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2021-07-09T18:21:17.980 | <p>I never understood mechanically how a vending machine prevents people from accessing goods through the dispensing panel. From what I understand it's not a locked door that unlocks only upon payment, dispenses goods and then re-locks. In my experience it seems to be a non-automated mechanical system powered by the weight of the dispensed goods when they fall into it.</p>
<p>I remember as a young child I would out of curiosity try to stick my hand into the panel and see if I can reach up but was unable to do so as my hand would be blocked by what felt like a literal wall. So I know there is some mechanism in place to ensure goods can only flow out and nothing can come in, but I don't understood mechanically how such a thing works.</p>
| |mechanical-engineering|machine-design|product-design| | <p>As others have mentioned, the door or flap can't prevent theft if someone is willing to put in the effort to steal. Same goes for some automated teller machines. As things stand the vending machine primary purpose is to process payment for the correct good that the buyer removes.</p>
<p>The most effective method to theft prevention I've seen is the security camera. While it does not prevent theft, it deters it and can help to easily file with law enforcement (some security companies do just that!). Usually someone brazen enough to steal in front of or vandalize a camera have other offences, and just goes to add charges when the person does something that gets them apprehended.</p>
<p>It is no wonder that some newer tech vending machines simply want a credit card authorization to open a cabinet and charge for anything removed- as detected via some other means than controlling a dispensing screw/clip.</p>
| 45052 | How do vending machines prevent people from stealing goods? |
2021-07-10T22:40:25.237 | <p>Is there a way to add a "fuzzy" texture to a body in solidworks?</p>
<p>Something like this:
<a href="https://i.stack.imgur.com/Z6kzK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z6kzK.jpg" alt="enter image description here" /></a></p>
<p>The purpose - to 3d print a body with a fuzzy surface to get better bonding</p>
<p>Any other ideas to improve the bonding surface will be great (fiberglass -> epoxy -> 3d printed stuff (PLA / PETG not sure which one to use yet)</p>
<p>(I prefer to do it in solidworks and not add the fuzzy feature in Cura)</p>
| |solidworks|3d-printing|carbon-fiber| | <p>Yes.</p>
<p>This has been a feature since 2019 <a href="http://help.solidworks.com/2019/english/solidworks/sldworks/c_3d_textures.htm" rel="nofollow noreferrer">http://help.solidworks.com/2019/english/solidworks/sldworks/c_3d_textures.htm</a></p>
<p>The only 'downside' is that the output is a mesh body not BREP, but this isn't an issue in its intended use case of 3D print files. For rendering you would just bump map the smooth surface.</p>
| 45065 | Fuzzy texture in soildworks |
2021-07-12T06:19:09.937 | <p>The question is about why are motorcycles not manufactured with handlebars that can turn to great degrees like a bicycle. In most motorcycles, you hardly have a steering angle of about 20-30 degrees, whereas on a bicycle, you can even go beyond 90 degrees. Is this done with the purpose of stability in mind, as motorcycles travel at far higher speeds than a bicycle, and sudden steering of the vehicle at such speeds can cause it to flip over. Why is this not done at least for motorcycles belonging to the 'streetbike' or 'naked' category, as they are meant to be ridden in cities, at least theoretically, where you might need to take tight turns.</p>
| |design|bicycles| | <p>The simplest answer to the question is "because the motorcycle gets in its own way"</p>
<p>Take a look at this picture of a trials motorcycle; it has perhaps the smallest of turning circles available in a motorcycle of a given length:</p>
<p><a href="https://i.stack.imgur.com/vhmih.png" rel="noreferrer"><img src="https://i.stack.imgur.com/vhmih.png" alt="enter image description here" /></a></p>
<p>And here's a picture of a typical bicycle steering/suspension setup:</p>
<p><a href="https://i.stack.imgur.com/y93Fd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/y93Fd.png" alt="enter image description here" /></a></p>
<p>On a motorcycle, it's fairly universal to have the suspension arms anchored in yokes in two places (the red clamps around the black tubes are part of the yoke - there is a yoke at the top, near the handlebars and another underneath, near the tire) for reasons of strength. In a bicycle it's common to have the suspension forks anchored in a single yoke underneath the frame tube in which it rotates. This means the steering can spin all the way round, 360 degrees, if it needs to (and other incidentals like cables aren't attached)</p>
<p>You can cast off the cables, or route them through the steering tube but you still won't evade this simple principle; the frame of the motorcycle gets in the way of the suspension tube and limits how much it can rotate.</p>
<p>Exceptions exist; this is a downhill bicycle:</p>
<p><a href="https://i.stack.imgur.com/kdaRr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/kdaRr.png" alt="enter image description here" /></a></p>
<p>It adopts the motorcycle style of suspension anchoring for reasons of strength, considering the use - high speed, heavy impact. You'll be able to swing these handlebars quite far too; the suspention tubes and frame are slim, but it won't spin 360 degrees.</p>
<p>This is a child's trials motorcycle:</p>
<p><a href="https://i.stack.imgur.com/aTckI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/aTckI.png" alt="enter image description here" /></a></p>
<p>Basically a bicycle with an electric motor; this machine certainly won't be suffering anywhere near the stress that a typical motorcycle would - you won't be doing 100mph on it and need to hit the brakes suddenly and hard, or be landing after a huge jump in a motocross race and be loading several hundred pounds of weight onto the front wheel.</p>
<p>For the stresses most motorcycle front suspension is subjected to you need that double anchor of above and below, and generally the bigger and heavier the motorcycle, the bigger all that componentry is going to need to be.. and the bigger it gets, the less far you can swing it before it bangs into something else. Notice how, as the components get bigger and more heavy duty because the bikes get heavier, the forks also have to be set further and further apart and more forwards, just to keep allowing a reasonable range of swing:</p>
<p><a href="https://i.stack.imgur.com/wYEgn.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/wYEgn.jpg" alt="enter image description here" /></a></p>
<hr />
<p>Side note; anything can be designed for any purpose. The owner of this machine clearly felt being able to turn on a dime was important and snapping the suspension in normal use unlikely:</p>
<p><a href="https://i.stack.imgur.com/FZgXz.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/FZgXz.jpg" alt="enter image description here" /></a></p>
<p>..but by and large engineering anything in this world is about achieving an ideal set of compromises for the task at hand and motorcycles and bicycles have rather different tasks at hand, different sets of physical stresses in operation and are thus engineered appropriately. The difference you've identified is a consequence of that and saying "they should be able to..." does also omit all the other things the "should" be able to do too (not snap when a 200 pound rider doing 60mph grabs the front brake but runs into a hole in the road anyway).</p>
<blockquote>
<p>Is this done with the purpose of stability in mind</p>
</blockquote>
<p>Unlikely to be the primary consideration. How two wheeled vehicles steer at speed isn't really a factor here. The process for steering at a constant speed is:</p>
<ul>
<li>When you want to turn right you (subconsciously) turn the steering to the left slightly</li>
<li>This causes the -cycle to start to fall over to the right</li>
<li>As the cyle falls over rightwards, you can then start turning the steering to the right to "catch it" - turning the steering to the right causes a -cycle to "fall over to the left"</li>
<li>If your -cycle is "already falling to the right" then this "steer to the right to cause the cycle to fall over to the left" means you can counteract the "falling over to the right" with some amount of "falling over to the left"</li>
<li>This means you can adopt a stable "describing a right hand arc" position where the amounts of "falling to the right" and "falling to the left" are balanced, the cycle is leaned over, it's steering is set pointing slightly right and it is not falling either left or right but is describing a right hand arc</li>
<li>You continue leaning into the direction of turn without falling off to the left or right of the cycle</li>
<li>If you need to turn more sharply to the right, you need to make the -cycle fall more to the right, so you actually again steer it a bit left, to induce a rightwards-fall, then as it falls you can feed in more and more "turn to the right" to achieve another balanced set of forces where you're leaned over more, you're describing a sharper turn, more centripetal force is trying to push the -cycle over to the left, more lean/gravity is trying to push the bike over to the right</li>
<li>When you want to straighten up out of a right turn you actually turn <em>more</em> to the right, which pushes the bike "over to the left" harder, causing you to become more upright. You can then, as you come upright, start feeding in turning the steering leftwards (back to being straight).</li>
<li>If you were leaned over to the right, in stable configuation, then turned more to the right and kept it that way you would very soon be thrown off the -cycle as it violently picks itself upright and then over to the left.</li>
<li>If you were leaned over to the right, in stable configuation, but your steering is turned all the way to the right and banging into the frame of the cycle you wouldn't be able to pick it upright from steering alone because you can't turn any more to the right to initiate a "fall over to the left"; you'd have to accelerate to cause an increase in centripetal force which would push the cycle upright</li>
</ul>
<p>One of the most significant skills in riding larger motorcycles at speed and flicking them between upright and leaned-over for fast, sharp corners is knowing exactly how much opposite steer to apply to cause the bike to quickly fall over to a lean angle that will go round the corner, and then how to quickly apply same-direction steer to get to a point where the falling over is halted, the bike goes round the corner and then the reverse operation picks the cycle upright again</p>
<p>Critically, none of this steering operation needs anywhere near 90 degrees of turn; the faster you go, the less you need. To a huge extent steering a cycle is not just about "pointing the front wheel in the same direction as the corner" - it is absolutely vital that the steering is also used to make the cycle fall over and your brain is doing a constant balancing act between the current speed, the rate of acceleration or deceleration (slower speeds need greater turns on the steering to cause a "falling over" in one direction or another), the amount of "fall to the left/right" and how much the steering is turned.</p>
<blockquote>
<p>where you might need to take tight turns</p>
</blockquote>
<p>If you have a cycle that can turn 90 degrees, and it is the rear wheel that is driven, you can achieve a scenario where the front wheel is turned 90 degrees and becomes a perfect brake for the motion of the cycle. There's no overall force that will encourage the front wheel to rotate one way or the other and indeed as a front wheel is turned more and more toward 90 degrees the force that must be applied by the back wheel to get things moving at all is greater and greater. A cycle with a wheel turned 90 degrees doesn't steer at all.</p>
<p>If the front wheel were driven, it wouldn't matter; the front wheel would easily rotate even at 90 degrees and the bike would pivot round its back wheel- the smallest turning circle it can achieve without travelling slightly backwards at the same time. You can achieve this with a cycle that only has a relatively small amount of steering angle by lying it down more, so that the contact point the front wheel makes with the floor reaches the point of the tire that is level with the axle when the wheel is upright. So long as you can turn a motorcycle steering far enough to the right that the frontmost point of the wheel is toughing the floor and nothing else is, you can drag that cycle round in a very tight turn. Of course you're using your body strength to counteract gravity and you're pulling it sideways to rotate the front wheel but it's a balanced set of forces achieving a turn all the same. In "normal operation" of a cycle though the balanced set of forces come from a different arrangement, so we get back towards that "because there's a set of compromises that must be met across all environments so that is what we do" which ultimately leads to an answer of "motorcycles don't turn as sharply as bicycles because they dont need to". You'll never try to flip one round in your hallway (tip for bicycle; don't steer to do that either - put the back brake on, walk backwards, then spin it round on its back tire and put it down)</p>
| 45083 | Why don't motorcycles have as good a turning radius as bicycles? |
2021-07-13T13:27:53.183 | <p>First off I'm not an engineer so please excuse the simple language I use to explain my question!</p>
<p>I have an arrangement whereby I have a steel plate suspended from basically a bridge. The plate is free to move up and down, guided in a channel by needle rollers. The vertical movement is controlled by an M4 threaded rod, with a load cell between the top of the bridge and the M4 nut, which is screwed in and out to effect movement. Hopefully the three images will convey what I'm doing.</p>
<p>The steel plate and its attachments weigh ~800g. If I suspend this outside of my test jig,I get a consistent mV output in line with what I expect to see.</p>
<p>However, if I raise the plate via the control nut, I can get an output that works out as 970g, and going down the reading can drop to 630g. It's not always this bad, but it was easy to re-produce this range, Note, readings are taken after leaving it to settle.</p>
<p>Plate appears to move very freely (just by feel), and rollers are well oiled (WD40).</p>
<p>Can anyone explain what is happening here, and how to overcome it, or allow for it?</p>
<p>Thanks!</p>
<p><a href="https://i.stack.imgur.com/T6aM7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T6aM7.jpg" alt="enter image description here" /></a></p>
<p>enter image description here</p>
<p><a href="https://i.stack.imgur.com/gXifa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gXifa.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/NuI38.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NuI38.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|measurements|bearings|friction| | <p>When you are raising the plate you are also raising the bearings and since they need to roll there is the frictional component pulling down on the bearings and pulling down on the plate.</p>
<p>When you lower the system you are pulling the bearings down and since they again need to roll there is the frictional component pulling up on the bearings and pulling up on the plate.</p>
<p>Adding a vibrating source, to jiggle the system, may help minimize the error. Another way to minimize error is to measure the load uni-directionally (measure the load in one direction only). That is, to lower the system, but then raise it slightly to bias residual loads in one direction. Thus, only measure the load while being raised slightly.</p>
<p>I would postulate, if you removed the bearings and allowed the plate to "clatter" between the guide edges the forces would compare favorably.</p>
<p>I'm not really sure why you would want to know the weight of the raising and lowering system unless you really expected it to change because of added mass during operation. If you can live with the frictional components so be it. If you expect service is to be required occasionally, then you might plan for occasional maintenance.</p>
<p>The load cell could provide the information of service being required, as the measured load separation between raising and lowering (as you have measured) expand.</p>
<p>I would venture that this is for controlling water flow in rectangular concrete channels. If dirt channels, your bearings are likely to get real dirty.</p>
<p>If you want to gage water flow over time, I would say, use a V-notch weir or a rectangular weir down stream of the "gate". The down stream level of the water should be below the bottom of the V or the "rectangle". Water flow is a function of the up stream level of water above the bottom of the V or "rectangle".</p>
<p>If the gate is partially submerged in water, then the weight of the gate will be diminished by the weight of displaced water by the volume of the partially submerged (wetted) gate.</p>
| 45102 | Inconsistant load cell readings - (static) fiction perhaps? |
2021-07-13T14:04:21.273 | <p>The test that led to the Chernobyl accident was to see if they could keep cooling water circulating long enough in the event of a power cut for the generators to kick in.</p>
<p>So, why didn't they just power their own circulation system? Why be dependent on the grid, and not just use some of the power they were outputting to run their own systems? Or at least have it as an insta-backup?</p>
<p>Even more directly, why not just link the turbine and the pump together mechanically with some gearing to govern the speed-ratios?</p>
| |nuclear-engineering| | <p>Your question stems partly from a common misunderstanding. The test was <strong>not</strong> meant to be conducted with the reactor running. In reality, the procedure called for the reactor to be SCRAMmed simultaneously with the turbine. So there was supposed to be no power available.</p>
<p>What happened was that <a href="https://chernobylcritical.blogspot.com/p/part-5-after-explosion.html" rel="nofollow noreferrer">there was a miscommunication</a>, and the shift manager waited for confirmation from the turbine department or the deputy chief engineer. Then, (I'm speculating here) the reactor operator apparently noticed that the automatic rods were descending and brought this to his attention, and he finally gave the command to shut down the reactor.</p>
| 45103 | Chernobyl - why didn't they power their own circulation system? |
2021-07-14T01:23:21.007 | <p>I need to draw the isometric view and top section view of the object given only the front and right side view.
<a href="https://i.stack.imgur.com/O7BT1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7BT1.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/kaTl2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kaTl2.png" alt="enter image description here" /></a></p>
<p>I understand on how other edges of these views correlate with each other. I just do not get on what to do with this surface (highlighted with red).</p>
<p><a href="https://i.stack.imgur.com/aJZBs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aJZBs.png" alt="enter image description here" /></a></p>
| |technical-drawing|drafting|drawings| | <p><a href="https://i.stack.imgur.com/XQKm6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XQKm6.png" alt="straight lines added" /></a></p>
<p>I see 4-solutions. I added straight lines (in GIMP) to represent canted faces. Either the original posted by NMech along with either or both surfaces as canted are also viable alternatives.</p>
<p>(As NMech noted, they could have curvature (except you might then have to show the center of the arc, unless they were b-splines(?).)</p>
| 45111 | Finding the isometric view of an object from the front and right side view |
2021-07-14T09:14:47.070 | <p>I have a T-shaped pipe that is connected to a vacuum pump. Along one side, there are 4 areas of 9 holes where air will get sucked in from the surroundings. It's used to pick up some small objects.</p>
<p>The problem is the 2 areas of holes that are closer to the center have a much higher intake flow/pressure compared to the 2 further areas, that it renders those 2 less-/ineffective.</p>
<p>Hole sizes and the general positions of the hole areas are fixed, so I'm thinking of balancing it by adjusting the number of holes at each area instead. <em>(edit: Currently can't use a flow regulator in this application, any other suggestions instead of changing number of holes are welcome as well)</em></p>
<p><em>I am not trying to achieve the exact same suction pressure for each area, but rather to have sufficient suction pressure at each area.</em> Will changing the number of holes in each area be able to balance the suction pressure? My guess was reducing the number of holes for the 2 middle areas or increase for the 2 end areas, but I'm not sure by how many. If you could provide a general idea/example on how to calculate this, or an alternative method to accomplish this, that would be great.</p>
<p><a href="https://i.stack.imgur.com/nDQ8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nDQ8S.png" alt="T-shaped Pipe" /></a></p>
| |mechanical-engineering|fluid-mechanics|design|vacuum-pumps| | <p>The two critical variables for flow rate will be resistance and diameter/area of flow.</p>
<p>The holes in the outer zones could be over drilled to increase their diameters.</p>
<p>If that is not an option then reducing the number of holes in the central zone and increasing the number of holes in the out zone would help.</p>
<p>Alternatively, as others have stated, you could cover some holes in the central zone with tape to block them. As an alternative to this, the resistance could be increased and hole diameters reduced by initially covering all the holes in the central zone with very sturdy tape which has good adhesive properties. After this, using a smaller diameter pin or spike, smaller diameter holes can be placed in the tape where the existing hole are.</p>
<p>If this is successful an adapter could be made for fit over the existing holes which has the holes with the diameters and number of holes required. The adapter could be screwed or clamped to the suction manifold.</p>
| 45115 | Balancing suction air flow at different areas on pipe |
2021-07-14T15:43:14.720 | <p>I need a little help with SolidWorks.</p>
<p>I need to assemble such a mechanism, which I am currently working on.</p>
<p>But I get a rather complicated and cumbersome algorithm:</p>
<ol>
<li>First, I build a "wireframe" from the necessary lines and other connections.</li>
<li>Then I build cylinders for rotary articulations. 2. Then inside them I build auxiliary surfaces, place circles on them and build a circular arc with their help.</li>
<li>I do the same with other arcs.</li>
<li>I build the platform separately. it is impossible to do everything in one picture.</li>
</ol>
<p>Maybe a more experienced person can tell me how I can build such a mechanism in a simpler way?</p>
<p>I would be glad and grateful for your help.</p>
<p><a href="https://i.stack.imgur.com/Jjttc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jjttc.png" alt="enter image description here" /></a></p>
<p><a href="https://www.youtube.com/watch?v=kN09M5QGGBk" rel="nofollow noreferrer">https://www.youtube.com/watch?v=kN09M5QGGBk</a></p>
| |solidworks|cad|solid-mechanics|software| | <p><a href="https://i.stack.imgur.com/XTp0v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XTp0v.png" alt="enter image description here" /></a></p>
<p>I tend to use SolidWorks assembly to trial and error these types of assemblies, then worry about how to make it work. Constrain (Angle and Distance) the crap put of components to start and remove them as components are interconnected.</p>
<p>Following the process I outlines in the comments:</p>
<p>Platform: 100mm.</p>
<p>Upper Arm: 105mm by 50mm, Angle 90°. The dimensions you will have to tweak at some point. Your algorithms. I started with 100mm by 50mm and it's not perfect.</p>
<p>Lower Arm: 50mm by 50mm, Angle 45°.</p>
<p>Mate all the lower faces. Place an angle mate of 120° on each of the lower arms. The angle mates will allow you to construct the mechanism. Mate faces to top plane.</p>
<p>Mate one of the lower arm origins to the assembly origin. Origin must be on mated faces.</p>
<p>Mate Upper Arms to Lower Arms. Place an angle mate of 90° on upper arm to lower arm. Make sure all upper arms are in the same direction.</p>
<p>Mate platform to upper arms. You can see I cheated and mated them to a point.</p>
<p>It's a <em>rather complicated and cumbersome device</em>, but this will get you going in the right direction. Remove all angle mates and you should have the basic mechanics.</p>
<p>Once you get that working, upper arm is identical, but lower connection cannot be in the same place, so bottom 1 will be at 50mm by 50mm at 45°. Next 60mm by 40mm at 45°. Last at 70mm by 30mm at 45°. The actual numbers will depend on your actual dimensions.</p>
<p>This has been solved on the web with 3D printed files <a href="https://www.youtube.com/watch?v=hj0oFC73A5c" rel="nofollow noreferrer">3D Print Timelapse Build: Spherical Parallel Manipulator</a>.</p>
<p><a href="https://www.thingiverse.com/thing:3951917" rel="nofollow noreferrer">Thingiverse project.</a></p>
| 45119 | Spherical Parallel Manipulator in SolidWorks |
2021-07-15T10:27:17.787 | <p>I hope these exist and someone can tell me what they're called. What I'm looking for is screws to mount a PCB where the top of the screw is normal, then there's a smooth length, then a nut with a spring held in between the top of the screw to the bottom of the bolt so that, for example, if you were to press on the PCB a bit it would move a bit vertically. I'm terrible at drawing things, so I'll try with text:</p>
<pre><code>===== <Top of screw
|\| <Spring on smooth part of screw/bolt
|\| <Spring on smooth part of screw/bolt
|\| <Spring on smooth part of screw/bolt
=== <Return to normal threading
[=] <Nut
</code></pre>
<p>Anyone know?</p>
| |terminology| | <p>It think you are looking for a spring-loaded heat sink attachment screw. These provide soft compression of a heatsink to a PCB. But you want the spring on the otherside so the mounted board can be pressed down. As far as I know, you can do that. But I'd ask the vendor to confirm it.</p>
<p><a href="https://i.stack.imgur.com/G6hW4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G6hW4.png" alt="enter image description here" /></a></p>
| 45129 | Screw/Bolt to mount PCB with spring |
2021-07-15T13:25:46.127 | <p>I am a control engineering student and I am studying the frequency response of a system, so the Laplace domain, the Bode plot, poles, zeros,etc. ...
I have clear grasp of their meaning, but if I think about the practice it is hard to grasp, at least for me.</p>
<p>For example, I would like to study the frequency response of a DC motor with Arduino, but really don't know from where to start.</p>
<p>So, <strong>how do I study the frequency response of a physical system with Arduino?</strong> (for example a DC motor)</p>
| |control-engineering|control-theory|signal-processing|frequency-response|nonlinear-control| | <p>The arduino does impose some limits. But a classical frequency identification is build from a few steps:</p>
<ol>
<li>As the other comments already suggest, the design of a suitable input. As mentioned, a sinusoidal input (I actually recommend multisine) is suitable to accurately identify the response of a few frequencies. White noise or band-limited white noise can identify an entire range of frequencies, but less accurate as the energy per frequency must be divided. The common sense here is that if you compute the frequency domain transformation of the input signal, the higher the magnitude at certain frequencies, the better the identification for these frequencies. On top of that, for physical reasons, I suggest to add a DC component (so a constant offset) to ensure the DC motor keeps spinning, this prevent you from measuring non-linear start-stop behaviour.</li>
<li>Sampling rate. The maximum frequency you can possible identify is equal to the Nyquist frequency - half the sampling rate. With identification, getting the cut-off frequency / resonance frequency is essential, so make sure you can measure fast enough.</li>
<li>Experiment duration. The longer you measure, the more detailled the frequency identification looks. This means you can identify lower frequencies, and have more point is the high frequency region.</li>
<li>Choice of Window and whether windowing should be applied. Windowing allows to get a more detailed frequency identification, whilst cutting the total identification length. So low frequencies are not very accurate, while higher, more desired frequency regions are.</li>
</ol>
<p>So that leaves us with the Arduino issues. Most notably, a problem between 2 and 3. I dont know which Arduino you use, but something like an Uno might be troublesome. See suppose you want to measure at 1kHz (which might be fair), the USB transfer protocols might harm this sampling rate and make it impossible to continuously send data. Therefore, it is advisable to store the data in the Arduino and send it afterwards. Which leaves us with point 3: memory. The arduino is fairly limited in memory so how long can you actually measure and still store it?</p>
<p>Hope these provided some insights. Any questions are always welcome.</p>
| 45133 | How do I study the frequency response of a physical system with Arduino? |
2021-07-15T22:51:22.287 | <p><a href="https://i.stack.imgur.com/kJuD7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kJuD7.png" alt="Triangle" /></a></p>
<p>If we imagine the load being placed at the top of the below triangle, I realise that the downwards force is transferred into one which goes in a 'south east' and 'south west' direction down to the end of the triangle.</p>
<p>What I am trying to understand however, is how we get the sideways movement from a straight down movement. Is this simply a product of the force needing to travel along a beam which can only transfer compression, therefore even though there is no 'input' sideways force, one comes from the downwards force only. Or is the sideways force created by the reaction force from the opposing 'leg' of the triangle?</p>
<p>Thank you!</p>
| |structural-engineering|civil-engineering|statics|stresses| | <p>The direction of Forces isn't necessarily along the connecting element. If that happens depends a lot on the constraints between the different elements. For example see the following image:</p>
<p><a href="https://i.stack.imgur.com/58pt2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/58pt2.png" alt="enter image description here" /></a></p>
<p>In the left column is a "welded" structure, while on the right column is a pin jointed structure (a basic truss) if you like.</p>
<p>On the top there are the shapes before deflection, while at the bottom are <span class="math-container">$\color{red}{\text{the shapes after deflection with red}}$</span> (superimposed on the original structure). (<strong>NOTE</strong>: on the no joints there is another case with the opposite curvature, but I left it for simplicity sake's).</p>
<p>AS you can see the behavior is totally different (at least initially). <em>Eventually</em> the jointed structure will also have a similar shape due to buckling. However, on the no-joint structure the curves will start to appear immediately.</p>
<h2 id="structure-with-joints-3v92">structure with joints</h2>
<p>I will focus on the structure with joints and why the forces indeed travel towards one direction. The direction of the force is parallel to the member.</p>
<p><a href="https://i.stack.imgur.com/WGunk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WGunk.png" alt="enter image description here" /></a></p>
<p>If I take a member - let;s say no 1 - then the only way that this element will be in equilibrium of moments and forces is when the forces are in the direction of the beam.</p>
<p><a href="https://i.stack.imgur.com/Qra3V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qra3V.png" alt="enter image description here" /></a></p>
<p>IF there are any forces perpendicular to the element axis then there is no equilibrium of moments of forces. Try it for any combination of forces (any of the yellow, green, red, purple that don't sum up to zero).</p>
<p><a href="https://i.stack.imgur.com/E8HXE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E8HXE.png" alt="enter image description here" /></a></p>
<p>As a result, the structure divides the force in a way that new forces are generated that cancel each other (horizontal forces on the top sides of 1 and 2. Those forces are such that the resultant force combined with the fraction of the vertical force that travels downwards is in the direction of the beam.</p>
<h2 id="a-word-of-caution-about-fluid-pressure-5umc">A word of caution about (Fluid) Pressure</h2>
<p>One final note, is that pressure from fluids is not the same thing. Although there are a lot of similarities with fluid pressure and material stress (similar units, definition), pressure does not have a direction (its a state quantity/function). Pressure is always normal to a surface. Therefore, although the result is the same (diversion of the flow, diversion of forces), the mechanism that happens is different.</p>
| 45139 | How exactly do forces transfer in different directions? |
2021-07-17T08:27:34.943 | <p>Here's where I'm struggling</p>
<p><a href="https://i.stack.imgur.com/cuXWn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cuXWn.jpg" alt="enter image description here" /></a></p>
<p>So I've tried to find the expression for deflection of a simply supported beam acted on by 4 point loads at varying distances.</p>
<p>First I balance out the vertical forces and then I balance out the moment about A. These two equations will get me the reaction forces on edges A and B.</p>
<p>Then I used the double integration method to find the expression for deflection of beam for a section x, 'x' distance away. The two constants can be found using the boundary condition that deflections at both edges (x=0 and x=L) is zero.</p>
<p><em>I've only used a check mark instead of finding the expression because I just want you guys to check my methodology because I seem to be making a mistake somewhere</em></p>
<h2 id="heading-jekf"><a href="https://i.stack.imgur.com/5bxco.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5bxco.jpg" alt="Heading" /></a></h2>
<p>Here I've used the same general expression to find the deflection for a single point load 'w' acting on the center of the beam.
Found the constant C1 using the condition that deflection is zero (y=0)at x=L.</p>
<p>The expression I get at the end is different from the expression I see mentioned in books or online. It's supposed to be 48 in the denominator and not 32.</p>
<p>If I go through each step it makes sense to me. But the answer is wrong. So I definitely have a fundamental misunderstanding of the method. I would appreciate it if someone could point this out.</p>
<p>All of this has to do with a larger problem I was working on but I noticed something in that problem which made me go back to all these basic derivations and notice that I've been doing something wrong.</p>
<p>Also I'm new to stackechange so I apologise if I messed up the format of the post or didn't conform to the rules of the forum.</p>
| |mechanical-engineering|structural-engineering| | <p><strong>I am writing this as an answer to the comment about the superposition method. I started writing it in my original answer, but it was too long and confusing, so I opted to do another answer.</strong></p>
<h2 id="superposition-method-etl3">Superposition method.</h2>
<p>Let's assume you have this problem.</p>
<p><a href="https://i.stack.imgur.com/eMwh1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eMwh1.png" alt="enter image description here" /></a></p>
<p>In that case you only need the following equation.</p>
<p><a href="https://i.stack.imgur.com/oKQEU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oKQEU.png" alt="Formulas for superposition" /></a></p>
<p>Essentially what you do is you apply the above equation when only <span class="math-container">$P_1$</span> is applied and you apply it again when only <span class="math-container">$P_2$</span> is applied.</p>
<h2 id="faf11d55-e160-4cf7-9ba2-39ba5a05bbd8-g6qc"><span class="math-container">$P_1$</span></h2>
<p>so starting with the case:</p>
<p><a href="https://i.stack.imgur.com/YzMV2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YzMV2.png" alt="enter image description here" /></a></p>
<p>so the application for this problems yields the deflection <span class="math-container">$y_1$</span> when only <span class="math-container">$P_1$</span> is applied.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 2$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 2<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>you can further substitute <span class="math-container">$a=\frac{L}{2}$</span>, and <span class="math-container">$b=L-\frac{L}{2}$</span>
and simplify</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 2$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 2<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(L-\frac{L}{2}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(L-\frac{L}{2}\right)^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(L-\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{\left(L-\frac{L}{2}\right)}\left(x-\frac{L}{2}\right)^3-\left(L^2-\left(L-\frac{L}{2}\right)^2\right)x -x^3\right)$$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(\frac{L}{2}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(\frac{L}{2}\right)^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{\left(\frac{L}{2}\right)}\left(x-\frac{L}{2}\right)^3-\left(L^2-\frac{L^2}{4}\right)x -x^3\right)$$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div><h3 id="a1de742-c307-4b58-b0ae-01e6cf0462aa-03p9"><span class="math-container">$P_2$</span></h3>
<p>Similarly you can apply for the second problem when only <span class="math-container">$P_2$</span> is applied. Now lets denote <span class="math-container">$y_2$</span> as the deflection of problem 2.</p>
<p><a href="https://i.stack.imgur.com/TFQTX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TFQTX.png" alt="enter image description here" /></a></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 4$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 4<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_2$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>if you check the equation is exactly the same. The only difference is that <span class="math-container">$a=\frac{L}{4}$</span>, and <span class="math-container">$b=L-\frac{L}{4}= \frac{3L}{4}$</span>. So substituting a and b and simplify ( I won't do it as thourougly)</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 4$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 4<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_2$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Pb}{6\cdot L\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$y_2$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(L-\frac{L}{4}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(L-\frac{3L}{4}\right)^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P\left(L-\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{L-\frac{L}{4}}\left(x-\frac{L}{4}\right)^3-\left(L^2-\left(L-\frac{L}{4}\right)^2\right)x -x^3\right)$$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$y_2$</span> (simpified)</td>
<td style="text-align: center;"><span class="math-container">$$ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$-\frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div><h3 id="putting-it-all-together-ar5j">putting it all together</h3>
<p>The total deflection will be equal to <span class="math-container">$y_1$</span> and <span class="math-container">$y_2$</span>. However there is a problem the a and b are different. To be clear I will rewrite the final forms of <span class="math-container">$y_1$</span> and <span class="math-container">$y_2$</span></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 2$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 2<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_1$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>and</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">Section</th>
<th style="text-align: center;"><span class="math-container">$0<x<\frac L 4$</span></th>
<th style="text-align: center;"><span class="math-container">$\frac
L 4<x<L$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;"><span class="math-container">$y_2$</span></td>
<td style="text-align: center;"><span class="math-container">$$ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$-\frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right)$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>So the solution is to break up the beam into three sections.</p>
<ul>
<li><strong>Leftmost section <span class="math-container">$0<x<\frac{L}{4}$</span></strong></li>
</ul>
<p>in that region the deflection will be equal to:</p>
<p><span class="math-container">$$y_{tot}(x) = \color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } 0<x<\frac{L}{4}$$</span></p>
<p><span class="math-container">$$y_{tot}(x) = \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right) \qquad\text{ when } 0<x<\frac{L}{4}$$</span></p>
<ul>
<li><strong>Middle section <span class="math-container">$\frac{L}{4}<x\frac{L}{2}$</span></strong></li>
</ul>
<p><span class="math-container">$$y_{tot}(x) = \color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } \frac{L}{4}<x\frac{L}{2}$$</span></p>
<p><span class="math-container">$$y_{tot}(x) = \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{4}<x<\frac{L}{2}$$</span></p>
<ul>
<li><strong>Rightmost section <span class="math-container">$\frac{L}{2}<x<L$</span></strong>
In the rightmost section the total deflection will be.</li>
</ul>
<p><span class="math-container">$$y_{tot}(x) =\color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } \frac{L}{2}<x<L$$</span></p>
<p><span class="math-container">$$y_{tot}(x) = \color{green}{\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)} + \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{2}<x<L$$</span></p>
<p>Or in another form:</p>
<p><span class="math-container">$$y_{tot}(x)= \begin{cases}
\color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right) \qquad\text{ when } 0<x<\frac{L}{4}\\
\color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{4}<x<\frac{L}{2}\\
\color{green}{\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)} + \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{2}<x<L
\end{cases}$$</span></p>
<h2 id="final-thoughts-mvkk">final Thoughts</h2>
<p>Although there is a lot of computation, here its much easier to repeat and program into a software. You don't need to think too much, you just need to be careful in the calculations.</p>
<p>You can also do the same with the deflection (although a better way IMHO, is to get the deflections and then differentiate.</p>
| 45161 | I seem to have a fundamental misunderstanding about the double integration method for calculating beam deflection |
2021-07-18T16:17:48.480 | <p>Hey guys I am trying to balance out the combustion equation for reactants: Octane + Air</p>
<p>with products: CO2, H2O, N2, and O2; and 120% oh excess air.</p>
<p>My balancing equation looks like this:</p>
<p><span class="math-container">$$ C_8H_{18} + 1.2* a(O_2 + 3.76 N_2) \rightarrow bCO_2 +cH_2O + dO_2 + eN_2 $$</span></p>
<p>My approach was as follows:</p>
<p>C: <span class="math-container">$b = 8$</span></p>
<p>H: <span class="math-container">$2c = 18$</span></p>
<p>O: <span class="math-container">$2b + c + 2d = 2*1.2a $</span></p>
<p>N: <span class="math-container">$e = 3.76*1.2a$</span></p>
<p>However, I think somewhere in my steps there's something I am not doing correctly as I am unable to determine 'a' with the system of equations I tried developing.</p>
| |mechanical-engineering|chemistry| | <p>you also need to take into account that the complete burn for octane is:</p>
<p><span class="math-container">$$2C_8H_{18} + 25\;O_2 \rightarrow 16\;CO_2 +18\;H_2O $$</span></p>
<p>Therefore <span class="math-container">$a$</span> should be equal to <span class="math-container">$\frac{25}{2}$</span>.</p>
| 45173 | Balancing Equation of Octane in combustion Chamber |
2021-07-18T20:44:37.623 | <p>You're basically replacing oil with oil in plastic form. Where is the gain?</p>
<p>Ok so we got this reply.</p>
<blockquote>
<p>Not sure if you're trolling but the point is that the windmill makes power as long as it exists and functions. So if it generates more power in its lifetime than the oil that took to make it, that's the gain. Is this not obvious? Its the same with solar cells</p>
</blockquote>
<p>So, a wind turbine weighs a thousand tons (or whatever). That's 10 gwh of oil. Does it produce that much energy?</p>
| |wind| | <p>Every once in a while questions like this come out of the woodworks, usually with a tone similar to other key-board warriors we have heard before. you don't know if it's a tease or a genuine deep misunderstanding.</p>
<p>In the spirit of improving my gravely lacking skills in being entertaining, I take a stab at this question.</p>
<p>If we compare burning a windmill to burning oil, burning the oil is a lot more efficient.</p>
<p>Also, the industry has many provisions to burn the oil as cleanly as possible whereas there have not been any attempts at burning the windmills efficiently!</p>
| 45177 | Wind turbines are plastic. How does that "work"? |
2021-07-19T03:09:05.627 | <p>I'm trying to get a safety factor of <span class="math-container">$n\geq2$</span> for the root of my snap fit beam (pic at bottom of the post). <br>
I have gone through the calculations to get the SF at the root's edge from the geometric and material properties (using a variety of plastics), and my SF seems unnaturally small. <br>
I don't have a lot of experience with this yet, so I wanted to check here to make sure I didn't forget something. <br>
If someone would be kind enough to check my work, I would greatly appreciate it. Is my work incorrect, or have I just overestimated the flexibility of my materials?
<br><br></p>
<p><strong>Find:</strong> Safety factor at the root edge, <span class="math-container">$n_e$</span> <br>
(I also found SF at the root center, <span class="math-container">$n_c$</span>, from transverse shear, but this isn't critical since stress from bending moment is zero there.)</p>
<p><strong>Geometric properties</strong> <br>
Beam length: <span class="math-container">$l=6.40~\text{mm}$</span> <br>
Maximum deflection: <span class="math-container">$y_{d,\text{max}}=1.20~\text{mm}$</span> <br>
Beam depth: <span class="math-container">$h=1.60~\text{mm}$</span> <br>
Beam width: <span class="math-container">$b=7.31~\text{mm}$</span> <br>
Root fillet radius: <span class="math-container">$r_f=0.48~\text{mm}$</span></p>
<p><strong>Material properties</strong> (ex.) <br>
SABIC LNP STAT-KON 5E003M: <span class="math-container">$S_y=50~\text{MPa}$</span>, <span class="math-container">$E=9060~\text{MPa}$</span> <br>
(this is a fairly brittle plastic, but I got very small SFs for many other plastics as well)</p>
<p><strong>Equations (beam root)</strong>
<span class="math-container">$$I=\frac{1}{12}bh^3~~~~~~~~~~~~~~P=\frac{6y_dEI}{x^2(x-3l)}=-\frac{3y_{d,\text{max}}EI}{l^3}~~~~~~~~~~~~~~y_{d,\text{max}}=y_{d,x=l}$$</span>
Equation for <span class="math-container">$P$</span> is based on the cantilever diagram at the bottom of the post.</p>
<p>Variable shear force and bending moment:
<span class="math-container">$$V(x)=P~~~~~~~~~M(x)=P(l-x)$$</span>
At <span class="math-container">$x=0$</span>:
<span class="math-container">$$V_{x=0}=-\frac{3y_{d,\text{max}}EI}{l^3}~~~~~~~~~M_{x=0}=-\frac{3y_{d,\text{max}}EI}{l^2}$$</span>
Normal stress at the root edge:
<span class="math-container">$$\sigma=\frac{My}{I}~~~~~~y=\frac{1}{2}h~~~~~~\sigma_\text{max,nom}=\frac{Mh}{2I}~~~~~~\sigma_\text{max}=K_t\sigma_\text{max,nom}$$</span></p>
<p>Stress concentration factor <span class="math-container">$K_t$</span> can be found by:
<a href="https://i.stack.imgur.com/38d8g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/38d8g.png" alt="enter image description here" /></a> <br>
From my design, <span class="math-container">$K_t$</span> turns out to be <span class="math-container">$\approx1.40$</span>.</p>
<p>Shear stress at root center:
<span class="math-container">$$\tau_\text{max,nom}=\frac{3V}{2A}~~~~~~~~~~~\tau_\text{max}=\tau_\text{max,nom}$$</span>
Von Mises stresses: <span class="math-container">$\sigma_c'=\sqrt{3\tau_\text{max}^2}$</span> at root center, <span class="math-container">$\sigma_e'=\sigma_\text{max}$</span> at root edge.</p>
<p>Safety factor:
<span class="math-container">$$n_c=\frac{S_y}{\sigma_c'}~~~~~~~~~~~~~~~~n_e=\frac{S_y}{\sigma_e'}$$</span></p>
<p><strong>Results (beam root)</strong> <br>
For 5E003M, <span class="math-container">$n_e=0.056$</span> and <span class="math-container">$n_c=0.725$</span>.</p>
<p>As you can see, these safety factors are terrible! As far as changing the material properties, I could decrease the width or depth of the beam, but the length must stay the same. <br>
Any insight would be appreciated, thanks! :D</p>
<p><br><br><br></p>
<p><a href="https://i.stack.imgur.com/ivmfU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ivmfU.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/GBknI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GBknI.png" alt="enter image description here" /></a></p>
| |beam|plastic| | <p>This is a suggested procedure to achieve your goal. I'll simplify your model to a cantilever beam with a constant cross-section as shown below, and assume the critical section is located at the root of the fillet.</p>
<p><a href="https://i.stack.imgur.com/vrdVt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vrdVt.png" alt="enter image description here" /></a></p>
<p><strong>The first step is to determine the maximum deflection <span class="math-container">$\delta_v$</span>.</strong></p>
<p><strong>The second step is to find out the equation for the deflection.</strong> For this exercise, I'll ignore the shear deformation, but apply the method developed by Timoshenko to account for "large deflection".</p>
<p>In his book, "Mechanics of Materials", co-authored with Gere, Timoshenko provided a table for ease of pinpoint the correct equation as shown below. In the table, column (m) is the numerical value of <span class="math-container">$\dfrac{PL^2}{EI}$</span>, and column (n) is the numerical value of <span class="math-container">$\dfrac {\delta_v}{L}$</span>. The equation of deflection is simply (m)/(n).</p>
<ul>
<li><p><span class="math-container">$(m)$</span> = <span class="math-container">$\dfrac{PL^2}{EI}$</span>, <span class="math-container">$(n)$</span> = <span class="math-container">$\dfrac {\delta_v}{L}$</span>, <span class="math-container">$\dfrac{(m)}{(n)}$</span> = <span class="math-container">$\dfrac{PL^2}{EI}$</span>/<span class="math-container">$\dfrac {\delta_v}{L}$</span>.</p>
</li>
<li><p>Let <span class="math-container">$\lambda = \dfrac{m}{n}$</span> and rearrange the terms,</p>
</li>
<li><p><span class="math-container">$\delta_v = \dfrac {PL^3}{EI \lambda}$</span>, or <span class="math-container">$P = \dfrac{EI \lambda\delta_v }{L^3}$</span>.</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/gEetJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gEetJ.png" alt="enter image description here" /></a></p>
<p><strong>The third step is to check shear stress:</strong></p>
<p>At this point, I'll conservatively introduce the form factor to account for the escalated shear stress due to shear deformation. For a rectangular section, the form factor <span class="math-container">$f_s$</span> = 6/5 = 1.2, so</p>
<p><span class="math-container">$\sigma = f_s*\dfrac{3P}{2A} \leq \dfrac{f_y}{n}$</span>, where <span class="math-container">$n$</span> is the desired safty factor.</p>
<p><strong>The fourth step is to determine the bending stress.</strong></p>
<ol>
<li><p>Assume elastic behavior, <span class="math-container">$\sigma_b = \dfrac{6M}{bh^2} \leq \dfrac{F_y}{n}$</span>, or</p>
</li>
<li><p>assume plastic behavior, <span class="math-container">$\sigma_b = \dfrac{4nM}{bh^2} \leq F_y$</span></p>
</li>
</ol>
<p><strong>The last step is to determine <span class="math-container">$P$</span>:</strong></p>
<p>Since <span class="math-container">$M = P*a$</span>, plug <span class="math-container">$M$</span> into the two equations above, you will get the <span class="math-container">$P$</span> that satisfies the limit of the bending stress with the desired safety factor. However, you also need to back-check/compare the <span class="math-container">$P$</span> derived from steps two (deflection) and three (shear stress) before making the conclusion.</p>
<p>If you still couldn't get satisfactory force with the desired safety factor, you will have to increase the depth/thickness of the member or adjust the deflection limit.</p>
<p>Hope this helps.</p>
| 45180 | Finding Safety Factor for Plastic Snap Fit Cantilever |
2021-07-19T11:19:48.940 | <p>I need to construct the auxiliary views with the given front and right side orthographic views. The view is in third angle projection.</p>
<p><a href="https://i.stack.imgur.com/t9uWo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t9uWo.jpg" alt="enter image description here" /></a></p>
<p>I observed that there is an oblique surface so that perhaps I need to construct here the secondary auxiliary view. However, I am confused in constructing those auxiliary views. I cannot make the oblique surface appear as an edge in the primary auxiliary view. I am stuck. Here, I already constructed the isometric view as my guide.
<a href="https://i.stack.imgur.com/oxfh1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oxfh1.jpg" alt="enter image description here" /></a></p>
<p>This is my outline for the construction of primary auxiliary view and I don't know what to do with the oblique surface and how to connect the points. My attempt is to use the fold-line method.</p>
<p><a href="https://i.stack.imgur.com/ilSkL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ilSkL.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|technical-drawing|drafting|drawings| | <p>You did pretty well, but @Transistor is correct that the horizontal line needs to be extended. Maybe it looks like this. However, I don't think the inclined plane is correct as there will be another line (shown in red) to connect the corners.</p>
<p><a href="https://i.stack.imgur.com/4P262.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4P262.png" alt="enter image description here" /></a></p>
| 45184 | How to Construct a Primary Auxiliary View given that an oblique surface with 6 sides |
2021-07-19T21:41:53.183 | <p>The task is to cool steel pieces in boxes with <strong>indoor air</strong> using a fan.
The number of boxes is <strong>6 pcs</strong>, each weighs about <strong>15 kg</strong>.
It is required to cool them <strong>from 90°C to 30°C</strong> with using <strong>30°C air in 9 min</strong>.</p>
<p>EDIT 2:</p>
<p>As I was advised, I tried to calculate the SHC:</p>
<p><a href="https://i.stack.imgur.com/iSsdk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iSsdk.png" alt="enter image description here" /></a></p>
<p>Thanks for any ideas</p>
| |thermodynamics|cooling| | <p>Specific heat capacity of steel is 0.466. In other words it takes 0.466 joules to raise the temperature of one gram of steel one degree Celsius. So how hot is the steel initially. In your case 90 degrees to get down to 30 you must remove from 90,000 grams, or 2.5 megajoules.</p>
<p>Forced air cooling for example of electronics electronic components use fan cooling. Makers of fans fans and compressors must consider the relationship between fan or impeller speed in rpm (revolutions per minute) and volumetric air flow in CFM (cubic feet per minute).</p>
| 45198 | How to calculate the volume of air that will cool steel? |
2021-07-20T05:27:53.827 | <p>Usually the governing equation for a thermometer is given by,</p>
<p><span class="math-container">$$\frac{m c_v}{h\;A_s}\dfrac{dT(t)}{dt}+ T(t) = T_\infty$$</span></p>
<p>Now for a problem the thermal conductivity of the thermometer bulb is also given. I am not sure if I have to account for the thermal conductivity in the problem. How would conductivity affect the given equation?
To be more precise what would be the effect on time constant for the system as my end goal is obtaining the time constant.</p>
| |instrumentation| | <p>The first governing equation you give above uses the <a href="https://en.wikipedia.org/wiki/Lumped-element_model" rel="nofollow noreferrer">lumped-capacitance assumption</a>; in other words, the thermal conductivity is assumed to be high enough, or the diameter small enough, or the external heat transfer slow enough, that the entire bulb is essentially at a single temperature <span class="math-container">$T(t)$</span>. This is a fantastically useful simplifying assumption—as long as the <a href="https://en.wikipedia.org/wiki/Biot_number" rel="nofollow noreferrer">Biot number</a> is suitably small.</p>
<p>It sounds like you've now been tasked with analyzing the situation without making this assumption. With spatial temperature variation now an option inside the sphere, we have the <a href="https://en.wikipedia.org/wiki/Heat_equation" rel="nofollow noreferrer">conduction heat equation</a></p>
<p><span class="math-container">$$k\nabla^2 T(r,t)=c\rho\frac{\partial T(r,t)}{\partial t};$$</span></p>
<p><span class="math-container">$$-k\frac{\partial T(r=r_0,t)}{\partial r}-h[T(r=r_0,t)-T_\infty]=0,$$</span></p>
<p>where <span class="math-container">$k$</span> is the thermal conductivity, <span class="math-container">$\nabla^2$</span> is the <a href="https://en.wikipedia.org/wiki/Laplace_operator#Three_dimensions" rel="nofollow noreferrer">Laplacian</a> (in spherical coordinates, where I've assumed axisymmetry), <span class="math-container">$T$</span> is the temperature, <span class="math-container">$r$</span> is the distance from the sphere center, <span class="math-container">$t$</span> is time, <span class="math-container">$c$</span> is the specific heat, <span class="math-container">$\rho$</span> is the density, <span class="math-container">$r_0$</span> is the sphere radius, and <span class="math-container">$h$</span> is the heat transfer coefficient at the surface.</p>
<p>The first equation describes conduction inside the sphere, and the second provides a boundary condition that you'll need for the solution (in addition to the initial conditions).</p>
<p>Multiple heat transfer textbooks and handbooks describe the details of the solution, including Incropera & DeWitt, <em>Fundamentals of Heat and Mass Transfer</em>, §5.6, Radial Systems with Convection. The full solution contains an infinite series with a variety of time constants; the first (dominant) term has a time constant proportional to the time over the <a href="https://en.wikipedia.org/wiki/Fourier_number" rel="nofollow noreferrer">Fourier number</a>, the Fourier number being <span class="math-container">$kt/c\rho r_0^2$</span>.</p>
<p>Does this all make sense?</p>
| 45206 | Effect of conductivity on time constant |
2021-07-21T07:53:20.440 | <p>I am doing a experiment on various gear combinations and am wondering How I can measure the efficiency (Torque loss) of each gear pair), I have a input shaft connected to the first gear and is also connected to a motor which drives it, this gear meshes with the second gear and thus the second shaft. each gear combo will be tested at the same conditions, driven at 500, 1000, 1500 and 2000 RPM each for 5 minutes. I am wondering however how do I measure the efficiency of each gear pair, I do not have access to a torque meter and I am unsure what to use as the output load. could anyone help with this? thanks</p>
| |mechanical-engineering|mechanical| | <p>Use a pulley with spooled string on it dropping a weight on the input shaft for input torque, and a pulley spooling up a string with a weight on the output shaft for load torque. With pulley radius and mass you can calculate input and output torques. You then need an RPM sensor on one of the shafts and you can use the gear ratio to calculate the RPM on the other shaft.</p>
<p>If you want to use a motor for input torque or generator for output torque for continuous operation, constant speed testing, or variable input/output torque convenience, you need torque sensors. For each motor or generator you use, you need a torque sensor. You cannot get around not using torque sensors if you want the convenience of using a motor for input torque or generator for load torque. You cannot use calculations to avoid a torque sensor.</p>
<p>The whole point of using a falling mass on the input shaft pulley and a rising mass on the output shaft pulley is that you don't need torque sensors because you aren't using motors and generators which have unknown electrical-mechanical efficiencies which also vary under load and RPM.</p>
<p>But this pulley mass setup is inconvenient in a number of ways. The most obvious is that that you cannot run continuously and you need more fall height to accelerate to higher RPMs if you want to test at those RPMs.</p>
<p>But the biggest weakest of this setup is that you mostly cannot run at constant speed. Unless you have enough fall height and the appropriate balances of masses, your falling and rising masses won't reach equilibrium where they achieve a constant speed. The problem then is that the weight of the masses (multiplied by pulley radius) is not actually the torque. The falling mass applying the input torque will apply less tension to the string than the weight of the mass while the rising mass acting as the load torque will have more tension on the string than the weight of the mass.</p>
<p>That means that you need some way to measure instantaneous acceleration. Since you already need an RPM sensor it might not be that big a deal but it does mean your RPM sensor needs to be able to give discrete readouts at sufficient bandwidth so you can calculate instantaneous RPM. Then you can calculate the string tension which will give you the actual force on the pulley which can be used to compute torque.</p>
<p>Alternatively, you can provide enough fall distance and select your input and output weights appropriately so that the system does reach constant speed, and at that point you can take the RPM measurement and use the weights and pulley diameter to calculate input and output power.</p>
<hr />
<p>If testing gear setups that are very similar where differences might be small, it's going to be pretty tough either way if the only difference is material. Even motor-generator setups with torque sensors can be very finnicky.</p>
<p>It should be easier to measure small differences in efficiency if run more power through the gears which would increase the losses. A difference of 2W is difficult to measure. A difference of 100W is much easier. So run more power through it.</p>
<p>Another thing you should really do is cascade the same gears setup to increase the losses which makes them larger easier to measure. So if you intended to test one pair of gears A-B, repeatedly cascade that pair with itself A-B-A-B-A-B-A-B-A-B which will make the losses larger and easier to measure.</p>
<hr />
<p>If you make a clutch with a lever arm (called a prony brake) you can connect the arm of the prony break to a fixed surface using a linear force scale 90 degrees to the arm. With arm length and the force scale you can calculate torque. So this would allow you to measure output torque without pulleys and string. I found prony brakes are surprisingly difficult to make though. It's difficult to get a consistent smooth interface with constant pressure without stiction so it runs smoothly for a stable reading.</p>
<p>But I know of no equivalent device to allow a linear force gauge to be used to measure the rotation of a motor for input torque.</p>
<p>Alternatively what you could do is just hang a weight off the output pulley and then slowly pour sand into a bucket hanging off the input pulley until the output weight starts to rise. That would give you a measure of the static friction rather than the dynamic friction of the gears when running, but it is much easier to obtain. Of course, you can also do the reverse where you hang a weight off the input shaft and remove sand from a bucket hanging off the output pulley until it starts to rise. If doing this, don't discount tapping the gearbox a bit to determine how close you are to the threshold. Friction torque does not not necessarily stay constant though in a gearbox and may vary with RPM and load torque (just like motors and generators do).</p>
| 45224 | How can I measure gear efficiency? |
2021-07-21T22:15:46.607 | <p>I'm trying to model a basic ball and beam system using Euler-Lagrange Equation. My system looks something like this
<a href="https://i.stack.imgur.com/vHDH8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vHDH8.png" alt="Source: Modelling and Control of Ball and Beam System using Coefficient Diagram
Method (CDM) based PID controller" /></a></p>
<p>I have come up with this final Euler-Lagrange Equation:</p>
<p><a href="https://i.stack.imgur.com/i3Ytp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i3Ytp.png" alt="enter image description here" /></a></p>
<p>Where <em><strong>J<sub>B</sub></strong></em> is the ball's moment of inertia, <em><strong>r</strong></em> is the radius of the ball, <em><strong>m</strong></em> is the mass of the ball, <em><strong>g</strong></em> is the acceleration constant, <em><strong>β</strong></em> is the ratio of <em><strong>d</strong></em> to <em><strong>L</strong></em>, <em><strong>r<sub>B</sub></strong></em> is the position of the ball along the beam, and finally <em><strong>θ</strong></em> is the gear angle.</p>
<p>The Euler-Lagrange Equation was acquired after finding the partial derivatives with respect to <em><strong>r<sub>B</sub></strong></em>.</p>
<p>My question is: How do I proceed with finding the transfer function? I have seen two research papers straight away cancelling the term with <span class="math-container">$\dot{θ}$</span> from the equation, changing the equation to Laplace domain, and finding <em><strong>r<sub>B</sub></strong></em> to <span class="math-container">$θ$</span>. I'm assuming that this was done due to an assumption, but I'm unable to figure out what this assumption is. Is this a correct way to do it?</p>
<p>Alternatively, can I proceed with leaving the <span class="math-container">$\dot{θ}$</span> term and changing it to the Laplace domain, and again find the transfer function from there?</p>
<p>Also, the research paper has proceeded with finding the transfer function from the Euler-Lagrange equation taken by finding the partial derivatives with respect to <em><strong>r<sub>B</sub></strong></em>. What about finding it with respect to <span class="math-container">$θ$</span>?</p>
<p>I'm a bit confused, so I'd appreciate some clarification. Thanks!</p>
| |control-theory|modeling|transfer-function| | <blockquote>
<p>How do I proceed with finding the transfer function?</p>
</blockquote>
<p>The concept of (Laplace transform) transfer functions are applicable for <strong>linear</strong> systems. So, you need to <a href="https://en.wikipedia.org/wiki/Linearization#Uses_of_linearization" rel="nofollow noreferrer"><em>linearise</em></a> your equations of motion. Linearisation uses the concept of <a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow noreferrer">Taylor series</a>.</p>
<p>In the example given in your question, that reduces to removing the term <span class="math-container">$\dot{\theta}^2$</span>. This comes from the assumption used for linearization.</p>
<blockquote>
<p>but I'm unable to figure out what this assumption is.but I'm unable to figure out what this assumption is.</p>
</blockquote>
<p>The assumption is that values of <span class="math-container">$\dot{r},\ \ddot{r},\ \theta,\ \dot\theta,\ \ddot\theta$</span> are <em>small</em> such that
<span class="math-container">$\dot{r}^2,\ \ddot{r}^2,\ \theta^2,\ \dot\theta^2,\ \ddot\theta^2, \ \dot{\theta}\dot{r},\ \theta\dot\theta$</span> etc.
are negligibly small compared to <span class="math-container">$\dot{r},\ \ddot{r},\ \theta,\ \dot\theta,\ \ddot\theta$</span>.</p>
<blockquote>
<p>Alternatively, can I proceed with leaving the θdot term and changing it to the Laplace domain, and again find the transfer function from there?</p>
</blockquote>
<p>Just to clarify, the papers (I think) are not cancelling <span class="math-container">$\dot{\theta}$</span> terms, they are cancelling terms which are the product of two or more "<em>small-valued</em>" variables; e.g. <span class="math-container">$\theta\dot{\theta}$</span> or <span class="math-container">$\dot{\theta}^2$</span> or <span class="math-container">$\theta^2$</span>.</p>
<blockquote>
<p>What about finding it with respect to θ</p>
</blockquote>
<p>I assume that the <em>input</em> to your system is <strong>defined</strong> as <span class="math-container">$\theta$</span>. If the input to the system was something else, like torque to the disc, then, you are right. The partial derivative with respect to θ does need to be taken.</p>
<h2 id="edit">edit</h2>
<blockquote>
<p>what if "θ˙^2" has a considerable value, such that the equation cannot be linearized. How would it be approached?</p>
</blockquote>
<p>It is better to follow a text book for knowing the proper way to do this. I will attempt to illustrate the specific case below. I follow the <a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow noreferrer">Taylor series</a> expansion as mentioned above.</p>
<p>Let the <strong>steady</strong> value of <span class="math-container">$\dot{\theta}$</span> be <span class="math-container">$a$</span>.
Then,
<span class="math-container">$$
\dot{\theta}^2_{(a+\delta_\dot{\theta})}
=
a^2 + \frac{2\dot{\theta}|_{a}}{1!} \delta_{\dot{\theta}}
+ \frac{2}{2!} \delta^{2}_{\dot{\theta}}
=
\color{blue}{a^2} +
\color{red}{
\frac{2 a}{1!} \delta_{\dot{\theta}}
}
+ \frac{2}{2!} \delta^{2}_{\dot{\theta}}
$$</span></p>
<p>The term in blue is part of the steady state and is usually not accounted for <em>deviations from steady state</em> which we want to analyse. The term in red is the linear part which we will use for further analysis of deviations from steady state. The variable <span class="math-container">$\dot\theta$</span> is now replaced by its <em>variation</em> from steady state value <span class="math-container">$a$</span>. The new variable which will appear in the linear analysis is denoted <span class="math-container">$\delta_{\dot{\theta}}$</span>. However, a lot of textbooks will not introduce a new symbol for this purpose. They will reuse the original symbol <span class="math-container">$\dot{\theta}$</span> and the reader needs to understand from context, if they are referring to the original variable or the variation from its steady state value. Note that the term is red is linear in the new variable <span class="math-container">$\delta_{\dot{\theta}}$</span></p>
<p>The variation <span class="math-container">$\delta_{\dot{\theta}}$</span> is assumed to be a small value. If the variation from steady state value is also not small, then linear analysis is not suitable and one needs to do non linear analysis and transfer functions cannot be used (IMO).</p>
| 45235 | Modelling of Ball and Beam System |
2021-07-22T12:49:09.490 | <p>The situation is we have to prepare a gas mix of air and CO<sub>2</sub> which has to be given as input from 0.1% to 10% CO<sub>2</sub> of air mix to the aqueous medium in a bioreactor?</p>
<p>I'm trying to build a simple and cheap DIY bioreactor. So how do we mix pure CO<sub>2</sub> from a CO<sub>2</sub> canister with air from a compressor or from a compressed air cylinder at the right flow rates so that we can achieve a desired air mix with CO<sub>2</sub> supplied at 1 LPM to 30 LPM?</p>
<p><em><strong>More particularly, what is the cheapest and best volumetric or mass flow meters and CO<sub>2</sub> detectors that can be sourced and put together for this purpose? Also, what are the parameters to consider for gas mixing calculation?</strong></em></p>
<p>The need is to have a satisfactory and approximate accuracy in terms of gas mixing and CO<sub>2</sub> detection. I have also added a simple diagram of this gas mixing system for reference.</p>
<p><a href="https://i.stack.imgur.com/PMXNf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PMXNf.jpg" alt="Gas mixing" /></a></p>
<p>It'd be great to get help from the community. Please do suggest if any other possible ways exist as well.</p>
| |mechanical-engineering|fluid-mechanics|airflow|process-engineering|prototyping| | <p>There are a lot of ways. To cut cost, reduce your expensive sensors to a bare minimum - probably just measuring what goes into the reactor.</p>
<p>Controlling by mass tends to be expensive compared to by volume or pressure. Ideal gas law is your friend here. Pressure times volume is proportional to number of particles times temperature. It's pretty inexpensive to bring two gasses to the same temperature, especially if you let them mix, but even if you just have their lines adjacent, they'll get there with enough time at sufficiently low flow.</p>
<p>In your diagram, if you theoretically just have valves instead of the mass flow meters, you could adjust them until you reach the desired output and then connect the reactor, wasting some gasses initially. From then, you would only adjust one valve at a time carefully and slowly bringing the setup closer to a target condition.</p>
<p>There's a more interesting method involving the ideal gas law I mentioned. You start with a "tank" of air (~.036% CO<sub>2</sub> by volume) of known volume and pressure. You add CO<sub>2</sub> from the cylinder until it reaches the percentage you want. By the ideal gas law (approximation), you can detect when you reach this desired percentage by measuring the temperature (pretty inexpensive) and pressure (need control that anyway)! You can control it with a pressure regulator like one for pumping tires. Final step is to dump this formulated mix to a desired location. This is where the "tank" being a pneumatic actuator instead helps - just pressurize the other chamber and move the mix onward either to the reactor (cheap) or a holding tank (you should just use it as the "tank" if that's the case, and add air or CO<sub>2</sub> based on the sensor). At the extra cost of more check valves, you can even use that other chamber as the start of the next mix (probably cheap enough to toss it though). With a programmable controller in charge of it all, you can have it adjust the target pressures on the fly based on the CO<sub>2</sub> sensor reading from the previous mix.</p>
<p>Should make for a good diy (or high school science lab) project.</p>
| 45249 | How to mix pure CO2 with air so that a bioreactor can be fed with air at 0.1% to 10% CO2? |
2021-07-22T18:51:39.700 | <p>I'm looking for a generic term/process that explains how we can make a closed- loop control system.</p>
<p>The typical elements of a closed loop systems are controller, actuator, plant and sensors.</p>
<p>If you consider a closed loop system with negative feedback loop, it feeds the error to the controller by subtracting the out-put from the desired value.</p>
<p>I'm interested to know how does exactly it happen in real world example?</p>
<p>As far as I know the feedback loop provides an electrical signal, the value that we provide ( the desired value ) is just a string of bits and bytes, so there should be something here right before the controller to do the subtraction operation.</p>
<p>How dose exactly this whole process happen ?</p>
<p>I know engineers design/simulate the whole thing in MATLAB/Simulink or ... . But I don't know the process that turns the simulation into a real ( physical ) final product.</p>
| |control-engineering|systems-engineering|feedback-loop| | <blockquote>
<p>I'm looking for a generic term/process that explains how we can make a closed-loop control system.</p>
</blockquote>
<p>There's a pretty good definition over on <a href="https://www.electronics-tutorials.ws/systems/closed-loop-system.html" rel="nofollow noreferrer">Electronics Tutorials</a>:</p>
<blockquote>
<blockquote>
<p><em>A Closed-loop Control System, also known as a feedback control system is a control system which uses the concept of an open loop system as its forward path but has one or more feedback loops (hence its name) or paths between its output and its input. The reference to “feedback”, simply means that some portion of the output is returned “back” to the input to form part of the systems excitation.</em></p>
</blockquote>
</blockquote>
<blockquote>
<blockquote>
<p><em>Closed-loop systems are designed to automatically achieve and maintain the desired output condition by comparing it with the actual condition. It does this by generating an error signal which is the difference between the output and the reference input. In other words, a “closed-loop system” is a fully automatic control system in which its control action being dependent on the output in some way.</em></p>
</blockquote>
</blockquote>
<blockquote>
<p>I'm interested to know how does exactly it happen in real world example?</p>
</blockquote>
<p>Here's an example which is purely mechanical. No electronics, no bits and bytes and no Matlab.</p>
<p><a href="https://i.stack.imgur.com/ohU6c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ohU6c.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. A mechanical governor. The vertical shaft is driven by the engine and the faster it goes the more the weights are thrown outward and upward (against gravity) causing the lever arm to reduce throttle. Source: <a href="https://en.wikipedia.org/wiki/Centrifugal_governor" rel="nofollow noreferrer">Centrifugal governor</a>.</em></p>
<p>These governors are used on steam and diesel engines such as on diesel generators which must run at constant speed to maintain mains frequency. Note that the feedback is negative. Increase in feedback causes the fuel to be reduced.</p>
<blockquote>
<p>The typical elements of a closed loop systems are controller, actuator, plant and sensors.</p>
</blockquote>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Element</th>
<th>Component</th>
</tr>
</thead>
<tbody>
<tr>
<td>Plant</td>
<td>The engine.</td>
</tr>
<tr>
<td>Controller</td>
<td>The governor and lever.</td>
</tr>
<tr>
<td>Actuator</td>
<td>The throttle valve.</td>
</tr>
<tr>
<td>Sensors</td>
<td>The centrifugal weights.</td>
</tr>
</tbody>
</table>
</div>
<p>Have a look at <a href="https://www.youtube.com/watch?v=CXogtrK4g5s&ab_channel=sparepartsuppliers" rel="nofollow noreferrer">Diesel engine speed governor explained</a> to see a US military training film for more on this.</p>
<p>If you can grasp the principle of operation of this it should be easy enough to translate the principles to an electronic control.</p>
| 45251 | How does feedback implementation look like in a real word closed- loop process? |
2021-07-24T00:00:43.940 | <p>I saw this today: <a href="https://i.stack.imgur.com/BYdKl.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BYdKl.png" alt="enter image description here" /></a></p>
<p>It's not the first time I see it, but I wonder what this is about. Why do they have three big what I assume are camera lenses? And what's the two smaller "spots" in the same area?</p>
<p>Are they for making some kind of "3D" view or what? If so, why three lenses instead of two?</p>
| |consumer-electronics| | <p>The answer depends on what phone this is exactly but here is my best guess for each spot:</p>
<ul>
<li><p>The three big lenses are to provide different focal lengths.</p>
</li>
<li><p>The white spot is an LED for flash.</p>
</li>
<li><p>The smaller dark spot is possible the LiDAR.</p>
</li>
</ul>
<p>The LiDAR would indeed help with 3d scanning and providing your phone with the data needed for augmented reality.</p>
| 45271 | Why do modern phone cameras apparently have three or even more different camera lenses in almost the same spot? |
2021-07-24T12:15:37.580 | <p>Recently I came across a FEM code of a linear deformation beam element, and it made me wonder what is the correct relation between shear stress and shear strain in a beam element.</p>
<p>In the beam element of the image below, it seems that shear strain is defined as <em>Shear Strain</em> = <span class="math-container">$\delta_{v}/L$</span>, with L being the length of the element and (if I am not wrong) the shear deformation, <span class="math-container">$\delta_{v}$</span>, be calculated from <span class="math-container">$$\delta_{v}=(\delta_{1}-\delta_{2})+(\phi_1+\phi_2)\frac{L}{2},$$</span> <span class="math-container">$\delta_{i}$</span> and <span class="math-container">$\phi_i$</span> being the vertical displacement and the rotations of the ends,</p>
<p>in this case, is the <em>Shear Strain</em>, the <strong>engineering shear strain, <span class="math-container">$\gamma$</span></strong> or is it the <strong>tensor shear strain, <span class="math-container">$\varepsilon_{xy}$</span></strong>?</p>
<p><a href="https://i.stack.imgur.com/aSArd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aSArd.png" alt="enter image description here" /></a></p>
<hr />
<p><strong>A comment on the answer:</strong></p>
<p>As stated by a comment below, the definition of shear may not be same for the beam elements as the assumptions in classic mechanics of materials. It seems to me that the definition of engineering strain and tensorial strain becomes vague when it comes to beam elements.</p>
<p>Usually beams have free faces without stress, so the equal shear stresses assumed for an infinitesimal element with small deformation concept is not applicable here. That is the shear stresses change along the height of the beam and thus we need a coefficient to account for these changes. A common solution is that the cross-section area is modified by a beam shear coefficient and a shear area is defined:
<span class="math-container">$$A_s = coef.A$$</span>
So now the shear stiffness, that depends on the area, is also modified by coef.</p>
<p>On less common cases where a similar element is used to simulate a volume (like a lattice network model), I see that they use 2G instead of G (for example see equation (22) in <a href="https://doi.org/10.1016/0029-5493(78)90217-0" rel="nofollow noreferrer">T. Kawai 1978</a>). This is like the idea that they are assuming that the Shear Strain is tensorial shear strain thus 2G should be used instead of G. But another interpretation can be that the coef. in this case is assumed equal to 2 since the beam does not have free surfaces.</p>
<p><a href="https://i.stack.imgur.com/xfcdA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xfcdA.png" alt="enter image description here" /></a></p>
<p>(above image adopted from <a href="http://solidmechanics.org/text/Chapter3_2/Chapter3_2.htm" rel="nofollow noreferrer">here</a>)</p>
| |structural-engineering|beam|finite-element-method|shear|deformation| | <p>Shear strain is defined as the angular deformation caused due to parallel or shearing force. In structural engineering, there are two cases of shear strains that are of particular concern. The occurrence of each depends on the loading that causes shear deformation as indicated below.</p>
<p><strong>1) Shear Strain due to Pure Shear</strong></p>
<p><a href="https://i.stack.imgur.com/6Sm4a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Sm4a.png" alt="enter image description here" /></a></p>
<p><strong>2) Shear Strain due to Shear Deformation</strong></p>
<p><a href="https://i.stack.imgur.com/Vjevq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vjevq.png" alt="enter image description here" /></a></p>
<p>Note, shear deformation becomes a concern when L/d (Span/beam depth) <span class="math-container">$\leq$</span> 10. Otherwise, it is usually ignored.</p>
<p><strong>Edit:</strong></p>
<p>For typical beams that follow the beam theory (Euler-Bernoulli or Timoshenko), as stated above, the shear deformation is very small thus usually ignored. The measured strain is, therefore, only consisted of the horizontal stretching and shortening of the extreme fibers in the direction of the longitudinal axial axis, which I think is your case.</p>
<p><a href="https://i.stack.imgur.com/4IMQJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4IMQJ.png" alt="enter image description here" /></a></p>
| 45277 | Is the calculated shear strain in a beam, engineering strain or tensorial strain? |
2021-07-24T22:59:20.190 | <p>I have a camera that uses a USB cable to connect to my computer. I want to extend the cable from 3ft to about 6ft. Originally I just created my own cable, soldering 4 26AWG wires to a blank USB connector and connecting the other side to the correct leads on the camera (this cable shown on right in the photo). The original wire (shown on the left in the photo) has slightly smaller wires probably either 28AWG or 30AWG.</p>
<p>When I plugged the new cable in, the camera would not come up and my computer said it was drawing too much power. So I tried just extending the original cable by just splicing additional wire before the connector. Once again I used they 26AWG wire.</p>
<p>With the extended original cable, the camera still does not show up, and it also does not give the warning that it is drawing too much power.</p>
<p>I am not sure what is happening. Given that the original cable extended does not work I am guessing that either the increase in wire gauge or the added length is causing an issue, but I do not understand why. If anyone has context on this it would be greatly appreciated!</p>
<p><a href="https://i.stack.imgur.com/tTu5Q.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/tTu5Q.jpg" alt="enter image description here" /></a></p>
| |electrical-engineering|electrical|wire| | <p>For the USB signal, you have to make a 90 Ohm characteristic impedance twisted pair. The characteristic impedance depends on the insulator and on the outer/core diameter ratio of the wires. Search online calculators on the net.</p>
<p>A simple way to twist wires:</p>
<ul>
<li>connect the two wires making a long wire</li>
<li>twist the wire using an electrical drill</li>
<li>put the wires parallel keeping the wires taut</li>
<li>let them twist together</li>
</ul>
<p>It is also possible to twist the two wires independently in the same direction, before letting them twist together.</p>
<p>Also, the max. length of USB signal twisted pair is around 5 m.</p>
| 45284 | Extending USB cable makes it not work |
2021-07-25T16:37:23.963 | <p>I would like to build a rack for my home with a pull up bar. I'm leaning towards a steel pipe with a diameter of 42.4 mm, but have no idea how to determine what wall thickness is needed so that the pipe doesn't bend under load.</p>
<p>This is a first sketch of the rack. Don't mind the rest of the structure, as most of it is subject to change. The idea is based around the bar, so any change to the bar would change the structure completely. You can think of it as though the bar rests on an idealized structure.</p>
<p><a href="https://i.stack.imgur.com/0ggjN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0ggjN.png" alt="First sketch" /></a></p>
<p>The bar is 2190 mm and the unsupported part is 2000 mm long. What I'm wondering is what wall thickness is needed for the bar to not permanently bend?</p>
<p>I found <a href="https://engineering.stackexchange.com/questions/40292/estimating-dynamic-load-from-human-body">this post</a> discussing what dynamic loads are expected in a similar scenario. One answer says to calculate with 8.9 kN for a person weighing 100 kg with 1 m long arms. I weigh about 70 kg and could travel about 0.60 m (from highest to lowest point). If I do the same calculations I get a load of about 5.5 kN.
This scenario would probably not be applicable though. The idea is to use it to build strength rather than power, so a static load of about 1.4 kN (2X bodyweight) might be more suited.</p>
<p>While writing this question I realized that I could also go for a 28 mm steel rod, if it would be cheaper or handle more load for the same price.</p>
<p>Also, I'm thinking it might be best not to fasten the bar in order to reduce stress and only limit its horizontal movement, so it cannot slide/roll out of position. Is that a good idea?</p>
| |mechanical-engineering|structural-engineering|materials| | <p>I will add my answer to the others although the results will not be much different.</p>
<p>The way I see it the following parameters are the most important:</p>
<ul>
<li><strong>The worst case scenario for the bending of the beam</strong>:</li>
</ul>
<p>Instead of the open handle (which is a four point bending), I would go for the close handle (which is three point bending). This is a worst loading scenario for a given force.</p>
<p><a href="https://i.stack.imgur.com/4aj1n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4aj1n.png" alt="enter image description here" /></a></p>
<p><strong>Figure 1: Three point bending load case with shear forces and bending moments</strong></p>
<ul>
<li><strong>Maximum force</strong>:</li>
</ul>
<p>The idea behind the calculation in <a href="https://engineering.stackexchange.com/questions/40292/estimating-dynamic-load-from-human-body">the question you are referring</a> is what would happen if someone pulled the chin up and then let go only to hang. IMHO, this is still a valid scenario in your case. Because the intended use is only for strength, that does not preclude an impact load (the equivalent in a car design would be to design cars only for transporting and for crashworthiness). Therefore, I would still go for the 5.5kN (which is BTW approximately 500 kg).</p>
<ul>
<li><strong>Use yield stress instead of uts</strong>:</li>
</ul>
<p>Now this is the hard part, since you probably will be able to get easier steels rod with better strength properties rather than steel pipes. The value would be depended on the surface/heat treatments that the rod has been subjected to, so you need to have an idea. In any case you can take a small value and compare.</p>
<h2 id="comparison-of-42.4mm-diameter-pipe-and-28-m-steel-rod-tijv">comparison of 42.4mm diameter pipe and 28 m steel rod.</h2>
<p>In order to compare those to, the most basic comparison is check for their second moments of area. Since you don't know the thickness I will assume its t. Then:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">pipe</th>
<th style="text-align: center;">rod 28</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">pipe 42.4mm with t</td>
<td style="text-align: center;">rod 28[mm]</td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$$\frac{\pi}{64}(d_p^4 - (d_p-2t)^4)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$\frac{\pi}{64}d_r^4 $$</span></td>
</tr>
<tr>
<td style="text-align: center;"><span class="math-container">$$I_p=\frac{\pi}{64}(42.4^4 - (42.4-2t)^4)$$</span></td>
<td style="text-align: center;"><span class="math-container">$$I_r = \frac{\pi}{64}28^4 $$</span></td>
</tr>
</tbody>
</table>
</div>
<p>The structure capable of enduring more bending is the one with the greater I. Therefore, <strong>pipe would be better than rod if</strong>
<span class="math-container">$$ I_p>I_r$$</span>
<span class="math-container">$$\frac{\pi}{64}(42.4^4 - (42.4-2t)^4) > \frac{\pi}{64}28^4 $$</span></p>
<p>After some calculations we obtain that <strong>the pipe will be stronger if the thickness is greater that <span class="math-container">$ t>1.15 [mm]$</span></strong>. (Of course, if you use so small a thickness, then the pipe would probably crumble due to localised bending). So in terms of geometrical resistance, the pipe will probably behave better although I would require it to have a thickness of at least 8[mm] (which I doubt you will have.</p>
<p>Additionally, as I mentioned earlier, it is probable that you will <em>easier</em> find a rod with better material properties.</p>
<h2 id="calculation-for-simply-supported-7mdn">Calculation for simply supported</h2>
<p>Because, I believe the 28[mm] rod is going to be easier on the grip, I will proceed with this calculation and provide the minimum yield stress for the beam.</p>
<p>I assume that the maximum force of F=5.5[kN] is correctly calculated and I will use that as maximum load.</p>
<p>Then the maximum bending for the 3 point bending will be equal <em>for a simply supported beam</em> would be equal to:</p>
<p><span class="math-container">$$M=\frac{Fl}{4}= \frac{5.5 [kNm]}{4}= 1.125[kNm]$$</span></p>
<p>Then maximum stress would be equal to:</p>
<p><span class="math-container">$$\sigma = \frac{M}{I_r}\frac{d_r}{2}=\frac{M}{\frac{\pi}{64}d_r^4}\frac{d_r}{2}=\frac{32 M}{\pi d_r^3}$$</span>
<span class="math-container">$$\sigma = \frac{1125[Nm]}{2.248\cdot 10^{-6} [m^3]}$$</span>
<span class="math-container">$$\sigma = \frac{1125[N]}{2.248\cdot 10^{-6} [m^2]}$$</span>
<span class="math-container">$$\sigma = 522 [MPa]$$</span></p>
<p>This is quite a high strength for yield (although you probably will be able to get a material with those values).</p>
<h2 id="improvement-with-fixed-ends-fs2n">improvement with fixed ends</h2>
<p>The above calculation is for simply supported beam. However, if you <strong>welded</strong> the additional 190mm of the beam to the structure, you would have a fixed beam with a concentrated load. In that case the loading condition would improve. I.e.</p>
<p><a href="https://i.stack.imgur.com/5wFua.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5wFua.png" alt="enter image description here" /></a></p>
<p><strong>Figure 2: Beam fixed at both ends (source: awc</strong></p>
<p>In that case the maximum bending moment would be <span class="math-container">$M_f= \frac{F l}{8} = \frac{M}{2}$</span>. So the bending moment would be <strong>only half of the simply supported case.</strong>. As such the minimum yield stress would need to be greater or equal to:</p>
<p><span class="math-container">$$\sigma_y \ge 261 [MPa]$$</span></p>
<p>(<strong>CAVEATS</strong>: there are some asterisks here, regarding with the rigidity of the rest of the structure and stress concentrations, however in general the welded version should perform better). This -hopefully- also answers your final question regarding constraining only horizontally.</p>
| 45292 | What thickness of steel bar/pipe for home gym? |
2021-07-26T14:07:17.613 | <p>CSR - Carbon composition resistors.</p>
<p>P1 :About tolerance resistors , what I have understood is that if a 220 Ω resistor has a color band of silver. Then , it’s tolerance = ± 22%. So , resistance of the CSR or the obstruction to the flow of electrons will be in the range of 242 or 198 Ω . Any one number value in this range.
From the values 198 to 248 ohm. What does this mean ? Like resistance is obstruction to the flow of electrons. Formula is <span class="math-container">$R= $$\frac{V}{I}$</span>. Now , the value of V and I should be such that resistance value is always between 198-242 ohm ? It cannot exceed or be lower than this range limit values.</p>
<p>Example in my textbook:
<a href="https://i.stack.imgur.com/C4PK5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C4PK5.jpg" alt="enter image description here" /></a>
P2: Also , in a CSR. Is it specified that from which direction the flow of electron is going to start. Left or right. Since online it says that resistors are blind to polarity.</p>
<p>I would like to confirm if for <span class="math-container">$1st$</span> part , my understanding is correct or not ?</p>
<p>For <span class="math-container">$2nd$</span> part , I am not getting what is the reason and the right answer.</p>
| |electrical-engineering| | <p><strong>From the values 198 to 242 ohm. What does this mean?</strong></p>
<p>If you measure the resistance of the resistor, it should measure a specific value in this range at room temperature, if it is within tolerance. This value can change as the resistor warms up. The value is determined by the manufacturing process.</p>
<p>A batch of resistors would have values falling around the color coded value and tolerance, with few measurements at the extremes. Values should follow a bell curve type of normal distribution. Center will be around 220Ω and no binning for tolerance (not cost effective to test for tolerance on mass produced components).</p>
<p>Actual value will depend on ambient temperature of circuit, aging (value drifts over life - PPM/year), humidity and use and abuse over life cycle.</p>
<p><strong>Is it specified that from which direction the flow of electron is going to start. Left or right.</strong></p>
<p>Resistors have no inherent polarity, but depend upon direction of current going through them and if you are using Conventional Current or Electron Flow (which may be a part of your confusion).</p>
<p>[Conventional Current] Current goes into positive side of resistor. [Electron Flow] Current goes into negative side of resistor.</p>
<p><strong>Now, the value of V and I should be such that resistance value is always between 198-242 ohm? It cannot exceed or be lower than this range limit values.</strong></p>
<p>A damaged resistor can be outside the range. It doesn't mean the circuit will not work or fail, just that the resistor is outside tolerance.</p>
| 45303 | Tolerance resistors in CSR |
2021-07-27T13:57:43.837 | <p>Please have patience with me as I have not looked at this sort of theory in many years (I am also very new to StackExchange).</p>
<p>A question asks the following:</p>
<blockquote>
<p>A sloping swimming pool two-thirds full of sea water provides the supply for a fixed installation protecting a risk situated above the pool.</p>
<p>The pool measures 12m long, 6m wide and 1m at the shallow end sloping to 2.5m at the deep
end. A pipe is placed so that its inlet is at the bottom of the deep end of the pool. A pump
imparts pressure to the water in the pipe such that a mercury manometer indicates a pressure
difference of 681mm Hg between inlet and outlet. The inlet of the pipe is 80mm diameter; the
outlet nozzle is 45mm in diameter and is 5m above the inlet. The inlet velocity of the water is
5m/s, the density of sea water is 1050kgm-3 and the density of mercury is 13600kgm-3.</p>
<p>Use Bernoulli’s theorem to calculate the velocity of the water through the outlet nozzle?</p>
</blockquote>
<p>By applying the continuity equation the following is found:</p>
<p><span class="math-container">$v_1A_1=v_2A_2$</span></p>
<p><span class="math-container">$v_2=15.8m/s$</span></p>
<p>However, the solution is given as <span class="math-container">$v_2=10m/s$</span></p>
<p>If I apply Bernoulli’s equation:</p>
<p><span class="math-container">$$
\frac{P_2 -P_1}{ρ}+\frac{1}{2}(v^2_2-v^2_1)+g(z_2-z_1)=0
$$</span>
<span class="math-container">$$
\frac{-90792.5}{1050}+\frac{1}{2}(v^2_2-25)+9.81(5)=0
$$</span></p>
<p><span class="math-container">$$
v_2=10m/s
$$</span></p>
<p>This is the answer that the solution gives but continuity is not met as shown above. Is this question ill-posed?</p>
<p>I am very confused why continuity is not met when Bernoulli’s equation is used and why the two methods produce different results. A similar question was asked <a href="https://engineering.stackexchange.com/questions/21882/can-the-continuity-equation-and-bernoulli-contradict-each-other?newreg=2742a82bf3fe47b3ae34f4bb3d22880c">here</a> but this was for a system without a pump and I am not sure if it is applicable.</p>
<hr />
<p>My question is: Can Bernoulli’s equation be used when the system has a pump or is it no longer valid due to energy not being conserved?</p>
| |mechanical-engineering|fluid-mechanics|bernoulli| | <p>No, there is a problem with the question. If the flow rate is set at the inlet, then that is the flow through the pipe, period. The pump doesn't add mass, just a motive force. Bernouli would apply if you were doing the flow calcs yourself. In this case, they gave you the answer at the beginning. The flow rate through the exit will be equal to the flow rate through the inlet no matter what the pump did or how high the exit is.</p>
<p>Bernouli probably doesn't work because they made up the pump head.</p>
| 45318 | Is Bernoulli’s equation applicable for systems with pumps? |
2021-07-27T14:09:23.253 | <p>Hey guys I am trying to model the performance of a hypothetical capacitor.</p>
<p>Capacitance depends on its geometry, but the charging and discharging of the capacitor depend on a resistive load within the circuit.</p>
<p>I am trying to see the performance of this capacitor when used to satisfy the energy demand of a household or power grid. However I am not sure as to how I can provide an approximate resistance value for a household or city load.</p>
<p>Thank you guys for your help.</p>
| |electrical-engineering|simulation|power-electronics|battery| | <blockquote>
<p>Capacitance only depends on its geometry, ...</p>
</blockquote>
<p>False. Capacitance is also determined by the dielectric (the insulator) between the plates and this can have a huge effect on the value obtained.</p>
<blockquote>
<p>... but the charging and discharging of the capacitor depend on a resistive load within the circuit.</p>
</blockquote>
<p>The load doesn't have to be purely resistive but we'll go with "resistive" for now.</p>
<blockquote>
<p>I am trying to see the performance of this capacitor when used to satisfy the energy demand of a household or power grid. However I am not sure as to how I can provide an approximate resistance value for a household or city load.</p>
</blockquote>
<p>Use <span class="math-container">$ P = \frac {V^2} R $</span> which you can rewrite as <span class="math-container">$ R = \frac {V^2} P $</span> where <span class="math-container">$V$</span> is the city supply voltage and <span class="math-container">$P$</span> is the demand of the city in watts (W) (and you'll have to apply the correct conversion if going from kW, MW or GW).</p>
<p>Energy is measured in joules (J) which are equal to watt-seconds (Ws) and 3600 Ws = 1 Wh (watt-hour). 3,600,000 Ws = 1 kWh.</p>
<p>The energy stored in a capacitor is given by <span class="math-container">$ E = \frac 1 2 C V^2 $</span>. Note that the capacitor voltage will start to fall immediately you start drawing charge from it and from the formula above you should be able to see that 75% of the energy will be gone when it falls to half voltage.</p>
<p>You now have enough information to start doing some calculations. Try this: What value of capacitance do you require to run a 1 kW load at, say, 200 V (chosen to make the maths easy) for 10 hours?</p>
<hr />
<p>Just checking, but you <em>are</em> aware that capacitor storage is DC and that city supply is AC, aren't you?</p>
| 45320 | Modeling a simple capacitor |
2021-07-27T18:29:20.143 | <p>I'm an intern at an electrical utility and I've been tasked with doing a proposal/cost analysis to evaluate whether or not starting a pilot with a company specializing in robotic inspections is worth it. The pilot, if it goes through, would consist of a tethered drone conducting an infrared/thermography scan and a LiDAR point-cloud (light detection and ranging) scan, in addition to a high-resolution RGB scan of the manhole, with permitting and de-watering as rather significant additional costs. The economics are not favorable, and there are some questionable aspects of the entire process, mainly</p>
<ol>
<li>The permitting, de-watering of the manhole, 4-gas monitor air sampling, and lid removal/re-installation is still necessary procedure for the process, even if a tethered drone as opposed to human personnel will make entry (thus an additional cost requiring two teams; the crew that removes the lid will still have to wait around for the drone inspection crew to finish, all the while still charging for the time spent doing nothing. The drone inspection company claims they have a solution to this by going to other manholes while one is being de-watered, but I don't know how this will actually work in practice.)</li>
<li>Union arguments/conversations about a future where they aren't in the picture or something of the sort</li>
<li>The fact that the data must be analyzed off-site, so a problem can't be remediated until the data is analyzed and then a return to the manhole structure is made.</li>
</ol>
<p>What I'm wondering is: How effective is LiDAR in practice? Does it enable inspections to be done faster? If so, how much faster? I know that underground inspections are prone to human error and missed information, but there is some information on LiDAR effectiveness that I can't seem to find when looking at research. Say if one has a manhole structure where a vault exists. Resolution of photos and ability to see 2nd or 3rd cable deep on a rack in the back office is critical to evaluate defects – particularly cables with cracks and no active oil leak. I don't see how a tethered vehicle is going to be able to position itself to find these defects. It makes it even more difficult to give an honest analysis because I haven't ever been inside a manhole myself.</p>
<p>If anyone could speak towards their experience with IR/thermography and LiDAR point-cloud scan effectiveness in manhole structures, your advice will be greatly appreciated.</p>
| |electrical-engineering|structural-engineering|infrastructure| | <p>There's nothing like kicking the dirt. I would advise actively participating in a manhole inspection, at least once so you know what it like. Do an inspection with an experienced person. By doing so you will have first hand experience of what is involved and what can and can't be easily seen or done via such inspections. Doing more such inspections would be better because there is always something missed & seen with each inspection. With more inspections you acquire more knowledge and experience. As a refresher do one inspection a year.</p>
<p>Scanning underground voids via laser technology is nothing new. It's been <a href="https://www.maptek.com/forge/june_2013/scanning_underground.html" rel="nofollow noreferrer">available to the mining industry</a> for 15 to 20 years, though it is little used because of the cost and having to amend schedules and work practices. In some instances it has been used to scan a void and then project digital photographs onto the inside surface of the scanned shape so as to produce a 3D "walk/fly through" representation of the void.</p>
<p>Such images are good at presenting an overall representation of the surface of a void. They won't tell you if a cable is red or blue and a small leak from the rear of a cable of pipe can be missed because the image is taken from one perspective. With human inspection, a person can reposition themselves, safely, to get a better view of something that could otherwise be missed.</p>
<p>Because of the way the technology works, making a lot of laser shots, a lot of data is produced that can take time to process. This could be alleviated by using fast computers and efficient optimized software. Some software can be very sloppily written, which can result in longer processing times. I question why it couldn't be done on site. If laptop computers are not good enough, then a trailer or van with high specification workstations could be used.</p>
<p>If the manholes have cables, pipes or ladders in them, they can produce shadow effects by being an obstruction to the laser beams resulting in parts of the wall of the shaft not being picked up. Falling water, even as a consistent stream of drops, may also be problematic.</p>
| 45324 | Effectiveness of LiDAR in manhole structures? |
2021-07-30T07:05:32.260 | <p>I have a sheet of metal that had grooves cut by laser in it; I need to analyze the depth of the grooves, and for that I think I will need to obtain a thin section slice of the metal and then put the sample under a microscope.</p>
<p>A microscope is not a problem to get, I have access to several. But my biggest issue here is how to obtain the section slice. Has anyone had experience doing something like that? I know that a microtome is used for obtaining very thin slices of biological samples, but not metals, I think. How do I slice the metal? Or is there perhaps a better, easier method of measuring the grooves' depth?</p>
| |materials| | <p>There may be non-destructive options, like laser profilometers or even optical microscopy that can extract Z axis data. Check the back side too.</p>
| 45345 | How do I get a thin section slice of grooved metal to analyze groove depth under a microscope? |
2021-07-30T11:50:54.687 | <p>This isn't mine, but I own one of these and I have no idea what it is. With the spike attached it looks like a marking gauge of some sort.</p>
<p><a href="https://i.stack.imgur.com/0v4vM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0v4vM.jpg" alt="enter image description here" /></a></p>
| |tools| | <p>It appears to be a base holding a scribe. Normally they are used to hold indicators (which are like touch sensors that measure how far the arm is deflected from their neutral position). You mount whatever you want on the post.</p>
<p>This particular style is apparently called a "surface gauge". I know them as universal bases, indicator bases, holders, etc. but those are apparently a different style. I've never seen one actually used with a scribe; Always an indicator.</p>
<p>Basically a base with an adjustable vertical post that you mount measuring devices to. The base is ground flat and you place it on a surface plate or machine table. You slide it around.</p>
<p>For example, you can check to see if two things are the same height by sliding an indicator between the two pieces and seeing of the dial reads the same, or if a horizontal beam/rod or surface is flat and parallel by sliding an indicator along the piece, or if something is round and by finding the high spot, zeroing the dial, then spinning it in the lathe or rotary table or indexing head and seeing if it the dial moves.</p>
<p>I am not sure what you are supposed to do with the scribe though. Seems super imprecise, has no give to climb onto higher surfaces...it just crashes and leaves marks, and no way to tell if you are touching a surface you have slid over or if it is imperceptibly lower. Way more useful with an indicator...and surface plate or machine table.</p>
<p>The base currently looks detached from the post and is just laid down and resting on top.</p>
<p><a href="https://i.stack.imgur.com/YSizI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YSizI.png" alt="enter image description here" /></a>
<a href="https://www.grainger.ca/en/product/p/STA57A?gclid=CjwKCAjwxo6IBhBKEiwAXSYBs9vS4mdogXc4W3YO7ABP4gNff7ve4xmEAf4NBTv-TPQ2MaCRPXG3FxoC-dgQAvD_BwE&cm_mmc=PPC:+Google+PLA&ef_id=CjwKCAjwxo6IBhBKEiwAXSYBs9vS4mdogXc4W3YO7ABP4gNff7ve4xmEAf4NBTv-TPQ2MaCRPXG3FxoC-dgQAvD_BwE:G:s&s_kwcid=AL!3645!3!483556496838!!!g!737429528392" rel="nofollow noreferrer">https://www.grainger.ca/en/product/p/STA57A?gclid=CjwKCAjwxo6IBhBKEiwAXSYBs9vS4mdogXc4W3YO7ABP4gNff7ve4xmEAf4NBTv-TPQ2MaCRPXG3FxoC-dgQAvD_BwE&cm_mmc=PPC:+Google+PLA&ef_id=CjwKCAjwxo6IBhBKEiwAXSYBs9vS4mdogXc4W3YO7ABP4gNff7ve4xmEAf4NBTv-TPQ2MaCRPXG3FxoC-dgQAvD_BwE:G:s&s_kwcid=AL!3645!3!483556496838!!!g!737429528392</a>!</p>
| 45348 | What's this tool? |
2021-07-30T22:28:19.913 | <p>I'm trying to compare a really tough polyurethane to baltic birch plywood to see if I can use it instead (a lot of subtractive processes to the plywood, want to eliminate it)</p>
<p>So what i'm looking for is how much weight it can carry per thickness and length
Currently i'm using 400mm X 100mm X 16mm blocks of plywood (then engrave it and so on)
And I know for a fact it can handle 100kg load with some bending but not breaking (weight in the middle)</p>
<p><a href="https://i.stack.imgur.com/upGFM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/upGFM.png" alt="enter image description here" /></a></p>
<p>This is the TDS of the material:
<a href="https://i.stack.imgur.com/Tf554.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tf554.png" alt="enter image description here" /></a></p>
<p>Is it possible to calculate?</p>
<p>Thanks!</p>
| |structural-engineering|structural-analysis|strength| | <p><strong>TL;DR: If the tensile strength for the other material is greater, then the material behaves better and you can use longer and thinner planks.</strong></p>
<h2>Assumptions</h2>
<p>For the sake of simplicity, I will assume that :</p>
<ul>
<li><p>the breadth of the plank <span class="math-container">$b$</span> is a constant for both cases, and only the thickness <span class="math-container">$h$</span> and the distance between supports <span class="math-container">$L$</span> is changing.</p>
</li>
<li><p>the load is a distributed load <span class="math-container">$w$</span> along the length of the material . In that case the distributed load <span class="math-container">$w $</span> is in units of <span class="math-container">$N/m$</span> and it includes the load on the shelf and the weight per meter of the plank.</p>
</li>
<li><p>That you are interested only in strength. i.e. when the plank breaks.</p>
</li>
</ul>
<h2>calculations</h2>
<p>In that case the bending moment is
<span class="math-container">$M = \dfrac {wL^2}{8} $</span></p>
<p>The second moment of area for a rectangular cross-section is
<span class="math-container">$$I = \frac{bh^3}{12}$$</span></p>
<p>Then the maximum stress in the beam is:</p>
<p><span class="math-container">$$\sigma_{max} = \dfrac {M }{I }\cdot \dfrac {h }{2 } = \dfrac{12 w L^2 }{8 bh^2}$$</span></p>
<p><span class="math-container">$$\sigma_{max} = \dfrac{3 w L^2 }{2 bh^2}$$</span></p>
<p>The maximum stress <span class="math-container">$\sigma_{max} $</span> needs to be less than the allowable stress for the material <span class="math-container">$\sigma_{all}$</span> in order to avoid the failure.</p>
<p><span class="math-container">$$\sigma_{max} \le \sigma_{all} $$</span>
I'll save you the analysis, you end up with</p>
<p><span class="math-container">$$\dfrac{L }{h}\le \sqrt{\dfrac{2 b}{3 w }\sigma_{all}} $$</span></p>
<p>For this case, since comparison is required, the limiting case is taken ie. :
<span class="math-container">$$\dfrac{L }{h} = \sqrt{\dfrac{2 b}{3 w }\sigma_{all}} $$</span></p>
<p>From here you can see that (<strong>for the uniform loading case</strong>)the length L and the thickness <span class="math-container">$h$</span> have a competing relationship. I.e. when the length increases, the thickness also needs to increase proportionally, in order to fail at the same breaking load.</p>
<h2>Bottom line</h2>
<p>Higher values of the ratio <span class="math-container">$\frac{L}{h}$</span> means that you can have longer beams with thinner planks.</p>
<p>Assuming the load <span class="math-container">$w$</span> and the breadth <span class="math-container">$b$</span> is constant, then you can see that the ratio <span class="math-container">$\frac{L}{h}$</span> is proportional to <span class="math-container">$\sqrt{\sigma_{all}}$</span>. i.e.:</p>
<p><span class="math-container">$$\frac{L}{h} \propto \sqrt{\sigma_{all}}$$</span></p>
<p>where for the material you are considering (from the TDS) <span class="math-container">$\sigma_{all}$</span> is the tensile strength is between <span class="math-container">$41$</span> and <span class="math-container">$46$</span> [MPa].</p>
<p>If the tensile strength for the other material is greater, then the material behaves better and you can use longer and thinner planks.</p>
<h2>Caveats</h2>
<p>This is a simplified analysis and there are a few assumptions for convenience (which -IMHO- are realistic and do not invalidate the analysis).</p>
<ul>
<li><p>The first one is the type of load (uniform distributed load)</p>
</li>
<li><p>The second is that the plywood behaves a bit differently that the polyurethane (one exhibits composite material behavior, while the latter behaves more like a uniform material).</p>
</li>
</ul>
| 45356 | Calculating break point (bend) of a material |
2021-08-02T14:06:20.560 | <p>We have a bit of wall sticking out of a flat wall</p>
<pre><code> ____
__________| |_______________
</code></pre>
<p>I have always called it the sticky outy bit of the wall and everyone knew what I meant. My boss told me I need to find out the proper name for a bit of wall that sticks out because it is a point of reference in a document. I can't keep on calling it the sticky-outy bit that I've called it for the last 3 years.</p>
<p>I've been looking at <a href="https://www.house-design-coffee.com/architectural-dictionary.html" rel="nofollow noreferrer">https://www.house-design-coffee.com/architectural-dictionary.html</a>. There is the word architrave but it is for a horizontal section below the roof.</p>
<p>There is also pilaster but all the examples I've seen have fancy decorations on it. A sort of fake column. Can it be called a pilaster if there are no decorations?</p>
<p>The other alternative is protrusion - can this be used for something that sticks out from ceiling to floor or just a bit of wall.</p>
<p><strong>Edit</strong></p>
<p>To answer @Fred's query - it is about 1ft (30cm wide) and about 4" (10cm) deep. It stretches from ceiling to floor. I've got no idea what it is for. Sorry, I cannot supply a photograph - it is in one of those restricted areas where cameras aren't allowed.</p>
| |building-design|terminology| | <p>If both the wall and the bump-out are made of masonry/bricks, it is usually called a pilaster, which was built integrally with the wall.</p>
<p>If the bump-out is concrete, it is called a column.</p>
<p>Both have structural significance in carrying gravity loads and resisting lateral loads.</p>
<p>However, if it is made of drywall material, then it is usually a decorative column, or to hide the steel column behind.</p>
<p>Note, you can check whether there is a horizontal beam, at the upper end, lays perpendicular to the bump-out to determine if this is a structural column, or decorative protrusion.</p>
| 45393 | What is this sticky outy bit called? |
2021-08-02T16:36:52.023 | <p>I need a little guidance to select a motor for a lifting mechanism for a weight of 3 kg. I am going to use two motors and both motors will have a gear (attached to the shaft) very close to each other and in between the gears, a thick rope will be placed so that the gears will bite into the rope and then move up. Each of the motors needs to lift 1.5 kg (hence a total of 3 kg).
To JUST counter the downforce of 2 kg (20 N) I need two motors of 3 kg-cm so if I use two motors of 5 kg-cm (@150 rpm) the payload will start moving up? Please tell me if I am right or wrong. Thank you.</p>
<p>my calculations:
2 kg</p>
<p>down force 2 X 9.8 = 19.6 N ~ <strong>20 N</strong></p>
<p>radius = 3 cm</p>
<p>torque = 20 X 0.03 = 0.6 N-m = 6.114 kg-cm [1 Newton meter is equal to 10.19 kg-cm].</p>
<p>i.e Each motor capable of giving <strong>3</strong> kg-cm.</p>
| |motors| | <p>Most of the motors has this curve, let's map the axis as an example:
<a href="https://i.stack.imgur.com/Al7Du.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Al7Du.jpg" alt="Motor curve" /></a></p>
<p>map in x's axis, 8.7 -> 5kgf (1.67kgf at 3cm radius) -> 1.5kgf would be 90%.</p>
<p>map in y's axis 100rpm -> 150rpm (take 100 as percentage)</p>
<p>Supposing increasing 1% in torque is a decreasing of 1% in velocity (with axis as percentages of the maximum)</p>
<p>Then you would have 10% of the 150rpm maximum, it is 15rpm, with 3cm would be 15*2<span class="math-container">$\pi$</span>*0.03/60= 0.04712m/s upward.</p>
<p>With your motor capacities of 3kgf you would be using 100% of the torque, then 0% of velocity.</p>
<p>This is a linear approximation, the motors have a dead zone so be sure that your percentage velocity isn't inside that dead zone (in which the motor wouldn't move on because of static friction), I had a pololu motor with 1.8kgf it was 20% dead zone as reference specially if it has weight on it.</p>
| 45394 | Selecting motors to lift 3 kg |
2021-08-02T21:44:35.793 | <p>I am reading a paper and the dimensions for some electrodes are given, but I was wondering if I was reading it properly:</p>
<p><strong>30 × 15 mm<sup>2</sup></strong>, is this the same as: <strong>(30 × 15)mm<sup>2</sup></strong> ?</p>
<p>To track the published paper I provide below its Digital Object Identifier (DOI) commonly used in online research papers. This number should lead you directly to the paper.</p>
<p>DOI: 10.1002/aesr.202000093</p>
| |electrical-engineering|unit| | <p>If you are referring to the article <a href="https://onlinelibrary.wiley.com/doi/10.1002/aesr.202000093" rel="nofollow noreferrer">A Structural Battery and its Multifunctional Performance</a> then the dimensions appear to be incorrectly written.</p>
<blockquote>
<p><strong>Structural Battery Full Cell Preparation</strong> (Section 4)</p>
</blockquote>
<blockquote>
<p>An illustrative overview of the structural battery composite full cells manufacture is shown in Figure 1. The negative electrode was made from a CF spread tow and the positive electrode was a commercially available LFP electrode foil. <strong>Both electrodes were cut in dimensions of 30 × 15 mm<sup>2</sup></strong>.</p>
</blockquote>
<p>I think this is an area of <strong>30 mm × 15 mm = (30 × 15) mm<sup>2</sup></strong>. Your second version is correct. Theirs is sloppy writing.</p>
| 45398 | Reading odd dimensions |
2021-08-03T04:18:40.570 | <p>I am sending my design for machining. In my design I am trying to assemble two metal parts with each other. In these holes I will be using m3 socket head bolt. I am confused what tolerance value should I put in the diameter of the hole.</p>
<p>Can anyone tell me how to calculate the tolerance for the diameter of the holes that will use standard screws such as m3/m6.</p>
<p><a href="https://i.stack.imgur.com/RERJa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RERJa.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>IF I understand your question correctly what you are looking for is the clearance hole diameter. I.e. how big to make the hole to be able to let an M3 bolt to pass through.</p>
<p>you can find a lot of tables like the following in the internet, where you can see that the clearance hole is between 3.2 and 3.9 mm <em>depending on the fit</em> that want to have.</p>
<p><a href="https://i.stack.imgur.com/MwgNY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MwgNY.jpg" alt="enter image description here" /></a></p>
| 45403 | How to choose diameter of hole that will fit m3 socket head bolt to assemble two parts? |
2021-08-03T07:53:06.787 | <p>How is the engine of a tractor different from a car?
So far, all I know, is, that a tractor has a rather high torque <span class="math-container">$ M $</span> and a car engine has rather high power <span class="math-container">$ P $</span>. But physically, they are quite connected: <span class="math-container">$ P = M * \omega $</span>.</p>
<p>Finally, I know, what matters is lastly what the engine can deliver to the tyres but nevertheless, I guess, the engines are quite different?</p>
| |mechanical-engineering|automotive-engineering| | <p>You optimize an engine for a tractor by looking at how it's used. This is not the same as looking at torque versus power, because in the end, it is really only power that is required. Torque can be gained through gears.</p>
<p>Tractors:</p>
<ul>
<li>Move rather slowly compared to road vehicles</li>
<li>do not need acceleration or drivability like a car</li>
<li>operate external equipment via a Power Take-off (PTO). This means you can't run the engine slowly since the driven equipment needs power regardless of the speed over ground of the tractor</li>
<li>are operated continuously for long periods. This affects operating cost (tends towards diesel engines because a diesel is more efficient) and design considerations for longevity.</li>
</ul>
<p>What you typically end up with is a diesel engine that is relatively small for the given size of the tractor, and is built with a very heavy duty rating, so it will last a long time. And very short gearing.</p>
| 45406 | How do you optimize an engine for a farm tractor? |
2021-08-04T21:38:54.120 | <p>Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about. What I imagine is the case in a real world situation is that I have a fluid that I want to cool down from temperature A to temperature B, and I want to find an exchanger that can do the job. What I know are the inlet temperatures and flowrates of both fluids.</p>
<p>Most calculations require both inlet and outlet temperatures to be entered, but if I'm looking to buy an exchanger, I can't know what the outlet temperatures are going to be.... I only know my inlet temperatures... so I'm a bit confused.</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heat-exchanger| | <p>Your specification is that you know the two inlet temperatures, <span class="math-container">$T_{h,i}$</span> (hot in) and <span class="math-container">$T_{c,i}$</span> (cold in). Interestingly, you also say that you <em>want to cool the hot stream to a defined temperature <span class="math-container">$T_{h,o}$</span></em>. So, you have at least three defined parameters for your "real world" case.</p>
<p>Let's take also as a given that you are specifying the fluids for the hot and cold fluids. This means, you are specifying whether the fluids are water, air, oil, steam, or whatever. When you do this, because you have the inlet temperatures of the fluids, you also specifying the starting points for the specific heat capacities of your fluids, <span class="math-container">$\tilde{C}_{p,h/c}$</span> (J/kg K). Take the specific heat capacities as constant over the heat exchanger to start.</p>
<p>Two approaches can be taken next. In one case, you also specify the mass flow rate of at least one stream. Let's suppose that, because you have specified that you want to cool the hot stream, you also specify how much flow rate <span class="math-container">$\dot{m}_c$</span> (kg/s) you need to cool.</p>
<p>Immediately, you see that you can calculate the required heat demand (W)</p>
<p><span class="math-container">$$ \dot{q} = \dot{m}_h\tilde{C}_{p,h}\left(T_{h,i} - T_{h,o}\right) $$</span></p>
<p>Now, you are in position to determine the mass flow rate and outlet temperature of the cold inlet stream.</p>
<p><span class="math-container">$$ \dot{q} = \dot{m}_c\tilde{C}_{p,c}\left(T_{c,o} - T_{c,i}\right) $$</span></p>
<p>Essentially, you are in a position to play around with the three parameters <span class="math-container">$T_{c,o}$</span>, <span class="math-container">$\dot{m}_h$</span>, and <span class="math-container">$\dot{m}_c$</span>. When you specify any one of the three, the other two are defined by the two heat balance equations.</p>
<p>But what about the TYPE and SIZE of the heat exchanger? The type is a choice such as shell-and-tube or complex with flows as parallel or cross-flow. The size is expressed as the area <span class="math-container">$A$</span> (m<span class="math-container">$^2$</span>).</p>
<p>Assume that you have solved the heat transfer equations. The simplest method to set the type and size is the log mean temperature difference (LMTD)</p>
<p><span class="math-container">$$\dot{q} = f\ U\ A\ \Delta_{LMTD} T $$</span></p>
<p>In this, <span class="math-container">$f$</span> is a factor based on the type of exchanger, <span class="math-container">$U$</span> is the <a href="https://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html" rel="nofollow noreferrer">overall heat transfer coefficient</a> (W/m<span class="math-container">$^2$</span> K), and <span class="math-container">$\Delta_{LMTD} T$</span> is the <a href="https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference" rel="nofollow noreferrer">log mean temperature difference</a> (K). The value of <span class="math-container">$U$</span> will depend on the construction of the exchanger, the type of exchanger, and the mass flow rates.</p>
<p>In some cases, the problem is open-ended. We are only given the inlet temperatures and the fluids. We are after both outlet temperatures as a function of flow rates, type of exchanger, and size of exchanger. The more respected approach in this case is the number of transfer units (NTU) method.</p>
<p><span class="math-container">$$ \dot{q}_{act} = \epsilon\ \tilde{\dot{C}}_{p,min}\left(T_{h,i} - T_{c,i}\right) $$</span></p>
<p>Here, <span class="math-container">$\epsilon$</span> is an effectiveness that depends on the type of exchanger, <span class="math-container">$U$</span>, and <span class="math-container">$A$</span>, while <span class="math-container">$\tilde{\dot{C}}_{p,min}$</span> (W/K) is the smallest value combining <span class="math-container">$\dot{m}$</span> (kg/s) and <span class="math-container">$\tilde{C}_p$</span> (W/kg K) for the fluids.</p>
<p>Summary details are provided effectively <a href="https://en.wikipedia.org/wiki/NTU_method" rel="nofollow noreferrer">at this Wikipedia link</a>. Full details and graphs are provide <a href="https://www.intechopen.com/chapters/54521" rel="nofollow noreferrer">at this link</a>. Finally, a comparison of LMTD and NTU methods is provide <a href="http://kb.eng-software.com/eskb/ask-an-engineer/theory-equations-and-calculated-results-questions/difference-between-the-effectiveness-ntu-and-lmtd-methods" rel="nofollow noreferrer">at this link</a>.</p>
| 45444 | How are heat exchangers normally calculated? |
2021-08-05T13:38:27.573 | <p>I am making a cylinder of polypropylene that must be pressurized (2 bar). The rough dimensions of the current design (A) are 7 cm tall, 5 cm in diameter, and wall thickness of 1-2 mm. The issue is that the end plate (disc) begins to bulge when the pressure is increased.</p>
<p>I want to make a new design that is stiffer. Is there a shape/pattern that is better in terms of preventing bulging than naively increasing thickness?</p>
<p>I want to maximize stiffness (prevent bulging) while minimizing the thickness of the end plate. I don't care about how much material I use. I.e. if the stiffness is the same solution B and C are equally good.</p>
<p>Constraints</p>
<ul>
<li>The end plate must be flat on the inside, but there are no constraints on the outside shape.</li>
<li>The final product must be injection moldable.</li>
</ul>
<p><a href="https://i.stack.imgur.com/MzjXQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MzjXQ.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|design|pressure|cad|plastic| | <p>This is a minor addition to Fred_dot_u's answer.</p>
<p>Regarding the radial stiffeners the best shape for the stiffeners (minimizing the deflection for a given weight would be a stiffener with the following shape (in green).</p>
<p><a href="https://i.stack.imgur.com/GHtBC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GHtBC.png" alt="enter image description here" /></a></p>
<p>This optimization only makes sense if you have large production numbers, otherwise you can use plain rectangular profile for the stiffeners if production is small.</p>
<p>It is best that the stiffeners are all connected in a central hub like the following.</p>
<p><a href="https://i.stack.imgur.com/9zXWz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9zXWz.png" alt="enter image description here" /></a></p>
| 45453 | How to design stiff plastic sheet |
2021-08-06T02:42:44.997 | <p>Reading through ASME Y14-5 I noticed a Note in para 2.6 "NOTE: When basic dimensions are used, there is no accumulation of tolerances. A geometric tolerance is required to create the tolerance zone. In this case, the style of dimensioning (chain, baseline, direct) is up to the discretion of the user. <strong>Locating features using directly toleranced dimensions is not recommended</strong>".</p>
<p>What does this mean? Aren't direct tolerances the best compared to to chain, or baseline? So why wouldn't someone want to locate a feature using a directly toleranced dimension, unless this has something to do with using a GD&T block (I'm self-learning GD&T at the moment, so I haven't gotten to those symbols yet).</p>
| |drafting| | <p>Note the phrase at the very beginning: "When <strong>basic</strong> dimensions are used..." So this paragraph isn't about regular size dimensions but onlybasic dimensions.</p>
<p>Basic dimensions aren't measured or inspected but are used as the <strong>theoretically perfect positions</strong> of features and/or other datums. So you essentially can't give them a tolerance, but you can tolerance the size and position of the feature (like a hole) at the location described by basic dimensions.</p>
<p><a href="https://i.stack.imgur.com/GTW48.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GTW48.png" alt="Example of Basic dimensions" /></a></p>
<p>Note the locations of the holes are basic, but the envelope dimensions of the part are directly toleranced as are the holes. Further there is geometric tolerancing on the holes to describe the allowable deviation from the theoretical perfect positioning set by the basic dimensions.</p>
<p><a href="https://www.gdandtbasics.com/reporting-basic-dimensions/" rel="nofollow noreferrer">Source</a></p>
| 45463 | Why is locating a feature using directly toleranced dimensions not recommended? |
2021-08-06T18:39:52.243 | <p>Friends, I am an EE/SW/Controls eng. Please, pardon me of perhaps incorrect terminologies.<br/>
I am working on a R&D project: A chemical lab equipment has a 2" long metal pipe enclosed in insulation material. "One end" of the pipe is attached to a heating element and a temperature sensor. My work is controlling the temperature of the pipe. Including mechanical engineers and a chemist, as a controls engineer I am, our group discusses what material the pipe needs to be, either aluminum or copper.<br/></p>
<p>I saw a few YouTube clips that demonstrate aluminum delivers heat faster than copper. In controls perspective, faster delivery means less delay and better control. Thus, my intuition tells to use an aluminum pipe, while others prefer copper for its higher thermal conductivity. <br/></p>
<p>Could someone explain how one is better than another for this case?</p>
<hr />
<p>Edited<br/>
The YouTube clip is <a href="https://www.youtube.com/watch?v=Sdpcjm2vmLY" rel="nofollow noreferrer">here</a>.<br/>
The pipe contains a section of <a href="https://en.wikipedia.org/wiki/Gas_chromatography" rel="nofollow noreferrer">Gas chromatography "Column"</a>, between the "Inlet + trap" to the "Oven". Ideally the temperature has to be the same as the Oven, which changes up to 15'C/sec. Otherwise, we try to keep it at a certain temperature(TBD) other than "Cold spot". The pipe can be a pair of aluminum plates with channels, I thought.</p>
| |materials|control-engineering|thermodynamics|heat-transfer|chemical-engineering| | <p>Simplified analysis to compare materials, for different limiting-case applications, given the same geometry:</p>
<ul>
<li>Want max heating ramp rate, with fixed-power electric heater -- minimize Volumetric Heat (i.e. specific heat * density)</li>
<li>Want max cooling ramp rate, with fixed-temperature coolant -- minimize (VolumetricHeat / ThermalConductivity)</li>
<li>Want best control performance at/near steady state -- maximize Thermal Conductivity (and possibly thermal mass, but depends on location and nature of noise in thermal load)</li>
</ul>
<p>Aluminum 6061-T6:</p>
<ul>
<li>density (g/cc) = 2.7</li>
<li>specific heat (J/gK) = 0.9</li>
<li>volumetric heat (J/ccK) = 2.43</li>
<li>thermal conductivity (W/mK) = 167</li>
<li>CTE (ppm/K) = 24</li>
</ul>
<p>Copper, pure:</p>
<ul>
<li>density (g/cc) = 8.9</li>
<li>specific heat (J/gK) = 0.385</li>
<li>volumetric heat (J/ccK) = 3.43</li>
<li>thermal conductivity (W/mK) = 385</li>
<li>CTE (ppm/K) = 17</li>
</ul>
<hr />
<p>Edit #1, from OP @jay<br/></p>
<ol>
<li>If the volume of the pipe was the same, aluminum would be more effective delivering the heat from one end to another. If the weight of the pipe was the same, copper would be more effective.<br/></li>
<li>Since, the length/shape of the pipe is constrained by the mechanical structure, the volume matters more.<br/></li>
<li>Thus, I would choose aluminum pipe, if I have good trust in my controller.<br/></li>
<li>I would choose copper pipe for less influence from disturbance (ex. from ambient), if the performance of the controller was excluded from the considerations.</li>
</ol>
<hr />
<p>Edit #2 from OP jay<br/></p>
<p>However, in practice and mechanical design perspective:<br/></p>
<p>For 2, if you are building from pipe/tube, there are a finite number of specific options wall thickness. So just make a table of the bulk heat capacity in J/K for the options you have given mechanical constraints. That will inform the remaining analysis, but is not necessarily the performance limiting factor. Often not.<br/></p>
<p>Regarding effect of controller, it depends on placement of heater and sensor.<br/></p>
<p>If you are not sure what you are doing, then a big thermal mass with a slow response might actually be more practical than something light and aggressively tuned.</p>
| 45474 | Aluminum or Copper, which one is better to control and maintain the temperature? |
2021-08-06T19:47:16.563 | <p>I've got a mechanism that uses a bimetallic strip; for aesthetic reasons I'd like to electroplate it with a thin layer of gold - how does this affect the relative displacement of a bimetallic strip in response to temperature? My intuition tells me that if the plating covers the entire assembly, it should all be subject to the same expansion, and shouldn't affect the relative displacement of the strip (accounting for the obvious stiffening effect of adding more material), right? Any advice is appreciated, I'm having trouble thinking about this intuitively.</p>
| |metallurgy|metals|temperature|thermal-expansion| | <h2>Why a bimetallic strip heating up is a bending problem</h2>
<p>When a bimetallic strip is heated by <span class="math-container">$\Delta T$</span>, from its original temperature, what happens is that one strip tends to grow longer and the other less. because they are bonded together, they end up to bend and rotate.</p>
<p><a href="https://i.stack.imgur.com/TbtiS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TbtiS.png" alt="enter image description here" /></a></p>
<p><strong>Figure 1: Diagram of a bimetallic strip showing how the difference in thermal expansion in the two metals leads to a much larger sideways displacement of the strip (Source: <a href="https://www.wikiwand.com/en/Bimetallic_strip" rel="nofollow noreferrer">wikiwand</a>)</strong></p>
<p>The bend happens because a moment couple is generated, because one strip applies a constraint to the other (i.e. the one with the higher expansion tries to pull the other one with the lower <span class="math-container">$\alpha$</span>, and vice versa)</p>
<p><a href="https://i.stack.imgur.com/63snk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/63snk.png" alt="enter image description here" /></a></p>
<p><strong>Figure 1: Moment Couple on a bimetallic (Source: <a href="https://www.calqlata.com/productpages/00045-help.html" rel="nofollow noreferrer">wikiwand</a>)</strong></p>
<p>So this ends up being a problem being a bending problem with a bending moment which can be calculated from the diagram above.</p>
<h2>Bending problem considerations.</h2>
<p>The main issues are how much does the second moment of area changes, and also how does the stresses change.</p>
<p>Just for reference, a bimetallic strip is between 0.5 and 10 mm (sometimes more), and gold electroplating is in the <a href="https://www.silvexinc.com/about-us/news/how-thick-gold-plating-electronics/#:%7E:text=Gold%20plating%20is%20the%20deposit,for%20making%20jewelry%20and%20coinage." rel="nofollow noreferrer">0.00025 – 0.005 mm </a>.</p>
<p>The second moment of area is affected by the plating -albeit not significantly) because the the gold plating will be further away from the neutral axis, therefore it will have the highest contibution <em>for a layer with that thickness</em>. However, overall because the thickness of the plating is at best 100 less, its effect would be at most about 5% of the overall contribution. If you take into consideration that the steel modulus is also greater than gold, that will probably reduce even further the overall effect.</p>
<p>The other part, is the distribution of the stresses. The presence of the gold plating would impose again another constrain (i.e. try to expand equally both top and bottom layers). That will create an opposite bending moment and the overall bending moment will be <strong>slightly</strong> less. However, again that will depend on the ratio of thickness of the electroplating and the bimetallic strip.</p>
| 45475 | In what way does plating a bimetallic strip affect its displacement? |
2021-08-06T20:46:23.030 | <p>My understanding is that thermal conductivity is responsible for the time until something will heat up. For example, if you hold a copper bar over a candle it will quickly become hot at the other end. If you instead hold a brass bar it will take a longer time, but the end result will be the same.</p>
<p>If this is correct (is it?) then why does thermal conductivity matter in a double pipe heat exchanger? It will take some more time for the pipe to heat up and start transferring heat, but once that happens it's all the same, is it not?</p>
| |mechanical-engineering|heat-transfer|heat-exchanger|heat| | <h2>Example</h2>
<p>Since the rest of the answer is too long I will start with an example.</p>
<p>Imagine you could build a heat exchanger from wooden pipes. If you run through the liquid you would observe very little efficiency (i.e. the hot liquid would remain hot and the cold cold). There would be too little heat exchange, because the wood would not allow heat to pass through the material at any significant rate.</p>
<hr />
<h2>combined conduction and convection</h2>
<p>You have to remember that the problem of heat transfer between the hotter fluid (h) and the cooler (c) is a combined conductivity and convection problem.</p>
<p><a href="https://i.stack.imgur.com/jGGy7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jGGy7.png" alt="enter image description here" /></a></p>
<p><strong>Figure Conducting wall with convective heat transfer (source: <a href="https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node123.html" rel="nofollow noreferrer">MIT thermodynamics</a>)</strong></p>
<p>Where:</p>
<ul>
<li><span class="math-container">$T_2$</span> is the hotter fluid</li>
<li><span class="math-container">$T_1$</span> is the cooler fluid</li>
<li><span class="math-container">$\delta_1$</span>, <span class="math-container">$\delta_2$</span> are the convective zones. I.e. the zone where the temperature in the fluid will change. This zone will change with the velocity and the viscosity and other geometric parameters</li>
</ul>
<p>In those problem sometimes, the concept of <strong>thermal Resistance</strong> is introduced. Thermal resistance shows what is the energy per area for a degree of temperature difference. It is important to note that as the Thermal resistance increases, less heat energy is transmitted per temperature degree difference <span class="math-container">$\Delta T$</span> (or <span class="math-container">$\Delta T_{LMTD}$</span> to connect to the previous question)</p>
<p>In the above case, the thermal resistance is:
<span class="math-container">$$R= \frac{1}{h_1A} + \frac{L}{A \color{red}{k}} + \frac{1}{h_2A}$$</span></p>
<p>where:</p>
<ul>
<li>L is the thickness of the wall (pipe)</li>
<li>A is the exchange area</li>
<li><span class="math-container">$h_1$</span> and <span class="math-container">$h_2$</span> the heat convectivity coefficient</li>
<li><span class="math-container">$\color{red}{k}$</span> heat conductivity coefficient</li>
</ul>
<p>As you can see the thermal resistance is inversely proportional to the thermal conductivity. This means that when <em>thermal conductivity of the pipe is high, then the thermal resistance is low</em>.</p>
<h2>Steady state of a heat exchanger</h2>
<p>The problem with heat exchanger is that they come into too many flavors. Usually you encounter the parallel and the counterflow (but there are several others). Below is the example of those types of heat exchangers. In this image you can see the temperature difference</p>
<p><a href="https://i.stack.imgur.com/TphLz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TphLz.png" alt="enter image description here" /></a></p>
<p><strong>figure: Temperature difference along the length of the heat exchanger for different types of flows</strong></p>
<p>In that image you can see immediately the problem. If you have a heat exchanger with a parallel flow you will have different temperature difference at different section of the heat exchanger (thus different heat transfer rates because the <span class="math-container">$\Delta T$</span> varies). (The same problem exists with the counterflow to a lesser extent.)</p>
<p>However, this is <strong>the temperature distribution of the steady state.</strong></p>
<h2>Transient state (short parenthesis)</h2>
<p>The transient state is (in the simplest case) when the heat exchanger starts its operation. Then the fluid that passes can be assumed to have a uniform distribution (i.e. does not change with location).</p>
<p>However, the temperature distribution will change with time, and it will approach the steady state solution. It is obvious that during the transient state the heat transfer will be greater because the temperature difference will be greater compared to the steady state. So the bottom line is that, <strong>at the steady state there is the least heat transfer compared to all the transient states that led to it</strong>.</p>
<h2>Putting everything together</h2>
<p>From the previous paragraphs, I was trying to highlight that <strong>the steady state will have a temperature distribution along the length of the heat exchanger that should remain constant (wrt to time).</strong></p>
<p>Therefore for a infinitesimally small portion along the length of the heat exchanger (with area A) the temperature difference can be assumed to remain constant wrt time. In that portion, the heat transfer rate can be assumed to be:</p>
<p><span class="math-container">$$\dot{Q} = \frac{\Delta T}{R}$$</span></p>
<p>However as we've seen before as the thermal conductivity (k) increases, then thermal resistance R decreases, and as R decreases the heat transfer rate increases. That means that:</p>
<blockquote>
<p>for increasing <strong>thermal conductivity (k)</strong>, there is more heat transfer for the same temperature difference.</p>
</blockquote>
<p>Since in the steady state the temperature difference is constant, the heat exchange will be dominated by the thermal resistance, and therefore it would be affected by changes in the thermal conductivity.</p>
<h2>Why is the effect not pronounced in metals</h2>
<p>The reason the heat conductivity effect is as not pronounced in metals is that because Thermal Resistance R is determined by the equation:</p>
<p><span class="math-container">$$R= \frac{1}{A}\left(\frac{1}{h_1} + \frac{L}{\color{red}{k}} + \frac{1}{h_2}\right)$$</span></p>
<p>Then, the lowest of the <span class="math-container">$h_1$</span>, <span class="math-container">$h_2$</span> and <span class="math-container">$\frac{k}{L}$</span> will have the dominating effect on the heat conductivity. Since usually <span class="math-container">$h_1$</span>, <span class="math-container">$h_2$</span> coefficients are significantly lower than <span class="math-container">$\frac{k}{L}$</span> (unless L becomes too great), any change of thermal conductivity is not affecting affecting significantly the thermal resistance. (this is exactly the point mart's second paragraph is trying to raise).</p>
| 45479 | Why does thermal conductivity of a pipe matter in a double pipe heat exchanger? |
2021-08-07T11:43:09.880 | <p>In internal flow, when a fluid flows inside a circular pipe, the most practical value of critical Reynolds number is taken as 2300. I was wondering if this number is also the same for non circular ducts? Does the critical Reynolds number of a duct will also be around 2300 for most practical cases?</p>
| |mechanical-engineering|fluid-mechanics|convection| | <p>To my understanding, if a flow has a different geometry, such as flow in a square duct, or over a turbine blade, transition will occur at different values of Re.</p>
<p>Additionally another very important point is that in circular tube:</p>
<ul>
<li>if the flow is undisturbed (i.e. very smooth pipe walls) then the transition to transient flow is at much greater Re numbers</li>
<li>if Re is less than 2000, then the flow will <strong>always be laminar</strong> even when it is disturbed.</li>
</ul>
| 45491 | Is this value of critical Reynolds number also valid for ducts? |
2021-08-07T15:20:50.213 | <p>I have previously done this question on math stack, and I think that maybe here is better.</p>
<p>I am studying the theory of discrete-time systems, comparing them with continuous ones in the control theory field.In particular, I have focused on the advantages and disadvantages to prefer sometimes the first over the latter.</p>
<p>Some papers say that nowadays, most control systems are implemented through digital devices (as DSP, etc.). So sometimes, it is preferable to design directly a discrete-time controller rather than a continuous one. Indeed the engineers prefer this choice.</p>
<p>I have also read that designing a continuous-time controller is not entirely avoided; indeed, many mathematicians prefer this choice. In that case, the idea is that a continuous-time is approximately a discrete-time system, with an arbitrarily small sampling time.</p>
<p>My question is: <strong>what are the disadvantages of choosing a too small sampling time above all on the control effort?</strong></p>
<p><strong>Note:</strong> the control effort is the amount of energy or power necessary for the controller to perform its duty.</p>
| |control-engineering|control-theory| | <p>From my -limited- experience the problem when dealing with control systems decreasing the sampling time <strong>beyond</strong> the optimum has the following adverse effects:</p>
<ul>
<li><strong>The cost of ADC goes up very fast</strong>: Usually when you need to control a motor or another physical system you need to take some measurements. Many times the output is analog and needs to be converted to analog. This Analog to Digital (ADC) conversion takes time. The faster you need data, the higher the cost. An example from a well known supplier at 50 kHz sampling rate for 1 differential channels is about 300 Euro, when you go up to 1.5 MHz the cost goes up to 2500 Euro. And although you can get even faster sampling rates, with dedicated DSP/FPGA computer cards, the total cost of development is the order of 5 digits.</li>
<li><strong>Gather more noise</strong>: again in my experience, when the control cycle duration goes down (ie. how many control cycles per second are carried out), and assuming the ADC sampling rate remains constant, you end up with only a few or maybe one data point per cycle. This has the effect that the inherent measurement noise affects the system (you get more disturbances). In most cases, that might not be a problem, but if it combines with common problems like poor grounding it can be life hell.</li>
<li><strong>Additional computational/programming effort</strong>: Increasing the ADC sampling rate, for a given control cycle rate, means that you have more data in each cycle. That means more processing. That processing can be as cheap as just a sum or a mean of the data, or it can have adverse effects when doing more complex dsp operations (e.g. power spectrum, of filtering). That can impose a burden on the CPU, that can sometimes affect the cycle time. Additionally depending on the hardware architecture (8-bit, 16-bit, etc) some programming effort might be required to avoid overflows during calculations.</li>
</ul>
| 45496 | Why prefer a discrete-time system because of a practical implementation instead of continuous one |
2021-08-10T13:47:02.373 | <p>I recently saw <a href="https://engineering.stackexchange.com/questions/45514/how-large-would-a-solar-panel-array-need-to-be-in-order-to-supply-all-the-world">this question</a> and I immediately thought that the discussion was too limited to current energy production, it simply assumes that the amount of energy we use today is the amount of energy we currently need. But the biggest part of the world population live in countries that do not produce enough energy to cover their demand.</p>
<p>So I was wondering how big should be a solar panel array in order to supply the world energy needs. But I guess that first one should determine what could be the energy need compared to the current production. Predicting real energy need, by the time the solar array would be ready, I think is impossible because it should take into account technological changes population growth and additional demands like desalination due to the exhaustion of fresh water. But at least a rough estimate could be made on the hypothesis of no population growth and simply the developing countries aligning to the developed ones. I saw this <a href="https://en.wikipedia.org/wiki/List_of_countries_by_energy_consumption_per_capita" rel="nofollow noreferrer">list</a> which gives an idea of how skewed is the production. So my question now is:</p>
<p>if the developing countries want to reach a per capita energy consumption similar to that of China, which seems reasonable, how much the current energy consumption would grow?</p>
| |energy| | <p>Energy use is not a problem. Efficiency and cleanliness should be the goal. Our planet will not even achieve the status of a type-I civilization until we harness all the energy of our sun. A type-II civilization harnesses the energy of its galaxy.</p>
<p><a href="https://futurism.com/the-kardashev-scale-type-i-ii-iii-iv-v-civilization/amp" rel="nofollow noreferrer">https://futurism.com/the-kardashev-scale-type-i-ii-iii-iv-v-civilization/amp</a></p>
| 45532 | If the developing countries consumed as much energy per capita as China how much the current energy need would grow? |
2021-08-10T15:57:03.593 | <p>Like the ones in a:</p>
<ul>
<li><p>cinema ticket selling point</p>
</li>
<li><p>purge contention chamber (sci-fi)</p>
</li>
<li><p>reception point of a public institution</p>
</li>
<li><p>police control point</p>
</li>
</ul>
<p>(where one can attend someone else, behind a security glass)</p>
| |terminology|architecture| | <p>'kiosk' is the first word that springs to mind</p>
<p>Other words depending on the case are:</p>
<ul>
<li>stand</li>
<li>stall</li>
<li>counter</li>
<li>booth</li>
</ul>
| 45533 | Correct term for "Cabin Window"? |
2021-08-10T16:17:49.280 | <p>I have built a little gearbox to reduce the RPM and increase the torque, that is composed of 8 gears that are all equal: they have 12 teeth "in the small side" that transfers the motion to the "big part" that has 36 teeth.
So at the end I have (1st gear with only 12t, 2nd-7th 12(small) 36(big), 8th just 36t to receive the motion from the 7th gear.</p>
<p>Can anybody explain how can I calculate the gear ratio in this case?</p>
| |gears| | <p>Image created in Inkscape by yours truly.
<a href="https://i.stack.imgur.com/mRYfb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/mRYfb.png" alt="reduction gearing" /></a></p>
<p>If I understand your description correctly, the diagram above should represent your question. Each "donut" is a pair of 12/36 gears joined to move simultaneously, while each donut drives the succeeding one with the ratios presented.</p>
<p>The text has been placed at the interface of each gear pair. The first circle drives the second one with a 1 to 3 reduction, while the center of the second gear drives the third. This is mechanically unsound, but suitable as a discussion diagram.</p>
<p>There are seven interfaces/reductions, which means your gear ratio is 1 to 3 raised to the 7th power. The final reduction is therefore 1 : 2187
If there is one more reduction, the ratio is 1 : 6561</p>
<p>If you did not have equal gears between each set, you would have to multiply the denominators of all the ratios together to calculate the final ratio.</p>
<p>From a mechanical standpoint, consider to create a planetary gear set. You can accomplish substantial reduction in a more compact design.</p>
| 45536 | How to calculate this gear ratio? |
2021-08-11T03:29:33.140 | <p>I have been reading about the elasticity of materials and most often in engineering there is always some ideal concepts which are the extreme cases. So is there any particular definition to what is a perfectly elastic and perfectly plastic material?</p>
<p><strong>Perfectly Elastic Material</strong></p>
<p>A lot of people say that from material science point of view a more elastic material means the material has greater resistance to elastic deformation eg steel being more elastic than rubber . With that definition ,a perfectly elastic material should then be defined as " a material which suffers zero deformation under any value stress (within elastic limit)". But in this case it becomes similar to the definition of a rigid body , which is "a body that suffers no deformation under stress".</p>
<p>Yet another possible definition could be that a perfectly elastic material is one which behaves as an elastic material over its entire stress-strain curve i.e behaves as an elastic material till fracture. Here then we could have perfectly linear elastic material and perfectly non-linear elastic material.</p>
<p><strong>Perfectly Plastic Material</strong></p>
<p>A perfectly plastic body could be defined as one which produces no restoring force for any value of stress applied. Thus a perfectly plastic body would always suffer permanent deformation for any value of load applied or in other words a perfectly plastic body would show plastic behaviour throughout the stress-strain curve.</p>
<p>If perfectly plastic and perfectly elastic are well defined ,how would their stress-strain curve look like?</p>
| |stresses|elastic-modulus| | <p>A perfectly elastic material is simply one that always returns to original shape after loading.</p>
<p>A perfectly plastic material is one that always plastically (permanently) deforms under load.</p>
<p>The shapes of the stress-strain curves can practically be whatever, as long as the material follows the "rules" above, then it will be considered those things.</p>
<p>These terms are used as estimations and for mathematical modeling and categorizations. There isn't truly any material that is either. All materials will behave between the two extremes at different points, more or less. But quartz is typically considered the most perfectly elastic material, and wax or putty is close to perfectly plastic. These materials have their own values for their engineering properties (quartz is very stiff, putty is soft), but a material that is either perfectly elastic or perfectly plastic does not have to have those same values. They could have any slope values eminating out from around the origin (0,0). It's just "difficult to make" a material in specific ways because physics/nature is just the way it is.</p>
<p>Most materials but not all (that we consider in engineering) are generally "elastic perfectly-plastic", meaning they have an elastic region and what can be estimated as a perfectly-plastic region. That being said, a stress-strain curve does not presuppose an elastic region-- the elastic region is marked/noted on the curve IF and only IF elastic behavior is observed. That is to say, you can (theoretically, hypothetically) have a curve that LOOKS like it has a typical elastic region, but indeed it does not return to it's initial shape after load is released. And as someone else mentioned, not all elastic regions are linear.</p>
<p>I believe something like cast-iron or graphite could be considered (fairly) perfectly elastic since it doesn't really have a plastic region to speak of. These are "brittle" materials, and their yield and ultimate tensile strengths are considered equal. So, their whole behavior (up until break) is "perfectly elastic".</p>
<p>And, just for completeness, a rigid body is one that doesn't deform (either elastically or in a plastic way) under load-- or, for estimation/relative purposes, deforms very little under load. A rigid body doesn't have to be elastic or plastic to be considered rigid (but, of course, it could be one or the other).</p>
| 45546 | Does perfectly elastic and perfectly plastic material make any sense? |
2021-08-11T12:33:05.133 | <p>Most of the sources that I had read about either talks about anelasticity or viscoelasticity, they don't compare both. From what I have read so far ,both anelastic and viscoelastic materials are the same. And they both show significant amount of time dependent strain component.
Is there any difference between these two?</p>
| |stresses|plastic|elastic-modulus| | <p>Anelastic materials have an element of time delay between stress and strain, and a time-dependent relaxation of strain. <a href="https://en.wikipedia.org/wiki/Anelasticity" rel="nofollow noreferrer">Wikipeidia link</a></p>
<p>But the viscoelastic materials have both properties of elastic and viscous materials when undergoing deformation.</p>
<p>Viscous materials resist shear flow. And have defusion of atoms under stress as compared to bond stretching along crystallographic planes.</p>
<p><a href="https://i.stack.imgur.com/lQo5o.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lQo5o.jpg" alt="stress strain curve" /></a></p>
<p>Stress–strain curves for a purely elastic material (a) and a viscoelastic material (b). The red area is a hysteresis loop and shows the amount of energy lost (as heat) in a loading and unloading cycle. It is equal to , where is stress and is strain.<a href="https://en.wikipedia.org/wiki/Anelasticity" rel="nofollow noreferrer">1</a></p>
<p><a href="https://en.wikipedia.org/wiki/Viscoelasticity" rel="nofollow noreferrer">Viscoelastic- Wikipedia link.</a></p>
| 45551 | Is there any difference between an-elastic and visco-elastic materials? |
2021-08-12T05:14:35.240 | <p>An open belt drive with flat belt has different angle of wrap on the larger and smaller pulley. I have to calculate a tension in the pulley. We generally use the formulae
<a href="https://i.stack.imgur.com/T5Ajp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T5Ajp.png" alt="Formulae" /></a></p>
<p>Here θ is the angle of wrap. Now the angle of wrap is different for larger pulley and smaller pulley in the belt drive. So which angle of wrap should be used in this equation?</p>
| |mechanical-engineering|pulleys|machine-design| | <p>It depends on the product of angle of wrap (θ) and coefficient of friction (μ) of the corresponding pulley. Whichever product is less that angle of wrap is considered.</p>
<p>In case the coefficient of friction is same for both the pulleys in open-belt drive system, angle of wrap of the smaller pulley is usually considered.</p>
<p>In case of crossed-belt drive system, angle of wrap is same for both smaller as well as larger pulley so the smaller coefficient of friction is considered in the formulae.<a href="https://i.stack.imgur.com/njk03.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/njk03.png" alt="enter image description here" /></a></p>
| 45563 | In an open belt drive, which angle of wrap should be taken for calculating tension in the belt? |
2021-08-12T12:24:55.433 | <p>My boss ordered a screw pump driving spindle from an alternative maker. That spindle, as expected, cannot turn with the two idle spindles inside the housing.</p>
<p>Now he is asking for the spindle dimensions so that he can fabricate a correct spindle at some workshop.</p>
<p>Ok, the height. But how to measure the epicycloid curve? How somebody measures a screw pump spindle?</p>
<p>i am inside a ship i only have a vernier and a micrometer.</p>
| |measurements|pumps| | <p>Just a general method to considering form, fit, and function when measuring any part:</p>
<p>Start with the overall outer dimensions and areas where it would interfere or interface with other pieces (fit). Since this is where the previous attempt failed, be sure to give it extra attention.</p>
<p>Then focus on the areas that allow the part to do its job (function). This could also include additional requirements for be where it interfaces with other moving parts or bears structural loads - such as special treatments or different materials.</p>
<p>Finally (form), if you were to sketch the main views of the part, knowing what the part is, what other dimensions/annotations would you need? For strange curves, assume some reasonable shape approximations (ellipse, circle in an alternate view, parabolas) and take at least two additional measurements to verify each approximation (if 3 points make a circle, take 5 and see how far the other two fall out of the circle determined by the 3). Use any existing specifications of similar parts for reference.</p>
| 45565 | How to measure a screw pump spindle? |
2021-08-12T12:48:45.400 | <p>This is to settle a debate.</p>
<p>I was working on a scaffold next to a building, as the wind blew, the scaffolding would sway. As I stood there I was trying to maintain my stability, but my movements seemed to be making the sway of the scaffolding worse.</p>
<p>I thought this was resonance, my brother maintains it is not. If it's not, is there a term for this?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|dynamics|vibration| | <p>There is a name for this phenomenon in the aeronautics world: <em>pilot-induced oscillation.</em> This occurs when the pilot's own response time, error condition measurement and system gain couple to the dynamics of the system and force it into divergent behavior. In short, it means the pilot's control inputs are in phase with the system dynamics, making the error condition worse instead of damping it out.</p>
| 45566 | A question on resonance |
2021-08-12T20:07:29.877 | <p>How does the fabricant of prestressed pre-tensioned beams calculate the equivalent pressure in the hydraulic jack (in bars) to the design prestress force (in kN)? Are there factors to multiply like a jack efficiency factor or are there some losses we must add before setting the right pressure to the jack?</p>
| |structural-engineering|civil-engineering|beam|concrete|prestressed-concrete| | <p>Just to give a very rough narrative.</p>
<p>Let's say we have a 20 ft long beam with strands of prestressed eccentricity of 1 ft. and our goal is to prestress this beam for 100 kips.ft moment at midspan.</p>
<p>So we have to provide 100 kips tension in the enter provided by the strands.</p>
<p>Ideally in a lossless situation we would need to apply 100kips tension at the jack.</p>
<p>But there are many factors that cause the loss of prestress which need to be counted for and added to the jack's force.</p>
<ul>
<li><p>Concrete beam will strain under the compression stress caused by the strands and cause relaxation. this depends on the quality of concrete and section properties.</p>
</li>
<li><p>Shrinkage of the concrete again depending on the concrete and the humidity of environment and some more factors.</p>
</li>
<li><p>Creep of the concrete. depends on concrete and environment, humidity, the beam will perform.</p>
</li>
<li><p>Mechanical losses at the jack bulkhead and material properties and details of the jack.</p>
</li>
</ul>
<p>The above relaxing factors can be complicated and sometimes are just assumed arbitrarily from past performances. Or sometimes there are codes by Caltrans, for example, specifying the loads and factors and equations that apply in each case.</p>
| 45571 | How to calculate the required prestressed jacking pressure to apply to a pre-tension concrete beam? |
2021-08-13T16:11:24.463 | <p>This is the basic requirement of the wooden door frame:</p>
<p><a href="https://i.stack.imgur.com/ugGro.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ugGro.jpg" alt="enter image description here" /></a></p>
<p>The requirement is to make sure that the door frame does not skew or deform.</p>
<p><strong>Questions:</strong></p>
<ul>
<li>What's the minimum number of wooden pieces to be added inside the frame, so that the frame of the door remains solid, and doesn't deform?</li>
<li>What's the principled approach to handle this? Because currently I mostly use <em>eyeballing</em> and intuitive thinking.</li>
</ul>
<hr />
<h1>1. My guesses so far</h1>
<h2>1.1. Extend from corners at 45 degrees</h2>
<p><a href="https://i.stack.imgur.com/MrXaf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MrXaf.jpg" alt="enter image description here" /></a></p>
<h2>1.2. Extend from corners to opposite, regardless of degree</h2>
<p><a href="https://i.stack.imgur.com/GIcES.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GIcES.jpg" alt="enter image description here" /></a></p>
<h1>2. Surprises</h1>
<p>I'm a bit surprised about how others do their frames, without angular pieces. For example, checkout this:
<a href="https://i.stack.imgur.com/QQHmY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QQHmY.png" alt="enter image description here" /></a></p>
<p>No angular pieces at all. What's wrong? Am I missing something? Or are the engineers missing something?</p>
| |structural-engineering|structural-analysis|wood|door| | <p><strong>One</strong></p>
<p>A gate like that will have a single diagonal from upper right to lower left. All you really need to do is turn the rectangle into triangles.</p>
| 45584 | What's the minimum number of required wooden pieces to make this door structually solid? |
2021-08-15T14:41:01.560 | <p>Hello I am designing a Aluminium fiber connector as shown below</p>
<p><a href="https://i.stack.imgur.com/f0Y88.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f0Y88.png" alt="enter image description here" /></a></p>
<p>I am tasked to maintain the object at a thermal equilibrium of 25 degrees Celsius. The object will be heated with 50W of power. I have been reading up about how much heat will a certain amount of power generate. <a href="https://sciencing.com/how-8643971-convert-wattage-degrees.html" rel="nofollow noreferrer">https://sciencing.com/how-8643971-convert-wattage-degrees.html</a>. Is this applicable to my situation?</p>
<p>After knowing the temperature increase due to the power source, how do I achieve a thermal equilibrium of 25 degrees Celsius? From my understanding, the thermal diffusivity must equal the rate of thermal conduction for it to be in equilibrium? I do not know how to relate the thermal diffusivity formula to thermal conduction.</p>
| |mechanical-engineering|thermodynamics| | <p>If 50 W is being dissipated in the device, then to maintain <em>any</em> constant temperature, for <em>any</em> material, you must remove 50 W. This is independent of the material properties.</p>
<p>If you choose passive cooling, then the device will heat up to an equilibrium temperature for which the heat loss to the environment is 50 W. Consider the simple <a href="https://en.wikipedia.org/wiki/Lumped-element_model" rel="nofollow noreferrer">lumped-capacitance model</a>, in which the material temperature is idealized as uniform (this is a reasonable assumption when the thermal conductivity is high). Assuming that heat is lost primarily through convection, then an energy balance gives <span class="math-container">$P=hA(T-T_\mathrm{env})$</span>, where <span class="math-container">$P$</span> is the power, <span class="math-container">$h$</span> is the convective coefficient, <span class="math-container">$A$</span> is the surface area, <span class="math-container">$T$</span> is the device temperature, and <span class="math-container">$T_\mathrm{env}$</span> is the ambient environmental temperature.</p>
<p>Perhaps <span class="math-container">$h=1\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$</span> for natural convection, and perhaps <span class="math-container">$T_\mathrm{env}=20^\circ \mathrm{C}$</span>. Then, for <span class="math-container">$P=50\,\mathrm{W}$</span> and a surface area of <span class="math-container">$A=0.1\,\mathrm{m}^2$</span>, say, the equilibrium temperature is projected to be <span class="math-container">$520^\circ\mathrm{C}$</span>. To get this down to <span class="math-container">$25^\circ\mathrm{C}$</span>, you might have to encase the device in a block of a high-thermal-conductivity material (such as copper) to give a surface area of <span class="math-container">$A=1\,\mathrm{m}^2$</span> and blow gas past it to obtain a higher forced-convection coefficient of <span class="math-container">$h=10\,\mathrm{W}\,\mathrm{m}^{-1}\,\mathrm{K}^{-1}$</span>, for example. Or you could use fins, a classic approach for increasing surface area. Alternatively, you could run coolant lines through the device or the encasing block with fluid at <span class="math-container">$<20^\circ\mathrm{C}$</span> to carry away the heat. But it seems clear that passive cooling alone won't be anywhere near sufficient.</p>
| 45606 | Thermal Conduction and Thermal Diffusivity |
2021-08-17T02:41:15.140 | <p>Let us say I simply design a low pass filter like (1/s+1) with the cutoff frequency as 1 rad/sec, when I implement it in real software, do I have to do the discrete-time realization? If not, what issues will I have? How to design a discrete low pass filter that have the same cutoff frequency?</p>
| |design|pid-control| | <p>For a great many engineering uses, you can approximate your idealized s-plane filter, into a z-plane equivalent, by one of the transformations mentioned in AJN's answer, using closed-form "cookbook" methods.</p>
<p>If the upper frequency is comfortably below Nyquist freq. (including saving room for the transition band where the roll-off happens), there will be few problems. Just as in analog, neither ideal differentiators nor integrators are really possible, since both have infinite response (at low and high f, respectively).</p>
<p>Start with Bilinear transform. Look up how to do the adjustment for "frequency warping". For most day-to-day uses, that should get you there.</p>
<p>Other notes</p>
<ul>
<li>In some implementations, there may also be issues with arithmetic precision (for instance with very-low frequency dynamics) so be mindful of that</li>
<li>Don't implement higher orders (e.g. 4-pole or 6-pole) directly, instead decompose into a series of "biquadratic" building blocks.</li>
<li>If implementing "integrators", and data originates from an integer form (e.g. ADC), try to prevent accumulated rounding errors.</li>
</ul>
<p>If you must get closer to Nyquist frequency, designing directly in the z-plane might be something to look at. Also check out the <a href="https://dsp.stackexchange.com/">DSP StackExchange</a>, there may be more questions of this type there.</p>
| 45631 | Discrete-time realization for continuous time domain controller/filter? |
2021-08-17T08:31:34.923 | <p>I was trying today to calculate the terminal velocity of a hailstones with increasing diameter and mass. I was trying to figure out if larger hailstones will have a higher impact velocity or lower.</p>
<p>I performed the calculation (I was a bit surprised by the end result), which I am presenting as a potential answer. I would be very happy to see if there are any improvements to this answer (if there are additional factors that I need to consider)</p>
| |applied-mechanics|dynamics|aerodynamics|drag| | <p>The terminal velocity <span class="math-container">$V_t$</span> will be reached when the drag coefficient is equal to the force of gravity:</p>
<p><span class="math-container">$$F_{drag} = mg$$</span>
<span class="math-container">$$C_D\frac{1}{2}\cdot \rho_{air} \cdot A \cdot V_{t}^2= m\cdot g$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$C_D$</span>: is the drag coefficient (for a sphere is 0.5)</li>
<li><span class="math-container">$\rho_{air}$</span> is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)</li>
<li><span class="math-container">$A$</span>: cross-sectional area of the hailstone assuming its a sphere <span class="math-container">$\frac{\pi d^2}{4} =\pi r^2 $</span></li>
<li><span class="math-container">$V_{t}$</span>: the terminal velocity of the sphere</li>
<li><span class="math-container">$m$</span>: the mass of the sphere (assuming its a sphere the volume is <span class="math-container">$\frac{4}{3}\pi r^3$</span>, and the density of the sphere is <span class="math-container">$\rho_{sphere}$</span>)</li>
<li><span class="math-container">$g$</span>: the acceleration of gravity</li>
</ul>
<p>Therefore:
<span class="math-container">$$C_D\frac{1}{2} \rho_{air} \cdot \left(\pi r^2\right)\cdot V_{t}^2= \frac{4}{3}\pi r^3 \rho_{sphere}\cdot g$$</span></p>
<p><span class="math-container">$$ V_{t}^2= \frac{4\cdot 2 \cdot\pi r^3 \rho_{sphere}\cdot g}{3C_D \rho_{air} \cdot \left(\pi r^2\right)}$$</span>
<span class="math-container">$$ V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere}}{\rho_{air} }\cdot r }$$</span></p>
<p>which indicates that the terminal velocity of a sphere increases proportionally to the square root of the diameter (radius).</p>
<hr />
<p>if the buoyancy is considered then:</p>
<p><span class="math-container">$$ V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere} - \rho_{air
}}{\rho_{air} }\cdot r }$$</span></p>
| 45632 | Does the terminal velocity of a sphere in free drop increase with increasing diameter? |
2021-08-17T14:07:51.373 | <p>Assume a sphere (hail) that is free dropping in air. At some the sphere will reach close to the terminal velocity. The question is:</p>
<ul>
<li>how long <span class="math-container">$t_p$</span> in seconds does it take before a percentage <span class="math-container">$p$</span> of the terminal velocity is reached</li>
<li>how far <span class="math-container">$t_p$</span> in m does the sphere travel before a percentage <span class="math-container">$p$</span> of the terminal velocity is reached</li>
</ul>
<hr />
<p>This is an extension to a <a href="https://engineering.stackexchange.com/questions/45632/does-the-terminal-velocity-of-a-sphere-in-free-drop-increase-with-increasing-dia">previous question</a>.</p>
<p>Although I found (eventually) the equation for terminal velocity, I couldn't find the time and the distance, so I opted for deriving them. I'm putting up the extended derivation for scrutiny, and/or alternative answers.</p>
| |applied-mechanics|dynamics|aerodynamics|drag| | <p>I suggest using the relationship below to derive the equations you are looking for.</p>
<p><span class="math-container">$F = ma$</span></p>
<p><span class="math-container">$W = mg$</span></p>
<p><span class="math-container">$D = \dfrac{C_d\rho V^2A}{2}$</span></p>
<p><span class="math-container">$C_d$</span> = Drag Coefficient (Shape dependent)</p>
<p><span class="math-container">$\rho$</span> = Atmospheric Density</p>
<p>The terminal velocity is reached when <span class="math-container">$W = D$</span>,</p>
<p><span class="math-container">$mg = \dfrac{C_d\rho V^2A}{2}$</span>, thus</p>
<p><span class="math-container">$V_t = \sqrt{\dfrac{2mg}{C_d \rho A}}$</span></p>
<p>Note, at this stage, <span class="math-container">$a = \dfrac{mg - D}{m} = 0$</span>.</p>
<p>Force equilibrium needs to be maintained throughout the fall:</p>
<p><span class="math-container">$F = ma = \dfrac{dV}{dt} = mg -D$</span></p>
<p><span class="math-container">$\dfrac{dV}{dt} = g(1 - \dfrac{C_d\rho A}{2mg})V^2$</span></p>
<p>From here, you shall be able to derive the equation for "<span class="math-container">$t$</span>" (see ref. 2 for derivation), and distance traveled "<span class="math-container">$s$</span>" at any given time.</p>
<p><strong>ADD:</strong> On approximately halfway through the linked wiki article (ref 2), at the bottom of "Derivation for Terminal Velocity", there is a boxed text that says "<strong>Derivation of the solution for the velocity v as a function of time t</strong>". To its far-right, click the link "show", the text will open to view.</p>
<p>References:</p>
<ol>
<li><p><a href="https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html" rel="nofollow noreferrer">https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html</a></p>
</li>
<li><p><a href="https://en.wikipedia.org/wiki/Terminal_velocity#:%7E:text=Based%20on%20wind%20resistance%2C%20for%20example%2C%20the%20terminal,more%20closely%20as%20the%20terminal%20speed%20is%20approached" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Terminal_velocity#:~:text=Based%20on%20wind%20resistance%2C%20for%20example%2C%20the%20terminal,more%20closely%20as%20the%20terminal%20speed%20is%20approached</a>.</p>
</li>
</ol>
| 45641 | How long does it take and how far does it travel a sphere in free fall before reaching a percentage of the terminal velocity? |
2021-08-19T03:23:14.190 | <p>I want to semi-automate the process of loading products (in the form of cardboard box) to trucks.</p>
<p><em><strong>Current process</strong></em> utilizes manual pallet jack to bring the pallet full of cases inside the trucks, then the box will be stacked manually, so in total there is three worker involved on the process.</p>
<p><em><strong>What I'm looking for</strong></em> is to <strong>try to increase the efficiency</strong> (more boxes handled in shorter amount of time), and my first idea is by removing the manual pallet transfer and changing it to transfer the boxes through a conveyor inside the truck. But, I'm still not sure if this is the most efficient way to do this. <strong>Is there any other way to increase the efficiency?</strong></p>
<p>The rules that needs to be obeyed on this improvement: <strong>no pallet inside the trucks, because the shipment needs to be non-palletized</strong></p>
<p>Other solutions that I have explored is the slip-sheet loading method, but this involves renting forklifts and could be much more expensive.</p>
| |mechanical-engineering|industrial-engineering| | <p>Solutions from cheapest to most expensive:</p>
<ol>
<li>Extendable non-powered conveyor. The first thing to do is stop building pallets so that you can take boxes off the pallet later. The first rule of waste reduction is avoid multiple handling actions. <a href="https://www.grainger.com/product/55KR72?ef_id=CjwKCAjwgviIBhBkEiwA10D2jzYainvP-Q2-iGsjt2elx_c_NRWEn40iIa6i8_Xs42pvsADOjQQWWRoC4skQAvD_BwE:G:s&s_kwcid=AL!2966!3!496359972664!!!g!437101822733!&gucid=N:N:PS:Paid:GGL:CSM-2295:4P7A1P:20501231&gclid=CjwKCAjwgviIBhBkEiwA10D2jzYainvP-Q2-iGsjt2elx_c_NRWEn40iIa6i8_Xs42pvsADOjQQWWRoC4skQAvD_BwE&gclsrc=aw.ds" rel="nofollow noreferrer">Example</a> These are extremely common.</li>
<li>Powered version of the above. <a href="https://www.gilmorekramer.com/more_info/best_flex_15_power_roller_conveyor/best_flex_15_power_roller_conveyor.shtml" rel="nofollow noreferrer">Example</a> Common for operations with a bit more cash and for unloading that trailer when it gets to its destination.</li>
<li>Semi-permanent extending conveyor. Usually attached to a powered inlet conveyor. <a href="https://www.gilmorekramer.com/more_info/ntl_drive_out_boom/ntl_drive_out_boom.shtml" rel="nofollow noreferrer">Example</a></li>
<li>Permanent extendable conveyer. These are typically attached to an incoming conveyor system that drops boxes to the right dock door. Big Box retailers typically have one dock door per store, fed by an automated system. <a href="https://www.gilmorekramer.com/more_info/tl2_heavy_duty_2_stage_gravity_truck_loader/tl2_heavy_duty_2_stage_gravity_truck_loader.shtml" rel="nofollow noreferrer">Example 1</a> <a href="https://www.gilmorekramer.com/more_info/maxxreach_telescopic_conveyor/maxxreach_telescopic_conveyor.shtml" rel="nofollow noreferrer">Example 2</a></li>
</ol>
<p>There are issues with extending conveyors, namely that you can no longer drive along the dock. You could theoretically also have robots move product onto a conveyor from a dropped pallet. The other option is live/walking floor trailers, but this really only works for dedicated lanes between your own facilities.</p>
<p>ETA: Please do not consider slipsheets. These will increase your damage rates, require more lift truck driver training, require expensive push-pull attachments, and you have to buy the slip sheets themselves, which is 100% waste. If you are stacking to the ceiling you will still have to throw cases onto the top of the slip-sheeted loads. Similarly, avoid clamp attachments unless you have a product that absolutely can take it.</p>
| 45663 | Options on 'semi'-automating the loading of non-palletized products into trucks |
2021-08-19T04:43:52.830 | <p>In effectiveness NTU method for heat exchangers, we have effectiveness of the heat exchanger defined as</p>
<p><span class="math-container">$$\epsilon = \frac{Q_{actual}}{Q_{max}}$$</span></p>
<p>Where <span class="math-container">$$Q_{max} $$</span> is the maximum rate of heat transfer in a counter flow heat exchanger with some prescribed inlet temperature and heat capacity rates.</p>
<p><span class="math-container">$$\text{when}\begin{cases}
C_H< C_C, & \dot{Q}_{max} = C_H(T_{h,i}- T_{c,i})\\
C_C< C_H, & \dot{Q}_{max} = C_C(T_{h,i}- T_{c,i})\\
\end{cases} $$</span></p>
<p>What will be the maximum rate of heat transfer that will be taken when heat exchanger is balanced i.e. <span class="math-container">$$C_h = C_c $$</span></p>
| |mechanical-engineering|heat-transfer|heat-exchanger| | <p>When the two heat capacities are the same, then Hot out = cold in, the same as the CH < CC. You can't get more heat out of the hot side than the delta between the hot entering temp and the cold entering temp, so this max occurs for all conditions where the cold side heat capacity is equal to or greater than the hot side.</p>
| 45664 | Maximum rate of heat transfer in a balanced counter flow heat exchanger in NTU method |
2021-08-19T12:25:23.100 | <p>I spotted this heavy machinery/loader attachment at a road reconstruction job site in Ontario:</p>
<hr />
<p><a href="https://i.stack.imgur.com/6QzrS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6QzrS.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/TgzqD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TgzqD.jpg" alt="enter image description here" /></a></p>
<hr />
<ul>
<li>The device a has horizontal tube that is cut off at an angle.</li>
<li>The tube is attached to a vertical mounting plate.</li>
<li>The tube portion appears to be custom built -- the angled cut through the tube is too rough to be a factory made.</li>
</ul>
<hr />
<p>Question:</p>
<p>What is the purpose of this device?</p>
| |civil-engineering|mechanical| | <p>It looks like it's being used to backfill narrow watermain trenches with gravel.</p>
<p><a href="https://i.stack.imgur.com/9wozH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9wozH.jpg" alt="enter image description here" /></a></p>
<hr />
<p><strong>Edit:</strong></p>
<p>It's also used for filling up an excavator bucket with gravel -- so that the excavator doesn't need to move out of position to go over to the gravel pile.</p>
<p><a href="https://i.stack.imgur.com/5FtVF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5FtVF.jpg" alt="enter image description here" /></a></p>
<p>It's hard to tell from the photo, but the excavator in the back is straddling a narrow trench. And the gravel pile is too far away for it to reach.</p>
<p>It seems like it would be an ordeal for the excavator to move away from its position over the trench every time it needed gravel. And in this particular location, there's not enough room for the loader to drive up and dump gravel directly into the trench.</p>
| 45670 | Heavy machinery/loader attachment with custom tube |
2021-08-19T15:35:04.483 | <p><a href="https://i.stack.imgur.com/Ex2uP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ex2uP.jpg" alt="enter image description here" /></a></p>
<p>My vehicle built in '08 suffered extreme damage to the dashboard, it permanently melted and is now sticky to the touch even when cool.</p>
<p><a href="https://i.stack.imgur.com/mSZ64.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mSZ64.jpg" alt="enter image description here" /></a></p>
<p>Presumably this is why so many vehicle owners use covers to block the sunlight when parked. I never liked the idea of these, they seem like a hassle that better engineering could fix.</p>
<p>Have there been recent manufacturing / vehicle design improvements across the automotive industry so future cars will be resistant to high temperatures? Or is this still a common problem? If so, what fundamental trait of car interior materials causes this issue to be either so hard or so expensive to solve?</p>
| |materials| | <p>I want to add to the answers above the other side of the story, the car's history of sun exposure and maintenance.</p>
<p>Some cars are parked mostly under the shade or indoors. Some are parked in harsh climates and exposed to extreme sunlight without even so much a window left a crack open to reduce the heat.</p>
<p>I have a car that was exposed mostly to the sun coming from the driver side over many years and the dashboard shows small cracks on that side.</p>
<p>Sometimes one uses the wrong chemicals to polish the dashboard, they may look nicer while the built-in UV resistive coat is being compromised.</p>
| 45675 | Why are vehicle interiors susceptible to damage from the sun when some plastics seem to be fine? |
2021-08-19T22:29:40.810 | <p>I saw a video showing a water bowl for a cat that used an upside down PET bottle to provide "fresh" water when cats consume some of the water in the bowl. I wanted to design something similar.</p>
<p>Here is a very very high level sketch:</p>
<p><a href="https://i.stack.imgur.com/zZISu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zZISu.jpg" alt="enter image description here" /></a></p>
<p>The idea is that an upside down water bottle is screwed into the top hole (on the left) and some sort of piping will lead the larger hole on the right to fill up with water. Naturally due to the pressure the water will overflow in the hole on the right.</p>
<p>My question is:
Is there a way to prevent the water from overflowing in the lower basin when a water bottle has much more water stored than the basin can hold? I am very unversed in those topics so I unfortuantely even failed to find the right terms to search for in google...</p>
| |fluid-mechanics|piping| | <p>The water level in the lower basin will not exceed the level of the bottom of the bottle opening. When the empty device is loaded with a full bottle (by keeping the bottle upright!) and then inverted, the basin will allow air into the bottle, which in turn allows water to enter the basin.</p>
<p>When the basin water level reaches the opening of the overturned bottle, air can no longer enter, which prevents more water from leaving.</p>
<p>As water evaporates or is consumed by the pet, the lowered basin water level again allows air to enter, repeating the cycle.</p>
<p>There is a bit of disparity between the bottom of the inverted bottle and the top level of the water, which is related to air pressure. For additional information and possibly some involved mathematics, consider to search for "physics of inverted bottle water dispensers." There is a <a href="https://physics.stackexchange.com/questions/88669/physics-of-the-inverted-bottle-dispenser">Physics SE related question</a> which nearly matches your question.</p>
| 45685 | Watering utility that does not spill |
2021-08-21T08:17:49.047 | <p>Hollow steel sections are used often in trusses:</p>
<p><a href="https://i.stack.imgur.com/hyVx6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hyVx6.png" alt="enter image description here" /></a></p>
<p>I've been taught that the advantage of a truss structure is that it can be stiff without any of the members being under bending moment. This is an advantage especially for the joints, which can now be designed nominally pinned.</p>
<p>In most structural codes, we categorize joints as rigid or pinned by comparing the stiffness of the joint to the stiffness of the connecting element. Eurocode for example gives an analytical method for calculating the stiffnesses of various beam-to-column end plate joints. However, I am interested in similar methods for hollow sections, especially RHS. Eurocode gives lot of info on how to calculate the axial resistance of these joints, but not their moment stiffness:</p>
<p><a href="https://i.stack.imgur.com/9zPnS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9zPnS.png" alt="enter image description here" /></a></p>
<p>So my question is, how do we design such a steel truss joint to be pinned or rigid? I am mostly interested in design to Eurocode, but if other codes have more on this topic I would be interested in them also.</p>
| |structural-engineering|steel| | <p>Assessing the stiffness of a hollow steel section joint typically involves evaluating the joint's structural behavior and calculating its stiffness characteristics. Here are some general steps to help you assess the stiffness of a hollow steel section joint:</p>
<p>Understand the joint type: Determine the specific type of joint you are working with, as different joint configurations have varying stiffness characteristics. Common types include welded connections, bolted connections, or a combination of both.</p>
<p>Gather information: Obtain relevant information about the joint, such as the dimensions and properties of the hollow steel sections involved, the connection type, and any additional components or reinforcement of <a href="https://keytoinfo.com/" rel="nofollow noreferrer">keytoinfo</a> present in the joint.</p>
<p>Analyze the joint geometry: Assess the geometry of the joint to understand how the hollow steel sections are connected. Consider factors such as the presence of welds or bolts, the arrangement of the joint, and any additional reinforcement used.</p>
| 45696 | How to assess the stiffness of a hollow steel section joint? |
2021-08-21T12:05:29.583 | <p>I live in an area that has an abundance of river rock and a shortage of crushed stone. Can river rock be used in place of crushed stone as a sub base for foundations, roads, etc. or does the smoothness cause problems when it comes to movement, compaction, settling, etc.?</p>
<p>My gut tells me that the "sharp" edges of crushed stone will result in better interlocking when compacted and thus less shifting/movement. However, I don't have any actual evidence/data to back this up.</p>
<p>Searching around the internet suggests that river rock is used predominately for decorative tasks, but most of the resources are also from the US where I believe river rock is significantly less available (more expensive) than crushed stone and it could just be that people only use it decoratively because it is not economical to use it as a sub base.</p>
| |structural-engineering|civil-engineering| | <p>Yes, it can be used in foundations and subbases of roadways if the rocks meet property and gradation requirements. However, crushed stone/gravel is often the choice for the base material.</p>
<p>Base course is a layer of the pavement structure immediately beneath the surface course. It typically consists of high quality aggregate such as crushed gravel, crushed stone, or sand that provides a uniform foundation support and an adequate working platform for construction equipment. Base may consist of unbound materials, such as gravel or crushed stone, or stabilized materials, such as asphalt-, cement- or lime-treated materials.</p>
<p>The subbase course is typically a <strong>granular borrow</strong> that is placed between the base and subgrade. It can be constructed as either a treated or untreated layer. Untreated or unbound aggregate subbase layers are characterized in a manner similar to the subgrade in pavement design. The material quality requirements of strength, plasticity, and gradation for subbase are not as strict as for a base. The subbase course must be better quality than the soil subgrade, the subbase is often omitted if soil subgrades are of high quality.</p>
<p>Depending on site conditions, subgrade improvements may also be performed. However, the role of different base and subbase layers and rationale for using different base types and layering are not well documented as many agencies specify standard or typical base and subbase layers based on historical performance and their own experiences. For example, it is unclear where and why a treated base should be used, or why one type of treated base is preferred over another.</p>
| 45697 | Can river rock be used as a subbase in place of crushed stone? |
2021-08-22T01:40:48.710 | <p>This is a toy for infants. My son was playing with it, and I started wondering how you could manufacture this. I can't think of any reasonably cheap way it could be done.</p>
<p><a href="https://i.stack.imgur.com/26etE.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/26etE.jpg" alt="spherical toy" /></a></p>
<p>I believe it's made of plastic and not silicone. It's flexible, but not elastic.</p>
<p>My first thought would be injection molding, but I can't see any way to get parts of the mold in and out of the sphere. My wife suggested that perhaps the mold had no interior surface and was rotated like they do with chocolate, but I don't think that's the case, because if you look carefully you'll see that the inside surface is rounded and not flat.</p>
<p>Anyway I'm stumped. How do you think they did it?</p>
| |injection-molding| | <p>OP injection molding tag is correct. OBall uses injection molding and plastic welding.</p>
<p>The OBall is the invention of David E. Silverglate.</p>
<p><a href="https://patents.google.com/patent/US20120028743A1/en" rel="nofollow noreferrer">Toy Ball Apparatus with Reduced Part Count</a></p>
<p><a href="https://i.stack.imgur.com/vireA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vireA.jpg" alt="OBall" /></a></p>
<p>Reduced image from <a href="https://www.kids2.com/products/10340-oball-classic-easygrasp-toy-redbluegreenyellow" rel="nofollow noreferrer">Kids II</a>.</p>
<p>It consists of four identical, flat, injection molded, pentagon and hexagon shapes with circular (or elipitical) holes, which are shaped and plastically welded into spheres.</p>
<p>Pentagon and hexagon edges are the same size and individual connected circles are only connected along one edge. The four shapes are clearly shown in colors above and from the patent. Solid lines on each part are hard connections, while dashed lines represent connections within the part. Patent shows how individual edges on parts are interconnected. Injection molding with a plastic (PETE [or PET] from Jim Clark link: <a href="https://treehozz.com/what-is-oball-made-of" rel="nofollow noreferrer">What is Oball made of?</a>) at a higher melting material and welded with a plastic at a lower melting material.</p>
<p>Figure 8 from the patent.
<a href="https://i.stack.imgur.com/gKWij.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gKWij.png" alt="enter image description here" /></a></p>
<blockquote>
<p>[0022] The shape and number of the mesh components 14 are designed in a manner that decreases manufacturing costs incurred using a process such as <strong>injection molding</strong>. Regarding the number of mesh components 14, it will be appreciated that when four mesh components 14 are utilized the production time may be significantly reduced when compared to a toy ball apparatus 10 having ten mesh components. The decreased production time may in turn decrease the toy ball apparatus's manufacturing cost.</p>
</blockquote>
<blockquote>
<p>[0029] FIG. 9 illustrates toy ball apparatus 10 in its assembled State, in which the plurality of mesh components 14a, 14b, 14c, and 14d have been coupled to enclose the closed Volume 20 and form the mesh 12, by joining adjacent mesh components 14 along their cooperative mating Surfaces 19 and securing the mesh components 14 together, for example, by <strong>plastically welding</strong> the mesh components 14 together along the cooperative mating Surfaces 19.</p>
</blockquote>
<hr />
<p>Each part has 3 pentagons and 5 hexagons.</p>
<p>Clearly Part B and C are the same. But A and D are obfuscated by the patent lawyers. Edge numbering indicates they knew all parts were the same. They just divided it to obscure the true inovation.</p>
<p>In all 4 parts, central hexagon has no writing on it. It is surrounded by 3 pentagons and 3 hexagons. The final hexagon with A1,A2,A3&A4 (or B1,B2,B3&B4, etc.) connections are always connected to hexagon with A16&A17 (or B16&B17, etc.) connections.</p>
<p>So tooling is for one part, which means true inovation was making ball from 4 identical flat injection molded parts.</p>
<p>SolidWorks drawing of ball made from 4 identical parts.</p>
<p><a href="https://i.stack.imgur.com/Pq3ir.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pq3ir.png" alt="enter image description here" /></a></p>
| 45701 | How was this toy made? |
2021-08-22T01:58:03.453 | <p>My question is how can all of these bricks be supported by the thin columns on either side? How can there be that much empty space under all of the bricks that the arrow is going through? Looking at it I am under the impression that it should collapse because there are no bricks underneath to support such a huge area. It also does not look like there is a beam or anything else directly under the rectangle of breaks so they should fall down.</p>
<p><strong>In the second image</strong>, there is a stone support to support the weight of the bricks above but the first image does not.</p>
<p>The <strong>third image</strong> looks like it should collapse as well.
<a href="https://i.stack.imgur.com/oMsY0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oMsY0.jpg" alt="brick building" /></a></p>
<p><a href="https://i.stack.imgur.com/d7XZw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d7XZw.jpg" alt="brick building with stone structural support" /></a></p>
<p><a href="https://i.stack.imgur.com/EiFTs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EiFTs.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|materials|structural-analysis|building-physics|construction-management| | <p>This was intended more a comment regarding the third image, (however I opted to put a</p>
<p>The confusing part for me is, the arch like section in the middle, and the differences between the style of the walls, and the ceiling (which appears to be wood). So it seems to me to be some type of renovation.</p>
<p>If you blow up the image, you will notice that the final row has a different orientation.</p>
<p><a href="https://i.stack.imgur.com/UYmHe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYmHe.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/OMBIA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OMBIA.png" alt="enter image description here" /></a></p>
<p>Although this might be to produce a visual effect, another interpretation is that the bricks clad to an underneath structure.</p>
<p><a href="https://i.stack.imgur.com/nvNJh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nvNJh.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/KpPvr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KpPvr.png" alt="enter image description here" /></a></p>
<p>Sometimes the brick weight also is not that great, and the end effect can look very convincing </p>
<p><a href="https://i.stack.imgur.com/oZR1J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oZR1J.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/pp4Lk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pp4Lk.png" alt="enter image description here" /></a></p>
<p>My point is that without a closer inspection its difficult to tell what is happening. I agree though that if that are indeed chemically bonded bricks without any underlying steel structure I wouldn't want to live in the third house.</p>
| 45702 | How are these bricks being supported? |
2021-08-22T13:26:56.260 | <p>The motor on my awning just bit the dust (again). This is the second time this has happened.</p>
<p>I'm looking to replace it with something that I can build myself, using the existing motor enclosure. You can see the exact motor I'm talking about below (amazon link). I just don't want to pay that much money for a replacement. I'd rather rip the burnt out one apart and buy a cheap motor to do the same thing.</p>
<p>What I'm struggling with is what motor to get that would handle a 20ft awning. And to fit in the existing enclosure, the diameter has to be 3cm.</p>
<p>I found the following motor, but I believe it may be way under powered to do it. Dimensions wise, it's like perfect for the enclosure that the fried motor is in.</p>
<p><a href="https://www.amazon.ca/gp/product/B00HDDXBEY/ref=ox_sc_act_title_1?smid=A1IQ6DRJX762AU&psc=1" rel="nofollow noreferrer">https://www.amazon.ca/gp/product/B00HDDXBEY/ref=ox_sc_act_title_1?smid=A1IQ6DRJX762AU&psc=1</a></p>
<p>Here are some pictures of the burnt out motor (removed from the enclosure):</p>
<p><a href="https://i.ibb.co/wBhFnsh/awning-motor2.jpg" rel="nofollow noreferrer"><img src="https://i.ibb.co/wBhFnsh/awning-motor2.jpg" alt="ddasd" /></a></p>
<p><a href="https://i.ibb.co/pnFyF58/awning-motor1.jpg" rel="nofollow noreferrer"><img src="https://i.ibb.co/pnFyF58/awning-motor1.jpg" alt="ddasd" /></a></p>
<p><strong>OEM Replacement motor:</strong></p>
<p>DOMETIC 3310423.209U Torsion Assembly</p>
<p><a href="https://www.amazon.ca/Dometic-3310423209U-Drive-Assembly-Awning/dp/B07L3JF78D/ref=asc_df_B07L3JF78D/?tag=googleshopc0c-20&linkCode=df0&hvadid=347072134503&hvpos=&hvnetw=g&hvrand=3888389780963375681&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9001077&hvtargid=pla-843687300391&psc=1" rel="nofollow noreferrer">https://www.amazon.ca/Dometic-3310423209U-Drive-Assembly-Awning/dp/B07L3JF78D/ref=asc_df_B07L3JF78D/?tag=googleshopc0c-20&linkCode=df0&hvadid=347072134503&hvpos=&hvnetw=g&hvrand=3888389780963375681&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9001077&hvtargid=pla-843687300391&psc=1</a></p>
| |motors| | <p>To answer your original question: it's unsure. It depends on what we calculate vs what the products say</p>
<p><strong>What the products say</strong></p>
<p>The original has, according to the label, 20 InLb.</p>
<p>The replacement is 2.33 kg per cm, which is 2,02 LbIn, which is 24InLb.</p>
<p>The replacement has 4 LbIn to space compared to the original. This is likely due to the lower RPM, increasing torque (much like a pulley system. The box will go up slowly, but you'll have an easier time despite it needing the same amount of power total.).</p>
<p><strong>What we can calculate</strong></p>
<p>This is most likely the case, as the original does have much more power in Watt (12V * 1.85A vs 12V * 0.098A). From that I would calculate:</p>
<ul>
<li><p>12V*1.85A = 22.2 W</p>
</li>
<li><p>22.2W with 35 revolutions per minute means 6.06NM or 4.48Lb-Ft.</p>
</li>
<li><p>12V*0.098A = 1.176W</p>
</li>
<li><p>1.176W with 25 revolutions per minute means 0.45NM or 0.33Lb-Ft.</p>
</li>
</ul>
<p>Unfortunately I can't take inefficiencies of motors in account, but it's clear the packaging and the calculations might not align. Maybe the LbIn and InLb have accidentally been switched or something.</p>
| 45713 | Help finding motor to replace awning motor (want to jerry rig something together) |
2021-08-23T21:02:51.863 | <p>Figure (a) shows the problem and (b) the free-body diagram. The question is to verify the reactions at C and E.</p>
<p>I understand the remainder of the free-body diagram but do not see how C and E are determined.</p>
<p>I tried force x perpendicular distance from pivot C but don't get the correct answer: (1500N x 3.5m) + (900 x 0.5) = 5700N</p>
<p>Any help appreciated.</p>
<p><a href="https://i.stack.imgur.com/lI4A2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lI4A2.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|statics| | <p>Globally the sum of moments must be zero to be stable. You can find one of the reactions from taking moments about one of the supports (C or E), then get the rest by satisfying the equilibrium requirements <span class="math-container">$\sum F_Y = 0$</span> and <span class="math-container">$\sum F_X = 0$</span>.</p>
<p><a href="https://i.stack.imgur.com/bHnfA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bHnfA.png" alt="enter image description here" /></a></p>
<p><strong>Assume roller support at C with restrain in X-direction only, and pin support at E.</strong></p>
<p>Say <span class="math-container">$\sum M_E = 0$</span>, let clockwise moment be positive, then</p>
<p><span class="math-container">$F_{CX}*1.5 - 1500*5 - 900*2 = 0$</span></p>
<p><span class="math-container">$F_{CX} = (1500*5 + 900*2)/1.5 = +6200$</span> - pointing to right (to produce +M as assumed).</p>
<p><span class="math-container">$\sum F_Y = 0$</span>, let the upward force be positive, thus</p>
<p><span class="math-container">$F_{EY} - 1500 - 900 = 0$</span></p>
<p><span class="math-container">$F_{EY} = 1500 + 900 = +2400$</span> - Acting upward.</p>
<p>The last, <span class="math-container">$\sum F_X = 0$</span>, let positive force pointing to the right, so</p>
<p><span class="math-container">$F_{CX} + F_{EX} = 0$</span></p>
<p><span class="math-container">$F_{EX} = -F_{CX} = -6200$</span> - pointing to left.</p>
<p>Note that I've changed the nomenclature of the reactions to be more consistent with common practice.</p>
| 45731 | Mechanics Statics: verifying reactions at supports |
2021-08-24T03:17:39.773 | <p>I am designing a heat sink, the diameter of the hole is 6mm. and the diameter of the fibre rod is 1.2mm.</p>
<p><a href="https://i.stack.imgur.com/DcxM0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DcxM0.png" alt="enter image description here" /></a></p>
<p>The material I am using is aluminium, with thermal conductivity of 230. The ambient temperature is 25 degrees celsius. Assume that the boundary condition on the fiber rod wall is at a temperature of 70 degrees celsius.</p>
<p>I am trying to determine the temperature of thermal equilibrium. I am facing trouble calculating the Convection Coefficient ( h ).</p>
<p>I first calculate the heat flux through the entire object through conduction, Using Fourier's Law, <span class="math-container">$\vec q = -k ( T_s - T_\infty) $</span>. <span class="math-container">$\vec q = 230 \times ( 70 - 25 )$</span> which gives <span class="math-container">$10350W/m^2$</span>.</p>
<p>Next, I calculate the convection from the surface to the ambient air, Using Newton's law of cooling. The heat Transfer equation gives <span class="math-container">$\vec q=h(T_s - T_\infty)$</span>. By re-arranging the equation gives <span class="math-container">$h=10350/ (70-25)$</span> . I feel that my calculations are inaccurate. Can anyone enlighten and help me with this? I know that there are much more to convection than this, such as nusselt number etc? Any help will be greatly appreciated!</p>
| |mechanical-engineering|thermodynamics|convection| | <p>I will try and go through the calculation. This is not a solution however, because what I'll be providing you is ok for flat plates/walls, and the irregular geometry that you have will need some numerical tools to simulate and get an reasonably accurate estimate.</p>
<hr />
<p>Assuming that you have a constant wall temperature of 70oC at the wall of the fiber wall, then the heat power (energy in the unit of time) <span class="math-container">$\dot{Q}$</span> ending up in the environment will be:</p>
<p><span class="math-container">$$\dot{Q} = \frac{\Delta T_{tot}}{R_{tot}}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\Delta T_{tot}= 70-25 [^oC]$</span> is the total temperature difference</li>
<li><span class="math-container">$R_{tot}$</span> is the total thermal resistance (see MIT <a href="https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node123.html" rel="nofollow noreferrer">lectures on combined conduction and convection</a>)</li>
</ul>
<p>The total thermal resistance can be found (for the flat wall case) as:</p>
<p><span class="math-container">$$R = \frac{L}{kA} + \frac{1}{h_{\infty}A}$$</span></p>
<p>where:</p>
<ul>
<li>k is the Thermal Conductivity 230 (W/m K)</li>
<li><span class="math-container">$h_{\infty}$</span> is the convective coefficient of air side.</li>
<li>L is the thickness of the aluminium</li>
<li>A is the cross-sectional area of the heat exchange.</li>
</ul>
<p>The reason you will need a numerical method (FEM or similar), is that the last two parameters (L and A) provide a problem, because:</p>
<ul>
<li>the thickness L of the aluminium is not constant</li>
<li>the cross-sectional area A is not constant.</li>
</ul>
<p>However for a simplified analysis you can calculate an average quantity for both L, and A and use that to get a ball park figure.</p>
<h2>convective heat transfer coefficient of the process</h2>
<p>The convective heat transfer coefficient is expressed in <span class="math-container">$\frac{W}{m^2.^oC}$</span>. This heat coefficient is very much depended on the velocity. The easiest way (avoiding <a href="https://en.wikipedia.org/wiki/Nusselt_number" rel="nofollow noreferrer">Nusselt</a> and <a href="https://en.wikipedia.org/wiki/Prandtl_number" rel="nofollow noreferrer">Prandtl number</a> which are required for a detailed calculation) the easiest way would be to use the following <a href="https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html" rel="nofollow noreferrer">equation (ref)</a>):</p>
<p><span class="math-container">$$h_\infty = 12.12 - 1.16 V + 11.6 \sqrt{V} \tag{eq.1}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$V$</span> is the velocity of air as it approaches the exchange surface.</li>
<li><span class="math-container">$h_\infty$</span> is the convectivity heat transfer coefficient in <span class="math-container">$\frac{W}{m^2. C}$</span></li>
</ul>
<p>Make a note, that although in your case is zero, you can use a fan to provide faster air and thus increase the convective coefficient and thus the overall heat transfer. (that is why computers use fans).</p>
<hr />
<h2>quick calculation of required power per area</h2>
<p>Assuming you have a 50 W source for the rod of <span class="math-container">$d_f=1.2 mm$</span> diameter, with a fiber length of <span class="math-container">$L_f=60mm$</span>, then the exchange area of the rod would be:</p>
<p><span class="math-container">$$A= \pi\cdot d_r \cdot L_f=226 mm^2$$</span></p>
<p>That means that per <span class="math-container">$m^2$</span> the energy is in the order of <span class="math-container">$ \frac{P}{A} = 221\frac{kW}{m^2}$</span> (which is quite a lot). You might need to increase the dimensions (also I don't know the fraction of the 50W that ends up on the heat sink, however even if its only 10%, you'd still have 22 <span class="math-container">$\frac{kW}{m^2}$</span>). Just to give a measure of reference direct sun on a clear sky is about 1 <span class="math-container">$\frac{kW}{m^2}$</span>.</p>
<hr />
<p>assuming the fibre is not touching then the <span class="math-container">$R_{tot}$</span> would be modified as follows (in red is the additional quantity):</p>
<p><span class="math-container">$$R =\color{red}{\frac{1}{h_{fr}A_{fr}}}+ \frac{L}{kA} + \frac{1}{h_{\infty}A}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$h_{fr}$</span> is the heat convectivity coefficient between the fibre rod wall and the hole</li>
<li><span class="math-container">$A_{fr} = \pi d_{fr} L_{fr}$</span> the exchange are between the fibre rod wall and the hole</li>
<li><span class="math-container">$d_{fr}$</span> the diameter of the fiber rod</li>
<li><span class="math-container">$L_{fr}$</span> the length of the fiber rod</li>
</ul>
| 45734 | Calculating Thermal Convection Coefficient |
2021-08-24T10:27:17.543 | <p>I need a mechanism to switch the vertical position of an assembly in as square a profile as possible - i.e. a <a href="https://www.youtube.com/watch?v=WHOrVsWOynE&list=PLG6OkRtGEdYr9bpowlDjsmyAighVRFOjN&index=14" rel="nofollow noreferrer">normal crank mechanism</a> produces a sine wave profile, to explain myself better.
I've seen some <a href="https://www.youtube.com/watch?v=iCwwbjUYaaM&list=PLG6OkRtGEdYr9bpowlDjsmyAighVRFOjN&index=7" rel="nofollow noreferrer">quick return mechanisms</a> (<a href="https://www.youtube.com/watch?v=56fJjeeEo38" rel="nofollow noreferrer">and this</a>) - these obviously travel slowly in one direction and 'quick return in the other - I need the same thing, as quickly as possible, and in both directions.</p>
<p>Any ideas?</p>
| |mechanical-engineering|mechanisms|linear-motion| | <p>Obviously you can not make a perfect square wave with any mechanism simply because they would need a infinite speed, acceleration, jerk (snap, crackle and pop). So no discontinuous functions and some finite raise time.</p>
<p>So from a design perspective there are many possible places to look at solutions. But the locations where I would concentrate my efforts are</p>
<ol>
<li>Cams</li>
<li>Linkages</li>
<li>Bistable mechnisms</li>
</ol>
<p>First cams. Like said the canonical cam to start out with is the modified scotch yoke with a dwell because its simple to understand and easy to reason with. (not because its the best possible cam to do this, but gives you some insight into cam design in this case). The cam you got in @NMechns answer is okay but read a bit further when i talk a bit on the details of cam design. So the 2 dwell scotch yoke looks as follows:</p>
<p><a href="https://i.stack.imgur.com/kUBIp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kUBIp.png" alt="enter image description here" /></a></p>
<p><strong>Image 1</strong>: A modified scotch yoke. OUter black line is follower inner surface. the dashed circle the path the cam takes.</p>
<p>I have added an dwell in both the horizontal and vertical of the yoke. Just to show they are possible, other dwells could be built. each dwell consists of a offset base circle. The two dwell is not the one you are looking for but im just showing it for completeness, image 2 shows its movement function and its relationship to a non modified scotch yoke.</p>
<p><a href="https://i.stack.imgur.com/Z69lU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z69lU.png" alt="enter image description here" /></a></p>
<p><strong>image 2</strong>: Graph if the x movement.</p>
<p>In reality you don't need the scotch vertical dwell you only need the horisontal one See image 3. In fact you never see the vertical dwell in practice because its problematic.</p>
<p><a href="https://i.stack.imgur.com/xXIv2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xXIv2.png" alt="enter image description here" /></a></p>
<p><strong>Image 3</strong>: Scotch yoke without the vertical dwell.</p>
<p>This should give the intuition needed to design the internal cam you need. If you need a faster raise time you can just slant the move lines a bit inward.</p>
<p><strong>The role and shape of the blends</strong></p>
<p>Remember when I said that you can not have discontinuities in your graph. Well if you do have discontinuities then your system is going to bounce. In general bad things happen if you bounce, either the cam follower loses contact or it causes a huge load in the rest of your system. The secondary result is that this causes the surfaces to erode reducing mechanism lifespan dramatically.</p>
<p>So for any high speed cam all of your shapes have to match:</p>
<ul>
<li>Position</li>
<li>Velocity (first derivative of position)</li>
<li>Acceleration (second derivative of position)</li>
<li>Jerk (third derivative of position)</li>
</ul>
<p>Now these functions are a bit tricky to work with. But literature has good candidates.</p>
<p>Other things to watch out are:</p>
<ul>
<li><p>the friction between cam and follower if you rely on sliding. Choosing the material pairs is important. Here the Scotch yoke is a good candidate since the central pin can be changed to a bearing so it can have genuine rolling instead of sliding.</p>
</li>
<li><p>the load can cause the cam to bounce. Which is why many cam systems have a spring. The scotch yoke (especially one devoid of the vertical dwell) does not have a problem with forces that move against the cam, but can have problems with something pulling the cam along at the same direction the cam is moving. If you have this problem put a inner race inside the yoke too this way it cant move too much(This also applies to NMechs cam too which i would modify to be a grove cam), or use a spring load.</p>
</li>
</ul>
<p><strong>Could a four bar do the job?</strong></p>
<p>Not alone. I tried to find a suitable four bar in my catalog of known paths. No suitable on hit my eye. Its still possible i don´t have easy access to a four bar curve fitter so who knows. But a six bar can do it.</p>
<p>However, you can take a bi-stable mechanism and drive with a crank rocker. Then you could do this. The plus side is that this could be really fast. I dont have time to sketch this out but imagine a rocker next to a electrical latching switch and the rocker has two pins that move it form state to another one when moving back and one when moving there.</p>
<p>Anyway i dont have much more time for this.</p>
| 45739 | Ideas for mechanism for quick reciprocating motion - a 'truly' mechanical switch |
2021-08-26T08:07:57.543 | <p>Thinking about the complications of building and maintaining turbine rooms installed at high altitudes in horizontal axis wind turbines (plus the difficulties related to it when, for example, a fire happens), I wonder if there is any working design where an axle, transmission belts... or any other means is employed to keep the turbine even at ground level. Too much power loss? Too expensive to build? Where does the complexity come from?</p>
| |turbines|wind-power|power-transmission|generator| | <p><strong>TL;DR:IMHO, the driver for keeping the electrical generation closer to the hub is for primarily for eliminating -as much as possible- moving parts for safety concerns. The secondary benefit is the reduced losses.</strong></p>
<hr />
<p>IMHO the problem is the distance and the complexity. For the following examples I will use as an example a Wind turbine with the following characteristics:</p>
<ul>
<li>Power : 2MW</li>
<li>hub height : 90 m</li>
<li>rotational velocity @max output: 20 rpm (usually between 9 and 19)</li>
<li>gearbox : planetary 1:100.</li>
</ul>
<p>The torque on the shaft of the wind blades is about 950 [kN.m].</p>
<p>After the gearbox, the rpm increase to 2000, and the torque drop to 9.5 [kN.m]</p>
<p>For arguments sake let's use those values.</p>
<hr />
<p>To transfer the energy from the hub to the ground you'd need either:</p>
<ul>
<li>Right angle gearbox and a shaft going the entirety of the hub height (Which can be several tenths of meters high - a 2MW wind turbine is about 80 meter height).</li>
<li>A chain</li>
<li>or belt to transmit the power.</li>
</ul>
<hr />
<h2>Right angle gearbox and a shaft</h2>
<p>Lets assume we'd make the shaft running down the tower from ordinary structural steel (yield stress 235 MPa). Then the diameter of the shaft (based on simple torsional stress calculation) would have to be :</p>
<p><span class="math-container">$$d = 2\cdot \sqrt[3]{\frac{M_t}{\pi\cdot \tau_{allow}}} = 60mm$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$M_t$</span> is the torsional moment (9.5 kN.m)</li>
<li><span class="math-container">$\tau_{allowed}$</span> is the allowable shear stress (assuming a VonMises yied Criterion 0.57*235 MPa= 133 MPa)</li>
</ul>
<p>That shaft (with a length of 90m) would weigh at least 2000kg (safety factor is still 1), and it would rotate at about 2000rpm. So you would need a lot of support along the length of the beam (in the form of bearing to allow for lateral support and avoid oscillations of the shaft).</p>
<h2>chain</h2>
<p>Assuming the biggest roller chain (in <a href="http://kcm.co.jp/pdf/eng_pdf/Group_Catalog_Power%20Transmission%20Chain.pdf" rel="nofollow noreferrer">this table from kcm</a>), (Chain no. 200), which has rollers 40 mm diameter, and weighs approximately 16 kg/m you'd need approximately 16*(90*2)= 2880 kg of chain.
Because the max allowable load of that chain is about 72kN, in order to transfer the 2 MW, you'd need to have that chain move at least at 28 m/s (~ 100 kph).</p>
<p>In this particular case you'd have to have a bearing at the top that is able to carry the weight of the chain 2880kg plus the additional dynamic loads.</p>
<hr />
<h2>Bottom line</h2>
<p>Any mechanical transmission would have large masses moving at high speeds along great lengths. Having rotating masses like that is an accident waiting to happen, and the results can be catastroptic.</p>
<p>So, IMHO, the driver for keeping the electrical generation closer to the hub is for primarily for eliminating -as much as possible- moving parts for safety concerns. The secondary benefit is the reduced losses.</p>
| 45762 | Is transmitting windmills' mechanical power to ground level that complex? |
2021-08-26T22:30:53.193 | <p>a quick question:</p>
<p>I want to convert between MMBTU and MWh to get at emission factors from energy.</p>
<p>I have the following information:</p>
<p>I use the data given in 40 CFR 98 (see <a href="https://www.epa.gov/sites/default/files/2015-07/documents/emission-factors_2014.pdf" rel="nofollow noreferrer">EPA source here</a>) by the EPA and gives me the Stationary Combustion Emission Factors by Energy content (in MMBTUs) for CO2, CH4 and N2O emissions. I convert the CH4 and N2O emissions using the global warming potentials (21 for CH4 and 310 for N20) to CO2eq.</p>
<p>This gives me 94.007 kgCO2eq/MMBTU for coal and 53.112 kgCO2eq/MMBTU for natural gas.</p>
<p>Now that I have CO2eq emissions by a common unit of energy, I use the information I have on a kWh to convert this to kg CO2eq emissions by kWh. A kWh contains 3412 BTUs or 0.003412 MMBTUs of energy. Therefore, a kWh of coal produces 0.3207519 kg CO2eq and a kWh of gas produces 0.1812181 kg CO2eq. (or Therefore, a coal produces 0.32 tCO2eq/MWh and a kWh of gas produces 0.181 tCO2eq/MWh ).</p>
<p><strong>BUT</strong>: I have also found information, like <a href="https://www.e-control.at/en/marktteilnehmer/oeko-energie/stromkennzeichnung/umweltauswirkungen" rel="nofollow noreferrer">here</a> where I get values for a kWh of 0.882 tCO2eq/MWh for Coal or 0.440 tCO2eq/MWh for Gas. These two pieces of information clearly don't square up!!</p>
<p>In essence, I have the information that a country produced 9,028,638 MWh of Coal electricity. With an emission factor of e.g. 0.91 tCO2eq/MWh that would be 8.76 MtCO2eq. But using the other methodology, 9,028,638 MWh would be 28,369,254 MMBTUs and so would be 28,369,254 * (94.007/1000) = 2.6 MtCO2eq</p>
<p>So clearly using the two methodologies, I get two very different results - but I have no clue where I'm going wrong...</p>
<p>Is the emission factor just so different in individual cases?</p>
<p>Thanks for the help!</p>
| |energy|energy-efficiency|emissions| | <p>IMHO, the difference is that you are trying to compare different things. I'll try to sum it up in the following table:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;"></th>
<th style="text-align: center;">coal</th>
<th style="text-align: center;">gas</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">EPA</td>
<td style="text-align: center;">0.32 tCO2eq/MWh</td>
<td style="text-align: center;">0.181 tCO2eq/MWh</td>
</tr>
<tr>
<td style="text-align: center;">e-control</td>
<td style="text-align: center;">0.880 tCO2eq/MWh</td>
<td style="text-align: center;">0.440 tCO2eq/MWh</td>
</tr>
<tr>
<td style="text-align: center;">ratio of EPA to e-control</td>
<td style="text-align: center;">0.36</td>
<td style="text-align: center;">0.41</td>
</tr>
</tbody>
</table>
</div>
<p>It appears to me that when you are using the data in the EPA report you are considering the <em><a href="https://en.wikipedia.org/wiki/Primary_energy" rel="nofollow noreferrer">primary</a></em> energy content (so its MWh of primary energy).</p>
<p>However, the values from the e-control report consider MWh of electricity generation. Therefore in order to produce 1 MWh of electricity you need to consume about 2.5 to 4 MWh of primary energy depending on the type of fuel and the efficiency of the generator.</p>
<hr />
<p>Although there is variability in the efficiency, the interpretation above is supported by the following data:</p>
<ul>
<li>According to <a href="http://needtoknow.nas.edu/energy/energy-sources/fossil-fuels/natural-gas/#:%7E:text=A%20gas%2Dfired%20plant%20was,as%20much%20as%2060%25%20efficient." rel="nofollow noreferrer">this</a> the efficiency of gas fired electrical generators is about 42%, and coal is about 33%.</li>
<li>According to <a href="https://en.wikipedia.org/wiki/Fossil_fuel_power_station#:%7E:text=Typical%20thermal%20efficiency%20for%20utility,(i.e.%20temperatures%20too%20low.)" rel="nofollow noreferrer">this</a> the efficiency of coal electrical generators is about 37%. (In "Basic concepts: heat into mechanical energy" it states <em>Typical thermal efficiency for utility-scale electrical generators is around 37% for coal and oil-fired plants</em>)</li>
</ul>
| 45781 | Converting between MMBTU and MWh |
2021-08-28T13:25:44.137 | <p>I was hoping someone might be able to help clarify where I might be going wrong with this question. I am being asked to essentially determine the maximum normal and shear stresses at the rigid support C due to load <span class="math-container">$F$</span> at the static weights.</p>
<p><a href="https://i.stack.imgur.com/w5xRk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w5xRk.png" alt="Fabric sample testing mechanism" /></a></p>
<p>I considered there would be a torsional shear stress induced by the torque, calculated by
<span class="math-container">$$ \tau_{xy} = \frac{Tr}{J} $$</span>
where <span class="math-container">$T = Fb$</span>, <span class="math-container">$r$</span> is the radius of the shaft (the point of maximum torsional shear), and <span class="math-container">$J$</span> is the polar moment of area.
There would also be vertical shear, maximised at the vertical centre of the beam, calculated using:
<span class="math-container">$$\tau = \frac{FQ}{Id} $$</span>
where <span class="math-container">$Q$</span> is the first moment of area for a semicircle, and <span class="math-container">$d$</span> is the diameter.</p>
<p>Where I'm having the most trouble is the bending moment at C. I chose to model the beam as a straight beam (which it clearly is not), supported (cantilever) at both ends with a load in the middle, giving a maximum bending moment of
<span class="math-container">$$M_{C} = \frac{FL}{8} $$</span>
from a table in my text, where <span class="math-container">$L = 2ac$</span>.
I am almost certain this is incorrect, so I was hoping someone might be able to provide some insight into where I am going wrong. I am assuming that the points that will experience the greatest stresses are most likely the top and bottom edges of the beam, where both bending and torsional stresses are maximised (and vertical shear stresses are zero). I know once I have the correct stress components I can use Mohr's circle to determine the principal stresses and maximum shear stresses that I would use in further analysis.</p>
<p>Full disclosure, this is a homework question, but it is a multiple choice quiz that allows multiple attempts and tells you the number correct, so I already have the "answer" to the question (which relates to material choices after determining the maximum stresses), so this is really only for my own understanding. Appreciate any help :)</p>
| |stresses|beam| | <p>The reactions at support "C" are as shown below. From here, you can calculate the normal stresses (due to bending) and shear stress (due to shear force and torsion).</p>
<p><a href="https://i.stack.imgur.com/pKa2T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pKa2T.png" alt="enter image description here" /></a></p>
| 45808 | Stresses due to combined loadings in a bent shaft supported at both ends |
2021-08-28T16:18:58.773 | <p>I am trying to understand the flow of an ideal, frictionless incompressible fluid through a pipe for these two cases.
<a href="https://i.stack.imgur.com/trFU9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/trFU9.jpg" alt="enter image description here" /></a></p>
<p>Note: Please consider Flow velocity in the tank is zero, and the fluid is incompressible.</p>
<p>Case 1: This is simple to understand. In this, I can calculate the flow velocity by applying Bernoulli's equation as the pressure difference is completely converted to kinetic energy.</p>
<p>i)So am I correct in understanding that the presence or absence of the pipe has no effect on the flow rate?</p>
<p>ii) Is the pressure at every point of the pipe the same as atmospheric pressure?</p>
<p>Case 2:</p>
<p>iii) Will the flow rate be the same as in case 1 or will the addition of the diverging section cause the flow rate to increase.</p>
<p>iv)Does the inclusion of the diverging section cause the flow to expand and so increase the static pressure (to atmospheric pressure) and if this is true will the static pressure before the expansion be less than atmospheric pressure?</p>
<p>Thank you</p>
| |fluid-mechanics| | <p><strong>Case 1</strong></p>
<p>(i) - Yes, correct.</p>
<p>(ii) - <span class="math-container">$P_2 = P_1$</span></p>
<p><a href="https://i.stack.imgur.com/xab2N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xab2N.png" alt="enter image description here" /></a></p>
<p><strong>Case 2</strong></p>
<p>(i) - Yes, the flow rate is the same for both cases. No, the flow rate is constant but the velocity varies due to change in cross section (<span class="math-container">$Q = v_1A_1 = v_2A_2$</span>).</p>
<p>(ii) - Again, let's write Bernoulli's equation:</p>
<p><span class="math-container">$P_2 = \dfrac{1}{2}\rho v_1^2 + P1 - \dfrac{1}{2}\rho v_2^2$</span>, since <span class="math-container">$v_2 < v_1$</span>, so <span class="math-container">$P_2 > P_1$</span>. The result is expected based on the Bernoulli's principle, that states, with constant datum, slower moving fluids create greater pressure (force) than faster moving fluids.</p>
<p>Due to the presence of head, both <span class="math-container">$P_1$</span> & <span class="math-container">$P_2$</span> are not likely less than <span class="math-container">$P_{atm}$</span>.</p>
| 45811 | Frictionless flow in pipes |
2021-08-29T19:52:14.263 | <p>this is a formula for torque and I was wondering for what P in this case stands for</p>
<p><a href="https://i.stack.imgur.com/ijQyf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ijQyf.png" alt="enter image description here" /></a></p>
<p>.
.
.</p>
<p><a href="https://i.stack.imgur.com/Msznc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Msznc.png" alt="this is how the construction should look like" /></a></p>
| |torque|solid-mechanics| | <p><span class="math-container">$$M_t = \frac{P}{2\pi\cdot n}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$M_t$</span> is the torque transmitted through the shaft (unit in SI: Nm).</li>
<li>P is the power transmitted through the shaft (unit in SI: W).</li>
<li><span class="math-container">$n$</span> are the rotations of the shaft in revolutions per minute (rps). (this is important for the units to be correct)</li>
</ul>
<hr />
<p>If you want to use <span class="math-container">$n$</span> with revolutions per minute (<strong>rpm</strong>), you should use the following formula</p>
<p><span class="math-container">$$M_t = \frac{60\cdot P}{2\pi\cdot n [rpm]}$$</span></p>
<p>(by square brackets next to a quantity I am presenting the units you should use in the equation)</p>
<hr />
<p>As mentioned elsewhere another common equation (which basically incorporates the conversion between rpm and angular velocity) is the following:</p>
<p><span class="math-container">$$M_t =\frac{P}{\omega}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\omega $</span> is the angular velocity (units in SI: rad/s)</li>
</ul>
<p>The relation of angular velocity with the rotations per second and minute correspondigly are <span class="math-container">$\omega = 2\pi \cdot n[rps]$</span> and <span class="math-container">$\omega = \frac{2\pi \cdot }{60}n[rpm]$</span></p>
| 45826 | torque for a machine shaft |
2021-08-30T06:58:18.833 | <p>Lets say we have a very large and deep circular footing for a tall tower such as an industrial chimney or a wind turbine, and for convenience we want to pour it not at once but in sections. Can this be done, provided that we have sufficient dowels between the pour sections? Or it must be a single pour?</p>
| |reinforced-concrete|foundations| | <p>Since the footing is "large and deep", <strong>both the vertical and horizontal construction or expansion joints can be utilized to partition the concrete pour.</strong></p>
<p>Note, other than the capability/capacity of the concrete plant and crew to produce and handle the concrete placement, the amount of <strong>heat generated during hydration</strong> needs to be evaluated, as it is usually the most important factor in determining the size and thickness of an individual pour, with the consideration of the available temperature control method employed.</p>
<p>Another factor affecting the thickness of the pour is the potential of aggregate separation when poured from the height. It can be difficult to vibrate the concrete uniformly and adequately, that usually the main cause for having honeycombs in the concrete.</p>
| 45836 | Pouring large circular footing in segments |
2021-08-30T17:24:36.650 | <p>I have been doing some experiments using a small brushed permanent magnet DC motor as a generator. I have it coupled to a similar (as in, exactly the same) motor which is powered by a PSU. thus driving the 'generator'. Note, this may look like a homework like question, but it's actually first steps in an actual real application.</p>
<p>In more detail, both motors are 12V, Current: 0.18A, Power Rating: 5.662W, Rotational Speed Max: 8,768rpm, Efficiency: 59.58%, Direction of Rotation: Anti-Clockwise</p>
<p><a href="https://ie.farnell.com/velleman-sa/mot3n/dc-motor-12vdc-180ma-11500rpm/dp/3498470?CMP=i-bf9f-00001000" rel="nofollow noreferrer">https://ie.farnell.com/velleman-sa/mot3n/dc-motor-12vdc-180ma-11500rpm/dp/3498470?CMP=i-bf9f-00001000</a></p>
<p><strong>Firstly, some questions relating to this motor, used as a motor.</strong></p>
<p><strong>Consumption:</strong> From looking at many similar sized motors while choosing this one, I came to the conclusion that the power rating in watts, seem to be the mechanical power output (putting the RPM and torque into a calculator site gives 5.69W). If 100% efficient it should thus consume 5.6W, as it's max 60% efficient it should really consume ~10W. Yet, the specs say 12V * 0.18a = 2.1W. What am I not understanding?</p>
<p>Also, the current consumption is indicated at about 1A by the PSU meter.</p>
<p><strong>Direction:</strong> The spec lists a direction for the motor. I thought DC motors went just as well in either direction (e.g. car window up/down). I took one of them apart expecting the brushes to be angled etc but looks the same in both directions to me. Why is a direction given, and what are the implications for generation (more on this later)</p>
<p><strong>Generation questions</strong></p>
<p><strong>Setup:</strong> the two motors are joined using a coupler, see pic. The driving motor is supplied 12V from the PSU and the generating motor is connected to some resistors to form a load (I'll just detail one value, but I tried 5/10/10/20 ohms)</p>
<p><strong>Results:</strong> With a 10 ohm load (9.8 actually) I measure 5.8V = 0.59A * 5.8V = 3.4W. This seems perfect as the driving motor should be putting in 5.6W (mechanical), * by the 59.58% efficiency = 3.3W (electric), which is what I'm getting</p>
<p>This simple view assumes I'm getting a nice flat DC waveform out, but I assume it's really like AC straight from a bridge rectifier (don't have a scope) so maybe RMS should come in somewhere?</p>
<p><strong>Input power:</strong> My main question, and getting back to the driving motor, is the readings I'm getting on the PSU/driving motor side. The PSU has a switchable meter reading volts or amps.</p>
<p>With the above test load, the driving motor is consuming about 1.2A at 12V, according to the PSU. However, using my multimeter, I read about 7v and 0.5 to 1.5A (all readings are jumping about wildly - RFI on cheap meter?) So, according to the PSU the driving motor is consuming 14W and 3.5 to 10W using the multimeter - both way more them the 12v* 0.18a = 2.1W specs. IS back EMF affecting my readings?</p>
<p><strong>Finally, another oddity</strong>. When reading the voltage across the load resistance, I saw that for some I got a stable reading while for others the meter digits were making no sense. But, if I reversed the driving motor supply polarity, the readings stabilized.</p>
<p><a href="https://i.stack.imgur.com/zldrP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zldrP.jpg" alt="enter image description here" /></a></p>
<p><em><strong>EDIT 23 Sept '21</strong></em></p>
<p>After spending time trying to get good readings with two multimeters, I bit the bullet and ordered a cheap PC based oscilloscope. So I'd like to update the question with these results</p>
<p><strong>Setup</strong></p>
<p>On the drive side I used Channel 1 to measure voltage at the motor, and CH2 to measure voltage across a 1.3R resistor (and thus calculate current via Ohm's Law). As the two probe grounds are connected, I had to use the motor/resistor junction as common GND for both channels and invert as required</p>
<p><a href="https://i.stack.imgur.com/IBy5M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IBy5M.png" alt="enter image description here" /></a></p>
<p><strong>Results on driving side</strong></p>
<p>I'm seeing an average of 11.75V across the motor and 1.0V across the resistor (=0.77A) giving a power in of 9W. The motor efficiency is 60% so this should give 5.4W mechanical output, which matches the rating of 5.662W</p>
<p><a href="https://i.stack.imgur.com/Khupm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Khupm.png" alt="enter image description here" /></a></p>
<p><strong>Output</strong></p>
<p>Not shown in the setup diagram is the fact the motor is coupled to another one as before with a 9.4R load. Output average is 6.7V giving 0.71A and thus 4.7W. Assuming the input side is correct, this means the motor is 87% efficient as a generator. Have I made a mistake somwhere?</p>
<p><a href="https://i.stack.imgur.com/jBjNh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jBjNh.png" alt="enter image description here" /></a></p>
| |electrical-engineering|motors| | <p>Power rating of a motor is the mechanical power out. Divide that by efficiency and you should roughly get electrical power in. Efficiency will be different if not at full-load. To get 6W mechanical @ 60% efficiency, you have to put 10W electrical in.</p>
<p>You have a problem. Driving motor correctly means generator is going in the incorrect direction, which may explain <strong>But, if I reversed the driving motor supply polarity, the readings stabilized.</strong> Small, cheap, dc motors can have brushes aligned to get more torque in one direction.</p>
<p><strong>This simple view assumes I'm getting a nice flat DC waveform out, but I assume it's really like AC straight from a bridge rectifier (don't have a scope) so maybe RMS should come in somewhere?</strong></p>
<p>Nope. You have a commutator, which the brushes ride on. You'd get DC with a ripple. You could add an inductor in series with output to filter ripple.</p>
<p>On your varying power consumption. <em>What size wires and how far is power supply from motor?</em> Your efficiency has to be applied to motor and generator. There is also a mechanical efficiency between the motor and generator. 10W means 6W (60%) from motor and 3.6W (60%) from generator.</p>
<p>Here you also have a problem. If your generator is at full-load, your motor will be overloaded. I suggest you graph out input and output (measure voltage and current at different loads).</p>
<hr />
<p>Most motors (DC and three-phase ac) can be used as generators with some being better than others.</p>
<p>Efficiency has to deal with losses (copper loss, hysterisis, windage, bearings), so they apply to motor or generator. Although the motor will be at 60%, generator will be different, probably slightly higher because copper losses will be less (half-load means half current so I2R losses will be 25% of full-load copper losses). You have a small motor, but theory still applies.</p>
<p>Your erratic readings are due to motor going in the wrong direction. As proof of concept, you have a genset.</p>
<p>If you drive a BLDC, you can get a three-phase generator.</p>
| 45845 | DC motor as generator - some questions |
2021-08-31T20:17:57.777 | <p>I'm trying to compute the total angular momentum of a 3D multi-body, pendulum-like (as in, each body is connected to another one), mechanical system. Let us consider, for a simpler case, a 2D double pendulum. With some tips, I hope to transfer this to the more difficult 3D case myself.</p>
<p><a href="https://i.stack.imgur.com/3P0vt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3P0vt.png" alt="enter image description here" /></a></p>
<p>Now, I am aware that angular momentum can be computed as
<span class="math-container">$$L = I\omega$$</span>
If we generalize and don't use the standard inertia equation for a beam rotating around its edge, for beam one, we can say
<span class="math-container">$$ L_1 = (I_{1,com} + m_1(\bar{x}_{1,com} - \bar{x}_{origin})^2 )\cdot \dot{\phi}_1 $$</span>
Now, issues arise when considering the second beam. I could simply add
<span class="math-container">$$ L_2 = (I_{2,com} + m_2(\bar{x}_{2,com} - \bar{x}_{origin})^2) \cdot \dot{\phi}_2 $$</span>
but I am not sure if this correct as I need to consider the beam rotating around the origin, as well as the beam rotating around 'itself'. So would I also need to add a term that is
<span class="math-container">$$ L_{2,self} = (I_{2,com} + m_2(\bar{x}_{2,com} - \bar{x}_{1,pinjoint})^2) \cdot \dot{\phi}_2 $$</span>
? It seems like I'm missing some fundamental understanding so I would really wish to not only obtain the answer but also some reasoning/explaining behind it.</p>
<p>In short, is it
<span class="math-container">$$ L_{total} = L_1 + L_2 $$</span>
is it
<span class="math-container">$$ L_{total} = L_1 + L_2 + L_{2,self} $$</span>
or is it 'something inbetween'....</p>
<p>I appreciate all effort!</p>
| |mechanical-engineering|beam| | <p>It is all relatively straightforward, albeit tedious. Do not try to compute things component by component in dynamics because you are going to be overwhelmed and make mistakes. Even the pure vector notation can be rather overwhelming.</p>
<p>The paradox here is that 2D systems are not necessarily simpler in terms of notation, but rather more complex. 2D systems are just projections of their 3D counterparts, and sometimes these projections can become obtuse.</p>
<h2>Kinematics</h2>
<p>First you need to establish the <strong>Kinematics</strong> describing the position and velocities of all important points as a function of the swing angles (the degrees of freedom).</p>
<p><sub>Note the <em><strong>boldface</strong></em> variables denote vectors, <em>italisized</em> variables are scalars, and upright variables are also scalar but pertaining to out-of-plane components.</sub></p>
<h3>Positions</h3>
<p><span class="math-container">$$\boldsymbol{r}_{1}=\begin{pmatrix}0\\
0
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\boldsymbol{c}_{1}=\begin{pmatrix}c_{1}\cos\varphi_{1}\\
c_{1}\sin\varphi_{1}
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\boldsymbol{r}_{2}=\begin{pmatrix}\ell_{1}\cos\varphi_{1}\\
\ell_{1}\sin\varphi_{1}
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\boldsymbol{c}_{2}=\begin{pmatrix}\ell_{1}\cos\varphi_{1}+c_{2}\cos\varphi_{2}\\
\ell_{1}\sin\varphi_{1}+c_{2}\sin\varphi_{2}
\end{pmatrix}$$</span></p>
<h3>Velocities</h3>
<p><span class="math-container">$$\dot{\boldsymbol{r}}_{1}=\begin{pmatrix}0\\
0
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\dot{\boldsymbol{c}}_{1}=\begin{pmatrix}-\dot{\varphi}_{1}c_{1}\sin\varphi_{1}\\
\dot{\varphi}_{1}c_{1}\cos\varphi_{1}
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\dot{\boldsymbol{r}}_{2}=\begin{pmatrix}-\dot{\varphi}_{1}\ell_{1}\sin\varphi_{1}\\
\dot{\varphi}_{1}\ell_{1}\cos\varphi_{1}
\end{pmatrix}$$</span></p>
<p><span class="math-container">$$\dot{\boldsymbol{c}}_{2}=\begin{pmatrix}-\left(\dot{\varphi}_{1}\ell_{1}\sin\varphi_{1}+\dot{\varphi}_{2}c_{2}\sin\varphi_{2}\right)\\
\dot{\varphi}_{1}\ell_{1}\cos\varphi_{1}+\dot{\varphi}_{2}c_{2}\cos\varphi_{2}
\end{pmatrix}$$</span></p>
<h2>Dynamics</h2>
<p>Now we move into the dynamics of the system by describing the momentum quantities</p>
<h3>Part Momentum</h3>
<p>For each part, linear momentum <span class="math-container">$\boldsymbol{p}_i$</span> and angular momentum <span class="math-container">${\rm H}_i$</span> of each body about the origin is described below</p>
<p><span class="math-container">$$\begin{aligned}\boldsymbol{p}_{1} & =m_{1}\dot{\boldsymbol{c}}_{1}\\
& =\begin{pmatrix}-m_{1}\dot{\varphi}_{1}c_{1}\sin\varphi_{1}\\
m_{1}\dot{\varphi}_{1}c_{1}\cos\varphi_{1}
\end{pmatrix}
\end{aligned}$$</span></p>
<p><span class="math-container">$$\begin{aligned}{\rm H}_{1} & ={\rm I}_{1}\dot{\varphi}_{1}+\boldsymbol{c}_{1}\times\boldsymbol{p}_{1}\\
& ={\rm I}_{1}\dot{\varphi}_{1}+\begin{pmatrix}c_{1}\cos\varphi_{1}\\
c_{1}\sin\varphi_{1}
\end{pmatrix}\times\begin{pmatrix}-m_{1}\dot{\varphi}_{1}c_{1}\sin\varphi_{1}\\
m_{1}\dot{\varphi}_{1}c_{1}\cos\varphi_{1}
\end{pmatrix}\\
& =\underbrace{\left({\rm I}_{1}+m_{1}c_{1}^{2}\right)}_{\text{parallel axis theorem}}\dot{\varphi}_{1}
\end{aligned}$$</span></p>
<p>Above the parallel axis theorem comes out naturally from the definition of angular momentum and the kinematics.</p>
<p><span class="math-container">$$\begin{aligned}\boldsymbol{p}_{2} & =m_{2}\dot{\boldsymbol{c}}_{2}\\
& =\begin{pmatrix}-m_{2}\left(\dot{\varphi}_{1}\ell_{1}\sin\varphi_{1}+\dot{\varphi}_{2}c_{2}\sin\varphi_{2}\right)\\
m_{2}\left(\dot{\varphi}_{1}\ell_{1}\cos\varphi_{1}+\dot{\varphi}_{2}c_{2}\cos\varphi_{2}\right)
\end{pmatrix}
\end{aligned}$$</span></p>
<p><span class="math-container">$$\begin{aligned}{\rm H}_{2} & ={\rm I}_{2}\dot{\varphi}_{2}+\boldsymbol{c}_{2}\times\boldsymbol{p}_{2}\\
& ={\rm I}_{2}\dot{\varphi}_{2}+\begin{pmatrix}\ell_{1}\cos\varphi_{1}+c_{2}\cos\varphi_{2}\\
\ell_{1}\sin\varphi_{1}+c_{2}\sin\varphi_{2}
\end{pmatrix}\times\begin{pmatrix}-m_{2}\left(\dot{\varphi}_{1}\ell_{1}\sin\varphi_{1}+\dot{\varphi}_{2}c_{2}\sin\varphi_{2}\right)\\
m_{2}\left(\dot{\varphi}_{1}\ell_{1}\cos\varphi_{1}+\dot{\varphi}_{2}c_{2}\cos\varphi_{2}\right)
\end{pmatrix}\\
& ={\rm I}_{2}\dot{\varphi}_{2}+m_{2}\left(\ell_{1}^{2}\dot{\varphi}_{1}+c_{2}^{2}\dot{\varphi}_{2}+\ell_{1}c_{1}\left(\dot{\varphi}_{1}+\dot{\varphi}_{2}\right)\cos\left(\varphi_{2}-\varphi_{1}\right)\right)
\end{aligned}$$</span></p>
<p>Again the above is based on the parallel axis theorem but since the kinematics include the motion of two bodies, it is not straightforward to apply, and the reason I discouraged going component by component, but instead insist on using <strong>at a minimum</strong> some vector notation.</p>
<p><sub>Note the 2D (projection) of the cross product operator <span class="math-container">$\boldsymbol{a}\times \boldsymbol{b} = a_x b_y - a_y b_x$</span></sub></p>
<h3>Total Momentum</h3>
<p>The combined momenta is described below as a linear algebra operation in terms of the motion variable(s) <span class="math-container">$\dot{\varphi}_i$</span>.</p>
<p><span class="math-container">$$\begin{aligned}\boldsymbol{p} & =\boldsymbol{p}_{1}+\boldsymbol{p}_{2}\\
& =\begin{bmatrix}-\left(m_{1}c_{1}+m_{2}\ell_{1}\right)\sin\varphi_{1} & -m_{2}c_{2}\sin\varphi_{2}\\
\left(m_{1}c_{1}+m_{2}\ell_{1}\right)\cos\varphi_{1} & m_{2}c_{2}\cos\varphi_{2}
\end{bmatrix}\begin{bmatrix}\dot{\varphi}_{1}\\
\dot{\varphi}_{2}
\end{bmatrix}
\end{aligned}$$</span></p>
<p><span class="math-container">$$\begin{aligned}{\rm H} & ={\rm H}_{1}+{\rm H}_{2}\\
& =\begin{bmatrix}{\rm I}_{1}+m_{1}c_{1}^{2}+m_{2}\ell_{1}^{2}+m_{2}\ell_{1}c_{2}\cos\left(\varphi_{2}-\varphi_{1}\right)\\
{\rm I}_{2}+m_{2}c_{2}^{2}+m_{2}\ell_{1}c_{2}\cos\left(\varphi_{2}-\varphi_{1}\right)
\end{bmatrix}^{\intercal}\begin{bmatrix}\dot{\varphi}_{1}\\
\dot{\varphi}_{2}
\end{bmatrix}
\end{aligned}$$</span></p>
<p><sub>Note the transpose <span class="math-container">${}^\intercal$</span> above which applies the dot product between the two vectors to return a scalar.</sub></p>
<p>Again the values above are summed about the origin. It is always a good idea to use either a fixed (nonmoving) reference point for the summations or use the center of mass. Otherwise, there are terms that need to be included. See <a href="https://ethz.ch/content/dam/ethz/special-interest/mavt/mechanical-systems/mm-dam/documents/Notes/Dynamics_LectureNotes.pdf#page=84" rel="nofollow noreferrer">Equation (3.80) in this excellent reference in dynamics</a>.</p>
<hr />
<h2>Appendix I</h2>
<h3>Cross Products</h3>
<p>There are 3 projections of the cross-product on the plane</p>
<p><span class="math-container">$$\begin{aligned}\boldsymbol{r}\times{\omega} & =\begin{pmatrix}r_{y}\\
-r_{x}
\end{pmatrix}\omega=\begin{pmatrix}r_{y}\omega\\
-r_{x}\omega
\end{pmatrix}\\
\omega\times\boldsymbol{r} & =\begin{pmatrix} & -\omega\\
\omega
\end{pmatrix}\begin{pmatrix}r_{x}\\
r_{y}
\end{pmatrix}=\begin{pmatrix}-r_{y}\omega\\
r_{x}\omega
\end{pmatrix}\\
\boldsymbol{a}\times\boldsymbol{b} & =\begin{pmatrix}-a_{y} & a_{x}\end{pmatrix}\begin{pmatrix}b_{x}\\
b_{y}
\end{pmatrix}=a_{x}b_{y}-a_{y}b_{x}
\end{aligned}$$</span></p>
<p>See that <span class="math-container">$\omega$</span> is a scalar, and <span class="math-container">$\boldsymbol{r}$</span>, <span class="math-container">$\boldsymbol{a}$</span>, and <span class="math-container">$\boldsymbol{b}$</span> are vector quantities.</p>
| 45872 | Angular momentum of a multi-body system |
2021-09-01T03:52:47.360 | <p>I'm presented with a problem requiring some analysis to determine if an antenna bracket mounted to a concrete blockwork wall is sufficiently strong enough to endure a wind load given the known surface area of two antennas mounted on it.</p>
<p>Excuse the terrible diagram:
<a href="https://i.stack.imgur.com/bDIc6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bDIc6.png" alt="enter image description here" /></a></p>
<p>The force due to the surface area of the antennas & wind velocity is assumed to be 288N at the end of the vertical post acting out of the page and attempting to 'lever' bolts B1:B3 out of the wall (pullout/tensile force).</p>
<p>Distance B1 to force = 1250mm, B2 to force = 912mm, B3 to force = 742mm</p>
<p>I've simplified the problem to assume B2 doesn't exist, B1 is the fulcrum and B3 is the 'load' in a class 2 lever system:
<a href="https://i.stack.imgur.com/mPp11.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mPp11.png" alt="enter image description here" /></a></p>
<p>For the simplified system I work out the tensile force acting on B3 to be in the realm of 709N:</p>
<p>(1250/(1250-742))*288N = 708.67N</p>
<p>Having an electrical background I'd be very interested to hear how a mechanical engineer would approach the problem and whether there is a better way (than my simplified system) to represent the forces on each of the 3x bolts.</p>
| |mechanical-engineering|civil-engineering|building-physics|bolting| | <p>For small equipment mounting, let me introduce two simple methods:</p>
<p><strong>1) Rigid Rotation Method</strong> (Classic)</p>
<p><a href="https://i.stack.imgur.com/mCZh8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mCZh8.png" alt="enter image description here" /></a></p>
<p><strong>2) Cantilever Method</strong> (Conservative approximation)</p>
<ul>
<li>Assume cantilever action, for which the applied load rotates about the point "A", and the rotation is restricted by the two bolts farthest from the point "A".</li>
</ul>
<p><a href="https://i.stack.imgur.com/sSmyv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sSmyv.png" alt="enter image description here" /></a></p>
<p><strong>Conclusion:</strong></p>
<p>I recommend using the <strong>Metod 2</strong> for its simplicity and embed safety. Also, I suggest adding a rod from the fixture to the wall to take care of the vibration induced by wind.</p>
<p><a href="https://i.stack.imgur.com/ojblK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ojblK.png" alt="enter image description here" /></a></p>
<p><strong>Please check my math for mistakes.</strong></p>
| 45875 | Tensile Force on Line of Bolts |
2021-09-01T21:39:00.430 | <p>I was wondering how I could get Ttmax (maximum torsional stress) or rather what the formula for it could be</p>
<p><a href="https://i.stack.imgur.com/t3YQ2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t3YQ2.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|torque|stresses| | <p>Tc/J where:</p>
<ul>
<li>T: internal torque in the shaft</li>
<li>c: radius of the circular rod</li>
<li>J: Polar moment of inertia (pi*d^4/32 for a solid circular shaft)</li>
</ul>
| 45884 | trying to find maximum torsional stress |
2021-09-01T21:55:44.773 | <p>I am building a small testing device consisting of a double acting pneumatic cylinder. I am using an existing pneumatic line running at 100 psi and I would like to control the force exerted by the piston on my test sample? My question is whether an electronic 5/3 direction control valve can be used to control the output force of the cylinder, the signal for this valve would be dictated by the readings from a load cell attached to the cylinder rod. This is a cyclic test so i need to ensure that the force remains within the allowable limits across all cycles.
In case this is not possible can anyone advice about a proper way of doing it? I don´t require high precision +-5% of variance is accepted. I don´t have a lot of budget so, easy to implement solutions are preferred.</p>
<p>Thank you!</p>
| |mechanical-engineering|control-engineering|pneumatic| | <ul>
<li>For a quantitative application, the electronic pressure regulator in Transistor's answer is probably the most appealing other than component cost. (It may still need another valve to move backwards, at least with the regulator shown).</li>
<li>You actually can implement hysteresis control with a simple 5/3 valve, since you have the load cell in the loop already. You <em>may</em> need to slow the ramp-up rate, with an orifice, a buffer tank, or both. The tradeoff in speed to realize the 5% accuracy may or may not be acceptable.</li>
<li>You can also potentially eliminate paying for the pressure sensor in the electronic regulator, with an electronic (solenoid) proportional valve designed for double acting cylinders. However I'm not sure if this would really save money, since the circuitry to drive it precisely is also a slight inconvenience.</li>
<li>The simple 5/3 (or 5/2) valve could be used combined with a mechanical regulator, if the load is always the same. Not sure if you will get your 5% with complete reliability.</li>
<li>The simple 5/3 (or 5/2) with direct 100psi can again be used, providing fixed end-to-end motion, but with an adjustable spring coupling to vary the load mechanically. Again presuming the load is always the same. Primitive, but perhaps the simplest of all.</li>
</ul>
| 45885 | Force control of pneumatic piston |
2021-09-02T08:52:21.387 | <p><a href="https://i.stack.imgur.com/keLNO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/keLNO.png" alt="Optical Fiber Heat analysis" /></a></p>
<p>I am conducting a thermal analysis of optical fibers. ( Software used AutoCAD Fusion 360)</p>
<p>The Inputs conditions are :</p>
<p>Let's assume an input power of 1250W, with an efficiency of 98%. 2% loss which becomes heat. (25W)</p>
<p>The internal heat of 25W at the inner core ( power loss ) gives 407.7 degrees Celcius.</p>
<p>Convection Coefficient of <span class="math-container">$10W/m^2K$</span> (Convection of still air)</p>
<p>The material of the rod I assume is glass with an emissivity of 0.9 ( Radiation ).</p>
<p>The temperature of the optical fiber reaches 407.7 degrees Celcius, which is inconsistent with real life. The temperature should be approximately 35 degrees Celcius? Why is this so, am I missing out on another method of heat loss? Any help will be greatly appreciated.</p>
| |optics|communication|photonics|fiber-optics| | <p>If you increase the length of the fibre, you will get more reasonable results. The losses -in my mind- need to be distributed along the entire length of the fiber .</p>
<p>Currently, you seem to model a very small portion of the fiber, and as such you have a very small area that need to transfer a seemingly small amount of energy. However, if that energy is going through a small surface area then the energy over area is high and therefore the end temperature difference will need to be high.</p>
<p>Another way you can approach it, is that you can you can adjust the heat losses. I.e. if you have a 1m of fiber, and you model 0.1 m , then instead of 25 W you can use 2.5 watts (provided everything else in the model is correctly setup).</p>
| 46893 | Heat loss in Optical Fiber |
2021-09-02T11:22:20.843 | <p>I am given the structure below, and supposed to find the forces occuring in <span class="math-container">$A, F, E, D, C$</span> and <span class="math-container">$B$</span>. The task hints of using symmetry, as several forces are equal in magnitude.</p>
<p><a href="https://i.stack.imgur.com/uEq6g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uEq6g.png" alt="enter image description here" /></a></p>
<p>Supposedly, <span class="math-container">$A=B$</span>, and <span class="math-container">$G=H$</span>, where the image below depicts the top beam, and these forces.</p>
<p><a href="https://i.stack.imgur.com/AJIS1l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AJIS1l.png" alt="enter image description here" /></a></p>
<p>My question is therefore <em>why</em> <span class="math-container">$A=B$</span> and <span class="math-container">$H=G$</span>. I have asked other students/T.As., and they say it is obvious or intuitive, but can not really <em>explain</em>. Can anyone help?</p>
| |mechanical-engineering|civil-engineering|statics|beam|deflection| | <p>You seem to have no trouble identifying that the structure itself is symmetric, only how that automatically tells you that <span class="math-container">$A = B$</span> and <span class="math-container">$G = H$</span>.</p>
<p>I'm going to take two approaches to this answer, one a bit blunter, the other a bit more visual.</p>
<hr />
<h2>The blunt approach</h2>
<p>One could argue that when the structure (including loading) is symmetric, the onus should be on you to give a reason why the reactions shouldn't be symmetric.</p>
<p>If the reactions aren't symmetric, then the internal stresses and deflection aren't symmetric. So why should a symmetric structure have asymmetric deflection? Why should its left span sag more (for example) than the right span?</p>
<p>Wouldn't that imply that the left span is more flexible or under greater load than the right span (which we know isn't the case, since it's symmetric)?</p>
<hr />
<h2>The visual approach</h2>
<p>Let's take a walk.</p>
<p>Initially, we are standing in a position such that we can see the structure exactly as shown in the problem statement, with A closer to us but slightly to our left and B in the distance, slightly to our right.</p>
<p>We then walk forwards and a bit to our right until we are exactly between the E and C supports and then we look at the rest of the structure. What we'll see is precisely what you drew: a horizontal beam that starts at A to our left, passes through G, then H, and ends at B. All those points have a 2-meter spacing between them, and the entire beam is under a distributed load <span class="math-container">$w$</span>.<sup>1</sup></p>
<p><a href="https://i.stack.imgur.com/ErNJH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ErNJH.png" alt="enter image description here" /></a></p>
<p>Let's then say we sit down and work out all the reactions somehow. Now let's imagine that you get that <span class="math-container">$A \neq B$</span> and <span class="math-container">$G \neq H$</span>. Specifically, let's say we get that <span class="math-container">$A > B$</span> and <span class="math-container">$G > H$</span>. Or, put another way, we get that (left and right defined from our current position between C and E):</p>
<p><span class="math-container">$$\begin{align}
\text{left-most reaction (A)} &> \text{right-most reaction (B)} \\
\text{left-inner reaction (G)} &> \text{right-inner reaction (H)}
\end{align}$$</span></p>
<p>We haven't thought about symmetry yet, so we just accept that result as correct and move on.</p>
<p>Now we stand up and walk around the entire structure until we are standing precisely between supports D and F, and we once again look at the rest of the structure. This time we see a horizontal beam that starts at B to our left, passes through H, then G, and ends at A. All those points have a 2-meter spacing between them, and the entire beam is under a distributed load <span class="math-container">$w$</span>.</p>
<p><a href="https://i.stack.imgur.com/B3W5d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B3W5d.png" alt="enter image description here" /></a></p>
<p>Now, do we need to sit down and calculate the results for this "new" beam? Of course not. It's obviously the same beam we calculated before, just from a different perspective.</p>
<p>But still, let's say we did go through the trouble of redoing our calculations. Assuming we did everything exactly the same way as the first time, we'd once again get that:</p>
<p><span class="math-container">$$\begin{align}
\text{left-most reaction} &> \text{right-most reaction} \\
\text{left-inner reaction} &> \text{right-inner reaction}
\end{align}$$</span></p>
<p>But this time that means that <span class="math-container">$B > A$</span> and <span class="math-container">$H > G$</span>...</p>
<p>So when we looked at the beam from one perspective (between C and E) we get a different result than we got from another (between D and F). That obviously makes no physical sense: the beam's true reactions don't care about where we are standing when we calculate them.</p>
<p>So we know that our result is incorrect. And in fact, we can ask ourselves: what's the only result that's independent of where we're standing? One where <span class="math-container">$A = B$</span> and <span class="math-container">$G = H$</span>.<sup>2</sup></p>
<p>This thought experiment only works when the beam is symmetrical. After all, when the beam is symmetrical, the change in perspective is simply an inversion of the labels. Why should beam A-G-H-B have different results than those from B-H-G-A?</p>
<p>If it isn't symmetric (say, if the spans are actually 1 m (A to G), 3 m (G to H) and 2 m (H to B), then our calculation standing between C and E will give us that:</p>
<p><span class="math-container">$$\begin{align}
\text{left-most reaction (A)} &< \text{right-most reaction (B)} \\
\text{left-inner reaction (G)} &< \text{right-inner reaction (H)}
\end{align}$$</span></p>
<p>and the result standing between D and F will be consistent:</p>
<p><span class="math-container">$$\begin{align}
\text{left-most reaction (B)} &> \text{right-most reaction (A)} \\
\text{left-inner reaction (H)} &> \text{right-inner reaction (G)}
\end{align}$$</span></p>
<hr />
<p><sup>1</sup> <sub>In the diagrams below, G and H should actually be springs; they've been marked as rigid supports to keep thing simple and because this is irrelevant to the question at hand.</sub></p>
<p><sup>2</sup> <sub>Obviously one where all reactions are equal would also be independent of where we're standing but would fail to satisfy the standard equilibrium equations.</sub></p>
| 46895 | Utilizing symmetry in statics problem |
2021-09-02T16:48:19.973 | <p>In light of the recent (Sept. 2021) power outages relating to storms and flooding, I've been thinking about a new version of a question that's been asked many times before: why not bury electrical cables <strong>when laying road?</strong></p>
<p>I believe this would address multiple issues of normal underground cabling:</p>
<ol>
<li><p><strong>No need to dig separate trenches</strong>. Since road is already ostensibly being laid on a suitable surface, we can simply place an undergound cabling shroud physically embedded in or directly adjacent to the road.</p>
</li>
<li><p><strong>Protection against erosion and other ground stresses</strong>. The road-embedded cabling benefits from the structural integrity of the road itself.</p>
</li>
<li><p><strong>Cables follow natural routes for power delivery.</strong> Since electricity is usually delivered to locations where there is already a road-based route leading to them, there's no need to construct new routes for underground/aboveground cabling.</p>
</li>
<li><p><strong>Easy access</strong>. Unlike a soil-buried cable, a road-buried cable could have a manhole-like detachable cover which wouldn't be swept away by erosion and could be physically anchored to road.</p>
</li>
</ol>
<p>This seems like it would remove the major costs of cable burying and add some marginal costs to road laying. The main downsides I can think of are:</p>
<ol>
<li><p>the need to coordinate road-laying and cable-laying</p>
</li>
<li><p>a solution to upgrade existing roads</p>
</li>
<li><p>potentially much longer paths for cable to follow (although overhead lines seem to follow roads anyway)</p>
</li>
</ol>
<p>Are there any other reasons why this wouldn't work?</p>
| |electrical-engineering|civil-engineering| | <p>The fundamental reason is that at high voltages, buried cables get capacitively coupled to the surrounding earth and the buried transmission line experiences more losses per mile of run than an overhead line.</p>
<p>The practical reason is that it is easier to repair cables when they are overhead than when they are buried, and when overhead, they do not require a (failure-prone) insulating jacket like they do when buried.</p>
| 46902 | Why not bury electrical cables when laying road? |
2021-09-02T19:40:16.677 | <p>Shortly I will have two 3/4" unkeyed shafts needing 3" type-A V-belt pulleys, one each. They'll each have their own motor driving them, and will need to run to 5000 RPM and possibly higher in the future. Is anyone making pulleys for keyless bushings? Or will vibration be reasonable using H-type bushings with compatible pulleys while ignoring the keyways?</p>
| |mechanical-engineering|mechanical|pulleys| | <p>Here are four shaft connection methods.</p>
<p><a href="https://i.stack.imgur.com/CRunW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CRunW.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. Image source: <a href="https://www.ptintl.com/four-shaft-attachment-methods/" rel="nofollow noreferrer">PTI International</a>.</em></p>
<p>I've used something like the Adaptor style successfully. The ones I used had a slot cut on the shaft clamp. The inner and outer ring are conical fit and as you tighten them up they pull together progressively squeezing on the shaft. They're a nice job.</p>
| 46909 | Bushed 3" V-belt pulley with unkeyed 3/4" shaft? |
2021-09-03T00:47:06.893 | <p><a href="https://i.stack.imgur.com/oVsOu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oVsOu.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jTT8O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jTT8O.png" alt="enter image description here" /></a></p>
<p>This assembly is extremely hard to mesh. Even, if I used an extremely finer mesh and took a long time to solve, it still keep failing. This model can be found at <a href="https://drive.google.com/file/d/1rbkwAJNMCYCoay7aoWPJKrbNC1anB0i0/view?usp=sharing" rel="nofollow noreferrer">https://drive.google.com/file/d/1rbkwAJNMCYCoay7aoWPJKrbNC1anB0i0/view?usp=sharing</a>.</p>
<p><strong>Would appreciate it if someone can help!!</strong></p>
| |solidworks|simulation|meshing| | <p>One important (and often neglected) parameter in FEM meshes is the <strong>aspect ratio</strong>. The aspect ratio is the measure of a mesh element's deviation from having all sides of equal length. A high aspect ratio occurs with long, thin elements.</p>
<p><a href="https://i.stack.imgur.com/tSzi7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tSzi7.png" alt="enter image description here" /></a></p>
<p><strong>figure: low and high aspect ratio (source <a href="https://blogs.solidworks.com/tech/2017/08/good-fea-mesh-heres-answer-yes-no-need-know-meshing-infographic.html" rel="nofollow noreferrer">Soidworks blogs</a>)</strong></p>
<p>Because having high values of aspect ratio means that the FEA accuracy reduces, therefore most mesh generators try to make -as much as possible- low aspect ratio elements.</p>
<p>The problem occurs with rods and sheets of metal. Both of them have (at least) one dimension that is significantly higher that the other. So for example for an element representing a sheet (there are a lot like that in the structure you are presenting), the through thickness of the shell imposes a <em>minimum length for <strong>solid</strong> elements</em>.This does the following:</p>
<ul>
<li>increases the aspect ratio,</li>
<li>uses only one (or two in the best case) elements in the direction of the thickness.</li>
</ul>
<p>Although the mesh generator will try to create a mesh with finer solid elements, there are practical problems. To explain it the following example is usually used.</p>
<p>Assume you want to model a solid cube with an edge 100 mm. Also for simplicity assume you will use cubic elements (not the basic tetrahedral). Then you can use</p>
<ul>
<li>1 cubic element, with edge equal to 100mm</li>
<li>8 cubic elements with and edge equal to 50mm</li>
<li>64 cubic elements with and edge equal to 25mm</li>
<li>512 cubic elements with and edge equal to 12.5mm</li>
<li>...</li>
</ul>
<p>if you wanted to model with 1mm elements you'd need 1000000 elements.</p>
<p><strong>So when you reduce the element size, the computational effort increases by a lot (and the memory requirements).</strong></p>
<hr />
<p>In your case, because you have a lot of sheets elements and gears with teeth (which I would remove for the purposes of the analysis)., the mesh generator tries its best for a good compromise.</p>
<p>The solution as jooja pointed is to use shell and beam elements where appropriate. This is still something that needs to be carefully implemented (and there are a lot of do's and don'ts but that's another matter).</p>
| 46913 | Mesh part failed in solidworks issue |
2021-09-03T10:36:46.450 | <p>What is the difference between <strong>nominal model</strong> and <strong>mathematical model</strong> of a real plant?</p>
<p>I am interested in understanding the difference since I want to understand why in some cases is not the best idea to proceed with the control design based on the model.</p>
| |control-engineering|modeling| | <p>This might give you an idea of what is the nominal model, and what it does.</p>
<p>"In continuous control problems we always have models. The idea that we are going to build a self-driving car from trial and error is ludicrous. Fitting models, while laborious, is not out of the realm of possibilities for most systems of interest. Moreover, often times a coarse model suffices in order to plan a nearly optimal control strategy. How much can a model improve performance even when the parameters are unknown or the model doesn’t fully capture all of the system’s behavior?</p>
<p>In this post, I’m going to look at one of the simplest uses of a model in reinforcement learning. The strategy will be to <strong>estimate a predictive model</strong> for the dynamical process and then to use it in a dynamic programming solution to the prescribed control problem. <strong>Building a control system as if this estimated model were true is called nominal control, and the estimated model is called the nominal model.</strong> Nominal control will serve as a useful baseline algorithm for the rest of this series.</p>
<p>From the above, you can see the difference between a nominal model (predictive) and a mathematical model (exact).</p>
<p><a href="http://www.argmin.net/2018/02/26/nominal/#:%7E:text=The%20strategy%20will%20be%20to%20estimate%20a%20predictive,the%20estimated%20model%20is%20called%20the%20nominal%20model" rel="nofollow noreferrer">http://www.argmin.net/2018/02/26/nominal/#:~:text=The%20strategy%20will%20be%20to%20estimate%20a%20predictive,the%20estimated%20model%20is%20called%20the%20nominal%20model</a>.</p>
<p>This paper address the mathematical model <a href="https://www.pearsonhighered.com/assets/samplechapter/0/1/3/6/0136156738.pdf" rel="nofollow noreferrer">https://www.pearsonhighered.com/assets/samplechapter/0/1/3/6/0136156738.pdf</a></p>
| 46920 | Nominal model and Mathematical model of a real plant |
2021-09-03T11:02:26.340 | <p>Two tanks with three continuos, redundant level sensors each need to be supervised so that when a certain total volume is reached, a specific actor is triggered. This could be done via the installed PLC system, however it is likely that a regulatory body will demand a hardware solution. However SIL compliance is not demanded. The important thing is that it's hard to alter the function or change parameters.</p>
<p>So I need a device that can do the following:</p>
<ul>
<li>Take average from three analog signals for tank 1, add offset</li>
<li>same for tank 2</li>
<li>add both values and give a digital signal to trigger if total exceeds a certain value</li>
<li>don't add value for tank 1 or 2 to the total, if a maintenance switch assigned to this tank is on</li>
</ul>
<p>Can this be done in a logic module? If so what specifications (not the specific model) do I need? If not, what's the device called and how to specify?</p>
| |electrical-engineering| | <p>All the major industrial PLC manufacturers make micro PLCs.</p>
<p><a href="https://i.stack.imgur.com/WbStU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WbStU.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. Siemens LOGO! base module with built-in HMI and keypad. A larger panel-mounted remote display is available.</em></p>
<p>The base unit has up to four analog inputs which are 0 - 10 V. You may require 3 × 2-channel 4 - 20 mA analog input modules.</p>
<p><a href="https://i.stack.imgur.com/kHgBa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kHgBa.png" alt="enter image description here" /></a></p>
<p><em>Figure 2. Most of the program requirements done for Siemens LOGO! using Logosoft Comfort and running in the simulator. Download over Ethernet. Double-click for full size view.</em></p>
<ul>
<li>The analog inputs go through an analog amplifier for scaling. The AM (analog markers) are required to terminate the analog signal as far as I can remember.</li>
<li>AVERAGE 1 takes sums the three analog signals and divides by 3.</li>
<li>The SUM block sums TANK 1 and TANK 2.</li>
<li>The COMPARE block compares against the setpoint. I've added some hysteresis.</li>
</ul>
<p>The only thing left for you is to add in some MUX (multiplexor) blocks to output a zero if the maintenance switch is on.</p>
<hr />
<p>I strongly recommend that you avoid hobby boards for industrial controls. PLCs are hardened for industrial environments, have proper isolation between inputs / outputs and control and are very reliable and have a large userbase.</p>
<p>If there is <strong>any danger to personnel or property</strong> then you need a fail-safe system or fail-safe backup system in conjunction with the PLC. Once you install this you will be liable.</p>
| 46921 | Hardware device to evaluate several sensors |
2021-09-03T11:11:32.020 | <p>I was having a discussion today, which led to the question</p>
<blockquote>
<p>why do heavy trucks still use air brakes?</p>
</blockquote>
<p>To my knowledge, it has been used for at least 40 years (I remember that as a kid), and apparently (I was told today but I haven't gotten around to verify it) they are still widely used.</p>
<p>From what I remember, one of the things I was cautioned was that if you repeatedly pressed on the air brakes, then after a while the air buffer would empty and it would take time to fill up (thus losing braking capacity).</p>
<p>Anyway, I wanted to know what are the benefits and disadvantages of air brakes compared to other technologies, e.g. hydraulic lines, or electrical system, (even KERS systems for more modern electrical vehicles).</p>
<p><strong>UPDATE</strong>: From the answers I understand that the main issue is reliability and "technical debt". I want to push a bit further and understand, what's stopping air brakes from being used in other vehicles. E.g. is it cost, performance, inability to accompany AC/DC drum rhythm?</p>
| |mechanical-engineering|automotive-engineering|braking| | <p>"Power" brakes, like the ones you have in your car, use air pressure to assist your pedal effort. In passenger cars the air pressure is actually negative: it relies on intake manifold vacuum. That vacuum is limited to about -15 PSIG: once all the air is drawn from a container, the difference between that pressure and the sea level atmosphere is all you can get. Air brakes use compressors to get unlimited (in principle) pressure which is needed in heavy truck brakes.</p>
<p>Brake geometry, disc/drum, has nothing to do with it.</p>
<p>The vacuum reservoir (I hate using that term to describe a continer of emptiness.) can be depleted by pumping the brakes while the engine is idling and that will increase the pedal effort - just like in air brakes.</p>
| 46922 | Why do heavy trucks use air brakes? |
2021-09-03T14:37:52.677 | <p>I have a second order system with the transfer function as: "1/(s^2 + 5.45s + 1)", I thought it should be an over-damped system, but when I use the damp function in MatLab, it returns the damp ratio zeta as 1 with 2 different natural frequency, what does that mean?</p>
<p>">> <span class="math-container">$display(G6)$</span></p>
<p><span class="math-container">$G6 = \dfrac{1}{s^2 + 5.45s +1}$</span></p>
<p>Continuous-time transfer function.</p>
<blockquote>
<blockquote>
<p><span class="math-container">$damp(G6)$</span></p>
</blockquote>
</blockquote>
<p>Eigenvalue Damping Frequency</p>
<p>-1.90e-01 1.00e+00 1.90e-01<br />
-5.26e+00 1.00e+00 5.26e+00</p>
<p>(Frequencies expressed in rad/seconds)"</p>
| |control-engineering|matlab| | <p>According to <a href="https://www.mathworks.com/help/control/ref/lti.damp.html#mw_774bba0d-fc75-4b75-a377-8ab21de1d8ee" rel="nofollow noreferrer">Matlab doumentation</a>, the damping is calculated <em>per pole</em>. and it is given by the formula</p>
<p><span class="math-container">$$
\zeta = -\cos(\angle s)
$$</span>
where s is the pole location. Since both poles of an over damped system are on the (negative) real axis, the damping is 1 for both poles.</p>
<p>For over damped systems, the poles are purely real and one cannot <em>naturally</em> pair the poles as in the case of complex conjugates. This may be why Matlab decides to report the damping <em>per pole</em>.</p>
| 46931 | Why matlab returns a damp ratio as 1 for a over damped system? |
2021-09-04T00:48:54.677 | <p>My Question was not well received in <a href="http://physics.stackexchange.com">physics.stackexchange.com</a> and I get a simple answer when I want to know the theory inside this question. I want to cook more efficient wasting less ( gas or electricty ) so my though is this when I boil the water cooking cassava the heat is 100° costant then the flame can be reduce for not waste energy, and the water still will be 100° otherwise if I am cooking at the flame is at the maxiximum temperature this food will be cooked faster?. so we have 2 scenarios:</p>
<p>Minimization Energy:</p>
<p>Flame Low = food ready in 30 minutes</p>
<p>Cook Faster:</p>
<p>Flame High = food ready in 10 minutes</p>
<p>Are this escenarios related like work and energy theorem?(e.g. if i reduce the flame to medium will be ready in 20 minutes ) I want to know the performance and the waste of energy in each scenario, <strong>which is better for save gas?.</strong></p>
<p>I take thermodynamics in my university but only law 0 and 1. and not too much of the 2nd law that is related to engineering. I know some basics of chemist process like matter&energy balance</p>
<p>This question is more for a professional of industrial process or chemical engineering so that's why I am posting here.</p>
| |thermodynamics|chemical-engineering| | <p>As long as the water is 100C and atmospheric pressure the food will cook at the same speed regardless of the heat applied to the pot. Any heat in excess of what's required to keep the water 100C simply converts water into steam.</p>
<p>An easy way to vastly improve the efficiency of boiling food is to boil with the cover on (even if the instructions say uncovered). You can usually keep the water boiling with the stove turned down as low as 20%. You just have to be careful as the pot will boil over if the heat is too high. I've been cooking food this way for years, and it definitely does not change the cooking time.</p>
| 46941 | More temperature after boil the water can cook more faster the food? |
2021-09-05T09:05:57.477 | <p>So I wanted to ask if a prefab can be built that looks like this house?</p>
<p><a href="https://i.stack.imgur.com/a3e3P.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a3e3P.jpg" alt="enter image description here" /></a></p>
<p>Also, can wool bricks be used in the house's construction?</p>
| |materials| | <p>Yes.</p>
<p>This can be delivered as a set of wooden-framed walls all with completed wiring and pipework pre-fitted, as well as insulation and final internal wall finishes. Only some joints need painting.</p>
<p>A friend had one delivered like that, build from the foundations was 2 weeks.</p>
<p>There are companies that do this and I won't tell you the one he used as it may not be local to you. You can easily search for others as there are several. Try looking at the trade shows - they always are present there.</p>
<p>As for insulation it can be any material as long as its performance specs meet building regulations. We used "wood wool" as well as lamb's wool in ours, but we specified 30cm of insulation. A no-brainer between a small construction cost increase or lower heating bills for 40 years...</p>
| 46963 | Can the house in the image be built as a prefab? Can wool bricks be used in its construction? |
2021-09-05T16:22:46.213 | <p>It is easy to find on Google that the tensile strength of graphene is 130,000 MPa. But what is it's compressive strength?</p>
| |materials| | <p>Graphene is an allotrope of carbon (same as diamond, graphite and fullerenes ), (allotrope to my mind is a fancy way of saying that the molecules can be arranged in different configurations). Graphene in particular arranges the molecules of carbon in a two dimensional hexagonal lattice.</p>
<p><a href="https://i.stack.imgur.com/UEWRD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UEWRD.png" alt="enter image description here" /></a></p>
<p><strong>Figure: hehagonal lattice of carbon (source <a href="https://www.graphene-info.com/graphene-structure-and-shape" rel="nofollow noreferrer">Graphene-info</a>)</strong></p>
<p>In essence it looks like very thin surface, very much like a thin sheet of paper, with a dimension of 1-2 [nm] (or 1-2 millionth of a mm). Therefore, it has a similar behavior to a piece of paper. I.e.</p>
<ul>
<li>the tensile force required for failure for a sheet of paper is well defined.</li>
<li>the compressive force required to fail (through crumbling) varies depending on the dimensions and the support. I.e. buckling becomes the dominant method.</li>
</ul>
<p>I.e. a longer sheet would fail under its own weight.</p>
<p>--</p>
<p>For comparison purposes, <em>Diamond</em>, has the following lattice. The lattice is face-centered cubic Bravais lattice</p>
<p><a href="https://i.stack.imgur.com/oUg0S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oUg0S.png" alt="enter image description here" /></a></p>
<p><strong>Figure: Diamond lattice(source <a href="http://butane.chem.uiuc.edu/pshapley/GenChem2/C2/3.html" rel="nofollow noreferrer">uiuc.edu</a>)</strong></p>
<p>The three dimensional nature of this structure allows the creation of 3d objects (with varying thicknesses).</p>
| 46970 | What is the compressive strength of graphene? |
2021-09-06T03:12:26.387 | <p>I have seen the different definitions of the velocity ratio for Gears and Gear trains. I have referred a book called Theory of machines by S. S. Ratan. In that book the definitions are as following:</p>
<ol>
<li>For gears, <strong>velocity ratio</strong> is ratio of angular velocity of follower to that of the driving gear.</li>
<li>For gear train, <strong>speed ratio</strong> is given as the ratio of speed of driving to that of the driven shaft.</li>
</ol>
<p>Why are these two definitions different (inverse of each other) as I think Velocity ratio and speed ratio are the same thing. Also if there is a difference, it will be helpful if a bit explanation is provided.</p>
| |mechanical-engineering|gears|machine-design| | <p><strong>TL;DR IMHO, this is something that think has its roots to older times, where the strength design of gears required tables and charts</strong></p>
<p>Like solarMike said, they are essentially the same thing. I don't think many people nowadays pay too much attention to this detail.</p>
<p>IMHO, the reason for the existence stems from the different need. Like you stated, (some authors) use (<strong>note that I'll start the opposite way</strong>):</p>
<blockquote>
<ol start="2">
<li>For gear train, <strong>speed ratio</strong> is given as the ratio of speed of driving to that of the driven shaft.</li>
</ol>
</blockquote>
<p>In that case, the big picture is important. I.e the entire drivetrain, and more specifically its kinematics. In that case the important is to define the angular velocities and torque on the various stages of the drivetrain (or maybe just at the end). Therefore, it that case, the speed ratio focuses on that.</p>
<blockquote>
<ol>
<li>For gears, <strong>velocity ratio</strong> is ratio of angular velocity of follower to that of the driving gear.</li>
</ol>
</blockquote>
<p>In this particular case, the gears themselves are the object of study (to be honest I think the velocity ratio is a poor name, and I have seen it as gear ratio). In that case, except for the kinematic characteristics there are also concerns about strength of the gears. (<em>I haven't seen S. S. Ratan's book but I bet that there is some calculation method for stress inside</em>).</p>
<p>Therefore, in the case of the <strong>gears</strong>, you isolate two components and you try to design them so that they a) perform kinematically as they should and <strong>b) the components themselves don't fail</strong>.</p>
<p>The second part is what's important. I've found several different methodologies for calculating gears (US, European, German, British etc). Almost every manufacturer has one. However, in the majority of cases the gear ratio plays a significant role(even though in more recent cases they avoid it as much as possible). So you have graphs like the following (this is just one example, I am sure that Ratan's book will probably have something similar):</p>
<p><a href="https://i.stack.imgur.com/Djlc0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Djlc0.png" alt="enter image description here" /></a></p>
<p><strong>Figure : Hardness ratio factor <span class="math-container">$C_H$</span> (source: shigley Mechanical Engineering Design 8th edition)</strong></p>
<p>On the x-axis, the reduction gear ratio <span class="math-container">$m_G$</span> is what you call velocity ratio. As you can see in this graph it is much easier to view the curves (rather than if they were between 0-1). So, <strong>because this is at component level (i.e. between the two gears)</strong>, which is the driving and which is the driven doesn't really matter. Instead what it <em>was</em> important was to put a lot all the relevant information without duplicating it. Therefore, they opted to define another ratio which essentially it was always greater than zero, so that the tables could be used without unnecessary duplication.</p>
| 46982 | Velocity or Speed Ratios in Gears and Gear Trains |
2021-09-06T07:55:26.717 | <p>I want to fabricate an aluminum shelf for my media center. 22 1/16” x 37 5/8” with a rectangular cutout 2 1/2” deep on one of the long edges of the sheet with 6” of material on either side of the cutout. (<a href="https://cad.onshape.com/documents/57e301ae0e78317e9d4c2972/w/6cfbee75e3f4ecfea2f3e3bb/e/bc438004f0aa4347176726bf" rel="nofollow noreferrer">CAD model</a>)</p>
<p><a href="https://i.stack.imgur.com/wiWFc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wiWFc.png" alt="enter image description here" /></a></p>
<p>The shelf will be supported by four shelf pins on the short edges near the corners.</p>
<p>The question I need to answer is how thick the aluminum needs to be to support my intended load (100 lbs of equipment, more-or-less in the center).</p>
<p>I've been playing around with <a href="https://www.simscale.com/projects/MatthewYo/metal_shelf" rel="nofollow noreferrer">SimScale</a> and (assuming I'm doing it right) I've determined that a tenth of an inch of aluminum is sufficient, but I'd like a better methodology for computing the <em>minimum</em> required thickness than "guess and check".</p>
<p>What is the right way to compute the required thickness of the sheet of aluminum? Modelling the problem as a rectangle secured on the short edges with a 100 lb load in the center seems like the first step, but coming from a CS/EE background I'm lacking the statics knowledge to enable me to properly set up the problem.</p>
| |aluminum| | <p>An adequate simplification would be to treat it as a simply supported beam (there is not much added benefit to treat it a plate).</p>
<p>For a simply supported beam with:</p>
<ul>
<li>Length L = 37 5/8''</li>
<li>width w = 22 1/16'' - 2 1/2'' = 19 9/16'' (the width is at its thinness point)</li>
</ul>
<p>Assuming the load is dead center:</p>
<p><span class="math-container">$$M=P\cdot \frac{L}{2} = 1881.25 [lbf.in]$$</span>
where:</p>
<ul>
<li>P is the load (100 lbf)</li>
</ul>
<p>The maximum stress that will be developed is:</p>
<p><span class="math-container">$$\sigma = \frac{M}{I}\frac{t}{2}$$</span></p>
<p>where:</p>
<ul>
<li>I is the second moment of area of the plate <span class="math-container">$I = \frac{w\cdot t^3}{12}$</span></li>
<li>t is the thickness</li>
</ul>
<p>the maximum stress needs to be less than the allowable stress:</p>
<p><span class="math-container">$$\sigma_{all} \ge \sigma = \frac{12 M}{w\cdot t^3}\frac{t}{2}$$</span></p>
<p>Therefore
<span class="math-container">$$t \ge \sqrt{\frac{6 M}{w\cdot \sigma_{all} }} = \sqrt{\frac{2 P\cdot L}{w\cdot \sigma_{all} }}$$</span></p>
<p>Assuming you use 6061 which has a yield stress of 40000 psi with a safety factor of 1, then the minimum resulting thickness is 0.1201 inches (which is consisted with your calculations).</p>
<p>For comparison purposes, assuming you applied a safety factor of N=2 (therefore the <span class="math-container">$\sigma_{all}=\frac{40000}{N}=20000 [psi]$</span>, the minimum thickness would be 1.7 inches.</p>
<h2>additional check with deflection.</h2>
<p>Its best to check with deflection.</p>
<p>The max deflection for a simply supported beam with a concentrated load is:</p>
<p><span class="math-container">$$d_{max} = \frac{P L^3}{48 EI}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$d_{max}$</span> is the max deflection. Usually, its defined as a fraction of the beam span length to avoid visible sagging (e.g. 1/250 L =0.1505'' ).</li>
<li><span class="math-container">$E$</span> is the elasticity modulus of aluminium = 10000 ksi</li>
</ul>
<p>So again (after some algebra):</p>
<p><span class="math-container">$$t = \sqrt[3]{\frac{12 P L^3}{48 E\cdot d_{max}\cdot w}}$$</span></p>
<p>If you apply the values, then the minimum thickness results in 0.3564 inches. <strong>As you can see its double the minimum thickness when you calculate based on stresses</strong>.</p>
<p>So, if you use 0.17 inches the shelf will not break, but it will have a noticeable by eye sag, which will not be pleasant.</p>
| 46985 | Computing the required thickness of an aluminum sheet to support a given load |
2021-09-06T09:17:40.437 | <p>Lets assume we have a mechanical element from uniform material that is supposed to work in a static load conditions. Moreover lets assume there is no gravity or gravity is negligible.
Are there any examples when reducing the element mass/volume actually gives an element more strength? I mean pure strength, no strength to weight ratio, the element cost is also irrelevant in this case.</p>
<p>I have mechanics lectures some time ago and if I remember correctly, the lecturer have provided an example of that. Unfortunately I cannot remind myself any details about that. Maybe the "trick" was that he reduced the mass simultaneously changing the shape a little bit in a way that some pressure concentration was dispersed?</p>
<p>Unfortunately I cannot produce or find any example of that now and I am not even sure if I remember this all correctly.</p>
<p>I will appreciate any help.</p>
| |structural-analysis|stresses|structures|strength| | <p>If the load is applied in flexure (i.e. bending), then a small amount of material a long way from the neutral plane can contribute more to the load-bearing capacity than a large amount of material near the neutral plane (that's what the second moment of area is all about). Similarly, if the load is applied in torsion then a small amount of material a long way from the axis about which the torsion is applied can contribute more to the load-bearing capacity than a large amount of material near the axis about which the torsion is applied (that's what the polar moment of area is all about).</p>
| 46991 | Mechanics: example of less mass/volume gives more strength |
2021-09-06T14:06:08.210 | <p>A machine member having a diameter of 50 mm and a length of 250 mm is supported at one end as a cantilever subjected to different forces as shown in the figure. Find the maximum stress at points A and B.
<a href="https://i.stack.imgur.com/Z1XPS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z1XPS.png" alt="enter image description here" /></a></p>
<p>I know to find axial stress will be equal to 15kn/area of crossection and bending moment due to 3kn force can be found as (3kn x 250 x 25)/moment of inertia. I'm not sure will it be polar moment of inertia or area moment of inertia and then how will I be able to achieve maximum stress. kindly looking for help, thanks in advance.</p>
| |mechanical-engineering|applied-mechanics|stresses|moments|bending| | <p>The maximum shear stress due to vertical load is</p>
<p><span class="math-container">$$\tau_{max} = \frac{VQ_{max}}{I_c b}=\frac{4V}{3A}=\frac{4*3kN}{3\pi25^2}$$</span>
And ot happens at the horizintal axis passing through the centroid.</p>
<p>The max shear stress due to torsion happens at the surface of the bar.</p>
<p><span class="math-container">$$\tau_{max} = \frac{2t_{max}}{\pi r^3}=\frac{2kN*1000mm/m}{\pi *25^3}$$</span></p>
<p>So the maximum shear stress is the sum of the two stresses and happens along the two sides of the bar.</p>
<p>As for the bending moment:</p>
<p>M=3kN*250mm=750kNmm
<span class="math-container">$$I_x = \pi r^4 / 4 \rightarrow S_X=I/c= \pi r^3/4 $$</span></p>
<p><span class="math-container">$$\sigma =m/S=750*4kNmm/(\pi*250^3)$$</span></p>
<p>This has to be added and subtracted from the normal tension of <span class="math-container">$\frac{15kN}{\pi 25^2}$</span></p>
| 47001 | Find the maximum stress at a point in a machine member |
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