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2019-07-09T15:24:30.927 | <p>I want to customise a piece of fabric that is 100% polyester (exclusive of decorations, wash at 30º) with vinyl. My idea is to use a vinyl cutter (such as Silver Bullet) to cut a pattern on the sheet, remove the background, transfer to a transfer sheet, and use a heat press to adhere the design to the fabric.</p>
<p>I have seen good results of this method on cotton and I am not sure if a heat press would work on polyester.</p>
<p>How can I transfer a digital design onto 100% polyester fabric?</p>
| |materials|heat-treatment| | <p>I got another similar shirt, 100% polyester, to see if it could work. I found the maximum temperature for polyester at <a href="https://www.thespruce.com/select-correct-ironing-temperature-for-fabrics-2146186" rel="nofollow noreferrer">this page</a>:</p>
<blockquote>
<h2>Iron Setting Temperatures in Celsius and Fahrenheit</h2>
<p>If your iron uses a different scale or you want to know more exacting temperatures for ironing different types of fabric, follow these guidelines:</p>
<p>...</p>
<ul>
<li>Polyester: 148 C/300 F</li>
</ul>
</blockquote>
<p>I found duration for heat pressing <a href="http://wiki.makespace.org/Equipment/TShirtPress" rel="nofollow noreferrer">at this page</a>:</p>
<blockquote>
<p>Suggested settings: temp=160 time=16 (seconds). Varies with different vinyls.</p>
</blockquote>
<p>So I used 20 seconds at 148ºC. I cut the design by hand or on a Silver Bullet Vinyl Cutter. The result works very well:</p>
<p><a href="https://i.stack.imgur.com/eJThN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eJThN.jpg" alt="First shirt with hand-cut design"></a></p>
<p><a href="https://i.stack.imgur.com/10jRc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/10jRc.jpg" alt="Second shirt with computer-cut design"></a></p>
| 29109 | Heat press of vinyl on 100% polyester fabric |
2019-07-09T18:16:13.357 | <p>I have a quick question about the force needed to accelerate/decelerate an irregular object. </p>
<p>Lets say I have a 3D object with the following general shape,
<a href="https://i.stack.imgur.com/YAZxr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YAZxr.png" alt="Gripper Assembly on a Fixed Beam"></a>
This system is in general a gripper (gray) holding a load (yellow) and fixed onto a translating horizontal beam (blue).</p>
<p>I was thinking, to calculate the stopping force of the load and gripper, I just would need to find the center of mass of the load and gripper and apply Newton's Second Law with the expected acceleration/deceleration?</p>
<p>Or does using a center of mass calculation with Newton's Second Law result in a large error from the actual system behavior?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|applied-mechanics|mechanical| | <p>As you said the basic deceleration is <span class="math-container">$$ \alpha =\frac{v_{initial}-v_{final}}{t}=\frac{v}{t}$$</span> </p>
<p>Of secondary order magnitude are things like the tendency of the load to tilt forward an swing up under negative acceleration and that leads to checking for:</p>
<ul>
<li><p>The stability and stiffness of the beam and its mass compared to the mass of load and bracket.</p></li>
<li><p>The play of the bracket so as it does not permit the stopping degenerate into vibration and shimmying of the bearings.</p></li>
<li><p>The hammer action of the load at the end of its course.</p></li>
</ul>
| 29114 | Stopping Force of Irregularly Shaped Object |
2019-07-10T18:07:39.903 | <p>Does there exist any type of Nut, that can be screwed on a threaded rod that can slipped in without having to go to the ends of the threaded rod? I can sort of imagine a Nut that was a <code>U</code> shape instead of <code>O</code> shape, or perhaps two <code>C</code> shapes that could interlock some how. Does anything like this exist?</p>
| |fasteners| | <p>try a sliding speed nut: <a href="https://i.stack.imgur.com/6SQkO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6SQkO.jpg" alt="enter image description here" /></a>
very easy to slide along threaded rod <a href="https://www.leevalley.com/en-ca/shop/hardware/jig-and-fixture-parts/101377-1-4-20-and-5-16-18-brass-speed-nuts" rel="nofollow noreferrer">https://www.leevalley.com/en-ca/shop/hardware/jig-and-fixture-parts/101377-1-4-20-and-5-16-18-brass-speed-nuts</a></p>
| 29127 | Fastening a Nut on threaded rod from the middle |
2019-07-11T15:23:33.400 | <p>I am reading the rocket propulsion elements book by sutton and I came across the following statement: "Because the propellant tank has to fly. its mass is at a premium and the tank material is therefore highly stressed."
I am just learning, and have no strong technical background, so I will appreciate the explanation!
Thanks! </p>
| |structural-engineering|aerospace-engineering| | <p>It means, the fuel tank's mass costs fuel to move it and of course that extra fuel has mass too and that mass needs more fuel. Premium here means extra fuel.</p>
<p>So there is a an effort to keep the mass of the fuel tank at the absolute safe minimum, hence allowing the tank material exposed to high stresses.</p>
| 29141 | What does it mean that the mass of a propellant tank is at premium? |
2019-07-11T20:46:13.530 | <p>I have a 2D vibration sensor that can only measure in the X and Y directions, and it is mounted on top of a motor that is fixed into place, as shown in a rough sketch below.</p>
<p><a href="https://i.stack.imgur.com/PzRNH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PzRNH.png" alt="rough depiction of vibration sensor on motor"></a></p>
<p>Would I be able to work out the vibrations in the Z direction based on a cross-correlation of just the vibration in the X and Y directions?</p>
| |mechanical-engineering|vibration| | <p>What vibrations are you interested in?</p>
<p>Are the vibrations somehow interlinked - what else beyond the question you placed to us do you know?</p>
<p>If you can constrain the question more, the situation may change. If you cannot, then @timwescott's answer is obviously the correct one.</p>
| 29150 | Would a cross correlation between vibrations in the X and Y directions help determine vibrations in the Z direction? |
2019-07-13T07:40:33.673 | <p>I am trying to find the transfer function of a plant that contains numerous equipment including a heater, several tanks, pumps, membranes etc.</p>
<p>I need to design 2 PID controllers: the first one to control the temperature of the stream entering one of the membranes, and the second PID controller needs to maintain the solution concentration in one of the tanks.</p>
<p>I have seen that there is a system identification tool in Matlab that I can use to find the plant transfer function. The problem is, I am not clear on exactly what the plant transfer function should describe. Should it represent the whole plant (e.g. all the tanks, pumps etc.) or only the part of the plant that I am interested in to be able to design the controllers (e.g. resulting in two different plant transfer function equations for each controller).</p>
<p>Thanks in advance.</p>
| |chemical-engineering| | <p>When you're working on systems design, the plant is the thing you're controlling, and it's up to you to figure out what that thing is.</p>
<p>Keep in mind that your plant may not be my plant -- in fact, if you're buying a sub-system from me, that sub-system may be your plant, while to me it is my closed-loop system.</p>
<p>For the purposes of your control loops, for each controller the plant transfer function you're interested in is the one that describes the relationship between whatever plant input that your controller affects, and whatever plant output that is affected by your controller.</p>
<p>So, for example, one of your "plants" has a heater command as an input, and a temperature as an output. The other one (I presume) has a flow rate as an input, and a solution concentrate of the desired reactant as an output.</p>
<p>Keep in mind that if these two processes (temperature and concentration) are linked, so that each one affects the other, then you may not be able to design each PID controller independently. In such a case you'd first need to worry about whether the coupling is strong enough to be an issue, and then what to do about it (which would be an answer to a different question).</p>
| 29177 | Identify plant transfer function to design PID controller |
2019-07-15T08:16:20.727 | <p>I have not taken any course in civil engineering. What I was taught was that a broad base would prevent something from toppling. So it makes me wondering instead of nailing the foundation of a building vertically deep down into the ground, if it would make sense to make multiple buildings share a horizontal foundation.</p>
| |structural-engineering|civil-engineering|structural-analysis|building-design|building-physics| | <p>Things topple when their center of mass moves past their supports.</p>
<p>Think of a square box: if you push it over slightly and let go, it'll fall back onto its original position. But if you push it past 45 degrees (when the center of the box would be exactly over the supporting edge), it'll topple onto one of its other sides.</p>
<p>Therefore, obviously, the wider the base, the harder it is for something to topple.</p>
<p>Unfortunately, that's not how earthquakes cause buildings to collapse. It's not that the earth moves the foundation so far that the building's center of gravity moves past it.</p>
<p>Earthquakes damage buildings by moving the base very quickly. The law of inertia (an object at rest wants to stay at rest) means that though the base moves, the rest of the building doesn't want to. And once the rest starts moving one way, the earthquake is then pulling the foundation the other way. In the worst case, this leads to a positive feedback loop, with each earthquake movement feeding the rest of the building's movement (the so-called "resonant frequency"). This causes significant stresses in the structure, which can damage or even destroy the building if it's improperly designed.</p>
<p>(For an example of how a low-energy earthquake at the resonant frequency is worth than a high-energy earthquake, see <a href="https://youtu.be/6JwEYamjXpA?t=768" rel="nofollow noreferrer">this video</a>)</p>
| 29190 | Can a horizontal foundation safeguard against earthquakes? |
2019-07-15T23:08:06.980 | <p>A box is being lifted by a crane. Sometimes the box will be completely full, and sometimes it will only be partially full (indicated by the gray shaded regions).</p>
<p><a href="https://i.stack.imgur.com/8TmRe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8TmRe.png" alt="enter image description here"></a></p>
<p>The question is, where is the axis of bending in both cases? Does it follow the center of mass? Or does it always follow the geometric center?</p>
<p>(Of course I hope the axis of bending will always be in the middle so it could take full advantage of the roof and floor to resist bending.)</p>
<p>Also, if it matters, the crane will actually lift the box at 4 points, not 2. And I moved them inwards from the edges just because I thought that would resist bending more, due to a shorter beam.</p>
<p>Once I know where the axis is, I can calculate the moment of inertia and use the deflection formula </p>
<p><span class="math-container">$$\delta = \dfrac{5qL^4}{384 E I}$$</span></p>
<p><strong>Update</strong></p>
<p>As advised, I did this for a U-shape, but also for the square, for comparison. I had to choose concrete numbers so I picked a 3x3x12 box with wall thickness of 1 cm.</p>
<p><a href="https://i.stack.imgur.com/hDqTL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hDqTL.png" alt="enter image description here"></a></p>
<p>For the square, I found the formula <a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="nofollow noreferrer">here</a>.</p>
<p><span class="math-container">$ I = a^4 - (a - 2t)^4 $</span></p>
<p><span class="math-container">$ I = 3^4 - (3 - 0.02)^4 = 2.13849584$</span></p>
<p>For the U-shape, I found the much more complex formula <a href="http://www.efunda.com/math/areas/squarechannel.cfm" rel="nofollow noreferrer">here</a> and took it about the y-axis since that diagram was vertical.</p>
<p><span class="math-container">$ I = \frac{2ta^3 + (a - 2t)t^3}{3} - (2ta + (a - 2t)t)\frac{2ta^2 + (a - 2t)t^2}{2a^2 - 2(a - 2t)(a - t)}$</span></p>
<p><span class="math-container">$ I = \frac{2ta^3 + at^3 - 2t^4}{3} - \frac{(2ta + ta - 2t^2)(2ta^2 + at^2 - 2t^3)}{2a^3 - 2(a^2 - 3ta + 2t^2)} $</span></p>
<p><span class="math-container">$ I = \frac{0.54000298}{3} - \frac{(0.0898)(0.180298)}{54 - 2(8.9102)}$</span></p>
<p><span class="math-container">$ I = apprx. 0.179553$</span></p>
<p>So now I can finally use the deflection formula with both I's. I'm using steel so E = 200 GPa. The distributed load, q, is for a box 1/3rd full of crushed iron ore which has a density of 2,500 kg/m<span class="math-container">$^3$</span>. q = 72,523 N/m.</p>
<p><span class="math-container">$\delta = \dfrac{5qL^4}{384 E I}$</span></p>
<p><span class="math-container">$\delta = \dfrac{5 * 72523 * 12^4}{384 * 200,000,000,000 * I}$</span></p>
<p><span class="math-container">$\delta = \dfrac{7,519,184,640}{76,800,000,000,000 * I}$</span></p>
<p><span class="math-container">$\delta = 0.00004578$</span> m for the square.</p>
<p><span class="math-container">$\delta = 0.0005452766$</span> m for the U-shape.</p>
<p>...I have to conclude I did something wrong. It is hard to imagine a box of that size, with just 1-cm thick walls, would deflect by less than a millimeter when holding up iron ore, even at 1/3rd full.</p>
| |mechanical-engineering|statics|deflection| | <p>As implicitly mentioned in <a href="https://engineering.stackexchange.com/a/29207/1832">@kamran's answer</a>, the neutral axis (which is what I understand by 'axis of bending') is constant regardless of the applied load.</p>
<p>That's because the neutral axis is a property of the structure resisting the load, and in this case the load does not help to "carry itself".</p>
<p>So the neutral axis will merely be equal to the position of the box's centroid. However, it is important to note the box should almost certainly be considered a U-shape. That's because the box's lid is unlikely to be sufficiently "bound" to the rest of the box to truly act monolithically with the rest. So you'd want to consider a lid-less box as your cross-section (though having to carry the weight of the lid, of course).</p>
<p>Unfortunately, U-shapes are very tricky to design under bending, due to the possibility of localized buckling of the "wings". This is especially problematic Given that your U-shape needs to be "belly-down" (that is, looking like a U, not an <span class="math-container">$\Pi$</span>), which means that the top of the beam (in this case, the wings) will be under compression. And given how low the center of gravity would be in this orientation, it'll be very high compression in the members the least equipped to withstand it.</p>
<p>I'd honestly recommend talking to an engineer to calculate this for you (especially given that there are myriad other aspects we haven't taken into consideration, such as wind and dynamic forces), or perhaps renting boxes made for this situation instead of trying to make your own. Another "solution" would be to use more than two support points on each side of the box. Using many supports would drastically reduce the applied load.</p>
| 29205 | Where is the axis of bending for a box lifted by a crane? |
2019-07-15T23:21:19.837 | <p>I’m a graduate structural engineer with 4 years experience, working in the UK. A family friend is planning on doing a house extension which will require some structural calculations for steel beams. My question is:
What will building control in England stipulate in terms of the structural engineer’s qualifications, liability insurance etc?
Do you need to be a chartered engineer with insurance to undertake structural calculations in England?
I think my experience is sufficient for what is a relatively simple calculation pack but I’m not sure what the legal stance is on this in England. </p>
| |structural-engineering|civil-engineering|licensure| | <p>IANAL, but the first step is to find whether you need planning permission for the work. Talk to your local council's planning department. They are usually helpful.</p>
<p>If you do need to submit a planning application, you will have to create drawings in a standard architectural format for inspection by anyone who may potentially be affected by the work - e.g. the owners of neighbouring properties. The planning department will notify them of the application to give them opportunity to comment on it before it is approved. These drawings don't specify the technical details of the construction, but show the changes to the appearance of the existing buildings.</p>
<p>Whether you need planning permission or not, the work needs to comply with the UK building regulations.</p>
<p>See <a href="https://www.gov.uk/government/collections/approved-documents" rel="nofollow noreferrer">here</a> for (detailed) official guidance on how to do that.</p>
<p>You may also want to contact the house insurers at the planning stage, rather than presenting them with a fait accompli at the end. They may wish to inspect the work in stages as it is carried out, or require copies of any planning department inspections. Failure to coordinate all the required inspections can cause significant delays to the work plan. </p>
<p>In general, the UK system doesn't care about the formal qualifications of the people involved, so long as the work meets the required standards. AFAIK there is no legal necessity for any form of insurance either on the work itself or the finished building - though it would be foolish not to have insurance against third party claims if there is any risk of a disaster damaging other people's property. </p>
<p>You will probably want some long-term guarantee of the work done should you want to sell the property - e.g. the National House-Building Council's standard guarantee is 10 years. That should be provided by the builder.</p>
<p>Since the builder will be guaranteeing the work, they will need to be satisfied that the technical specifications for the work are sensible, rather than just "doing whatever you tell them" regardless of the risks involved.</p>
<p>You may also require final inspections if you have made major changes to the water, gas, or electricity installations.</p>
<p>The bottom line of all this is that "doing the calculations" is often trivial compared with "understanding how the system works" - which is the expertise that professionals working in the field have.</p>
| 29206 | Do you need to be a chartered engineer to submit structural calculations to building control in the UK? |
2019-07-16T02:53:27.460 | <p>Everything I've read on the Rankine cycle indicates that the feedwater pump is necessary for the cycle to work. However the steam cycle used in traditional household boiler / radiator systems seems to challenge this.</p>
<p>The cycle for the household boiler seems to be a closed loop cycle with just a heat source (boiler) and cold source (radiators). But those two pieces are sufficient to have steam flow through the radiators, condense, and then return to the boiler.</p>
<ul>
<li>Is my understanding of the boiler / radiator cycle correct?</li>
<li>If so, and if the system was fitted with a turbine, could it produce any amount of power?</li>
<li>If it produced any amount of power, how inefficient would it be compared to a Rankine cycle?</li>
<li>Is there a name for the type of cycle in a boiler / radiator system?</li>
</ul>
| |thermodynamics|heat-transfer|heating-systems| | <p>Your heating system is not a Rankine cycle. In fact it's not considered a thermodynamic cycle at all because all it does is transfer heat. If you add a steam turbine you could possibly produce a tiny amount of mechanical power, but it would be very limited because the temperature and pressure are so low.</p>
<p>But if you do that you'll quickly discover the problem - steam will be leaving the system in order to power the turbine. That means you'll run out of water, so you have to condense the steam and pump the water back in, or vent it to the atmosphere and pump new water in. In either case, you'll need the feedwater pump.</p>
| 29208 | Steam cycle for a boiler / radiator system vs Rankine cycle |
2019-07-16T04:19:39.043 | <p>I was looking at laminar flow filters to aid in my mycology work I'm doing. Most filters use an aluminum extrusion with fins. I figured the options out there are pretty expensive for what you get and figured I could just 3d print my own straightener. Question is what shape do I use? I have seen honeycomb used and would be an easy option, but it really got me thinking about what shape would be optimal for this. We all know that when streamlining a positive body that a teardrop is close to optimal, but putting teardrops in a grid would yield a weird negative shape in the pattern. Any thoughts or ideas on this would be greatly appreciated, I plan on open sourcing the final design. Thanks</p>
| |fluid-mechanics|airflow|aerodynamics|airfoils| | <p>Just doing a quick search, this NASA paper (<a href="https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19810020599.pdf" rel="nofollow noreferrer">https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19810020599.pdf</a>) indicates that no particular geometry has significant advantages for producing flow uniformity. I would stick to a well-tested geometry that is the simplest to produce with a 3D printer. Also, keep in mind the purpose of using honeycomb is to remove turbulence (vortices) from the flow via reducing the hydraulic diameter. Thus, so long as you don’t inflict a large pressure drop through the honeycomb (if you care about pressure drop), any shape with an appreciable reduction in flow hydraulic diameter will reduce the turbulent intensity. This paper seems to be useful as well: <a href="https://conservancy.umn.edu/bitstream/handle/11299/108725/pr338.pdf?sequence=1" rel="nofollow noreferrer">https://conservancy.umn.edu/bitstream/handle/11299/108725/pr338.pdf?sequence=1</a></p>
| 29211 | Laminar flow hood/air straightener |
2019-07-17T17:13:33.640 | <p>When I'm looking at images / videos about the load supported by reinforced concrete slabs, I always see a compression at the top of the slab in every point that varies from the center to the edges:</p>
<p><a href="https://i.stack.imgur.com/f4eoZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f4eoZ.png" alt="enter image description here"></a></p>
<p>But isn't there a tension on the slab near the edges caused by the fact that the slab is supported by the walls?</p>
<p>I mean, why I always see the first and not the second schema below?</p>
<p><a href="https://i.stack.imgur.com/kqkmK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kqkmK.png" alt="enter image description here"></a></p>
| |structural-engineering|structural-analysis|reinforced-concrete| | <p>From your image, we can clearly see supports on two edges of the slab. It seems reasonable to assume this is your entire model (that is, that you haven't created a view where the rest of the structure is "hidden").</p>
<p>That being the case, the slab will behave entirely according to how you've defined the supports. By the slab's behavior, it seems you've defined the supports as pinned, not restricting rotations. This being the case, the slab is free to rotate around the supports and does so, such that no negative bending moment is generated.</p>
<p>In a real structure, however, the slab's behavior is affected by the stiffness of its supports.</p>
<p>Slabs are usually supported in one of two ways: either by walls or by beams (which are themselves supported by columns)<sup>1</sup>.</p>
<p>When slabs such as yours (which is only supported on two opposing edges) are supported by walls, the entire wall-slab-wall structure behaves like a column-beam-column frame. Just as the beam tries to deflect by rotating its supports (the tops of the columns) and the columns resist, generating negative bending moment on the beam, the walls resist the slab's rotation, generating the same negative moment.</p>
<p>Obviously, the columns/walls are flexible, so there is rotation at the beam/slab's edges. Just not as much as there would be if the connection were pinned.</p>
<p>When a slab is supported by a beam, however, is does basically behave as if it were a simple pinned support. That's because the slab rotating is felt by the beam as a rotation around it's own axis. That is, the beam feels torque (as well as bending moment from the vertical force transferred from the slab, of course). And beams are very flexible to torque, especially the types of beams commonly used in non-infrastructure construction. So the slab basically feels it can transfer its load to its support (the beam) and that it can rotate freely around that support: the very definition of a pinned support.<sup>2</sup></p>
<hr>
<p>That being said, there's a catch.</p>
<p>I've so far only discussed a single slab being supported by walls and/or beams.</p>
<p>But what if a wall/beam is supporting not just one slab, but two? That is, you have a slab to either side of the support.</p>
<p>Well, in that case, the other slab's stiffness must also be taken into consideration. Making matters worse, two side-by-side slabs will usually be designed under the same loading conditions. So if one slab under load is trying to rotate a wall/beam one way, the other slab will be trying to rotate it the other way, most likely canceling that rotation out entirely.</p>
<p>So if two slabs share a supporting wall or beam, they are usually treated as being fully fixed on that edge.</p>
<hr>
<p><sup>1</sup> <sub>Slabs can also be directly supported by columns, but that's much less common and orders of magnitude harder to calculate.</sub></p>
<p><sup>2</sup> <sub>There is a slight effect from the beam's vertical deflection at midspan, but it is negligible and usually ignored.</sub></p>
| 29235 | Why not a reverse bending in the slab? |
2019-07-17T21:22:16.607 | <p>I have a quick question about the parallel axis theorem.</p>
<p>Let's say I have a box-like object rotated about the following pivot point</p>
<p><a href="https://i.stack.imgur.com/lYgRd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lYgRd.png" alt="enter image description here"></a></p>
<p>I was wondering, by applying the parallel axis theorem, would the moment of inertia about the pivot point be the following, </p>
<p><span class="math-container">$$M = I_{cm} + M_d^2+M_h^2$$</span></p>
<p>where <span class="math-container">$I_{cm}$</span> is the moment of inertia about the box's centroid and <span class="math-container">$M$</span> is its mass.</p>
<p>Or, does the parallel axis theorem not hold true for two axis shifts?</p>
| |mechanical-engineering|structural-analysis|applied-mechanics|statics|torque| | <p>Moment of inertia is based on both a point of reference <strong><em>and</em></strong> an axis of rotation.</p>
<p>So the answer to your question depends on the axis you're asking about.</p>
<p>For the moment of inertia around the horizontal axis, you only care about the vertical shift (and visa-versa for the vertical axis, of course):</p>
<p><span class="math-container">$$I_{xx} = I_{cm} + M\cdot h^2$$</span></p>
<p>For any other axis, things get much more complicated, as suggested by <a href="https://engineering.stackexchange.com/a/29242/1832">@kamran's answer</a>.</p>
| 29238 | Parallel Axis Theorem for x and y Shift |
2019-07-16T22:08:15.610 | <p>I am wondering if boiling water that is trapped inside of a semi-enclosed pipe would be a low-cost method of creating an underwater thruster that would function as a silent marine propulsion system.</p>
<p>Please reference the following conceptual drawing:</p>
<p><a href="https://i.stack.imgur.com/dgFFf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dgFFf.png" alt="enter image description here"></a></p>
<p>This drawing is showing a cross-sectional side view of a conceptual underwater thruster. There would be a copper pipe that would be partially positioned within a metal cylinder and within this metal cylinder there would be an AC induction heating element that would partially surround the closed end of the copper pipe. The other end of the copper pipe would be located outside of the metal cylinder and it would be open-ended. This heating element should create a rising water temperature gradient within the closed end of the copper pipe. </p>
<p>The AC induction heating element would be turned on and would remain on so that it would continually boil the water. This will cause the water pressure (psi) inside the pipe to be higher that the ambient water pressure outside of the pipe/thruster and this imbalance should cause the thruster to move forward as indicated on the drawing. This thruster could be mounted externally to a ship or submarine's hull or it could be installed internally in the stern area of either vessel.</p>
<p>The main premise of this idea is that the boiling water will not be able to exit the copper pipe so there should be continual high water pressure inside the pipe and thus continual thrust. Also, there should always be water inside of the pipe because there should be a continual circulating flow of warm water exiting the copper pipe while colder outside water enters it.</p>
<p>Could boiling water inside a semi-enclosed pipe be a low-cost method for a silent marine propulsion system?</p>
<p><strong>EDIT</strong></p>
<p>Based on the answers and comments I have received to original design above, I have redesigned this 'Boiling Water Thruster', shown in the new drawing below:</p>
<p><a href="https://i.stack.imgur.com/FIUNJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FIUNJ.png" alt="enter image description here"></a></p>
<p>I believe this new underwater thruster design should provide silent propulsion for the ship/submarine, although the thrust would be low compared to propeller driven craft.</p>
| |fluid-mechanics|thermodynamics|design|propulsion| | <p>It can kinda work. Ish.</p>
<p>Your design is based, at its core, on the idea that warm water rises. Warm water only rises in cold water, so I am sceptical of the efficacy of simple thermal gradients. Nevertheless, bubble pumps have been in commercial use for many years, so if you get it just right I suppose it's possible. </p>
<p>However, you have some serious mass flow problems in your design. If I understand correctly, in your design you intend the flow path to be something like this:
<a href="https://i.stack.imgur.com/1P3aT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1P3aT.png" alt="naive pipe pump"></a></p>
<p>Now, this system is pretty hard to analyze simply. As the essential components are the pumping action and the flows, I think it will be easier to see on a simpler system that is effectively the same (the symbol is a generic pump):
<a href="https://i.stack.imgur.com/dM5Ho.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dM5Ho.png" alt="equivalent mass flow system"></a></p>
<p>(let right be positive)
I'm going to play a little loosy-goosy with mass and mass flow rate so as to not complicate the analysis with a whole lot of derivatives. </p>
<p>I assume you know that <span class="math-container">$F = ma$</span>? In this system, there are two forces on the submarine: the force from sucking in the inlet water, which pushes to the left, and the force from pushing out the effluent water, which pushes to the right. Ok, so it is more useful to use the momentum forms of Newton's laws here:
<span class="math-container">$$ F = \frac{dp}{dt}, \Delta p_{net} = 0$$</span></p>
<p>This means he change of the momentum of the craft is equal to the change of momentum of the water. At the inlet pipe, the water is moving into the sub at some velocity <span class="math-container">$+v_{in}$</span>, and some mass <span class="math-container">$m_{inlet}$</span> is coming in. As <span class="math-container">$p = mv$</span> and the momentum imparted to the sub is exactly opposite to that imparted on the water, so <span class="math-container">$$\Delta p_{in, sub} = -\Delta p_{in, water} = -V_{in} \cdot m_{in}$$</span> It is similar at the outlet pipe, so <span class="math-container">$$\Delta p_{out, sub} = -\Delta p_{out, water} = +V_{out} \cdot m_{out}$$</span></p>
<p>Now, unless you are somehow generating water inside the craft, <span class="math-container">$m_{in}$</span> <em>must</em> equal <span class="math-container">$m_{out}$</span>. Therefore, the total momentum imparted on the craft is:
<span class="math-container">$$ \Delta p_{tot} = \Delta p_{in, sub} + \Delta p_{out, sub} = V_{out} \cdot m_{out} - V_{in} \cdot m_{in} = (V_{out} - V_{in})\cdot m_{water}$$</span></p>
<p>And there's the key problem with the design! If <span class="math-container">$V_{out} = V_{in}$</span>, then $\Delta p = 0} and the craft cannot accelerate, no matter how fast you pump the water. </p>
<p>Ok, now for some fluid mechanics. How can we make <span class="math-container">$V_{out} \ne V_{in}$</span>?
Well, the mass <span class="math-container">$ \dot{m}$</span> flowing through an orifice is proportional to area, velocity, and density:
<span class="math-container">$$ \dot{m} = A \cdot V \cdot \rho $$</span>
because we know <span class="math-container">$\dot{m}_{in} = \dot{m}_{out}$</span>, then
<span class="math-container">$$ A_{in} \cdot V_{in} \cdot \rho_{in} = A_{out} \cdot V_{out} \cdot \rho_{out}$$</span>
Now while the would change from inlet to outlet if they are different temperatures, water only has a density change of 3% over its entire liquid range, so I am going to disregard that, so:
<span class="math-container">$$ A_{in} \cdot V_{in} = A_{out} \cdot V_{out} $$</span>
Thus the only way that we can cause the craft to accelerate is to have a smaller outlet orifice than inlet (I'll let you work that one out). </p>
<p>If we move back to your design, I can't see how you could possibly enforce that in a simple tube. The inlet and outlet streams are almost completely uncontrolled (they touch! you're going to have Rayleigh waves out the wazoo!) and while you could maybe do <em>something</em> (put a mesh to restrict flow on the bottom?) I can't see an efficient one.</p>
<p>Also, boiling is loud. Have you ever boiled a pot of water? It's certainly not silent. You would also need to get power from somewhere. If it's from a nuclear reactor (as you mention in a comment) those are also <em>really</em> loud.</p>
<p>If you're dead set on this bubble pump thing, I feel it would be much more feasible to simply put the inlet at the front:</p>
<p><a href="https://i.stack.imgur.com/c4qrz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c4qrz.png" alt="bubble pump with inlet at front"></a></p>
| 29239 | Could boiling water inside a semi-enclosed pipe be a low-cost method for a silent marine propulsion system? (redesigned) |
2019-07-18T17:20:48.043 | <p>I am trying to test a piece of furniture based on a particular code, the code states that i can either drop a bag (with a specific shape) of 57 kg 1.2 in above the seat or I can use other methods as long as they can be shown to be equivalent. My company is interested on using vertical pneumatic cylinders instead since the test needs to be repeated 100 000 times.</p>
<p>My question is: If i manage to get a piston that produces the equivalent of 57kg in force and i adjust its speed to match the speed at contact for a free falling bag (0.77 m/s), would this produce equivalent effects on my test sample?</p>
<p>I am not sure if i am overlooking or neglecting something. Any advice will be appreciated</p>
<p>Thanks </p>
| |pneumatic|product-testing| | <p>It's not a question of force, its momentum m*v in a non-elastic collision that remains constant. So you have to measure the mass of your pneumatic moving rod and make sure it complies with Newton's law in a way that its momentum is equal to the momentum of the bag at the time it touches the sofa. </p>
<p><strong>Edit</strong></p>
<p>After OP's comment, I added detail subscripts to identify moving parts, to clarify I mean the momentum of the pneumatic punch rod should be equal to the momentum of the bag at its final velocity just before it collides with the sofa.</p>
<p><span class="math-container">$$m_{rod}*v_{rod}= m_{bag}*v_{bag}$$</span>. </p>
<p>This will deliver the same impact to the sofa. As you said in your comment kinetic energy will not preserve.</p>
| 29248 | Emulate impact forces using pneumatic piston |
2019-07-19T23:49:48.930 | <p>I am having a house built and the builder found organic soils during the excavation. They removed the soils and then filled the cavity with 1x3 stone (see picture). This is the corner where the sump crock is supposed to be and I am concerned that the digging of the hole for the sump will unsettle the stone. The builder doesn't think so. My question is should I be concerned about this or will disturbing the stone undermine the foundation.
<a href="https://i.stack.imgur.com/UhA1e.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UhA1e.jpg" alt="enter image description here"></a>
Thank you.</p>
| |civil-engineering|soil|foundations| | <p>If the excavation has a slope of roughly 3 vertical to 1 horizontal, that's about 70 degrees with horizontal. An unloaded slope in stones, gravel or sand will not be stable at an angle of more than approximately 40 degrees with horizontal, so it is not possible to make the excavation without disturbing the soil. That angle is called the angle of internal friction and is considered a material parameters that describes the quality of the soil. The value of 40 degrees is an estimate of mine that I am guessing might be applicable to your stone fill, and it assumes a well-compressed fill of decent quality stones. And furthermore, it does not include any safety factors, so the slope of the excavation should be further reduced to about 35 degrees with horizontal.</p>
<p>Instead of moving the sump crock away from the foundation, you could consider installing a small retaining wall.</p>
| 29264 | 3 x 1 stone under footing and sump crock |
2019-07-20T15:58:41.947 | <p>Suppose I want to implant a bio-chip that consumes a lot of power into my body, for example, a full-fledged CPU, and I don't want to lug around a heavy battery like the ones used in pacemakers, is it possible to be in Rome and do as the Romans do, i.e. draw and utilize the sugar inside my blood as fuel, like a huge living and breathing body cell or an organ like the liver? Let's assume I can put this device where there's a lot of blood flow, say the jugular vein.</p>
| |biomedical-engineering| | <p>One possible way to do this <em>could</em> be a <a href="https://en.wikipedia.org/wiki/Microbial_fuel_cell#Power_generation" rel="nofollow noreferrer">microbial fuel cell</a>. </p>
<p>This has, apparently, been tested ex-vivo: <a href="https://www.researchgate.net/publication/236071090_A_pacemaker_powered_by_an_implantable_biofuel_cell_operating_under_conditions_mimicking_the_human_blood_circulatory_system_-_battery_not_included" rel="nofollow noreferrer">A pacemaker powered by an implantable biofuel cell operating under conditions mimicking the human blood circulatory system - battery not included</a></p>
<p>Another approach, which also works at least ex-vivo, is called Abiotic Biofuel Cell: <a href="https://onlinelibrary.wiley.com/doi/abs/10.1002/elan.201400440" rel="nofollow noreferrer">Pacemaker Activated by an Abiotic Biofuel Cell Operated in Human Serum Solution</a></p>
<p>Keywords here are biofuel cell and microbial fuel cells, knowing these you can google the rest yourself.</p>
| 29274 | Is it possible to power bio-chips with blood sugar? |
2019-07-21T15:21:13.787 | <p>To follow up with my <a href="https://engineering.stackexchange.com/questions/29190/can-a-horizontal-foundation-safeguard-against-earthquakes">previous question on the seismic design of buildings</a>, I would like to ask if it makes sense to support buildings with strong thick 'crisscrossed' metallic frames.</p>
<p>I have the impression that an 'X-structured design' provides good structural integrity, as commonly seen in construction sites. Would it be wise to construct buildings out of a metallic 'skeleton' that is formed throughout by intersecting the diagonals of squares?</p>
| |structural-engineering|civil-engineering|structural-analysis|building-design|building-physics| | <p>earth quake is basically random vibration of the earth in several different ways, S wave, P wave etc. which is a complex and different subject.</p>
<p>X bracing is a way of making the structure stiff, so in short buildings combined with other measures is mostly an effective method.</p>
<p>But in high risers and tall building the critical thing is to mitigate the huge shears that X bracing imparts on the foundation and also make sure the building structure vibration caused the earth quake does not resonate with the vibration of site.</p>
<p>So in tall buildings the strategy is to design the building for earth quake and wind fluttering in a way that the natural frequency of the building is not close to the frequency of earthquake or wind turbulence.</p>
<p>That is why in tall buildings they sometimes use moment frame as opposed to X bracing. These are classified as passive methods.</p>
<p>In some critical structures they insulate the building from the foundation by using rollers and shock absorbers. </p>
<p>Or in some new projects the have active smart earthquake resisting computer controlled hydraulic jacks and braces, which make the building sway, opposing and breaking the impulse of the earthquake. </p>
| 29288 | Can a crisscrossed metallic skeleton resist earthquakes for buildings? |
2019-07-21T18:21:32.613 | <p>I am currently designing a setup for an experiment which involves loading a weight to its extension arm, and by design this structure is not bolted to any surface so its free to move. I need to simulate whether this structure will tip over because of the torque caused by this weight. I looked at some FEA softwares such as patran but I can only get static stress analysis, what kind of technique/software should I use to simulate this kind of motion in computer.</p>
<p>Thanks in advance<a href="https://i.stack.imgur.com/tsZiC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tsZiC.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-analysis| | <p>@kamran described how to analytically find the location of the C.G in your case. However, according to your comments I understand you want to play a little and gain experience with engineering programs. Am I Right?</p>
<p>Following @alephzero and @ Wasabi♦ comments, here you can find a way for implementing their approach on ANSYS WORKBENCH:</p>
<p>A simplified model was created - containing two structures having different geometries. Each of these geometries is a one piece item made of a single material. This was done just for simplicity, of course you can apply the same approach on any super complicated assembly.
To make the examination even easier each item was simply supported to the ground as shown in the following screenshot:
<a href="https://i.stack.imgur.com/X9jou.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X9jou.png" alt="enter image description here"></a></p>
<p>Now, what you would like to do is to examine the support reactions. As long as the center of gravity is located between the two pin joints (no tip over), the two reactions will push the structure upwards. This is the case of the left scenario as shown on the following screenshot.
In case the center of gravity location falls out of the lower base limits, one of the reactions will be directed downwards - which means a tip over in your case. This is illustrated on the right side of the following screenshot:</p>
<p><a href="https://i.stack.imgur.com/1Ss9g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Ss9g.png" alt="enter image description here"></a></p>
<p>As already discussed, Using a FEA tool just for analyzing the center of gravity location is quite a wasteful procedure. Generally, examining the reactions is used as a sanity checks for the finite element model.</p>
| 29295 | Structure analysis |
2019-07-21T19:26:01.390 | <p>I am designing a PID controller for a system that requires several step changes in the reference variable (concentration of solute in a tank) over a set period of time. I can find the plant transfer function experimentally, by applying a step change in the input (flow rate) and measuring the resulting changes in concentration over time. </p>
<p>What is not clear to me is: does the plant transfer function change each time I apply a different step change in the input variable (e.g. from 0.1 L/hr to 0.08 L/hr then after a while from 0.08 L/hr to 0.05 L/hr), or is the plant transfer function fixed for a given system (regardless of changes in the reference/input variable)? Thanks in advance.</p>
| |control-engineering|chemical-engineering|pid-control| | <p>Not enough detail, but in your case very possibly not.</p>
<p>A transfer function is a description of a <em>linear system</em>, using the Laplace transform (or the z transform if you're sticking to sampled time).</p>
<p>The basic problem with the above is that most real-world systems are nonlinear. We get around that by holding one eye closed and <em>pretending</em> that our system is linear (if we're measuring its response), or by finding its linearization around an operating point (if we're designing from a mathematical model).</p>
<p>I'm not up on the chemical engineering details, but I suspect that there are all sorts of nonlinearities in your system.</p>
<p>So, if the apparent transfer function that you extract from your measurements at one flow rate is considerably different than the one you extract at a significantly different flow rate, then yes, your linearized model of your nonlinear system's behavior at that different flow rate is, indeed, probably different.</p>
<p>If you've got the math chops for it, write out the differential equations that describe the reaction -- if they're nonlinear, then you can (A) expect that the transfer function is only an approximation, and (B) you can linearize them yourself around various operating points and see if the gains are different.</p>
| 29297 | Does the plant transfer function of a system change with variations in the reference variable? |
2019-07-22T00:50:52.853 | <p>Why Bode plots are on jw axis when determining gain and phase margin?
Usually it is explained as 1+G(s)H(s) should not be zero and margins represent how far away from this we are. But for showing this ("distance" to G(s)H(s) = -1, i.e. 0dB, -180 degree phase combination) we use bode plot which are GH values on jw axis. I would rather expect it to be on the whole s plane.</p>
| |control-theory| | <p>The idea behind Bode plot analysis is that you want to know how close the poles of the closed-loop transfer function are to the stability boundary. If a pole is <em>on</em> the stability boundary, then the closed-loop gain would calculate as infinite.</p>
<p>In the case of Laplace-domain analysis, the stability boundary is the <span class="math-container">$s = j \omega$</span> line -- any pole that has a positive real part is outside of the stability region, and the system is unstable.</p>
<p>When you do an open-loop Bode plot, you are concerned with how close the open-loop gain gets to -1, because if it were <em>exactly</em> -1 then there would be one or more poles sitting right on the stability boundary. The degree to which the open-loop gain "misses" -1 is a measure of the degree to which the resulting closed-loop poles are far away from the stability boundary.</p>
<p>Note that a drawback of the Bode plot is that you only know how far the closed-loop poles are from the stability boundary -- not whether they are actually stable. For that you need to apply some intuition, or you need to make a Nyquist plot, and know how many unstable zeros the system has.</p>
<p>Note, too, that for a sampled-time system expressed in the z domain you can do a Bode plot by tracing <span class="math-container">$z = e^{j \omega \theta}$</span> for <span class="math-container">$0 \le \theta \le \pi$</span>, because that's the stability boundary in the z domain.</p>
| 29299 | Gain and phase margin |
2019-07-22T10:16:43.477 | <p>I have a very noisy air conditioner that have a 10kHz peak at 75 dB, the rest (fan/compressor/pump) are under the regulation limit. </p>
<p>I would like to understand what is the measurement spectrum. I think 10kHz is out of the measurement spectrum, but I am not sure. </p>
<p>What is the procedure to measure household appliances noise level?</p>
| |acoustics|regulations|noise| | <p><strong>What is the measurement Spectrum?</strong> </p>
<p>From BS EN 60704-1:2010+A11:2012 Page 10, Section 4.1 Paragraph 3</p>
<blockquote>
<p>The preferred noise emission quantity is the A-weighted sound power
level, LWA, in decibels</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/8SEM8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8SEM8m.png" alt="Acoustic Weighting Curves"></a></p>
<p><strong>It is <em>not</em> correct to assume that a peak at 10kHz is outside the measurement range</strong> - you can reduce your 75dB measurement by -2.5dB according to the curve shown above.</p>
<p><strong>What is the procedure?</strong></p>
<p>Page 11 Paragraph 1</p>
<blockquote>
<p>According to this standard, two principal methods exist, the direct
method and the comparison method, as described in 4.2 and 4.3 below.
These two methods can be used alternatively.</p>
</blockquote>
<p>And then subsequently in Page 11 Section 4.2</p>
<blockquote>
<p>The direct method can be used only for measurements in qualified test environments
according to ISO 3744 for free field conditions over reflecting plane(s), and according to ISO
3743-2 for special reverberation test rooms. </p>
</blockquote>
<p>and Page 11 Section 4.3</p>
<blockquote>
<p>The comparison method for measurement is explicitly described in ISO
3743-1 and in ISO 3743-2.</p>
</blockquote>
| 29306 | Sound level emissions (IEC 60704-1) spectrum? |
2019-07-23T11:57:11.807 | <p>I recently saw a tweet about <a href="http://www.esa.int/spaceinimages/Images/2009/02/Hypervelocity_impact_sample" rel="nofollow noreferrer">Hypervelocity Impact Sample</a> and someone ask how is it possible to shoot a projectile at that speed and what happened to the used projectile when hitting the shield.</p>
<p>I found that it was probably shooted by a <a href="https://www.nasa.gov/centers/wstf/site_tour/remote_hypervelocity_test_laboratory/two_stage_light_gas_guns.html" rel="nofollow noreferrer">Two Stage Light Gas Guns</a> but I didn't found anything about the projectile.</p>
<p>I only found this:</p>
<blockquote>
<p>They got their name because hypervelocity impact produces local
pressures in the target and projectile that greatly exceed the
material strength of these materials, the material behaves as if it
has no strength, i.e., like a fluid, or hydrodynamic behavior</p>
<p><a href="https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090010053.pdf" rel="nofollow noreferrer">Handbook for Designing MMOD Protection </a></p>
</blockquote>
<p><a href="https://i.stack.imgur.com/y2Uzg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y2Uzg.png" alt="HYPERVELOCITY IMPACT SAMPLE"></a></p>
<h3>What happened to the projectile ?</h3>
<p>Does the projectile melt with the shield ?
<del>At this speed, does the projectile melt before hiting the shield ?</del></p>
<p>EDIT: deletion of the last question, test done in vacuum.</p>
| |materials|aerospace-engineering| | <p>When the collision velocity of a projectile is fast and the target hard and unyielding the impact produces forces that are greater than the compressive yield strength of the projectile and target. </p>
<p>So the projectile will collapse onto itself in plastic deformation while penetrating into a deepening crater in the target while converting its huge kinetic energy into heat.</p>
<p>This heat will liquify the projectile and crater into molten metal which can produce a cone of ejected red hot metal reflecting back and leaving traces of burn lines radiating out.</p>
<p>The impact force,F , must be such that: </p>
<p><span class="math-container">$$ \frac{F}{A_{projectile}} = m\frac{\delta v}{\delta t}/A= m\frac{v_{projectile}}{ \text{very small t}}/A> F_{\text{projectile yield strength}} $$</span></p>
<p>This is the first rough explanation. Weird things happen in a fast collision. Extensive tests have been done by artillery and shell manufacturers and designers of armory shields. Albeit the data is proprietary trade secret.</p>
| 29325 | What happened to the projectile in Hypervelocity Impact tests? |
2019-07-23T12:22:58.157 | <p>I am wondering the following. What are the typical cycletimes (or timesteps), that are used in real industry-standard pid-controllers (for example for assembly-robots on a car-production-line). I didnt find anything specific in the literature, just the theory. Therefore I am asking here. Often, these information is not published by the vendors of these pid-servo-controllers, so maybe someone has some experience here. Thanks.</p>
| |control-engineering| | <p>Without being at all facetious, it's the cycle time the engineer in charge of the control loop felt was best.</p>
<p>It depends on the size of the motor, the requirements of the loop, the sophistication of the computing hardware, and how the motor is controlled. I've written motor control loops with a cycle time of 100Hz, driving a current command to a transresistance amplifier. I've written motor control loops with a cycle time of 10kHz, that repeated at each cycle of the PWM signal to the H-bridges of a class D amplifier, and that controlled the motor current cycle by cycle. I've done quite a bit in between.</p>
<p>In general, you look at the fastest mode of the system that you're going to control (i.e., motor current, or motor speed, etc.), and you sample faster than that. The rule of thumb is 10 times, but depending on the precision you need, and the difficulty of attaining high sampling rates, you may sample lower than that or much, much higher (5 is a practical lower limit, there is no practical upper limit, although the required numerical precision of your controller goes up linearly, or possibly as the square of the sampling rate).</p>
| 29326 | Typical cycletime of an industry pid controller for servomotors |
2019-07-25T15:37:01.610 | <p>I want to buy a "Swiss File #4".</p>
<p>I visited some local stores but noone knew what that was.</p>
<p>I found this nice set of files. Can someone read what it says in the top?
Is some of these file equivalent to "Swiss File #4"?</p>
<p>If yes, i will buy this instead.</p>
<p><a href="https://i.stack.imgur.com/x5lN9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x5lN9.jpg" alt="enter image description here"></a></p>
| |tools| | <p>In the US, the wording you are looking for is "Needle File Cut 4", I believe.</p>
<p>Amazon search: <a href="https://www.amazon.com/s?k=needle+file+cut+4&ref=nb_sb_noss_2" rel="nofollow noreferrer">https://www.amazon.com/s?k=needle+file+cut+4&ref=nb_sb_noss_2</a></p>
<p>A Diamond file is not the same, having industrial diamond particles embedded into the surface, rather than teeth cut into the steel.</p>
<p><a href="https://en.wikipedia.org/wiki/File_(tool)#Diamond_files" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/File_(tool)#Diamond_files</a></p>
<p>It is possible that the files in your photo would work well for your application, but they are <strong>not</strong> equivalent to a Swiss #4.</p>
| 29363 | Swiss File #4 Alternatives |
2019-07-26T12:46:02.810 | <p>When I was in school, photosynthesis was taught as this amazing biological process that took light and water and turned carbon dioxide into oxygen and carbon. It was implied that this was limited to the domain of plants. </p>
<p>Have we industrialised photosynthesis so that we can shine light on a test tube and have it brake cardbon dioxide into oxygen?</p>
<p>My question is: <strong>Is there an industrial process that makes use of photosynthesis?</strong></p>
| |chemical-engineering|chemistry| | <p>Thanks to @am304 who has not yet posted a full answer - but answered in a comment. </p>
<p>In the American Chemical Society Journal <a href="https://www.acs.org/content/acs/en/pressroom/presspacs/2015/acs-presspac-april-29-2015/artificial-photosynthesis-could-help-make-fuels-plastics-and-medicine.html" rel="nofollow noreferrer">we read</a>:</p>
<blockquote>
<p>The groups developed a stand-alone, nanowire array that captures light and with the help of bacteria, converts carbon dioxide into acetate. The bacteria directly interact with light-absorbing materials, which the researchers say is the first example of “microbial photoelectrosynthesis.” Another kind of bacteria then transforms the acetate into chemical precursors that can be used to make a wide range of everyday products from antibiotics to paints.</p>
</blockquote>
| 29384 | Is there an industrial process that makes use of photosynthesis? |
2019-07-26T17:00:09.800 | <p>Please refer to the following youtube video (pause it as soon as it starts)...</p>
<p>"<strong>USS William D Porter, the Unluckiest Ship in the Navy</strong>" by <strong>The History Guy</strong>
<a href="https://youtu.be/f9Gb4PakFTU?t=162" rel="nofollow noreferrer">https://youtu.be/f9Gb4PakFTU?t=162</a></p>
<p>There is a running clock super-imposed over the film which gives a HH:MM:SS:FF (where FF is the number of frames elapsed this second, 0-29).</p>
<p>How did 1940's technology achieve this?</p>
<p>I suppose I must concede that this counter may have been added years or even decades later, but the clocks on display here (there are several clips within this video that feature different styles of clocks) seem to "blend in" with the old footage which leads me to believe that they were added (perhaps in post-processing) during the war period.</p>
<p>Given the astonishing ingenuity of wartime engineers I think this would be a relatively trivial task for them, but I would like to know how it was done.</p>
| |engineering-history| | <p>If you right click on the YouTube video and set the playback speed to the minimum, if is obvious that the video has been converted from 24 FPS cine film to 30 FPS video for TV broadcasting. Every fourth frame in the original is repeated.</p>
<p>The sequence numbers correspond to the video frames (0-29 in each second) not to the original cine frames, and therefore must have been added when the cine film was converted. There is no indication when or why that was done, but first commercially successful video recorder with helical scanning heads (the Ampex 1000) was developed in 1953-54. It is unlikely that any video conversions made before that date would be in a format that was still playable, and they would probably have been of lower quality than the YouTube video. </p>
<p>To summarize, the time stamps could have been added at any time in the last 60 years or so, but were not part of the original film footage.</p>
<p>At a superficial look, the video seems to simply repeat every 4th frame, rather than using the so-called <a href="https://en.wikipedia.org/wiki/Three-two_pull_down" rel="nofollow noreferrer">3:2 pulldown</a> interpolation that was used for telecine broadcasting. That fact might limit the time when the conversion might have been done, but of course "obsolete" equipment sometimes continues to be used long after there are "better" alternatives available.</p>
<p>If the conversion was done for historical archiving, the lack of interpolation may be a deliberate choice to preserve the original content of each film frame as accurately as possible, in case someone wants to examine it frame by frame for research purposes.</p>
| 29386 | WW2 technology - How was this on-screen frame counter added to old film-media footage? |
2019-07-27T01:09:27.750 | <p>My group made an RCA video/audio to RF modulator for a school project. The video is a NTSC composite video signal. Here is a clip of our tv receiving the output of the project on NTSC channel 2:</p>
<p><a href="https://drive.google.com/open?id=1-1xRHI8razjYvvptTsPsPBHt8T0z-yi5" rel="nofollow noreferrer">https://drive.google.com/open?id=1-1xRHI8razjYvvptTsPsPBHt8T0z-yi5</a></p>
<p>In the video, we use an usb-to-hdmi adapter from Raycue, a hdmi-to-composite video/audio adaptor from Ozvavzk, and a tv from Thomson Consumer Electronics (model 19GT241).</p>
<p>I made measurements of our audio and carrier signals using a Siglent SDS 1202X-E Oscilloscope and a $12 active probe (which is listed on eBay as "RF Active Probe 0.1-1500 MHz").</p>
<p>I measured the audio carrier signal to be 55.2 MHz and I measured the video carrier signal to be 59.2 MHz. I did a FFT of the two signals - the audio's bandwidth looks narrow like a dirac-delta function and the video's bandwidth is about 500 KHz wide.</p>
<p>According to the "Pan-American television frequencies" article on Wikipedia, if we are to transmit on channel 2, then our audio and video carrier signals should be tuned to 59.75 MHz and 55.25 MHz, respectively. I can't find information that supports the carrier frequencies of our project are alternative ones for channel 2. Why is our tv playing audio and video if the carrier frequencies in our RF project are not tuned according to the article?</p>
| |electrical-engineering|rf-electronics|radio| | <p>The old analog signal TVs had a wide frequency range allocated to each channel.
Channel 2 for instance occupied everything between 54 MHz and 60 MHz.</p>
<p>Because of effects of temperature and other conditions on the hardware (e.g. resistors and capacitors), the transmitter and the receiver could drift slightly from the exact frequency.
To compensate, the TV had a "fine tuning" control allowing one to adjust it to match the actual frequency. The set you were using either had AFT (Automatic Frequency Control), or you manually used the fine tuning yourself.</p>
<p>Wikipedia's "<a href="https://en.wikipedia.org/wiki/Pan-American_television_frequencies" rel="nofollow noreferrer">Pan-American television frequencies</a>" has this table, showing that your 55.2 and 59.2 were well within the channel 2 frequency range:</p>
<p><a href="https://i.stack.imgur.com/8MPMH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8MPMH.png" alt="VHF low-band"></a></p>
| 29393 | Why is the tv playing if the carrier frequencies in our RF project appear unsupported? |
2019-07-27T09:25:30.723 | <p>When applying the moment distribution method also known as the Hardy Cross method, how do I take into account the self-weight of columns? In the case of the beams, I can just apply the weight load laterally. However, for the column the load is axial, therefore causing 0 fixed end moment even when there is a difference in the support bending moment reaction. </p>
| |structural-engineering|structural-analysis| | <p>I'm not sure what system of columns and beams you're describing, but the answer appears to be in the question: the self-weight of the columns has no effect on the support bending moment reactions. You will use it in reactions at the column footings, but otherwise ignore it.</p>
| 29396 | Self Weight in Moment Distribution (Hardy Cross) |
2019-07-27T21:44:30.367 | <p>In automatic transmissions some parts need to be moving with the engine in one moment and in the other they need to be stopped. Why don't transmisisons then use detent clutches to accomplish this?</p>
| |mechanical-engineering|transmission| | <p>Detent clutches use mechanical interference to force the halves of the clutch into hard engagement. The wear surfaces in this case are metal-on-metal which works well when the engagement duty cycle is low, as for example in the overrunning-clutch mechanism used in motorcycle starters. </p>
<p>However, for purposes of shifting gear ratios in a car transmission, you do not want hard engagement because it jolts the car whenever a shift is made. Furthermore, the duty cycle of engagement and disengagement is high in a car transmission, which means the engagement surfaces will wear themselves out of tolerance before the desired service lifetime of the transmission as a whole has been met.</p>
| 29405 | Why aren't detent clutches used in car transmissions? |
2019-07-28T17:10:07.897 | <p>Something simple is tripping me up in a design, and the problem can be reduced to a simple 'seesaw' setup: a rigid beam with pin joint and a weight hanging from each end. I have attached an image with dimensions and values of interest:</p>
<p><a href="https://i.stack.imgur.com/4sGcH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4sGcH.png" alt="Beam and Fulcrum"></a>. </p>
<p>The pin joint is not centered in the beam so that one end is further from the joint than the other. The beam has been rotated to an angle, and the length of each side's projection onto the horizontal is known. My question is this: In analyzing this problem I end up showing that the torque induced by each weight is equal but the reaction induced at <span class="math-container">$F_2$</span> by <span class="math-container">$F_1$</span> does not equal <span class="math-container">$F_2$</span>, and vice versa. Have I made a mistake in my analysis, and if so where?</p>
<hr>
<p>Known: <span class="math-container">$A=30\ deg;\ R_1=10\ inch;\ R_2=5\ inch$</span></p>
<p>Given: <span class="math-container">$F_1=10\ lbs;\ F_2=20\ lbs$</span></p>
<p>Determine the vertical reaction force <span class="math-container">$F_{1,2}$</span> induced at <span class="math-container">$F_2$</span> by <span class="math-container">$F_1$</span>:</p>
<p>1) <span class="math-container">$F_{1,tangent}=F_1\cos(A)$</span></p>
<p>2) <span class="math-container">$L_1=R_1/\cos(A)$</span></p>
<p>3) <span class="math-container">$\tau_1=F_{1,tangent}L_1=F_1R_1=100\ in.lbs$</span></p>
<p>4) <span class="math-container">$L_2=R_2/\cos(A)$</span></p>
<p>5) <span class="math-container">$F_{1,2,tangent}=\tau_1/L_2=100\cos(A)/R_2$</span></p>
<p>6) <span class="math-container">$F_{1,2}=F_{1,2,tangent}\cos(A)=15\ lbs$</span></p>
<p>Using this result of <span class="math-container">$F_{1,2}=15$</span> and going backwards yields <span class="math-container">$\tau_2=75\ in.lbs\neq\tau_1$</span>. So it seems like I made an error, but I can't find it. </p>
<p>To look at it another way, wouldn't this result suggest <span class="math-container">$F_2$</span> cannot be greater than <span class="math-container">$15\ lbs$</span> without breaking equilibrium. Yet when <span class="math-container">$F_2=20$</span> the torque induced is <span class="math-container">$\tau_2=100=\tau_1$</span>.</p>
<p>And so I keep going in circles. I could really use some help! Thank you for your time.</p>
| |mechanical-engineering|statics| | <p>As @Kamran stated:</p>
<p>F1xL1=F2xL2
Therefore:
F2=[F1XL1] ÷L2</p>
<p>Radial calculations are a mechanical assumption when gravity is acting upon all equally and therefore discounted too. This means discounting angle A also from the equilibrium equation. </p>
| 29411 | Simple Fulcrum Analysis |
2019-07-29T15:11:06.377 | <p>I general, if you model a system (pendulum for example) and you do not include damping into the system (maybe caused by friction): Is it generally possible to tune a PI or PID controller for an second-order-dynamical system that is undamped? Because without friction/damping no energy can leave the system, and therefore I am wondering, if PID-control can generally work for these systems or if this will lead always to oscilations or unstable control?</p>
| |control-engineering| | <p>Yes, as long as the plant is controllable, observable, and doesn't vary too much from the model.</p>
<p>In terms of pole placement, a PID controller with a settable bandwidth on the derivative adds two poles, and has four settable parameters. The system has four poles (two from the plant, two from the PID controller), so in theory, not only can you stabilize the system, you can put those poles wherever you want.</p>
<p>In practice, you'd want to use a more robust design technique.</p>
| 29428 | PID-control of undamped system |
2019-07-31T12:20:32.483 | <p>I am trying to determine the temperature of an underground concrete vault buried 10m underground. The vault has the following dimensions: 2640mm x 5995mm x 2640mm with a wall thickness of 250mm. The transformer has the following dimensions: 2083mm by 1702mm by 1257mm. It has heat losses of 2080W to 11000W. How can I determine the temperature inside the vault? How would adding mechanical ventilation change the temperature inside this vault? </p>
| |mechanical-engineering|electrical-engineering|thermodynamics|heat-transfer|temperature| | <h1>Assumptions</h1>
<p>Assume steady state and neglect all radiation terms. Assume the surroundings to the concrete wall is an infinite heat sink at a fixed (constant) temperature). Finally, neglect the heat transfer that occurs at the corners of the concrete box (set the internal and external areas of the box to be equal as far as heat transfer is concerned).</p>
<h1>Formulations</h1>
<p>The energy balance from the transformer to the inside air of the concrete wall is</p>
<p><span class="math-container">$$ P = h_i A_T (T_T - T_{ai}) $$</span></p>
<p>In this, <span class="math-container">$P$</span> is the power supplied, <span class="math-container">$T_T, T_{ai}$</span> are the transformer and internal air temperatures, <span class="math-container">$h_i$</span> is the convection coefficient inside the box, and <span class="math-container">$A_T$</span> is the area of the transformer. This equation says that power generated by the transformer is convected away to the inside air of the container. The convection coefficient <span class="math-container">$h_i$</span> depends on how fast the air is moving (stagnate/free convection to forced convection). This gives you one equation with two unknowns (the temperatures).</p>
<p>The energy balance from the air in the box to the concrete wall is</p>
<p><span class="math-container">$$ P = h_i A_T (T_T - T_{ai}) = h_i A_w (T_{ai} - T_{wi}) $$</span></p>
<p>In this, <span class="math-container">$T_{wi}$</span> is the temperature of the internal wall and <span class="math-container">$A_w$</span> is the area of the wall. This equation says that power generated by the transformer is convected away to the inside air of the container that is then convected to the box wall. It is one equation with two unknowns (the temperatures).</p>
<p>The energy balance on the concrete wall without vents is</p>
<p><span class="math-container">$$ k \frac{(T_{wi} - T_g)}{w} = h_i (T_{ai} - T_{wi}) $$</span></p>
<p>In this, <span class="math-container">$k$</span> is the thermal conductivity of concrete, <span class="math-container">$T_g$</span> is the (constant) ground temperature, and <span class="math-container">$w$</span> is the wall thickness. This equation says that heat flow to the inside walls from the air is conducted away through the walls. It gives one more equation with the same two unknowns (the temperatures).</p>
<p>The energy balance on the concrete wall with vents is</p>
<p><span class="math-container">$$ k (1 - f_v) \frac{(T_{wi} - T_g)}{w} + h_v f_v (T_{wi} - T_g) = h_i (T_T - T_{wi}) $$</span></p>
<p>In this, <span class="math-container">$f_v$</span> is the fractional area of the wall that contains vents and <span class="math-container">$h_v$</span> is a pseudo coefficient for convection-type heat transfer through the vents. It also gives one more equation with the same two unknowns (the temperatures).</p>
<h1>Solution</h1>
<p>In all cases without or with vents, you can combine the system to obtain two equations and two unknowns. The system can be solved.</p>
<p>A starting point is to solve the flux equation without vents to find <span class="math-container">$T_{wi}$</span>. </p>
<p><span class="math-container">$$ P = k A_w \frac{(T_{wi} - T_g)}{w} $$</span></p>
<p>This says that the energy from the transformer goes through the walls by conduction.</p>
<p>Use the value of <span class="math-container">$T_{wi}$</span> and the equation <span class="math-container">$P = h_i A_w(T_{ai} - T_{wi})$</span> to find <span class="math-container">$T_{ai}$</span>. Continue with the heat transfer from transformer to air to find <span class="math-container">$T_T$</span>. Finally, use these first guesses in the equation with vents to re-calculate the temperatures.</p>
<p>A comparable problem is given <a href="https://engineering.stackexchange.com/questions/29117/simple-thermal-analysis-for-sealed-aluminum-box-with-3-7-watts-of-heat-generatio/29279#29279">at this link</a>.</p>
| 29439 | How do I determine the temperature inside a transformer vault? |
2019-07-31T20:24:04.600 | <p>I am having difficulty with two questions from a book I am studying about statics. I have seen the answers but it did not help me understand how the answers are derived. I want some advice on how to think about these types of problems and whether there are any tricks which help to solve them. </p>
<p>Question 1 (1st image);</p>
<p>A small probe is gently forced against a circular surface with a vertical force F (the red vector). Determine the n- and t-components of this force as functions of the horizontal position S.</p>
<p><a href="https://i.stack.imgur.com/gdjqU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gdjqU.png" alt="enter image description here"></a></p>
<p>Question 2 (2nd image);</p>
<p>Determine the x- and y-components of the tension T which is applied to point A of the bar OA. Neglect the effects of the small pulley at B. Assume that R and alpha are known.</p>
<p><a href="https://i.stack.imgur.com/p4k0Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p4k0Y.png" alt="enter image description here"></a></p>
| |statics| | <p>In fact, these are just geometry problems. </p>
<p><strong>The answer to the first question:</strong> </p>
<p><span class="math-container">$$\begin{align}
F_n &= F\dfrac{(r^2 - s^2)^(1/2)}{r} \\
F_t &= F\dfrac{s}{r}
\end{align}$$</span></p>
<p>Figure shows the details of the solution.</p>
<p><a href="https://i.stack.imgur.com/ATgJd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ATgJd.png" alt="Solution of the first question"></a></p>
<p>If you think that way, you can solve the second question.</p>
| 29445 | Need help in deriving components of forces when they depend on lengths and angles. (2 questions) |
2019-07-31T21:49:46.793 | <p>I've been studying this topic for a couple of days now. I started by asking some questions regarding output Torque and gear ratios relations.
Now I came with the whole picture and some calculations that I been working on. My goal is to find out if I'm in the right direction, and also point-out relations between equations that I'm not sure of.
To make this more clear and clean, I divide the system into three sections; A, T and B.
<a href="https://i.stack.imgur.com/ZHtiM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZHtiM.png" alt="enter image description here"></a></p>
<p>In section A its the electric motor and worm/spur torque transmission.</p>
<p>On another topic with a lot of help, I realise that the power provides by the motor is crucial to generate the output torque and angular velocity that your system requires. No matter the radius, ratios or N of teeth of the spur; if the power isn't enough, the gate won't move at the desired velocity.</p>
<p>So, the problem in question:
My goal is to find out if this system will be able to move a gate of 600 kg with a velocity of 0.33 m/s, and how much time it will take to reach that velocity.</p>
<p>The technical data of the motor is:</p>
<p>P = 1/2 hp ; RPM = 1450 ; 220V/50Hz</p>
<p><strong>SECTION A:</strong></p>
<p>P = 0.5 hp = 373 (kg * m2 /s3)
Pin = Tin * win</p>
<p>win = 1450 rpm = 151.84 (rad/seg)</p>
<p>Tin = 373 (kg * m2 /s3) / 151.84 (rad/seg) = 2.456 N*m</p>
<p>This, of course, is in an ideal situation, without considering losses. (how can I add an estimation of electric motor losses?)</p>
<p>The worm/spur system has a gear ratio of 23:1. </p>
<p>Considering the efficiency of the system of 80%, we have:</p>
<p>Tout = 23 * (0.8 * Pin) / win = 45.2 N*m </p>
<p>Now, Tout also can be calculated by Tout = 0.8 * Pin / wout</p>
<p>angular velocity applied on the spur is: wout = (0.8 * Pin) / Tout</p>
<p>wout = 0.8 * 373 (kg * m2 /s3) / 45.2 N*m = 6.6 rad/sec --> 63 rpm</p>
<p>To consider real circumstances, its the moment of inertia has something to do within this case? I mean, is this Tout enough to accelerate the spur to 63 rpm??</p>
<p><strong>SECTION T:</strong></p>
<p>In this section, I'm wondering what happens with force transmitted in the distance "d" to the pinion. I'm not sure if I need to consider losses in this section.</p>
<p><strong>SECTION B:</strong></p>
<p>Here I have a lot of question about what's happening.
The Pinion has 17 teeth and is module 4.</p>
<p>From gear design, I understand that module, and axial pitch determines the Lead and lead angle of the gear. This has a direct relation with the surface contact (involute) in the system and frictional forces.</p>
<p>Is the torque Tout the same as calculated for the worm/spur system? What about the radius of the pinion? If I increase the radius, the Torque will increase as well?
The radius of the pinion is = 0.025 (m)</p>
<p>Tout = F * r </p>
<p>F = 45.2 N*m / 0.025 m = 1808 N</p>
<p>Does this mean that the force applied to the rack is 1808 N? Is this force enough to accelerate until 0.33 m/s?</p>
<p>F = m * a = 600 kg * 9.8 m/s2 = 5880 N ; here I need to consider a friction factor according to the wheels and ground. (0.1)</p>
<p>F = 5880 N * 0.1 = 588 N; this means that I need to generate 588 N force to move the gate, but what about the acceleration? if the gate has 4 m, how much time will take to reach the 0.33 m/s</p>
<p>I need a last push to understand this system entirely.</p>
| |mechanical-engineering|gears|torque| | <p>O.K.
After your <a href="https://engineering.stackexchange.com/questions/29360/torque-and-rpm-calc">last question</a> we still have a couple of things to figure out. let's dig into it.</p>
<p>Similarly to my last answer, let's start our analysis at the load section (the gate). Its free body diagram may look as follows:</p>
<p><a href="https://i.stack.imgur.com/q8Khy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q8Khy.png" alt="enter image description here"></a> </p>
<p>The pinion, from the rack & pinion mechanism generates the pushing force. As you correctly noticed, there is a friction resisting force. Last time we decided to conservatively define the friction coefficient as 0.1. It means that the total friction force would be 600 [kg] times 9.8 [m/sec^2] * 0.1 = 588 [N]. The net force, i.e. the difference between the pinion pushing force to the friction force is what drives the gate and causes it to accelerate. This difference divided by the mass results in the gate acceleration.
For instance, let's say you would like your gate to reach its final velocity (0.33 m/sec) after 4 seconds. Assuming constant acceleration, it would end up in 0.33 [m/sec] over 4 [sec] = 0.0825 [m/sec^2].
The net force for acceleration the gate should be 0.0825 [m/sec^2] times 600 [kg] = 49.5 [N].
This means that the pinion force is 49.5 [N] + 588 [N] = 637.5 [N].</p>
<p>Got it? Don't confuse between M<em>g which is the weight of the gate to M</em>a which is the force needed to drive at a acceleration.</p>
<p>Regarding the pinion radius - we already discussed it last time. The rack & pinion mechanism does not multiply the input torque by a factor (like the worm gear), it just converts the torque into a force. This conversion factor is the pinion radius. </p>
<p>Regarding the inertia - It should be consider only when dealing with acceleration. Your constant speed calculation shall not take it into account.
I guess the pinion inertia would be negligible relatively to the reflected gate inertia. Try to compare between the two after you get all the basic ideas.</p>
| 29446 | Reduction System for Gate-opener motor |
2019-08-01T01:54:50.350 | <p>It appears to be a waste of electricity to keep a bank of refrigerators wide open with no doors or covers whatsoever on them. </p>
<p>They seem to be just blowing cold air out into the supermarket.</p>
<p>They are not turned off at night when the store closes. Even the "green" Whole Foods-type places do this. </p>
<p>But is it really that inefficient to operate them that way?
What is the relative efficiency between a refrigerator with a door and one without a door in a supermarket?</p>
<p><a href="https://img-s-msn-com.akamaized.net/tenant/amp/entityid/CCiQxr.img?h=416&w=624&m=6&q=60&u=t&o=f&l=f&x=750&y=399" rel="nofollow noreferrer"><img src="https://img-s-msn-com.akamaized.net/tenant/amp/entityid/CCiQxr.img?h=416&w=624&m=6&q=60&u=t&o=f&l=f&x=750&y=399" alt="Fridge without door"></a></p>
<p><a href="https://i.stack.imgur.com/a2zso.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a2zso.jpg" alt="Fridge with door"></a></p>
| |electrical-engineering|refrigeration| | <p>The doorless models (this was the photo in the question when I answered it) aren't as inefficient as they appear to be. <a href="https://i.stack.imgur.com/2wYGT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2wYGT.jpg" alt="Original photo."></a></p>
<p>Whenever someone opens a regular fridge door, its cold air pours out onto the floor and warm air replaces it inside the fridge. If the doors are opened frequently, or held open for significant time, there is a lot of waste.</p>
<p>On the other hand, the doorless models are designed to have a laminar flow of cold air from top to bottom. Most of the cold air emitted at the top is sucked back in at the bottom, with far less mixing with the surrounding warm air than one might think.</p>
<p>So for products where temperature control isn't critical, and where many people are likely to take the product or to spend a long time looking and deciding, the doorless models are a good choice.</p>
<p>(They're also good from a marketing perspective, as they provide an excellent view of the product and don't have doors that fog up, but that's not the question.)</p>
| 29448 | Relative inefficiency of doorless fridges at supermarkets |
2019-08-01T16:46:17.797 | <p>I'm dispensing around 1uL of volume with 1% accuracy. Although I'm confident in my pump, I'd like to be able to routinely confirm that I'm getting the exact volume I'm requesting every time. </p>
<p>How can I reliably measure volume at this small scale? I've tried a precision scale, but the evaporation rate for the fluid is quick and I'm losing non-negligible volumes while waiting for the scale to settle. </p>
| |chemical-engineering|sensors|pumps|flow-control|microfluidics| | <p>This is a challenging measurement. In my opinion best done with a very good balance. Fully shielded from thermals and vibration, obviously. Use a small container with small opening diameter to limit evap to what will hopefully be under 10% of the final measurement (and we will correct for it more later). A clean 2mL glass bottle is good. Fill bottle halfway, with 1mL of a low-surface-tension liquid, like 50% alcohol (does not need to be same the liquid being pumped). Ensure process fluid is degassed. Submerge outlet tube (but do not let it touch the bottle!) and immobilize the tube. Small tube outside diameter better to reduce surface tension force. Building a small fixture and attaching it to the enclosure of the balance is a good idea to protect the delicate tube from vibrations/movement/keep it from touching bottle, and if you're going to need a large number of measurements, also to facilitate the process of changing the bottle. Close the door to balance enclosure and wait a little bit for everything to come to thermal equilibrium. Program the balance to go into a mode that continuously outputs unfiltered data, so that you can see the evaporation rate and background-subtract it. Do not "TARE" the balance at any point, do that in post-processing. Some programming is required to do it right. Pump a little to let the fluid flow also come to thermal equilibrium. stop flow, to read the evaporation rate. then restart the flow or dispense process, and take your measurements. No touching the balance or table at any point. Take note of the ambient temperature.</p>
| 29457 | How can I confirm nL Volume Measurement for Pump and Sensor Calibration |
2019-08-03T15:09:14.393 | <p>A friend and I were in a discussion and can't seem to find the answer to the question above. I said it was possible and he said it isn't and now we are both confused so could anyone help here?</p>
<p>Also would it be possible to spin carrier and sun gear while holding ring gear stationary?</p>
| |mechanical-engineering| | <p>No, if you hold the sun gear the carrier will turn at slower speed than the ring gear.</p>
<p>You could have a setting where the ring gear is turning at the same speed as the input speed but then the carrier gear will turn slower or else it will break, because of the fact that is being spinned by the ring gear around the sun gear. </p>
<p>And as to the second part of you question, again if you hold the ring gear the sun gear and carrier will spin at different speeds.
hear is a youtube clip, <a href="https://www.bing.com/videos/search?q=planetry%20gear%20animation&docid=608052271535164301&mid=7CADA6E2BE3B0EB5CBDB7CADA6E2BE3B0EB5CBDB&view=detail&FORM=VIREHT" rel="nofollow noreferrer">planetary gear</a></p>
| 29478 | In a planetary gearset is it possible to move both carrier and ring gear at the input speed while holding sun gear stationary? |
2019-08-04T11:05:00.500 | <p>This is from a design entrance that I am preparing for. I know it's a little different than a theoretical projection drawing that engineering books usually cover. It is more practical and visualization based question. I have tried really hard to find some resources to understand the approach to solve this kind of questions. I didn't find any. So, if someone decides to put it on hold. I would like to request them to attach some resource in the comment.</p>
<p><a href="https://i.stack.imgur.com/SndKZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SndKZ.png" alt="enter image description here"></a></p>
<blockquote>
<p>What would be the number of surfaces of the common part?</p>
</blockquote>
<p>My initial assumption was I need to first visualize the shape of the common part then count the number of surfaces. But, as people have told that is not necessary. Still, I would also like to know</p>
<blockquote>
<p>How to visualize and draw the common part?</p>
</blockquote>
<p>I have tried to solve this myself but the best I can come up with this. I would also like to mention that I am not an engineering student. But for my exam asks some questions from engineering drawing. I have a book for it written by someone from my country but that does not cover any union kind of question. I would be very much obliged if you show me the proper method to do it.</p>
<p><a href="https://i.stack.imgur.com/KVg8v.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVg8v.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|drafting|drawings|machine-design| | <p>I don't want to spoil the fun of solving this for yourself so I offer the following suggestions:</p>
<ol>
<li>Redraw the assembly with all edges, including the hidden edges, shown as dotted lines.</li>
<li>Draw solid lines on all the intersections.</li>
<li>Join up any vertices which form edges not on the outer surfaces of the assembly.</li>
<li>Count the faces.</li>
</ol>
<p>Step 3 is the only tricky bit. In your example the uppermost edge of the horizontal element will form one such edge.</p>
<p>Post a sketch in an update to your question and we'll see if you've understood it.</p>
<hr>
<p><a href="https://i.stack.imgur.com/0SNS7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0SNS7.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Intersection of two prisms. The green lines are edges that already exist on the isometric view. The orange lines will be new edges created by the intersection operation.</em></p>
| 29487 | visualization of intersection of two objects |
2019-08-04T16:37:27.427 | <p>So the reason why torque converter is needed is because it disconnects engine from transmission. It has also extra benefit of multiplying torque but that is not 100% necessary. From what I have seen all automatics have some sort of constant input which is always connected to the shaft. Would it be possible to design an automatic transmission which wouldn't have constant input and therefore wouldn't need torque converter?</p>
| |mechanical-engineering|transmission| | <p>The original automatics did not have torque converters ; they mostly had fluid couplings . Chrysler in particular , had more than a half dozen different ones in the various cars. They were more like "gear changers" with an automatic clutch ; some still included a clutch pedal. Performance was very poor by todays standards. And there were overdrives that would make one shift . The GM Hydromatic was the first good automatic ( 1939 Olds and 1940 Cadillac as I remember) . It had a fluid coupling that had limited torque converter function - I forget the percentage. It could shift hard enough to burn rubber in the first 2 shifts ( with the V-8 engines). </p>
| 29490 | Is it possible to design automatic transmission without torque converter? |
2019-08-04T17:06:01.293 | <p>I'm learning a bit of stick welding since I keep needing it for certain projects. Total beginner.</p>
<p>The latest thing I decided to make is a table similar to this one: <a href="https://i.stack.imgur.com/5FZS8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5FZS8.jpg" alt="enter image description here"></a></p>
<ul>
<li><p>The material is <strong>2mm</strong> thick <strong>40x40mm</strong> square tubing - don't know the exact metal type, should be "regular" steel that was a bit corroded from staying in the steel yard.</p></li>
<li><p>I tried using <strong>2mm</strong> and <strong>1.6mm rutile electrodes</strong>. </p></li>
<li><p>I tried the <strong>1.6mm at 55~ Amps AC</strong> and the <strong>2mm at 60-65 Amps AC</strong>. With higher or lower Amps, my electrodes started sticking.</p></li>
<li>The angle I'm holding the electrode at is similar to this one: <a href="https://i.ytimg.com/vi/qTaQf0iPH-g/maxresdefault.jpg" rel="nofollow noreferrer">https://i.ytimg.com/vi/qTaQf0iPH-g/maxresdefault.jpg</a></li>
</ul>
<p>What is happening that on straight flat sections I can spark and keep a straight line and don't burn a hole in metal. The weld is strong and consistent. </p>
<p>When trying to weld the two pieces together, on the outside and inside angles (see red circle) I keep burning through and bigger and bigger gaps keep opening because of this.</p>
<p>On the inner corners I have not had this issue and I have successfully welded thicker metals with 2mm and 2.5mm electrodes without issues.</p>
<p><strong>EDIT</strong>: Edited for clarity that I'm not actually welding without melting metal :)</p>
| |metals|welding| | <p>Welding always melts steel, you mean you are "burning through", too much penetration. Wire feed welding -MIG is much easier to use than stick . MIG , also identified as flux core with or without gas and solid wire with gas - is the best choice for your job. Stick is difficult to learn by trial and error and more difficult with thin base metal as you have. Stick is not so bad with base metal of 1/2 in. and turn the amperage up high enough to prevent "sticking". Apparently you have the best electrode , 6013, that is the high titania/rutile type. Reverse polarity will give the lowest penetration ; more of the heat goes to melting the electrode and less heat into the work. </p>
| 29491 | Arc-welding 2mm thick steel at an angle without burning-through |
2019-08-05T04:25:32.550 | <p>My intuition on how a twist drill bit operates is that the cutting is done by the end of the drill bit and that the fluted section carries away material as it's cut.</p>
<p>When I look at twist drill bits, though, it appears that the edge of one of the flutes is sharpened. What purpose does this serve? Does this edge actually cut material?</p>
| |mechanical-engineering|machining|tools| | <p>The edge is sharp to make sure the side of the hole is left smooth minimizing the cutting marks and to make sure that the resulting diameter is correct.</p>
<p>However, if a precise diameter is required then a reamer is usually used to finish the hole.</p>
| 29498 | Why is the fluted edge of a twist drill bit sharp? |
2019-08-05T05:57:12.993 | <p>I noticed that the screw-on of my water tap often loosens itself up after I rotate the tap left and right enough times. Every time I rotate the tap left, it loosens the screw a bit, then if I rotate the tap right, the screw does not tighten itself to the same amount. </p>
<p>This, repeated enough times, loosens the screw enough so that the tap would leak.</p>
<p>I thought it might be solved if the screw would have a wavy thread. So, for example, instead of gently sloping downwards by 1 mm every revolution, it would slope up 1 mm for the first half, and down 2 mm for the second half. This way, the screw would lock itself in a local "peak" in the thread, and not loosen itself no matter if I rotate the tap left or right.</p>
<p>Is there such a kind of screw? If so, what are they used for? If not, is it because they have some kind of mechanical weakness?</p>
| |mechanical-engineering|civil-engineering| | <p>This idea doesn't work, because screwing the parts together will tend to strip the threads before the parts are fully squeezed together.</p>
<p>Instead, engineers use different general methods to get the threads to lock and not unscrew themselves in the manner you experienced, as follows:</p>
<p>1) using screws or nuts with plastic inserts in them which interfere with the threads in the mating part. These are called <em>nylock</em> fasteners or <em>lock nuts</em>. </p>
<p>2) using threads whose <em>diameter</em> increases progressively along the length of the threaded portion, so as to generate progressively more interference as the parts are screwed together. these are called <em>tapered</em> or <em>pipe threads</em>. </p>
<p>3) inserting a <em>lock washer</em> between the parts with teeth that bite into the nut or screw head that discourage the parts from unthreading. </p>
<p>4) assembling the part with a glue applied to the threads that sets up and prevents the parts from being unscrewed. This glue is called <em>thread locker</em>. </p>
| 29500 | Why is the thread of the screw linear? Can it be wavy? |
2019-08-05T07:45:57.040 | <p>I am designing a hydrophone, from the piezo buzzer. To hold the circular piezzo plate, I have designed a mechanical fixture. I need to use a washer/o-ring to make it leak-proof. I am confused about which one to use.</p>
<p>Can you please explain the difference between washer and o-ring. What is the difference between them?</p>
| |mechanical-engineering|design|sensors|vibration|seals| | <p>In simple words, Washers are flat o-rings</p>
| 29503 | What is the difference between washer and O-ring, what are their advantages of one over the other? |
2019-08-05T14:12:39.347 | <p>I read about how an LCD screen works <a href="https://www.explainthatstuff.com/lcdtv.html" rel="nofollow noreferrer">here</a>, and it made me wonder what actually happens when the screen shuts off. Based off of this, I can think of several possibilities:</p>
<ol>
<li><p>Electricity is supplied to all the pixels/crystals, thus making them all appear dark. While this fits with the on/off explanations given in that article, it seems like a significant waste of energy to use extra energy when <em>not</em> in use.</p></li>
<li><p>Electricity is <em>not</em> supplied to all the pixels, which would light them all up, but the backlight shuts off, so the screen appears dark.</p></li>
<li><p>Electricity is maintained to the pixels that were receiving electricity when the screen was on, but the backlight shuts off so the screen appears dark.</p></li>
</ol>
<p>I would be curious to know if different phases of the off screen are different (such as a off-screen-like screen saver versus a sleep mode or being shut off, etc.).</p>
| |computer-hardware| | <p>There's two different "shut off" methods: showing a black screen and actually powering off. To show a black screen, the backlight stays on but none of the pixels are energized. Light that gets polarized from the first filter doesn't change orientation (because the liquid crystals aren't energized) and so that light gets blocked by the second filter. </p>
<p>Turning off also de-energizes the pixels, but it also turns the backlight off. This is obviously more energy efficient, but also makes for darker black colors because the polarizing filters only stop <em>most</em> of the light. </p>
<p>Most displays will have a single light source for the backlight, but some displays, like edge-lit displays, can control <em>regions</em> of light and other displays use individual LEDs with each pixel. Both methods can vary the intensity of the backlight and this is sometimes called "high dynamic range" displays. You can read more about that <a href="https://www.cnet.com/news/led-local-dimming-explained/" rel="nofollow noreferrer">here</a>. </p>
| 29510 | What happens when an LCD screen shuts off? |
2019-08-05T14:59:31.017 | <p>What FEA software is capable of simulating static studies on orthotropic materials such as wood? I must be able to define mechanical properties in X,Y and Z. I would also like to have access to failure criteria such as tsai-hill or tsai-wu and probably even other custom criteria. My company uses solid edge simulation at the moment but i have not found a way to account for orthotropic materials, I also tried solidworks simulation on my side and I was able to perform most of what i need but i want to consider any additional options before i make a recommendation to my company to shift to a new software.
Any hints or advice would be greatly appreciated.
Cheers!</p>
| |finite-element-method|solidworks|wood| | <p>Any of the major FEA systems (Nastran, Abaqus, Ansys, Adina, Comsol, etc) can do this. If you have a fairly limited requirements (e.g. only linear statics), look for the vendor with the cheapest entry-level or student-level version of the code which includes orthotropic material models.</p>
| 29511 | FEA software for Orthotropic materials |
2019-08-05T16:02:57.770 | <p>I'm revising for the UK LGV theory test. One of the questions goes something like this:</p>
<blockquote>
<p>Which axle configuration is most effective at preventing petrol tankers from rolling over?</p>
<ul>
<li>A. Tandem axles with double wheels</li>
<li>B. Tandem axles with air suspension</li>
<li>C. Tri-axles with single wheels</li>
<li>D. Tri-axles with double wheels</li>
</ul>
</blockquote>
<p>The correct answer is apparently C, with the following explanation:</p>
<blockquote>
<p>The type of suspension fitted to a vehicle will influence its resistance to ‘roll-over’. Modern tri-axle semi-trailers fitted with single wheels on each side extend the tracking width available, making this the most stable configuration.</p>
</blockquote>
<p>I have the following questions:</p>
<ul>
<li>Why are single wheels better than double wheels?</li>
<li>Is this specific to a tri-axle setup, or does it also apply to twin-axles or single-axles?</li>
<li>Is this specific to trailers, or does this apply to rigid vehicles as well?</li>
<li>If single wheels are better, why do so many vehicles have double-wheels?</li>
</ul>
| |mechanical-engineering|automotive-engineering| | <p>Track width is one of the most important factors in stability and resistance against rollover in cars and trucks. </p>
<p>That's why we see in race cars they even extend the wheels outside of the car body, or all street legal fast sport cars have flared out fenders to accommodated a wide track width stance.</p>
<p>In a double wheel axle the width of the wheelbase is from the center of the left double wheel to the center of the right double wheel which is approximately 20% shorter than a single wheel width, even though it has the benefit of distributing the wight over larger area of the pavement, hence applying less pressure on the road.As has been mentioned in @Chuck's answer. </p>
<p>The other reason is when there is two wheel at each end when the truck is urged to roll as by a turn or a bump, the outside wheel's tire starts to compress into a more flattened shape but the the tire next to it winch is under much less torque because the fact that it is not far enough will bulge out a bit and exacerbate the situation on the outermost tire.</p>
<p>If you watch some videos of monster truck races which some have double-wheels you'll see this effect. </p>
| 29514 | Why are tri-axle trailers with single wheels more stable than double wheels? |
2019-08-05T21:10:25.673 | <p>What are some materials that allow air to pass but not, or to a lesser extent, water vapor? </p>
<p>Here are two materials I am aware of:</p>
<ul>
<li><strong>Teflon</strong> is a material that allows water vapor to pass, but not air. Water itself can't pass.</li>
</ul>
<p>What are some other sourceable substances which have this feature? What are some defining terms and measurements used to identify this property in a material, and how can the rate of water vapor / air transfer be manipulated in such a substance?</p>
| |materials| | <p>I just check on Google and I find the following results</p>
<ul>
<li><a href="https://www.nitto.com/eu/es/about_us/brand/promotion/innovation/temish.html" rel="nofollow noreferrer">TEMISH (TM) of Nitto </a></li>
<li>Silicone (yes, you read that right) (fixed for spelling mistake) - Apparently, the American company GE tested it out in 1950s
<a href="https://i.stack.imgur.com/chybt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/chybt.jpg" alt="enter image description here"></a></li>
<li><a href="https://www.dupont.com/brands/tyvek.html" rel="nofollow noreferrer">Tyvek (R) of DuPont</a></li>
<li><a href="https://www.gore-tex.com/" rel="nofollow noreferrer">Gore-Tex (R)</a></li>
</ul>
<p>The stuff above can happen due to something called "Molecular Sieve", find quick reading <a href="https://en.wikipedia.org/wiki/Molecular_sieve" rel="nofollow noreferrer">here</a>. Basically, the material concerned is porous ("with holes"), but these holes are very, very, very small, at the atomic level. The holes can block bigger molecules/atoms (like water - H2O), but allow air (say oxygen - O2) to pass through. Any material with sufficient small holes can "block" the water, and "unblock" air.</p>
<p>Of course, said is easier than done, because the water molecules have the size of nearly 3 angstroms (2.75 if you want more precise value), so... yeah. Building material like that is <em>possible</em>, but <em>very hard</em>.</p>
| 29523 | What is a material that allows air to pass but not water vapor? |
2019-08-06T16:21:23.787 | <p>I need to move a machine, approx weight of 900Kg
I will put 3 tubes under the machine and roll the machine on the tubes
Holidays season, all industrial suppliers are closed around me and only have access to 2 kinds of tubes:
20mm diameter and wall thickness 1.5mm
or
25mm diameter and wall thickness 1.5mm</p>
<p>Is the thickness enough to support the weight?
Material is construction steel (we call it S235 here but I don't know the international name)
i would be interested in the formula to make the calcultion also</p>
<p>Thank you</p>
| |steel| | <p>It depends on the dimensions and geometry of your load and the evenness of the track. At first look it looks quite inadequate.</p>
<p>But as a basic rule the smaller the diameter the stronger is the tube. </p>
<p>If you're really out of alternative options, you may want to fill the smaller size tube with rich cement grout and use a 3/4 inch thick plywood between your load and rollers, such that you distribute the load to a wider contact surface.</p>
<p>Make sure that when there is 2 rollers left under the load they share the load equally. </p>
| 29530 | Steel tube wall thickness for moving a machine |
2019-08-06T20:25:54.707 | <p>If I understand it correctly in a CO<sub>2</sub> laser engraver/cutter the laser itself is fixed in position and mirrors are used to redirect the laser to the material.</p>
<p>Is there some mechanism used to keep the focal point on the material or is the focal point set to the midpoint of the machine's range? Wouldn't the beam be wider at the edges of the machines travel?</p>
| |lasers|cutting| | <blockquote>
<p>... the laser itself is fixed in position and mirrors are used to redirect the laser to the material.</p>
</blockquote>
<p>Yes. This system is called "flying optics". The delicate laser is fixed in position and only the mirrors and lens move about. This simplifies the mechanical setup and reduces the inertia and required motor sizes.</p>
<blockquote>
<p>Wouldn't the beam be wider at the edges of the machine's travel?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/hUUjI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hUUjI.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Divergence specification for Syntrad's <a href="https://www.synrad.com/sites/default/files/2019-05/ti60%20Data%20Sheet%20US%204.10.19%20LoRes.pdf" rel="nofollow noreferrer">ti60</a>.</em></p>
<p>The beam divergence can be quite small. The Synrad unit above has < 7 mrad which would give a divergence of up to 7 mm/m of beam length. The optical path (mirrors, etc.) at maximum range would have to have large enough mirrors and the final collimation lens diameter would have to cope with the enlarged beam. The other course of action would be to correct for the divergence at source. </p>
<p><a href="https://i.stack.imgur.com/axh1n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/axh1n.jpg" alt="enter image description here"></a></p>
<p><em>Figure 2. See <a href="https://www.scienceabc.com/innovation/what-is-a-laser-collimator.html" rel="nofollow noreferrer">Science ABC</a> for example.</em></p>
| 29535 | Laser engravers and focal points |
2019-08-06T22:58:54.010 | <p>Hello dear engineering community.</p>
<p>I have a specific question about multiphase flows. I was wondering if anybody could help me with it, please.</p>
<p>Assume, I have a one dimensional water-air flow. Air is the dispersed phase. Let's say, it's a vertical flow. Water phase flow is incompressible.</p>
<p>QUESTION: if volume fraction (either of liquid or gas phase) is time dependent, then does it mean that that the flow is necessarily transient (either liquid or gas, or the entire mixture flow)? Or it can be steady?</p>
<p>Thank you in advance.</p>
| |fluid-mechanics|multiphase-flow| | <p>I think I've a partial answer to my question.</p>
<p><em>1) Proof that liquid phase flow is steady if liquid volume fraction is time independent.</em></p>
<p>Continuity equation for incompressible liquid phase: <span class="math-container">$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$</span> Let's say that volume fraction is not time dependent and only spatially dependent; and assume that velocity is still spatially and time dependent. Then we'll have:<span class="math-container">$$\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$</span>
<span class="math-container">$$\frac{d \alpha_L}{dx}·v_L+ \frac{\partial v_L}{\partial x}· \alpha_L=0$$</span>
<span class="math-container">$$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{\partial v_L}{\partial x}·\frac{1}{v_L}$$</span>
Now we have that the entire left hand side is only spatially dependent, whereas the entire right hand side is assumed to be both spatially and time dependent. But it can't be. If we know for sure that the left hand side is not time dependent and is equal to the right hand side, then the right hand side must also not be time dependent: <span class="math-container">$$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{d v_L}{d x}·\frac{1}{v_L}$$</span>
<strong>Therefore, it follows that if liquid volume fraction is not time dependent then the liquid flow is steady.</strong></p>
<p>The same discussion should work for the gaseous phase (but I am not sure about that).</p>
<p><em>2) Proof that liquid phase flow is transient if liquid volume fraction is time independent.</em></p>
<p>Continuity equation for incompressible liquid phase: <span class="math-container">$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$</span>
Let's say that volume fraction is both time and space dependent. Then assume that velocity can be only spatially dependent. Prove that it can't be. Let's do further math manipulations with liquid continuity:
<span class="math-container">$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L+\frac{d v_L}{d x}·\alpha_L=0$$</span>
<span class="math-container">$$\frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)=-\frac{d v_L}{d x}$$</span>
Now, even if both of <span class="math-container">$\alpha_L$</span> partial derivatives are constants and with the assumption that <span class="math-container">$v_L=f(x)$</span>, the left hand side is still time dependent because <span class="math-container">$\alpha_L=f(x,t)$</span>.</p>
<p>Let's write an integral formulation: <span class="math-container">$$v_L=-\int \frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)dx$$</span>
From the integral formulation it is explicitly seen that since the integrand is time dependent, the result will also be time dependent. It means that if <span class="math-container">$\alpha_L=f(x,t)$</span>, <span class="math-container">$v_L$</span> cannot be a function of space only.</p>
<p><strong>Therefore, it follows that if liquid volume fraction is time dependent then the liquid flow is transient.</strong></p>
<p>The same discussion should work for the gaseous phase (but I am not sure about that).</p>
| 29539 | How to infer from the volume fraction function whether a flow is transient or steady? |
2019-08-07T00:53:51.917 | <p>I am trying to verify windfarmer generation estimates manually. To do so I am trying to manually calculate the estimated energy production (kWh) of a wind turbine. The information I have available is the </p>
<ol>
<li>Hourly weather data (wind speed, direction & air density at Hub Height,
and temperature) </li>
<li>Power Curve of wind Turbine</li>
<li>Thrust Coefficient of wind turbine</li>
<li>Temperature de-ration curve of wind turbine</li>
<li>Turbine specifications (blade radius, swept area, cut in & cut out speeds, Cp, Rated Power, etc)</li>
</ol>
<p>How would I use this data to estimate the potential annual generation?</p>
<p>I tried multiplying hourly wind speeds * the specific value in the power curve, then de-rating the result based on the given temperature, However it seems to significantly overestimate the answer. </p>
<p>Unfortunately I cannot share the data, but all I am looking for is a formula or calculation methodology. Any help be greatly appreciated!</p>
| |electrical-engineering|energy|renewable-energy|wind-power| | <p>For an approximate estimate, use</p>
<p><span class="math-container">$ P = \frac{1}{2}\rho C_p A u^3 $</span></p>
<p>where <span class="math-container">$\rho$</span> is the air density, <span class="math-container">$C_p$</span> is the power coefficient (read off the power curve that you have, where it will depend on the wind speed), <span class="math-container">$A$</span> is the swept area of the rotor, and <span class="math-container">$u$</span> is the wind speed.</p>
<p>If you have hourly data, calculate this for each hour. That will give you a power in Watts (or kW, or MW), so you can then assume that that is the average output over the hour centred on the time of the weather measurements, and hence that the energy generated in that hour is the same number of Wh (or kWh, or MWh).</p>
<p>You could try to manually include temperature de-rating (I don't know exactly how this works, but if you have a curve you could apply it), but if your aim is to check that the Windfarmer result is sensible, this ought to get you pretty close.</p>
| 29542 | How to estimate wind generation using power curve? |
2019-08-08T05:10:38.863 | <p>I purchased a pair of noise canceling headphones today and I was just wondering how exactly they are able to cancel out background noise? </p>
| |mechanical-engineering|electrical-engineering|manufacturing-engineering|process-engineering|computer-engineering| | <p>Briefly they take the waveform of the ambient noise then produce the inverse waveform which is then generated. The two waveforms effectively cancel out leaving a much quieter ambient sound level.</p>
<p>The important part is to be able to separate the wanted "noise" (music or voices) from the unwanted noise.</p>
<p>A search will give you several results that will provide much more detail.</p>
| 29564 | How exactly do noise canceling headphones work? |
2019-08-08T14:21:26.017 | <p>I am a relatively recent BSME graduate working on reproducing the results of a technical paper to strengthen my ANSYS CFX skills. The paper is "Effect of Section Thickness and Trailing Edge Radius on the Performance of NACA-65 Series Compressor Blades in Cascades at Low Speeds" by Herrig, Emery, and Erwin. My goal is to reproduce pressure coefficient vs axial chord plots for one of the test cases using ANSYS CFX. </p>
<p>Problem Statement/Givens from Paper: The test was performed in a low-speed porous wall wind tunnel with an air average inlet speed of 95 ft/s. The particular airfoil geometry I am examining is an NACA 65-(12)12 (5 inch chord) at an angle of attack of 23.5 degrees with a solidity of 1.5. I made some assumptions not stated in the paper for my CFX model. I applied a free slip wall in place of the porous walls to capture the effect of boundary layer shedding. I also assumed the outlet static pressure was 1 atm and the inlet temperature is 15C. All other solid surfaces are applied as adiabatic walls with a no-slip condition. I am using the default k-epsilon turbulence model, isothermal heat transfer, and neglecting buoyancy effects. The pressure coefficient is to be determined along a mid-span plane on the surface of an airfoil far from the end walls (the third airfoil of the 5 I put in my cascade model). <a href="https://i.stack.imgur.com/Cr7AW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cr7AW.png" alt="Geometry produced in ANSYS"></a></p>
<p>The Problem: The provided results from the paper have a minimum pressure coefficient of 0. My results, however, have a minimum ~1.05. I noticed all the pressure coefficient figures in the paper have a minimum at 0 and no negative values. </p>
<p>My Attempt at a Solution: I created a polyline at the intersection of the center blade surface and the mid-span plane. I wrote expressions in CFX Post to determine the area-averaged density, inlet velocity (although this is redundant), and static pressure. Then, I created a new variable to calculate the pressure coefficient. The calculation being performed is (P_static_local-P_static_inlet)/(1/2*rho_static_inlet*velocity_inlet^2). I've attached the results from the paper and my produced results. My trend seems to be mostly correct (save the waviness from producing the airfoil profile myself), however, the problem of the y-axis scale still remains. <a href="https://i.stack.imgur.com/CXaCS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CXaCS.png" alt="Results from Paper"></a> <a href="https://i.stack.imgur.com/MryAz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MryAz.png" alt="Results from my Simulation"></a>. To complicate things further, the definition of the pressure coefficient they provide in the paper uses a local total pressure instead of a local static pressure, which changes the shape of the plot completely. <a href="https://i.stack.imgur.com/TFJ4Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TFJ4Z.png" alt="Variables from Paper"></a> <a href="https://i.stack.imgur.com/Bm5rk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bm5rk.png" alt="My results using their definition of pressure coefficient"></a></p>
<p>Does anyone know of a standard used to adjust the minimum pressure coefficient to 0? This is the only sort of logical explanation I can conjure at the moment. Of course, if anyone sees any glaring errors in my methodology please comment and let me know. </p>
<p>Thanks for your time</p>
| |fluid-mechanics|aerospace-engineering|aerodynamics|cfd| | <p>In the report <span class="math-container">$P$</span> is the total pressure and <span class="math-container">$p$</span> is the static pressure. See the page in the report before the one you attached in the OP.</p>
<p>With that definition, <span class="math-container">$S$</span> will be zero at the stagnation point near the leading edge of the airfoil, and <span class="math-container">$S$</span> will always be positive, as in their plots.</p>
<p>The "calculation being performed" that you give for the pressure coefficient in the OP using two static pressures is not the same as the paper's definition.</p>
| 29569 | Pressure Coefficient with a Minima of 0 on Compressor Airfoils in Cascade |
2019-08-11T05:47:59.973 | <p>I'm thinking of big halls, especially those that use chandeliers for light. The idea's simple: build a parabolic ceiling that uses mirrors, place the chandelier at the focus, and illuminate the entire hall with light. Since the light would be coming from the ceiling, it wouldn't be possible to "stand in someone's light" either, which would be great for e.g. exams. One would also be able to use just one light source for the entire hall. Why hasn't this been done?</p>
<p>I feel like I'm missing some obvious reason but I can't place it. One guess could be that big parabolic mirrors are expensive, but we have such mirrors elsewhere (e.g. for telescopes, science museums). It could be that such a hall would look really bright if one looks upwards (at the ceiling), but that could be mitigated by, e.g., placing a one-way mirror between the ceiling and ground level that allowed light down but not up. What am I missing?</p>
| |structural-engineering| | <p>In the case of a mirrored paraboloid with a lamp at the focus, all the light rays would shine directly downward. Any object would have a sharply defined shadow directly below it; people would look sinister with dark eye sockets and nostrils. This is known as <em>hard</em> light.</p>
<p>Contrast that with a light-colored ceiling (of any shape), which reflects light in many directions. This creates <em>soft</em> light, and is more pleasant to be in.</p>
| 29596 | Why don't we have parabolic ceilings with mirrors & a light at the focus? |
2019-08-11T10:23:17.000 | <p><a href="https://i.stack.imgur.com/bWPP8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bWPP8.png" alt="enter image description here"></a></p>
<p>So i need to calculate the angular velocity of the dark disk, <span class="math-container">$w^*_3$</span>.
In the master solution they state it is : <span class="math-container">$w^*_3 = w_3 + \overline{w_3}$</span>
with :</p>
<p><span class="math-container">$w_3 = \frac{|\dot{r}|}{4a}$</span> </p>
<p><span class="math-container">$|\dot{r}| = v_a + \overline{w_3}\cdot a$</span></p>
<p><span class="math-container">$|\dot{r}| = v_b - \overline{w_3}\cdot 2a$</span></p>
<p><span class="math-container">$ \Rightarrow|\dot{r}| = 2a\cdot (w_S+w_R)$</span></p>
<p><span class="math-container">$\overline{w_3} = \frac{1}{a}(|r|-v_1)=\frac{1}{2a}(v_b-|r|)=\frac{1}{3a}(v_b-v_a)$</span></p>
<p><span class="math-container">$ \Rightarrow \overline{w_3} = 2w_r-w_s$</span></p>
<p><span class="math-container">$ \Rightarrow w_3^* = \frac{5}{2}w_r-\frac{1}{2}w_s$</span></p>
<p>So, the question I have is that if I understood correctly, <span class="math-container">$w^*_3$</span> is the angular velocity of the disk itself (in its coordinate system) if you look at it from the disks center. So shouldn't be <span class="math-container">$\overline{w_3}$</span> already be my <span class="math-container">$w^*_3$</span>? Because it is the angular velocity with which my disk is actually spinning?(in it's coordinate system)</p>
| |mechanical-engineering|structural-engineering|mathematics| | <p><span class="math-container">$\omega ^*_3$</span> is the rotation rate of the disk in the frame pictured (equal to <span class="math-container">$2 * pi *$</span> frequency point <span class="math-container">$P$</span> will appear in the 9 o'clock position). It is the sum of the orbit rate of <span class="math-container">$\omega _3$</span>'s disc center about the system center in the frame pictured (<span class="math-container">$\omega_3$</span>), plus the rotation rate about it's own axis in a reference frame where this axis is fixed (<span class="math-container">$ \bar\omega_3$</span>). The first term considers the contact points with <span class="math-container">$A$</span> and <span class="math-container">$B$</span> fixed and the system rotating with disk <span class="math-container">$\omega_3$</span>, so one turn of the center is one rotation of the disk, and point P makes one revolution. The second term considers <span class="math-container">$\omega_3$</span>'s axis fixed and the contact points free to run. </p>
<p>If we take <span class="math-container">$\omega_S$</span> as given in each case, we get two different (linear) relations for <span class="math-container">$\omega_R$</span> and <span class="math-container">$\omega_3$</span>. So <span class="math-container">$\omega_R = C_1*\omega_S$</span> and <span class="math-container">$\bar\omega_R= C_2*\omega_S $</span> This lets us set up a general solution <span class="math-container">$\omega^*_3 = f(\omega_S,\omega_R)$</span>, where <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> are solved using <span class="math-container">$\omega_S$</span> and then they are used to compute <span class="math-container">$\omega^*_3$</span></p>
| 29597 | Question regarding calculation of angular velocity connected to coordinate transformation |
2019-08-11T18:08:52.150 | <p>a long time back at school, I had a difficult excercise which was named X, Y and Z bodies.
They were drawings where both the front view and the top view were fully given. We had to find the view from the left plus the body in a sketch drawing.
Unfortunately, I don't remember the solutions anymore and I don't find out them anymore.
Could anyone help?
As an example, I added the Y body with below link.</p>
<p><a href="https://i.stack.imgur.com/4GIoN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4GIoN.png" alt="Y body"></a></p>
<p>Ah, yes: the forms are squares and the middle of the Y is the middle of the squares - in both views.</p>
<p>Thanks for your help and hints in advance!</p>
<p>Best
Tom</p>
| |technical-drawing| | <p>The bottom right line is meant for transferring lines between the top view and the Left view. Same with the Front and Left vertical line. Ideally, the picture would also include dotted lines to indicate any hidden geometry inside or under the view, but let's assume that there aren't any hidden geometries - since there aren't any hidden lines.</p>
<p>It would help indexing the corners in each view, to transfer the geometry properly, and knowing where each line crosses one from another view. Indicate also the "bottom" corners on the front view as "under" the upper corners in the top view. </p>
<p>Draw straight lines out from the points from the top view, until you reach the bottom right line. Switch directions upwards, to project it to the left side view. Project the same point from the front view as well, and where the two lines intersect from the same points in the left view plane, is where you have the location of that point in the left view.</p>
<p>I can work through this exercise precisely, with steps when I get home from work in a few hours :)</p>
| 29601 | Strange CAD excercises |
2019-08-12T16:24:47.973 | <p>I have this old squirrel cage induction motor (nameplate below) that I just replaced the bearings on. When it runs, there is a rattling noise that gradually gets louder and then softer, repeating every ~5 seconds. <a href="https://photos.app.goo.gl/oRZagiXwXN5VyAJo7" rel="nofollow noreferrer">Here's a video of it in action.</a> I don't think the motor made this noise before replacing the bearings, but I'm not sure. (I just bought it and the only time I ran it was briefly at the seller's house.) Does anybody know what might be the source of this rattle? I've put pictures of the motor and the rotor below, as well. Thanks!
<a href="https://i.stack.imgur.com/UqxBv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UqxBv.jpg" alt="motor"></a></p>
<p><a href="https://i.stack.imgur.com/HC6gH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HC6gH.jpg" alt="nameplate"></a>
<a href="https://i.stack.imgur.com/G83ig.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G83ig.jpg" alt="rotor"></a></p>
| |mechanical-engineering|motors| | <p>I listened to the recording and I suspect there's a loose bearing in the motor. Here's how to test it:</p>
<p>With the motor OFF, grab the end of the shaft and pull up and down on it as hard as you can. There should be NO wiggle at all in the shaft when you do this. </p>
| 29615 | What is causing a pulsating, low-frequency vibration in an induction motor? |
2019-08-12T16:27:03.933 | <p>I am using Nema17 12 v motors with A4988 drivers for a project that drives an XY stage. I am programming the pulses to the STEP pin of the drivers in C and I can control the frequency and the shape of each pulse. I was wondering if anyone can tell me (or point me to a reference about) what is the relationship between pulse SHAPE and performance of the motor (as distinct from pulse FREQUENCY). By 'performance' I mean any effect on torque or the reliability of converting the step signal into a rotor step. For example, what would be the difference (for the same frequency) of applying these pulse shapes to the STEP pin of the driver:
<code>
a) -_____-_____-_____-_____ etc. (most of the cycle the signal is 'off'/'logical zero')
b) ---___---___---___---___ etc. (equal time for 'on' and 'off')
c) -----_-----_-----_-----_ etc. (for most of each cycle the signal is 'on'/'1')
</code>
I am currently using form (b) for my pulses (as most examples I have seen do this) but I was wondering what advantage/disadvantage there may be for changing this. Many thanks in advance. </p>
| |stepper-motor| | <p>With this driver, I think your changes will have no effect whatsoever.</p>
<p>A rising edge on the STEP line simply tells the driver to advance the motor firing sequence. It will advance and then hold and wait for the next step. How long you hold the signal high will not change anything.</p>
<p>Here's a a chart from the datasheet. Notice that there are no vertical lines at the STEP falling edges. You can also see that when the edge occurs, nothing changes in the output waveform.</p>
<p><a href="https://i.stack.imgur.com/QDBK6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QDBK6.png" alt="output current vs step signal chart"></a></p>
<p>There <em>are</em> settings you can change that will change the pulse shape that the motor sees. For example micro-stepping and decay settings, but these are set elsewhere. </p>
| 29616 | Effect of step pulse shape on stepper motor function |
2019-08-13T06:26:01.760 | <p>I have a Proportional solenoid valve that uses a 0-30VDC for control (opening the orifice) I only have capability to provide 0-5V or 0-10V. I need to control the valve using my PC. Is there any DC Power supply device that I can control using PC to provide 0-30V (that I can change continuously as required during operation)? I have tried a voltage multiplier circuit. I am getting the required voltage but the Transistors (BC547) are burning out quickly due to over-current/heating.</p>
<p>I am not a electrical/electronics engineer hence don't know if there is a better transistor or how to choose one.</p>
<p>Hence, searching for the PC controllable DC Power supply. if available.
<a href="https://i.stack.imgur.com/eIUsU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eIUsU.png" alt="This is the circuit I have tested. But the BC547 transistor is burning out after 5 min."></a></p>
| |valves|electrical|circuit-design|electronics| | <p>I have found a good device that served my purpose very well.</p>
<p><a href="https://www.scientificindia.com/products/programmable-3-channel-power-supply-1.aspx" rel="nofollow noreferrer">https://www.scientificindia.com/products/programmable-3-channel-power-supply-1.aspx</a></p>
<p><a href="https://i.stack.imgur.com/pXnBm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pXnBm.png" alt="enter image description here"></a></p>
<p>This type of device is also available from manufacturers like Siglent outside India.</p>
<p>Thank you for all the responses.</p>
| 29626 | PC Controllable DC Power supply (0-30V) |
2019-08-13T14:31:54.100 | <p>In HYSYS it is possible to create multiple lists of components using their proper fluid package to better calculate the properties of very different molecules.</p>
<p>Here I defined two lists:</p>
<p><strong>List 1</strong>
<a href="https://i.stack.imgur.com/Ft2uf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ft2uf.png" alt="List 1 HYSYS"></a></p>
<p><strong>List 2</strong>
<a href="https://i.stack.imgur.com/GCH8j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GCH8j.png" alt="List 2 HYSYS"></a></p>
<p>I also defined the fluid packages I need and when I go to simulation and create a stream, I can only specify the composition for the first list of components as shown after. <strong>How should I do if I want to use the second list of components?</strong></p>
<p><a href="https://i.stack.imgur.com/mWfTd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mWfTd.png" alt="Stream HYSYS"></a></p>
| |chemical-engineering|software| | <p>You would need to create a whole new property package to use List 1 or 2 even if the EoS is the same as the initial one. There are a few times when this is impossible such as in Sulphur Recovery (Sulsim) or some Acid Gas properties. Also you should know you can share component lists with two property packages because there are times your reactor requires NRTL but your distillation tower requires CPA or it works fine with PR because you have to have the Efficiencies already measured out.</p>
<p>If you want to change component lists within the stream and keep the data from stream 1... You better create 2 streams one with Basis 1 and the other with Basis-2 each one with its own component list. </p>
<p>Edit: Oh also to change the basis you do it in the Fluid Package cell in Worksheet/Conditions (Perhaps this was the answer to your question)</p>
| 29631 | How to use a specific list of components in a stream in ASPEN-HYSYS? |
2019-08-15T17:16:52.173 | <p>The horizontal Force <span class="math-container">$A_x$</span> is easy but how did they get the vertical forces <span class="math-container">$A_y$</span> & <span class="math-container">$B_y$</span>? I tried using similar triangles for the Force and length of the sides, somehow I can't manage to solve it. I wonder where I am making a mistake. </p>
<p><a href="https://i.stack.imgur.com/eVoUm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eVoUm.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|statics|mechanical| | <p>There are multiple ways to solve this problem but I use the moment equilibrium around the points A and B. </p>
<p><span class="math-container">$$A_y 3l-q_0 4l^2=0$$</span></p>
<p>You can find the <span class="math-container">$A_y$</span> reaction force from the equation above and use it again in the moment equilibrium around A this time: </p>
<p><span class="math-container">$$-4q_0l^2-B_y 3l=0$$</span></p>
<p>You can also skip the second part, and use your intuition, since the structure is in equilibrium, the vertical forces should cancel each other out, otherwise the structure accelerates.</p>
| 29651 | How do I find the vertical reaction forces? |
2019-08-16T07:53:30.473 | <p><a href="https://i.stack.imgur.com/6fAuR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6fAuR.png" alt="illustration of wall"></a></p>
<p>Consider a high and wide <strong>concrete wall</strong> of say <strong>30 cm thickness</strong>, covered with a sheet of <strong>vapor barrier</strong> on one side, and a sheet of say <strong>30 cm thick</strong> and effective <strong>insulation</strong> on the other side.</p>
<p>The <strong>vapor barrier</strong> completely blocks water vapor. The <strong>insulation</strong> lets water vapor through and leads away any <strong>condensed</strong> water downwards where it is <strong>drained away</strong>. The concrete lets vapor vapor through, albeit slowly.</p>
<p>The air on the vapor barrier side (the inside of a hypothetical building) is <strong>25°C with 50% Relative Humidity</strong>, and the air on the insulation side (the outside of a hypothetical building) is <strong>10°C with 100% RH</strong>. See illustration above which shows the wall from above.</p>
<p>According to some vendors of basement insulation, a setup like this will cause the concrete wall to become relatively dry due to vapor pressure towards the insulation in which the water vapor will condense and allow the water to drip down and then be drained away.</p>
<p>But I can't make myself understand how this process works from a physical point of view. I thought an infinite source of water vapor spreads out until all air is saturated with water, so I would expect the wall to end up having 100% RH.</p>
<p>Will the wall end up relatively dry, at say 70% RH? If yes, how exactly?</p>
| |building-physics|vapor-pressure| | <p>After some research, I managed to come up with an answer myself. I will wait with accepting it to allow time for feedback.</p>
<p>Let's first calculate the RH of the concrete. First, we need to calculate the temperature distribution.</p>
<h2>Temperature distribution</h2>
<p>We will assume the following thermal conductivity values:</p>
<ul>
<li>concrete: 1,5 W/mK</li>
<li>insulation: 0,04 W/mK</li>
</ul>
<p>this gives us R (thickness divided by thermal conductivity) values of:</p>
<ul>
<li>concrete R<sub>c</sub> = 0,3 m / 1,5 W/mK = 0,2 m²K/W</li>
<li>insulation R<sub>i</sub> = 0,3 / 0,04 = 7,5 m²K/W</li>
</ul>
<p>The flow of heat from the warm side to the cold side is given by</p>
<p>q = ΔT / ΣR = (25 - 10) / (0,2 + 7,5) ≈ 2 W/m²</p>
<p>To calculate the temperature drop over the concrete which we can call ΔT<sub>c</sub>, we use the above formula but solve for ΔT and only use R<sub>c</sub> for ΣR, which gives us</p>
<p>ΔT<sub>c</sub> = R<sub>c</sub> × q = 0,2 * 2 = 0.4 K = 0.4 °C</p>
<p>For demonstrative purposes, let's round this up to 1 °C. So we have:</p>
<ul>
<li>temperature in concrete right next to the vapor barrier = 25 °C</li>
<li>temperature between concrete and insulation = 24 °C</li>
<li>temperature at insulation outside = 10 °C</li>
</ul>
<h2>Water vapor distribution</h2>
<p>To be able to calculate RH in the concrete, we need to know how the water vapor is distributed. Water vapor diffusion occurs so that concentration of water molecules becomes evenly distributed. Temperature only affects the speed of the diffusion<sup>*</sup>. Since RH is 100% outside of the insulation at 10 °C, we know that water vapor content is 9,40 g/m<sup>3</sup>.</p>
<p>If the air in the concrete has more than 9,40 g/m<sup>3</sup> water molecules, the water molecules with diffuse in the direction of the insulation, since the concentration of water vapor is lower there. And since the water vapor content is fully saturated, it will condense and flow away.</p>
<p>If the air in the concrete has less than 9,40 g/m<sup>3</sup>, water molecules will diffuse into the concrete, but only up to 9,40 g/m<sup>3</sup>.</p>
<p>To calculate relative humidity in the concrete, we need to know maximum water vapor capacity of air at 24 °C and 25 °C which is 21,75 g/m<sup>3</sup> and 23,02 g/m<sup>3</sup> respectively. So the RH in the concrete becomes 9,40/21,75 ≈ 43% and 9,40/23,02 ≈ 41%.</p>
<p><sup>*</sup> Temperature itself does have a small impact of diffusion, but in the domain of building physics, it is negligible.</p>
<h2>Answer summary</h2>
<blockquote>
<p>Will the wall end up relatively dry, at say 70% RH? If yes, how
exactly?</p>
</blockquote>
<p>The RH of the wall will become ≈ 40% because water vapor diffusion is driven by difference in water molecule concentration, not difference in RH.</p>
| 29663 | Can temperature gradient keep concrete wall below 100% RH in 100% RH atmosphere? |
2019-08-16T09:24:26.633 | <p>I need to write a non-technical product requirements document (PRD) for a new machine that I want designed.</p>
<p>I need to ensure that I have included all the salient information concisely and to the point.</p>
<p>I have searched online for template PRDs but all of the documents I have come across, assume that the product is software so do not fit the needs of a PRD for a mechanical device.</p>
<p>Can someone outline here, the structure and sections required for a PRD for a mechanical device (e.g. an engine), failing that, a link to such a document would suffice.</p>
| |mechanical-engineering|design|product-engineering| | <p>Before we start, just a word of caution that an engine is not only a mechanical device, they come with their own controllers (FADECs), and most of the time they have to integrate with some software (even when they operate through a throttle servo).</p>
<p>Alright, I'll suggest two methodologies, both of them are applicable to many scenarios.</p>
<ol>
<li>The Flexible Way</li>
<li>The Well-Specified Way</li>
</ol>
<p>I've personally gone through both methodologies in aerospace and defense projects.</p>
<p><strong>Flexible way</strong> can be used when you don't know the specifics of what you are doing, e.g. you want to outsource the expertise of designing and integration of the subsystem. You can do this when you have good business partnerships with the subcontractor (who will develop and supply the Product). Several of the specifications of the design will eventually change during the project. Specifications may change because of the bigger system or because of the Product itself.</p>
<p>In Flexible way, the Work Description can be more important than the Technical Specifications (=PRD, as you call it). So it's two documents you have to consider: </p>
<ol>
<li>Work Description: HOW will they develop the PRODUCT (including how
will they interact with you)</li>
<li>Technical Specifications: WHAT will the PRODUCT do, or What
properties will it have.</li>
</ol>
<p>For the flexible way, writing a brief document will suffice. Key Performance Parameters, and Key Characteristics/Properties could be sufficient. So the structure of the document can be as follows:</p>
<ol>
<li>General Specifications</li>
<li>Performance Requirements</li>
<li>SubPart-A Requirements (Assume that you know the main part breakdown of the Product)</li>
<li>SubPart-B Requirements</li>
<li>Interface/Integration Requirements</li>
<li>Other Requirements</li>
<li>Appendix (e.g. CAD data for interface, technical drawings, draft interface control document, etc.)</li>
</ol>
<p><strong>The Well-Specified way</strong> is used when the Product to be developed is so important that you cannot risk changing the specifications on the road. It can be a mission critical subsystem or a performance critical item. For sake of completeness I'll refer you to the Defense Acquisition template. <a href="http://acqnotes.com/acqnote/careerfields/system-requirements-document" rel="nofollow noreferrer">ref</a></p>
<p>Structure of a System Technical Specification document is as follows:</p>
<ul>
<li><ol>
<li>Scope/Introduction 8</li>
</ol></li>
<li>1.1. System or Subsystem Identification 8</li>
<li>1.2. System or Subsystem Overview 8</li>
<li>1.3. Document Overview 8</li>
<li><ol start="2">
<li>Applicable Documents 8</li>
</ol></li>
<li>2.1. General 8</li>
<li>2.2. Government Documentation 9</li>
<li>2.2.1. Government Specifications, Standards, and Handbooks 9</li>
<li>2.2.2. Other Government Documents, Drawings, and Publications 9</li>
<li>2.3. Non-Government Publications 9</li>
<li>2.4. Order of Precedence 10</li>
<li><ol start="3">
<li>System or Subsystem Requirements 10</li>
</ol></li>
<li>3.1. Required States and Modes 11</li>
<li>3.2. System of Subsystem Functional Requirements 11</li>
<li>3.2.1. System or Subsystem Function 11</li>
<li>3.3. System External Interface Requirements 11</li>
<li>3.3.1. Interface Identification and and Diagrams 11</li>
<li>3.3.2. Project Unique Identifier of Interface 12</li>
<li>3.4. System Internal Interface Requirements 12</li>
<li>3.5. System Internal Data Requirements 12</li>
<li>3.6. Adaption Requirements 12</li>
<li>3.7. Safety Requirements 13</li>
<li>3.8. Security and Privacy Requirements 13</li>
<li>3.9. System Environmenet Requirements 13</li>
<li>3.10. Computer Resource Requirements 13</li>
<li>3.10.1. Computer Hardware Requirements 14</li>
<li>3.10.2. Computer Hardware Resource Utilization Requirements 14</li>
<li>3.10.3. Computer Software Requirements 14</li>
<li>3.10.4. Computer Communications Requirements 14</li>
<li>3.11. System Quality Factors 14</li>
<li>3.12. Design and Construction Contraints 15</li>
<li>3.13. Personnel-Related Requirements 15</li>
<li>3.14. Training-Related Requirements 15</li>
<li>3.15. Logistics-Related Requirements 15</li>
<li>3.16. Other Requirements 16</li>
<li>3.17. Packaging Requirements 16</li>
<li>3.18. Statutory, Regulatory, and Certification Requirements 16</li>
<li>3.18.1. Statutory Requirements 16</li>
<li>3.18.2. Regulatory Requirements 16</li>
<li>3.18.3. Certification Requirements 16</li>
<li>3.19. Precedence and Criticality of Requirements 16</li>
<li>3.20. Demilitarization and Disposal 17</li>
<li><ol start="4">
<li>Verification Provisions 17</li>
</ol></li>
<li>4.1. Verification Methods 17</li>
<li>4.1.1. Demonstration 17</li>
<li>4.1.2. Test 17</li>
<li>4.1.3. Analysis 17</li>
<li>4.2. Inspection 17</li>
<li>4.3. Special Verification Methods 17</li>
<li><ol start="5">
<li>Requirements Traceability 18</li>
</ol></li>
<li>5.1. Traceability to Capability Document or System Specification 18</li>
<li>5.2. Traceability to Subsystems Requirements 18</li>
<li><ol start="6">
<li>Appendix 18</li>
</ol></li>
<li>6.1. Appendix A: Acronyms and Definitions 18</li>
<li>6.2. Appendix B: Key Performance Parameters/Key system Attributes 18</li>
<li>6.3. Appendix C: Requirements Traceability Matrices 18</li>
<li>6.4. Appendix D: Verification Matrices 19</li>
</ul>
| 29665 | What are the essentials of a good PRD (product requirements document) for a new mechanical device |
2019-08-16T09:50:23.217 | <p>I know there are some aspects we can try to apply this surface area to volume ratio to say physical chemistry in order to see reactivity of chemicals. However I am very confused in my structural engineering elective course, it states that surface area to volume of high rise tower like structures is very important. I do see that the SA:V ratio of certain shapes in general give a very interesting and neatly simplified expression once computed, but I do not see what the application of SA:V ratio is when building structures of towers (Eiffel tower CN Tower etc...).</p>
<p>Can anyone help me explain the significance of SA:V ratio when building high rise structures (I would appreciate technical details as engineering is not my major of study i want to understand alittle better)? </p>
| |structural-engineering| | <p>In high rise design surface area to volume ratio is a significant factor in the amount of energy used to keep the building air conditioned and useable for its design function. The larger the surface the more heat transfers either from the building to outside in the winter or vice versa in the summer.</p>
<p>At the first glance it would seem a cube would have the least surface to volume ratio and be an ideal shape.</p>
<p>However a cube has a big portion of its usable floor are near its core, far away from the exterior windows which provide natural lighting and ventilation. So the core of the building needs continuous lighting and ventilation. which adds to the energy consumption.</p>
<p>The optimal building form factor and SA/V is one which would address all these concerns and find a balance between an energy efficient shell and energy needed for lighting and ventilation, etc.</p>
<ul>
<li><p>Location and angle of orientation of the building with respect to sun rays, wind direction and snow drift.</p></li>
<li><p>The climate. Moderate Climate gives flexibility to have more surface to take advantage of natural light.</p></li>
<li><p>The extreme conditions the building envelope will likely be exposed to.</p></li>
</ul>
| 29666 | Surface Area to Volume Ratio |
2019-08-16T10:28:43.187 | <p>I'm wondering whether there are any notable losses in a typical automobile clutch when fully engaged.</p>
<p>Only losses that come to mind:</p>
<ul>
<li>material twisting with torque changes</li>
<li>slipping due to imperfect friction</li>
</ul>
<p>Are these even significant and are there any more? What is a typical energy efficiency of a fully-engaged clutch?</p>
| |mechanical-engineering|automotive-engineering| | <p>The effects the OP mentioned do exist and would probably be measurable in controlled conditions in a test rig. However they are negligible, because they only dissipate a tiny amount of energy <em>as the torque through the clutch changes</em> with driving conditions. </p>
<p>Nearly all of the strain energy in the components that is stored as they twist up under torque load is recovered when the load is removed at they return to their undeformed shape. The amount of energy lost in one cycle of loading and unloading (independent of how long the cycle takes) is of the order of 0.001% to 0.01%, for metal components.</p>
<p>If the clutch is <em>continuously</em> slipping as it transmits power, it will quickly fail because that is not how it is designed to operate.</p>
<p>The analogous effects in bolted joints between components can dissipate a significant amount of energy and affect the mechanical damping of the system, but only if the load in the joint is continually changing at high frequency - e.g. something like a cylinder head bolt where the loads are changing in sync with the engine RPM, hundreds of times per second. </p>
<p>Any energy losses in the clutch will be tiny compared with losses in the gearbox, for example. </p>
| 29667 | Efficiency of a fully-engaged friction clutch |
2019-08-17T13:45:32.907 | <p>Nuclear power always require some way of converting the energy from nuclear to electrical or mechanical to be useful. For example, in a nuclear power generation facility, this is done through heat/steam which drives mechanical generators. What is the mechanism used to convert nuclear energy to rocket power?</p>
| |rocketry|nuclear-engineering|nuclear-technology| | <p>In <a href="https://en.wikipedia.org/wiki/Nuclear_thermal_rocket" rel="nofollow noreferrer">Nuclear Thermal Rockets</a> (NTRs) </p>
<blockquote>
<p>the heat from a nuclear reaction replaces the chemical energy of the propellants in a chemical rocket. In an NTR, a working fluid, usually liquid hydrogen, is heated to a high temperature in a nuclear reactor and then expands through a rocket nozzle to create thrust. The external nuclear heat source theoretically allows a higher effective exhaust velocity and is expected to double or triple payload capacity compared to chemical propellants that store energy internally.</p>
</blockquote>
| 29687 | How does a nuclear powered rocket engine work? |
2019-08-19T09:28:15.440 | <p>Currently I'm doing a personal project, but I've some doubts about the calculation process. Well, I've a platform (2.30m length x 0.32m width). This platform is supported by 4 wheels with a diameter equals 120mm. This platform loads a motorbike. The maximum mass (platform + motorbike) that it will move is about 400kg. The surfaces contact are rubber with concrete (coefficient of friction 0.45). Rolling coefficient I think is 0.02). I'm uncertain wether to put 2 driven wheels or 4. The platform must travel a distance of 7 m. For the process I propose a speed of 0.1 m / s.
Well, How can I calculate the required torque to move this platform with a motorbike?</p>
| |mechanical-engineering|mechanical|industrial-engineering| | <p>I put an answer but it don't reads correctly.. I put there:</p>
<p>My calculations are:
Total Weight (plaform + motorbike) = 4905 N
Radius of wheel = 0,06 m
Desired top speed = 0,1 m/s
Desired acceleration time = 3 s
Maximum incline angle = 5 degree
mu (rubber-concret) = 0,45
Crr = 0,02</p>
<p>To determine the Total Tractive Effort (TTE):
TTE = RR + GR + FA where:
RR - Force necessary to overcome rolling resistance
GR - Force required to climb a grade
FA - Force required to accelerate to final velocity</p>
<p>RR = 4905 x 0,02 = 98,1 N
GR = 4905 x sin 5º = 427,5 N
FA = 4905 x 0,1 / (9,81 x 3) = 16,67 N
TTE = 542,27 N</p>
<p>Tw (wheel torque) = TTE x Rw (radius wheel) x RF (resistance factor)
Tw = 542,27 x 0,06 x 1,1 = 35,79 Nm</p>
<p>Are there correct? Thanks</p>
| 29710 | How to calculate required torque |
2019-08-20T10:56:54.983 | <p>The Blast Furnace flue dust contains high rates of Zn (Zinc). I'd like to know the reason behind this. The chemical reaction of Zn inside blast furnace dust. Why it's rates becoming high?</p>
| |chemical-engineering|metallurgy|chemistry| | <p>Not likely unless they use a very unusual ore. Blast furnaces generally do not use scrap. Scrap likely contains zinc ( die castings) . Electric arc remelt furnaces melt a high percentage of recycle containing zinc. There may still be open hearths or other furnaces used to melt recycle. So, furnaces other than blast furnaces are likely the source. Depending on the system , dust from different types of furnaces may be collected in one facility which may not be differentiated , leading to the idea that zinc is coming from a blast furnace. Generally the mills remove as much zinc as possible from the recycle as I understand it causes deterioration of the refractories. After it is a few hours old ,zinc oxide is innocuous and is used in skin creams. </p>
| 29740 | Why do Blast Furnace flue dust generates high rates of Zn (Zinc)? |
2019-08-21T13:04:19.003 | <p>What is the difference between both of these terms? Especially in configuration and application?</p>
| |electrical-engineering|control-engineering|microelectronics| | <p>In pure control theory, everything is a mathematical entity. So a controller is a mathematical entity that realizes a certain transfer function (or a certain time-dependent behavior, if there are multiple inputs or outputs involved).</p>
<p>A microcontroller is a term for a chip that contains a microprocessor core, memory, and peripherals, that can run stand-alone but for needing to be mounted to a board. In other words, it's what would have been built onto a 200mm x 200mm square board in 1978, and would have cost \$200 in parts (in 1978 dollars), -- only microcontrollers hawe ten to one hundred times more of just about everything, and you can get them starting at about \$0.50 in quantity.</p>
<p>To either confuse things or clarify: </p>
<p>You can take a microcontroller and you can <em>implement</em> a controller on it, using the microcontroller's peripherals, and make it part of a <em>physical</em>, real-world control system. This will rapidly lead to you being whacked upside the head with the difference between theory and practice, when you first turn your system on expecting good stable behavior from your plant.</p>
<p>But you cannot take a controller from control theory and implement a microcontroller inside of it. You could use some of the same language and implement a <em>model</em> of a microcontroller (with much cursing and head-scratching from both your local control theorists and your local computer scientists). But you couldn't realize a physical microcontroller.</p>
| 29753 | Controller vs Micro controller? |
2019-08-21T20:02:51.323 | <p>We only have one adjusted FWD modulus value for HMA layers, but I can only see input options in the form of a grid of frequency and temperature.</p>
<p><a href="https://i.stack.imgur.com/UICQb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UICQb.png" alt="enter image description here"></a></p>
| |pavement| | <p>The FWD modulus results are an input for the existing asphalt layers. One value can be input and a grid is not necessary, as shown in your screenshot above. </p>
<p>You should evaluate the backcalculated values for the asphalt layers and make sure the data you have is independent of temperature. The dynamic modulus of any asphalt layer is dependent on temperature and if the deflections are measured over a wide enough range of temperature, the backcalculated values can be grouped by time of day and temperature, as long as temperature is included in the FWD data file. These are engineering decisions and need to be made by the pavement engineer.</p>
| 29758 | How can I use backcalculated modulus results from FWD testing to represent my existing condition using Pavement ME Design? |
2019-08-21T20:50:41.003 | <p>So far, I have only been able to model dry layers as I've only been able to represent the FDR layer with one modulus value.</p>
| |pavement| | <p>Please refer to the Manual of Practice for the Mechanistic-Empirical Pavement Design Guide (MEPDG). The following is an extract from that manual:</p>
<blockquote>
<p>Full-Depth Reclamation (In-Place Pulverization of Conventional
Flexible Pavements) - Cold in-place recycling of the HMA and existing
aggregate base layers, and hot in-place recycling of HMA. Cold
in-place recycling as a rehabilitation strategy is considered
reconstruction under the MEPDG design/analysis process and would be
defined as a new flexible pavement. Hot in place recycling as a
rehabilitation strategy is considered mill and fill with an HMA
overlay of the existing flexible pavement. The thickness of the hot
in-place recycled material is considered part of the HMA overlay, as
well as the thickness of the milled material. Full-depth reclamation,
however, was not included in the global calibration of AASHTOWare
Pavement ME Design.</p>
<p><strong>Mechanistic-Empirical Pavement Design Guide - A Manual of Practice 2015</strong> <strong>(AASHTO)</strong></p>
</blockquote>
<p>How you treat or simulate the FDR layer is an engineering decision. It depends on how the FDR layer is stabilized and designed. </p>
<p>If you are adding an emulsion to the reclaimed layer, it can be simulated as a conventional asphalt layer and the pavement can then be simulated as new construction. If an emulsion is used and the layer is simulated as an asphalt layer for new construction, you need to determine whether the default or local calibration coefficients for fatigue cracking apply to this FDR layer. </p>
<p>This decision will depend on how that FDR layer with emulsion added is designed and constructed. Some State agencies model the FDR layer as a high strength granular layer with a representative resilient modulus when a low amount of asphalt binder (4 percent or less) is used as the stabilizing material (Virginia for example). </p>
| 29760 | How can we model a full depth reclamation (FDR) layer with emulsion (wet FDR layer) in a pavement design? |
2019-08-22T20:40:47.137 | <p>How are the integrated climatic model as discussed in <a href="http://hdl.handle.net/2142/95718" rel="nofollow noreferrer">Enhanced Integrated Climatic Model Version 2.0</a>, and the thornthwaite moisture index (TMI) used to calculate erosion, suction, and frost heaving in Pavement Mechanistic-Empirical Design?</p>
<p>Specifically interested in how precipitation affects pavement performance within Pavement Mechanistic Empirical Design (PMED) software.</p>
| |pavement| | <p>Erosion, frost heaving, and suction are all ultimately caused by precipitation on the pavement. Precipitation is used in the following models:</p>
<ul>
<li><p>The Enhanced Integrated Climatic Model (EICM) uses precipitation to calculate the monthly Thornwaite Moisture Index (TMI).</p></li>
<li><p>The TMI is used in calculating the equilibrium suction for the base and subgrade layers. The suction values are used to estimate water content within each sublayer which impacts the resilient modulus values.</p></li>
<li><p>The flexible roughness model uses the mean annual rainfall to calculate the frost and swell potential for the site factor parameter in the model.</p></li>
<li><p>The continuously reinforced concrete pavement (CRCP) model uses annual rainfall for the erosion calculation</p></li>
</ul>
<p>The Pavement ME Design software does not currently model precipitation infiltration into the pavement system from the surface. It assumes that the pavement system has adequate drainage.</p>
<p>Resources:</p>
<ol>
<li><p>Mechanistic Empirical Pavement Design Guide – A Manual of Practice</p></li>
<li><p>NCHRP 1-37A Documentation – Main document and double letter appendices</p></li>
<li><p>NCHRP Report 602 – Calibration and Validation of the Enhanced Integrated Climatic Model for Pavement Design </p></li>
</ol>
| 29767 | What is the influence of precipitation on pavement performance in both jointed plane concrete pavements (JPCP) and asphalt concrete pavements? |
2019-08-23T10:10:12.880 | <p>I would like a tig welder. It has to be able to weld 8mm thick steel flat bar so we can fix the front gate if it breaks. </p>
<p>I also want to be able to do stainless and aluminium for projects.</p>
<p>It seems my budget can only get me at most a 200A machine.. My budget is £500</p>
<p>so would a 200A machine be enough?</p>
<p>also why do people say tig without argon isn't good?</p>
| |welding| | <p>200 A is plenty of power for that relatively thin material. TIG stands for " tungsten inert gas" , without argon there is no inert gas so it is not TIG ( I don't know what it is). You need TIG for aluminum , although it requires practice. MIG is usually the first choice for steel and stainless . However , for MIG the "inert gas" can be replaced with CO2 or flux core wire.</p>
| 29777 | 200A tig welder enough to weld 8mm thick steel flat bar/stainless steel sheet/tubing |
2019-08-23T13:48:54.123 | <p>Can we build a building that could last thousands of years, if it was abandoned, in a humid continental climate? </p>
<p>Is this even possible with present knowledge? How would such a structure be built? What foundation can actually withstand seasonal temperature changes, and precipitation, without requiring repairs after a couple decades at most?</p>
<p>It doesn't need to be very large, assume the size of a simple home, so 1500 square feet.</p>
| |structural-engineering| | <p>In Mexico's hot and humid Yucatan peninsula are many examples of exactly what you are asking about, some dating back to 400 BC.
Many are abandoned in the middle of the rain forest, while others have been tidied up for tourists.</p>
<p>The most famous is "<a href="https://en.wikipedia.org/wiki/El_Castillo,_Chichen_Itza" rel="nofollow noreferrer">El Castillo, Chichen Itza</a>", which has been declared <a href="https://en.wikipedia.org/wiki/World_Heritage_Site" rel="nofollow noreferrer">a UNESCO World Heritage Site</a>:</p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/5/51/Chichen_Itza_3.jpg/1920px-Chichen_Itza_3.jpg" alt="Chichen Itza"></p>
<p>This one is only a thousand years old, but it could obviously continue surviving for many more centuries or millennia.</p>
<p>We <em>can</em> build structures that will last for thousands of years, but we generally choose not to.</p>
<p>EDIT: <strong>How is it able to not fall apart like our structures? We construct a concrete foundation, its cracking almost immediately.</strong> – J. M. Becker</p>
<p>The Coliseum in Rome was built over 2000 years ago, and most of it is still standing. Much of it is made from concrete that has endured all this time. The Romans expected it to last "forever".</p>
<p>Consider the equivalent of the anthropic principle. The ancient societies <em>did</em> build many flimsy structures, but the only examples that remain of their work for us to see are the few that <em>were</em> well built.</p>
<p>But there's another factor to it. Today's society is very different from what it used to be.</p>
<p>The Mayans built their structures as permanent monuments. They were intended to last. (Egyptian pyramids are nearly 5000 years old, but the original question asked for a "humid climate", so I didn't use them as an example.)</p>
<p>Today's structures are built to make a profit for their builders, and to provide a quick and <em>modern</em> building for their occupants. In a few years they will be <em>old fashioned</em> and replaced by another <em>trendy</em> building. There's no point in spending a fortune building something that will endure.</p>
<p>Our current society and economy is based on this. I use tools that were made 50 to 100 years ago, and they are still in good condition. Today, unless I'm willing to spend an incredible price for high quality manufacturing, anything I buy now will be cheaply made in China and will need to be replaced in a year or two.</p>
<p>In today's dollars, years ago I could buy a tool for \$20–30, depending upon the quality I wanted. Today I can buy a tool of similar quality for \$150, or a cheap version for $5. There's nothing in between. Most people choose the \$5 version, which means that the good quality versions must be made in small quantities, and so end up costing much much more.</p>
<p>Household refrigerators that were made 50 years ago are still running fine (energy inefficient though), while most made 10 years ago are now scrap.</p>
<p>When it comes to houses and office buildings, the situation isn't much different.</p>
| 29780 | Can we build a building that could last thousands of years, if it was abandoned, in a humid continental climate? |
2019-08-23T18:12:48.510 | <p>When you're putting up a new building, you have to check the soil underneath for stability. How deep do you typically have to check? What's the extent of your "due diligence" when ensuring that a building won't collapse because of unstable ground? I imagine there might be different rules in different jurisdictions, but is there a typical, ballpark depth? </p>
<p>(Relatedly: Has a building ever collapsed because of unstable conditions below the checked area?)</p>
<p>I'm a total newbie, and I apologize for the many flaws you will be able to spot in this question.</p>
| |structural-engineering|structural-analysis|building-design|building-physics|construction-management| | <p>Depends on the structure. You go down until you have capacity to carry the load plus a safety factor. If you don't want the 150percent safety factor be prepared to do a core sample for every casson for the entire depth. If that allows you to go down 150ft instead of 225 on 20 supports it's worth it. LOCAL factors rule, ask around for a good old boy driller or excavator and call him. He'll tell you what's under your site. Plan for that and adjust accordingly. He'll be within 10-20 percent on depth of soil transitions 200 feet down.</p>
| 29786 | How deep do you have to check for problems underneath a building site? |
2019-08-24T21:16:28.750 | <p>I have a 12" flow meter that uses a float on a stem and whenever gas is supplied, the initial flow spikes the float to the top and gets wedged. The flow meter is plumbed downstream of a control valve and ball valve that is being used to regulate the flow at 40 cfm. The supply side is helium that can range between 60 - 80 psig and the outlet of the meter is plumbed to a chamber at half atmosphere, ~ 400 Torr.</p>
<p>I'm aware that a ball valve should not be used to fine tune flow, that's why I need some clarification to properly correct this issue. I attached an image to help better explain my current setup.</p>
<p><a href="https://i.stack.imgur.com/vaB0e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vaB0e.png" alt="enter image description here"></a></p>
| |flow-control| | <p>The flow meter needs to be installed such that it is always under pressure (in normal operation). That is it should be after the regulator but before the on/off valve. That way you will get no surge of gas entering an empty pipe and pushing up the float. It is quite a common problem.</p>
| 29801 | Proper Flow Meter Installation |
2019-08-25T05:54:59.197 | <p><img src="https://i.stack.imgur.com/Wod2e.jpg" alt="enter image description here"></p>
<p>The rotor of fan is shown as a series of circles arranged in a circular manner. From youtube videos, I came to know that this is sort of a top view and the circles actually represent conductors through which current 'can' flow into the screen or come out of the screen (towards the viewers).
I wonder as to what happens to the current that arises in these conductors (due to the induction of EMF).</p>
<p>Where do the electrons flow to ? There has got to be some path, which makes that path a conductor, and hence susceptible to be affected by lorentz force. </p>
<p>So why isn't that shown ?
(Diagrams would be highly helpful)</p>
| |motors| | <p>Each of the circles or conductors are actually paired with another, probably diametrically opposite but this is not always the case.</p>
<p>So, if the current is coming towards you on one circle then it will be going away from you on its pair.</p>
<p>That is the path you are looking for as the linking connections are not shown as that diagram is a section drawn through the rotor.</p>
| 29805 | Why doesn't the flow of current through the conductors in the rotor of a fan affect it? |
2019-08-25T07:25:26.817 | <p>Tractor carries more than his own weight on its trolley. The linkage between the tractor and the trolley is far away from the tractor. So, the whole weight of the trolley is on the back of the tractor. But why does not a tractor do wheelie with more weight on his behind?</p>
<p>To make it understandable, here are some photos I designed:</p>
<p>Tractor with empty trolley:</p>
<p><a href="https://i.stack.imgur.com/eZlS5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eZlS5.png" alt="enter image description here" /></a></p>
<p>Tractor with weight on its trolley must be like this but it doesn't happen.
<a href="https://i.stack.imgur.com/w0bkc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w0bkc.png" alt="enter image description here" /></a></p>
<p>What is the reason for that?</p>
| |automotive-engineering|car| | <p>Because the rear wheels are larger so the hitch connection point is below the centre line of the rear axle.</p>
<p>However, if there is enough power then the tractor can do a wheelie anyway, have a look a tractor pulling competitions...</p>
| 29807 | Why does a tractor not do wheelie with a lot of weight on the trolley? |
2019-08-26T07:08:24.350 | <p>According to <a href="https://www.youtube.com/watch?v=P13Mau2VUWw" rel="nofollow noreferrer">What is Prestressed Concrete? - Practical Engineering</a>, one way to reduce deflection in the steel rebar is to pre-stress the rebar by applying compressive stress in the member before service which is done by tensioning the steel rebar.</p>
<p>My question is: Why does applying compressive stress to member decrease deflection?</p>
<p>especially with relation to the formula:
<span class="math-container">$y=\frac{FL^3}{48EI}$</span></p>
<p>where</p>
<ul>
<li><span class="math-container">$y$</span> is deflection</li>
<li><span class="math-container">$F$</span> is force applied to member at <span class="math-container">$L/2$</span></li>
<li><span class="math-container">$L$</span> is length of beam</li>
<li><span class="math-container">$E$</span> is elastic modulus</li>
<li><span class="math-container">$I$</span> is second moment of area</li>
</ul>
| |concrete|deflection|prestressed-concrete| | <p>Concrete is strong in compression but weak in tension. In tension, it will crack at a much lower stress than the allowable compression stress.</p>
<p>A reinforced concrete beam without any prestressing will generally have compression in concrete and tension in the rebars. If there are any significant tension stresses in the concrete, the concrete will crack and the rebars will take over, transferring the tension instead of the concrete. Since the part of the concrete which is cracked can no longer transfer any tension stresses, it will not contribute to the second moment of area.</p>
<p>If the concrete beam had a large compression force from prestressing, there would be no tension stresses and therefore also no cracking. That means the entire concrete cross section contributes to the second moment of area.</p>
<p>So in terms of the deflection formula, you quoted in the question, the answer is that the second moment of area is different, because we avoid the cracking.</p>
| 29824 | Why does rebar in reinforced pre-stressed conrete reduce in deflection if compressive stress is applied to member? |
2019-08-26T07:22:18.513 | <p>When driving a car with manual transmission on start you slowly release the clutch so the car doesn't jerk. So clutch slippage here is not that bad if I understand this correctly?</p>
<p>And also couldn't dog clutch or synchromesh units be used as a mean of disconnecting the engine from the transmission? Then perhaps clutch could be used further down in a DCT to change between gears?</p>
<p>Thanks in advance.</p>
| |mechanical-engineering|mechanisms|mechanical|transmission| | <p>Clutches in car transmissions are designed on purpose to allow slip transiently, to make gear changes and standing starts smooth. This is not a bad thing. But they are not designed to slip all the time, which would wear them out quickly and generate lots of heat. </p>
<p>It is possible to use the synchronizers in a <em>motorcycle</em> transmission to "bang" through the gears without using the clutch, but this is because the motorcycle weighs a lot less than a car and the synchronizers are stout enough relative to the weight of the bike to take this sort of punishment- for a while, anyway.</p>
<p>But this is decidedly <em>not</em> the case for cars and trucks, where this sort of abuse will ruin the synchronizers and eventually break the teeth off the gears.</p>
| 29825 | Is it ok for clutch to slip in DCT in certain situations? |
2019-08-26T13:23:17.763 | <p>"1 TR is equivalent to the rate of heat transfer needed to produce 1 U.S. ton (2000 lbs) of ice at 32F from water at 32F in one day, i.e., 24 hours."</p>
<p>While defining 1 Ton of Refrigeration (TR), why doesn't some specific value of pressure come into definition?</p>
<p>Isn't freezing water (to ice) related to pressure?
-->If yes then why do we go away with it while defining 1TR ?</p>
| |thermodynamics|refrigeration| | <p>The values for Standard Pressure are 1 atm or 101325 Nm-2 or 760 mm Hg or 101.325 kPa.
This pressure generally use in defining standards . Also for ideal condition the value of pressure taken into account is 1 atm . Hence it is to obvious to mention the value of standard there . </p>
| 29832 | 1 Ton of Refrigeration |
2019-08-27T13:26:18.880 | <p>ESD-safe tweezers have a black coating on the handle. If I am grounded to a work bench, and normal tweezers are already conductive, then what is the point of the conductive black coating? Screw drivers sometimes have handles which may be insulators, but tweezers are all metal. Are they not as conductive as I am expecting? Maybe an oxide layer? </p>
| |electrical-engineering|electrical| | <p>The tweezers that you are describing have what's known as a Static Dissipative coating. This coating <em>is</em> conductive, but it has a very high resistance.</p>
<p>What this means is that they are able to dissipate any differential charge slowly/safely, without there being a very low impedance path to ground that might produce an arc.</p>
<p>Put simply, the tweezers will not prevent static discharge, but they will limit the peak current at which this occurs, which can be enough to save some (but not all) components from destruction.</p>
| 29846 | What's the point of ESD-safe tweezers? |
2019-08-27T14:38:59.983 | <p>Firstly i am new to hydraulics, i want to build a hydraulic circuit that powers 4 cylinders at each leg of a Quad-Wheeled-Robot. The idea is to move the Chassis of the Robot up and down in all possible ways to avoid obstacles. <strong>I did design a system but i am not sure if it has everything i need is there anything that i need to add to it?.</strong></p>
<p><strong>Below is a picture of the simple design using only one actuator cylinder.</strong> </p>
<p><strong>It has the following specs:</strong> </p>
<ul>
<li>Dynamic Load: 4000 N bis 6000 N</li>
<li>Motor: 12V BLDC 1000W (Feedback)</li>
<li>Pump: min 80 to max 150 Bar @ 3 to 5 L/min</li>
<li>Tank: 5 l</li>
<li>Control Valve: 4x Proportional 4/3 Servo-Control Valves <strong>(closed Center)</strong></li>
<li>Zylinder: Stroke 200mm (Piston/Bore 32mm;
Rod:16mm)</li>
<li>Check Valve and Unloading valve (if i need one)?
Also i have added a Cooling and Filter.</li>
</ul>
<p><strong>Updated Version all return lines going through Cooler</strong></p>
<p><a href="https://i.stack.imgur.com/ReSri.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ReSri.png" alt="enter image description here"></a></p>
| |hydraulics|valves|actuator| | <p>If the first VMA 1.025 is acting as your pressure relief valve, you might be OK, depending on duty cycles/pressures/flowrates/etc...You do still build some heat flowing over that relief valve, but like above, depends on your system. I had a similar issue in a design recently so I feel the need to mention it. If you go with a dump, put the valve right after the pump, dump into a T/Y fitting before the radiator, just add a check valve before the T/Y on the other incoming side of the T/Y. With a center position, P and T blocked, you'll be building pressure/heat upstream of that valve with nowhere to go,not so much an issue in constant operation, but may be evident when the machine idles. If intermittent a great deal between operation and not, I would maybe cycle off the pump. </p>
<p>I like the updated diagram, watch your filter though....you can filter your return oil coming out of your radiator into the tank, but your pump should not be puling oil through a filter, your going to want a strainer for that. I know, pretty similar names, but your strainer has much larger holes, never want to starve a pump for oil.</p>
| 29847 | Help with simple hydraulic circuit to use for a four legged Robot |
2019-08-29T14:20:26.567 | <p>One of my suspension bolts came undone while I was mountain biking recently, sadly I didn't notice. I landed a big drop with only 2-3 threads left engaged, and unsurprisingly it's ruined them.</p>
<p>The rest of the threads (another 20 or so) are fine. It's a blind hole.</p>
<p>I have the right size tap to cut the thread, but I don't know if that's the right thing to do. Don't I risk being "out of sync" with the 20 good threads and causing problems with them?</p>
<p>I feel there should be a tap that you insert deep into the bolt hole, and it then somehow expands into the known-good threads, before being able to be wound out.</p>
<p>If that does exist, please let me know what it's called and where I can get one.</p>
<p>If it doesn't, please give me advice on how best to fix this hole with the existing tap that I have is <a href="https://www.parktool.com/product/frame-tap-tap-10" rel="nofollow noreferrer">Frame Tap</a> </p>
<p><a href="https://i.stack.imgur.com/vixDS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vixDS.jpg" alt="Frame Tap"></a></p>
| |threads|bolting| | <p>The product that Jonathan R Swift created for himself does exist in commercial form, for 10, 12 and 14mm spark plug threads. Can save having to Helicoil a head, including all the labor to pull it, etc. Sold by NAPA, Amazon, and others.</p>
<p><a href="https://www.napaonline.com/en/p/SER3689" rel="nofollow noreferrer">https://www.napaonline.com/en/p/SER3689</a></p>
| 29885 | Fixing a blind bolt hole when the first 2-3 threads are ruined? |
2019-08-30T12:49:41.963 | <p>from the <a href="http://T.%20Belytschko,%20L.%20Schwer:%20Large%20Displacement,%20Transient%20Analysis%20Of%20Space%20Frames;%20International%20Journal%20For%20Numerical%20Methods%20In%20Engineering;%20Vol.%2011,%2065-84%20(1977)" rel="nofollow noreferrer">paper in this link</a> , which is the Belytscko beam. I could not figure out if it is a 3D beam or a 2D one. </p>
<p>if you don't have access to that paper, please check the <a href="http://www.lstc.com/pdf/ls-dyna_theory_manual_2006.pdf" rel="nofollow noreferrer">page 73 on this link</a>.</p>
<p><a href="https://i.stack.imgur.com/XsWZb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XsWZb.png" alt="Beam schematic"></a></p>
<p>what confused me was the schematics on the original paper and the second link. in the beginning the general movements start from 3D but I don't see all the DOFs in the above image</p>
| |structural-engineering|structural-analysis|finite-element-method| | <p>It is a 3D beam (6 degrees of freedom at each node). The figure is 2D just to make things simpler.</p>
| 29895 | Is Belytschko beam defined for 3D or 2D? |
2019-09-02T02:47:03.250 | <p>I intend to make an oscillating motion from a simple computer DC fan (say 5V). </p>
<p>My question is: could a signal generator with a sinusoidal signal of (say voltage 5V and frequency of 5 Hz) be used to power such a DC fan to produce the oscillating motion ?</p>
<p>Second question:
it is related to <a href="https://engineering.stackexchange.com/questions/18427/oscillations-in-motor-control-systems">Oscillations in motor control systems</a>.
Will there be a damage for a low frequency and low voltage input to the fan ?</p>
<p>For additional info, the load is lightweight, and the mechanical rotation is about ±45 degrees.</p>
| |electrical-engineering|motors|prototyping| | <p>Yes and no, maybe.</p>
<p>Computer fans are specialized, cheap brushless motors, with their own built-in controllers. The ones that I took apart 30 years ago had 4-pole coil assemblies and an unknown number of poles in the magnets, and a 3-transistor circuit that actually made the rotation happen.</p>
<p>You should be able to convert one to a galvonometer by disassembling it and driving the coils directly, in two pairs, in opposition. Whether this will give you enough rotation is an unknown to me -- it depends on whether the magnet has two poles or six (either of which should work, for a fan).</p>
<p>As long as you don't drive the thing with too much current you shouldn't damage anything -- the two things to worry about are overheating the coils and demagnetizing the magnets.</p>
| 29919 | oscillation motion from DC motor (computer fan) |
2019-09-03T02:14:03.697 | <p>I know that the simple way to measure the voltage stored in a lead-acid battery is to simply measure the positive and negative using a voltmeter.</p>
<p>In my case, I think that my battery has a builtin charge controller and it is sealed like the picture below. How can I effectively measure it without opening?</p>
<p><a href="https://i.stack.imgur.com/0x7Hc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0x7Hc.png" alt="asure "></a></p>
<p><a href="https://i.stack.imgur.com/Gj8xr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gj8xr.jpg" alt="enter image description here"></a></p>
<p>If there is no way to measure it, how do I open this thing? I tried to open it using flat screwdriver and it leaves a dented mark.</p>
<p><a href="https://i.stack.imgur.com/Ix9wX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ix9wX.png" alt="enter image description here"></a></p>
| |battery| | <p>It is almost <strong>impossible to accurately know the value of charge stored in a battery</strong>. The amount of charge a battery can hold varies with battery life, battery use, charging method and environmental issues.</p>
<p><strong>Niel's</strong> explains nicely how to know if the battery is fully charged (with liquid electrolyte filled batteries) and <strong>Tim's</strong> tracking of the battery charge state by watching the battery voltage, and the current, likewise.</p>
<p>However, both are <strong>only indicators of how much charge is in the battery. Not of "How much charge the battery can hold"</strong> and, more to the point how long it will take to discharge.</p>
<blockquote>
<p>Battery Lifetime <a href="https://www.pveducation.org/pvcdrom/lead-acid-batteries/characteristics-of-lead-acid-batteries" rel="nofollow noreferrer">Characteristics of Lead Acid Batteries</a></p>
<p>Over time, battery capacity degrades due to sulfation of the battery and shedding of active material. The degradation of battery capacity depends most strongly on the interrelationship between the following parameters:</p>
<p>the charging/discharging regime which the battery has experienced
the DOD of the battery over its life
its exposure to prolonged periods of low discharge
the average temperature of the battery over its lifetime</p>
</blockquote>
<p>Measuring voltage is pointless as this will decrease as soon as load is applied. Even a nearly useless lead acid battery can, under certain circumstances, indicate a high voltage.</p>
| 29931 | How can I measure the amount of voltage stored in a sealed lead-acid battery? |
2019-09-03T20:11:37.337 | <p>Soil compaction can occur through heavy vehicle traffic. I attached a picture (see below). There are three samples approx. to 5m distance to each side of a paved road. Vehicle traffic is sprodically (few times in a year, e.g. at harvest).</p>
<p>Do you think at the sampling points in the soil (0-30-cm depth) compaction could have occurred through vehicle traffic? The blue area is untilled, i.e. more aggregate stability. The grey part is tilled, i.e. soil aggregate structure is loose.* </p>
<p><a href="https://i.stack.imgur.com/GfTVQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GfTVQ.png" alt="road runs between two fields"></a></p>
<p>*The thing is, I noticed very high accumulation of soil organic matter under the blue area which also had lots of vegetation one yr prior to sampling -- unlike the grey area which has been cropland for long time. It is said that compaction can suppress mineralization of soil organic matter because of lower aeration in soil which inhibits decomposer communities.</p>
| |pressure| | <p>30cm depth over 5m distance is approximately a cone with an angle of 3.5 degrees. This is outside the 45 degrees angle commonly assumed for influence of compaction or surcharge load. It means it could not benefit from traffic.</p>
<p>If anything this zone could have been negatively impacted. </p>
<p>According to Terzaghi this will by the bulging area of compression influence plane. </p>
| 29946 | Does this soil get compacted by occasional heavy vehicle traffic? |
2019-09-04T09:55:21.567 | <p>A room 10"x10"x17" lxbxh respectively to be heated 397F if a heater is placed inside the room (2 kw) how many heater are to be required to heat the room from room temperature to 397 F in 1 hour.</p>
| |thermodynamics| | <p>You miss a piece of key information: the "beginning" temperature of the room. Oh, and for the love of God, what is wrong with SI units that you Americans never use them?</p>
<p>Converting units</p>
<pre><code>10"=0.254m
17"=0.4318m
397F=202.778C=475.778K
</code></pre>
<p>Ok, from converting units makes me wonder the sanity of yours and your teacher. But considering that this is an Engineering SE, and we Engineers are never known to be sane in the first place...</p>
<p>Assumption 1: The heater is just a single point - A room barely reaches 0.5m (just around your thigh level) will not be enough to put any regular heater in. Therefore, just assume this for a calculation exercise like this</p>
<p>Assumption 2: Humidity is ignore</p>
<p>Assumption 3: Ideal gas</p>
<p>Assumption 4: Begin temperature as normal, or 25C, 298K</p>
<p>The temperature change is
<span class="math-container">$\Delta T=475.778-298=177.778K$</span></p>
<p>The heat capacity of air: according to <a href="https://www.ohio.edu/mechanical/thermo/property_tables/air/air_cp_cv.html" rel="nofollow noreferrer">this site</a>, the heat capacity changes according to temperature. The same goes for density from <a href="https://en.wikipedia.org/wiki/Density_of_air" rel="nofollow noreferrer">a reference source</a></p>
<p>But, writing them in full would be long and tidious (and I doubt such is the case for a simple question like this), so just assume the 2 values would not be changed, therefore, <span class="math-container">$c=1.003$</span> and <span class="math-container">$\rho =1.225$</span></p>
<p>The mass of the air is
<span class="math-container">$m=\rho * V = 1.225*0.254*0.254*0.4318=0.034126kg$</span></p>
<p>The total heat energy required (assumption number 6: No heat loss) is
<span class="math-container">$Q=m*c*\Delta T=0.034126*1.003*177.778=6.085kJ$</span></p>
<p>For one hour heating, the net heat output power is</p>
<p><span class="math-container">$P=Q/t=6.085/3600=1.69*10^{-3}kW=1.63W$</span></p>
<p>So, this is significantly smaller than the heat output of 2kW of the heater. Something is wrong, either with my series of assumption, or with the way this question is given.</p>
| 29961 | How many heaters needed |
2019-09-04T20:43:07.697 | <p>This question, in short, is: Can a gas be compressed with high pressure liquid?</p>
<p>As a scenario, imagine a tank full of atmospheric gas at STP. Could something like a pressure washer pump (in this case water, but generally any) liquid into the tank thereby compressing the gas inside the tank? Pressure washers often reach pressures that are quite high (easily 2000+ psi depending.) This seems like a cheap (if inefficient) way to get high pressure gas.</p>
<p>I haven't been able to find any material on this idea, and I find it highly unlikely that I am the first person to think this. What this indicates to me is that this is NOT a good idea but the reason is not obvious to me.</p>
<p>If the reason happens to do with the choice of gas/liquid, would this be solved with different materials? For instance, if the water is a problem for some reason could something like ln2 be used to compress nitrogen gas in some sort of specialized pump that could handle those type of temperatures?</p>
| |compressed-air|compressors| | <p>This is common in any farm water system. A pressure tank is on a side branch of the line from the well to the farm water system. At the cut-in pressure the well pump starts, and pushes water into the tank until the system reaches the cut out pressure. Typically the two pressures are about 20 psi apart. There is some adjustment available.</p>
<p>Water would absorb air under pressure, so some mechanism is in place to regulate the amount of air in the tank. This was usually some combination of a float valve that released excess air, and a bleeder valve in the top section of well pipe that led some air back in the pipe so when the pump started, a slug of air was included. I'm sure there are other ways. </p>
<p>Failure of either of these would result in a waterlogged tank. The well pump would cut it, almost immediately pressureize the system, cut off, and repeat as often as several times a second. Hard on pump.</p>
<p>More commonly now there is either a bladder or a diaphragm in the tank to separate the water and the air. These require no additional components. Usually the tank is pressurized to about half the max pressure of the system.</p>
<p>I have two in my farm system, one at the house and one near the far end of the irrigation line 1500 feet away. A third of a mile of moving water has considerable momentum. My solenoid operated valves were taking a beating when turning the water off. I was popping connections from the pressure surge. Adding a 25 gallon pressure tank meant that the surge would just overpressurize the tank somewhat. </p>
<hr>
<p>Water compressing air is used in hydraulic rams. The momentum of a pipe full of water is halted with a 'clacker' valve. The moving water loses it's momentum rushing into a air containing side chamber. That air pressure then is used to move a small amount of the water to a higher elevation. A well tuned system (Air volume, large in stream pipe, and small diameter discharge pipe have approximately equal volumes) can pump water to surprising elevations. More info here: <a href="https://en.wikipedia.org/wiki/Hydraulic_ram" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hydraulic_ram</a></p>
<hr>
<p>water pumped air can also be used as a way to generate energy. Consider a vessel with an open bottom and two one-way valves on top. When a wave passes it compresses the air in the vessel. A valve opens and that compressed air goes to a storage tank. The wave falls, that valve shuts, and another one opens admitting air into the top of the tank. </p>
<p>The compressed air can then be used to operate a turbine. The efficiencies of this are low. 6 foot waves would only generate about 3 psi. There are better ways to extract energy from waves.</p>
| 29974 | Compressing gas with liquid |
2019-09-04T20:46:43.340 | <p>I have a pellet smoker, and I am customizing it for tighter temperature control. Currently, I am controlling how much material is burning to control temperature using a PID loop. I want to achieve tighter control by also slowing down or speeding up the fan in response to temperature changes. I'm looking to find an algorithm that would allow these two to work together.</p>
<p>A pellet smoker works by using an augur to feed wood pellets into a burn chamber. Temperature is controlled by turning the augur on or off to control the amount of material that is burning in the burn chamber, and varying the speed of a fan to control how much air is flowing to the burn chamber.</p>
<p><a href="https://i.stack.imgur.com/jqFtA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jqFtA.png" alt="Traeger Pellet Smoker example"></a></p>
<p>Currently, I am controlling temperature with a PID loop controlling only the augur motor. The fan is running at a fixed speed of 90% of its maximum.</p>
<p>This leads to an acceptable temperature range, but I would like to achieve tighter temperature control by also using a PID loop on the fan speed rather than running it at a fixed speed.</p>
<p>I can control the fan with a completely independent PID loop. But I was hoping someone can point me towards and algorithm that would get them working together since they are pretty tightly coupled, and the fan influences temperature faster, but also doesn't work at all if the augur isn't feeding enough material.</p>
| |control-engineering|pid-control| | <p><strong>Controlling temperature using PIDs on two separate controllers</strong></p>
<p>Currently, I am controlling how much material is burning to control temperature using a PID loop. I want to achieve tighter control by also slowing down or speeding up the fan in response to temperature changes. I'm looking to find an algorithm that would allow these two to work together.</p>
<hr>
<p><strong>So what is the real question are you looking for an algorithm to use as a second separate controller?</strong></p>
<p>There is one very basic problem with this approach. One controller will influence the other controller so you will have an unstable control cycle. For example as the fan reduced speed to reduce the temperature the argue speed would increase to raise the temperature back to it's original point, the fan would then reduce speed further to compensate.</p>
<p>This is a common problem in air conditioning where both humidity and temperature control are needed one fights the other for dominance. Although the example you have given is more complicated some of the basics remain the same. You cannot use two independent PID controls. A special set of circumstances exist in which one control (I will call it a channel) needs to have priority control.</p>
<p><strong>What is required is a two channel PID controller with a priority control option.</strong></p>
<p>To be honest I have never come a cross an algorithm for this that works successfully except in a dedicated controller like Eurotherm or someone similar would manufacture. 20 years ago they cost over 2000 usd. (you are paying for the development costs). Today I have no idea of the costs. Algorithms for PLC's from Siemens and the likes are next to useless and my old company asked a university to estimate a time and cost to develop such an algorithm They estimated 2 years at which point the cost became irrelevant and the idea was scraped.</p>
<p><strong>Your problem is infinitely more complexed</strong> than trying to control Temperature and Humidity. They are just two independent states that are physically related. So you can control each and just give priority to one. What you are trying to do is <strong>create 2 channels which are both based on temperature</strong> so even though one channel could be given priority it will still produce conflict with the second channel and the result would be a totally unstable control cycle. I think even Einstein would have considered this a considerable challenge.</p>
<p>The only way forward that I can see is to decide on an alternative control strategy that provides two independent control channels, or scrap the idea.</p>
<p>However there is some light at the end of the tunnel If you had addressed this question to a cooking web site they may have given you better advice.</p>
<h2><strong>A smoker with little airflow</strong>, like some electrics, <strong>creates a larger stagnant bubble of air protecting the meat from smoke</strong>. <strong>Rapid, random air movement breaks up the bubble. For example, a strong convection fan.</strong></h2>
<p><a href="https://genuineideas.com/ArticlesIndex/srasmokeparticles.html" rel="nofollow noreferrer">Without the smoke, you might as well be grilling.</a></p>
| 29975 | Controlling temperature using PIDs on two separate controllers |
2019-09-04T20:49:12.270 | <p>A client has asked whether a container can be placed on an existing unreinforced concrete foundation of depth 100 mm. The grade of the concrete is unknown. How can the capacity (in kPa) of this plane concrete foundation be evaluated?</p>
| |civil-engineering|foundations| | <p>In order to determine the capacity, you will need to know a couple of things.</p>
<ol>
<li>Dimensions of the pad and container</li>
<li>Bearing capacity of the soil under the slab. You would normally receive this from a geotechnical report.</li>
<li>The strength of the concrete. This can be determined either through core testing or something called a "schmidt hammer". The schmidt hammer can give you a ball park idea of the strength of the concrete. However you need to calibrate it against something of a known strength first.</li>
<li>You will want to check the concrete pad for crushing, shear. You local building code should provide guidance on capacity checks for both bearing and shear.</li>
<li>You will also want to look at unbalanced loads from the container, and potentially the container not being centered on the pad.</li>
</ol>
<p>Note that many parts of the world will consider plain concrete (unreinforced) to be non-structural in nature. That does not mean that it can't carry a load, just that its relatively weak compared to the structural loading reinforced concrete tends to take. </p>
| 29976 | Capacity of plain concrete foundation on grade [Civil UK] |
2019-09-05T00:44:32.847 | <p>I have lubricated the ball-bearings of two hand spinners (the anti-stress toys) and contrarily to my expectation the time they keep on spinning has been divided by four. </p>
<p>One fact I have noticed, the rolling bearings now makes almost no more noise at all: they were really noisy before.</p>
<p>I am trying to explain myself why/how the energy in the two hand-spinners would dissipate so much faster ? It is contrary to my intuition.</p>
<p>Any idea ?</p>
<hr>
<p>When I say divided by 4, it is an estimation:</p>
<p>I have an anti-smoking anti-stress technique that involves a short meditation, and I was using two hand-spinners, one in each hand for a duration of roughy 2 minutes, a bit more, (repeated numerous time for one year or so). </p>
<p>But after lubricating them this afternoon with WD-40, the duration falls to 30 seconds at most.</p>
| |bearings| | <p>The lubricant you put into the spinner bearings has both viscosity and surface tension, and it coats the rotating parts inside the bearing. Therefore, as the spinner bearing revolves, it churns the lube and makes and breaks air/lube interfaces- which dissipates energy and makes the spinner slow down. </p>
<p>This energy dissipation mode is absent in the case of a dry bearing, in which case the main energy dissipation mode will be radiating away the noise generated when the bearing parts rattle around as the bearing spins. </p>
<p>This illustrates the fact that in a roller- or ball-bearing, the purpose of the lubricant inside them is not to reduce friction (the rolling elements do that) but to inhibit wear caused when the balls or rollers skid against the bearing races instead of rolling smoothly. The lubricant adds a little bit of frictional drag but the tradeoff is worth it.</p>
| 29978 | How/why lubricating a rolling bearing would make them spin less time (dissipate energy faster)? |
2019-09-05T05:56:27.740 | <p>During truss fabrication, the truss member are usually connected by welding does behave like fixed joints not pin joint so how will that behave like truss without pin connection?</p>
| |mechanical| | <p>The geometry of the trusses are usually long narrow members carrying axial loads,connected to each other in triangles. And the connections are designed to let the axis of truss members pass trough a common center on the connection plate or connection bolted bracket. </p>
<p>Because of this geometry there is no moment at the connections, even if they are rigidly connected. Therefore the connection can be assumed to be a pin connection.</p>
<p>There may be very small tertiary moments due to heat expansion or deflection but those are well within the tolerance range of the truss.</p>
<p>That doesn't mean if a truss will be designed for extreme temperatures or large deflections the effect can be ignored. </p>
| 29984 | Practical consideration of truss |
2019-09-05T09:52:38.810 | <p>For a DC motor, speed of the armature is inversly proportional to current.</p>
<p>But at the same time, we know that torque of the armature is directly proportional to the armature current. Since speed can be increased by applying torque, this means that speed is directly proportional to armature current.</p>
<p>How is this possible ?
Where have I gone wrong ?</p>
| |motors| | <p>For a DC motor, speed of the armature is inversely proportional to current.</p>
<p>But at the same time, we know that torque of the armature is directly proportional to the armature current.</p>
<p>Since speed can be increased by applying torque, this means that speed is directly proportional to armature current.</p>
<p><strong>How is this possible ? Where have I gone wrong ?</strong></p>
<hr />
<p><strong>Where you have gone wrong 1</strong>
Firstly your question is poorly defined you should clearly state what type of equipment you are referencing. There are numerous types of D.C. Motors and variations on how the excitation can be achieved. Therefore I will ignore differential compound D.C. Motors.</p>
<hr />
<p><strong>Where you have gone wrong 2</strong> is your statement Since <strong>speed can be increased by applying torque</strong>. <a href="http://lancet.mit.edu/motors/motors3.html" rel="noreferrer">Torque speed curve</a></p>
<p>In fact <strong>this is the exact opposite</strong> of what happens. <strong>As speed increases the torque reduces.</strong> As the the load is reduced and torque approaches zero the armature speed reaches maximum. because <strong>AS THE SPEED INCREASES THE BACK EMF INCREASES AND THEREFORE THE CURRENT REDUCES.</strong> so the statement <strong>speed of the armature is inversely proportional to current</strong> is correct.</p>
<blockquote>
<p><a href="https://www.electronics-tutorials.ws/inductor/inductor.html" rel="noreferrer">Tutorials</a> emf = inductance x rate of current change” and a circuit has an inductance of one Henry will have an emf of one volt induced in the circuit when the current flowing through the circuit changes at a rate of one ampere per second.</p>
<p>One important point to note about the above equation. <strong>It only relates the emf produced across the inductor to changes in current</strong> because if the flow of inductor current is constant and not changing such as in a steady state DC current, then the induced emf voltage will be zero because the instantaneous rate of current change is zero, di/dt = 0.</p>
</blockquote>
<p>Therefore it is obvious that as the speed of the armature increases so the rate of change of flux will increase and so the back emf also increase.</p>
<hr />
<p><strong>However</strong> if you were to say that the <strong>speed of a Electric locomotive can be increased by applying torque</strong> that is correct, at least in the initial stages of movement.</p>
<hr />
<p><strong>Where you have gone wrong 3</strong></p>
<p>Your statement torque of the armature is directly proportional to the armature current is also incorrect <a href="https://www.yourelectricalguide.com/2017/09/dc-motor-characteristics-and-applications.html" rel="noreferrer">electrical guide</a></p>
<p>In fact <strong>T α Iaφ</strong> Torque α Armature current x Field flux and thus <strong>the equation T= I? will depend on what type of motor you are referencing.</strong></p>
<blockquote>
<p>On light loads, the <strong>torque produced by the series motor is proportional to the square of armature current</strong> and hence curve drawn between torque and armature current up to magnetic saturation is a parabola. But after magnetic saturation flux φ is independent of excitation current and so torque is proportional to Ia and hence characteristics become a straight line.</p>
</blockquote>
<p>Whilst</p>
<blockquote>
<p>The flux of a <strong>shunt motor</strong> is practically constant. Therefore, <strong>T α Ia</strong> Torque produced is proportional to the armature current</p>
</blockquote>
| 29985 | Contrradicting answers for relation of speed of a motor with current |
2019-09-07T12:01:49.800 | <p>From the finite element theory, what is remained a bit unclear for me, is that how do we know the location of gausian points where the integration is done for each specific type of element ? it is not indicated in many FEM books, but if someone could explain it here , it would be fantastic</p>
| |mechanical-engineering|applied-mechanics|finite-element-method| | <p>If you are using standard Gauss quadrature, from the formulation, you determine the coefficients (weights of Gauss location) and their locations. You cannot arbitrarily decide their locations. Here are some examples: <a href="http://edwilson.org/BOOK-Wilson/G-inter.pdf" rel="nofollow noreferrer">http://edwilson.org/BOOK-Wilson/G-inter.pdf</a></p>
| 30017 | How are the location of Gausian points determined in Finite element theory? |
2019-09-09T12:35:28.827 | <p>This site:(<a href="https://www.motioncontroltips.com/faq-what-are-dc-shunt-motors-and-where-are-they-used/" rel="nofollow noreferrer">https://www.motioncontroltips.com/faq-what-are-dc-shunt-motors-and-where-are-they-used/</a> ) explains the speed regulation of DC shunt motor as follows: If the load on the motor is increased, the armature rotation slows and back EMF is reduced, since back EMF is proportional to speed. With less back EMF voltage and a constant supply voltage the net voltage increases. The increase in net voltage results in an increase in armature current. Since torque is proportional to armature current, torque also increases. Finally, this increased torque allows the motor to increase its speed and compensate for the slowdown due to loading. Hence, a DC shunt motor is able to self-regulate its speed, and can be referred to as a constant speed motor.</p>
<p>Considering that torque is inversly related to speed for a DC shunt motor, wouldn't this entire explanation be wrong as the last line mentions increase in speed due to increase in torque ?</p>
| |motors| | <p>Considering that torque is inversely related to speed for a DC shunt motor, wouldn't this entire explanation be wrong as the last line mentions increase in speed due to increase in torque ?</p>
<hr />
<p>torque is inversely related to speed</p>
<p>This means</p>
<blockquote>
<p>The <strong>maximum Torque</strong> is applied at <strong>Zero speed</strong>, because torque is inversely related to speed.</p>
</blockquote>
<p>However</p>
<blockquote>
<p><strong>as speed increases, Torque reduces</strong></p>
</blockquote>
<p>until</p>
<p><strong>The required design speed and it's associated value of Torque are reached.</strong></p>
<p>If additional load is applied <strong>speed reduces</strong></p>
<blockquote>
<p><strong>as speed reduces, Torque increases</strong></p>
</blockquote>
<p>Which then takes us back to where we started. The increased Torque will accelerate the speed, similar to what it did on starting, only in this case we are not increasing from zero or stall speed. But some point slower than the designed speed of the D.C. Shunt motor.</p>
<blockquote>
<p><strong>as speed increases, Torque reduces</strong></p>
</blockquote>
<p>until</p>
<blockquote>
<p>The required speed and it's <strong>New associated value of Torque are reached</strong>.</p>
</blockquote>
<p>Therefore if you relate this to the explanation. And view them as <strong>steps in a process not simultaneous events</strong>,</p>
<p><strong>The explanation is correct.</strong></p>
<blockquote>
<p>this increased torque</p>
<p>allows the motor to increase its speed and compensate for the slowdown due to loading</p>
</blockquote>
| 30046 | Isn't this explanation of speed regulation in a DC shunt motor wrong? |
2019-09-09T12:46:40.397 | <p>I'm planning a thought project for measuring the C02 in a room (at office namely), and communicating its results to a computer. </p>
<p>But I'm blocked at the early stage: which sort of sensor and which is the mechanism to measure the CO2 concentration in air ?</p>
<p>I saw devices for sale, not communicating with computers, only LCD screens, and there were no indications on what is the system/mechanism used.</p>
<p>Any idea ?</p>
| |gas| | <p>I have used NDIR sensors, such as these:
<a href="http://www.alphasense.com/index.php/products/ndir-safety/" rel="nofollow noreferrer">http://www.alphasense.com/index.php/products/ndir-safety/</a></p>
<p>It's based on a light source which is flashed at a regular frequency and reflected on a series of mirrors before ending on a receiver. The sensor provides 2 signals: one "reference" signal, which remains constant regardless of the % of CO2, and one "active" signal which changes with the % of CO2 in the air. You have to calibrate the sensor with different % of CO2, and you can compensate for temperature and pressure variations.</p>
| 30048 | What kind of sensors to be used to measure the carbon dioxyde concentration? |
2019-09-09T14:56:34.357 | <p>I’m trying to find out if it’s possibly to form a cylinder or tube with a diameter of 1 inch using some kind of curling die for a press brake. I’ve seen dies that make small 360 degree bends like that but they always have additional length of sheet metal extending from the tubular section. Does anyone know if it’s possible to form just a small tube without additional material extending out? This tubing would be used for handles to lift something, similar to a handle bar on a bicycle or something. </p>
| |mechanical-engineering|materials|design|steel|industrial-engineering| | <p>You can do - the process called "bumping". You can see how it works - with the backgauge of the press brake you push the metal and perform a lot of bends.</p>
<p>This is the process to make light poles in industry. You will never get small diameters as 1 inch.</p>
| 30054 | Press brake tooling for 360 tube? |
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