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2020-03-21T11:37:35.150 | <p>If you let a color TV display something from an unplugged display port, it displays snow, right? Randomly varying black or white pixels on the whole screen.</p>
<p>My guess was that the television was decoding the noise from the unplugged wires as a video signal.</p>
<p>Assuming this, it makes sense that B&W TVs make B&W snow (they would decode the noise as being a B&W video signal). But then <strong>color TVs</strong> should make <strong>colored snow</strong>, right? Shouldn't they decode the noise as being a color video signal?</p>
| |electrical-engineering|signal-processing| | <p>The question was well answered by "Transistor" on Mar 21 at 14:31. The only part missing was that the other colour tv systems of the past, known as PAL (Phase Alternate Line) and SECAM (Séquentiel Couleur à Mémoire) worked in a very similar fashion, both had a colorburst signal at 4.43 MHz, superimposed on a blanking portion of a horizontal synchro pulse. PAL with phase alternating between 135° and 225° from line to line. SECAM did not care about the phase since it used FM modulation for colour information.
The important part is that all the analog tv receivers had a circuit designed to detect the presence of color signal, with noise immunity at the top of the requirements list.
So when the signal is very week or noisy (or missing) the analog color tv set reverts back into a black and white mode.</p>
| 33737 | Why do color TVs make black and white snow? |
2020-03-21T14:13:01.080 | <p>How does the sine part i.e. <span class="math-container">$\dfrac{\ell\sin(\theta)}{2}$</span> appear in the calculation of shear flow (<span class="math-container">$q$</span>)? I mean, if I'm not mistaken, the sine part is the distance from the x-axis to point D, but how is it derived? </p>
<p><a href="https://i.stack.imgur.com/dyg1G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dyg1G.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|beam|solid-mechanics| | <p>The calculation you are showing concerns the shear force in the top flange. The shear stress is calculated at a horizontal distance of <span class="math-container">$l$</span> from B. To use Zhuravskii's shear stress formula, you need the first moment of area of that part of the top flange with length <span class="math-container">$l$</span>. It has an area of <span class="math-container">$tl$</span> and its centroid has a distance to the <span class="math-container">$X$</span>-axis of the distance from the axis to B, <span class="math-container">$|Y_B|$</span>, plus the distance from B to the center of the bit of flange measured parallel to the <span class="math-container">$Y$</span>-axis, i. e. <span class="math-container">$\frac{1}{2}l\sin\theta$</span>. So that gives a term of <span class="math-container">$tl\left(|Y_B|+\frac{1}{2}l\sin\theta\right)$</span></p>
| 33738 | Shear flow for unequal leg channel |
2020-03-22T20:14:15.250 | <p>Consider we have a second order system (plant) described by the following transfer function:</p>
<p><span class="math-container">$$ P(s) = \frac{b_0}{s^2+a_1*s+a_2} $$</span></p>
<p>which is controller by a PD-controller:</p>
<p><span class="math-container">$$ C(s) = K_p+K_ds $$</span> </p>
<p>In order to study the impact that the external noise has to the system, I have come across the so called noise sensitivity function which is described by the below equation:</p>
<p><span class="math-container">$$ N(s) = \frac{C(s)}{1+P(s)C(s)} $$</span> </p>
<p>However, deriving <span class="math-container">$\ N(s) $</span> produces an improper transfer function with degree of numerator being <span class="math-container">$\ 3 $</span> and of denominator being <span class="math-container">$\ 2 $</span>. This results, for example, to the imcapability of obtaining a step response in order to study the behaviour. Is the definition of the noise sensitivity function wrong or am I missing something ? </p>
| |control-engineering|control-theory|transfer-function|noise| | <p>This is because your controller is also improper. A common way to correct this is to add a low-pass filter to the derivative (effectively making it a high-pass filter). It can be noted that the PD controller could then also be seen as a lead-lag filter.</p>
<p>However, if you are only interested in the step response you could also use that a step is the integral of an impulse. So you could add an integrator to the system, which raises the order of the denominator, and use impulse instead of step on the causal modified system.</p>
| 33758 | Noise sensitivity transfer function |
2020-03-23T20:33:36.107 | <p>I should have produced a custom designed bottle, but metal molds are some expensive and because we don't require too many products (we just need 100-200 piece), I need a way to reduce the cost. </p>
<p>So I thought that, maybe I can use wooden mold for this. But I don't have experience about that. Is there anyone who has tried/used this method or anyone who has knowledge about that? Is it correct method?</p>
<p><strong>Specs</strong>: </p>
<ol>
<li>Bottle thickens can be same as any water bottle (you can see examples <a href="https://www.google.com/search?q=5%20Liter%20Natural%20Rectangular%20bottle&sxsrf=ALeKk00WlBGtRl2qDDokZT2_JJ0tzVzVpA:1584994438295&source=lnms&tbm=isch&sa=X&ved=2ahUKEwixpNyztLHoAhUTj3IEHS_8C6AQ_AUoAXoECGMQAw&biw=1532&bih=673" rel="nofollow noreferrer">in this search</a>).</li>
<li>Color can be clear or white (clear is preferred)</li>
<li>Some inlet and outlet necks needed (so mold needs to be some complex)</li>
<li>Bottle will be 5 to 7 liters with the shape of like rectangular prism</li>
<li>Material will be derivative of polyethylene probably</li>
</ol>
| |product-engineering|molding| | <p>You will have some of the same considerations of creating a mold when using wood as when using metal. </p>
<p>Underhangs/undercuts are to be avoided.
Corners are to be radiused, no sharp edges.
Alignment of mold halves must be ensured.</p>
<p>For your listed considerations, it doesn't appear that a wood mold, especially one made from hardwood would be a problem. With such a small item count, one can expect that you also do not need full automation. Manual processing of the bottle making will certainly save money.</p>
<p>Consider that you'll need someone to create the mold halves for you. If you have an existing positive model, it can be used to create the tool paths for a CNC router, although the size you've referenced would mean a router with a substantial thickness capacity. Hobby CNC routers are often limited to around 100 mm or so.</p>
<p>Finding someone to make the molds in wood will be your most challenging component.</p>
<p>Added below in response to comment:</p>
<p>Manual processing means opening the mold by hand, inserting the envelope of plastic, closing and clamping the mold and activating the injection of air. Placing the mold into a heated environment is also included in this sequence, as is removing it and de-molding the part.</p>
<p>This is to contrast an automated processing system where the mechanicals are constructed to be mostly hands-off. Push the button and everything happens at hundreds of parts per hour.</p>
<p>With respect to the type of wood, you'll likely have to experiment. Close-grained hardwoods will be strongest and most heat resistant as well. You'll have to test for heat based on the melting point or glass-transition temperature of your selected plastic. As with metal molding, pre-heat the mold prior to inserting the plastic, use appropriate safety gear.</p>
<p>Mold lifetime and other durability considerations will again depend on your selection of wood, as well as local availability of this material. As you've noted, you have CNC capability, which means a dozen molds aren't going to be much more expensive than one, in terms of material cost.</p>
| 33776 | Can wooden mold be used for blow molding? |
2020-03-23T22:35:20.363 | <p>My generator (a Fermont MEP-803A, 4 cyl diesel, 10kW rated) can run in three distinct modes:</p>
<ul>
<li>Single-phase 120V, in which one terminal is hot, and another is neutral</li>
<li>Single-phase 120V/240V, in which two terminals are hot, and another is neutral. The difference between the terminals is 240V, and each one to neutral is 120V.</li>
<li>Three-phase 208Y, in which three hot terminals are each 208V apart from each other, and each one is 120V from the neutral terminal.</li>
</ul>
<p>I need to run a bunch of lights that all need 120V. So the obvious thing would be to just use the 120V mode. But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals.</p>
<p>My thinking, based on my coarse understanding of generators, is that in three phase operation, the torque should be more homogeneously distributed over the course of a single turn of the engine, so there should be less vibration and more equality in effort among the cylinders. In short, I'm guessing that three-phase generator operation would have the same advantages three-phase motors have over single-phase motors.</p>
<p>Is my thinking correct here? If not, why not? Conversely, are there any disadvantages to three-phase operation over single-phase operation?</p>
| |torque|vibration|energy-efficiency|generator|diesel| | <p>While the other answers here are compelling, they're only half the truth. If you consult your manual you'll find that the generator load windings are reconfigured when using the mode selection switch to one of three output modes.</p>
<p><a href="https://i.stack.imgur.com/IheLc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IheLc.png" alt="enter image description here"></a></p>
<p>In either mode, the full output of the generator is available, either as 104A@120V(1ph, one circuit), 52A@120/240(1ph/split, two circuits), or 34A@120/208(3ph, three circuits). </p>
<p>If you select position #2 (1ph @ 120V) then you can place your entire load between L3 and LO without worrying about balancing anything.</p>
<p>The only reason to select three-phase mode is if you actually have a three phase load to drive. It has no other effect on the available power of the generator, but using it in that mode does require you to manually balance your single phase loads. Choose the mode that fits your load. If you only need 120V 1ph service and you need all 104A for single phase 120V loads then choose that mode. </p>
<p>In fact, if you run single phase loads in 3-phase mode or in 120/240 split phase mode then you <strong>DO</strong> have to worry about keeping the phases reasonably balanced yourself by evenly distributing loads to the L1/L2/L3 terminals (in 3-phase mode), or to the L1/L3 terminals (in 120/240V split phase mode).</p>
<blockquote>
<p><strong>Operator's manual, 2-18</strong> </p>
<p>When using single phase connections, always attempt to balance loads between terminals(do not connect all loads between one terminal and LO). Failure to observe this caution can result in damage to generator set.</p>
</blockquote>
<p>If you have a 3ph load you need to power, obviously you need to select three-phase mode, and a 3-phase load will naturally be balanced if you're not adding any single phase loads on top of it. If you're only driving single phase loads, save yourself the headache and just use mode position #2 (1ph, 120V). That way all your loads simply go between LO and L3 and you don't have to worry about balancing anything.</p>
<p>As always, the lesson is - <a href="https://greenmountaingenerators.com/manuals-and-support/mep-803a-manuals/" rel="nofollow noreferrer">Read your manuals</a>.</p>
| 33778 | Three-phase versus single-phase operation of a generator |
2020-03-24T03:53:05.707 | <p>I have a well with a 1.5 HP pump. The pump is rated for 240V input. At the moment, the only nearby available power is 4-wire 208Y 3phase. I am considering connecting two of the three phases to the well pump, so it receives 208V where it is expecting 240V.</p>
<p>To what extent, and by what mechanism, will it hurt the pump if it is powered by 208V instead of 240V? Are there warning signs I can observe (like something electrical I can measure, or something I can observe about the flow of water) that would indicate whether this lower voltage is causing problems?</p>
<p>More generally, when an AC motor says it's rated for a certain voltage, without specifying a range, how tight is that calibration? Like, what is the "probably not much worse" voltage range? My intuition says "ohh eee ah around 15%" but I'm realizing I have no actual basis for that sentiment.</p>
| |electrical-engineering|motors|torque|pumps|current| | <p>The 1.5 HP motor is built to draw a specific amperage at full load at 240V. Because power equals voltage times current, a lower than rated voltage will draw more amps through the motor. So if your supply is 208, your current will increase by a ratio of 240/208, or about 15%. Low voltage/high amps is a common failure mode for hand power tools being used with undersized extension cords.</p>
<p>If the motor is under a light load, you will likely never notice. But if it is running at full load, you might find it overheating, tripping overloads, or just generally failing early.</p>
| 33782 | Running a well pump rated for 230V on 208V power |
2020-03-24T15:39:31.853 | <p>I'm new to Control Systems. I've been given this block diagram and I would like to find the overall transfer function. How can I do it?
I was thinking about simplifying the diagram by placing in parallel C(s) and R(s), but I don't think it is the right approach.</p>
<p><a href="https://i.stack.imgur.com/e1WI7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e1WI7.png" alt=" "></a></p>
| |control-engineering| | <p>This is actually quite simple when one understood the concept, I'll try to make the start and you can solve the rest from there on your own.</p>
<p>You want to get a function that describes <span class="math-container">$y$</span> in terms of <span class="math-container">$v$</span>, so you just work your way back from the end. Let's call the edge that goes into the <span class="math-container">$G(s)$</span> block <span class="math-container">$x$</span>.
Then we can write for the last part of the block diagram: <span class="math-container">$$ y = G(s) \cdot x$$</span>
Now we can move on further through the diagram, by replacing <span class="math-container">$x$</span> with what it actually is, a sum of the edges that come out of <span class="math-container">$C(s)$</span> and <span class="math-container">$R(s)$</span>. From here it's mostly rinse and repeat. </p>
<p>Eventually you will have a function that contains only your input and, in case of a feedback like here, also your output, which you can then solve for <span class="math-container">$y$</span>. This yields the transfer function.</p>
| 33790 | Find the transfer function of a basic block diagram |
2020-03-25T12:12:04.293 | <p>The stiffness of a uniformly loaded cantilever beam is given as follows:</p>
<p><a href="https://i.stack.imgur.com/sBiyI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sBiyI.png" alt="enter <imag{e description here"></a></p>
<p>I am <strong>trying to derive the stiffness</strong>, but I am facing a problem. I would be glad, if someone could help me out.</p>
<p><strong>What I have done so far:</strong> </p>
<p>From my understanding, the stiffness <span class="math-container">$k$</span> is defined as:
<span class="math-container">$$k = \frac{F}{\delta}$$</span>
where <span class="math-container">$F$</span> is the force that moves the beam at a certain point by <span class="math-container">$delta$</span></p>
<p>Hence, I calculate the tip deflection and the equivalent point load at the tip and use those two values to get the stiffness.</p>
<p>I know that the deflection curve is:</p>
<p><span class="math-container">$$y(x) = \frac{p}{24EI}(6L^2x^2-4Lx^3+x^4)$$</span></p>
<p>The tip deflection becomes: <span class="math-container">$$\delta_{tip} = y(L) = \frac{p}{24EI}(6L^4-4L^4+L^4) = \frac{pL^4}{8EI}$$</span></p>
<p>For the equivalent force at the tip, I do the following:</p>
<ol>
<li><p>I calculate the equivalent point load</p></li>
<li><p>Then I calculate the equivalent tip load from the point load (moment = 0 around fixed end)</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/vfd4h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vfd4h.png" alt="enter image description here"></a></p>
<p>But then I end up with the following stiffness:</p>
<p><span class="math-container">$$k=\frac{\frac{1}{2}pL}{\frac{pL^4}{8EI}} = \frac{4EI}{L^3}$$</span></p>
<p><strong>which is wrong...</strong> </p>
| |mechanical-engineering|statics| | <p>You came so close to the answer, you ran right past it!</p>
<p>As you've said, the stiffness coefficient is defined as force over distance: how much force you need to apply to deflect an object (a beam or a spring, for example) in a given way a certain distance.</p>
<p>As you correctly derived, the deflection at the point of a beam under a distributed load is</p>
<p><span class="math-container">$$\delta = \dfrac{pL^4}{8EI}$$</span></p>
<p>Now, that's actually basically the answer right there, actually. It's saying that if you apply a distributed load <span class="math-container">$p$</span> over a beam, it will deflect in a given way a certain distance <span class="math-container">$\delta$</span>. That's really close to our definition of a stiffness coefficient above, except that here we're talking about a distributed load instead of a force.</p>
<p>So, how can we convert this distributed load to a force? You actually tried too hard. All we have to do is calculate the total load <span class="math-container">$P$</span> created by the distributed load, which you correctly derived as</p>
<p><span class="math-container">$$P = pL$$</span></p>
<p>And we're done. There's no need to perform any sort of transformation to the force's location or anything. If we just plug that into the deflection equation, we get</p>
<p><span class="math-container">$$\begin{align}
\delta &= \dfrac{PL^3}{8EI} \\
\therefore \dfrac{P}{\delta} &\equiv k = \dfrac{8EI}{L^3}
\end{align}$$</span></p>
<p>Huzzah.</p>
<p>The thing to remember here is that stiffness coefficients are unique to the exact loading and boundary configuration given, and to the specific point they are calculated for. If you convert a distributed load into a concentrated one and calculate the resulting deflection, you'll get a different stiffness coefficient (after all, a beam under distributed loading deflects differently from one under a concentrated load). If you use that same beam with the same loading, but calculate the deflection at midspan, you'll get another stiffness coefficient. None of these coefficients are wrong, they're just correct for different cases.</p>
<hr>
<p>So calculating the stiffness coefficient is the easy step: calculate the deflection at the desired point, sum up the total force applied throughout the entire structure, and then divide the deflection by that.</p>
| 33802 | Derivation of the stiffness of a uniformly loaded cantilever beam |
2020-03-25T14:45:22.367 | <p>Before I start, sorry if this doesn't make sense English is not my first language. </p>
<p>In my country electricity is nationalized so there is only one energy company and one grid. When you build new house you just connect your house cables to the gird.</p>
<p>My question is in country/region where there are multiple companies providing energy, does each company have its own grid/cables/wires reaching homes? </p>
<p>And what happens if there is a new company, does it have to build it is own grid?</p>
<p>Are there multiple transmission lines for each company?</p>
| |electrical-engineering|electrical| | <p>In the UK, the grid is divided into the transmission network (high voltage; long distance) and the distribution networks (lower voltage networks; connects directly to the consumer and small generators). The transmission network is owned by National Grid (in England and Wales) and the distribution networks are owned by a different companies that operate in different regions.</p>
<p>Your electricity supplier doesn't need to own any physical infrastructure. They buy ahead of time the amount of generation they expect their customers to need on the electricity market. Physical location of customers and generators doesn't matter; the grid is treated like a collective "pool" of power and you have to put in as much in as your customers will take out. Wind forecasting allows wind farms to sell generation ahead of time.</p>
<p>National Grid then look at the generation profile and make adjustments to ensure that circuits are not overloaded, generation matches demand and various other constraints are met. Engineers in the control room can see prices for each generator to increase or decrease generation. The price for fossil fuel generators to decrease generation is usually negative because they save fuel but still get paid for the electricity they would've generated. For wind farms it is positive because they loose their renewable energy incentives.</p>
<p>Electricity flows into and out of the transmission network are all metered. A company called Elexon does the accounts with those readings. As a generator or electricity supplier, if your generation or demand didn't match what you contracted for, you pay money according to the difference. This money pays for National Grid to balance the system as above.</p>
<p>I don't know how Elexon work out how much your electricity supplier's customers consumed given that suppliers share the distribution networks. I assume it must be based on estimates for households and meter readings for large consumers.</p>
<p>Essentially owning infrastructure (power lines, cables etc.) is a separate business from being an electricity supplier.</p>
| 33804 | How do electric companies supply electricity to homes in countries where energy is privatized? |
2020-03-26T15:58:37.037 | <p>Currently I'm designing a tracked vehicle for a school project. It has to carry a weight and be able to overcome some obstacles which are about 3/4th the height of the tank. What I'm stuck at, is calculating the required torque for the motors.
<a href="https://i.stack.imgur.com/NsYWO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NsYWO.png" alt="Basic layout with variables. The upper right wheel is going to be powered."></a></p>
<p>This is the basic layout of the tank with some variables. The wheel in the upper right is going to be powered.
Given that there will be enough traction between the tracks and the obstacle, how can I accurately calculate the required torque for the motor of this tracked vehicle to climb the obstacles?</p>
<p>Thanks in advance!</p>
| |mechanical-engineering|torque|electric-vehicles| | <p>The max torque at the time of the climbing is</p>
<p>Let's call the power wheel radius .r.</p>
<p>T=mg*sin(theta)*r</p>
<p>After the track passes the hump of non powered wheel the tank will flip back, because it's CG falls back outside of the track footprint. </p>
| 33835 | Required torque of climbing tracked vehicle |
2020-03-27T04:51:43.590 | <p>I want to design a robust control system using the internal model design specifications. The block diagram is the one shown below:</p>
<p><a href="https://i.stack.imgur.com/nbo1B.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nbo1B.jpg" alt="enter image description here"></a></p>
<p>I am trying to obtain the transfer function <span class="math-container">$\ \frac{Y(s)}{R(s)} $</span> in order to acquire the characteristic polynomial of the closed loop system but for some reason I am stuck. This is what I have done so far:</p>
<p><span class="math-container">$$ E_{a}(s) = R(s) - Y(s) $$</span></p>
<p><span class="math-container">$$ U(s) = E_{a}(s)G_{c}(s) - KX(s) $$</span> ( <span class="math-container">$\ U(s) $</span>: control input to the process <span class="math-container">$\ G(s) $</span>)</p>
<p><span class="math-container">$$ Y(s) = U(s)G(s) = E_{a}(s)G(s)G_{c}(s)-KX(s)G(s) $$</span></p>
<p><span class="math-container">$\ $</span></p>
<p><span class="math-container">$$ Y(s) = R(s)G(s)G_{c}(s) - Y(s)G(s)G_{c}(s) - KX(s)G(s) $$</span>
<span class="math-container">$\ $</span></p>
<p><span class="math-container">$$ Y(s) = \frac{G(s)[R(s)G_{c}(s)-KX(s)]}{1+G(s)G_{c}(s)} $$</span></p>
<p>From this point, I really don't know how to continue. Obviously the term that confuses me is <span class="math-container">$\ X(s) $</span>. So, now how should I proceed and obtain the overall transfer function ?</p>
| |control-engineering|control-theory|transfer-function|robust-control| | <p>Well, after a while this is what I finally came up with. I have tested the results in simulation environment as well as on the real system and the produced control behaviour was what I expected. Firs of all, I rewrote the block diagram in the form shown below:</p>
<p> <a href="https://i.stack.imgur.com/F55SR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F55SR.png" alt="enter image description here"></a></p>
<p>The whole procedure of the transfer function is shown below:</p>
<p>The state variables of the plant process are <span class="math-container">$\ x_1, \ x_2 $</span> and they are equal to: </p>
<p><span class="math-container">$$ x_1 = y \rightarrow X_1(s) = Y(s) $$</span>
<span class="math-container">$$ x_2 = \dot{y} \rightarrow X_2(s) = sY(s) - y(0) $$</span></p>
<p>Considering zero initial conditions, the transfer function of the inner loop is (simple feedback system):</p>
<p><span class="math-container">$$ H(s) = \frac{G_p(s)}{1+k_1G_p(s)+sk_2G_p(s)} $$</span> </p>
<p>And now it all comes down to the outer loop being again a simple feedback loop:</p>
<p><span class="math-container">$$ T(s) = \frac{G_c(s)H(s)}{1+G_c(s)H(s)} $$</span></p>
| 33844 | How to obtain transfer function of control diagram with Internal Model Control? |
2020-03-28T10:46:17.077 | <p>I bought a 200x90 sq.cm. bed and assembled it. The bed frame is assembled out of hollow iron rods. After using for a few days I started having back pain. So I checked the bed and found that some of those rods got bent in the middle and that made the whole frame uneven. I want to strengthen it, however I don't have much stuff at hand, and we are in a lockdown due to coronavirus. </p>
<p>I have not thrown out the wooden planks from my older bed. So I thought putting them on the bars will distribute the load and may help it. My question is what is the pattern that will provide maximum strength? I arranged them in a way (see picture) and it already fills quite solid. I will appreciate any feedback. For information, I don't have an electronic drill machine or any such thing; also I don't know how to use them.</p>
<p>Thank you for reading, it is my first post here. :) Note: I also posted this question in home improvement stack exchange, just fyi.
<a href="https://i.stack.imgur.com/uouMA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uouMA.jpg" alt="enter image description here"></a></p>
| |statics|stability| | <p>I would take the wooden slats from the old bed and cut them so they can stand on end and be trapped between the steel tubing and the floor, thereby supporting the load that the steel tubing by itself cannot. If you wish, you can cut a round notch in one end of the slat so as to capture the tubing; this will help prevent the slats from popping out of position. </p>
| 33860 | Strengtheing a weak bed frame |
2020-03-28T22:19:47.577 | <p>If I already have a small DC axial 2m3/h fan blowing air at something, what will happen if I add a larger more powerful 6m3/h DC axial fan blowing from behind that one (i.e. its the larger fan first, then then smaller, then the object to blow air at)? Will the smaller one just hinder the airflow from the first, or are their outputs cumulative? </p>
<p>Many thanks!</p>
| |fluid-mechanics|airflow|compressed-air| | <p>It depends on their aerodynamic properties.</p>
<p>Jet engines are multiple fans as well, but they can and do different functions. </p>
<p>Some work together to increase the pressure and reduce the speed of stream.</p>
<p>Some even drive energy from the stream of air.</p>
<p>In your case similar situations can happen.</p>
<p>The combination can increase the flow or the pressure. Or it may cause the entire stream to rotate in the cylinder and stall the system. </p>
| 34865 | Cumulative effect of two axial fans |
2020-03-29T04:16:28.707 | <p>I am doing a Finite Element analysis of a continuum. In my problem definition I use these units:
mm for length, kPa for applied stress, and MPa for material stifness (Young's modulus). </p>
<p>With these units, I get my output in: kPa for stress and micron for displacement, which is what I prefer.</p>
<p>Now I have to add Mass Density information to my problem. To benchmark my setup, I am trying to reproduce an old textbook example, in which the Mass Density is given in units of:
<span class="math-container">$$
\\lb-sec^2 / in^4\,
$$</span>
What would be the equivalent of this in my adopted SI unit system.</p>
<p>My impulse is to use:</p>
<p><span class="math-container">$$
\\kN / mm^3\,
$$</span></p>
<p>Any suggestions or advice will be appreciated!</p>
| |mechanical-engineering|finite-element-method|dynamics|experimental-physics|unit| | <h1>First Part</h1>
<p>Simplifying the
<span class="math-container">$$
1 \ lbf = 1 \ slug .ft/sec^2
$$</span>
then
<span class="math-container">$$
1 \ lbf \ sec^2 / in^4 = 1 \ (slug .ft/sec^2 ). (sec^2/in^4) = 12 \ slug /in^3
$$</span></p>
<p><span class="math-container">$$
12 \ slug / in^3 * (14.6 kg / 1\ slug)*(in/25.4 \ mm)^3 = 0.0107 \ kg/mm^3
$$</span>
Note that that each <code>slug</code> equals to 14.6 <code>kg</code> so <code>(1 slug / 14.6 kg)</code> equals to <code>1.0</code> and any expression multiplied with 1.0 will still matematically equal to original expression.</p>
<p>then</p>
<p><span class="math-container">$$
1.0 \ kg/mm^3 = 93.53 \ lbf \ sec^2 / in^4
$$</span></p>
<p>we name above as eq1</p>
<hr />
<h1>Second Part</h1>
<p>Changing shape of the target formula:</p>
<p><span class="math-container">$$
1 \ kN / mm^3 = 1 \ kN / mm^3 * (1000 \ N/1 \ kN) * (1 \ kgf / 10 \ N) = 100 kgf/mm^3
$$</span></p>
<p>again note that</p>
<p><span class="math-container">$$
(1000 \ N/1 \ kN) = (1 \ kgf / 10 \ N) = 1.0
$$</span></p>
<p>Also i took <span class="math-container">$ kgf = 10 N $</span> which is approximate, correct amount is <span class="math-container">$ kgf = 9.81 N $</span>
finally
<span class="math-container">$$
kN / mm^3 = 100 \ kgf/mm^3
$$</span></p>
<p>or</p>
<p><span class="math-container">$$
1.0 \ kgf/mm^3 = 0.01 \ kN / mm^3
$$</span></p>
<p>we name above as eq2</p>
<hr />
<h1>Third part</h1>
<p><code>kg</code> and <code>kgf</code> units are not consistent units on the paper, but looks equal on non scientific conversations. For example someone can say this chair mass is 10kg or it's weight is 10kgf, both refers to same chair with same mass and weight. so if we conventionally assume: <code>1 kgf = 1 kg</code>, eq1 and eq2 can be combined like this:</p>
<p><span class="math-container">$$
kgf/mm^3 = 0.01 \ kN / mm^3 = 93.53 \ lbf \ sec^2 / in^4
$$</span>
(Note: base on comments, <code>1 kgf</code> equals <code>9.8 N</code>, but we assumed <code>1 kgf = 10 N</code></p>
<h1>Final part</h1>
<p>so</p>
<p><span class="math-container">$$
1.0 \ kN / mm^3 = 9353.0 \ lbf \ sec^2 / in^4
$$</span></p>
<p>or</p>
<p><span class="math-container">$$
1.0 \ lbf \ sec^2 / in^4 = (0.00010691) \ kN / mm^3
$$</span></p>
<p>Which are the answers</p>
<hr />
<h1>Notes</h1>
<p>I am not sure about the assumption of <code>1 kgf = 1 kg</code> in third part, but the rest should not have problem. Anyways this answer could be considered as a suggestion.</p>
| 34868 | Unit Conversion for Mass Density |
2020-03-29T09:27:56.253 | <p>Apologies if I have done something wrong, first time posting on a stack exchange website.
<br>
<br>
Four forces act perpendicular to the surface of a circular slab (assume mass of slab is 0 and middle of slab is at 0,0,0)(all forces in -y direction)
<br>
Force 1 is located at 5,0,0 and has a magnitude of 77.4kN
<br>
Force 2 is located at 0,0,5 and has a magnitude of 48.6kN
<br>
Force 3 is located at 0,0,-5 and has a magnitude of 24.4kN
<br>
Force 4 is located at -5,0,0 and has a magnitude of 97.6kN
<br>
What is the coordinates of the resulting force (x,y,z) to two decimal places?
<br></p>
<p><a href="https://i.stack.imgur.com/KXhYo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KXhYo.jpg" alt="Visual Representation of Question"></a></p>
<p><br>
<br>
My working currently and my initial answer
<a href="https://i.stack.imgur.com/IKmS5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IKmS5.jpg" alt="Working and answer"></a></p>
<p><br>
<br>
The given answer is (0.49,0,-0.41) and I don't know where I went wrong.
<br>
<br>
After talking to my lecturer, she said the question was supposed to be determining the z coordinate only, which is why it would say the answer was correct when 0.49 was put first, meaning that the answers I calculated were correct.</p>
| |statics|moments| | <p>We start by the forces along the x-axis about D and call the resultant distance from D, X.</p>
<p><span class="math-container">$\Sigma M_{about\ D}=0 \rightarrow X*(97.6+24.4+77.4+48.6 ) \\= 73*5+77.4*10 \quad \rightarrow X=\frac{365+774}{248}=4.59$</span> </p>
<p><span class="math-container">$ 5-4.59=0.41$</span> X is 0.41 meters to the left of the origin <span class="math-container">$ \ X=-0.41m \quad $</span> as per the book.</p>
<p>The same steps apply to the Y-axis.
Please check my numbers. also, I assumed the radius 5m not 6m as shown, because you asked 5m.</p>
| 34871 | Resultant force coordinate for parallel 3d forces acting on a slab |
2020-03-30T02:09:27.780 | <p>I'm interested in printing a working two degree of freedom planetary gear set. I searched around and found a paper from University of Maryland in 1988 (<a href="https://i.stack.imgur.com/vim9v.png" rel="nofollow noreferrer">source</a>). I am having difficulty translating the information into a real-world model. </p>
<p>The goal is to make a system that allows two independent inputs to affect the speed and torque of a single output. As the paper points out, a differential system doesn't quite make this possible.</p>
<blockquote>
<p>Although mechanical systems with multiple inputs or multiple outputs have existed for many years, generally, they have been used as a series of one-DOF devices rather than as true multi-DOF mechanisms. For example, the automotive bevel-gear differential, a two-DOF mechanism with one input and two outputs, is made-up of a one-DOF gear train in series with its gear box. </p>
</blockquote>
<p>I tried analyzing a standard differential system but I don't believe you can alter it to use two independent inputs. The six link systems described in the paper seem to be the right direction, and so I would like to understand the graph.</p>
<p>Here is the figure from the paper:</p>
<p><a href="https://i.stack.imgur.com/vim9v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vim9v.png" alt="Excerpt from paper"></a></p>
<p>And here is my attempt to interpret graph 6-1-1:</p>
<p><a href="https://i.stack.imgur.com/K9PkF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K9PkF.png" alt="enter image description here"></a></p>
<p>I think this is clearly incorrect because elements 5 and 6 have no effect on the system. Could anyone help me interpret the graph?</p>
| |mechanical-engineering|gears| | <p>After help from jko I was able to put together a working model for a six link epicyclic gear system. It is essentially three standard planetary gear mechanisms tied together, each with a sun, four planets on a carrier, and a ring gear. I color coded the important interconnections:</p>
<p>1) Blue: The left and right planet arrangements share a common carrier. This double-sided carrier is also the main axis of rotation. The two parts making up this carrier mesh and their rotation is coupled.</p>
<p>2) Yellow: The middle and right planet arrangements share a common sun gear.</p>
<p>3) Red: The left sun gear is also the middle carrier.</p>
<p><a href="https://i.stack.imgur.com/fNTNs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fNTNs.png" alt="6 Link Mechanism"></a></p>
| 34885 | Gear train using two inputs and one output. Understanding six link epicyclic gear topology - Interpreting displacement graphs |
2020-03-30T04:22:19.970 | <p>I might be wrong here. The solid modeling done in CAD like Catia, NX, Solidworks,etc can be taken as input in CNC and the desired shape and size of piece can be obtained.The CAD itself convert the model into G -code.So why do we need to learn CNC programming if CAD can do it? Stay safe.</p>
| |mechanical|cnc| | <p>Since certain projects for certain tasks can be composed speedier than the time it takes to create an identical CAD document.</p>
<p>Likewise, utilization of factors makes the program more helpful ie a program to create a chamber can be composed to accept width as the controlling contention while delivering numerous CAD records for different various measurements requires significant investment.</p>
<p>On the off chance that you work a CNC machine, you will learn g-code in the end. Regardless of whether it's simply a subset of the g code. The reasons are various however generally it comes down to:</p>
<p>You have to arrange the machine and apparatuses every once in a while. At that point it may be functional to have the option to compose short pieces to change instruments, work organizes, and so on I mean this isn't truly enchantment it's more like stating load instrument number 16. You get this actually rapidly.</p>
<p>You might be in a circumstance where you have to troubleshoot code created by the postprocessor of your CAD/CAM application. Or on the other hand, you have to check a program. In these cases, you should have the option to peruse the g-code at least cursorily.</p>
<p>You should have the option to include a stop in the program to tidy up a drag between activities, make computerization changes between programs, and so on</p>
<p>You may need to change your postprocessor.</p>
<p>You may need to roll out a little improvement to the program. Sure you can return to the scoundrel application. Yet, that may mean leaving the shop floor opening the CAD or you may very well sort in the change without even a moment's pause.</p>
| 34887 | What is a purpose of learning G-code? |
2020-03-30T18:37:49.287 | <p>I'm programming a drawing plugin for autocad to draw cross-section in beams and ribs. I need to know if it is worth it to add (to my plugin) the feature to add more than two rebar rows, or if I'll be covering the most use cases with two.</p>
| |structural-engineering|civil-engineering|autocad| | <p>I have seen 4 layers at top and 6 layers in the bottom of some deep beams in a high-rise hospital building.</p>
| 34897 | How common is it to use more than two Rebar rows in a beam (bottom or top rebar)? |
2020-03-30T22:16:46.363 | <p>Now that Medtronic has open sourced (<a href="http://newsroom.medtronic.com/news-releases/news-release-details/medtronic-shares-ventilation-design-specifications-accelerate" rel="nofollow noreferrer">WIP</a> reportedly for its completeness) the design of a ventilator, the following question to experts in medical engineering with specialization on lung ventilators.</p>
<p>What could be blocking obstacles, if any, for a typical DIY shop with say mid-range equipment to replicate such a device following these specs?</p>
| |medical-devices| | <p>A typical DIY shop would fall over the electronics immediately. This is more a piece of electronics than a mechanical device. First of all, the shop would need the source of the chips which as far as I can tell aren't even referenced. This is a set of assembly drawings, not complete product specifications. It tells you how to build a circuit board, but certainly not the processor at the heart of it. Second, the shop would need a software arm to write the software for the product, which is also not given. What is given is a set of software <em>Requirements</em>, which is very different from software design specifications.</p>
<p>Maybe a manufacturer of similar devices could use this to aid their design and manufacturing process, but a DIY shop has no chance.</p>
| 34900 | Complexity of Medtronic respirator vs. DIY |
2020-03-30T23:15:32.087 | <p>On <a href="https://en.wikipedia.org/wiki/Buckling" rel="noreferrer">Wikipedia</a>, buckling is defined as follows:</p>
<blockquote>
<p>In engineering, buckling is the <em>sudden</em> change in shape of a structural component under load such as the bowing of a column under compression or the wrinkling of a plate under shear. If a structure is subjected to a gradually increasing load, when the load reaches a critical level, a member may suddenly change shape and the structure and component is said to have buckled. </p>
</blockquote>
<p>Why do we define buckling as a <em>sudden</em> change in shape? This definition implies that a column is perfectly straight until we apply a load over a critical limit, after which the column suddenly curves sideways. But in real life, columns are not exactly straight and loads are not applied exactly to the center lines of columns so there is a bending moment on the column (and therefore everywhere in the column) for <em>any</em> load, not just a load above a certain limit.</p>
<p>Classical Euler buckling assumes a perfectly ideal column, and for that buckling appears as a bifurcation solution after a critical load is reached (and not before), but a real life column is not ideal and any eccentricity on the column or the load means that a bending moment is technically present for any load, however small.</p>
<p><strong>So are real life instances of buckling not strictly speaking buckling according to this definition?</strong></p>
| |structural-engineering|buckling| | <p>In the real world, and even in most moderately complex numerical simulations, you are right: columns are not perfectly straight and any load applied will be slightly off center and apply a moment in all situations. </p>
<p>As an exercise take a ruler and push the ends together with your hands. At some point the center of the ruler will bow out in one direction or the other. If you maintain that same force the ruler will remain in the same shape (static equilibrium). Think about this from an energy perspective: while pushing on the ends of the ruler when it is straight there is zero internal energy being stored in the ruler because no displacement occurs. When the stick suddenly flexes your hands are now closer than they were before and you have applied force over that distance, this necessarily means you have done work on the system. That work is conserved as energy stored elastically in the ruler. </p>
<p>The ruler could be bent to the same state purely by applying moments on either end. During the initial phase of loading, by pure compression, before buckling the amount of moment needed to to get to this same state is offset by the increasing compression force. Eventually enough compression occurs that the additional moment needed approaches zero. You can derive theoretical limits to this for perfect geometries to your hearts content. In the real world due to a variety of factors this limit may be significantly less if you actually tested it to its limit. </p>
<p>Some of the potential deformed shapes that are possible can have huge deformations and wild shapes. If your structure was rubber you might be able to see that in the real world. For things like a steel I-beam buckling it will "buckle" according to these theories up until it yields and becomes plastic deformation.</p>
<p>Yes in the real world pure buckling according to the theories exists albeit with some correction factors needed for meaningful engineering decisions .. and overbuilding everything is always the best option. </p>
| 34902 | Definition of buckling as a sudden deformation |
2020-03-31T07:40:50.193 | <p>I was looking at the following form of the Euler equations:</p>
<p><span class="math-container">$$\frac{\partial \vec{v}}{\partial t}+(\vec{v} \cdot \nabla) \vec{v}+\frac{1}{\rho} \operatorname{grad}(p)=\vec{k} \quad \Longleftrightarrow \quad \frac{\partial v_{i}}{\partial t}+\sum_{j=1}^{3} \frac{\partial v_{i}}{\partial x_{j}} v_{j}+\frac{1}{\rho} \frac{\partial p}{\partial x_{i}}=k_{i}, \quad i=1,2,3$$</span></p>
<p>But I don't understand the notation of the gradient (<span class="math-container">$(\vec{v} \cdot \nabla) \vec{v}$</span>)
Because I only know the gradient for scalar valued functions from <span class="math-container">$\mathbb{R^n} \rightarrow \mathbb{R}$</span></p>
<p>It looks to me as if the gradient of a vector field is defined as the Jacobian. Is this correct?</p>
| |fluid-mechanics| | <p>The expression <span class="math-container">$(\vec{v}\cdot\nabla)\vec{v}$</span> is called convective derivative of the flow velocity and is called convective acceleration.</p>
<p>Using Einstein notation : <span class="math-container">$\vec{v} = v_i \hat{e}_i$</span>, where <span class="math-container">$v_i$</span> represents the velocity component along <span class="math-container">$\hat{e}_i$</span> direction; <span class="math-container">$i = 1,2,3$</span>. (Note: <span class="math-container">$i$</span> is a dummy index and we can represent it by any other letter as well).</p>
<p>Similarly, <span class="math-container">$\nabla = \frac{\partial}{\partial x_i} \hat{e}_i$</span>; <span class="math-container">$i = 1,2,3$</span></p>
<p>Let us study the expression term by term:</p>
<ul>
<li><p><span class="math-container">$\frac{\partial \vec{v}}{\partial t} = \frac{\partial v_i}{\partial t} \hat{e}_i$</span></p></li>
<li><p><span class="math-container">$(\vec{v}\cdot\nabla)\vec{v} = (v_j \hat{e}_j \cdot \frac{\partial}{\partial x_j}\hat{e}_j) v_i \hat{e}_i = \sum_{j=1}^3 v_j \frac{\partial v_i}{\partial x_j} \hat{e}_i$</span></p></li>
<li><p><span class="math-container">$\frac{1}{\rho} \text{grad}(p) = \frac{1}{\rho} \nabla p = \frac{1}{\rho} \frac{\partial p}{\partial x_i}\hat{e}_i$</span></p></li>
<li><p><span class="math-container">$\vec{k} = k_i \hat{e}_i$</span></p></li>
</ul>
<p>Now, let us gather all terms along <span class="math-container">$\hat{e}_i$</span> direction, we get
<span class="math-container">$$\frac{\partial v_i}{\partial t} + \sum_{j=1}^3 v_j \frac{\partial v_i}{\partial x_j} + \frac{1}{\rho} \frac{\partial p}{\partial x_i} = k_i,\hspace{3mm} i = 1,2,3$$</span></p>
<p>Hope this answers your question!</p>
| 34911 | Euler-Equations in gradient form |
2020-04-01T12:26:46.433 | <p>Sometimes engines are described as being intended for a particular orientation, e.g., "The 4.6 L; 278.6 cu in (4,565 cc) LD8 (VIN "Y") is a transverse V8 for front-wheel drive cars." What exactly does that entail? Why would an engine function differently with transverse instead of longitudinal mounting?</p>
| |automotive-engineering| | <p>Your question asks, how do longitudinally and tranversely oriented engines differ. Allow me to explain.</p>
<p>The primary mechanical differences relate to MOUNT positioning and HARMONIC BALANCE. The timing cover may differ as well as valve covers, intake and exhaust manifolds, and connections for hoses and wiring harnesses. In this case, it has nothing to do with the angle of the cylinder orientation (as another poster had commented).</p>
<p>Although you only asked about the engine, I am compelled to emphasize that drivetrain pairing is a significant consideration with chassis dynamics and performance. Generally, transverse orientations are paired with transaxles while longitudinal motors are paired to transmissions. However, there are exceptions to every rule and the late 70's Cadillac Eldorado and early 90's Acura Legend are two interesting examples of longitudinal front wheel drive vehicles you can examine for further study.</p>
<p>If you have a more specific inquiry on this subject matter, I am happy to help.</p>
| 34931 | How do automotive engines differ for longitudinal vs. transverse applications? |
2020-04-02T09:31:10.243 | <p>I have a project in which I have to use a up to 2.5mm diameter needle with a 90 degree (L shape) angle similar than the one in the following picture:
<a href="https://i.stack.imgur.com/4QWTL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4QWTL.jpg" alt="enter image description here"></a></p>
<p>I say similar because there is only the tip of the needle is only 1mm since the 90 degree angle</p>
<p>I need to find some kind of material I can get a wire of that goes inside the needle that will bend so it moves back and forward throughout the L shape of the needle and strong enough so it can push a springed metal present at the 2 or 3 mm from the tip of the needle.</p>
<p>I have the dilemma that if the material is too stiff it won't easily go back and forward because of the L shape but if it is not enough it won't have enough strength to make the spring to compress on the tip end.</p>
<p>I thought about a wooden snake alike solution but since the wire must be so thin I can't find anything to make it of.</p>
<p><a href="https://i.stack.imgur.com/ttpPr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ttpPr.jpg" alt="enter image description here"></a></p>
<p>Maybe you have a better idea. </p>
<p>Thank you so much in advance</p>
| |materials|stiffness|material-science| | <p>Here is an idea: try using a bundle of very thin wires and fuse the ends together at the tip. You could prototype this with small gauge, multi-strand, electrical wire. Solder the ends together so you have a solid tip for pressing. You can buy stainless steel wire down to 50 microns very inexpensively and fuse the ends together if corrosion is a concern. </p>
<p>You will get the combined axial stiffness of the bundle because the fused tip along with the constraint of the needle tip will keep the wires bundled together while it pushes. The wires will be able to turn the corner more easily because they are able to move relative to one another in bending. It's important that you do not fuse both ends or you will loose some of this advantage. </p>
<p>You could try something like this: <a href="https://www.mcmaster.com/34235t27" rel="nofollow noreferrer">https://www.mcmaster.com/34235t27</a> (nylon coated)
or this: <a href="https://www.mcmaster.com/3461t63" rel="nofollow noreferrer">https://www.mcmaster.com/3461t63</a> (uncoated)</p>
<p>They are braided bundles of 133 wires with an OD of 0.94 mm. I would suggest the uncoated version for most flexibility. </p>
| 34942 | Looking for thin flexible and strong material |
2020-04-02T15:49:11.770 | <p>What is the cause of turbojets'/gas turbines' high power to weight ratio compared to piston engines? They certainly use a generally less efficient thermodynamic cycle. </p>
| |mechanical-engineering|turbomachinery|jet| | <p>Turbojets and turbofans support extremely large mass flow rates of air & fuel through them, relative to their weight. (The majority of the work performed by burning that fuel is taken up by running the compressor stage of the engine and what's left over generates thrust.) In comparison, piston engines cannot support similar fuel burn rates relative to their weight.</p>
| 34946 | What gives gas turbine/turbojet engines such high energy density? |
2020-04-02T20:49:50.670 | <p>Problem 3.6 in Khalil's Nonlinear Control: <em>Use given Lypunov candidate function to prove that the origin is exponentially stable</em>. The system is
<span class="math-container">$$\dot{x}=\begin{bmatrix}x_2\\-h(x_1)-2x_2\end{bmatrix},\hspace{0.5cm}h(x_1)=x_1\left(2+\frac{x_1^2}{1+x_1^2}\right)$$</span>
and the given Lyapunov candidate function is
<span class="math-container">$$V(x)=\int_0^{x_1}h(s) ds+\frac{1}{2}\left(x_1+x_2\right)^2$$</span></p>
<p>To prove exponential stability, we need to find four positive constants, <span class="math-container">$k_1,k_2,k_3,a$</span>, such that
<span class="math-container">$$k_1\lVert x \rVert^a \le V(x) \le k_2\lVert x \rVert^a$$</span>
and
<span class="math-container">$$\dot{V}(x)\le -k_3\lVert x \rVert^a$$</span>
We can write the Lyapunov function more explicitly as
<span class="math-container">$$V(x)=\frac{3x_1^2}{2} + \frac{1}{2}\left(x_1+x_2\right)^2 - \frac{1}{2}\ln(1+x_1^2)$$</span>
and its time derivative as
<span class="math-container">$$\dot{V}(x)=-\left(\frac{x_1}{\sqrt{2}} + \frac{x_2}{\sqrt{2}}\right)^2 - \frac{3x_1^2}{2}-\frac{x_2^2}{2}-\frac{x_1^4}{1+x_1^2}$$</span>
It is relatively clear that we can use <span class="math-container">$a=2$</span>, and that there exists some <span class="math-container">$k_1,k_2,k_3$</span> that will hold.
But how do we find suitable values for the <span class="math-container">$k_i$</span>'s ?</p>
| |control-engineering|control-theory|mathematics|nonlinear-control| | <p>You don't necessarily have to find <em>exact</em> constants <span class="math-container">$k_1,k_2,k_3$</span>, only need to show that there exists some <em>positive</em> constants. In your example above, I can say that </p>
<p><span class="math-container">$$
\dot{V} \leq -\bigg(\frac{x_1 + x_2}{\sqrt{2}}\bigg)^2 = -\frac{1}{2}\|x\|^2,
$$</span></p>
<p>so <span class="math-container">$k_3 = 1/2$</span> would work. For the function <span class="math-container">$V(x)$</span> itself, we know that it is positive definite. Note that the log term is <em>always</em> negative so </p>
<p><span class="math-container">$$
V(x) \leq \frac{3}{2}x_1^2 + \frac{1}{2}\|x\|^2 = 2\|x||^2 - \frac{3}{2}x_2^2 \leq 2\|x\|^2,
$$</span></p>
<p>so <span class="math-container">$k_2 = 2$</span> would work. For <span class="math-container">$k_1$</span>, you can just pick an arbitrarily small positive constant that satisfies the lower bound.</p>
| 34954 | Bounds to prove exponential stablity for given Lyapunov function |
2020-04-02T21:43:21.617 | <p>Wikipedia has a good <a href="https://commons.wikimedia.org/wiki/File:Gearbox_4gears.gif" rel="nofollow noreferrer">animation</a> of a manual transmission, but something irks me.</p>
<p><a href="https://i.stack.imgur.com/QSMcf.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QSMcf.gif" alt="enter image description here"></a></p>
<p>If I understood correctly, in the animation, the input shaft is in green, and the output shaft is in cyan. The dog clutches (synchromesh in modern cars) are therefore between the free gears and the output shaft. This forces both the engine and the gear assembly to change speed in a gear shift. If the synchromesh were between the input shaft and the free gears, only the input shaft (therefore the engine) would need to change speed during a gear change. I don't see anything after a quick Google that suggests things are different in many other gearboxes.</p>
<p>What is the reason for this that I'm apparently missing? Acting as a flywheel is the only that crosses my mind.</p>
| |car|transmission| | <p><a href="https://i.stack.imgur.com/EMEQi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EMEQi.png" alt="Honda B Series"></a></p>
<p>Image from <a href="https://www.j-k-tuning.com/Transmission/BTrans2.php" rel="nofollow noreferrer">https://www.j-k-tuning.com/Transmission/BTrans2.php</a></p>
<p>You can put them anywhere you want to (and people do). But it is easier to put them on big gears than little ones. If your four speed was working the other way around with the green as an output shaft, all the gears would have to have the inverse ratios, and you would have all your syncros on little gears.</p>
<p>In the image above, you can see both shafts carry shift forks in this model.</p>
| 34956 | Why does the synchromesh in manual transmissions act on the output shaft? |
2020-04-03T11:39:15.897 | <p>I'm planning to drive a <a href="https://www.ebay.com/itm/New-Hiwin-MGN9H-Linear-Guides-MGN-Series-Linear-Bearings-30mm-to-1190mm-Long-/172789096756" rel="nofollow noreferrer">small linear guide</a> (4cm long) with a servo in a setup like this:</p>
<p><a href="https://i.stack.imgur.com/Pswbu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pswbu.png" alt="enter image description here"></a></p>
<p>I wonder if the ball bearings inside would sustain too much stress due to the vertical forces exerted by the links whilst driving it linearly. AFAIK ball bearings can support reasonable radial load, but I'm aware that these guides are usually driven by lead screws instead. Is it a good idea to drive the guide the way presented above?</p>
| |mechanical-engineering|bearings|linear-motion|servo| | <p>If you limit the retracted angle of the arm to <span class="math-container">$\ a<60^{\circ} $</span> it should be ok.</p>
<p>That would limit the maximum vertical load at two times the horizontal push. </p>
<p><span class="math-container">$ F_y = \ \leq 2F_x$</span></p>
| 34965 | Linear guide ball bearings |
2020-04-03T15:04:59.580 | <p>I'm making a Z (coffee table) linkage with two interlocking C channels and a couple of steel linkage bars. While moving, they aren't load bearing but when extended, they will take a weight of around 300lbs. I need to figure out whether 3mm T66 aluminium can take that weight without tearing.</p>
<p>The entire weight will be spread across two bolts (10mm with bushing) and the top half of that so that's Pi x 10mm diameter / 2 because the weight is all on the top half x 2 bolts - around 31.4mm x 3mm thick = 94.2mm squared. I've found <a href="https://www.makeitfrom.com/material-properties/6060-T66-Aluminum/" rel="nofollow noreferrer">tables of properties</a> for T66 which give figures in MPa (megapascals?) but that's where my half-remembered high school physics gives out.</p>
<p>How do I calculate the maximum load across a given area from these properties?</p>
<p>Edit: To include sketch.
<a href="https://i.stack.imgur.com/nahfP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nahfP.png" alt="Unwin rails"></a>
The block to the left is the (fixed) seat base, along the top are Unwin rails along which the seat can slide. When not needed, the linkage around the 4 bolts can rotate it away into the floor. When they are raised (as shown), the seat+occupant, for which I am allowing 300lbs) can sit directly over the hinged section. Ignoring the rail overlap, that means that the full weight is supported by the top of the two round pivots (actually, since it's made from C channel and there are one of these on each side, that's 2x2x2x10mm bushings in 3mm aluminium).</p>
<p>Hope this makes the question clearer.</p>
| |aluminum|material-science| | <p>I confirm that in your case the yield stress is an adapted limit value.
However, there is a unit error in your computation. You forgot to convert the area in m<sup>2</sup> which means that the results is not 362 Pa but 362x10<sup>6</sup> Pa = 362 MPa.
Assuming that your reasoning is correct to compute the effort, your are above the limit of 170 MPa.</p>
| 34969 | Calculating tensile strength for aluminium |
2020-04-04T06:05:26.650 | <p>I am reading a textbook on fundamentals of vibrations and it has a very brief introduction to shock spectra. It mentions that </p>
<blockquote>
<p>For comparison purposes, it is customary to use the response of an
undamped single degree of freedom system to obtain the shock spectrum.</p>
</blockquote>
<p>So is it that we always find the maximum response of an undamped single DOF system to get the shock spectrum? If yes, then how does it help us, because can't the response of the structure which I am analysing (say a building) be widly different from that of a single DOF system? For example, consider two shocks. The shock which seems "stronger" (higher maximum response in general) when using a single DOF system may actually be weaker when using my structure. </p>
<p>Also, just saying an undamped single DOF system seems vague because response to the same shock may differ for different single DOF systems. Is there a specific single DOF system that is used here (like the mass-spring system)?</p>
| |structural-engineering|structural-analysis|vibration|shock| | <p>Not necessarily, many shock response spectra are created for systems under the impulse of a shock with damping.</p>
<p>However many codes allowing seismic design using the response spectra with an estimated 100-year or whatever design earthquake require the assumption of the three spectra <span class="math-container">$ S_d, \ S_v, \, and \ S_a $</span></p>
<p>related as:</p>
<p><span class="math-container">$ S_v= \omega S_d\\ S_a= \omega^2 S_d$</span></p>
<p>And this is valid only for a single degree of freedom system with no damping.</p>
| 34988 | Is shock spectrum always obtained for an undamped single degree of freedom system? |
2020-04-05T08:47:18.937 | <p>See video:
<a href="https://www.youtube.com/watch?v=75Z7-gmd_qk&t=226" rel="nofollow noreferrer">https://www.youtube.com/watch?v=75Z7-gmd_qk&t=226</a></p>
<p>This does not make sense to me. As far as I got it, bottom pipe is isolated from top 2 and just controls valve of the top 2 pipes. So, when air from bottle goes into left "liar" trough top 2 tubes of the right one it should just stop there. There is no way air can get to the middle one and also return to the right one to close/open it.</p>
<p>I hope somebody can explain this to me, because I'm starting to loose my mind a bit over this.</p>
<p>Edit:</p>
<p>I used Transistor's schematic to illustrate my frustration better:
<a href="https://i.stack.imgur.com/s2FC3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s2FC3.png" alt="enter image description here"></a></p>
<p>See, in the video it's clearly seen that input air pipe goes only into A cylinder's input pipe. B and C does not receive air at all! Also it seems that B and C are closed initially, while A is open. And I can't really see any springs that you mentioned either. This is what really frustrates me. How B and C can operate if they are not even connected to air supply?</p>
<p>Edit:</p>
<p><a href="https://i.stack.imgur.com/ctBXY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ctBXY.png" alt="enter image description here"></a></p>
<p>OK, I'm feeling really embarrassed now. Are this tubes actually fused together? The problem is, Theo holds his finger over it almost all the duration of operation and it really seems that this 2 tubes are separate... In this case, Transistor's schematic is correct and my problem is solved.</p>
| |fluid-mechanics|pneumatic| | <p>o.k. i think you'r just confused because they have different tops, and the tubs are hard to follow, but they are just all connected to the bottle, as i draw in my first answer
<a href="https://i.stack.imgur.com/w8BPY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w8BPY.png" alt="enter image description here"></a></p>
| 35006 | How does 3 "liars" in Strandbeests work? |
2020-04-05T09:00:42.563 | <p>(<a href="https://drive.google.com/drive/u/0/folders/1G6pZJOcN0G8Zk7YzqGfa1C8KK2aCzxId" rel="nofollow noreferrer">solidworks files here</a>)</p>
<p>I've made a small assembly thet mooves smoothly manually, but when I try to moove it the same way is motion study, it collides is a strange way... what can be the cause of that?<a href="https://i.stack.imgur.com/A5Xl9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A5Xl9.png" alt="manually"></a></p>
<p><a href="https://i.stack.imgur.com/xocd0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xocd0.png" alt="enter image description here"></a></p>
| |mechanical-engineering|cad|solidworks| | <p>General points to note: The "Standard" mates are more robust than the Mechanical mates, and should be used wherever possible. A cam should have one continuous face, you can use the fitspline tool to create this.</p>
<p>See below for how I fixed your model:</p>
<p><a href="https://i.imgur.com/bJk9p4V.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/bJk9p4V.gif" alt="fix Cam Screencap"></a></p>
| 35007 | Solidworks motionstudy/motion study causes parts to "jump" |
2020-04-05T19:08:48.483 | <p>V is a rotation factor to account for the difference in ball rotations for outer ring rotation vs. inner ring rotation.</p>
<p>V = 1 for inner ring rotation
V = 1.2 for outer ring rotation</p>
<p>From Shigley's Mechanical Engineering Design - Ninth Edition</p>
<p>What is the reason behind this?</p>
| |mechanical-engineering|design|bearings| | <p><a href="https://i.stack.imgur.com/zk36C.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zk36C.gif" alt="enter image description here"></a></p>
<p><em>Figure 1. Source: Wikipedia's <a href="https://en.wikipedia.org/wiki/Ball_bearing" rel="nofollow noreferrer">Ball-bearing</a>.</em></p>
<p>The distance rolled on the inner ring is the same as the distance rolled on the outer ring. Due to the difference in circumference the ball will pass the same point on the inner ring about twice as often as it passes the same point on the outer ring (for these dimensions).</p>
| 35015 | Why rotation factor of bearing for inner ring is 1 and for outer ring is 1.2? |
2020-04-06T06:57:23.340 | <p>Problem statement: 2 kg of air at 500 kPa expands adiabatically in a closed system until its volume (V) is doubled and it's temperature become equal to that of surrounding which is at 100 kPa, 5°C. For the process determine</p>
<ol>
<li>maximum work </li>
<li>change in availability</li>
<li>irreversibilty</li>
</ol>
<p>For air take <span class="math-container">$c_v= 0.718\text{ kJ/kg K}$</span>, <span class="math-container">$u = c_v T$</span> and <span class="math-container">$PV=mRT$</span> where <span class="math-container">$R= 0.287\text{ kJ/kg K}$</span></p>
<p>In the solution, maximum work is solved using state equation including entropy which I have no doubt about. But my doubt is: </p>
<ol>
<li>why can't we use integration <span class="math-container">$p\partial V$</span> for an adiabatic process to find maximum work here?</li>
<li>as per the question, final temperature is same as that of surrounding and final volume is 2 times of initial volume also it is an adiabatic process. Final pressure using <span class="math-container">$PV^{1.4}=\text{constant}$</span> and <span class="math-container">$PV=mRT$</span> are giving different result for the final pressure of the system.</li>
</ol>
| |thermodynamics| | <p>1) The maximum mechanical work of a process is not the same as the actual work done on/by the system. The former is calculated using the assumptions:</p>
<ul>
<li>reversible</li>
<li>ideal gas</li>
</ul>
<p>In the reversible case,</p>
<p><span class="math-container">$$ \Delta U = 0 \rightarrow w = -q\ \ \mathrm{IUPAC\ form,\ for\ Clausius\ form\ } w = q $$</span></p>
<p>Next, the starting equation for work is <span class="math-container">$w = -\int p_{ext}\ dV$</span>. Only when we say that the process is reversible can we state that <span class="math-container">$w = -\int p\ dV$</span>.</p>
<p>2) <strong>For a reversible adiabatic processes with ideal gases</strong>, we can substitute into the mechanical work using <span class="math-container">$pV^\gamma = $</span>constant. An adiabatic reversible process is also always isentropic. An isentropic process is not always a reversible adiabatic one. An isentropic process also does not always mean that we are using an ideal gas. In summary, we cannot say that we can use <span class="math-container">$pV^\gamma = $</span>constant for the work of any generic isentropic process. This is a shortcut that ignores the three founding rules for the substitution (an adiabatic, reversible process on an ideal gas).</p>
<p>Finally, regardless of whether the real process is reversible or irreversible, we can always write</p>
<p><span class="math-container">$$\Delta U = \int T\ dS - \int p\ dV $$</span></p>
<p>All terms in the above are state functions, and the change <span class="math-container">$\Delta U$</span> does not care about the path.</p>
| 35025 | Maximum work from the process |
2020-04-06T10:34:13.687 | <p>So to the point. In our project (Robotics project) we mostly use sheet metal and PLA printed parts to build. But we want to level up and be ready for an advanced run and even low-volume production of the product. Right now we're in the last days of an advanced prototype (Phase II) . My question is, what do you guys recommend to level up the design and start to be ready for low-volume production (15 - 20 pieces)? What are the tips and tricks when designing a more finished product? For example, The two things we're considering to change are CNC machining the plastic parts that have high loads on them, and for other 3D printed parts we are considering using different technology than FDM printing. Right now plastic injection molding is not an option, we are not in mass production. Carbon composites are not an option too. I will be glad to hear from you about more ways of bringing the product closer to an industrial one and start thinking about low-volume production. Any article/guide will be great!</p>
| |machining|robotics|industrial-engineering|product-engineering|production-technology| | <p>I believe your primary goal when transitioning from single prototypes to small batch production should be to improve manufacturability, and ensure that your design is scalable.</p>
<p>Examples of parts/systems that may need to be revised:</p>
<ul>
<li><p>Parts that were made without drawings, for example by match drilling, or "freestyling"</p></li>
<li><p>Parts that are not easily obtainable in quantity</p></li>
<li><p>Parts/systems that have known issues, which were manageable for a single prototype.</p></li>
<li><p>Systems that require excessive manual labor for manufacture. For example hand soldered cable harnesses may need to be replaced with off-the-shelf options.</p></li>
<li><p>Parts that are functional, but not aesthetically pleasing may need to be replaced</p></li>
</ul>
<p>I would recommend focusing less on methods, and more on results. Instead of asking "how do I level up", I would ask: "if I started making 20 of these right now, what problems would I run into".</p>
<p>To address the issue of 3D printed components. First ask yourself what the problem with them is, is it Aesthetics? Strength? Time? Cost? There is nothing inherently wrong with using 3D printed parts on a product.</p>
<p>There are other options, but they will all be more expensive than 3D printing (for 10-20 units) so you will probably want to have a good reason for choosing something else. Some alternatives:</p>
<ul>
<li>CNC, good looking and strong parts, but very expensive.</li>
<li><p>Laser cutting / water jet. Almost as cheap as 3d printing, strong materials, looks good. But the parts will need to be re-designed to be 2D</p></li>
<li><p>Resin molding. Not economical, in my experiance this is only used to simulate what an injection molded part will look like.</p></li>
<li><p>Other 3D printing. There are a lot of types, such as sintered plastic/metal, resin, etc that may be able to solve problems that your FDM PLA prints are having.</p></li>
</ul>
| 35028 | From advanced prototype to low-volume production |
2020-04-06T13:04:49.780 | <p>The picture below shows a veritcal <a href="https://en.wikipedia.org/wiki/Joint_(building)" rel="noreferrer">construction joint</a> of an approximately 50 years old reinforced concrete industrial structure. The expansion joint was likely originally sealed with a <a href="https://www.raygrahams.com/products/138890-flexcell-expansion-joint-2440mm-x-100mm-8-x-4.aspx" rel="noreferrer">flexcell</a> which due to age and water ingress has deterioriated and come away from the joint exposing a peforated metal pipe.</p>
<p>My question is does any one know why the pefortated pipe was installed there originally approximately 50 years ago? Does it have a drainage purpose? The joint should have been <a href="https://en.wikipedia.org/wiki/Waterstop" rel="noreferrer">waterstopped</a>. Is this a second line of defence? If so where do you expect it to drain to? Has anyone seen a similar detail to this before?</p>
<p><a href="https://i.stack.imgur.com/J1mOC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/J1mOC.png" alt="perforated pipe within construction joint"></a></p>
| |civil-engineering|concrete| | <p>Yes, it is a hydrostatic pressure relief pipe (part of the drainage).</p>
<p>Behind all the retaining walls, swales, subdivision's low land lots, you need to have a trench or channel filled with aggregate and a perforated pipe leading to surface runoff drainage as a minimum.</p>
<p>The Aggregate mus be stratified from 1.5" near the pipe to sand 18" out.</p>
<p>I suspect in your photo corrosion has washed the aggregate out.</p>
| 35030 | Why is there this perforated pipe inside this vertical construction joint in a reinforced concrete wall? |
2020-04-07T00:30:46.100 | <p>I want to manually measure the dimension of a pattern printed on a flat surface, with an accuracy of about 0.1mm. For simplicity sake, let's say this is a black rectangle, printed on a white background, and i want to measure its width and height. The dimension can be between 20cm and 80cm.</p>
<p>I have tried with a vernier caliper, but placing the caliper accuratly enough on the rectangle borders is impossible, especially since the caliper jaws do not lie flat on the surface.</p>
<p>What kind of measuring instrument could i use? Is there something dedicated to this task, or a way to use a standard instrument to do that?</p>
| |mechanical-engineering|measurements|distance-measurement| | <p>I had a plastic ruler at one time in my life and it looked something like Figure 1.</p>
<p><a href="https://i.stack.imgur.com/3Yotv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Yotv.png" alt="enter image description here"></a>
<em>Figure 1. Overview.</em></p>
<p><a href="https://i.stack.imgur.com/gQMHF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gQMHF.png" alt="enter image description here"></a></p>
<p><em>Figure 2. Close up of precision grid.</em></p>
<p><a href="https://i.stack.imgur.com/9RJ9X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9RJ9X.png" alt="enter image description here"></a></p>
<p><em>Figure 3. In-use.</em></p>
<p>How to use:</p>
<ul>
<li>Align the right edge to the nearest whole cm less than the width of the item to be measured.</li>
<li>In the precision section, find the closest intersection between a sloping line and a horizontal line.</li>
<li>Add the three numbers. In this case it's 30 + 5 + 0.7 mm. = 35.7 mm.</li>
</ul>
<p>I drew this in a few minutes on a CAD program. Vertical spacing is 2 mm. Print it off on laser-compatible transparent film on a decent laser printer and you should be in business. If you mirror the image before printing you will print on the "back" of the film and will have no parallax error when the film is pressed on the part. To make a longer scale you can tape several together.</p>
| 35042 | Measuring a pattern size on a flat surface |
2020-04-07T07:51:32.163 | <p>When double-checking/confirming results produced by structural analysis software, many engineers recommend verifying and spot-checking key or critical members if the entire structure is not analysed. </p>
<p>However, for rigid jointed structures with sway, it is not possible (at least in my understanding) to analyse individual members. This means that hand calculation methods such as moment distribution or slope deflection cannot be applied directly. </p>
<p>How would I go about analysing those members in a sway frame?</p>
<p>Also, to what storey of an indeterminate structure would you analyse purely by hand calculations without computer software when considering code requirements?</p>
| |structural-engineering|civil-engineering|structural-analysis| | <p>In my professional experience, I've found few opportunities to perform literal "by hand" verification.</p>
<p>Unless you're dealing with simple structures (or structures which can be simplified as such), it gets quite complicated and if you perform any mistakes you won't know if it's because of the model or your "by hand" calculation. And I personally have a hard time identifying (non-obvious) mistakes in my own calculations when I'm not even sure they're there.</p>
<p>That being said, the hard part of "by hand" verification of non-trivial structures is the actual structural analysis. Thankfully, the world has many different computer programs that do this exceedingly well.</p>
<p>So I check my work in one structural analysis program by creating another model in another program. This other model, however, is usually much simpler: hell, my tool of choice was <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, which only does 2D frame analysis. So if I was trying to double-check a 3D program, I'd have to recreate the model in Ftool using slices, manually calculating the applied loads (usually using <a href="https://debug.pi.gr/default.aspx?ch=557" rel="nofollow noreferrer">simple distribution areas</a>), considering the support conditions (it's much easier to be explicit and intentional about your supports and hinges in 2D than 3D), etc. It's important not to get too detailed in this model: don't try to model a complex support condition using master-slave relations or whatever; make the best simplification you can: fully hinged or fully fixed (or if absolutely necessary, do both and get an average of the results).</p>
<p>And then I let Ftool do its job and give me the internal forces, which I can then compare to the 3D model.</p>
<p>If they're close enough, great. It's much harder to get the boundary conditions wrong in a 2D model (where you explicitly define what they are at a single point, while the 3D model may require you to apply conditions to multiple points to represent the same effect, giving you room for error), so if they're close but actually wrong, it's because you explicitly messed up (i.e. in both models you forgot that that one beam there is actually hinged to the column, not fixed... a mistake you likely would've made even if doing the verification fully by hand). Getting someone else to take a look at your 2D model should further reduce the odds of that happening. And the odds of you having made different mistakes in each model that cancel out to give similar wrong results are infinitesimal.</p>
<p>If they're different, it's far easier to double-check that you didn't mess up your 2D model, and then compare the results to see what's similar and what's different, helping you identify your mistake (while doing it by hand risks a screw up in one element of the stiffness matrix leading you to get everything wrong).</p>
<p>This offers an efficient verification which is quite close to "by hand" -- you are still the one defining the structure's loads, supports and dimensions, in a format (2D) where it's much harder to accidentally mess up the model -- while still being efficient and honestly less error-prone than literally doing it by hand.</p>
| 35047 | Analysis Of Statically Indeterminate Structures By Hand |
2020-04-07T10:34:28.383 | <p>I'm going to buy Thomas vacuum cleaner with Aqua filter and use it to sand walls a bit with grinder. And after finishing the decoration, use it as an ordinary cleaner. May construction plaster, putty sanding harm it? Or it is OK to use the household vacuum for small amount of grinding?</p>
| |grinding| | <p>Actually I've bought household Thomas Multi Clean X10 Parquet vacuum with Aqua filter and rent the industrial cleaner Hilti VC-40UM with Hilti DGH-130 concrete grinder. There was about 2 full buckets of dust, I cleaned the filter for Hilti about 10 times during work of single small 3x5 meters room. So I can say that household cleaner won't last after such an action. In addition to that Aqua filter passes about 20% of dry dust through. Then it is being caught by sponges and microfiber filter before the engine. I suppose 20% dust from 2 buckets would be enough to kill those filters</p>
| 35050 | Can I use household vacuum cleaner with AquaFilter as construction/industrial one |
2020-04-07T11:51:16.720 | <p>Hello all I have a magnet in my car to stick my phone to the dashboard. But I changed phone and now my new phone needs some metal stuck to the case in order to attract to the magnet.</p>
<p>I read online these metals are called ferromagnetic (iron, steel, and others).</p>
<p>I wonder if any of you fine folks can suggest me a place I can butcher some from something I might already have at home (or can be easily found). At the moment we have the lockdown and most shops are shut. I can't figure it out on my own haha.</p>
<p>Just need like an inch or 2 square piece of metal that will stick to my magnet in the car. Sadly, I don't have great tools, but might be able to borrow some rudimental hand tools.</p>
<p>Thanks for reading and any help!</p>
| |magnets| | <p>A steel washer would work well.</p>
<p>Find one about 1" in diameter and that should be sufficient.</p>
<p>If you have a file or sanding tool then round the edges off.</p>
<p>Make sure the surfaces you are gluing are "keyed" ie scratched to provide a better joint.</p>
| 35052 | Cheap / Easy source of small amount of ferromagnetic metal |
2020-04-07T16:43:55.950 | <p>The reinforced concrete beam (approximate dimensions 36" x 18") shown in the image below has a row of circular holes at 3/5ths of the height of the cross-section at approximately 300 mm centres. </p>
<p>My question is why does this concrete beam have these holes?</p>
<p>My first thought is that these holes were due to the <a href="http://www.siteright.net/tie-bar-and-tie-bar-connector-p-88.html" rel="nofollow noreferrer">tie bars</a> providing lateral support to the formwork vertical walls. However, they appear to be too frequent for that to be true.</p>
<p><a href="https://i.stack.imgur.com/LasbW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LasbW.jpg" alt="row of circular holes through concrete beam"></a></p>
| |civil-engineering|beam|reinforced-concrete| | <p>I've not seen them before but I think that they are surely penetrations to run wiring. They on top for reasons including</p>
<ol>
<li>close to the lid where wiring would be normally attached between the beams</li>
<li>in a place of lower bending stress, near the compression zone, not in the tension zone were reinforcing would be located or concentrated</li>
</ol>
<p>I suppose that they could be for air flow, a venting of sorts.</p>
<p>In any case, I believe that they are not there as a design feature of the beam. That is, these holes serve no purpose for the beam itself.</p>
| 35057 | Why is there a row of circular holes in this concrete beam? |
2020-04-08T13:16:52.060 | <p>I need to calculate whether a polycarbonate sheet can withstand a certain maximum force. The force is an impact force on any one point of the sheet at any given time. All I have are the Young's modulus and the dimensions of the sheet. Any ideas how I can go about calculating the maximum force the sheet can withstand? This polycarbonate plate is placed vertically and held on all edges by a frame.</p>
<p>Thanks!</p>
| |structural-engineering|structures|stresses| | <p>For impact force calculations you need some more information such as the shear modules and the material properties like toughness and tensile yield and shear yield.</p>
<p>Also you need to define a physical object that will impact the sheet.</p>
<p>If the object is much more rigid than the sheet and pointed it will go through like a bullet.</p>
<p>Imagine an impact force has caused your sheet to behave like trampoline table.</p>
<p>You will see a cone shape deflection under the load opening up to a wide sloped flange depression.</p>
<p>On the very sharp bottom of this cone most of the stress is tensile due to cylindrical elongation of this segment.
And some crushing at the center.</p>
<p>But as we move up most of deformation is due to punching shear on concentrically widening rings with a bit of biaxial bending moment.</p>
<p>The military shield and armoire manufacturers have done extensive research on the competition between the pentration of mortar and resistance of shields.</p>
<p>If you are limiting your limit to elastic range, your force has to have a nominal minimum diameter and you can assume an arbitrary dynamic factor of 3 and check Roark's handbook of engineering formulas for some static forces and edge support conditions. </p>
| 35068 | Calculate whether a polycarbonate plate can withstand a certain force |
2020-04-08T16:14:32.267 | <p>I have been working on a question, and my lecturer has hinted me with this equation that doesn't include forces due to weight?</p>
<p><span class="math-container">$C_{M_x}=C_{M_{ac}}+C_L(x/c−x_{ac}/c)$</span></p>
<p>What he told us was that due to the derivation of the equation we can ignore weight with this equation? However when taking the moment of course one must take weight into account correct? Is it included in the lift coefficient?</p>
| |aerospace-engineering|aerodynamics| | <p>These are the dimensionless coefficients. The equation computes the <span class="math-container">$C_{m_x}$</span> at some point not at the aerodynamic center by adding the <span class="math-container">$C_{m_{ac}}$</span> to a (nondimensional) force times a (nondimensional) distance. Note that the <span class="math-container">$C_{m_{ac}}$</span> is independent of angle of attack. To the first order, it is a function of the camber line. But the <span class="math-container">$C_l$</span> is a function of angle of attack. So the <span class="math-container">$C_{m_x}$</span> varies with angle of attack.</p>
<p>If you are looking for where the weight is "hiding", changes in weight are reflected in changes to <span class="math-container">$C_l$</span> in level flight.</p>
| 35069 | Why is it when taking the moment coefficient relative to c.o.p., the weight is ignored? |
2020-04-09T06:53:49.637 | <p>I have continuous A-weighted sound pressure level (SPL) values at 1 minute time resolution (<code>LAeq,1min</code>). I would like to calculate <code>LAeq</code> SPL at <code>half-hourly</code> and <code>hourly resolutions</code> using these values. Can someone suggest on how this could be achieved? </p>
| |acoustics|noise| | <p>You can find from various sources online or on textbooks that <span class="math-container">$L_{eq}$</span> is calculated as</p>
<p><span class="math-container">$$ L_{eq} = 10 \log_{10} \left(\frac{1}{T} \int_{0}^{T} 10^{L \left( t \right)/10} dt \right) \tag{1}\label{1}$$</span></p>
<p>where <span class="math-container">$T$</span> is the measurement duration and <span class="math-container">$t$</span> represents time. Please note that <span class="math-container">$10^{L \left( t \right)/10} = P \left( t \right)$</span>, with <span class="math-container">$P \left( t \right)$</span> the power of the sound. Thus, equation \eqref{1} can also be written as</p>
<p><span class="math-container">$$ L_{eq} = 10 \log_{10} \left(\frac{1}{T} \int_{0}^{T} P \left( t \right) dt \right) \tag{2} \label{2}$$</span></p>
<p>We can now solve for the part inside the parentheses. This would result to</p>
<p><span class="math-container">$$ L_{eq} = 10 \log_{10} \left( \frac{1}{T} \int_{0}^{T} P \left( t \right) dt \right) \implies \frac{L_{eq}}{10} = \log_{10} \left( \frac{1}{T} \int_{0}^{T} P \left( t \right) dt \right) \implies \\ \implies 10^{L_{eq}/10} = \frac{1}{T} \int_{0}^{T} P \left( t \right) dt \implies T \cdot 10^{L_{eq}/10} = \int_{0}^{T} P \left( t \right) dt\tag{3}\label{3}$$</span></p>
<p>Since you can find the power integral corresponding to each <span class="math-container">$L_{eq}$</span>, all you have to do then is to sum the power up and divide by the total duration and convert to dB. One step at a time. Summing up the integrals of equation \eqref{3} you get (where the limits of the integrals are in seconds)</p>
<p><span class="math-container">$$ \int_{0}^{60} P \left( t \right) dt + \int_{60}^{120} P \left( t \right) dt + \int_{120}^{180} P \left( t \right) dt + \ldots + \int_{3480}^{3540} P \left( t \right) dt + \int_{3540}^{3600} P \left( t \right) dt = \int_{0}^{3600} P \left( t \right) dt$$</span></p>
<p>where we have used the known equality from calculus (from right to left)</p>
<p><span class="math-container">$$ \int_{a}^{c} f \left( x \right) dx = \int_{a}^{b} f \left( x \right) dx + \int_{b}^{c} f \left( x \right) dx, ~~~~~ a < b < c \tag{4} \label{4}$$</span></p>
<p>Now, in order to calculate the corresponding one-hour <span class="math-container">$L_{eq}$</span> you have to first divide with the total duration (<span class="math-container">$3600$</span> seconds this is) and then convert to deciBel. So, for our numbers this is</p>
<p><span class="math-container">$$ L_{eq, tot} = 10 \log_{10} \left( \frac{1}{3600} \int_{0}^{3600} P \left( t \right) dt \right) $$</span></p>
<p>where you will use the result from the power integral summation in the logarithm.</p>
<h2>Notes</h2>
<p>There are some things to note here.</p>
<ol>
<li>You most probably won't have the power function but you will acquire the integral values directly from the measured <span class="math-container">$L_{eq}$</span> values with the aid of equation \eqref{3}.</li>
<li>You most probably be working in the digital domain, where all integrals should be converted to sums. So, for example, equation \eqref{2} should be written
<span class="math-container">$$ L_{eq} = 10 \log_{10} \left(\frac{1}{N} \sum_{n = 1}^{N} x \left(\left[ n \right] \right)^{2} \right) \tag{5} \label{5}$$</span>
where <span class="math-container">$N$</span> is the number of samples in your signal vector, <span class="math-container">$n$</span> is the sample index in the range <span class="math-container">$\left[ 1, N \right]$</span> and <span class="math-container">$x \left[ n \right]$</span> is the sample values. Please note that they are squared since power is proportional to the square of the signal amplitude. This should be used in case you have to calculate the power from time-series measurements.</li>
<li>In a similar manner you can calculate the half-hour <span class="math-container">$L_{eq}$</span>, or any other duration for this purpose.</li>
<li>Please keep in mind that in this approach it is assumed that the measured <span class="math-container">$L_{eq}$</span> values are taken in consecutive time intervals and this is the sum of the power integrals results from equation \eqref{4}. If the noise you measure is (wide sense) stationary then the results should be fairly close (in engineering accuracy) whether the measurements are taken in consecutive intervals or not. Otherwise, you may encounter significant deviations.</li>
</ol>
<p>I assume you can do the coding part of this fairly simple algorithm (only exponentials, multiplication, addition and logarithms are involved, which are readily available in pretty much every programming language) on yourself. If not, I believe you could ask in either <a href="https://dsp.stackexchange.com/">Signal Processing SE</a>, <a href="https://scicomp.stackexchange.com/">Computational Science SE</a> or some other programming related SE site.</p>
| 35076 | How do I calculate hourly LAeq sound pressure level using 1-mint LAeq values? |
2020-04-09T22:03:05.040 | <p>I want to calculate the improvement in flow rate that can be achieved with changing the tubing size when the tubing is sub-millimeter size. </p>
<p>I originally approached this using a simplification of Poiseuille's law which says <span class="math-container">$\Delta Q=\Delta r^4$</span></p>
<p>However, lets say I'm changing from 0.5 to 1.5 millimeter tubing. That gives <span class="math-container">$\Delta r=0.050$</span>, which when raised to the 4th power is a tiny number (6.25E-05). This seems wrong intuitively, based on the <em>relative</em> increase in size (i.e. new tube is 300% of the original tube) </p>
<p>If I convert to micrometers first, the result jumps to 6.25E06, which seems really really large. </p>
<p>What am I missing here, because this doesn't seem like the right approach. </p>
<p>Thanks!</p>
| |fluid-mechanics|flow-control| | <p>You made a small mistake if we are going to use Poiseuille's law.</p>
<p><span class="math-container">$ Q_{final}= Q_{initial}\Delta r^4 \rightarrow \Delta Q =(\frac{r_{final}}{r_{itial}})^4$</span></p>
<p>Plugging your numbers,</p>
<p><span class="math-container">$ Q_{final}=Q_{initial} (0.75/0.25)^4 \\ =Q_{initial}3^4=81Q_{initial}$</span></p>
| 35087 | Relative improvement in flow from diameter change with sub-milliimeter tubing |
2020-04-10T01:49:10.320 | <p>I am looking at designing a rack and pinion system that will be laser cut out of some plastic. While the cad system that I am using has a spur gear generator, I was looking at the design of the rack gears. I thought that it needed an involute type profile, but many of the references that I see have a flat profile instead of an involute type profile. For example, this metal rack gear from McMastercarr has a seemingly flat profile.</p>
<p><a href="https://i.stack.imgur.com/jm7vq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jm7vq.png" alt="mcmastercar" /></a></p>
<p>Is this common? Is the reason that it has a flat profile due to the fact that the radius of the base circle for racks is theoretically infinite?</p>
| |mechanical-engineering|design|gears| | <p>I found my old Machine Design (Machine Design by Norton) textbook, and this is how it puts it:</p>
<blockquote>
<p>If the diameter of the base circle of a gear is increased without limit, the base circle will become a straight line. If the "string" wrapped around this base circle to generate the involute were still in place after the base circle's enlargement to an infinite radius, the string would be pivoted at infinity and would generate an involute that is a straight line. </p>
</blockquote>
| 35090 | Does the rack for a rack and pinion often have or need to have the involute profile? |
2020-04-11T03:33:00.617 | <p>I am designing a home CNC mill for machining nonferrous metals. I plan to use linear guides to support either a Y-axis table or the 2 legs of a Y-axis moving gantry. </p>
<p>Linear guides are used to provide low friction linear motion along one axis while resisting forces in other directions (both rotational and translational). They come in many sizes as shown below, and must be mounted to very flat surfaces and adjusted for parallelism with respect to a "reference" guide. </p>
<p><a href="https://i.stack.imgur.com/VhW3R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VhW3R.png" alt="enter image description here"></a></p>
<p>I have a number of identical linear guides at my disposal that are 12mm in width which is on the smaller end of the spectrum (although they are lower in height profile). I am curious if using more than 2 linear guides (3 or 4 all in parallel) is beneficial (provided that I mount them all in parallel). Are there any drawbacks I should think about when considering using more than 2 linear guides to guide something?</p>
<p>The reason I ask is (1) I don't want to waste a linear guide if this is a "weakest link" situation, and (2) using 2 linear guides is by far the most common I've seen in commercial designs for linear stages.</p>
<p><a href="https://i.stack.imgur.com/bxauJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bxauJ.png" alt="enter image description here"></a></p>
<p>Someone might ask "if you're worried about rigidity why dont you go buy some bigger stronger linear rails"? I guess that's also part of my same question - could using multiple smaller linear guides additively achieve the same effect as a bigger stronger linear guide (while at the same time maybe being lower profile which I think should decrease the moment arm for the roll axis)?</p>
<p><a href="https://i.stack.imgur.com/ZwCBV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZwCBV.jpg" alt="enter image description here"></a></p>
<p>Here are some diagrams from newports website if they help for reference: <a href="https://www.newport.com/t/manual-positioning-basics" rel="nofollow noreferrer">https://www.newport.com/t/manual-positioning-basics</a></p>
<p>straightness deviation:
<a href="https://i.stack.imgur.com/vy62g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vy62g.png" alt="enter image description here"></a></p>
<p>tilt rigidity:
<a href="https://i.stack.imgur.com/eH0OP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eH0OP.png" alt="enter image description here"></a></p>
<p>load capacity:
<a href="https://i.stack.imgur.com/ChG3r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ChG3r.png" alt="enter image description here"></a></p>
| |linear-motion| | <p>The idea is mostly sound. </p>
<p>I have a load F that is too high for my linear rails....so instead of beefing up my rails, I increase the load sharing across 3 or more rails, thus dividing it.</p>
<p>In reality, I have actually seen such things....but only on poorly built machines as a half solution.</p>
<p>In general one <em>can</em> do anything they want....but that doesn't mean it's the correct way to do it.</p>
<p>Linear rails are specifically designed for uses of 2 rails for a specific type of load and/or torque, in various orientations and environments within specific tolerances and drift or alignment tolerances. </p>
<p>Beyond this, trying to align 3, or four rails along a linear axis accurately, without them binding or causing premature wear on the rails or carriages themselves would be tedious and difficult.</p>
<p>But it is possible...you just wouldn't see anyone do it. </p>
| 35107 | Can I use more than 2 linear guides to support a linear stage? |
2020-04-12T19:15:31.987 | <p>I was recently reading some of Leonardo da Vinci's <a href="https://en.wikisource.org/wiki/The_Notebooks_of_Leonardo_Da_Vinci/XIII" rel="nofollow noreferrer">notes on architecture</a>. One passage on arches states:</p>
<blockquote>
<p>An Experiment to show that a weight placed on an arch does not
discharge itself entirely on its columns; on the contrary the greater
the weight placed on the arches, the less the arch transmits the
weight to the columns. The experiment is the following. Let a man be
placed on a steel yard in the middle of the shaft of a well, then let
him spread out his hands and feet between the walls of the well, and
you will see him weigh much less on the steel yard; give him a weight
on the shoulders, you will see by experiment, that the greater the
weight you give him the greater effort he will make in spreading his
arms and legs, and in pressing against the wall and the less weight
will be thrown on the steel yard.</p>
</blockquote>
<p>There are two parts to my question:</p>
<ol>
<li>What are the physical principals behind Leonardo's statements on the distribution of an arch's load to its columns?</li>
<li>How does his experiment demonstrate these principals?</li>
</ol>
<p>On 1, I interpret his scenario as in the below diagram. There is some arch which is supported at is ends by two columns <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span>. There are additionally two abutments <span class="math-container">$A_1$</span> and <span class="math-container">$A_2$</span> to the sides of the arch to counter the arch's horizontal thrust. A downward force <span class="math-container">$F$</span> is applied to the apex, which is directed through internal compression forces to the arch's ends, where we see <span class="math-container">$F_1$</span> being applied to <span class="math-container">$A_1$</span> and <span class="math-container">$C_1$</span> (and analogously for the other side). My understanding is that no matter how large <span class="math-container">$F$</span> is, the vertical components of <span class="math-container">$F_1$</span> and <span class="math-container">$F_2$</span> must both be <span class="math-container">$F/2$</span> in order to satisfy the static equilibrium constraints.</p>
<p>Then, how do I reconcile this analysis with Leonardo's claim "the greater the weight placed on the arches, the less the arch transmits the weight to the columns"? Is Leonardo assuming a different setup than I have? Or is one of us wrong?</p>
<p><a href="https://i.stack.imgur.com/g2F44.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g2F44.jpg" alt="Arch Diagram"></a></p>
| |statics|building-physics|architecture| | <p>I think he may have been thinking two things,</p>
<ul>
<li><p>Friction.</p></li>
<li><p>Large deformations</p></li>
</ul>
<p>If we follow him in his path of organizing the gradual transfer of vertical reactions to tangential friction forces from the wall taking most the vertical load, he must have been thinking of a gradual change in the mechanism of the arch from a regular arch to an arch with a plastic hinge formed on the apex.</p>
<p>That would make it possible for the vertical forces to be supported entirely by a triangular force mechanism with the external force on the top and a cracked or flexed apex and to friction forces on the two lateral bottom sides as the tangential components of the friction force.</p>
<p>Let's assume at the ultimate case when the apex is fully cracked or turned into a plastic hinge, a triangular truss substituting the arch with the two base angles at 45 degrees and the load P and the friction factor of 0.8 just to illustrate his idea.</p>
<p>The horizontal reaction at the walls</p>
<p><span class="math-container">$F_h= P/2 \quad \text{and the vertical component of friction, Fr} \quad F_r= 0.8P$</span> </p>
<p>So for an angle of 45 degrees, the wall takes 80% of the load.</p>
| 35132 | What does Leonardo da Vinci mean in his statement on arches? |
2020-04-14T03:04:51.080 | <p>I have a worm gear set with no drawings. I need to model them. The only things I know are
1) The Gear ratio is 30, and the worm gear only has one thread. It is a single start as opposed to a multi start worm.
2) The diameter of the worm is 0.7 Inches, and the length of the worm is 0.852 Inch
3) The diameter of the gear is 1.007 Inches, and it has 30 teeth. </p>
<p>I do not know the pressure angle, diametrical pitch or gear module. </p>
<p>What other piece of missing information would I need to know? How do I go about figuring out the diametrical pitch? How do I figure out the module.</p>
| |gears| | <p>You can use angle gages to find pitch angle. Looks like a feeler gage but for angles.</p>
<p>You use thread measuring wires with a micrometer and equations to find thread diameters.</p>
| 35153 | How to I determine the dimentions for a worm gear? |
2020-04-14T08:32:35.230 | <p>I am reading a report titled ``<em>Definition of safe separation criteria for external store and pilot escape capsules</em>'' in which the author uses following expressions.</p>
<p>From Newton's second law:
<span class="math-container">$$ z \left( t\right) = \frac{1}{m} \int \int\limits_{0}^{t}F \left( \tau \right)d\tau dt$$</span></p>
<p>From Taylor Series,</p>
<p><span class="math-container">$$ F \left(\tau \right) = F \left( 0 \right) + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n}}{n!}$$</span></p>
<p>Substituting the Taylor series expansion in first equation, we get</p>
<p><span class="math-container">$$ z \left( t\right) = \frac{1}{m} \left[ \frac{F \left( 0 \right) t^{2}}{2!} + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n+2}}{\left(n + 2 \right)!}\right] + w \left( 0 \right) t + z \left( 0 \right)$$</span></p>
<p>Questions:</p>
<ol>
<li>I am familiar with differntial equation form of Newton's Second law,
i.e., <span class="math-container">$F = m \frac{d^{2}z}{dt^{2}}$</span> but I don't know how the
author obtained integral form, especially the limits of integration.</li>
<li>Since the integral with respect to <span class="math-container">$\tau$</span> is a definite integral, why the author has added constant of integration when integrating with respect to <span class="math-container">$\tau$</span>.</li>
</ol>
| |applied-mechanics| | <h3>Review of Terms</h3>
<p><code>w(t) -> v(t) -> velocity at time t</code></p>
<p><code>z(t) -> x(t) -> position at time t</code></p>
<p><code>w(0) -> initial velocity</code></p>
<p><br/></p>
<h3>Derivation</h3>
<p>We can compute <span class="math-container">$w(t)$</span> from force, since <span class="math-container">$\frac{dw}{dt} = F(t) / m$</span>, so <span class="math-container">$$w(t) - w(0) = \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$</span></p>
<p>Which means <span class="math-container">$$w(t) = w(0) + \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$</span></p>
<p>We can compute <span class="math-container">$z(t)$</span> from velocity: <span class="math-container">$$z(t) - z(0) = \int_{0}^{t} \left( w(\tau) d\tau \right)$$</span></p>
<p>Where we substitute the above integral expression for <span class="math-container">$w(t)$</span>: <span class="math-container">$$z(t) - z(0) = \int_{0}^{t}\left( w(0) + \int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$</span></p>
<p>Since <span class="math-container">$w(0)$</span> is constant, we can compute its integral in closed form: <span class="math-container">$w(0) * t$</span>.</p>
<p><span class="math-container">$$z(t) = z(0) + w(0)t + \int_{0}^{t}\left(\int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$</span></p>
<p>And you can use that Taylor expansion to write an expression for this integral.</p>
<p><br/></p>
<h3>Conclusion</h3>
<ol>
<li><p>The integral form is the double-integral of the differential equation. </p></li>
<li><p>The integration constants you are seeing come from the fundamental theorem of calculus — they’re the initial conditions. Recall the classic kinematic equation: <span class="math-container">$z(t) = z(0) + w_0t + \frac{a_0t^2}{2}$</span> for the same expression, where acceleration is known to be constant. <span class="math-container">$w_0$</span> is <span class="math-container">$w(0)$</span>. The body’s initial velocity applies a linear offset to its motion</p></li>
</ol>
| 35158 | Converting Newton's Second Law Differential Form to Integral Form and Integration Limits |
2020-04-14T10:31:12.003 | <p>According to Wikipedia <a href="http://en.wikipedia.org/wiki/Heat_exchanger" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Heat_exchanger</a>, plate exchangers typically serves lower pressure fluids than shell and tube heat exchanger, why?</p>
<p>Tubes in plate exchangers are simpler and have larger diameter, seems to be more robust to hold higher pressure fluids.</p>
| |heat-transfer|heat-exchanger| | <p>Because the plates are usually very thin for their dimensions and tubes tend to have more strength for the pressure loading.</p>
<p>I would check better references such as Simonson: Heat transfer for more thorough detail instead of wiki...</p>
| 35162 | Why do plate exchangers typically serve lower pressure fluids than shell and tube heat exchanger? |
2020-04-14T13:51:39.843 | <p>I'm trying to spec a stepper motor that would allow me to position a tube, as well as spin the tube continuously in either direction. Ideally at a maximum of 60rpm, but variable speed which be ideal.</p>
<p>The tube would be held in the middle. It is 1035mm long and has a 42mm diameter and weighs 1kg.</p>
<p>What are the calculations I would need to be able to work this out please?</p>
<p>Regards</p>
| |mechanical-engineering| | <p>The rotation at 60rpm creates air drag. It is kind of turbulent and not linear, due to rotation, but we just consider the simple version.</p>
<p><span class="math-container">$Fd = 1/2C_d ρ v^2 A \quad \text{for a bar moving into wind perpendicular to long axis}$</span></p>
<p>because of rotation the speed hence the drag changes from 0 to <span class="math-container">$(1.035/2 *2 \pi)/s$</span> and an average of Va= 1.035pi/s.</p>
<p>Assuming a <span class="math-container">$Cd=0.6 and \rho= 1.2kg/m^3 $</span> and assuming average drag = 1/2 max Fd</p>
<p><span class="math-container">$F_d= 1/2*0.6*1.2*1.01*0.042 \pi^2=0.6/4*1.2*0.042*1.01 \pi^2 *0.50015=0.0376kg \quad \text{at 60RPM}$</span> </p>
<p>This force works at 2/3*L/2, So torque is,</p>
<p><span class="math-container">$ \tau= 2/3* 0.5015*0.0375= 0.0126kgm=0.123 Nm$</span></p>
<p>This torque should be enough to position the bar and rotate it at a maximum of 60 RPM. you can add more if you need a more agile system.
check my numbers, please.</p>
| 35167 | Torque Required for spinning tube |
2020-04-14T19:06:58.613 | <p>Cabins have just been installed on an asphalt car park. The portable cabins weigh 4000 kg with 4 no. 300 mm square spreaders under each foot.</p>
<p>The car park is constructed from:
1" thick rolled asphalt
2" thick bitumen macadam
6" thick granular subbase
well compacted ground</p>
<p>The effective pressure on the asphalt is:
4000 kg = 40 kN / (4 × 0.3²) = 111 kN/m²</p>
<p>The question is whether this load is likely to cause permanent deformation of the asphalt leading to problems with ponding of water in the future.</p>
<p>I know this is a complicated question depending on the semi-solid nature of the asphalt and its rheological characteristics. The temperature for the duration of the portable cabins is not expected to exceed 30°C. The cabins are expected to be in place for approximately a month.</p>
<p>Any guidance, rules of thumb or a point in the right direction would be appreciated.</p>
| |civil-engineering|highway-engineering| | <p>It depends on the tar and bitumen creeping properties.</p>
<p>Some asphalt perform well under transient loads but slowly creep under steady load. Hence the concrete is material of choice for runways.</p>
<p>Although the max 30 sounds like a mild temperature, some asphalt which has been exposed previously to high temps tend to let the aggregate sink and tar float. </p>
<p>Bottom line, in the absence of lab tests, one has to try to check for any signs of pavement deterioration most visible around the posts penetration or intersection of the curbside and street asphalt and the potential depression under street lane painting. </p>
| 35171 | Permanent uneven deformation of asphalt car park under portable cabins |
2020-04-14T20:04:35.317 | <ul>
<li>This question is specifically about stationary or vehicle mounted crossbows, not man-portable ones</li>
<li>The projectile size would measure about 1.8 - 2 meters in length, so the crossbow would have corresponding proportions</li>
<li>The materials would be wood and metal/metal compounds, no plastics</li>
<li>The goal is to have a siege engine capable of cocking itslef, like a recoil operated firearm would</li>
</ul>
<p>The easiest concept I have in my head for this would be as follows:</p>
<ol>
<li>There's a carriage riding on two rails that carries the projectile</li>
<li>The carriage has two holes on the sides</li>
<li>A wire is threaded through these holes</li>
<li>At the front of the crossbow two springs are placed around the wire</li>
<li>When the carriage gets released and propells the projectile it hits the springs and gets pushed back into it's resting position</li>
</ol>
<p><a href="https://i.stack.imgur.com/PqYPq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PqYPq.jpg" alt="Recoil operated crossbow"></a>
<em>(simple sketch of it)</em></p>
<ol>
<li>Could the cocking springs provide enough energy to push against the limbs and get the carriage back into it's resting position?</li>
<li>Are there any pitfalls with a system like this?</li>
<li>Is there a more efficient way to build a recoil operated crossbow?</li>
</ol>
| |mechanical-engineering|springs|coil-spring| | <p>To make my comment into an actual answer, this design is not feasible because the energy balance doesn't add up. All the energy that you have available is stored in the limbs being pulled back. Now you need to do three things with it:</p>
<ol>
<li>Accelerate the projectile</li>
<li>Reload the crossbow</li>
<li>Overcome friction in the system</li>
</ol>
<p>The problem is, that the energy required for reloading is exactly the same as you started out with, but you still want to accelerate your projectile, so both together doesn't work. Not to mention friction losses of the sliding action and in the springs, so even without projectile it would probably not work.</p>
<p>The reason automatic firearms can do this, is that they tap the high pressure gas in the barrel, basically using some of the chemical energy in the gunpowder. </p>
| 35173 | Are recoil operated crossbows/ballistae feasible? |
2020-04-15T02:47:36.213 | <p>I am trying to fabricate a very small elliptical nozzle; if possible, 10um x 20um. I am almost certain that this cannot be achieved with regular machining. Ideally, the material would be corrosion resistant (e.g. stainless, brass) although that is not absolutely necessary. </p>
<p>The depth of the hole would depend on the material. I need to have small deflection with a pressure differential of 80kPa. Therefore, a stronger material like SS would maybe have 1-2mm of depth. </p>
<p>The tolerance is less important than the smoothness of the edges. A hole that is a couple of microns off spec is better than one that has edges with roughness of 1-2um. </p>
<p>Could it be done with laser cutting? Are there any other methods that could achieve such feature size? Thanks.</p>
<p>Edit: added material and tolerance requirements.</p>
| |machining|lasers|cutting| | <p>Do you need that for production at scale? If no, a focused ion beam could easily give you the accuracy you seem to need, but 2mm in depth maybe borderline...</p>
| 35181 | Smallest possible elliptical hole |
2020-04-15T12:25:48.707 | <p>I'm working on 3d printing a planetary gear for LEGO. I've got a constrained set of planetary + sun gears and was looking into designing a custom ring gear.</p>
<p>I've read a little bit into the design of gears but am currently confused. This is what I've found.</p>
<pre><code>Sun gear: 20 teeth, OD=21.3mm
Planet gear: 12 teeth, OD=13.38mm
</code></pre>
<p>In both cases, module (m) = 1 (for the gears to mesh the module has to be the same?)</p>
<p>Pressure angle = 20 degrees.</p>
<p>I was just wondering how I would go about designing such a ring gear in CAD, i.e. figuring out the number of teeth, tooth size, height etc. Thanks in Advance.</p>
| |mechanical-engineering|gears|cad| | <p>As jko suggests, one would search for tutorials based on your selection of CAD software. Fusion 360 has a gear generator as to others. I purchased a <a href="https://woodgears.ca/gear/index.html" rel="nofollow noreferrer">gear generator program</a> to create SVG files for laser cutting. It also allows one to adjust the parameters to meet requirements.</p>
<p>I used your figures for outside diameter in the program, which resulted in a variety of parameters:</p>
<p>12 tooth gear:
11.886 pitch diameter</p>
<p>20 tooth gear:
19.810 pitch diameter</p>
<p>shaft spacing 15.848</p>
<p>The 20 degree pressure angle is taken as a constant.</p>
<p><a href="https://i.stack.imgur.com/S4ATT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S4ATT.png" alt="sun and planet gear"></a></p>
| 35187 | Calculating and Designing a Ring Gear for LEGO |
2020-04-15T14:33:09.817 | <p>This is in continuation/related to the following question asked in the below link</p>
<p><a href="https://engineering.stackexchange.com/questions/23741/eulerian-and-lagrangian-descriptions-of-velocity">Eulerian and Lagrangian descriptions of velocity</a></p>
<p>According to my understanding Lagrangian approach is an approach where you track the individual object of which you want to take measurement of for eg in a fluid flow you track a particular fluid particle and measure the physical quantity of that particular fluid particle.(leading to pathlines)</p>
<p>But in the Eulerian Approach, you basically try to measure the physical properties around a particular area/surface.</p>
<p>In application to continuum mechanics in structure, it is mentioned that the initial/reference position of a body is called as the Lagrangian Configuration and the current position(Mapped from Initial Configuration) is called as Eulerian Approach. </p>
<p>Q1) I am not able to understand why is that so? Is it because of the fact the in Lagrangian Configuration/Initial Condition the body is supposed to be rigid and it becomes easier to understand rigid body motion and when there is no rigid body motion ie when deformation starts to occur we shift to Eulerian Approach to study deformation?</p>
<p>Q2) If the answer to the above question is yes, why is the Eulerian approach used to study deformation and if not why do we use two approaches??</p>
<p><strong>EDIT 1:</strong></p>
<p>Thanks for your reply. I have been using this particular reference. <a href="http://web.mit.edu/abeyaratne/Volumes/RCA_Vol_II.pdf" rel="nofollow noreferrer">http://web.mit.edu/abeyaratne/Volumes/RCA_Vol_II.pdf</a> in which Eulerian and Lagrangian configurations are explained.(Page 6, 1.3).
Here it describes Eulerian configuration as </p>
<blockquote>
<p>"the representation (1.6) which deals with the positions of the particles in the deformed configuration, (the configuration in which the physical quantity is being characterized,) is called the Eulerian or spatial description." </p>
</blockquote>
<p>and for lagrangian</p>
<blockquote>
<p>"If a reference configuration has been introduced we can label a particle by its position
x = χref(p) in that configuration, and this, in turn, allows us to describe physical quantities in Lagrangian form."</p>
</blockquote>
<p>This is why I am getting confused with Eulerian and Lagrangian.
Is there any specific reason so as to why its named like that?</p>
<p><strong>EDIT 2:</strong></p>
<p>Adding on more reference for the same with the same query(Why are these configurations called Eulerian or Lagrangian ? )</p>
<p><a href="https://i.stack.imgur.com/52Fdl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/52Fdl.png" alt="enter image description here"></a></p>
| |mechanical-engineering|mechanical| | <p>I think it is quite not clear for you what is an Eulerian and Lagrangian description.<br>
As you briefly said at the beginning of your post, in the Eulerian approach, you consider a small volume that is fixed is time and space, and study what happens to this volume: what enters or goes out (convection), if there are some forces applied to it, if it is heated or not etc.<br>
In the Lagrangian approach, you consider a material particle an follows its evolution. That is why for example there is no convection term in a Lagrangian description of the motion, since there is no material motion to consider within a material particle.<br>
Lagrangian and Eulerian approaches are thus two different way to describe the evolution of physical quantities, and the equations that result are different.<br>
In solids mechanics, the Lagrangian description of the motion is very widely used,
while the Eulerian approach is usually preferred in fluid mechanics.</p>
<p>EDIT:<br>
I went through the chapter you mention in the reference you sent. It is a very good book, but quite involved if you start studying continuum mechanics. Section 1.3 is very clear about what is the difference between the Eulerian and Lagrangian <strong>approaches</strong> (or description, as written in the book): They are two different "mindsets" to describe the evolutions of physical quantities.<br>
Either you decide that you decide that you sit in a specific location, and look what happens around you. You observe that particles are moving, the temperature is changing etc. You observe the evolution of the physical quantities from that specific location point of view. This is an <strong>Eulerian approach</strong>.<br>
Otherwise you decide to follow a specific particle and move with it. You position is changing with time, and you observe the evolution of the physical quantities from that specific particle point of view. This is a <strong>Lagrangian approach</strong>. </p>
<p>I do not know the reference you are using, but having quite extensively studied continuum mechanics, these are the most common appellations for the various configurations: </p>
<ul>
<li>Initial configuration = reference configuration = undeformed configuration </li>
<li>Current configuration = deformed configuration </li>
</ul>
<p>EDIT:<br>
I was quite surprise you wrote about "Eulerian configuration" and "Lagrangian configuration" since I never encountered this terminology before. Actually, they are not in the reference you provided neither. The terms that I introduced (reference/current/(un)deformed) are the one that are used as well.<br>
You have however Eulerian and Lagrangian <strong>coordinates</strong>, as introduced in the snippet you added in your post. The Lagrangian coordinates <span class="math-container">$\mathbf{X}$</span> are measured in the undeformed configuration, and the Eulerian coordinates <span class="math-container">$\mathbf{x}$</span> are measured in the deformed configuration. You thus have the displacement <span class="math-container">$\mathbf{u}$</span> of a material point, which is the difference between its current position and its original position, that is given by <span class="math-container">$\mathbf{u} = \mathbf{x}-\mathbf{X} $</span>. </p>
<p>Be careful : in your post you say that the current configuration is the Eulerian approach. "Approach" here means a way to describe the motion. "Configuration" relates to the state of the body at a certain time.<br>
When you decide to write your equations that describe the motion of the domain you want to study, you chose one or the other approach. You cannot switch from one to the other. </p>
<p>I think that the questions you asked result from a misunderstanding of these notions, which, I agree, are quite complex.</p>
| 35191 | Eulerian and Lagrangian --configuration-- (continuum mechanics) |
2020-04-18T10:05:16.423 | <p>Tunnel boring machines are a common tool in deep tunnel construction. I am curious if they have ever been mass produced in specific types. What are the possible problems in (a possible) mass production of such machines?</p>
| |structural-engineering|tunnels| | <p>One of the main problem with tunnel boring machines is they are required for a specific tunneling job: a certain sewer tunnel, a one off underground rail tunnel, etc. Once the job has been completed the tunnel boring machine may no longer be required. In a particular location there may not a need to continually excavate tunnels. Tunnel boring machines aren't easily mobile. Logistics are required to move them from site to site.</p>
<p>Tunnel boring machines are made to construct a tunnel of a specific diameter. They may also be constructed to deal with specific types of ground conditions: clay soils, sand, highly stressed rock, etc. A machine built to perform for a specific set of conditions may not be applicable to other ground conditions.</p>
<p>For a given set of tunneling conditions, tunnel boring machines may not be the most appropriate device to use. In near surface tunnels in sandy soils, a cut and cover approach may be more appropriate technically and economically.</p>
<p>Tunnel boring machine manufacturers don't want to have a yard of unsold machines waiting to be purchased. It's better for them financially to make machines as required and to the specifications of the client.</p>
| 35243 | Have tunnel boring machines been mass produced? What are the difficulties in mass producing a tunnel boring machine? |
2020-04-18T15:37:02.557 | <p>Reading about shearing stresses in beams, it is commonly stated that it is necessary to load a section through its shear center, so no twisting is applied to the beam. This is specially important in thin-walled open section, that have very weak torsion strength.</p>
<p>But how do you desing a beam in such a way? For example: if I'm designing a floor beam with a C-shape section, how do I enforce that the shearing forces act through the shear center?</p>
<p>Thanks!</p>
| |mechanical-engineering|structural-engineering|shear|cross-section| | <p>For a C channel with the following dimensions:</p>
<p>b = width of the flange measured to the center of the web, inches or mm</p>
<p>c = Distance to the shear center, inches or mm</p>
<p>h = Height of section measured to the center of flanges, inches or mm</p>
<p>I = The second moment of area, in^4 or mm^4</p>
<p>t = flange thickness, inches or mm</p>
<p><span class="math-container">$ c=\frac{b^2h^2t}{4I_x}$</span></p>
<p>If we apply load on this offset falling outside of the channel opposing the flanges there will be no twisting torque.</p>
<p>But usually in the construction C channels are rarely used as beams, the favorite section fo steel is I beams, or similar shapes because of their symmetry, a large relative I and the fact that they have been designed as beam and already include many requirements to be used as a beam by codes.
However, C channels in most cases (if ever) are used in situations where they are laterally braced by the floor slabs or roof sheathing so the torque created by loading, not through the shear center is resisted by those members.</p>
| 35253 | How do you ensure that a beam is loaded through the shear center? |
2020-04-19T19:46:11.170 | <p>I have a few questions regarding the physical interpretation of the information presented on a Bode plot. I believe it is one of those cases where I "don't know what I don't know", so please bear with me if I meander.</p>
<p>1) The magnitude plot of a Bode diagram represents the ratio of a system's output and input amplitudes, in dB. Given any system, is there a threshold dB which represents "too much", or is it a system-by-system assessment which must be made by the designer? (Note: I believe I know the answer to be the latter, but want to be sure). Could you provide a contrasting example, such as a filter design vs. something with vastly different technical requirements?</p>
<p>2) What is the physical interpretation of a magnitude plot having infinite gain at <span class="math-container">$0$</span> rad/s (for example, the bode plot of a pure integrator)? When an undamped second order system hits resonance, you get a theoretically infinite-dB magnitude response, which is obviously bad. So why does this logic not apply to very low frequencies? Is the infinite gain of a pure integrator simply a mathematical construct?</p>
<p>3) In controller design, what is the physical interpretation of phase margin and gain margin? I understand the underlying mathematics which brings about those particular properties, but not what they represent. For example, the gain margin is the amount of gain which can be added to the system prior to it going unstable, but why would you want more gain? Is it simply that more controller gain allows it to better track the input signal? </p>
<p>4) How does the gain of the system as represented by the magnitude bode plot relate to the gains of the controller? I feel as though much of my confusion stems from this distinction, where large bode plot magnitudes are not necessarily desirable but large controller gains always seem to be so.</p>
<p>Thank you very much in advance.</p>
| |control-engineering|control-theory|stability|frequency-response|feedback-loop| | <p>I apologize in advance for the length of the text but as you can understand these concepts can't be easily defined in few words. I will do some research about the second question and come back to update the answer.</p>
<p><strong>1.</strong> The magnitude bode plot of a system indeed represents the ratio you mentio. However, there is not something like a general "good" requirement for the gain ratio of any given system. Each system represents a different procedure which includes different characteristics. The gain ratio may come from a design specification. Although, the output of any system is very often needed to track some reference either time dependent (like a trajectory or a simple sinusoidal) or constant (a step input of any amplitude). This means you want your gain to be equal to <span class="math-container">$1 \ (0dB)$</span> ideally in order to have zero steady state error. Consider the following first order lowpass filters with transfer functions (they are randomly chosen for demonstration purposes):</p>
<p><span class="math-container">$$L(s) = \frac{1}{s+15}$$</span>
<span class="math-container">$$F(s) = \frac{15}{s+15}$$</span></p>
<p>The first one has DC gain <span class="math-container">$K_1 = 0.0667$</span> which means there is much loss of the input. The second one has DC gain <span class="math-container">$K_2 = 1$</span> which is the ideal in general and also what you want by a filter (at least very often). The bode plots of these two filters are the following:</p>
<p><a href="https://i.stack.imgur.com/ftuNy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ftuNy.jpg" alt="enter image description here"></a> </p>
<p>Notice that while the pase plots are identical the magnitude plots have significant differences. The plot for <span class="math-container">$F$</span> is the ideal case where the gain at low frequencies is <span class="math-container">$0dB$</span> while the gain of <span class="math-container">$L$</span> at low frequencies is very much undesirable. For a more complex example I place here the bode plot of a closed loop system which corresponds to my diploma thesis and I have designed. The closed loop system consists of the following parts:</p>
<p><span class="math-container">$$T(s) = \frac{b_0}{s^2+a_1s+a_2} \rightarrow Plant \ Dynamics$$</span></p>
<p><span class="math-container">$$C(s) = K_p + K_d\frac{15s}{s+15} \rightarrow PD-Controller$$</span></p>
<p><span class="math-container">$$F(s) = \frac{15}{s+15} \rightarrow Lowpass \ Filter$$</span></p>
<p><a href="https://i.stack.imgur.com/u4VFr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u4VFr.jpg" alt="enter image description here"></a></p>
<p>See that at low frequencies the gain is about (<span class="math-container">$0dB$</span>) with too litle variation and approaches the ideal case. Generally the engineer is responsible for tuning the system in order to achieve certain behaviour, which in many cases dictates that the DC gain should be <span class="math-container">$0$</span>.</p>
<p><strong>3.</strong> The purpose of control systems is to be able to control real mechanical systems and in general procedures which take place in the real world. In order to be able to do so, it is almost always great to have a mathematical model which represents the procedure you are trying to control and force certain behaviours. However, since the real world is not perfect you will never (and believe me never) be able to obtain a mathematical model which represents perfectly the real world procedure. First of all, it is almost always prefered to have a linear model (because we have a bunch of design tools for linear models and they are much simpler than non-linear) although the real world is full of non-linearities which means you only approach the real procedure. You will also need sensors to measure data that correspond to the system's behaviour and measurements have noise. Most importantly though is that there are always disturbances and the so called unmodeled dynamics (dynamics that can't be modeled but we know they exist) that influence your system's behaviour and have not been taken into account when designing your controllers and perfoming simulations since the simulations perfect. All these things (linearization, disturbances, noise, unmodeled dynamics) influence your system and add phase and gain to your system which you are not able to take, as said before, into account when designing your controllers. This means that the more gain and phase margin the more robust your systems to these variations is. For example, if you have a gain margin of <span class="math-container">$5dB$</span> and there are some variations that may add up to <span class="math-container">$4.5dB$</span> gain, your system will remain stable regardless of these variations but if these variations add up to <span class="math-container">$6dB$</span> gain your system most probably will go unstable. The same goes for phase margin as well. That does not mean you make your system "infinite" robust because you pay a cost for more robustness but I will leave it to you to earch for robust control (very interesting subject). I suggest you also to look for the so-called disk margin which can tell you many great things about your system including phase and gain margin. </p>
<p><strong>4.</strong> The issue regarding large controller gains is very much debatable. Generally, the larger the controller gains are the more aggresive and fast your system is. For example, if you have a closed loop system with third poles, you want your third pole to be like <span class="math-container">$10$</span> times larger than the two dominant poles in order not to affect the behaviour which is dictated by the dominant poles. This means that the gains of the controller will surely be very large in order to achieve this. However, when controlling a real system you need actuators (dc motors, servo motors, stepper motors etc) which feed the control signal produced by your controller to the real system. These actuators always come with certain limitations. They can produce a certain amount of power, force or torque which are generated by the controller. The larger the controller gains the more torque, for example, is required by the actuator yet the actuator can only provide a certain amount of torque. You need to be very careful because if you do not take into consideration these limitations you can damage the actuator or even the system itself. You always need to saturate the control signal and the more saturation the control signal undergoes the worst your produced results regarding the system's behaviour may be because there are phenomena like integral windup that may come up. </p>
| 35266 | Physical interpretation of a Bode plot |
2020-04-19T20:34:29.227 | <p>What are the degrees of freedom of an internal hinge in a plane beam?</p>
<p>I understand that the bending moment=0, does this mean there is no slope either?</p>
<p>Or does it have both degrees of freedom of deflection (V) and slope (θ)?</p>
| |beam|stiffness| | <p>Let's start talking not about hinges, but supports. Specifically, why do supports generate the forces (including bending moments, if applicable) they do (or don't)?</p>
<p>Think of a simply supported beam under a downward force. Its supports generate upward reaction forces. Why? Because if they didn't, the system would be unbalanced, and we wouldn't have a structure but a mechanism accelerating according to <span class="math-container">$F = ma$</span>. More specifically, though, the supports create their reactions because the specific nodes they are attached to would move otherwise. For example, you can create structures where the deflection at a given unsupported point is zero. See the example below, where the deflection at midspan is zero and vertical reactions at each support.</p>
<p><a href="https://i.stack.imgur.com/7OXre.png" rel="noreferrer"><img src="https://i.stack.imgur.com/7OXre.png" alt="enter image description here"></a></p>
<p>If you then put a support at the midspan, that support's reaction will be zero; after all, it's not having to resist that point's movement.</p>
<p><a href="https://i.stack.imgur.com/MjDw2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MjDw2.png" alt="enter image description here"></a></p>
<p>This is a trivial example of a fundamental relationship between reactions and deflections (including rotations): boundary conditions create reactions when they are inhibiting a deflection. If a given boundary condition doesn't create a reaction, it doesn't resist that deflection.</p>
<p>So, as you've stated, hinges have zero bending moment. What does that tell us? That hinges don't resist rotations. They also don't have force reactions (though axial and shear forces aren't necessarily zero at hinges; they transferred through the hinges, not absorbed), so we also know hinges don't resist deflections.</p>
<p>So, to sum up: hinges can have non-zero deflections and non-zero rotations (specifically, the slope on one side of the hinge can be different from the other side).</p>
| 35268 | Degrees of freedom of an internal hinge in plane beam? |
2020-04-20T01:25:12.033 | <p>Diesel exhaust is hot, and there's a lot of it.</p>
<p>Diesel radiators also get pretty hot.</p>
<p>In a typical diesel engine, which one is dissipating more heat?</p>
<p>EDIT: In response to a request for specificity below, I am specifically interested in heat losses on five different gensets I operate. The engines are a 0.7L Kubota D722 (no turbo), two Lister Petter LPW4 engines (also no turbos), a 2.216L Perkins 404D-22 (also no turbo), and a 4.5L John Deere 4045TF151 (turbocharged). These engines are run at 1800 RPM (except for the kubota, which is 3600 RPM), under roughly half of rated continuous load.</p>
<p>That is my specific situation, but I am looking for a general understanding of the matter, not just situation-specific advice.</p>
| |heat-transfer|energy-efficiency|heat-exchanger|engines|diesel| | <p>They are very nearly equal for typical four-stroke non-turbo diesels under load. A turbo diesel under load should have slightly more radiator loss than exhaust loss.</p>
<p>At the bottom is a link to the technical spec sheet for a Cat 3412 powered genset. It's a probably a bit bigger than what you had in mind. It is a turbo with aftercooler (A/C in the doc below). Since it is setup for stationary power, the turbo is big, heavy, slow and efficient, and the aftercooler has a substantial heat rejection.</p>
<p>All industrial engines will have sheets like this. They can be customized by the mfg for unusual operating conditions such as high altitude or arctic conditions. This lets the dealer's installers get everything sized and arranged right on the second try (and they just love to fly to northern Alaska to fix stuff in the winter).</p>
<p>For over-the-road diesels, the heat rejection for internal jacket water and exhaust would be closer together. Radiators are draggy and heavy and you design vehicle engines to minimize them. You can reject more heat from surfaces in a vehicle.</p>
<p>Cat has a nice engineering primer on engine cooling - <a href="http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199" rel="noreferrer">http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199</a></p>
<p>Vehicles often use air-to-air aftercooling and oil cooling, so the aftercooler and oil cooler heat doesn't end up in the radiator. Most larger engines use jacket water for aftercooling and oil cooling, so you need to add those values to the internal jacket water heat to size the radiator.</p>
<p><a href="http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199" rel="noreferrer">Cat 3412 genset technical data sheet</a> (see section titled heat rejection on page 2)</p>
<p>I'll try to dig up a similar sheet for mobile equipment engines later. Or someone else can ;) </p>
| 35273 | In a diesel engine, does more heat leave the engine through the tailpipe or the radiator? |
2020-04-20T12:28:44.880 | <p>I understand that for fully developed flow inside a tube (with constant properties), Nusselt number doesn’t change in axial direction but it is surprising that it is also independent of Reynolds number which means that the rate of heat transfer (for constant pipe diameter and fluid properties) doesn’t change with mass flow rate (and velocity).
So could someone please give an explanation for this.</p>
| |fluid-mechanics|heat-transfer|heat-exchanger| | <p>TLDR: at any given cross-sectional slice along the length of the pipe, the Reynolds number only affects the overall amplitude of the lateral temperature variation, not its detailed functional form. Since the Nusselt number involves the quotient of the local radial temperature gradient at the wall by the average radial temperature gradient between wall and central axis, the Nusselt number depends only on the details of the functional form of the lateral temperature variation, not on its overall amplitude. Hence, Nusselt number doesn't depend on Reynolds number.</p>
<p>Long and algebra-heavy version: The fully-developed laminar flow in the pipe has the Poiseuille velocity profile</p>
<p><span class="math-container">$$u = 2\langle u\rangle\left(1-\left(\frac{2r}{D}\right)^2\right)$$</span></p>
<p>where <span class="math-container">$\langle u\rangle$</span> is the mean velocity, <span class="math-container">$r$</span> the radial co-ordinate of a cylindrical polar co-ordinate system aligned with the pipe, and <span class="math-container">$D$</span> the pipe diameter.</p>
<p>In terms of Reynolds number,</p>
<p><span class="math-container">$$u = 2\frac{\nu}{D}\,\textit{Re}\,\left(1-\left(\frac{2r}{D}\right)^2\right)$$</span></p>
<p>where <span class="math-container">$\nu$</span> is the kinematic viscosity of the fluid.</p>
<p>The temperature field is governed by the steady, axisymmetric advection-conduction equation</p>
<p><span class="math-container">$$u\frac{\partial\!\! T}{\partial\! z} = \kappa\frac{1}{r}\frac{\partial\!\!}{\partial\!r}\left(r\frac{\partial\!\! T}{\partial\! r}\right)$$</span></p>
<p>where <span class="math-container">$\kappa$</span> is the thermal diffusivity of the fluid, and the <span class="math-container">$\partial^2\! T/\partial\! z^2$</span> term has been neglected on the basis of a geometrical scaling argument that temperature varies more rapidly along the radial direction than along the axial direction. Looking for solutions separable in these cylindrical polar co-ordinates, one finds</p>
<p><span class="math-container">$$T = T_0+T_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-\frac{24}{D^2}r^2\right)$$</span></p>
<p>where <span class="math-container">$T_0$</span> and <span class="math-container">$T_1$</span> are arbitrary constants, <span class="math-container">$\textit{Pr} := \nu/\kappa$</span> is the Prandtl number of the fluid, and <span class="math-container">$z$</span> is the axial (i.e. along-pipe) co-ordinate.</p>
<p>The heat flux per length into the pipe wall is</p>
<p><span class="math-container">$$q = -\mathrm{\pi} kD\left(\frac{\partial\!\! T}{\partial\! r}\right)_{r = D/2}$$</span>
<span class="math-container">$$ = 12\mathrm{\pi} kT_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-6\right)$$</span></p>
<p>where <span class="math-container">$k$</span> is the thermal conductivity of the fluid. The temperature difference between wall and axis is</p>
<p><span class="math-container">$$\Delta\! T := \left(T\right)_{r = 0}-\left(T\right)_{r = R/2}$$</span>
<span class="math-container">$$= T_1\left(\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z\right)-\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-6\right)\right)$$</span>
<span class="math-container">$$= T_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z\right)\left(1-\exp\left(-6\right)\right)$$</span></p>
<p>The Nusselt number is defined as</p>
<p><span class="math-container">$$\textit{Nu} := \frac{q}{\mathrm{\pi}k\Delta\! T}$$</span>
<span class="math-container">$$= \frac{12\exp\left(-6\right)}{1-\exp\left(-6\right)}$$</span></p>
<p>The dependence on Reynolds number (and everything else) has cancelled.</p>
| 35279 | Why Nusselt number for laminar flow in a pipe is independent of Reynolds number |
2020-04-21T05:23:24.093 | <p>In FEM is symmetry technically a boundary condition?</p>
<p>For example using the FEM on a symmetrically loaded and supported beam. The FEM can be simplified by using symmetry.</p>
<p>However, is this symmetry just implemented as a boundary condition or is it technically one? Because it's not actually constraining the movement given that the result is the same with or without it.</p>
<p>So my own impression is it's not a boundary condition?</p>
| |finite-element-method|stiffness| | <p>The symmetry itself is not a boundary condition. It is a property of your system which means that both the geometry and the load are symmetric with respect to an axis or a plane. It allows to reduce the computation to a downsized domain, which leads to considerable computational time saving.<br>
I guess you are using a FEA software and manually reduced the mesh. You then have to tell the software that some of the boundaries are not the physical boundaries of the domain you study, but are part of the plane of symmetry. Under the hood, the software applies some boundary conditions that are consequences of symmetry.<br>
For example, if you have a planar symmetry, no displacement on the axis perpendicular to this plane is allowed, whereas only the rotation around this normal axis is permitted. So to answer your question, in "FEA software" language, symmetry is a boundary condition for the reasons I just explained, but this is not very rigorous.<br>
The fact that you get the same results with and without the symmetry BC may be due to various reasons, one of which may be chance, if your load does not "put into use" the degrees of freedom blocked by the symmetry.</p>
| 35294 | FEM - is symmetry a boundary condition? |
2020-04-22T02:35:42.087 | <p>I'm studying Root Locus method and I still confused. The question below exemplifiques my doubt.</p>
<p>Determine the <span class="math-container">$K$</span> gain so that the dominant roots have a damping factor equal to <span class="math-container">$0.5$</span>, where </p>
<p><span class="math-container">$$G(s) = \frac{K(s^2+5s+9)}{s^2(s+3)}$$</span></p>
<p>I have successfully draw the root locus from this <span class="math-container">$G(s)$</span>, but I can't figure out how I could calculate the gain <span class="math-container">$K$</span>.</p>
| |control-engineering|control-theory| | <p>An analytical solution.</p>
<p>The closed-loop system is <span class="math-container">$G(s)/(1+G(s))$</span> and its poles are those of <span class="math-container">$1+G(s)=0$</span>.</p>
<p>In this case that is <span class="math-container">$$k \left(s^2+5 s+9\right)+(s+3) s^2=0 \ \ \ \ (1)$$</span>.</p>
<p>For general third-order system with a pair of complex dominant poles, the poles are the roots of <span class="math-container">$(\alpha +s) \left(s^2 + 2 \zeta s \omega _n+\omega _n^2\right)=0$</span>. Here <span class="math-container">$\alpha$</span> is the real pole, <span class="math-container">$\zeta$</span> is the damping factor, and <span class="math-container">$\omega _n$</span> is the natural frequency. In this case <span class="math-container">$\zeta=0.5$</span> and hence the equation becomes </p>
<p><span class="math-container">$$(\alpha +s) \left(s^2 + s \omega _n+\omega _n^2\right)=0 \ \ \ \ (2)$$</span></p>
<p>If we equate the coefficients of (1) and (2) we will have 3 equations and 3 unknowns of which <span class="math-container">$k$</span> is one, and hence can be computed. </p>
<pre><code>Expand[s^2*(s + 3) + k*(s^2 + 5*s + 9) - (s^2 + 2*(1/2)*wn *s + wn^2)*(s + a)]
eqns = Thread[CoefficientList[%, s] == 0]
sols = Solve[eqns, {a, k, wn}, Reals]
k /. sols
</code></pre>
<blockquote>
<p>9 k + 5 k s + 3 s^2 - a s^2 + k s^2 - a s wn - s^2 wn - a wn^2 - s wn^2</p>
<p>{9 k - a wn^2 == 0, 5 k - a wn - wn^2 == 0, 3 - a + k - wn == 0}</p>
<p>{{a -> 3, k -> 0, wn -> 0}, {a -> 9/2, k -> 9/2, wn -> 3}}</p>
<p>{0, 9/2}</p>
</blockquote>
<p><span class="math-container">$k=0$</span> is an artifact of the calculations. The answer turns out to be <span class="math-container">$k=\frac{9}{2}$</span>.</p>
| 35303 | Determine the gain K from a root locus |
2020-04-22T11:26:42.417 | <p>I am reading a paper titled <em>Conditions for safe separation of external stores</em> by EE Covert<a href="https://arc.aiaa.org/doi/abs/10.2514/3.44728?journalCode=ja" rel="nofollow noreferrer"> [Link]</a>. In Section 1 <em>Introduction</em> of the paper, the author has given an expression for <span class="math-container">$d$</span>, the relative distance between some point <span class="math-container">$r_{p}$</span> on the store and <span class="math-container">$r_{q}$</span> on the aircraft as:</p>
<p><span class="math-container">$$d = d \left( 0 \right) + \int\limits_{0}^{t} \left\{ \left[ R \right] \cdot \left( U + \omega \times r_{p}\right) - \left( V_{A/C} + \Omega \times r_{q} \right)\right\} \mathrm{d}t $$</span></p>
<p>where,</p>
<ol>
<li><p><span class="math-container">$ V_{A/C} $</span>: aircraft translational velocity in aircraft body coordinates.</p>
</li>
<li><p><span class="math-container">$ \Omega $</span>: aircraft rotational velocity in aircraft body coordinates.</p>
</li>
<li><p><span class="math-container">$ r_{q} $</span>: position vector between aircraft center of gravity, and point of interest in aircraft.</p>
</li>
<li><p><span class="math-container">$ r_{p} $</span>: position vector between origin and point on store.</p>
</li>
<li><p><span class="math-container">$ \left[ R \right] $</span>: rotation matrix to transfer store velocities to aircraft coordinate frame.</p>
</li>
<li><p><span class="math-container">$ U $</span>: store linear velocity in store coordinates.</p>
</li>
<li><p><span class="math-container">$ \omega $</span>: store angular velocity in store coordinates.</p>
</li>
</ol>
<p>I want to derive this expression, but I don't know where to start.</p>
| |applied-mechanics|dynamics| | <p>To derive these EOM, you have to understand that there are two coordinate systems defined here, namely the store coordinate system (<span class="math-container">$X,Y,Z$</span>) and the aircraft coordinate system (<span class="math-container">$X',Y',Z'$</span>). Suppose the store has a linear velocity <span class="math-container">$U$</span> in its coordinate system, meaning <span class="math-container">$U := U_x\hat{i}' + U_y\hat{j}' + U_z\hat{k}'$</span>. In the <em>aircraft</em> coordinate system, the velocity will be different, because the aircraft is translating and rotating as well. In general, the velocity at a point <span class="math-container">$Q$</span> in some reference frame, given the velocity at point <span class="math-container">$P$</span> is given by</p>
<p><span class="math-container">$$
\vec{v}^Q = \vec{v}^P + \vec{\omega} \times \vec{r}^P,
$$</span>
where <span class="math-container">$\omega$</span> denotes the angular velocity of that reference frame. This just comes from regular dynamics. Applied to this problem, the velocity of the origin (see Figure 1 in the paper) given the velocity of the store is </p>
<p><span class="math-container">$$
U_0 = U + \omega\times r_p
$$</span>
However, this is still in the reference frame of the <em>store</em>. In order toconvert this to the reference frame of the aircraft, you have to change reference frames. To do this, you have to multiple by the <em>direction cosine matrix</em> <span class="math-container">$R := R_{A/S}$</span>, which rotates the store frame to the aircraft frame, so in total we get</p>
<p><span class="math-container">$$
^AU_0 = R \ ^SU_0 = R(U + \omega\times r_p).
$$</span>
This is just the first term in the equation in the paper, because it <em>only</em> considers the velocity of the store. However, the aircraft is moving too! So, you apply the same procedure to get the second term, except you have to add in a negative sign. This gives you the full equation that you are trying to understand.</p>
| 35308 | Equation governing relative motion between aircraft and store |
2020-04-22T13:03:16.960 | <p>Consider the borehole record below (in particular the <em>Water Depth</em> column).</p>
<p>There is a white inverted triangle at a depth of 3.20 m. What is the meaning of this symbol?</p>
<p><strong>The white inverted triangle indicates the depth at which water was struck.</strong></p>
<p>There is a black inverted triangle at a depth of 2.50 m. What is the meaning of this symbol?</p>
<p><strong>The black inverted triangle indicates the depth to which the water rose after 20 minutes.</strong></p>
<p>My new question is why does the water rise above where it was originally struck? What is this implication of this? Is the location of the black inverted triangle the <em>true</em> location of the water table within the soil? Why was water struck lower than this?</p>
<p><a href="https://i.stack.imgur.com/69cLL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/69cLL.png" alt="borehole record"></a></p>
| |civil-engineering|geotechnical-engineering| | <p>The water table in some locations may vary during the day or over time. For this reason, the borehole drilling process includes waiting periods to see if there are any changes to the water table.</p>
<p>The depth at which the borehole initially finds the water table may be incorrect, an artifact of the drilling process itself. Therefore, if any changes to the water table level are observed, boring must be halted for a short while (10+ minutes) to let the table stabilize (and this must be included in the report).</p>
<p>I've found a <a href="https://www.rbkc.gov.uk/idoxWAM/doc/Other-1448588.pdf?extension=.pdf&id=1448588&location=volume2&contentType=application/pdf&pageCount=1" rel="nofollow noreferrer">nice report</a> (pdf) stored by the UK government which just so happens to explain this on page 7. It also has nice examples of borehole records (i.e. page 8) which indicate differing water levels over time.</p>
<blockquote>
<h2>GROUND WATER</h2>
<p>The depth at which ground water was struck is entered on the Borehole Records. However, this observation may not indicate the true water level at that period. Due to the speed of boring and the relatively small diameter of the borehole, natural ground water may be present at a depth slightly higher than the water strike. Moreover, ground water levels are subject to variations caused by changes in the local drainage conditions and by seasonal effects. When a moderate inflow of water does take place, boring is suspended for at least 10 minutes to enable a more accurate short term water level to be achieved. An estimate of the rate of inflow is also given. This is a relative term and serves only as a guide to the probable flow of water into an excavation.</p>
<p>Further observations of the water level made during the progress of the borehole are shown including end of shift and overnight readings and the depth at which water was sealed off by the borehole casing, if applicable.</p>
<p>Whilst drilling through granular soils, it is usually necessary to introduce water into the borehole to permit their extraction. When additional water has been used a remark is made on the Borehole Record and the implications are discussed in the text.</p>
</blockquote>
<p>What I'm going to say next is based on what I've found in Brazilian codes; I'm not 100% sure if this also applies elsewhere, but I believe it does.</p>
<p>Once the borehole has reached the requested depth, Brazilian codes (NBR 6484:2001 6.5.5-6) require the removal of water from the borehole in an attempt to lower the measured water table as much as possible. After waiting at least 12 hours, a new water table depth measurement is made and registered in the report.</p>
<p>I'm unsure if this was the exact procedure used in your report, but it seems clear that the two marks indicate the water table observed at two different moments. The report really should come with a legend with this information (especially since water table measurements should come with timestamps).</p>
| 35309 | What is the meaning of these two inverted triangle water table symbols in a borehole record? |
2020-04-25T17:08:50.427 | <p>I'm reading company training materials and I don't understand a few terms. I'm hoping that some of you might be able to decode. The company receives product and delivers it to work sites. I believe these are forklifts or ?order pickers?. An excerpt from the slides:</p>
<blockquote>
<p>"if your branch has a personal man up device like the Little Joey or JLF"...</p>
</blockquote>
<p>Can you help me to understand the terms used in that excerpt? I assume Little Joey is referring to <a href="https://bigjoeforklifts.com/pages/products" rel="nofollow noreferrer">https://bigjoeforklifts.com/pages/products</a> but I couldn't find "JLF"; perhaps a trade name.</p>
| |manufacturing-engineering| | <blockquote>
<p>I'm reading company training materials and I don't understand a few terms.</p>
</blockquote>
<p>So I'm sure that <a href="https://engineering.stackexchange.com/a/35373/1633">alephzero's answer is probably correct</a>, but you're reading <em>company training materials.</em> Acronyms and jargon could mean different things in different fields (sprints and MVP, for example), but I would assume you're reading training material to learn something about your company.</p>
<p>You should find someone in your company you feel comfortable approaching and ask them about this. </p>
| 35370 | Are these "parts pickers"? "JLF" and "Little Joey" |
2020-04-25T21:30:30.650 | <p>Trains where originally designed to keep the train on track. At the time trains where pulled by horses and used to transport coal (and such) out of mines. The two functions have very different requirements:<br>
- Keep the train on track<br>
- Push the train forward<br>
How effective would a solution be to push/pull the train forward that pulls on a serrated track?<br>
Or some other surface that provides far more grip/traction?</p>
| |rail|transportation|propulsion|power-transmission| | <p>Trains do sometimes use different means for propulsion: <a href="https://en.wikipedia.org/wiki/Rack_railway" rel="noreferrer">Rack railways</a> use cogs, and <a href="https://en.wikipedia.org/wiki/Funicular" rel="noreferrer">Funiculars</a> use ropes</p>
<p>I think the reason these are uncommon is that the available traction of smooth tracks is more than sufficient for all but really steep slopes (>10% ). Train engines may be limited in both torque and power when it comes to moving a heavy train up a steep slope, so the lack of traction is only part of the problem. That's why trains tend to avoid slopes, or if absolutely necessary go up in S curves or zigzags.</p>
<p>Racks have hugely greater maintenance costs, and hugely greater friction resistance, so they would be avoided except for special circumstances (very steep slope, very slow speed, and special engine/light load).</p>
<p>Even in mines, I think it is more common to use a combination of smooth tracks and elevators - it's worth dumping ore from one to the other rather than laying down rack rails.</p>
| 35376 | Why do trains use the same wheels for staying on track and propulsion? |
2020-04-25T22:47:41.087 | <p>A satellite launcher has a unit feedback system, whose TF global open loop is given by:</p>
<p><span class="math-container">$$G_c(s)G(s) = \frac{K(s^2-4s+18)(s+2)}{(s^2-2)(s+12)} $$</span></p>
<p>a) Draw the root locus for this function</p>
<p>b) Determine the range values of <span class="math-container">$K$</span> that make this system stable.</p>
<p>I don't know where start to evaluate a), because the TF has the same number of zeros and poles so, in this case, there's no branches and asymptotes?</p>
<p>In the item b), I achieve the following expression to evaluate the Routh–Hurwitz criterion:</p>
<p><span class="math-container">$$s^3(1+K)+s^2(12-2K)+s(-2+10K) -24 + 36K = 0$$</span></p>
<p>However, when I finished the Routh–Hurwitz table and evaluated the inequalities, don't seems correct with the root locus provided by MATLAB.</p>
| |control-engineering|control-theory|matlab| | <p>Suppose we have a third order polynomial in the form :</p>
<p><span class="math-container">$$ s^3+a_2s^2+a_1s+a_0 = 0$$</span></p>
<p>There is nice caveat for third order systems which is derived from the Routh-Hurwitz stability criterion. In order for this polynomial to be stable the following three conditions have to be met (trying to derive the Routh-Hurwitz table will be a total mess for this particular system):</p>
<ul>
<li><span class="math-container">$a_2 > 0$</span></li>
<li><span class="math-container">$a_0 > 0$</span></li>
<li><span class="math-container">$a_2a_1 > a_0$</span></li>
</ul>
<p>The characteristic polynomial of the third order system is:</p>
<p><span class="math-container">$$ (K+1)s^3+(12-2K)s^2+(10K-2)s+36K-24=0 $$</span></p>
<p>which by considering the fact that <span class="math-container">$K>0 \ (\Rightarrow K+1>1>0)$</span> can be rewritten:</p>
<p><span class="math-container">$$ s^3+\frac{12-2K}{K+1}s^2+\frac{10K-2}{K+1}s+\frac{36K-24}{K+1}=0 $$</span></p>
<p>The above requirements for this particular polynomial are:</p>
<ul>
<li><span class="math-container">$\frac{36K-24}{K+1} > 0 \ \Rightarrow \ K > 0.6667 $</span></li>
<li><span class="math-container">$\frac{12-2K}{K+1} > 0 \ \Rightarrow \ K < 6$</span></li>
<li><span class="math-container">$\frac{(12-2K)(10K-2)}{(K+1)^2} > \frac{36K-24}{K+1} \ \Rightarrow \ K\in[0 \ 2]$</span></li>
</ul>
<p>Taking these into consideration we conclude that the gain <span class="math-container">$K$</span> should lie somewhere in between the interval:</p>
<p><span class="math-container">$$ 0.6667 \ \le \ K \ \le \ 2 $$</span></p>
<p>If you indeed try the values <span class="math-container">$0.6666$</span> or <span class="math-container">$2.01$</span> for <span class="math-container">$K$</span> you will see that your system goes unstable. For your information, there is a same caveat for the second order polynomials of the form:</p>
<p><span class="math-container">$$ s^2+a_1s+a_0 = 0$$</span></p>
<p>This polynomial is stable if only and only if <span class="math-container">$a_1,a_0 > 0$</span>.</p>
<p>Now, regarding you root locus of your open loop function, it is somewhat challenging to derive it since there is some complexity going on. You can always use some software to obtain it. This is the root locus from MATLAB:</p>
<p><a href="https://i.stack.imgur.com/weAK8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/weAK8.jpg" alt="enter image description here" /></a></p>
<p>Below is the root locus of the closed loop system for a specific value of <span class="math-container">$K=2$</span>. Notice that the marks for the closed loop poles are indeed located on the imaginary axis which means that the system is critically stable (not strictly stable).</p>
<p><a href="https://i.stack.imgur.com/eBqr6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eBqr6.jpg" alt="enter image description here" /></a></p>
<p>The system becomes strictly stable for values of the gain <span class="math-container">$K$</span> which lie in the interval: <span class="math-container">$(0.6667 \ 2)$</span>. For <span class="math-container">$K=1$</span> the root locus of the closed loop system becomes:</p>
<p><a href="https://i.stack.imgur.com/IXhXj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IXhXj.jpg" alt="enter image description here" /></a></p>
<p>And as a last test the root locus for <span class="math-container">$K=0.6667$</span> of the closed loop system also includes one pole of the closed loop system on the imaginary axis which implies again that the system is critically stable and not strictly stable:</p>
<p><a href="https://i.stack.imgur.com/QMVu0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QMVu0.jpg" alt="enter image description here" /></a></p>
<p>As you see the stability of the system is very well stated throught the Routh-Hurwitz criterion. There are some rules of thumb in order to obtain the root locus of a system, such as that the poles of the system "go" towards the zeros of the system. However, I encourage you to try and obtain some on your own and have some software package to check them. Check also these series on how to draw them by hand. They are really good.</p>
<p><a href="https://www.youtube.com/playlist?list=PLUMWjy5jgHK3-ca6GP6PL0AgcNGHqn33f" rel="nofollow noreferrer">https://www.youtube.com/playlist?list=PLUMWjy5jgHK3-ca6GP6PL0AgcNGHqn33f</a></p>
| 35378 | Root Locus and Routh–Hurwitz stability criterion |
2020-04-26T01:55:01.080 | <p>Fused quartz glass has an extremely low expansion coefficient. I think I once read on the internet that it can go from 1500°C into water without cracking from thermal shock. Sometimes people drip what's left over into water to put it somewhere but that type of glass probably wouldn't crack. Also, it can get heated to a really high temperature before it starts softening to a noticeable amount. I think it might also be stronger than soda lime glass. It's probably also extremely insoluble in water unlike soda lime glass. It can also be put in an annealer so that it will cool very slowly and be free of stress after it cools. So why isn't that the normal regular type of glass we use?</p>
| |mechanical-engineering| | <p>Because compared to ordinary glass, it is expensive to make, and difficult to process. For these reasons it is used only in those applications where its thermal properties are needed. </p>
| 35383 | Why don't we regularly use fused quartz glass? |
2020-04-26T14:16:36.670 | <p>Consider the design of the bracket shown in the figure below (it is assumed that the anchors are sufficient and are not considered below).</p>
<p>My question concerns the detailing of the welded connection detail and the overall strength of the bracket. The design principle is that the welded connection shall have a capacity equal to atleast the moment capacity of the plate. Note that the bracket behaves as a cantilever so the point of maximum moment is the connection itself.</p>
<p>Steel partial factor <span class="math-container">$= 1.15$</span></p>
<p>Permissible stress in steel <span class="math-container">$= \frac{275 \text{ N/mm}^2}{1.15} = 239 \text{ N/mm}^2$</span> </p>
<p>Elastic modulus <span class="math-container">$= \frac{bd^2}{6} = \frac{100 \times 10^2}{6} = 1666 \text{ mm}^3$</span></p>
<p>Moment capacity <span class="math-container">$ = 239 \times 1666 = 398174 \text{ Nmm} = 398 \text{ Nm}$</span></p>
<p>Load <span class="math-container">$F$</span> partial factor <span class="math-container">$= 1.6$</span></p>
<p>Maximum load <span class="math-container">$F_\text{max} = \frac{398 \text{ Nm}}{1.6 \times 0.5 \text{ m}} = 122.5 \text{ N} = 12.5 \text{ kg}$</span></p>
<p>The connection is then designed so that it is atleast as strong as the plate it is joining. There are three options shown in the picture below:</p>
<ol>
<li>closed corner (3 passes with a 6mm leg length E42 electrode)</li>
<li>open corner (3 passes with a 6mm leg length E42 electrode)</li>
<li>semi-open corner (2 passes with a 6mm leg length E42 electrode)</li>
</ol>
<p>My first question is which of the three corner details is preferred for this design? Will all three details provide at least the moment capacity of the plate? Could you recommend an even better detail?</p>
<p>My secondary (bonus) question is whether my maths above is correct and that the bracket should conservatively be capable of supporting 12.5 kg (assuming a suitable anchoring detail)?</p>
<p><a href="https://i.stack.imgur.com/S498a.png" rel="noreferrer"><img src="https://i.stack.imgur.com/S498a.png" alt=""></a></p>
| |structural-engineering|welds| | <p>I haven't done a complete check of your calculations but I do have two comments:</p>
<ul>
<li>You describe the thickness as 10mm yet use 100mm in the calculation of the section modulus, resulting in a capacity 100 times too great.</li>
<li>As Chuck has noted in a comment, you haven't accounted for the shear stresses. (With an elastic stress distribution, it is possible this won't impact the result, though.)</li>
</ul>
<p>Your details 1 and 2 are one-sided fillet welds, and doing one-sided fillet welds like this is generally considered poor practice. They have three big problems:</p>
<ul>
<li>It is very difficult to ensure full quality in the innermost fiber of the weld, which will have a significantly larger stress than the rest of the weld. Therefore, initial cracking/yielding in those fibers is likely, and that will weaken the weld.</li>
<li>Pitting corrossion is likely to occur if there is any possibility of
humidity, resulting in a reduced lifetime.</li>
<li>With a 6mm leg length, you only have a throat thickness of about 4mm, resulting in the section modulus of the weld being
<span class="math-container">$\frac{4^2}{10^2}=16\%$</span> of the section modulus of the plate, so you
should not expect anything close to full capacity.</li>
</ul>
<p>Your detail number 3 is a possibility but since the plates have an overlap of only 4mm (40% of the thickness), you should not assume full strength without a more detailed capacity calculation of the weld.</p>
<p>There are two obvious principles for doing full strength welds:</p>
<ul>
<li>A butt weld with throat thickness equal to the plate thickness with a backing run.</li>
<li>A butt weld with throat thickness equal to half the plate thickness (5mm) on one side combined with a fillet weld with the same throat thickness on the other.</li>
</ul>
| 35396 | How should the welded corner of this bracket be detailed? |
2020-04-27T03:19:13.903 | <p>I am very confused with something regarding bikes...</p>
<p>How is it that a bike wheel continues to spin (either from momentum or from going downhill) when the rider is not pedalling, without also spinning the pedal?</p>
<p>Similarly, what mechanism allows the addition of this spin from momentum with to the spin from the pedalling that makes the wheels spin faster than they would with just one or the other?</p>
<p>I've literally searched all over the internet for this and have found nothing, so I'd really appreciate the help.</p>
| |mechanical-engineering|gears|mechanisms| | <p>Basically several ratchets, below animation shows when you are not pedalling. In the other direction the ratchet will engage and provide force.</p>
<p>If you lift a bike and get the wheel spinning or just leading the bike, sometimes the pedals will turn very slowly due to the friction of ratchets.</p>
<p><img src="https://i.stack.imgur.com/XJ244.gif" alt="Ratchets]"></p>
<p>Image borrowed from <a href="https://www.notubes.com/technology/neo-durasync-speedsync" rel="noreferrer">https://www.notubes.com/technology/neo-durasync-speedsync</a> (no affiliation)</p>
| 35411 | How do bike wheels keep turning when the rider stops pedalling? |
2020-04-27T14:50:16.507 | <p>I am curious to know if a Jacob's Ladder would produce enough hot air to keep a hot air balloon afloat, or in other words, that it could be a substitute for a standard gas burner used for hot air balloons which carry humans. I am not actually planning to build one and use one for a real hot air balloon, it's just a scientific curiosity that I've had for a long time that I would now like to get an answer to from the Engineering community.</p>
<p><a href="https://i.stack.imgur.com/h9pPl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h9pPl.jpg" alt="enter image description here"></a></p>
<p>I am thinking that one could use several car batteries connected in series as the power source and a cable from these batteries would be connected to a high voltage transformer which would provide the high voltage for the Jacob's Ladder.</p>
<p>Not being an engineer, I have no idea how large the Jacob's Ladder would have to be in proportion to the hot air balloon, nor do I have any idea as far as how long the two rods would have be or what their thickness would have to be to handle the high voltage spark and heat from the plasma.</p>
<p>Also, I understand that the Jacob's Ladder could not inflate a hot air balloon by itself, one would still have to use a standard hot air balloon inflater/fan for doing that, yet once it's inflated, the Jacob's Ladder would then be turned on and would produce the hot air needed to make the balloon rise and keep it afloat.</p>
<p>Could a Jacob's Ladder produce enough hot air for a hot air balloon?</p>
<p><strong>EDIT</strong></p>
<p>Based on the comments and answer I've received about this idea, I am now thinking that instead of using car batteries, one could bring on board a small gas-powered electric generator and connect that to the high voltage transformer. A small, lightweight model like the one shown below may be an ideal model to use:</p>
<p><a href="https://i.stack.imgur.com/EWwxx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EWwxx.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering|design| | <blockquote>
<p>I am thinking that one could use several car batteries connected in series as the power source and a cable from these batteries would be connected to a high voltage transformer which would provide the high voltage for the Jacob's Ladder.</p>
</blockquote>
<p>Since we're talking about a balloon, you need energy to heat the air, and that energy will be stored in something like a propane tank (or car batteries as you suggest) then converted to heat via a burner, a resistor, or Jacob's ladder.</p>
<p>What you need to know is how much energy per kilogram you can store. This is called <strong>specific energy</strong> for a battery. This is the amount of energy (in Joules or Watt.hours) that can be stored in 1 kilogram of battery.</p>
<p><a href="https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery" rel="noreferrer">Lead-Acid battery</a>: 30-40 Wh/kg</p>
<p><a href="https://en.wikipedia.org/wiki/Energy_density" rel="noreferrer">Propane:</a> 50 MJ/kg, or 13800 Wh/kg (1 Wh = 1W*3600s = 3600J).</p>
<p>Plutonium-239: 23,222,915,000 Wh/kg (about 23 Twh)</p>
<p>You should consider the complete system's specific energy. For propane you have to add the weight of tanks and burner, which will weight about as much as the propane. So the total weight doubles, which means the specific energy halves, so in the end, rounding down you get say about 6 kWh/kg specific energy. It's pretty good.</p>
<p>This is still way better than what batteries can do, which is why the method you would use to turn electric energy into heat doesn't actually matter for the feasibility of the project. It's the energy storage that matters. This is also why we don't have electric rockets. Specific energy of rocket fuel is a very important parameter since it has to lift and accelerate its own weight.</p>
<p>For Plutonium-239 it would be the other way around: the energy storage would be tiny, but the "burner" would be a problem, if you want to climb in the balloon without getting radiation sunburn then you will need a huge amount of lead and concrete shielding which means it will only be a popular solution for something that doesn't fly, like an aircraft carrier (unless <a href="https://en.wikipedia.org/wiki/Supersonic_Low_Altitude_Missile" rel="noreferrer">you don't care</a>).</p>
<p>Also there is a problem with the <a href="https://en.wikipedia.org/wiki/Hot_air_balloon#Burner" rel="noreferrer">power of a hot air balloon burner</a>, which is at least 2 MW (Megawatts). It isn't that expensive to make a bigger burner, but 2MW of electricity is really huge.</p>
<p>As for the other part of the question "could you use an electric arc to heat stuff" then yes, it is done in <a href="https://en.wikipedia.org/wiki/Electric_arc_furnace" rel="noreferrer">arc furnaces</a>. Problem is, the temperature of the arc is high enough to melt pretty much everything, which is why it's a great choice for a furnace to melt steel... but in open air at high power your electrodes would quickly melt and react with ambient oxygen and burn. Carbon is a popular material for <a href="https://en.wikipedia.org/wiki/Arc_lamp" rel="noreferrer">arc lamp electrodes</a> as it withstands high temperatures, but it burns in air, so it can't be used to heat air.</p>
| 35417 | Could a Jacob's Ladder produce enough hot air for a hot air balloon? |
2020-04-27T20:25:51.490 | <p>I am building a soapbox (go kart) and I am planing on what materials I could use for the frame and the chassis. I have thought about using bamboo or steel. </p>
<p>Steel is normally the go to material when building the chassis and frame. However, I am starting to think bamboo may be a better choice. My reasoning is that: </p>
<ul>
<li>Bamboo is less dense than steel, this makes the structure lighter and could enable it to go faster.</li>
<li>It is cheaper than steel. This leaves room for prototyping and allows me to iterate my way to perfection.</li>
<li>More flexible than steel this could be good as the go kart will jump over bumps and slides. </li>
<li>No welding would be needed.</li>
<li>Depending on the type of steel used, bamboo can have a higher tensile strength than steel. </li>
<li>More eco-friendly.</li>
</ul>
<p>Also, it is possible to make a bamboo composite that produces a stronger bamboo based material.</p>
<p>Could you let me know your thoughts on what I should use: steel, bamboo or something else?</p>
| |structural-engineering|materials|automotive-engineering|steel| | <p>I think the only way to decide this is to make a preliminary design of your cart - twice, once with steel, once with bamboo. You don't need to engineer every detail, but you need to have close look at the critical issues:</p>
<ul>
<li>How to connect individual members (with the tools and skills <em>you</em> have access to)</li>
<li>Cost</li>
<li>Weight, center of gravity</li>
<li>rigidity or stiffness of the whole structure and individual members</li>
<li>How does bamboo fail vs. steel?</li>
<li>Safety or which material is more likely to impale the rider in case of accident</li>
</ul>
<p>In my experience, some issues only show up in manufacturing/building, but many issues will become apparant the moment you make a decent drawing of your design. </p>
| 35433 | Soapbox - GoKart material - Bamboo or steel? |
2020-04-27T22:03:18.510 | <p>I'm working on a design to feed Airsoft BB's through a funnel. A simple gravity fed system (ie. I don't want to force them out at speed) but every now and again the BB's get jammed at the end of the funnel.</p>
<p>I would like to use a motor to constantly keep the BB's moving and prevent jams.</p>
<p>The video below highlights the issue:</p>
<p><a href="https://youtu.be/eq6Sbfp9BUQ" rel="nofollow noreferrer">https://youtu.be/eq6Sbfp9BUQ</a></p>
<p>I would welcome any tips to improve the design/make it functional.</p>
<p>At the moment my thought is to add ridges to the inside of the funnel, this combined with the mixing action should shift them around enough to prevent the jam?</p>
| |mechanical-engineering|fluid-mechanics|manufacturing-engineering|mechanical|pipelines| | <p>What I've seen in commercial applications of this nature involve a vibrating device, which dislodges the stuck balls. It causes random movement within the block or pack of balls and helps to disorganize them enough to allow movement.</p>
<p>From the video clip, you'd want something to move the block in the center out of the way, even just a tiny amount.</p>
<p>Have you considered an oscillating motor to rock the spinning motor through a small angle, similar to an inverted windshield wiper travel? The pivot could be at the end of the gantry holding the spinning motor causing it to rock back and forth, perhaps only a couple of millimeters at the spinner end.</p>
<p>Consider also that an increase in size of the exit will improve the process. For "single file," one needs only to be smaller than twice the size of the ball.</p>
| 35436 | Airsoft BB Funnel Design - Electric funnel jam solution? |
2020-04-29T05:43:47.200 | <p>I devised a setup, and I show it in the diagram below. Suppose I have these specifications:</p>
<ol>
<li>Water in the bottle is 5 liters and it's initial temperature is 28°C.</li>
<li>The bottle capacity is 6 liters.</li>
<li>The tank water is at an initial temperature of 35°C. </li>
<li>The air inlet fan AND the vapor exhaust have a CFM of 1000 (one is in negative I guess). The circular vent cross-sectional radius is 1ft.</li>
<li>The cloth surface area inside the enclosed space is <span class="math-container">$1m^2$</span> (I don't know if this is realistic considering the bottle size. If not, just use a suitable one in calculations.)</li>
<li>Ambient temperature outside enclosed space is 35°C.</li>
<li>Enclosed space volume is <span class="math-container">$5.8m^3$</span></li>
<li>(Don't ask about humidity inside enclosed space, as it is supposed to be constant since one fan will cause vapor, the other takes it out. For calculations, use a random, suitable, humidity %).</li>
</ol>
<p>To bring the water temperature down to 3°C, how long will I have to run the inlet and exhaust fans? Please don't close this outright. It took a lot of energy to make that diagram, so if anything is missing, just comment and I'll update.</p>
<p><a href="https://i.stack.imgur.com/eQAUz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eQAUz.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <p>Your proposed device is an adaptation of a <a href="https://en.wikipedia.org/wiki/Coolgardie_safe" rel="nofollow noreferrer">Coolgardie Safe</a>, devised in the 1890s. To be effective, a continual supply of moving air is required to keep the items with the devices cool.</p>
<p><a href="https://i.stack.imgur.com/S0YzV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S0YzV.jpg" alt="enter image description here"></a></p>
<p>Your proposed device has the water tank on the bottom and relies on capillary action for the cloth to absorb the water. A Coolgardie Safe has the water tank on top and relies on gravity to wet the cloth. Also your device has the item inside draped by the cloth, whereas the Coolgardie Safe has the cloth attached to the frame of the device, creating a large cooling zone.</p>
<p>Depending on atmospheric humidity levels and levels of airflow, <a href="https://lifeonspringcreek.com/2017/02/10/coolgardie-safe/" rel="nofollow noreferrer">reducing the temperature by 3</a> <span class="math-container">$\sf^\circ$</span>C, could take between one hour to and 1.75 hours.</p>
<p>Reducing the temperature to 3 <span class="math-container">$\sf^\circ$</span>C won't happen because of dew point consideration.</p>
<p>Your proposed device and the Coolgardie Safe operate on the same principles as evaporate air conditioners/coolers. They do not have the capability to achieve such low temperatures.</p>
| 35452 | How effectively will this evaporation setup cool down the water? |
2020-04-29T16:08:46.733 | <p>This question is specifically for felt tip pens and marker pens that contain a felt inner tube. The parts of the pen are:
- the body case
- the felt nib
- the felt inner tube</p>
<p>I do not understand is why the ink within the felt inner tube moves up into the nib and then onto the surface in contact with the nib. Why does the ink not stay in the inner tube?</p>
<p>Please comment if you think any other tags should be applied to this question.</p>
| |fluid-mechanics|liquid| | <p>the fibers forming the <em>pen filler</em> are less densely packed than the fibers forming the <em>nib</em>, which extends into the filler a significant distance. This creates a <em>capillarity gradient</em> wherein the nib has a greater affinity for the ink than the filler does, and tends to saturate itself at the expense of the filler. The capillarity gradient guarantees that 1) almost all of the ink in the filler gets used and 2) that the nib doesn't go dry before the ink in the filler us used up. </p>
| 35464 | In marker pens, why does the ink travel to the nib? |
2020-04-30T15:26:05.450 | <p>What is the difference between base shear and pseudo lateral load in seismic analysis of buildings or are they the same?</p>
| |structural-engineering|structural-analysis| | <p>Base shear is the <strong>result</strong> of any type of analysis, including pseudo lateral load and seismic analysis. </p>
| 35484 | What is the difference between base shear and pseudo lateral load in seismic analysis of buildings? |
2020-04-30T17:14:34.663 | <p>My Question:</p>
<p><a href="https://i.stack.imgur.com/x6i9N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x6i9N.png" alt="enter image description here"></a></p>
<p><span class="math-container">$$R1 + R2 - 800 - 300 \cdot 4 = 0$$</span>
<span class="math-container">$$R1 + R2 - 2000 = 0$$</span>
<span class="math-container">$$R1 + R2 = 20000kN -- - (i)$$</span>
<span class="math-container">$$ - R2 \cdot 8 + 300 \cdot 4 \cdot 6 + 800 \cdot 1.5 = 0 $$</span>
<span class="math-container">$$300\cdot4\cdot6 + 800\cdot1.5 = R2 \cdot8$$</span>
<span class="math-container">$$R2 = \frac{8400}{8} = 1050kN$$</span>
<span class="math-container">$$R1 = 2000 - r2$$</span>
<span class="math-container">$$R1 = 2000-1050$$</span>
<span class="math-container">$$R1 = 950kN$$</span></p>
<p><span class="math-container">$$R1 = 950kN$$</span>
<span class="math-container">$$R2 = 1050kN$$</span></p>
<p>Free Body Diagram:</p>
<p><a href="https://i.stack.imgur.com/CyqfR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CyqfR.jpg" alt="enter image description here"></a></p>
<p>Is the calculations and free body diagram correct?</p>
| |mechanical-engineering| | <p>Your answer is indeed correct (neglecting the weight of the beam).
<a href="https://i.stack.imgur.com/OzAob.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OzAob.png" alt="enter image description here"></a></p>
<p>I will recommend this site which i use quite often to solve this sort of problems (it saves me a ton of effort and time): <a href="https://clearcalcs.com/freetools/beam-analysis/au" rel="nofollow noreferrer">https://clearcalcs.com/freetools/beam-analysis/au</a></p>
<p>Cheers!</p>
| 35488 | Reaction forces at R1 and R2 and draw free body diagram |
2020-05-01T01:32:38.493 | <p>I've noticed that aluminum squeeze tubes have a fold at the end. The fold is not the seal of the tube, as I can unfold and see the seal on the very end of the tube once unfolded. <a href="https://en.wikipedia.org/wiki/Tube_(container)#Materials" rel="nofollow noreferrer">Wikipedia</a> states that </p>
<blockquote>
<p>Aluminium tubes generally have the far end folded several times after the contents have been added. The tube is thus hermetically sealed and nearly germ-free due to the high temperatures during the production process. The inside of the tube can be coated to prevent the material from reacting with the contents.</p>
</blockquote>
<p>I don't understand why the 2nd sentence follows from the first, especially in my case where the seal is not the fold, rather it looks like the seal was made, then it was folded. Shouldn't the seal be enough to hermetically seal the tube?</p>
<p>Why are aluminum tubes folded at the end? How does the folding relate to hermetically sealing the tube?</p>
| |aluminum|seals| | <p>The tube is sealed in the following manner:</p>
<p>The aluminum from which the tube is formed is laminated or otherwise coated with a thin layer of plastic on its inside, which is easily heat-fused together and which thereby seals off the material inside from oxygen on the outside. The outer surface is also typically laminated or otherwise coated with another type of plastic to which paint will easily stick, in support of labeling the tube. </p>
<p>The heat-fused end can be split open if the tube is squeezed hard enough. For this reason the end of the tube is folded over on itself several times to make it harder for the heat-fused end to burst open when squeezed. </p>
| 35495 | Why are aluminum tubes folded at the end |
2020-05-01T12:47:03.490 | <p>I'm looking into fitting ball-type inflatable equipment (eg bike tires, yoga balls, basketball ball, etc) to a 2/4 silicone tube. This will be to measure the pressure inside the inflatable, so the other end of the tube goes to a pressure sensor.</p>
<p>The top valve seems to have a <a href="https://en.wikipedia.org/wiki/Schrader_valve" rel="nofollow noreferrer">schrader valve</a> compatible threading.</p>
<p>The bottom end seems to be called descriptively (hose fitting, air pipe fitting, tube connector, pneumatic fitting, etc), are there any DIN or other names for this?</p>
<p>How is each end named of how does one goes about finding such component?</p>
<p><a href="https://i.stack.imgur.com/9kE6c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9kE6c.png" alt="enter image description here"></a></p>
| |pneumatic| | <p>I searched for "Schraeder Hose Barb".</p>
<p><img src="https://i.stack.imgur.com/B852j.png" alt="enter image description here">!</p>
| 35502 | What is this pneumatic connector called? |
2020-05-02T12:35:06.753 | <p>The airflow needed for a system to sustain itself at 80°C with 25°C ambient temperature can be calculated with this formula:</p>
<pre><code>Airflow = (P * t) / (ΔT * D * SHC) [m3/s]
</code></pre>
<p>Where</p>
<pre><code>P = Power [watts]
t = Time [seconds]
ΔT = Difference in Temperature [°C or K]
D = Density of Air [kg/m3]
SHC = Specific Heat of Air [J/(kg*K)]
</code></pre>
<p>For a 105W CPU to remain at 80°C the airflow needs to be:</p>
<pre><code>Airflow = (105 * 1) / (55 * 1.2 * 1000)
Airflow = 0.00159m3/s or 5.27m3/h
</code></pre>
<p>This means any small 80mm fan typically used for case mods to improve airflow by a little:</p>
<p><a href="https://i.stack.imgur.com/SNbRt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SNbRt.jpg" alt="enter image description here"></a></p>
<p>Can cool a 3900X beast CPU that requires an air cooler with a beefy heatsink and a large fan:</p>
<p><a href="https://i.stack.imgur.com/mKLxd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mKLxd.jpg" alt="enter image description here"></a></p>
<p>How can I further improve this formula to take into account the <strong>thermal resistance</strong> of the heat source and some <strong>fluid dynamics</strong> inside the case when it's simplified to just a box?</p>
<p>I know this is more of a thesis amount of work that's needed to be done but I'm only looking at a very rough estimate.</p>
| |fluid-mechanics|thermodynamics|heat-transfer|airflow| | <p>I usually work out the equation from first principles to solve for <em>t</em>.</p>
<p><span class="math-container">$$ t = \frac {\Delta T \times m \times SHC} P$$</span></p>
<p>where <em>t</em> is in seconds, <em>ΔT</em> is in K or °C, <em>m</em> is in kg and <em>SHC</em> is J/kg·K.</p>
<p>In your case you've got:</p>
<p>= SHC of Air ~= 1000 J/kg·K.
- Density of Air ~= 1.2 kg/m<sup>3</sup>.
- Airflow ~= 0.035 m<sup>3</sup>/s.
- Mass flow = airflow × density = 0.035 × 1.2 = 0.042 kg/s
- Power ~= 265 watts</p>
<p>Temperature difference = Power / (Specific Heat of Air * Density of Air * Airflow)</p>
<p>Rearranging we get</p>
<p><span class="math-container">$$ \Delta T = \frac {P \times t} { m \times SHC} = \frac {265 \times 1} { 0.042 \times 1000} = 6.3° \text C $$</span></p>
<p>This tells us that if the temperature of the PC is stable then the air will exit the box 6.3°C warmer than it goes in. Does this match your experimental readings?</p>
<p>You can rearrange the equation to calculate mass flow for a particular <em>ΔT</em>.</p>
<hr>
<p>From the comments:</p>
<blockquote>
<p>Also, the temperature difference should have no unit attached to it.</p>
</blockquote>
<p>That's not true. The difference between any two measurements will have the units of the original measurements. The difference between two lengths will be in metres, the difference between two times will be in seconds and the difference between two temperatures will be in °C or K (or °F if you're from some of the British colonies).</p>
| 35518 | How can I calculate the airflow needed for a fan to cool a system? |
2020-05-03T10:48:01.573 | <p><a href="https://i.stack.imgur.com/1SeKK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1SeKK.png" alt="enter image description here"></a></p>
<p>A Spherical buoy has a diameter of 2m and weighs 3kN. It is designed to be used in a variety or circumstances by it being floated in water and anchored to the floor with a chain.
Draw the free body diagram and determine the following given thta the volume of a sphere is, </p>
<p><span class="math-container">$$Volume =\frac{\pi D^3}{6}$$</span> </p>
<ol>
<li><p>The upthrust or buoyant force (Fb) in sea water.</p></li>
<li><p>The tension in the chain (T)</p></li>
</ol>
<p>Assume at 4Degrees the specific gravity of sea water is 1.042, g = 9.81m/s^2</p>
<p>1.</p>
<p>Equation</p>
<p><span class="math-container">$$Pf Vd g$$</span></p>
<p><span class="math-container">$$Fb = PfVdg = (1.042kg/m^3)(4.188m^3)(9.81m/s^2)$$</span>
<span class="math-container">$$ = 42.8N$$</span></p>
<p>Is this correct? Trying to learn this. I think it is correct. But wanted to check so i understand how to do it correctly. </p>
<p>Then i am going to find question 2.</p>
| |mechanical-engineering|fluid-mechanics|hydrostatics| | <p>A Buoy by definition has to float to be visible to the ships to alert the to the submerged obstacles or the depth datum.</p>
<p>Assuming the weight of the sphere and the tie, W the sphere sinks such that the buoyancy of the part under the water, called a sphere cap, is equal to the wight of the sphere plus approximately the weight of the tie.</p>
<p>Assuming the water <span class="math-container">$ \rho=1 $</span></p>
<p>In a calm sea the buoy will set at H, submersion height such that:</p>
<p><span class="math-container">$ V=W= \frac{\pi H^2}{3}(3r-H) $</span></p>
<ul>
<li><p>r the radius of the sphere</p></li>
<li><p>H is the height of the submersion (Cap)</p></li>
</ul>
<p>they usually anchor the buoy to the floor of the sea with an anchor which is many times heavier the sphere buoyancy even if fully submerged so the maximum tension on the tie is the full buoyancy minus the weight of the sphere.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/VN4Y7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VN4Y7.jpg" alt="buoy and displaced water"></a></p>
| 35524 | Archimedes Principle Question |
2020-05-03T15:21:03.300 | <p><a href="https://i.stack.imgur.com/4Satg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Satg.png" alt="enter image description here"></a></p>
<p>Gabrielli-Von Karman graph plots the specific power, that is, power required to move certain weight, against speed. It gives an idea how much power is required to move certain mass at a certain speed using a certain vehicle.</p>
<p>There are some vehicles whose curves go down at some intervals, such as helicopter, racecar and jet fighter. How does this happen? Curving down implies that we need <em>less</em> power at higher speeds. Why would this be true, especially for aircrafts as we know air resistance is highly proporsional to speed?</p>
| |power| | <p>There is a good reason for airplanes and I guess it can be extended to some race cars with spoilers which function similar to flaps and air-brakes.</p>
<p>An airplane's thrust demand graph (power needed to keep it going) is divided into two parts of the drag curve as per the attached graph.</p>
<p>At slow speeds such as take-off or landing the angle of attack is higher and some flaps are deployed adding to the drag to the advantage of more lift, so there is a lot of "Incidental" drag ( as if when you hold you hands out of a moving car window in a tilt), as the speed increases the angle of attack is reduced and the "Parasite" drag which is basically the friction of the stream of air in which the plane is flying increases. That is why we get the total drag ( power demand ) sloping down in midrange speeds and then go back up.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/EAuwU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EAuwU.jpg" alt="gray curve."></a> </p>
| 35530 | How can a curve in Gabrielli-Von Karman graphs curve down? |
2020-05-04T13:47:39.803 | <p>I have a fan on a PC case running at 100% speed making 40dB/A of noise (sound pressure). Then I add a second fan right next to it running at 100% speed making 30dB/A of noise.</p>
<ol>
<li><p>What is the combined noise level?</p></li>
<li><p>What happens if add and extra 50dB/A fan?</p></li>
<li><p>And then 10 more 30dB/A fans?</p></li>
</ol>
<p>Forget about space, treat them as sources coming from the same point in 3D space.</p>
<p><strong><em>EDIT</em></strong></p>
<p>The total signal level from sources with different strengths can be calculated as:</p>
<pre><code>Lt = 10 log ((S1 + S2 ... + Sn) / Sref)
</code></pre>
<p>Where</p>
<pre><code>Lt = total signal level (dB)
S = signal (signal unit)
Sref = signal reference (signal unit)
</code></pre>
<p><a href="https://www.engineeringtoolbox.com/adding-decibel-d_63.html" rel="nofollow noreferrer">Source</a></p>
<p>I can't directly use this formula because it requires the sound power of fans instead of the sound pressure manufacturers such as <a href="https://noctua.at/en/products/fan" rel="nofollow noreferrer">Noctua</a> typically provide. How can I convert it? What will I need?</p>
| |mechanical-engineering|acoustics|audio-engineering|waves| | <p>This is a pretty straightforward calculator: <a href="https://www.engineeringtoolbox.com/adding-decibel-d_63.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/adding-decibel-d_63.html</a></p>
<p>Basically noise level is dictated by your 2 loudest sources, so 50 dB + 40 dB (assuming they are at or near the same location) would result in about 50.5 dB.</p>
| 35546 | How can I calculate the total signal level (noise) of PC fans? |
2020-05-04T19:01:00.000 | <p>I know this might not be the most ideal place to ask this question, but I couldn't think of any place better.</p>
<p>I recently purchased a high vacuum chamber setup and have been disassembling it for transport. I removed the hot cathode ion gauge, and then on further inspection, I realized it used an unfamiliar mounting flange. It is a gendered connection. Does anyone know what type it is?</p>
<p><a href="https://i.stack.imgur.com/9iZfL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9iZfL.jpg" alt="male flange with brass? gasket"></a></p>
<p><a href="https://i.stack.imgur.com/kay8A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kay8A.jpg" alt="same side"></a></p>
<p>I was weary of trying to remove what looks like a brass gasket. It's pretty thin.</p>
<p><a href="https://i.stack.imgur.com/sbcOk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sbcOk.jpg" alt="female end - no CF style knife edge"></a></p>
<p><a href="https://i.stack.imgur.com/3fzHF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3fzHF.jpg" alt="same side as above"></a></p>
<p>No Conflat style knife edges, but the edge on the first step is fairly sharp, and might ramp upwards (hard to tell)</p>
<p><a href="https://i.stack.imgur.com/uUeEB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uUeEB.jpg" alt="dimensions"></a></p>
| |vacuum| | <p>I have some experience with high vacuum so thought I put in my two cents along with what I was able to find on this particular flange. My first inclination was this is a custom unit and I have since checked with my vacuum community and they are of the same opinion.</p>
<p>Here is an exact quote from the most serior person in our group, "Looks custom. I would think an o-ring would fit over the shoulder in that thin groove and then get compressed when the two flanges are bolted together".</p>
<p>I have come across several unique flanges, especially when it comes to cathode gauges mounting to the chamber side. Here are a couple I just pulled off the shelf in the lab:</p>
<p><a href="https://i.stack.imgur.com/BvsAj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BvsAj.png" alt="enter image description here" /></a></p>
<p>These will likely never see service again and are kept for the extremely unlikely event that they <em>might</em> be useful to someone down the road who could have them machined for their particular application.</p>
<p>Here are some photos of some connections that are currently installed on my 80's era General Ionex Tandem Accelerator:</p>
<p><a href="https://i.stack.imgur.com/X0Q35.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X0Q35.png" alt="enter image description here" /></a></p>
<p>The main point I am making is even though conflats and KFs are ubiquitous today, odd configurations were common a few decades ago and it is usually to deal with attaching the cathode gauge(s) to a custom vacuum chamber. In my experience these unusual pieces work quite well so long as the are in good condition, cleaned and installed properly with good and correctly sized o-rings.</p>
<p>If you need help locating a new o-ring - brass, copper, gold or otherwise, PM me; although it looks like yours is probably fine (assuming there wasn't also a rubber unit, but that is unlikely)</p>
<p>Good luck with your new system, and hope this helps!</p>
| 35550 | Need help identifying this vacuum chamber flange |
2020-05-04T22:00:56.977 | <p>I am analyzing a built-up C-channel for bending purposes (2 point loads of 1200 N each @ 30 cm symmetrically from the CL). The flanges are made of a different material than the web, they are also of a different thickness. The two materials have very different elastic and strength properties. Assume a perfect bond between the flanges and web.</p>
<p><a href="https://i.stack.imgur.com/2Hh2W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Hh2W.png" alt="enter image description here"></a></p>
<p>So far i have used the "equivalent section" trick to convert the flanges' dimensions as if they were made of material 2 and to find the stresses in both the web and flanges, after this i ran an FEA on the beam and the stresses are very close to what the theory predicts.</p>
<p>The next and hardest step in the analysis is to figure out the deflection of the beam due to the bending load. While the "equivalent section" trick is helpful to find the maximum stresses, i'm not sure it can be used to find the actual deflection. Additionally, i need to figure out the <strong>shear center</strong> of the section and all formulas i have seen so far assume a uniform thickness in the web and flanges which is not my case.
Furthermore, since the shear center will be located somewhere to the left of the section, there will be a resultant torque on the section which will induce a shear stress in both the flanges and web. While Roark's formulas for stress and strain provide guidance on how to calculate torsional shear stresses in C-channels, it generally assumes uniform material properties which is not the case of this channel. The materials are a ton weaker in shear than they are tension/compression so this is actually my main concern.</p>
<p>Can anybody recommend any resources, tools or tricks to solve this problem? I have access to my college's mechanics of materials books and even in the advanced edition i could't find anything that might help me. In theory i could simply use FEA software for this section but i will have nothing to verify the results against and I always like to at least have an analytical ballpark number to double check.</p>
<p>Any help would be greatly appreciated. Thank you fellas.</p>
| |mechanical-engineering|structural-analysis|stresses|solid-mechanics| | <p>Assuming your sketch has been drawn to scale, it won't be easy to make a hand calculation of the location of the shear center. The challenge is not the use of two different materials but that the usual assumption of a thin-walled cross section won't be very accurate.</p>
<p>If you need the accurate location of the shear center, you will pretty much have to use a FEM with 3D elements. The shear flow in a thick-walled cross section is too complicated a topic to be worth the bother.</p>
<p>To calculate the approximate location using the approximation of a thin-walled cross section, you can use the following approach:</p>
<ul>
<li><p>Calculate any cross section parameters you need (area, first moment of area and section moment of area) for the transformed cross section, which is very similar to using an equivalent section, but not quite. That is, you pick one material as the reference material (e. g. number 2) and multiply the contribution of the other material (1 in that case) with the ratio of E for the two materials. The difference is that you don't adjust the width or thickness of material 1. The transformed cross section does not have geometric representation as a cross section of a single material.</p></li>
<li><p>Assume the cross section is loaded in pure shear, i. e. a vertical shear load placed in the shear center.</p></li>
<li><p>Use Zhuravskii's shear stress stress formula to calculate the sum of horizontal shear in each flange in material 1. Clearly there will also be some horizontal shear in material 2, but based on the approximation of a thin-walled cross section, we're assuming it is a small contribution.</p></li>
<li><p>There will also be some vertical shear in the flanges in material 1, but we'll assume it's a small contribution. Then the centroid of vertical shear will be located in the centroid of the web.</p></li>
<li><p>The horizontal shear forces and the vertical will both contribute to a torsional equilibrium about the shear center but with opposite signs. The only unknown in that equation is the location of the shear section, so solve for that.</p></li>
</ul>
| 35551 | Bending behavior of built-up C-Channel |
2020-05-04T22:54:32.063 | <p>I have came across this problem and I couldn't figure it out </p>
<p>I have used the following equation </p>
<p><span class="math-container">$$\begin{align}
y &= 6 - \dfrac{2}{3}x \\
V &= 26 - \dfrac{2}{3}x^2 \\
M_x &= 26x-\dfrac{2}{9}x^3
\end{align}$$</span></p>
<p>for the triangular part of it but the answers provided with the question are not matching mine</p>
<p><a href="https://i.stack.imgur.com/dFj2r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dFj2r.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/n9lDS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n9lDS.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/91aXr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/91aXr.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|civil-engineering|statics| | <p>One easy way of drawing the shear and moment diagram is to separate the loading, draw the diagrams, and then superpose them.</p>
<p>Let's call the uniformly distributed load W1 and the triangular load W2.</p>
<p>In the diagram, this load and its shear and moment are shown in blue color.</p>
<p>The top part is the W1 loading and its shear is shown as a blue rectangular on the second row. and its moment which is a triangle on the third row in blue.</p>
<p>The fourth row is the triangular loading that stars from 0 at 6 meters and ends at 4 kN at the support. In the diagram, this load and its shear and moment are shown in tan.</p>
<ul>
<li>The max shear of the W1 is simply</li>
</ul>
<p><span class="math-container">$$ V_1= W_1L=2*10 =20kN \ \text{at support}$$</span></p>
<ul>
<li>The moment of W1 is</li>
</ul>
<p><span class="math-container">$$ \ M= -\frac{W_1L^2}{2}=2*10^2/2=-100kNm$$</span></p>
<ul>
<li><p>The max shear of W2 is the area of the shear diagram</p>
<p><span class="math-container">$$ \ V_2=4*6/2 =12kN$$</span></p>
</li>
<li><p>The max moment of W2 is at the support and is the area of the shear</p>
<p><span class="math-container">$$ M= -W_2L^2/6= -4*36/6=-24kNm $$</span></p>
</li>
</ul>
<p>Therefore we have the max shear at the support</p>
<p><span class="math-container">$$ V_{max}= V_1+V_2= 20+12=32kN$$</span></p>
<p>And max moment again at the support</p>
<p><span class="math-container">$$ M_{max}=-24+(-100)=-124kNm$$</span></p>
<p>.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/EM4tE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EM4tE.jpg" alt="enter image description here" /></a></p>
| 35554 | A statics problem containing a distributed triangular load and a linear load |
2020-05-05T03:30:13.507 | <p>when shopping on digi-key or other electronics distributor, I can see the Torque - Holding spec, given in oz-in and mNm. </p>
<p>I understand Nm in a torque context when the force is applied as a lever at a distance from the rotation point (for instance when applying 1 N force on a wrench, 20 cm from the rotational point, would give 0,2Nm). </p>
<p>But what does it mean when we are talking about motor torque? It seems to me that it is like opposite: the motor is applying force at the rotational point, so the distance is always 0.</p>
<p>Does it mean motor specs are always measuring the force the motor is applying 1 meter away from it? I am trying to have a good intuition of what Nm mean when looking at motors, because tutorials online only talk about wrenches or other cases when the force is applied <strong>away</strong> from the ration point.</p>
<p>Thanks a lot!</p>
| |motors|torque| | <p>Let's say you have a motor with torque rating of 10 Nm and it came in a box with just a shaft.</p>
<p>The rating just means if you attach a pulley with R =1 meters you get ten Nm torque. But if you attach a 0.2 meters pulley you get 5*10=50 Nm torque. Of course your belt runs 5 times slower.</p>
<p>Shaft torque/ demand torque = radius of your pulley in meters. </p>
| 35560 | Basic intuition for motor torque |
2020-05-05T10:54:33.883 | <p>Consider the following extract from a British reinforced concrete drawing from 1970:</p>
<p><a href="https://i.stack.imgur.com/a0oHF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a0oHF.png" alt=" "></a></p>
<p>It states that <em>'REINFORCEMENT TO BE G.K 60. UNLESS STATED OTHERWISE'</em></p>
<p>What is the meaning of GK 60? What property of the steel reinforcing bar is it referring to? To which standard does it pertain?</p>
| |civil-engineering|reinforced-concrete|engineering-history| | <p>Gk and GK are references to a part of Eurocode design formulae.</p>
<p>The gamma attribute (not mentioned in your question) is the factor of safety whereas GK is the load factor. So I would, in my capacity as a UK civil engineering student, take GK to be the gross load at which catastrophic failure would occur. GK is the weight of the structural mass × gravitational force acting on it. Usual factor of safety is 1.25 to 1.4 but most often 1.35</p>
| 35564 | In a reinforced concrete detail drawing what does 'reinforcement to be GK 60' mean? |
2020-05-05T11:20:09.510 | <p><a href="https://i.stack.imgur.com/lrEwB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lrEwB.png" alt="enter image description here"></a></p>
<p>A crane is used to lower a vertical cylindrical pillar into a reservoir. The pillar has a mass of 4 tonnes and is submerged to a depth of 2.5m. It has a diameter of 1.1m.</p>
<p>The specific gravity of fresh water is <span class="math-container">$$1125kg/m^3$$</span>, and <span class="math-container">$$9.81m/s^2$$</span>.</p>
<p>Use a free body diagram to show the forces acting on the pillar when submerged and calculate the upthrust, the tension in the rope in air and the tension in the cable when submerged.</p>
<p>I have this for Upthrust:</p>
<p><span class="math-container">$$ Fbuoy = Pf \cdot Vd \cdot g$$</span>
<span class="math-container">$$\pi r^2h$$</span>
<span class="math-container">$$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$</span>
<span class="math-container">$$1.125 \cdot 2.375 \cdot 9.81 = 26.2N$$</span></p>
<p><span class="math-container">$$Upthrust = 26.2N$$</span></p>
<p>Tension in air</p>
<p><span class="math-container">$$T = w - Fb$$</span>
<span class="math-container">$$\pi r^2h$$</span>
<span class="math-container">$$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$</span>
<span class="math-container">$$Volume = 2.375m^3$$</span>
<span class="math-container">$$ w = m \cdot g$$</span>
<span class="math-container">$$w = 4 \cdot9.81 = 39.24kN$$</span>
<span class="math-container">$$w = 39.24kN$$</span>
<span class="math-container">$$Fb = Pf \cdot Vd \cdot g$$</span>
<span class="math-container">$$1.29 \cdot 2.375 \cdot 9.81 = 30N$$</span>
<span class="math-container">$$T = w - Fb$$</span>
<span class="math-container">$$T = 39.24 - 30 = 9.24N$$</span></p>
<p><span class="math-container">$$Tension in air = 9.24N$$</span></p>
<p>Tension in water</p>
<p><span class="math-container">$$T = w - Fb$$</span>
<span class="math-container">$$\pi r^2h$$</span>
<span class="math-container">$$3.14 \cdot .55^2 \cdot 2.5 = 2.375m^3$$</span>
<span class="math-container">$$Volume = 2.375m^3$$</span>
<span class="math-container">$$ w = m \cdot g$$</span>
<span class="math-container">$$w = 4 \cdot 9.81$$</span>
<span class="math-container">$$w = 39.24kN$$</span>
<span class="math-container">$$Fb = Pf \cdot Vd \cdot g$$</span>
<span class="math-container">$$Fb = 1.125 \cdot 2.375 \cdot 9.81$$</span>
<span class="math-container">$$T = w - Fb$$</span>
<span class="math-container">$$T = 39.24 - 26.2 = 13.04N$$</span></p>
<p><span class="math-container">$$Tension in the cable when sumerged = 13.04N$$</span></p>
<p>Are these correct?</p>
<p>If not could someone help?</p>
<p>I have been trying to complete Archimedes Principle but just getting a little stuck on the calculations. </p>
| |mechanical-engineering| | <p><a href="https://i.stack.imgur.com/mgCL4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mgCL4.png" alt="enter image description here"></a></p>
<p>Hi. Here is what the Free-body diagram should look like. When the cylinder is suspended in the air, the weight of the pillar must equal the vertical tension in the rope. When lowered, the bouyant force pushes the pillar upwards with a force equal to the weight of the volume of water displaced. This bouyancy force reduces the tension in the rope by the amounts shown.</p>
<p>Cheers!</p>
| 35565 | Archimedes Principle |
2020-05-05T14:12:14.863 | <p>I can't even see how can I solve this truss question
I'm an engineering student, our professor assigned us this truss problem.
I have found an answer saying that it is a zero member.</p>
<p>I have tried but I couldn't even see where should I start from.</p>
<p><a href="https://i.stack.imgur.com/9v5EW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9v5EW.png" alt="question"></a></p>
<p>update [answer attempt]</p>
<p>i am working on this problem but i couldn't find any value once i reach a joint with 2 unknowns </p>
<p><a href="https://i.stack.imgur.com/hNYiP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hNYiP.jpg" alt="1"></a></p>
<p>like this joint for example [B]
[<img src="https://i.stack.imgur.com/6mbNZ.jpg" alt="2">]<a href="https://i.stack.imgur.com/6mbNZ.jpg" rel="nofollow noreferrer">3</a></p>
<p>i have even used the momentum equation but i can't go anymore
<a href="https://i.stack.imgur.com/PqfDm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PqfDm.jpg" alt="4"></a></p>
| |civil-engineering|statics| | <p><a href="https://mathalino.com/reviewer/engineering-mechanics/problem-430-parker-truss-method-sections" rel="nofollow noreferrer">https://mathalino.com/reviewer/engineering-mechanics/problem-430-parker-truss-method-sections</a></p>
<p>i have the found the answer here</p>
<p>it is a similar question and it is solved in details </p>
<p>good luck </p>
<p>also take a look on the simple model made by wasabi it will help you for sure </p>
<p>he used ftool for it </p>
| 35566 | I can't even see how can i solve this truss question |
2020-05-06T17:54:26.250 | <p>Is there some kind of graph or relation between <em>laser power</em> and <em>efficiency</em>. What I'm trying to find out is whether the efficiency of a laser decreases with increase in power or is it that there is no definite relation. </p>
<p>For example, which would be better; using <em>10 nos</em> <strong>100 mW</strong> laser diodes concentrated to one point or a <em>single</em> <strong>1 W</strong> laser diode.</p>
<p>Found these two laser diodes:
<a href="https://i.stack.imgur.com/VCGUE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VCGUE.png" alt="Laser Diode Stats"></a></p>
<p>Here are the links to the datasheets of the same, <a href="https://www.frlaserco.com/uploads/files/20150413155503_ML520G73.pdf" rel="nofollow noreferrer">ML520G73</a> <a href="https://www.frlaserco.com/uploads/files/20171222143549_ML562G84-01.pdf" rel="nofollow noreferrer">ML562G84</a></p>
<p>So which will be more efficient using <strong><em>one</em> 2.5 W</strong> laser diode or <strong><em>five</em> 0.5 W</strong> laser diode (the five are set up to point to the same location). And what would be the advantages or disadvantages of both setups?</p>
<p>Thinking of making a laser engraver, but had this idea of using multiple low powered laser diodes instead of a single high powered one. But in a dilemma of what's a better choice. I don't feel complicated in using a multiple lasers instead of a single one.</p>
| |energy-efficiency|lasers| | <p>As I already mentioned in my comment, I would recommend to take a single laser source rather than five seperate laser sources, for multiple reasons:</p>
<ol>
<li>The effort to align the beams of the five diodes on the same spot is out of proportion, when you can instead just get a single beam out of one diode.</li>
<li>The 2.5 W diode that you have been looking at has just as much of a linear current-power behaviour as the 0.5 W diodes, so you can also adjust the power there, if needed.</li>
<li>In case you want less power than the minimum threshold power of the 2.5 W diode: try to avoid reducing the power at all. For pulsed laser manufacturing processes (engraving is one of them), you run the process at 100 % power and influence the power-per-processing-length via the axis speed. That way the laser always operates at its optimal pulse form. The pulse form and peak power are only specified for 100 % power by the manufacturer (that's a chapter for itself, feel free to make a new question if you're interested).</li>
</ol>
<p><strong>Finally, a disclaimer for this kind of questions about lasers:</strong>
The laser sources you picked here are class 4 lasers. I assume from your question (I may be wrong) that you don't have much experience with lasers. Be extremely careful with what you are doing, as they can cause permanent damage, not only to your eyes but also your skin, or fire and explosion hazard. <strong>Follow proper procedures for laser safety and make sure that you don't put yourself or anyone else in danger.</strong></p>
<p>I keep my old answer for the mechanism behind power-efficiency relations here in case it still helps someone:</p>
<p><strong>TL-DR:</strong> There are saturation effects in the laser-active medium that can lead to the laser process running more inefficient the more power we try to "suck out" of the laser-active medium.</p>
<p><strong>Long version:</strong> This is a pretty interesting question, maybe on first glance one might think that, as long as we can cool the losses, there would be no decrease in efficiency with increasing the power. However, what might come the closest to what you ask is the <strong>saturation effect</strong> of the laser-active medium, especially in the case of optically pumped lasers. To understand what is happening here, we need to look a bit deeper into the mechanisms behind the laser process.</p>
<p>For simplification, consider a rod laser with cross sectional area <span class="math-container">$A$</span> and length <span class="math-container">$L_L$</span>, the equations also work for disk and fiber lasers, as they are just extreme cases of rod lasers. The maximum power that we can extract from this laser rod is given by
<span class="math-container">$$P_{max \ extract.}=g_{kl}(v)I_sAL_L$$</span>
This equation can be derived from the equations describing the population inversion of the laser levels. For your question, the two remaining variables in the equation are important, <span class="math-container">$g_{kl}(v)$</span> is the <strong>small-signal gain coefficient</strong> and <span class="math-container">$I_s$</span> is the <strong>saturation intensity</strong>.
The saturation intensity, again, is described by
<span class="math-container">$$I_s = \frac{hv}{\sigma_{ou}\tau_o},$$</span>
so it is a function of the energy of the emitted photon <span class="math-container">$hv$</span> (-> output wavelength of our laser), the cross section <span class="math-container">$\sigma_{ou}$</span> (-> probability that a photon is emitted) and the mean life time <span class="math-container">$\tau_o$</span> (-> average time until a particle in the upper laser level spontanously emits a photon) and thus depends on values that are <strong>inherent to the laser-active medium</strong>.</p>
<p>To return now to your question, there is a saturation mechanism in our laser-active medium that occurs when the intensity of the laser beam that gets amplified in the laser-active medium <strong>reaches the saturation intensity</strong>.
In this case, the <strong>gain coefficient</strong> <span class="math-container">$g$</span> falls down to only half of the small-signal gain coefficient <span class="math-container">$g_{kl}$</span>. What a small gain coefficient means, is, that the population inversion, an absolutely necessary condition for the laser process, gets less and we can not amplify our laser beam as much as we would like to. If the intensity increases even more, the gain coefficient gets even lower. At this point, we see an effect that gets pretty close to your question, a lowered overall efficiency due to higher power.</p>
<p>A graph to illustrate this behaviour is shown below:</p>
<p><a href="https://i.stack.imgur.com/8JXbc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8JXbc.jpg" alt="enter image description here"></a>
Source: Graf, Thomas; Laser - Grundlagen der Laserstrahlerzeugung (<em>Fundamentals of laser beam generation, translation by me</em>); Springer 2015 </p>
| 35596 | Laser Power vs Efficiency |
2020-05-06T21:54:47.140 | <p>I am stuck on the calculation of forces on a certain system involving parallel springs.
The schematic of the system is as follows:</p>
<p><a href="https://i.stack.imgur.com/CkzY7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CkzY7.jpg" alt="enter image description here"></a></p>
<p>The lower plane is fixed and the upper plane movable. There is a symmetric 2x2 matrix of springs between the upper and lower plane. All springs (same size and spring constant) are rigidly attached to the respective planes. The upper plane has a rigid extension (not shown in the image) where forces can be applied such that the upper plane moves/rotates.
In a parallel setup of springs with a symmetric force acting in -z direction I know that I can just add up the spring constants. However what if the force is not parallel to the springs? In case of a single spring I would model the bending of the spring as beam, but what about a spring array? </p>
<p>Is there a way to compute the forces/ torques acting on the upper plane when I know the relative position (rotation and translation) of the upper plane to the lower plane?</p>
<p>Thank you in advance!</p>
| |mechanical-engineering|applied-mechanics|modeling|springs| | <p>Found this answer <a href="https://study.com/academy/answer/a-rigid-plate-of-side-2a-is-supported-on-four-identical-coli-springs-each-of-stiffens-k-at-the-corners-a-b-c-d-as-shown-in-the-figure-below-a-vertical-force-p-is-now-applied-at-a-point-with-the-coord.html" rel="nofollow noreferrer">here</a>. Feel free to check it out for more details.</p>
<p><a href="https://i.stack.imgur.com/ZaHpk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZaHpk.jpg" alt="enter image description here"></a> </p>
<blockquote>
<p><p><b>Given Data</b>: </p> <ul> <li>The stiffness of each spring is k
</li> <li>The total force on the plate is P </li> <li>The length of
the plate is 2a </li> <li>The coordinate axis are x and y </li>
<li>The reaction force on spring A is <span class="math-container">${R_A}$</span> </li> </ul> <ul>
<li>The reaction force on spring B is <span class="math-container">${R_B}$</span> </li> </ul> <ul>
<li>The reaction force on spring C is <span class="math-container">${R_C}$</span> </li> </ul> <ul>
<li>The reaction force on spring D is <span class="math-container">${R_D}$</span> </li> </ul> <p><br> </p>
<p>The expression for force equilibrium, </p> <p><span class="math-container">${R_A} + {R_B} +
{R_C} + {R_D} = P $</span> </p> <p><br> </p> <p>The expression for the moment
is </p> <p><span class="math-container">$\begin{align*} {R_A} \times 2a + {R_B} \times 2a - P
\times \left( {a - x} \right) &amp;= 0........\left( 1 \right)\\ {R_A}
+ {R_B} &amp;= \dfrac{{P\left( {a - x} \right)}}{{2a}}........\left( 2 \right)\\ {R_B} \times 2a + {R_C} \times 2a - P \times \left( {y + a}
\right) &amp;= 0.......\left( 3 \right)\\ {R_B} + {R_C} &amp;=
\dfrac{{P\left( {y + a} \right)}}{{2a}}........\left( 4 \right)
\end{align*}$</span> </p> <p><br> </p> <p>The expression due to the
uniformity, </p> <p><span class="math-container">$\begin{align*} {R_A} + {R_C} &amp;=
\dfrac{P}{2}.......\left( 5 \right)\\ {R_B} + {R_D} &amp;=
\dfrac{P}{2}.......\left( 6 \right) \end{align*}$</span> </p> <p>Substituting
the value of <span class="math-container">${R_C}$</span> in equation <span class="math-container">$\left( 3 \right)$</span> from equation
<span class="math-container">$\left( 5 \right)$</span> </p> <p><span class="math-container">$\begin{align*} {R_B} \times 2a + \left(
{\dfrac{P}{2} - {R_A}} \right) \times 2a - P \times \left( {y + a}
\right) &amp;= 0\\ {R_B} - {R_A} &amp;=
\dfrac{{Py}}{{2a}}.........\left( 7 \right) \end{align*}$</span> </p> <p>By
solving equation <span class="math-container">$\left( 2 \right)$</span> and equation <span class="math-container">$\left( 7 \right) $</span>,
</p> <p><br> </p> <p>Expression for the reaction force in spring A is
</p> <p><span class="math-container">${R_A} = - \dfrac{{Py}}{{4a}} + \dfrac{{P\left( {a - x}
\right)}}{{4a}}........\left( 8 \right)$</span> </p> <p><br> </p>
<p>Expression for the reaction force in spring B is </p> <p><span class="math-container">${R_B} =
\dfrac{{P\left( {a - x} \right)}}{{4a}} +
\dfrac{{Py}}{{4a}}........\left( 9 \right)$</span> </p> <p>Expression for the
reaction force in spring C is </p> <p><span class="math-container">${R_C} = \dfrac{P}{2} -
\dfrac{{Py}}{{4a}} + \dfrac{{P\left( {a - x}
\right)}}{{4a}}........\left( 10 \right) $</span> </p> <p>Expression for the
reaction force in spring D is </p> <p><span class="math-container">${R_D} = \dfrac{P}{2} -
\dfrac{{P\left( {a - x} \right)}}{{4a}} +
\dfrac{{Py}}{{4a}}........\left( 11 \right)$</span> </p> <p>Thus equation
8,9,10,11 are the expression for the forces in A,B,C,D
respectively.</p></p>
</blockquote>
| 35600 | Calculate forces/ torques on a parallel spring array |
2020-05-07T00:45:27.463 | <p>Im in the process of building an espresso machine from readily available parts. I have a solenoid pump that supplies pressurized hot water to the coffee grounds. I want to be able to tune the pressure and flow of the system.</p>
<p>I can tune the max line pressure using an over-pressure valve, but I'm less sure about how I can reduce the flow-rate and maintain whatever pressure I set at the OP valve.</p>
| |fluid-mechanics|pumps|pipelines|valves| | <h1>Short answer:</h1>
<p>I suspect that so long as your pump is a positive displacement pump (ex: the "vibratory pump" or "rotary vane pump" as described in <a href="https://clivecoffee.com/blogs/learn/the-pump-the-heart-of-your-espresso-machine" rel="nofollow noreferrer">this</a> webpage about espresso pumps) and is sized to output the flowrate you desire, then you only need a discharge valve or an appropriately designed flow restriction device (ex: a restriction orifice or your "pressurized filter basket").</p>
<p>If you need to reduce flowrate of a fixed-speed positive displacement pump yet maintain a controlled pressure, read on about installing a recycle line.</p>
<h1>Long answer:</h1>
<p>Controlling mass flowrate and pressure through a pressure vessel (an espresso machine) is possible. I'm not an expert in espresso machines but I am somewhat familiar with the equipment needed to carry out the chemical engineering unit processes of solid-liquid extraction (ex: making espresso); I'm used to larger machinery.</p>
<h2>Pump</h2>
<p>You need a pump that can is capable of providing more than the maximum desired flowrate at more than the maximum desired pressure. A quick search for espresso pumps brings up <a href="https://clivecoffee.com/blogs/learn/the-pump-the-heart-of-your-espresso-machine" rel="nofollow noreferrer">this page</a> about rotary and piston pumps, both <a href="https://www.engineeringtoolbox.com/classification-pumps-d_55.html" rel="nofollow noreferrer">positive displacement</a> types. These pumps tend to output constant flowrate no matter the output pressure (they may still output less due to seal leakage and power limits of the motor).</p>
<p>If your pump isn't a positive displacement type, there are engineering methods that use "<a href="https://www.engineeringtoolbox.com/pump-system-curves-d_635.html" rel="nofollow noreferrer">pump curves</a>" as <a href="https://engineering.stackexchange.com/users/20490/drew">Drew</a> described to take into account quirks about other pump types (ex: centrifugal). However, no matter the type of pump used, as long as sufficient pressure and flowrate capacity in the pump is available, I know of three methods for controlling the pressure and flowrate through the system:</p>
<ol>
<li>Speed control</li>
<li>Discharge control</li>
<li>Recycle control</li>
</ol>
<p>You will need at least two of the three.</p>
<h2>1. Speed control</h2>
<p>"Variable Frequency Drive" or, VFD, is a way to adjust the speed of the pump's rotating elements. Slower generally means less flow rate and pressure.</p>
<p>For your application, I imagine this would be an electronic module that adjusts the frequency and voltage of the alternating current electricity that powers your device. In the U.S. it is probably a frequency of 60 Hertz (Hz) at 120 volts of alternating current (VAC). I am not finding many hits on searches for variable speed espresso pumps so perhaps this type of control isn't popular. This doesn't surprise me since adjusting frequency of power requires complex electronics.</p>
<p>Using a VFD is like adjusting your pump size without having to buy a new pump.</p>
<h2>2. Discharge control</h2>
<p>This type of control involves restricting flow leaving the pressure vessel with an adjustable valve. Generally, when the valve is more closed, an increase in upstream pressure and reduction in flowrate through the system results.</p>
<p>The "pressurized filter basket" you mentioned in a comment on another answer sounds like it may fulfill the role of a discharge control. However you decide to restrict discharge flow, the discharge valve position (or orifice sizes of the basket?) will have to be adjusted in tandem with the other control method you select (Speed control or Recycle control) in order to achieve a desired pressure and flow through the pressure vessel.</p>
<h2>3. Recycle control</h2>
<p>This method of control involves sending a portion of the flow leaving the pump or pressure vessel back into the inlet of the pump.</p>
<p>In your application, this may be achieved by installing two tees:</p>
<ol>
<li>A tee on the high pressure piping at the vessel or pump outlet</li>
<li>A tee on the low pressure piping at the pump inlet.</li>
</ol>
<p>Then, you connect the two tees with a length of pipe containing a flow-restricting valve; this is a "recycle line". When the pump runs, some amount of fluid continuously recirculates if the recycle line valve is open. As the valve closes less fluid is permitted to recycle and the flow of fluid out of the system increases (provided the discharge control valve (if present) is somewhat open).</p>
<p>Recycling fluid wastes some energy since fluid is pressurized by the pump and then depressurized at the recycle valve without leaving the system. The wasted energy heats the fluid somewhat. Recycling fluid might be desired if a limited supply of water is available or more extensive leaching of the coffee grounds bed is desired.</p>
<h2>Maintaining pressure and flowrate</h2>
<p>As I mentioned before, if you have a pump that can output more than the desired flowrate at more than the desired pressure, then two of the three control methods will be necessary to maintain pressure and flowrate within the pressure vessel. This is the result of what in chemical engineering is known as a "<a href="https://en.wikibooks.org/wiki/Introduction_to_Chemical_Engineering_Processes/Multiple_Components_in_Multiple_Processes" rel="nofollow noreferrer">degrees of freedom analysis</a>". You will notice this phenomenon if you set up your system in a <a href="https://en.wikipedia.org/wiki/Category:Chemical_engineering_software" rel="nofollow noreferrer">process simulator</a> (ex: DWSIM (<a href="https://en.wikipedia.org/wiki/Free_and_open-source_software" rel="nofollow noreferrer">FOSS</a>), VMGSIM (proprietary)) and fix pressure and flowrate of the pressure vessel.</p>
<h2>Recycle and Discharge control example</h2>
<p>For example, let's say you employ discharge control along with recycle control. Your positive displacement pump runs at a fixed speed. Fluid is pumped from the pump to a the pressure vessel. The fluid then flows out of the vessel to a tee at which point the flow splits: one stream goes to the discharge control valve and one stream goes to the recycle control valve. How much each valve restricts flow depends upon the desired pressure and flowrate setpoint of fluid passing through the vessel.</p>
<h3>Manual Control</h3>
<p>At this point you could manually adjust the recycle and discharge valves to be more open or more closed and monitor the resulting pressure (ex: with a pressure gauge) and flowrate through the system (ex: with a flowmeter).</p>
<p>A procedure for reaching your target pressure and flowrate might be the following:</p>
<ol>
<li>Completely open the recycle valve.</li>
<li>Completely open the discharge valve.</li>
<li>Start the pump. Flowrate may be lower than desired and vessel pressure may be lower than desired.</li>
<li>Slowly close the recycle valve until the flowrate rises to the desired setpoint. This step will cause the vessel pressure to increase.</li>
<li>Slowly close the discharge valve until the vessel pressure rises to the desired setpoint. This step will cause the flowrate to decrease.</li>
<li>Repeat step 4.</li>
<li>Slowly open the discharge valve until pressure falls to the desired setpoint. This step will cause the flowrate to increase somewhat.</li>
<li>Slowly open the recycle valve until the flowrate falls to the desired setpoint.</li>
<li>Repeat steps 5 through 8 until both the desired discharge flowrate and vessel pressure setpoints have been achieved.</li>
</ol>
<h3>Automatic Control</h3>
<p>It is possible to automate what is achieved by the manual control procedure I described by connecting both the recycle and discharge control valves to computer-controlled actuators and connecting the computer to a pressure transmitter and a flowmeter.</p>
<p>The computer would run two "control loops": a pressure control loop and a flow control loop. Often in industry these will be "<a href="https://en.wikipedia.org/wiki/PID_controller" rel="nofollow noreferrer">PID</a>" control loops (short for "proportional" "derivative" and "integral"). A control loop is a computer program that makes a small adjustment to valve position, waits for a certain amount of time, reads the resulting change in a measured variable (ex: pressure), calculates a new small adjustment, repeating the process. Even though each control loop only responds to a single variable, the system as I've described has two degrees of freedom and therefore two control loops will converge upon the appropriate valve positions.</p>
<p>However, I suspect in your application the valve positions will probably be similar from each batch of espresso to the next, especially if the pressure drop caused by the coffee grounds themselves is similar for each batch. If so, then you may not need automatic control to achieve the pressure and flowrate control you desire.</p>
<h3>Restriction Orifice</h3>
<p>If you found that the quantified flow restriction required by each valve throughout each batch didn't change batch-to-batch, then each valve could be replaced with an <a href="https://en.wikipedia.org/wiki/Discharge_coefficient" rel="nofollow noreferrer">appropriately sized</a> "restriction orifice". I imagine this is what espresso machine manufacturers would do to control temperature and pressure while at the same time reducing the cost of each machine.</p>
<p><a href="https://i.stack.imgur.com/692z7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/692z7.jpg" alt="BORDA INLETS AND ORIFICES"></a>
Examples of piping restriction orifices. Image by <a href="https://commons.wikimedia.org/wiki/File:BORDA_INLETS_AND_ORIFICES_-_NARA_-_17427362.jpg" rel="nofollow noreferrer">NASA</a> released into the public domain.</p>
<h2>Speed and Discharge control example</h2>
<p>If, for example, you perfectly sized your pump to always output the required flowrate no matter the downstream pressure (which is probably nearly the case for positive-displacement pumps), then you would only need a discharge control valve to maintain the desired vessel pressure and flowrate.</p>
<p>Sizing a positive displacement pump correctly in your application would have the same effect as adjusting the speed of your pump motor to achieve a target flowrate, assuming that pressure drop through the coffee grounds does not change throughout a batch.</p>
<h2>Summary</h2>
<p>I suspect that so long as your espresso pump is a positive displacement pump and is sized to output the flowrate you desire, then you only need a discharge valve or an appropriately designed flow restriction device (ex: a restriction orifice or your "pressurized filter basket").</p>
| 35602 | Divide Flow and Maintain Pressure |
2020-05-07T01:04:52.957 | <p>According to the 1D models of compressible fluid flow, the effects of <a href="https://en.wikipedia.org/wiki/Fanno_flow" rel="nofollow noreferrer">pipe wall friction</a> and the effects of <a href="https://en.wikipedia.org/wiki/Rayleigh_flow" rel="nofollow noreferrer">heat addition from the environment</a> both drive a flow toward Mach 1 (i.e. the speed of sound). Whether the flow starts off supersonic (M > 1) or subsonic (M < 1), the M = 1 condition is the maximum entropy point. Wall friction will actually cause a subsonic flow of gas through a pipe to accelerate up to M = 1 (this is still counter-intuitive to me). So why then don't we find that most fluid flows through a pipe ultimately end up with sonic flow velocities?</p>
<p>I also do not understand what happens after the flow reaches M = 1. The textbooks I've skimmed generally say "if there is still pipe length left or heat addition past the point of M = 1, then the inlet conditions must spontaneously change such that the flow reaches M = 1 at the end of the pipe." But they do not explain what these changes are or what mechanism enforces them. What if I am controlling the inlet conditions to be a certain pressure? Then what happens at the end of a rough walled pipe if the flow reaches M = 1 in the middle? Or what if I forcibly add more heat past the M=1 point (in an initially subsonic flow) where the inlet pressure is fixed? Physically, what will happen?</p>
<p><strong>TL;DR</strong>: Entropy is maximized when a fluid flow velocity through a pipe reaches the speed of sound. So why aren't all our natural gas pipelines carrying gas at the speed of sound? Or why not even our water pipes?</p>
| |compressible-flow| | <p>These are good questions to ask yourself.</p>
<p>The most succinct answer to your questions is: Mass Balance and Energy Balance.</p>
<blockquote>
<p>Wall friction will actually cause a subsonic flow of gas through a
pipe to accelerate up to M = 1 (this is still counter-intuitive to
me).</p>
</blockquote>
<p>Gas bulk velocity increases but mass flow rate remains the same.</p>
<p>In a gas transmission pipeline friction causes the pressure of gas to drop as the gas travels down the pipeline. This pressure drop causes a decrease in gas density and a decrease in gas temperature. For steady state flow, the mass flux past any point along the length of the pipeline <strong>must</strong> be equal to all other points. The same cannot be said for velocity since "conservation of velocity" isn't a thing. Here is an equation for gas velocity <span class="math-container">$v_{gas}$</span>, gas volumetric flowrate <span class="math-container">$\dot{V}$</span>, pipe cross-sectional area <span class="math-container">$A_{pipe}$</span>, gas mass flowrate <span class="math-container">$\dot{m}_{gas}$</span>, and gas density <span class="math-container">$\rho_{gas}$</span>:</p>
<p><span class="math-container">$$v_{gas}(x)=\frac{\dot{V}_{gas}}{A_{pipe}}=\frac{\dot{m}_{gas}}{\rho_{gas}(P,T) \cdot A_{pipe}}$$</span></p>
<p>If the mass flowrate <span class="math-container">$\dot{m}_{gas}$</span> is held constant and the pipe is rigid (<span class="math-container">$A_{pipe}$</span> held constant), then a decrease in gas density will cause the velocity to increase proportionally. Eventually this increase in gas velocity will continue until <span class="math-container">$M=1$</span>, available pressure drop falls to zero, or the gas reaches the end of the pipeline (and consumed by the customer).</p>
<p>If the mass flowrate <span class="math-container">$\dot{m}_{gad}$</span> is not held constant, then it may be that <span class="math-container">$M=1$</span> is never reached.</p>
<blockquote>
<p>The textbooks I've skimmed generally say "if there is still pipe
length left or heat addition past the point of M = 1, then the inlet
conditions must spontaneously change such that the flow reaches M = 1
at the end of the pipe." But they do not explain what these changes
are or what mechanism enforces them.</p>
</blockquote>
<p>As someone who worked with pipeline models in the oil and gas industry, here's my take. The book seems to be talking about an edge-case scenario where you purposefully let pipeline pressure fall until you the gas mach number <span class="math-container">$M$</span> manages to reach <span class="math-container">$1$</span>. It seems to be a way of indirectly explaining the nature of <a href="https://neutrium.net/fluid_flow/choked-flow/" rel="nofollow noreferrer">choked flow</a> in few words.</p>
<p>Here are more than a few words:</p>
<p>The changes to inlet conditions would be to gas temperature, pressure, or flow rate in order to raise the speed of sound or decrease bulk gas velocity until the <span class="math-container">$M=1$</span> point hits the end of the pipe.</p>
<p>The mechanism that enforces this conservation of mass and energy. In my experience when flow through a pipe reaches <span class="math-container">$M=1$</span>, no further increase in mass flow rate of a given gas composition is possible unless the gas temperature is increased. The point at which <span class="math-container">$M=1$</span> is like the event horizon of a black hole: no information can be transmitted downstream via pressure (sound) beyond that point.</p>
<p>I believe this point-of-no-return for pressure information at a <span class="math-container">$M=1$</span> point in a pipeline is because if you try to pump gas molecules through a pipe beyond their speed of sound, the excess energy dumped into the pipeline by the pump to perform this acceleration goes into raising the gas temperature instead of the bulk gas velocity in a single direction along the pipe. If small supersonic regions where <span class="math-container">$M>1$</span> of the gas do appear, they are quickly dissipated by turbulence due to the presence of the nearby pipe walls, preventing other groups of gas molecules from achieving <span class="math-container">$M>1$</span>.</p>
<p>For example, let's say I have a positive displacement pump that pushed a constant mass flow rate of gas into a pipeline and I disabled all high discharge pressure and temperature shutdown safety mechanisms. Let's say the pipeline's primary outlet valve were closed due to a downstream equipment problem. At some point a pressure relief valve along the pipeline opens and begins venting gas to atmosphere. As pipeline pressure builds up, mass flow rate through the relief valve increases. Increased mass flow rate means increased gas bulk velocity which means increased <span class="math-container">$M$</span> at the relief valve. <span class="math-container">$M$</span> would rise until <span class="math-container">$M=1$</span>. However, at <span class="math-container">$M=1$</span>, additional increases in relief valve's inlet pressure cannot increase the mass flow rate any further for the reason I stated earlier. Some increase in mass flow rate would be achieved due to increased temperature caused by the adiabatic heating of the pipeline gas (as the runaway pump continued to shove gas into the pipeline) but this does not raise <span class="math-container">$M$</span> since the speed of sound rises with rising temperature (both temperature and speed of sound are functions of <a href="https://www.physicsforums.com/threads/speed-of-sound-and-density-relation.616698/post-3975464" rel="nofollow noreferrer">the root mean square of gas molecule velocities</a>). Eventually the flow rate through the pressure relief valve (hopefully one designed to handle the flow of the runaway pump) would stabilize with <span class="math-container">$M=1$</span> at its orifice (the "end of the pipe").</p>
<p>The principle I'm trying to get at is that the system will adjust in response to whatever variables you fix. If you try to fix more variables than degrees of freedom you have available in your system, then you'll find that you cannot achieve control of one of the variables.</p>
<blockquote>
<p>What if I am controlling the inlet conditions to be a certain
pressure? Then what happens at the end of a rough walled pipe if the
flow reaches M = 1 in the middle? Or what if I forcibly add more heat
past the M=1 point (in an initially subsonic flow) where the inlet
pressure is fixed? Physically, what will happen?</p>
</blockquote>
<p><span class="math-container">$M=1$</span> requires energy to sustain. I know this because I have worked near 15 MMSCFD (million standard cubic feet per day, or 17,658 standard cubic meters per hour) relief valves going off: you can hear its scream for miles (the extreme turbulence of gas from within the piping). In a pipeline, that energy is provided by pressure drop. This pressure drop can be provided by increasing upstream pressure with a pump/reservoir or by decreasing downstream pressure ("backpressure"). Heating a section of piping may increase pressure.</p>
<p>But let's explore a situation that attempts to hit all your question's theoretical requirements. Let's say we have a new empty pipeline at atmospheric pressure. Let's say we decide to violently fill it with gas from a set of compressors and heat exchangers so powerful that we can instantly and continuously maintain a controlled inlet gas temperature and pressure. It will be a violent affair but let's imagine this crazy boundary condition for this dynamic model. The pipeline is initially quiet and uniformly at <span class="math-container">$1 atm$</span>. We start the compressors and the inlet pressure and temperature are instantly on target at the mouth of the pipeline. The gas immediately enters the pipeline causing a high pressure wave of gas to flow at its speed of sound. This speed of sound varies with the front's temperature. The front temperature is much cooler than the inlet temperature due to the <a href="https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect" rel="nofollow noreferrer">Joule-Thompson effect</a>. All the while the compressors and heat exchangers stubbornly maintain a constant inlet pressure and temperature, injecting enormous quantities of heat and mass, although the mass flow rate decreases gradually. At some point in time there is a situation where <span class="math-container">$M=1$</span> is located midway along the pipeline length. No gas has been able to travel downstream of this point since pressure information cannot be transmitted downstream through the <span class="math-container">$M=1$</span> point. Pressure downstream of <span class="math-container">$M=1$</span> remains at atmospheric pressure. Adding heat past the <span class="math-container">$M=1$</span> point doesn't do anything but warm empty pipe.</p>
<blockquote>
<p>So why aren't all our natural gas pipelines carrying gas at the speed
of sound? Or why not even our water pipes?</p>
</blockquote>
<p>Because natural gas pipeline companies want to maximize mass flowrate while minimizing cost within government regulations. Mass flowrate is higher if gas density within the pipeline is higher. Gas density for natural gas nearing <span class="math-container">$M=1$</span> is low and can cause problems such as <a href="https://www.youtube.com/watch?v=M7h59Tg9DS4" rel="nofollow noreferrer">hydrate formation</a> for long runs of pipe.</p>
<p>As for water pipes, liquid water flowing at high speeds <a href="https://wiki.pengtools.com/index.php?title=Erosional_velocity" rel="nofollow noreferrer">erodes</a> piping. It is also incompressible so you don't get much benefit at all in terms of increased density for transferring it at higher pressures. Transporting water in its vapor phase for distances long enough for pressure-drop-induced <span class="math-container">$M=1$</span> situations to occur is rare and expensive (especially since steam systems require insulation and boilers to prevent condensation back to its liquid state).</p>
| 35603 | Why don't all gas pipe flows end up at sonic speed? |
2020-05-07T22:20:40.413 | <ol>
<li><p>Why haven't car manufacturers caught up with Tesla's automobiles?</p></li>
<li><p>E.g. why haven't the <a href="https://old.reddit.com/r/OutOfTheLoop/comments/4czrt7/whats_difference_between_the_tesla_model_3_and/d1p5pn0/" rel="nofollow noreferrer">Chevy Bolt, Nissan Leaf</a>, <a href="https://old.reddit.com/r/OutOfTheLoop/comments/2sipxv/why_is_tesla_the_only_electric_car_company_people/cnr342z/" rel="nofollow noreferrer">VW e-Golf</a> driven (pun intended) TSLA out of business?</p>
<p>I list TSLA's advantages that don't appear grueling or covert or confidential to mimic, like <a href="https://old.reddit.com/r/OutOfTheLoop/comments/2sipxv/why_is_tesla_the_only_electric_car_company_people/cnq2af2/" rel="nofollow noreferrer">its exterior appearance</a>. Yet u/skogoa wrote that <a href="https://old.reddit.com/r/OutOfTheLoop/comments/2q8m0j/what_is_so_innovative_about_tesla_motors/cn49ohn/" rel="nofollow noreferrer">"Tesla is very good at marketing. Their technology isn't all that special but they have managed to build quite a lot of hype."</a></p></li>
</ol>
<p><a href="https://old.reddit.com/r/OutOfTheLoop/comments/2q8m0j/what_is_so_innovative_about_tesla_motors/cn3vhct/" rel="nofollow noreferrer">i_start_fires. 62 points 5 years ago</a></p>
<blockquote>
<p>Well, they're the only company producing cars that can run 250+ miles on electric power alone. They are single-handedly building a network of charging stations in the US making cross-country electric travel a viable reality for the first time. They recently released their entire patent portfolio into the public domain, allowing any company to use their electric car, battery, and charger designs for free.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/OutOfTheLoop/comments/1l4it0/whats_the_deal_with_tesla_cars/" rel="nofollow noreferrer">Turtlecupcakes. Aug 25 2013.</a></p>
<blockquote>
<ul>
<li><p>Their range nearly competes with gas (2-300 miles, I believe)</p></li>
<li><p>They look like high-end vehicles, not some pile of plastic that runs on batteries.</p></li>
<li><p>The entire in-car dashboard and computer system is built from the ground up so that software can control nearly every single motor, display, relay, and so on. This means that all sorts of functionality can be added at any point through firmware updates. Most existing car manufacturers just build new cars on the old computer systems that were basically hardwired to perform specific function. (So even though the AC is digitally controlled, the only way to change its behavior is to basically take the car apart, pull out the microcontroller, and reprogram it.)</p></li>
</ul>
</blockquote>
<p><a href="https://old.reddit.com/r/OutOfTheLoop/comments/1l4it0/whats_the_deal_with_tesla_cars/" rel="nofollow noreferrer">NiceTryNSA. 26 points. Aug 26 2013.</a></p>
<blockquote>
<p>Recently, they beat out all other cars PERIOD in safety ratings, getting the highest safety rating ever awarded to a vehicle. On the corner-load test, Telsa actually broke NHTS's $250,000 machine.</p>
<p>Also: C&D Car of the Year, and Consumer Reports gave it the highest score ever given to a car: a 99/100.</p>
</blockquote>
| |automotive-engineering| | <p>The other car manufacturers have nothing to gain from destroying Tesla, and nothing much to lose by letting it potter along in its own little niche market.</p>
<p>Tesla seems to have made the brain-dead strategic decision to play "not invented here" with everything. Don't collaborate with battery manufacturers, build your own battery plants. Don't use existing dealerships for sales and service, build your own network from scratch. Build your own proprietary national (or international) charging station network. And so on, and so on.</p>
<p>Patents don't have any value unless you can use the technology efficiently. As a wannabe global car manufacturer Tesla is hopelessly inefficient compared with the competition. That's one reason they have never succeeded in making a profit consistently for a whole year. In terms of "vehicles produced per employee per year", they are a factor of three behind the industry leaders. If they fixed that problem at their current output level, and kept their current (top of the market range) prices, they would be making \$1bn to \$2bn profit a year, instead of numbers hovering around zero. (That estimate ignores any one-off costs involved in firing about 30,000 employees, which is what they would need to do play in the same league as the competition). Those profit figures would still be peanuts compared with GM's \$14bn or Toyota's \$50bn in 2019.</p>
<p>This inefficiency issue is even more strange, because logically an EV should be a much simpler product to manufacture than a conventional car. </p>
<p>From the point of view of GM or Toyota, a partnership with Tesla would be like an international Olympic squad choosing to become dependent on "athletes" who can just about run 100 meters in 30 seconds, and manage a high jump of 2 feet six inches.</p>
<p>Of course the Tesla fanboys don't care about any of this. So long as Musk remains a god, they will buy. What happens when Musk either discovers he isn't immortal, or gets bored with cars and takes personal leadership of his plan to colonize Mars by taking a one-way trip there, is "tomorrow's problem". </p>
| 35620 | What's so special about Tesla's all-electric automobiles, compared to other car manufacturers'? |
2020-05-08T13:47:41.643 | <p><a href="http://www.refinerlink.com/blog/US_PADD_Overview/" rel="nofollow noreferrer">Crash Overview of U.S. PADDs and Why They’re Important | RefinerLink</a></p>
<blockquote>
<p><a href="https://i.stack.imgur.com/BR3rt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BR3rt.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\color{red}{[1.]}$</span> Although 49% of US refining capacity resides in PADD III, notice how 58% of domestic crude is produced in PADD III and 55% of net crude imports also go to PADD III. <span class="math-container">$\color{red}{[2.]}$</span> This implies that crude oil must move out of PADD III to other PADDs to remain <span class="math-container">$\color{forestgreen}{balanced}$</span>. As a matter of fact, roughly 1.4 Million barrels a day of oil moves from PADD III to PADD II and PADD IV.</p>
</blockquote>
<ol>
<li><p>I don't grasp the implication. How does <span class="math-container">$\color{red}{[1]}$</span> above imply <span class="math-container">$\color{red}{[2]}$</span>?</p>
</li>
<li><p><span class="math-container">$\color{forestgreen}{balanced}$</span> with what? Why must the PADDs' crude oil be balanced?</p>
</li>
</ol>
| |petroleum-engineering| | <p>The total crude throughput is domestic production plus net crude imports.</p>
<p>All of the crude gets refined.</p>
<p>So if 49% of the total crude is refined in PADD III, and more than 49% of domestic production is in PADD III, and more than 49% of net imports arrive at PADD III, It doesn't matter what the ratio of imports to domestic production is, the total amount has to be more than 49%. So crude has to be shipped out of PADD III to refiners in other PADDs.</p>
| 35631 | Why do 58% of domestic crude produced in PADD III and 55% of net crude imports also go to PADD III imply that crude oil must move out of PADD III? |
2020-05-08T17:37:17.367 | <p>If I were to remove all the air from a canister of WD40 so that it was a vacuum inside, and reinforced the walls so that it won't collapse in from atmospheric pressure, what would happen when I pull the trigger? Will it now suck like a vacuum cleaner?</p>
| |fluid-mechanics|pressure|air| | <p><a href="https://i.stack.imgur.com/s0I6N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s0I6N.png" alt="enter image description here"></a></p>
<p><em>Figure 1. A typical aerosol valve. Source: <a href="https://en.wikipedia.org/wiki/Aerosol_spray" rel="nofollow noreferrer">Wikimedia</a>.</em></p>
<p>An Internet image search will show that most aerosol valves are of the type shown in Figure 1. Here it should be clear that if the differential pressure across the valve is reversed (pressure inside < pressure outside) that the valve will tend to open. The internal spring opposes the opening and so the ΔP at which the valve opens depends on the spring pressure. </p>
<ul>
<li>If the spring force is high enough then the valve may not even open at ΔP = -1 bar and your vacuum could be maintained.</li>
<li>If the spring force is lower then the pressure inside the can will settle at that ΔP that just balances the spring force.</li>
</ul>
<blockquote>
<p>... what would happen when I pull the trigger? Will it now suck like a vacuum cleaner?</p>
</blockquote>
<p>Once you depress the valve the pressures will will begin to equalise at a rate determined by the ΔP, the resistance of the air path and the viscosity of the air. Since the rate is proportional to ΔP the pressure equalisation will follow an exponential curve rather than linear.</p>
| 35633 | Will an empty spray can maintain a vacuum? |
2020-05-08T20:05:16.733 | <p>Why is this overdefined? Can't I translate eveything yellow horizontally to the right?</p>
<p><a href="https://i.stack.imgur.com/r2XKO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r2XKO.png" alt="enter image description here"></a></p>
| |solidworks| | <p>The coincident relation at the bottom of the line on the right will prevent you from moving it horizontally. It is overdefined because your <code>20.75</code> dimension is probably too short or too long to reach the corner that the right-hand yellow line is coincident too. Your line is also vertical, meaning that it cannot move only the upper point of the yellow line to satisfy the dimension.</p>
<p>Hope this helps!</p>
| 35638 | Why is this overdefined? Can't I move right vertical line to the rright? |
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