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2023-08-27T20:50:30.877 | <p>I'm designing a small tea strainer in the shape of a cylinder that goes into a glass bottle, which is also shaped like a cylinder. Currently, I'm still designing it in Fusion 360, but I'm facing one particular challenge.</p>
<p>You see, the circumference of the strainer and the inner glass wall of the bottle aren't that far apart. In other words, the strainer is supposed to fit the bottle as closely as possible. One problem that I can foresee is that whenever water enters the strainer and exits from the sides, the cohesive and adhesive forces of water will cause droplets to stick to the outer wall of the strainer and inner wall of the bottle.</p>
<p>What would be an appropriate solution here? Would water still build up if I added small vertical ridges to the strainer to encourage the droplets to gather and flow downwards? How far apart would the glass and stainless steel have to be in order for the problem to be negligible?</p>
| |steel|cad|tolerance|liquid|product-design| | <p>You are close. Use the ridges to establish a uniform gap or standoff between the strainer and the vessel walls, and then size the gap width (that is, the ridge height) so gravity forces will overcome surface tension forces in the gap. This condition will allow <em>two-phase flow</em> within the gap i.e., the tea can run out the bottom of the gap while air burbles its way into the gap at the same time.</p>
| 56142 | How to prevent water droplets from staying on stainless steel / glass surfaces? |
2023-08-30T11:58:55.817 | <p>I need to measure a machine for a liquid of which I know the viscosity (89 Kinetic Viscosity) but I don't have the liquid at hand. Is there a clever way of increasing viscosity of water to 89?</p>
| |liquid|viscosity| | <p>There are a whole class of chemicals called <em>thickening agents</em> used in the processed food industry which when dissolved in water, increase its viscosity. Some are used to thicken liquid foods so people who have difficulty swallowing can consume them; these are available in drug stores.</p>
| 56184 | How to raise viscosity of water |
2023-09-01T12:34:26.967 | <p>I am simulating heat transfer through a 1-dimensional wall with convection on both sides. For the wall, I implemented a finite difference model like this:</p>
<p><span class="math-container">$$
u(x, t + ∆t) ≈ u(x, t) + c [u(x + ∆x, t) − 2u(x, t) + u(x − ∆x, t)]
$$</span></p>
<p>where <span class="math-container">$c = α \frac{∆t}{(∆x)^2}$</span> and <span class="math-container">$α$</span> is the thermal diffusivity.
The implementation of this model works fine if I look at only the wall and assume that the temperatures on both ends of the wall are known at all times.
Now I would like to extend the model to include convection on both sides of the wall. It looks like the best way to do it is to use convective boundary conditions, as shown <a href="https://qdotsystems.com.au/boundary-conditions-for-the-heat-conduction-equation/" rel="nofollow noreferrer">here</a>:</p>
<p><a href="https://i.stack.imgur.com/ww4Hg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ww4Hg.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$T_{\infty}$</span> and <span class="math-container">$h$</span> are known on both sides, but variable over time. <span class="math-container">$k$</span> is known and constant.</p>
<p>I am struggling to find a starting point for the implementation of these boundary conditions as an extension of the above model.
Anything that helps me get started will be highly appreciated!</p>
<p>Intuitively, I would guess that for <span class="math-container">$x = 0$</span>, the above equation would have to look something like this:</p>
<p><span class="math-container">$$
u(0, t + ∆t) ≈ u(0, t) + c [u(0 + ∆x, t) − 2u(0, t)] + a * T_{\infty}
$$</span></p>
<p>where <span class="math-container">$a$</span> is a function of <span class="math-container">$h$</span> and <span class="math-container">$∆t$</span>. And for <span class="math-container">$x = L$</span>, an analogous equation will have to be used. Could that be right?</p>
| |thermodynamics|heat-transfer|simulation|numerical-methods|convection| | <p>Let us assume that you divide the space into <span class="math-container">$n$</span> equal partitions, such that <span class="math-container">$\Delta x=\frac{L}{n}$</span>. There are <span class="math-container">$n+1$</span> nodes, denoted by <span class="math-container">$x_i$</span>, where <span class="math-container">$i=(0,1,....n)$</span>. The target is to find <span class="math-container">$u_i=u(x_i)$</span> at each time. Let <span class="math-container">$u_i^j$</span> represent the temperature value at <span class="math-container">$x_i$</span> at time step <span class="math-container">$j$</span>. You don't mention the initial condition, so I will assume that initially everything is at ambient <span class="math-container">$T_{\infty}$</span>, such that <span class="math-container">$u_i^0=T_{\infty} \forall i$</span>. The target is to develop a time march scheme to get <span class="math-container">$u^{j+1}_i$</span> given <span class="math-container">$u_i^j$</span>.
Let us observe the BC on the left end. We have <span class="math-container">$\frac{\partial u}{\partial x}(x=0)=-b(T_{\infty}-u(x=0))$</span>. Discretising it, we get <span class="math-container">$u_{1}^j-u_{-1}^j=-b(T_{\infty}-u_0^j)$</span>. Note that I have absorbed <span class="math-container">$\Delta x $</span> in <span class="math-container">$b$</span>. Here <span class="math-container">$u_{-1}$</span> corresponds to a ghost point, a point in space which does not exist, but is assumed for mathematical convenience (remember the left edge is <span class="math-container">$x_0$</span>). Similarly, you can write a discretised equation for the right wall. Let us look at your time stepping scheme now</p>
<p><span class="math-container">$u_i^{j+1}=u_i^j+c(u_{i+1}^j-2u_i^j+u_{-1}^j)$</span></p>
<p>Start at <span class="math-container">$j=0$</span>. You know the right hand side for <span class="math-container">$j=0$</span> (our initial condition), when i=1,2.....(n-1). Thus you get <span class="math-container">$u_i^{1}$</span> when <span class="math-container">$i=1,2,3...(n-1)$</span>.</p>
<p>When <span class="math-container">$i=0$</span>, you end up needing <span class="math-container">$u_{-1}^0$</span>, which is not given by the initial condition (note that <span class="math-container">$u_{-1}^0\neq T_{\infty}$</span>, this equality is only valid for physical points and not ghost points), but it can be easily obtained from the BC! Notice that <span class="math-container">$u_{-1}^0=u_1^0+b(T_{\infty}-u_0^0)$</span>.</p>
<p>You will have to repeat this procedure again for <span class="math-container">$i=n$</span>, where you will need another ghost point <span class="math-container">$u_{n+1}$</span> whose value can be obtained in terms of physical points by using the right wall BC.</p>
<p>Now you know <span class="math-container">$u^1_i \forall i \in (0,n)$</span>.</p>
<p>Repeat for as many time steps as needed</p>
| 56197 | Implementing convective boundary conditions for 1-dimensional finite difference computation of heat equation |
2023-09-01T14:05:11.703 | <p>First of all I understand there are so many variables so I'm going to try to be as explicit as possible:</p>
<ul>
<li>I have various chargers which state they can deliver 3A @ 5V</li>
<li>I have an iPhone 12 Pro Max (I am not sure what is the default expected charging voltage, 5V maybe?)</li>
<li>I am using the charger shown below which has 2 types of slots: 1 type (the 3x green slots) with 3.1A and 1 other type (the 1x orange slot) with QuickCharge 3</li>
<li>I am using a charging cable with 2.4A max capability with visual display of charging power (it shows me the watts)</li>
</ul>
<p>Now:</p>
<ul>
<li>if I plug the cable in the green slots (with 3.1A) the display shows approx 5W which I assume is 1A @ 5V</li>
<li>if I plug the cable in the orange slot (QuickCharge 3) the display shows approx 12W which I assume is 2.4A @ 5V</li>
</ul>
<p>Why is this happening? More specifically, why don't I have 12W/2.4A charging with the green slots too? I also tried multiple chargers which advertise themselves as 2-3A charging but all I get is the slow 5W charging.</p>
<p><a href="https://i.stack.imgur.com/vqgcE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vqgcE.jpg" alt="output info" /></a></p>
<p><a href="https://i.stack.imgur.com/1YB5Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1YB5Y.jpg" alt="output slots" /></a></p>
| |power| | <p>Quick Charge is a standard for chargers and power supplies to negotiate and provide voltages greater than 5V. In addition, when the power supply is providing the control of the voltage or amperage, the charger can potentially bypass some of its miniaturized, internal control circuitry that was built around the dumb 5V coming in; there's a difference between 5V, and 5V with the potential to command it down to less than 4V within some time per a standard.</p>
<p>Your assumption that the 5W is 1A @5V is probably about right. Your assumption that 12W is 2.4A@5V is probably wrong; QuickCharge 3 goes up to a potential of 22V (although yours may not; manufacturers probably aim a bit lower to make sure they are in spec).</p>
<p>You can read more about voltages and QuickCharge on the wikipedia article and its cited sources <a href="https://en.m.wikipedia.org/wiki/Quick_Charge" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Quick_Charge</a></p>
| 56198 | Charger not charging phone at advertised power, why? |
2023-09-04T04:24:24.327 | <p>I'm trying to find an accurate price estimate for an electron beam evaporator for a research project. The intended application is for use in large-scale manufacturing.</p>
<p>How does one usually go about finding this information? Are there any recommended reliable resources?</p>
| |materials|manufacturing-engineering|production-technology| | <p>You can not get this information on the internet easily since these are costly machines which are not typically bought off amazon.
The best way to get this information is to ask a manufacturer (eg Kurt Lesker, CHA etc) for a quote. They will ask your application and recommend devices which will suite you.
You can expect them to cost from a few hundred thousand USD to about a million USD.</p>
| 56221 | Price of e-beam evaporator |
2023-09-04T09:56:02.870 | <p>So, this might be a somewhat strange need, as I don't seem to see many resources on this. I'm trying to create a device where I can set an amount of force for it to generate, and when it would behave as such:</p>
<ol>
<li>If there is no or only a small opposing force, it would drive forward in the direction of the force.</li>
<li>If there is a constant force being applied that is equal and opposite to the force being generated, it would stall indefinitely without damage.</li>
<li>If there is an opposing force larger than the generated force, it would be pushed back, while still providing the constant force as a resistance, therefore reducing the effective backward force; and it should be able to handle all this without damage.</li>
<li>All this only need to occur in the range of no more than 5 cm, and the simpler and smaller the system is, the better.</li>
</ol>
<p>I was wondering if a linear motor, considering that they are also called a force motor, can handle such a demand, and if so, what type? There's also the concern that such a small linear motor can't be found anywhere (or might be prohibitively expensive).</p>
<p>Another thought that I had was to use a torque motor (with some gears to translate torque into linear force, of course), but I couldn't find a definitive source saying that those can indeed provide a constant torque even when being pushed back by the load, I just see that they provide high torque at low speed. Also, on the topic of torque motors, do DC torque motors exist, or are they AC only?</p>
<p>Of course, if you have a better idea for achieving this need of a constant force no matter what's actually happening in terms of movement and outside load, please let me know!</p>
<p>Finally, please do tell me if you think a different community would be a better place for this question.</p>
<p>Thank you all for your time!</p>
| |motors|linear-motion|linear-motors| | <p>If the changes are slow enough, a simple pulley with counterweight might do the job. Constant force hangers are used very commonly for piping, where you can find interesting designs using combinations of springs and lever arms. One of the simplest designs is <a href="http://www.piping-engineering.com/spring-support-piping.html#constant-effort-spring" rel="nofollow noreferrer">here</a>.
<a href="https://i.stack.imgur.com/M8PMf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M8PMf.jpg" alt="enter image description here" /></a></p>
<p>Similar effect is also achieved in <a href="https://www.completeguidetoarchery.com/how-does-a-compound-bow-work/" rel="nofollow noreferrer">compound bows</a>, that are optimised to resist the middle part of the draw with the maximum force in order to store maximum energy (the force also drops at the end of the draw, so you can hold the fully drawn bow with a lots of energy with smaller effort than traditional bow).</p>
| 56226 | Method to achieve a constant force no matter the actual velocity when under load |
2023-09-06T16:49:05.353 | <p>The time-averaging operator definition gives us</p>
<p><span class="math-container">$$\overline{u_iu_j}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}u_i(x,t)u_j(x,t)dt.$$</span></p>
<p>From my understanding, this should mean that the Reynolds stress tensor, <span class="math-container">$\overline{u_iu_j}$</span>, is a function of space only, and not of time.</p>
<p>But the <a href="https://en.wikipedia.org/wiki/Reynolds-averaged_Navier%E2%80%93Stokes_equations#Equations_of_Reynolds_stress" rel="nofollow noreferrer">Reynolds Stress Equations</a> contain partial derivatives of the stress tensor with respect to time, i.e. <span class="math-container">$\frac{\partial \overline{u_iu_j}}{\partial t}$</span>.</p>
<p>Since the stress tensor is time-averaged (and hence independent of time), why doesn't it follow that these time-derivative terms are zero?</p>
| |fluid-mechanics|turbulence| | <p>This answer just depends on how you define your averaging. The Reynolds stress transport equation defines it abstractly, so it can have both time and space variation.</p>
<p>The equation you give is for time averages, but you can also consider space averages. For example, for a homogeneous turbulent flow, it makes sense to define the averages as space averages that are averaged throughout the entire domain. In that case, the Reynolds stresses and other statistics vary with time. Space averaging can also be done in a periodic manner called "pattern averaging" too.</p>
<p>There are also "ensemble averages" that allow for both time and space variation. How? Well, you average over different "realizations" of the flow. That is, you collect data from one "run" (either experimentally or numerically), and you then repeat the test as many times as you see fit. Given slight differences in the initial conditions, the data are different in each run. You then average over the "ensemble" of runs and form an ensemble average and then varies in both space and time.</p>
<p>This is why the Reynolds stress transport equation can have both time and space variation, since it considers averaging in a very abstract form. For some applications, you can consider the time derivatives to be zero, but for others, they may not be. It all just depends on your application and how you define your averages.</p>
| 56250 | Why are Reynolds Stresses time-varying? |
2023-09-06T20:49:08.290 | <p>In "Theory of Elastic Stability" by Timoshenko and Gere, the authors say for a beam in pure torsion,</p>
<blockquote>
<p>For a beam of thin-walled open section it can be assumed with
reasonable accuracy that the shearing stress at any point is parallel
to the corresponding tangent to the middle line of the cross section
and is proportional to the distance from that line.</p>
</blockquote>
<p>What is meant by the "tangent to the middle line"? A cross section would be a two dimensional shape. For an I-beam, it would be the shape of an I. But I can see two reasonable definitions of what a middle line could be, since an I has two axes of symmetry. And even then, I'm not sure what a tangent to the line is, other than the line itself.</p>
| |beam| | <p>I found a manuscript <a href="https://s28490.pcdn.co/wp-content/uploads/2019/06/asm-B840.pdf" rel="nofollow noreferrer">https://s28490.pcdn.co/wp-content/uploads/2019/06/asm-B840.pdf</a> that clarifies the situation.</p>
<p><a href="https://i.stack.imgur.com/9igOn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9igOn.png" alt="enter image description here" /></a></p>
<p>The "middle line" mentioned in the quote is not strictly a mathematical line, but may be a curve, or in the case of an I beam or T beam, a join of more than one line segments, which describes the shape of the cross section if the width of the walls were shrunk down to zero.</p>
| 56251 | Shearing Stress in a Beam with Torsion |
2023-09-07T06:21:45.650 | <p>Budget ~$5k</p>
<p>Current setup:
2x 20k cfm water-based smoke machines facing each other 6' apart blowing across the top or bottom of a fan or set of fans. I was hoping there could be a way to get a vortex that rose in the sky using a series of fans. I've searched youtube and google to find someone who has done this because I feel like I've seen it before, but they've been no help. This is for a work project in defense.</p>
<p>Limitations:
<em>I can't use heat to get the smoke to rise higher due to fire risk.</em></p>
<p>Specific recommendations on: types of industrial fans to use, many small fans vs a few big fans, configuration/setup, all given the budget above would be very helpful. While googling, I found industrial high-velocity fans like <a href="https://www.industrialfansdirect.com/collections/loading-dock-truck-cooling" rel="nofollow noreferrer">this</a>, but they claim to only project <=100' each, which would be underwhelming for me. I know this a fluid dynamics problem, so even recommendations on a setup that would expedite testing for height would be much appreciated.</p>
<p><strong>Thank you!</strong></p>
| |fluid-mechanics|airflow|air| | <p>Your objectives and constraints are not clear, but based on your comment that you "merely looking to extend the 'throw range' of these fans", you may want to consider using an <a href="https://en.wikipedia.org/wiki/Air_vortex_cannon" rel="nofollow noreferrer">air vortex cannon</a>. The best engineering solution, however, might be to see if you can mitigate the fire risk so you can use real smoke.</p>
<p><strong>Air-Vortex Cannon</strong></p>
<p>As @Phil-Sweet notes in their comment, simply trying harder to force a fluid through a fluid doesn't usually extend the range much because the fluid jet disintegrates because of instabilities. Suitably generated <a href="https://en.wikipedia.org/wiki/Vortex_ring" rel="nofollow noreferrer">toroidal rings</a>, however, can propagate stably for considerable distances. Check out:</p>
<ul>
<li><a href="https://www.youtube.com/watch?v=fxZu_Y_m8i4" rel="nofollow noreferrer">Wham-O Air Blaster commercial</a></li>
<li><a href="https://www.youtube.com/watch?v=IyAyd4WnvhU" rel="nofollow noreferrer">blowing down a straw house</a> on the BBC</li>
<li>this <a href="https://www.youtube.com/watch?v=3zPNFFiM-pQ" rel="nofollow noreferrer">HMS Engineering giant vortex cannon</a></li>
<li><a href="https://www.youtube.com/watch?v=ayaiArVkpA4" rel="nofollow noreferrer">blowing out a candle</a> 180 feet away</li>
<li>and the many other <a href="https://www.youtube.com/results?search_query=vortex+cannon&sp=CAM%253D" rel="nofollow noreferrer">vortex cannon videos on YouTube</a>.</li>
</ul>
<p>There are at least a couple of potential issues. First, such a cannon is pulsed not continuous. Second, the air vortex can propagate long distances, but the water/glycol "smoke" may become invisible long before the toroid reaches its maximum range.</p>
<p><strong>Buoyant Gases</strong></p>
<p>You could entrain your water/glygol smoke in a buoyant gas, but I think they are all either dangerous/flammable/explosive (e.g. hydrogen and methane), or too valuable and expensive (e.g. helium and neon).</p>
<p><strong>Mitigate the Fire Risk</strong></p>
<p>Since we know that actual smoke from a fire can rise extremely high in the atmosphere, your best approach might be to try to figure out how the fire risk can be mitigated sufficiently to allow using actual smoke or hot-air driven particulates, e.g. rent a barge and do your experiment in the middle of a large lake.</p>
| 56256 | I'd like to create a column of smoke that goes as high as possible. Is there a fan or configuration of fans that is best? |
2023-09-08T23:55:57.637 | <p>I am trying to develop a program in Python that runs a few different basic analyses for optimization like minimizing compliance and buckling. I want to use machine learning to try to develop a fast (once programmed) multi-objective optimizer. To do this I need a way to calculate the critical buckling load of whatever arbitrary geometry it is optimizing, so that the agent can learn to minimize this value.</p>
<p>I have searched what feels like everywhere and cannot find any method of calculating the critical buckling load that is not just using a massive prebuilt software like Abaqus or Fusion 360. The problem with using scripting for those services is that in order to run this computationally intensive program and train the machine learning model, I need to use cloud computing and cannot use a service like Microsoft Azure with another commercial program like Abaqus (as far as my understanding goes).</p>
<p>Does anyone know if it is currently possible and reasonable to program an FEA to determine the critical buckling load of an arbitrarily shaped object?</p>
<p>I currently represent the object as a 3D grid of voxels, and expect to run it at a resolution of roughly 30x30x30 for now.</p>
| |finite-element-method|buckling| | <p>Representing the geometry using voxels seems very inefficient to me. Although voxels could be translated to solid brick elements in FEA, buckling analysis is usually more important for thin shell and beam structures, so you would need a way to extract shell and beam elements from the voxels.</p>
<p>Critical buckling load is very sensitive to small changes to geometry and calculating it precisely can be very difficult. The most precise analysis involves initial imperfections of the geometry, large deformations and nonlinear material model, which is quite a challenge even for professional FEA packages.</p>
<p>Maybe you could use just the simplest analysis, the <a href="https://physics.stackexchange.com/a/384454/270075">eigen buckling</a> in combination with static analysis (which is not very precise) and focus only on frame structures (represented using beams instead of voxels of course).</p>
| 56263 | Is there a way to program a buckling analysis of an arbitrary shape, using a standard programming language/libraries (not Abaqus)? |
2023-09-09T21:48:33.830 | <p>Sorry I am not an Engineering professional or student. But I had a question regarding US Nuclear weapons. Specifically, is it possible that Nuclear weapons we created in the 50s still work?</p>
<p>The major doubts I have (and pleases add "In the 75 years since these weapons were developed" to each doubt:</p>
<ol>
<li>Radioactive warheads and decay: would all this time affect the electronics in a bomb and or the ability for the core to go nuclear?</li>
<li>Starter explosive: As I understand it, a nuclear bomb's reaction is activated by a conventional explosive. If so, do we believe that even with decay to the conventional explosive, that it's still enough to ignite the nuclear reaction?</li>
<li>Delivery systems & ICBMs: Do the ICs on these weapon systems still function? Does anyone know how to repair or perform maintainance on them? Have they been replaced? Ditto with control systems.</li>
<li>Control Systems: I can't imagine these systems have been replaced - it seems way too risky, or? So how have they been maintained, and does anyone know how to use them anymore?</li>
</ol>
<p>In short, it seems to me like the bombs we built 75 years ago were built using technology... from 75 years ago. How in the world have they been able to maintain this (assuming that they have maintained it)?</p>
| |mechanical-engineering|electrical-engineering|control-engineering| | <ol>
<li><p>Electronics in today's sense didn't exist. The implosion explosives used a special sort of squib designed at Los Alamos. <a href="https://en.wikipedia.org/wiki/Exploding-bridgewire_detonator" rel="noreferrer">https://en.wikipedia.org/wiki/Exploding-bridgewire_detonator</a>. There was a battery and a few switches and that was about it.</p>
</li>
<li><p>The high explosives used in the lenses were a bit more carefully prepared than what was seen in a typical shell-filling process. They were individually bullet proof and quite well sealed and storage was decent. They can be considered viable for a good, long time. There are examples Comp B with far less care existing for 70 years. But the point that they need a very high degree of uniformity as a set in order to work is a bit of issue.</p>
</li>
<li><p>Yes - they still work, and I know how to repair them. I was the chief of engineering for Minuteman at Minot From '89 to '93. That was long enough ago that some of the old hands that designed the things in the early '50s were still around - the Paperclip guys and the old Boeing Autonetics guys that designed the flight and ground computers. As far as IC's, The early Minuteman system didn't have any. By Minuteman III, they had a four-bit register, of which there was about a dozen in the ground computer. Minuteman's computer systems represented about 50% of the planet's total floating point operation capability during the several years of buildout. You can buy a computer that works a million times faster today, but you can't buy one that does what these old rack systems did. You would need about a million times more lines of machine code to duplicate the 2k program, and the telemetry wouldn't be as good. I was one of the last people to get detailed training on the system computers at the field level. Depot obviously had people. The command and control systems have been upgraded to deal with changings threats. The ground control hardware proved to be pretty robust. There has been some concern about encryption over the years.</p>
</li>
<li><p>Within the missile fields, site to site comms is done via HICKS cable. It's the same as deep-sea phone cable. It has a pressurized armored jacket around the wires. The site terminations have lighting protection (via an ESA vault) and then it's just ordinary amplifiers and receivers. They all have real-time fault monitoring via message error counts and replacing them is routine maintenance work. So this is a matter of choosing to do adequate maintenance, which hasn't always been the case.</p>
</li>
</ol>
<p>Odds and ends - these systems were designed to be maintained by 18-year-old kids out of high school that had a knack for mechanicing. They are very modular and seldom require more than simple remove-and-replace maintenance (Okay, some of it isn't exactly simple, but you can train for it). Everything went back to depot for repair work.</p>
<p>As to reliability, when I was at Minot, we had all the missiles on Strategic Alert all the time (unlike today). There are 150 silos. On any given day, one would be getting pulled apart for depot rework, one would be in the middle of depot rework, and one would be getting put back together. These sites belonged to depot and weren't my problem. So we had a baseline of about 147 silos on strategic alert. Our launch-ready rate was about 99.4% when I left in '93. That is computed on a minute-by-minute basis. The 0.6% was inop due to failures or a maintenance team being on site. We could expect 146 ready out of 147 given 20 minutes notice, and probably 148 given 24 hours notice.</p>
| 56269 | Nuclear weapon doubts |
2023-09-11T17:12:34.027 | <p>I'm in the market for a new heating and air conditioning unit for my home. Most offer variable speed blowers and the more expensive models are equipped with BLDC motors, which seem overkill. I am having a difficult time seeing the advantages to variable speed blowers other than perhaps to reduce noise.</p>
<p>I figure the goal is to heat up the interior as fast as possible by heating the coils as hot as possible and forcing as much air as possible over them. Of course, you don't want to turn your exhausts into hot air guns. Am I missing something here?</p>
| |thermodynamics|heat-transfer|heating-systems| | <p>ordinary blower fans are AC induction motors with a separate winding for each fixed speed (low, medium, high). they are very quiet, cheap to build, efficient, and are well-established in practice.</p>
<p>However, it is possible to do the trick with a <em>single set of windings</em> if you have a variable frequency AC power source, for example, or with a BLDC-type motor. By building the motor with a single set of windings, the motor can be made smaller and lighter (which means less expensive). Then you trade off the cost savings for a lighter, cheaper motor against the cost increase of the VFAC drive board or the BLDC drive board.</p>
<p>Note that the air molecules themselves don't care what kind of motor is used to power the fan.</p>
| 56291 | Whats the advantage of variable speed furnace blowers? |
2023-09-12T01:49:35.600 | <p>I have had a long-time interest and fascination with railgun technology and for a long time I have wanted to build a small-scale, simple railgun as a scientific experiment. This railgun is designed to use a capacitor as the projectile.</p>
<p>I want to point out that I am not an electrical engineer nor do I have a background in working with electronic components nor with building things with electronical components. I do have a self-taught understanding of basic circuit design and how some basic electrical components work, such as batteries, resistors, diodes, and capacitors.</p>
<p>I am seeking design advice on the best type of capacitor that I should buy for this simple railgun. I have recently been doing a lot of online research on capacitors, and at this point, I am rather confused by all the various types, sizes, and specifications.</p>
<p>My goal with this simple railgun is to try to launch a small capacitor that will travel at least six meters in distance. I am planning to use four D-size batteries to charge up the capacitor.</p>
<p>I am thinking that to achieve this goal I will need to use a low-weight capacitor, say one weighing 3 ounces or less, yet one that will produce enough current for a long enough time to accelerate the capacitor to the point that it travels at least six meters in distance.</p>
<p>Below is a conceptual design drawing I made to illustrate the working principle of this simple railgun:</p>
<p><a href="https://i.stack.imgur.com/9MLMk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9MLMk.png" alt="enter image description here" /></a></p>
<p>What would be the best capacitor to use for a railgun that uses a capacitor as the projectile?</p>
| |electrical-engineering|materials|design|power-electronics|electromagnetism| | <p>There are new-technology capacitors available now that can store large amounts of energy; they are called <em>ultracapacitors</em> and have capacities of order ~farads.</p>
<p>Before you go out to buy some, however, I have the following recommendations.</p>
<p>First of all, go find you some data on a capacitor's energy storage capacity on a <em>per gram basis</em>, for each major type of capacitor on the market.</p>
<p>Then calculate the projectile "muzzle velocity" which will yield a 6-meter flight length, in the earth's gravity. From this, derive the energy release required on a <em>per gram basis</em> to accelerate a projectile to that velocity.</p>
<p>Now, assuming a 6-volt charge in all cases, you should be able to equate the required energy release per gram for launch to the energy stored in the capacitor per gram and see if it is possible to self-launch a capacitor projectile 6 meters.</p>
| 56299 | What would be the best capacitor to use for a railgun that uses a capacitor as the projectile? |
2023-09-17T10:51:50.103 | <p>I am presently studying fluid statics from Munson's textbook. I do not understand part (b) of example 2.6. The full question is as follows:</p>
<blockquote>
<p>GIVEN The 4-m-diameter circular gate of Fig. E2.6a is located in the inclined wall of a large reservoir containing water (<span class="math-container">$\gamma = 9.8 \mathrm{kN/m^3}$</span>). The gate is mounted on a shaft along its horizontal diameter, and the water depth is <span class="math-container">$10\mathrm{m}$</span> above the shaft. <br>
FIND Determine <br>
a) the magnitude and location of the resultant force exerted on the gate by the water and <br>
b) the moment that would have to be applied to the shaft to open the gate.</p>
</blockquote>
<p>with the associated diagram</p>
<p><a href="https://i.stack.imgur.com/QW5rx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QW5rx.png" alt="enter image description here" /></a></p>
<p>I had a few questions about this problem. Firstly, what is the role of the shaft? Is it used to hold the circular gate in place? Can the gate rotate freely about the shaft in the absence of the stop? The author writes the following in his solution to the problem</p>
<blockquote>
<p>The moment required to open the gate can be obtained with the aid of the free-body diagram of Fig. E2.6c. In this diagram <span class="math-container">$\mathcal{W}$</span> is the weight of the gate and <span class="math-container">$O_x$</span> and <span class="math-container">$O_y$</span> are the horizontal and vertical reactions of the shaft on the gate.</p>
</blockquote>
<p>It appears that his convention for <span class="math-container">$x$</span> and <span class="math-container">$y$</span> is different than the coordinate system he has used. That aside, I understand where <span class="math-container">$O_y$</span> comes from; I guess it is the reaction force to the weight <span class="math-container">$\mathcal{W}$</span>. However, it is not clear to me where <span class="math-container">$O_x$</span> arises from. Is it a reaction force to the horizontal component of <span class="math-container">$F_R$</span>? It doesn't seem like that to me. Next, the author writes</p>
<blockquote>
<p>We can now sum moments about the shaft <br>
<span class="math-container">$\sum M_c = 0$</span></p>
</blockquote>
<p>I assume this implies an equilibrium condition at which point the shaft just opens. I also assume we discount the presence of the stop since it, in the current state of the gate, only prevents anticlockwise rotation of the gate.</p>
<p>Any kind of inputs and explanations would be most appreciated.</p>
| |fluid-mechanics|torque|moments|water-pressure| | <ol>
<li>Yes, the gate can rotate freely about the shaft. It is clear that the lower part of the gate will experience more force than the part of the gate above the shaft (because of depth), and therefore, it will have the tendency to rotate counter clockwise. The stop prevents this by applying a net force.</li>
<li>Remember that water applies hydro-static force, which means that the force is normal to any face it is acting on. Therefore, the force applied by water on the gate is normal to gate, and has both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> components. Since the force has both these components, the reaction from the shaft has to balance these, and thus <span class="math-container">$O_x$</span>. (From the diagram, see also that <span class="math-container">$O_y$</span> is not only the weight, but also a component of the hydro-static force)</li>
<li>Yes you are right, when you apply that moment, you are neglecting the stop. Or equivalently, think of the moment which is needed to open the gate in clockwise direction (stop does not prevent CW rotation).</li>
</ol>
| 56352 | How do I determine the moment of force that would have to be applied to the shaft to open the gate in this question on fluid statics? |
2023-09-18T13:22:49.833 | <p>Living in the north, I am wondering if a sitting car battery's chemical reactions give off enough heat to justify insulating it. That is, will it result in significantly more power available to start a cold car?</p>
<p>[I did ask that this be moved to the engineering forum]</p>
| |thermodynamics|temperature| | <p>Note that batteries, including lead-acid and Li-ion batteries, are most susceptible to overheating while charging. Any jacketed insulated battery should be equipped with temperature-regulating charge control.</p>
<p>Temperature-regulated charge control is standard for good Li-Ion systems, but not common for Lead-Acid automotive start batteries.</p>
| 56367 | Practical application: questioning the value of insulating a car battery |
2023-09-20T10:10:39.250 | <p>I'm trying to expand my theoretical knowledge on turbomachinery, but I cannot decipher the equation derivation steps in the book I'm using. I attached the derivation steps bellow as an image.</p>
<p>I understand the eq (51), mass through the inlet ring is the same as the mass through the outlet ring. The part that boggles my brain is how the eqs (52) and then (53) were derived.</p>
<p>Perhaps someone could help me with some pointers to understand this completely.</p>
<p><a href="https://i.stack.imgur.com/QtvOR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QtvOR.png" alt="enter image description here" /></a></p>
| |turbomachinery| | <p>It looks like Eq. (52) is a general statement of torque and angular momentum transfer across the control volume boundary. Presumably, that was discussed previously. It is not derived from Eq. (51) which is a conservation of mass, the flow in equals the flow out.</p>
<p>In <span class="math-container">$\rho_1 \vec c_1 d\vec S_1$</span> (and subscript 2 case) the vector multiplication is understood to use the component of <span class="math-container">$\vec c_1$</span> perpendicular to vector area <span class="math-container">$d \vec S_1$</span> to obtain volume flow through the surface.</p>
<p>Eq. (53) removes the integration symbols of Eq. (52), perhaps we are looking at a small part of the rings A and D. In Eq. (52), <span class="math-container">$F_B dA$</span> is a force perpendicular to the radius r, so <span class="math-container">$rF_B dA$</span> is an increment of torque (also called moment of force). Eq. (53) shows the small increment of torque <span class="math-container">$dM$</span> equal to <span class="math-container">$rF_B dA$</span>. Looking at Eq. (51), <span class="math-container">$\rho_1 \vec c_1 d\vec S_1$</span> is an increment of mass flow so call it <span class="math-container">$d \dot m$</span>, and the same amount at exit (subscript 2). The right side of Eq. (52)
now becomes the right side of Eq. (53).</p>
<p>Excuse me if I explain too much, <span class="math-container">$c_u$</span> is the component of <span class="math-container">$c$</span> in the plane perpendicular to the axis x and perpendicular to the radius r. Some books will use a <span class="math-container">$\theta$</span>, but u is common because it is in the wheel speed direction and U is the wheel speed, that is, the speed of the blade at a given radius.</p>
| 56375 | Euler's turbine equation derivation |
2023-09-20T13:07:33.730 | <p>I am confused by Axial loading mentioned in the figure below. The long black line is a piece of rubber passing over different rollers. And at the point "Axial Load" a weight is attached. The rubber piece itself is drawn back and forth.</p>
<p><a href="https://i.stack.imgur.com/Z4J5k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z4J5k.png" alt="enter image description here" /></a></p>
<p>I believe this load attached is acting perpendicular to the axis of the rollers and the rubber piece movement so it's a radial load. Or am I wrong in my interpretation?</p>
| |forces| | <p>As Abel said, this is axial to the rubber since the rubber is thin enough that you can assume that the force is passing through the centreline.
Also, for the rollers, yes the force is radial.
If there were friction, then there would be a tangential component to the rollers as well.</p>
| 56376 | Is this axial load or radial load |
2023-09-20T13:40:12.060 | <p>In the research paper "I. Krakovský et al., A few remarks on the electrostriction of elastomers, 1999" the authors stated the equations of the principle stresses that are produced in an isotropic dielectric, as follows:</p>
<p><span class="math-container">$$\sigma_{xx} = \sigma_{yy} = \frac{1}{2} \varepsilon_0 \varepsilon E^2 \left(1 + \frac{a_2}{\varepsilon}\right)$$</span></p>
<p><span class="math-container">$$\sigma_{zz} = - \frac{1}{2} \varepsilon_0 \varepsilon E^2 \left(1 - \frac{a_1+a_2}{\varepsilon}\right)$$</span></p>
<p>where <span class="math-container">$\sigma_{xx}$</span>, <span class="math-container">$\sigma_{yy}$</span>, and <span class="math-container">$\sigma_{zz}$</span> are the principal stresses. <span class="math-container">$\varepsilon_0$</span> is vacuum permittivity and <span class="math-container">$\varepsilon$</span> is the relative permittivity. <span class="math-container">$E$</span> is the electric field, and <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span> are two parameters describing the change of dielectric properties of material in shear and bulk deformation.</p>
<p>I would like to compare them with other equations of stress from another paper, but in the other paper the stress tensor is stated in its tensor form containing all the elements including shear stress, so without the principle stresses presentation.</p>
<p>Thus, I would like to know how to convert the principles stresses back and get the full stress tensor, so that I could see if both papers have same stress equations.</p>
| |stresses|insulator| | <p>Given a set of principal stresses, it is not possible to get a unique state of stress. This is because the stress tensor has 6 independent components, whereas there are only 3 principal stresses.
However, given a state of stress, you can calculate the unique principal stresses.
Your best bet would be to calculate the principal stresses from the other paper and match it with these. To do that, you would just have to get the eigen-values of the stress tensor.</p>
| 56378 | Calculating stress tensor from the principle stresses |
2023-09-21T22:04:05.820 | <p>I'm making a game where you have to hit targets, kind of like whack-a-mole except the targets aren't actively actuated. I'd still like an element of mechanical "feel" to it, so I'm aiming to put the targets on a rod with a spring, so that when you whack it, it goes down, triggers a limit switch so the game knows you've hit it, and it springs back up. See below for a rough model. The intention is to use an off-the-shelf shaft, spring and shaft collar.</p>
<p><a href="https://i.stack.imgur.com/g8fhZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g8fhZ.png" alt="A simple spring mechanism" /></a></p>
<p>In terms of scale, the target is going to be somewhere in the range of cue ball to grapefruit, and made from a light material (e.g. plastic, resin, etc), so the spring doesn't have to be particularly strong. (I realize this is slightly vague, but it's TBD and I'm only tasked with making the mechanism.)</p>
<p>It needs to withstand a few hundred whacks per day for a couple of months. The mallet will be coated in thick foam but wielded by members of the public.</p>
<p>Would using linear bearings in the brackets or a linear bearing pillow block as the whole bracket be overkill?</p>
<p>Is there an off-the-shelf component I could use for the bracket otherwise, or would I be better to have a metal bracket fabricated? I've hunted around for a while, but I don't know what to call that kind of bracket.</p>
<p>I've also considered 3D-printing a bracket, but I'm not sure if it will be strong enough to put up with repeated whacks. Might I get away with it if I designed enough bracing into the bracket?</p>
<p>The limit of the travel can be set by a fixed surface (e.g. the cabinet) so theoretically the bracket primarily needs to withstand the force applied to it by the spring under compression, but if the target is whacked diagonally, the bracket must withstand the lateral force too.</p>
<p>Another option I have considered is laser cutting 1/2" ply to make a box-jointed housing, but not sure if the wood against the metal will make a kind of jittery movement. I'm imagining kind of like below, but with box joints and probably an extra sort of lintel under the horizontal pieces.</p>
<p><a href="https://i.stack.imgur.com/xL2VP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xL2VP.png" alt="A different housing" /></a></p>
<p>I have time and resources to try a couple of simple prototypes, but I can't feasibly whack it thousands of times to see if it gives.</p>
| |mechanical-engineering|mechanisms|metalwork| | <p>I'd probably make the brackets out of a low-friction plastic like nylon. You <em>could</em> 3D print them, but they are (or at least can be) simple enough that there may not be much point.</p>
<p>Assuming you're only making a relatively small number, I'd probably just get rectangular nylon bar stock, drill a suitable-size hole and cut it to length (and probably a couple screw holes to mount it). No need for the L-shaped mounting base--just use thick enough material to mount as rectangle. Something along this general order:</p>
<p><a href="https://i.stack.imgur.com/xFx0e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xFx0e.png" alt="enter image description here" /></a></p>
<p>That'll use a little more material than the L-shaped bracket you've drawn, but the savings in fabrication time will more than compensate. Given the application, it doesn't seem as if the slight increase in friction from a larger bearing surface is at all likely to matter.</p>
<p>At the opposite extreme, if you were going to be making millions of these, it would probably make sense to get molds made, and have essentially the entire mount injection molded. This has much higher initial cost (molds are pretty expensive) but reduced unit cost.</p>
<p>As far as dealing with lateral forces, I'd mostly try to design it to minimize the ability to exert lateral force in the first place. If you look at a typical whack-a-mole kind of game, the button that sticks up won't be a hemisphere. It'll have a much larger radius, so the user has no way to hit it that produces significant lateral force. The "button" that sticks up is also a tight enough fit in the hole that you can't really move it very far laterally in any case.</p>
<p>You could also use a shaft that's flexible enough (e.g., probably also plastic) that most lateral force would simply flex the shaft rather than being transmitted to the brackets.</p>
| 56400 | Making a basic sprung mechanism that can withstand moderate percussive force |
2023-09-22T14:46:13.563 | <p>I need to know whether neoprene and HNBR (Hydrogenated Nitrile Butadiene Rubber) seals can tolerate pH11 atmosphere. The seals would be in touch with a fertilizer which is pH11.</p>
| |materials|material-science| | <p>Per <a href="https://fluidpowerjournal.com/demystifying/" rel="nofollow noreferrer">https://fluidpowerjournal.com/demystifying/</a> HNBR is</p>
<p>Recommended for:</p>
<ul>
<li>Hot water and steam</li>
<li>Oils and fuels</li>
</ul>
<p>Not recommended for:</p>
<ul>
<li>Polar solvents</li>
<li>Strong acids</li>
<li>Chlorinated hydrocarbons</li>
</ul>
<p>If I had a choice, I'd try for something more akin to pvc or ptfe...</p>
| 56403 | Need to know tolerance of certain rubbers to pH 11 Bases |
2023-09-22T18:01:45.853 | <p>What would be a good way to actively control the loudness of an acoustic piano note over time after the note has already been struck by the hammer, such as by actively changing the amplitude of a piano string's oscillations over time instead of having it naturally die down?</p>
<p>features wanted:</p>
<ul>
<li>Ideally, the sound quality remains the same, and the only discernable difference between the sound produced by this method and a natural acoustic piano sound is the fact that the amplitude of a natural piano note dies down over time while the amplitude produced by this new method can follow any arbitrary function of time.</li>
<li>minimal change or add-on to an actual acoustic piano, minimal cost, complexity, power used, and minimal moving parts for less maintenance.</li>
<li>Additionally, since I have heard that the presence of harmonics and exact waveform of a note is different between different musical instruments, I'm thinking it might be more "faithful" to the original sound of a piano to not generate the waveform artificially but amplify the piano string with its own preexisting oscillations after it has been struck by a hammer. This would be like having a speaker and microphone right next to each other so that the microphone picks up the speakers sounds and amplifies the speaker's own preexisting oscillations in a usually very bad feedback loop. Would this be a good idea, and if so, how could it be implemented?</li>
</ul>
<p>A couple existing ideas:</p>
<ul>
<li>use electromagnetism, perhaps pass AC electricity through the string and immerse it in a magnetic field like in a microphone. Problems/questions: would this produce enough force to make a difference? Can you take advantage of coils of wire instead of a straight wire? Wouldn't the wire heat up and thus change pitch (not wanted)? Or have the wire be magnetized and have changing magnetic field to cause it to oscillate?</li>
<li>mechanical wave oscillator, just attach one end of the string to some sort of oscillating mechanism. Problems/questions: what would this look like?</li>
<li>just have a computer digital reproduction of a piano sound whose amplitude can be controlled. Problems/questions: kind of a cop-out, as the sound wouldn't be from a "real" piano string and soundboard.</li>
<li>somehow change the soundboard itself?</li>
</ul>
| |mechanical-engineering|electrical-engineering|vibration|electromagnetism|acoustics| | <p>What you describe has been on the market for over 40 years for electric guitar use; it is called an <em>E-Bow</em>. It contains a sense coil, a drive coil, a battery, and an amplifier circuit inside a case you can hold in your hand. When held near a vibrating metal guitar string, the vibrations are picked up by the sense coil, amplified by the amp, and sent into the drive coil which forces the string's vibrations to increase in amplitude. To modulate the strength of the coupling, the guitarist changes the spacing between the E-Bow and the string.</p>
| 56407 | Controlled oscillation of a piano string by electromagnetic, mechanical, or other means |
2023-09-23T10:02:25.283 | <p>In this blueprint I was not able to find where the height of the point of the cone angle should be.
<a href="https://i.stack.imgur.com/4nChx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4nChx.jpg" alt="enter image description here" /></a></p>
<p>Here's the original blueprints:<br />
<a href="http://library.centerofthewest.org/digital/collection/WRAC/id/398/rec/8743" rel="nofollow noreferrer">http://library.centerofthewest.org/digital/collection/WRAC/id/398/rec/8743</a>
<a href="http://library.centerofthewest.org/digital/collection/WRAC/id/2664/rec/8855" rel="nofollow noreferrer">http://library.centerofthewest.org/digital/collection/WRAC/id/2664/rec/8855</a>
<a href="http://library.centerofthewest.org/digital/collection/WRAC/id/397/rec/8742" rel="nofollow noreferrer">http://library.centerofthewest.org/digital/collection/WRAC/id/397/rec/8742</a></p>
<p>How can I find the height of this point in relation to origin?</p>
| |metallurgy|technical-drawing|drawings| | <p>The height of the center can be found from the .208R+.003 dimension. The lowest point of this radius is dimensioned vs a nearby flat face with .025±.002.</p>
<p>The reason the center is not dimensioned directly from any flat face will be to do with the design intent and tolerance stackups. The dimension that <em>matters</em>, is the distance from the flat to the surface of the cone (.025±.002) - not the distance to the centre.</p>
<p><a href="https://i.stack.imgur.com/GOf9H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GOf9H.png" alt="marked up print" /></a></p>
| 56421 | Find height of cone point in blueprint |
2023-09-25T14:53:29.453 | <p>To test crash barriers, we crash cars into them at a specific speed and angle. The car must hit a designated impact point within plus or minus X inches.</p>
<p>I'm wondering if it is possible to build a system to automate the steering process, using sensors (or whatever else) without access to the test car's on-board ECUs. The first idea which came to mind was a computer vision system which controls the vehicle's steering wheel mechanically - essentially a mechatronics system.</p>
<p>Are there any existing systems that do this that I can take a look at? I think the hardest part may be the accuracy, since the car must be steered into a specific spot on the test barrier. The speed is taken care of by another system.</p>
| |mechanical-engineering|automotive-engineering| | <p>Yes, we have steering robots that attach a motor to the steering wheel. Trouble is that interferes with the airbag, so an important part of the test is invalid. The proper way around this is to correlate your FEA to the ACTUAL test path, and then run it again at the perfect test path. Crash FEA is good enough these days that we can use it to calibrate the airbag sensors. Rails can't be used as they provide an unrealistic boundary condition.</p>
| 56434 | Automated steering system for crash testing vehicles |
2023-09-26T08:56:13.357 | <p>I am trying to calculate the maximum torque a cylindrical body can withstand when clamped by 2 set screws. The cylinder is placed in a housing with a sliding fit against housing surface, and the clamping force produced by the set screws are represented by F1 and F2.</p>
<p>I would like to verify that my below analysis is correct, would appreciate any comments:</p>
<p><a href="https://i.stack.imgur.com/U7ISH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U7ISH.png" alt="enter image description here" /></a></p>
<p>For example, if we apply 20Nm torque to the cylinder, the force tangent to the surface of the cylinder is T=F<em>R, 20Nm=F</em>0.05m, F=400N.</p>
<p>Using a formula for estimated force produced from an M4 screw torqued to 1Nm, the force from each set screw is estimated to be ~850N. Assuming the coefficient of static friction between the set screw (steel) and the cylinder (brass) to be 0.5, the friction force from the set screws is 850N as shown above, and the cylinder should not turn.</p>
| |torque| | <p>Your analysis seems correct. The cylinder should not turn.</p>
| 56442 | What is the maximum torque a cylinder can withstand without turning when clamped by 2 set screws? |
2023-09-26T22:13:32.173 | <p>Air flow (eg in a duct) has a dynamic pressure, and when the air flow goes through e.g. a 90°-bend the bend has a dynamic-loss-coefficient.<br />
The dynamic-loss-coefficient is used to calculate the total-pressure-loss by the formula:<br />
Total Pressure loss = C * Velocity Pressure<br />
(Velocity pressure is a different name for dynamic pressure. Both Total Pressure loss and velocity pressure are in Pascal, while dynamic-loss-coefficient C doesn't have a unit.)</p>
<p>For most bends/structures C is less than 1.<br />
<em>What I don't understand is what a C greater than 1 actually represents/means in the real world?</em><br />
(Does it mean the air turns backwards?)</p>
<p>(Because if you subtract the total pressure loss from the dynamic pressure it would be negative.)</p>
<p>(Since this also has practical relevance there are several organizations empirically determining/collecting these values (e.g. ASHRAE and SMACNA), and in most cases C<1, but there are quite a few cases where C>1. E.g. an <a href="https://i.imgur.com/Ai8CX4F.png" rel="nofollow noreferrer">example</a> of an fitting by SMACNA and an <a href="https://i.imgur.com/w6YoDbC.jpg" rel="nofollow noreferrer">example</a> by ASHRAE. (Edit: <a href="https://i.imgur.com/Tt0NKCv.png" rel="nofollow noreferrer">Heres</a> an image of the definitions and formulas used by SMACNA.)<br />
But the examples are just for illustration and don't matter in themselves, i'm just interested in what happens to the airflow when C>1.)</p>
<p>Regards</p>
| |fluid-mechanics|airflow|hvac| | <p>The <span class="math-container">$C$</span> coefficient simply quantifies the pressure loss of a restriction as a scale multiple or fraction of the dynamic pressure <span class="math-container">$\frac 1 2 \rho v^2$</span>. But the dynamic pressure is usually much less than the total pressure of the flow.</p>
<p>For example, air flowing at 10 m/s at a static pressure of 10 atm has a dynamic pressure of</p>
<p><span class="math-container">$$ \frac 1 2 \left( \frac{10}{1} 1.22 ~\rm{kg/m^3} \right)(10 ~ \rm{m/s})^2 = 610 ~\rm{Pa ~ or} ~0.006 ~\rm{atm} $$</span></p>
<p>which is far less than the static pressure of 10 atm. So your flow could lose many times the dynamic pressure going over restrictions and still have plenty of pressure left at the end. There is nothing special about the loss coefficient <span class="math-container">$C$</span> being greater than one.</p>
| 56451 | What does a dynamic-pressure loss coefficient of more than 1 mean (for eg air going through bend)? |
2023-09-27T09:16:24.157 | <p>I am doing some experiments in <a href="https://en.wikipedia.org/wiki/Fractional_freezing" rel="nofollow noreferrer">fractional freezing</a> with a modified <a href="https://www.walmart.com/ip/Arctic-King-7-Cu-ft-Chest-Freezer-Black/589721123" rel="nofollow noreferrer">residential deep freezer</a>. I have programmed my own PID controller on an Arduino setup such that it will not cycle the refrigeration pump on and off too frequently to cause damage. This has worked okay but I would like to control the temperature more precisely.</p>
<p>I am open to any control ideas, but one I have been contemplating is running the single phase motor/compressor unit with a <a href="https://rads.stackoverflow.com/amzn/click/com/B09SLV71Y4" rel="nofollow noreferrer" rel="nofollow noreferrer">single phase VFD</a>. Definitely wont have as much turn-down as 3 phase motor but should give me enough controllable range to make it work. Thoughts? Alternatives?</p>
| |electrical-engineering|control-engineering|refrigeration| | <p>Probably the simplest thing is to use a hot gas bypass valve and run it off the controller.</p>
<p><a href="https://hvacknowitall.com/blog/the-hot-gas-bypass-valve-explained#google_vignette" rel="nofollow noreferrer">https://hvacknowitall.com/blog/the-hot-gas-bypass-valve-explained#google_vignette</a></p>
| 56454 | Methods for precission control of a freezer cooling output |
2023-09-27T12:06:35.543 | <p>Browsing the internet yesterday I came across this very esoteric paper which purports to describe the construction of a device which uses the electrolysis of water into hydrogen and oxygen (a highly endothermic process) with KOH as a catalyst to cool down a room as a replacement for an Air Conditioner: <a href="https://www.researchpublish.com/upload/book/Electrolysis%20Air%20Cooler-3057.pdf" rel="nofollow noreferrer">https://www.researchpublish.com/upload/book/Electrolysis%20Air%20Cooler-3057.pdf</a></p>
<p>The enthalpy of the reaction is +285.83 kJ/mol, and while most of the energy input comes from the electricity being supplied, a fair portion (around 20%) comes from heat in the surroundings at standard temperatures so the basic chemistry/physics is there:</p>
<blockquote>
<p>As stated, splitting a mole of liquid water to produce a mole of hydrogen at 25°C requires 285.8
kJ of energy—237.2 kJ as electricity and 48.6 kJ as heat.</p>
</blockquote>
<p>Reference for this: <a href="https://www.nrel.gov/docs/fy10osti/47302.pdf" rel="nofollow noreferrer">https://www.nrel.gov/docs/fy10osti/47302.pdf</a></p>
<p>The authors of the paper describe the construction of such a device as well as a test showing that it is indeed capable of reducing temperatures inside a room (the paper itself is not well written so a lot of things are unclear, but they definitely did get the idea to work )</p>
<p>I found the idea quite intriguing, especially given that it generates hydrogen which can be stored to be used later as a fuel. I was not able to find any other references to this idea on the internet, which makes me suspect there is something else going on which makes the idea impractical for why other people haven't independently come up with the same thing and developed it further but I can't think of anything except for the hydrogen gas generated being an explosive hazard (but that can be stored/vented away so shouldn't be a big issue).</p>
<p>So, is this idea viable, or is there something big I'm missing?</p>
| |electrical-engineering|thermodynamics|chemical-engineering|cooling|electricity| | <p>You have referenced a process with a Co-efficient Of Performance of 0.25. According to Wikipedia, "Most air conditioners have a COP of 2.3 to 3.5"</p>
<p>You suggest that, if the energy could be recovered from the separated hydrogen and oxygen, you could get an improved COP.</p>
<p>If what you want to do is "cool a room", electrolysis of water is an extremely ineffective way of doing so, even with the referenced catalyst.</p>
<p>There are other similar cooling systems that work on chemical association/dissociation rather than phase change. It's not a crazy idea. And, as you suggest, hydrogen generation is currently an area of high interest. But using this process for room cooling is not something that, at present, would interest anybody who actually wanted to cool a room.</p>
| 56456 | Using the electrolysis of water to cool a room |
2023-09-29T07:28:41.783 | <p><a href="https://i.stack.imgur.com/qcbid.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qcbid.jpg" alt="enter image description here" /></a></p>
<p>As in the picture, we have a room with vents at the bottom to let cold air in and at the top to let hot air out. Normally the air flow is limited by the temperature gradient (correct me if I'm wrong). If we place a hot object in the vent on top, won't that create a faster draft? That way we could get very high "wind" speed indoors to replace fans.</p>
<p>Is this possible, and if yes, what's the limit for air flow per second?</p>
| |mechanical-engineering|structural-engineering|fluid-mechanics|thermodynamics| | <p>I take a stab at this very unfocused but interesting question.</p>
<ul>
<li>Yes a hot object like a small flame near or even inside the vent would accelerate the airspeed and potentially help move the heat from the room if the air pressure shock it creates would ride behind the temperature gradient flow, pushing it out like a rocket nuzzle. Also the heat it radiates or converts is blown away by the flow faster than it gets a chance to be absorbed by the duct and walls or even backfire.</li>
<li>such a system would work under a limited set of controlled parameters such as initial velocity, temperature, and laminar flow. It may need priming to get to those conditions and people and furniture in the room or opening a door may break the cycle!</li>
<li>The design of the vent and stack has to accommodate this (rocket nuzzle) action.</li>
</ul>
| 56473 | Will hot object near ceiling vent create more draft? |
2023-09-29T16:31:35.030 | <p>I have a guitar and also the rectangular cardboard box which it was delivered in. I want to make a more durable carry case for the guitar using the cardboard delivery box. It seems a shame to throw away a perfectly good box which is already the correct size and I don't want to buy a guitar case just yet. Dimensions of the box are 100x45x17 centimeters. The cardboard is made from two layers of corrugated material sandwiched in three layers of flat. I find from research that the cardboard needs to be sealed first with pva/wood glue after which fine woven glass cloth and then polyester resin can be applied. In previous experiments (when I didn't understand the necessity to seal it first) the cardboard warped and destroyed the shape of my project which was a tea tray. My question is how can I stop the cardboard from warping? If successful with the glassfibre I then want to apply a final color coat( I think this is called a gel coat).</p>
| |composite-resin| | <p>Consider that the typical multi-layer corrugated cardboard material has all the flutes running parallel/in the same direction. In order to minimize the warping you are experiencing, it would be necessary to create additional layers by using adhesive to bond "cross-grain" cardboard to the existing surface. You'd have to pick an adhesive, perhaps via experimentation that would not create warping while being applied or while curing.</p>
<p>PVA/wood glue is water based and will indeed warp while creating the new layers. Contact cement is not water based (unless you buy that particular product!) and will provide strong bonds between layers.</p>
<p>Additionally, gel coat is usually applied as the first layer within a mold, excluding ambient atmosphere during the curing process. Be sure to select a gel coat product which provides for this aspect of your build. For outer skin application, the terms "<a href="https://support.jamestowndistributors.com/hc/en-us/articles/360022132074-What-s-the-difference-between-gelcoat-with-wax-and-without-wax-" rel="nofollow noreferrer">gel coat with wax</a>" will provide suitable results in a search. Useful information at the linked site.</p>
| 56475 | Stop cardboard from warping when impregnating with polyester resin |
2023-09-30T23:00:52.843 | <p>I suspect/hope I'm not kidding myself too much with the physics of this; it seems to stack up in my mind; but can we not relieve tension on, say, a steel cable running from a space station in orbit around the Earth by propelling the station towards the Earth using ion thrusters, thereby relieving tension? Bear in mind that this idea came about after I began working on a design for a new space elevator consisting of about 100 stages of about 1km each stretching to the Karman line; I was planning on using turbofan jet engines and dirigibles up to the end of the stratosphere, and then rings of rocket engines after that to stabilise the whole thing and relieve tension; now I'm thinking it might just be good enough to simply use ion thrusters in space to counteract the tension on the cable. Is there any reason why that wouldn't work?</p>
| |civil-engineering|aerospace-engineering| | <p>The tension on the cable is due to centrifugal<sup>1</sup> force.</p>
<p>So, to help (at all) you need something that produces more thrust downward than its mass will produce in centrifugal force upward.</p>
<p>I'm pretty sure no current ion thruster even comes close to that. For example, NASA's X3 ion thruster weighs about 500 pounds, and produces about 1.2 pounds of thrust.</p>
<p>You'd also pretty much need to deliver power through the cable as well, which makes the problem even more difficult. The X3 thruster needs 250 amps of electricity to deliver its 1.2 pounds of thrust. In normal situations, to carry 250 amps, you'd use a wire around 00 to 0000 AWG (depending on distance). The distance you're dealing with is <em>much</em> greater than either of those is designed for though. To keep cabling halfway reasonable, you'd almost certainly want to run (much) higher voltage over the cable and step it down at the thruster to get the necessary current. Oh, but that (you can probably guess it) is going to add its own weight/mass, leading to more centrifugal force to overcome.</p>
<p>Until or unless ion thrusters have <em>drastically</em> higher thrust/mass ratio, this seems like a losing proposition. Even with drastically improved thrust/mass ratio, power delivery is likely to remain problematic as well.</p>
<p><strong>Reference</strong></p>
<p><a href="https://newatlas.com/x3-hall-thruster-test-record-nasa/51869/" rel="nofollow noreferrer">https://newatlas.com/x3-hall-thruster-test-record-nasa/51869/</a></p>
<hr />
<sup>
1. For the moment, ignoring phrasing details about centrifugal vs. centripetal force.
</sup>
| 56497 | Can we use ion thrusters in space to relieve tension on a space elevator cable? |
2023-10-01T03:20:26.030 | <p>Between working on my and my girlfriend's bikes, I've come across different style compression ferrules (aka sleeves/olives) used in the hydraulic brake systems and would like to know:</p>
<ol>
<li>What is the purpose of the grove in the middle?</li>
<li>Would it be safe to use a standard brass hydraulic ferrule (straight or curved body) as long as it's rated for similar or higher pressure? (I would continue to use a barbed insert rather than switch to smooth.)</li>
</ol>
<p>(<a href="https://www.worldwidecyclery.com/products/trp-hydraulic-hose-kit-for-5-0mm-hose" rel="nofollow noreferrer">example 1</a>; <a href="https://www.jensonusa.com/Shimano-BH90-Hose-Bolt-and-Fitting-Unit" rel="nofollow noreferrer">example 2</a>; <a href="https://rads.stackoverflow.com/amzn/click/com/B08JJM76WH" rel="nofollow noreferrer" rel="nofollow noreferrer">example 3</a>; <a href="https://www.pexuniverse.com/brass-compression-sleeves" rel="nofollow noreferrer">example 4</a> "standard")</p>
<p>To be clear, these ferrules and inserts go over/inside single or dual-layered PVC tubing.</p>
| |hydraulics| | <p>I think the feature acts like a lock so it fits into the brake well like a seal. These hoses don't need crimping, according to this video. <a href="https://youtu.be/z5qIrpGR3yY?si=uP8OnMla9kn24HwP" rel="nofollow noreferrer">https://youtu.be/z5qIrpGR3yY?si=uP8OnMla9kn24HwP</a></p>
<p>As for your question on safety, I'm not sure if will fit in the first place.</p>
| 56499 | Compression ferrules for bicycle hydraulic brakes |
2023-10-01T07:35:47.790 | <p>Mechanical engineering student here (the professor is not being helpful). When a single alloying element is added to a normal iron-carbon steel, the resulting changes in eutectoid temperature and composition are relatively easy to determine using graphs such as the following:</p>
<p><a href="https://i.stack.imgur.com/Rj1nJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rj1nJ.png" alt="enter image description here" /></a></p>
<p>How would you predict the changes if multiple alloying elements were added instead of just one? I understand that some elements tend to stabilize ferrite versus austenite (you can tell by the relative slopes of the upper graph), but this only provides a qualitative answer. Is there any way to calculate exactly what the changes would be? Thanks.</p>
<hr />
<p><sub> Image source: <a href="https://www.tf.uni-kiel.de/matwis/amat/iss/kap_9/illustr/s9_2_1.html" rel="nofollow noreferrer">https://www.tf.uni-kiel.de/matwis/amat/iss/kap_9/illustr/s9_2_1.html</a> </sub></p>
| |steel|metallurgy|material-science| | <p>The effect of multiple alloy elements is not as simple as single elements. The addition of one element changes how the material interacts with another and each element combination has a different effect. Additions of Mn, for example, will change how much Ni or C the matrix can solubilize, and that should change the behavior of your upper plot.</p>
<p>You could qualitatively approach this problem, for example, by using Carbon Equivalent, Chromium Equivalent or Nickel Equivalent Formulae to understand different phase diagrams, like the Schaeffler Diagram, shown below.
[<img src="https://i.stack.imgur.com/RNuXk.png" alt="Schaeffler Diagram, used primarily for study of stainless steels. Source: ][1]" /></p>
<p>The method most used nowadays in alloy design and research is that of CALPHAD, or CALculation of PHAse Diagrams. Experimental data is modeled and chemical activity models calculate phase stability criteria, yielding exactly what your question tries to understand. Via a CALPHAD software (like Thermocalc), one could generate plots or tables to understand the effect of many different variables, including alloying elements.</p>
| 56502 | Effect of multiple alloying elements on steel's eutectoid properties |
2023-10-03T11:01:15.503 | <p>L.S.</p>
<p>Currently for my course in Continuum Mechanics, I am asked to calculate the force potential in the configuration as shown in the picture. First I will introduce the full problem, then show my calculations. I hope somebody can help me and show my error as it is hard to get in contact with my professors.</p>
<p><a href="https://i.stack.imgur.com/fCJ9B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fCJ9B.png" alt="Mechanism of two solid bars." /></a></p>
<blockquote>
<p>The mechanism is made up of two solid bars, the long one has a length <span class="math-container">$L$</span>, the short one a length of <span class="math-container">$L/2$</span>. The reference system used is <span class="math-container">$\{\mathbf{e}_x, \mathbf{e}_y\}$</span>. One end is locked along <span class="math-container">$\mathbf{e}_x$</span> and <span class="math-container">$\mathbf{e}_y$</span>, the other end is only allowed to slide along <span class="math-container">$\mathbf{e}_x$</span>. At the connection between the two bars, a torsion spring with a stiffness <span class="math-container">$\kappa$</span> is acting. The spring is associated with an angle <span class="math-container">$\gamma$</span> and is unstretched for an angle <span class="math-container">$\gamma_0$</span>. A force <span class="math-container">$P$</span> is applied to the mechanism and a mass <span class="math-container">$m$</span> is hanging from
the end of the long bar. Under the action of the forces, the
mechanism moves and its movements are described using an
angle <span class="math-container">$\beta$</span>.</p>
</blockquote>
<p>Now my task is to compute the force potential <span class="math-container">$\mathcal{F}$</span> associated with forces on the mechanism as a function of <span class="math-container">$L$</span>, <span class="math-container">$m$</span>, <span class="math-container">$\beta$</span>.</p>
<p><span class="math-container">\begin{equation}
\mathcal{F} = -W_E
\end{equation}</span>
<span class="math-container">\begin{equation}
W_E = W_P + W_{mg} = \int \mathbf{P} \cdot \text{d}\mathbf{r}_P + \int m\mathbf{g} \cdot \text{d}\mathbf{r}_{mg}
\end{equation}</span>
<span class="math-container">\begin{equation}
= \int_{\beta_0}^{\beta} \mathbf{P} \cdot \frac{\text{d}(\mathbf{r}_P)}{\text{d}\beta}\text{d}\beta
+ \int_{\beta_0}^{\beta} m \mathbf{g} \cdot \frac{\text{d}(\mathbf{r}_{mg})}{\text{d}\beta}\text{d}\beta
\end{equation}</span></p>
<p>From the picture I can make out the following position- and forcevectors:</p>
<p><span class="math-container">\begin{equation}
\mathbf{P} = [-P, 0]^T
\end{equation}</span>
<span class="math-container">\begin{equation}
\mathbf{g} = [0, -g]^T
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
\mathbf{r}_P = L[\cos(\beta), 0]^T
\Longrightarrow
\frac{\text{d}(\mathbf{r}_P)}{\text{d}\beta} = L[-\sin(\beta), 0]^T
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
\mathbf{r}_{mg} = L[\cos(\beta), \sin(\beta)]^T
\Longrightarrow
\frac{\text{d}(\mathbf{r}_{mg})}{\text{d}\beta} = L[-\sin(\beta),\cos(\beta)]^T
\end{equation}</span></p>
<p>Now if we put this together, we now get</p>
<p><span class="math-container">\begin{equation}
\mathcal{F} = -W_E = -\left(\left( PL\int_{\beta_0}^{\beta} \sin(\beta) \text{d}\beta \right)+
\left( -mgL\int_{\beta_0}^{\beta} \cos(\beta) \text{d}\beta \right)\right)
\end{equation}</span></p>
<p>If we evaluate the integrals we get</p>
<p><span class="math-container">\begin{equation}
PL\int_{\beta_0}^{\beta} \sin(\beta) \text{d}\beta = -PL \cos{\beta} \vert_{\beta=\beta_0}^{\beta}
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
-mgL\int_{\beta_0}^{\beta} \cos(\beta) \text{d}\beta = -mgL\sin{\beta} \vert_{\beta=\beta_0}^{\beta}
\end{equation}</span></p>
<p>And this gives us
<span class="math-container">\begin{equation}
\mathcal{F} = PL (\cos(\beta) - \cos(\beta_0)) +
mgL(\sin(\beta) - \sin(\beta_0))
\end{equation}</span></p>
<p>However the answer the textbook gives me is</p>
<blockquote>
<p><span class="math-container">\begin{equation}
\mathcal{F} = PL (\cos(\beta) - \cos(\beta_0)) -
mgL(\sin(\beta) - \sin(\beta_0))
\end{equation}</span></p>
</blockquote>
<p>I believe I have made a mistake in the definition of the position vectors <span class="math-container">$\mathbf{r}_p$</span> and/or <span class="math-container">$\mathbf{r}_{mg}$</span>, but I can't find my mistake.</p>
| |mechanical-engineering| | <p><span class="math-container">$\vec{g}=(0,-g)$</span></p>
<p><span class="math-container">$\frac{d\vec{r}_{mg}}{d\beta}=L(-\sin\beta, \cos \beta)$</span>.</p>
<p><span class="math-container">$\vec{g}.\frac{d\vec{r}_{mg}}{d\beta}=-gL\cos \beta$</span>.</p>
<p><span class="math-container">$\mathcal{F}=-W_E= \underline{+} m\int \vec{g}.\frac{d\vec{r}_{mg}}{d\beta} d\beta $</span></p>
| 56529 | Virtual work of two forces, how do you know whether they are in the same direction? |
2023-10-04T00:49:46.643 | <p>There is a system with two controllers in series having the transfer functions G<sub>1</sub>(s) and G<sub>2</sub>(s) which is separated by a summing point through which the disturbance D(s) enters the system. The output from the 2nd controller is fed back through another element having the transfer function H(s).
The closed loop transfer function C<sub>D</sub> is given by the formula C<sub>D</sub>/D(s) = G<sub>2</sub>(s) / (1 + G<sub>1</sub>(s) * G<sub>2</sub>(s) * H(s)).</p>
<p>Can someone help me figure out how we get to this final formula.<a href="https://i.stack.imgur.com/O2pvA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O2pvA.png" alt="enter image description here" /></a></p>
<p>I included the diagram too. It is from the book Modern Control Engineering by Katsuhiko Ogata.</p>
| |control-engineering|control-theory|transfer-function| | <p>Assume <span class="math-container">$R(s) = 0$</span>. Now the left most summing junction can be removed from the diagram <strong>taking care to shift the -ve sign to the remaining summing junction</strong>.</p>
<p>The above assumption is valid since transfer functions are derived for SISO systems. So, the other inputs are allowed to be assumed zero (or constant as the case may be).</p>
<p>Then redraw the diagram with <span class="math-container">$G_2$</span> in <span class="math-container">$\text{forward}$</span> path and <em>both</em> <span class="math-container">$H(s)$</span> and <span class="math-container">$G_1(s)$</span> in <span class="math-container">$\text{feedback}$</span> path (with negative feedback).</p>
<p>Now the closed loop transfer function is given by
<span class="math-container">$$
\frac{\text{forward}}{1 + \text{forward} \cdot \text{feedback}}
$$</span></p>
| 56534 | Transfer functions of a closed loop system subjected to a disturbance |
2023-10-07T13:30:06.963 | <p>I am wondering if the interaction between the magnetic field created by a conical-shaped solenoid and the Earth's magnetic field will result in this conical-shaped solenoid being propelled through the Earth's atmosphere.</p>
<p>I am planning to build such a solenoid and then I plan to mount it on a small raft which I will build to see if the raft will be propelled across the water by this solenoid.</p>
<p>This solenoid will be a hollow cone made out of plastic and will have an electrified wire wrapped around it. I am planning to use a plastic cone in which the larger end will be 41cm in diameter and the smaller end will be 10cm in diameter. The power source will be either a battery pack placed on the raft, or perhaps via wires coming from a DC power supply.</p>
<p>I am thinking that when this conical-shaped solenoid is pointed towards the North Pole, with the smaller opening of the cone pointed towards the North Pole, the Earth's magnetic field within the interior of this solenoid will interact with the magnetic field of the solenoid. Due to this interaction, the solenoid (and the raft) will be propelled towards the North Pole. This working principle is based off of the interaction of two current carrying wires shown in the drawing below.</p>
<p><a href="https://i.stack.imgur.com/lV8nU.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lV8nU.gif" alt="enter image description here" /></a></p>
<p>This drawing is from: <a href="https://www.physics.louisville.edu/cldavis/phys111/notes/magn_forces_curr.html" rel="nofollow noreferrer">https://www.physics.louisville.edu/cldavis/phys111/notes/magn_forces_curr.html</a></p>
<p>Changing the polarity of the solenoid should result in the solenoid be propelled towards the South Pole.</p>
<p>I created a drawing which shows a cross-sectional view of how the magnetic field of this solenoid and the Earth's magnetic field should interact with each other:</p>
<p><a href="https://i.stack.imgur.com/Y7q47.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y7q47.png" alt="enter image description here" /></a></p>
<p>Before I put in the time and effort to build this test model, I would like to know if this working principle is a sound one or if this test model will not work due to me misunderstanding something.</p>
<p>Would a conical-shaped solenoid be propelled by its interaction with the Earth's magnetic field?</p>
| |electrical-engineering|design|electromagnetism|propulsion| | <p>I think you'll be very disappointed.</p>
<p><a href="https://i.stack.imgur.com/AUPnf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AUPnf.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. Image source: <a href="https://scienceclass.github.io/5-minute-magnetic-compass.html" rel="nofollow noreferrer">Science Class</a>.</em></p>
<p>Your coil will form a very weak magnet since it has an air core. (A good solenoid will use an iron core as it has a permeability 1000 times that of air.)</p>
<p>Place your solenoid on a boat and it will rotate so that it's north pole will point north. (The Earth's north pole is actually a magnetic south pole! That's why the magnet's north pole points in that direction.) Once that happens the turning moment will fall to zero and the forces will be balanced with the same force pulling north and south.</p>
<p>I think you'll be making a battery powered compass - and a very weak one.</p>
| 56564 | Would a conical-shaped solenoid be propelled by its interaction with the Earth's magnetic field? |
2023-10-08T17:23:49.307 | <p><em><strong>Background:</strong></em> Read <a href="https://www.tomshardware.com/news/dollar3-sata-ssd-vs-dollar5-nvme-ssd-which-is-the-better-deal" rel="nofollow noreferrer">this article on Tom's Hardware</a> that mentions SSDs, and during the process notes:</p>
<p>"[The] Silicon Motion's SM2263XT PCIe 3.0 SSD controller [...] adheres to a dual-core configuration sporting the <em><strong>Arm 32-bit Cortex-R5 architecture</strong></em>"</p>
<p>Reading <a href="https://en.wikipedia.org/wiki/ARM_Cortex-R" rel="nofollow noreferrer">the ARM Cortex-R article on Wikipedia</a> it sounds like the processor is just like any other processor, except "cores are optimized for hard real-time and safety-critical applications. Cores in this family implement the ARM Real-time (R) profile, which is one of three architecture profiles"</p>
<p>From my admittedly outsider perspective, that sounds like a normal CPU. Apparently the SSD is a dual core like my desktop used to be.</p>
<p><em><strong>Question:</strong></em> Can the cores on an SSD like this be accessed and used for general computation? (Not necessarily this specific SSD, just generally)</p>
<p><em>Note:</em> Not sure if this is the correct StackExchange. Dithered between CS, Super User, Internet of Things, and Hardware Recommendations, yet none seem to have a computer hardware or hardware engineering, or similar.</p>
| |consumer-electronics|computer-engineering|embedded-systems|computer-hardware| | <p>Yes, but if you have to ask, no.</p>
<p>SSD controllers run firmware contained in flash memory: it is the same kind of memory used for the SSD. If you know how to re-flash the controller, and know what i/o has been attached to that controller, you can communicate with it and use it.</p>
<p>This is not information that is normally publicly available.</p>
<p>But some controllers have been used to emulate two or more devices (both in production and in research projects). That is, not just an SSD.</p>
<p>PCIe is a general-purpose communication channel used for all kinds of things. So that SSD has a general-purpose communication channel, and that SSD is controlled by the controller which is able to re-map the SSD pages, so <em>even if</em> the actual i/o options are extremely limited, it can be programmed to communicate by mapping file data.</p>
<p>Since this is a demonstrated mal-ware channel, you can expect that restrictions are in place: at best it will be difficult to learn about, at worst there will be some kind of fuse locks and signature in place.</p>
| 56568 | Can the Processor Cores on SSDs be Used for General Computation? |
2023-10-08T22:35:03.607 | <p>I am figuring out the state space equations of a connected 2-tank system. For the 1st tank, the state space equation is as follows :
<code>dh̃₁/dt = (-K₁/ρA₁) h̃₁ + (1/ρA₁) m̃ᵢ</code> where h̃₁= h - h̄, m̃ᵢ = m - m̄ and K₁ = (k/2√h̄).
No problems here. Perfectly clear. What is tripping me up is the equation for the 2nd tank.</p>
<p>First, I do a mass balance.
<span class="math-container">$$ρA₂dh₂/dt = k√h₁ - k√h₂$$</span>
Then we perform linearization using the Taylor series expansion and ignore the terms with the 2nd derivative and higher.<br />
So, we get <code>ρA₂dh₂/dt = (k√ħ₁) + (k/2√ħ₁) (h₁ - ħ₁) - (k√ħ₂) - (k/2√ħ₂)(h₂ - h̅₂)</code></p>
<p>Next, we get
<code>ρA₂dh₂/dt = (k/2√ħ₁) (h̃₁)- (k/2√ħ₂)(h̃₂)</code></p>
<p>My question is simple, what happened to the k√ħ₁ and - (k√ħ₂) terms which were there earlier?</p>
<p><a href="https://i.stack.imgur.com/EdjTE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EdjTE.png" alt="enter image description here" /></a></p>
| |control-engineering|control-theory|modeling| | <p>They got cancelled.</p>
<p>At equilibrium when <span class="math-container">$dh₂/dt = 0$</span>, the equation for the second tank is <span class="math-container">$0 = k√ħ₁ - k√ħ₂$</span>.</p>
| 56571 | State space representation of a 2 tank system |
2023-10-09T15:56:59.780 | <p>I'm trying to perform 2D analysis for a steady laminar incompressible flow through a horizontal channel with a height of H (m) and length of 2H. The boundary conditions is of the form:</p>
<p><span class="math-container">$u=\frac{3}{2}U_{avg}\left(1-\frac{2y}{H}\right),v=0$</span> at <span class="math-container">$x=0$</span>,</p>
<p><span class="math-container">$\frac{\partial u}{\partial x}=0,\frac{\partial v}{\partial x}=0$</span> at <span class="math-container">$x=2H$</span>,</p>
<p>with no-slip condition at the walls. Is it possible to implement these boundary conditions in Fluent? I think that UDF may help but I have no experience dealing with it. Thanks in advance, I will appreciate any help.</p>
| |fluid-mechanics|ansys|fluent| | <p>I am not sure if you have already found a way or not, but for others just posting the answer. You can always refer the Fluent UDF user guide and samples.
These are few additional examples which you can refer:</p>
<p><a href="https://www.learncax.com/knowledge-base/blog/by-category/cfd/writing-a-user-defined-function-udf-in-ansys-fluent" rel="nofollow noreferrer">https://www.learncax.com/knowledge-base/blog/by-category/cfd/writing-a-user-defined-function-udf-in-ansys-fluent</a></p>
<p><a href="https://www.southampton.ac.uk/%7Ezxie/SESS6021/udf_tut1.pdf" rel="nofollow noreferrer">https://www.southampton.ac.uk/~zxie/SESS6021/udf_tut1.pdf</a></p>
<p>Thanks</p>
| 56582 | Variable inlet velocity implementation in ANSYS Fluent |
2023-10-10T07:44:18.973 | <p><a href="https://en.wikipedia.org/wiki/Masonry_dam" rel="nofollow noreferrer">A Masonry Dam</a> is build out of individual stones or bricks. The Wikipedia link says that it is "sometimes" joined with mortar, and has a cite for it but the link is 404 dead.</p>
<p>Is that really true? How could a masonry dam not require mortar? The dam needs to be water-tight against a wall of water, which has larger pressures at the bottom. I don't see how the individual blocks could be so tight and perfectly-fitting to stop water at high pressures.</p>
<p>I was unable to find a single example of a masonry dam that explicitly does not use mortar. I guess I'm missing something. Is there some other common seal mechanism besides mortar?</p>
| |structural-engineering|masonry|hydroelectric-dams| | <p>A "face" coating that provides the waterproof layer.</p>
<p>There are many materials used, from polymers to asphalt, cement etc</p>
<p>The material is chosen as it has to resit water movement - driven by wind etc</p>
| 56587 | How can a masonry dam not require mortar? |
2023-10-10T14:05:03.573 | <p>On many websites, I read that double-acting reciprocating compressors are more efficient than single-acting compressors.
These sources say it's because they complete two compression cycles on each crank shaft turn.</p>
<p>Why should the number of compression cycles per crank shaft turn matter for energy efficiency?
That should give me double the amount of compressed fluid at double the energy expense, isn't it?</p>
<p>These sources also mention that a double-acting reciprocating compressor uses a crosshead between the piston and the rod. I could imagine that this crosshead can absorb the typical piston thrust and thus reducing the piston friction. But does it make such a big difference that double-acting pistons are one of the "most efficient ways" to provide a compressed fluid?</p>
<p>Two example websites claiming the above statements:</p>
<p><a href="https://www.titusco.com/equipment/compressors/reciprocating-compressors/#1678981011717-c80d8bbf-c693" rel="nofollow noreferrer">Link1</a></p>
<p><a href="https://blog.exair.com/2017/09/19/about-double-acting-reciprocating-air-compressors/#:%7E:text=The%20double%20acting%20compressor%20compresses,of%20air%20compressor%20very%20efficient." rel="nofollow noreferrer">Link2</a></p>
| |mechanical-engineering|compressed-air|compressors|compressed-gases| | <p>You will find losses in a single acting compression such as leakage, even with a seal as the pressure difference is high, any leaked fluid in a dual acting compressor is less of a loss as it leaks into something that will have work done to it.
You would also find inertial losses with a single acting compressor but with a dual acting compressor these are two for one assuming they don't change.
As for your reasoning you are right if the systems weren't lossy then the single and dual acting compressors should be the same.</p>
| 56591 | Why are double-acting reciproating compressors more efficient than single-acting ones? |
2023-10-10T19:52:46.453 | <p>I am watching some videos on control theory by <a href="https://www.youtube.com/@johnrossiter2323/playlists" rel="nofollow noreferrer">John Rossiter</a>, and they are very good. So far I have watched most of the videos on classical control, including transfer function, feedback, etc.</p>
<p>One topic that keep arising is the importance of integrators to help reduce the offset of the system to zero. That is, in order to reduce the difference between the target level and the actual level of the system to zero, an integrator is required in the simple first and second order systems provided.</p>
<p>But I am not clear on what is meant by an integrator. The presentation in the videos is about the Laplace transform structure of an integrator as a pole located at zero. That is, the inclusion of a <span class="math-container">$\frac{1}{s}$</span> in the transfer function. The integrator can be included in the controller or the plant itself. I understand the math easy enough.</p>
<p>However, I am not clear on what is the physical intuition behind an integrator? Like is the integrator some sort of mechanical component of a control system, or some sort of
design aspect. I suppose I am trying to find some examples of systems that contain integrators, so that I can understand what action the integrator is performing.</p>
| |control-engineering|control-theory|pid-control| | <p>In the physical sense, an integrator adds up your input over time.</p>
<p>Electrical Current * time -> (change in) Electrical Charge</p>
<p>Fluid Flow * time -> (change in) Volume</p>
<p>Heat Flow * time -> (change in) Temperature (for a given thermal mass)</p>
<p>Object Velocity * time -> (change in) Object Position</p>
<p>Force * time -> (change in) Momentum</p>
<p>You might see this in the "plant" of a control loop. For example if you're controlling the fluid volume in a tank (system output), there might be an intermediate input that is the fluid flow. In practice you might have another thing like a valve which controls your flow, so this is a little bit of an example.</p>
<p>You might also see an integrator in the "compensator" of the control loop. In that case, it's purely mathematical, and if we want to have intuition, we should think about the control loop in a purely mathematical sense. The integrator gives you "lots of response at low frequency and no response at high frequency". The bode plot representation is a nice way to approach this.</p>
| 56597 | What is the physical intuition behind an "integrator" in control theory |
2023-10-10T20:15:29.727 | <p><a href="https://i.stack.imgur.com/GdLm1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GdLm1.png" alt="Problem solution" /></a></p>
<p>See image... why is it that in part 1 the answer is negative but in part 2 it is added instead of subtracted?</p>
<p>I thought because I got a negative in part 1 I would have to subtract in part 2 but the solution shows otherwise.</p>
| |mechanical-engineering|materials|civil-engineering| | <p>In Part 1 the rotation is clockwise WRT rigid disk B if we chose plane B as reference and look at it in such a way that the steel assembly is coming out of that plane. This viewpoint is used in line 1 of your notes to calculate the <span class="math-container">$\phi_{C/B}$</span>.</p>
<p>However, if we look from behind B toward A the <span class="math-container">$\phi_{C/B}$</span> is counterclockwise and positive.</p>
<p>This is the view used for total rotation(line 2 of your notes). Hence both rotations are counterclockwise and additive!</p>
| 56598 | Rotation of a tube |
2023-10-10T20:16:59.043 | <p>I am reading the book "Model-Based Fault Diagnosis Techniques" by Steven X. Ding. Here, they describe a system with the equation <span class="math-container">$$ ẋ = (A + ΔA_F)x +(B + ΔB_F)u+E_f f$$</span> Then, <span class="math-container">$$ ΔA_F = A_i θ{_A}{_i} $$</span> is given</p>
<p>Then, they proceed with finding the partial derivative of <span class="math-container">$ẋ$</span> with respect to <span class="math-container">$θ{_A}{_i}$</span> for which they get the answer as <span class="math-container">$ A \frac {∂x}{∂θ{_A}{_i}} +A_ix $</span>.</p>
<p>I tried doing it myself, and this is how I do it.</p>
<p><span class="math-container">$$ \frac {∂ẋ}{∂θ{_A}{_i}} = A \frac {∂x}{∂θ{_A}{_i}} + A_i x \frac {∂θ{_A}{_i}}{∂θ{_A}{_i}} + A_i θ{_A}{_i} \frac {∂x}{∂θ{_A}{_i}} $$</span></p>
<p>Which can be simplified as</p>
<p><span class="math-container">$$ \frac {∂ẋ}{∂θ{_A}{_i}} = A \frac {∂x}{∂θ{_A}{_i}} + A_i x + A_i θ{_A}{_i} \frac {∂x}{∂θ{_A}{_i}} $$</span></p>
<p>Essentially, the last term is surplus and I am not sure how I can get the answer as in the book.</p>
<p>Just to add some context, will the product rule not apply for the last term ? As <span class="math-container">$x$</span> and <span class="math-container">$θ{_A}{_i}$</span> are both functions of <span class="math-container">$θ{_A}{_i}$</span>.</p>
| |control-engineering|control-theory|partial-differential-equations| | <p>I agree with you, and I think this is simply an inconsequential mistake from the book, since in the proof, this result is only an intermediate step, and it is later evaluated at <span class="math-container">$\theta_{A_i} = 0$</span>.</p>
| 56599 | Partial derivative of change in state with respect to multiplicative faults |
2023-10-12T02:14:28.263 | <p>I am designing a front panel for an electronic device with PCB-mounted components on it. I need to provide precise CAD geometry for drill holes, etc. to the third party that will mill / manufacture the panel out of 2mm thick aluminum stock.</p>
<p>The device includes two PTV09A-2 (side mount) potentiometers. The potentiometer has a main shaft that obviously involves a drill hole in the front panel. The body also contains two small circular bumps that I think should align with holes in the panel to prevent the pot body from rotating if the shaft/knob is rotated to the stops, and prevent mechanical stress on the pins connecting the pot to the PCB.</p>
<p>Looking at the drawing from the datasheet below, however, I cannot find any dimensioning that tells me how I should align drill holes for these relative to the central shaft, or what hole diameter is appropriate for them... all I can tell is that they are 1.2mm tall. (Bumps circled in green below.) I do notice that in the profile (top) view of both variants, the anti-rotation bumps have a dashed line going through the middle; but I don't understand what that means or if it implies something about their alignment.</p>
<p>I can measure the diameter of the bumps easily enough with calipers, but trying to establish center-to-center offset from bumps to the pot shaft with a reasonably tight tolerance is harder. Am I missing something that would give me a clue to their relative location?</p>
<p><a href="https://i.stack.imgur.com/RRtwz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RRtwz.png" alt="mechanical drawing of PTV09A-1 and PTV09A-2 potentiometers" /></a></p>
<p>Full datasheet at <a href="https://www.bourns.com/docs/Product-Datasheets/PTV09.pdf" rel="nofollow noreferrer">https://www.bourns.com/docs/Product-Datasheets/PTV09.pdf</a></p>
| |cad|technical-drawing| | <p>You're correct. The dimensions aren't there.</p>
<p>I suspect that the bumps may be used in the manufacturing assembly process and aren't intended for use in mounting. They may even be designed as spacers / stand-offs to keep a minimum distance between the face of the pot and the front panel. I notice that the "w/Metal Threaded Bushing" versions (which <em>are</em> designed for panel mounting) don't have the bumps.</p>
<p>Make the solder pads for the body lugs good and meaty - and maybe double-sided - and you should be fine!</p>
| 56611 | What are the positions/diameter of the anti-rotation bumps on this mechanical drawing? |
2023-10-13T06:21:05.277 | <p>I have a basic question regarding the calculation of mass flow rates of a channel with multiple outlets. In the scenario depicted in the image below, we have a single inlet with a mass flow rate (<em>m1</em>) of 0.87 kg/s, and I'm trying to determine the mass flow rate (<em>m2</em>) at each of the six outlets.<a href="https://i.stack.imgur.com/ZcPVY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZcPVY.png" alt="enter image description here" /></a></p>
<p>Considering mass conservation,<code>M1 = M2</code>.</p>
<pre><code> M1 = density * area of all the outlets * velocity
0.87 (kg/s) = 1.18 * 10^-9 (kg/mm^3) * 6 * 1 (mm^2) * velocity
velocity = 122881355.932 (mm/s)
velocity = 122,881.355 (m/s)
</code></pre>
<p>Now, to find the velocity for one outlet, I divide the total velocity by the number of outlets:</p>
<pre><code>Velocity per Outlet = 122, 881.355 / 6 = 20,480.22 m/s
</code></pre>
<p>Therefore, the mass flow rate <code>M2</code> for one outlet is calculated as:</p>
<pre><code>m2 = 1.18 (kg/m^3)∗ 1 * 10^-6 (m^2) ∗ 20,480.22 m/s = 0.0241666 kg/s
</code></pre>
<p>Is this calculation correct?</p>
<p>Thanks in advance.</p>
| |fluid-mechanics|pipelines| | <p>In theory, 1 outlet will have 1/6 of the total mass flow, BUT flow effects will change that.</p>
<p>If you need to have a precise measurement at each exit then calibrated orifices or pitot tubes may be needed.</p>
| 56619 | How to calculate mass flow rate at the Multiple Channel Outlets? |
2023-10-14T21:55:53.557 | <p>What does this F (similar to mathematical function symbol) means in blueprints?</p>
<p>Year of blueprint 40s.</p>
<p><a href="https://i.stack.imgur.com/VhP9K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VhP9K.jpg" alt="enter image description here" /></a></p>
| |metallurgy|technical-drawing|drawings| | <p>I think it means finishing the surface. The finish symbol indicates the surface must be machined and finished. <a href="https://en.wikipedia.org/wiki/Engineering_drawing_abbreviations_and_symbols" rel="nofollow noreferrer">source Wiki</a></p>
<p>The tick line in the middle indicates the direction of the machining!</p>
| 56639 | What does this F symbol means in blueprints? |
2023-10-14T23:04:36.747 | <p>I am trying to design a digital control system of which output depends on the difference between 1 input signal 1 tick ago and the output 1 tick ago ,and 1 another input signal.</p>
<p>The output <span class="math-container">$y(n)$</span> = input <span class="math-container">$i(n)$</span> <span class="math-container">$\cdot$</span> <span class="math-container">$b(n)$</span> and <span class="math-container">$b(n)$</span> = input <span class="math-container">$d(n-1)$</span> <span class="math-container">$-$</span> <span class="math-container">$y(n-1)$</span> Here is what I have come up with so far:</p>
<p><a href="https://i.stack.imgur.com/gMGNW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gMGNW.png" alt="" /></a></p>
<p>I have made the time delay part.The inverse Z transform of <span class="math-container">$z^{-1}$</span> is <span class="math-container">$\delta(n-1)$</span> so I have created the time delay in both <span class="math-container">$D(s)$</span> and the feedback loop.However I need to somehow put I(s) in the overall system so that it does what I want but so far I have been unsuccesfull on doing so.How should I keep on?</p>
<hr />
<p>Here is what I have come up with:</p>
<p><a href="https://i.stack.imgur.com/gMGNW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gMGNW.png" alt="" /></a></p>
<p>Note:I have added a steady signal to the summer</p>
| |control-engineering|control-theory| | <p>First, recognize that your system, as given, is not linear time-invariant, because of the multiplication by <span class="math-container">$i(n)$</span>. Depending on how you treat <span class="math-container">$i(n)$</span> and it's effects, your system is either nonlinear (if you take <span class="math-container">$i(n)$</span> as a regular old input) or time-varying (if you take <span class="math-container">$i(n)$</span> as a predefined signal, and make it part of the system's time dependence).</p>
<p>Either way, using <span class="math-container">$z$</span> notation on the signals is misleading. Even for non-linear systems, using <span class="math-container">$\frac 1 z$</span> inside a block to denote a unit delay is pretty common; that's what I do <em>with the expectation that the reader will still understand it's not an LTI system</em>. Here's what I would draw, using <span class="math-container">$\Sigma$</span> to denote a summing junction, and <span class="math-container">$\Pi$</span> to denote a product junction. It's pretty much the same except I've left out the <span class="math-container">$A$</span> term which you did not mention in your question.</p>
<p><a href="https://i.stack.imgur.com/58J6p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/58J6p.png" alt="enter image description here" /></a></p>
| 56641 | Creating a recursive control system |
2023-10-18T12:06:42.173 | <p>I'm trying to to teach myself to design gears for a hobby project, and I want to make a similar design. I typically use Autodesk inventor, but I was wondering if there was something better for this sort of thing. <a href="https://i.stack.imgur.com/x9tr5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x9tr5.png" alt="gears with teeth missing" /></a></p>
<p>This is from <a href="https://engineering.stackexchange.com/questions/11204/is-there-a-mechanism-that-can-turn-circular-motion-into-alternating-circular-mot">this</a> question.</p>
| |design|gears|cad| | <p>It is conjecture based on my experience with the 3D modeling program known as <a href="https://openscad.org/" rel="nofollow noreferrer">OpenSCAD.</a></p>
<p>The colors of the components and placement are consistent with the program, as well as the background color. There is usually a set of axes visible with reference measurements, but those are easily disabled, allowing for a screen shot of this nature.</p>
<p>From the gallery option in the linked site:</p>
<p><a href="https://i.stack.imgur.com/JVTvj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JVTvj.jpg" alt="openscad sample image" /></a></p>
| 56677 | What software was this designed in? |
2023-10-19T14:36:33.050 | <p>In the AISC's <em>Steel Construction Manual</em> Table 8-4 "Coefficients C for Eccentrically Loaded Weld Groups", it mentions the Special Case where the load is not in plane with the weld group. <a href="https://i.stack.imgur.com/KUvvD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KUvvD.png" alt="Special Case" /></a></p>
<p>However, this Special Case is not mentioned again for the remainder of the weld groups in the rest of the chapter's tables on other weld groups. For example:
<a href="https://i.stack.imgur.com/tIMTL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tIMTL.png" alt="No Special Case" /></a></p>
<p>Why wouldn't the out of plane case be applicable to the other weld groups?. I hesitate to apply the case to other groups when the manual doesn't explicitly state it's valid. The manual as a whole is rather explicit, so I believe that if the special case is not mentioned for the other groups, it's not valid to use it for them.</p>
| |structures|welds|aisc| | <p>An AISC consultant engineer responded to this in an email. The answer is no, the special case only applies to Table 8-4 geometry. My interpretation of the reasoning is as follows.</p>
<p>These tables are for loads in the same plane as the weld group. The geometry of two closely spaced parallel welds is approximately in plane with the load regardless of the orientation. This is something of an exception (hence the "special case") of the geometry and I believe that if the perpendicular distance between the welds in the tables increased much more than shown (i.e. k > 2) then this special case would also be invalid for this weld group geometry.</p>
| 56694 | Is the AISC's Steel Construction Manual's out of plane "special case" valid for all weld groups? |
2023-10-21T06:54:30.453 | <p>I have noticed that the exhaust flames for rockets can be either everywhere, like in <a href="https://www.livemint.com/lm-img/img/2023/07/16/1600x900/Chandrayaan-3-was-launched-from-the-Satish-Dhawan-_1689468375924_1689497193513.jpg" rel="nofollow noreferrer">this image</a> or it can be hardly noticeable, like in <a href="https://twitter.com/AmitShah/status/1715599841316004074?ref_src=twsrc%5Etfw%7Ctwcamp%5Etweetembed%7Ctwterm%5E1715599841316004074%7Ctwgr%5Edef270d92b222e7c9dbee008a8580583d88c441f%7Ctwcon%5Es1_&ref_url=https%3A%2F%2Fwww.hindustantimes.com%2Findia-news%2Fgaganyaan-mission-live-updates-isro-space-missions-news-today-21-october-2023-101697846634364.html" rel="nofollow noreferrer">this video</a>. I am just curious, what makes this difference? Is one better than the other?</p>
| |rocketry|jet| | <p>The side boosters in the first image are solid fuel rockets, which generate a lot of particles (soot and unburned fuel) in their exhaust plumes. These solid particles glow orange and white hot, creating a "fiery" plume. Kerosene liquid fuel engines like the Saturn V first stage or Falcon 9 first stage also generate sooty flames.</p>
<p>In contrast, the center engines in your first image, and main engine in your second image, are hydrogen or clean burning methane fueled. These generate much less soot, so what you see is the color of the gaseous combustion combustion itself (usually pink for hydrogen and blue for methane).</p>
<p>Regarding the size and shape of the exhaust plume, this depends on the exit pressure. Some engines (usually first stage like the outer boosters in your first image) are optimized for near-sea level thrust. So they try to make the nozzle exit pressure match or slightly exceed sea level pressure, making the plume roughly cylindrical or spread out slightly.</p>
<p>In contrast, engines that are optimized for higher altitude flight like the Space Shuttle main engines, Artemis core stage engines, and seemingly the center engines in your first example, have a lower exit pressure. This causes the plume to contract inward at sea level from the higher atmospheric pressure around it. If you observe high altitude flight photos of the Saturn V, or any upper stage in vacuum, the plume looks different and very spread out.</p>
| 56708 | Why are the rocket exhausts so different from rocket to rocket? |
2023-10-24T14:19:27.090 | <p>I am currently in a discussion with a so called '9/11 truther' who claims there is no adequate explanation for the collapse of the central core columns of the WTC towers. (Sorry to beat a dead horse)</p>
<p>Based on my viewing of a documentary, I argued that the composite floors added a 'stiffening' effect (i presume engineers call this 'diaphragmic action'?), and that because the floors collapsed first, this compromised the structural integrity of the columns and caused the main columns to collapse quickly after the floors.
This documentary is can be seen here and the explanation is at 45:10.
<a href="https://www.youtube.com/watch?v=DgwlFY-4Txc" rel="nofollow noreferrer">https://www.youtube.com/watch?v=DgwlFY-4Txc</a></p>
<p>I'm however not 100% whether this explanation is satisfying enough. It seems logical enough for the perimeter columns, but is it also true of the core? Is the idea of of your building core being stabilised by diaphragmic action of the floors something that is common or should a framed tube building core be designed completely independant of any stabilising effect this action might provide?</p>
<p>My discussion partner claims that it is absurd for a building core to be dependant on the floors like this. It might be a tall order, but I wonder if someone experienced in this subject can provide some clarity.</p>
<p>Just for context, I am not an engineer, but a draughtsman in the construction industry.</p>
| |structural-engineering|building-design| | <p>Yes, it was. The structure was designed on the basis of the pipe-in-pipe principle. Concrete core housing elevators and stairs get their lateral strength and stability from the perimeter columns which act together like a pipe with a huge moment of inertia.
For a hollow square tubing with the wall thickness, T, and base of <span class="math-container">$B\cdot B$</span>, the moment of inertia which is a measure of lateral stiffness is</p>
<p><span class="math-container">$$I\propto T\frac {B\cdot B^3}{12}$$</span></p>
<p>So the outer pipe is more effective in resisting lateral force by a power of four.
<span class="math-container">$$\frac{I_{outer}}{I_{core}}=(\frac{B_{outer}}{B_{inner}})^4$$</span>
The slabs played a dual task of supporting the floor load and shear transfer beams to connect the outer pipe to the inner pipe.
The softening of the slab trusses and yielding due to the heat was one of the major components of the collapse.</p>
| 56731 | Was the stability of the WTC1&2 core columns dependant on the diaphragmic action of the building's floors? |
2023-10-24T18:08:23.943 | <p>Inspired by <a href="https://electronics.stackexchange.com/questions/685779/electromagnetic-pendulum-pulse-driver-having-trouble-getting-circuit-running">this question</a>, I am looking for a low-friction pivot for a clock pendulum. The design uses magnetic repulsion to keep the pendulum going. A magnet is attached to the bottom of the pendulum and a coil is fixed just below it. My pivot now is a nail and a hole in a piece of wood, almost anything will be better.</p>
<p>I don't need help with the electronics or magnetics, I need to reduce the friction so it will last much longer with batteries.</p>
<p>The pivot needs to constrain the motion to a single plane so I can continue to use magnetic repulsion. The voltage generated by the coil as the magnet on the pendulum passes by triggers the coil repulsion circuit.</p>
<p>I have a toy that uses a rolling pivot, but this requires two points of contact to keep it centered. I could possibly put the two points much closer together to meet my needs.</p>
<p>I have seen designs that use two threads, but I really want something that looks like a clock pendulum.</p>
<p>Note that the pendulum is just for show, it won't be used as the timebase. The pendulum will weigh an ounce or two, although I have seen cool clocks where most of the clock was on the pendulum, this would weigh more.</p>
<p>What other pivot types are available? Nothing too exotic, something I can buy cheaply or build myself.</p>
<p><a href="https://i.stack.imgur.com/hSXr6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hSXr6.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/TV6y2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TV6y2.jpg" alt="enter image description here" /></a></p>
<p>[my pictures]</p>
| |friction| | <p>I did some research and found that the most accurate pendulum clock ever made used a knife edge on agate. We can probably assume that low friction is necessary for accuracy. I'm not sure that I have the skills and tools to build something like this. I am decent at woodworking, not so much with metal. It would also be rather delicate and easily upset.</p>
<p>Of course, ball-bearings are often used, but these are usually optimized for heavy loads, and they are often lubricated with oil. Then I thought of something designed for light loads, fidget spinner bearings! A bought a high-quality bearing and modified my pendulum design to use it.</p>
<p>Unfortunately, I didn't get as much improvement as I expected. It seems to have about 3X less friction than a nail on wood. Maybe they are optimized for high speeds.</p>
<p>I improved the electromagnetic portion of the design to get about another 3X improvement. I am now in the ballpark of my goal (run for one year on 2 AA batteries).</p>
| 56733 | Pivot Point with Low Friction |
2023-10-25T14:40:46.317 | <p>This is a question I've had on my mind for a while. I was recently looking at machine in operation: it was a machine which had a bunch of vertical spindles holding yarn bobbins, rotating at a high speed, side-by-side. If someone wanted to analyse the dynamics of one of these spindles, they would need some sort of model for the torque due to drag acting on it.</p>
<p>Maybe, data on a large number of drag coefficients is available, and this is what's used, I thought initially. But then looking more closely, surely the streams generated would be merging into another. These bobbins aren't independant of each other in that sense. Quick googling revealed this is called interference drag.</p>
<p>This made me curious about how a mechanical engineer that's designing a machine, who isn't necessarily an aerodynamics expert, goes about analysing the dynamics of situations like this. In many situations, parts of a machine must be moving in complicated ways/trajectories, and at high speeds. Applying the drag equation here seems inappropriate since it appears that <span class="math-container">$C_D$</span> is experimentally determined for an object moving in a straight path. Also, like above, the streams from the different moving parts would be merging into each other</p>
<p>So how then, in practice, do mechanical engineers do such analyses when they're designing machines? Any part of the machine is moving at high enough speeds that you would need to account for drag.</p>
| |mechanical-engineering|machine-design| | <p>In many cases, probably a majority, they ignore air drag, and simply design the torque capability of the engine as enough to deliver the required output, plus some sizable margin to overcome sources of friction.</p>
<p>That is a general engineering principle. If you have a system that is difficult or impossible to model, you model it as a simple system and add margin (or safety factors) to the capability. Then all you have to do is show by estimation that the unknown factors are too small to compromise the result.</p>
<p>For example if I'm designing a table, I may not be able to model the stress distribution of every object that could be put on the table, nor the complex transient loading that occurs when items are dropped on it and the table vibrates. But I do know that the force necessary to arrest the fall will not be more than a completely elastic collision from height <span class="math-container">$H$</span> per the following formula:</p>
<p><span class="math-container">$$F= 2 \frac Mt \sqrt{2gH}$$</span></p>
<p>(see <a href="https://physics.stackexchange.com/a/313909/313823">here</a>)</p>
<p>Let's say this value is twice the weight of the item.</p>
<p>So if I put a weight limit on the table of 400 lb, I would design the table structurally to withstand at least 800 lb. The goal is to calculate the worst case loading scenario that will envelop all the unknown conditions, and design for that condition, plus margin.</p>
<p>For your specific question about air drag, that is going to be insignificant for the motions of mechanisms. It is more relevant for vehicles in motion.</p>
| 56748 | How do mechanical engineers, in practice, account for drag/air resistance when designing machines? |
2023-10-25T18:41:02.230 | <p>I am designing a device which clicks when rotated so as to wake up an accelerometer, allowing the device to run on low power and wake to take a reading.</p>
<p>I have access to SLA and FDM 3D printers. When I design a disc with a notch in the sla printer the material is degraded over use and the click diminishes. The fdm printer suffers from accuracy for the dimensions used.</p>
<p>The outer disc is 3cm in diameter in which there is another disc that is 2.5cm in diameter. The requirement is to use the space in the middle to use any materials to create a satisfying click.</p>
<p>I have used</p>
<ol>
<li>notches which suffer from degradation</li>
<li>A spring mounted in the outer and inner disc which pushes out a small piece of plastic but this keeps getting caught.</li>
<li>Mounted a small push button switch in the inner ring but due to the square shape of the button part and rolling motion I get no click.</li>
<li>Printed leaf spring but this does not have the potential energy to create the click.</li>
</ol>
<p>Any ideas would be a great help?</p>
| |mechanical-engineering|3d-printing| | <p>I assume you need quite a click to activate an accelerometer and 3D printers are not known for making very wear resistant components.</p>
<p>Making it simply an electrical contact wake-up might be in order then with a gentler spring for tactile feel. You can use a hammer punch to punch out copper or aluminum foil flat rings to embed inside channels in your 3D printed part to send the requires electrical currents around.</p>
<p>The other alternative is to put in bearing surfaces into your 3D printed parts so they abrade against each other and not the 3D printed plastic. For example, a bunch of sockets for ball bearings, with perhaps one spring loaded ball bearing opposite to click against.</p>
| 56755 | How to engineer a click from small concentric discs |
2023-10-26T19:43:32.863 | <p>I am trying to find a proof as to why we can read the velocity error constant <span class="math-container">$K_v$</span> of a closed-loop directly from the Bode diagram. That is, why the following is true: <span class="math-container">$K_v = \frac{|L(j\omega)|}{\omega}$</span> and therefore <span class="math-container">$K_v = |L(j\omega)|_{\omega=1}$</span>. Here <span class="math-container">$K_v$</span> is the velocity error constant of the closed-loop transfer function of a system with loop transfer function <span class="math-container">$L(s)$</span>.</p>
<p>For the position error constant it is relatively obvious that we can read it from the low-frequency asymptote since by the very definition of <span class="math-container">$K_p$</span> we take the limit as the frequency approaches zero: <span class="math-container">$K_p=\lim_{s\rightarrow 0} L(s)= \lim_{\omega\rightarrow 0} L(j\omega)$</span>. However, the same reasoning is not readily applicable to <span class="math-container">$K_v$</span>.</p>
| |control-theory| | <p>The static velocity error constant is obtained from the initial -20db/decade segment of the Bode plot or the extension of that segment.</p>
<p>It can be obtained as either the intersection of the segment or its extension with the <span class="math-container">$\omega=1$</span> vertical line or the <span class="math-container">$0$</span> dB horizontal line.</p>
<p>Here are a couple of examples.</p>
<p><a href="https://i.stack.imgur.com/h4eqr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h4eqr.png" alt="enter image description here" /></a></p>
<p>For the proof that it can be obtained as the intersection of the segment and the <span class="math-container">$\omega=1$</span> line, let's start with the definition <span class="math-container">$k_v=\lim_{s \to 0}\ s \ L(s)$</span> and let <span class="math-container">$\omega_\epsilon$</span> be a very small value of <span class="math-container">$\omega$</span> which is much smaller than 1 and from which we can compute <span class="math-container">$k_v$</span>.</p>
<p>That is, <span class="math-container">$k_v= |j\ \omega_\epsilon L(j \ \omega_\epsilon)|$</span></p>
<p>Consider the -20db/decade segment. It starts say from <span class="math-container">$\omega_\epsilon$</span> where the magnitude is <span class="math-container">$|L(j \ \omega_\epsilon)|$</span>. When that segment reaches <span class="math-container">$\omega=1$</span>, the frequency has increased by a factor of <span class="math-container">$\frac{1}{\omega_\epsilon}$</span>. So the magnitude should have decreased to</p>
<p><span class="math-container">$$20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10} \frac{1}{\omega_\epsilon}$$</span></p>
<p><span class="math-container">$$=20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10}| \frac{1}{j \ \omega_\epsilon}|$$</span></p>
<p><span class="math-container">$$=20 \log_{10} |{j \ \omega_\epsilon} L(j \ \omega_\epsilon)|$$</span></p>
<p><span class="math-container">$$= 20 \log_{10}k_v$$</span></p>
<p>The other proof, that it can be obtained from the intersection of the segment and the <span class="math-container">$0$</span> dB line is similar. Let's assume that the frequency when it intersects the <span class="math-container">$0$</span> dB line is <span class="math-container">$\bar{\omega }$</span>.</p>
<p><span class="math-container">$$20 \log_{10} |L(j \ \omega_\epsilon)| - 20 \log_{10} \frac{\bar{\omega }}{\omega_\epsilon}=0$$</span></p>
<p><span class="math-container">$$ \log_{10} \frac{|L(j \ \omega_\epsilon)|\omega_\epsilon}{\bar{\omega }}=0$$</span></p>
<p><span class="math-container">$$\bar{\omega } = |L(j \ \omega_\epsilon)|\omega_\epsilon=|L(j \ \omega_\epsilon)||j \ \omega_\epsilon|= |j \ \omega_\epsilon L(j \ \omega_\epsilon)|=k_v$$</span></p>
| 56769 | Why is it true that the velocity error constant can be read from the Bode diagram at the frequency w=1 rad/s? |
2023-10-29T11:52:18.653 | <p>I know structural steel is considered to have an elastic behavior and after a certain deformation, it has a plastic behavior.</p>
<p>How about gold metal?</p>
| |mechanical-engineering|structural-engineering|solid-mechanics|elasticity| | <p>Gold has a Young's modulus of 79 GPa which is very similar to silver, but significantly lower than iron or steel.</p>
<p><a href="https://www.totalmateria.com/page.aspx?ID=CheckArticle&site=ktn&LN=NO&NM=230#:%7E:text=The%20Young%27s%20modulus%20of%20elasticity,lower%20than%20iron%20or%20steel" rel="nofollow noreferrer">https://www.totalmateria.com/page.aspx?ID=CheckArticle&site=ktn&LN=NO&NM=230#:~:text=The%20Young%27s%20modulus%20of%20elasticity,lower%20than%20iron%20or%20steel</a>.</p>
| 56797 | Does gold have elastic behavior? |
2023-10-31T07:40:23.273 | <p>Recently, it occured to me that there is a lot of "unused" space on flat-top buildings, and so I wondered, whether it wouldn't be a reasonable ideas to put (vertical) wind turbines on buildings, especially high-rise ones, where one could expect a lot of wind.</p>
<p>However, I couldn't find any example of such a project. Perhaps I just didn't search thoroughly enough, or perhaps there are good reasons why this isn't a good idea after all. Can someone enlighten me as to the practicality of my idea?</p>
| |aerodynamics|wind-power| | <p>It's more likely that solar panels would be installed on the roofs of buildings, including tall buildings. Solar panels have no moving parts and if there is failure associated with the panels the worst that will happen is the inverter catches fire.</p>
<p>Wind turbines on the other hand, rely on moving parts to generate electricity. In the unlikely event that bits fall off a wind turbine, such as blade or two, or the wind turbine was not properly secured to the building, the bits could fall to the ground kill people and cause damage to other property.</p>
<p><em><strong>Edit</strong></em>
Additionally, for wind turbines to be placed on top of buildings, the buildings would need to be constructed to successfully deal with the vibrational loads the wind turbine would place on the building.</p>
| 56832 | Wind turbines on high-rise buildings? |
2023-10-31T13:53:40.587 | <p>In robotics, I've seen quite a few people make use of cycloidal drives. They seem robust, compact and have a relatively simple design, while still achieving high reduction ratios.</p>
<p>Yet outside robotics, they don't seem to be that popular- often times, other high reduction gearing is used e.g. worm drives.</p>
<p>I was wondering why this is and what advantages/disadvantages cycloidal drives have over their worm gear counterparts.</p>
<p>Maybe it's ease of manufacturability, cost or something else I'm missing.</p>
| |motors|gears|torque| | <p>Your question is so open ended that its hard to answer. But lets try. The reason is almost certainly cost, manufacturing capability and legacy reasons.</p>
<p>A worm gear is very easy to make in volume. This may sound a bit weird, since if all you have is a cnc mill then it seem that a cycloid is a no brainer. A worm gear can be made by hobbing or power sikiving. While they require a specialized tooling once you have the tool its extremely simple to make these parts. So for a worm gear you need 4 bearing journals and two features with special tooling. This is relatively simple to do with even non CNC equipment in a dedicated setup.</p>
<p>The worm drive is also a very old invention. This means there are old dedicated factory setups that can churn out worm drives very cheaply. The result of this is that all kinds of weird things have been done. I have seen ultra cheap worm drives that have been stamped out of sheet stock. While technically possible with cycloid drives too, I have not encountered one.</p>
<p>Then there is the thing that worm drives change the direction of the axis of the output. This is a benefit in some applications. Likewise many machinery dont need to rotate both ways so much of the must have features in robotics like backlash might be of no concern to the application at hand.</p>
<p>And finally there is also some cultural inertia, if you have a design that works why adapt it to a new form factor. Consider that the tuners of guitars and ukuleles are worm gears! Could you manufacture as small cycloidal gears, for less than 10 cents a piece?</p>
<p>Beverything except robotics is surely a big area so there can be many other reasons too.</p>
| 56837 | Worm drive vs Cycloidal Drive |
2023-11-02T05:01:54.527 | <p><a href="https://i.stack.imgur.com/Nz758.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nz758.png" alt="enter image description here" /></a></p>
<p>I have to write 0xA5 to the DAC. So should I send 0x00100A50 or 0x001A5000?
The command bits are 0x0 as i just want to write to the buffers. Say I want to send to channel 1 with channel Address 1 (D23 to D20 & the Don't Care bits are taken as 0. I am having this confusion because I am not familiar with left aligned binary data. If 0x001A5000 is the correct one, How can I write 0xA500 to the DAC?</p>
| |electronics|digital-communication|adc| | <p>What the left aligned data really means is that for the 70004, you have 14 bits Data followed by 2 bits of what is likely "Don't care" and for 60004, you have 12 bits Data followed 4 likely "Don't care" bits (adjacent to the mode bits). This "left aligned" mention allowed TI to fit all the info for three different devices into a single table.</p>
<p>Take care to also get the rest of the bit and byte order correct, as well as order of bits on the line.</p>
<p>Easiest way to read TI doc here is as a 32-bit entity. We'll suppose this uint32 serves that purpose:</p>
<pre><code>uint32_t spicmd;
void SetRW(uint8_t RW){
//Treating don't cares as part of RW since it's easier to read in hex
spicmd &= 0xFFFFFFF;//remove old bits
spicmd |= ((uint32_t)RW) << 28;//set new bits; extra bits go over msb so no need to remove
}
void SetCommand(uint8_t RW){
//Treating don't cares as part of RW since it's easier to read in hex
spicmd &= 0xFFFFFFF;//remove old bits
spicmd |= ((uint32_t)RW) << 28;//set new bits; extra bits go over msb so no need to remove
}
//... The part you are interested in: depends on which dac you have
void SetDatDAC80004(uint16_t DAT){
//80004 is 16-bit so nothing so no shift
spicmd &= 0xFFF0000F;//remove old bits
spicmd |= ((uint32_t)DAT) << 4;//set new bits
}
void SetDatDAC70004(uint16_t DAT){
//70004 is 14/16 bits so shift 2; rest is same as 80004
//alternatively, you could just have treated the lsb of your DAC as 4 times what you would for the 80004
SetDatDAC80004(DAT << 2);
}
void SetDatDAC60004(uint16_t DAT){
//60004 is 12/16 bits so shift 4; rest is same as 80004
//alternatively, you could just have treated the lsb of your DAC as 16 times what you would for the 80004
SetDatDAC80004(DAT << 4);
}
//SPI is not inherently 32 bits so make sure your 32 bits goes out in the correct order
//Supposing you have the ability to send a spi byte
extern void QueueSPIByte(uint8_t byte);
void QueueSPICmd(){
QueueSPIByte(spicmd >> 24);
QueueSPIByte(spicmd >> 16);
QueueSPIByte(spicmd >> 8);
QueueSPIByte(spicmd);
}
</code></pre>
| 56869 | Left Aligned Binary Data |
2023-11-02T16:21:44.537 | <p>Should I worry about galvanic corrosion if using an aluminium shaft inside a stainless steel bearing in something that's going to be washed down frequently?</p>
<p>Considering I will press the shaft into the bearing, is there any way to guard against this? If pressed together in a dry environment, and then I add a bead of silicone sealant at the mating surface (see diagram), will this protect it? <a href="https://i.stack.imgur.com/jyG6W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jyG6W.png" alt="enter image description here" /></a></p>
<p>Thanks</p>
| |steel|aluminum|corrosion| | <p>My bigger concern would be the bearing surviving the frequent washdown lol.</p>
<p>A bead of silicone like you mentioned will help with galvanic corrosion, but as others have mentioned will not fully eliminate the issue. The silicone would help by increasing the distance between the two conductive surfaces and reducing the duration and frequency a conductive water solution would complete the circuit touching both the aluminum and stainless steel at the same time. Galvanic corrosion only takes place if there is a completed circuit involving an electrolyte.</p>
<p>One method that may work in this case would be to send off the shaft to have it professionally anodized. This creates an aluminum oxide layer that is not electrically conductive. Then provided this coating is not scratched away when pressed onto the bearing (might need some heat on the bearing or cold on the shaft to make that happen), it should electrically insulate it from the bearing while providing the same mechanical properties. You would then want to add the silicone like you mentioned because there are still pin holes in this coating and is only truly non conductive when dry in that protected area under the bearing.</p>
<p>Other mitigation options would be paint, oil, shields to deflect direct spray, washing with water that has less TDS like reverse osmosis water, or you treat the aluminum as sacrificial and find a way to easily replace it.</p>
| 56872 | Risk of corrosion putting aluminium shaft in a stainless steel bearing with frequent washdowns? |
2023-11-04T00:06:54.703 | <p>Is it possible to generate 10 GHz frequency RFs at 400 V amplitudes while only using 10-100 W order power? I tried looking up how 10s of GHz frequencies are generated online and found next to nothing except communication products that had no spec sheets available.</p>
<p>The main requirements I have are that</p>
<ul>
<li>I want to generate 10 GHz,</li>
<li>have the waves transmit over a distance of 10 cm,</li>
<li>while having at least 200 V amplitude at the receiver point</li>
<li>and only use about 10-100 Watts but ideally fractions of Watts.</li>
</ul>
<p>Is such an experimental/instrument set up even physically possible? I know that the medium the waves will transmit through, also affect the decay envelop of the wave function and therefore, the decay of the wave's power, so let's just assume that these waves will propagate through an atmosphere in the 100s of Pascals.</p>
<p>Lastly, the emitter's material is non-specific (Because I do not know what they should be for such high frequencies), but the geometry of the emitter should be a ring. If someone can give me a rough order magnitude of power requirements for these waves and/or the resources so I can start reading into RF/communication technologies and do these calculations on my own with the formulas used in that field I would really appreciate it.</p>
| |telecommunication|waves| | <p><strong>UPDATE</strong> From clarifying the specs in comments: the poster wants a region with a high electric field generated by as low an input power as possible.</p>
<p>You want standing waves, not propagating waves; therefore you want a resonator, not an antenna. Either make the device you have into a resonator, put a resonator in it, or put it as a whole into a resonator.</p>
<p>Since there seem to be a bunch of specific constraints on geometry which are not described in the question, I can only offer very general guidance:</p>
<p>(a) build a resonant structure</p>
<p>(b) figure out all of your geometry requirements, in the following format: volume where you need your field; volumes of space where you can't have anything; where you could have metal, and where you have to have metal</p>
<p>Now the problem becomes simple: you need to make a resonant cavity which fits that geometry. For a rectangular metal box, basically two walls have to be spaced some multiple of wavelength/2 apart. But it also doesn't have to be rectangular, or metal, or a closed structure - many more ideas here: <a href="https://en.wikipedia.org/wiki/Microwave_cavity" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Microwave_cavity</a></p>
<p>Easy geometries are box, parallel plates (which may be shorted together at one edge or two opposite edges), and cylinder. If possible, it may be easier/more efficient to use whatever the device you're working with is made of as part of the resonator (especially if it's already metal), but if not, just put the whole resonator inside it.</p>
<p>A section of transmission line or waveguide with shorted or open ends is also a resonator (and may also have open sides, eg in <a href="https://en.wikipedia.org/wiki/Air_stripline" rel="nofollow noreferrer">air stripline</a>). A box can be thought of as a waveguide section with shorted ends. Open resonators will leak through the open sides - two facing parabolic reflectors is probably the least-leaky open resonator possible (like parallel plates but curved).</p>
<p>Resonators can be built out of dielectrics: <a href="https://en.wikipedia.org/wiki/Dielectric_resonator" rel="nofollow noreferrer">dielectric resonator</a>, these tend to be much more compact but more lossy than ones built of a conductive material.</p>
<p>Anything you put in your resonant structure (eg the probe) will detune it somewhat, and perturb the field (could be a lot, if it's metal) so you probably need to design something that works both with and without it.</p>
<p>The "ring emitter" described in the question would only work well to the extent you can build a resonant structure that incorporates it (eg: the two ends of a cylinder)</p>
<p><strong>Original answer</strong> For any kind of electromagnetic radiation (plane waves) in vacuum, the formula connecting the peak electric field strength <span class="math-container">$E_0$</span> and the average energy flux per unit area (intensity) <span class="math-container">$I$</span> is:</p>
<p><span class="math-container">$E_0 = \sqrt{ \frac{2 I}{c \epsilon_0} }$</span></p>
<p>where <span class="math-container">$c$</span> is the speed of light and <span class="math-container">$\epsilon_0$</span> is the dielectric permeability of vacuum (<a href="https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/16%3A_Electromagnetic_Waves/16.04%3A_Energy_Carried_by_Electromagnetic_Waves#mjx-eqn-16.31" rel="nofollow noreferrer">source</a>)</p>
<p>There are several missing pieces in the specs you describe.</p>
<p>To sum up the specs: frequency 10GHz, input power 10-100W, amplitude 200-400V, distance 10cm</p>
<p>The missing parts:</p>
<ul>
<li>the area (cross-section) of the beam</li>
<li>the size of the receiver (ie antenna)</li>
<li>what is in between the transmitter and receiver?</li>
<li>are you trying to create a electrical field, or are you trying to transmit power?</li>
</ul>
<p>The cross sectional area is unspecified in your case, but let's assume 10x10cm = 100 cm^2.</p>
<p><span class="math-container">$I = 100 \ \mathrm{W} \ / \ 100 \ \mathrm{cm}^2 = 1 \ \mathrm{W}/\mathrm{cm}^2$</span></p>
<p><span class="math-container">$E_0 = 2745 \ \mathrm{V}/\mathrm{m} = 274.5 \ \mathrm{V}\ / \ 10 \ \mathrm{cm}$</span></p>
<p>You can see why the area and the size of the receiver matter: (a) the total power is not important, just the power transmitted per unit area, and (b) the electric field is volts/meter, not just "volts" (so a larger receiver would pick up a larger potential difference across it). However, as long as the receiver is the same size as the width of the beam, the resulting voltage across the receiver ends up being the same - for a narrower beam with the same power the intensity is larger, <span class="math-container">$E_0$</span> in V/m is larger, but it is multiplied by the smaller size of the receiver.</p>
<p>This makes all kinds of assumptions, eg the transmission happens through free space in vacuum, is roughly a plane wave of uniform intensity with a certain beam cross-section (*of course it can't be exactly a plane wave if it has a finite cross-section), and the receiver is perfectly absorbing and the same size as the beam. The receiver in this case would normally be called an "antenna" - there is a huge variety of designs that work at 10GHz, from patch antennas to parabolic to dielectric lens.</p>
<p>In practice, unless you really need the transmission to be through free space, you would probably use a <a href="https://en.wikipedia.org/wiki/Waveguide" rel="nofollow noreferrer">waveguide</a>, and the other end of the waveguide may be reflecting or absorbing or somewhere in between. Looking up some waveguide equations would help you calculate the electric field in a waveguide.</p>
<p>A lot of both waveguides and coax cables (and the things that they connect to) are impedance matched to 50 ohms. The (rms) voltage expression is very simple in that case, sqrt(100W * 50 ohm) = 70.7 V.</p>
<p>If you are just trying to create a large field, you want standing waves, ie waves going back and forth inside a waveguide. This is also similar to what happens in a microwave oven. You can think of this as the equivalent of shining a flashlight through a small hole into a tin can or between parallel mirrors; the light inside would bounce off the walls multiple times, and be much brighter than if the walls were black.</p>
<p>It would be easier to answer the question if you describe what you're actually trying to do: transmit information? transmit energy? create a field? something else?</p>
<p>Lastly, a safety note: 100W at 10GHz is enough to cause serious injury; it can focus down to a small spot (including by accident) which would heat up <em>very</em> fast. The way to think about it is somewhere in between radio and a laser beam, with some of the same safety issues as a 100W laser. You don't sound like you're familiar with this, so probably the first piece of advice is to find someone who knows how to work with high power microwaves to help you, and make sure anything you do is safe. Read about maximum permissible exposure limits, get a power meter, shield your setup, test at low powers first, and generally treat it with a healthy amount of caution.</p>
<p>See also:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Standing_wave_ratio" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Standing_wave_ratio</a></li>
<li><a href="https://en.wikipedia.org/wiki/Wave_impedance" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Wave_impedance</a></li>
<li><a href="https://en.wikipedia.org/wiki/Impedance_of_free_space" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Impedance_of_free_space</a></li>
</ul>
| 56885 | Generating 10 GHz Frequencies with Reasonable Power Requirements |
2023-11-04T08:33:53.200 | <p><a href="https://en.wikipedia.org/wiki/Scroll_compressor" rel="nofollow noreferrer">Scroll compressors</a> are efficient, reliable, and a fill a pressure/flow niche in the array of available compressor technologies. However, I have never seen one used the other direction for generating power from a compressed air source, steam, or power cycle. Any thoughts on why this might be the case? Any examples of one being used as a prime mover?</p>
| |mechanical-engineering|turbines|compressors| | <p>Looks like "scroll expander" was the term I was looking for.</p>
<p>Still looks to be a pretty rare application, but here are some instances:</p>
<ol>
<li><a href="https://airsquared.com/products/scroll-expanders/e15h022a-sh/" rel="nofollow noreferrer">Air Squared commercially available scroll expander</a></li>
<li>Here is a good <a href="https://atami.oregonstate.edu/biblio/development-small-scale-scroll-expander" rel="nofollow noreferrer">thesis from Oregon State University</a> (<a href="https://ir.library.oregonstate.edu/downloads/v405sd97c?locale=en" rel="nofollow noreferrer">direct pdf link</a>).</li>
<li>Some short <a href="https://www.sciencedirect.com/topics/engineering/scroll-expander" rel="nofollow noreferrer">Science Direct Excerpts</a></li>
</ol>
| 56888 | Can a scroll compressor be used to generate power? |
2023-11-05T00:49:51.557 | <p>Question about a home my wife and are looking at purchasing. The garage had an office/apartment on the second floor, and the seller disclosed that the floor had sagged and been fixed with shoring beams (I may have the wrong terminology).</p>
<p>We snapped some pictures and it looks like they are permanent rated beams, but I don’t know how to tell if they’re properly installed. Any advice would be greatly appreciated. <a href="https://i.stack.imgur.com/KZylt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KZylt.jpg" alt="image of a few of the beams" /></a></p>
| |structural-engineering|foundations|home-improvement| | <p>No, they are not okay. Even as a short-term solution for a couple of days. If the building department knows about this it will red-tag the garage (prohibit entering it) until they pull a permit and do it right.</p>
<p>Even their insurance company would suspend the policy if they knew of illegal construction.</p>
<p>Beams have to be calculated. posts should be calculated and rest on a proper foundation with a proper baseplate. The fasteners should be code-approved. As it is if somebody inadvertently bumps gently into the columns with a car trying to park, the building will collapse on their head.</p>
<p>This is the textbook example of what a dangerous job, done by unprofessionals, is.</p>
| 56895 | Shoring beams properly installed? |
2023-11-08T21:49:38.830 | <p>I posted this question in physics, but they advise me to post it here so here I am.</p>
<p>I was looking for the formula to calculate the burden in bench blasting and I found this <a href="//www.oricaminingservices.com/uploads/Blastec/Background%20to%20surface%20part%20calculations.pdf" rel="nofollow noreferrer">document</a>. The problem is I have to calculate the concentration of the charge first with the unit kg/m. I know that some materials are given but I need the formula to calculate other materials in which I have their energetic values calories/g calories/cm<sup>3</sup> ..</p>
| |energy|mining-engineering| | <p>The charge concentration, in kilograms per meter of drill hole, serves two purposes. The first is being able to determine the amount of explosives needed to blast a given amount of rock. This is needed when ordering explosives from the supplier or when taking out enough explosives from the magazine.</p>
<p><span class="math-container">$M_{Explosives} \ = \ I_b \cdot N \cdot (L_H \ - \ L_S)$</span></p>
<p>Where:</p>
<ul>
<li><span class="math-container">$M_{Explosives}$</span> is the mass of explosives required [kg]</li>
<li><span class="math-container">$I_b$</span> is the charge concentration [kg/m]</li>
<li><span class="math-container">$L_H$</span> is the length of the drill holes [m]</li>
<li><span class="math-container">$L_S$</span> is the length of the stemming [m]</li>
<li><span class="math-container">$N$</span> is the number of drill holes</li>
</ul>
<p>The other purpose of the charge length is, if one knows the energy equivalent of each type of explosive [J/kg], for its given density of explosive, it becomes possible to calculate the amount of energy expended in the blast by the explosives used.</p>
| 56939 | Blasting - how to calculate the charge concentration Ib (kg/m)? |
2023-11-09T16:26:05.690 | <p>What happens in this scenario - a modern electrical motor, three phase, squirrel cage, that is not specifically designed for operation via frequency inverter, <em>is</em> in fact operated on an FI? How long will the motor last, compared to 50Hz only operation? How will it fail (if at all)?</p>
| |electrical-engineering|motors| | <p>In my experience there are three main failure modes for motors on a VFD. These modes of failure are more likely when the motor is not rated "Inverter Duty". However note that when used properly on a high quality VFD, a general purpose motor will usually last LONGER on a VFD than without due to smoother acceleration, current protection, voltage protection, torque limiting options, and mechanical harmonic frequency avoidance.</p>
<ol>
<li><p>Insufficient cooling. Heat is the primary killer of motors in my experience. Generally an induction motor is cooled by a small impeller mounted to the shaft. At 50Hz(full speed Europe) or 60Hz(full speed US) this provides sufficient cooling for the windings and bearings. When you continuously operate on a VFD at low RPM there is not sufficient cooling. Even inverter-duty motors with industrial grade safety factors should not be operated below 20Hz without supplemental fan cooling. For a non rated motor it would be better to not drop below 30Hz without supplemental cooling. Even with supplemental cooling you would never want to use a general purpose motor for very slow control with vector or sensor-less vector control; it would most certainly burn up.</p>
</li>
<li><p>Electrical noise and voltage spikes. Engineers minimize electric motor winding insulation thickness to get more power in a smaller lower cost package. Since general duty electric motors have less insulation, the windings can arc and fail from voltages spikes due to the fast IGBT switching in the VFD. Inverter-grade motors have thicker insulation resulting in a larger form factor, higher cost, but better protection (but not invincible) to voltage spikes. The best way to reduce this is to buy a high quality VFD (I recommend Yaskawa). Another way to help reduce these spikes is to add a line filter and/or ferrite beads on the lines.</p>
</li>
<li><p>Eddy currents. This failure mode is less common but on lower quality VFDs (Allen Bradley lol), the electrical noise that is generated from the IGBTs switching on and off at high frequency can cause a charge to build up on the rotor of the motor that then discharges through the bearings quickly destroying them and causing them to burn up even on industrial grade motors. One solution is to have new ceramic bearings pressed into the motor for $$$. The best solution is just to by a good VFD in the first place (Yaskawa). Another possible option that we did not explore is adding a grounding graphite brush to the output shaft.</p>
</li>
</ol>
<p>A good article on these topics from <a href="https://www.automationdirect.com/products/motors/index" rel="nofollow noreferrer">automation direct.</a></p>
| 56945 | What are the issues with using a general purpose motor on a variable frequency drive? |
2023-11-10T08:35:24.003 | <p>I have a borewell dug in my farm and the depth of the borewell is around 195 feet. I have applied for a Solar pump subsidy and I am about to get the equipment with following details:</p>
<ol>
<li>4.7kW solar panels</li>
<li>Controller</li>
<li>5 HP 12 stage submersible water pump</li>
<li>Borewell diameter - 6 inch</li>
</ol>
<p>As I understand from the technician, this is just sufficient lift the water from the borewell. But what I plan is to lift to the highest elevation in the farm so that the water can be fed by gravity in the farm as and when needed.</p>
<p>In the snapshot from Google maps I have calculated the distance to 'push' the water:</p>
<p><a href="https://i.stack.imgur.com/9yzvv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9yzvv.jpg" alt="enter image description here" /></a></p>
<p>So now the details are as follows:</p>
<ol>
<li>Depth of borewell = 195 feet ≈ 200 feet</li>
<li>Horizontal distance to push water = 333 mtr = 1092 feet ≈ 1200 feet</li>
<li>Number of pipe joints = Around 8 to 10</li>
<li>Altitude difference between borewell point and highest elevation land = 68 feet ≈ 70 feet</li>
<li>Planned elevated tank construction's highest elevation point = 20 feet</li>
<li>Desired flow rate - I do not know it exactly in litres per minute but I do know that 5 HP 12 stage output is around 1 lakh litres per day, so I am ok to forgo around 50% of this to push the water to my desired location.</li>
</ol>
<p>So the pump should vertically lift a total of 290 feet and should horizontally push a distance of 1200 feet along with around 3 joints in between.</p>
<p>I would like to know if the provided equipment is capable of handling lift and push of water for these distances and elevation, if not, what is the capacity I should be looking at.</p>
<p>Please let me know if the above details are sufficient enough. I need some scientific guidance and best practices as against the technicians guidance because they recommend out of their experience and not out of precise calculation as such.</p>
<p><strong>Edit</strong>: I received the pump and solar PV specifications from technician, attached are the details:</p>
<p><a href="https://i.stack.imgur.com/jtDJv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jtDJv.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/D5hoi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D5hoi.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/lfOUX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lfOUX.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/cWsG4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cWsG4.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/fBg9G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fBg9G.jpg" alt="enter image description here" /></a></p>
<p>Kindly guide me with the maximum lift for the water to higher altitudes.</p>
| |pumps|solar-energy|photovoltaics|solar| | <p>To determine if the provided solar pump system is capable of meeting your requirements, we need to consider a few factors, including the total dynamic head (TDH) and the pump's performance curve.</p>
<p>The TDH is the total equivalent height that a fluid is to be pumped, taking into account friction losses in the pipe. It includes the vertical lift, the friction loss due to the horizontal distance, and any additional pressure needed for the system to function.</p>
<p>Here are the components of the TDH in your case:</p>
<p>Vertical lift from the borewell: 200 feet
Elevation difference to the highest point: 70 feet
Height of the elevated tank: 20 feet
Friction loss due to pipe length and joints: This will depend on the type and diameter of the pipe, as well as the flow rate.
The approximate total vertical lift needed is the sum of the vertical lift from the borewell, the elevation difference, and the height of the elevated tank:</p>
<p>Total Vertical Lift = 290 feet
</p>
<p>(Total Vertical Lift=200feet+70feet+20feet=290feet)</p>
<p>The friction loss in the piping due to the horizontal distance and joints can be estimated using the Hazen-Williams equation or other similar hydraulic calculations, but you'll need to know the type of pipe, the diameter, and the flow rate.</p>
<p>To ensure the pump can handle the TDH, you would typically look at the pump's performance curve, which shows the relationship between the flow rate (in gallons per minute or liters per minute) and the height it can pump to at that flow rate. You mentioned that the 5 HP 12 stage pump can pump approximately 1 lakh liters per day. Assuming continuous operation over 24 hours, that's about:</p>
<p>100000 liters / (24 hours × 60 minutes) ≈ 69.44 liters per minute</p>
<p>However, you also mentioned you are willing to forgo around 50% of this flow rate to achieve the desired lift, which would be around 34.72 liters per minute.</p>
<p>Without the specific pump curve, it's impossible to give a definitive answer. If you can provide the make and model of the pump or the pump curve, it would be easier to evaluate.</p>
<p>In the absence of a pump curve, as a general rule, each stage of a submersible pump can raise water by a certain number of feet depending on the design. You would need to know how many feet each stage can lift and then check if 12 stages are sufficient for 290 feet. If not, you might need a pump with more stages or higher horsepower.</p>
<p>Additionally, it's crucial to consider the efficiency of the solar panel system and whether it can consistently provide the required power for the pump, especially during peak usage times.</p>
| 56953 | Pump capacity calculation |
2023-11-12T08:35:46.127 | <p>I read an experiment which demonstrated that metal filings fired at a surface using air blasts charges the particles, surface and airstream. I understand this is triboelectricity, resulting from friction.</p>
<p>For reference: <a href="https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1929.0004" rel="nofollow noreferrer">https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1929.0004</a></p>
<p>Experiments show that the charge increases if</p>
<ul>
<li>airstream is faster</li>
<li>airstream is hotter</li>
<li>particles are finer (greater surface area)</li>
<li>particles are shot at surface obliquely</li>
</ul>
<p>What if we use concentrated sunlight to heat air (e.g. to 300° C) and blow that air at high speeds to fire fine sand on some surface and continuously do so to generate electricity? Why isn't this feasible? Is there some physical limit that comes into play which makes this idea irrelevant?</p>
<p>EDIT 1: I'm frustrated as to why someone is assuming I meant to make some perpetual mumbo jumbo machine. It's not! This is a conversion of heat (in solar) to electricity. The device being so simple, I think it should be useful for charging cell phones and such, with lack of efficiency being a non-issue, since solar is mostly wasted anyway!</p>
<p>I am using the concentrated sunlight to generate a temperature (and thus, pressure) gradient in the setup, so that airflow is generated. There is no "pump" in my modified system. I am trying to heat air to generate "chimney" effect airflow.</p>
<p>Note: Generally I don't like talk about efficiency when it comes to simple easy to make, use and maintain devices like this. Unless it's too low, don't bring it up.</p>
| |electrical-engineering|materials|thermodynamics|experimental-physics| | <p>This seems to be your question:</p>
<blockquote>
<p>What if we use concentrated sunlight to heat air (e.g. to 300° C) and blow that air at high speeds to fire fine sand on some surface and continuously do so to generate electricity? Why isn't this feasible?</p>
</blockquote>
<p>It isn't feasible because you have to generate power to blow this air around to capture some tiny electric charge potential. You are converting energy too many times to be useful: solar to heat, solar to pressurized/moving air, sand charges to electricity. You may not like talking about efficiency, but efficiency is what makes things feasible. A machine that mostly heats up air and sand can certainly be built, but these things have to make money. Make a statement to investors that you don't care about efficiency and they'll throw you out of the room.</p>
| 56976 | Electricity generation from sand particles in airstream hitting surface |
2023-11-13T14:56:35.630 | <p>I want to evaluate the efficiency of a heat exchanger. From measurements I know all four temperatures <span class="math-container">$T_1$</span>, <span class="math-container">$T_2$</span>, <span class="math-container">$t_1$</span> and <span class="math-container">$t_2$</span> as well as the mass flows of both fluids <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> and if necessary I would also able to retrieve e.g. pressure measurements.</p>
<p>Now after some research it seems there are multiple ways to calculate a value to attribute to a heat exchange process.</p>
<p>On a very simple level I found the thermal efficiency</p>
<p><span class="math-container">$$
\eta_{thermal} = \frac{T_1 - T_2}{T_1 - t_1}.
$$</span></p>
<p>Then there is the so called effectiveness
<span class="math-container">$$
\epsilon = \frac{q_{act}}{q_{max}}
$$</span></p>
<p>where I should be able to calculate <span class="math-container">$q_{act}$</span> by</p>
<p><span class="math-container">$$
q_{act} = \dot{m} c_p (T_1 - T_2)
$$</span></p>
<p>I am not 100% sure this is correct though, and I don't know which temperature I would need to use for <span class="math-container">$c_p$</span> in this equation.</p>
<p>Then the next question is how do I calculate q_{max}? Could I use the same equation just with e.g. the inlet temperature of the hot fluid <span class="math-container">$T_1$</span> and the outlet temperature of the cold fluid <span class="math-container">$t_2$</span>? But which mass flow would I then use for the calculation?</p>
<p>Finally there are more sophisticated methods like NTU and LMTD for which I am not sure if I actually need them, if I in my case simply want to calculate an efficiency/effectiveness of a heat exchanger.</p>
<p>So what is the correct way if I basically want to compare a single heat exchanger during it's operating life? Efficiency? Thermal or another equation? Or effectiveness? And do I need a method like NTU or LMTD to calculate it?</p>
| |thermodynamics|heat-transfer|heat-exchanger| | <p>Effectiveness using <span class="math-container">$\epsilon = \frac{q_{act}}{q_{max}}$</span> seems to be most general. It assumes comparing 2 cases and I would choose comparing cases where inlet temperatures and flowrates are fixed for both of them.</p>
<p>When calculating <span class="math-container">$q_{act}$</span> for the case where you already know all the inlet/outlet temperatures, you can use enthalpy difference between inlet and outlet multiplied by mass flow instead of approach with <span class="math-container">$c_p$</span>:</p>
<p><span class="math-container">$$q_{act} = \left(h_C\left(T_{C,out}\right)-h_C\left(T_{C,in}\right)\right)\cdot \dot{m}_C = \left(h_H\left(T_{H,in}\right)-h_H\left(T_{H,out}\right)\right)\cdot \dot{m}_H$$</span></p>
<p>However, calculating <span class="math-container">$q_{max}$</span> might be trickier, because you need to imagine, that the heat exchange between both fluids is perfect at any point in the exchanger, and this I think can manifest in 2 ways:</p>
<ul>
<li>In cocurrent heat exchanger, each fluid enters with different temperature, but they will reach the same temperature at the outlet, i.e. <span class="math-container">$T_{C,out}^* = T_{H,out}^*$</span>.</li>
<li>In countercurrent heat exchanger, different inlet temperatures, <span class="math-container">$T_{C,in}$</span> and <span class="math-container">$T_{H,in}$</span>, are also given at the start and the outlet temperatures will be: <span class="math-container">$T_{C,out}^* = T_{H,in}$</span> and <span class="math-container">$T_{H,out}^* = T_{C,in}$</span>;</li>
</ul>
<p>Knowing all the inlet/outlet temperatures, you can also use the enthalpy approach for calculating <span class="math-container">$q_{max}$</span>:</p>
<p><span class="math-container">$$q_{max} = \left(h_C\left(T_{C,out}^*\right)-h_C\left(T_{C,in}\right)\right)\cdot \dot{m}_C = \left(h_H\left(T_{H,in}\right)-h_H\left(T_{H,out}^*\right)\right)\cdot \dot{m}_H$$</span></p>
<p>Expressing efficiency, flowrates can cancel out, so the expression is simplified:</p>
<p><span class="math-container">$$\epsilon = \frac{h_C\left(T_{C,out}\right)-h_C\left(T_{C,in}\right)}{h_C\left(T_{C,out}^*\right)-h_C\left(T_{C,in}\right)} = \frac{h_H\left(T_{H,in}\right)-h_H\left(T_{H,out}\right)}{h_H\left(T_{H,in}\right)-h_H\left(T_{H,out}^*\right)}$$</span></p>
<p>Some heat exchangers might involve different setups, but I think it is always possible to imagine situation with perfect heat transfer between the fluids.</p>
| 56988 | What is the correct way to calculate the efficiency of a heat exchanger? |
2023-11-14T00:46:46.853 | <p>Suppose there is hot water supply at fixed flow rate and inlet temperature around 40~60C, also there is cold air to be heated by the water at fixed flow rate and inlet temperature around -10~20C.</p>
<p>Heat exchange system 1: use water coils to exchange heat directly.</p>
<p>Heat exchange system 2: first use plate heat exchanger to transfer from the incomming hot water to secondary water, then use
water coils to transfer secondary water to air.</p>
<p>Both system have the same water coil. The only thing varies is the design of the plate heat exchanger subsystem.</p>
<p>System 2 has the advantage of isolation and disadvantage of extra pipes/pump/tank.</p>
<p>What about the heat exchange capability comparison between them?</p>
<p>Suppose the heat exchange capability of system 1 is <span class="math-container">$P_1$</span> (i.e. the power at which heat is taken away from the incomming hot water), can there be an estimation about the maximum heat exchange capability that system 2 can reach? e.g. <span class="math-container">$P_2 \le 0.5P_1$</span>, <span class="math-container">$P_2 \le 1.5P_1$</span>, etc?</p>
| |heat-transfer|hvac| | <p>Question: In a residential setting, how does the heat exchange capability of a cascaded heat exchange system (System 2) compare to a direct heat exchange system (System 1), given the provided parameters?</p>
<p>Answer:</p>
<p>Comparing System 1 and System 2:</p>
<p>System 1: Direct heat exchange using water coils.
System 2: Cascaded heat exchange involving a plate heat exchanger and water coils.
Key Considerations:</p>
<p>Efficiency of Direct vs. Cascaded Heat Transfer: System 1's direct approach typically offers higher efficiency due to fewer stages of heat transfer, thus less thermal resistance. System 2, with its additional stage (plate heat exchanger), introduces more potential points for heat loss.</p>
<p>Impact at Residential Scale: The efficiency difference between these systems is less pronounced in residential applications compared to larger, industrial ones.</p>
<p>Complexity of System 2: The added components in System 2 (extra pipes, pump, tank) mean more opportunities for heat loss, but also offer the benefit of isolation.</p>
<p>Rough Estimation of Heat Exchange Capability:</p>
<p>If System 1's heat exchange capability is denoted as P1, then System 2’s capability (P2) is likely to be somewhat lower, considering the additional components and stages involved.
Estimated Range: A reasonable estimation might place P2 in the range of 0.7P1 to 0.9P1. This suggests a decrease in heat exchange capability for System 2 compared to System 1, but not a significantly drastic one, especially in a residential context.
Conclusion:</p>
<p>While System 2’s cascaded approach does introduce complexity and potential points for efficiency loss, in a residential setting, this decrease in efficiency might not be as significant as it would be in industrial applications. The estimated range reflects a balance between the typical efficiencies of residential-grade heat exchangers and the inherent losses due to the additional components in System 2. It's important to note that this is a general estimation and actual performance can vary based on specific system design and operational conditions.</p>
| 56993 | Can cascaded heat exchange system transfer more heat? |
2023-11-15T20:36:58.770 | <p>I want a system to move something in the opposite direction to the driven thing, but twice the distance. I'm thinking the below system of pulleys. I just wonder if it will actually work or will it just stick and not want to move? And then how to avoid that happening?
<a href="https://i.stack.imgur.com/1ntD4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1ntD4.png" alt="enter image description here" /></a></p>
| |motors|friction|pulleys| | <p><a href="https://i.stack.imgur.com/nQTph.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nQTph.png" alt="enter image description here" /></a></p>
<p>Yes.</p>
<ul>
<li>If the motor moves to the right by distance <span class="math-container">$x$</span> then the inner left pulley most move the same distance.</li>
<li>This will increase the length of the fixed end by <span class="math-container">$x$</span> and the length of the top left string by <span class="math-container">$x$</span> and will require <span class="math-container">$2x$</span> of string to be drawn from the bottom.</li>
<li>The load will move by <span class="math-container">$2x$</span>.</li>
</ul>
| 57017 | Will this pulley system work? |
2023-11-17T00:33:08.160 | <p>This is my first post here. I explain the big picture of my current circumstances, I am a software engineer currently making a platform for evaluating process equipment in plants. I do not really know the technical details since it is not my area of expertise and I apologize if this question does not make any sense.</p>
<p>Basically, the engineer who used to help me build this software found these formulas to calculate effectiveness of a shell and tube heat exchanger:</p>
<p><span class="math-container">$$
Cmin = m_1*C_{p 1}
$$</span></p>
<p><span class="math-container">$$
Cmax = m_2*C_{p 2}
$$</span></p>
<p>Then we'd have to choose which one of the two is the minimum and the maximum. Then we'd have to do this:</p>
<p><span class="math-container">$$
C = \frac{Cmin}{Cmax}
$$</span></p>
<p>However, the engineer said that when there is a phase change Cmax would be considered as infinity and therefore C would equal 0.</p>
<p>If C equals 0 then that would mean the formula that would be used for effectiveness would be the following:</p>
<p><span class="math-container">$$
\epsilon = 1-e^{-NTU}
$$</span></p>
<p>And NTU would be:</p>
<p><span class="math-container">$$
NTU= \frac{U*A}{Cmin}
$$</span></p>
<p>First of all, is this logic true? If not, what is the right logic? And how should I calculate Cmin in the NTU formula? And would this still be true even if there is phase change in both sides of the sell and tube exchanger?</p>
<p>Again, I'm sorry if I wasn't clear enough, this is not my area of expertise...</p>
| |chemical-engineering|heat-exchanger| | <p>First of all, I feel sorry about that I did not see your question earlier.<br />
Your reasoning is suitable for most cases, but since you are not a chemical or mechanical engineer we can forget about messy details and pointless scenarios.<br />
Your inference that if
<span class="math-container">\begin{equation}
c=0
\end{equation}</span>
Then for all heat exchangers
<span class="math-container">\begin{equation}
\epsilon=1-e^{(-NTU)}
\end{equation}</span>
is absolutely correct. Also you can calculate <span class="math-container">$C_{min}$</span> by comparing following two:
<span class="math-container">\begin{equation}
C_{hot}=\dot{\mathbf{m_{hot}}}*C_{p,hot}\\
C_{cold}=\dot{\mathbf{m_{cold}}}*C_{p,cold}
\end{equation}</span>
Where <span class="math-container">$\dot{\mathbf{m_{cold}}}$</span> is mass flow rate of cold fluid and <span class="math-container">$\dot{\mathbf{m_{hot}}}$</span> is mass flow rate of hot fluid, notation is the same for specific heats.
<span class="math-container">$C_{min}$</span> is the smaller one from <span class="math-container">$C_{hot}$</span> and <span class="math-container">$C_{cold}$</span>. Mass flow rate can be calculated by many ways. For example one can calculate mass flow rate if cross sectional area, density of the fluid and velocity of fluid is known, by the formula for steady flow <span class="math-container">$\dot{\mathbf{m}}=\rho*v*A_{c}$</span>.<br />
The specific heats are obtained from tables for most cases, so you will not calculate them.<br />
For the last question I have less to say. It's a troublesome question to develop a clear answer, for me. But I believe we would use some other techniques to analyse the situation with phase change in both sides, or we would divide the analysis of process in order to account for all physical phenomena respectively, or maybe we would at least add some extra terms for phase change. Also I think the answer of this question would be case-specific. Therefore it would be best for you to consult the engineer you work with.</p>
| 57036 | How do I calculate effectiveness and NTU of a heat exchanger when there is phase change? |
2023-11-21T00:01:49.527 | <p>I am using a 1/2 hp sump pump to provide flood coolant to a CNC milling machine. The sump pump has a 1.5" outlet. The milling machine already has a 1/2" adapter and manifold which connect to Loc-Line hoses that each have a ball-valve for on-off.</p>
<p>Where is the best place to put the 1.5" -> 0.5" reducer? I'm debating whether I should do that at the sump outlet, or if I should run a length of pipe (PVC or other) close to the sump's outlet diameter and then use the reducer closer to the manifold. Does it make a difference?</p>
| |fluid-mechanics|pumps|pipelines| | <p>I would add a by-pass valve back to the sump.</p>
<p>That will allow the flow to be controlled from max to min.</p>
| 57078 | Where should I restrict the flow of a pump? |
2023-11-22T19:45:19.920 | <p>Currently, I'm trying to understand the Two-Ray Ground-Reflection Model.
During my research, I came across the following simplification expressing the received power as follows (taken from <a href="https://en.wikipedia.org/wiki/Two-ray_ground-reflection_model" rel="nofollow noreferrer">Wikipedia</a>):
<a href="https://i.stack.imgur.com/1V8bp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1V8bp.png" alt="enter image description here" /></a></p>
<p>The part that is not clear to me is <span class="math-container">$e^{-j\Delta\phi}$</span>:
How is that derived from <span class="math-container">$e^{-j2\pi l/\lambda}$</span> and <span class="math-container">$e^{-j2\pi (x + x\prime)/\lambda}$</span>?</p>
<p>I know that <span class="math-container">$\phi$</span> can be calculated with <span class="math-container">$\phi = 2\pi d/\lambda$</span>, where <span class="math-container">$d$</span> is the distance between the transmitter and the receiver.</p>
<p>In the above equation, the ground-reflected path is expressed as <span class="math-container">$(x + x\prime)$</span> and <span class="math-container">$l$</span> is the direct line-of-sight path.</p>
<p>Moreover, the approximation only holds if the signal is "narrow band relative to the inverse delay spread <span class="math-container">$1/\tau$</span>, so that <span class="math-container">$s ( t ) \approx s ( t − τ )$</span>".</p>
<p>Why is that important?</p>
<p>Thank you for your answers in advance!</p>
| |telecommunication|wireless-communication|signal-processing|waves|communication| | <blockquote>
<p>The part that is not clear to me is e−jΔϕ: How is that derived from e−j2πl/λ and e−j2π(x+x′)/λ?</p>
</blockquote>
<p>Attempting to take <span class="math-container">$e^{-j2\pi l/\lambda}$</span> as a common factor we get
<span class="math-container">$$
\left|e^{-j2\pi l/\lambda}
\right|^2
\left|
\frac{\sqrt{G}}{l}
+\Gamma(\theta)\frac{e^{-j2\pi (x+x\prime)/\lambda}}{(x+x\prime) (e^{-j2\pi l/\lambda})}
\right|^2
$$</span></p>
<p>On the above, you might have to use the simplifications <span class="math-container">$\frac{e^a}{e^b} = e^{a-b}$</span> and <span class="math-container">$|e^{j\ c}|=1$</span>.</p>
<blockquote>
<p>Moreover, the approximation only holds if the signal is "narrow band relative to the inverse delay spread 1/τ, so that s(t)≈s(t−τ)".
Why is that important?</p>
</blockquote>
<p>They might be interested in finding out the locations or angles for which the signal and its reflected version <strong>destructively interfere</strong> and result in low signal strength. If a signal <span class="math-container">$s(t)$</span> and its time delayed version <span class="math-container">$s(t-\xi)$</span> needs to cancel out, for all values of <span class="math-container">$t$</span>, then the assumption above (<span class="math-container">$s(t)\approx \color{red}{-}^{\dagger}s(t-\xi)$</span>) is required. The word <strong>narrow band</strong> is simply used to indicate that the signal needs to looks like a sinusoidal wave (which has the required property of periodicity mentioned above).</p>
<p><sup><span class="math-container">$\dagger$</span> The minus sign may be provided by the act of reflection. I am not sure.</sup></p>
| 57103 | Two-Ray Ground-Reflection Model: Understanding the Difference in Phase Offsets at the Receiver |
2023-11-23T14:30:46.753 | <p>The Janka hardness test is a test used to measure wood's resistance to denting. The test involves pressing a steel ball that is 0.444" in diameter halfway into a piece of wood and measuring the force required.</p>
<p>The 0.444" ball intentionally has a 1 sq cm circular diameter but I have two questions about this method:</p>
<ol>
<li>Generally why is a steel ball used rather than just a flat 1 sq cm anvil? It seems like the ball might give a better representation of the wood strength across a larger cross section of the wood but is there a well defined reason for this?</li>
<li>Is the force required really even measurable in pounds per square centimeter (odd combination of units there I know)? Considering there will potentially be some additional friction and compressive forces at play. i.e.: white oak has a janka hardness of 1360 lbf or 616 kg. Would it be reasonable to say it takes around 8774 PSI to dent white oak since 616 kg per sq cm is equal to 8774 lbs per square inch?</li>
</ol>
<p>In regards to solar mike's comment about side friction. I get that a sphere will have very little to no side friction once pressed in halfway but wouldn't there be additional friction prior to that point? The area of the cross section of the ball is 1 sq cm but the area of the sphere is 4 sq cm so the half being pressed in the wood is actually 2sq cm. I'm unsure what part this plays as the area of a nail for example isn't nearly as important as the area of the tip of the nail. The side may provide some additional friction but ultimately only the cross section of the nail is penetrating the wood.</p>
| |force-measurement| | <p>The steel surface needs to start and remain parallel to and tangent to the wood surface. This is automatic and easy to re-create with a spherical steel surface. The first part is unnecessarily difficult with a flat surface: the second part is impossible with sharp corners.<br />
Both conditions can be met with parabolic/hyperbolic/thumb shaped probe: when considering adhesion forces, non-spherical probes may be used, but steel ball bearings and spherical probes are a standard item.</p>
| 57123 | Benefits of a steel ball in janka hardness test |
2023-11-26T17:07:53.197 | <p>In gear geometry there are 3 main angles, which can be misleading:</p>
<ol>
<li>Profile angle</li>
<li>Pressure angle</li>
<li>Operating pressure angle</li>
</ol>
<p>Further for simplicity I would restrict the area of interest to "spur gears". It is quite obvious that for standard mating gears these above mentioned angles are equal, but how do they differ when a profile shift occurs?</p>
<p>We can distinguish 2 types of corrections:</p>
<ol>
<li><span class="math-container">$V_0$</span> shift <span class="math-container">$\longrightarrow \sum x = 0$</span></li>
<li><span class="math-container">$V_+ V_-$</span> shift <span class="math-container">$\longrightarrow \sum x \neq 0$</span></li>
</ol>
<p>So how these angles:</p>
<ul>
<li>are geometrically defined?</li>
<li>change during profile shift?</li>
<li>can be calculated?</li>
</ul>
| |gears|terminology|geometry| | <p>In spur gears, the profile angle, pressure angle, and operating pressure angle usually align for standard gears. But with profile shift, things change.</p>
<p>The profile angle, which is the slope of the gear tooth relative to the gear base circle, remains constant even with a profile shift. The pressure angle, the angle at which gear teeth transmit force, and the operating pressure angle, the pressure angle during gear operation, are the ones affected by profile shift.</p>
<p>In a V0 shift (∑x=0), the gears are modified but the center distance remains the same, whereas in a V+V− shift (∑x≠0), the center distance changes. These shifts alter the effective working pressure angle: a positive shift increases it, and a negative shift decreases it.</p>
<p>Calculating these angles post-shift involves gear tooth geometry and the involute function. It's complex and often handled by specialized gear design software, but can be done manually through detailed geometric analysis if necessary.</p>
| 57153 | Profile and pressure angle definition in gear geometry |
2023-11-28T11:20:40.050 | <p>I am dealing with a European contractor, and often I see a division symbol (÷) in their documents, screenshot below. Could somebody explain what that means? P.S. Hope this is the right stack for it. If not, kindly suggest where to post it instead.</p>
<p>Thanks in advance.</p>
<p><a href="https://i.stack.imgur.com/J7bL5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J7bL5.png" alt="Screenshot of welding proceedure" /></a></p>
| |welding| | <p>It seems that in Polish , which the above is written in , the division sign is sometimes used to denote a range of values. <a href="https://academic-accelerator.com/encyclopedia/division-sign" rel="nofollow noreferrer">source</a></p>
| 57164 | Division symbol in welding proceedure |
2023-11-30T03:17:16.827 | <p>The specification for car hitches routinely state the <em>tongue weight</em>, but do not specify how far that weight can/should be applied. What am I missing? Shouldn't the maximum <em>bending moment</em> be the one to be specified?</p>
<p>It matters because I identified one hitch manufacturer who states</p>
<blockquote>
<p>This tow hitch is rated at 1588 kg (3,500 pounds) towing and 238 kg (525 pounds) tongue
weight.</p>
</blockquote>
<p>That is terrific, because every other hitch builder for this vehicle only handles 91 kg (200 pounds) (the so-called class I).</p>
<p>Nominally, a 238 kg (525 lb) hitch is amply sufficient to handle the load from a 2-bike rack (24 kg; 52 lb), a 2-bike add-on (20 kg; 44 lb), and four mountain bikes (13.5 kg; 30 lb; each), for a total of 98 kg (216 lb).</p>
<p>But the third and fourth bikes, as well as the add-on rack, are so far from the hitch I am concerned the torque would be excessive.</p>
<p>How do I relate tongue weight to weight x distance?</p>
<p><a href="https://i.stack.imgur.com/BnrRl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BnrRl.png" alt="hitch tongue weight" /></a></p>
| |torque|forces|force-measurement| | <p><a href="https://i.stack.imgur.com/yFzV7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yFzV7.png" alt="hitch" /></a></p>
<p>you need to jack up the car and check the structural layout of the hitch. Usually, there are two arms welded to the sides of the hitch receptor reaching to the left and right side of the mainframe of the car bolted to the frame of the car through two rows of bolts on each arm.</p>
<p>You measure the moment of 4 bicycles times their distance from the hitch, say</p>
<p><span class="math-container">$$M = 200 lbs. * 1ft= 200lbs.ft$$</span></p>
<p>Half of this goes to each side bracket and induces tension and shear on front bolts and compression and shear on rear bolts, in addition to 500/4= 125lbs tension per bolt.
<span class="math-container">$$T= C =100/d$$</span></p>
<p>My estimate is a hitch that is rated 500lbs. tongue load can handle this because it has already been designed to take the moment of 500 * D, the distance of the receiver to the brackets. That distance is many times bigger than d.</p>
| 57181 | How do I relate tongue weight to weight x distance? |
2023-12-05T04:53:27.023 | <p>What are those holes for? (Those are for CO<sub>2</sub>.)</p>
<p>Here are a few examples :</p>
<p><a href="https://i.stack.imgur.com/R34Nt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R34Nt.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/FvS0f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FvS0f.jpg" alt="enter image description here" /></a></p>
| |piping|compressed-gases|connectors| | <p>I believe that is called a weep hole for the upper fitting, intended to help detect abnormal leakage. Since the fluid should be contained in the internal sections, the fluid should not be present mid thread deep if seals are functioning. By routing it outside where one might observe or to a point that has a sensor, it gives indications that the sealing system has failed.</p>
<p>See for example
<a href="https://www.dynexhydraulics.com/coned-threaded-hydraulic-fittings-a-primer." rel="nofollow noreferrer">What's Up With That Tiny Hole</a></p>
<p>On the lower valve, I can't say for certain that it isn't some type of additional sense or vent port that has to do with affecting the valve based on whether the device is properly connected, over pressurized, etc or not- a bit more active a variant of the humble weep hole where what I had called a leak before might instead indicate a proper connection.</p>
<p>A hole can serve different functions on each part. Observing only one end might not tell us the hole story.</p>
| 57219 | What is this hole in some gas related threads |
2023-12-07T05:44:05.457 | <p><em>Confiteor</em>, I have only a rudimentary knowledge of how machines work. I'll be discussing fossil fuel engines only.</p>
<p>First off, I have a block of iron, B. I find a pan balance. I place B in one of the pans. I then place X in the other pan. X > B. The arms will move in such a way that you can infer, <em>B can't lift X</em>. Now remove X, place Y in its place. Y = B. Equilibrium!!!</p>
<p>Conclusion: If X > B then B can't lift X (we're talking about <em>weight</em> here).</p>
<p>My question is, with a car, the weight of the fuel (even at full tank) is much, much, less than the weight of the car + fuel. and yet ... a fraction of the fuel "lifts" the car.</p>
<p><em>Quomodo</em>?</p>
| |energy|machine-elements|fuel| | <p>the gasoline contains a tremendous amount of stored <em>chemical energy</em> per pound of weight. the engine in the car burns the gasoline to release the stored chemical energy as heat and convert that into shaft work, which turns the wheels of the car.</p>
| 57241 | How do cars move, it seems inexplicable (to me)? |
2023-12-07T18:17:11.057 | <p>I'm looking for a bolt without a thread that I can attach to an object using a kind of press nut. Whats the name of these kind of bolts and nuts? Can't find any information about it?</p>
<p><a href="https://i.stack.imgur.com/qyiDu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qyiDu.jpg" alt="Bolt with press nut" /></a></p>
| |mechanical-engineering| | <p>One name it goes by is a PEM pin.</p>
<p>In addition to the various washer names given by Solar Mike, it might also be called various forms of push-on retaining clips/rings (whereas a normal retaining clip expects a groove).</p>
| 57244 | Looking for bolt with press nut |
2023-12-07T22:08:24.883 | <p>I am reading Modern Compressible Flow by John D. Anderson Jr and I am going through the integral form of the continuity equation, I am struggling to wrap my head around the fact that said form has two integrals associated with it. I have come across the expression before, but I just hadn't stopped to think about why it has two integrals, any help will be appreciated. My impression has been that when we use the infinitesimal fluid element approach where the fluid element is on a blob (random geometric shape), we use a single integral to show that we are summing the fluid elements over that blob but with the two integrals, that's throwing me off. <a href="https://i.stack.imgur.com/Sk386.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sk386.png" alt="The expression represents the mass flow going into the control volume" /></a></p>
| |fluid-mechanics| | <p>If you are doing surface integrals starting from single dimensional infinitesimals, then you need two integrals because areas are two dimensional. If your integral is formed such that it already happens to start with an infinitesimally thick slice of the surface then you don't need two integrals because part of the work is already finished.</p>
<p>Essentially, with two integrals you are building up from linear dimensions, to a slice, and to an area. The first integral integrates any one a slice of the surface. The second integral integrates all the slices to form the entire surface.</p>
<p>It progresses the same way to volume integrals, where the surface you have after two integrals is actually a cross section of the volume. Then the third integral adds up all the cross sections to get you the volume.</p>
| 57248 | Why does a surface integral have two integrals? |
2023-12-08T11:20:37.877 | <p>I want to make an ultrasonic cutter for home use projects, something like <a href="https://www.youtube.com/watch?v=2-im2AgUe9w" rel="nofollow noreferrer">this</a> or there are many online videos of industrial ultrasonic cutters in e.g. the food industry. A typical design of an ultrasonic cutter is below on the left (a). It has a booster (a shaped metal cylinder whose length has been tuned to match the resonance of the oscillator), and then a sonotrode (to do the same thing I believe). The sonotrode has a particular profile of its own.</p>
<p>What is the purpose of the wedge shape of the sonotrode profile and the slots?</p>
<p>If I replace both with a single flat sheet of metal (as in (b)), tuned to the resonance of the oscillator as will the professional version be - what will I lose in terms of performance from the professional version?<a href="https://i.stack.imgur.com/AOuKQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AOuKQ.png" alt="enter image description here" /></a></p>
| |acoustics|cutting|ultrasound| | <p>The purpose of the various tapers and bulges along the length of the booster and sonotrode is to 1) extract the greatest amount of <em>energy</em> being fed into it by the oscillator, and then 2) to yield the greatest amount of oscillation <em>amplitude</em> right at the point of contact between the sonotrode and the workpiece.</p>
<p>A piece of sheet metal is too light and too flexible to permit either of these conditions from being met.</p>
| 57255 | Can I use a single flat sheet of metal as an ultrasonic horn? |
2023-12-08T15:51:36.010 | <p>I'm trying to detach the handle from my meat grinder. I'm not sure what this kind of lock is called but it had a crescent/half moon shape that locks against the grinding (clockwise) motion. In order to detach it I would have to twist in the counterclockwise direction. But the thing is the piece the handle is attached to is the swirly metal piece and I'm not sure if I can stop that could lock it in place while I turn it counterclockwise. I tried putting the thing in the freezer but the lever also shrank with the semicircle locking part. I do get my hands on dry ice sometimes and I'm thinking that I could maybe use some gloves and put that in direct contact with the semicircle, but I'm wondering if that will somehow transfer the temperature over to the lever anyways since they're in direct contact with each other. If anyone knows the name of the locking mechanism that might help me google better. Thanks<img src="https://i.stack.imgur.com/1RIQk.jpg" alt="the cursed beast" /></p>
| |mechanical-engineering|solid-mechanics| | <p>Because the joint created by the handle and the auger bit is solid metal, the threading is, as you've noted, held in place by the rotation.</p>
<p>It is common that a sharp blow on the handle in the correct direction will cause the threads to release. Use a rubber mallet or wood mallet or other dead blow mallet, one that will not peen the metal in the handle and leave unsightly marks.</p>
<p>You may be able to combine the mallet method with a wrench applied to the semi-circle end, but you may create metal damage in so doing.</p>
| 57259 | Trying to detach a detachable handle on a meat grinder |
2023-12-08T21:14:03.243 | <p>I would like to know if there is any special coating or special paint that can be applied to the tread of an automobile tire, strongly bonding to it, and which is strong enough and durable enough to be driven on. The purpose of this would be to extend the life of the tires that you bought for your car.</p>
<p>I would imagine that this special coating or paint has the characteristics of being able to adhere/grip the road in any weather condition and is durable enough to last at least a few thousand miles before being worn off of the tire's tread.</p>
<p>Is there any special coating or paint that will strongly bond to an automobile tire's tread and can be driven on?</p>
| |materials|automotive-engineering|chemical-engineering|elastic-modulus| | <p>To add a bit to DKNguyen's answer:</p>
<p>The wear resistance of rubber as used in tire manufacture scales with its hardness. Hardness is varied by controlling the amount of carbon black milled into the elastomer base. More carbon yields a harder composition. Hard rubber wears more slowly than soft rubber, so a coating of hard rubber on the tread surface of a tire will make it last longer than a soft one- but it also does not grip the road as effectively as soft rubber. What to do?</p>
<p>On motorcycle tires, the solution is to make the tire out of three different rubber compositions. The center "stripe" of the tire surface is made from hard rubber, so when you are blasting down the interstate in a straight line on a warm, sunny day you are rolling on the most wear resistant part of the tire, and you get long life.</p>
<p>Partway up on the <em>sidewall</em> of the tire, where you are rolling during a turn, a softer compound is used to yield more traction so you will not skid on a cold rainy day, and farther up than that the softest compound is used to give you even better grip in a sharper turn while "canyon carving".</p>
<p>Almost all the way up on the sidewall is a ridge that protrudes away from the body of the tire. This is called the "chicken strip" and by thickening the soft rubber into a ridge, that rubber can deform more easily than the soft rubber covering the bulk of the sidewall and furnish even greater grip in the steepest possible turn- beyond which portions of the underside of the motorcycle will begin to scrape the road, which lifts the tire out of contact with the pavement. A <em>low-side crash</em> is the result.</p>
<p>It is called the chicken strip because if you fail to "chicken out" of a steep turn prior to running out of sidewall tread, it will dig into the pavement and save your life at the last moment.</p>
| 57261 | Is there any special coating or paint that will strongly bond to an automobile tire's tread and can be driven on? |
2023-12-09T16:38:40.060 | <p>I have obtained the transfer function for a two-tank hydraulic system. I wish to obtain the time response for which I need to perform the Inverse Laplace Transform and have the function with natural frequency and damping factor. But I am lost, I don't know how to do it because I can't find a way to rearrange the polynomial to use the Laplace Transform Tables. <a href="https://i.stack.imgur.com/kevJE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kevJE.png" alt="The function" /></a></p>
| |transfer-function| | <p>Using the method of partial fractions as shown <a href="https://en.wikipedia.org/wiki/Laplace_transform#Impulse_response" rel="nofollow noreferrer">here</a> in Wikipedia.</p>
<ol>
<li>Factorise the denominator polynomial to degree one and two polynomials. In your example it is already done.</li>
<li>Do partial fraction expansion to bring the transfer function into the form
<span class="math-container">$$
H_2(s) = \sum_i\frac{a_i}{s+b_i}+\sum_j\frac{cs+d_i}{s^2+es+f}
$$</span></li>
<li>According to linearity property, the inverses of each term in the sums can be added together to get the result.</li>
<li>To find inverses of the individual first order and second order transfer functions, use the forward or inverse Laplace transform tables. e.g. <a href="https://en.wikipedia.org/wiki/Laplace_transform" rel="nofollow noreferrer">here</a></li>
</ol>
<p>Result will be available as a sum of exponentials and sinusoids.</p>
<p>The above method will work for most rational transfer functions.</p>
| 57269 | How would you take the Inverse Laplace Transform of this function? |
2023-12-11T10:43:55.233 | <p>The transition towards climate neutrality is a big political topic in Germany. One claim that is often made but typically only handwaivingly explained (if at all) is that we need (green) hydrogen.</p>
<p>I do understand that some processes might need hydrogen. And that hydrogen should be produced in a climate-neutral way.</p>
<p>But are there processes that currently use e.g. fossil gas which can/should be replaced with hydrogen?</p>
<p>The main example I found is <strong>steel production</strong>. However, I see that electric arc furnace (EAF) and induction furnace exist. My guess is that they are well-suited for scrap metal, but not for ore. Is that guess right?</p>
<p>If so, why are they not suited for ore? Or is it just the case that gas is so much cheaper than electricity?</p>
| |steel|metallurgy| | <p>It's almost certainly true that an electric arc furnace isn't adequate for steel production, at least not by itself.</p>
<p>It depends on what phase of steel making you care about though. An electric arc furnace is most often use to melt down scrap, which you then purify (e.g. with the Basic Oxygen Process), then add whatever you need to get the alloy you want.</p>
<p>You seem to care about starting from ore though. In that case, you're starting from iron ore (e.g, taconite) which is almost always some form of iron oxide.</p>
<p>So, the main impurity you need to remove from iron ore to get iron is oxygen. In a traditional blast furnace, you use carbon not only as something to burn to get heat, but also as a chemical reactant.</p>
<p>Simplifying the chemical process involved<sup>1</sup>, you're basically doing FeO<sub>2</sub> + C => Fe + CO<sub>2</sub>. You do have to get the iron ore hot before the oxygen will unbind from the iron and bind to the carbon instead.</p>
<p>Regardless of the precise details of the chemical process, the main point is that you need to do more than just get iron ore hot. Rather, you need to heat the iron ore in an oxygen-deprived atmosphere in the presence of something else that will absorb the oxygen from the iron. Traditionally, carbon has been used for both the burning to produce heat and as the material that absorbs oxygen from the iron ore.</p>
<p>So, while it's obviously possible to get the iron ore hot with an electric furnace, you still need to add something to absorb the oxygen being released from the iron ore. Oxygen binds to lots of other materials, so there's no shortage of candidates. The difficulty tends to be that it binds fairly strongly to most other elements, so you frequently have to expend a lot of energy to get the element in reasonably pure (oxygen-free) form.</p>
<p>Hydrogen is attractive in this respect. Although it does consume energy, it's pretty easy to separate water into hydrogen and oxygen via electrolysis. In fact, you can do it at extremely small scale with a glass of water and a AA battery.</p>
<p>I'm not sure whether you care, but hydrogen does pose some pretty serious engineering problems of its own though:</p>
<ol>
<li>even in liquid form, it's pretty low density, so you need large tanks of hydrogen to store enough to do much.</li>
<li>Designing hydrogen tanks is seriously non-trivial for at least a couple reasons:
<ul>
<li>many metals get brittle when exposed to reasonably pure hydrogen</li>
<li>hydrogen molecules are small enough that they leak through many materials</li>
</ul>
</li>
</ol>
<p>The difficulty here isn't really with finding elements that will combine with oxygen. The difficulty is much more with finding elements that combine with oxygen...gently enough, so to speak. If you wanted to use solely electricity, you could (for one example) purify buaxite (aluminum ore), which is normally done with an electrical process.</p>
<p>Then grind your aluminum into a fine powder, and heat alongside iron oxide. The oxygen will combine with the aluminum. The problem is that it does so rather...violently. The process is exceedingly exothermic--in fact, when you do this with a small quantity of iron oxide and aluminum, it's called a "thermite grenade". So while this works (in one respect) you'd need to do some pretty "special" engineering to keep it from destroying the furnace. It's not really a practical choice for steel production.</p>
<hr />
<ol>
<li>If memory serves, the real process happens in two steps. First, at relatively low temperature you get CO<sub>2</sub> + C => 2CO. Then at higher temperature you get: FeO<sub>2</sub> + 2CO => Fe + 2CO<sub>2</sub>. Caveat: I'm going from memory on the chemical reactions involved, so it's probably still a bit wrong. The real point isn't the precise reaction, but the fact that you're using the carbon not only as fuel, but also as a reactant, so replacing it means not only getting ore hot but using some other reactant as well.</li>
</ol>
| 57293 | Can electric furnaces be used for steel production from ore? |
2023-12-12T06:32:37.650 | <p>I would like to know what are the tools and the techniques to create a map as precise as possible of a place, typically interior, and also, in certains reasonable way, quickly.</p>
<p>That is, a 2D representation of the room and the things inside. I don't care about pillows or small ornaments, or even chairs, but big things like cupboards or fixed things like toilets. I don't care about the details, a rectangle for a bath tub would be good.</p>
<p>Also, as a corollary question, does some AI exists such as if you take some photos of a room from different positions (maybe on top of a ladder), it can triangulate the objects and create a 2D map from above (aside real measures, but only relative ones)?</p>
<p>Sorry if this isn't much in topic, it's impossible put these questions in search engine and obtain relevant results (like almost always nowadays). Also, it should be evident, I come from a different background. Thanks very much.</p>
<p>EDIT 2023-12-26: thanks everyone, it seems I can only accept a single answer, but all of them were useful.</p>
| |civil-engineering| | <p>Yes, 3d models from 2d shots has been available for decades now, but try <a href="https://lifehacker.com/magicplan-creates-a-floor-plan-with-your-phone-s-camera-1614766960" rel="nofollow noreferrer">https://lifehacker.com/magicplan-creates-a-floor-plan-with-your-phone-s-camera-1614766960</a> for what you are actually talking about. I have not tried it.</p>
| 57300 | Creating a map of a place |
2023-12-12T13:10:45.673 | <p>I am a high school science research student and am investigating the implementation of regenerative braking on roller coasters. Ive read the papers that were available on this topic and its said that usually around 70% of the coasters original kinetic energy is up for recapture.</p>
<p>My first problem is the formula, ive seen a couple motors using specific formulas that calculate the energy recovered but I don't have a specific motor to draw an equation from.</p>
<p>My second problem is the data required to calculate the energy recovered. What general stats do I need to find out the energy recovered such as velocity, mass, incline etc.?</p>
<p>There aren't many papers on this topic so anything pointing me in the right direction would help me significantly, thank you!</p>
| |regenerative-braking| | <p>My starting point would be to consider the conversion of energy between forms. If we model the rollercoaster car as a mass starting at its highest point, initially with zero speed but teetering on the edge of its first descent, then in an idealised system with no energy loss externally it would be be able to roll all the way around its track, returning to the exact same position and speed. This would involve a conversion of gravitational potential energy (which we can calculate, based on the mass of the car and the height difference in the track) into kinetic energy and then back again.</p>
<p>Now suppose that we want to stop and let passengers in and out at a point that isn't the highest, we need some way of taking energy out of our simple system (in order to deccelerate at the bottom) and then putting it back in (in order to accelerate and get back to the top).</p>
<p>Next we can think about the reasons this model is not realistic, these would include: friction at the wheels, air resistance, resistance in the wiring, losses in the motor, losses in the battery, etc. All these represent energy lost from the system (typically as heat). It also may not be feasible to recover all the kinetic energy into the electrical system (e.g. for capacity or safety reasons) and we may need to add mechanical brakes (which will convert all their kinetic energy into heat).</p>
<p>With the energy conversions quantified and the losses understood we can next start to think about how far from perfectly efficient each conversion might be. You may be able to find some typical % efficiency numbers for each step or loss mechanism. Ultimately, for a closed loop like a rollercoaster there will be an amount of energy lost per lap, and this will be equal to the amount you will need to add in order to complete the lap.</p>
| 57305 | How to determine the energy recovered by regenerative braking on a roller coaster |
2023-12-12T14:17:55.587 | <p>I am looking for a way to create prototype high temperature circuits capable of going up to 500 °C. I know one way is to generate the CAD file and send it to a PCB manufacturer. Another way is to get copper plated alumina wafers and just cnc/etch them myself.</p>
<p>I was wondering if there was a way to do this by adhering copper foil onto a high-temperature non-conductive wafer. It would not be as precise as having it PCB manufactured, but it would be 10 times cheaper and faster. The highest temperature capable foil that I found has been in the 200 °C range. Is there a high temperature capable adhesive that I can combine with regular copper foil? Is there a pre-made high temperature adhesive foil somewhere? Are there any other techniques that I am missing?</p>
| |temperature|circuits|adhesive| | <p>Another approach that may be practical for prototypes if your circuit design is not too tiny or complex is to make each separate element of the circuit in thin copper, either wire or sheet that is thick enough for practical hand work and mechanically fasten all of it together on a substrate sheet of Macor machineable ceramic (good to 1470F)with drilled holes in it that can accomodate machine screws or other methods like silver solder (which comes in various melting temperatures from around 1000 to 1300F so you can work in steps). Or you could use cheaper plain ceramics and drill the holes with diamond tools. Or find a friend that does ordinary pottery and have then make you a flat ceramic plate with the holes already in it. Do some research in this direction and real quick you will know more about it than I do.</p>
| 57306 | Alternatives to ceramic PCB |
2023-12-15T01:48:53.987 | <p>I recently bought <a href="https://rads.stackoverflow.com/amzn/click/com/B09B73X57M" rel="nofollow noreferrer" rel="nofollow noreferrer">a lathe</a> of the A site that was quite clearly from China. It claims to be 8" x 23". It came with a turret style tool-post which, frankly, I don't want to deal with shims so I bought a SHARS <a href="https://rads.stackoverflow.com/amzn/click/com/B087QMBF32" rel="nofollow noreferrer" rel="nofollow noreferrer">OXA</a> and <a href="https://rads.stackoverflow.com/amzn/click/com/B087QMBF32" rel="nofollow noreferrer" rel="nofollow noreferrer">AXA</a> quick-change tool-post (free returns), and I'm trying to decide which to install. I think I'm going to keep the larger one (AXA) and I was drawing up a way to mount the post to the lathe compound when I realized that there is a "nubbin" on the compound <a href="https://i.stack.imgur.com/iZpok.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iZpok.jpg" alt="compound_nubbin" /></a> where the turret style post's bolt attached to the cross slide.</p>
<p>I am trying to figure out what the best plan is do I:</p>
<ol>
<li>Grind off the nubbin, and increase the hole size on the cross slide from M8 to M12 so I can mount the AXA tool post directly?</li>
<li>Grind off the nubbin, and mount the toolpost mount plate to the cross slide and then the AXA tool post to that?</li>
<li>Grind off the nubbin, give up on the AXA and go with the OXA which use M8, and mount the tool post directly to the currently existing hole?</li>
<li>Leave the nubbin and shim around it so that the OXA tool post mounts above the existing hole?</li>
</ol>
<p>My concern is primarily why does the "nubbin" exist is it for strength? or just to give the turret style post something to rotate around? I am heartened that the mount plate that came with the AXA tool post is the exact thickness of the cross slide above the dovetail, with the nubbin removed.</p>
<p><a href="https://i.stack.imgur.com/Gg5vJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gg5vJ.jpg" alt="enter image description here" /></a></p>
<p>This might be largely "opinion" based, but I'm hoping I can get some real engineers feedback on the various methods and their effects on the lathe system.</p>
| |machining|machine-design| | <p>[Disclaimer: I'm much closer to being a "real machinist" than a "real (mechanical) engineer".]</p>
<p>I'm going to guess that the stock toolpost is one of the "4 way turret" ones, vaguely like this:</p>
<p><a href="https://i.stack.imgur.com/fsASe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fsASe.png" alt="enter image description here" /></a></p>
<p>Unlike this one, however, in your case the "nubbin" fits into a counterbored hole in the bottom of the toolpost.</p>
<p>I think it's primarily to provide a bearing surface, so as you rotate the toolpost, it maintains its position. For example, if you rotate it a full turn, and tighten it back in place, your tool should still be pretty close to the same position relative to the work piece (within a thousandth or so).</p>
<p>As to what to do...I'd prefer to counterbore the bottom of the toolpost over grinding off the nubbin. Leaving the nubbin intact means it can continue to accomplish what it was put there to accomplish in the first place, and most QC toolposts won't be adversely affected by counterboring the bottom of the mounting hole. To maximize its benefit, you want your counterbore to be a nice, tight fit over the "nubbin".</p>
<h3>Image Source</h3>
<p><a href="https://www.penntoolco.com/precise-revolving-tool-post-turret-with-swing-of-range-6-to-12-rtp-001/" rel="nofollow noreferrer">https://www.penntoolco.com/precise-revolving-tool-post-turret-with-swing-of-range-6-to-12-rtp-001/</a></p>
| 58351 | Proper way to mount a toolpost to a lathe |
2023-12-15T11:28:34.463 | <p>I am an HVAC engineer.
I know that in the refrigeration cycle the expansion valve controls the refrigerant flow rate but if flow through a positive displacement compressor is fixed, how does it do that?</p>
| |hvac|refrigeration| | <p>You will find that the mass flow rate is the same between both but the volume flow rate will change because of the valve. The pump has a fixed mass flow rate at a given condition but the valve changes the condition. If the valve is of the thermostatic type then this follows.
"The purpose of the thermostatic valve is to control the rate at which the refrigerant passes from the liquid line into the evaporator and to keep the pressure difference between the high and low pressure sides of the refrigeration system."<a href="https://www.sciencedirect.com/topics/engineering/expansion-valve#:%7E:text=An%20expansion%20valve%20is%20basically,orifice%20is%20now%20a%20low" rel="nofollow noreferrer">reference</a></p>
| 58360 | Refrigerant flow |
2023-12-16T20:33:02.527 | <h2>electrostatic whiteboards</h2>
<p>I have a bunch of these special dry erase sheets. Someone gave them to me as a gift about a year ago; I see they are still available for purchase on the Internet — you can find them if you look for something like <em>«static cling dry erase sheet»</em> or <em>«electrostatic whiteboard»</em>; they are available from different companies and, as far as I can tell, there is no single standard name for them. I am going to call one such an <em>«electrostatic whiteboard»</em> further in this question.</p>
<p>So, an electrostatic whiteboard is a flexible rectangular sheet, about 0.5 mm thick and about 1 m² in surface. Since it is light, transparent and warm to the touch, I assume it is made out of some kind of a plastic. The magic of it is that, put against a wall, it sticks with enough strength to hold its own weight and then some, but remains easy to peel off, stick again and so on, without any visible changes to either the wall or the electrostatic whiteboard.</p>
<p>Sadly, after about a year of being stuck to a wall and serving me faithfully, my electrostatic whiteboards slid down one after another and lost their stickiness. Even though they are not that expensive, I loathe to buy new ones since it is not environmentally conscious. Rather, I should like to restore the electrostatic whiteboards I already have to their former glory.</p>
<h2>my theory</h2>
<p>My understanding is that electrostatic whiteboards come strongly charged with static electricity. They are made of a strong electrical insulator, so it takes a long time for local surplus or deficit of electrons to <em>«escape»</em> to the surface and dissipate. The wall, meanwhile, is covered with paint that is, I guess, dielectric — so, upon coming near it, the electrically charged electrostatic whiteboard polarizes the surface of the wall into carrying the opposite charge and sticks to it thanks to the electrostatic force. However, over time electrons move through the material of the electrostatic whiteboard and it loses its charge. No charge — no stickiness.</p>
<h2>my question</h2>
<p>So, how do they charge electrostatic whiteboards at the factory?</p>
<p>I tried rubbing mine against the carpet but it does not help. I tried rubbing a thin polyethylene bag against the carpet and it does stick to the wall, but only weakly — it falls down under its own tiny weight within a minute. So, rubbing sheets of plastic against the carpet does not seem to be an efficient way to electrically charge them.</p>
<p>Should I take a more fluffy carpet? Should I rub more intensely, or for longer? Do I need to charge both sides of a sheet? Should I apply strong enough electricity that the electric charge soaks through the whole depth of the sheet despite its very low electron mobility?</p>
<p>Is my theory overall right or wrong?</p>
| |electrical-engineering|materials|statics| | <p>Despite its name, a "static cling" or "electrostatic" whiteboard sticks to a wall because of molecular adhesion and air pressure, not static electricity, so trying to recharge them will not revive them. What might help is carefully cleaning both the wall and the wall-facing side of the whiteboard.</p>
<p>This is because when you place the whiteboard against a wall, areas of the sheet that touch the wall "stick" via <a href="https://en.wikipedia.org/wiki/Intermolecular_force" rel="nofollow noreferrer">intermolecular forces</a>.
These are very short-range, but the area of the sheet is large and these weak molecular forces are very strongly reinforced by air pressure.
Sea level <a href="https://www.digikey.ca/en/resources/conversion-calculators/conversion-calculator-pressure" rel="nofollow noreferrer">air pressure over just one square centimetre</a> is sufficient to support a kilogram weight, so it doesn't take much direct contact area for air pressure and friction to hold the whiteboard sheet in place.
Any dirt, dust, film, wrinkles, roughness, distortion, … that interferes with direct wall-whiteboard contact will reduce stickiness.</p>
<p>Static electricity is not, however, completely irrelevant. One form of molecular adhesion is caused by the different <a href="https://en.wikipedia.org/wiki/Electron_affinity" rel="nofollow noreferrer">electron affinities</a> of different materials. When you peel the sheet away from the wall, you are likely to peel away some electrons or leave them behind, charging the sheet relative to the wall.
This <a href="https://en.wikipedia.org/wiki/Triboelectric_effect" rel="nofollow noreferrer">contact electrification</a> may cause some temporary electrostatic forces, but any macroscopic electrostatic charge should not last long, since ions in the air move to neutralize the charge. This is why if you rub a balloon in your hair and stick it to a wall, it will typically fall down within a few hours.</p>
<p>Static cling decal companies explain this much better than any static cling whiteboard vendor I can find:</p>
<ul>
<li><a href="https://www.letteringonthecheap.com/blog/2016/02/03/the-science-of-static-cling-decals/" rel="nofollow noreferrer">The Science of Static Cling Decals</a></li>
<li><a href="http://www.plctx.com/blog/an-introductory-guide-for-static-cling-labels/" rel="nofollow noreferrer">An Introductory Guide For Static Cling Labels</a></li>
</ul>
<p><a href="https://www.tekra.com/resources/tek-tip-white-paper/tek-tip-static-cling-versus-low-tack-adhesive-window-applications" rel="nofollow noreferrer">One cling film sheet manufacturer says</a> that their product becomes less sticky with time because "the effect of the plasticizers relax". (I am not exactly sure what this means, but my guess is that the sheet becomes less flexible so it can't mould itself to the surface to make good contact.) So you may be out of luck if your whiteboards have aged beyond their expected lifetime.</p>
| 58377 | How do I electrically charge a sheet of plastic strongly enough that it firmly sticks to the wall? |
2023-12-18T12:22:23.660 | <p>I have a client in Belgium that has 16 DM400-M10-54HBB on his roof. On Saturday the 2nd of December 2023, the PV output was 3.95 kW, according to the log of the Huawei SUN2000 4.6KTL-L1 5 kVA inverter. This means 247 W per panel.</p>
<p><a href="https://i.stack.imgur.com/UKMAi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UKMAi.png" alt="Power on 2/12/2023" /></a></p>
<p>This is the house, with the panels directed to the south-east:</p>
<p><a href="https://i.stack.imgur.com/80RzO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/80RzO.jpg" alt="PV panels" /></a></p>
<p>However, the irradiance was only 250-300 W/m<sup>2</sup> that day, according to multiple measurements accross the country (clear blue skies according to my naked eye). This is to be expected because we only get around 1100-1200 W/m<sup>2</sup> in summer and now it is winter.</p>
<p><a href="https://i.stack.imgur.com/OUne9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OUne9.jpg" alt="irradiance in Beerse" /></a></p>
<p><a href="https://i.stack.imgur.com/s3dAy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s3dAy.png" alt="irradiance in Stabroek" /></a></p>
<p>However, according to the datasheet, there is approximately 600 W/m<sup>2</sup> needed to produce this amount of power (yellow line is drawn by me to indicate the produced peak power on that day). How is this possible?</p>
<p><a href="https://i.stack.imgur.com/9HKo6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9HKo6.png" alt="Datasheet power curves" /></a>
Datasheet source: <a href="https://www.elektrototaalmarkt.nl/amfile/file/download/file/3868/product/41786/" rel="nofollow noreferrer">https://www.elektrototaalmarkt.nl/amfile/file/download/file/3868/product/41786/</a></p>
<p>The data in the Huawei logs seems accurate..</p>
| |electrical-engineering|solar-energy|photovoltaics| | <p>The answer from fromwastetowind is correct that solar irradiance is published for a flat horizontal surface. I entered the time, latitude, and logitude for Brussels into the <a href="https://gml.noaa.gov/grad/solcalc/azel.html" rel="nofollow noreferrer">NOAA ESRL Solar Position Calculator</a>. This gave a solar elevation of 16.87 degrees. Not knowing your customer's panel angles I just assumed a perfect angle 90 degrees to the incoming radiation (the best possible production). The best way to think about this calculation is that 0.29m^2 of solar radiation is spread out over the 1m^2 that the pyronometers measured.</p>
<p>The 861 watts per meter squared is a reasonable number because the radiation has to travel through more atmosphere at the high angle. And the roughly 600 watts per meter squared you figured from the spec sheet is reasonable for the not perfect angle on the fixed array.</p>
<p><a href="https://i.stack.imgur.com/4Zd7jm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Zd7jm.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/biUa0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/biUa0.png" alt="enter image description here" /></a></p>
<p>Also note that in some cases mixed cloud cover (like shown in the OP photo) can cause an increase in local irradiance. The light that was headed to other locations on the earth's surface gets diffused and a portion of that light can end up at the solar array being analyzed. This can be a big problem for closed loop solar hot water systems and additional safety factory has to be considered when sizing the thermal load dump. <a href="https://www.pv-magazine.com/2020/07/14/solar-causes-highest-power-peaks-under-mixed-cloud-conditions/" rel="nofollow noreferrer">Article on mixed cloud cover irradiation spikes</a>.</p>
| 58392 | PV solar panel is producing more than is possible according to its datasheet? |
2023-12-18T20:37:34.520 | <p>I have a bucket that I want to fill with water only up to a certain level, the bucket is indoors, the first thing I thought of was a siphon like the one in a toilet but would prefer a smaller mechanism. I am not very well versed in these things, so is there something that I can affix to a hose which will stop the water flowing past a certain level?</p>
| |fluid-mechanics|hydraulics|fluid|flow-control| | <p>Try and get ahold of a gasoline pump handle, a so-called "fill nozzle". Those shutoff when enough of the nozzle becomes submerged. Uses a clever mechanism of air pressure and levers so no electronic sensors required. Can even be locked open so you don't have to be there. What more could you ask for?</p>
<p>The moving gasoline produces a venturi effect that sucks in air through a narrow channel that runs along the water nozzle. So long as this air channel remains unobstructed (for example, by gasoline plugging it because the gasoline level has risen high enough), it causes a pressure change on a membrane against atmospheric pressure that is coupled to a spring loaded latch.</p>
<p>This spring loaded latch, the hand lever, and the valve that allows gasoline to flow are all coupled to the same lever at three different points, respectively. The fulcrum of this lever is actually the spring loaded latch and it serves to anchor the lever so the hand lever can pull open the gasoline valve. However, when the air channel is blocked off, the pressure on both sides of the membrane are such that the spring loaded valve cannot properly latch. No latching means the fulcrum isn't anchored at any point which means the entire lever is floating and the hand lever has no leverage to pull the gasoline valve open.</p>
<p><a href="https://www.youtube.com/watch?v=fT2KhJ8W-Kg" rel="nofollow noreferrer">https://www.youtube.com/watch?v=fT2KhJ8W-Kg</a></p>
| 58398 | How can I stop water flow in a hose after a certain water level? |
2023-12-19T23:54:38.573 | <p>What is the meaning of the term 'recoverable ore' when used in contexts relating to mining and the extraction of resources/minerals?</p>
<p>For example, several academic papers use the term 'recoverable ore' but it is typically not defined.</p>
<p><a href="https://link.springer.com/article/10.1007/s11004-013-9478-x" rel="nofollow noreferrer">Example paper 1.</a>
<a href="https://link.springer.com/chapter/10.1007/978-3-319-39264-6_22" rel="nofollow noreferrer">Example paper 2.</a>
<a href="https://link.springer.com/article/10.1007/s11053-013-9225-5" rel="nofollow noreferrer">Example paper 3.</a></p>
| |terminology|mining-engineering| | <p>To begin with, the titles and abstracts of the papers you refer to do not mention recoverable ore. They mention recoverable reserves or recoverable resources.</p>
<p>Resources, reserves and ore are not the same thing.</p>
<p>Generally, geologists discovery a mineral deposit and they will do an assessment that will determine the amount of mineralization or metal in the deposit. They do not apply mining and economic parameters to their analysis, they just determine a total mass of material according to three categories: measured, indicated and inferred. The result of their assessment is called a <strong>resource</strong>.</p>
<p>An <strong>inferred resource</strong> has the least degree of confidence. It's usually based on wide spaced drilling were the drill holes are least 50 meters apart.</p>
<p>An <strong>indicated resource</strong> has a good degree of confidence that it is as interpreted by geologists. It's based on closer spaced drilling, usually 20 or 25 meters. The complexity of the deposit also plays a factor and a more complex deposit may require closer spaced drilling.</p>
<p>A <strong>measured resource</strong> has the highest degree of confidence. It is based on drilling that is spaced at most 20 to 25 meters apart, the same as for an indicated, but the geologists have additional information that gives them greater confidence regarding the deposit. Such a classification usually applies to resources for a deposit that is being mined. For an underground mine, not only does the minerlization need to have closer spaced drilling, it also needs to be exposed on at least threes side by underground development, such as drives (drifts or tunnels) and raises. Closely drilled mineralization in the walls of a pit can also be classified as being measured. However, even with such exposure, pit or underground, if the deposit is complex of the geologists have doubts, such as resource can be downgraded to indicated status.</p>
<p>The conversion of resources to reserves and hence ore status, is done by mining engineers. Note, only measured and indicated resources can be converted to reserve status. Indicated resources have insufficient confidence to considered.</p>
<p>Mining engineers analyze the resource and geological models given to them by geologists. For open pit mines they use an open pit optimizer to obtain an initial pit shell to which they will design a pit that is practical to mine. Economic parameters which include mining costs, the value of the commodities being mined, processing costs and recoveries and marketing costs are applied to the material within the pit design to determine what is ore and can be categorized as part of the reserve. Material that is on no value will be classified as waste and mineralization that is marginal may be classified as low grade and stockpiled separately. If commodity prices improve later, such material may be recovered from the stockpile and processed. If this happens, such material is then reclassified as ore, and part of the reserve. For underground mines the mining engineers will complete a design for both development and stopes and then apply economic parameters to the stopes and development to determine which stopes, and hence, which parts of the mineral deposit are economic to mine. They are then classified as being part of the reserve and hence ore.</p>
<p>In short: geologists defined resources (size, shape, orientation and distribution of mineral deposit with no economics applied), mining engineers define reserves (economics applied: mine design, mineral processing and commodity value).</p>
<p>To your question, recoverable resources are parts of the resource that have the potential to be mined for a profit. Recoverable reserves, or ore, are parts of the reserve that can be mine profitably.</p>
| 58416 | What is the meaning of the term 'recoverable ore'? |
2023-12-20T01:40:19.120 | <p>I have found a part in an assembly which has been saved internally, while the rest are external. How can I turn this part into an external one? I need all the references to stay though.</p>
<p>Also, should a SLDPRT file be showing up in the project folder if it is an internal part? I'm wondering if it's possible that it got saved externally somehow without getting rid of the "^assemblyname" bit.</p>
| |cad|solidworks| | <ol>
<li>To save a Virtual Part externally, just right click it in the history tree, and click "Save Part(in External File)". If you see "Make Virtual" in the same position in the menu, then the part is already saved externally.</li>
</ol>
<p><a href="https://i.stack.imgur.com/Q5T7Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q5T7Z.png" alt="Save Part(in External File)" /></a></p>
<ol start="2">
<li>Using the process above should remove the ^assemblyname, but there are other ways you could have created a .SLDPRT file. If you have a virtual part, click "open part", and then File, Save As, you can save it. There will be a popup warning that looks like the image below. You can see that this includes the ^assemblyname. If you have ...^assemblyname.SLDPRT in your folder, the most likely scenario is that you did this, and your part isn't virtual at all any more, it's just got that filename.</li>
</ol>
<p>: <a href="https://i.stack.imgur.com/ysAIo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ysAIo.png" alt="virtual component references warning" /></a></p>
<ol start="3">
<li>If the above is true, and you just need to rename the part to remove ^assemblyname (without breaking references), then you should right click on the part using windows explorer, and rename using the tool in the SOLIDWORKS sub-menu. Do this when the parts/assemblies are not open.</li>
</ol>
<p><a href="https://i.stack.imgur.com/mwihE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mwihE.png" alt="SW Rename Sub Menu" /></a></p>
| 58418 | How can I make an internal part of a Solidworks assembly become external? |
2023-12-22T21:13:55.950 | <p>When the system is first order:</p>
<p><span class="math-container">$1/(1+s\tau)$</span></p>
<p>Why is the bandwidth: <span class="math-container">$f_{-3db}=1/(2\pi \tau)$</span>?</p>
<p>Where did the <span class="math-container">$2\pi$</span> come from? Why isn't it simply <span class="math-container">$1/\tau$</span>? The unit of <span class="math-container">$2\pi$</span> is related to radian. But the unit of <span class="math-container">$f_{-3db}$</span> is <span class="math-container">$[1/second]$</span> so unit of <span class="math-container">$1/(2\pi\tau)$</span> does not seem to equate to Hz</p>
| |control-engineering|signal-processing|circuit-design| | <p>Unit of frequency is <em>not</em> really <span class="math-container">$1/\mathrm{second}$</span>.</p>
<p><span class="math-container">$$
1\mathrm{Hz} = 1 \frac{\mathrm{cycle}}{\mathrm{second}}=360\frac{\mathrm{degrees}}{\mathrm{second}}=2\pi\frac{\mathrm{radians}}{\mathrm{second}}
$$</span></p>
<p><em>cycle</em>, <em>degrees</em>, <em>radians</em> are kind of <em>unitless</em> since they all represent <em>a fraction</em> of a whole.</p>
<p><sup>The <em>unit</em> of <code>radians</code> is kind of <em>unitless</em> since angle in radians is found by <span class="math-container">$\frac{\mathrm{arc length}}{\mathrm{radius}}$</span></sup></p>
<p><em>Natural</em> unit of frequency is <code>radian/s</code>.
<span class="math-container">$$
\left|\frac{1}{1+s_{\text{(-3 dB)}}\tau}\right| = \frac{1}{\sqrt2}\\
\implies 1+\omega_{\text{(-3 dB)}}^2\tau^2=2
\qquad (\text{letting }s = j \omega)\\
\implies \omega_{\text{(-3 dB)}} = \frac{1}{\tau}
$$</span>
whre <span class="math-container">$\omega_{\text{(-3 dB)}}$</span> is the bandwidth frequency. Its unit is <code>radians/second</code>. To convert to <code>hertz</code>,
<span class="math-container">$$
f_{\text{(-3 dB)}}=\frac{\omega_{\text{(-3 dB)}}}{2\pi}
$$</span></p>
| 58441 | Converting time constant to bandwidth of 1st order system |
2023-12-25T16:27:44.490 | <p>I have a spindle which is about 200 mm in diameter. I want to indicate where the axis of the spindle in pointing. I can attach any laser pointer etc. on the circumference of the spindle.</p>
<p>Is there any optical method using prisms, mirrors etc. to indicate the center of the spindle on the bed, irrespective of the height and angle of the spindle head?</p>
| |mechanical-engineering| | <p>This could be done using a pair of line-generating lasers. Mounting them 90° apart on the spindle would give a '+' crosshairs on the bed of the machine. The crosshairs would rotate as the spindle rotates.</p>
<p><a href="https://i.stack.imgur.com/ZtSeS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZtSeS.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. Laser crosshair generator. (Image is original.)</em></p>
<p>Something similar is used on many drill presses and for simplicity these are mounted on the press frame rather than on the spindle.</p>
| 58452 | Showing where the axis of a spindle is pointing |
2023-12-26T19:22:00.410 | <p>Well, while searching for dielectric elastomer actuators I can find that they normally have efficiencies around 80% or more (<a href="https://escholarship.org/content/qt7kf261zf/qt7kf261zf_noSplash_e480d01e957c9dd7a9e06b13b7356899.pdf" rel="nofollow noreferrer">source</a> page 38), they are cheap and easy to build.</p>
<p>But the more I search about them, the more I notice that I can't find projects more complex than proofs of concept and others uses, like microfluidic pumps.</p>
<p>Why is this lack of interest on actually using this class of actuators on more elaborated robots? What I'm missing?</p>
<p><a href="https://i.stack.imgur.com/8exqC.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8exqC.gif" alt="Artificial muscles using dielectric elastomers" /></a></p>
<p>This is the only example that I could find (<a href="https://www.pnas.org/doi/full/10.1073/pnas.1815053116" rel="nofollow noreferrer">source</a>).</p>
| |electrical-engineering|robotics|actuator| | <p>Preface: I am not an expert in Dielectric Elastomeric Actuators, but as a mechanical engineer I operate a small prototyping business that brings new technologies to market. Hopefully this can give you a manufacturer's perspective into some of the difficulties that I see in developing and profiting from this technology.</p>
<p>Currently, Dielectric Elastomeric Actuators (DEAs) are likely not commercially available because of a combination of several of the following factors:</p>
<ol>
<li>DEAs have a dangerously high working voltage. Even on a 2022 <a href="https://onlinelibrary.wiley.com/action/downloadSupplement?doi=10.1002%2Fadma.202106757&file=adma202106757-sup-0001-SuppMat.pdf" rel="nofollow noreferrer">improved 10nm MIT design</a> they still show voltage ranges between 300V and 1800V. While these voltages can be made safe for consumer equipment such as in florescent lights, the engineering, fail-safes, certifications, and liabilities make for a high barrier to entry into the market. If the voltage can be brought below 48V (perhaps with thinner layers), the safety risk and electrical certification is much less stringent.</li>
<li>DEAs require complex manufacturing including spin coating, vacuum bubble removal, nano meter level precision, thin graphene or carbon nanotube electrodes, unknown electrode application techniques, electrical connection to the electrodes, high-voltage sinusoidal IGBT driver circut, dielectric breakdown detection for safe failure, and lots of other details unknown at this time. As graphene and carbon nanotube production and applications mature there will be less manufacturing hurdles to overcome.</li>
<li>DEAs require prestretch to prevent <a href="https://softroboticstoolkit.com/book/dea-modeling" rel="nofollow noreferrer">Electromechanical instability</a>. Having two different displacements for a given voltage can cause the actuator to unpredictably move between the two creating very high accelerations and damage the elastomer and or electrode causing failure of the actuator. I did not find a satisfactory explanation of the reason for the initial force/voltage peak, but my understanding is that the elastomer chains are not initially lined up and lining them up ahead of time ensures that there is not a high force required for an initial displacement. While prestretching simple enough to do in theory, this is another mechanical engineering hurdle to overcome with every physical implementation of the actuator. In the future this hurdle could potentially be addressed with chemistry and cure processes to create directional cross-linking in the unloaded state.</li>
<li>Already proven <a href="https://en.wikipedia.org/wiki/Amplified_piezoelectric_actuator" rel="nofollow noreferrer">amplified piezoelectric actuators</a> fill many of the niche uses for DEAs. Piezo materials suffer from a lot of similar issues such as high voltage and preload required. The big difference is that this technology has already been developed and many of the engineering hurdles have been overcome. While being "soft" may provide a sufficient niche in many future applications, currently an off-the-shelf piezo actuator fills almost any need for a DEA.</li>
<li>In rapidly developing fields it is always important to consider that progress is slowed by <a href="https://patents.google.com/?q=(dielectric%20elastomer)&oq=dielectric%20elastomer" rel="nofollow noreferrer">patents</a>. Both by the researchers taking time to acquire patents, everyone having to perform a detailed patent searches to know if they can participate, and everyone avoiding key patent claims that restrict the best methods. Once a viable path to profit is realized, then patents accelerate development by encouraging the investment of capital, but until that point they are antagonistic. 3d printing exploded in popularity and development only after key patents expired.</li>
<li>Just speculating here, but a lot of the scientists that would be experts in further developing DEAs are being pulled towards developing other highly financially incentivized technologies such as new batteries, fuel cells, gas sensors, ultra capacitors and other leading edge electrode technologies.</li>
<li>There are actually <a href="https://softroboticstoolkit.com/actuators" rel="nofollow noreferrer">lots of competing soft actuator technologies</a> that I was unaware of. These dilute the focus on advancing one technology like DEAs and may ultimately have superior characteristics for applications that are not shadowed by piezo actuators.</li>
</ol>
<p>References:</p>
<ol>
<li>The <a href="https://www.electronicdesign.com/technologies/analog/article/21213099/electronic-design-elastomeric-actuators-lift-aerial-micro-robot" rel="nofollow noreferrer">Elastomeric Actuators Lift Aerial Micro-Robot</a> article
DKNguyen posted in the comments.</li>
<li>The bottom of the Elastomeric Actuators Lift Aerial Micro-Robot article also has a <a href="https://onlinelibrary.wiley.com/action/downloadSupplement?doi=10.1002%2Fadma.202106757&file=adma202106757-sup-0001-SuppMat.pdf" rel="nofollow noreferrer">“Supporting Information” link</a> that includes the technical details of how they were able to improve the performance.</li>
<li>First article from the OP: <a href="https://escholarship.org/content/qt7kf261zf/qt7kf261zf_noSplash_e480d01e957c9dd7a9e06b13b7356899.pdf" rel="nofollow noreferrer">Dielectric Elastomers for Actuation and Energy Harvesting</a></li>
<li>Second article from the OP: <a href="https://www.pnas.org/doi/full/10.1073/pnas.1815053116" rel="nofollow noreferrer">Realizing the potential of dielectric elastomer artificial muscles</a></li>
<li><a href="https://softroboticstoolkit.com/about" rel="nofollow noreferrer">Soft Robot Toolkit</a> looks like an excellent resource for developing prototypes.</li>
<li><a href="https://en.wikipedia.org/wiki/Dielectric_elastomers" rel="nofollow noreferrer">Dielectric Elastomers Wikipedia Article</a></li>
<li><a href="https://en.wikipedia.org/wiki/Soft_robotics" rel="nofollow noreferrer">Soft Robotics Wikipedia Article</a></li>
</ol>
| 58459 | Why aren't Dielectric Elastomer artificial muscle actuated robots more common? |
2023-12-27T15:36:14.733 | <p>Trying to figure out if this has a mechanical advantage on moving a 600 pound car faster with less effort than just two using 2 gears<a href="https://i.stack.imgur.com/gSrFK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gSrFK.jpg" alt="enter image description here" /></a></p>
| |gears| | <p>The only advantage of the lower arrangement is that it allows the motor sprocket to be offset from the 12:72 tooth final drive chain.</p>
<p>The 14:14 ratio changes nothing other than introduce additional friction losses on the chain and shaft bearings as well as some additional weight and complexity.</p>
| 58465 | What is the mechanical advantage of this chain drive arrangement? |