CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2018-01-07T22:23:57.260 | <p>I'm building a machine that operates over 20 orders of magnitude with logarithmic process characteristics. I have tried regular PID and also all sorts of variations of PID and did not get good results. What other alternatives to PID I have?</p>
<p>I start with 21% oxygen in a closed vessel (at high temperature) and pump it out ion by ion down to log10(pO2)=-20 of original amount. I have sensitive enough sensor, pump and power supply for it. This part all works great if I set pump currents by hand or babysit the process. To go up in concentration I just lower the pump current and leakage takes care of adding oxygen.</p>
<p>But as you see, in context of trying to automate the control with PID, what works well at one region is completely unsuitable at another region. It does not help that the full loop of measurements takes about 10 seconds. </p>
<p>I have limited the current used to 500mA, and at worst conditions the current needed to cope with the leakage is 700µA, so the range for the current is not so huge. Nor is the range for sensor voltage, which is about 100mV to -1V.</p>
<p>Maybe I could control pump voltage instead of pump current. Current results directly to change in amount of oxygen (which is logarithmic). The resistance of the pump depends on temperature, and most of all the oxygen gradient, and would yield much more linear and automatically regulated response.</p>
| |pid-control| | <p>An alternative to PID is Model Predictive Control (MPC) <a href="https://en.wikipedia.org/wiki/Model_predictive_control" rel="nofollow noreferrer">[1]</a>, <a href="https://www.sciencedirect.com/science/article/pii/S0957417417306000" rel="nofollow noreferrer">[2]</a>, <a href="http://www.uni-obuda.hu/users/szakala/SMC%202016%20pendrive/1139_smc2016.pdf" rel="nofollow noreferrer">[3]</a>, <a href="https://web.stanford.edu/class/ee364b/lectures/mpc_slides.pdf" rel="nofollow noreferrer">[4]</a>.</p>
<p>As your system is highly nonlinear, I would recommend using Nonlinear Model Predictive Control (NMPC) <a href="http://control.ee.ethz.ch/~apnoco/Lectures2015/07-Nonlinear%20Model%20Predictive%20Control.pdf" rel="nofollow noreferrer">[5]</a>, <a href="http://www.springer.com/gp/book/9780857295002" rel="nofollow noreferrer">[6]</a>, <a href="https://pdfs.semanticscholar.org/4238/9407bc489b9d42a3a3d9a00cc45e44dc79ea.pdf" rel="nofollow noreferrer">[7]</a>.</p>
<p>Model predictive controller, predicts the future dynamic of the system and takes action before hand. This controller performs an online optimization at each sampling time and handles the system constraints explicitly. </p>
| 18769 | Tricks for controlling process with logaritmich characteristics over 20 orders of magnitude |
2018-01-08T04:56:52.193 | <p>I've come across this phenomenon several times and never found any real reason for this.</p>
<p>When I consider a cylindrical object like a rod, I need to calculate several things. One of those is the stress tensor in cylindrical coordinates represented by $r$, $\theta$, and $z$ directions.</p>
<p>Now, in many books and in my lecture notes there's always this one line that says:</p>
<p>$$ \frac{\partial\bullet}{\partial\theta}=0$$</p>
<p>I know that when compressing a steel rod (assuming there is little to no friction at the contact points) the steel rod should get compressed in the $z$-direction and expand in the radial direction.
<strong>Why is there no stress in the circumferential direction?</strong></p>
<p>I've looked in many places and I hope that I didn't accidentally miss the other SE question with respect to this specific topic.</p>
| |mechanical-engineering| | <p>1) axisymmetry requires symmetry of the <em>entire problem</em>. The body and the loading and the material properties. For the bending case the boundary condition is not axi-symmetric and so you can not use an axisymmetry assumption to simplify the governing equations.</p>
<p>2) zero circumferential stress <em>does not follow</em> from axisymmetry. It happens to be so for this simple problem. You have zero transverse stress in a square bar in axial load too.</p>
<p>As a trivial example consider a cylinder under uniform pressure P over the entire surface. All normal stress components are equal to P, so clearly sigma_theta is not zero.</p>
| 18773 | Why is the partial derivative of stress in the circumferential direction equal to zero in axial symmetric bodies? |
2018-01-08T13:25:25.440 | <p>I am used to the metric system where everything is simple and consistent. I can do quick conversions between metric and imperial in my head, but especially when it comes to volumetric measures, I cannot be sure if British or US Customary units are used. Here I refer to units such as gallons, pints, etc. not cubic-linear units.</p>
<p>Questions arising from this confusion is how to know which unit is being used.</p>
<p>In technical publications, will the author or publisher note the applicable units somewhere?</p>
<p>For everyday use, is there a convention that says U.S. gallons or otherwise must be specified, or is it up to the whims of the manufacturer? If not, how can I reduce my chances of making an error if I cannot ask or measure?</p>
| |measurements| | <p>According to a quick scan of <a href="https://en.wikipedia.org/wiki/Imperial_units#Current_use" rel="nofollow noreferrer">Wikipedia</a>, no country actually officially uses imperial units any more. There are plenty of <em>casual</em> uses of imperial (for example, I live in the UK and measure my height in feet and inches, my weight in stone and pounds, and my milk and beer in pints) but no <em>technical</em> uses that I know of.</p>
<blockquote>
<p>In technical publications, will the author or publisher note the applicable units somewhere?</p>
</blockquote>
<p>In technical publications from the US, US Customary units will be used. In technical publications from the UK, metric units will be used. </p>
<p>So this should work for the vast majority of cases. i.e. pints, gallons etc. are US ones. The main caveat is the publication date: the further back in time you go, the more likely it is that imperial units could have been used (see the Wikipedia article for further detail).</p>
| 18780 | Is there a way to distinguish between U.S. Customary and British Imperial volumetric units? |
2018-01-09T21:39:39.337 | <p>From an engineering perspective, what limits the maximum speed you can reach with a regular car? I understand that some of the faster cars are for safety reasons limited to not run faster than say 250 km/h, but that's not my question.</p>
<p>I can think of several reasons, but not sure which of these is relevant:</p>
<ol>
<li>Is the limit set by some part (which?) breaking if I increase the rpm, as suggested by red marks on rpm meters?</li>
<li>Or is it rather that you cannot get in fuel fast enough to keep increasing the rpm?</li>
<li>Or is it that friction/drag increases as you speed up and the engine cannot overcome this as it can only generate a maximum amount of force/torque? If yes, what does this amount of torque/force depend on?</li>
</ol>
| |mechanical-engineering|car| | <p>To find the purely theoretical limit depending totally on limits of the rolling resistance of the tire and air drag. So here is the work.</p>
<h2>Formulas</h2>
<ul>
<li><span class="math-container">$F_d = \dfrac{1}{2}pA\mu_d v^2$</span>: this is the formula for drag force.</li>
<li><span class="math-container">$f_s = \mu_s mg$</span>: this is the formula for static friction force.</li>
</ul>
<p>NOTE: I am using values of a Tesla model 3.</p>
<h2>Known values</h2>
<ul>
<li>A = 0.23 m<sup>2</sup></li>
<li>g = 9.80665 m/s<sup>2</sup></li>
<li>m = 1611 kg</li>
<li><span class="math-container">$\mu_s$</span> = 0.7</li>
<li><span class="math-container">$\mu_d$</span> = 0.23</li>
<li>p = 1.2754 kg/m<sup>3</sup></li>
</ul>
<h2>Assumptions</h2>
<ul>
<li>Car is perfectly perpendicular to the force of gravity.</li>
<li>Car is at sea level.</li>
<li>Conditions are ideal.</li>
<li>The power of the engine is infinite.</li>
</ul>
<h2>Results</h2>
<p><span class="math-container">$V_{theo} = 1280.9\text{ mph}$</span></p>
| 18806 | What limits the speed of a car? |
2018-01-10T20:32:45.410 | <p>Assume you can use any whistle type, is there a way to choose the pressure point at which a whistle makes noise? For example I have a whistle that starts making noise at 2 psi but I would like to lower that to 1 psi. Would I change the area of the inlet, outlet? what feature would you change to accomplish this?</p>
| |fluid-mechanics|pressure|airflow| | <p>The most serious and comprehensive information about whistles is provided by the folks who work on pipe organs: expensive instruments which endure critique by serious listeners, and which require maintenance where they are installed because of their size. A pipe organ's <strong>Flue Pipes</strong> are in fact no different from whistles.</p>
<p>A whistle is a noisemaker that uses no reed. In this sense, we might say that a brass player's lips and a singer's vocal cords are <em>reeds</em>, while a flute player is playing a kind of <em>whistle</em>.</p>
<p>A whistle has two components: a resonator and an air supply. The resonator's opening may be called its mouth, and the air supply must flow freely past the resonator's mouth in order for it to work.</p>
<p>The air flows over the resonator's <em>mouth</em> and sucks out a bit of pressure from the resonator by the Bernoulli principle. Once this pressure is lower inside the resonator, the free air supply flow changes course and starts packing into the resonator, increasing its pressure. This sends a high pressure wave through the resonator which may either bounce back (for a closed resonator) or depart (for an open resonator).</p>
<p>At some moment in time while the air is flowing into the resonator, the pressure inside increases enough to divert the airflow away from the mouth of the resonator. Once the flow is diverted away from the resonator, it begins to remove air from inside and lower the pressure, creating another pressure wave inside. Now the cycle repeats.</p>
<p>The sound you hear originates both from the opening at the top of the resonator and an opening at the bottom (for open resonators) where there is this fluctuating air pressure.</p>
<p>The sound is at a tuned pitch (or frequency) that is the result of the pressure wave bouncing from top to bottom, bottom to top in the resonator. The frequency is a simple matter of length with flue pipes, but I can't even explain what's going on in an ocarina.</p>
<blockquote>
<p>How can I reduce the pressure of my whistle?</p>
</blockquote>
<p>In order to make a sound, your instrument must initiate the aforedescribed cycle, and also convert the air source's energy into acoustic energy. Improving either (by initiating sound earlier or converting energy more efficiently) may enable your instrument to use less pressure. Here are some conditions which may affect either, in order of my guess for the most important first:</p>
<ul>
<li><strong>Velocity of air supply</strong> (high velocity = earlier initiation of tone)</li>
<li><strong>Cross-sectional area of resonator near its mouth</strong> (smaller cross-sectional area = earlier initiation of tone)</li>
<li><strong>The resonator being stopped or open</strong> (stopped = earlier initiation of tone)</li>
<li><strong>Volumetric flow of air supply</strong> (high flow = more volume)</li>
<li><strong>Length along the flow direction of the resonator "mouth"</strong> (must be optimized)</li>
<li><strong>Length across the flow direction of the resonator "mouth"</strong> (must be optimized)</li>
<li><strong>Distance that the air supply travels freely</strong> (must be optimized)</li>
<li><strong>Laminarity of the air supply</strong> (unknown effect by me whether more or less sound)
<ul>
<li>You might increase the laminarity by passing the air supply through a long, straight tube, for instance.</li>
</ul></li>
<li><strong>Intended pitch of the resonator</strong> (higher pitch = more sound, to a point)</li>
<li><strong>Length of the resonator</strong> (shorter length = more sound, to a point)</li>
<li><strong>Atmospheric condition</strong> (ambient wind may "kill" the free flow of the air supply)</li>
<li><strong>Cross-sectional area of the resonator's other opening for open resonators</strong> (bigger rear opening = more volume)
<ul>
<li>There's a reason tubas, gramophones, Alpenhorns, and megaphones have a flared end: the smaller volumetric velocity produced by the large cross-sectional area enables more sound energy to transfer into the open air, making the best use of the player's effort. This is called <strong>impedance matching</strong>.</li>
</ul></li>
</ul>
<p>As suggested elsewhere, dividing your airflow among multiple resonators may solve your issue. Good luck!</p>
<p><a href="http://hardmanwurlitzer.com/pipes/" rel="nofollow noreferrer"><em>Further reading on pipe organ flue pipes ("whistles")</em></a></p>
| 18834 | Whistle Design - Optimizing sound initiation pressure and volume |
2018-01-11T09:34:56.407 | <p>I have little/no experience with pipe networks and have been searching for a methodology to calculate the discharge from multiple outlets of an oil pipe network with a single inlet. As far as i can tell I have too many unknowns to use something like the Hardy Cross method (and often i don't have loops, just multiple branches and outlets as shown below), and Hardy-Cross seems to be used to find flows in pipes with known inflows and outflows, which I don't have. </p>
<p>A good example may be something like this: </p>
<ul>
<li>Inlet flow rate is known </li>
<li>Fluid Properties are known</li>
<li>Geometry of all pipes/bends/outlets are known</li>
<li>The height above a datum of all pipes and outlets are known</li>
<li>Flow is each pipe branch is UNKNOWN</li>
<li>Discharge from each outlet is UNKNOWN (ext. pressure can be assumed ambient)</li>
<li>I have target outflows, and could modify pipe length/diameters to obtain them</li>
</ul>
<p>See my very crude diagram: <a href="https://i.stack.imgur.com/Z5Bog.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z5Bog.png" alt="enter image description here"></a></p>
<p><strong>Can anybody suggest an analytical method I can use to calculate the outflows and/or optimise the pipe lengths to obtain a set of target outflows?</strong> </p>
<p>A worked example of a similar problem would also be appreciated. </p>
<p>The person who tried to solve this problem before me just divided the inflow up based on the area of the outlets, without considering losses, bends or even the lengths of the pipes anywhere in the network. I have been searching for a while with limited resources for a worked example or solution method for this. </p>
<p>I will probably write a code to solve these problems if I find an appropriate methodology. (I'm aware commercial software <a href="https://www.cadreanalytic.com/cadreflo.htm" rel="nofollow noreferrer">does exist</a> that could calculate this, but I don't have access). </p>
| |mechanical-engineering|fluid-mechanics|civil-engineering|pipelines| | <p>You have 9 unknown values you are seeking (the flow at each of 9 outlets), and one that you need to find in the process (the total pressure drop) From the given information, you can assemble equations that you can solve using linear algebra methods.</p>
<p>The sum of the flows at the outlets will equal the inlet flow: </p>
<p>$$Q_t = Q_a + Q_b + Q_c + Q_d + Q_e + Q_f + Q_g + Q_h + Q_j$$</p>
<p>You can also set up equations based on the fact that the pressure drop will be equal along any path from a given node to an outlet. </p>
<p>Pressure drop in any segment can be found using many methods. In this case, the Darcy-Weisbach equation would probably be most useful:</p>
<p>$$h_f = f\cdot\dfrac{L}{D}\cdot\dfrac{v^2}{2g}$$</p>
<p>where: </p>
<ul>
<li>$h_f$ = head loss (m)</li>
<li>$f$ = friction factor </li>
<li>$L$ = length of pipe work (m) </li>
<li>$d$ = inner diameter of pipe work (m) </li>
<li>$v$ = velocity of fluid (m/s) </li>
<li>$g$ = acceleration due to gravity (m/s²) </li>
</ul>
<p>Equivalent length information for fittings can be found here: <a href="https://www.engineeringtoolbox.com/resistance-equivalent-length-d_192.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/resistance-equivalent-length-d_192.html</a></p>
<p>An excel template for using this method can be found here:
<a href="https://www.engineeringtoolbox.com/equivalent-pipe-length-method-d_804.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/equivalent-pipe-length-method-d_804.html</a></p>
<p>Your example is much more complex than the example given at that page (9 paths vs. 2), but the spreadsheet is scaleable.</p>
| 18840 | How Can I Determine Multiple Discharges from a Pipe Network? (Knowing the Inlet Conditions) |
2018-01-11T21:08:51.743 | <p>These items: <a href="https://www.parktool.com/product/dial-indicator-gauge-set-for-ts-2-2-and-ts-2-truing-stands-ts-2di" rel="nofollow noreferrer">https://www.parktool.com/product/dial-indicator-gauge-set-for-ts-2-2-and-ts-2-truing-stands-ts-2di</a> are mechanical gauges with roller tips that are intended to measure deviations from true in bicycle rims. </p>
<p>What is this kind of instrument called in general? </p>
<p>In particular, what are good keywords to look for devices like this the output of which can be read electronically?</p>
| |measurements|tools| | <p>In machine shops, the purely mechanical types are called dial indicators. For ones that can be read electronically, the classic ones are called LVDT's or Linear Variable Displacement Transducers. There are many types of these in use today as what are called proximity sensors which measure gap widths or clearances without having to touch the object being measured. </p>
| 18858 | What are the electronic readout versions of Dial Indicators called? |
2018-01-12T18:06:59.283 | <p>I have following assembly:
<a href="https://i.stack.imgur.com/lv1cl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lv1cl.png" alt="Problematic assembly"></a>
and I have following <code>Dowel holes</code> misaligned:
<a href="https://i.stack.imgur.com/KOr3H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KOr3H.png" alt="Problematic dowel holes"></a>
Now, both sets of holes (for both parts) were created using <code>Linear Sketch Pattern</code> tool and now I want to mate these parts "through" these dowel holes sets. The first option is to recheck <code>Linear Sketch Pattern</code> parameters for both sets, however is it possible to automatic creation of holes at exact position from first part?</p>
| |solidworks| | <p>Perhaps I'm not fully understanding you but it seems like what you are trying to do is create the dowels and the dowel holes in the same action; this is fairly straightforward.</p>
<p>Presumable you used <strong>Extruded Cut</strong> to make these holes, what you want to do is replace that feature with an <strong>Extruded Boss/Base</strong> using the same sketch and be sure to uncheck the <strong>Merge Entities</strong> box so that the extrusion is a separate body from your main part.</p>
<p>You need to <strong>Select all bodies</strong> in the scene and create an <strong>Intersect</strong> feature. Once again make sure that <strong>Merge Entities</strong> is turned <strong>off</strong>.</p>
<p>The final step is to use <strong>Insert>Feature>Split</strong> to export each of the bodies in the scene as individual parts so that you can combine them in assembly.</p>
<p>I hope this was the answer you were looking for</p>
| 18869 | Solidworks 2017 Dowel holes misaligned |
2018-01-13T06:47:25.473 | <p>I have a wheel which has a radius of 0.1m that needs to spin at 10000RPM. The outer surface of this wheel is pressed up against another free spinning wheel like a football or tennis ball launcher.</p>
<p>I am feeding metal powder through these two wheels, which I need to travel at 104.72m/s (basically the wheel needs to spin at 10000RPM while the powder is feed through). The powder is fed at a rate of 0.013 kg/s. </p>
<p>If I could work out the force on the wheel then I could work out the torque and hence power, however that is where I am not sure on what to do. The powder is being fed from almost standstill to a wheel traveling at over 100 m/s. It would seem the acceleration is almost instantaneous, but I know that's not the case. How would I work something like this out? Would I need the frictional coefficients of the wheels and powder?</p>
<p>(Assume this is operating in a vacuum and there is no air resistance)</p>
| |dynamics|acceleration|tribology| | <p>You actually don't need to know the force on the wheel to work on the power. Just look at the kinetic energy of the powder. Kinetic energy = $(1/2)mv^2$. For 0.013 kg of powder, the kinetic energy is $(1/2)(0.013)(104.72)^2=71$ Joules. You need to provide that much Joules every second, so the power is just 71 Joules/second = 71 Watts. </p>
<p>Of course, this assumes that the bearings in your wheels are frictionless. They are not. It will take some amount of power just to spin the wheels with no powder. This will depend on how good your bearings are. That's something that you could easily measure if you have the wheels already. If not, probably just round up to something like 100W and that should be more than enough. </p>
| 18877 | Power needed to accelerate metal powder through two spinning wheels |
2018-01-13T19:12:30.520 | <p>I've seen the terms "SCADA boxes", "SCADA out-of-the-box", "SCADA turnkey systems" used in various articles but without a definition. I'd like to know whether the terms are synonymous and if not what is the "generally accepted" concept of a "SCADA box"?</p>
| |control-engineering|industrial-engineering| | <p>The plant I used to work at had a Supervisory Control And Data Acquisition (SCADA). They had all kinds of transducers measuring everything in sight and we could sit in the control room and check on things all over the plant. </p>
<p>(<em>warstory</em>)
One night (when I should have been working) I was poking around and noticed the temperature of the Feedwater pump A bearing was high and just then, the call came in from the field that the Feedwater pump B was burning up and we needed to take it down. </p>
<p>I showed the high temp on the A pump to the operator on duty and he checked with the field operator who confirmed it was Feedwater pump B so he cut Feedwater B. The operator in the field immediately called back with "A, the A pump, not the B pump." The good news was we were able to bring B back up without too much delay and the plant had no difficulty.
(<em>end warstory</em>)</p>
<p>At this point, I have no recollection of being able to actually control anything with the SCADA system, such as turn on equipment, it was all for data logging. How did that differ from the Plant Monitoring System (PMS)? Again, I don't recall except to say that it allow monitoring all kinds of things that were not available to the PMS.</p>
<p>The SCADA system had these Remote Terminal Units (RTUs) out in the field that would interface with the transducers and send the data back to the system that was used for display. </p>
<p>I am suspicious of anybody that says they have a turn-key SCADA system. I expect they will sell you their off-the-shelf display software and the RTUs (or whatever their modern equivalents are) and a then charge you for the (possibly huge amount of) time spent on configuration. I would rather sign up to buy an "off-the-shelf" SCADA than try to cook something up in-house but it is hardly turn-key.</p>
<p><em>But that is not your question.</em></p>
<p>From what I can find on the web, it appears the SCADA-boxes you refer to are the modern equivalent of RTUs. They receive the data from the transducers, package it up and send it back to the main data logging and display system via some sort of network. The out-of-the-box and turnkey systems are the same thing. You buy display and data logging software from the vendor and either play hell configuring it yourself or pay them big bucks to do the configuration for you.</p>
| 18884 | What are SCADA Boxes? |
2018-01-14T12:24:06.590 | <p>By finite element type I specifically mean element geometry. </p>
<p>I am aware of Beam/Strut, Quad, Tri, Tet & Hex but are there other types, eg a hexagonal prism, rhombihedral etc.. ?</p>
| |finite-element-method|meshing| | <p>Certain geometries can benefit from polyhedral elements or elements with edge degrees of freedom. I can think of three main developments in that direction:</p>
<p>1) Voronoi cell finite elements, e.g., <a href="https://www.sciencedirect.com/science/article/pii/0045794994904359" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/0045794994904359</a></p>
<p>2) Isogeometric elements, e.g., <a href="https://www.sciencedirect.com/science/article/pii/S0045782504005171" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/S0045782504005171</a></p>
<p>3) Finite elements based on exterior calculus, e.g., <a href="https://www.cambridge.org/core/journals/acta-numerica/article/finite-element-exterior-calculus-homological-techniques-and-applications/1A2AEB067BCA561D9ED6D674026539B9" rel="nofollow noreferrer">https://www.cambridge.org/core/journals/acta-numerica/article/finite-element-exterior-calculus-homological-techniques-and-applications/1A2AEB067BCA561D9ED6D674026539B9</a></p>
<p>These elements are still being actively designed, but attention today is mostly on isogeometric and exterior calculus-based elements; primarily based of meshing issues and the need to solve multiphysics problems and to avoid element locking.</p>
| 18895 | Are there other finite element types besides the usual? |
2018-01-15T05:32:06.287 | <p>In 802.11n 2.4Ghz, the spectrum is varied from 2400Mhz to 2500Mhz and consists of 13 (or 14) channels, each channel having 20Mhz band width.
When I search google for best channel, all websites say 1,6 and 11 are the best channels since they won't overlap. But from my study, i found the following.</p>
<pre><code>channel 1 : 3 [2, 3, 4]
channel 2 : 4 [1, 3, 4, 5]
channel 3 : 5 [1, 2, 4, 5, 6]
channel 4 : 6 [1, 2, 3, 5, 6, 7]
channel 5 : 6 [2, 3, 4, 6, 7, 8]
channel 6 : 6 [3, 4, 5, 7, 8, 9]
channel 7 : 6 [4, 5, 6, 8, 9, 10]
channel 8 : 6 [5, 6, 7, 9, 10, 11]
channel 9 : 6 [6, 7, 8, 10, 11, 12]
channel 10 : 6 [7, 8, 9, 11, 12, 13]
channel 11 : 5 [8, 9, 10, 12, 13]
channel 12 : 4 [9, 10, 11, 13]
channel 13 : 3 [10, 11, 12]
</code></pre>
<p>Channel 1 overlaps with 3 channels, Channel 6 overlaps with 6 channels and finally channel 11 overlaps with 5 channels.</p>
<p>So from my study, I conclude that channel 1,2,3,11,12,13 have less overlap interference.</p>
<p>Am I correct?</p>
| |rf-electronics|wifi|waves| | <blockquote>
<p>Q: "Why channel 1, 6 and 11 are best for WiFi ... ?"</p>
</blockquote>
<p>The are three major sources of interference: Co-Channel, Adjacent-Channel, and non-WiFi.</p>
<ul>
<li><a href="http://community.arubanetworks.com/t5/Technology-Blog/An-Intro-to-Co-Channel-Interference/ba-p/273248" rel="nofollow noreferrer">Co-Channel interference</a> is simply where other WiFi equipment is using the same channel. and you need to share timeYour equipment waits until the channel is clear before trying it's luck, if there's enough competition or a fault in the equipment you get a collision; meaning you must wait and try again.</li>
</ul>
<p>When setting up <strong>your own</strong> equipment on <strong>different channels</strong> and seek to not interfere <strong>with yourself</strong> you would choose channels 1, 6, 11, and 14 (if it's available in your area).</p>
<p><a href="https://i.stack.imgur.com/gtMWR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gtMWR.png" alt="2.4 GHz channel widths"></a></p>
<p>Avoiding interference from other people's WiFi involves using a WiFi APP on your cellphone (or test equipment) to determine which channels are least congested. If you enable <a href="https://en.wikipedia.org/wiki/Channel_bonding#Wi-Fi" rel="nofollow noreferrer">channel bonding</a> you might double your transmission speed but you're more likely to double your collisions if anyone else is nearby.</p>
<ul>
<li><p>Adjacent channel interference occurs when a nearby channel is used by either you or your neighbor. If you use channel 1 and everyone else uses channel 6 or higher then your in luck.</p></li>
<li><p>Interference can also occur from non-WiFi sources such as cordless phones or microwaves.</p></li>
</ul>
<p><a href="https://i.stack.imgur.com/lfhar.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lfhar.png" alt="Sources of interference"></a></p>
<p>Most people only use a single WiFi channel in their home. In that case you're not going to want to stick with 1,6,11 but instead survey the channels and attempt to find one that is least used (all of the time).</p>
| 18904 | Why channel 1, 6 and 11 are best for Wifi communication 802.11n? |
2018-01-15T14:26:47.413 | <p>I own this vaporizer: <a href="https://www.dynavap.com/dynashop/vaporizers/omni/omnivap/xl-titanium" rel="nofollow noreferrer">(product link)</a>
<a href="https://i.stack.imgur.com/7LQmC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7LQmC.png" alt="Dynavap vaporizer"></a>
It has no electrical parts. To use it, you take off the cap at the left, put some "herbs" in it to smoke (vaporize), and put the cap back on. Then you heat the cap with a torch lighter until it says click. The click indicates that the cap (made of stainless steel) is at the right temperature. Then you can inhale from the right part and hot air with vaporized herbs will enter your body. It will click again when it has cooled down. <a href="https://youtu.be/i86qSoFolVs" rel="nofollow noreferrer">(video)</a></p>
<p>How does this click work? Can I make this myself with a CNC machine?</p>
<p>The click system is at the end of the cap.</p>
<p><a href="https://i.stack.imgur.com/R6alB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R6alB.jpg" alt="Cap"></a></p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <p>Flashers in your car that makes your blinkers blink used to work on the same principle. If you lightly bend a thin metal sheet, it will resist more to bending in the perpendicular direction. You can demonstrate it by holding a sheet of paper with one hand. It won't hang down if you push your thumb to curve the paper somewhat while holding it. But lightly push the other end of the paper down with your other hand, and eventually it will flip over and hang down.</p>
<p>Your vaporizer will probably work with that principle, but with a small metal sheet. And it is probably fitted in a location that compresses it, just like you pushed the other end of the paper.</p>
<p>That resistance to flipping over decreases as the temperature of the metal increases. So when heated, at some point, it will become weak enough to snap; creating the snapping sound.</p>
| 18910 | How does this mechanical heat click work? |
2018-01-16T13:47:01.747 | <p>I am a newbie at this!! </p>
<p>For an experiment, I have a diesel generator that releases hot flue gases at 150-160 Degree Celcius. </p>
<p><strong>I want the output gas to be at less than 75 degrees for a duration of ~ 1 hour without 'circulating' water. I'd like to do this with static water tank/bath</strong> (because 1 hr is usually the maximum time of operation at once for a generator.)</p>
<p>-The location I am in is where ambient temperature is ~ 18-30 degree celcius.
-Exhaust diameter is 4 inches</p>
<p>I want to be able to design a kind of heat exchanger: </p>
<ol>
<li><p>Can I just fill a tub of water, and divide the output exhaust into 4/5 tubes, take them through a tank of water? (like MUX-DEXMUX)</p></li>
<li><p>What's the amount of water that'd be required for the setup to be able to give that much heat reduction, for ~ an hour? </p></li>
<li><p>What sort of geometry should I build to not have much back-pressure on the gas flow.</p></li>
<li><p>What'll likely be the best material to build these pipes?</p></li>
</ol>
<p>What's the best direction/references/designs to look into for design such, water bath/non-circulating type heat exchangers? </p>
<hr>
<p>My experiment till now. <a href="https://i.stack.imgur.com/f1Hoy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f1Hoy.jpg" alt="Like this i took a spiral tubing, through a bucket of water and it almost worked. However this isn't a optimized design as it'd cause backpressure due to spirals"></a></p>
| |mechanical-engineering|heat-transfer|automotive-engineering|hvac|thermal-conduction| | <p>alright! After all the comments and what was available this is what we ended up designing<a href="https://i.stack.imgur.com/Do1fh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Do1fh.jpg" alt="DIY heat exchanger"></a></p>
<p>The inlet flue gas at 140 degrees was 65-75 degrees at outlet after an hour of continuous usage, with outdoor temperature as 25 degrees.</p>
| 18929 | Rapidly cool a flue gas without water circulation |
2018-01-16T22:16:14.180 | <p>I've learned that a high viscosity means that the fluid is more thick. <a href="https://en.wikipedia.org/wiki/Viscosity#Dynamic_(shear)_viscosity" rel="nofollow noreferrer">Dynamic Viscosity</a> thus represents the <em>resistance</em> to shear forces. Presumably, higher resistances are represented by higher numbers in the <a href="http://www.viscopedia.com/viscosity-tables/substances/iso-viscosity-classification/" rel="nofollow noreferrer">relevant ISO spec</a>.</p>
<p>Thus I'm puzzled about the unit of <a href="https://en.wikipedia.org/wiki/Viscosity#Kinematic_viscosity" rel="nofollow noreferrer">Kinematic Viscosity</a> (which is basically Dynamic Viscosity modulated by a measure of the density of the fluid). The unit is:</p>
<blockquote>
<p>m^2/s</p>
</blockquote>
<p>But how can this unit be understood intuitively? The higher the viscosity, the lesser the flow-rate and thus less m^2 fluid should flow per second in whatever measurement setup they are using.
Looking at the unit for kinematic viscosity, it seems that a higher value means that the fluid has a <em>lower</em> viscosity (i.e. is less thich), because, well, more square meters of fluid seems to flow per second.</p>
<p>How can the unit of kinematic viscosity be understood intuitively, when seen in relation to the definition of viscosity?</p>
| |machining| | <p>Let's start from the definition of dynamic viscosity, which is the proportionality constant in Newton's linear relationship between shear and velocity gradient: shear = (dynamic) viscosity x velocity gradient. Thus, dynamic viscosity has dimensions of stress per velocity gradient. Physically, it is a measure of how much stress a fluid experiences for a certain velocity gradient. Kinematic viscosity is defined as the dynamic viscosity divided by fluid density. Thus, kinematic viscosity is a measure of the stress a fluid experiences for a certain velocity gradient per unit of its density.</p>
| 18939 | Is there an intuitive explanation for the Kinematic Viscosity unit of mm^2/s? |
2018-01-17T15:04:57.377 | <p>I have some doubts about a quick-release mechanism that I'm working on and I'd like to run it by you. </p>
<p>I have hollow tube in which there's a spring-loaded rod. Part of it is made wider and there's a metal round pin that restrains the entire mechanism as shown below:
<a href="https://i.stack.imgur.com/xsGhI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xsGhI.png" alt="enter image description here"></a></p>
<p>The spring rate is 1.23 N/mm its free length is 135mm but it is compressed to 35 mm. </p>
<p>I'm using a servo quick release mechanism similar to this:
<a href="https://i.stack.imgur.com/OSL4Z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OSL4Z.jpg" alt="enter image description here"></a></p>
<p>It is driven by a servo with a stall torque of 9.8kg/cm</p>
<p>My question is, would the lateral force the rod is exerting on the pin be a problem for the motor? E.g. the friction between the pin and the hole of cylinder being so great that the motor won't be able to move? Is there a way to mitigate this? </p>
| |mechanical-engineering|springs|servo| | <p>The force which is needed to pull the pin out, will depend on the friction of the pin and the neighbouring material, and the surface at which the friction works. Let's assume a friction factor $\mu$ of 1, though it's probably less for plastics. That means that theoretically, just as much force is needed to remove the pin, as the force put on the pin by the spring, which is 123N(or 12.5kg) in your case, assuming the springrate is constant. If the servo has an arm of 2 cm, then the maximum force the servo will be putting on the pin, is half it's stall torque, so 4.9kg. That wouldn't be enough to remove the pin. However, if the friction factor, or the arms length is lower, it may be enough.</p>
<p>But I'd rather be concerned that the pin doesn't get damaged after a few times, as your servo pulls it out of the tube under force. That used to be the one lifelimiting factor in the BB guns I had as a kid. Those used the same construction as you show here.</p>
<p>I'd advice you to use a solenoid to pull the pin out of it's hole, it does so with enough force, and much faster than a servo, greatly reducing the stress on the pin. You can easily find a solenoid strong enough to pull the pin out.</p>
| 18948 | Quick release for spring-loaded rod |
2018-01-18T15:08:19.843 | <p>I have been playing OpenRails Train SImulator recently and I thought i made my own train(which i did), then the physics part came up. It required me to fill in a lot of parameters, like braking, sanders, and of course, engine. Now the engine part is confusing me, it requires me to enter:</p>
<p>Mass(KG), Max Power(kW), Max Force(kN), Continuous Force(kN), and Max Velocity(mph).</p>
<p>Now the Mass, MaxForce and MaxContinousForce is easy to find, but what about the rest? In order for me to know more about what i am doing, i would like to know how the speed in the game is calculated using all these values? Because i thought the speed of the train was only calculated using the MaxForce and the Mass value, but all these values seem to have some connection as if one of those listed values is changed, speed is different. So how would I calculate the speed of a train using these parameters(What is/are the formula('s)? And how are the horsepower values and force values related to each other in terms of final speed?(As again, i though kW didn't matter in speed)</p>
<p>Can someone explain me?</p>
<p>Thanks!</p>
<p>Edit:
So mainly what i want to know is:
Is the calculation of speed in general affected by horsepower/kW(does it have anything to do with speed/force at all?)</p>
| |mechanical-engineering|acceleration| | <p>Let's put aside this specific game, and just say we want to calculate the speed of some generic vehicle. I think that will answer what you are really trying to get at. Our vehicle is going to obey Newtons\'s second law $F=ma$, force = mass times acceleration. So, let's add up all of the forces that act on the train. The main ones will be the driving force from the engine, and then rolling resistance and air drag as pointed out by @SolarMike. Let's ignore rolling resistance to simplify the problem, and just look at the others. So </p>
<p>$F_{engine} + F_{drag} = ma$. </p>
<p>Now, we know from <a href="https://en.wikipedia.org/wiki/Drag_(physics)" rel="nofollow noreferrer">wikipedia</a>, that $F_{drag} \propto v^2$, where $v$ is the velocity. To simplify the discussion, I'll lump all of the constants into just one, so $F_{drag}=-Cv^2$, where the minus sign is indicating that the drag force is in the opposite direction of the applied engine force. So at this point we have:</p>
<p>$F_{engine} - Cv^2 = ma$. </p>
<p>Now, what is the formula for the engine? Well, it's complicated and will depend on your specific engine. If you have a transmission, it will depend on what gear you are in. But two important points: it is dependent on speed, and above a certain speed it will start to decrease with speed. i.e. as you go faster and faster, the engine is producing less torque. You can find some real engine curves <a href="http://web.itu.edu.tr/~sorusbay/ME422/LN11.pdf" rel="nofollow noreferrer">here</a>, but for this discussion, let's oversimplify it a lot and just assume a linear relationship $F_{engine}=A-Bv$. So now we have:</p>
<p>$A-Bv - Cv^2 = ma$. </p>
<p>So, imagine your train starts at v=0. The left hand side is positive, so acceleration is positive and your train speeds up. Now it has some velocity. The left hand side is still positive, but less than before. The engine is producing less force, and the air drag is acting against us. So the train continues to accelerate, but not as much. Finally, at some speed, the left hand side will be zero. The air drag will have increased to the point where the force produced by the engine exactly equals the air drag. At this point, acceleration is zero. The train has reached its top speed.</p>
| 18961 | How is train speed calculated in OpenRails? |
2018-01-19T02:14:15.010 | <p>I am not able to understand the relation between ride frequency and wheel deflection of a car, as per given in Race car vehicle dynamics book by milliken. Is it something to relate with harmonic motion of a forced oscillator? I still don't get why there is a 188 in the numerator.</p>
<p><a href="https://i.stack.imgur.com/GzjM0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GzjM0.png" alt="enter image description here"></a></p>
| |mechanical-engineering|design|dynamics| | <p>This is a log-log plot in which $log(\omega)$ is plotted against $log(x)$. </p>
<p>For example, the frequency for $x=5$ is $\frac{188}{\sqrt{5}}=84.0762$ and the frequency for $x=10$ is $\frac{188}{\sqrt{10}}=59.4508$. </p>
<p>On the log scale, the tick marks show the original values, but the points $\{5,84.0762\}$ and $\{10,59.4508\}$ are actually plotted at $\{\log (5),\log(84.0762)\}=\{1.60944,4.43172\}$ and $\{log(10),\log(59.4508)\}=\{2.30259,4.08515\}$.</p>
<p><a href="https://i.stack.imgur.com/9ZYvd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ZYvd.png" alt="enter image description here"></a></p>
| 18965 | Relation between ride frequency and wheel deflection of a car |
2018-01-19T02:47:58.097 | <p>(Please move this question to the appropriate exchange site if it is not this one; I originally posted this <a href="https://travel.stackexchange.com/questions/108460/how-do-locks-for-shoe-lockers-in-sentos-work">here</a>, and a comment suggested that I move it here.)</p>
<p>In Danganronpa it appears that the lock at 9:27 in this <a href="https://www.youtube.com/watch?v=qOVU8UZRrHc" rel="nofollow noreferrer">video</a> has a weird shape, and I cannot figure out how it might work. Using Google, it appears that these are locks for shoe lockers in <a href="https://en.wikipedia.org/wiki/Sent%C5%8D" rel="nofollow noreferrer">Sentōs</a>, but I haven't been able to find a picture showing how they work.</p>
<p>A comment on the original place I put this question pointed me to this <a href="https://bramanswanderings.com/2014/04/03/japanese-locker/" rel="nofollow noreferrer">website</a>, which says "The wooden key was interesting as it had grooves in it. If the pattern of the grooves match the pins in the lock then the key will go all the way down and trip the catch on the lock." However, this seems a little suspicious because then it looks like you could open any lock by using a small stick.</p>
<p>After some more googling I found this <a href="https://www.ebay.com/i/323002263864?rt=nc" rel="nofollow noreferrer">auction</a> which has pictures of inside of the lock (and also pictures of the key, which are different from those in the game; I am now guessing that the key shape in the game probably does not exist, but only because I cannot come up with a lock mechanism for it). It does not explain how it works, though, and I am still interested in this.</p>
<p>How do they work? More specifically,</p>
<ol>
<li><p>Is the picture of the lock in the video realistic?</p></li>
<li><p>Is the description from that <a href="https://bramanswanderings.com/2014/04/03/japanese-locker/" rel="nofollow noreferrer">website</a> accurate, and if not then how does it work?</p></li>
</ol>
| |design|mechanisms| | <p>The key in the game is not a realistic representation of keys used in traditional Japanese shoe lockers - it has lateral groves besides normal lengthwise groves used in these locks. They are usually low-security locks that act on principle similar to warded locks (only fixed obstructions blocking a mismatching key; a key that is significantly slimmer will just work) - and similarly, easy to bypass. Possibly there exist more secure variants; I haven't encountered any.</p>
<p>A lock like in the video would be fairly easy to implement as a variant of a <a href="https://www.youtube.com/watch?v=ehA8WUX4xaY" rel="nofollow noreferrer">dimple lock</a> - a lock where the key has holes on the flat surface and engages the tumblers with its sides instead of edges. </p>
<p>Instead of drill holes, groves can be used to engage the tumblers at various locations to various depth; to avoid necessity of turning, and thanks to the size of the key, the tumblers could be considerably bigger and act directly as the latches on the door. Still, this sort of key, made of wood, would sooner or later get dented by the tumblers, failing eventually. So, a lock like this is possible, is not very practical, and it's highly dubious if it's used in reality.</p>
| 18966 | How do locks for shoe lockers in Sentōs work? |
2018-01-19T16:38:36.030 | <p>There is a large drilling crane or something, which is connected by a giant hose to a skip full of water (central skip, blue). It looks like a jacuuzi but I suspect they are using it for something else.</p>
<p><a href="https://i.stack.imgur.com/g3NBm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g3NBm.jpg" alt="construction_site_jacuuzi"></a></p>
<p>Perhaps it is not a jacuuzi? What is it for if not?</p>
| |civil-engineering| | <p>It is likely to be tanks for <a href="https://en.wikipedia.org/wiki/Drilling_fluid" rel="noreferrer">drilling fluid</a> (aka 'mud'). </p>
<p>This is a slurry of various substances, such as bentonite clay suspended in oil or water which serves various functions. In particular it provides cooling and lubrication for the drill head and helps to clear debris created by drilling. </p>
<p>When drilling into pressurised oil and gas wells it also provide a hydrostatic head which resists the well pressure. </p>
<p>Typically it will be pumped down the centre of the drill pipe and returns up through the gap between the drill pipe and the bore. </p>
<p>Drilling fluid is usually made up on site as needed by mixing dry ingredients with oil or water plus any additional additives. </p>
| 18973 | What is this construction site jacuuzi for |
2018-01-19T22:25:24.197 | <p><a href="https://i.stack.imgur.com/0PicO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0PicO.jpg" alt="enter image description here"></a>I know if a rod is hinged at its two ends and if there is no load act upon it along its length then the forces exist only along its length but i don't see how this characteristic comes into play ? </p>
| |civil-engineering|beam|torque|power-transmission| | <p>If you imagine a square or rectangular shape made from rods joined with points at the corners. If those pins act as hinges then the whole shape can easily deform, effectively acting as a 4 bar link. </p>
<p>On the other hand if the shape is a triangle then even if the corners are hinged then it can't be deformed without stretching one of the bars. Also, whichever way you try to deform it one of the bars will always be in tension. So the stiffness of the structure doesn't depend on the rigidity of the joints. </p>
<p>This is the fundamental reason why breaking up a structure into triangular sections tends to be an efficient way to make a stiff, lightweight structure. </p>
<p>This approach also tends to be fairly straightforward to design and analyse. </p>
| 18980 | Why in the construction of power transmission towers , the rods are hinged oblique? |
2018-01-20T05:01:34.453 | <p>I have a room of computer rack that with 2750 watt power needs, which according to <a href="https://www.enviromon.net/advanced-server-btu-calculator/" rel="nofollow noreferrer">this site</a>, I need 9000 BTU/h AC. I'm deciding if I should put a vent in the room or not. When there's a ventilation, it will look like this:
<a href="https://i.stack.imgur.com/1zJIV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1zJIV.png" alt="Scenario 1"></a>
Note that there are two wind turbine ventilators on the roof. Also the attic is cool because of the good air circulation.
Question for this scenario:
1. Will the AC and the grill vent vanquish the humidity from the air outside, which is warm and humid?
2. Will dust enter the room via vent despite the fact that the airflow is upward to the wind turbine ventilator?
3. Is 9000 BTU an optimum number? Is it too much or too little? I'm thinking most of the heat is drawn to the vent.</p>
<p>The second scenario will look like this:
<a href="https://i.stack.imgur.com/BAVMe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BAVMe.png" alt="Second scenario"></a>
In this scenario, instead of a ventilation with a chimney, there is a ducting that caries heat, which escapes to the attic and goes to the wind turbine vents. Question:
Will this give faster heat reduction whereas there is no way humid air will come to the room?</p>
<p>regards</p>
| |thermodynamics|heat-transfer|airflow|heat-treatment| | <p>Additionally, switch rooms and IT rooms tend to just recirculate the room volume with some small DX cooling. It is not viewed as an occupied space, hence no need for extract for heat rejection, you would need some provision for smoke extract though.</p>
<p>Also another design consideration, exposing an IT space to external environment is not a good idea if you are likely to get high/very low RH.</p>
<p>Thats my 5 cents.</p>
| 18981 | Room with computer racks, with or without ventilation |
2018-01-21T01:36:48.313 | <p>I am designing a mock-up aircraft throttle, like one of these:<a href="https://i.stack.imgur.com/CMNbG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMNbG.jpg" alt="enter image description here"></a></p>
<p>In an airplane, it it slides forward and backward to control the thrust of the engines. The problem is, I am not sure how to go about adding more friction to the sliding motion. If I were to just design a sliding handle, If you let go of it, it wouldn't be held in place, and it would slide back down. I need something sort of "reverse bearing" that will add friction to the rotation of the handle, so that when you let go of the handle, it wont slide around-- it will just stay in the same position until it is moved again.</p>
| |mechanical-engineering|friction| | <p>In Piper Cherokee airplane, the power throttle assembly is just a bushing and shaft with an adjustable pressure handle to increase the grip.
It has worked on my airplane for 41 years, no problems.</p>
<p>It is a really simple mechanism.
Here are some photos from Ebay. they sell it for $85. <a href="https://www.ebay.ie/itm/Piper-Cherokee-PA28-151-161-Control-Quadrant-Assy-P-N-65850-00-/112844567379?hash=item1a460f7353" rel="nofollow noreferrer">Cherokee quadrant throttle</a></p>
| 18991 | Mechanical device that creates friction when rotating? |
2018-01-21T02:31:21.780 | <p>I hear a lot about open loop and closed loop Model Predictive Control (MPC).</p>
<p>What is the difference between an open loop MPC and a closed loop MPC?</p>
<p>Any block diagram demonstrating the difference is also appreciated.</p>
| |control-engineering|control-theory|optimal-control| | <p>I think the other answer is not complete. Model predictive control (or 'receding-horizon control' is a technique in which a predictive system model is used to evaluate a sequence of future control inputs; of all such control input sequences, an optimization algorithm chooses the best one. Typically, the <em>first</em> input of the sequence is implemented. After a certain amount of time, the process is repeated to find a new control input.</p>
<p>When the predictive model is deterministic, this is straightforward. Now, when the predictive model is <em>uncertain</em> (e.g., stochastic or adversarial), one often differentiates between <em>open-loop MPC</em> and <em>closed-loop MPC</em>:</p>
<p><strong>Open-loop MPC</strong> considers only a single, fixed sequence of future control inputs; this sequence must yield good performance under all possible realizations of the uncertainty. In other words, when constructing our initial plan, we ignore the fact that we're going to re-plan (with more information) soon. We plan as though we were committed to our input sequence (even though we aren't).</p>
<p><strong>Closed-loop MPC</strong> considers the <em>effect of feedback</em> (or <em>recourse</em>). This is the fact that, before the controller makes future decisions, it will have more information available than it does now. For this reason, the controller must optimizes over control <em>policies</em>, not just inputs. These problems are quite often intractable.</p>
<p>I'll add one more: <strong>Certainty-equivalent MPC</strong> replaces the uncertain predictive model with a deterministic one (usually by replacing all random quantities with an average or nominal value). At this point, there's no uncertainty, and we simply precede by optimizing over input sequences, as in open-loop MPC. This is often simply referred to as <em>MPC</em> (without a qualifier.</p>
<p>Check out <a href="https://web.stanford.edu/class/ee364b/lectures/stoch_mpc_slides.pdf" rel="noreferrer">Boyd's MPC description here</a>.</p>
| 18992 | Open loop versus closed loop Model Predictive Control |
2018-01-21T04:30:08.513 | <p>I am working on a model city.</p>
<p>Suppose I have utility poles A, B and C (brown) that go around a sharp corner. Looking from above, how do I connect the poles with power lines (gray, solid)? </p>
<p>Are the crossarms of utility pole B rotated correctly with respect to the direction of the lines? Do the crossarms of pole B the correct length, or do they need to be longer than normal? Are there any other problems I should keep in mind?</p>
<p>Thanks.</p>
<p><a href="https://i.stack.imgur.com/lBiUb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lBiUb.png" alt="enter image description here"></a></p>
| |civil-engineering|power| | <p>You can easily see that the wires don't allign(parallel) with the dotted line which would be the line they should follow to maintain the same mutual distance. To calculate the widht the mast should have, you can take the width of the first mast, and divide that by $cos(\alpha)$ where $\alpha$ is <em>half</em> the angle the middle mast makes in degrees.</p>
<p>In reality(at least in my country), the masts in sharp corners like those of yours, are lower and stronger than normal masts. Normal masts directing wires in straight lines mostly only get loaded perpendicular to the ground. Masts in corners, colloquially called 'little soldiers' in my country, are build thicker and shorter since they need to take on lateral forces much more. They are not necessarily as wide or wider than normal masts. I believe they are also build at an angle relative to the ground, further increasing resistance to lateral forces.</p>
| 18993 | Connecting power lines around corners |
2018-01-21T05:11:17.203 | <p>This is a plot of Yaw moment vs Lateral acceleration of my vehicle. I know that it can give me the analysis of trimmed conditions during cornering, yet unable to properly quantify it. What all can I achieve out of this plot. Thanks<a href="https://i.stack.imgur.com/1HjbI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1HjbI.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|automotive-engineering| | <p>What are you trying to find in the data?<br>
Under normal circumstances, ie. no drifing or such the following can be assumed; at the same vehicle speed, lateral acceleration is inherent to yaw speed, and the yaw speed is inherent to the corner radius. So yaw acceleration, or yaw moment as you can also define it, is created when steering into a corner. Theoretically, it is 0 the moment you keep the steering wheel at any steady position.</p>
<p>The above is theoretic, and based on formulas and basis physics. In automotive world, the highly dynamic reality ofter differs vastly from the theory we've learned. Explaining the V shape would be very hard to explain with only formulas and theoretic blahblah. The below is what my intuition tells me happens in reality:</p>
<p>You don't suddely steer, and keep the steering wheel in position like a step response, but you gradually steer in, creating yaw moment during steering since the corner radius decreases. Yaw moment is higher in slow sharp corners, since you turn the steering wheel fast. Those sharp corners are made at a <em>relatively</em> high speed, making for high lateral acceleration, explaining the upper right and left corners of your V. At the highway, you would steer relatively gently and slower, since situations are more dangerous at higher speeds, and your tyres aren't able to pull the same stunts at higher speeds without breaking out.
This explains why the upper corners meet in a mutual point, creating a V shape.</p>
<p>The fact that the upper left corner is higher likely indicates that your turns in one direction(probably right turns) are made faster than the other direction. Probably caused by the fact that you can oversee corners better, and that you have the right of way, and have to slow down less.</p>
| 18994 | Interpretation of the testing data of my vehicle |
2018-01-21T11:52:47.427 | <p>I am working on a problem at the moment and would appreciate some of your expertise regarding this issue. If you see the image below, I am trying to calculate heat transfer from a fluid entering a vessel/tank on the grey cuboid within. To begin with, I am focussing on side C.</p>
<p>I would like to understand how we would work out the Velocity of the fluid at point C when water is pumped in at A (intake), and, when discharging from point (B), i.e A flow, B return.</p>
<p>The overall problem I am trying to solve is how long it would take to heat the cuboid from convection heat transfer (forced + also natural). At the moment I am calculating (Re) and need to determine velocity.</p>
<hr />
<p>It is not a home work question, it is for general research. There is no actual question I’m working from.</p>
<p>I would like to understand how to work out velocity at C when there is a flow into A and out of B. This will allow me to work out Re at C.</p>
<p><a href="https://i.stack.imgur.com/ReR7Q.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ReR7Q.jpg" alt="Example" /></a></p>
| |mechanical-engineering|heat-transfer|convection| | <p>Assuming incompressibility you can work with volume-flow to calculate (a first guess) the velocity, $u$ with the following equations.</p>
<p>Probably you have some volume or mass $\dot{m}$ flow entring at (A).
This mass-flow is entering the box through a opeing with a certain cross-section $A_\mathrm{A}$. The relation is:</p>
<p>$u_\mathrm{A} = \frac{\dot{m}}{\rho\, A_\mathrm{A}}$</p>
<p>The same relation should also be valid for the exit B.</p>
<p>$u_\mathrm{B} = \frac{\dot{m}}{\rho\, A_\mathrm{B}}$</p>
<p>For the velocity around the cube we will now assume a constant velocity $u_\mathrm{C}$ which is of course not really the case but it is a start.</p>
<p>Here the equation has the same structure you just need to put in the correct area ($A_\mathrm{C}$) which is the area of the box minus the area of the cube.</p>
<p><a href="https://i.stack.imgur.com/xrq99.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xrq99.png" alt="enter image description here"></a></p>
<p>$u_\mathrm{C} = \frac{\dot{m}}{\rho\, (A_\mathrm{box} - A_\mathrm{cube})}$</p>
| 18998 | Convection Heat Transfer |
2018-01-21T12:56:51.393 | <p>I have tried searching for this, but I am unable to find a good answer. There is all sorts of stats about the longest bridge in the world, the highest, the longest span, but I think one of the most difficult things is the depth of the pier.</p>
<p>From my internet search I found a San Francisco article that says the deepest pier in the world is on the <a href="https://www.extranomical.com/HTML_PAGES/OAKLAND_BAY_BRIDGE.htm" rel="nofollow noreferrer">Bay Bridge at 242 feet</a> (~74m). This was built in the 1930's, so does this feat of engineering still remain at the deepest, or has it been superseded?</p>
<p>I initially wondered about this because of Boris Johnson refloating the idea of the idea of a road bridge between England and France. At 26 miles, that would not be a world breaking engineering challenge, but it seemed to me the depth of the piers. The Strait of Dover deepest point is 68m, so this does seem like a feasible project from a engineering point of view, but I think the question remains whether it is an efficient use of resource. Particularly when the Channel Tunnel utilisation is only running at 48% or so.</p>
| |civil-engineering|bridges| | <p>Piles of the Padma Bridge are 128 meter (420 ft) long. Each pier has 6 - 7 piles.</p>
<p><a href="http://www.padmabridge.gov.bd/mainbr.php" rel="nofollow noreferrer">http://www.padmabridge.gov.bd/mainbr.php</a></p>
| 19004 | What is the deepest bridge pier in the world? |
2018-01-21T17:18:38.250 | <p>I've been trying to find actual figures for the braking power of cars (in order to determine things like stopping time). As I understand it this comes in two parts: the maximum force that the brakes can apply on the tires, and the maximum force the tires can apply on the road.</p>
<p>I'd love to find figures for particular cars (or brake systems or tires) but I would be fine with any figures for actual cars under reasonable conditions. It's easy to find math problems of this sort online but I don't expect their figures to have much in common with the real world.</p>
<p>If someone could point me in the right direction (car data sheets, third party testing, government regulations, etc.) I'd be very much obliged.</p>
| |mechanical-engineering|automotive-engineering|energy| | <p>To elaborate on my comment; obviously generated braking force is defined by how hard you push the braking pedal. The required force to lock the wheels (given that ABS is disabled) depends on how much traction the tyre has. Let's say our tyre has a constant friction coefficient of 1. That's a simplified situation where the tyre has as much traction force, as there is force on the tyre by the vehicle's weight. </p>
<p>This allows you to decelerate(or accelerate) with a maximum of 1G or 9.81m/s2, before your tyres start slipping. That means that you'd get to a stop in 2.83s when travelling 100km/h. You can see from this figure, that most supersports cars are limited in their 0-100 time by the tyres. They use as wide tyres as they can on the driven wheels, to maximise traction. Heating up the tyres and lowering their pressure, and a slick surface on dry hot asphalt further increases their traction. But it doesn't allow you to accelerate to 2G or something.(except for crazy top fuel dragsters) But that's why i'm rather sceptical about Tesla's promised 0-100 figures for the announced roadster.</p>
<p>In reality, a friction coefficient is not close to constant and way more complex. In reality, your tyres are <em>always slipping,</em> which is the main reason they wear out. It gets worse in corners, acceleration, braking etc. but it's always present in a certain amount.</p>
<p>Also, traction sharply decreases once the wheel starts slipping by a certain amount. I believe that was usually 20% or something. You only have kinetic friction at that point, which is always lower than static friction. That's the reason ABS works, it constantly regains static friction by releasing the brakes briefly. I hope this gives you an better understanding of how and why tyres work.</p>
| 19009 | How much braking force do cars generate? |
2018-01-21T17:26:12.227 | <p><a href="https://www.moneyweb.co.za/news-fast-news/drought-hit-cape-town-at-point-of-no-return-tightens-water-targets/" rel="noreferrer">Cape Town, South Africa is experiencing an extremely serious drought</a>, with estimates putting April 21, 2018 as the day that the city will be forced to shut off all municipal water supplies and begin water rationing, except to essential services such as hospitals.</p>
<p>One possible solution is to delay this date by turning the supply off and on for scheduled periods of time over the next few months. However, experts have said that this is not a solution that can be implemented, because the pipes aren't supposed to deal with this variation of pressure and leaks and other infrastructure damage will occur.</p>
<p>Is there a way that a city could implement a water rationing system where water is shut on and off periodically, while minimizing leaks and damage to infrastructure?</p>
| |civil-engineering|water-resources|pipelines|piping|infrastructure| | <p>It is difficult to provide a useful answer to this question without additional information. For example, what is the system pressurization - operating psi/bars? Having installed ductile iron water mains, and having pressure tested new installations to destruction (resulting from hairline cracks in new pipes), I can confirm that the effects can potentially be devastating to infrastructure - especially roads and possibly other foundations/buried structures. In addition to the obvious negatives associated with traffic disruption and reinstatement, there will also be substantial water losses should this happen.</p>
<p><a href="https://i.stack.imgur.com/MTMo7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MTMo7.jpg" alt="Road collapse due to pipe failure."></a></p>
<p>Where HDPE pipe is used, one of the predictors of service life is the cyclic nature of the pressurization. Cyclic pressurization may not lead to catastrophic failure, as it could for ductile iron, but it will certainly shorten the lifespan of the network.</p>
<p>What is the nature of the network design in terms of control points? Can the network even be controlled? The valves, tees and joint connections will likely be the most fragile part of the system.</p>
<p>Nonetheless, in my view the best chance of implementing such a programme successfully would be to partially shut down the network on a rotational schedule in such a way that essentially the entire network would have a chance to stay pressurized - or at least ensure that the portion of the network that is exposed to risk would be minimized. This would require endpoint valves however. It could be possible, depending upon the nature of the network, to install these at strategic locations.</p>
<p>Whether such an approach is economic, or would provide any benefits at all in terms of a reduction of water consumption, is another question. My guess is it may not, but then again a rationing programme may not be enforceable either.</p>
| 19010 | Is there a way to reduce leaks and infrastructure damage when turning water on and off at the scale of city? |
2018-01-22T16:05:30.047 | <p>I am currently working on my dissertation project using abaqus, my model is an arch bridge consisting of a shell and solid elements, there are concrete and steel materials also, I have checked all of the constraints, and I found that there are some constraints giving the error messages 'time increment required is less than specified' and 'too many attempts made for this increment', specifically, those constraints happened in shell-solid tie constraints. the other constraints between solid to solid were fine.</p>
<p>Can anybody help me resolve these problems?</p>
<p><a href="https://i.stack.imgur.com/SqnMu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SqnMu.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/AZqpV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AZqpV.jpg" alt="enter image description here"></a></p>
| |structural-engineering|finite-element-method|abaqus| | <p>Try to increase the iteration count in each step. You can follow this video to solve this issue.
<a href="https://www.youtube.com/watch?v=mQ09lt8yRtc&t=2s" rel="nofollow noreferrer">https://www.youtube.com/watch?v=mQ09lt8yRtc&t=2s</a></p>
| 19029 | How to Solve this error message on Abaqus 'time increment required is less than specified' and 'too many attempts made for this increment'? |
2018-01-22T19:36:41.477 | <p>I've been trying to find the value of thermal resistivity for some linear disc resistors. I know that Thermal Resistivity is the reciprocal of Thermal Conductivity but the thermal conductivity on the data sheet is given in W/cm^2.C/cm which is the units for the heat transfer coefficient per cm not conductivity if the "period" is actually a multiplier and not just a period? Is this some convention I am unaware of on data sheets like this? I do have the thickness of the resistor disc in use so I think I can get the actual conductivity by multiplying the value given by the thickness squared? </p>
| |materials|thermodynamics|thermal-conduction| | <p>They are missing parentheses, as noted by agentp, but it should be said that <code>W/(cm^2*C)/cm</code> is actually a relatively useful way of describing Thermal conductivity.</p>
<p>Let's re-format that as:</p>
<p>$$Conductivity =\frac{Power\ Transfer\ Rate}{ \left(\frac{Surface\ Area\ *\ Temperature\ Difference}{Thickness} \right)}=\frac{W}{ \left(\frac{{cm}^2 *\Delta °C}{cm} \right)}$$</p>
<p>The reason this is useful, is that it incorporates the values that you may already know about your system, in the units that you are likely to use them. Namely,</p>
<ul>
<li>Rate of Power Transfer through the material - ($W$)</li>
<li>Surface area over which Power Transfer is occuring - ($cm^2$)</li>
<li>Thickness through which Power Transfer is passing - ($cm$)</li>
<li>Temperature difference between hot and cold sides of material - ($°C$)</li>
</ul>
<p>Setting the units out as they have done means that you can intuitively input the values you have, combined with the published conductivity values, to find the Power Transfer Rate that you are most likely looking for</p>
<p>$$Power\ Transfer\ Rate = Conductivity*\left(\frac{Surface\ Area\ *\ Temperature\ Difference}{Thickness} \right)$$</p>
<p>It is clear that, for a fixed conductivity, doubling the surface area (dividing by a larger number) will mean that you will experience double the power transfer.
Likewise, doubling the thickness will halve the power transfer.</p>
<p>This is slightly more difficult to see when using the SI units <code>W/(m.K)</code></p>
| 19032 | Thermal Conductivity? Units On a Data Sheet |
2018-01-22T21:20:57.873 | <p>I know how to solve it by moment line methode but i want to use singularity function here and i'm very confused. My strategy consist of writing the correspond singularity function for each element but i have absolutely no clue how can i write the right singularity function for the inverse ramp distributed load at the right side? <a href="https://i.stack.imgur.com/QmeIH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QmeIH.png" alt="enter image description here"></a></p>
| |mechanical-engineering|statics|beam| | <p>Are you familiar with the <em>Dirac delta function</em>, <em>Heaviside step function</em> and the <em>Ramp function</em>?</p>
<p>You could use Laplace Transforms together with those three handy functions:</p>
<p><strong>Dirac Delta function</strong></p>
<p>$$\delta(x)= \left\{\begin{array}{l}+\infty\qquad&x=0\\0&x\neq0\end{array}\right. $$
The <em>Dirac Delta function</em> is equal to infinity, at $x=0$, and $0$ everywhere else. Respectively $\delta(x-a)=+\infty$ if $x=a$.</p>
<p>With the <em>Euler-Bernoulli beam theory</em>, where
$$ EI\frac{d^4w}{dx^4}=q(x) $$
a static interpretation of the Dirac function is, that a point load $Q$ at point a represents a distributed load with infinite density, thus can be written as $q(x)=Q\cdot \delta(a-x)$</p>
<p><strong>Heaviside step function</strong></p>
<p>The step function is the integral of the <em>Dirac Delta function</em>
$$ H(x)=\int_{-\infty}^x\delta(s)ds $$
or
$$ H(x)= \left\{\begin{array}{l}1\qquad&x\geq0\\0&x<0\end{array}\right. $$
Thus, a distributen load, with magniuted $q$, between $x=a$ and $x=b$ can be written as $q(x)=q\left[ H(x-a)-H(x-b) \right]$</p>
<p><strong>Ramp function</strong></p>
<p>The ramp function again is the integral of the Heaviside function.
$$ R(x)= \left\{\begin{array}{l}x\qquad&x\geq0\\0&x<0\end{array}\right. $$
Thus, a distributied load , starting at $x=a$, with a slope $m$ can be written as $ q(x)=m\cdot R(x-a) $</p>
<p>The following distributed load, for example:
<a href="https://i.stack.imgur.com/hRJ7X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hRJ7X.png" alt="enter image description here"></a></p>
<p>This load can be written as:
$$ q(x)=\frac{q}{b-a}\left[R(x-a)-R(x-b)\right]-\frac{q}{d-c}\left[R(x-c)-R(x-d)\right] $$
<em>please see <a href="https://i.stack.imgur.com/H0xuR.jpg" rel="nofollow noreferrer">attached file</a> for the derivation</em></p>
<p><strong>Further explanation of the Dirac Delta function</strong></p>
<p>Take, for example the point force $Q$. That force can be approximated by a distributed load $Q=q\cdot dx$, therefore $q=\frac{Q}{dx}.$ Now, as a point load acts on an interval $dx\to0$, we can say $\lim_{dx\to0} q=+\infty$</p>
<p>This is also represented, if you look at a shear force diagram. We know, that $V'(X)=q(x)$, and we know that a point load $Q$ (external force or support) at point $a$ causes a "jump" in the shear force diagram, according to its magnitude, this can be represented by the Heaviside function. $V=Q\cdot H(x-a)$, now
$$V'(x)=q(x)=(Q\cdot H(x-a))'=Q\cdot H'(x-a)$$
As explained above, the derivative of the <em>Heaviside function</em> is the <em>Dirac delta function</em>, therefore
$$ q(x)=Q\cdot \delta(x-a) $$</p>
| 19033 | How to deal with inverse distributed load ramp starts and ends somewhere along the beam? |
2018-01-23T10:06:42.480 | <p>I've Googled the hell out of this and it doesn't seem to be, i.e.</p>
<ul>
<li>It's structured like a standard stud (not reinforced top)</li>
<li>It doesn't run along a floor joist</li>
<li>The uprights are pretty badly cut and not snug</li>
</ul>
<p>But it looks to me like the remains of an attic support structure (it an attic conversion/extension into a dormer-like structure) and to my untrained eye, it seems like a lot of roof (80" from nearest edge to back wall) to support itself.</p>
<p>I was planning on building a stud wall on the outside edge and running a support beam along the length to be safe but is there any way I can be sure that the support isn't even needed?</p>
<p><a href="https://i.stack.imgur.com/AitHD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AitHD.jpg" alt="Wall I want rid of"></a></p>
| |building-design| | <p>I have seen many of these DIY jobs by home owners that has been defying any sense of logic. Not only this could be a part of support of the roof by cutting the diagonal 2x4s that supported the rafters. It may not be even sitting on a solid member of framing. </p>
<p>I would try to get a read of the framing both of roof and floor of attic.</p>
<p>An experienced Carpenter could help.</p>
<p>I have a small probe that can send it's camera through a 1/4 inch hole and has a few LED lights to light up dark places. This probe connects to your droid phone and you can record what you see. I bought it for $28 from Amazon, it is really a handy tool.</p>
| 19037 | Is there any way I can be sure a wall is NOT load bearing? |
2018-01-23T20:13:07.880 | <p>I would like to independently check a vendor's silo design for proper explosion venting. Worst case, it is essentially cylinder filled with a stoichiometric mixture atomized sugar dust and atmospheric oxygen initially at atmospheric pressure. The roof of the cylinder is designed to detach from the structure with, lets assume, zero force. The roof also has some amount of mass per square inch.</p>
<p>Lots of different ways to go about it, but initially to skip transient calcs I was thinking I could calculate the worst case by assuming 100% of oxygen reacts instantly. Put that energy into the ideal gas law and calculate the final pressure if unvented. Not sure if this will truly be the max pressure or if the transient shock wave pressure would exceed this?</p>
<p>The minimum pressure would be the psi required to lift and accelerate the roof up. The vented roof pressure loading should be pretty easy, again not considering transients.</p>
<p>Any thoughts on how to better roughly calculate wall pressures without cfd? Thanks!</p>
| |structural-engineering|fluid-mechanics|combustion|pressure-vessel| | <p>You have two pressures which are important here. One pressure (the one you are considering and that @Bart has calculated) is caused by the chemical reaction, i.e. instead of sugar dust and air you have some new gases (carbon oxides and water and probably nitrogen oxides) at elevated temperature. This is called the constant volume equilibrium pressure.</p>
<p>Additionally there is a flame front travelling through your explosive mixture that causes a jump in pressure (and density and temperature). This is the transient explosion pressure, and this pressure is much larger.</p>
<p>If you want to estimate the explosion pressure, for a first order guess you can calculate the <a href="https://en.wikipedia.org/wiki/Chapman%E2%80%93Jouguet_condition" rel="nofollow noreferrer">Chapman-Jouguet pressure (CJ pressure)</a>. There is a <a href="https://www.youtube.com/watch?v=2_9q0f6Lnvk" rel="nofollow noreferrer">good youtube video</a> that explains how to do that. Note that this is a 55 minutes lecture video.</p>
<p>Alternatively, if you do not want to do the calculation by hand, you can use <a href="https://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htm" rel="nofollow noreferrer">NASA's CEA software (Chemical Equilibrium with Applications)</a>. This will probably take a similar amount of time to use.</p>
<p>Or, you just use a number from <a href="http://www.dustexplosion.info/dust%20explosions%20-%20the%20basics.htm" rel="nofollow noreferrer">this</a> (8.5 bar) or <a href="https://www.fauske.com/blog/bid/316347/kst-and-pmax-tests-for-combustible-dust-who-or-what-are-they" rel="nofollow noreferrer">this</a> (8.5 bar) or <a href="https://www.crystalsugar.com/media/277431/Powdered-Sugar.pdf" rel="nofollow noreferrer">this</a> (9 bar) website, which all seem reasonable.</p>
<p>Note that you will have higher pressures in edges and especially corners due to concentration of shock waves. For this to calculate you need special CFD software.</p>
<p>And yes, 8.5 bar are enough to blow away the roof.</p>
| 19049 | How to calculate the pressure loading of a vented silo explosion? |
2018-01-23T21:29:25.167 | <p>I have read about the application of sewage sludge incinerators<sup>1</sup> but they do not have a net-positive power-generation.
<a href="https://i.stack.imgur.com/jjGGe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jjGGe.png" alt="enter image description here"></a>
Is there a way to generate net-power from sewage sludge?
What are the processes to do so?</p>
<hr>
<p>1: German: <a href="http://www.berlin-klimaschutz.de/de/projekte/energieerzeugung-durch-klaerschlammverbrennung" rel="nofollow noreferrer">www.berlin-klimaschutz.de/...</a></p>
| |power|waste-water-treatment| | <p>In order to generate net-positive power from sewage-sludge the chemical energy contained in the the sludge needs to be big enough in order to:</p>
<ol>
<li>Dry the sludge to an acceptable level</li>
<li>Overcome all process inefficiencies</li>
</ol>
<p>In a study from Dieter O. Reimann<sup>1</sup> the necessary energy content to maintain incineration of different wast types was calculated.
<a href="https://i.stack.imgur.com/0eP96.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0eP96.png" alt="enter image description here"></a>
Reimann concludes:</p>
<blockquote>
<p>To achieve self-propagating incineration, a minimum calorific value in the sewage sludge app. 2,500 kJ/kg above the energy required for drying should therefore be aimed at.</p>
</blockquote>
<p>An overview of the properties of sewage-sludge shows that even in dried form sewage-sludge is under the minimum value of 2,500 kJ/kg.
<a href="https://i.stack.imgur.com/Gpsyl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gpsyl.png" alt="enter image description here"></a></p>
<p>A energy-content of 20MJ/kg for <em>dried</em> sewage-sludge is reported by Gerhardt et al.<sup>2</sup>:
<a href="https://i.stack.imgur.com/JyH6k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JyH6k.png" alt="enter image description here"></a></p>
<p>This means there exist the possibility to produce electrical-power using sewage-sludge if the water-content is low enough. Which could be achieved by a solar-thermic-process<sup>3</sup>.</p>
<h2><a href="https://i.stack.imgur.com/TRvEe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TRvEe.jpg" alt="enter image description here"></a></h2>
<p>[1]: <a href="http://ec.europa.eu/environment/archives/waste/sludge/pdf/workshoppart4.pdf" rel="nofollow noreferrer">Problems about sewage sludge incineration</a>, founded by the <a href="http://ec.europa.eu/environment/index_en.htm" rel="nofollow noreferrer">European Union</a><br>
[2]: <a href="https://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/43_1_DALLAS_03-98_0197.pdf" rel="nofollow noreferrer">Fuel Characteristics Of Sewage Sludge...</a><br>
[3]: <a href="https://www.isah.uni-hannover.de/36.html?&tx_tkforschungsberichte_pi1%5BshowUid%5D=545&tx_tkforschungsberichte_pi1%5Bbackpid%5D=62&L=1&cHash=65d685bfd16ebb33ee7ceb807b8529e1" rel="nofollow noreferrer">Universität Hannover - ISAH</a> </p>
| 19052 | Is it Possible to Generate Power From Sewage Sludge? |
2018-01-25T08:53:03.320 | <p>Need help with this.</p>
<p>Part of a sea defense consists of a section of a concrete wall 4 meters high, and 6 meters wide.
<a href="https://i.stack.imgur.com/3M6jM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3M6jM.png" alt="enter image description here"></a></p>
<p>a) Calculate the resultant thrust on the section when the sea reaches a height of 3 metres relative to the base of the section.</p>
<p>b) Calculate the overturning moment experienced by the section about point A.</p>
<p>This is my answer. Have i done this correctly? I am not sure if i am on the right track with this or not. Especially the overturning moment.</p>
<p>a) Thrust</p>
<p>$Density of water = 1000kg/m^3$</p>
<p>Width = 6m</p>
<p>Height of fluid = 3m</p>
<p>$f = pg \frac{h} {2} A$</p>
<p>Ft = 1000 x 9.81 x 1.5 x (6x3)</p>
<p>Ft = 264,870 N</p>
<p>b) Overturning Moment</p>
<p>$M = Ft \frac{h} {3}$</p>
<p>$M = 264.870 x \frac{3} {3} = 264.870 Nm$</p>
| |mechanical-engineering|hydrostatics| | <p>Usually with this kind of problems what I would do is to analyse what effect is caused by what source. In your specific problem, there is water under gravitational load. The gravitational field imposes a <a href="https://en.wikipedia.org/wiki/Vertical_pressure_variation" rel="nofollow noreferrer">vertical pressure distribution</a> inside the water. This pressure acts on the wall.</p>
<p>Thus the first answer I would give is the distribution of pressure $p$ as a function of the gravity of earth $g$, the density of water $\rho$ and the vertical coordinate $z$ (I would choose $z=0$ at sea level and positive values of $z$ going downward, but you can choose a different coordinate system). This gives you a function $p(z)$. This is not directly asked for, but is required to answer the two questions.</p>
<p>In the second step, since the <a href="https://en.wikipedia.org/wiki/Influence_line#Distributed_loads" rel="nofollow noreferrer">load is distributed</a>, I would integrate this distributed load (force per unit length) along the height:
$$
F=\int_0^h w \cdot p(z) \, \mathrm{d}z
$$
with width $w=6\,\mathrm{m}$ and height $h=3\,\mathrm{m}$. The distributed load is the product of pressure and wall width, because the pressure does not vary with width. If it did, I would have to integrate not only along the height axis $z$, but also along the width axis.</p>
<p>Finally, to calculate the momentum, I would integrate the local <a href="https://en.wikipedia.org/wiki/Torque" rel="nofollow noreferrer">torque</a>, with is the product of the distributed load $w \cdot p(z)$ and the lever arm length $(h-z)$:
$$
M=\int_0^h w \cdot p(z) \cdot (h-z) \, \mathrm{d}z
\quad\mbox{.}
$$
Now you can calculate everything and check if my results are the same as yours. :-)</p>
| 19088 | Hydrostatic Pressure |
2018-01-25T09:24:17.987 | <p>I am currently working on abaqus modeling, I am trying to find a quick way to get the strain and stress values shown in the output file.</p>
<p>What do I need to put in the .inp file? Does anyone know how to resolve this?</p>
| |structural-engineering|finite-element-method|abaqus| | <p>just guessing you might be seeking ascii human readable reporting, you can put in the inp file, (after your first <code>*STEP</code> line.)</p>
<pre><code> *EL PRINT,ELSET=setname
E
S
</code></pre>
<p>this will dump the data into your ".dat" file. This is primarily useful if you want data for a small element set (like one particular element). If you need the whole model data do yourself a favor and learn to directly read the output database (odb) file.</p>
| 19089 | How to obtain strain and stress values by editing abaqus input file? |
2018-01-25T18:17:14.813 | <p><a href="https://i.stack.imgur.com/20Gcp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/20Gcp.jpg" alt="enter image description here"></a></p>
<p>I am currently making a corroded model in Abaqus, but not all locations are suffering from corrosion. As shown in the picture there are some parts (the blue parts) that do suffer from corrosion, so since the picture is only part of the overall bridge structure, I am wondering whether there exists many ways to make finer mesh size in corroded locations only without creating a partition?</p>
<p>Please comment if you know how to resolve this problem.</p>
<p>Thank You</p>
| |corrosion|abaqus|meshing| | <p>I have solved this problem
I used partition and adjust the mesh seed size </p>
| 19096 | is there any other ways to refine mesh size in particular place without making a partition? |
2018-01-26T21:33:17.913 | <p>I understand bevel gears are used in situations were a 90 degree change is required, but I can't visually understand why. Why can't we just use standard spur gears at 90 degrees, what do we actually benefit from using a bevel?</p>
<p>Essentially I'm asking how the engagement between gear teeth react if we just used spur gears for a 90 degree change, rather than bevel gears.<br>
<a href="https://i.stack.imgur.com/724Qq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/724Qq.png" alt="enter image description here"></a></p>
| |gears| | <p>One of the key design characteristics of a gear is its tooth profile. This is typically based on what is called an Involute Curve, and the shape of the tooth is designed to maintain consistent force through the tooth at its pitch line during rotation and to minimize backlash within the series. The interweaving of the teeth themselves is referred to as the gears "meshing" and the curvature of the teeth allow them to clear one another as the gear rotates. </p>
<p>This Wikipedia illustration give a good visualization of how the teeth actually "mesh": <a href="https://upload.wikimedia.org/wikipedia/commons/c/c2/Involute_wheel.gif" rel="nofollow noreferrer">https://upload.wikimedia.org/wikipedia/commons/c/c2/Involute_wheel.gif</a></p>
<p>I'd also take a look at <a href="http://Geargenerator.com" rel="nofollow noreferrer">http://Geargenerator.com</a> which give a great animation of gear tooth interaction and allows you to vary the tooth parameters to see how the shape of the teeth interact. </p>
<p>Now, reconsidering your original question, you can easily see that simply putting two flat spur gears at a right angle to one another would defeat the entire concept of proper gear "mesh" and in fact the teeth would bind because they wouldn't "fit" into one another. </p>
<p>A set of bevel gears solves this problem by placing the teeth on an angle so that they face one another and mesh properly!</p>
| 19113 | Why do we use bevel gears? |
2018-01-26T21:44:03.120 | <p>I'm reading this article on afterburners:</p>
<p><a href="https://www.airspacemag.com/flight-today/how-things-work-afterburners-18481403/" rel="noreferrer">https://www.airspacemag.com/flight-today/how-things-work-afterburners-18481403/</a></p>
<p>It says in the article:</p>
<blockquote>
<p>A typical jet engine uses only about half the oxygen it ingests, leaving a large amount of potential energy. The afterburner, which is a long extension at the back of the engine, combines much of the remaining oxygen with jet fuel, squirted into the high-speed exhaust stream from the engine’s turbine, and ignites the mixture.</p>
</blockquote>
<p>I've searched many places but I've never really understood why the engine only uses about half of the oxygen it ingests. Instead, the afterburner is needed to burn more oxygen. But why is not possible to design jet engines that burn more oxygen instead of adding another separate component, the afterburner? Why do jet engines only burn about a half of the oxygen ingested?</p>
| |combustion| | <p>There are some parts to your question:</p>
<ul>
<li>Part 1: The highest temperature is achieved with a air-fuel-ratio below 1.</li>
<li>Part 2: The temperature limit of available materials cooling processes is even lower.</li>
<li>Part 3: Given an upper temperature limit the thermodynamic-cycle of a jet-engine can only deliver more power by re-heating the exhaust-gases.</li>
<li>Part 4: Every design-process is a trade-off between multiple factors. For the after-burner two factors are efficiency and complexity of the engine.</li>
</ul>
<p><strong>Part 1: Combustion Reactions</strong><br>
Depending on the <a href="https://en.wikipedia.org/wiki/Air%E2%80%93fuel_ratio" rel="nofollow noreferrer">air-fuel-ratio</a> the combustion-temperature will vary.
The following figure shows the relation. Observe that the maximum power (think temperature) is not achieved for a air-fuel-ratio of 1 (, i.e. <a href="https://www.engineeringtoolbox.com/stoichiometric-combustion-d_399.html" rel="nofollow noreferrer">stoichiometric combustion</a>, $\lambda = 1$). This is due to dissociation of the combustion products.<br>
<a href="https://i.stack.imgur.com/tebUR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tebUR.png" alt="enter image description here"></a></p>
<p><strong>Part 2: Temperature Limit of Materials</strong><br>
Due to the high mechanical loads on the rotating turbine parts (aero- and centrifugal-loads) the maximum allowable temperature is below the theoretical maximum limit. The following figure shows the how the development of new materials did allow for higher temperatures in the jet-engine<sup>1</sup>.
<a href="https://i.stack.imgur.com/P94ji.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P94ji.png" alt="enter image description here"></a></p>
<p><strong>Part 3: Thermodynamic Cycle</strong><br>
The basic (idealised) thermodynamic cycle of a jet-engine is called <a href="https://en.wikipedia.org/wiki/Brayton_cycle" rel="nofollow noreferrer">Brayton-Cycle</a>. The following figure shows this cycle.
<a href="https://i.stack.imgur.com/JCL33.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JCL33.png" alt="enter image description here"></a>
The <a href="https://en.wikipedia.org/wiki/Enthalpy" rel="nofollow noreferrer">enthalpy</a>-difference between state 5 and 6 is what is converted into thrust. Given the temperature limits the only way to increase the power output of the thermodynamic cycle (vertical distance between state 5 and 6) is to re-heat the turbine-exhaust.
<a href="https://i.stack.imgur.com/lwAz6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lwAz6.png" alt="enter image description here"></a>
Observe that the temperature limit for stationary parts is higher, this is also due to the lower operating hours of the after-burner.</p>
<p><strong>Part 4: Engineering considerations, trade-offs</strong><br>
Usually the additional thrust produced by the after-burner is not needed throughout the whole mission. Therefore it is acceptable to sacrifice efficiency for mechanical simplicity.
<a href="https://i.stack.imgur.com/bYVoF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bYVoF.png" alt="enter image description here"></a>
The shown <a href="https://de.wikipedia.org/wiki/Eurojet_EJ200" rel="nofollow noreferrer">EJ-200</a> also show the engine stations. It is easily seen the difference in mechanical complexity between stations 4-5 and stations 5-6.</p>
<p>Conclusion:</p>
<blockquote>
<p>But why is not possible to design jet engines that burn more oxygen instead of adding another separate component, the afterburner? </p>
</blockquote>
<p>It is theoretical possible to design an engine which would burn all oxygen. However, until now this was not necessary.</p>
<blockquote>
<p>Why do jet engines only burn about a half of the oxygen ingested?</p>
</blockquote>
<p>Because fuel is burned for maximum heat not to deplete oxygen. The maximum temperature is limited by material and dissociation.</p>
<hr>
<p>[1]: <a href="http://www.virginia.edu/ms/research/wadley/high-temp.html" rel="nofollow noreferrer">engineering.virginia.edu</a></p>
| 19114 | Why afterburners are needed instead of burning more fuel inside the turbine? |
2018-01-27T19:01:16.817 | <p>I was given this problem:</p>
<p>A current of 1,400 Amps flows in a copper bus bar. The bus bar is 0.6 cm wide by 9 cm long. You can assume the current flows only on the surface of the bus bar (so you can ignore the “depth” of the copper bar). Find the average electron velocity in the copper bus bar.</p>
<p>Copper has a free electron density of 1029 electrons/cubic meter.</p>
<p>Charge C = 1.6022e-19 Coulombs</p>
<p>Current = i = charge times velocity</p>
<p>n = amounts of charge per volume</p>
<p><strong>However because Im asked to ignore the third dimension of the copper bar
Im not sure if Im solving this problem correctly</strong></p>
<p>$$\begin{align}
i &= \frac{dq}{dt} \ldots \frac{n(volume)q}{dt}=\frac{n{A(V_dt)q}}{dt} \\
i &= \frac{n*A*V_dq}{dt} \\
V_d&=1.618E^{-8}um/s
\end{align}$$</p>
| |electrical-engineering|circuits| | <p>I would argue that your free electron density isn't correct. </p>
<p>Unless I have misunderstood your question, the equation you need is,</p>
<p>$$v_e = \frac{I}{nAq}$$
$$v_e = \frac{1400}{1029\times 5.4\times10^{-6} \times 1.6\times10^{-19}}$$
$$v_e = 1.572\times10^{25} \text{ m/s}$$
$$v_e = 1.572\times10^{31} \text{ $\mu$m/s}$$</p>
<p>$n$ should be about $8.5\times10^{28}$, though, which makes the actual answer:</p>
<p>$$v_e = \frac{1400}{8.5\times10^{28}\times 5.4\times10^{-6} \times 1.6\times10^{-19}}$$
$$v_e = 0.01906 \text{ m/s}$$
$$v_e = 1.906\times10^{-8} \text{ $\mu$m/s}$$</p>
| 19127 | Find average electron velocity on copper bus bar(um/s) |
2018-01-27T23:05:59.587 | <p>I've been trying to get this sweep to work for a while now. It works with a solid circle. It doesn't work with an annulus. I can provide more info if necessary. I made sure the order of the features matched what I've seen in other threads about this issue but that doesn't seem to help. Every time I try to build it, I get the error "The intermediate profile # 2 could not be solved."
<a href="https://i.stack.imgur.com/uea3z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uea3z.png" alt="enter image description here"></a></p>
| |solidworks| | <p>You are missing a huge number of guide curves in this scenario. I've taken a screenshot showing what happens when you have an internal and external diameter guide curve for just one 'tooth' on your profile - the other vertices aren't guided, attempt to go in a straight line, and produce self-intersecting geometry that can be previewed, but not built.</p>
<p><a href="https://i.stack.imgur.com/Xwb2j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xwb2j.png" alt="Two of 8 guide curves"></a></p>
<p>I am, therefore, going to suggest another way to model your part, since adding all of those curves would be exceptionally time consuming...</p>
<p>Given that your "Flow Axis" is a straight line, this would make <em>much</em> more sense if it were built using a revolve, rather than a sweep. This gives you full control over the cross section.</p>
<p>I have modelled your part using the following procedure:</p>
<ol>
<li>Create a 'master sketch', showing the internal profile, external profile, and the root of the grooves. I've assumed that the wall thickness should scale down as the diameter reduces (what would happen if the sweep/guide curve procedure you were trying had 'worked'), but if you wanted constant wall thickness, then simply edit the 'master sketch' to use 'offset' curves from the outer diameter, rather than the curves shown.</li>
<li>Revolve the inner area, the solid segment of the body through 360 degrees</li>
<li>Revolve the outer area, the 'ribs', through 8 degrees (my geometry is an arbitrary guess, of course)</li>
<li>Circular Pattern the rib revolve into 24 instances</li>
</ol>
<p><a href="https://i.stack.imgur.com/EK7Fb.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EK7Fb.gif" alt="Demo Gif"></a></p>
<p><a href="https://i.stack.imgur.com/Lo3Uf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lo3Uf.png" alt="Finished Item"></a></p>
| 19129 | Sweep with guide curve in SolidWorks |
2018-01-28T00:56:34.527 | <p>I have this pickup truck that I love dearly. I often change the aerodynamic configuration of the truck when I use my camper shell, or roof top tent, etc.</p>
<p>Is there a reasonably easy way to create a mesh model of my truck from images so that I can use ANSYS to model the aerodynamic effects?</p>
<p>Thanks!</p>
| |automotive-engineering|aerodynamics|mechanical| | <p>What you seek is called 3D photogrammetry and is a relatively exciting segment of 3D modeling. You can use those terms to search with your favorite search engine and be overwhelmed with options. I found a link to a summary performed a bit more than a year ago, which might be more useful to you.</p>
<p><a href="https://pfalkingham.wordpress.com/2016/09/14/trying-all-the-free-photogrammetry/" rel="nofollow noreferrer">https://pfalkingham.wordpress.com/2016/09/14/trying-all-the-free-photogrammetry/</a></p>
<p>The author of the summary took photos of a couple different models to present reasonable challenges to the software available and posted his results.</p>
<p><a href="https://i.stack.imgur.com/Tm4y0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tm4y0.jpg" alt="3d model"></a></p>
<p>The image above came from the linked site. You'd be learning to use image processing software to clean up the results in order to accomplish your goal, but it's time well spent, considering the image file you'd have in return.</p>
| 19131 | Scanning and modeling my personal vehicle for aerodynamic simulation |
2018-01-28T15:06:10.623 | <p>I am a chemical engineering student learning about material balance, so the problem below will not be a ‘real’ one , but just an exercise for me to understand better. </p>
<p>This system below is not a reaction.</p>
<p>My question is: can I say that the overall material balance for this system is:</p>
<p>$$F1 + F2 + F3 = P1 + P2 + P3$$</p>
<p><a href="https://i.stack.imgur.com/vICgQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vICgQ.jpg" alt="enter image description here"></a></p>
<p>I have this doubt because Feed 3 comes in after Product 1 and Product 2 have been removed. Does this matter at all? </p>
| |chemical-engineering| | <p>I'm assuming R is a recycled material that gets reprocessed from product stream P1. </p>
<p>Given the fullness of time and the complete reprocessing of R so that at the end of production, no R is present then you can say that F1 + F2 + F3 = P1 + P2 + P3.</p>
<p>However, processes and processing plants rarely operate that way. Usually there will be an amount of R within the system. To account for this F1 + F2 + F3 = P1 + P2 + P3 + R.</p>
<p>With material balance, conservation of mass is paramount - what goes in must come out.</p>
<p>Regarding your concerns about F3 being added after P1 and P2 have been removed, F3 is added to the residual from the P1 and P2 process to produce P3. Your concern would be valid if there was no residual feeding into the second process to which F3 is added.</p>
<p>You need to consider the system as whole when it comes to material balance. There are three inputs/feeds and three products, with some of the feed recirculating.</p>
| 19134 | Concept on material balance |
2018-01-29T01:22:41.297 | <p>I have this data sheet for a stepper motor:</p>
<p><img src="https://i.stack.imgur.com/b1bu0.png" alt="stepper motor data sheet"></p>
<p>The pitch diameter of the pinion attached to it is listed as "Ø6". What unit is this? Looking at the <a href="https://en.wikipedia.org/wiki/https://en.wikipedia.org/wiki/%C3%98" rel="nofollow noreferrer">wikipedia page for Ø</a> it seems like this is just a nonce symbol indicating that the measurement in question is a diameter, but then the question still remains, what unit is it? Does it mean 6 millimeters? Is there some convention about default units in data sheets I'm missing?</p>
| |mechanical-engineering|motors|gears| | <p>Engineering drawings come in two flavours metric and north american (mainly US only). Metric drawings are using mm as a unit, even in architecture where a mm is a slightly small unit. North american drawings are using inches as their base unit. If they would use anything else then the unit would read next in the dimension (this is incredibly rare). </p>
<p>Now, the sheet states that the unit is mm in the title block. The title block applies to the whole drawing. Other much weaker hints are the A4 size paper (ISO size indicates mm), projection rotation rule (image is drawn in first angle, US almost exclusively use third) and scale info. Although you really shouldn't measure distances on drawings since the numbers are what counts and there can be reproduction errors and/or dimensions not in scale.</p>
<p>Other possible hints would be invoking any ISO standards vs ASME standards which here are not present (for example a general tolerance). Also simply assuming mm* would work for the bulk of drawings made for any other purpose than US consumption. Besides in this case inhes would sound incredibly big.</p>
<p>* Although, inches are much more common in electronics than mechanical drawings, so best not guess anything.</p>
| 19139 | Meaning of O-slash in data sheet about gear diameter? |
2018-01-29T14:46:57.397 | <p>I'd like a simple explanation of the function of <strong>by pass<a href="https://i.stack.imgur.com/nFU5c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nFU5c.png" alt="enter image description here"></a></strong>,</p>
<p>here is an image of the project am working on and if it can help clarify.</p>
| |mechanical-engineering|pipelines|petroleum-engineering| | <p>“By-pass” generally means to take an alternative path or route around something, so could be a village or a valve or a flow meter.</p>
<p>Sometimes meters have a by-pass arrangement with 3 valves so that they can be replaced without stopping the flow.</p>
| 19144 | What is the use of a " by pass " |
2018-01-29T15:48:13.467 | <p>I have done some research about the subject but I can't find the difference between Temperature Element (<strong>TE</strong>) and Temperature Indicator (<strong>TI</strong>)
<a href="https://i.stack.imgur.com/k0zhA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k0zhA.png" alt="enter image description here"></a></p>
<p>Here is an image of the P&ID of the project am working on if it can help clarify my question.</p>
| |mechanical-engineering|electrical-engineering|instrumentation| | <p>TI = Temperature Indicator; Shows temperature local (in pipe, canal, wall, …)
TT = Temperature Transmitter; Remote monitoring temp. (0-10V, 4-20mA, ...)
TE = Temperature Element; Remote monitoring temp. by resitance (Pt100, NTC 5k, ...)</p>
<p>TTI = Temperature Transmitter Indicator; Remote monitoring temp. with display</p>
<p>PT = Pressure Transmitter
TA = Temperature Alarm
TAh = Tempreature high Alarm
FI = Flow Indicator
QTrh = Quality Transmitter relative humidity</p>
| 19147 | What is the difference between TE and TI? |
2018-01-29T18:46:39.440 | <p>I am trying to find the torque exerted on the shaft of a capstan and bow drive. In particular, the system I'm looking at has two wire rope pulleys, and two idler pulleys, and is used to drive a carriage as seen Below:
<a href="https://i.stack.imgur.com/B2ndK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B2ndK.png" alt="enter image description here"></a></p>
<p>In this case the drive pulley consists of two separate pulleys tensioned against one another, with the end of two wire ropes terminated in the pulleys. </p>
<p>Naively, I would assume that the force F is equal to the Torque T * the radius r of the drive pulley. </p>
<p>Are there any additional effects due to a multiple number of wraps on the pulley (one side reels out, while the other side reels in, keeping the total angle of wraps equal throughout the motion)?</p>
<p>Does the capstan effect play a large part with this mechanism? From what I understand the capstan effect should cause less tension at the wire rope stops embedded in the drive pulley, but not effect F. </p>
<p>Thanks!</p>
| |torque|pulleys|power-transmission| | <p>In theory, the force is just the torque, divided by the radius.
In practice, wrapping the rope a few times around the pulley will rob a little torque from it, but it may even be unnoticeable. At the same time, friction between rope and pulley will increase, thus allowing you to transfer more force to the rope before it starts slipping over the pulley. It also helps to tension the rope to reduce slipping(much like a derailleur on mountainbikes), but again it'll cost you a little torque. So only consider it if the rope keeps slipping.</p>
| 19150 | Torque Required to Drive Capstan and Bow |
2018-01-30T05:24:00.817 | <p>Why the closed loop dominant poles of the root locus can show the response of the system? Wont it neglect the effect of the closed loop zeros?</p>
<p>As I read on the books, root locus method deal with the closed loop poles. It sketch the locus of the close-loop poles under an increase of one open loop gain(K) and if the root of that characteristic equation falls on the RHP. It means the close loop pole fall into RHP and make system unstable. But the exercises and examples also treat Root locus as a method of designing compensation or gain to fulfill the system requirement like damping factor and setting time. Isn't the root locus method for stability only?</p>
<p>I thought the system response should include the closeloop zero. Did I miss something? </p>
<p>Also, can Nyquist plot also used to show the system response and to design the system? As I thought its some how same as Root locus, with frequency being the varying K(gain).</p>
| |control-engineering|control-theory| | <p>Late answer here. Zeros can have a significant effect on CL (closed loop) response. If they are within the bandwidth of the loop, they would cause overshoot in the step response. The CL stability is not affected as discussed elsewhere.</p>
<p>It may be related to note, that it's often easy to remove unwanted zeros from the CL transfer function.</p>
<ul>
<li>LHP zeros can be cancelled or approximately-cancelled directly at the input.</li>
<li>It's also possible to accomplish the equivalent result by shifting zeros into the feedback path, and poles into the forward path. It can be a useful alternative to direct cancellation, in case the unwanted CL zero was created by a controller zero that can change. This will be shown below:</li>
</ul>
<p>Consider a loop with the familiar CL transfer function <span class="math-container">$\displaystyle\frac{F}{1+FG}$</span></p>
<p><a href="https://i.stack.imgur.com/Igsmo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Igsmo.png" alt="basic-loop-fg" /></a></p>
<p>Let's break F and G into numerator and denominator (that is, zeros and poles).</p>
<p><span class="math-container">$F:=\displaystyle\frac{N_F}{D_F}$</span> and <span class="math-container">$G:=\displaystyle\frac{N_G}{D_G}$</span></p>
<p>Thus</p>
<p><span class="math-container">$\displaystyle\frac{F}{1+FG}=\frac{\frac{N_F}{D_F}}{1+\frac{N_FN_G}{D_FD_G}} = \frac{N_FD_G}{D_FD_G+N_FN_G}$</span></p>
<p>To avoid CL zeros (or more commonly, just to eliminate the stray one that falls into the CL bandwidth), we simply need to not place the unwanted zero into <span class="math-container">$N_FD_G$</span>. We can usually do this while still obtaining the desired <span class="math-container">$(1+FG)=(D_FD_G+N_FN_G)$</span>, by simply moving the "troublesome" zero out of <span class="math-container">$N_F$</span> and into <span class="math-container">$N_G$</span>. (And less commonly, out of <span class="math-container">$D_G$</span> and into <span class="math-container">$D_F$</span>).</p>
<p>The root locus analysis is completely unaffected.</p>
| 19157 | Control theory: Closed loop zeros, Root locus and its dynamic response |
2018-01-30T12:02:18.263 | <p>A contractor delivered three components, all are supposed to be in 1.4571. Part A is actually done in <a href="http://www.m-woite.de/en/materials/14404.shtml" rel="nofollow noreferrer">1.4404 / (X2CrNiMo17-12-2)</a> (according to pretty conclusive documentation). Part B is allegedly done in <a href="http://www.m-woite.de/en/materials/14571.shtml" rel="nofollow noreferrer">1.4571 / (X6CrNiMoTi17-12-2)</a>, but the material looks like part A (shiny, silvery). Part C meanwhile is also allegedly done in 1.4571, but looks different (silvery gray). "Allegedly" means that for the documentation for parts B and C is far less conclusive.</p>
<p>So, from someone who has encountered both steels: Would I expect them to look the same?</p>
<p>Or, to rephrase my question: Part A and C look the same but, according to the documentation provided, should be different materials - does this tell me something is amiss?</p>
| |mechanical-engineering|materials|steel| | <p>Not familiar with Russian specs but these look like ( in USA names) 316 L and 316 ; If so , you could not tell them apart by appearance. Also , if they have similar composition tolerances to USA specs you may not be able to separate them with chemical analyses ; 316 may have identical composition to 316 L. </p>
| 19169 | Do the steels 1.4404 and 1.4571 look the same? |
2018-01-30T16:24:56.690 | <p>I am looking for a way to double check readings I am getting through a mass flow meter that appear to be reading lower than expected. </p>
<p>Once flow starts in the system, the pressure regulator at the supply tank drops in pressure readout. This drop should be due to the dynamic pressure of the flow of gas through the regulator so I was thinking it may be possible to correlate the mass flow meter's readings with the calculated mass flow resulting in the drop in pressure at the regulator. I started by using the ideal gas law (low enough pressures) to calculate the density of the gas at the static pressure read by the regulator.</p>
<p>$$\rho=\frac{PM}{RT}$$</p>
<p>Then I took the pressure difference read from the tank regulator between the static no-flow to flowing as the dynamic pressure.</p>
<p>$$P_d=\frac{1}{2}\rho v^2$$</p>
<p>Using the above equation I calculated the velocity of the fluid flow, and then just used the cross sectional area to calculate the mass flow.</p>
<p>$$\dot{m}=\rho A_c v$$</p>
<p>When calculating the density I am getting an unreasonable value for fluid velocity so I was wondering if this procedure makes sense?</p>
| |fluid-mechanics|compressible-flow| | <p>If I am not mistaken, you are looking at the drop in pressure right at a pressure reducing regulator when flow initiates, and wanting to infer mass flow from the value of this drop. If this is true, I would like to add that there are many reasons why a pressure regulator would "droop" other than the velocity head effect. Regulators have springs that compress when the fill valve opens, and there are several friction losses that together comprise the flow performance curve. Many designs have "lock up" that presents a significant offset between the static and flowing conditions. Have I interpreted your question correctly?</p>
| 19174 | Using pressure drop to correlate mass flow |
2018-01-31T00:37:42.570 | <p>In order to build a lighting system for a microscope (with a digital camera) I want to make a very small diameter laser beam (parallel, rather than convergent). If possible around 1 um diameter. </p>
<p>How could I build such a laser using a common diode laser? </p>
| |optics|lasers| | <p>If you want a small illumination area on the object plane, start by focussing your source onto a 1-micron pinhole, then relay that pinhole (a single biconvex lens will suffice) to the object plane. Be aware that:<br>
1) rather little light will make it through the pinhole<br>
2) due to diffraction effects, you will get a center spot and surrounding rings of illumination. As Jack B pointed out in his answer (for a parallel beam), you can't do better than this. </p>
| 19180 | How can I build a micrometer diameter laser beam from a common laser diode? |
2018-01-31T08:31:33.413 | <p><img src="https://i.stack.imgur.com/VbOaj.jpg" alt="enter image description here"></p>
<p>I want to have a star dot above each bar chart. I don't know how to present that index. Please anyone help me to resolve this problem,</p>
| |statistics| | <p>Quite seriously: <em>Friend Don't Let Friends Use Excel</em>. Learn to use Matlab/Octave, or python, or R, or just plain Gnuplot.<br>
No matter what tool you use, since "significance" is a separate variable, you need to convert your collection of "*" , "**" etc to numeric values and plot the desired symbols at the matching category locations. </p>
| 19185 | How do I include statistical significant index in bar chart in excel? |
2018-01-31T22:50:57.180 | <p>If trying to achieve a 1:4 gear ratio, will a planetary gear system help reduce the loss of torque? Or is it essentially no different than the loss of torque when using a large drive gear to rotate a small driven gear?</p>
| |mechanical-engineering|gears|torque| | <p>Apart from frictional losses, which should be fairly low for a 1:4 ratio, the torque/speed ratio is inescapable ie if you increase the angular speed by a factor of 4 the torque must be reduced by a factor of 4. </p>
<p>Power is torque x angular velocity so if this was not the case you would get more power out than you put in, which as we know is not possible. </p>
<p>Planetary gears can be used to produce quite high ratios in a compact package and also have the ability to change ratios by locking different combinations of input/output/carrier eg in automatic transmissions but their fundamental principal of operation is the same as any other type of gear or indeed any type of mechanical advantage. </p>
| 19199 | Gear acceleration and maintaining torque |
2018-02-01T13:41:04.173 | <p>In theory of bending of beams, often these terms are used. There are no clear definition present. Most of them contradicting. Which of the property(bending moment and shear force) is zero in which case? </p>
<p>The definition of <a href="https://en.wikipedia.org/wiki/Contraflexure" rel="nofollow noreferrer"><code>contraflexure</code></a> is defined as the point of zero bending moment. Then why have another term, i.e <code>inflection point</code>? </p>
<p><em>NOTE:The wikipedia link has no references.</em></p>
| |mechanical-engineering|structural-engineering|applied-mechanics|beam| | <p>From what my teacher taught me, both are the same point.It is called contraflexure in the bending moment diagram (where the bending moment changes sign, i.e hoging to saging or the other way) .Here bending moment neednot be zero, it just changes sign(it neednot change gradually, it can also make a sudden jump).It is called the point of inflection in the deflection curve also called elastic curve (the point where the deformation changes from concave to convex or the other way)
<img src="https://i.stack.imgur.com/6pF1p.jpg" alt="enter image description here"></p>
<p>Refer this question I asked in Math stack exchange
<a href="https://math.stackexchange.com/questions/3095917/can-point-of-inflection-occur-at-a-point-where-second-derivative-doesnot-exist">https://math.stackexchange.com/questions/3095917/can-point-of-inflection-occur-at-a-point-where-second-derivative-doesnot-exist</a> , for better understanding of point of inflection. For furthur reference use these links
<a href="https://math.stackexchange.com/questions/1021582/definition-of-point-of-inflection">https://math.stackexchange.com/questions/1021582/definition-of-point-of-inflection</a>
<a href="https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist">https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist</a>
<a href="https://math.stackexchange.com/questions/688503/what-is-inflection-point">https://math.stackexchange.com/questions/688503/what-is-inflection-point</a></p>
| 19206 | Are the points of inflection and contraflexure the same? |
2018-02-01T18:50:43.967 | <p>I have an extractive distillation column set up in AspenPlus. Methylcyclohexane (MCH) is expected to come out from the top (in the distillate stream) and the bottoms stream is mostly Toluene/Phenol, with a bit of MCH. The feeds are MCH+Toluene (at stage 8 out of 25) and Phenol (at stage 18 out of 25).</p>
<p>Temperature - 100 C. Pressure - 1 atm.</p>
<p>When I use the N.R.T.L (non-random two-liquid) model for the simulation, I get weird results. However, when I use UNIFAC, I get what I expect.</p>
<p>I thought NRTL was supposed to take into account deviations from ideality (caused by the phenol group?), so I don't know what's throwing it off.</p>
<p>Any suggestions?</p>
| |thermodynamics|chemical-engineering|modeling| | <p>At moderate temperatures and pressure NRTL should be a good option provided that the binary interaction parameters are correct. </p>
<p>You need to check:</p>
<ol>
<li><p>Whether the binary interaction parameters for your three compounds are in Aspen Plus database. If not, try to find them in literature or, use UNIFAC to determine only the unknown parameters (most process simulators allow you to do that).</p></li>
<li><p>If the three binary interaction parameters are in the Aspen Plus database, check whether they represent the experimental behaviour of your system. In particular, try to reproduce binary Pxy and/or Txy diagrams. Otherwise, try to reproduce any experimental bubble or dew point information you might have.</p></li>
</ol>
<p>Of course, if you don't have good binary interaction parameters, UNIFAC being totally predictive is a good choice. As you know, UNIFAC divides a molecule into groups which contribute differently to the Gibbs free energy. The compounds you mention can be decomposed into fairly standard groups (you can check the list <a href="http://www.ddbst.com/published-parameters-unifac.html#ListOfMainGroups" rel="nofollow noreferrer">here</a>) and UNIFAC results should be reliable.</p>
<p>Finally, I have to say that, though uncommon, it may happen that process simulators contain errors in interaction parameters (NRTL, Wilson, UNIQUAC) or similar that lead to poor predictive capabilities. I did work in a project (mostly confidential) where a widespread process simulator completely failed to reproduce the VLE behavior of something similar to 2-butanol in water/methanol solution because the interaction parameters were wrong. We detected it comparing with experimental data of Txy and we regressed and corrected the parameters.</p>
| 19214 | Failure of N.R.T.L method to model Phenol / Methylcyclohexane / Toluene system in AspenPlus. Why? |
2018-02-02T08:56:23.410 | <p>I asked a <a href="https://engineering.stackexchange.com/questions/19168/flow-in-a-t-junction-with-pipes-of-different-diameters">previous question</a> about pipe t-junctions with pipes of varying diameters, and it got me wondering... </p>
<p><strong>How are losses handled and calculated in pipes with holes drilled directly into them?</strong> Flow runs perpendicular to the hole and most of the theory I can see for flow through a pipe exit assumes the pipe and orifice are coaxial. Or for an orifice plate directly in the pipe flow, neither of which seem appropriate.</p>
<p>I'm guessing some theory must exist since common objects like lawn sprinklers contain these pipes with many holes drilled into them. I have been thinking about a pipe with one hole drilled in it of small diameter. Essentially we have a free jet exiting perpendicular to the main pipe flow. </p>
<p><a href="https://zavax.files.wordpress.com/2015/09/vlcsnap-2015-10-05-14h19m29s85.png" rel="nofollow noreferrer"><img src="https://zavax.files.wordpress.com/2015/09/vlcsnap-2015-10-05-14h19m29s85.png" alt="enter image description here"></a></p>
| |mechanical-engineering|fluid-mechanics|civil-engineering|pipelines|fluid| | <p>So, I attempted to solve a simplified problem using Bernoulli, it is an approximation though, since it doesn't account for the change in flow direction and subsequent turbulence caused after the hole. I rearranged equations for volumetric flow rate and head loss. <strong>Please let me know if you spot any errors.</strong> </p>
<p><a href="https://i.stack.imgur.com/AYuO7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AYuO7.png" alt="enter image description here"></a></p>
<p><strong>Note:</strong> the last two lines rearrange from an earlier form of the equation to find pressure drop, and <strong><em>SG</em></strong> refers to specific gravity of the fluid. </p>
| 19228 | Losses through holes drilled in a pipe |
2018-02-02T10:47:11.907 | <p>I would like to compare the shape of the frequency spectrum of a plastic and aluminium component. Now if we assume that the plastic and aluminium component only differ in the material, while the mass and stiffness stay the same (I know, large assumption), how would generally differ the shape of the frequency spectrum from each other? Obviously, the damping plays a major role here and would be much higher in case of plastic, therefore the response amplitude would be lower, the resonance zone wider and a shift (which is many times neglected) towards a lower frequency would be made... <strong>Could I in a very simplified version, just for the sake of understanding, draw it like this?</strong>
<a href="https://i.stack.imgur.com/QbU0S.png" rel="noreferrer"><img src="https://i.stack.imgur.com/QbU0S.png" alt="enter image description here"></a>
Another question would be the temperature dependence of the elastic modulus. If the temperature rises, the modulus is reduced which also affects the eigenfrequency. In case of aluminium not so much, but in case of plastics, the difference can be enormous. My understanding is that in case of elevated temperatures, the damping increases; in case of plastic much more than in case of aluminium. <strong>Could I visualize this as:</strong>
<a href="https://i.stack.imgur.com/Ap5XC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Ap5XC.png" alt="enter image description here"></a></p>
<p>Any constructive answers are much appreciated!
Kind regards,</p>
<p>Luka</p>
| |mechanical-engineering|structural-engineering|materials|frequency-response|eigenvalue-analysis| | <p>In the perfectly elastic spectrum, response to the <strong>dampening</strong> coefficient of the second order differential equation of vibration behaves very similar to how you have shown, but not quite. It is well documented by reviewing how a system responds when subjected to a forced time-varying load at different speeds. Notice how it doesn't get wider for the frequency shifts as the dampening increases:</p>
<p><a href="https://i.stack.imgur.com/RfEYv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RfEYv.png" alt="enter image description here"></a></p>
<p>$$\begin{gather}
x(t) = X_0 \sin(\omega t + \phi) \\
X_0 = \dfrac{KF_0}{\sqrt{\left(1-\dfrac{\omega^2}{\omega_n^2}\right)^2+\left(2\varsigma\dfrac{\omega}{\omega_n}\right)^2}} \\
\phi = \tan^{-1}\left(\dfrac{-2\varsigma\dfrac{\omega}{\omega_n}}{1-\dfrac{\omega^2}{\omega_n^2}}\right)
\end{gather}$$</p>
<p>However, as time goes on, the natural frequency of a material experiencing <em>creep</em> will vary. While creep adds to the dampening, it adds some other time-dependent side effects. From <a href="https://en.wikipedia.org/wiki/Creep_(deformation)" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>Creep (sometimes called cold flow) is the tendency of a solid material to move slowly or deform permanently under the influence of mechanical stresses. It can occur as a result of long-term exposure to high levels of stress that are still below the yield strength of the material. </p>
</blockquote>
<p>As a result of creep, the natural frequency of the curve can become wider over time, but that's somewhat speculative until the actual creep mechanism is understood and the governing PDEs are resolved. However, this could give results similar to your temperature plot.</p>
| 19230 | Elastic (Young's) Modulus relation to the Eigenfrequency |
2018-02-02T17:40:26.280 | <p>What is the meaning of this statement "In a reciprocating engine the flow of intake charge takes place through the intake valve opening which is varying during the induction operation. <strong>Also</strong>, <strong>the</strong> <strong>maximum</strong> <strong>gas</strong> <strong>velocity</strong> <strong>through</strong> <strong>this</strong> <strong>area</strong> <strong>is</strong> <strong>limited</strong> <strong>by</strong> <strong>the</strong> <strong>local</strong> <strong>Sonic</strong> <strong>velocity</strong>"</p>
| |mechanical-engineering|engines|car| | <p>If the local velocity of the inlet charge through that area hits mach 1, the flow is "choked" by the presence of a normal shock wave. I have never heard of flow rates high enough to induce choked flow in an intake runner tube, however; this seems unlikely. </p>
| 19233 | What is Inlet-valve Mach index? |
2018-02-03T07:47:07.043 | <p>I have made bridge model in Abaqus,
I also have done experimental study and I have the load-displacement data from my experiment,</p>
<p>I am now going to compare the results from Abaqus and the results from experiment</p>
<p>From this step, I want to compare the load-displacement curves between experiment and analysis in Abaqus.</p>
<p>how to get this information from abaqus to excel, please help me resolve this problem, I really appreciate it</p>
<p>Thanks</p>
| |finite-element-method|abaqus| | <p>I found the answer, I input the separated input files to solve the maximum line number limit in dat file</p>
| 19238 | How to get all forces and displacements values in abaqus and export it to excel? |
2018-02-03T16:28:53.570 | <p><a href="https://i.stack.imgur.com/I4vHx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I4vHx.jpg" alt="This is the question "></a></p>
<p>I tried to solve this question I arrived at this diagram but taking the moments about the center of the beam I arrived at x=L/2. Can someone tell me what I need to do to get the correct answer?</p>
<p><a href="https://i.stack.imgur.com/xuNpn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xuNpn.jpg" alt="Diagram"></a></p>
<p><a href="https://i.stack.imgur.com/zw6mX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zw6mX.jpg" alt="This is my answer"></a></p>
| |mechanical-engineering|statics| | <p><strong>Second Attempt:</strong></p>
<p>After reading your comments, I approached the problem using a different process.</p>
<p><strong>Step 1:</strong>
Find the y-forces using $\Sigma$$F_y = 0$.</p>
<p>$(2\cdot Tsin(\theta))$ - $(W\cdot L)$ $=0$</p>
<p>$2Tsin(\theta)$ = $WL$</p>
<p><strong>Step 2:</strong> Treat the beam as a collection of beams.</p>
<p>I decided to break the beam into three parts. The outside edges can be thought of as cantilever beams that are attached by the supports.</p>
<p><strong>Step 3:</strong> Making the first cut at the midpoint to the left support.</p>
<p>Using the formula for bending moment of a simple beam under a uniformly distributed load, we can determine that $M_{mid} = \frac{WS^2}{8}$</p>
<p><strong>Step 4:</strong> Making the second cut at the left sling.</p>
<p>Using the formula for bending moment of a cantilever beam under a uniformly distributed load, we can determine that $M_{cantilever} = \frac{Wx^2}{2}$</p>
<p><strong>Step 5:</strong> Setting equations equal and solving.</p>
<p>From left end, to get the sum of moments to equal zero, $M_{cantilever} = \frac{M_{mid}}{2}$</p>
<p>Therefore:</p>
<p>$\frac{WS^2}{16} =\frac{Wx^2}{2}$</p>
<p>$WS^2=8Wx^2$</p>
<p>$S^2=8x^2$</p>
<p>$S = 2\sqrt{2}x$</p>
<p><strong>Final:</strong></p>
<p>As you can see, the center span $S = 2\sqrt{2}x$. Because $S = L-2x$, you can substitute to find the equation $L-2x = 2\sqrt{2}x$</p>
<p>$L = 2\sqrt{2}x + 2x$</p>
<p>$L \approx 4.8284x$</p>
<p>$x \approx \frac{1}{4.8284L} \approx 0.2071L$</p>
<p>_</p>
<p>Where $x$ is the distance from a support on the beam to the left end of the beam.</p>
<p>Where $S$ is the span of the beam.</p>
<p>Where $L$ is the length of the beam.</p>
<p>To clarify, the beam of length $L$ is divided into three sections; the left cantilever of length $x$, the center span $S$ of length $L-2x$, and the right cantilever of length $x$.</p>
<p>Where $Tsin(\theta)$ is y-component of the tension force.</p>
<p>Where $W$ is the weight of the beam.</p>
| 19243 | Beams question help |
2018-02-04T08:33:58.437 | <p>I have a vapor system that requires a 50 psi feed of DI water. Without getting fancy, is there a simple commercial off-the-shelf part I can buy, fill with water and hand-pressurize? Or something with an NPT inlet & outlet that I could pressurize with nitrogen or some other inert gas? At maximum, I think I'd need a 10 gallon reservoir - something that can sit on a lab bench.</p>
<p>The alternative is to slap a small pump in the reservoir, but that seems like overkill. </p>
| |pumps| | <p>A well storage pressure tank may serve your purpose. There are reverse osmosis tanks that are designed to handle up to 100 psi (per Amazon item post) and ordinary well tanks with internal bladders that provide at least your specified 50 psi.</p>
<p>Bladder tanks will be larger than the water capacity, as half the volume contains a rubber bladder. Air pressure fills the bladder, pressurizing the water as the well pump fills the water volume portion. When the pump reaches cut-off pressure, the bladder is compressed and maintains that pressure as the water is drawn off.</p>
<p>Without a bladder tank, as your tank empties, the pressure would drop.</p>
<p>If you will be hand-pressurizing the tank, you are the cut-off switch and can adjust via air-pressure gauge on the air pump or fill nozzle of the bladder.</p>
<p>A quick search discovered bladder well storage tanks as small as 2 gallon, 4 gallon, and 14 gallon which makes it more likely one will find a 10 gallon capacity.</p>
| 19248 | Simplest Way to Make A Pressurized Water Reservoir |
2018-02-04T19:19:59.343 | <p>I'm quite curious about the 3-ring release system used in parachutes, as described in this Wikipedia article:</p>
<p><a href="https://en.wikipedia.org/wiki/3-ring_release_system" rel="noreferrer">https://en.wikipedia.org/wiki/3-ring_release_system</a></p>
<p>I wonder why is this system so popular? I've been told it's nearly ubiquitous in the industry, but even after Googling and asking some skydivers I haven't been able to find out what makes it better.</p>
<p>More specifically, the Wikipedia article mentions that each of the loops multiply the mechanical advantage of the small cord loop that is secured by the yellow wire. Could somebody describe in detail how this multiplication of mechanical advantage takes place, and why is the number of ring chosen to be 3 instead of say, 5?</p>
<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/3Ring_release_animation.gif/220px-3Ring_release_animation.gif" rel="noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/3Ring_release_animation.gif/220px-3Ring_release_animation.gif" alt="3-ring release system"></a></p>
| |mechanical| | <p>Before the "3 ring circus," as we jumpers called it, were the Capewell variations. Here are some pics:<a href="https://www.flickr.com/photos/43867826@N07/sets/72157622676844920/" rel="nofollow noreferrer">https://www.flickr.com/photos/43867826@N07/sets/72157622676844920/</a></p>
<p>The "shot and a half" required moving a cover to access the release mechanism. "One shots" released by just moving the cover. </p>
| 19251 | Mechanical advantage of the 3-ring release system |
2018-02-05T12:41:03.837 | <p>I tried the question but I didn’t get the correct answer, can anyone tell me what mistake I made?</p>
<p><a href="https://i.stack.imgur.com/s8EpC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s8EpC.jpg" alt="Question"></a></p>
<p><a href="https://i.stack.imgur.com/DKaXC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DKaXC.png" alt="enter image description here"></a></p>
<p>$$\begin{align}
I &= \int y^2\text{d}A \\
&= 25^2\cdot10^{-6}\cdot200\cdot50\cdot10^{-6} \\
&= 25^2\cdot10^{-8} \\
\sigma &= \dfrac{My}{I} \\
&= \dfrac{10^3\cdot25\cdot10^{-3}}{25^2\cdot10^{-8}} \\
&= 4\cdot10^6
\end{align}$$</p>
| |mechanical-engineering|statics| | <p>While I agree with <a href="https://engineering.stackexchange.com/a/19268/1832">AndyT's philosophy</a> of simply using the solved equations, it is important to learn the origin of these solutions, so here we go.</p>
<p>The mistake you made was that you didn't perform the integration at all. You simply got the equation for the moment of inertia, ignored the integration sign, set $y$ as half the section's height, and replaced $\text{d}A$ with the section's area. That's not integration.</p>
<p>This is:</p>
<p>$$\begin{align}
I &= \int_A y^2\text{d}A \\
&= b\int_{-\frac{h}{2}}^{\frac{h}{2}} y^2\text{d}y \\
&= b \left.\dfrac{y^3}{3}\right|_{-\frac{h}{2}}^{\frac{h}{2}} \\
&= \dfrac{bh^3}{3}\left(\dfrac{1}{8}+\dfrac{1}{8}\right) \\
&= \dfrac{bh^3}{12}
\end{align}$$</p>
<p>Once you have the correct moment of inertia, calculating the stress is straightforward.</p>
| 19258 | Beam question to find the maximum stress |
2018-02-05T17:39:18.730 | <p><a href="https://i.stack.imgur.com/hZ0Sk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hZ0Sk.jpg" alt="Question"></a>
I have got 19.6 as the answer for (a) but the book tells 14 is the answer. Can someone tell what mistake I have done or is there in the book.
<a href="https://i.stack.imgur.com/Atefp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Atefp.jpg" alt="This is my solution "></a></p>
| |mechanical-engineering|dynamics| | <p>You omitted the inertia of the mass of $0.5 kg$. The total inertia of the system is the inertia of the disk plus the inertia of the mass.</p>
<p>$$ I = \frac{M R^2}{2} + m R^2 $$</p>
<p>The total torque about the axle is $T= m g R$. Then use $I \alpha =T$ to get the equation</p>
<p>$$ \left(\frac{M R^2}{2} + m R^2\right)\alpha=m g R$$</p>
<p>This gives</p>
<p>$$\alpha=\frac{2 g m}{R (2 m+M)}=\frac{(2) (9.8) (0.5)}{(0.2)((2)(0.5)+2.5)}=\frac{9.8}{0.7}=14$$ </p>
| 19263 | Torque and angular acceleration |
2018-02-06T23:17:41.103 | <p>From what I understand most types of steel are strong both in tension and compression, yet they are still usually significantly stronger in tension. Suppose a common I-beam supported on both ends and with a load on top of it; the bottom flange experiences tension, while the top one compression, yet since steel is somewhat stronger in tension, why would the bottom flange be as thick as the top one? Is it not done because it's easier to manufacture symmetrical ones or is there some other reason?</p>
<p>Is it possible to use assymetrical I-beam to save weight and steel in that case or is my thinking completely off? </p>
| |structural-engineering|steel|beam| | <p>Academically, I believe you are correct in that you could potentially optimize an I-beam to have a higher failure-load to mass ratio. You would have to play with the <a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="noreferrer">area moment of inertia</a> equations to check for sure. <a href="https://en.wikipedia.org/wiki/T-beam" rel="noreferrer">T-beams</a> would be the extreme of what you are proposing. They are less common, but composite steel/concrete T-beams are used extensively in parking garages, with additional steel in the lower tensile section.</p>
<p>It is important to note that many structures are design-limited by maximum deflection rather than structural failure. Additional area on the tensile member will reduce this defection. It would be interesting to see how deflections would compare between a symmetrical and the hypothetical failure-load optimized beam.</p>
<p>Symmetrical is superior for torsional loads or compound loads where the flange loading may be variable. Symmetrical is also superior for columns because it reduces eccentricity and the potential for buckling for a given mass/length.</p>
<p>Being able to use a beam in multiple situations is important because structural steel is a commodity. The price of a particular style of I-Beam is inversely proportional to how much it is used in industry in your region. This is often a bigger factor than the cost of the additional steel in the beam or cost of the additional mass in your design.</p>
<p>Additionally if your structure requires lots of trapezoidal shapes, you want to use symmetrical material so you waste less material.
Right-side-up > Upside-down > Right-side-up > Upside-down...</p>
<p>Also, for every symmetrical beam, it saves the engineers an orientation specification in the drawings, and saves the fabricators a potential error.</p>
<p>Here is a good <a href="http://publications.lib.chalmers.se/records/fulltext/182546/182546.pdf" rel="noreferrer">I-beam thesis</a> to research further.</p>
| 19279 | Why are steel I-beams mostly symmetrical? |
2018-02-07T14:13:03.557 | <p>I am trying to solve this but I am stuck; I have watched loads of YouTube videos but still don't understand how to complete it:</p>
<p>A mass of $m=0.12 kg$ of air has an initial temperature of $T_1=500°C$ and pressure of $p_1=0.8 MPa$. If the air is expanded according to the law $pV^{1.2} = c$ to a final volume of $90\ litres$, determine</p>
<p>i) its initial volume, $V_1$</p>
<p>ii) its final pressure, $p_2$</p>
<p>iii) its final temperature. $T_2$</p>
<p>For air, take $R_{specific} = 287 Jkg^{-1} K^{-1}.$</p>
<hr>
<p>I have got these equations which i believe i need to use.</p>
<p>$$pV = nRT$$</p>
<p>$$n = \frac{p V}{R T}$$</p>
<p>Are these the correct equations to use?</p>
<p>I also think the fixed amount of gas is the constant?</p>
<p>I have been looking at Boyle's law. And Charles' law.</p>
<p>Any help appreciated.</p>
<p>Thanks.</p>
| |mechanical-engineering|thermodynamics| | <p>You can also solve the problem without knowing the average molar mass air, which is why R(specific) is given.
Multiplying R(specific) with mass will give you the number of moles multiplied by the universal gas constant.
i.e mR(specific)=nR
Proceed on the usual lines to get the answer after this.</p>
| 19288 | Applications of Thermodynamics |
2018-02-07T16:06:19.180 | <p>I'm curious about which scenarios one would use a force function when using a PLC programming interface. Examples pertaining to process engineering, manufacturing, or industrial settings would be ideal.</p>
| |manufacturing-engineering|process-engineering|instrumentation| | <p>A "force" in a PLC is the act of forcing a memory bit on or off from the PLC programming software. It is used exclusively for testing. Forcing a bit on or off over-rides all other PLC logic. You can think of it like both setting it and making it read only. "set" a bit is also an option with most PLCs and is also useful for testing. It differs from "force" in that it changes the value only once and does not over-ride any logic.</p>
<p>For example if I wanted to test if a light has failed, I can "force" its output bit in the PLC software to determine if it has failed or if it is an issue with the logic.</p>
<p>If I want to test a if some latching logic works, I would "set" the trigger and see if it responds how I expected.</p>
| 19292 | Real world applications where one would use a force function in PLC programming? |
2018-02-07T16:24:03.087 | <p>This may be an obvious question, but here goes anyhow.</p>
<p>If you have two pulleys (different diameters) with infinitely strong bearings and shafts, does increasing the belt tension have any effect on the torque required to turn the system? I.e. Is a super tight belt is harder to turn than a weakly tensioned belt (assuming no belt slip in either case)?</p>
<p>Ignoring the cost and wear implications of doing this as increasing the tension will obviously wear out your belts and bearings significantly faster.</p>
<p>I've done a force balance, and it seems as though, no, everything balances out. If you increase the tension on a belt you're increasing the force on both sides and the torque on the pulley input seems to be independent of the belt tension. When the system is moving at a constant speed under a constant load, belt tension doesn't seem to factor in (again ignoring the loads on the bearings and shaft)</p>
<p>However from a practical point of view it seems to make intuitive sense that if you crank up the belt tension it'll be harder to turn them. Is my intuition wrong, or is there something I've over looked?</p>
| |design|pulleys| | <p>From reading your answers (provided as comments) to the answers given, the “best solution” to increasing the load capacity is to change to multi-belt pulleys: 2 and up to 8 are used in some situations - belts can provide a “softness “ compared to gears or a shaft.</p>
| 19293 | Belt tension to rotational force |
2018-02-08T11:34:52.863 | <p>When I am drawing things (Assembly) in SolidWorks, on some parts which I work on, I can select the edge of a circle and then measure the dimension to a line, like shown below:</p>
<p><a href="https://i.stack.imgur.com/Cg60g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cg60g.png" alt="Measuring Edge of Circle"></a></p>
<p>But the problem is that on some others I simply can't. It won't allow me to select the Edge of a cirlce when I hold down the SHIFT key. If I "insist" and drag the measurement to an edge, it jumps to the other end, like shown below:</p>
<p><a href="https://i.stack.imgur.com/DbeHY.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DbeHY.gif" alt="Opposite Edge selected"></a></p>
<p>I think I'm doing everything right: Doubleclick Sketch to edit it, select a line, then hold SHIFT and then try selecting the edge of a circle. I already searched on the internet around, but no one had really a solution for it. The closest I came was <a href="https://forum.solidworks.com/thread/19072" rel="nofollow noreferrer">this</a>.</p>
<p>Why does it work on some projects and on some others it don't? </p>
<p>Any help is greatly appreciated.</p>
<p>EDIT: I'm using SolidWork 2016 Premium x64 (SP 0.0)</p>
| |measurements|solidworks| | <p>You can achieve this simply and quickly by adding a point to the circumference of your circle. I like this method because it's easy to see exactly where the dimension is going to, and it's robust in the case where the line become non-horizontal etc.</p>
<p><a href="https://i.stack.imgur.com/5PuQG.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5PuQG.gif" alt="point on circle demo"></a></p>
<p>You can achieve this more 'officially' by using the 'minimum' arc condition in the 'leader' tab of the dimension properties. I don't like this method, because relying on manually altering dimension properties doesn't sit well in my personal workflow. You may prefer it, though.</p>
<p><a href="https://i.stack.imgur.com/6yXIB.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6yXIB.gif" alt="dimension options demo"></a></p>
<p>A more pertinent point, however, is that it's not good practice to dimension a part like this, since that's not how it will be manufactured. If possible, the dimensions in your sketches should match with those that would be desirable on the 2D manufacturing drawing.</p>
| 19307 | Selecting edge of a circle in Solidworks |
2018-02-08T11:39:28.003 | <p><a href="https://i.stack.imgur.com/7bcPB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7bcPB.png" alt="internal hinge"></a></p>
<p>We have a beam with 1 internal hinge just like the picture. We know the internal hinge can not transfer moments between AB and BD. We know the 30 kN force makes some moments on BD & AB. If B (internal hinge) can not transfer moments between the two elements, then why does the 30 kN (which is applied on BD) cause moments in AB?</p>
<p>I don't know if i'm clear enough but the 30 kN force makes some moments on AB while we know that no moments can transfer between AB & BD.</p>
| |structural-analysis|structures|beam|moments| | <p>The moment at the hinge is sought:</p>
<p><span class="math-container">$$M=-R_1\times X_1+W_2\times X_2-W_3\times X_3+W_4\times X4$$</span></p>
<p>where the sign of the moment is <span class="math-container">$+CCW$</span>:</p>
<p><span class="math-container">$$\begin{align}
P_1&=\text{reaction left}\\
P_2&=\text{weight left}\\
P_3&=\text{weight right}\\
P_4&=\text{reaction left}
\end{align}$$</span></p>
<p><span class="math-container">$X_1$</span> through <span class="math-container">$X_4$</span> are the absolute values of the distances from the hinge.</p>
<p>Moment at the hinge is the sum of the moments about it.</p>
| 19308 | Transfer of moments in beams with internal hinges |
2018-02-08T15:31:11.583 | <p>How can I calculate thrust in Newton from mass flowrate and velocity? </p>
<p>Please note,I have mathematical skills of 7 year old,the most complex task I can do is multiply and divide,I dont understand any equations.</p>
| |airflow|thrust| | <p>Well the best learning happens when there is a need, and the internet can answer almost any question. Easiest way to understand physics is to use <a href="https://en.wikipedia.org/wiki/SI_derived_unit" rel="nofollow noreferrer">SI derived units</a> and always carry your units through your equation.</p>
<p>velocity = 5 m/s
<br>mass flowrate = ?kg/s
<br>volumetric flowrate = 5m^3/s
<br>Thrust: ? Newtons = kg*m/s^2</p>
<p>I asked google what the <a href="https://www.google.com/search?q=mass%20of%20cubic%20meter%20of%20air&oq=mass%20of%20cubic%20meter%20of%20air" rel="nofollow noreferrer">mass of a cubic meter of air</a> was:
<br>density = 1.293 kg/m^3</p>
<p>The density lets us convert your volumetric flow into mass flow:
<br>(looks like googles calculator carries units now which is cool)</p>
<p><a href="https://www.google.com/search?q=5m%5E3%2Fs%20*%201.293%20kg%2Fm%5E3%20%3D&oq=5m%5E3%2Fs%20*%201.293%20kg%2Fm%5E3%20%3D" rel="nofollow noreferrer">5m^3/s * 1.293 kg/m^3</a> = 6.465 kg/s
<br>mass flow = 6.465 kg/s</p>
<p>Now we just multiply mass flow and velocity and confirm our units work out:
<br><a href="https://www.google.com/search?ei=eql8WvnjOpC8jwPgv4egCQ&q=6.465%20kg%2Fs%20*%205%20m%2Fs%20%3D&oq=6.465%20kg%2Fs%20*%205%20m%2Fs%20%3D" rel="nofollow noreferrer">6.465 kg/s * 5 m/s</a> = 32.325N</p>
<p>So the thrust is 32.325N or <a href="https://www.google.com/search?ei=O6p8WtW7GaW4jAPQ9JzADQ&q=32.325%20Newton%20in%20pound%20force&oq=32.325%20Newton%20in%20pound%20force" rel="nofollow noreferrer">7.27 pounds force</a></p>
| 19316 | How to calculate thrust from mass flowrate and velocity |
2018-02-08T16:58:55.533 | <p>Been brushing up on my control systems knowledge and I keep seeing that a P controller WILL NOT have 0 error and that doesn't make sense to me.</p>
<p>Imagine a tank problem, the setpoint is 50 cm, you open the drain valve and the level goes down to 49. Controller output goes up and you start putting water into the tank. Controller output does not go to 0 until the level is back to 50. So while this may take a while if your Kp is low, it will get there. </p>
<p>I understand that steady-state error is not measured at infinite time but the idea that a P controller will not reach the SP is odd to me. </p>
| |mechanical-engineering|control-engineering|control-theory| | <p>Take a linear system:</p>
<p><span class="math-container">$$\dot{x}=Ax+Bu$$</span></p>
<p>and assume, either that there is a constant affine force <span class="math-container">$d$</span> affecting your system (in your example, the valve left open which drains your tank)</p>
<p><span class="math-container">$$\dot{x}=Ax+Bu+d$$</span></p>
<p>or that you want to stabilize your system outside of the origin at <span class="math-container">$x=r$</span>. In the second case you can introduce the change of variable <span class="math-container">$z=x-r$</span> and obtain</p>
<p><span class="math-container">$$\dot{z}=Az+Bu+Ar$$</span></p>
<p>taking <span class="math-container">$Ar=d$</span> we notice that the two situations are similar. Both situation will generate a steady state error for a proportional controller.</p>
<p>For the rest of the answer I will demonstrate without loss of generality the steady state error in the first case. Say that the pair <span class="math-container">$(A,B)$</span> is stabilizable. It means there exists a matrix <span class="math-container">$K$</span> such that <span class="math-container">$u=Kx$</span> makes the system exponentially stable at the origin when <span class="math-container">$d=0$</span> by imposing that <span class="math-container">$(A+BK)$</span> is Hurwitz. Indeed the state trajectory is then given by:</p>
<p><span class="math-container">$$x(t) = e^{t(A+BK)}x_0$$</span></p>
<p>which provides <span class="math-container">$$\lim_{t\to +\infty}x(t) = 0$$</span></p>
<p>However, when <span class="math-container">$d\neq 0$</span>, the state trajectory becomes:</p>
<p><span class="math-container">$$x(t) = e^{t(A+BK)}x_0+ \int_{0}^{t}e^{(t-s)(A+BK)}d \,\mbox{d}s \\
= e^{t(A+BK)}x_0+ \int_{0}^{t}e^{s(A+BK)}\mbox{d}s \, d\\
= e^{t(A+BK)}x_0+ (e^{t(A+BK)}-I_n)(A+BK)^{-1}d $$</span>
finally providing
<span class="math-container">$$\lim_{t\to +\infty}x(t) =- (A+BK)^{-1}d$$</span></p>
<p>Of course <span class="math-container">$(A+BK)$</span> being Hurwitz, <span class="math-container">$\ker [(A+BK)^{-1}] = \{0\}$</span>, and since <span class="math-container">$d\neq 0$</span>, this demonstrates the existence of a steady-state error.</p>
<p>Note that in some cases, taking <span class="math-container">$u=Kx+f$</span> with a well chosen <span class="math-container">$f$</span> will compensate this steady state error and avoid the use of a proportional integral (PI) controller. This solution is however less robust than using a PI controller.</p>
| 19318 | Why does a proportional controller have to have a steady state (offset) error? |
2018-02-08T17:39:28.363 | <p>I want to mount a small engine on a shaft of a large engine and rotate them both in the same direction, will the RPMs be a sum of the their RPMs or will the resulting number be a multiplication?</p>
| |mechanical-engineering|gears| | <p>It's definitely not multiplication! Imagine you had a motor spinning in your hand at 1000rpm, and you decided to turn it over once every 30 seconds (2rpm). Would the output shaft be spinning at 2000rpm, or 1002rpm?</p>
| 19321 | Does angular velocity of a body rotating on top of another rotating body add up or multiply? |
2018-02-08T20:10:32.683 | <p>I recently learned the basics of automotive clutches, more specifically friction clutches. Here is a simplified version of their operation explained to me (the car starts at rest): The engine and the transmission are connected via friction plates. When the clutch is pressed down, the plates are separated. When to clutch is released slowly, the plates press harder together and the friction between them causes the car, via transmission, to gain speed.</p>
<p>But what confuses me is this: The car weights many tonnes, and the engine spin very fast. When the plates are coming into contact with each other, they are slipping until the car begins to move. Isn't this causing a huge amount of heat to be produced? And most of all, how are the friction plates manufactured so that they don't "smoothen out" very quickly, reducing the friction coefficient between them? I understand that they <strong>do</strong> wear out and need to be replaced sometimes, but considering the high torque and speed of the engine, you would think they'd wear extremely quickly.</p>
<p>Also, isn't the force needed to press the plates together enormous? Considering the large mass of the car, how kind of mechanism is used inside to car the press the plates together? And how can a person then, using only one foot, be able to again separate these plates? I assume the clutch pedal is connected to some kind of hydraulic system to amplify the force?</p>
<p>I've Googled a lot and watched several Youtube videos on the matter but they all simply show that friction holds the plates together and pressing the clutch releases them.</p>
| |mechanical-engineering|car|transmission| | <p>To answer your questions:</p>
<ol>
<li>Yes, a lot of heat gets produced. Most of it will sink into the flywheel, but the clutch does get very hot. That's why it's good to keep the revs low when feathering the clutch. Don't pull away revving more than 1200rpm. Better to keep it at 1000, also with a petrol car. That may require some practice, though. It's not hard at all to burn the clutch and flywheel. Your flywheel then will look like this:</li>
</ol>
<p><a href="https://i.stack.imgur.com/WM8oH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WM8oH.jpg" alt="burned flywheel" /></a></p>
<p>If you'd run your fingers across it, you'd feel a crater-like surface. Chips of
metal just melted and sprung off the flywheel, that's how hot it can become.
You'd get this when trying to pull away with a caravan on a steep hill for too
long, or when your clutch is incorrectly tuned (issue on older cars).
The glue/resin in the clutch can melt and cause a glazed surface on the clutch, reducing its friction.</p>
<p>It's not that hard to get a car moving. Try to push a car when it's in neutral, one person is adequate to get it moving. And the clutch has it much easier, being helped by a transmission and a final gearing; after the clutch, the transmission with a gearing of about 3 to 4 (in 1st) and a differential also with a gearing of about 3 to 4 is used to amplify torque. So the torque of the engine is amplified roughly 12 times.</p>
<ol start="2">
<li>They achieve smooth operation by carefully designing the lever ratios in the pedal and clutch arm, and by using diaphragm springs. Older clutches used conventional springs, newer clutches since the 60s use diaphragm springs. Those are springs with a progressive-like behaviour. With older coil spring clutches, you feel a constant force when depressing the clutch pedal. It kind of tires your leg in traffic jams. With modern ones, you feel a 'wobble' in the pedal. At the end of its travel, the spring is pushed with much greater force. It makes it possible to use minimal travel and relatively weak leg strength to control massive amounts of force. I won't elaborate about that here, but you could ask about it in a separate question.</li>
</ol>
<p><a href="https://i.stack.imgur.com/CKSBz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CKSBz.jpg" alt="diaphragm vs convo" /></a></p>
<ol start="3">
<li><p>It's not a problem if the clutch has a smooth surface, better even. It won't wear down as quickly. If it's flat, that doesn't mean it doesn't create friction. It gets pushed hard against the flywheel and pressure plate, creating enough friction for operation. The material of a clutch is designed to withstand wear by friction, and it gets used relatively little. The brakes are used far more and are loaded much more, and thus they wear out much quicker.</p>
</li>
<li><p>Yes, the force is indeed great. Consider a car that uses a clutch with a mean radius of 12cm, and the width of the padding on the clutch is 3cm. That gives us a surface of about 0.023m2. Assume the clutch has to transmit 100Nm of torque. That means the force on the clutch padding will be about 833N.</p>
<p>If our clutch has a friction coefficient of 0.8, we need to push on the clutch with a force of <span class="math-container">$F=833/0.8=1042N$</span> That's more than 100kg. In reality, the clutch has to be able to withstand much more. We're lucky to have levers and hydraulics.</p>
</li>
<li><p>To clean up your main confusion: The pedal travel is relatively long with roughly 25cm, considering the pressure plate only backs away a few mm. The force of your leg is indeed amplified by a lever and hydraulics, the same way you're able to jack up the car with your arms. If your foot travels 25cm, and the clutch only travels 2.5mm, that means the travel is reduced a factor of 100. It also means the force is amplified by a factor of 100. 1kg of pressure by your leg translates in 100kg force on the clutch.</p>
</li>
</ol>
| 19326 | Friction clutch and the force required |
2018-02-09T04:12:14.733 | <p>Currently I am doing a comparison results between analytical results and measurement results, I compare stress and strain values in 32 specified locations, unfortunately I found that there are some locations does not give a good agreement.
and my professor want me to think about the reason, instead of modifying the model,
does anyone come up with an idea about a good reason for this problem?
Thank you
Cindy</p>
| |structural-engineering|civil-engineering|data| | <p>Two possibilities come to mind:</p>
<p>One, the assumptions / theory / model is not describing / modelling the situation correctly,</p>
<p>Or,</p>
<p>Two, the source data is incorrect / has an error at those locations - perhaps the sensors are not mounted the same or the mounting points have been damaged / moved or the calibration for each sensor has drifted.</p>
| 19331 | What is a good reason if the data from analysis and measurement does not give a good agreement? |
2018-02-09T08:16:36.603 | <p>In that graph, I want to show the two values and the error percentage from the comparison between two values. I am using microsoft excel.</p>
<p>can anyone give me any idea which best graph for this case?
Thank you</p>
| |statistics|data| | <p>Based on the discussion in comments, you probably want to use a bar chart combined with a line for the error. Have a look at <a href="https://support.office.com/en-us/article/create-a-combination-chart-cf2c969a-b11e-4a62-bb8f-da021177452f" rel="nofollow noreferrer">Create a combination chart</a> for more details.</p>
<p>In your case, and using dummy data, it would probably look something like this:</p>
<ul>
<li>Data in Excel</li>
</ul>
<p><a href="https://i.stack.imgur.com/OCYoN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OCYoN.jpg" alt="enter image description here"></a></p>
<ul>
<li>Resulting combination graph</li>
</ul>
<p><a href="https://i.stack.imgur.com/RoyJX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RoyJX.jpg" alt="enter image description here"></a></p>
| 19336 | What is the best graph type to show a comparison value between two cases? |
2018-02-09T13:21:52.213 | <p>I am learning about gears, hobs and things of this sort. The main book is the machinist handbook which does a pretty good job of explaining things but things are in imperial scale.</p>
<p>This causes me a lot of confusion, especially when dealing with gear cutting blades where imperial uses DP and metrics uses Modules.</p>
<p>Are there any recommended books as thorough as the machinist handbook about gears but one that uses metric system instead?</p>
| |gears|machining|cutting| | <p>There is an English translation of "Tabellenbuch Metall", which is the german equivalent. This uses metric DIN standards. <a href="https://www.amazon.co.uk/Mechanical-Trades-Handbook-Ulrich-Fischer/dp/3808519134" rel="nofollow noreferrer">https://www.amazon.co.uk/Mechanical-Trades-Handbook-Ulrich-Fischer/dp/3808519134</a> </p>
| 19340 | Gear cutting books in metric system |
2018-02-12T04:31:33.320 | <p>I own a warehouse that needs to push 25,000 CFM. I currently have an old warehouse exhaust fan that doesn't push anywhere close to this. The diameter of the fan blade is about 36 inches, but the frame in which the fan sits is 41 x 41 inches. I've done a bit of looking and can't seem to find a fan that is 41 inches or less in diameter that pushes 25,000 or more CFM. So, I really have two questions:</p>
<ol>
<li>Does a fan exist that would push 25,000 CFM which that could fit into the 41 x 41 opening (probably means close to a 36 in diameter fan blade); and</li>
<li>Is there a mathematical function that relates RPMs of a fan, the number of blades, and size of the blades to CFM (or a similar function such that when you plug in the size and speed of the fan, you can find the CFM)?</li>
</ol>
<p>Thanks.</p>
| |airflow|hvac|cooling| | <p>I did some quick math. Call the opening 3 ft sq or 9 ft^2. Dividing 25,000 ft^3 by 9 ft^2 will give you the third dimension of 2,777 ft, which would need to travel in a minute. This is about 31.5 mph wind speed. I did a quick search of measured wind speed for fans, they were approximately 4 to 8 mph. Probably not for 25,000 CFM.</p>
| 19376 | Do 36 inch diameter fans exist that push 25,000 CFM? |
2018-02-12T06:43:26.063 | <p><a href="https://i.stack.imgur.com/mwCGL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mwCGL.png" alt="enter image description here"></a></p>
<p>I am aware that Absolute pressure is zero-referenced to a perfect vacuum and Gauge pressure is zero-referenced to ambient air pressure. So, Absolute pressure is equal to Gauge pressure plus atmospheric pressure.</p>
<p>However, the Part B of this problem, confused me on what the absolute pressure under piston B is, which comes from the fact that the system is open on both piston A and B. In this case, is the (Weight of block)/(Area of piston), Gage or Absolute pressure? </p>
| |thermodynamics| | <p>It doesn't seem that the test creator relies on your ability to differentiate between absolute and relative (or gauge) pressure. It's not mentioned anywhere, other than in B where he wants you to give the absolute pressure, which is impossible to answer, since no ambient pressure is given. </p>
<p>It's not really possible to give an answer at all since the described situation is not static. If the mass is increased to $150\ kg$, it will launch piston 1 rapidly in the air. Furthermore, in a static situation, the pressure below B is/must be the same as below piston A or anywhere else in the fluid. The whole question gives me the idea that the creator of the test doesn't really know what he's talking about.</p>
<p>I think he just wants to hear what the relative pressure in the fluid would be, if the mass on piston B is $150\ kg$, and piston A is frozen. Then you would come to
$3316\ kgf/m^2$ or $32.1\ kPa$ relative, assuming a gravitational acceleration of $9.81\ m/s^2$.</p>
<p>The absolute pressure doesn't matter anyway, since only relative pressure will be able to do any work. You can mention that the ambient pressure should be given to be able to give an absolute pressure, but in my experience, smartassery is rarely rewarded in school. Although the teacher is just plain wrong here and he should accept his fault...</p>
| 19381 | What are the differences between Gauge and Absolute pressure? |
2018-02-13T08:32:16.463 | <p>Does structural metal tubing (e.g., either as <a href="https://en.m.wikipedia.org/wiki/Hollow_structural_section" rel="nofollow noreferrer">HSS</a>, or in <a href="https://en.m.wikipedia.org/wiki/Space_frame" rel="nofollow noreferrer">space frames</a>, or even motorcycle/bicycle frames) ever come with any small metal rods or columns <em>inside</em> of the tubing itself?</p>
<p>While it might add some weight (and/or manufacturing complexity), could it also significantly improve the rigidity and load-bearing ability of the tubing?</p>
<p>*Edit: a more "biological" visualization of this could be similar to that of a bird bone's hollow structure (see <a href="https://qph.fs.quoracdn.net/main-qimg-e3a99fcd31cbfd69c3cbf2192210df5d" rel="nofollow noreferrer">this image</a>), but with possibly far fewer crossbeams (to save on weight), perhaps just one or two small "spokes" or crossbeams every now and then?</p>
| |structural-engineering|structural-analysis|beam|metals|columns| | <p>I ride dirt bikes as a hobby and the frame of the bike is a mix of tubes, patch stiffener plates welded over it, reinforcing sleeves and what not. It is designed to be light and tough and take a lot of punishment!
Here is the frame of Suzuki RMz450, a very agile and powerful dirt bike:</p>
<p><a href="https://i.stack.imgur.com/xI1ht.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xI1ht.jpg" alt="RMZ450 2012 Frame."></a></p>
<p>But in design of tubular trusses I have never had to be concerned with local buckling or concentrated stress mid length of member.</p>
<p>Because they are designed mostly to have a joint at every possible load or support point! Basically they undergo only tension or compression stresses.</p>
| 19393 | Support columns inside metal tubing |
2018-02-13T08:48:40.017 | <p>-- <em>This issue might be due to a bug in SolidWorks 2014 SP2</em> --</p>
<p>I'm trying to route a water hose from one location to another without passing through a keepout area (the red box shown).</p>
<p>I am simply sweeping a circle (red sketch) along a Projected Curve defined by two orthogonal sketches, one on TOP (green curve) and one on FRONT (blue curve).</p>
<p><a href="https://i.stack.imgur.com/EjB08.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EjB08.png" alt="enter image description here"></a></p>
<p>It's very nearly working, apart from that I cannot seem to persuade Solidworks to complete the Projected Curve. It always seems to stop short of the final location.</p>
<p>Each of the two control curves begin and end with straight line segments that are tangent to the spline that governs the path. I need these to get a nice, straight controlled entry/exit from the flat surfaces.</p>
<p>The control points begin and end at the same locations in space (projected onto their respective planes).</p>
<p>What am I doing wrong, that is preventing this curve from being completed?</p>
| |solidworks|technical-drawing| | <p>I'd recommend using a 3D Sketch Spline for creating your hose - this will more accurately model the natural shape that the hose would assume given the boundary conditions of perpendicular to the grey transparent block at the in/out positions. Let me know if you have any questions regarding this method:</p>
<p><img src="https://media.giphy.com/media/3ohs4vhSqZtKssDF1m/giphy.gif" alt="Gif showing how to use 3D Sketch Spline for Hose Creation[1]"></p>
| 19394 | Solidworks - Projected Curve is incomplete |
2018-02-13T15:46:51.520 | <p>I am a beginner in control system for hydraulic servo systems. The reason why I haven't started yet to do simulations of hydraulic servo system is because I have no information about the components I want to use.</p>
<p>For exampe. To do a simulation of a hydraulic system, you need to know the dynamical characteristics of the components you are using. </p>
<p>If you look at this <a href="https://home.uia.no/geirh/PDF/C39.pdf" rel="nofollow noreferrer">document</a>, you sure know what I'm talking about.</p>
<p>Let's a assume that you are building a system like that and you want to simulate it in MATLAB.</p>
<p>Well, OK. First you need to know what components you want to have. That's easy, just follow the picture. Then when you have found all the components you need, you also need to know the dynamics such as gain factors, damping, leackage etc.</p>
<p>The problem that all suppliers I have visit, don't show their specification inside the data sheet of the products they are selling. </p>
<p>Example, if you want to do a transfer function of a <a href="https://docs-emea.rs-online.com/webdocs/12c4/0900766b812c44d2.pdf" rel="nofollow noreferrer">hydraulic pump</a> and you go to Bosch Rexroth to buy a hydraulic pump. All they going to give you are static equations e.g formulas.</p>
<p>So that led me to the question: If I want to have dynamical information of hydraulic pump, hydraulic cylinders, electrical motors. What supplier should I visit then? </p>
| |control-engineering|control-theory|hydraulics|simulation|servo| | <p>The dynamics you’re looking for are largely governed by how you connect the different components in your design. Although the hydraulic pump you select is important, the equations provided in the supplier catalog should (at least) be sufficient for an initial design analysis (i.e. make an assumption with respect to pump’s efficiency by using the provided charts or google for typical/sane values). Depending on the required detail level of the model (which is determined by the design specifications) additional details might or might not have to be included in the model.</p>
<p>With respect to your last question; in my experience the level of detail you’re looking for is generally not available in supplier catalogs. Mainly because it is not needed for sizing a pump, motor, etc. However, if you know what additional information you need, try giving the supplier a call. They might be willing/able to help ...</p>
| 19401 | Suppliers who write dynamical specification inside their data sheet? |
2018-02-13T16:26:52.227 | <p>I need to design some custom involute gears for a project that I am working on and this is my first dive into gears.</p>
<p>I worked out how to calculate the size of gears by the required numbers of teeth and module.</p>
<p>This gear dimensions:
Module = 1mm
Teeth = 36
Clearance = 0.1
Pitch Diameter = 72mm
Outside Diameter = 74mm
Whole Depth = 2.1mm</p>
<p>As you can see it's a tiny gear and I'd like to be able to make the blade to cut this gear myself.</p>
<p>With a standard pressure angle of 20 degrees, how would I calculate the angle of the cutting surface to make this gear profile?</p>
| |gears|machining|cutting| | <p>This is more of comments than an answer.
module = pitch circle diameter / Number of Teeth. A gear with 1mm module and 36T would have a 36mm pitch diameter, not 72"
Also, companies such as <a href="https://us.misumi-ec.com" rel="nofollow noreferrer">https://us.misumi-ec.com</a> sells 1mm module 36T gears with selectable bores sizes.</p>
| 19403 | Calculating the angles for gear cutters |
2018-02-14T14:53:12.283 | <p>I have laminar flow in a tube. Consider the tube to be 0.2 m long and with an average velocity of 0.05 m/s. The analytical expression for my transfer function is: $E(t)= \frac{\tau^2}{2*t^3}$ for $t$>=$\frac{\tau }{2}$ and $E(t)=0$ for $t < \frac {\tau }{2}$. $\tau$ is the mean residence time. In this case: $\tau$=0.2m/0.05m/s=4 s. I want to convolute this with an exponential equation: $E_2=(1-exp(\frac{-t}{2.55}))$. This equation descripes the magnetization of a particle in a static magnetic field.</p>
<p>I want to get the average magnetization of the particles at the outlet of the tube. So I thought I would do the following: $E_{out}(t)=E(t)*E2(t)$. I would then take the value at t=4s. I have the results from CFD which gives the following: <a href="https://i.stack.imgur.com/TOxPu.png" rel="nofollow noreferrer">CFD result</a>, the red line is the magnetic field and the green line is $E_{out}(t)$. I'm interested in the value at x=0,2m which is 0.7942.</p>
<p>My results of the convolution are totally different:<a href="https://i.stack.imgur.com/7bCdc.png" rel="nofollow noreferrer">Convolution result</a></p>
<p>What am I doing wrong? I'm a little bit confused. Has anyone an idea of how to approach this problem. </p>
<p>Best regards,</p>
<p>Gesetzt</p>
| |chemical-engineering|cfd|matlab| | <p>The average magnetization exiting the tube in steady state should just be the product not the convolution.
$$\int_0^\infty E(t)\, E2(t) \, dt$$
$$\int_{\frac{\tau}2}^\infty \left(1-e^{-t\frac{t}{k}}\right)\, \frac{\tau^2}{2\,t^3} \, dt$$
$$-\frac{\tau^2 Ei\left(-\frac{\tau}{2 k}\right)}{4 k^2} - \frac{\tau \, e^{-\frac{\tau}{2 k}}}{2 k} + e^{-\frac{\tau}{2 k}}-1$$
Where $Ei$ is the <a href="http://mathworld.wolfram.com/ExponentialIntegral.html" rel="nofollow noreferrer">Exponential Integral</a></p>
<p>This looks like it comes out to a bit less than your CFD results:
<a href="https://i.stack.imgur.com/leK16.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/leK16.png" alt="Plot of fraction of flow magnetized"></a></p>
<p>The x axis in this case is the average residence time which should be proportional to the length along the tube, making our plots comparable.</p>
| 19422 | Convolution with RTD laminar flow |
2018-02-15T00:42:26.087 | <p><a href="https://i.stack.imgur.com/j18gP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j18gP.jpg" alt="enter image description here"></a></p>
<p><sup>Source: <a href="https://www.motherearthnews.com/organic-gardening/low-tunnel-construction-mini-hoop-house" rel="nofollow noreferrer">Low Tunnel Construction: How to Build a Mini Hoop House</a> </sup></p>
<p>I have a small greenhouse that I want to keep as <strong>cold</strong> as possible in the spring. I plan to cover the greenhouse with a <strong>polyethylene tarp</strong> to shield it's contents from the sun.</p>
<p>However, I'm not sure what colour of tarp would be the most effective at shielding heat.</p>
<p>I suspect that light coloured tarps would allow for solar gain. And I wonder if a black tarp might also retain warmth by absorbing heat from the sun.</p>
<p><strong>Question:</strong></p>
<p>Are there specific colours of material that are known to let in the least amount of heat?</p>
| |thermodynamics|heat-transfer|environmental-engineering|thermal-insulation| | <p>White reflect the greatest amount of solar radiation. I build and maintain Ice Runways for the National Science foundation, and any dirt or grease matter gets on the surface the solar gain on that item melt a hole in the surface of the runway and weakens the landing surface. We do everything possible to keep the surface white so the "albedo effect" reflects the greatest amount of solar energy to maintain structural integrity of the runway in the summer. <a href="https://nsidc.org/cryosphere/seaice/processes/albedo.html" rel="nofollow noreferrer">https://nsidc.org/cryosphere/seaice/processes/albedo.html</a></p>
| 19429 | What colour of tarp will let in the least amount of heat? |
2018-02-15T16:21:18.343 | <p>I drew a quarter of my element and i want to generate the whole part by using 'pattern mirror'. The actual element is totally symmetrical (as you can see in the first image sketched by yellow curves) but in my final results (the second image) that ain't so and I have absolutely no clue why!
<a href="https://i.stack.imgur.com/MRorS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MRorS.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/xDbjS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xDbjS.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|solidworks| | <p>When you have created a 3D part then you don't have to go to sketch circular pattern for an easy approach. Instead, you can directly go to feature toolbar there you will find and a ‘Linear Pattern’ below which there will be an arrow for a drop-down menu from there select 'Circular Pattern'. Or you can go to <strong>Insert>Pattern/Mirror> Circular Pattern</strong> and then you can perform the same step by selecting features you want for the circular pattern.</p>
<p>If you are doing it from ‘Sketch’ then you have to go to ‘Circular Sketch Pattern’ from the drop down of Linear Sketch Pattern. Then select sketches of features, not the features and then you can create a circular sketch pattern. And thereafter extruding the circular sketch pattern will give you complete part.</p>
| 19439 | I need some help with 'pattern mirror' tool in Solidworks |
2018-02-15T16:56:51.317 | <p>When reconditioning of a pneumatic cylinder, which fit between piston and cylinder is required? and how much tolerance between mating parts is required?. Will surface finish inside of cylinder play a major role?.</p>
| |pneumatic| | <p>General comment, this is a direct answer to the question. Most pneumatic cylinder, especially bores of 2" or less are non repairable. Actually I have never seen a pneumatic cylinder repaired, but I think the procedure is to get a new seal kit. I suggest contacting the manufacture of the cylinder.</p>
| 19440 | Cylinder Reconditioning |
2018-02-15T22:32:43.210 | <p>I was always curious why lead was chosen as the default for metal acid batteries. This article describes lead-acid battery operation and there are plenty of resources like this on how lead acid batteries work, but I could not find any resources that explain "why lead?"</p>
<p>What about lead makes it the standard for metal-acid batteries?</p>
<p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/electric/leadacid.html" rel="nofollow noreferrer">Lead Acid Battery on HyperPhysics</a></p>
| |electrical-engineering|chemical-engineering|battery| | <p><a href="https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery" rel="nofollow noreferrer">Lead acid batteries</a> has been around a long time and is easy to manufacture. They are rechargeable, recyclable, and reasonably safe. <a href="https://en.wikipedia.org/wiki/VRLA_battery" rel="nofollow noreferrer">AGM or Absorbent Glass Mat</a> lead acid has the added benefit of being sealed.</p>
<p>The reason they are so common is because of the high <a href="https://en.wikipedia.org/wiki/Comparison_of_commercial_battery_types" rel="nofollow noreferrer">watt-hour/$ ratio</a>:</p>
<ul>
<li>Lead acid 6.77–17.41</li>
<li>Alkaline 0.48</li>
<li>Lithium 2.75</li>
</ul>
<p>They are the battery of choice for <a href="https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery#Applications" rel="nofollow noreferrer">car starters</a>, <a href="https://en.wikipedia.org/wiki/Uninterruptible_power_supply" rel="nofollow noreferrer">UPS (uninterruptible power supplies</a>), commerical/industrial battery backup systems, off-grid power systems, electric forklifts, emergency lighting, etc.</p>
<p>Per watt-hours in use they are definitely the most common. Per unit, it would most likely be alkaline, followed by lithium and lithium ion.</p>
| 19445 | Why lead was chosen as the metal for lead-acid batteries? |
2018-02-16T01:01:10.570 | <p>I have this pneumatic cylinder that I wish to connect to a wooden door in order to make it kinda like a Star wars door. I'm wondering what the best way to connect the cylinder to the door would be. </p>
<p>The cylinder has a threaded rod at the end. Looks like the one attached. I was wondering if there are components like a bracket that can be screwed onto the door and has a threaded loop to which the Rod could be screwed to?<img src="https://i.stack.imgur.com/nIaxI.png" alt="enter image description here"><img src="https://i.stack.imgur.com/XX4L5.jpg" alt="enter image description here"></p>
| |pneumatic| | <p>Based on the threads at the end of your sample cylinder, you would be seeking a female threaded rod end:</p>
<p><a href="https://i.stack.imgur.com/RJLQA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RJLQA.jpg" alt="female threaded rod end"></a></p>
<p>There are a multitude of variations of this product, almost always including the words "rod end" with different modifiers. You could have a forked threaded rod end.</p>
<p><a href="https://i.stack.imgur.com/I1dCS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I1dCS.jpg" alt="forked threaded rod end"></a></p>
<p>It would not have to be female threads, of course. Depending on the work load, you may not have to have the bearing insert of the first image, although it allows for longer life and easier alignment in use.</p>
| 19447 | How to connect a pneumatic cylinder to a door? |
2018-02-16T16:49:15.080 | <p>I recently outfitted a crew to build scaffold in preparation for an upcoming turnaround at our plant. Among the tools requested were canvas buckets for manually hauling material and tools up and down scaffolds. In sourcing these I went through a reliable well established industrial supplier with a know brand. I ordered bags rated for 100 lbs which above the maximum load for manual hoisting permissible on our site.</p>
<p>Several days after arrival on site one of these bags failed with a 27 lb load in it. Nobody was hurt but it could have easily turned out differently. We immediately pulled all bags from use. In investigating it was discovered that when used outside in winter conditions that the plastic parts of the bag bottoms become brittle. No temperature restrictions were offered in any documentation related to these bags and these tools were being used for the purpose they were designed for.</p>
<p>I feel awful that a decision I made put people at risk. Is there some aspect of due diligence in tool selection that a reasonable engineer would have been expected to take that I had missed? </p>
| |tools|specifications| | <p>As a busy engineer, I would say you did your due diligence. There are always more variables than we are given time to consider. The higher the risk of failure/injury, the more time we spend on it, but there are limits. I often try to have another engineer or manager look at my work so there is another set of eyes on it. Also, taking time to test new tools/equipment is a good when possible. In hindsight this would have addressed the issue. Sometimes you can do "field testing" by just testing the smallest portion of the system at a time. This could be like first giving out one bag to a worker that you know will give your prompt and reliable feedback before your give out all of them.</p>
<p>Our company just started requiring a <a href="https://en.wikipedia.org/wiki/Hazard_analysis" rel="nofollow noreferrer">hazard analysis</a> for all capital projects and major process changes. A small change like bags may have flown under the radar, but a procedure like that might be something worth implementing.</p>
<p>On the other side of things, remember that the workers are not helpless. They consciously or subconsciously know to test their tools before they use them. They should also be wearing proper PPE like hardhats and steel toe boots to reduce risk in general. It also helps to let these guys know that they have a say in what tools they use. If they feel something is unsafe they need to make everyone aware. We recently implemented a <a href="http://docs.healthandsafetyhub.co.uk/Vennsys/Media_library/vennsys-media-take-5-meter-reader-booklet.pdf" rel="nofollow noreferrer">take 5 safety program</a> that is designed to assist with this.</p>
<p>In addition to this, please let the manufacturer know (not just the vendor). You will save more people than just the people at your facility. If they are of any credibility they will refund or replace your bags too.</p>
| 19458 | Potentially fatal near miss: Due diligence in tool selection |
2018-02-17T16:45:37.907 | <p>In the textbook : <em>Electric Power Principles</em>, in Chapter 1 </p>
<p>on the working of a pv cell : </p>
<blockquote>
<p>The source current Is is the result of absorption of photons in
sunlight that cause separation of valence electrons from their atoms.
The resulting hole/electron pairs fall across the high field gradients
present at the diode junction. Because any voltage resulting from
this current tends to forward bias the actual junction, the voltage
available is limited</p>
</blockquote>
<p>I don't understand the train of thought leading to a limited voltage, and I'd appreaciate a better explaination or a reading recomendation. </p>
| |electrical-engineering|power-electronics|power-engineering|photovoltaics| | <p>The simplest explanation I can give is that a forward biased diode will start to transmit a high level of current with only a small increase in voltage, once a certain threshold is reached. This means that if you increase the rate at which electrons are generated with more sunlight, then instead of the voltage increasing in line with this, the current will go up instead. There are numerous resources explaining the mechanisms behind this elsewhere on Electronics SE, and the wider internet.</p>
<p>If you would like a higher voltage for a particular project, you can simply connect multiple solar cells in series.</p>
| 19472 | Working mechanism of photovoltic cells |
2018-02-19T06:59:13.030 | <p>there
I installed SINO 942 tracker in my bike, it works well, but then my bike's self start has stopped working (it works when I ride bike for around an hour), so
I think the tracker is eating up battery when the bike is parked for long hours.</p>
<p>I am already thinking about installing a manual switch, or a rechargable
12 V battery (so it wont need to connect to main battery), but all of these
are manual and regularly to-be-interacted-with solutions.</p>
<p>Please suggest best solution I can use to rectify this issue.</p>
| |electrical-engineering| | <p>You can go with your second battery, but use a split-charge system so whenever the bike is running the second battery gets charged.</p>
<p>An example of a split charge system / relay is <a href="https://www.ebay.co.uk/itm/160477-0-727-10-EQUIV-12V-100A-100-AMP-HEAVY-DUTY-SPLIT-CHARGE-RELAY-CAMPER-VAN/201605769301?hash=item2ef0a3e855:g:0f4AAOSwgsxaaEY8" rel="nofollow noreferrer">here</a></p>
| 19479 | Efficient solution for Bike Tracker |
2018-02-19T10:59:39.357 | <p><strong>Brief</strong> I am actually of 17 not even in college but I have dream for making my project/dream/aspiration/or So to come true so I am thinking of making a device that performs a function say like Thermometer measures Temperature, now working and doing the function is not a concern of now (a Question for me), the thing is if I got the data of Reading of final Temp (assume it has performed its temperature calculation and given output [Yep its digital]) so now it gave me some value x .</p>
<hr>
<p><strong>Problem</strong> I want to get this value x into my server or website (maybe I am unable to express it but I want to get it to me through Internet) and store that value for data of <strong>Person P</strong> so , I am not sure how I would achieve that. Assume all calculation and functions are done and are independent of device now I just need to know how to store that value for use? Would it require something like connecting it to a wifi card or similar and making an OS (Till now this has came across my mind ) </p>
| |electrical-engineering|telecommunication| | <p>Just as Mahendra says, in terms of hardware you'll need your sensors connected to something such as an arduino or raspberry pi and have an internet connection. If using a pi you have both wired/wireless options ready out of the box. </p>
<p>Are you planning on hosting your own server or buying server space? Either way linux would be an easy way to go. Watch <a href="https://youtu.be/vzojwG7OB7c" rel="nofollow noreferrer">this video</a> to set up your own server, it shows you how to install Apache, PHP and MySQL. If buying some webspace, you'll need to buy a domain name from a registrar such as GoDaddy or 123reg and then buy the space from somewhere such as hostinguk.net, anyone that hosts linux with PHP and sql. Once you've done both, link your domain with the webspace.</p>
<p>Now you have everything, you can program your hardware to read the sensor input and send it to your websites database. There are countless guides on all of this online, this answer should point you in the right direction atleast.</p>
| 19482 | How do I get data from my Hardware to server or so? |
2018-02-19T22:52:08.987 | <p><a href="https://i.stack.imgur.com/YwsLy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YwsLy.jpg" alt="enter image description here"></a></p>
<p>I have <em><a href="https://en.wikipedia.org/wiki/Prunus" rel="nofollow noreferrer">prunus</a></em> pits that I <a href="https://en.wikipedia.org/wiki/Stratification_(seeds)" rel="nofollow noreferrer">stratify</a> in a refrigerator during winter. </p>
<p>Normally, the seeds germinate in springtime, which is ideal.</p>
<p>However, often some of the seeds <strong>germinate prematurely</strong>. When this happens, it's necessary to grow the seeds/plants indoors during winter until they can be planted outdoors (after the <a href="https://en.wikipedia.org/wiki/Growing_season" rel="nofollow noreferrer">last frost date</a>). <strong>This is not ideal.</strong></p>
<p>Rather than let the premature pits grow, an alternative would be to slow germination down (or even prevent it) by keeping them at <strong>just above freezing</strong>. It's my understanding that the colder plants are, the slower they grow (up until freezing, of course--the plants are too young to go dormant for sub-zero temperatures).</p>
<p>So, the question becomes, "Where in my house is the temperature at just above freezing?". </p>
<p>Typically, refrigerators are at a temperature of around 6°C. From experience, I've found that 6°C is too high; the plants grow and eventually die due to lack of light.</p>
<p><strong>How can I store the seeds at 1°C in a household setting?</strong> </p>
<p><a href="https://i.stack.imgur.com/rEgU3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rEgU3.jpg" alt="enter image description here"></a></p>
| |temperature|environmental-engineering|refrigeration| | <p>The temperature control in a fridge is pretty crude. The evaporator is in the freezer and the compressor is controlled with an electric thermostat. The fridge portion however relies on convection from the freezer. The temperature control from the fridge is sometimes nothing more than a slider that establishes how quickly the convected air moves. There are likely better methods employed in more expensive units so you will have to look at your specific fridge. In any case, the temperature can fluctuate quite because the designers were not concerned with a little temperature fluctuation harming food.<br>
<a href="https://i.stack.imgur.com/5Tdip.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Tdip.gif" alt="enter image description here"></a></p>
<p>Also, if your freezer has an auto defrost cycle this will also increase the temperature for <a href="https://en.wikipedia.org/wiki/Auto-defrost#Mechanism" rel="nofollow noreferrer">15 minutes every day</a> (or whatever the timer settings are).</p>
<p>Before we go changing things, it might be helpful to get a <a href="https://rads.stackoverflow.com/amzn/click/B01HH7YD2Y" rel="nofollow noreferrer">wifi temperature logger</a> or something similar so you can see how much the temperature fluctuates and how much modifications improve it.</p>
<p>Assuming this fridge is 100% dedicated to this project; I would try adjusting the fridge temperature to target your 1°C with its current crude control. Fill the rest of the fridge with gallon jugs of 3/4 full of tap water(room for freezing) to help stabilize the temperature. If the freezer is not loaded, also fill it with 3/4 filled gallon jugs of water. This increased thermal mass will help stabilize the temperature. It also helps ensure the temperature does not dip much below 0°C if controls are not ideal.</p>
<p>If that doesn't fully address it, figure out how the <a href="http://www.appliance411.com/faq/howdefrostworks.shtml" rel="nofollow noreferrer">auto defrost cycle</a> works on your specific refrigerator/freezer and disable it. This <a href="https://www.repairclinic.com/Shop-For-Parts/a4/Refrigerator-Parts" rel="nofollow noreferrer">parts site</a> has pretty good schematics of most models. Of course, now you will have to remember to do it manually before and after or your freezer/fridge will run all the time and waste power.</p>
<p>If that doesn't get the temperature stable enough you can replace the passive convection temperature control system with an <a href="https://rads.stackoverflow.com/amzn/click/B004YTSB7C" rel="nofollow noreferrer">ac fan</a> and <a href="https://rads.stackoverflow.com/amzn/click/B01KNXETWS" rel="nofollow noreferrer">pid controller</a>. The fan turns on when it needs more cold air from the freezer and turns off when it is cold enough. Definitely more work, but it will be able to hold the temperature much closer than with other methods.</p>
| 19494 | Store seeds at 1°C |
2018-02-20T09:32:41.430 | <p>Shear stress is introduced in statics as $\frac{P}{A}$.</p>
<p>However, the maximum shear stress with neutral axis is different. It is $\frac{3P}{2bh}$ (Where $b$ is width and $h$ is height)</p>
<p>What is the difference between these concepts?</p>
| |mechanical-engineering|structural-engineering|stresses|experimental-physics|shear| | <p>The difference is in the assumptions. The first "shear stress" is simply assuming the stress is uniform over the cross section. As a result, we see a shearing force $P$ shearing a rod of uniform cross section $A$, and due to the assumption or uniform shear stress across the cross section, we have the resulting $\frac{P}{A}$ shear stress.</p>
<p>When we abandon this assumption of uniform shear stress, and we use the assumptions for beam shear, we find "maximum shear stress". Here, we find the shear stress develops parabolically across the cross section, and develops a maximum sheer stress at the center line of the cross section, for symmetric cross sections. For a rectangular cross section, this maximum sheer stress happens to be $\frac{3P}{2bh}$, but for other cross sections the value differs. </p>
<p>In general, the value is $$\frac{PQ}{Ib}$$</p>
<p>where $b$ is the thickness of the cross section at the cut, $Q$ is the first moment of the area above the cut referenced to the centroid of the area: $$\int^{h_t-c}_{h_c-c} x dA$$</p>
<p>Where $c$ is the height of centroid from the bottom of the cross section $$\frac{\int^{h_t}_{0} x dA}{\int^{h_t}_{0} dA}$$, $h_t$ is the height of the top of the area from the bottom of the cross section, $h_c$ is the height of the cut from the bottom of the cross section. $I$ is the second moment of the area above the cut referenced to the centroid:</p>
<p>$$\int^{h_t-c}_{h_c-c}{x^2}dA$$</p>
<p>(Note different uses of the reference point other than the bottom of the cross section will result in easier formulas for Q and I). A <a href="https://engineering.stackexchange.com/questions/19100/what-is-the-difference-between-tau-vq-it-and-tau-v-a?rq=1">related post</a> may be able to help you with more visuals.</p>
| 19508 | What is the difference "maximum" shear stress and shear stress |
2018-02-20T09:45:31.730 | <p>I was looking into using <a href="https://www.bosch-motorsport.com/content/downloads/Raceparts/en-GB/51865867208058251.html" rel="nofollow noreferrer">Lambda Sensor LSU 4.9</a> to know the lambda value. From the datasheet, its looks though this sensors cannot be used without a special IC such as Bosch lambdatronic LT4. Can we get any information out of the sensor by connecting only 12v to the heater+ and heater-. What would be the minimum circuit to get some information from this sensors?</p>
| |sensors| | <p>No you can't. The LSU4.9 is a wideband sensor, and so differs from the old school narrow-band zirconia sensors that just produced a voltage. The wideband incorporate an oxygen pump along with a zirconia element to make it a wideband. That pump must be controlled for proper working, as well as the temperature which must be kept within limits. You can't just put 12V on the heater, it will destroy the sensor. The heater has to be throttled. And you can't just only use the zirconia element within the sensor to use it as a narrowband.</p>
<p>Driving the 4.9 requires a quite sophisticated construction. The LT4 uses the CJ125 IC, and even without the LT4, you will need that IC. The CJ125 IC is specially designed and made by Bosch to allow users to use the LSU sensors. And it is quite adequate for that purpose. </p>
<p>Some manufacturers succeeded in making their own circuitry without the CJ125, some with more success than others. But unless you're capable of producing your own IC's, you'll have to use the CJ125. And even then correctly using it is certainly not easy. A lot of peripheral hardware is needed to use the CJ125. At that point, you could best just use the LT4 or a similar tool. Innovate and AEM are manufacturers that also offer tools to read out the 4.9. Both use the CJ125 IIRC.</p>
<p>14Point7 offers a cheap PCB(~$40) that closed-loop heats and reads the sensor, and gives you the lambdavalue and temperature in either an analog voltage or via I2C communication. I have used it and it works fine. I think it's the closest you'll come to successfully reading the 4.9 yourself.</p>
| 19509 | Can lamda sensor 4.9 be used without the special LSU-IC? |
2018-02-20T12:57:22.833 | <p>So I have laser cutter. And I am thinking about buying a control unit (<a href="http://www.topwisdom.com.cn/en/product.asp?dlm=11" rel="nofollow noreferrer">this one</a>, to be exact). </p>
<p>But my concern here is the following: If the input DXF file contains splines, they will be converted to lines (G01) and arcs (G02 or G03), since these are all the entities G-Code has.</p>
<p>Since have absolutely no knowledge about motors or controlling them, the question is as follows:</p>
<ul>
<li>Does the control unit from the link do the same? Does it convert DXF to G-Code?</li>
<li>If yes, is there a control unit that would quide my laser cutter over a spline from the DXF and NOT on some series of lines and arcs?</li>
<li><a href="http://www.topwisdom.com.cn/en/product.asp?dlm=11" rel="nofollow noreferrer">This webpage</a> says that only Z axis is controlled with PID. Why? How are other axis controlled. Does this mean I could in general choose between PID and G-Code?</li>
</ul>
| |control-engineering|motors|pid-control| | <p>Not really an answer, but I can't comment, so:</p>
<blockquote>
<p>Does this mean I could in general choose between PID and G-Code?</p>
</blockquote>
<p>PID and G-code are not two interchangable approaches to control a CNC-machine, they serve different purposes.</p>
<p>G-code is basically the programming language to tell the CNC-machine where to drive with the axes, when to do stuff like turn on the laser, turn on cooling water, turn on process gas, whatever your machine can do. </p>
<p>PID is a (simple) closed-loop control algorithm, in case of the control unit you linked, it ensures that the z-axis is at the position it is supposed to be, even under the influence of some disturbance. </p>
| 19513 | Are all motors controlled with G-Code? |
2018-02-20T16:21:19.840 | <p>I'm translating a Russian text about a drug product manufacturing process. The text mentions briefly the vial stoppering machine, in which rubber stoppers are first fed into the hopper and then directed towards vials, and a <strong><a href="https://www.multitran.ru/c/m.exe?l1=2&l2=1&s=%EF%F3%E0%ED%F1%EE%ED" rel="nofollow noreferrer">пуансон</a></strong> drives the stopper into the vial. Multitran <a href="https://www.multitran.ru/c/m.exe?l1=2&l2=1&s=%EF%F3%E0%ED%F1%EE%ED" rel="nofollow noreferrer">provides multiple options</a> for how to translate this "пуансон". It's basically some shaft-like member.</p>
<p>I googled and found a <a href="http://www.findpatent.ru/patent/231/2319659.html" rel="nofollow noreferrer">description of some stoppering machine in Russian</a>, and it has a diagram, with "пуансон" under number 60:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/75iq5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/75iq5.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>The text there says that the axle (59) rotates, making the cam (58) rotate and push this.. plunger\piston\punch member (?) towards the vial and drive the rubber stopper into the vial. </p>
<p>I'm unsure which word to use for this part. </p>
| |terminology| | <p>A camshaft is rotating a cam that is part of the shaft. It seems the cam acts on a roller, fixed on a push-rod, and so pushes down the push-rod in the guider.
I think the push-rod pushes the rubber down with a punch cap.</p>
<p>I think push-rod is the word you're searching for if you're talking about part 60.
Plunger may be a synonym, but I think of a kind of pump when hearing that term.</p>
| 19516 | What do I call this shaft that drives the rubber stopper into the glass vial in a vial stoppering machine? |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.