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2021-02-17T20:13:18.840 | <p>Overview</p>
<p>I'm planning a rack for solar panels on the roof of a cargo trailer. I will mount the panels to a pair of 15 series 80/20 aluminum extrusion rails. I will mount the rails to the roof of the trailer using aluminum angles. Here's a view from a Sketchup model giving a rough idea:</p>
<p><a href="https://i.stack.imgur.com/Eo1B0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Eo1B0.png" alt="enter image description here" /></a></p>
<p>Specifics</p>
<p>My question revolves around mounting the set of aluminum angles to the trailer's steel roof beams. Here is a rough model showing the relevant parts:</p>
<p><a href="https://i.stack.imgur.com/ITJ63.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ITJ63.png" alt="enter image description here" /></a></p>
<p>The aluminum angle is 1/4" 6061-T6 with 2" legs cut to 2" wide. The steel beam in the trailer roof is 1" wide, 1-1/4" tall, and about 1/8" thick. Between the angle and the beam is the thin aluminum skin of the trailer.</p>
<p>The interior of the trailer is finished and prevents any access to the beams from the interior so I can't use through-bolts to attach the angle to the beam.</p>
<p>So my current plan is to drill and tap, from the outside, the top of the beam. Then I can attach the angle to the beam using two bolts that screw into the top of the beam. I plan to use 1/4"-28 bolts. I also plan to use VHB tape between the aluminum angle and the aluminum roof skin.</p>
<p>Question</p>
<p>Finally, the actual question. Since the steel beam is only 1/8" thick, is the choice of using 1/4"-28 bolts strong enough? Or would 1/4"-20 be better? My thought is finer threads gives me a better grip over courser threads.</p>
<p>Side question (that may be off-topic) - is there a better, stronger way to attach the aluminum angle to the steel roof beams than screwing in bolts given that the steel is only 1/8" thick but I can't use through-bolts?</p>
| |mechanical-engineering|strength|threads| | <p>Rivnuts are great, until they aren't. Get one that doesn't have a good grip and spins when you want to remove the bolt and it's all over.</p>
<p>Your idea of using 1/4-28 threads is a good one, especially in steel and especially with the backup of VHB tape. With sufficient VHB tape, you could almost get away with tape only, but the risk factor is high.</p>
<p>One-eighth of 28 is 3.5, which is the number of threads you'll have engaged in the steel using that tap. With a quarter-twenty tap, it's only 2.5, but it is steel and if you don't lay into the wrench and strip the threads.</p>
<p>Thread locker will be good for additional reliability.</p>
| 40430 | Securing aluminum angle to steel tubing without through-bolts |
2021-02-17T21:33:57.667 | <p>I am doing some temperature measurements for reliability testing, over an expected range of 16C to 45C. I have an NIST traceable thermocouple thermometer (S220-T8), however it does not come with thermocouple probes.</p>
<p>Without a calibrated thermal reference, how can I be sure that K-type thermocouples purchased from Amazon are accurate? Shouldn't thermocouples be subject to unit-to-unit variability and thus require some sort of voltage-to-temperature characteristic that is reported? I am finding that sometimes even "name-brand" companies are not providing such a datasheet.
I understand that they work based on something called the Seebeck effect, and maybe the Voltage/Temp relation is intrinsic to the metal alloy used. As long as the little pellet of metal is "K-type" and is in thermal equilibrium with the thing being measured then maybe that is good enough?</p>
<p>I have found an NIST traceable thermocouple probe for $100. I assume this probe will be within some spec or that they will provide the voltage to temperature relation. Is it advisable to purchase one of these NIST traceable probes and then use it to check the error of the remaining probes?</p>
| |temperature|thermal-conduction| | <p>Go to Omega.com to learn all you want to know about thermocouples (TC). I'll give you here a few basics, directed at your questions.</p>
<p>TC science is well understood and standardardized. Every type TC is made with it's own precise metal alloy and all TC of the same type will output a voltage uniquely related to the temperature of the TC junction, within accepted accuracy. Review "The Basic Laws of Thermocouples." The insulators on each TC type are color coded and each Type have their appropriate accuracy within appropriate temperature ranges. Omega used to provide a table of Temperature/Voltage data increments in their hardcopy catalogues, so my guess is that they have this data on their website. You can use such a chart to calibrate your TC, but you'd need an accurate Temperature measuring device and an accurate voltmeter, and it's really not worth the bother.</p>
<p>The TC industry is extremely well controlled, and so it's highly unlikely that any TC wire you buy will be "inaccurate." So if you buy only the wire itself, no need to worry. TC wire is very cheap, coming in all Types and diameters. It's very easy to make a TC junction from the wire itself. All you have to do is fuse the two separate wires (one + the other -) together by a hot acetylene torch, or by an electronic "TC maker" that uses a discharge from a capacitor. This is essentially fool proof. If the wires are fused, the TC is a TC. You can also imbed (drill a hole and "peen" it in) one wire in say the metal piece you want to measure the temperature of and the other wire a short distance away, and the reading will give you the average temperature of the space between. The TC Law that assures this is the Law of a Common Junction.</p>
<p>You mention "probes," but you didn't say what kind of probe geometry you're using. In general, TC wires are often mounted inside probes, and there, the manufacturing quality is important. There are "surface probes" to measure the temperature of flat surfaces, "point probes" that are used to stick inside things, and others with variously different probe materials suitable for different environments.</p>
<p>TC don't age, unless the probes become corroded or broken, but the basic TC itself never wears out or loses calibration, as long as the TC welded junction is intact.</p>
<p>That's about all I can do for you here, but please go to Omega.com and learn more.</p>
| 40433 | How can I trust a K-type thermocouple purchased from a no-name company? |
2021-02-18T10:21:12.120 | <p>Having a cast aluminium enclosure with the 3mm wall thickness. now I need to mount this to a mounting plate which fixed in vertical walls. can I have some idea on how should I choose the appropriate fastener?</p>
<ol>
<li>Cast aluminium enclosure</li>
<li>mounting plate
for now, I have placed a 5M countersink bolt</li>
</ol>
<p><a href="https://i.stack.imgur.com/uzyLd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uzyLd.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/7lwhA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7lwhA.png" alt="enter image description here" /></a></p>
<p>Thanks in advance</p>
<p><a href="https://i.stack.imgur.com/O7vv2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7vv2.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|design|aluminum|fasteners| | <p>The easiest way to maintain water resistance is to not puncture the enclosure.</p>
<p>It looks like the box has holes extending through the cover down to the bottom, I'm guessing these are specifically for mounting? (The 2 large holes). Use those if you can.</p>
<p>The next best option would be to attach the box so that the bottom is against the plate, and screws come in from the outside into the screw holes that extend from the top of the box to the bottom.</p>
<p><a href="https://i.stack.imgur.com/vOAjQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vOAjQ.png" alt="box screw placement diagram" /></a></p>
<p>Another option is to attach the plate with screws from the inside. Place the holes so the inner surface around the hole is flat, and then use sealing screws.</p>
<p><a href="https://i.stack.imgur.com/BvVyP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BvVyP.png" alt="sealing screw" /></a></p>
<p>Depending on how water proof you need the box to be, it might also be viable to just mount the box using angle brackets so the screw holes are facing the ground. Although not fully waterproof, water would be unlikely to get in that way.</p>
| 40438 | Mounting methods for thin wall cast aluminum |
2021-02-18T13:12:57.107 | <p>I am responsible for a brick kiln in the Punjab province of Pakistan. My government requires all brick kilns to be upgraded to match recent new standards, which include the addition of a blower near the chimney. We consulted with local engineers who presented several different blower designs and conflicting arguments regarding which design is best.</p>
<p>Here are the two main designs most of them agreed upon, with the fan being <strong>52 inches in diameter and having 6 blades</strong>:</p>
<h1>Blower Design #1</h1>
<p><a href="https://i.stack.imgur.com/5isHF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5isHF.png" alt="Diagram of a centrifugal blower showing a 24-inch outlet diameter, with the following gaps measured between fan and housing every 90 degrees starting immediately before the outlet: 18 inches (top); 12 inches (left); 8 inches (bottom); 2 inches (right)." /></a></p>
<h1>Blower Design #2</h1>
<p><a href="https://i.stack.imgur.com/ALuNe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ALuNe.png" alt="Diagram of a centrifugal blower showing a 24-inch outlet diameter, with the following gaps measured between fan and housing every 90 degrees starting immediately before the outlet: 24 inches (top); 18 inches (left); 10 inches (bottom); no dimension specified (right)." /></a></p>
<p>Which design will work better? I also have my doubts about the fan blades being straight. Would straight blades be appropriate for this use? What about forward or backward curved blades?</p>
<p>We will mount the blower at the base of the chimney, pulling hot air from cooking bricks and blowing it into the chimney. The goal is to improve the efficiency of the coal-burning process in the kiln. We refer this as a "zigzag kiln."</p>
<p>Here is a sketch of a similar blower configuration. Our blower will be placed above the walls instead of in front of the walls, but the airflow otherwise should be the same:</p>
<p><a href="https://i.stack.imgur.com/WFTM7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WFTM7.png" alt="Sketch of blower drawing air from an area on the left and blowing into a separate chimney area on the right" /></a></p>
| |mechanical-engineering|airflow|combustion|emissions| | <p>Normally fans are not custom built by a company that does not build fans. The concept of a centrifugal fan is not particularly complicated, but it is difficult to achieve target flow rate, pressure, and efficiency if your company does not do it on a daily basis. If you are not able to purchase a fan, another option is to copy someone else's simple straight blade design with known fan curves.</p>
<p>Answering your specific question: The top design with the two inch gap on the left is better at handling trash (foreign material in the air stream). The bottom design with the small gap on the right will have a better efficiency because it wont recirculate as much air around the housing. The style of the fan (blade shape) is really a minimal problem compared to knowing the flow rate, and pressure required. This <a href="https://www.nrel.gov/docs/fy03osti/29166.pdf" rel="nofollow noreferrer">NREL article</a> has some good information on centrifugal fan blade shape, but it basically simplifies to particulate handing reliability vs efficiency.</p>
<p>For clarification of the application (for myself) I attached some references below for a "continuous moving fire zig-zag brick kiln"</p>
<ul>
<li><a href="https://breathelife2030.org/wp-content/uploads/2016/09/12.pdf" rel="nofollow noreferrer">Introduction to Brick Kilns</a></li>
<li><a href="https://www.ccacoalition.org/sites/default/files/resources/Bricks-SEA.pdf" rel="nofollow noreferrer">FACTSHEETS ABOUT BRICK KILNS IN SOUTH AND SOUTH-EAST ASIA</a></li>
<li><a href="https://nepis.epa.gov/Exe/ZyPDF.cgi/P100EF3D.PDF?Dockey=P100EF3D.PDF" rel="nofollow noreferrer">Reducing Black Carbon Emissions in South Asia</a></li>
<li><a href="https://www.mdpi.com/2073-4433/10/3/107/pdf" rel="nofollow noreferrer">Zigzag Brick Kilns in Nepal</a></li>
<li><a href="https://blog.biokeram.com/kiln-firing-history-development-and-types-of-kilns" rel="nofollow noreferrer">Kiln firings: History, development and types of Structural kilns</a></li>
</ul>
<p><a href="https://i.stack.imgur.com/1UUTa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1UUTa.png" alt="brick kiln image from GKSPL introduction to brick kilns" /></a></p>
<p>Other than coal fired brick industry rules-of-thumb (unknown to me), here are some ways of estimating specifications. This is potentially how the engineers sized those 52 inch diameter fans above.</p>
<p>I would estimate the airflow requirements based on stoichiometric fuel combustion and excess oxygen requirements. If you know how much fuel is added and how much excess O2 is needed, engineers can calculate the flue exhaust flow rate that this fan will need to provide.</p>
<p>Temperature of the flue of the current operation should be known, but note that forced draft modification may change this temperature. Temperature of the air affects the density of the air. The lower the density, the larger the fan needs to be to transfer the same mass flow.</p>
<p>If the new zig-zag stacking configuration has been used with natural draft, you can estimate the minimum pressure requirements by measuring the pressure at the base of the stack compared to atmospheric; and add in the buoyancy pressure of the hot flue. This is a difficult one to estimate because the flow requirements and duct geometry are unknown.</p>
<p>I would recommend over-sizing the fan and motor by at least 50% (which the engineers have likely already done), then when commissioning, adjust pulley diameters to reduce the speed of the fan to the optimum operating flow rate. This also gives the ability to adjust the pressure/flow if other brick packing and temperature configurations are used in the future. A variable frequency drive, makes this even easier if it is available. I do not recommend using a damper as it greatly reduces the efficiency of the system. Note that if the electric motor is has less than 40% of its rated amps after optimization, it should be replaced with a smaller motor to allow the motor to be fully loaded and operate more efficiently.</p>
<p>Once the engineers have all of that information they just need to select a fan that meets the requirements of airflow and pressure for a given temperature. There are lots of good fan manufactures in every country and they will produce a fan that meets your needs cheaper than you can design and build it. <a href="https://www.alibaba.com/product-detail/Factory-direct-industrial-centrifugal-fan-blower_1600076266590.html" rel="nofollow noreferrer">Alibaba</a> is a good resource if you cant find any.</p>
<p>As side note, the forced draft will greatly change the heating and cooling rate of the bricks. The airflow may need to be reduced from optimum fuel use in order to reduce cracking. Also forced draft may structurally load areas of the kilns with vacuum pressure that may not have had any load in the past (baffles, doors, walls, roof, etc). Even 500 pascals can cause significant loads over large surfaces. It would be a good idea to have the engineers evaluate that scenario before commissioning. Also, unintended air (tramp air) may leak through cracks and access doors creating cold zones that were not a problem in the past. Make sure everything is well sealed.</p>
| 40440 | How do I choose a blower design for an induced draft brick kiln? |
2021-02-18T18:52:00.727 | <p>my father has a sports court construction business and I work with him infield and in the office. He usually almost always builds the sports courts using post-tension concrete with 4000 PSI concrete. He also includes turndowns in all the courts he builds. My question to this would be: Is it necessary or beneficial to use a turndown for the slab? The reason he uses post-tension is to avoid cracks which really does work marvelously, but is it necessary to have a turndown when there is no load on the edges of the slab? Can it actually be a negative component of such type of slab since if there is some type of soil erosion on the edges of the slab where the turndown is, this will cause an uneven load distribution of the slab itself on the edges, even though they're tensioned by the PT cables? Or is a turndown the right thing to do in this case? I'm not a civil engineer or anything but wanted to know if anyone had any insight on what would be best in this situation. Thank you!</p>
| |reinforced-concrete| | <p>In general, the edge beam (turndown) serves two purposes - prevent the tendency of curling at edges of the slab, and prevent loss of soil under the slab due to erosion. Ideally, the turndown shall be sit on the compacted granular material, that is free to drain, and will not influenced by the freeze-thaw cycles.</p>
| 40444 | Turndown for a Post-Tension Concrete Court |
2021-02-18T19:52:42.153 | <p>I'm reading a document describing the procedure to do a random vibration test. In it, a PSD (Power Density Curve) curve is defined, and there's a note at the bottom of the curve:</p>
<pre><code>Total acceleration effective: 50 to 500 Hz = 2.0G
</code></pre>
<p>What does it mean? Thanks.</p>
| |vibration| | <p>For a discrete system with <span class="math-container">$N$</span> degrees of freedom, the displacement can be expressed as
<span class="math-container">$$
\mathbf{u}(t) = \sum_{i=1}^N \boldsymbol{\Phi}_i \Gamma_i s_i(\omega_i, t)
$$</span>
where <span class="math-container">$\boldsymbol{\Phi}_i$</span> is the mode shape, <span class="math-container">$\Gamma_i$</span> is the participation factor, and <span class="math-container">$s_i$</span> is a function of the modal frequency, <span class="math-container">$\omega_i$</span>.</p>
<p>The second derivative of the displacement, <span class="math-container">$\ddot{\mathbf{u}}$</span> (with respect to time, <span class="math-container">$t$</span>) gives the <strong>total</strong> acceleration. Some corrections are needed when all the <span class="math-container">$N$</span> modes are not known.</p>
<p>In your case, the modes chosen are only between 50 Hz and 500 Hz. Therefore, the term <strong>effective</strong> has been used.</p>
<p>To get the <strong>g</strong> value, the acceleration (<span class="math-container">$\ddot{\mathbf{u}}$</span>) is divided by the acceleration due to gravity.</p>
| 40448 | What's the meaning of 'Total acceleration effective'? |
2021-02-20T05:20:45.780 | <p>I am attempting to 3D print my own Quad lock mount and require a spring. I've taken some measurements but don't know how to search for where I can buy more springs. Dimensions:
12mm height, 20mm OD. Thanks</p>
| |springs|3d-printing|coil-spring| | <blockquote>
<p>Dimensions: 12mm height, 20mm OD.</p>
</blockquote>
<p>You need more information than that to select or design a spring. Typically you need to specify several of the following:</p>
<ul>
<li>The free length.</li>
<li>The force required at L<sub>1</sub>. (Remember the spring exerts zero force when in its uncompressed state.)</li>
<li>The force required at L<sub>2</sub>, typically the maximum load and not quite fully compressed.</li>
<li>The <em>rate</em> expressed in N/mm or colonial measurement units.</li>
</ul>
<p>I only had to work through this once and realised that to get a reasonably consistent force over a range of positions that quite a long spring was required.</p>
<p>Search for online compression spring calculator should get you something to work with and once you have a specification you can look at standard catalog parts.</p>
| 40461 | Sizing a spring |
2021-02-20T17:23:57.140 | <p>I'm looking at a hobby micro servo, which has a 5mm splined shaft output. Specifically, this is a Tower Pro model SG90 or clone. Close-up photos of the shaft below.</p>
<p>I've found online diagrams for both a 30 degree Flat Root Spline shaft, and a 30 degree Filet Root Spline. Can someone please explain the difference between these spline designs?</p>
<p><a href="https://i.stack.imgur.com/zF6lV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zF6lV.jpg" alt="Micrograph of splned output shaft" /></a></p>
<p><a href="https://i.stack.imgur.com/o3Bqb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o3Bqb.jpg" alt="Closeup of spline teeth" /></a></p>
<p>I'm adding a drawing approximating the photos (and a hub) of a 21 tooth spline with a 5mm pitch circle and 45 degree pressure angle. (To simplify, I've not used involute teeth.)</p>
<p>I have filleted the exposed corners of the teeth at 0.06 mm. Is it these that make it a fillet root spline?</p>
<p>I've also added a fillet to the empty spaces joining the faces.</p>
<p><a href="https://i.stack.imgur.com/uXow4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uXow4.png" alt="detail drawing of spline" /></a></p>
<p>P.S. here's a link to my <a href="https://cad.onshape.com/documents/ca0cae54ca707e3d00ffdb59/w/b70f9fe24bf5ee92396a8648/e/f0e98e690a99cf50ba90bc92" rel="nofollow noreferrer">in-progress CAD model</a>. Feedback welome.</p>
| |mechanical-engineering|meshing| | <p>According to ISO 4156-1:2005, the following definitions apply:</p>
<blockquote>
<p>3.6
fillet root spline</p>
<p>spline having a tooth or space profile in which the opposing flanks are connected to the root circle (D<sub>ei</sub> or D<sub>ie</sub>) by a <em><strong>single</strong></em> fillet.</p>
<p>3.7 flat root spline</p>
<p>spline having a tooth or space profile in which <em><strong>each</strong></em> the opposing flanks are connected to the root circle (D<sub>ei</sub> or D<sub>ie</sub>) by a fillet.</p>
</blockquote>
<p>[emphasis added]</p>
<p>The distinction is quite subtle, as even flat root splines have fillets.</p>
<p>From what I now understand, the photos in my question are fillet root splines. However, the diagram is sort of a hybrid. I'll try to draw a second diagram to illustrate the difference and add it here.</p>
| 40470 | Flat root spline vs filet root spline: what's the difference |
2021-02-20T23:23:08.853 | <p>So I'm not actually an engineer, I'm a novice computer programmer but I think in order to solve a problem I have I'm going to need some help from the world of engineering.</p>
<p>I remember in high school they were teaching a civil engineering calculus class and one thing they spent a lot of time on was how to figure out the perfect curve for the perfect slope to transition the road to a bridge smoothly. My computer program is a digital paint and photo software which may include 3D sculpting and modeling later on. There is an existing tool that I don't know a ton about but it does something kind of similar called a <a href="https://en.wikipedia.org/wiki/B%C3%A9zier_curve" rel="nofollow noreferrer">Bézier curve</a> and basically, it lets you find smooth curves. But I'm looking for something else as well, I think.</p>
<p>Here's a rough picture of what I'm trying to do. Imagine this is someone's face. It is showing a mid-tone and a shadow. The green circle is where an AI-powered healing stamp will attempt to conceal a blemish. But there's a problem. There's both shadow and midtone in the same area as the stamp, and we're basically inventing contours in the skin. The robotic way to do it would be to just B-line between both instances of shadow, but maybe it would look more organic if there was a smooth transition between the external shadow's slope going in and the artificial shadow that continues.</p>
<p>So, what math would you use to figure that out, and does anybody know of any good place to learn it? Could I do it with just Calculus I under my belt?</p>
<p><a href="https://i.stack.imgur.com/hHGvp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hHGvp.jpg" alt="A diagram showing an abstracted part of a face in two examples, one where the shadow is cutting across the healing stamp's surface linearly, the other where a smooth curve is connecting them." /></a></p>
| |civil-engineering|mathematics|bridges| | <p>I think you are referring to ‘vertical curves’ for the slope transition of roadway to bridge. Also known as parabolas.</p>
<p>The graphic transition can be accomplished with a spline. Bezier is a spline. See Wiki: <a href="https://en.wikipedia.org/wiki/Spline_(mathematics)" rel="nofollow noreferrer">Spline</a></p>
| 40478 | What's the name of the calculus used to make smooth transitions in roads? |
2021-02-21T05:08:22.413 | <p>I have a measure of a quantity, x, and I know the standard deviation of the measure, STDx.
I want to compute the inverse of x, which is y= 1/x, but then what would happen to the standard deviation?
How can I compute STDy?
Many thanks in advance.</p>
| |measurements| | <p>The standard method, as set out in the BIPM <a href="https://www.bipm.org/utils/common/documents/jcgm/JCGM_100_2008_E.pdf" rel="nofollow noreferrer">Guide to the Expression of Uncertainty in Measurement</a>, is to work out the derivative of <span class="math-container">$y$</span> with respect to <span class="math-container">$x$</span> (in this case, <span class="math-container">$\mathsf{d}y/\mathsf{d}x = -1/x^2$</span>), evaluate it at the modal value <span class="math-container">$x_{\star}$</span> of <span class="math-container">$x$</span> (i.e. at the peak of the probability density distribution over <span class="math-container">$x$</span>), then estimate the standard uncertainty <span class="math-container">$\sigma_y$</span> of <span class="math-container">$y$</span> from the standard uncertainty <span class="math-container">$\sigma_x$</span> of <span class="math-container">$x$</span> by <span class="math-container">$\sigma_y = \sqrt{\left(\left(\mathsf{d}y/\mathsf{d}x\right)_{x = x_{\star}}\right)^2\sigma_x^2}$</span>.</p>
<p>The GUM sort of presents that formula as a first principle in itself, but actually, it emerges from the leading-order Laplace's-method approximations to the integrals that define the mean and standard deviation of a probability distribution over a continuous variable. Working in terms of leading-order Laplace's-method approximations means that any skew, kurtosis, etc. in the probability distribution over <span class="math-container">$x$</span> has no effect on the result. (However, if the modal value of <span class="math-container">$x$</span> is exactly zero, things get a bit more complicated, because in that case, the mode is no longer the leading-order Laplace's-method approximation to the mean.)</p>
| 40482 | Error propagation in reciprocal of a measurement |
2021-02-21T10:22:59.040 | <p>My professor told us that If we spray water in dry air the process is on an Isenthalpic line of the Mollier diagram. Is it always true? if yes, can you explain me why?</p>
| |air|air-quality| | <p>Let us presume that the process being considered is vaporization of liquid water to a gas state in dry air. As long as the temperature and pressure of the liquid water is constant during the process, vaporization of the liquid to gas in dry air will not change the molar enthalpy of the remaining liquid.</p>
<p>In those cases where for example the temperature of the liquid water cools down as some of the liquid vaporizes, then the molar enthalpy of the liquid water will change (decrease) as some of it evaporates.</p>
| 40489 | Is this an isenthalpic process? |
2021-02-23T02:09:42.757 | <p>Can I use <strong>220-12v,1.5 Amps</strong> Stepdown transformer along with 12V 2A <strong>(<a href="http://www.eicsemi.com/DataSheet/KBP200_10.pdf" rel="nofollow noreferrer">KBP200</a>)</strong> bridge rectifier to build a charger for 12v, 7.5AH battery or I should use transformer with Higher output, maybe 15V</p>
| |electrical-engineering|design|power-electronics|electronics| | <p>You will need more than a 220V:12V transformer.</p>
<p>Fullwave bridge rectifier output:</p>
<p><span class="math-container">$$ V_{DC} = 0.637\ V_{MAX} = 0.9\ V_{RMS}$$</span></p>
<p>This is the generic formula (textbooks), which gives you 10.8V. Higher current may cause < 10.8V. Lower current > 10.8V. As in: the actual voltage depends on the load current.</p>
<p>From <a href="https://www.electronics-tutorials.ws/diode/diode_6.html" rel="nofollow noreferrer">Full Wave Rectifier</a>.</p>
<p><a href="https://i.stack.imgur.com/tSmSu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tSmSu.png" alt="enter image description here" /></a></p>
<p>The peak voltage of a bridge rectifier has two diode losses (Forward voltage of diode depends on current).</p>
<p>If you do the ideal math:</p>
<p><span class="math-container">$$V_{MAX} = V_{RMS} \ \sqrt {2} - 2 V_{F\ diode} = 12V \times \sqrt {2} - 2 \times 0.7V = 15.6V$$</span></p>
<p>which seems enough to make a 12V charger, but that is peak voltage (top of ripple), while actual voltage depends upon load current. Essentially if there is no load current, then maximum DC voltage is 15.6V.</p>
| 40533 | Power Electronics Project |
2021-02-23T13:49:34.177 | <p>We were learning about superpositions of potential flows.</p>
<p>so,</p>
<p><span class="math-container">$$\Psi_{doublet} = \Psi_{source} + \Psi_{sink}$$</span></p>
<p>where: <span class="math-container">$\Psi_{source} = \frac{Q'}{2\pi}\theta_1$</span>, and <span class="math-container">$\Psi_{sink} = \frac{-Q'}{2\pi}\theta_2$</span></p>
<p>I then expect: <span class="math-container">$\Psi_{doublet} = \frac{Q'}{2\pi}(\theta_1 - \theta_2)$</span></p>
<p>but final form we were given was, <span class="math-container">$$\Psi_{doublet} = \frac{-Q'}{2\pi}(\theta_1 - \theta_2)$$</span></p>
<p>I have no clue as to where that extra minus sign is coming from.</p>
<p>Thank you guys kindly for your time and help.</p>
| |mechanical-engineering|fluid-mechanics|fluid| | <p>Seems to me that the source in this case is <code>θ2</code> since the sign convention normally is <code>-Q</code> for inward flow. This site <a href="https://www.ecourses.ou.edu/cgi-bin/eBook.cgi?doc=&topic=fl&chap_sec=07.3&page=theory" rel="nofollow noreferrer">https://www.ecourses.ou.edu/cgi-bin/eBook.cgi?doc=&topic=fl&chap_sec=07.3&page=theory</a> provides an example where <code>θ2</code> is the source angular displacement.</p>
| 40543 | A Doublet's stream function |
2021-02-23T15:07:32.110 | <p><a href="https://i.stack.imgur.com/zoo2b.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zoo2b.jpg" alt="enter image description here" /></a></p>
<p>So , I was trying to solve it. Then , I notice that acc of 2kg mass has 2effects. One acc from the force to left and one from the Question , it says acceleration to be up. Then , the 2kg mass would go diagonally and not straight .</p>
<p>I am not getting how to solve this Question further . Is it possible to solve this Question from inertial frame.</p>
| |mechanical-engineering|applied-mechanics|acceleration| | <p>This is a simple static problem, with the assumption that all contact surfaces are frictionless. In the static position (no motion), the cable has a tension equal to 2g. Then, in order to move the 2kg mass upward, there is an additional tension that equals to 2a, so at motion under given acceleration, the total tension in the cable will be massx(g+a), that is T = 2x(9.81+4) = 27.62N. Since the tension is constant throughout the cable, the force on the 10kg block must equal to -27.62N, the minus sign indicates the push force is in direction opposite to the tension.</p>
<p>A final check needs to be made to ensure the 3kg mass is adequate to resist the tension without displacement, that is easily proved, since the weight of the mass 3x9.81 = 29.43N > 27.62N, check!</p>
| 40550 | How to solve this Question regarding compound accelerations |
2021-02-24T12:05:00.077 | <p>Consider the following steelwork drawing:</p>
<p><a href="https://i.stack.imgur.com/U9fKv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U9fKv.png" alt="enter image description here" /></a></p>
<p>The location of the 6 no. M12 bolts are indicated with a circle and cross.</p>
<p>However, it is not clear whether the shank of the bolts should point into or out of the page. Without explicitly drawing the nut and bolt, what is a shorthand way to indicate the direction of the bolt shank and location of the nut?</p>
<p><strong>EDIT: I've done some digging and it seems there is no short hand for this. It is dealt with via an explicit drawing of the washer/nut and the use of drawing notes. I'll keep the question open in the hopes someone knows a shorthand way to notate which side the nut is to go on</strong></p>
<p><a href="https://i.stack.imgur.com/Vi8hQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vi8hQ.png" alt="enter image description here" /></a></p>
| |structural-engineering|technical-drawing| | <p>In structural construction, if it doesn't matter, it is left up to the contractor to put the bolts in however makes it easier for them.</p>
<p>If it matters for some reason (clearance, aesthetics, etc.) then there should be a note. Drawing a bolt head on the drawing and making it clear that it is not a nut and will be noticed by the contractor is hard. It is easier to add a general note or a call out that says, "Bolt heads shall face ... ".</p>
<p>I had a project where we were modifying an existing riveted structure. The client wanted the modifications to be as invisible as possible. For this project, we had to specify twist-off bolts with button heads. The button heads were exposed which made it harder for the contractor to get the twist-off wrench into tight spaces. This all was covered with one general note that said that the bolt head was to be on the exposed surface.</p>
| 40568 | How to indicate bolt direction on a drawing? |
2021-02-24T18:15:48.013 | <p>What key physical, chemical and mechanical properties of carbon ceramic brakes make it better compared to other brakes for example metallic brakes made of cast iron ?</p>
<p>I know that they are much more light weight and can handle much higher temperatures. But what other properties make them better than other brakes ?</p>
| |mechanical-engineering|materials|material-science| | <p>Well mass and high-temperature behavior are really the key benefits with carbon/ceramic brakes.</p>
<p><strong>High temperature:</strong>
Their high temperature properties make them more dependable during heavy use as they barely suffer from brake fade, they are also not prone to warping due to carbon fibers extremely low coefficient of thermal expansion (CTE). The CTE is actually negative :)</p>
<p><strong>Weight:</strong>
Reducing the vehicles un-sprung/un-dampened mass improves handling.
Reducing the total mass is generally desired in most vehicles, the reasons are many.
Reducing the wheels moment of inertia improves handling.</p>
<p><strong>Wear:</strong> They will last much longer than steel discs. However considering their much higher cost, compared to steel discs, they are still more expensive in the long run.</p>
<p><strong>Chemical:</strong> Another benefit but I would not call it key is that they are basically inert. They will not corrode. No more rusty discs.</p>
<p><strong>Price:</strong> They are expensive compared to steel, however compared to carbon/carbon discs they are cheap and still offer many of the benefits.</p>
| 40573 | Why carbon ceramic brakes |
2021-02-25T09:54:28.120 | <p>I have a Mitutoyo Sj-210, that plots automatically a Firestone curve. I have another device that gives me the SRM values and ect but no curve. I have no idea how the Sj-210 does it despite being useful. How would i plot a Firestone curve manually. What would I have to do on Excel to take the values I have and plot them.</p>
| |diagram| | <p>In excel what you should do to generate the curve is the following:</p>
<ol>
<li>take all the measurements for the shaft in one column (name it PROFILE)</li>
<li>Sort the data column in ascending/descending order (doesn't really matter)</li>
<li>Add another column (name it INDEX) with ascending integer (1,2, 3,4 etc)</li>
<li>then add another column (name it CDF) and divide the value in the same row on the INDEX column with the maximum value in the INDEX COLUMN. (You can multiply that by 100% but that is up to you).</li>
</ol>
<p>Then the Abbot firestone curve can be created by creating a scatter plot diagram with CDF on the X axis and PROFILE on the y-axis.</p>
| 40585 | How does one plot an Abbott Firestone curve |
2021-02-25T13:32:48.037 | <p>I know that you can have metallic and ceramic composite brakes.</p>
<p>My question is can you have disc brakes for a car that consists of polymetric materials or would this be impossible due to its low melt temperature ?</p>
| |mechanical-engineering|materials|material-science| | <p>I'm not an expert, but the disk in a disk brake system needs to absorb enourmous amounts of heat to dissipate the energy of motion. It then needs to transfer that heat to the air. Polymers generally are poor heat conductors and have relatively low heat capacity. Add to this lower strength and you have a poor choice for disk material. As Jim Clark mentions in his comment thermosetting polymers don't melt and can have higher degradation temperatures, but they will burn. All in all, it is hard to justify any polymer over something cheap and effective like cast iron.</p>
| 40589 | Polymer based disc brakes |
2021-02-26T19:30:23.907 | <p>Im trying to get my stepper to spin faster then 1000 pps, and my torque curve on my stepper's datasheet is showing it being capable of that(<a href="https://www.omc-stepperonline.com/download/17HS19-2004S1_Torque_Curve.pdf" rel="nofollow noreferrer">https://www.omc-stepperonline.com/download/17HS19-2004S1_Torque_Curve.pdf</a>) but when i try to spin it faster then 1000 pps, it vibrates violently and does't move.</p>
<p>I'm giving a drv8825 24V 2A max as the datasheet specifies, and the DRV8825 max pulses is 250 khz, which is significantly more then 1000 pps, so I think its a code problem.</p>
<p>My code:</p>
<pre><code>int main() {
xEN = 0;
yEN = 0;
xDIR = 1;
yDIR= 0;
for(int i = 0; i < 1000; i++){
xSTP = 1;
wait(0.001);
xSTP = 0;
wait(0.001);
}
}
</code></pre>
<p>I've read stepper motors need to be accelerated towards higher RPM, but im not sure how to do that. Like, how would I achieve 3000 pps?</p>
<p>EDIT: This seems to spin incredibly quickly and loudly but anything in the middle of this seems to just vibrate violently</p>
<pre><code>int main() {
xEN = 0;
yEN = 0;
xDIR = 1;
yDIR= 0;
for(int i = 0; i < 1000; i++){
xSTP = 1;
wait(0.0005);
xSTP = 0;
wait(0.0005);
}
}
</code></pre>
| |electrical-engineering|stepper-motor| | <p>Welp, it appears NMech above this is right about using the chip's clock as the basis for the pulses. I don't quite fully understand the code yet, but using code from <a href="https://www.exploreembedded.com/wiki/LPC1768:_Timers" rel="nofollow noreferrer">https://www.exploreembedded.com/wiki/LPC1768:_Timers</a> for prescalar microsecond code and <a href="https://www.electronicshub.org/how-to-use-timer-in-lpc1768/" rel="nofollow noreferrer">https://www.electronicshub.org/how-to-use-timer-in-lpc1768/</a> for the rest.</p>
<p>I set the prescalar to be based on microseconds and used the second link's Tim0 code to make a timer that stops once it reaches the specified amount of microseconds and now I can pulse past 1000 pps</p>
<pre><code>int main() {
xEN = 0;
xDIR = 1;
Timer0_init(); //<---code for this in second link
while(1){
xSTP = 1;
delayus(333); //code for prescalar in first link
xSTP = 0;
delayus(333);
}
}
</code></pre>
| 40609 | When I try to spin my stepper motor faster then 1000 pps, it vibrates and doesn't move |
2021-02-28T17:11:17.977 | <p>I am trying to repair a <strong>plissee blind</strong>, that does only fold up until half, then stops, no matter how high you push it.
It has a metal-thing holding three metal spirals that can be connected to the plastic drum in the middle. The middle plastic drum has two square-shaped holes where the metal spirals can be connected (see pictures, it's hard to describe).
It also has a kind of shuttle that is connected to the two strings pulling the blind.
The shuttle is connected to the metal-thing via another string. <strong>The metal thing should strongly pull the shuttle to the right, so the blind will be kept up</strong> (this is the problem).
If you pull the blind down, the shuttle should move left and should be kept left by the weight of the blind.
So far I understand it. But now I have no clue how I can make the metal-thing have that strong pull to the right.
I can pretty much guess how the left and the right metal spiral should be connected to the middle plastic drum, because I could see some dirty marks on the metal thing, caused by the metal spirals moving around.
But I have no clue what to do with the middle metal spiral. Even theoretically, I have no clue about its function, because however I put it, it seems to "do nothing".</p>
<p>Is there anybody out there with a better technical understanding to imagine that spiral spring in practise? I am thankful for every hint or resource, as I could find absolutely nothing online (at least in German; I am missing the right research words in English, as these seem to be kind of special terms).</p>
<p><a href="https://i.stack.imgur.com/Dwe9j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dwe9j.jpg" alt="whole spring" /></a></p>
<p><a href="https://i.stack.imgur.com/6jnqF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6jnqF.jpg" alt="connection to the middle plastic drum" /></a></p>
<p><a href="https://i.stack.imgur.com/ngFfw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ngFfw.jpg" alt="metal basis" /></a></p>
<p><a href="https://i.stack.imgur.com/At3D9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/At3D9.jpg" alt="spring inside the blind rail" /></a></p>
<p><strong>Edit</strong>:
<a href="https://i.stack.imgur.com/F0lVt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F0lVt.jpg" alt="left or right spiral only have one connector" /></a>
<a href="https://i.stack.imgur.com/mknIs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mknIs.jpg" alt="left or right spiral and their spindle" /></a></p>
| |springs| | <p>I came to the conclusion that one of the metal spirals must have broken in two, because I researched vendors' websites of these springs (e.g. <a href="https://www.hunterspringandreel.com/products/springs/constant-torque-spring-negator/spring-motor-assemblies" rel="nofollow noreferrer">these datasheets</a>) and I don't think there exist any three-spirals-with-three-spindles springs.
As I could only find industry-level vendors of these springs, and none that claimed to be specifically suitable for this plissee blind, I gave up.</p>
<p>For others researching this: it is called <em>constant torque spring</em> :)</p>
| 40632 | Repairing blind in bathroom: How do spiral springs with 3 revolving drums work? |
2021-03-02T10:49:28.023 | <p>See below for an extract from a general arrangement drawing from the 1980s. It makes reference to a '9" rocker slab' which appears to be a slab supported by steel beams.</p>
<p><a href="https://i.stack.imgur.com/iM3lF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iM3lF.png" alt="enter image description here" /></a></p>
<p>Does anyone have any more information on the construction details of a rocker slab?</p>
| |civil-engineering|reinforced-concrete| | <p>Found this <a href="https://www.academia.edu/9773741/CONTAINER_TERMINAL_AND_INTERMODAL_RAIL_YARD_OPERATIONAL_AREA_CONSIDERATION_FOR_PAVEMENT_DESIGN" rel="nofollow noreferrer">reference</a> , a white paper from a firm "Moffatand Nichol," which says</p>
<blockquote>
<p>Localized areas found between structure and external service yard can
be designed as a form of articulation or ‘rocker’ slab, which as the
name suggests, allows for normal movements irrespective of on-going
settlements. While rocker slabs may be a solution for particular
applications, they can never provide an operationally friendly
solution for large storage areas or service yards.</p>
</blockquote>
<p>My take, skimming some other sites' documents which mention but don't strictly define them, is that a "rocker slab" is a single-cast piece of concrete which rests on supports at both ends (and possibly the sides), thus joining two separate parts of framing. Quite possibly without being bonded in place (as the slab is a preform). Local bolts would be required to anchor the slab.</p>
| 40655 | What is a 'rocker slab'? |
2021-03-03T01:42:56.750 | <p>I have a large DC Motor I ripped out of a treadmill and I'm curious if the back emf generated from manually rotating the motor is a DC output or an AC output. Is it AC only when it's from the power grid and other sources treated as DC? How can I always tell?</p>
<p>For context, the motor is a 2.65Hp, 21.4A, P.M.D.C Motor</p>
<p>Thank you!</p>
| |motors|electrical| | <p>For Permanent Magnet DC motor, even if you rotate it with the constant rate the back EMF you'll get externally from the motor connector pins ought to be more <em>similar</em> to a rectified AC.</p>
<p><a href="https://i.stack.imgur.com/KhqeSm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KhqeSm.png" alt="enter image description here" /></a></p>
<p>However, I wouldn't expect it to be exactly sinosoidal.</p>
| 40675 | Back EMF from a motor |
2021-03-03T18:21:02.453 | <p>Hello I am modeling a vehicle that has a seperately excited DC motor as the power plant utilizing the constant-torque and constant-power regions for traction. The acceleration is governed by the following equation
<span class="math-container">$$
M \frac{dv}{dt} = F_t -(\beta_1+\beta_2v^{2}+\beta_3v)
$$</span></p>
<p><a href="https://i.stack.imgur.com/oAyVk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oAyVk.png" alt="velocity profile" /></a></p>
<p>The problem with my model is that the <span class="math-container">$\beta_1$</span> component, the rolling resistance, is a constant value and causes the velocity to go negative for the first 0.7 seconds when the model is executed . Any suggestion on how to fix this issue?</p>
| |simulink| | <p>I solved the issue with a switch function block, it not the best solution but it works for my basic simulation. Thanks again<a href="https://i.stack.imgur.com/A0twe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A0twe.png" alt="function block" /></a></p>
| 40696 | Modeling Vehicle Acceleration - Simulink |
2021-03-03T20:49:02.520 | <p>In continuum mechanics, why does local derivative and convective derivative only apply to spatial descriptions?</p>
<p>Is it because in the material description, the motion focuses on the particles, and not on fixed spots in space, and in each particle's motion the time and space are simultaneously changing, so we can't perform derivative w.r.t time and space separately?</p>
| |fluid-mechanics|solid-mechanics| | <p>Yes, your second paragraph is a pretty good natural-language explanation.</p>
<p>The way I'd have put it is as follows:</p>
<p>A partial derivative is a derivative with respect to (something), calculated on the assumption that (something else) is held constant. Specifically, the "local derivative" is defined as the derivative with respect to time, calculated on the assumption that the location in space under observation is held constant; the "material derivative" is defined as the derivative with respect to time, calculated on the assumption that the identity of the parcel of fluid under observation is held constant; and the "convective derivative" is defined in some texts as being the same thing as the material derivative, and in other texts as being the difference between the material derivative and the local derivative. The material derivative of velocity is the same thing as the "acceleration" in Newton's second law, so when one writes Newton's second law for a fluid (which is what the Navier-Stokes momentum equation is) in terms of the material derivative of velocity, the "mass times acceleration" side of the equation appears as a simple, single term; whereas when one writes Newton's second law for a fluid in terms of the local derivative of velocity, one has to add something on to the local derivative of velocity to obtain the acceleration, hence ending up with two terms on the "mass times acceleration" side of the equation.</p>
<p>(A little linguistic note of caution: the verb "derive" does not mean "compute a derivative".)</p>
| 40700 | Why does local derivative and convective derivative only apply to spatial descriptions? |
2021-03-04T05:09:31.230 | <p>So I've been trying to find out how to determine what proportion of fluid will flow down a given path when a main line branches into two lines at a 90 degree T-Junction and I can't find anything helpful.</p>
<p>Lets just assume water at room temperature in steady-state conditions.</p>
<p>It seems that pressure losses will have an effect on proportion of distribution. If the two paths have similar designs, ie; the same energy losses, the flow will split roughly 50:50. However, if one path has more energy losses than the other, then < 50% of fluid will go down that path. So there is a relationship between the proportion of pressure losses in the line to the proportion of flow down that path. What is this relationship? Is there a numerical model to determine this?</p>
<p>Also, intuitively, at a 90 degree T-junction, momentum/inertia should have an effect. Obviously the fluid would prefer to continue on straight rather than take a 90 degree turn, especially at high speeds. Do we consider this when determining flow division? Is there a numerical model to help me determine it's effect on how much flow will go down a given path?</p>
<p>Any answers or resources guiding me in the right direction, or numerical models to help me determine flow distribution at a T junction would be greatly appreciated!</p>
<p>Thank you in advance.</p>
| |mechanical-engineering|fluid-mechanics|civil-engineering|pipelines| | <p>In flow problems, it's generally a good idea to work backwards in realtion to the flow: Find where each branch is ending and the pressure at this point (height, atmospheric pressure, <span class="math-container">$p_{out 1}, p_{out 2}$</span> ). Then, find the relation between pressure loss and flowrate for each branch (<span class="math-container">$\Delta p_1(Q_1), \Delta p_2(Q_2)$</span>).</p>
<p>You know the the pressure at the junction is the same for both flows (<span class="math-container">$p_{in}$</span>), and that <span class="math-container">$p_{in}-\Delta p = p_{out}$</span>. So it follows: <span class="math-container">$p_{out 1}+\Delta p_1(Q_1)=p_{out 2}+\Delta p_2(Q_2)$</span>. Knowing <span class="math-container">$Q_{total}=Q_1+Q_2$</span> you can find the flows.</p>
<p>You won't solve this analytically, only numerically.</p>
<p>As for your second question, consider this (admittedly a bit cryptic) diagram:</p>
<p><a href="https://i.stack.imgur.com/UthFi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UthFi.png" alt="enter image description here" /></a></p>
<p>(<a href="https://www.schweizer-fn.de/zeta/abzweig/abzweig.php#t_stueck_trennung" rel="nofollow noreferrer">from here</a>)</p>
<p>The red dashed line is the <span class="math-container">$\zeta$</span> value for flow straight through the pipe, the lines above are for branching flows (at different sizes of the off-branch). You see that the <span class="math-container">$\zeta$</span> value is always higher for branching flows, so your hunch in effect correct.</p>
| 40709 | How do you determine flow distribution at a dividing T-Junction? |
2021-03-04T05:58:32.437 | <p>Can anyone please explain this question to me? I am okay with answering a gear assembly with 2 gears at a time. But the third one here is confusing me.<a href="https://i.stack.imgur.com/G24nM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G24nM.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|gears|torque|solid-mechanics| | <p>There a few steps in this problem.</p>
<ul>
<li>The first one is to figure out what is the torque that each shaft is subjected.</li>
<li>The second one is to determine the twisting angle.</li>
</ul>
<p>In this you are making the assumption that the disks do not deform.</p>
<p>So the regarding the torque (if you know the basics about gear assemblies this is obvious):</p>
<ul>
<li>the short rod (L/2) is subjected to T torque</li>
<li>the long rod (L) is subjected to <span class="math-container">$\frac{3}{2} T$</span> torque</li>
</ul>
<p>The magnitude of the twist is given by:</p>
<p><span class="math-container">$$\Delta \theta = \frac{M_{t,i} L_{i}}{G_i J_i}$$</span></p>
<p>where <span class="math-container">$J_i =\frac{\pi d^4}{32}$</span></p>
<p>therefore the twist is :</p>
<ul>
<li>for the short <span class="math-container">$$\Delta\theta_s = \frac{T L }{2 \;G \;J}$$</span></li>
<li>for the long <span class="math-container">$$\Delta \theta_L = \frac{3 T L }{2\; G \;J}$$</span></li>
</ul>
<p>However because of the gear ratio from long to short being 3/2, twisting one unit on the long (DE) will result in 3/2 unit rotation on the short.</p>
<p>Therefore the final twisting of point A is:</p>
<p><span class="math-container">$$\Delta\theta_A=\Delta\theta_s +\frac{3}{2}\Delta \theta_L $$</span>
<span class="math-container">$$\Delta\theta_A=\frac{T L }{2 G \;J} +\frac{3}{2}\frac{3 T L }{2 G J} $$</span>
<span class="math-container">$$\Delta\theta_A=\frac{T L }{G J} \left (\frac{1}{2}+\left(\frac{3}{2}\right)^2\right)$$</span>
<span class="math-container">$$\Delta\theta_A=\frac{11}{4}\frac{T L }{G J} $$</span></p>
<p>Finally substituting <span class="math-container">$J_i =\frac{\pi d^4}{32}$</span> (if everything went ok)</p>
<p><span class="math-container">$$\Delta\theta_A=88\frac{T L }{G \pi d^4} $$</span></p>
| 40710 | What is the angle of twist in end point A? |
2021-03-04T10:45:18.860 | <p><a href="https://i.stack.imgur.com/WJnJr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WJnJr.jpg" alt="keyclip" /></a></p>
<p>I've tried to search for the name of the connecting mechanisms on these clips but I can't find it. I'm trying to build something where two parts are connected on one axis (diameter less than 5 mm) but each moves independently. So I need a connecting mechanism which doesn't let them come apart but which is also small enough to fit in that hole. What are some mechanisms that would allow this?</p>
| |design|machine-elements| | <p>IMHO, they are both variations of the <strong>pin joint</strong> or [<strong>revolute joints</strong>] (<a href="https://en.wikipedia.org/wiki/Revolute_joint" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Revolute_joint</a>) or <strong>pivot joints</strong>.</p>
<p><a href="https://i.stack.imgur.com/zEhFc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zEhFc.png" alt="enter image description here" /></a></p>
<p>It comes in many types and flavours that is difficult to keep track of.
<a href="https://i.stack.imgur.com/CktbQm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CktbQm.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/gT3tvm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gT3tvm.png" alt="enter image description here" /></a></p>
<p><strong>Update:</strong></p>
<p>Regarding how you stop the pin to fall out over time there are many ways. Interference fit or other types of frictional connections are <strong>not</strong> an option if you are after a freely rotating joint. Usually you will implement one of the following (not exhaustive)</p>
<ul>
<li>some type of safety pin or split pin</li>
</ul>
<p><a href="https://i.stack.imgur.com/HGwMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HGwMf.png" alt="enter image description here" /></a></p>
<ul>
<li>using rivets</li>
</ul>
<p><a href="https://i.stack.imgur.com/aAVFN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aAVFN.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/eZ9vG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eZ9vG.png" alt="enter image description here" /></a></p>
| 40718 | What are the names of these connection mechanisms? |
2021-03-04T12:50:30.817 | <p>I've been playing around with 3D printing some compliant mechanisms, mostly to get precise linear motion, and I've noticed that many commercial compliant mechanisms are often designed like this:
<img src="https://www.thorlabs.de/images/xlarge/17681-xl.jpg" alt="XY Flexture" /></p>
<p>Where the flexible parts are only thin by their joints. Is there any benefit to this over simply keeping it thin across the whole flexible part? All my designs use flexible parts which are thin all the way, granted I'm mostly working in PLA so I need the parts to be long and thin to handle the stresses without breaking, but what is the reason behind keeping those parts thick in the middle?</p>
| |design|3d-printing|flexures| | <p>I'll start from the particular example you are asking about. It is a positional adjustment device. It is heavily used in laser optical tables to provide micrometer adjustments. And it falls in the category of compliant mechanisms like you suggest.</p>
<p>There are a few requirements for these devices (most have been already pointed by others). IMHO, the main two are</p>
<p>a) resistance to vibration (the added mass serves this purpose, because it lowers the natural frequency).</p>
<p>b) <strong>spring back</strong> repeatedly to original position. This is a very important aspect for these types of devices. They do not rely on the screw to set the position, they passively press onto the screw, thus minimizing any backlash that might be present.</p>
<p>Now regarding why <strong>compliant mechanism need to have thick sections</strong>:</p>
<p>In my understanding is that for the same force levels, the deformation of thin sections provides the range of movement, while the deformation of thick parts is negligible, thus allowing to maintain a functional shape.</p>
<p>If the thickness was everywhere small/flimsy, you would have a structure that would deform everywhere significantly. That may not always be a problem. The problems occur when you are trying to move/displace this structures a lot. Then their mode of displacement becomes unpredictable. (you might have experienced the problem when you are trying to use a tape measure in non straight position like the following)</p>
<p><a href="https://i.stack.imgur.com/C0SMxm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C0SMxm.jpg" alt="enter image description here" /></a></p>
| 40721 | Why are many compliant mechanisms only flexible in the joints? |
2021-03-04T16:30:56.830 | <p>I am investigating the flow and heat transfer over a cylinder in a wind tunnel. I want to record the data that I need to calculate the flow characteristics, But how can I make sure that the steady-state condition is achieved? There are very small changes in measured parameters like velocity and we know that there is no change anywhere with time in steady-state condition.</p>
| |fluid-mechanics|heat-transfer|wind-tunnels| | <p>The main issue you have is not the wind tunnel air flow but the heat transfer.</p>
<p>Generally the air flow in a wind tunnel can very quickly reach a steady state (for both open and closed wind tunnels). However, the problem I believe is with the heat transfer in the cylinder.</p>
<p>Depending on how you setup your experiment it will depend.</p>
<p>For example, if you are using an open wind tunnel, then the temperature of the air during the day will fluctuate. That will change the heat transfer gradient, and you will not (trully) have a steady state.</p>
<p>Another parameter that affects is if the cylinder is heated internally to a temperature or if its just idly sitting there. In the first case (internally heated) you will have a steady state when the external temperature of the cylinder stops to change temperature. On the other hand (not heated cylinder) you will have a steady state, when the cylinder <em>internally</em> has the same temperature with the air flowing around it.</p>
<p>As you can see, there are many variations, and you need to provide more information for a more useful answer.</p>
| 40729 | How can we determine that the flow in a wind tunnel reaches the steady-state condition? |
2021-03-04T18:19:29.207 | <p>I have an engineering competition coming up and one of the things I need to build is structure with both compression and tension members. The structure is scored based on the load held/weight ratio, and I already have the design I'm looking for but I need ways to reduce weight.</p>
<p>I was thinking about comparing the strength of a hollow rectangular section and a solid one with the same cross-sectional area. I think that I understand that for members under tensional stress, the strength of the two should be equal because the cross section of the two members is the same.</p>
<p>Would this be different for compression members? i.e., would one be able to carry more weight if one used a hollow member instead of a solid one?</p>
<p>EDIT: Another question: If I'm making this in a structural simulation, does the dimensions of the member matter if the cross-section is still the same? I guess you can assume here that the member is undergoing pure compression parallel to the length of the member.</p>
| |mechanical-engineering|structures| | <p>In compression, except for very short members the member fails by buckling under Euler's formula:</p>
<p><span class="math-container">$$P_c= n\frac{ π^2 E I}{ L^2} $$</span></p>
<p>Where n is a factor based on the column's end connections.</p>
<p>Therefore we see that with the same area of a section if we can increase its I, second area moment, we increase the compression force the member can safely take.</p>
<p>The I of hollow square or rectangular or circular section is very close to a section with the same size. Alternatively, the same section area if used in a hollow section creates a lrger I than a solid section.</p>
<p>Lets use a square of sides a as an examole</p>
<p><span class="math-container">$$I=a^4/12 $$</span></p>
<p>but for a hollow section where the void is a square with sides a/2<span class="math-container">$$ I=a^4/12-a/2^4/12= \ 15/16 *a^4/12$$</span></p>
<p>Removing 25% of section area leads to only 1/16 less I.</p>
| 40736 | Hollow (rectangular) sections versus solid sections |
2021-03-04T19:32:27.397 | <p>Sorry I'm having a migrane and can't think rn, which is why I am posting such a basic question for confirmation...</p>
<p><a href="https://i.stack.imgur.com/CM01m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CM01m.png" alt="Sprockets on rotating beam" /></a></p>
<p>The above diagram shows two sprockets on a beam which is attached to a driven gear.
The Red sprocket doesn't rotate (it is fixed). The orange sprocket is at the other end of the beam and can spin freely. Theres a chain between them (not drawn)</p>
<p>Relative to the gear/beam, the red sprocket is rotating in reverse.</p>
<p>So would I be correct in saying that the orange sprocket will turn with the opposite angular velocity of the Driven Gear?</p>
| |gears|mechanisms| | <p>If the sprockets have the same number of teeth both the red and orange sprockets will remain in the same relative orientation. The writing on the orange will remain horizontal.</p>
<p>If red has <em>N</em> teeth then the chain will advance <em>N</em> links per rotation. If orange has <em>n</em> teeth then it will rotate <em>N/n</em> teeth relative to the red.</p>
<p>So rotating the beam one revolution clockwise will result in the the orange sprocket rotating anti-clockwise relative to the beam by N/n turns.</p>
| 40741 | Sprokets on rotating body |
2021-03-04T21:29:15.093 | <p>I cannot seem to visualise why the area of the upstream face of control volume is 2pir∂r, can someone pls clarify on why is this the case?</p>
<p><a href="https://i.stack.imgur.com/5G7bO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5G7bO.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/bjtnW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bjtnW.png" alt="enter image description here" /></a></p>
| |fluid-mechanics|geometry| | <p>Because of the very thin layer of <span class="math-container">$\delta r$</span>.</p>
<p>if we peel it off and stretch it like a very thin rectangular with <span class="math-container">$L=2\pi r$</span> and thickness as <span class="math-container">$\delta r, \ $</span> its area will be <span class="math-container">$A=l*\delta r= 2\pi r*\delta r$</span></p>
| 40747 | Why is the area of upstream face of control volume a stated? |
2021-03-05T13:12:38.940 | <p>In continuum mechanics, the definition of trajectory or pathline is the locus of the positions occupied by a given particle in space throughout time.</p>
<p>And streamlines are a family of curves which for every instant in time are the velocity field envelopes.</p>
<p>For a stationary velocity field, trajectories and streamlines coincide..</p>
<p>How is this possible? Even if the velocity didn't depend on time, and was only changing with the change of position, it doesn't make sense for the streamline and trajectory to coincide because they hold different physical meaning.</p>
| |fluid-mechanics|solid-mechanics| | <blockquote>
<p>How is this possible? Even if the velocity didn't depend on time, and
was only changing with the change of position, it doesn't make sense
for the streamline and trajectory to coincide because they hold
different physical meaning.</p>
</blockquote>
<p>The only thing that represents a single physical thing is a trajectory. Pathlines, streaklines, and streamlines can all be thought of as aggregations of trajectories that were assembled using different rules. They are traces in space, and somebody decided to make life easy and define them in a way that they are all congruent for a constant velocity field.</p>
<p>A trajectory is a point in space and time. If you start with a velocity field, you need an initial point and time to get started. The trajectory point is then found by integrating to the final time.</p>
<p><span class="math-container">$$P\small{(x_0\,,y_0\,,t_0\,,t_1)}\normalsize{ =} \int_{t_0}^{t_1} \mathbf{V}(\small{x(t)\,,y(t)\,,t}\,\normalsize{)\,dt}$$</span></p>
<p>If <span class="math-container">$\mathbf{V}$</span> is a stationary field, several things are simpler. <span class="math-container">$\mathbf{V}$</span> doesn't have a time parameter, and the bounds of the integral can be shifted by any constant we like.</p>
<p><span class="math-container">$$P\small{(x_0\,,y_0\,,t_1)}\normalsize{ =} \int_{0}^{t_1} \mathbf{V}(\small{x(t)\,,y(t)}\,\normalsize{)\,dt}$$</span></p>
<p>A pathline is a collection of trajectory points that all have the same starting point <span class="math-container">$P$</span>, with <span class="math-container">$t_1$</span> being anything.</p>
<p>A streamline is basically the same thing, but we travel across the field infinitely fast. No time elapses. But this makes collecting data points from the integral difficult. So some clever person realized that the same curve can be gotten by parameterizing the curve differently. Instead of using time, we can use distance as the variable of integration. And instead of integrating the velocity, we integrate the normalized velocity.</p>
<p>So <span class="math-container">$s$</span> is the new variable of integration, and it is related to <span class="math-container">$t$</span> by <span class="math-container">$ds=|\mathbf{V}\small{(x\,,y)}\normalsize{|\,dt}$</span></p>
<p>And a streampoint becomes <span class="math-container">$\displaystyle P\small{(x_0\,,y_0\,,s_1)}\normalsize{ =} \int_{0}^{s_1} \mathbf{v}(\small{x(s)\,,y(s)}\,\normalsize{)\,ds}$</span>
where <span class="math-container">$\displaystyle\mathbf{v}=\frac{\mathbf{V}}{|\mathbf{V}|}$</span></p>
<p>So what we are doing is reparameterizing the integral to get rid of an infinity problem and provide nice bounds on the definite integral. This parameterization is different at each point in the field, but it all comes back together when you integrate over the field. Different parameterizations, same shaped curve.</p>
<p>And that's okay, because with pathlines, streakline, and streamlines, we don't keep an index of the parameter values that produced a given point the way we do with a trajectory. A trajectory should retain this timestamp info.</p>
<p>Now if you want to test your calculus chops, assume <span class="math-container">$\mathbf{V}$</span> isn't stationary, and <span class="math-container">$ds=|\mathbf{V}\small{(x\,,y\,,t)}\normalsize{|\,dt}$</span> and figure out what the real restrictions on <span class="math-container">$\displaystyle\mathbf{v}=\frac{\mathbf{V}}{|\mathbf{V}|}$</span> are such that streamlines and pathlines are the same. A constant <span class="math-container">$\mathbf{V}$</span> is overly strict, there are certain <span class="math-container">$\mathbf{V}(t)$</span> for which this is true as well.</p>
| 40757 | How is it possible for trajectory and streamline to coincide in case of stationary velocity field? |
2021-03-05T13:38:37.177 | <p>There are many places I see a knob used to set the vertical position of an object (like in a microscope stand), but how does it not just simply slide down due to gravity once the knob is released. It's not like they first have to pull/push out the knob before turning it so as to lock it first. Instead once you turn it, it just remains in that position. I believe it's a rack and pinion mechanism, but I don't understand how it locks itself in place after it's been set.</p>
| |mechanical-engineering|mechanisms| | <p><a href="https://i.stack.imgur.com/5j9lT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5j9lT.png" alt="table lifting mechanism (thang010146 youtube)" /></a></p>
<p>I've found this method on YouTube (Channel: thang010146), but as NMech shows there are a few other methods like the lead screw and the worm drive screw jack.
But all look like they work with the same principle of using the friction of the thread to stop sliding. So we just have to match the friction to the max load capacity we want.</p>
<p><a href="https://i.stack.imgur.com/I9kzr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I9kzr.png" alt="worm drive screw jack" /></a></p>
| 40758 | vertical rack and pinion self-stopping |
2021-03-05T22:46:57.537 | <p>I designed this ball feeder for my robot however I'm experiencing frequent ball jams as the ball is stuck against the wall of the hole entry or just a ball being moved left and right not entering the.</p>
<p>I was thinking of maybe adding a wall to one side of the opening to guide the ball into the opening. However since the balls are on top of the moving disc they could fall on th other side and get stuck there.</p>
<p>I should mention that this is implemented using a single servo motor so only 0-180* movement is possible.</p>
<p>I'm open to any suggestion either ne design or improvement.</p>
<p><a href="https://i.stack.imgur.com/ANn9F.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ANn9F.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/B5dup.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B5dup.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/43Yfv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/43Yfv.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|motors|design|servo| | <p><a href="https://i.stack.imgur.com/XSzLc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XSzLc.jpg" alt="enter image description here" /></a></p>
<p><strong>I think you need change the structure, as show image of ball feeder.</strong></p>
| 40770 | Improving ball feeder design |
2021-03-05T23:44:22.610 | <p>I am looking to calculate the force that would be required to pull a single axle 12000 lb box trailer with a 72 sq ft frontal area up a 10% grade at 80 mph.</p>
<p>It has been quite a long time since I've done any engineering and I want to make sure I'm not way off. I am only looking to find a comfortable upper limit for potential force required to pull any trailer. My numbers estimate a heavy trailer going up a very steep grade faster than one probably should.</p>
<p>For this example I'm estimating 1200 LB of strain from pulling the weight of the trailer and another 1200 LB from the drag of the wind. Is this a fair rough estimate or am I way off?</p>
| |mechanical-engineering| | <p>The general equation of an object moving uphill is FT >= FR, in which FT is the thrust produced by the object, and FR is the resistance force that impede the movement of the object. FR includes,</p>
<ol>
<li><p>Downward pull of gravity, F1 = mgsin(\theta), the angle \theta = slope (10 degrees)</p>
</li>
<li><p>Friction of tires on pavement, F2 = mgcos(\theta)(\mu), \mu is the friction coefficient (it varies, but can be conservatively taking as 1)</p>
</li>
<li><p>Aerodynamic drag, F3 = C(\rho)AV^2/2, in which,</p>
</li>
</ol>
<ul>
<li><p>C is drag coefficient, it is about 0.71 for truck trailer.</p>
</li>
<li><p>\rho is density of the air, it can be assumed as 24x10-4 slugs/ft^3 at sea level.</p>
</li>
<li><p>A is the frontal surface of the truck trailer.</p>
</li>
</ul>
<p>so FR = F1 + F2 + F3. Note that if your are only interested in the force required to toll the trailer, you shall set mg in the equation equal to the weight of the trailer (12000 lb), otherwise, mg is the total weight of the truck and trailer.</p>
<p>Depends on the condition of your truck, you need to ensure it has adequate power to drive uphill. Good luck.</p>
| 40771 | Upper limit of force required to pull a trailer |
2021-03-05T23:59:10.167 | <p>Hi so I have this senior design project involving cooling human skin as it is irradiated by an infrared LED. Human skin not only is super absorbent to IR, but that IR goes THROUGH the thickness of it really well. There are some very long path lengths of IR photons within the human body.</p>
<p>In my heat transfer class, we only saw radiation as dumping all its energy right at the very surface of some block of mystery meat or whatever. In that class we only cared about absorption in terms of how much radiation isn't reflected. We then treated all the radiation that got absorbed as being "stopped" by that surface essentially. After that the heat from this interaction would be distributed by conduction within the receiving block thing.</p>
<p><strong>Million dollar question</strong>: What is happening to that topmost surface temperature if absorbed radiation is going THROUGH the surface? So now my deeper layers are being heated by both conduction and direct radiation?</p>
<p>The end goal is to figure out the surface temperature of this irradiated spot of skin, then cool it by cold air convection until a goal surface temperature is reached. Any diagrams in answers would be super helpful. I am a visual creature. Part of my trouble is photons are hard to visualize and I keep picturing bullets stopped by walls when I know photons don't have mass and it's a poor analogy. It's really tripping me up.</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heating-systems|heat-exchanger| | <p>Photons either pass through a given material or they don't.</p>
<p>If they pass through, it's as if the material isn't there as far as they're concerned (ignoring refraction).</p>
<p>If they get absorbed, then they dump all their energy into the material. This might excite the incident atom such that it then re-emits the photon with the same or different wavelength, such that the effective "heating" of the material is just the difference between the energy given by the incoming and outgoing photons.</p>
<p>What you need to remember is that a beam of light is composed of an almost infinite number of photons. So whenever you talk about light going through a material and heating it, you're talking about a few different groups: photons that went through the material completely undisturbed (i.e. with the same energy as before passing through) and which don't contribute at all to heating the material, photons which are entirely absorbed by the material without any "output" photons and therefore heat the material by their energy, and intermediary photons which are absorbed but also emit photons themselves and therefore heat the material less.</p>
<p>And if it is possible for light to pass through a material, you'll get an iterative process: some fraction of the photons (let's say 70%) will be absorbed by the material's surface, and the rest (30%) will pass through. On the second layer, 70% of the photons which managed to pass through will be absorbed and 30% will pass through, rinse and repeat until you get entirely past the material. (Real life obviously doesn't really have "layers", but it's a useful mental image). Only some fraction of the absorbed photons will emit photons of their own in random directions; these outgoing photons will then also have to gamble to see how many "layers" they can pass through before being absorbed.</p>
<p>So, to answer your million dollar question: the surface temperature will be increased by the fraction of the photons which <em><strong>don't</strong></em> pass through. It will also have some "inner" heating from "outgoing" photons emitted by inner layers which have absorbed some of the photons which managed to pass through the surface, but that will likely be a very minor complement.</p>
| 40772 | Does a photon impart energy on a surface (to raise its temp) if it goes THROUGH that surface? |
2021-03-06T11:40:23.993 | <p>I'm having trouble reading several dimensions on this technical drawing.</p>
<p>First, I don't know how big is the diameter of the marked cylinder.
M24 is a threaded hole in my opinion and that cannot say anything about the cylinder.
<a href="https://i.stack.imgur.com/Zo48U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zo48U.png" alt="enter image description here" /></a></p>
<p>And what is the diameter of the rotor ? Is it 40 ? if yes then why they didn't put any diamater mark next to the number 40 ?
And I cannot find out the angle of the central key relative to slots?
<a href="https://i.stack.imgur.com/wDtO0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wDtO0.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|technical-drawing| | <p><a href="https://i.stack.imgur.com/9M2Yh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9M2Yh.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. The cylinder is 24 mm diameter with an <strong>outer</strong> thread of M24 x 2 (pitch) extending to 16 mm from the end.</em></p>
<blockquote>
<p>And what is the diameter of the rotor? Is it 40? if yes then why they didn't put any diamater mark next to the number 40?</p>
</blockquote>
<p>Yes, 40 mm. It should be marked with a Ø symbol or "DIA".</p>
<blockquote>
<p>And I cannot find out the angle of the central key relative to slots?</p>
</blockquote>
<p>It's not there in the extract you have given. I would assume 45° - but that could be an expensive assumption.</p>
| 40781 | Ambiguous technical drawing sizes |
2021-03-06T17:56:02.703 | <ol>
<li><p>"<a href="https://www.youtube.com/watch?v=3l6EcBg9EPA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=3l6EcBg9EPA</a>" [@24:30 - 25:10] He says something about the load of a threaded interface being on only one small part of the spiral. To me this seems a bit odd right?</p>
</li>
<li><p>How much contact area is there really between a screw and a nut? "<a href="https://engineering.stackexchange.com/questions/13547/where-is-the-contact-point-between-screw-spindle-teeth-and-nut-teeth/13562#13562">Where is the contact point between screw spindle teeth and nut teeth?</a>" - Here one answer suggests that it's dependant on the thread type and tolerances which makes sense, but I couldn't find any reference values in my machining books, besides this link which briefly mentions a contact area of 30-35% without further explaining it "<a href="https://fastfixtechnology.com/rail/the-benefits-of-lockbolts/" rel="nofollow noreferrer">https://fastfixtechnology.com/rail/the-benefits-of-lockbolts/</a>"</p>
</li>
<li><p>Can you even talk about "contact area" or is it like with the teeth of gears more like a contact line?</p>
</li>
<li><p>Isn't the contact area almost 100% along the whole thread because of the elongation of the screw when torqued down correctly?</p>
</li>
<li><p>Why doesn't this value apear when calculating screws, as I would imagine it to be important when considering surface pressure between the two threaded parts(e.g. what material to use)?</p>
</li>
</ol>
<p>I guess it's important to mention that I was thinking about machined threads not self tapping screws or something like that.</p>
| |threads| | <p>When you see the distribution of the load in a thread, it usually involves only the first few threads.</p>
<p><a href="https://i.stack.imgur.com/wfkqN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wfkqN.png" alt="enter image description here" /></a></p>
<p>That is easier to understand when you consider that the thread is under load. The load is gradually transferred to the plate through the bolts. The fact that the core of the bolt transfers that load, creates a small elongation/compression (enough to develop pressure onto the thread).</p>
<p>As the load is gradually transferred to the plate, the remaining threads are less and less loaded. Eventually there is no pressure on the threads, because - like you pointed out - machined threads are expected to fit without interference when they are unloaded.</p>
<p>That is why the contact area is defined in the first few threads, and also, when you are doing calculation on the thread (when you are using a hard bold with a softer material), you tend to calculate that most of the load will be on just on thread so that the the contact area is even less.</p>
<ul>
<li>Regarding the 5th question: depending on the calculation that you are using the contact area comes into play. However, most of the time, this is not required (i.e. in most cases performing that calculation will yield a result that is not a constrained). The reason is that bolts have been developing and improving for many years. As a result, the final form of bolts nowadays has a ratio of thread size to thread core, which (most of the time) makes sure that the bolt does not cut across the core cross-section. (Probably the worst scenario is to have a bolt break inside the thread). And although threads are not the top priority, the ratio of strength is very close to one.</li>
</ul>
<p>Keeping in mind that bolts are usually designed with a safety factor, and that a portion of the core carries the total of the load, while a single thread carries about 35 to 50% of the load at any time, you can understand that the failure of threads is not usual (unless there are material of uneven hardness involved).</p>
| 40789 | What is the contact area between threads and how does it change? |
2021-03-06T18:00:19.317 | <p>Could engines run on 1 cylinder while idling to save fuel? And then either each "idle" session alternate a different cylinder so wear and tear is even, or change the active cylinder periodically throughout an idle session (i.e every 1 minute idle, change the singular active cylinder). I realize that this is probably pointless with modern start / stop features on many newer engines, but why wasn't my idea possible before we came up with the tech to enable modern start / stop on engines?</p>
| |mechanical-engineering|engines| | <p>If you shut off the ignition for the cylinder you wish to disable, the action of the piston rising and falling in the cylinder (with no combustion) will tend to load the combustion chamber with motor oil, which then fouls the spark plug- so that when you want to turn that cylinder back on, it refuses to fire.</p>
<p>That puts the driver in a perilous situation, if (s)he were trying to pass another car on the highway. You are cruising along with cylinders idled, then you pull out to pass, step on the gas- and instead of the idled cylinders suddenly "waking up", the engine stumbles, bucks, misfires and smokes. Yikes!</p>
| 40790 | Could engines run on 1 cylinder while idling to save fuel? |
2021-03-06T19:39:42.287 | <p>I would like to start by saying I am not an engineer, so I apologize if this question seems simpleminded.</p>
<p>I am making a manual tarp reel from pvc pipe, and I am trying to decided whether to use Schedule 40 or Schedule 80. I need to know which one will sag, deflect?, the least amount under its own weight. The pipe will be <strong>10ft long</strong> and in a fixed-fixed position.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;">Type</th>
<th style="text-align: left;">Outer Diameter</th>
<th style="text-align: left;">Inner Diameter</th>
<th style="text-align: left;">Weight</th>
<th style="text-align: left;"><em>E</em></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;">Schedule 40</td>
<td style="text-align: left;">1.315”</td>
<td style="text-align: left;">1.049”</td>
<td style="text-align: left;">.32 lbs/ft</td>
<td style="text-align: left;">400,000 psi</td>
</tr>
<tr>
<td style="text-align: left;">Schedule 80</td>
<td style="text-align: left;">1.315”</td>
<td style="text-align: left;">.957”</td>
<td style="text-align: left;">.41 lbs/ft</td>
<td style="text-align: left;">400,000 psi</td>
</tr>
</tbody>
</table>
</div>
<p>When I used <span class="math-container">$\frac{wL^4}{384EI}$</span> , I got 4.683 for Sch 40 and 5.233 for Sch 80. Is this even the correct formula? The Sch 40 doesn’t visually seem to sag 4 inches.</p>
<p>Any help and explanations are much appreciated! Thank you!</p>
| |mechanical-engineering|structural-engineering|civil-engineering|beam|deflection| | <p>Here's my solution in GNU Octave. Everything is in inches. The formula you used (fix-fix support) gives
0.412" deflection for S-40 and 0.437" deflection for S80.</p>
<pre><code>clc, clear
format long
E = 400000 ;
D_1 = 1.315;
d_1 = 1.049;
d_2 = 0.957;
w_1 = 0.32/12;
w_2 = 0.41/12;
L = 10*12;
I_1 = pi/64 * (D_1^4 - d_1^4);
I_2 = pi/64 * (D_1^4 - d_2^4);
fprintf('I_1 = %f\n', I_1)
fprintf('I_2 = %f\n', I_2)
disp("Pin-Pin Support")
d_1 = w_1 * L/2 /(24*E*I_1) *(L^3 - 2*L*(L/2)^2 + (L/2)^2 )
d_2 = w_2 * L/2 /(24*E*I_2) *(L^3 - 2*L*(L/2)^2 + (L/2)^2 )
disp("Pin-Roller Support")
d_1 = 5*w_1*L^4 / (384*E*I_1)
d_2 = 5*w_2*L^4 / (384*E*I_2)
disp("Fix-Fix Support")
d_1 = w_1*(L/2)^2 / (24*E*I_1) * (L - L/2)^2
d_2 = w_2*(L/2)^2 / (24*E*I_2) * (L - L/2)^2
</code></pre>
<p>Result</p>
<pre><code>I_1 = 0.087343
I_2 = 0.105609
Pin-Pin Support
d_1 = 1.655542171849186
d_2 = 1.754296885580493
Pin-Roller Support
d_1 = 2.060840877820563
d_2 = 2.183772056739203
Fix-Fix Support
d_1 = 4.121681755641125e-01
d_2 = 4.367544113478407e-01
</code></pre>
| 40794 | Deflection of a fixed-fixed pvc pipe under self weight |
2021-03-06T22:54:23.353 | <ul>
<li>I have a tubing system where I need to precisely control <code>relative humidity</code>. All my sensors are giving me temperature <code>[°C]</code> and relative
humidity <code>[%]</code>.</li>
<li>Right now I have simple PID controller, and it's input is hooked up to the <code>relative humidity</code> values.</li>
</ul>
<p>But I had an idea: wouldn't be my controller more precise if I would feed into the input <code>absolute humidity</code> values and not <code>relative humidity</code>?</p>
<p>My though process:</p>
<ol>
<li><a href="https://en.wikipedia.org/wiki/Humidity#Relative_humidity" rel="nofollow noreferrer">Relative humidity</a> is dependent on temperature, partial pressure of water vapor and equilibrium vapor pressure of water. So it is dependent on 3 values.</li>
<li><a href="https://en.wikipedia.org/wiki/Humidity#Absolute_humidity" rel="nofollow noreferrer">Absolute humidity</a> on the other hand stated by Wikipedia is not dependent on the temperature around it.</li>
</ol>
<p>If I would compute <code>absolute humidity</code> from <code>relative humidity</code> and <code>temperature</code> and feed it into PID input. Would I achieve greater precision?</p>
<hr />
<p>I don't know if computing the <code>absolute humidity</code> from <code>relative humidity</code> and <code>temperature</code> would transfer the dependency of temperature or it would eliminate it.</p>
| |control-engineering|dynamics|airflow|pid-control|heating-systems| | <p>Creating an output from 2 inputs won't have more precision; it will have 2 sources of error.</p>
<p>If you are trying to control relative humidity, then using that as your control variable only makes sense.</p>
| 40802 | Is controlling humidity more precise when calculating with relative or absolute humidity? |
2021-03-07T00:53:38.857 | <p><a href="https://i.stack.imgur.com/D8DQM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D8DQM.jpg" alt="Question" /></a></p>
<p>I have a question that goes thus. <strong>At what angle <span class="math-container">$\theta$</span> must the <span class="math-container">$500N$</span> force be applied in order that the resultant <span class="math-container">$R = 1000N$</span>. For this condition what will.ne the angle <span class="math-container">$\beta$</span> between <span class="math-container">$R$</span> and horizontal</strong></p>
<p>I got</p>
<p><span class="math-container">$ \theta =97.9^\circ$</span> and <span class="math-container">$\beta=34.1^\circ$</span></p>
<p>But I'm not sure that's rightly done. I need help with verification of result</p>
<p><img src="https://i.stack.imgur.com/KRJbX.jpg" alt="my attempt" /><img src="https://i.stack.imgur.com/sza39.jpg" alt="enter image description here" />
<em>Here's my solution</em></p>
| |mechanical-engineering|statics|solid-mechanics| | <p>Since we are working on force vectors. You can should always verify your answers graphically with direction in mind. The diagram below is an example utilizing the original sketch and using parallelogram method. Note that in which the vectors (500N & 800N) shall be drawn to correct length and direction, thus the resultant R can be directly measured off</p>
<p><a href="https://i.stack.imgur.com/d5sj6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d5sj6.png" alt="enter image description here" /></a></p>
<p>In addition, the angle between the resultant force and the horizontal x-axis can be solved by triangular method (solving the right triangle).</p>
<p>[![enter image description here][2]][2]</p>
<p>Please let me know, if the solves above contain mistakes.
[2]: <a href="https://i.stack.imgur.com/lMfVD.png" rel="nofollow noreferrer">https://i.stack.imgur.com/lMfVD.png</a></p>
| 40805 | Resultant of coplanar forces |
2021-03-07T13:50:33.967 | <p>I'm trying to find out the angle/ length of the oblique line, or the solid angle of the conic part.</p>
<p>This is what I've tried so far:
According to the pythagorus theorem the length of the line is equal to:
<span class="math-container">$$\sqrt{(50-22)^2+14^2}$$</span>
but obviously this method gives a contradictory results, because as you can read, the line makes a 45° angle with the vertical axis, so its length should be equal to:
<span class="math-container">$$\sqrt{14^2+14^2}$$</span></p>
<p>I make somewhere a mistake which I obviously don't see it.</p>
<p>The next question is about the dimension 15 ( marked in red ).
I spend an awful amount of time to figure it out, but still cannot say anything.</p>
<p>At the end I want to know if this is something I'll confront a lot in real life mechanical engineering ? I mean a really confusing old drawings of parts sometimes with wrong/incomplete dimensions ?
<a href="https://i.stack.imgur.com/wriuF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wriuF.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|pumps|technical-drawing| | <p>50 and 22 refer to diameters of the conical bit. Usually in axisymmetric objects you denote diameters. It is the same as the 36 and 45 diameters (which are more easily understood).</p>
<p>So it is:</p>
<p><span class="math-container">$$\sqrt{\left(\frac{50-22}{2}\right)^2 +14^2}$$</span>
<span class="math-container">$$\sqrt{\left(14\right)^2 +14^2}$$</span></p>
<p>So there is no contradiction.</p>
<p>Regarding the 15, it actually refers to the distance of the highlighted small channel at the underside of the conical bit.</p>
<p><a href="https://i.stack.imgur.com/OQmo3.png" rel="noreferrer"><img src="https://i.stack.imgur.com/OQmo3.png" alt="enter image description here" /></a></p>
| 40815 | Can you help me to interpret technical drawings? |
2021-03-07T21:44:05.370 | <p><a href="https://i.stack.imgur.com/ziqgn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ziqgn.png" alt="problem on centroid " /></a></p>
<p>I was studying this example but I didn't understand how the author arrived at <strong>(A₁ - A₂)G₁G</strong></p>
<p>Please I need enlightenment</p>
| |statics|centre-of-gravity| | <p>The equation of finding centroid of an area is <span class="math-container">$$\bar{x} = \sum\left(\frac{A_i \times d_i}{\sum A_i}\right)$$</span>, "i" is the numerical label of part of an area that makes up the entire area, in here <span class="math-container">$A_1 = 6 \times 8 =48$</span> (ideally it should be <span class="math-container">$A$</span>, or <span class="math-container">$A_0$</span>, but let's stick to the givens). And "d" is the distance of the centroid of the ith partial area measured to the centroid of the centroid of the entire area, that is marked as "<span class="math-container">$G_1$</span>" here.</p>
<p>Now let's identify the partial areas of interest, and its distance to the centroid. As indicated, <span class="math-container">$A_2 = 3$</span>, and <span class="math-container">$d_2 = -x$</span> (negative because it is to the left of the centroid axis); and let's call <span class="math-container">$A_3 = A_1 - A_2 = 48 - 3 = 45$</span>, and <span class="math-container">$d_3 = 1$</span>. So,</p>
<p><span class="math-container">$\sum A_i = A_2 + A_3 = 3 + 45 = 48 $</span></p>
<p><span class="math-container">$\sum(A_i \times d_i) = 3 \times (-x) + 45 \times 1 = -3x + 45 $</span></p>
<p>To maintain equilibrium of the remaining area, <span class="math-container">$\sum(A_i \times d_i)$</span> must equal to <span class="math-container">$A_1 \times d_1$</span> (sum of the parts must equal to the whole). <span class="math-container">$D_1$</span> is the distance of the centroid of the entire/original area measured to its centroid, <span class="math-container">$G_1$</span>, thus <span class="math-container">$d_1 = 0$</span>, so</p>
<p><span class="math-container">$\sum(A_i \times d_i) = -3x + 45 = A_1 \times d_1 = 0 $</span></p>
<p>now solve for x, <span class="math-container">$x = \frac{45}{3} =$</span> +15 mm (the positive solution indicates the original assumption <span class="math-container">$d_2 = -x$</span> is correct, the area stays to the left of the original centroid, <span class="math-container">$G_1$</span>).</p>
<p>The problem is testing your understanding on "the first moment of area method", which is described as,</p>
<p>"The first moment of area of a shape, about a certain axis, equals the sum over all the infinitesimal parts of the shape of the area of that part times its distance from the axis [Σ ( a × d )]. First moment of area is commonly used to determine the centroid of an area."</p>
<p>As indicated above, it is most useful in finding the centroid of an irregular shape/area, by summing the area of the parts, and summing the area-moments (A x d) about a reference axis, then the distance of the centroid is obtained through dividing the sum of area-moments by the sum of the areas. Finally, you shall keep one important concept in mind, the area moment of an area/shape about its centroid axis is always zero.</p>
<p>Sorry for the lengthy write up. Hope you got the idea without trouble.</p>
| 40827 | Centroid and Center of Gravity |
2021-03-08T02:05:33.500 | <p><a href="https://i.stack.imgur.com/G3Lfi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G3Lfi.png" alt="enter image description here" /></a></p>
<p>So this is as far as I have gotten. The formulas written in green are the formulas that I have to find/ derive but I am hitting a wall. I know I am supposed to integrate but I am not sure where to go from here to get to the end formulas. If someone could please help explain/ walk me through this, I would really appreciate it.</p>
| |structural-engineering|structural-analysis|structures|moments| | <p><a href="https://i.stack.imgur.com/ro7Sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ro7Sn.png" alt="Translational" /></a></p>
<p><a href="https://i.stack.imgur.com/ulSzM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ulSzM.png" alt="Rotational" /></a></p>
<p><a href="https://i.stack.imgur.com/iwr6I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iwr6I.png" alt="Algebra" /></a></p>
<p>The equations found do match the equations above, they are just written differently but will yield the same outcome.</p>
| 40832 | How to derive fixed end moment of a beam fixed at both ends with a point force not in the center |
2021-03-08T15:51:08.817 | <p>What are some examples of real polymer products that can be be produced by injection moulding ? I thought plastic bottles may be one but I think that is done mainly using injection blow moulding</p>
| |materials|material-science| | <p>cell phone housings and trim, Lego blocks and all other similar types of plastic toys, camera housings, plates, bowls, and dishes of plastic as well as plastic knives, forks and spoons, automobile dashboards, PC and tablet housings, refrigerator drawers, plumbing pipe, fittings, fixtures and sinks, electrical enclosures and conduit fitments, booster seats for children, just to name a few.</p>
| 40843 | Injection moulding |
2021-03-08T17:08:01.270 | <p>Let us we are moving in a car. There is a wall in front of us, we need to decide can we go around it or not. It is known that width of the wall is <span class="math-container">$w$</span> and our speed is <span class="math-container">$v = const$</span>.</p>
<blockquote>
<p>What is a simple approach to set formally conditions upon which it is safe to go around?</p>
</blockquote>
<p>I understand that there are many possible details such as traction, weight of car etc. and will be glad even for the most simplistic analysis. If you can provide a source, that too is great. Thanks.</p>
| |control-engineering|applied-mechanics|kinematics|geometry| | <p>Each car depending on its handling has a maximum safe turning speed <span class="math-container">$v_{max}$</span>, and radius <span class="math-container">$r$</span>. Let us say the current speed <span class="math-container">$v<v_{max}$</span>, then your turning angular velocity is <span class="math-container">$$\omega= v/r.$$</span> We need to go an arc of <span class="math-container">$\pi/2$</span> so the time it takes to turn is <span class="math-container">$$t=\pi/2\omega= \pi r/2v$$</span> which will give the decision distance <span class="math-container">$x$</span> as <span class="math-container">$x=t\cdot v$</span>.</p>
<p><em>Edit</em></p>
<p>Above was for a wall wider than the cars cornering turn, if it is less, then the arc is smaller and we have <span class="math-container">$\theta= arccos(r-\text{car-width})$</span>.</p>
| 40847 | To stop or to go around? |
2021-03-08T17:21:38.757 | <p>I am trying to analyse a wing structure in Ansys static structural. In most examples I have been through the lifting pressure is uniformly applied under the bottom of wing or it is simplified to point loads. I have 2 questions regarding this,</p>
<ol>
<li>Is it okay to apply pressure loads on bottom wing as - (lift*load factor/wing area)</li>
<li>How can I apply pressure loads accurately taking into account coefficient of pressures at the wing
surface?</li>
</ol>
<p>EDIT 1: the wing is divided in 3 parts, the middle parts has a spar box which acts as a slot for the spar extensions of outer wings. knowing the stresses at this junction is critical</p>
| |structural-analysis|aerospace-engineering| | <p>You can not replace an uniform load with concentrate load without losing truth on stress and deflection. But, if the support reaction is the sole concern, then "yes", you may.</p>
| 40848 | Is it okay to replace pressure with point loads for structural analysis? |
2021-03-09T08:13:54.493 | <p>Having a serious issue with mechanical vibration in my home (I call it "mechanical" because its nature is not from sound, but rather from some sort of a device - for clear image whats happening, please refer <a href="https://engineering.stackexchange.com/questions/40462/reducing-vibration-coming-from-the-floor-while-at-sleep">this question</a> - tried almost all ways as suggested there, with no obvious effect). And I need to sleep ASAP...</p>
<p>In other words - I need to create a very isolate bed environment. Rebuilding floor is not an option. Need something what I can add up to the current state of the apartment. Probably most radical way (construction wise) could possible be to create a second floor (for bed area only), attaching it to the walls (by making holes in walls).</p>
<p>But the actual idea I am looking for - is hanging bed. The question here - would it work? Could someone do some calculations whatsoever, to get a better picture of what result that could bring? F.ex. I imagine, for some level of vibration, some vibration force could endup making bed to swing (what sounds as a less damaging effect to me, while at sleep). Also, maybe there is a chance for vibration to exhaust while traveling via that construction?</p>
<p>Could someone please give some advice?</p>
<p><a href="https://i.stack.imgur.com/c7Bvi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c7Bvi.png" alt="enter image description here" /></a></p>
<p>EDIT:</p>
<hr>
<p>Now thought of maybe a better design. See image below. Idea - to place next to walls (to the right and to the left of the bed) some solid material (like very solid wood, metal etc.) which then is jointed via the ceiling. And attaching ropes to hang a bed. In this design, I really think vibration would not travel to the bed.</p>
<p>Anyone could please comment this? You help is needed so much, people...</p>
<p><a href="https://i.stack.imgur.com/hwO8b.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hwO8b.jpg" alt="enter image description here" /></a></p>
| |vibration|sound-isolation| | <p>Springs and dampers are the classic method to reduce vibration, but don't expect the solution to be easy. Vibrations is a 4000 level ME course.</p>
<p>If the problem is just the floor, then hanging the bed is probably the easiest option. I would recommend 4 ropes or chains anchored into the ceiling joists above the bed.</p>
| 40866 | Creating an isolate bed environment for sleeping (to reduce mechanical vibration from floor) |
2021-03-09T17:35:09.693 | <p>I'm not sure this is the right stack exchange community to be asking this, but I have a little issue. There's a great microscope-camera from Carl Zeiss (AxioCam HRc) which I got for free and would like to use. It originally came with a PCI interface board and a driver CD; the PCI interface board is built into a computer which runs Windows NT and the driver CD is missing. To make matters worse, Zeiss doesn't support this model anymore.</p>
<p>So I wanted to know if it's possible to just remove the PCI interface board and attach it to my Windows 10 computer, and also to somehow take the driver from Windows NT and install it on Windows 10? I know that Windows has always been big on backwards-compatibility, but I'm way out of my depth when it comes to drivers, interfaces, etc. and so I don't even know if and how I can do this. Any help would be highly appreciated!</p>
| |computer-engineering| | <p>I googled and found this site. read their instructions on installation carefully and do not download junkware if any is packed with your zip file.</p>
<p><a href="https://www.driverscape.com/download/carl-zeiss-axiocam-hr" rel="nofollow noreferrer">AxioCam driver</a></p>
| 40872 | Microscope Camera Windows NT |
2021-03-09T18:37:09.157 | <p>I am looking to combine a pellet stove water heater with a solar water heater.
First of all I wold like to apologise for the ugly drawing, though I think it represents my idea well enough.
The whole thing will be a closed loop cycle or two cycles rather.
What we are looking to do is use whatever energy we can collect from the solar panels in the winter time and use it to aid the heating of the boiler/buffer vessel. The main source of energy will be the pellet stove.
My question is what can we do in order to use the solar energy first so we don`t end up stealing energy from the boiler.
P.S. The drawing is just the direction of my thoughts.
Please feel free to butcher it :)
<a href="https://i.stack.imgur.com/XbcMy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XbcMy.jpg" alt="enter image description here" /></a></p>
| |solar| | <p>Thanks a bunch for the input everybody.
I`ll probably mix and match as from what you guys suggested as I see fit after a bit extra research.
FYI the domestic tank I am using has two separate coils for heating the water in it and as an addition it has an electrical heating element so there is no threat of legionaires. The water in it will come from the public water supply for now.</p>
| 40877 | Combining solar water heating with pellet stove water heater |
2021-03-10T03:37:28.533 | <p>I am designing a wing for a Vertical Takeoff and Landing aircraft. I am trying to choose a wingtip winglet for my aircraft. One of the option is a aft-swept wingtip winglet. However, this structure would cause the horizontal loading of my wing to increase, according to my lecture notes. I am wondering whether this effect could strengthen my wing?</p>
<p>Here is a picture of my aircraft, the red crossed component is a rotation mechanism for my wings
<a href="https://i.stack.imgur.com/OWmY0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OWmY0.png" alt="enter image description here" /></a></p>
| |pressure|stresses|beam|aircraft-design| | <p>The image below is from the <a href="https://en.wikipedia.org/wiki/Wingtip_device" rel="nofollow noreferrer">wiki winglet page</a> shows the <a href="https://en.wikipedia.org/wiki/Wingtip_vortices" rel="nofollow noreferrer">wingtip vortices</a> with and without wiglets. It is producing lift to oppose wingtip vortices and as a result adds a horizonal force component (Think Newton equal and opposite).</p>
<p><a href="https://i.stack.imgur.com/2rzOE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2rzOE.png" alt="enter image description here" /></a></p>
<p>Obviously the results of this force are highly geometry dependent, as winglet could be up, down, or both. In the case of the photo, the upward winglet is pushing air outward and consequently has a resultant force oriented towards the aircraft. This creates a load and a moment that must be structurally supported by the wing and root of the wing; just as if it were a cantilever light pole opposing gravity. This moment is in addition to the much greater lift forces on the wing that are more or less operating in the same direction. Basically there is more load, so you need to build the wing stronger to accommodate it. Putting it into perspective though, a properly designed winglet will have a very small force (say less than 5%) relative to the lift of the wing.</p>
<p>The wing will not stiffen because of the winglet or this force. Excluding the geometric complexities of a tapering wing, it will somewhat follow the <a href="https://en.wikipedia.org/wiki/Deflection_(engineering)#End-loaded_cantilever_beams" rel="nofollow noreferrer">cantilever beam deflection equation</a> where the deflection is proportional to load.</p>
| 40888 | If an airfoil underwent horizontal loading, would it stiffen the airfoil? |
2021-03-10T09:47:27.623 | <p>I would like to clarify if the equations that I got from this figure are correct.</p>
<p><a href="https://i.stack.imgur.com/buWKS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/buWKS.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$$-T_{CD}sin(30) + T_{DE}cos(0) = F_x = 0$$</span></p>
<p><span class="math-container">$$T_{CD}cos(30) - \frac{W}{2} = F_y = 0$$</span></p>
<p>and</p>
<p><span class="math-container">$$-T_{ED}cos(0) + T_{EG}sin(10) = F_x = 0$$</span></p>
<p><span class="math-container">$$T_{EG}cos(10) - \frac{W}{2} = F_y = 0$$</span></p>
| |statics| | <p>if</p>
<ul>
<li>the points D and E are pivot joints, or equivalently if all the connecting elements are wires/cables,</li>
<li>GE does not change in length</li>
</ul>
<p>then you've written the right type of equations.</p>
<p><em><strong>Some minor notes:</strong></em></p>
<ul>
<li>The equilibrium is <span class="math-container">$\sum F_x $</span> not <span class="math-container">$F_x$</span> (and <span class="math-container">$\sum F_y $</span>) so for example:</li>
</ul>
<p><span class="math-container">$-T_{CD}sin(30) + T_{DE}cos(0) = F_x = 0$</span></p>
<p>is properly written:</p>
<p><span class="math-container">$$-T_{CD}sin(30) + T_{DE}cos(0) = \mathbf{\sum F_x} = 0$$</span></p>
<ul>
<li>Usually the symbol <span class="math-container">$T$</span> in this context is used to denote the tension on a wire in that case, regarding the tension of the cables in GE, I would have preferred to write:</li>
</ul>
<p><span class="math-container">$$-T_{ED}cos(0) + \mathbf{\color{red}2\cdot}T_{EG}sin(10) = \sum F_x = 0$$</span></p>
<p><span class="math-container">$$\mathbf{\color{red}2\cdot}T_{EG}cos(10) - \frac{W}{2} = \sum F_y = 0$$</span></p>
<p>The reason for <span class="math-container">$\mathbf{\color{red}2}$</span> is that you have two cables between GE and therefore the tension on each cable will be <span class="math-container">$T_{EG}$</span>.</p>
| 40890 | Equations of Static Equilibrium |
2021-03-11T09:24:11.113 | <p>I want to optimize the below function with GA(genetic algorithm):
<span class="math-container">$$
\min_{l_{x_1},l_{x_2},l_d}||S_QS(z)||_2
$$</span></p>
<p><span class="math-container">$S_Q$</span> and S(z) are defined as:
<span class="math-container">$$
S_Q=\Biggm[\matrix{6.6549 &-0.806&6.883\cr
-0.806 &12.146 &6.818\cr
6.883 &6.818 &29.071}\Biggm]
$$</span>
<span class="math-container">$$
S(z)=\Biggm[\matrix{z-1-l{x_1}&-0.05&-1 \cr
-l{x_2}&z-1&-1\cr
-l_d &0 &z-1}\Biggm]^{-1}
$$</span>
where z is the z-transform variable</p>
<p>I write below code and using optimization toolbox in MATLAB</p>
<pre><code>function z=objective_function(o)
lx1=o(1);
lx2=o(2);
ld=o(3);
Sq=[6.6549,-0.806,6.883;
-0.806,12.146,6.818;
6.883, 6.818,29.071]
z=tf('z')
s=[z-lx1-1,-0.05,-1;-lx2,z-1,-1;-ld,0,z-1]
z=norm(Sq*inv(s));
end
</code></pre>
<p>but do not work and give me that error (The "norm" command cannot compute the H2 norm of improper discrete-time models)</p>
| |matlab| | <p>Taking the norm of a system is something different that taking the norm of a matrix. Luckily, matlab has you covered:
<a href="https://nl.mathworks.com/help/control/ref/lti.norm.html" rel="nofollow noreferrer">https://nl.mathworks.com/help/control/ref/lti.norm.html</a></p>
<p>This page even shows an example in discrete time. However, what I think is missing in your example is that you should provide a sampling time. As the norm is calculated for every value of <span class="math-container">$z$</span> (I believe it is something like the integral of the trace of the absolute value of the system for every <span class="math-container">$e^{jw}$</span>), the sampling time should be included. I'd recommend first testing if <code>z=tf('z', -1);</code> works, but otherwise specify a generic sampling rate.</p>
| 40916 | Optimization with GA (genetic algorithm) |
2021-03-11T21:21:26.300 | <p>I have a solution to a problem on the mechanics of materials.</p>
<p>For the last step to find the deflection Vmax I am not sure how equation (15) was simplified to get the answer shown below:</p>
<p><a href="https://i.stack.imgur.com/PqGn1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PqGn1.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|solid-mechanics| | <p>You have to review how the "Macaulay Method", or "Step Function", works. As in the note, that "-a" in the middle term is less than the given value (x=0), so the term drops out. And the result only consists of the first term and the third term.</p>
<p>Form od the Step Function:</p>
<p><a href="https://i.stack.imgur.com/CS99X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CS99X.png" alt="enter image description here" /></a></p>
| 40925 | Rearranging to find Vmax |
2021-03-12T01:24:43.130 | <p>As the title of this post says, how do the contraints increase the stiffness. For example, I was doing an exercise where I had to calculate the deflection on a rubber block with and without deflection and they mentioned that "the constraint increases stiffness by a factor of 1.33 (for v=0.5)" where v is the Poisson's ratio. So, where does the 1.33 come from and what are the effects of the constraints.</p>
<p>I understand that imposing constraints will make it harder to stretch/bend, but I do not understand how they came up with 1.33</p>
<p>Thanks!</p>
| |materials|material-science|stiffness| | <p>Let me see if I understood it correctly:</p>
<p>You have a rubber block under a uniaxial load (compression or tension). That block may or may not be constrained on one pair of sides (the load is applied along the block's x-axis and the block can't deform in the y-axis but can in the z-axis, for example).</p>
<p>If this is the case, then the 1.33 is easy enough to explain.</p>
<p>Let's start by remembering that the Poisson ratio describes how big perpendicular deformations are when under axial loads.</p>
<p>Given that, let's think about the unconstrained case: you have a rubber block under some load. You calculate the axial deformation (along the x-axis, let's say) and it comes out as <span class="math-container">$\epsilon_x = \epsilon$</span>. We also know the rubber's Poisson ratio, so we immediately know that <span class="math-container">$\epsilon_y = \epsilon_z = -\nu\epsilon$</span>.</p>
<p>But, oh, sorry, actually we've been doing the constrained version this entire time, sorry! So we actually know that <span class="math-container">$\epsilon_y = 0$</span>. So, how can we fix our mistake? Well, we can apply forces along the y-axis which will resist that <span class="math-container">$\nu\epsilon$</span> deformation. The value of that force doesn't matter to us, only the knowledge that it generated a "counter-deformation" of <span class="math-container">$\nu\epsilon$</span>.</p>
<p>But wait, if we're applying an axial load along the y-axis which causes a deformation of <span class="math-container">$\nu\epsilon$</span>, then we also have a simultaneous deformation along the x-axis! And the value of this deformation will be equal to <span class="math-container">$-\nu^2\epsilon$</span>.</p>
<p>So, the force applied along the x-axis caused a deformation of <span class="math-container">$\epsilon$</span>, which caused a transversal deformation of <span class="math-container">$-\nu\epsilon$</span> along the y-axis, which caused a reaction from the walls which caused a counter-deformation of <span class="math-container">$\nu\epsilon$</span> (so that the final deformation along the y-axis is zero, as expected), which then causes a deformation of <span class="math-container">$-\nu^2\epsilon$</span> along the x-axis. Fittingly for a rubber block, there was a lot of bouncing back and forth here.</p>
<p>So, at the end of the day, our deformation along the x-axis is
<span class="math-container">$$\begin{align}
\epsilon_x &= \epsilon - \nu^2\epsilon \\
&= \epsilon(1 - \nu^2) \\
\end{align}$$</span></p>
<p>And since Young's modulus is defined as <span class="math-container">$E = \dfrac{\sigma}{\epsilon}$</span>, we end up getting that the "effective Young's modulus" in this case is</p>
<p><span class="math-container">$$\begin{align}
E_{x,\ \mathrm{nominal}} &= \dfrac{\sigma}{\epsilon} \\
E_{x,\ \mathrm{effective}} &= \dfrac{\sigma}{\epsilon_x} \\
&= \dfrac{\sigma}{\epsilon(1 - \nu^2)} \\
&= \dfrac{E_{x,\ \mathrm{nominal}}}{1 - \nu^2}
\end{align}$$</span></p>
<p>And for <span class="math-container">$\nu = 0.5$</span>, that gives us:</p>
<p><span class="math-container">$$\begin{align}
E_{x,\ \mathrm{effective}} &= \dfrac{E_{x,\ \mathrm{nominal}}}{1 - \left(\dfrac{1}{2}\right)^2} \\
&= \dfrac{E_{x,\ \mathrm{nominal}}}{\left(\dfrac{3}{4}\right)} \\
&= \dfrac{4E_{x,\ \mathrm{nominal}}}{3} \\
&\approx 1.33E_{x,\ \mathrm{nominal}}
\end{align}$$</span></p>
<p>And since stiffness is proportional to Young's modulus, that 1.33 factor simply carries over to the stiffness.</p>
| 40929 | How do the constraints imposed on a material increase the stiffness? |
2021-03-12T12:42:36.527 | <p>I am a high school student learning to analyze aircraft models in Autodesk CFD for the first time.
My current project is analyzing a supercritical airfoil on Autodesk CFD (that was designed in Autodesk Fusion360).</p>
<p>Can I get help with the following :
I researched several turbulence models(<a href="https://knowledge.autodesk.com/support/cfd/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/SimCFD-UsersGuide/files/GUID-E9E8ACA1-8D49-4A49-8A35-52DB1A2C3E5F-htm.html#:%7E:text=Turb.%20model%20%20%20Turbulence%20Model%20%20,model%20wi%20...%20%206%20more%20rows" rel="nofollow noreferrer">https://knowledge.autodesk.com/support/cfd/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/SimCFD-UsersGuide/files/GUID-E9E8ACA1-8D49-4A49-8A35-52DB1A2C3E5F-htm.html#:~:text=Turb.%20model%20%20%20Turbulence%20Model%20%20,model%20wi%20...%20%206%20more%20rows</a>) and I chose the SST k omega turbulence model, with a 250 m/s inlet velocity, 0 gauge pressure , unknown outlet and a compressible flow , I am getting a lift to drag ratio of approx. 4.3.I think the lift to drag ratio is too low. I am suspecting the analysis conditions are to blame. Are these the correct input parameters? Am I missing anything else?
Thank you.
Here is a screenshot of the analysis:<a href="https://i.stack.imgur.com/w1bsv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w1bsv.png" alt="I ran the airfoil analysis at angle of attack=5degrees
" /></a>
The following image is my mesh along with the convergence plot after working with the suggestions.
<a href="https://i.stack.imgur.com/Tt5Um.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tt5Um.jpg" alt="This is my mesh along with the convergence plot after working with the suggestions." /></a></p>
| |fluid-mechanics| | <p>Welcome to Engineering.SE</p>
<p>Regarding your question, in CFD analysis so many things can go wrong. You have a good start with some right inlet and initial conditions, but the problem you're simulating isn't an easy one, and requires taking care of so many details.</p>
<p><strong>1 - Let's start with the boundary conditions</strong>
<a href="https://i.stack.imgur.com/0Htvf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Htvf.png" alt="enter image description here" /></a></p>
<p>If you're simulating a wind-tunnel-like situation or a free stream over a body, you should not keep the non-slip conditions for the walls (except the airfoil of course), this will affect your results greatly and it's adding the complexity of non-developed internal flow, so we need to avoid that. I am not expert in AutoCAD CFD, but you need to make upper and lower walls as "slip" walls.</p>
<p><strong>2 - Convergence</strong>
<a href="https://i.stack.imgur.com/BG7wG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BG7wG.png" alt="enter image description here" /></a>
Judging from the curves of residuals, I don't think you've achieved numerical convergence, so I wouldn't count on any of the results you obtained. A typical converged solution curves would look similarly like <a href="https://www.simscale.com/knowledge-base/how-to-check-convergence-of-a-cfd-simulation/" rel="nofollow noreferrer">this</a>:
<a href="https://i.stack.imgur.com/4Fzzd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Fzzd.png" alt="enter image description here" /></a></p>
<p><strong>3 - Geometry of the problem is too small</strong></p>
<p>Box-wind-tunnel simulations can give quite good results, but the typical 2D airfoil set up would look like the following <a href="https://www.mdpi.com/1996-1073/12/3/488/htm" rel="nofollow noreferrer">picture</a>, with a freestream <span class="math-container">$U_{\infty}$</span> boundary condition or simple inlet-outlet setup. Also there are the mesh refinements around the airfoil (all of the simulation "drama" happens right there!).</p>
<p>And, you did not show us your final mesh, so that can also be one of your simulation problems.</p>
<p><a href="https://i.stack.imgur.com/qc8jv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qc8jv.png" alt="enter image description here" /></a></p>
<p><strong>4 - Turbulence</strong></p>
<p>The choice of <span class="math-container">$\text{SST} \ k-\omega$</span> is good, but what about the wall treatment? what is the max <span class="math-container">$y^+$</span> around the airfoil? Flow around airfoil involves adverse pressure gradients, so I would make my <span class="math-container">$y^+$</span> as low as it takes to capture the viscous layer.</p>
<p>You also should try different turbulence models, and in your case I would give Spalart–Allmaras model a try.</p>
<hr />
<p>I hope I haven't scared you off, please investigate the above points we discussed, and let me know if anything isn't clear to you.</p>
| 40935 | Confused with supercritical airfoil transonic condition analysis |
2021-03-12T12:50:02.553 | <p>This is an ABEC7 bearing with a 12 mm hole and a 19 mm external diameter:</p>
<p><a href="https://i.stack.imgur.com/D2nV8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D2nV8.jpg" alt="enter image description here" /></a></p>
<p>It has a tightly mounted adapter (the black part), which in turn has a 6mm hole and two external diameters: 12 mm (partially hidden in the bearing, partially visible) and 15.5 mm (fully visible). The adapter is required for a common 5 mm axle of two bearings. The wider parts are on the outside. The narrower parts are fixed into a common wheel.</p>
<p>Unfortunately, the adapter is made of aluminium and I would want one which is made of steel. Thus the questions:</p>
<ol>
<li>What is the proper technical name of the adapter?</li>
<li>Can it be bought "off the shelf"?</li>
<li>If not, can it be easily ordered from a typical metal workshop?</li>
</ol>
<p>EDIT: I was asked for a precise drawing and so I attach it. As seen, I need two versions, both with an increased internal diameter. I could adapt, however, an off-the-shelf part into any of these versions with a drill, apart from the 16/18 mm outer diameter, which is better but not necessary.</p>
<p><a href="https://i.stack.imgur.com/tcvHB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tcvHB.png" alt="enter image description here" /></a></p>
| |bearings| | <p>Regarding the name, I don't think that there is an official technical name, but then again I'm not a native speaker (nor can I speculate on the function of this shaft which in some cases might lend it a name).</p>
<p>However, if you go to any workshop with a lathe you should be able to order one of those. The main thing that you need to determine (apart from diameters, lengths and hole sizes) is the interference fit (since its tightly mounted).</p>
<p>You can always take the old part and ask them to replicate it (since I surmise that you might not be adequately technically inclined to calculate and provide the necessary tolerances for the interference fit).</p>
| 40936 | How this adapter inside the bearing is called? How to find/make a replacement? |
2021-03-12T14:49:07.567 | <p>I'm sure everybody has come across a case like the following:</p>
<p><a href="https://i.stack.imgur.com/S6D22m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S6D22m.png" alt="enter image description here" /></a></p>
<p>My experience is that this corrosion must act as an insulator(/or shortcircuit). I have managed -in some cases- to clean by scrubbing and using alcohol based cleaners. However that is not always the case.</p>
<p>Could someone suggest the best way to restore an adapter (if its possible).</p>
<p>If possible, it would be nice - along with or instead of commercial products that maybe available in only certain parts of the world- to have a solution/substance that is appropriate. If the requirement is just, scrub harder then that's also ok.</p>
| |battery|corrosion|energy-storage| | <p>The classic cleaner for removing corrosion products is a mild acid at elevated temperature. For contacts made of steel or chrome-plated steel, warm phosphoric acid works very well. This is an ingredient used in popular "rust remover" products you can buy in hardware stores.</p>
<p>If you use anything like this, it is imperative that you take great care to prevent the rust remover from getting into the guts of the device, where it will promptly wreak havoc. Also important is the need to thoroughly rinse the parts with water to remove all the rust remover when the job is done. Finally, water must be excluded from the cleaned parts, otherwise corrosion will continue. Spray a little bit of WD-40 onto the cleaned parts to prevent this, and wipe off any excess.</p>
| 40938 | How do you clean/restore corroded battery terminals? |
2021-03-12T16:35:13.933 | <p>Please to understand what I mean, see the picture below :</p>
<p><a href="https://i.stack.imgur.com/kkzGc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kkzGc.png" alt="Laser cutting speed chart" /></a></p>
<p>I don't understand why it is not a linear curve (or almost), because at first sight it should be no ?</p>
<p>If you need 10 sec to cut a length of 1mm thickness, why not 20 sec for a 2mm and 30sec for 3mm... I can't figure this out !</p>
<p>If you have an answer, please don't hesitate to share your experience with me !</p>
| |manufacturing-engineering|lasers|cutting| | <p>To see why this graph can only have a hyperbolic shape, it can help to take a look at the energy balance of the laser cutting process with absorbed laser power <span class="math-container">$P_A$</span>.</p>
<p><span class="math-container">$$P_A = \underbrace{vtw}_{V_M}\cdot \rho \cdot \underbrace{(c_p \Delta T_P + h_M + \xi h_E)}_{\text{process energy}}+\underbrace{P_L}_{\text{losses}}$$</span></p>
<p>The term labelled "process energy" doesn't bother us further here, it describes the energy we need to put in to bring the material from environment temperature to process temperature and is <strong>constant for a given material</strong>. Same with the power losses <span class="math-container">$P_L$</span>, maybe a little unintuitive, but experiments showed that they depend mainly on the thermal conductivity of the material, so we approximate them as <strong>constant as well</strong>. The material's density <span class="math-container">$\rho$</span> is of course also constant.</p>
<p>What remains? <span class="math-container">$V_M$</span>, the molten volume per time, which consists of cutting velocity <span class="math-container">$v$</span>, the sheet thickness <span class="math-container">$t$</span> and the width of our cut <span class="math-container">$w$</span>. The width <span class="math-container">$w$</span> depends on the laser beam diameter, so <strong>we also consider <span class="math-container">$w$</span> constant</strong>.</p>
<p>In the end, only <span class="math-container">$v$</span> and <span class="math-container">$t$</span> are left. We consider everything in the energy balance except those two constant, so let's claim:</p>
<p><span class="math-container">$$ v \cdot t = const.$$</span></p>
<p>A product being constant? That is just the equation of a hyperbola with the axes as asymptotes.</p>
| 40943 | Why laser cutting speed suddenly decrease when cutting sheets from 1 to 3mm thickness then stabilize? |
2021-03-12T18:00:51.697 | <p>I'm trying to create a block or anything else in soildworks that I can move and see the compression of the lower wire further from the arm</p>
<p>Something like this</p>
<p><a href="https://i.stack.imgur.com/bJsNa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bJsNa.png" alt="exemple" /></a></p>
<p>The example show just one joint but I want it for a long array of joints
But I have no idea how to go about it</p>
<p>Can you give me a direction on how to do it?</p>
| |solidworks|robotics|wire| | <p>In answer to the addition question in the comments -</p>
<p>In the case you are describing, the cable should be anchored parallel to the beam. The length of the wire required to limit motion is self-evident from being a straight line - although you should ensure that you have compensated for stretch in the wire, and flex of its mounting under the forces involved.</p>
<p>Then, it's the same method to create a flexible cable that you can use to approximate a possible bent state</p>
<p><a href="https://i.stack.imgur.com/w75AL.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w75AL.gif" alt="dynamic cable spline" /></a></p>
<p>Of course, you can get more complicated with it by adding additional construction lines to the style spline:</p>
<p>Which best represents the behaviour of your specific cable is something you will have to test experimentally, really.</p>
<p><a href="https://i.stack.imgur.com/mHyPy.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mHyPy.gif" alt="dynamic cable spline" /></a></p>
| 40944 | How to model robotic wires behavior in solidworks |
2021-03-14T09:47:02.077 | <h3>Introduction</h3>
<p>A water storage power plant pumps water up to a reservoir when there is more power available than needed and lets the water flow down when power is needed.</p>
<p>For pumping water up, electricity is needed.</p>
<h3>Concept / Idea</h3>
<p>I'd like to create a water storage power plant, where the electric pump is replaced by the sun (see also image below):</p>
<ul>
<li>The sun would evaporate water on the ground.</li>
<li>The water would rise up to a reservoir (as gas).</li>
<li>The water can cool down at the reservoir and becomes a fluid again.</li>
<li>The water is ready to flow down to produce energy.</li>
<li>The whole circle would take place within an vacuum (or near-vacuum) environment so water boils by the sun alone.</li>
</ul>
<p><a href="https://i.stack.imgur.com/e2FXQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e2FXQ.png" alt="system" /></a></p>
<h3>Question</h3>
<ul>
<li>Is something like this feasible? Where are the pitfalls?</li>
<li>What is the professional terminology for this concept? (it's difficult to search without one)</li>
<li>Are there any publications/researches regarding this concept? (hopefully some, that usual human beings can understand)</li>
</ul>
| |power|water-resources|power-engineering| | <p>I suggest that this is a needlessly complex way to store solar energy.</p>
<p>The typical objective of a water energy storage system is to pump the water up with excess energy during low demand. This system is just using sunlight to move the water (you've just described rain, really). Almost certainly it would be better to just produce electricity from the sun and then pump the water. That way you can get the raw power from the system during the afternoon without the rest of this system.</p>
<p>This reminds me of attempts to make Stirling engines for large energy production. The energy density just isn't high enough for the capital investment.</p>
| 40970 | Sun as water pump in underpressure water power circle |
2021-03-14T17:39:28.870 | <p><a href="https://i.stack.imgur.com/95Y5j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/95Y5j.jpg" alt="enter image description here" /></a></p>
<p>It is often said for I beams that the flange carries most of the <em>moment</em> subjected to the beam, while the web carries most of the <em>shear force</em>. The latter is obvious from pictures like above, showing the distribution of shear stress.</p>
<p>But why does the flange carry most of the moment? The normal stress on a beam is linearly proportional to distance from the neutral axis. So why is it assumed that the flange carries the bending moment?</p>
| |structural-engineering|stresses|beam| | <p>What is really proportional to the distance from the neutral axis (let's denote it <span class="math-container">$z$</span>) is the strain.</p>
<p><a href="https://i.stack.imgur.com/vSxBs.png" rel="noreferrer"><img src="https://i.stack.imgur.com/vSxBs.png" alt="enter image description here" /></a></p>
<p>For pure bending and a symmetric cross-section the equation for strain is given by
<span class="math-container">$$\epsilon(z) = \frac{z}{\rho}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\rho$</span> or <span class="math-container">$R$</span> is the radius of curvature of the beam</li>
</ul>
<p><a href="https://i.stack.imgur.com/n2ZlE.png" rel="noreferrer"><img src="https://i.stack.imgur.com/n2ZlE.png" alt="enter image description here" /></a></p>
<p>Because in the linear region the constitutive equation is <span class="math-container">$\sigma(z) = E\cdot \epsilon(z)$</span>, the development of stress at distance z is proportional to the strain.</p>
<p>Because stress in a small area <span class="math-container">$\Delta A$</span> is defined as <span class="math-container">$\sigma(z) =\frac{\Delta F}{\Delta A}$</span>, in that small area <span class="math-container">$\Delta A$</span> which is at distance z you are getting a Force :</p>
<p><span class="math-container">$$\Delta F(z) = E\cdot \Delta A\cdot \epsilon(z)$$</span></p>
<p>Those forces produce bending moments the further away they are from the neutral axis (about which the cross-section rotates), the contribution of the bending moment is greater.</p>
<p><span class="math-container">$$\Delta M(z) = \Delta F(z) \cdot z= E\cdot \Delta A\cdot \epsilon(z)\cdot z$$</span></p>
<p>So at a distance (z) any small area <span class="math-container">$\Delta A$</span> will produce a bending moment equal to</p>
<p><span class="math-container">$$\Delta M(z) = \frac{E\cdot}{\rho} \Delta A\cdot z^2$$</span></p>
<p>However, as you can see at the flanges you get more area at the same (more or less distance) (e.g. <span class="math-container">$z=\pm\frac{A}{2}$</span> for the image below).</p>
<p><a href="https://i.stack.imgur.com/9K6iW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9K6iW.png" alt="enter image description here" /></a></p>
<p>So, although the stresses and strain (in the linear region) follow a linear distribution wrt to the distance from the neutral axis, you can see that the actual bending moment eventually is proportional</p>
<ul>
<li>to the square of the distance z ,</li>
<li>to the area at distance z from the neutral axis.</li>
</ul>
<p>Therefore, because the flanges have a significant area, further away from the neutral axis they have greater contribution to the bending moment which resists bending.</p>
| 40977 | Why is it said that the flange of an I beam carries most of the moment? |
2021-03-15T03:32:19.067 | <p>In electrical i usually use cross-references to identify their symbols locations, I am still yet a lot to discover P&ID. So my question is, does P&ID have this such of thing?</p>
<p><a href="https://i.stack.imgur.com/FvbZm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FvbZm.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/uXbos.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uXbos.png" alt="enter image description here" /></a></p>
| |electrical-engineering|process-engineering|piping|instrumentation|pi-diagram| | <p><strong>Physical location</strong><br />
It depends. Ideally, every actutator(including hand operated valves) and sensor should have a unique identifier (Don't know the english term, the german terms are AKZ or KKS). These usually show up in the wiring diagram too. The identifying scheme changes from plant to plant, it is IME unusual to indicate physcial location - usually it's pureyl functional (example: everything to do with the emergency flare starts with 7123, if you are lucky these valves and sensors etc are also nearby the the flare)</p>
<p>Normally, you have a list for all sensors, actutators, occasionally for manually operated valves anda list for all sensors. The physical location should be apparent from the descriptor of the part, it can (but this is unusual IME) be marked in drawings. Usually the location in the P&ID is sufficient to find the item in the drawing (example: first valve in flow direction on the pressure side of a pump).</p>
<p>you can also indicate physical location in a P&ID by drawing boxes around components within the same room or building. A lot depends on how complicated the plant is.</p>
<p>When BIM is used, it is possible to integrate the P&ID with the physical drawings, so that each item and pipeline in the drawing is matched with the P&ID.</p>
<p><strong>Cross references</strong><br />
Typically a P&ID has far fewer pages than a wiring diagram (often only one), One uses arrows and text boxes that connect different P&ID pages or diagrams.</p>
<p><a href="https://i.stack.imgur.com/j5HdA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j5HdA.png" alt="enter image description here" /></a></p>
<p>This, from the DIN ISO EN 10628-1 is actually a poor example - there's no clear indication (drawing number etc.) where the arrow connects.</p>
| 40986 | Does P&ID have cross-reference same as electrical drawing? |
2021-03-15T20:30:53.443 | <blockquote>
<p>In <strong>Lagrangian or Material description</strong>, the physical properties are
described in terms of the material coordinates and time. It focuses on
what is occurring at a fixed material point (or particle) labeled by
its material coordinates as time progresses.</p>
</blockquote>
<p>My confusion: But the material surface (defined below) is written in terms of material coordinates only in case of the reference configuration and there is no time parameter. I understand that in reference configuration it is logical to not be dependent on time, but in the definition of material description, time is supposed to be included.</p>
<blockquote>
<p>In <strong>Eulerian or Spatial description</strong>, the physical properties are
described in terms of the spatial coordinates and time. It focuses at
a fixed point in space as time progresses.</p>
<p>A <strong>material surface</strong> is a mobile surface in the space constituted
always by the same particles.</p>
<p>In the <strong>reference configuration</strong>, the material surface is defined
in terms of the material coordinates as f(X,Y,Z) = 0, where the set
of particles (material points) belonging to the surface are the same
at all times.</p>
<p>In the <strong>spatial description</strong>, it is defined as f(x,y,z,t) = 0. The
set of spatial points belonging to the surface depend on time, and
the material surface moves in space.</p>
</blockquote>
| |fluid-mechanics|solid-mechanics| | <p>One way to think about this is to imagine that the body is made up of spheres and you're sitting on one of them. As the body deforms, all you can observe is the distance between you and neighboring spheres; but not how you're positioned in the surrounding space (with respect to a global coordinate system that's fixed in time).</p>
<p>The problem with a Lagrangian descriptions is that if the deformation is too large and you no longer can see your original neighbors, the description fails.</p>
| 40993 | Confusion in the Lagrangian description of Material Surface in Continuum Mechanics |
2021-03-16T00:46:03.917 | <p><em>Question</em> How do I calculate the ideal travel distance of the ball screw per revolution of the stepper / per step for the following hardware (I believe this is called the lead)?</p>
<p><em>Background</em> I have a pre-built <a href="https://www.amazon.co.uk/gp/product/B0899PC9NR/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1" rel="nofollow noreferrer">100mm Ball Screw Drive Linear Slide purchased from Amazon</a>.</p>
<p>The specifications of this assembly are as follows.</p>
<ul>
<li><p>Material: Aluminum</p>
</li>
<li><p>Boundary dimension: 40mm</p>
</li>
<li><p>Shaft diameter: 12mm / 0.47inch</p>
</li>
<li><p>Rated vertical load: 40KG</p>
</li>
<li><p>Rated horizontal load: 60KG</p>
</li>
<li><p>Accuracy: 0.01mm</p>
</li>
<li><p>Effective stroke: 100mm / 3.9inch</p>
</li>
<li><p>Ball screw length: 165mm</p>
</li>
<li><p>Sliding block: 60 x 80 x 30mm / 2.36 x 3.15 x 1.18inch</p>
</li>
<li><p>Pitch-row width: 68mm (max)</p>
</li>
<li><p>Pitch-row length: 60mm</p>
</li>
<li><p>Screw: M6</p>
</li>
<li><p>Slide table size: 190 x 40 x 20mm / 7.48 x 1.57 x 0.79inch</p>
</li>
<li><p>Weight: Approx. 1432g</p>
</li>
</ul>
<p>The stepper motor included with this module is 42BYGH48.</p>
<p><a href="https://i.stack.imgur.com/FrJ4c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FrJ4c.jpg" alt="enter image description here" /></a></p>
<p>Looks very similar to the ones defined <a href="https://mecheltron.com/en/product/42bygh-stepping-motor" rel="nofollow noreferrer">here</a> with the same wire out colours. Looking into this I found multiple sources refer the desired number this as the <em>lead</em> (<a href="https://www.machinedesign.com/mechanical-motion-systems/linear-motion/article/21834347/the-fundamentals-of-ball-screws" rel="nofollow noreferrer"><em>lead</em> is the linear distance traveled by the nut or screw after one full rotation</a>). Page 2 of <a href="https://tech.thk.com/en/products/pdf/en_b15_069.pdf" rel="nofollow noreferrer">this source</a> talks about this but I don't really follow where these numbers have come from.</p>
<blockquote>
<p>Can anyone suggest a good source to help explain to me what I need to calculate this for myself? What values am I missing from my setup to calculate this?</p>
</blockquote>
<p>Sorry, I suspect I'm thinking about this all wrong and look like an idiot! had a long day at work and just spent an hour digging around and made no progress! I was certain someone here could point me to exactly what I needed!</p>
| |gears|stepper-motor|linear-motion|linear-motors| | <p>In general the ideal travel per 1 revolution of the <strong>motor</strong>, will be depended on two things:</p>
<h2>screw dimensions</h2>
<p>the screw dimensions (in your case M6) and more specifically the pitch (see image below)</p>
<p><a href="https://i.stack.imgur.com/PE0X9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PE0X9.png" alt="enter image description here" /></a></p>
<p>Although its usually forgotten ISO 261 defines a coarse and a fine pitch. See <a href="https://en.wikipedia.org/wiki/ISO_metric_screw_thread" rel="nofollow noreferrer">wikipedia</a>. The pitch should be written on the designation of the bolt <strong>M6-1.00</strong> (coarse) or <strong>M6-0.75</strong> (fine). If its omitted then its assumed to be the coarse.</p>
<p>So, technically, for one rotation <strong>of the shaft(/screw)</strong> you get a travel displacement of 1 pitch. (this can get complicated when you have a multi-start screw but I won't get into that).</p>
<h2>Gear ratio</h2>
<p>If you notice in the previous section I highlighted <strong>one revolution of the shaft(/screw)</strong>. The reason is that (instead of directly coupling the shaft to the motor) you might intervene some sort of gearbox, or pulley or chain.</p>
<p>In that case, one rotation of the motor results in a multiple of rotations of the shafts. The multiple is defined as gear ratio, and it has multiple definitions. The most basic one is:</p>
<p><span class="math-container">$$i = \frac{n_{out}}{n_{in}}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$i$</span>: Gear ratio</li>
<li><span class="math-container">$ n_{out}$</span>: rpm at output (e.g. shaft)</li>
<li><span class="math-container">$ n_{in}$</span>: rpm at input (e.g. motor)</li>
</ul>
<h2>bottom line</h2>
<p>So 1 rotation of the motor, would result in 1 rotation of the shaft. Or equivalently</p>
<p><span class="math-container">$$\text{distance travelled} = i \cdot \text{pitch}$$</span></p>
| 40995 | How do I calculate the ideal travel distance of the ball screw per revolution of the stepper or per step for the following hardware? |
2021-03-16T05:03:29.707 | <p>(I feel it is necessary to mention that I am not an electrical or mechanical engineer; I just like to build fun things.)
I've already got a good explanation of the concepts involved in a variable field magnetic motor, but how is it achieved? It surely can't just be a DC motor with a retractable stator, right?</p>
<p>Here's an example of what I'm trying to make: <a href="https://www.youtube.com/watch?v=olnXy-J48SY" rel="nofollow noreferrer">https://www.youtube.com/watch?v=olnXy-J48SY</a></p>
<p>It really does look like a simple DC motor, but every one of these I've seen has that brown padding at the top of each armature loop. Is it something vital to the contraption?</p>
<p>I've scoured the internet, but there isn't a whole lot that goes further into this than the concepts involved. I realize this may force you to dead-reckon, but I'm desperate for information by now.</p>
| |mechanical-engineering|motors|electromagnetism|magnets| | <p>Although the previous answers address the OP's question, I am adding an answer since they do not explain the mechanism of the variable field magnetic motor.</p>
<p>The variable field magnetic motor is usually built as a brushless DC motor (BLDC). The mechanism works by moving the rotor out of line with the stator so the effective magnet strength is lower. This increases <a href="https://en.wikipedia.org/wiki/Brushless_DC_electric_motor" rel="nofollow noreferrer">velocity constant</a> <span class="math-container">$K_v$</span> because the motor has to run faster to produce enough back-EMF to counter-match the supply voltage. It is the permanent magnet equivalent of varying field current in a shunt-wound motor. Indeed, motor speed could be raised by simply increasing voltage, but this also increases <em>material</em> (iron) losses due to hysteresis and eddy currents. Field weakening reduces the magnetic field strength as it raises rpm, reducing the loss at light loading. However it also reduces torque and output power, so it is more suited to applications which require high efficiency at both low rpm & high torque and high rpm & low torque.</p>
<p><a href="https://www.youtube.com/watch?v=6_41btVawMc" rel="nofollow noreferrer">Here</a> is a video showing the <em>increasing speed</em> of the rotor with decreasing cross sectional interface between the curved surfaces of the rotor and stator.</p>
| 40996 | What Goes Into Making a Variable Field Magnetic Motor? |
2021-03-17T13:38:12.197 | <p>I own a CNC machine (Shapeoko 3 XXL) which I use to make projects that require good 90° angles, and I have been through several steps of improvement/calibration of the machine that are described in the documentation. However the rails of the machine's frame are nearly impossible to "square" below a deviation of about 1mm/meter or 2mm/meter. Besides that, the rails may be slightly different and some unknown factors can play a role an reduce precision of the machine. I suppose however that repeatability is far better than actual precision.</p>
<p>So I have though of something: why not measure the difference between the CAD and the result on several points on the xy plane of the machine and apply a correction to recalculate the CAD model, or even better, just the Gcode?</p>
<p>It makes so much sense that I thought I would easily find a computer program that already has been designed to apply such corrections on the Gcode, but I didn't fin any result on any search engine. Have you ever heard of such a project?</p>
| |cad|cnc| | <p>After searching for solutions online I decided to write my own code to do this task.</p>
<p>The python script is available here: <a href="https://github.com/studioadrienlucca/studioadrienlucca/blob/gcode_corrector/Gcode3stepsCorrector.py" rel="nofollow noreferrer">https://github.com/studioadrienlucca/studioadrienlucca/blob/gcode_corrector/Gcode3stepsCorrector.py</a></p>
<p>This script assumes that you have a Gcode file that uses G0, G1, G2, G3 commands as well as plane changes G17, G18, G19</p>
<p>It assumes also that you have made a 60x60 cm square on your machine and measured the sides and diagonals of the real output.</p>
<p>The script will revrite your Gcode using only G1 commands, with parameters associated to arc linearization.</p>
<p>Unfortunately I haven't had the time yet to switch my system ti Python 3 so it is 2.7. The code is a bit messy, but it works.</p>
<p>Anyone owning a Shapeoko 3 and Fusion 360 sould be able to benefit directly from this tool that will automatically correct Gcodes. The Z axis is not corrected because I assume the errors are very small in Z. I also assumes that the bed of the machine has been surfaced parallel to the spindle.</p>
| 41016 | Software correction of CAD or Gcode based on machine measurements |
2021-03-17T17:49:44.297 | <p><a href="https://i.stack.imgur.com/JMKER.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JMKER.png" alt="Bevel gear combo to transfer hand torque to a pinion to drive a rack" /></a></p>
<p>I'm trying to design a little device which is hand-driven and which involves turning a bevel gear pair. I want a small gear pair but I don't want it to be weak. The forces involved will be small (it's attached to a pinion along which will slide a rack < 1 kg), but of course I still need to check the safety factors involved. So here's my question:</p>
<p>Since it's hand-driven, what torque value can I enter for the strength-calculations? I search for "typical torque values human hand" (and variations on this), but I keep receiving maximum human turning strengths which are definitely not going to turn up with this device. The dial has a ~3 cm diameter.</p>
| |mechanical-engineering|design|torque|strength| | <p>You can easily measure this by dangling a weight from a string and winding that around a pulley of known diameter.</p>
<p>This arrangement will produce a fixed torque on the pulley, which you can adjust by changing the weight.</p>
<p>Now all you have to do is attach a knob to the the pulley and you can feel what a given torque is like. Adjust the weight until the torque required to turn the knob feels like a reasonably maximum to you. From there you can do a bit of math and figure out what that torque is.</p>
| 41017 | Good estimates for torque generated while turning a dial |
2021-03-18T10:36:44.520 | <p>I've just begun to learn pneumatics and I'm really confused where the pneumatic cylinder falls under in the categorization of what kind of mechanical motion it is suppose to produce. I've seen one source say that it is rotary and for analogy the motor is its linear counterpart. But I haven't seen any other sources to back this up definitely. I understand that it might be possible for there to be both types.</p>
<p>What is the classification of the one that's most commonly discussed when like someone is new to pneumatics?</p>
| |motors|mechanisms|pneumatic|linear-motion| | <p><a href="https://i.stack.imgur.com/hlxcq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hlxcq.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. A <a href="https://vane-actuator.kinetrolusa.com/2019/05/kinetrol-vane-actuators.html" rel="nofollow noreferrer">Kinetrol rotary actuator</a>.</em></p>
<p>This type of rotary actuator is also known as a vane type. This one has a 90° action and is popular for controlling ball or butterfly valves.</p>
<p><a href="https://i.stack.imgur.com/XH8td.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XH8td.png" alt="enter image description here" /></a></p>
<p><em>Figure 2. A rack and pinion style rotary actuator. Image source: <a href="https://www.hydraulicspneumatics.com/technologies/cylinders-actuators/article/21885216/pneumatic-rotary-actuators-back-to-school" rel="nofollow noreferrer">Hydraulics Pneumatics</a>.</em></p>
<p>These convert linear motion into rotary. The angular rotation is limited only by the length of the rack and actuating cylinders.</p>
<p><a href="https://i.stack.imgur.com/CcStl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CcStl.png" alt="enter image description here" /></a></p>
<p><em>Figure 3. Standard air cylinders are linear actuators. There is no rotational motion.</em></p>
<p>These are most common and most simple. Variations such as clamping cylinders use some additional mechanics to impart a rotation to the shaft as it retracts.</p>
| 41030 | Is a pneumatic cylinder in general a linear or rotary actuator? |
2021-03-18T12:56:59.757 | <p><a href="https://www.alibaba.com/product-detail/dc-brushless-centrifugal-circulation-small-water_1764477502.html?spm=a2700.galleryofferlist.normal_offer.d_title.10875798tE00rA" rel="nofollow noreferrer">Here</a> is the small <span class="math-container">$12 \; V$</span> pump that I am looking at. The manufacturer called the product as <em>DC brushless <em>centrifugal</em> circulation small water pump</em> but under its description it mentions that it is a <em>single stage pump</em>.</p>
<p><a href="https://i.stack.imgur.com/Ows1r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ows1r.png" alt="enter image description here" /></a></p>
<blockquote>
<p>I wonder what the difference is? Is it a misleading description or title or do those terms are used interchangeably?</p>
</blockquote>
| |motors|pumps| | <p>Single stage and centrifugal are describing two different properties.</p>
<p>A single stage pump is a pump that only has one acting impeller stage doing work on the fluid. If you put two pumps in series, you could say that you have a two-stage pumping setup. Some pumps are built with multiple stages - this is usually for high pressure applications.</p>
<p>A centrifugal pump is the style of impeller / casing. Centrifugal pumps use an impeller similar to what is shown below, and they rotate in a way that utilizes <a href="https://en.wikipedia.org/wiki/Centrifugal_force" rel="nofollow noreferrer">centrifugal force</a> to do work on the fluid, in this case creating pressure and driving the flow.</p>
<p><a href="https://i.stack.imgur.com/31obq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/31obq.jpg" alt="enter image description here" /></a></p>
<p>You could have a single stage centrifugal pump, a multiple stage centrifugal pump, a single stage NON-centrifugal pump, or a multistage NON-centrifugal pump setup. Below is a generic example of a multiple stage centrifugal pump - note that there are multiple impellers, and they don't have to be completely separate pumps - each of these increases the pressure supplied to the next impeller, so as to increase the pressure of the next impeller's discharge.</p>
<p><a href="https://i.stack.imgur.com/Yo6et.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yo6et.jpg" alt="enter image description here" /></a></p>
<p>For the vast majority of applications (especially water), single stage centrifugal pumps sized correctly cover most applications.</p>
| 41031 | What is the difference between a single-stage pump and a centrifugal pump? |
2021-03-18T19:24:07.097 | <p>My thermodynamics knowledge is extremely rusty, so I would appreciate some guidance on how to solve this problem.</p>
<p>As illustrated <a href="https://i.stack.imgur.com/0Yfz3.png" rel="nofollow noreferrer">here</a>, I have a process pipe (steel) with a gas flowing through it (blue), with air flowing over the pipe (orange) perpendicular to the process gas flow.</p>
<p>I need to calculate T2 of the process gas, or more specifically (but perhaps more complicated), I need to know if the internal surface of the pipe reaches a certain temperature at any point along its surface. However, T2 of the process gas at the end of the control volume would suffice.</p>
<p>From my research, I know that I need to calculate the Reynold's number of the gas flow inside the pipe, then use the correct Nusselt number correlation for the flow regime and pipe geometry - all well and good. This then gives me my heat flux from the gas to the inner surface of the pipe.</p>
<p>From the other side of the problem, because the ambient air has a velocity, I also need to calculate the Reynolds number of the air moving over the pipe (using the free stream velocity and the pipe outside diameter), I think I would then use something called the <a href="https://en.wikipedia.org/wiki/Churchill%E2%80%93Bernstein_equation" rel="nofollow noreferrer">Churchill-Bernstein equation</a> to get a Nusselt number for the air flow over the pipe, which similarly gives me the heat flux from the air to the pipe.</p>
<p>However, the temperature of the process gas is >> ambient air temperature, so I would expect a negative heat flux between the gas and the pipe wall.</p>
<p>Now I'm stuck. I'm positive there is an element of conduction through the pipe wall here, but I just don't know how to apply these separate elements of the problem to get the answer.</p>
<p>I am assuming here that the process is steady-state, so I'm not interested in what happens during the process start up. Does that mean I can ignore the effects of conduction through the pipe?</p>
| |mechanical-engineering|heat-transfer|thermal-conduction|convection| | <p>Stand at any center point along the axis of the pipe where the temperature is <span class="math-container">$T$</span>. Assume that you can obtain <span class="math-container">$h_i$</span>, the internal convection heat transfer coefficient, <span class="math-container">$k$</span> the thermal conductivity of the pipe with thickness <span class="math-container">$w$</span>, and <span class="math-container">$h_o$</span> the external convection heat transfer coefficient. Assume an external temperature of <span class="math-container">$T_\infty$</span>. The heat transfer rate <span class="math-container">$\dot{q}$</span> is defined by these equations with <span class="math-container">$A_i$</span> as the internal wall area and <span class="math-container">$A_o$</span> as the external wall area:</p>
<p><span class="math-container">$$ \dot{q} = U A_i (T - T_\infty) $$</span></p>
<p><span class="math-container">$$ U^{-1} = h_i^{-1} + wk^{-1} + A_i(h_o A_o)^{-1} $$</span></p>
<p>The term <span class="math-container">$U$</span> is the <a href="https://en.wikipedia.org/wiki/Heat_transfer_coefficient" rel="nofollow noreferrer">overall heat transfer coefficient</a>. In this example, <span class="math-container">$U$</span> is based on the internal area of the pipe. You can make use of the information at <a href="https://www.thermopedia.com/content/1007/" rel="nofollow noreferrer">this link</a> to change between inner or outer area (as well as to include fouling). Note that for cylinders, <span class="math-container">$A_i/A_o = R_i/R_o$</span>, the ratio of inner to outer tube radius.</p>
<p>This gives you the heat transfer rate from the center of the pipe to the ambient air. The value of <span class="math-container">$T$</span> is a function of distance <span class="math-container">$z$</span> along the pipe. The energy balance for the gas flow through an infinitesimally thick circular plane at <span class="math-container">$z$</span> is</p>
<p><span class="math-container">$$ \dot{m}\tilde{C}_p\ dT = -d\dot{q} = U\ 2\pi R_i\ dz\ (T_\infty - T)$$</span></p>
<p>where <span class="math-container">$\dot{m}$</span> is the mass flow, <span class="math-container">$\tilde{C}_p$</span> is the specific heat capacity, and <span class="math-container">$dT$</span> is the temperature change of the gas as it goes through the plane at <span class="math-container">$z$</span>. Note that if <span class="math-container">$T > T_\infty$</span> the gas temperature decreases as it passes through the plane as expected.</p>
<p>A useful transformation at this point is to define <span class="math-container">$\theta = (T - T_o) / (T_\infty - T_o)$</span> where <span class="math-container">$T_o$</span> is the temperature of the gas entering the tube and to define <span class="math-container">$Z = z/L$</span> where <span class="math-container">$L$</span> is the pipe length. With this, the governing equation becomes</p>
<p><span class="math-container">$$ d\theta = \beta \left(1 - \theta\right) dZ $$</span></p>
<p><span class="math-container">$$ \beta = \frac{2\pi R_i L U}{\dot{m}\tilde{C}_p} $$</span></p>
<p>Note that when <span class="math-container">$T_o > T_\infty$</span>, as we move from the inlet to the outlet, <span class="math-container">$z$</span> increases and <span class="math-container">$\theta$</span> goes from 0 to 1 (increasing).</p>
<p>In practice, <span class="math-container">$U$</span> may depend on <span class="math-container">$z$</span> and <span class="math-container">$\tilde{C}_p$</span> may depend on <span class="math-container">$T$</span>. A first approximation will neglect these variations. The dimensionless solution is</p>
<p><span class="math-container">$$ \theta = 1 - \exp\left( -\beta Z \right) $$</span></p>
<p>A plot of <span class="math-container">$\theta$</span> versus <span class="math-container">$Z$</span> for a value <span class="math-container">$\beta = 1$</span> is shown below. By example, at a location 50% down the pipe, we obtain <span class="math-container">$\theta \approx 0.4$</span>, meaning <span class="math-container">$T \approx 0.4(T_\infty - T_o) + T_o$</span>. When the gas enters at 100<span class="math-container">$^o$</span>C and the ambient is 20<span class="math-container">$^o$</span>C, the gas would be 52<span class="math-container">$^o$</span>C at the 50% position down the pipe. The position <span class="math-container">$Z = 1$</span> is the end of the pipe, where <span class="math-container">$\theta \approx 0.63$</span>.</p>
<p>Correlations for <span class="math-container">$h$</span> will give first approximations, the value of <span class="math-container">$h_o$</span> is dependent on radial position, and the dependence of <span class="math-container">$k$</span> and <span class="math-container">$\tilde{C}_p$</span> on temperature may not be negligible. The simplest approach for quick and dirty confidence is to define <span class="math-container">$\theta_{limit}$</span> in your system, estimate the range of <span class="math-container">$\beta$</span> you can expect, and use an interactive graphing tool such as GeoGebra to slide through that <span class="math-container">$\beta$</span> value to see whether the <span class="math-container">$\theta_{limit}$</span> is ever exceeded along the pipe for the estimated ranges. The maximum <span class="math-container">$\theta$</span> is at <span class="math-container">$Z = 1$</span> (the end of the tube). So your first problem is simply to determine whether you exceed this criterion:</p>
<p><span class="math-container">$$ \beta > -\ln\left(1 - \theta_{limit}\right) $$</span></p>
<p>Essentially, you want to stay below a limiting value of the ratio of heat transfer from the pipe to enthalpy flow of gas through the pipe, either by increasing the mass flow of gas through the pipe or by reducing the air flow over the tubes.</p>
<p>After that, you may also want to analyze whether the sides of the tubes that are upstream in the external air flow could be subject to cool below your limit because the local <span class="math-container">$h_o$</span> is potentially higher than you set in your first estimate using an average <span class="math-container">$h_o$</span>. You can walk backwards along the tube from its exit point because that is where the "cool spots" will most likely first appear.</p>
<p><a href="https://i.stack.imgur.com/pmXy9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pmXy9.png" alt="temperature plot" /></a></p>
| 41044 | How to calculate heat loss across a process pipe with perpendicular air flow |
2021-03-19T01:30:32.897 | <p>I major in chemical engineering, and I'm solving a example of unit operations.
The textbook is Warren L. McCabe, Unit Operations of Chemical Engineering 7th ed.
Here is example 24.1.
<a href="https://i.stack.imgur.com/X8h4O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X8h4O.png" alt="enter image description here" /></a></p>
<p>And this is its solution.</p>
<p><a href="https://i.stack.imgur.com/nXmnM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nXmnM.png" alt="enter image description here" /></a></p>
<p>From this, it says that the density of air is 29/369 × 492/620.
But it omitted the untis, so I can't understand the equation.
I assume that this equation is probably ρ = PM/RT, since molecular weight of air is 28.97 g/mol when I searched for it.
But the numbers 369, 492, 620 are still mystery. What is those?</p>
| |chemical-engineering|unit| | <p>If you look at Text equation 1.29, the 492/620 is Temperature in Rankine.</p>
<p><span class="math-container">$$T°F = T°R - 459.67$$</span></p>
<p>Ice point is 491.67°R. 160°F = 619.67°R.</p>
<p>If you do the math, it does not equal 0.0641, because the 369 is wrong. The math to get 0.0641 means 369 should be 359.</p>
<p>1 lb-mole of an ideal gas at 0°C and 1 atm occupies 359 <span class="math-container">$ft^3$</span> absolute.</p>
<p>The molecular weight of air is 29 lb/lb-mole.</p>
<p><span class="math-container">$$\frac {29 lb/lb-mole}{359 ft^3/lb-mole} \times \frac {492°R}{160°F + 460°R} = 0.0461 lb/ft^3$$</span></p>
| 41046 | How can I calculate the density of air? |
2021-03-19T04:16:15.707 | <p>Does any know what kind of water pump the French Firefighting Robot Colossus used to Help Save Burning Notre Dame Cathedral?</p>
<p>What kind of water pumps do fire fighting vehicle use or fire fighting robots use?</p>
| |pumps|robotics|fire| | <p>This is not a definitve answer, but since I went to the trouble to look it up I thought i might as well post what I found: Pretty sure there's no onboard pump</p>
<p><a href="https://i.stack.imgur.com/KTyUN.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KTyUN.png" alt="enter image description here" /></a></p>
<p>As you can see this is a normal firehose, not a suction hose with reinforcing spirals. this is relevant because it means water has to be supplied to Colossus at pressure, from a fire engine with a pump.</p>
<p>Next, let's look at the <a href="https://static1.squarespace.com/static/5df36a06aac1414fe3532950/t/6006ea0ee40b7c3f2c279ea1/1611065879941/Colossus+-+EN+.pdf" rel="noreferrer">published specs</a> - no mention of a pump or water tank:</p>
<p><a href="https://i.stack.imgur.com/3yGva.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3yGva.png" alt="enter image description here" /></a></p>
<p>6 batteries, 48V and 27 Ah translates to roughly 7.7 kWh. Smallest fire pumps I found where around 20 kW - I don't think it makes sense to power a fire pump from a battery.</p>
<p>Water has to be supplied anyway, so at most Colossus would carry a small booster pump. I sum I think the best approach for Colossus and similar vehicles is simply to have a water supply at sufficient pressure, and not carry an onboard tank or pump.</p>
| 41051 | Does any know what kind of pump "Colossus" used for saving the Notre Dame Cathedral? |
2021-03-19T06:16:03.013 | <p>While learning about I-beams I became curious to know if there is any relation between the flange length and web thickness of I-beams or if its arbitrary.</p>
| |mechanical-engineering|civil-engineering|solid-mechanics| | <p>The answer is "Yes", and the direct source to find such relationship is from the technical publishing, that offer steel design tables/charts showing the most optimum beam/column sections for certain load with respect to the length of the beam column.</p>
<p>The image below is an example of a chart comparing moment capacity (indicated on the left axis) of beam sections with respect to the unbraced length indicated on the bottom. While there are many sections will satisfy either constrain, there is only one that will satisfies both constrains with the least "weight", which often is the focus of optimization.</p>
<p>You can also make your own calculation to find the most optimum cross section for the constrains you have in mind. For instance, You want a section can resist a set of loads, moment and shear, yet having the least weight. In realizing the stress formulas, <span class="math-container">$f_b = My/I$</span>, and <span class="math-container">$f_v = V/A_w$</span>, and the facts that <span class="math-container">$I$</span> (moment of inertia), and <span class="math-container">$A_w$</span> (effective area of the web for shear) must be kept constant to yield the constant stresses due to the given set of loads, now is the simple mater by writing the equations for <span class="math-container">$I$</span>, <span class="math-container">$A$</span> and <span class="math-container">$A_w$</span> (<span class="math-container">$A$</span> is area of the entire cross section), and plug in varies combination of <span class="math-container">$b_f$</span>, <span class="math-container">$t_f$</span>, <span class="math-container">$d$</span>, <span class="math-container">$t_w$</span> that will yield the constant <span class="math-container">$I$</span> and <span class="math-container">$A_w$</span>, yet the <span class="math-container">$W = r \cdot A$</span> is the minimum.</p>
<p>Hope this helps.</p>
<p><a href="https://i.stack.imgur.com/WnTa4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WnTa4.png" alt="enter image description here" /></a></p>
| 41053 | What should be the relation between flange length and web thickness for an optimum I - beam? |
2021-03-19T11:18:49.347 | <p><a href="https://i.stack.imgur.com/sQN3G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sQN3G.jpg" alt="7" /></a></p>
<p>In the picture above we have the expression of stretch with respect to strain tensor, but in case of infinitesimal strain theory, we take stretch equal to 1+T.E.T</p>
<p>I don't understand why by taking Taylor series expansion, the answer would be 1+ x, with x= T.E.T, with my calculation, the first derivative is 1/radical(1+2x) not x</p>
| |fluid-mechanics|mathematics|solid-mechanics| | <p>Taylor is straightforward:
<span class="math-container">$$
\sqrt{1+2x}
=\left.\sqrt{1+2x}\right|_0
+\left.{d \over dx}\sqrt{1+2x}\right|_0x
+O(x^2) \\
=1
+\left.{d \over dx}{1 \over \sqrt{1+2x}}\right|_0x
+O(x^2) \\
=1
+x
+O(x^2) \\
$$</span></p>
<p>Note that Taylor holds for matrices variables under some conditions.</p>
<p>ps.If you can handle it, you also have the Generalized <a href="https://en.wikipedia.org/wiki/Binomial_theorem" rel="nofollow noreferrer">Binomial Expansion</a> for complex numbers (not matrices)
<span class="math-container">$$
(x+y)^r=\sum_{k=0}^\infty\left(\array{r\\k}\right)x^{r-k} y^k \ , \ \left(\array{r\\k}\right)={(r)\cdots (r-k+1)\over k!}\\
\sqrt{1+2x}=(1+2x)^{1/2}
=1+\left(\array{1/2\\1}\right)(2x)+O(x^2)\\
=1+{1/2 \over 1!}2x+O(x^2)=1+x+O(x^2)
$$</span></p>
| 41056 | Stretch in Infinitesimal strain theory |
2021-03-19T13:51:53.980 | <p>I was looking for some element's half-life, on the following website:</p>
<p><a href="http://nucleardata.nuclear.lu.se/toi/listnuc.asp?sql=&A1=214&A2=214&Z=88" rel="nofollow noreferrer">http://nucleardata.nuclear.lu.se/toi/listnuc.asp?sql=&A1=214&A2=214&Z=88</a></p>
<p>For 214,88-Ra, the half-life is shown to be: 2.46 s 3</p>
<p>I am unsure what the "3" stands for, is it "<span class="math-container">$\cdot 10^{-3}$</span>"?</p>
<p>Thank you for your help and time guys.</p>
| |energy|nuclear-engineering|nuclear-technology|unit| | <p>The half life of Radium 214 is 2.46[s]. You can check half times at the following <a href="https://en.wikipedia.org/wiki/List_of_radioactive_nuclides_by_half-life" rel="nofollow noreferrer">wikipedia article</a>.</p>
<p>I am not sure what the 3 stands for, but I suspect that this is some sort of annotation/footnote that was not properly inserted in the webpage.</p>
| 41058 | Half-life Unit Notations |
2021-03-19T16:18:29.643 | <p>Suppose we have <span class="math-container">$n$</span> carnot engines connected together labelled as <span class="math-container">$C_1,C_2...C_n$</span>, now suppose we feed an energy of <span class="math-container">$E_1$</span> into <span class="math-container">$C_1$</span>, now by the second law of thermodynamics, it must be that efficiency can't be <span class="math-container">$1$</span> , so some heat is dumped to environment, let's say we 'direct' the heat dump <span class="math-container">$E_2$</span> into <span class="math-container">$C_2$</span>, and we keep doing this procedure till we reach the <span class="math-container">$n-1th$</span> engine. Now, the heat dump of this engine into n+1th engine must be very small in magnitude ( say we have enough engines such that this happens), then this if we look at the at the total efficiency of the process:</p>
<p><span class="math-container">$$ e= \frac{Q_{in-sys} - Q_{out-sys} }{Q_{in-sys} }$$</span></p>
<p>The question I have is how to achieve the step where we direct the heat loss from the <span class="math-container">$ith$</span> engine to the <span class="math-container">$i+1th$</span> engine. If there was a way to do that I Think this would be a viable method for maximizing heat engnine output.</p>
| |thermodynamics| | <p>I'd say it's a fairly common exploit to squeeze what extra you can out of your heat source. The <em>Titanic</em> had stacked engines. The center shaft was driven by a blow-down turbine fed from the exhausts from the port and starboard engines.</p>
<p>The P 51 had a <a href="https://en.wikipedia.org/wiki/Meredith_effect" rel="nofollow noreferrer">Meredith effect radiator</a>.</p>
| 41062 | Is 'stacking' engines a viable way to maximize efficiency? |
2021-03-19T18:09:02.887 | <p>I am interested in knowing how much weight a piece of cardboard can hold in proportion to it's size.</p>
<p>Suppose we have independent variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span> corresponding to length and width and that these two vectors are in <span class="math-container">$\mathbb{R}^2$</span>. Then I think (could be wrong) the formula would be</p>
<p><span class="math-container">$$f(x,y) = \lambda_1 x + \lambda_2 y $$</span> where <span class="math-container">$\lambda_1 = \lambda_2$</span> if the piece of cardboard has square dimension.</p>
<p>However I am kind of stuck from here because I don't know how to get the <span class="math-container">$\lambda$</span> scalars to complete the formula. I tried looking online and there doesn't appear to be a general formula (surprised because cardboard is so common in our lives). Does anyone have any insight into figuring out this problem?</p>
<p>edit: the weight is assumed to be evenly distributed</p>
| |structural-engineering|solid-mechanics| | <p>Put a modest floor layer on the bottom (another piece or two of cardboard), and the failure location (with careful handling, no bouncing resulting in concentrated load) becomes the bends at the edges, so I would guess the max load should be proportional to the perimeter.</p>
| 41067 | How much weight can a cardboard of size $n \times m$ hold? |
2021-03-20T01:25:31.860 | <p>I am looking for a way to mirror the swept cut shown in the figure below about the object's midplane, which is parallel to the right plane. Can you do this without reference geometry, for example is there a way to select the midplane of the object without creating a new plane?</p>
<p><a href="https://i.stack.imgur.com/uQYnI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uQYnI.png" alt="enter image description here" /></a></p>
| |solidworks| | <p>No - there needs to be a physical plane or face to select for the mirror to be defined.</p>
<p>You can define the reference plane to be the mid-plane between two opposing faces, or to be on a central axis etc. such that the position of the mirror plane dynamically updates if/when the geometry changes.</p>
| 41074 | Alternative approach to reference geometry to mirror a swept cut |
2021-03-20T16:47:41.747 | <p>I have a peristaltic pump that was having some leaks because of connections between the tubing so I decided to replace it all with a single piece of tubing. Upon doing this though, I observed that the tubing on the outlet side of the pump was being forced upwards.</p>
<p>My peristaltic pump comes with a base that is made out of plastic. I want to glue something to the outlet side of the pump that will create some friction on the outside of the tubing keeping it from moving. I have some M20 nuts that are just large enough to fit around tubing and might offer the proper resistance if I can adhere them to the plastic.</p>
<p>However, the metal has a black coating on it that might cause some issues, and I am pumping mineral oil so I need to use some glue that will resist that for prolonged use while the pump is on. I'm fine with the nuts being permanently affixed to the plastic. I was wondering if anyone on this forum has some ideas for glues I can use to do this.</p>
<p><strong>Update</strong>
Here is a picture of the pump from Amazon. Looks like I got the last one. Anyway, the plastic is the black frame around the cylindrical motor.
<a href="https://i.stack.imgur.com/Nq0tj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nq0tj.jpg" alt="pump" /></a></p>
| |metals|plastic|adhesive|tubing| | <p>3M makes some unbelieveably strong "doublesticky" indoor/outdoor elastomeric tape that is used to do things like attach chrome trim to the outsides of car bodies. Home Depot sells it in rolls and squares; I do not know if it is mineral oil-resistant but it is worth a try.</p>
| 41080 | Adhesive to connect metal with a coating to plastic that resists mineral oil |
2021-03-20T18:38:52.927 | <p>I'm trying to design and simulate a simple knurling device. <a href="https://i.stack.imgur.com/uJJAv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJJAv.png" alt="enter image description here" /></a></p>
<p>The shaft ( blue part ) has a flat surface at the other end, where the small grey pin connect the green body to the shaft. I guess you can see it clearly in this view:</p>
<p><a href="https://i.stack.imgur.com/iAmhk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iAmhk.png" alt="enter image description here" /></a></p>
<p>My first question is, why we do even need to give the surface such a flat shape?</p>
<p>I think maybe it's easier to bore a hole, but I'm far from being sure.</p>
<p>And is there any other flat surface on the other side ( opposite side ) or there is only one flat side ?</p>
| |mechanical|machining|technical-drawing|machine-design| | <p>Yes, it would be easier to drill the hole but there are other reasons.</p>
<ul>
<li>It's makes the shaft the same width as the red bearing also clamped by the green sides.</li>
<li>The shaft will experience torque when the knob at the top is tightened. A hole though a round shaft would have high pressure on the curved edge at a point on each side some little distance away from the green faces. Holding the flattened shaft between flat plates gives a spanner-like grip on the shaft.</li>
</ul>
| 41082 | Shape of the shaft of knurling device |
2021-03-21T08:16:09.640 | <p>After heating at 950°C and holding the temperature for 3 hours there is a soaking time that is provided for 12 hours. What is the soaking time being referred to here? A detailed reply to this will be very welcome.</p>
| |mechanical-engineering|materials|metallurgy|metals|heat-treatment| | <p>At that high temperature , there are some things like carbides and intermetallics going into solid solution. Also some chemical homogenization ( which is similar to the solution). To a small degree the temperature is becoming more uniform ; although the surface has reached temperature the core of a thick component takes longer to reach temperature ; Usually addressed by typical ASME requirement by " one hour per inch of thickness". For a lower temperature like 500 C ,it would be time for stress relieving to develop , as in PWHT. That high temperature sounds like a solution anneal for a ferrous superalloy, followed by a fast cool. However 3 + 12 hours is exceptionally long in my experience.</p>
| 41093 | What is meant by soaking time in heat treatment? |
2021-03-21T10:36:15.953 | <p>I disassembled an iPod Nano 2 yesterday and was surprised to find a round brass-coloured plate glued to the back. I didn't find any mention of it at the teardown over at ifixit.com, as they didn't cut the case in half. The back of the logic board has two contacts pressed against this plate.</p>
<p>Since it just seems to be a piece of metal connected to the case and is quite discoloured, I'm reminded of galvanic anodes attached to chips to protect them from corroding. Does anyone know why this plate is there?</p>
<p>I'm a physics teacher and this seems like an opportunity to teach my students something.</p>
<p><a href="https://i.stack.imgur.com/7wUXI.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/7wUXI.jpg" alt="" /></a>
<a href="https://i.stack.imgur.com/72MpR.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/72MpR.jpg" alt="" /></a></p>
| |metallurgy|metals|consumer-electronics| | <p>To enlarge upon carloc's answer, you can actually see the contact studs for making connection to the piezo disc in the lower image. they are the gold-plated spring doots in the little green PC board.</p>
| 41095 | Unknown metal-piece connected to the case of an iPod Nano 2 |
2021-03-21T21:41:30.750 | <p>I am attempting to determine which material has been used in a spring. I know that the spring is ferrous. However, I would like to know whether the spring would have been heat treated.</p>
| |springs| | <p>Springs are steel because it is : 1- cheap, 2- available and, 3 - it makes excellent springs. It may contain various alloys and carbon levels. It may be cold drawn or heat-treated ( quench and tempered ) . Very likely the spring is cold drawn carbon steel because that is the cheapest. The main concern for most springs is hydrogen stress cracking . Very little hydrogen will cause this because of the high strength and high stress. So some protective coating may be applied , as simple as paint. Recently, hydrogen stress cracking caused a torsion spring in my garage door to fail . The present weather of cold nights and warm days caused moisture to condense on the cold spring , that introduced enough hydrogen ( few ppm ) from corrosion to fracture the spring . The spring was 25 years old and the original coating of oil/ grease was gone. I suggest oiling garage door springs as I have now done , to resist this corrosion / cracking.</p>
| 41106 | Common spring materials |
2021-03-21T21:44:03.747 | <p>From this <a href="https://www.nuclear-power.net/glossary/neutron-moderatoraverage-logarithmic-energy-decrement/" rel="nofollow noreferrer">https://www.nuclear-power.net/glossary/neutron-moderatoraverage-logarithmic-energy-decrement/</a>, I have found that the relationship between atomic weight and neutron energy loss can give the number of collision between a neutron and nucleus to slow down (average logarithmic energy decrement.)</p>
<p>I was wondering if this can be a good factor of selecting materials for fast and thermal reactor. I have not found the any relevancy between the relationship mentioned above and the material selection of the reactors.</p>
<p>Can anyone please explain how the relationship between atomic weight and neutron energy loss can be relevant to the selection of materials between fast and thermal reactor? Or, can anyone give me any link/sites to know about it more?</p>
| |materials|material-science|nuclear-engineering| | <p>Slowing a neutron down in a reactor in called <strong>moderation</strong> and the material that is responsible for moderation is the <strong>moderator</strong>.</p>
<h1>Thermal Reactors</h1>
<p>In a thermal reactor, moderation is essential as neutrons tend to be born with a lot of energy (fast) but thermal reactors relying on slow (also called thermal) neutrons to perform fission. To more efficiently slow down the neutron, as you note, we want to minimize the atomic mass of the moderator material--the mainstream thermal reactor concepts of the world are all based on moderators with <span class="math-container">$A \le 12$</span>.</p>
<p>However, this is counterbalanced against other neuronic and materials factors.</p>
<h2>Neutronic factors</h2>
<p>In addition to scattering, neutron-nucleus interaction can also result in the nucleus capturing the neutron. To be an effective moderator, this has to be minimized; we need to look at the <strong>cross-sections</strong> of our would-be moderator to make sure they're not strong absorbers. Boron, for example, is fairly light but is a neutron vacuum cleaner.</p>
<p><span class="math-container">$^{1}H$</span>, with <span class="math-container">$A=1$</span>, is of course the best choice from a scattering perspective at slowing down neutrons, but it also has a non-negligible absorption cross-section. <span class="math-container">$^{2}H$</span> (deuterium) with <span class="math-container">$A=2$</span> is twice as heavy but doesn't suffer from the absorption problem. Very pure carbon (e.g., graphite) with <span class="math-container">$A=12$</span> also has a fairly low absorption cross-section.</p>
<h2>Materials properties</h2>
<p>Even if we've picked some materials with good neutronic properties, we need practical materials to put in a reactor. The material must be stable at high temperatures, be able to be produced in high-purity forms in sufficient quantities, resist the effects of radiation, etc.</p>
<p>Hydrogen or deuterium gas are explosive in the presence of oxygen, aren't the best materials for heat transfer; a high heat capacity can be a nice property since we can then use the moderator as coolant. The answer is to use water or heavy water (water with deuterium instead of <span class="math-container">$^{1}H$</span>) as the moderator, usually under pressure, and put up with the absorption cross-section of oxygen and the high-energy gamma ray that is produced in the process (thankfully with a short half-life).</p>
<p>Graphite is very stable to very high temperatures, but obviously can't serve as the coolant either.</p>
<h2>Economics</h2>
<p>Water is abundant and cheap, and can be purified relatively efficiently.</p>
<p>High-purity graphite is not cheap, but not as expensive as heavy water.</p>
<p>Heavy water is very expensive.</p>
<p>However, heavy water's neutronic properties mean that is can be used without having to enrich the uranium in the reactor, which can save money on fuel.</p>
<p>Light water can be used as coolant and moderator, but the uranium has to be enriched to make a practical power reactor.</p>
<p>Graphite is used for gas reactors, where compressed gas is used for heat transfer. Gas reactors under development have the potential to reach very high temperatures but still achieve "walk-away" safety, meaning that they are inherently very safe.</p>
<p>Light water reactors have been the dominant players, but heavy water reactors (e.g., the CANDU family) and gas reactors (e.g., the AGR) are in profitable service today.</p>
<h1>Fast Reactors</h1>
<p>Fast reactors do not require moderation, as they are based around using fast neutrons.</p>
<h1>Summary</h1>
<p>For thermal reactors, the moderation efficiency (basically, maximizing the "energy loss fraction") is one factor in choosing a moderator material that has to be considered alongside other factors: other neutronic factors, material properties, economics. The only three materials that have been chosen for production power reactors are light water, heavy water, and graphite.</p>
<p>For fast reactors, the ability of materials to efficiently slow-down neutrons is not a design concern.</p>
| 41108 | Relevancy of (neutron energy loss and atomic weight) and the material selection of fast and thermal nuclear reactor |
2021-03-22T12:00:58.117 | <p>Why does a decrease in dislocation density result in decrease in yield strength and increase in ductility?A slightly detailed explanation is what I am looking forward to.</p>
| |mechanical-engineering|materials|manufacturing-engineering|metallurgy|material-science| | <p>Dislocations travel through a metal crystal while it is being deformed; it is in fact dislocation movement which makes metal crystals yield and deform at loads far below that which you could calculate on the basis of interatomic forces in a perfect lattice.</p>
<p>These mobile dislocations tend to pile up against one another and become entangled as the material deforms; these pileups inhibit further deformation and the material becomes more resistant to deformation than it was originally. This furnishes the basis for <em>cold work</em> as a strengthening mechanism.</p>
<p>If there are not many dislocations in the crystal to start with, then those which are present find it easy to move, and it is easy to form new ones from things like stacking faults within the crystal. This makes the crystal ductile- at least until dislocation pileup sets in.</p>
| 41115 | Why does decrease in dislocation density result in higher ductility? |
2021-03-23T04:18:32.680 | <p>The following question is from a past paper from Further Mechanics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts
<br/>
<BR/> Part 1:</p>
<blockquote>
<p>An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.
<br/><br> Find the distance of the centre of mass of the object from the end of the cylinder that is not attached
to the cone.</p>
</blockquote>
<p>I first roughly sketched the object as such</p>
<p><a href="https://i.stack.imgur.com/nsFr1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nsFr1.png" alt="sketch" /></a></p>
<p>Then finding the <span class="math-container">$\bar{x}$</span> of each part seperately:
<br/>
<br/>
For the cone I used the standard result of the centre of mass being <span class="math-container">$\frac{1}{4}r$</span> away from the base:</p>
<p><span class="math-container">$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$</span>
<br/></p>
<p>For the cylinder I derived <span class="math-container">$\bar{x}$</span> via integration:</p>
<p><span class="math-container">$$y=3r$$</span></p>
<p><span class="math-container">$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$</span></p>
<p><span class="math-container">$$\because dV=y^2 \pi dx $$</span></p>
<p><span class="math-container">$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$</span></p>
<p><span class="math-container">$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$</span></p>
<p><span class="math-container">$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$</span></p>
<p><span class="math-container">$$\therefore \bar{x}_{cylinder}=2r $$</span></p>
<p>Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was</p>
<p><span class="math-container">$$\bar{x}=\frac{11}{4}r$$</span></p>
<p>However the next part has had me at a complete loss for the better part of the day, it states that:</p>
<blockquote>
<p>Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.</p>
</blockquote>
<p>I tried coming up with a rough sketch for this also</p>
<p><a href="https://i.stack.imgur.com/wMXbx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wMXbx.png" alt="sketch 2" /></a></p>
<p>However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.</p>
<p>The condition for this question given in the marking scheme is that</p>
<p><a href="https://i.stack.imgur.com/WRdWl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WRdWl.png" alt="markscheme" /></a></p>
<p>Can someone explain what this means and why this is the case?</p>
| |applied-mechanics|dynamics| | <p>The CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then</p>
<p><span class="math-container">$$ \bar{X}= \frac{2r*m+5r*m/3}{4/3m} =r*11/4 $$</span></p>
<p>The cone side length (like the sharp tip of a pencil) is <span class="math-container">$5r$</span>, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees.</p>
<p>The CG is <span class="math-container">$5.25r$</span> from the tip of the cone and at rotation to rest on the side of the cone will be at <span class="math-container">$$5.25r*cos36.87= 4.19r<5r$$</span></p>
<p>I let you prove the condition.</p>
<p>It is well within the footprint of the pencil shape so it is stable.</p>
| 41139 | Condition for equilibrium of a rigid body on a horizontal surface |
2021-03-23T19:32:49.793 | <p>A pulley system has mass moment of inertia of 1.4kgm^2. If 8kg block is released from rest from S=1, by using energy approach what is angular velocity at S=2? Radius of pulley is 250mm. Note S is from the pulley centre.</p>
<p>This is what I did:</p>
<p><span class="math-container">$$T_1+V_1=T_2+V_2$$</span>
<span class="math-container">$$0+ 8(g)(1)= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} Igw^2 - 8(g)1$$</span></p>
<p>Then rearranged for W and got 12.85rad/s?? Am I on the right tracks please? <img src="https://i.stack.imgur.com/5NyDc.jpg" alt="enter image description here" /></p>
<p>The mass is at the bottom and it is originally 1m away. The mass then moves down to 2m away from the pulley from this original position of 1m away.</p>
| |dynamics| | <p>This is in the right track but not entirely.</p>
<p><span class="math-container">$$T_1+V_1=T_2+V_2$$</span></p>
<p>I found two problems</p>
<ul>
<li>one with the initial position. Let's assume position is h=0 initially, and the mass travels by S=1m. So the initial energy can be thought of as zero.</li>
<li>the other is with an added g in the rotational energy of the pulley. It should be <span class="math-container">$\frac{1}{2} Iw^2 $</span> instead of <span class="math-container">$\frac{1}{2} Igw^2 $</span>.</li>
</ul>
<p>Below I am showing the differences.</p>
<p><span class="math-container">$$0+ 8(g)(1)= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} Igw^2 - 8(g)1$$</span></p>
<p><span class="math-container">$$0+ \color{red}{0}= \frac{1}{2}(8)w^2(0.250)^2 + \frac{1}{2} \color{red}{I w^2 } + 8\cdot(g)\cdot \color{red}{(-1)}$$</span></p>
<p>The rest is exactly like you've done already.</p>
| 41153 | Pulley with mass dynamics question |
2021-03-24T17:36:54.807 | <p>How do you reverse the direction of a fillet like that highlighted below?</p>
<p><a href="https://i.stack.imgur.com/Jj73O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jj73O.png" alt="enter image description here" /></a></p>
| |solidworks| | <p>During the selection process, select the two faces (instead of selecting an edge).</p>
<p>This should create the desired outcome.</p>
| 41168 | Reverse direction of fillet |
2021-03-25T10:26:23.437 | <p>How can I obtain the below formulas of infinitesimal strain in cylindrical coordinates using matrix calculation given the first formula? I find it hard to study them because I still don't know how to derive them.</p>
<p><span class="math-container">$$
\epsilon_{ij}=\frac{1}{2}\left(u\otimes\nabla+\nabla\otimes u\right)\\
\,\\
\begin{align}
u\otimes\nabla
&=\begin{bmatrix}u_r\\u_{\vartheta}\\u_z\end{bmatrix}\begin{bmatrix}\dfrac{\partial}{\partial r}&\dfrac{1}{r}\dfrac{\partial}{\partial\vartheta}&\dfrac{\partial}{\partial z}\end{bmatrix}\\\\
&=\begin{bmatrix}\dfrac{\partial U_r}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_r}{\partial\vartheta}&\dfrac{\partial U_r}{\partial z}\\\\\dfrac{\partial U_{\vartheta}}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_{\vartheta}}{\partial\vartheta}&\dfrac{\partial U_{\vartheta}}{\partial z}\\\\\dfrac{\partial U_z}{\partial r}&\dfrac{1}{r}\dfrac{\partial U_z}{\partial\vartheta}&\dfrac{\partial U_z}{\partial z}\end{bmatrix}\end{align}
$$</span></p>
<p>Above, I show my try in deriving the first part of the tensor, but I didn't know how to derive the second part.</p>
<p><span class="math-container">\begin{align}
\varepsilon_{ij} &= \frac{1}{2} (U_{i,j} + U_{j,i})\\
\varepsilon_{rr} & = \cfrac{\partial u_r}{\partial r} \\
\varepsilon_{\theta\theta} & = \cfrac{1}{r}\left(\cfrac{\partial u_\theta}{\partial \theta} + u_r\right) \\
\varepsilon_{zz} & = \cfrac{\partial u_z}{\partial z} \\
\varepsilon_{r\theta} & = \cfrac{1}{2}\left(\cfrac{1}{r}\cfrac{\partial u_r}{\partial \theta} + \cfrac{\partial u_\theta}{\partial r}- \cfrac{u_\theta}{r}\right) \\
\varepsilon_{\theta z} & = \cfrac{1}{2}\left(\cfrac{\partial u_\theta}{\partial z} + \cfrac{1}{r}\cfrac{\partial u_z}{\partial \theta}\right) \\
\varepsilon_{zr} & = \cfrac{1}{2}\left(\cfrac{\partial u_r}{\partial z} + \cfrac{\partial u_z}{\partial r}\right)
\end{align}</span></p>
| |mathematics|solid-mechanics| | <h1>differential derivation</h1>
<p>The displacement vector is <a href="https://en.wikipedia.org/wiki/Infinitesimal_strain_theory#Strain_tensor_in_cylindrical_coordinates" rel="nofollow noreferrer">defined as</a> :</p>
<p><span class="math-container">$$\vec{u} = u_{r}\mathbf{\vec{e}_r} + u_{\theta}\mathbf{\vec{e}_\theta}+u_{z}\mathbf{\vec{e}_z} $$</span></p>
<p>by derivation you can obtain all the relevant strains.</p>
<h1>geometrical derivation</h1>
<p>One way to derive the strain tensor is from geometry.</p>
<p>The diagonal (normal) components <span class="math-container">$\epsilon_{rr}$</span> , <span class="math-container">$\epsilon_{θθ}$</span> , and <span class="math-container">$\epsilon_{zz}$</span> represent the change of length of an infinitesimal element. The non-diagonal (shear) components describe the change of angles.</p>
<p><a href="https://i.stack.imgur.com/LOOU6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LOOU6.png" alt="enter image description here" /></a></p>
<h2>normal strains</h2>
<h3>normal strain in radial direction <span class="math-container">$\epsilon_{rr}$</span></h3>
<p>The change of length in the radial dimension is only due to displacement in the radial direction.</p>
<p><span class="math-container">$$\epsilon_{rr} = \frac{ \left\{ u_r + \frac{\partial u_r}{\partial r} d r - u_r \right\} }{ d r} = \frac{\partial u_r}{\partial r}$$</span></p>
<p><a href="https://i.stack.imgur.com/eiAzL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eiAzL.png" alt="enter image description here" /></a></p>
<h3>normal strain in circumerential direction <span class="math-container">$\epsilon_{\theta\theta}$</span></h3>
<p>For the circumferential strain <span class="math-container">$ \epsilon_{θθ}$</span>, there are two sources :</p>
<ul>
<li><p>due to radial displacement:
<span class="math-container">$$\epsilon_{\theta\theta,r} = \frac{(r+u_r) d \theta - r d\theta}{r d \theta} = \frac{u_r}{r}$$</span>
i.e. if there is rotation and change in r, then the arc is greater for greater r, and smaller for smaller r</p>
</li>
<li><p>to circumferential displacement
<span class="math-container">$$\epsilon_{\theta\theta,c} = \frac{u_{\theta} + \frac{\partial u_{\theta}}{\partial \theta} d \theta - u_{\theta} }{r d \theta} = \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta}$$</span></p>
</li>
</ul>
<p>This is the "rigid rotation" of a segment at the same radius.</p>
<p><a href="https://i.stack.imgur.com/iu4Oj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iu4Oj.png" alt="enter image description here" /></a></p>
<p>So the total strain in the circumferential direction is:</p>
<p><span class="math-container">$$\epsilon_{\theta\theta,r} = \frac{u_r}{r}+ \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta}$$</span></p>
<h3><span class="math-container">$\epsilon_{r\theta}$</span></h3>
<p>In the z direction, there is not much of a surprise. The derivation is exactly the same as in the cartesian coordinates. If the infinitesimal element increases by <span class="math-container">$u_z$</span>, then the strain is:</p>
<p><span class="math-container">$$\epsilon_{zz} = \frac{\vartheta u_z}{\vartheta z}$$</span></p>
<h2>shear strains:</h2>
<h3><span class="math-container">$\epsilon_{r\theta}$</span></h3>
<p>This is equivalent the shear strain in xy coordinates in the sense that there is a change in shape of the component.</p>
<p><a href="https://i.stack.imgur.com/EaCgc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EaCgc.png" alt="enter image description here" /></a></p>
<p>Again there are two components:</p>
<ul>
<li>for a change in angle with the side of the element parallel to the radial direction
<span class="math-container">$$\frac{1}{2} \left[ \frac{\partial u_{\theta}}{\partial r} - \frac{u_{\theta}}{r} \right] $$</span></li>
<li>for a change in angle with the side of the element parallel to the circumerferntail direction
<span class="math-container">$$ \frac{1}{2r}\frac{\partial u_r}{\partial \theta} $$</span></li>
</ul>
<p>The total is:
<span class="math-container">$$\epsilon_{r \theta} = \frac{1}{2} \left[ \frac{\partial u_{\theta}}{\partial r} - \frac{u_{\theta}}{r} + \frac{1}{r}\frac{\partial u_r}{\partial \theta} \right]$$</span></p>
<h3><span class="math-container">$\epsilon_{r z}$</span></h3>
<p>The strain on r,z of a infinitesimally small element can be derived more or less like the xz direction.</p>
<p><a href="https://i.stack.imgur.com/6rxjn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6rxjn.png" alt="enter image description here" /></a></p>
<p>The new element has the same volume, but the angle between the edges initially parallel to r, and z have changed. For infinitesimally small angles:</p>
<p><span class="math-container">$$ \epsilon_{rz} = \frac{1}{2} \left( \frac{\partial u_r}{\partial z}+ \frac{\partial u_z}{\partial r}\right)$$</span></p>
<h3><span class="math-container">$\epsilon_{r \theta}$</span></h3>
<p>Similarly, for the shear strain in <span class="math-container">$r\theta $</span>, if you imagine the plane which is perpendicular to the <span class="math-container">$rz$</span> plane at a specific radius (See image below for a representation), then the strain is again obtained as you would obtain any shear strain in a cartesian system.</p>
<p><a href="https://i.stack.imgur.com/2ALGB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2ALGB.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$\epsilon_{\theta z} = \frac{1}{2} \left( \frac{\partial u_z}{r \partial \theta} + \frac{\partial u_{\theta}}{\partial z}\right)$$</span></p>
| 41178 | Infinitesimal Strain Tensor in Cylindrical Coordinates |
2021-03-25T16:05:50.807 | <p>Is it because of some ferrite stabilizer or something?A detailed explanation to this is most welcome.</p>
| |mechanical-engineering|materials|metallurgy| | <p>Depends what you mean by "manganese steel". If you mean a low alloy like AISI 1100 types , you are above the curie temperature or you are measuring wrong . If you mean a Hadfield type ( 11 to 14 % Mn) it cooled quickly enough to be mostly retained austenite . It certainly has nothing to do with ferrite stabilizing as that would be magnetic. Because railroad applications are the main use of Hadfield types , I suspect that is what you have. I suggest getting a book that covers Hadfield type alloys for details. These steels are relatively uncommon.</p>
| 41189 | Why is manganese steel non magnetic in as cast condition? |
2021-03-25T18:53:35.830 | <p>Recently, I visited industry where they produce steel coils (rebar + plain coils). At the end of the production process, I discover that rebar coils are getting sprayed by water:</p>
<p><a href="https://i.stack.imgur.com/zO6Q2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zO6Q2.jpg" alt="enter image description here" /></a></p>
<p>Whereas plain coils are not getting sprayed by water. What's the purpose of spraying water on rebar coils but not on plain coils?</p>
| |rolling| | <p>The rebar may be getting some hardening depending on temperature and chemical composition. If they are hot enough to show any color ( roughly 1200 F) it is definitely some hardening. Because the transformation from high temperature austenite takes some time ( depending on chemistry ) ; some hardening could develop at lower temperatures. The rebar coils would only need to be straightened and have the anchor pattern rolled into them to be finished. The other coils are likely to have significant further processing such as wire drawing or cold heading so they need to be less hard = more ductile so are cooled slowly.</p>
| 41193 | What's the purpose of spraying water on rebar coils but not on plain coils? |
2021-03-26T06:20:19.417 | <p>What is a Time Temperature curve basically? I have come across this a couple of times while dealing with heat treatment in my industry. A detailed reply to this is most welcome.</p>
| |mechanical-engineering|materials|manufacturing-engineering|metallurgy|material-science| | <p>You left off a "T" ; it is "time, temperature, transformation". It is a basis to evaluate hardenability of alloy steels. A sample is austenitized, then rapidly cooled to some temperature ,and held at that temperature for a specific time . Then rapidly cooled to room temperature and the microstructure examined . And after one finishes the one year course on steel metallurgy , one can determine what has occurred. Continuous cooling curves are more reflective of the real world but much more difficult to construct so not often used. USSteel had developed the most comprehensive list of TTT curves ; title - "Isothermal Transformation Diagrams " , pub. 1963, Pittsburgh , PA.</p>
| 41205 | What does time temperature specifically give us? |
2021-03-26T23:25:57.230 | <p>In rotational mechanical systems, there is a common ratio used when transferring quantities to different shafts.</p>
<p><a href="https://i.stack.imgur.com/G25uc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G25uc.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ \frac{N_D}{N_S} $$</span></p>
<p>Let's generalize number of teeth into number of a physical parameter associated with a connection.</p>
<p>The ratio then represents the number of the physical parameter of the destination over the source.
This ratio is used in different manners depending on the type of quantity.</p>
<p><strong>Rotational Mechanical System</strong></p>
<ol>
<li><p><strong>Gears</strong> link connections.</p>
</li>
<li><p>The <strong>loop</strong> connection is the connection.</p>
</li>
<li><p>The number of <strong>teeth</strong> is the physical parameter.</p>
</li>
<li><p>For <strong>torque</strong>, you multiply by the ratio.</p>
</li>
<li><p>For <strong>angle</strong>, you divide by the ratio.</p>
</li>
<li><p>For <strong>impedance</strong>, you multiply by the square of the ratio.</p>
</li>
</ol>
<p>Now get this, in the transformer this concept also applies. Now I know there are analogous quantities such as torque and voltage, angle and current and the two impedances for rotational mechanical and electrical networks. But I didn't notice at once that even this concept of transferring and referring quantities to connections it is not originally part of is also the same.</p>
<p><strong>Transformer in Electrical System</strong></p>
<ol>
<li><p><strong>Windings</strong> link connections.</p>
</li>
<li><p>The <strong>current</strong> connection is the connection.</p>
</li>
<li><p>The number of <strong>turns</strong> is the physical parameter.</p>
</li>
<li><p>For <strong>voltage</strong>, you multiply by the ratio.</p>
</li>
<li><p>For <strong>current</strong>, you divide by the ratio.</p>
</li>
<li><p>For <strong>impedance</strong>, you multiply by the square of the ratio.</p>
</li>
</ol>
<p>I haven't seen any source say it blatantly as I did, but if you are familiar with the lessons you can verify if my concepts are correct. My question is, is there an overarching concept that covers why these two systems have a similar behavior? Are there formal terminologies and would such concept be extendable to other systems provided that they would also have analogs to the quantities mentioned above.</p>
| |systems-engineering| | <p>This is well known feature called the <a href="https://en.m.wikipedia.org/wiki/Mechanical%E2%80%93electrical_analogies" rel="nofollow noreferrer">Mechanical - Electrical analogies</a>. As name implies there are in fact several of these spanning several domains, not only does it apply to transformers and gears it applies to spirngs, heat machines etc.</p>
<p>This was of great importance back in the day of analogue computers as this allowed you to simulate mechanical systems with electrical components. This is still used, the electrical engineers quite readily simulate heat loads with their circuit simulators. I have even seen the reverse done in a multibody simulator where a mechanical system simulated curcuitry, though it was more of a joke.</p>
<p>Simply, yes a gear is a transformer. Simply both are forms of a energy conservation used to exhange one linked property to another. Where the mechanical system is converting Force and Speed the electrical system is converting voltage and current. Similar systems exist also in the thermodynamic, acoustic and hydeaulic domains.</p>
<p>It isnt really suprising though, the technical requirements for all the systems are equivalent. If say one of the domains would come up with a super useful new basic use pattern all the others would seek to find the equivalent in case ot was useful.</p>
| 41224 | Transformers and systems with gears have similar computation when having quantities refer to different parts. What is the general concept? |
2021-03-27T07:14:26.747 | <p>Its very basic question. But there is no data all over internet about it.</p>
<p>Let's see how we calculate mass flow rate through pipe.</p>
<p>We take velocity of fluid and c/s area of pipe. Then we find volume flow rate using continuity equation. After that we multiply it by density So far so good.</p>
<p><span class="math-container">$$\frac{mass}{sec} = density \times area \times velocity$$</span></p>
<p>But which density?</p>
<p>The liquid is saturated (partly water and partly vapour with some dryness fraction)</p>
<p>Now if we see the density of water and water vapour at saturation point (100 celcius), there is a lot much difference.</p>
<ul>
<li><p>for water its around 1000</p>
</li>
<li><p>for vapour its around 0.598</p>
</li>
</ul>
<p>Both will give different answers.</p>
<p>This question is still haunting me even after my engineering is completed.
Over the years, I have seen numericals using them both without any reasonable explaination.
Help me understand this please!
Let's say dryness fraction is 0.6</p>
| |pipelines|steam|flow-control| | <p>If you interested in the mass rate of a partially full pipe at steady state , then things are very simple.</p>
<p><a href="https://i.stack.imgur.com/jhqob.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jhqob.png" alt="enter image description here" /></a></p>
<p>The total mass would be the sum of the mass rate</p>
<ul>
<li>in the liquid cross-section</li>
<li>in the vapour cross-section</li>
</ul>
<p>So total mass rate :</p>
<p><span class="math-container">$$\dot{m}_{total} = \dot{m}_{liquid} + \dot{m}_{vapour}$$</span>
<span class="math-container">$$\dot{m}_{total} = \rho_{liquid}\times A_{liquid} \times v_{liquid} + \rho_{vapour}\times A_{vapour} \times v_{vapour}$$</span></p>
<p>Just for comparison purposes, lets assume that the pipe is half full. In that case <span class="math-container">$A_{liquid} = A_{vapour}=\frac{\pi}{2} r^2 $</span>. (Also, in most cases the velocity of the vapour and the liquid should not be all that different but lets not make that assumption here).</p>
<p>In that case you see that the mass rate is predominantly from the liquid. i.e.</p>
<p><span class="math-container">$$\dot{m}_{total} = (\rho_{l}\times v_{l}+\rho_{v}\times v_v) \times A_{liquid}$$</span></p>
<p><span class="math-container">$$\dot{m}_{total} = (1000\times v_{l}+0.6\times v_v) \times A_{liquid}$$</span></p>
<p>So you see even if you calculated only the mass rate of the liquid, you would only incur less than 0.1% error in the mass rate.</p>
<h2>Other cases</h2>
<p>Of course if the flow is not steady state, and you've got gas bubbles and pockets (see below), that is different ball game.</p>
<p><a href="https://i.stack.imgur.com/XYkW5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XYkW5.png" alt="enter image description here" /></a></p>
<p>However, again if you could find a characteristic time that the flow can be considered periodic, you can define a control volume and again derive the same equation. If you are meticulous enough you should be able to derive the fluctuation of the mass rate within the characteristic time.</p>
| 42231 | Calculate mass flow rate of saturated fluid through pipe |
2021-03-27T16:26:03.080 | <p>I'm new to electronics and I have a problem with timers and pulses.</p>
<p>I have a 24 V input signal that turns on every now and then. On every rising edge of this signal (from 0 to 24 V), I want to output a 24 V signal for 100 ms. (No retrigger.)</p>
<p>I've been looking into LTC6993-1 and 555 timers, however, to my understanding both of these require a constant power source to operate (which I want to avoid) and they also seem rather complex compared to the problem.</p>
<p>I wonder if there is an easier approach to accomplishing this without having a constant power source and just using the 24 V input signal?</p>
| |electrical-engineering| | <p><a href="https://i.stack.imgur.com/sTXW0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sTXW0.png" alt="enter image description here" /></a></p>
<p><em>Figure 1 (a) Simple. (b) Still simple.</em></p>
<p>Figure 1a might work. It relies on the relay being a little slow to energise. Meanwhile you get an output pulse until RLY1 picks up.</p>
<p>If that's too quick then replace RLY2 with a 12 V relay, add in R1 with a value equal to the coil's resistance and add in some capacitance to cause a pick-up delay. The time constant will be approximately <span class="math-container">$ \frac {R_1 C} 2 $</span> farads.</p>
<p>If you want more precise timing then you'll have to go electronic. 24 V is standard industrial control voltage and you'll find many, many timers available for that.</p>
| 42242 | Output pulse on rising edge |
2021-03-27T18:13:41.157 | <p>I am working to develop a deep reinforcement learning model for balancing an inverted double pendulum.</p>
<p>To do so, I needed to derive the <a href="https://nbviewer.jupyter.org/github/mughees-asif/dip/blob/master/deep-learning-dip.ipynb#euler-lagrange" rel="nofollow noreferrer">equations of motion</a>, via the Euler-Lagrange (EL) method and now need to linearise the EL equations to get the local stability near the equilibrium point, when:</p>
<p><span class="math-container">$$\theta(t)=0\:\:\:\:\:\:\:\:\:\:\:\phi(t)=0$$</span></p>
<p>i.e., both pendulums are in the vertically upright position; <strong>vertically stable</strong>.</p>
<p>After reading several papers, the linearisation most commonly used, as described <a href="https://nbviewer.jupyter.org/github/mughees-asif/dip/blob/master/deep-learning-dip.ipynb#3.6.-Linearisation-and-Acceleration-" rel="nofollow noreferrer">here</a> is:</p>
<p><span class="math-container">$$sin(\theta(t))\approx\theta(t)\:\:\:\:\:\:\:\:\:\:\:sin(\phi(t))\approx\phi(t)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$$</span></p>
<p>Using these linearisations, the <a href="https://nbviewer.jupyter.org/github/mughees-asif/dip/blob/master/deep-learning-dip.ipynb#3.6.-Linearisation-and-Acceleration-" rel="nofollow noreferrer">final acceleration equations</a> each have a term that is multiplied by either <span class="math-container">$\theta(t)$</span> or <span class="math-container">$\phi(t)$</span>, which does not even make sense.</p>
<p>For example, one of the terms in the <span class="math-container">$x$</span>-acceleraion equation is:</p>
<p><span class="math-container">$$-M^{2} g l^{2} 4I + 4 M l^{2} - M l \theta(t)+...$$</span></p>
<p>As seen, the last term <span class="math-container">$M l \theta(t)$</span> is a multiplication of the mass, length and angle. How can a value just be multiplied by an angle?</p>
<p>Instead, if using, <span class="math-container">$\theta(t)=0$</span> & <span class="math-container">$\phi(t)=0$</span>, then:</p>
<p><span class="math-container">$$sin(\theta(t))=sin(0)=0\:\:\:\:\:\:\:\:\:\:\:sin(\phi(t))=sin(0)=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$$</span></p>
<p>Using the above linearisations, there are no values in the final acceleration equations that are being multiplied with just an angle?!?</p>
<p>Can someone please explain why the papers have used the linearisations stated in <span class="math-container">$(1)$</span> instead of <span class="math-container">$(2)$</span>, or provide some links to the relevant literature? Intuitively, the linearisations of <span class="math-container">$(2)$</span> correlate with the initial idea of vertically stabilising the pendulums?</p>
<p>Thank you.</p>
| |dynamics|kinematics| | <p>This answer comes late, but I hope it is still useful.</p>
<p>The second set of linearization linearize the problem to triviality. You are basically constraining the pendulae to not move. A more accurate terminology would be you are keeping them constant (by constraining), therefore eliminating the dynamics completely. If the system cannot move, then there is no balancing possible, even if the language might make you think otherwise, as balancing <em>consists</em> of making sure it doesn't move.</p>
<p>Notice that if <span class="math-container">$$\sin(\theta)=\sin(\phi)=0$$</span> then <span class="math-container">$$\theta=\phi=0 \pmod {2\pi}$$</span> and <span class="math-container">$$\dot{\theta} = \dot{\phi}=0 $$</span></p>
| 42247 | Linearisation of the Equations of Motion of an Inverted Double Pendulum |
2021-03-27T20:21:35.630 | <p>I'm looking to build a fish-tank stand for a decently large fish tank (300 gal). I'm looking at building it out of A513 steel because that is easily available locally. However, I am having trouble finding a good yield stress to build my design on. A513 does not have a maximum yield stress requirement, so the measured results I find on the internet are all over the place!</p>
<p>Is there a reasonable estimate for maximum yield strength for A513, short of going to the company providing them the steel and asking them which specific supplier they are using, and then getting data from that supplier?</p>
| |steel|yield-point| | <p>It is steel , if it is magnetic it is stronger than you need. I see your frustration , internet numbers from 30,000 psi up , or $ 65 for an ASTM Spec. ( I discarded all my ASTM books because every spec is reviewed every 5 years and none of mine were newer than 1995 . Specs may not be changed but you do not know ). Look at it another way ; your 300 gallon tank will weigh about 3000 lb with water ,gravel, etc. So if you have the lowest strength A -513 , you need 0.1 square inch of 30,000 psi steel to hold it up. Having built and modified several stands , I think having diagonal bracing or some other way to stabilize the tank from a side force is the major concern.</p>
| 42250 | Reasonable expected yield strength of A513 steel |
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