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2015-03-18T14:52:43.163 | <ol>
<li>Honey dippers are often made of wood. Is there a property of honey that drives this?</li>
<li>Honey dippers are shaped like ovoids with lateral grooves. Is there something about the viscosity of honey that caused this decision? Is this design somehow superior to using a spoon?</li>
</ol>
<p><a href="http://commons.wikimedia.org/wiki/File:Wooden_honey_dipper_(5536641494).jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JOelXm.jpg" alt="honey dipper"></a></p>
| |fluid-dynamics|design| | <p>Honey dippers work by using the viscosity of the liquid to raise a small amount of the liquid from a reservoir. Spoons work by trapping the small amount of liquid in a gravity well. Spoons therefore will work even when the liquid has low viscosity. The honey dipper method only works with appreciable amounts of liquid when that liquid has high viscosity.</p>
<p>Given enough time, the liquid will flow off the honey dipper. The viscosity lengthens that time, allowing you to move and turn the instrument to compensate for the slowly creeping flow before it separates and drips - until you want it to.</p>
<p>The material only needs to allow the thick liquid to adhere, so something hydrophobic or that resists adhesion, like teflon, would be a bad choice. In the specific case of the liquid being honey, the material of course has to be food-safe. Wood was the obvious choice long ago because it was available, cheap, and the technology for working it well established. Other materials available 100s of years ago would have worked too, like walrus tusk or ivory, but those were much more expensive. Metals are not a good choice due to corrosion and possible chemical reactions. Still, the same metals used for cutlery should have worked as well, but wood was much cheaper, and unlike the cutlery, the higher strength was not needed.</p>
<p>The shape is to provide a large surface area for the volume. This traps more of the liquid in the slow-flowing boundary layer.</p>
| 2131 | What drives the honey dipper design? |
2015-03-18T19:18:08.730 | <p><strong><em>Scenario:</em></strong><br>
Suppose the emergency brake button in the cab of a train is not working. Would the emergency brake cords / buttons in the passenger cars still work?</p>
<p>This question is inspired by <a href="https://scifi.stackexchange.com/q/84144/23802">this question on the SciFi StackExchange</a>, asking why the passengers in the <a href="https://www.youtube.com/watch?v=GYOYewO_Veg" rel="nofollow noreferrer">train fight scene from the movie SpiderMan 2</a> did not simply pull the emergency brakes.</p>
<p>In that particular scene, a R46 City Subway Car was rendered "unstoppable" by ripping out the speed control lever in the car, which also happened to disable the emergency brake button (see 16 seconds into the video).</p>
<hr>
<p>With movies, we suspend our disbelief in order to enjoy the story that is presented. But the above SciFi question got me to thinking about how emergency brake systems are designed for trains.</p>
<p>Given trains significant mass and momentum when moving, it seems like there would be multiple, redundant safety systems to provide braking capability for the train.</p>
<p><strong><em>My Question:</em></strong><br>
Is there a common safety design used for the braking system of trains? </p>
<p>Does that design account for portions of the system failing and allowing other portions to compensate for the failed components? (ie. would the passenger car emergency brakes still work?)</p>
| |mechanical-engineering|rail| | <p>I was trained to operate diesel electric locomotives for the Canadian Pacific Railway about 20 years ago. I'm not sure if the technology has been updated, likely not that much. The other posts about how the brakes work are correct. Air is pumped from the locomotives to fill reservoirs on each car to be able to release the brakes. Any large loss of pressure along the train applies the brakes. From the locomotive there are 3 ways to initiate emergency breaking. Press the engineer emergency brake button or let the timer expire on the engineer emergency brake button. (setup so if something happens to the engineer it will automatically stop the train). There is a also an emergency brake lever for the conductor to pull on his side of the cabin. The last way is change the direction of the throttle form forward to reverse or reverse to forward. All of these will release the air pressure and cause all the brakes to be applied both the train car brakes and locomotive brakes. </p>
| 2133 | If the emergency brake in a train is broken, do the passenger car brakes still work? |
2015-03-19T18:21:05.420 | <p><a href="http://en.wikipedia.org/wiki/Jewel_bearing" rel="nofollow noreferrer">Jeweled bearings</a> use a metal spindle with a jewel lined pivot hole (pictured below). They have been important in the manufacture of mechanical timepieces since the early 1700's. Their importance lies in the fact that they can be machined to very high accuracy, they have low and predictable friction, and their hardness gives them very low wear. They are known for operating in sealed environments for decades with no servicing. </p>
<p>With the advent of synthetic crystal manufacturing in the early 1900's jeweled bearings became the standard for every important bearing in mechanical timepieces. Every quality mechanical watch manufactured these days, for instance, has at least 17 jeweled bearings for the 'going train'. </p>
<p>$\hspace{150px}$<a href="http://en.wikipedia.org/wiki/File:Watch_jewel_bearing_and_capstone.svg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wDmS2m.png" alt="Jeweled bearing from wikimedia commons"></a></p>
<p>My questions are:</p>
<ul>
<li>Are there any other common applications for jeweled bearings besides mechanical timepieces?</li>
<li>If so, what are the largest jeweled bearings used for?</li>
</ul>
| |mechanical-engineering|friction|bearings| | <p>Compasses. </p>
<p>"The business part of a compass is out of sight. This is a set of tiny magnets glued under the compass card and surrounding the jeweled pivot that supports the center of the card." -The Annapolis Book of Seamanship, by John Rousmaniere</p>
| 2147 | Are jeweled bearings commonly used for applications other than mechanical timepieces? |
2015-03-20T02:28:52.070 | <p>I'm creating a light up dancing pole. This requires a translucent material that is strong enough to support the weight and force of a human being, even when it's hollowed out to so LEDs can go inside. The material must also be reasonably affordable, so something like diamond is out of the question. I figured that <strong>polycarbonate</strong> was my best bet. </p>
<p>I did some searching and found a polycarbonate tube with the right dimensions (8ft length, ~45mm outer diameter). Here are the exact specs:</p>
<ul>
<li>Length: 8 ft </li>
<li>Outer Diameter: 1.750 in.</li>
<li>Inner Diameter: 1.500 in.</li>
<li>Wall Thickness: 0.125 in</li>
<li>Weight: 0.332 lb/ft</li>
</ul>
<p>My question is whether this pole can support the torque of an average-strength 150lb human dancer. Not someone who is using all their bodily strength to destroy the thing, but apply force at different angles and forces.</p>
<p>It will be secured to the ground by bolting it to a weighted platform. My concern is whether the pole could potentially break.</p>
| |mechanical-engineering|materials| | <p>Given the variables involved and the relatively low cost of polycarbonate tubing, I'd be tempted to just buy a length and test it. Use a fairly chunky dancer to build in a factor of safety.</p>
<p>Polycarbonate isn't brittle and so catastrophic failure is unlikely. However, it scratches easily, which might degrade its light transmission over time. Stress corrosion cracking can be a long-term problem but I don't imagine any corrosive materials would be involved in the application you describe.</p>
<p>If the top of the pole can't be built-in, think about guy-lines ('shrouds' in yacht-speak), possibly with a spreader to give more room for the dancer.</p>
| 2149 | Can polycarbonate dance pole support the torque of a dancer? |
2015-03-20T14:51:32.553 | <p>One of the places I frequent recently had the curb and sidewalk rebuilt by the city. Previously, there was a roughly continuous transition from the street to the rest of the private drive next to the building. </p>
<p>After the rebuild, the sidewalk was leveled out which created two sloped areas coming from the street. The problem with this new design is that even at reasonable speeds, vehicles are somewhat bottomed out from the up-down-up motion created by the two slopes.</p>
<p>If it matters, this is a transition point from a public street and sidewalk onto private property for a non-residential building.</p>
<p><em>This is what the entryway looks like now</em>:<br>
<img src="https://i.stack.imgur.com/9MGik.png" alt="double slope picture"></p>
<p><em>What it used to look like</em>:<br>
<img src="https://i.stack.imgur.com/RGjvV.png" alt="previous version of entryway"></p>
<p>Please note, I have exaggerated the slopes in the illustrations in order to clarify what I'm asking about.</p>
<p><strong>Is there a common term used in roadway design that describes the double slope?</strong> I would like to be able to research the term to see if there was any code violation. At a minimum, I'd like to use the correct terminology when I ask the city about rectifying the problem. </p>
| |civil-engineering|terminology|roadway| | <p><strong>Name</strong></p>
<p>I'm not sure that it has a specific name. It is just, "the way driveways are when there is a sidewalk." </p>
<p><strong>Geometry</strong></p>
<p>Like most things related to roadway design in the US:</p>
<ol>
<li>It has been studied. There is a well named report available called, <a href="http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_w151.pdf" rel="nofollow noreferrer">Geometric Design of Driveways</a>.</li>
<li>It can vary based on local codes. See report above.</li>
</ol>
<p>The report linked above studied driveway geometry requirements from around the US and tried to do some modeling as well. The graphical results are shown below:</p>
<p><img src="https://i.stack.imgur.com/eqBKC.png" alt="Driveway Recommendations"></p>
<p>The conclusion that I got from skimming through the report was that lots of driveways have bad geometry; drive slow.</p>
<p>Unless there is a local code that the drive is in violation of, it might not be something that the city is willing to fix.</p>
| 2157 | Term for double slope from street |
2015-03-21T01:23:41.477 | <p>Magnetic bearings work by levitating a rotating shaft so that it is not in contact with its supports. This greatly reduces the friction of the system.</p>
<p>In all of the literature that I have seen on magnetic bearings, the bearings are described as "low friction" and not "no friction". </p>
<p><a href="https://en.wikipedia.org/wiki/Magnetic_bearing#Design">Wikipedia</a></p>
<blockquote>
<p>... they do not suffer from wear, have low friction ...</p>
</blockquote>
<p><a href="http://www.synchrony.com/knowledge/how-magnetic-bearings-work.php">Synchrony</a></p>
<blockquote>
<p>... they do not suffer from wear, they have low friction ...</p>
</blockquote>
<p><a href="http://www.steorn.com/bearings/performance/">Steorn</a></p>
<blockquote>
<p>... A low-friction bearing ...</p>
</blockquote>
<p><a href="http://www.calnetix.com/resource/magnetic-bearings/advantages-magnetic-bearings">Calnetix</a></p>
<blockquote>
<p>... extremely low friction and wear ...</p>
</blockquote>
<p>It would seem that there wouldn't be any friction since the magnets are keeping the rotating shaft from touching anything.</p>
<p>Where does the friction come from in magnet bearings?</p>
| |mechanical-engineering|friction|magnets|tribology| | <p>There are two types of losses in magnetic bearings, windage and electromagnetic losses. Windage or aerodynamic loss is the dissipation of rotational energy due to the viscosity of air or other gases trapped between the rotating and stationary portions of the system. These effects are more significant in high speed applications and clearly non-existent in a vacuum sealed chamber. The electromagnetic losses can come from hysteresis and/or eddy currents in the magnets. These effects have to do with a variation in the magnetic flux due to the rotating ferromagnetic materials. Below are some references I have found useful. </p>
<hr>
<p><strong>References:</strong></p>
<ul>
<li><a href="http://rads.stackoverflow.com/amzn/click/3642004962" rel="nofollow">Magnetic Bearings: Theory, Design and Application to Rotating
Machinery</a></li>
<li><a href="http://www.femm.info/dmeeker/pdf/01333140.pdf" rel="nofollow">Effect of Magnetic Hysteresis on Rotational Losses in Heteropolar
Magnetic Bearings</a></li>
</ul>
| 2162 | What causes friction in magnetic bearings? |
2015-03-23T03:59:28.190 | <p>In Ernest Cline's novel <a href="http://en.wikipedia.org/wiki/Ready_Player_One" rel="nofollow noreferrer"><em>Ready Player One</em></a>, the main character lives in the "stacks" - a dystopian vision of what a trailer park may look like in the future. The "stacks" are primarily composed of trailerhouses stacked</p>
<blockquote>
<p>... at least fifteen mobile homes high (with the occasional RV, shipping container, Airstream trailer, or VW microbus mixed in for variety). In recent years, many of the stacks had grown to a height of twenty units or more.</p>
</blockquote>
<p>and</p>
<blockquote>
<p>The trailers on the bottom level rested on the ground, or on their original concrete foundations, but the units stacked above them were suspended on a reinforced modular scaffold, a haphazard metal latticework that had been constructed piecemeal over the years.</p>
</blockquote>
<p>There's also a great picture from the cover which helps visualize:</p>
<p><img src="https://i.stack.imgur.com/TJXsY.jpg" alt="Ready Player One cover art"></p>
<p><strong>I'm curious how feasible this actually is.</strong> Can a typical mobile home support that much weight? Can standard steel scaffolding provide enough support? Can the <em>ground</em> support that much weight without a better foundation than typical for a motor home?</p>
| |civil-engineering|structures| | <blockquote>
<p>Can a typical mobile home support that much weight?</p>
</blockquote>
<p>No, most mobile homes are just flimsy wooden constructions where emphasis is on weight saving. If you want to stack cheap housing blocks I suggest stacking steel old shipping containers. They also come preequipped with anchor points in the corners.</p>
<blockquote>
<p>Can standard steel scaffolding provide enough support?</p>
</blockquote>
<p>Standard haphazard steel scaffold, probably not. Properly engineered lattice work yes.</p>
<blockquote>
<p>Can the ground support that much weight without a better foundation than typical for a motor home?</p>
</blockquote>
<p>No, the higher you go the higher the tipping momentum is, this will put more force on one side. This could be stabilized with tension wires between the individual towers. This will widen the effective base. (and provide a place for people to hang their laundry.)</p>
| 2180 | How high can you stack mobile homes? |
2015-03-23T13:36:47.700 |
<p>It's a movie trope that snipers are spotted because someone catches a glimpse of a reflection of the scope. It seems to me that this is a serious design flaw.</p>
<p>I also know that a good number of scopes have "lids" to protect against dirt, etc., when not in use.</p>
<p>My thought is this: Why is the front-most (nearest the target) lens/glass so close to the end of the pipe (as in ╞═════╪╡)? Why isn't that piece of glass hidden <em>way</em> inside the pipe (as in ╞═════╪════╡)? That <em>extra</em> bit of pipe could then block out any light/reflections apart from what is in the view of the shooter. It would also make the scope longer, but that can't be <em>that</em> much of a downside, can it?</p>
<p>If there <em>are</em> such scopes, how common are they? Or, put the other way around, given that reflections are undesired, how is this actually dealt with <em>outside of the movies</em>?</p>
| |optics| | <h3>The lenses which make up the scope will always reflect some light (~0.5%).</h3>
<p>All optical materials will reflect some amount of light. This is due to the fact that light travels slower inside of the medium than it does in the surrounding air. The slowing down of a wave always results in some reflection; in optics it is governed by the <a href="http://en.wikipedia.org/wiki/Fresnel_equations" rel="nofollow noreferrer">Fresnel equations</a>, but there are analogous reflections in <a href="http://en.wikipedia.org/wiki/Signal_reflection" rel="nofollow noreferrer">electronic circuits</a> and <a href="http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html" rel="nofollow noreferrer">waves traveling along a string</a>.</p>
<p><a href="http://en.wikipedia.org/wiki/File:Partial_transmittance.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vZ3FG.gif" alt="Partial Transmission"></a></p>
<p>There are <a href="http://en.wikipedia.org/wiki/Optical_coating" rel="nofollow noreferrer">techniques which use destructive interference between multiple reflections</a> at successive boundaries to give optical devices very, very low reflectivities at a specific wavelength. However, for a rifle scope it is important that the reflectivity be low for the entire visible range. Browsing the <a href="http://www.edmundoptics.com/technical-resources-center/optics/anti-reflection-coatings" rel="nofollow noreferrer">coatings offered by Edmunds</a> shows that a good coating can reduce the reflectivity down to about 0.5% across the entire visible range. </p>
<h3>The lens tube cannot simply be extended without degrading the image quality.</h3>
<p>There is a lot of information on hoods on <a href="https://photo.stackexchange.com/search?q=lens+hood">photography.SE</a> (<a href="https://photo.stackexchange.com/questions/399/why-are-some-lens-hoods-petal-shaped-and-others-not">here</a> and <a href="https://photo.stackexchange.com/questions/26752/could-i-get-a-lens-hood-that-suits-all-lenses/26753#26753">here</a> for instance), but I'll briefly summarize. The length of the hood is limited by the field of view. Essentially, light from every point in the scene needs to be able to reach <strong>every part of the front lens.</strong> I.E. it is not enough for the light to reach only the center portion of the lens. </p>
<p>For this reason, simply extending the tube beyond the front of the objective lens is not an option because light from the edge of the scene would not reach the edge of the lens. The hood must have a radius which is larger than the objective lens itself before it can be extended; otherwise you are throwing away light from part of the scene. Pay attention next time you see a professional photographer (or enthused amateur) with a lens hood and you will immediately notice this. The most effective hood is actually a cone with an angle set by the field of view which will start to make the scope very large and bulky.</p>
<p><a href="https://www.flickr.com/photos/colinsd40/4179924452/" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vis3vm.jpg" alt="Lens Hood from Flikr"></a></p>
<h3>The reflection is less noticeable than you might imagine.</h3>
<p>Since the objective lens of the scope is a focusing lens, the outer surface is curved like a diverging mirror. This means that the light redirected toward the enemy sniper is significantly less intense than it would be for a flat mirror. For a flat mirror the entire surface of the mirror redirects light from the sun to the enemy's eye, while a curved mirror has a much smaller area which gives the correct angle between the enemy's eye and the sun. </p>
| 2184 | Must rifle scopes have revealing reflections? |
2015-03-24T14:14:22.033 | <p>I was wondering, what is the strongest : a rectangle or an arc (same width/height) ? <img src="https://i.stack.imgur.com/4JPEA.png" alt="screenshot"></p>
<p>you have to imagine the screenshot in 3D : you have 4 walls and a paved layer (the rectangle) or a dome</p>
<p>Reason why the <em>rectangle</em> would be stronger : same size as the dome but more raw materials</p>
<p>Reason why the <em>dome</em> would be stronger : the vertical pressure is directed to the walls -> <a href="http://en.wikipedia.org/wiki/Keystone_(architecture)" rel="nofollow noreferrer">Keystone</a></p>
<p>Assumptions :</p>
<ul>
<li>the 4 "walls" are infinitely strong</li>
<li>the dome / rectangle has a weight negligible relatively to the force applied</li>
<li><b>The dome / rectangle is not infinitely strong, it can bend / break given the vertical pressure</b></li>
</ul>
<p><b>EDIT</b></p>
<p>I am making a plastic box and people will walk on it (box size is size of a foot). for some reasons walls must be thin and the box must be pretty thin too and we wondered what shape is stronger / is less likely to bend in its center</p>
| |structural-engineering| | <p>The answer to your question is very simple. </p>
<p>If the arch and the rectangle are made of the same material, and the rectangle height and width are equivalent to the arch rise and span, and they are loaded with the same load in the same place, and the weight of the material is neglected, the rectangle will always have at least as as much load capacity as the arch, and could have more depending on the material properties. Always. </p>
<p>This is a result of the principle of superimposition. If you overlay your arch on top of your rectangle, you'll see that the rectangle, in a manner of speaking, <em>contains</em> the arch: </p>
<p><img src="https://i.stack.imgur.com/b3yiC.png" alt="enter image description here"></p>
<p>The extra material above and below the arch, given your assumptions (the most important of which, aside from using the same material and dimensions, is ignoring self weight of the material), cannot negate the load carrying capacity of the arch "inside" of it. Therefore, the rectangle will have at least as much capacity as the arch. </p>
| 2204 | Mechanically, is an arc stronger than a rectangle? |
2015-03-24T23:51:19.043 | <p>I am mounting solar panels on top of a recreational vehicle (RV). To reduce drag, I was planning on building a small ramp in front of the panels to deflect wind before it hits the flat panel and mounts. Here is a rough sketch (red is brackets, black is panel, purple is ramp):</p>
<p><img src="https://i.stack.imgur.com/IKLCI.png" alt="Solar Panel"></p>
<p>Someone told me that there would be MORE drag if I did it this way than if I didn't put the deflector/ramp. I will do the work even if there will be VERY LITTLE difference between drag avoided by putting in a ramp and not having a ramp at all, but I definitely don't want to do it if it will INCREASE drag.</p>
<p>I think the space under the panel will be minimal. I am going to try to get them as close as possible to the roof. The panels are about 2" thick and the space between will be roughly 1/4-1/2 inch.</p>
<p>How would adding a ramp in front of the panel affect drag on the vehicle? Would it increase, decrease or stay the same?</p>
| |automotive-engineering|airflow|solar-panel|drag| | <p>The concept of what you're doing is sound, and as Russell McMahon notes the efficiency gains could be significant enough to justify the change.</p>
<p>I'd strongly suggest that you consider adding a ramp to the back edge as well. Drag force is very sensitive to the downstream (rear) end of a body as well You get some positive pressure at the front of the vehicle, but you also get strong negative pressure at the rear. This is made much worse by the development of turbulence and flow separation.</p>
<p>Conceptually, what you want to do is reduce the cross-section at the back end to reduce the area over which that negative pressure is applied. You can do that by tapering the back end (a reverse ramp in your case). The only trick is that you have to do so pretty gradually to avoid a phenomenon called flow separation. this is basically the air stream tumbling over itself and generating vortexes.</p>
<p>So go ahead with the ramp in the front, but add a very shallow (~10 degrees) ramp on the back as well. That'll give you the best improvement for minimal investment. Also, try to get the ramps to fit up right to the edges of the solar panels. You want the path that the air follows to be as smooth as possible.</p>
<p>For some experimental evidence and great pictures, check <a href="http://www.ara.bme.hu/oktatas/letolt/Vehicleaerodyn/Vehicleaerodyn.pdf">this presentation</a></p>
<p>p.5 shows small improvements in rounding the front of a van but they quickly level off (no benefit for making the front end smoother)</p>
<p>p.11,17,18 shows flow separation at the rear</p>
<p>p.18 (upper right) shows an optimal point for reducing the read x-section. Tapering too sharply undoes the benefit of reducing area.</p>
| 2207 | How would adding a wind deflector in front of roof-mounted solar panels affect drag on an RV? |
2015-03-25T16:48:06.667 | <p>I have a PCB of approximately 4" x 4". It has a single long spiraling trace on it. The trace is 50 mil 4 oz copper, so nominally it's good for something like twenty amps before it overheats. But this board overheats much faster. My supposition is that the windings being so close to each other has a compounding effect on the heat.</p>
<p>Now, presumably I could increase the surface area of the PCB, and thereby dissipate the same power with less temperature rise. My question is, how does one calculate such? What is the relationship between the power dissipated by my copper plane, its temperature rise, and its surface area? Assume still air.</p>
| |electrical-engineering|heat-transfer|temperature|power| | <p>Assumptions:</p>
<ul>
<li>The copper side with the traces is modeled as a sheet of copper rather than traces.</li>
<li>The body is thin enough that thermal conductivity within the body is unimportant, and the entire device is considered to be at a uniform temperature.</li>
<li>Only the two broad surfaces contribute to the heat loss, the sides are neglected.</li>
<li>The surroundings, including the air and radiative syncs, are at a uniform temperature $T_s$</li>
<li>Thermal coefficients: <a href="http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html" rel="noreferrer">$\epsilon_{cu}$=0.78</a>, <a href="http://www.ti.com/lit/an/snva183b/snva183b.pdf" rel="noreferrer">$\epsilon_{pcb}$=0.50</a>, <a href="http://www.thermal-wizard.com/tmwiz/convect/natural/hup-isot/hup-isot.htm" rel="noreferrer">$h_{up}=7.25\ \frac{\text{W}}{\text{m}^2\text{K}}$</a>, <a href="http://www.thermal-wizard.com/tmwiz/convect/natural/hdp-isot/hdp-isot.htm" rel="noreferrer">$h_{down}=3.63\ \frac{\text{W}}{\text{m}^2\text{K}}$</a></li>
</ul>
<p>Under these assumptions we can estimate the temperature of the board by simply equating heat flows. The heat coming in per unit time is from <a href="http://en.wikipedia.org/wiki/Joule_heating" rel="noreferrer">Joule heating</a> from the current running through the copper and is given by
$$
q_{in}=I^2R.
$$
The heat flowing out has two escape mechanisms; <a href="http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html" rel="noreferrer">radiative heat transfer</a> to the surroundings which is given by
$$
q_{rad}=\epsilon\sigma A(T^4-T_s^4)
$$
and <a href="http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html" rel="noreferrer">convective heat transfer</a> to the air which is given by
$$
q_{conv}=hA(T-T_s).
$$</p>
<p>Now we just equate the heat flows
$$
\begin{align}
q_{in}&=q_{out}\\
I^2R&=A\left[\sigma(T^4-T_s^4)(\epsilon_{cu}+\epsilon_{pcb})+(T-T_s)(h_{up}+h_{down})\right]
\end{align}
$$
we can rearrange this to look like a quartic equation
$$
\sigma(\epsilon_{cu}+\epsilon_{pcb})T^4+(h_u+h_d)T -
\left[\frac{I^2R}{A}+\sigma(\epsilon_{cu}+\epsilon_{pcb})T_s^4+(h_u+h_d)T_s\right]=0.
$$</p>
<p>This isn't easy to solve analytically, but Mathematica would have no problem. I put it into Python and numerically found the minimum of the absolute value. I assumed a total resistance of $R=1\Omega$ so your actual results may vary. The results are shown below. </p>
<p><img src="https://i.stack.imgur.com/QmWPd.png" alt="Board Temperature"></p>
<p>Here is the Python code also:</p>
<blockquote>
<pre><code>import scipy.optimize as opt
import numpy as np
import matplotlib.pyplot as plt
# Define the function with some extra variables
def tempOpt( t, ts, i):
sigma = 5.67e-8
ec = 0.78
ep = 0.50
hu = 7.25
hd = 3.63
r = 1
a = 0.0103
e = ec + ep
h = hu + hd
out = sigma*e*t**4 + h*t - i**2*r/a - sigma*e*ts**4 - h*ts
return out
# Decide the ranges for the current and temperature
currents = np.linspace( 0.1, 20, 50)
temps = [0, 20, 40, 60, 80, 100]
# Calculate the value
resDict = dict()
for temp in temps:
resVec = np.zeros( np.shape( currents))
cnt = 0
for current in currents:
# Define a new function of one variable and a minimum at zero
# Don't forget to convert between Kelvin and Celcius
def tempNow( t):
return abs( tempOpt( t+273.15, temp+273.15, current))
# Find the minimum with a reasonable guess
guess = temp + 30
resVec[cnt] = opt.fmin( tempNow, 30)
cnt += 1
# Store results in dictionary
resDict[temp] = resVec
# Plot
plt.figure(1)
plt.clf()
for temp in temps:
plt.plot( currents, resDict[temp], lw=2)
plt.xlabel('Current (A)')
plt.ylabel('Board Temp ($^\circ$C)')
leg = plt.legend( temps, loc=2)
leg.set_title( '$T_s$ ($^\circ$C)')
plt.title('Board Temperature (R=1$\Omega$)')
</code></pre>
</blockquote>
| 2217 | How can I calculate the temperature rise of a circuit board for a given power dissipation and area? |
2015-03-27T04:21:19.403 | <p>Why is it that a second order system with an additional pole or zero can be approximated into the general second order system for analysis?</p>
<p>This was asked during our lecture and I am still looking for the answer.</p>
<p>In my reading I've only learned that when an additional pole or zero is added to the system, the step response of this system starts to resemble the step response of the original system. I am not sure if this is the answer to the question.</p>
| |control-engineering|control-theory| | <p>I think there must be some other restrictions placed on your professor's comment. Adding a pole to a general second order system makes it a third order system and the dynamics can change dramatically. </p>
<p>One example I can think of in which adding a pole or zero does not change the system dynamics dramatically is when the pole or zero is at much higher frequency than the pair of poles of the second order system. In this case, the dynamics only change at very short timescales. The step response, for example, would only change at the very beginning and the damping or oscillatory behavior later in the response would remain essentially the same. </p>
<p>At any rate, I wouldn't fret too much about making approximations of such simple systems. Once you are actually working with them in the field you will likely use numerical software to understand them, and such software has no need to make any approximations for a third order system. </p>
| 2228 | Second order system with additional pole |
2015-03-27T06:25:38.760 | <p>If you look at the design documentation in our projects (Biogas and waste-to-energy plants), you find out what we plan to build but not what other options were considered and why they were discarded. This information rests only in the engineers head. Occasionally, a plant will be in a conceptual stage for years, sometimes switching between engineers. So I find myself revisiting an option that a coworker considered and ultimataly discarded last year. </p>
<p>What is a good way to track design decisions, the 'whys' and 'why-nots'? I'd strongly prefer an approach that's tried, tested and used elsewhere in industry. </p>
<p>Additional information: My company will be pursuing ISO 9001 certification, an approach that fits with this QA regime would be prefered.</p>
| |design|project-management| | <p>Others have made a lot of good comments about documentation in general, but I want to suggest a particular class of software that will help tremendously.</p>
<p>I find <a href="https://en.wikipedia.org/wiki/Revision_control" rel="nofollow">version control software</a> to be excellent for these sorts of issues. It's not too common outside of software development, unfortunately, but it's only a matter of time before it catches on in other fields.</p>
<p>Here's an example: My research group has a <a href="https://en.wikipedia.org/wiki/Apache_Subversion" rel="nofollow">Subversion</a> repository that most people use (A popular alternative software is <a href="https://en.wikipedia.org/wiki/Git_%28software%29" rel="nofollow">git</a>). Using version control helps ensure continuity of the research projects after someone leaves, because most of the files relevant to a project are there, along with a detailed log of what work they did at what time, and (if they do it right), their rationale. I'll elaborate.</p>
<p>Let's say you have a folder on your computer that contains all of the files</p>
<p>Everything is automatically dated when it's committed (the version control term for what others might call "uploading"). The commit message acts as a short memo (as Mahendra Gunawardena suggested, you might want longer and more official memos for bigger decisions). You describe the change to the project files, say "We decided that approach X is not feasible. We are going to try Y instead. Deleted the files associated with approach X and created a new CAD model for approach Y."</p>
<p>Then, later, if you decide that approach X is actually the right thing and you want to dig up the deleted files, it's not hard. You can roll back any part to older version with ease and continue from there.</p>
<p>There are some additional benefits to this approach. First, it makes sharing files with others pretty easy. They just check out the repository (a version control term which could either mean the current version, or all versions) and then regularly update it (there are commands to do that). Basically, everyone is synchronized. Second, you can get a backup (older versions) for basically free. With that being said, this doesn't mean you don't need to keep normal backups, just that you have another option for restoring accidentally lost files. I highly recommend backing up the entire repository as well.</p>
| 2229 | What's a professional way of tracking design history in plant engineering? |
2015-03-29T18:19:15.250 | <p>In order to reduce operational cost and improve the safety and reliability of aircraft's the aerospace industry and government is increasingly evaluating new ideas such structural heath monitoring systems for aircraft's. Such a system will enable continuous monitoring, inspection and detection of damage to the aircraft with minimal human involvement. This will also help the airline industry to improve aircraft maintenance scheduling process. </p>
<p><img src="https://i.stack.imgur.com/m5IwE.jpg" alt="enter image description here"></p>
<p><sub>Figure: Early Diagnosis for Microcracks in Aircraft </sub></p>
<p>Currently <a href="http://www.delta.com/content/www/en_US/about-delta.html" rel="nofollow noreferrer">Delta airlines</a> and an aircraft manufacture has teamed up with <a href="http://www.sandia.gov/" rel="nofollow noreferrer">Sandia Laboratory</a> to install sensors in aircraft to help understand flaw detection. Research also indicates that ultrasonic sensors are most widely used sensing element.</p>
<p><img src="https://i.stack.imgur.com/aR4GL.gif" alt="enter image description here"></p>
<p><strong>Question:</strong><br>
Which part of an aircraft can most benefit from structural health monitoring (SHM) system? Is it the fuselage, wings, landing gear, flaps, rudder, elevator, vertical stabilizer or another component? Why is this component the prime candidate? </p>
<p><strong>References:</strong></p>
<ul>
<li><a href="http://phys.org/news/2014-09-in-flight-sensor-health-safer-flights.html" rel="nofollow noreferrer">In-flight sensor tests a step toward Structural Health Monitoring for safer flights</a></li>
<li><a href="http://www.compositesworld.com/articles/structural-health-monitoring-composites-get-smart" rel="nofollow noreferrer">Structural health monitoring: Composites get smart</a></li>
<li><a href="http://www.ndt.net/search/docs.php3?id=4595&content=1" rel="nofollow noreferrer">Early Diagnosis for Microcracks in Aircraft</a></li>
<li><a href="http://www.cnn.com/2015/12/01/asia/air-asia-crash-report/index.html" rel="nofollow noreferrer">Air Asia QZ8501: Pilot response to equipment malfunction caused crash</a></li>
</ul>
| |electrical-engineering|aerospace-engineering|sensors| | <ul>
<li>First you need a part that <strong>exceeds its fatigue limit under expected stresses</strong> because you need slow coalescence of cracks occurring uniformly in the measured area, not a local defect which might occur outside the sensor placement and resistant to modelling based on measurements from other pieces of the part. Steel/titanium etc parts stressed below their fatigue limit won't tell you anything.</li>
<li>Parts that are easy to access can't lower maintenance costs as much as <strong>difficult to reach</strong> structures.</li>
<li>Since you're trying to lower the cost without compromising safety of the procedures, the logical step is to put a monitor on the parts expected to show signs of fatigue first and for which fatigue is well understood. You want agreement with the model, proving that you can design the sensor, not that the sensor happened to find or not find crack growth.</li>
</ul>
<p>Based on these considerations, problems I would expect are of interest are:</p>
<ol>
<li>The fuselage. Compression cycles are a different fatigue mechanism than time spent in the air, requiring modelling and checks.</li>
<li>Landing gear attachment point. Landing hardness and its effects on the fatigue of connecting structures can potentially not match airframe life <em>or</em> duty cycles, requiring checks.</li>
<li>Wings. Wings encounter a large number of cycles and require access panels to measure internal structures. Fatigue in the middle of a large beam is probably relatively predictable.</li>
<li>Engine pylon attachment points. Compressor stalls, vibration in the engine, and awkward landings/maneuvers can all result in uncommon forces that would be difficult to treat based on empirical data, requiring checks</li>
<li>The tail surfaces that are far off the ground are a good place to perhaps save time on lifts.</li>
</ol>
<p>Wings and fuselage are going to be best to demonstrate the sensor engineering principles. Pylons and landing gears (excepting any steel parts) would present the most insight perhaps.</p>
| 2240 | Which part of an aircraft can most benefit from a structural health monitoring (SHM) system? |
2015-03-30T06:08:09.233 | <p>This one should be easy to answer by looking at the thermodynamics, but I find it not to be. Say we have a natural gas stream, ~700 N∙m<sup>3</sup>/h, that we want to compress from 16 bar to 250 bar. What will the power consumption of the compressor be?</p>
<p>The thing that stymies me is that the power uptake is not path independent, but mostly depends on how many cooling steps you have in between. Or so it seems to me. You don't need to do the math for me, I just provided the numbers as an example. I would appreciate a detailed explanation of how to arrive at an approximate real world value.</p>
| |mechanical-engineering|gas|compressors| | <p>Maybe I didn't get the problem it sounds a little bit simple, but as I see you would have to calculate using a PVT correlation (Peng-Robinson for exemple) the power your transformation requires in a ideal system (ΔS=0, the power required for this transformation is path independent) and then you can divide the value you got by the efficiency of your compressor, this will change with your equipment, but 0,8 is a good value, having more cooling stages will give you more efficiency, with less colling stages you will have a lower value for efficiency.</p>
<p>You can use the following equations with a PVT correlation to calculate the power required by the transformation:</p>
<p>dU=CvdT+[T*(∂P/∂T)−P]dV</p>
<p>dS=(Cv/T)dT+(∂P/∂T)dV</p>
<p>dH=dU+d(PV)</p>
<p>Cv=Cp-R</p>
| 2244 | Typical power consumption of large compressor |
2015-03-31T11:24:20.767 | <p>I had a <a href="http://www.omegaeng.cz/ppt/pptsc_eng.asp?ref=PX26&Nav=preb02" rel="nofollow">PC Mountable Wet/Wet Differential Pressure Sensor</a>
I used it to measure a pressure difference , I don't know how to convert its output to pressure ! </p>
| |electrical-engineering|pressure|sensors| | <p>Well, it looks like your sensor is +/-5psid (based on the link you provided), so looking at the datasheet, we have (assuming a 10VDC excitation) 16.7mV at 1psi and 50mv at 5psi. Assuming the sensor is linear (the datasheet states that linearity is +/-0.25% of full scale), you have the following relationship:</p>
<pre><code>P = V * 0.1201 - 1.006
</code></pre>
<p>Where <code>P</code> is expressed in psi and <code>V</code> in mV. This does not take into account any possible hysteresis (0.2% of full scale according to the datasheet).</p>
<p>If you want an accurate reading, you should probably do your own calibration by applying a series of known pressures (measured with a calibrated instrument, e.g. manometer) and reading the sensor output.</p>
<p><strong>EDIT</strong></p>
<p>How did I arrive at the relationship? It's fairly easy, you're fitting a straight line to two points, so you're looking for a slope and an offset:</p>
<pre><code>1psi = 16.7mV * slope + offset
5psi = 50mV * slope + offset
</code></pre>
<p>That's 2 linear equations with 2 unknowns (<code>slope</code> and <code>offset</code>), which is easily solved to give the values mentioned above. That's really basic maths though, I would expect any engineer or even technician to know how to do that, which is why I hadn't included the details on how to calculate the values in the first place.</p>
| 2260 | Px26-015 pressure sensors output |
2015-03-31T20:08:01.130 | <p>...or what type of motor is used there?</p>
<p>I found this type of motor - usually powered with low-voltage AC (~12V), but at times with 230V, in several appliances that require very slow rotation and sometimes a fair momentum - a color-shifting lamp, the microwave plate, an ice cream mixer...</p>
<p>The funny property of it is it picks the start direction at random and keeps spinning in that direction until switched off - but I never faced a situation when it would get stuck in the "unstable balance" position.</p>
<p>So, what is this type of motor and why does it behave that way?</p>
| |electrical-engineering| | <p>Had a similar problem with my electrolux microwave turntable rotating when door was opened and stopped when closed. Also while rotating you could force it into the opposite direction. After checking the 3 safety microswitches which were found ok. Have noticed that mains polarity, live and neutral swapped influences this. The mains sockets i have are European so the plug can be inserted any way around, not like US or UK type. This has supprised me that some kitchen equipment can be polarity sensitive.</p>
| 2265 | Why does the microwave plate start in a random direction? |
2015-03-31T20:39:13.563 | <p>What general specifications are there in building a storm shelter in the American Midwest? I believe the major considerations are tornados and severe thunderstorms, as well as flooding along river valleys. </p>
| |structural-engineering| | <p><a href="https://www.fema.gov/" rel="nofollow">FEMA (Federal Emergency Management Agency)</a> publishes some specifications for the design of safe rooms.</p>
<p>Per <a href="https://www.fema.gov/safe-rooms" rel="nofollow">their guidance</a>:</p>
<ul>
<li><a href="https://www.fema.gov/media-library/assets/documents/2009" rel="nofollow">FEMA P-320</a> is for residential safe rooms</li>
<li><a href="https://www.fema.gov/media-library/assets/documents/3140" rel="nofollow">FEMA P-361</a> is for community and small business safe rooms</li>
</ul>
<p>And according to <a href="https://www.fema.gov/safe-rooms/frequently-asked-questions-tornado/hurricane-safe-rooms#Q18" rel="nofollow">their FAQ</a>, there are restrictions regarding potential flood risks.</p>
<blockquote>
<p>Per FEMA P-361, flood hazards should be considered when designing a residential safe room. Flood loads acting on a structure containing a safe room are strongly influenced by the structure’s location relative to the flood source. Tornado or hurricane residential safe rooms should be located outside of the following high-risk flood hazard areas: </p>
<ul>
<li>Flood hazard areas subject to high-velocity wave action (Zone V areas) and Coastal A Zones </li>
<li>Floodways </li>
<li>Any areas subject to storm surge inundation associated with any modeled hurricane category, including coastal wave effects</li>
</ul>
<p>More information on these siting restrictions can be found in FEMA P-361. </p>
</blockquote>
<p>You should also look at <a href="https://www.fema.gov/safe-rooms/frequently-asked-questions-tornado/hurricane-safe-rooms#Q20" rel="nofollow">their answer</a> regarding other applicable standards:</p>
<blockquote>
<p>FEMA P-361 provides the design criteria used with common building codes and standards to design a building. This means that the underlying building code (such as the International Building Code or International Residential Code) applies and building construction must comply with the many items that are regularly governed by code requirements. Standards such as ASCE 7 also apply but are used in conjunction with the safe room design criteria described in FEMA P-361 to produce a structure capable of resisting loads much higher than those for normal buildings.</p>
</blockquote>
<hr>
<p>So you are correct that the major considerations are tornadoes and severe thunderstorms. Their considerations for flooding are fairly common sense as they can be summarized with "don't build where it will flood", but they certainly provide more guidance than that.</p>
<p>It's also worth noting that the guidelines are applicable outside of the North American Midwest.</p>
| 2266 | Building a storm shelter |
2015-04-01T20:22:55.213 | <p>I am trying to figure out what kind of device is being used to stop and start the flow of liquid inside this device. Below is a link to the device. It looks too small to be a solenoid or a pump. How else would they do it?</p>
<p>I heard somewhere it may be a Venturi system but how do you electronically control a Venturi system that is that small?</p>
<p><a href="http://www.wunderbar.com/dispensing/beverage-dispensing/liquor-dispenser/skyflo" rel="nofollow">http://www.wunderbar.com/dispensing/beverage-dispensing/liquor-dispenser/skyflo</a></p>
| |electrical-engineering|valves|liquid|actuator| | <p>I'm going to cheat by referencing the patent. ;)</p>
<p>Automatic Bar Controls owns the Wunderbar site and has a patent on the product. The patent is <a href="https://www.google.com/patents/US8925769?dq=8925769&hl=en&sa=X&ei=GZUcVZjgFYGqNteEgdAE&ved=0CB4Q6AEwAA">US008925769</a>. The actual detail drawings don't seem to be available there though, so you have to get it straight from the US Patent Office <a href="http://pdfpiw.uspto.gov/.piw?PageNum=0&docid=08925769&IDKey=47085CEF0D39&HomeUrl=http%3A%2F%2Fpatft.uspto.gov%2Fnetacgi%2Fnph-Parser%3FSect2%3DPTO1%2526Sect2%3DHITOFF%2526p%3D1%2526u%3D%2Fnetahtml%2FPTO%2Fsearch-bool.html%2526r%3D1%2526f%3DG%2526l%3D50%2526d%3DPALL%2526S1%3D8925769.PN.%2526OS%3DPN%2F8925769%2526RS%3DPN%2F8925769">listing</a>.</p>
<p>The name of the patent is: <strong>Wireless spout and system for dispensing</strong>.</p>
<blockquote>
<p>Embodiments of the present invention provide a pouring device for a container for the dosing of liquid. The pouring device has various features intended to ease use, including an improved removal and attachment system and a system to identify and visually illustrate selected pour sizes. In a specific embodiment, the pouring device has a colored light or LED indicator system that allows the user to quickly and easily confirm the selected pour size. Embodiments may also include one or more features that ease attachment and removal of the pouring device to a liquid container, such as a replaceable cork system.</p>
</blockquote>
<p>The specific part is Claim 1:</p>
<blockquote>
<p>(c) an electrically operated valve disposed within the spout housing for selectively clamping the conduit so that a registerable amount of liquid is dosed;</p>
</blockquote>
<p>That seems to answer the question of how it works. It physically clamps off a tube to stop the flow of fluid. More specifically from the description in the patent:</p>
<blockquote>
<p>The silicone tube can be easily squeezed to stop the liquid flow. This can be done by a stepping motor, a motor with gearbox, or any other motor or appropriate mechanism.</p>
</blockquote>
| 2280 | What type of liquid flow control could this device use? |
2015-04-03T16:09:37.653 | <p>I know that the rubber on car and truck tires wear, and the road concrete wears out. I wondered:</p>
<p>While steel is hard and elastic, it still causes friction (interaction
between molecules) and therefore abrasion.</p>
<p>Let's say we have on average 20-30 wagons with four axles with a full load of 50-60 tons on each wagon and a traction vehicle. How many train passes can a railway rail endure before it must be replaced? What is the equivalent time frame in which this number of passes is achieved on average?</p>
| |civil-engineering|materials|rail|transportation| | <p><em>This is an admittedly North American response.</em></p>
<p><strong>MGT</strong></p>
<p>In the US, how much traffic goes over a given track in a year is measured in Million Gross Tons (MGT) e.g. 1 MGT = 2 000 000 000 lbs [spaces instead of commas to be world-friendly]. This is a measure of the total weight of cargo and vehicles but not necessarily the number of individual trains.</p>
<p><strong>Rail Life</strong></p>
<p>The life of a typical railroad rail is between <a href="http://www.aar.com/pdfs/Part1_WRI_Management.pdf" rel="nofollow noreferrer">1300 MGT and 380 MGT</a> depending on if the rail is on a straight (tangent) track or in a curve. There is more friction in a curve. Also, the rail on the low side of a super-elevated curve wears faster than the rail on the high side.</p>
<p><strong>Friction</strong></p>
<p>Because of this great difference in life of the rail, areas that have lots of curves employ machines that <a href="http://www.aar.com/pdfs/Part4_FrictionControl_SFK.pdf" rel="nofollow noreferrer">add lubrication to the rail</a>. This lubrication is applied after the locomotives pass so that their traction capability is not reduced. Some photos of these <a href="http://www.lbfoster-railtechnologies.com/Friction_Modifiers.asp#mstto=" rel="nofollow noreferrer">machines</a> are below:</p>
<p><img src="https://i.stack.imgur.com/F8gSa.png" alt="LB Foster machine"></p>
<p><img src="https://i.stack.imgur.com/UZui2.png" alt="LB Foster Rail Lubrication"></p>
<p><strong>Number of Trains</strong></p>
<p>In the US, the largest/heaviest trains carry only one commodity. A typical example of this is a so called "coal unit train". This is a 100-car train loaded with coal where each car weighs 286,000 lbs. Assuming an average of 500 MGT of life:</p>
<p>$$ \text{Number of Trains} = \frac{500*1e6}{(100*\frac{286,000}{2000})} \approx 35 000 \text{ trains}$$</p>
<p>A high capacity line might have 25 trains a day on one track, so the life span in years would be:</p>
<p>$$ \frac{35000}{25*365} \approx 4 \text{ years}$$</p>
| 2295 | How many train passes can railway tracks endure? |
2015-04-03T17:39:29.163 | <p>A lot of research has been devoted to creating electrical devices that emulate biological sensors, including:</p>
<ul>
<li>Visual: Cameras, color/light intensity sensors</li>
<li>Auditory: Microphones, ultrasonic sensors</li>
<li>Tactile: Pressure sensors, temperature sensors</li>
<li>Balance: Gyroscopes, accelerometers</li>
</ul>
<p>However, I have yet to find a comprehensive sensor/processing algorithm to detect and interpret odors. Certainly, there are "olfactory" sensors which are dedicated to a specific purpose, such as carbon monoxide detectors, and other hazardous gas detectors. But I have yet to find a general purpose sensor/processing algorithm that can readily detect and interpret odors within the range and resolution of a human nose. </p>
<p>Do such sensors/algorithms exist? If so, what are they and how do they work? If not, what are the primary obstacles to developing them?</p>
| |electrical-engineering|sensors| | <p>Odor assessment is usually performed by human sensory analysis using <a href="http://en.wikipedia.org/wiki/Chemoreceptor" rel="nofollow">chemosensors</a>:</p>
<blockquote>
<p>A chemoreceptor, also known as chemosensor, is a sensory receptor that
transduces a chemical signal into an action potential.</p>
</blockquote>
<p>Recently I have also heard of <a href="http://www.phonearena.com/news/Breakthrough-sensor-to-give-our-smartphones-a-sense-of-smell_id48089" rel="nofollow">a sensor from Honeywell that could potentially be used in smart phones</a>. These sensors are also called <a href="http://en.wikipedia.org/wiki/Electronic_nose#Working_principle" rel="nofollow">electronic noses</a>:</p>
<blockquote>
<p>Bio-electronic noses use olfactory receptors - proteins cloned from
biological organisms, e.g. humans, that bind to specific odor
molecules. One group has developed a bio-electronic nose that mimics
the signaling systems used by the human nose to perceive odors at a
very high sensitivity: femtomolar concentrations.</p>
<p>The more commonly used sensors for electronic noses include</p>
<ul>
<li>metal–oxide–semiconductor (MOSFET) devices - a transistor used for amplifying or switching electronic signals. This works on the
principle that molecules entering the sensor area will be charged
either positively or negatively, which should have a direct effect on
the electric field inside the MOSFET. Thus, introducing each
additional charged particle will directly affect the transistor in a
unique way, producing a change in the MOSFET signal that can then be
interpreted by pattern recognition computer systems. So essentially
each detectable molecule will have its own unique signal for a
computer system to interpret.</li>
<li>conducting polymers - organic polymers that conduct electricity.</li>
<li>polymer composites - similar in use to conducting polymers but formulated of non-conducting polymers with the addition of conducting
material such as carbon black.</li>
<li>quartz crystal microbalance - a way of measuring mass per unit area by measuring the change in frequency of a quartz crystal resonator.
This can be stored in a database and used for future reference.</li>
<li>surface acoustic wave (SAW) - a class of microelectromechanical systems (MEMS) which rely on the modulation of surface acoustic waves
to sense a physical phenomenon.</li>
</ul>
</blockquote>
| 2297 | Sensors / processing algorithms to emulate a human's sense of smell |
2015-04-03T21:31:26.793 | <p><strong>Given:</strong></p>
<p>My thermodynamics text reads as follows:</p>
<p>In SI units, the force unit is the newton ($N$), and it is defined as the force required to accelerate a mass of $1\cdot kg$ at a rate of $1\cdot\frac{m}{s^2}$. In the English system, the force unit is the pound-force ($lbf$) and is defined as the force required to accelerate a mass of $32.174\cdot lbm$ (1 slug) at a rate of $1\cdot\frac{ft}{s^2}$. That is...</p>
<p>$$1\cdot N = 1\cdot kg\times1\cdot\frac{m}{s^2}$$</p>
<p>$$1\cdot lbf = 32.174\cdot lbm\cdot\times1\cdot\frac{ft}{s^2}$$</p>
<p><strong>Question:</strong></p>
<p>For all practical purposes, such as at STP conditions or close to it like when we have a rounded off sea-level acceleration due to gravity of $32.2\frac{ft}{s^2}$ $(101\cdot kPa)$, can I just think of the $lbf$ in the following way...</p>
<p>$$W=1\cdot lbf=1\cdot lbm \times 32.174\cdot\frac{ft}{s^2}$$</p>
<p>and that for the weight of an object having a mass of $1\cdot kg$ (also at sea-level) in SI units as...</p>
<p>$$W=9.81\cdot N=1\cdot kg\times9.81\cdot\frac{m}{s^2}$$</p>
<p>Yes or no and why?</p>
| |mechanical-engineering|thermodynamics|international| | <p>Here's how I like to think of it. lbf is the force acting upon the mass. This is what, for example, your bathroom scale is measuring. lbm is the actual mass of the object. So the F =m*a in English units, lbf = lbm * a (aka gravity 32.2 ft/s2).</p>
<p>That's at least how I've always looked at it.</p>
| 2300 | Pound-force (lbf) vs Pound-mass (lbm) |
2015-04-03T23:18:11.360 | <p><strong>Given:</strong></p>
<p>A problem in my thermodynamics text is stated as follows...</p>
<p>Determine the mass and the weight of the air contained in a room whose dimensions are $V=$ $15ft$ x $20ft$ x $20ft$. Assume the density of the air is $\rho=0.0724\cdot\frac{lbm}{ft^3}$.</p>
<p><strong>My Solution:</strong></p>
<p>First find the mass...</p>
<p>$$m=\rho\times V$$</p>
<p>$$m=0.0724\cdot\frac{lbm}{ft^3}\times 6000\cdot ft^3$$</p>
<p>$$=434.3\cdot lbm$$</p>
<p>Now find the force acting on the air due to gravity. This is the weight of the air assumed at sea-level...</p>
<p>$$W=m\times g$$</p>
<p>$$W=434.3\cdot lbm\times32.174\cdot\frac{ft}{s^2}$$</p>
<p>$$=13976\cdot lbf$$</p>
<p><strong>Question:</strong></p>
<p>I find it hard to believe that in an average size room the air weighs a whopping $14,000\cdot lbf$. Did I do something wrong in my calculations or is this correct? If this is correct perhaps we earthlings living on the surface of the earth are the real <em>extremophiles</em>.</p>
| |mechanical-engineering|thermodynamics|pressure| | <p>A pound force is defined as the force required to accelerate a slug at 1 ft/s^2.
The density of air is $\rho = 0.0724 \ lb_m/ft^3 = 0.0724/32.2 \ slugs/ft^3$</p>
<p>The weight of the air is
$\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ slugs\ ft/s^2 = 434.4 lb_f$ </p>
| 2302 | Mass and weight of air in a room |
2015-04-03T23:56:48.727 | <p><strong>Given:</strong></p>
<p>The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the upward force, in $N$, that a $70\cdot kg$ man would experience in an aircraft whose acceleration is $6\cdot g’s$.</p>
<p><strong>My Solution:</strong></p>
<p>I account for the normal acceleration due to gravity by subtracting $1\cdot g$ from the upward acceleration of the aircraft. This is what we would see if we had drawn a free body diagram (not shown)...</p>
<p>$$F=ma=m\times(6g-1g)=70\cdot kg\times5\times9.81\frac{m}{s^2}=3433.5\cdot N$$</p>
<p><strong>Solution in Text:</strong></p>
<p>However the solution in the text does not do this...</p>
<p>$$F=ma=m\times6g=70\cdot kg\times6\times9.81\frac{m}{s^2}=4120.2\cdot N$$</p>
<p><strong>Question</strong></p>
<p>Why is the net upward acceleration not $5\cdot g's$?.</p>
| |aerospace-engineering| | <p>Well, the concept of g force measurement goes by measuring the net acceleration considering all effects. That is assume an object in free fall it will have a 0g acceleration, meaning Force= M(g-a), where a is pseudo acceleration=g. Hence F = M*A = M(g-g), A=The acceleration in g. Therefore, A=0g, which is freefall acceleration.</p>
<p>From <a href="https://en.m.wikipedia.org/wiki/G-force" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth's surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth's surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free-fall. The upward contact-force from the ground ensures that an object at rest on the Earth's surface is accelerating relative to the free-fall condition. (Free fall is the path that the object would follow when falling freely toward the Earth's center). Stress inside the object is ensured from the fact that the ground contact forces are transmitted only from the point of contact with the ground.</p>
<p>Objects allowed to free-fall in an inertial trajectory under the influence of gravitation only, feel no g-force acceleration, a condition known as zero-g (which means zero g-force). This is demonstrated by the "zero-g" conditions inside a freely falling elevator falling toward the Earth's center (in vacuum), or (to good approximation) conditions inside a spacecraft in Earth orbit. These are examples of coordinate acceleration (a change in velocity) without a sensation of weight. The experience of no g-force (zero-g), however it is produced, is synonymous with weightlessness.</p>
</blockquote>
| 2303 | Aircraft force in g's experienced by pilot |
2015-04-04T04:30:51.033 | <p><strong>Given:</strong></p>
<p>A problem in my thermodynamics text reads as follows...</p>
<p>A vacuum gauge connected to a tank reads $5.4\cdot psi$ at a location where the barometric reading $P_{Hg}= 28.5\cdot in$. Determine the absolute pressure in the tank. Take $\rho_{Hg} = 848.4\frac{lbm}{ft^3}$.</p>
<p><strong>My Solution:</strong></p>
<p>The formula for a vacuum gauge is as follows...</p>
<p>$$P_{vac}=P_{atm}-P_{abs}$$</p>
<p>Rewrite as...</p>
<p>$$P_{abs}=P_{atm}-P_{vac}$$</p>
<p>Determine atmospheric pressure but first convert $ft$ units to $in$ since that is how we traditionally define pressures in English units...</p>
<p>$$P_{atm}=\rho g h$$</p>
<p>$$\rho=848.4\cdot\frac{lbm}{ft^3}\times\frac{ft^3}{(12in)^3}=.4910\cdot\frac{lbm}{in^3}$$</p>
<p>$$g=32.174\cdot\frac{ft}{s^2}\times\frac{12in}{ft}=386.0\cdot\frac{in}{s^2}$$</p>
<p>$$P_{atm}=.4910\cdot\frac{lbm}{in^3}\times386.0\cdot\frac{in}{s^2}\times28.5\cdot in=5402\cdot\frac{lbm}{in\cdot s^2}$$</p>
<p><strong>Question:</strong></p>
<p>The atmospheric pressure obtained cannot be right because the scalar value $5402$ is no where near $14.7$ and the units obtained ($\frac{lbm}{in\cdot s^2}$) look completely wrong. Shouldn't they be $\frac{lbf}{in^2}$? Where did I go wrong? After this is solved I know how to obtain the absolute pressure in the tank so I shall stop here.</p>
| |mechanical-engineering|thermodynamics|pressure| | <p>$ \rho = 26.3 \ slugs/ft^3 $</p>
<p>$\rho g = \gamma= 847\ lb_f/ft^3$ "specific weight"</p>
<p>$ ft^3 = (12 in)^3$</p>
<p>$ P = \rho g h = \frac{847 lb_f}{(12 in)^3}\cdot 25.8 in$</p>
<p>The issue is the difference between a lb mass and lb force.</p>
| 2306 | Determine abs pressure in a tank where the atm and vac pressures are given |
2015-04-04T07:24:34.600 | <p><strong>Given:</strong></p>
<p>A problem in my thermodynamics text reads as follows...</p>
<p>The barometer of a mountain hiker reads $13.8$ $psia$ at the beginning of a hiking trip and $12.6$ $psia$ at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of $.074\cdot\frac{lbm}{ft^3}$ and take $g = 31.8\cdot\frac{ft}{s^2}$.</p>
<p><strong>My Solution:</strong></p>
<p>Our solution begins by solving for the change in air pressure and using this to determine elevation climbed...</p>
<p>$$\Delta P=P_1-P_2$$</p>
<p>$$=(13.8-12.6)\frac{lbf}{in^2}=1.2\frac{lbf}{in^2}$$</p>
<p>and...</p>
<p>$$\Delta P=\rho g h$$</p>
<p>rewrite as...</p>
<p>$$h=\frac{\Delta P}{\rho g}$$</p>
<p>But first we needed to convert $ft$ units to $in$ since that is how we traditionally define pressures in the English system...</p>
<p>$$\rho=.074\frac{lbm}{ft^3}\times\frac{ft^3}{(12in^3)}\times.03108\frac{slugs}{lbm}=1.33\times10^{-6}\frac{slugs}{in^3}$$</p>
<p>$$g=31.8\frac{ft}{s^2}\times\frac{12in}{ft}=381.6\frac{in}{s^2}$$</p>
<p>We can now solve for $h$...</p>
<p>$$h=\frac{\Delta P}{\rho g}=\frac{1.2\cdot lbf}{1.33\times10^{-6}\frac{slugs}{in^3}\cdot381.6\frac{in}{s^2}}=2364\frac{lbf\cdot in^2\cdot s^2}{slugs}$$</p>
<p><strong>Answer in text:</strong></p>
<p>$$h=2363\cdot ft$$</p>
<p><strong>Question:</strong></p>
<p>The scalar values are almost identical but the units just don't fit. Where did I go wrong?</p>
| |mechanical-engineering|thermodynamics|pressure| | <p>I'm glad we got rid of that obsolete system of obscure units decades ago. The metric system is so much easier.</p>
<p>From the definition of the <a href="https://en.wikipedia.org/wiki/Slug_%28mass%29">Slug</a>, </p>
<p>${1 Slug = 1 {lb}_f.{s}^2/ft}$.</p>
<p>Substituting that into the units you got, the ${slugs}$, ${lb}_f$ & ${s}^2$ go & ${ft}$ comes in.</p>
<p>The only mistake you made was to exclude ${in}^2$ in your final calculation.</p>
<p>Pressure in US units is pounds per <em>square inch</em>, not pounds as you had it. Correct for this and you end up with units of feet in the final calculation.</p>
| 2311 | Determine elevation from the difference in air pressure |
2015-04-05T06:48:07.117 | <p>I was recently on the UCLA campus (University of California Los Angeles) and I saw the Bunche Hall. I love the design and the architecture of the building. However I saw a little (what I would call) a closet with no door on the side but it was just three walls with a roof, extremely small, underneath the stairs, outdoors, a bit below the floor level. There is a small alley and then you take two stairs down and the closets face a fall.</p>
<p>I don't even know what to call it. My question is, from architectural/structural engineering point of view, what is this building device? Why is it there? What is its purpose? What is it even called? The two closets reminded of telephone booths but there was nothing but dirt, cobwebs, and leaves in there. Were they once used for something? I am posting the pictures below.</p>
<p><img src="https://i.stack.imgur.com/OQ4Qx.jpg" alt="enter image description here">
<img src="https://i.stack.imgur.com/P3sd2.jpg" alt="enter image description here">
<img src="https://i.stack.imgur.com/SiPDQ.jpg" alt="enter image description here">
<img src="https://i.stack.imgur.com/PRdEV.jpg" alt="enter image description here">
<img src="https://i.stack.imgur.com/ZlF3T.jpg" alt="enter image description here">
<img src="https://i.stack.imgur.com/AjQwR.jpg" alt="enter image description here"></p>
| |civil-engineering|structural-engineering|building-physics|building-design|architecture| | <p>I do not know, I've never been there and I have no other information other than what is in your question. However, ...</p>
<p>I'd say that best guess, until somebody who knows more presents, is your suggestion of now-unused telephone booths. </p>
<p>On the wall are two plates. These are at the same height in each case and consistent with connecting a telephone. Other equipment may be equally possible (eg: ATM machines, but seem less likely). More details about these plates and what is under them is probably the best clue apart from finding somebody who knows. </p>
<p><img src="https://i.stack.imgur.com/leCzv.jpg" alt="enter image description here"></p>
<p>There are other wall marks suggesting mountings and possibly a shelf on the right hand wall in the right cubicle, but high up marks in the left hand one are not so obviously relevant,</p>
<p>The lights were almost certainly identical (diffuser missing at right) indicating downwards illumination in both cases. </p>
<p><img src="https://i.stack.imgur.com/7LQE8.jpg" alt="enter image description here"></p>
<p>Reality? I don't know.
But, if I <em>had</em> to bet your life on it, I'd guestimate telephones. </p>
| 2319 | What's the purpose of these small closet-like-rooms-without-a-door? |
2015-04-05T09:58:48.483 | <p>I know that in case of wedge we have a 2D behavior which creates a discontinuity across the shock, but I am unable to understand why this discontinuity is not present in the case of a conical surface. I have a vague idea that the continuity equation ensures this as, due to increase in the flow area, the velocity must increase.</p>
<p>But still it would be great help if I could get a proper explanation, I couldn't find one in books and net.</p>
<p>I would also like to know which 2D constraints are actually relieved which causes the 3D relaxation effect.</p>
<p><img src="https://i.stack.imgur.com/PPd6l.png" alt="enter image description here"> </p>
<p><strong>EDIT:-</strong> I think i need to reiterate the question, I wanted to ask why does the supersonic flow bend toward the surface in case of being obstructed by a cone unlike in case of wedge, and as per its reasoning i have also read that the flow is more <strong>free</strong> in case of cone, due to presence of 3 dimensions, commonly known as 3D relaxation, so it would be nice if someone could comment upon it too.</p>
<p><img src="https://i.stack.imgur.com/VJHP9.png" alt="enter image description here"></p>
| |mechanical-engineering|fluid-dynamics|shock| | <p>The answer to this question is complicated since the flow is fundamentally different.</p>
<p>When supersonic flow hits a <strong>wedge</strong> it is abruptly turned by one <em>oblique shock</em>. After this all streamlines are parallel since in this 2D-flow-scenario the geometry/flow-area does not change anymore.</p>
<p>When supersonic flow (think of a stream tube/cylinder) hits a <strong>cone</strong> a differential volume element of the stream tube will constantly increase in size while it follows the cone surface. This changes the pressure and is called <em>relaxation</em>. Due to this volume increase of the stream tube the <strong>cone flow</strong> has two features/regimes which the <strong>wedge flow</strong> cannot have:</p>
<ol>
<li><p>The stream lines after the shock are curved (bent) so that they align with the cone geometry.</p>
</li>
<li><p>Certain flow-settings will result in a <em>sonic-line</em> within the flow downstream of the shock. This means that the supersonic flow after the oblique shock is decelerated until it is subsonic see figure below taken from <a href="http://naca.central.cranfield.ac.uk/reports/1955/naca-report-1242.pdf" rel="noreferrer">Naca Report 1242</a>.</p>
<p><img src="https://i.stack.imgur.com/nbgXE.png" alt="Naca Report 1242" /></p>
</li>
</ol>
| 2321 | Explanation of supersonic flow differences between a 2D wedge and a 3D cone |
2015-04-05T11:59:37.073 | <p>What are the construction procedures that allow the mitigation of potential problems due to water in a retaining wall?</p>
<p>To be more specific: I'm talking about reinforced concrete walls.</p>
<p>The cases of water are: if I'm studying a reinforced concrete wall such that the soil is not saturated, and it rains so that the soil is saturated or if the water table level rises.</p>
<p>What are possible solutions? I'm looking for general solutions.</p>
| |civil-engineering|retaining-wall| | <p><strong>Design</strong></p>
<p>In design, the presence of water behind the wall can have two effects. It can cause additional force to be applied to the wall from hydrostatic pressure, or its presence can cause the soil characteristics to change. </p>
<p>Changing soil properties has the most effect on clayey soils. Relatively dry clay will have cohesion and reduce the force on the wall. Wetter clay will lose this cohesion and apply more load to the wall. </p>
<p>The way that the wall is designed will effect how the water needs to be handled. If the wall was designed to for the effects of water all the way up the wall, then nothing may need to be done. If it was designed for only a certain height of water, then this must be assured.</p>
<p><strong>Removing water</strong></p>
<p>Depending on the volume of water that is expected and the source, a few different methods of water removal can be employed:</p>
<ul>
<li>Weep hole</li>
<li>Perforated drain pipe</li>
<li>Surface water diversion</li>
<li>Dewatering wells</li>
</ul>
<p><strong>Weep holes</strong> are small holes through the face of the retaining wall. These allow any water that is trapped behind the wall to slowly drain out. They are best for small volumes of water. They are often installed in all walls as a minimal level of assurance that water will not be trapped behind the wall.</p>
<p><strong>Perforated drain pipes</strong> are usually installed at the base of a wall in combination with free draining fill or geotechnical fabric. They can drain larger volume of water, but the connection details and where the outflow is located need to be investigated.</p>
<p><strong>Surface water diversions</strong> may include ditches or paved areas that keep surface water from running down a slope and being contained behind the wall. This method helps where ground water is less of a concern when compared to runoff.</p>
<p><strong>Dewatering wells</strong> are only an option that is considered where the wall must be built below the groundwater level and where the force from the water can not be designed for. The wells have pumps that lower the ground water elevation in the area. These can pump large quantities of water, but they require constant monitoring and power.</p>
| 2322 | What are the solutions to water problems in a retaining wall |
2015-04-06T03:33:21.427 | <p>I am trying to come up with a way to restrict beer flow using a valve that does not cause foam. I have found that a solenoid valve causes a change in the direction the beer must flow and this sharp change through a different size opening creates foam quickly. I have thought about a ball valve or pinch valve but I need it to be electric, work very quickly and have a small form factor. </p>
<p>Does anyone know if anything like what I created in this video exist? <a href="https://youtu.be/iE6qe7045Ck" rel="nofollow">https://youtu.be/iE6qe7045Ck</a></p>
<p>The idea would be electromagnets that cause the metal pieces to slide in one direction or the other opening and closing the flow of beer. The idea is keeping the opening in the valve the same diameter as the inner diameter of the tubing I would be using (3/8"). It would need to handle at most 14 psi and at least 12 psi. (Average beer lines are 9-10psi)</p>
| |mechanical-engineering|electrical-engineering|fluid-mechanics|valves| | <p><a href="http://www.globalspec.com/learnmore/flow_control_flow_transfer/valves/pinch_valves_multiturn" rel="nofollow">Look at pinch valves</a>: A small piece of tubing that can be pinched shut pneumatically or mechanically (solenoid). When open, you have the full diameter of your tubing. The only part that touches the medium is the tube, and you don't have any sliding seals. The latter part is important, because you don't want to mess up your beer with sealant grease and you don't want $CO_{2}$ to bubble along the seals. Also it seems you are looking for a DIY solution, so a simple build will be your friend.<br>
I don't know about foaming. You could test by having beer flow through tubing and pinch it with your fingers.</p>
| 2329 | Does this exist? -Sliding Fluid Restrictor -Solenoid Valve -Ball Valve |
2015-04-06T09:38:22.343 | <p>I'm currently looking around for a new bike and I've noticed lots of load-bearing carbon parts (frames, cranksets, forks etc.) with an interesting finish like this:</p>
<p><img src="https://i.stack.imgur.com/Cj8gy.jpg" alt="Carbon crankset"></p>
<p>and I was wondering <strong>what this type of carbon cloth is called</strong> and <strong>why it's used</strong>. I've seen it used on parts (e.g. forks) where the forces are predictable (laterally during cornering or longitudinally during braking) and thus unidirectional or woven cloth would appear to be the best option from a specific stiffness perspective. It does look pretty cool and thus presumably has marketing value but other than that, I'd appreciate any ideas!</p>
<p>Additional image:</p>
<p><img src="https://i.stack.imgur.com/tBtZc.jpg" alt="enter image description here"></p>
| |mechanical-engineering|materials| | <p>The forged composite material used by Lamborghini and Calloway (a.k.a. "forged carbon") has a higher modulus and tensile strength than other composite materials like glass fibers and even traditional carbon fiber composites made with twill or satin weaves. The "randomness" of the fiber orientation in the "forged carbon" composite produces a stronger part in terms of flexion resistance, Young's Modulus and tensile strength. The method used by Lamborghini for things like driving compartment shells and certain structural and suspension components is comprised of pre-fabricated sheet material made of a vinyl ester resin and short length (i.e. chopped) carbon fibers which are pressed (where the "forged" term comes into play) into molds under pressure and heat (similar to how pre-preg is used). The small-random fibers also allow for more complex shapes to be formed as compared to traditional carbon fiber twill fabrics.</p>
<p><a href="http://www.lambolab.org/wp-content/uploads/03research/pub/05chop/2011-ASC-montreal-forged-suspens-ICE.pdf" rel="nofollow noreferrer">http://www.lambolab.org/wp-content/uploads/03research/pub/05chop/2011-ASC-montreal-forged-suspens-ICE.pdf</a></p>
| 2333 | Forged carbon fibre finish |
2015-04-06T21:45:19.850 | <p>The ISO 9223 standard indicates that there should be no corrosion at temperatures below 0 degrees celsius. Independent researchers have however proposed to lower the minimun temperature stated in the standard to lower values in order to account for the actual corrosion observed in Nordic climates. What is the explanation for the observed corrosion at temperatures below the freezing point?</p>
| |materials|corrosion| | <p>As Dan mentioned in the comments, the presence of salt (and other pollutants) can alter the freezing point of water. </p>
<p>The <a href="http://www.iso.org/iso/catalogue_detail.htm?csnumber=53499" rel="nofollow">ISO 9223 abstract</a> states:</p>
<blockquote>
<p>key factors in the atmospheric corrosion of metals and alloys. These are the temperature-humidity complex, pollution by sulfur dioxide and airborne salinity.</p>
</blockquote>
<p>All of which are factored into the "time of wetness", which can increase in the presence of salinity (and other pollutants, acknowledged in the ISO standard itself) as found in observations made in Antarctica published in <em><a href="http://au.wiley.com/WileyCDA/WileyTitle/productCd-0470080329.html" rel="nofollow">Uhlig's Corrosion Handbook</a></em>, (p. 330):</p>
<blockquote>
<p>..that in the presence of marine salt, liquid water monolayers could form under ice layers resulting in high corrosion rates in temperatures well below 0C</p>
</blockquote>
<p>Sulfur dioxide is also identified in the ISO abstract, this a common pollutant from vehicles, industry etc. which according to can chemically react with water vapour, forming sulfuric acid, the basis of <a href="http://www.epa.gov/acidrain/what/" rel="nofollow">acid rain and acid snow</a>.</p>
| 2339 | How does corrosion occur at a temperature below 0 degrees Celsius? |
2015-04-06T23:20:08.780 | <p>What are the recommended or relevant standards for US city planning with regards to depth ranges for various utilities such as water, natural gas, and power, as well as for subway or rail, etc...?</p>
<p>For example, water and gas lines should be buried well below the frost line to prevent damage from ground movement. What standard(s) dictate those depths?</p>
<p>I am primarily interested in standards applicable to major cities such as Atlanta, Boston, Chicago, Dallas, Denver, New York, San Diego, Seattle, Washington, etc... My assumption is that there are enough similarities between the cities that a standard would have been developed.</p>
| |civil-engineering|design|standards| | <p>First, if you care about a certain city, look up the information. This should be information that is readily available. Big cities are usually go about publishing this information online.</p>
<p>Tunnels (subway or vehicle) are different enough that they don't have standard depths. They are designed to be far enough below ground to miss utilities and basements, unless they can't be for some other reason. </p>
<p>That being said, below is the information for a random location that I found first (<a href="http://www.co.washington.tx.us/default.aspx?Washington_County/Utility%20Rules" rel="nofollow">Washington County, Texas</a>):</p>
<blockquote>
<p>Depth of Underground Lines - The depth of underground lines shall be as specified herein for each type of utility. Where placements at such depths are impractical or where unusual conditions exist, the department shall specify other protection as may be appropriate in lieu of the depth of bury required for the particular utility line. Any and all buried utility lines will be placed at a minimum depth of 36". A ny deviation from the specified depth must be requested in writing and approved before Commissioners Court.</p>
<ul>
<li>High pressure gas and liquid petroleum lines will be constructed no less than forty-eight inches (48") lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway. </li>
<li>Fiber Optic lines will be constructed no less than forty-eight inches (48") lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway.</li>
<li>Communications cable will be constructed no less than thirty-six inches (36") lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway.</li>
<li>Water Lines will be constructed no less than thirty-six inches (36") lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway; crossings to be encased.</li>
<li>Underground Power line crossings and longitudinal shall be encased (placed in conduit) and buried a minimum of thirty-six inches (36") under roadway ditches, and sixty inches (60") below the pavement surface.</li>
<li>Cable television and copper cable communication lines shall have a minimum depth of cover twenty-four inches 24") under ditches or 18 inches beneath the bottom of the pavement structure, whichever is greater.</li>
</ul>
</blockquote>
<p><strong>Warning! - Do not rely on these for digging!</strong> </p>
<p><a href="http://www.call811.com/default.aspx" rel="nofollow">Call 811 to have utilities marked before you dig!</a></p>
| 2340 | How deep are utilities typically buried? |
2015-04-07T11:30:24.463 | <p><img src="https://i.stack.imgur.com/EYwcJ.png" alt="enter image description here"></p>
<p>From the previous examples I've seen, there is a connection rod, for example, between a and c and then another rod from c to one of the pins. However, in this case, the rod ac is directly connected to another pin? I assume there is no relative velocity in this case. Is this assumption correct? </p>
<p>EDIT: By pins, I was referring to O2 and O8. I've just realised that C is a slider. I thought it was a rigid 'pin' like O2 and O8. Sorry for the confusion. </p>
<p>So I think I know how to approach this problem now. Va is perpendicular to OA. Vc is a horizantal line from O2. Vc/a is perpendicular to the rod AC starting from Va.Then I need to intersect Vc and Vc/a to find the magnitude of Vc. Is this approach correct? </p>
<p>EDIT 2: I'm running out of time. I quickly sketched this to check if I'm on the right track.</p>
<p><img src="https://i.stack.imgur.com/uQbYW.png" alt="enter image description here"></p>
| |mechanical-engineering|applied-mechanics| | <p><a href="https://i.stack.imgur.com/1I0VP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1I0VP.jpg" alt=""></a></p>
<p>I have had the same exercise in school, maybe it helps, you just know $B= 1/2 AC$, and you know C is in horizontal line from 0.</p>
<p>If you want the acceleration diagram let me know, but it is just the same.</p>
| 2343 | How to draw the velocity and acceleration diagrams if there is no connecting rod between the pins? |
2015-04-08T02:45:00.753 | <p>I want to build a small ice cannon that uses pressurized air (either through an air compressor or a CO<sub>2</sub> cartridge) to shoot small ice pellets at my coworkers. I plan to use a thermoelectric cooler to freeze the water but I don't understand how to spec it.</p>
<p>For example, what are the key parameters of <a href="https://www.sparkfun.com/products/10080" rel="nofollow noreferrer">this product</a> that will let me know if the cooler will achieve what I'm hoping it to?</p>
<blockquote>
<h2>Thermoelectric Cooler - 40x40mm</h2>
<p>COM-10080 In Fritzing Library</p>
<p><strong>Description:</strong> Thermoelectric coolers (TEC or Peltier) create a temperature differential on each side. One side gets hot and the other
side gets cool. Therefore, they can be used to either warm something
up or cool something down, depending on which side you use. You can
also take advantage of a temperature differential to generate
electricity. The thermal tape listed below works very well to attach
heat sinks to the hot side.</p>
<p>This Peltier works very well as long as you remove the heat from the
hot side. After turning on the device, the hot side will heat quickly,
the cold side will cool quickly. If you do not remove the heat from
the hot side (with a heat sink or other device), the Peltier will
quickly reach stasis and do nothing. We recommend using an old
computer CPU heatsink or other block of metal to pull heat from the
hot side. We were able to use a computer power supply and CPU heatsink
to make the cold side so uncomfortable we could not hold our finger to
it.</p>
<p><strong>Features:</strong></p>
<ul>
<li>40 x 40 x 3.6mm</li>
<li>lmax - 7A</li>
<li>Umax - 15.4V</li>
<li>Qcmax - 62.2W</li>
<li>Tmax - 69C</li>
<li>1.7 Ohm resistance</li>
<li>127 thermocouples</li>
<li>Max Operating Temp: 180°C</li>
<li>Min Operating Temp: -50°C</li>
</ul>
</blockquote>
<p>If I want to freeze 10 mL of water from room temperature within a certain time period, how would I spec out the cooler? What are the base equations that govern how long it takes to freeze the water?</p>
| |mechanical-engineering|thermodynamics|cooling| | <p>Calculate latent heat Q= ML , using amount of water you want to freeze . Divide by time you want to want the cannon to be made. You Get cooling capacity . Add losses to this.</p>
<p>Make trials of TEC from calculator below.
<a href="https://viveksilwal.wordpress.com/2015/04/15/28/" rel="nofollow">https://viveksilwal.wordpress.com/2015/04/15/28/</a></p>
<p>Once this is done, Install a software Ansys Icepak, you can learn it in a week, there you can simulate TEC easily. Analyse your system there.</p>
<p>Although thermoelectric is very less efficient , but I think a good heat sink and tec combination can make you freezed balls.</p>
| 2355 | How do I spec out a thermoelectric cooler? |
2015-04-08T21:53:54.640 | <p>In some cars, I have noticed the rear wheels are located at the extreme rear of the vehicle. However, I have noticed that the rear wheels of buses are always located about 1/4th of the way forward from the rear. What is the reason for this?</p>
| |mechanical-engineering|automotive-engineering| | <p>Most busses in the U.S. use a rear engine/rear drive layout, with the engine located behind the rear axle for ease of maintenance. There is no drive shaft per se, because the engine/transmission/rear axle are integrated.</p>
| 2363 | Why are rear wheels not placed at the extreme rear of a bus? |
2015-04-09T17:45:20.373 | <p>Most people have had the experience of moving the tab of an <a href="https://en.wikipedia.org/wiki/Beverage_can#Stay-on-tab" rel="nofollow noreferrer">aluminum can</a> back and forth until it breaks off. It usually only takes a few complete back and forth motions before the tab breaks off.</p>
<p><img src="https://i.stack.imgur.com/6PljV.jpg" alt="Aluminum can with stay tab"></p>
<p><strong>What is the root cause of the tab breaking off?</strong></p>
<p>The possible causes seem to be:</p>
<ul>
<li>A fatigue fracture.</li>
<li>An overstressing of the metal.</li>
<li>A result of plastic deformation.</li>
</ul>
<p>But which one is it?</p>
| |mechanical-engineering|materials|mechanical-failure|fatigue| | <p>Almost everyone is partly right. You can fail the ring-pull by simple overload in one 'cycle' or you can accumulate plastic strain in three or four cycles. This wouldn't normally be considered even low-cycle fatigue but I don't know that there's a lower limit on how many cycles Paris' Law can be applied.</p>
| 2382 | Why do beverage can tabs break when bent? |
2015-04-09T17:46:10.807 | <p>I was thinking about this <a href="https://www.kickstarter.com/projects/voltera/voltera-your-circuit-board-prototyping-machine/description" rel="nofollow">kickstarter project</a> where they print circuit boards, and later they show it printing solder paste. I've seen other liquid or paste dispensing systems like this too, maybe for medical or chemical uses.</p>
<p>My question is how do you make a system that dispenses a known amount of paste like that? I get that I can run a motor and put pressure on top of a syringe and goo will come out of it. Maybe a stepper with some gearing? </p>
<p>Is there a feedback mechanism? It seems too small to have some kind of flow meter there. Are these kinds of things just open loop and you calculate the volumes ahead of time and just crank down the motor accordingly?</p>
<p>You must have to get it to some baseline first though, like a home position where paste is at the bottom of the nozzle.</p>
<p>Just curious how these things work.</p>
| |mechanical-engineering|fluid-mechanics| | <p>I do this type of things quite a lot at work, and yes they are typically syringe-actuated device, where you either apply a pneumatic pressure inside the syringe barrel for a certain amount of time (such as this <a href="http://www.loctite.co.uk/loctite-4087.htm?nodeid=8802622210049">glue dispensing machine</a>) or move the piston inside the syringe by a certain amount at a certain speed, using some kind of electrical drive (stepper motor + gearing) and a lead screw type mechanism (like this <a href="http://www.syringepump.com/NE-1000.php">syringe pump</a>).</p>
<p>Typically, there is no feedback mechanism, they operate open-loop, but are calibrated beforehand, by dispensing onto a balance and using gravimetry to work out the volume dispensed per unit of movement of the plunger.</p>
<p>As you say, you have to make sure, the system is primed before use so you would typically dispense a little bit to waste before using it for its designed purpose.</p>
<p>The challenge with viscous liquids (and therefore pastes and assorted goos), is you have to control the speed of the plunger quite carefully, otherwise you may over-dispense, and you may cavitate the pump if you refill too quickly. You would typically dispense and refill quite slowly in these cases.</p>
| 2383 | controlled depositing of goo with a syringe |
2015-04-10T13:47:48.143 | <p>I've been studying the design of turbopumps and their impellers, all the design equations and what-not; this question concerns the contours of the impeller vanes. I'm going to attempt to enter aerospace engineering in the future, so I figured I might as well start getting exposure to the concepts now.</p>
<p>I'm currently attempting to design an impeller in CAD to test my knowledge of the equations as well as get better at CAD, but one of the facets of the design rather baffles me.</p>
<p>As you can see here (<a href="http://my.fit.edu/~dkirk/4262/Liquid%20Rocket%20Engine%20Design.pdf" rel="noreferrer">source</a>, page 208):</p>
<p><img src="https://i.stack.imgur.com/NLDxC.png" alt="Impeller radial contour"></p>
<p>The calculation of the radial contour of the impeller (that is, concerning $\beta_1$ and $\beta_2$ should be relatively easy given the other content in the paper, but radial impellers in turbochargers and other such equipment almost always appear to have a "lip" or forward facing top edge near the eye of the volute or casing inlet.</p>
<p>Here is an example of what I'm referring to:
<img src="https://i.stack.imgur.com/iCYFO.jpg" alt="CAD-modeled impeller"> </p>
<p>You can tell that the vanes are backwards-curved, but near the top of the impeller vane closer to the eye, the curvature reverses and flares forward to approach near-flat slope.</p>
<p>My questions are:</p>
<ul>
<li><p>What is the purpose of this axial contour (forgive me if that isn't the proper term)?</p></li>
<li><p>Are there specific design parameters associated with that contour? The Rocketdyne text above didn't seem to make any mention of the axial profiles of the impeller, other than the diagram to the left describing the dimensioning of the shrouds.</p></li>
<li><p>Where can I find an explanation and source of design information for how to design the contour? </p></li>
</ul>
| |mechanical-engineering|fluid-dynamics| | <p>I should have a little more overview than I have of pumps but I don't. What you refer to seems to be called an <strong>entrance vane</strong> or an <strong>impeller guide vane</strong>. The purpose is to help the water gain the required rotational velocity:</p>
<blockquote>
<p>... in connection with a centrifugal impeller it is desirable to provide an <strong>axially extending entrance-vane section having vanes curved such that their radius of curvature gradually decreases in the direction of the fluid flow there through</strong>. <a href="http://www.google.com/patents/US2384265" rel="nofollow noreferrer">Source</a> [Patent, 1945]</p>
</blockquote>
<p>The water approaches the impeller with a flow direction along the axis. Without curving it inwards, the water would have a very abrupt change from going straight to moving with the blade, with a considerable waste of energy.</p>
<p>However, by curving the blades inwards, you 'scoop' the water up. The water will be pushed along the the slope of the vain and will help turn the axial motion that it had to rotational, increasing the pump efficiency.</p>
<p>This design is not easy to machine and hence only used where necessary, like rocket engines where you need a good deal of pressure. Most applications are just not worth the trouble.</p>
<blockquote>
<p>With this construction, the entrance vane section comprises a hub portion having a plurality of vanes extending radially therefrom, with each of the vanes having a rather complicated twisted profile. Obviously, such an entrance vane section is difficult to cast and/or machine. Accordingly, it is an object of this invention to provide a novel construction of such an impeller to facilitate its fabrication.</p>
</blockquote>
| 2398 | Contours of radial-flow impeller vane |
2015-04-11T16:55:05.160 | <p>I want to pull some weight horizontally and for that i have settled for a linear actuator.</p>
<p><img src="https://i.stack.imgur.com/Sao5w.gif" alt="enter image description here"></p>
<p>The actuator should pull some weight as it moves back and forth as the gif shows. </p>
<p>There are linear actuators i could find but they are too bulky with heavy motors and a lot of lubrication and the noise they make is not music to my ears.</p>
<p>I found electrical actuators to be a perfect fit but i do not know how they work for instance this actuator: <a href="http://www.kollmorgen.com/en-us/products/linear-actuators/rodless-actuators/r4-series/" rel="nofollow noreferrer">R4 Series</a> </p>
| |mechanical-engineering|linear-motors| | <p>For a minimum of moving parts, you can get linear stepper motors. With the right kind of controller (microstepping), the motion can be very smooth and quiet.</p>
<p>For example, I have in the past worked with linear actuators from <a href="http://www.nipponpulse.com/" rel="nofollow">Nippon Pulse</a>, driven by a motor controller from <a href="http://www.technosoftmotion.com/en/" rel="nofollow">Technosoft</a>. Very precise and powerful, but also very expensive!</p>
| 2409 | How does a rodless electrical linear actuator work? |
2015-04-12T05:46:26.693 | <p>I understand how a circle (or sphere if you want 3D) is the best shape for holding a vacuum inside a container, but what if you wanted to have a large positive pressure <em>inside</em> of the container instead of outside? Would a circle/sphere still be the best shape as far as material required to hold X amount of pressure? </p>
| |mechanical-engineering|pressure|geometry| | <p>It's all about the fact that spherical shell would not experience bending if the load is evenly distributed over the surface (which is the case when pressure is the only load). The very idea of using shells is to create them of such shape that stresses due to bending would be low, and the load bearing capacity of the shell would be provided by the tensile forces (or by compression) arising in the shell (tensile/compression forces lead to much lower stresses than bending moments do). An interesting example devoted to load bearing capacity of shells is presented on the following webpage (it is devoted to so-called "shallow" shells, which have low curvature, i.e. they closely resemble plates, but can withstand much larger loads, exactly because compression in the material plays much bigger role than bending in this case):</p>
<p><a href="http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm" rel="nofollow">http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm</a></p>
| 2419 | Is a circle the strongest 2D shape for containing internal pressure? |
2015-04-12T05:59:16.773 | <p><strong>When we apply a vertical shear force Sy and the structure is symmetric about x axis, then why is it logical to have the position of shear center at the location of intersection of line of action of shear force and x axis?</strong></p>
<p>If we go by basic definition then the moment about the shear center due to shear forces should be <strong>0</strong>, so the line of action of external shear force is logical, but when we consider moment about x- axis, rather than getting cancelled, <strong>I <em>think</em> they get added</strong>.</p>
<p><strong>eg. This is the situation.</strong>
<img src="https://i.stack.imgur.com/s4xHw.png" alt="enter image description here"> </p>
<p><strong>Now look at the final solution depicting the shear flows</strong>
<img src="https://i.stack.imgur.com/1P7mO.png" alt="enter image description here"></p>
<p><strong><em>So as I suggested, instead of moments being equal and opposite, I think they exactly equal in magnitude and direction. So where am I getting a wrong interpretation?</em></strong></p>
<p><strong>EDIT:-</strong> I tried to keep the question general and explained accordingly, to be specific the question is as follows:-</p>
<p><em>The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination
of direct stress carrying booms and shear stress only carrying walls. If the
section supports a vertical shear load of 10 kN acting in a vertical plane through booms
3 and 6, calculate the distribution of shear flow around the section.</em></p>
<p><em>Boom areas: B1 =B8 =200mm2, B2 =B7 =250mm2, B3 =B6 =400mm2, B4 =
B5 =100mm2.</em></p>
<p>x is horizontal and y is vertical</p>
<p><strong>It is a closed section beam. An idealized version of an airplane wing</strong></p>
| |mechanical-engineering|structural-engineering|aerospace-engineering|structures| | <blockquote>
<p>why is it logical to have the position of shear center at the location
of intersection of line of action of shear force and x axis?</p>
</blockquote>
<p>That statement isn't logical. I think you have misunderstood how the shear centre is defined. </p>
<p>The shear centre is the point such that an applied force passing through the S.C. does not cause any <strong>rotation</strong> of the section. In other words, if you apply a shear force through the shear centre, it does not cause any <strong>torsion</strong> in the beam but only <strong>bending</strong>. </p>
<p>The position of the S.C. depends only on the <strong>geometry</strong> of the beam section, not on the applied loads.</p>
<p>You can apply a shear force at any point on the beam. If the force does not pass through the S.C, you need to replace it by an equal shear force through the S.C, plus a moment about the S.C. You can then find the deflections and stresses due to bending (caused by the force through the S.C.) and torsion (caused by the moment about the S.C.) separately, and add them together to get the total deflections and stresses.</p>
<p>In your aircraft wing example, the shear forces caused by the aerodynamic lift and drag will act through the <strong>centre of pressure</strong> of the wing, and the centre of pressure is usually <strong>not</strong> the same point as the shear centre. The aerodynamic forces will therefore cause a combination of bending and twisting in the wing. </p>
| 2420 | Logic behind location of shear centre |
2015-04-12T18:09:56.273 | <p>I have been searching for this answer for awhile. I've read numerous texts and even watched some lectures online, but often times this is never explained and just given. The viscous stress term in the Navier-Stokes equations looks like</p>
<p>\begin{equation}
\nabla \cdot \tau = \nabla \cdot \mu \left(\nabla\vec{u} + (\nabla\vec{u})^T\right)
\end{equation}</p>
<p>Now the term $\nabla \cdot \mu \nabla\vec{u}$ is easy enough to understand as it is just velocity diffusion, but I have a hard time coming up with a physical interpretation of the term $\nabla \cdot \mu (\nabla\vec{u})^T$. After I expanded this term I ended up with</p>
<p>\begin{equation}
\nabla \cdot \mu (\nabla\vec{u})^T =
\begin{pmatrix}
\frac{\partial}{\partial x} \nabla \cdot \vec{u} \\
\frac{\partial}{\partial y} \nabla \cdot \vec{u} \\
\frac{\partial}{\partial z} \nabla \cdot \vec{u}
\end{pmatrix}
\end{equation}</p>
<p>which seems to imply that this effect is not present in a divergence-free velocity field, but I still can't come up with or find any physical intuition about what this term actually means. Does anyone understand what this term physically represents?</p>
| |mechanical-engineering|fluid-dynamics|fluid-mechanics| | <p>I agree with @sturgman one should not look at individual parts but try to understand it in ints context.</p>
<p>Looking at the very basic version of the Navier-Stokes-Equation (using <a href="http://en.wikipedia.org/wiki/Einstein_notation" rel="nofollow">Einstein-Notation</a>):</p>
<p>$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \rho k_i + \frac{\partial}{\partial x_i} \left( -p + \lambda^*\frac{\partial u_k}{\partial x_k}\right) + \underbrace{\frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right)}_{\nabla \,\cdot\, (\eta \, \left[ (\nabla\vec{u})+(\nabla\vec{u})^\mathrm{T}\right])} $$</p>
<p>The underbraced part in its original can be rewritten.</p>
<p>$$ \frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right) = \eta \left( \frac{\partial^2u_i}{\partial x_j \partial x_j} + \frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial{x_k}} \right]\right)$$</p>
<p>Which leads to:</p>
<p>$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \underbrace{\rho k_i}_{\text{I}} - \underbrace{\frac{\partial p}{\partial x_i}}_{\text{II}} +\underbrace{(\lambda^* + \eta)\frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial x_k}\right]}_{\text{III}} + \underbrace{\eta \left[ \frac{\partial^2u_i}{\partial x_j \partial x_j}\right]}_{\text{IV}}$$</p>
<p>In symbolic notation this should look like:</p>
<p>$$\rho\frac{\mathrm{D}\vec{u}}{\mathrm{D}t} = \rho\vec{k} - \nabla p + (\lambda^* + \eta)\nabla(\nabla\cdot\vec{u}) + \eta\nabla\cdot\nabla\vec{u}$$</p>
<p>Part $\text{III}$ is not always shown like this depending on the way the Newtonian stress tensor was introduced. Since $\lambda^*$ is a fluid property which is very hard to measure but varies only little, <em>Stokes Hypothesis</em> sets it to $-2/3 \eta$ (Which is technically only true for monoatomic gases).</p>
<p>Part $\text{III}$ describes a feature of a fluid where the atomic structure of the fluid-molecule can absorb energy, it is sometimes referred to as pressure-viscosity. Whereas Part $\text{IV}$ describes the resistance of the flow when sheared, part $\text{III}$ describes the resistance of a fluid-volume when it is "isobarically" expanded or compressed.</p>
| 2423 | What is the physical interpretation of the second term in the viscous stress tensor in the Navier-Stokes equations? |
2015-04-13T02:50:23.227 | <p>Consider a balloon, it is a flexible, expanding pressure vessel made of synthetic rubber. I am trying to design a system that will use hemispheric pockets of neoprene to grip an object, much like the one found in this journal:</p>
<blockquote>
<p>H. Choi and M. Koç, 'Design and feasibility tests of a flexible gripper based on inflatable rubber pockets', International Journal of Machine Tools and Manufacture, vol. 46, no. 12-13, pp. 1350-1361, 2006. (Behind a paywall, abstract available <a href="http://libra.msra.cn/Publication/6847499/design-and-feasibility-tests-of-a-flexible-gripper-based-on-inflatable-rubber-pockets" rel="nofollow">here</a>)</p>
</blockquote>
<p>Although I am not entirely certain where to start, I am dealing with a high pressure (8 bar) and need to work on a small scale (max radius of 40mm).</p>
<p>Values I have recieved from equations found online suggests thickness values in meters and radius expansion reaching the hundrends of meters.</p>
<p>What equations do I need to calculate a solution?</p>
<p>Are neoprene, latex and chloroprene viable materials to use or should I find alternatives?</p>
<p>Is a hemisphere that best pocket shape I can use?</p>
<p>Additionally, how can I calculate the amount of expansion?</p>
| |pressure|thermodynamics|compressed-air| | <p>This is fully doable. This is the in the higher range as bicycle tires (although with the outer tire as well and not the inner tube alone).</p>
<p>Alright, here goes:</p>
<blockquote>
<p>What equations do I need to calculate a solution?</p>
</blockquote>
<p>The pressure in a tube can can be modelled by the following equations:</p>
<p><em>Hoop Force:</em> </p>
<p>$$\sigma_H = \frac{pr}{t}$$</p>
<p><em>Axial Force</em></p>
<p>$$\sigma_A = \frac{pr}{2t}$$</p>
<blockquote>
<p>Is a hemisphere that best pocket shape I can use?</p>
</blockquote>
<p>That is a design question, although when inflated it should ideally be cylindrical to prevent extra stress. This is a bike tyre picture from Wikipedia:</p>
<p><img src="https://i.stack.imgur.com/I9Nsj.png" alt="enter image description here"></p>
<blockquote>
<p>Are neoprene, latex and chloroprene viable materials to use or should I find alternatives?</p>
</blockquote>
<p>Neoprene I imagine will leak air, although slowely. Latex and chloroprene should work. I would suggest that you get bike parts to do this if you are trying to build a prototype.</p>
<blockquote>
<p>Additionally, how can I calculate the amount of expansion?</p>
</blockquote>
<p>Your system is a little dynamic here so it will get bigger as you fill it with pressure.</p>
<ul>
<li><p>Rewrite the equations above with $\epsilon E $ to replace $\sigma$.</p></li>
<li><p>With a constant pressure and E-modulus you should be able to get the final strain.</p></li>
</ul>
| 2427 | How to design a hemispheric flexible pressure vessel? |
2015-04-13T06:33:37.157 | <p>I am trying to find some information regarding the design of plastic hinges that can move freely but also hold a position in any degree, like in this little camera: <a href="https://www.kickstarter.com/projects/podolabs/podo-the-first-stick-and-shoot-camera">https://www.kickstarter.com/projects/podolabs/podo-the-first-stick-and-shoot-camera</a></p>
<p>I want to know the considerations of design and if there's any kind of reference.</p>
| |mechanical-engineering|plastic| | <p>Here is an idea for you if you can't find an off-the-shelf product that fits your need. There are plastic coolant hoses that can be bent and stay in any position desired to direct the coolant flow for machine tool applications. The hoses are built from articulated plastic ball and socket joints. The hoses have the detent property due to the tight fit(friction) of the ball inside the socket. So just make your own hinge using two of these hose sections as pivot points and attach a plastic plate between the two pivots. It will be a little "bulky" but if you have no size constraints it won't matter. You will have a nice hinge that will stay where you position it, just like the coolant hoses themselves.</p>
| 2429 | Information regarding plastic hinges that can hold the position |
2015-04-13T13:19:19.540 | <p>I have to draw two custom pipes, 1 1/4" & 2", with on one end male tapered pipe thread and on the other end parallel pipe thread. The connections are Parallel female thread with a tapered male thread, with Teflon tape as sealant.</p>
<p>I'm familiar with the metric tread, but the British/ Whitworth thread system is new for me. After a search on the internet and the standards: DIN 3858 / DIN ISO 228, i have the following questions:</p>
<ul>
<li>According to DIN 3858, tapered thread largest available size is 1 1/2". Why are there couplings on the market that are larger? In my case 2".
<ul>
<li>Is the designation of "Pipe thread DIN 3858 - R 2" correct?</li>
</ul></li>
<li>What is the minimal thickness of the pipes when it concerns tapered thread? Are there guidelines for this?
<ul>
<li>For the 2" connection i selected a OD = 60.3 mm, t = 2 mm. (probably too thin)</li>
<li>For the 1 1/4" connection i selected a OD 42.4 mm, t = 2.6 mm.</li>
</ul></li>
</ul>
| |mechanical-engineering|threads| | <p>After some discussion with an experienced college and a continued search, I'am able to answer my own question.</p>
<p>DIN 3858 is valid for type C stud ends as in DIN 3852-2 and type Z tapped holes with parallel internal thread as in DIN 3852-2. Example Type C stud <a href="http://mdmetric.com/4100/Chapter%20D%20UK.pdf" rel="nofollow">-click-, page 19</a>.</p>
<p>In my case i'm looking for pipe threads and not thread on a stud. For this case EN 10226-1:2004 is valid. Also a references was available in DIN 3858. This guidance show the full pipe thread range until 6". This explains directly the bigger available sizes.</p>
<p>The wall thickness questions can be answered with some common sense. The wall thickness must be larger than the complete thread heigth incl the height created by the taper. Say roughly 2 time thread thickness.
From the steel pipe catalog, i even found some process pipes specifically for 2" & 1 1/4". 60.3 x 3.65 & 48.3 x 3.25. These fit with properly with the thread and my selection was wrong in conclusion. These were thin wall pipe suitable for compression couplings.</p>
<p>It took me some time, but some good lessons are learned for future work.</p>
| 2433 | British tapered pipe thread; Wall thickness? & Thread designation tpd? |
2015-04-13T14:49:51.510 | <p>Are there any industry standards (<a href="http://en.wikipedia.org/wiki/Specification_(technical_standard)" rel="nofollow noreferrer">Engineering Specifications</a>) for the minimum / maximum space required for the electrical box that holds a regular AC wall plug?</p>
<p><img src="https://i.stack.imgur.com/dr0USs.jpg" alt="Type B grounded duplex outlet"></p>
<ul>
<li>The problem I am trying to solve is quite simple - are / is there any local (USA, Canada, or EU) industry standards that require, or set, dimensions in the Height, Width, and Depth (in inches or centimeters) of the space required to be used by the electrical box for an AC wall plug as shown below:<br>
<img src="https://i.stack.imgur.com/0SVhHm.jpg" alt="enter image description here"></li>
</ul>
| |electrical-engineering|standards|specifications| | <p>The normative reference for the design of wall boxes for electrical work in the US is NEMA OS-1 for metal boxes and NEMA OS-2 for nonmetallic boxes. This, in turn limits the size that components to go into these boxes (including outlets) may be.</p>
<p>If you don't want to buy the standard, you can find a good bit of information from it posted from various manufacturers, <a href="http://www.emersonindustrial.com/en-US/documentcenter/EGSElectricalGroup/products_documents/commercial_products/commercial_box/switch_outlet_box_covers/switch_outlet_box_covers/AEC_MC007_Switch_and_Outlet_Boxes_and_Covers.pdf" rel="nofollow">here</a> is one example.</p>
<p>Page A3 of <a href="http://www.hubbell-rtb.com/literature_pdf/Hubbell-RTB_Catalog.pdf" rel="nofollow">this document</a> explains the limitations on how full the boxes can be, as dictated by NEC 314.16.</p>
<p>As you can see, there are many different sizes and configurations of these boxes, but they have fairly standard front-facing features so that various outlets, switches, or other devices can attach uniformly. When you buy an industrial outlet, the cut sheet will usually tell you how deep of a box it requires, since that's the most variable dimension between boxes.</p>
| 2435 | Industry standard for AC wall electrical box space |
2015-04-13T16:57:14.100 | <p>Many states require that a PE applicant have some number of years of "progressive engineering experience". Are there details as to what qualifies? For example, if an engineer is working in sales and not design, I would tend to think that would not count. But that's just my impression. Are there defined rules? General guidelines or standards of practice?</p>
| |licensure| | <p>The state of California has some specific guidance as to what is considered experience. (<a href="http://www.bpelsg.ca.gov/applicants/faq_eng.pdf" rel="nofollow noreferrer">California FAQ</a>)</p>
<blockquote>
<p>Qualifying engineering work experience is that experience in the appropriate branch of engineering which has been gained while performing professional level engineering tasks under the direction of a person authorized to practice in the branch of engineering in which the applicant is seeking licensure. There is no limit to the amount of such qualifying experience which will be accepted by the Board, provided that the experience meets the other requirements indicated herein. Applied engineering research is considered to be an engineering task, which may constitute qualifying experience.</p>
<p>Work in management, proposal writing, contract administration, estimating, sales, and other peripheral areas, however, is presumed to contain little or no element of qualifying experience, and therefore an applicant must provide a detailed explanation of what portions of such work are actually qualifying and why the Board's presumption is not correct, if the applicant expects to obtain any credit for this type of work. Such peripheral experience will then be evaluated on a partial credit basis as applicable to each applicant's particular situation. Thus, the actual credit allowed may range from near zero to a substantial amount.</p>
<p>Subprofessional work such as work normally performed by a drafter or a technician is not qualifying. Nor is construction inspection qualifying. However, work as a field engineer may be qualifying.</p>
</blockquote>
| 2437 | What constitutes progressive engineering experience? |
2015-04-13T16:27:56.593 | <p>Given the information on swords <a href="http://quantumchymist.blogspot.co.uk/2014/02/why-arent-more-blades-made-of-titanium.html?m=1" rel="nofollow">here</a>:</p>
<blockquote>
<p>When you need to place a great deal of strength into a thin blade, steel remains the best all around choice. Titanium blades will be simultaneously weaker and softer than similar steel blades.</p>
</blockquote>
<p>And the material (LDHEA) described <a href="http://www.tandfonline.com/doi/full/10.1080/21663831.2014.985855#tabModule" rel="nofollow">here</a>:</p>
<p>Summarized here from the Applications section of the Wikipedia article on Scandium, as:</p>
<blockquote>
<p>An alloy called Al20Li20Mg10Sc20Ti30 has been shown to be as strong as titanium, light as aluminium, and hard as ceramic.</p>
</blockquote>
<p><strong>In combination with other materials, how could LDHEA be used to make a good shortsword?</strong></p>
<p>Perhaps steel for the flat and LDHEA for the edge and point?</p>
| |materials|metals| | <p>As a HEMA practitioner and a physicist, I can tell you that different swords need different properties and it is unlikely one material will be the end-all for a sword material. Depending on the sword, you may actually want different parts of the blade to have different properties. <a href="https://www.google.com/search?q=rapier&oq=rapier&aqs=chrome..69i57j0l5.795j0j7&sourceid=chrome&es_sm=91&ie=UTF-8&safe=high">Rapiers</a>, for instance, are much less rigid than hand-and-a-half swords (or "<a href="http://en.wikipedia.org/wiki/Longsword">longsword</a>").</p>
<p>As I said earlier, different properties are desired for different parts of the blade. For instance, a high hardness on your edge is great, because you can deliver more cuts before it dulls. However, high hardness in your flat does not really help you, because it usually means brittleness, and your blade <em>will</em> come into contact with opponents' blades and potentially break or shatter, leaving you more-or-less defenseless.</p>
<p>Lighter blades are not always better. Sure, a lighter blade can <em>ideally</em> be moved faster, but the human body can only move so fast and benefits only so much from it. Cutting swords, however, need a certain degree of heft to them to enable their cuts. Trying to cut someone with a rapier, for instance, is usually laughable because it simply lacks the same momentum as a tulwar, katana, longsword, or any other sword. (Rapiers have a tendency to simply bounce off in such attempts to cut.) Also, some weight to your sword makes it harder to deflect, which is another problem rapiers classically encounter when attempting to cut.</p>
<p>It seems that the ideal blade, without regard to form, is made of a layering of both hard and soft metals, as seen in <a href="http://en.wikipedia.org/wiki/Wootz_steel">wootz</a> or <a href="http://en.wikipedia.org/wiki/Damascus_steel">damascus</a> steel. Many blacksmiths, when manipulating material properties for a sword, attempt to get a hard edge and soft flat. This material seems promising, but we need to know more about it. </p>
<p>How brittle is it? Dr. Carl Koch, the senior author for that paper, <a href="http://nextbigfuture.com/2014/12/new-alloy-is-as-light-as-aluminum-as.html">says</a> "we think it’s tougher – less brittle – than ceramics." That bodes well for it as a sword material. At the very least, it would make a good material for the edge and points of swords. As noted earlier, the flexibility of this material may eliminate it for certain swords or recommend it for others.</p>
<p>A lighter material lets your sword be larger, which means you could (in theory) have swords that were formerly limited to <a href="http://images2.fanpop.com/image/photos/8700000/The-Same-Sword-Brothers-inuyasha-8711240-448-243.jpg">anime and video games</a>. You also run into the problem of simply larger size not being what you want. For instance, a <a href="http://en.wikipedia.org/wiki/Zweih%C3%A4nder">two-handed</a> sword is not used in the same way as a <em><a href="http://en.wikipedia.org/wiki/Messer_(weapon)">Messer</a></em>.</p>
<p><strong>In review</strong>, it seems like this material is promising, and <em>could be good</em>, but hardness and strength-to-weight are not the only properties which make a good sword material. Toughness (brittleness) is an important factor which we do not know.</p>
| 2440 | How do I use this " Low-Density, High-Hardness, High-entropy Alloy" to make a good shortsword? |
2015-04-13T21:21:38.260 | <p>Here is <a href="http://s.hswstatic.com/gif/gear-worm.jpg" rel="noreferrer">an image of a worm gear</a>. What is english word for mechanical element (two of them will be necessary for a worm gear) in which the worm shaft will "lie" and rotate?</p>
<p>Here is one photo of the element where it is part of a housing; I need it as separate mountable element (I will mount it on plate):</p>
<p><img src="https://i.stack.imgur.com/Htdevm.jpg" alt="unknown element in worm gear housing"></p>
| |mechanical-engineering|gears| | <p>Not only does the shaft have a "bearing" surface but the ends of the shaft also need something to "bear" against due to the "thrust" of the shaft in either direction. So I would add the term "thrust bearing" to the mix some where in there ... this can be as simple as adding a single steel ball for a bearing surface to each end of the shaft to deal with the worm gear's lateral movement unless you have dealt with this in some other way. </p>
| 2447 | What is the mechanical element that holds a worm gear shaft in place called? |
2015-04-15T13:33:17.973 | <p>Is there a method for plating steel with stainless steel? </p>
<p>If so, is it chemical, electrical, or electrochemical? </p>
<p>I did a quick search on the internet but was unable to find a service. I'm interested in applying a food safe finish to something that would otherwise be cost prohibitive to make out of solid stainless steel.</p>
| |materials|steel| | <p>Someone mentioned allclad plating with stainless steel. That is not quite correct. They take a pair of thin stainless steel pots, then position them to have a 1/8 inch gap, and pour aluminum into the void. Since the melting point of aluminum is so low compared to stainless steel, this works quite well, and the stainless pots do not deform.</p>
| 2478 | Is stainless steel plating possible? |
2015-04-16T06:48:03.313 | <p>I have an aluminum square tube (diameter 1/2") that I would like to attach to a flat aluminum surface which has a circular hole of smaller diameter in it for water to flow through at low pressure. Would welding a flange to the square tube (with screw holes for going into the surface) be the best option, or are there better ways for making this kind of attachment? This is not a high pressure system so I don't think there will be any complications there.</p>
| |mechanical-engineering| | <p>If it needs to be removable, then yes, attaching a flange will be the best choice. If so, you'll need to put some sort of seal between the flange and the existing aluminum surface. The faying surfaces of aluminum will not, on their own, form a good seal. There are many types of seals, but an elastomeric gasket or O-ring will probably be your best bet. If you use an O-ring, it is best to put a groove in one or both of the mating parts so it is properly supported.</p>
<p>As for making the watertight aluminum-to-aluminum connection itself, brazing will probably be your best bet. Soldering is possible, but won't be as strong or reliable. Welding is possible, although a 1/2" tube is likely to have a thin wall, so you'd need a skilled TIG welder to make the joint both strong and watertight. Brazing is the best compromise because it will be reasonably strong, and doesn't take too much skill to make watertight. There will also be less of a chance of melting through the wall or collapsing the tube.</p>
<p>One thing that can make this kind of joint easier is running the tube through the flange, so it sticks out maybe 1/4" from the other side of the flange. This reduces the risk of deforming the tube as you heat it, and it's easy to grind or machine the extra tube off when you're done welding/brazing. If you can keep the gap reasonably small between the tube and the thickness of the flange, you'll also get some help from capillary action to make your weld stronger and more watertight.</p>
<p>Do keep in mind that some alloys of aluminum are significantly effected by heat, and heat as low as 400 degrees Fahrenheit can soften the aluminum permanently. If strength or hardness is important to you, you'll want to pay attention to what alloys you're using.</p>
| 2494 | Best way to attach a square tube to a flat surface? |
2015-04-16T07:57:00.897 | <p>Recently I observed installing new storm sewage along a city street. Pipes have plain surface (not corrugated) and are made of some plastic looking like HDPE (polyethylene) and their diameter is around 0,5 meters and the walls are around 40 millimeters thick. That's a lot of material. I assume the same structural strength could be achieved if the pipes were made corrugated and with much thinner walls. I saw corrugated plastic pipes for storm sewage once and so they surely exist.</p>
<p>Why would plain surface pipes be preferred over corrugated pipes for this scenario?</p>
| |civil-engineering|piping| | <p>You are correct in your assumption that corrugations give pipes strength and hence such pipes can have thinner walls that smooth walled pipes made of the same material when used for the same application.</p>
<p>Pipelines are made to transport fluids or slurries. When it comes to the flow of fluids or slurries in a pipe, one of the import considerations is the smoothness, or roughness, of the internal wall of the pipes. Smooth walled pipes, or as you called them “plain pipes” have a small surface roughness on their internal walls. The corrugations of corrugated pipes are a very large roughness. This will increase the resistance to flow within the pipes and increase the turbulence of flow in the pipe. Increased turbulent flow, particularly of slurries, can lead to increased wear of pipes. The smoother the flow in pipes the better.</p>
<p>Sometimes, the flow of sewage in pipes is not a continuous steady stream. As the spasmodic flow subsides the sewage fluid and solids will collect in the valleys of the corrugations. If the corrugated pipes are made of steel, this will lead to a higher probability of the pipes corroding more quickly in these locations. The other result will be the presence of decomposing sewage and fluid trapped in the corrugations which will lead to the creation of a habitat for harmful bacteria and the production of noxious gases. The result of which would be an increased risk to public health. To prevent this, such pipes will need periodic flushing.</p>
<p>With smooth walled pipes there will be less accumulation of sewage based items in the pipe as it will tend to form a narrower line along the center of the base of the pipe. Corrugations run transverse to the pipe, allowing for more material to collect in bottom of the pipe.</p>
<p>The wall thickness of any pipe will be determined by the strength properties of the material from which the pipe is made and the pressures that the pipe must withstand; both internal and external pressure. The deeper a pipe is buried the greater the soil/ground pressure that it must withstand. HDPE is a weak material, a wall thickness of 40 mm for a 500 mm diameter pipe means it will have to withstand high pressures but it also helps the pipe keep its shape.</p>
| 2496 | Why prefer plain pipes over corrugated pipes for storm sewage and culverts? |
2015-04-16T12:26:33.823 | <p>I have a 8-pole synchronous machine (as a generator), which is rated to produce 1000 kVA. Its synchronous reactance is 0.4 Ohms.
It is connected to a diesel engine with the grid frequency 60 Hz.
The load is resistive type, 500 kW, 480 V.
I need to calculate the speed, power factor and power angle.</p>
<p>I know the following:</p>
<ul>
<li>Speed can be calculated by: $n = 120 \cdot \frac{f}P$ , where $f$ is frequency (60 Hz) and $P$ are poles (8) ... and we get the result 900 rpm.</li>
<li>Power factor can be calculated by: $\mathrm{PF} = \cos(\mathrm{PA})$, where $\mathrm{PF}$ is power factor, $\mathrm{PA}$ is power factor angle (is that the same as power angle?)</li>
</ul>
<p>So, how can I calculate power factor and power angle?</p>
| |mechanical-engineering|electrical-engineering| | <p>If the load is purely resistive, the power factor is 1 and the power angle is zero by definition.</p>
| 2500 | Synchronous machine power factor and angle |
2015-04-16T23:36:50.343 | <p><a href="http://en.wikipedia.org/wiki/Chemical_oxygen_generator">Chemical oxygen generation (COG)</a> is used in commercial airliners to supply emergency oxygen to passengers and crew if the plane is depressurized. Weight for weight it produces far more oxygen than compressed oxygen tanks.</p>
<p>I read the original paper on COG, several sites from oxygen candle manufacturers and the US Federal Aviation Administration document on oxygen equipment (riveting I assure you). However, most of the rest of my knowledge is from my own background as a physicist; I want to know how these methods compare from an engineering perspective. </p>
<p>Besides reduced weight, are there any other advantages of COG in aircraft? Does it have added safety benefits when compared to high pressure gaseous oxygen storage? The oxygen candle burns extremely hot when it goes off but I'd imagine high pressure gas canisters could be far more dangerous and prone to exploding if damaged.</p>
| |aerospace-engineering|aircraft-design|chemical-engineering|compressed-gases| | <p>One advantage these chemical systems may have over compressed tanks is how well they are suited for intermittent use and long-term storage (as opposed to regular, continuous operation). The <a href="https://www.faa.gov/regulations_policies/handbooks_manuals/aircraft/amt_airframe_handbook/" rel="nofollow">US FAA Aviation Maintenance Technician Handbook</a> points out this advantage <a href="https://www.faa.gov/regulations_policies/handbooks_manuals/aircraft/amt_airframe_handbook/media/ama_Ch16.pdf" rel="nofollow">on page 16-5</a>:</p>
<blockquote>
<p>Sodium chlorate chemical oxygen generators also have a long shelf
life, making them perfect as a standby form of oxygen. They are inert
below 400 °F and can remain stored with little maintenance or
inspection until needed, or until their expiration date is reached.</p>
</blockquote>
<p>Compressed gas cylinders, on the other hand, require regular maintenance and testing whether in use or in storage. Pressure vessels are prone to metal fatigue and corrosion over extended periods. Exposure to extreme temperatures can accelerate fatigue or cause an already-weakened cylinder to fail; COG systems, in comparison, are somewhat less vulnerable to temperature extremes.</p>
<p>As part of an emergency system, an oxygen cylinder would experience infrequent use and mechanical fatigue would be less of a concern than corrosion.* Corrosion is driven mainly by the reactivity and partial pressure of the stored gas and by the presence of moisture in the cylinder. Proper selection of fittings such as valves and regulators is critical to minimize the potential for a galvanic reaction between the fitting and the cylinder. </p>
<p>Keep in mind that pressure vessels support a <em>difference</em> of pressure between the inside and the outside of the vessel. During a decompression event, the outside pressure drops rapidly while the inside pressure stays constant, leading to a rapid increase in stress on the vessel. This has two implications: first, the capacity of the tank must be reduced to account for the lower atmospheric pressure at operating altitude; second, a worn tank is most likely to explode just when you need it the most—not at all a desirable quality in a safety device. (COG systems do not, to my knowledge, explode, though <a href="http://lessonslearned.faa.gov/ll_main.cfm?TabID=4&LLID=10&LLTypeID=2" rel="nofollow">they can start fires when handled improperly</a>.)</p>
<p>This leads to strict inspection and maintenance requirements for compressed gas cylinders. They have to be rotated in and out of service every so often to go through requalification testing and potentially reconditioning or condemnation. For some idea of how complicated these requirements are, take a look at <a href="http://www.aviationpros.com/article/10387217/high-pressure-gas-cylinders-maintenance-requirements" rel="nofollow">this 2003 Aircraft Maintenance Technology article</a>.</p>
<p>All of this, of course, represents a cost to the operator—service contracts, training, etc. In <a href="http://www.slate.com/articles/business/moneybox/2014/12/cheap_airlines_why_americans_will_suffer_worse_service_on_flights_in_order.html" rel="nofollow">an industry that competes mainly on ticket prices</a>, cost is king, and reduced maintenance costs would be an attractive selling point.</p>
<p>It's also nice to be able to relate the safety system cost directly to the occasions when you need to deploy the system. Going back to the initial point about suitability for intermittent, emergency applications: You're paying maintenance costs for that cylinder whether there's a depressurization or not. The cost of the oxygen itself is probably minor in comparison. So reducing the frequency of incidents by an order of magnitude may not change the cost very much. On the other hand, only paying for the COG system when you use it means the airline can reduce its costs by reducing depressurization incidents. This isn't so much an engineering concern, and I am speculating a bit here; the point is, businesses often don't make decisions in the same way that engineers do, and that's something to keep in mind when comparing design alternatives.</p>
<hr>
<p>* You might think that pure oxygen is not a corrosive gas (perhaps because we breathe oxygen) but the definition of "corrosive" can vary a bit with context. For example, <a href="http://www.alspecialtygases.com/files/Design_and_Safety_Handbook_3001.5.pdf" rel="nofollow">the Air Liquide Design and Safety Handbook (p. 2)</a> uses this definition:</p>
<blockquote>
<h3>Corrosive Gases</h3>
<p>These are gases that corrode material or tissue on contact, or in the
presence of water. ... Due to the probability of irritation and damage
to the lungs, mucous membranes and eye tissues from contact, the
threshold limit values of the gas should be rigidly observed. Proper
protective clothing and equipment must be used to minimize exposure to
corrosive materials.</p>
</blockquote>
<p>The emphasis here is on personal safety and health impacts and in that context, oxygen isn't a corrosive gas. But in the context of storing a gas in a pressurized cylinder, the emphasis is on the potential for the stored gas to react chemically with the walls or components in a way that weakens them, and in that context, pure oxygen definitely counts as "corrosive."</p>
| 2507 | What are the advantages of chemical oxygen generation over compressed tank storage? |
2015-04-17T04:34:05.567 | <p>I am designing a system where a beam is acted on by a horizontal force. I intend to hold the beam in place with a set of screws, although I am uncertain how much force they will be able to withstand.</p>
<p>What method can I use to determine the holding force of screws?</p>
| |mechanical-engineering| | <p>Many cities have either their own code on allowable load for nails and bolts or have adopted a uniform building code. </p>
<p>In Los Angeles for example the building department has published these codes on their site, or give them to you as a hand out for free. </p>
<p>American Wood Council has many pages of guidelines and information on their site.<a href="http://www.awc.org/codes-standards/calculators-software/connectioncalc" rel="nofollow">their calculator</a> The allowable load for wood connections depends on many factors:<br>
wood structural rating, its humidity content, species, storage history as well as mechanical strength and type of fastener (screw, bolt, lag bolt, prefabricated fasteners and kits) and the mode of their failure( shear, bending moment, pull out, etc..) and alos the function of the connection, there are load factors for seismic loads or shear loads, etc. </p>
<p>For nails there are concerns as to the ductility of nail, its size and length, the distance between nails, the grain of wood and its angle and many more factors. All of these cases have been extensively tested and tabulated and are available from your City or the hardware store or the manufacturers site.<br>
A deputy structural engineer inspecting the construction site is responsible for checking the nailing and proper application of their use and will ask the builder to correct or remove not complying nails, fasteners or members and re-installing them correctly.<br>
You have to discuss your case with an engineer in your building department. In most cases it is free consultation.</p>
| 2510 | How to calculate screw pullout strength? |
2015-04-17T09:11:43.427 | <p>Electronic cigarette elements that heat the fluid are usually termed "atomizers." They are usually just some nichrome wire or similar wrapped around a wick that becomes saturated with the nicotine fluid. </p>
<p>What is the process that "vaporizes" the fluid? Does the element just heat the fluid until it evaporates? Or something else, e.g., is the majority of fluid sort of spattered off the wick due to a smaller proportion of it violently evaporating? Or some other process/phenomenon? Is the term "atomizer" a scientific term in this context, or just something adopted for electronic cigarettes?</p>
| |thermodynamics|vaporization| | <p>Vapor is something that forms over any liquid and does not require the liquid to be above its boiling temperature. The atmosphere is a great example; it contains about <a href="https://www.google.com/search?q=how+muych+water+vapor+in+the+atmoshpere&oq=how+muych+water+vapor+in+the+atmoshpere&aqs=chrome..69i57j0.7306j0j7&sourceid=chrome&es_sm=93&ie=UTF-8#q=how+much+water+vapor+in+the+atmosphere&spell=1" rel="nofollow noreferrer">37 billion gallons</a> of water in the form of vapor, but the temperature of (most of) the atmosphere is well below the boiling point of water. The goal of a vaporizer is simply to increase the rate of this process. </p>
<p>To understand how it does this you need to know a little bit about <a href="http://en.wikipedia.org/wiki/Vapor_pressure" rel="nofollow noreferrer">vapor pressure</a>. A liquid at a given temperature will have some of its molecules which are moving fast enough to escape the surface of the liquid and end up in the vapor phase above the liquid. If you put the liquid in a sealed container, this process will continue until the <em>partial pressure</em> of the vapor reaches the vapor pressure of the liquid. What is important about this process is that it is only the pressure of the substance itself (i.e. partial pressure) which matters, even if the rest of the air above the substance is at a higher pressure than the vapor pressure, the liquid will still vaporize until its partial pressure reaches the vapor pressure. The vapor pressure increases with increasing temperature until it reaches the atmospheric pressure at the boiling point of the liquid. </p>
<p><a href="http://en.wikipedia.org/wiki/File:Vapor_pressure.svg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1QeuJ.png" alt="Vapor pressure illustration and plots"></a></p>
<p>With that background, we can now discuss how the vaporizer in an e-cigarette works. The amount of ambient vapor (in the surrounding air) of the nicotine solution in e-cigarettes is very low so turning it into vapor is simply a matter of letting it sit for long enough. The amount of time it would take is too long for a smoker to wait though so they use two tricks to speed up the process. 1) The fluid is heated so that the vapor pressure increases, and 2) a wick is used to drastically increase the surface area between which the air and fluid interact. </p>
<p>The heating of the fluid has the desirable side effect of making the cloud of vapor visible too. The best way to understand this is to think about a boiling pot of water. The water is converting to a gas at the surface and rising away from the pot, but you can't see water in its gaseous stage. You can, however, see steam because the water gas is mixing with the cool air above the pot and small droplets of water are condensing, much like a cloud. The same thing happens in an e-cigarette. Local heating in the vaporization area heats the air along with it and produces a vapor which you can't see. When this vapor mixes with the cool air as you inhale, it condenses to tiny droplets which give it the desirable appearance of smoke. So, when you blow out after puffing you are actually blowing out a little nicotine filled cloud. </p>
<p>As for the term <a href="http://en.wikipedia.org/wiki/Atomization" rel="nofollow noreferrer">atomizer</a>, it does have a semi-scientific definition. An atomizer is something which creates a fine suspension of liquid droplets in a gas. This is exactly what I described to you above; the vaporizer turns the liquid to vapor which then becomes a suspension as the vapor saturated air cools. </p>
| 2511 | Are electronic cigarette "atomizers" just evaporating the fluid? |
2015-04-17T15:08:22.973 | <p>The design of a retaining wall commonly involves determining the lateral earth pressure using either Rankine theory or Coulomb theory. Both theories involve mobilising the shear resistance of a triangular wedge of soil extending for a considerable distance away from the base of the wall.</p>
<p>In the case of a double-walled cofferdam, such as the one in the picture below, the short distance between the two walls would prevent such failure wedge from extending all the way down to the bottom. In which case, how does one go about determining the earth pressure from the sand fill material in between the two walls?</p>
<p><img src="https://i.stack.imgur.com/9Ddfe.jpg" alt="double-walled cofferdam"></p>
| |civil-engineering|geotechnical-engineering|soil| | <p>From what I read, you are looking at the pressure the sand between the sheet piling exerts on them. In this case, I see two possibilities: (1) <strong>log-spiral analysis</strong> or (2) elastic analysis of <strong>Boussinesq</strong>.</p>
<p><strong>Log Spiral Analysis</strong></p>
<p>The log spiral analysis assumes that soil pressure is mobilised by a soil mass that follows the shape of a log spiral curve. This is commonly used for braced trench excavations, and the curve of the mass must intersect the surface at the perpendicular. The analysis is non-determinate, so a trial and error graphical (scaled) method is recommended, but we have worked out a computer based algorithm that does this trial and error process computationally.</p>
<p>In this case though, in your trial and error analysis, you can consider that the curve must be forced to occur within the geometrical limits of the distance between piled walls. So it could represent a realistic condition.</p>
<p>Log spiral is suggested as applicable to all passive soil retention problems. I think this assumption would be applicable to your situation, but this is something that should be verified.</p>
<p><strong>Boussinesq Elasticity Theory</strong></p>
<p>Boussinesq theory can be used to look at lateral (and vertical) pressure problems where deformation does not occur. In your case deformation will likely occur, but assuming that it cannot will produce higher stresses/pressures than are expected (there is no relaxation under the theory) so it will be a conservative result.</p>
<p>Also there is the assumption of an elastic half space within Boussinesq theory. As your system is restricted by hydrostatic pressure, it could be considered to <em>behave</em> as an elastic half space. But more information would be required.</p>
<p><strong>Other Considerations</strong></p>
<p>A very good, comprehensive, but dated information source is the <strong>Steel Sheet Piling Design Manual (1984)</strong>. Cellular cofferdams and pressure analysis is included, however, and a copy can be viewed at scribd.com <a href="http://www.scribd.com/doc/27226946/Steel-Sheet-Piling-Design-Manual" rel="nofollow noreferrer">here</a>.</p>
<hr>
<p>In the photo provided there is no doubt going to be construction traffic travelling along the region between the piles. I have used Boussinesq (as modified) specifically for this purpose on previous projects, to ensure that the structure can withstand these loadings. This is another very important issue to be studied - it will require the analysis of the specific equipment, track patterns and loadings - essentially the equipment manufacturers data. Your analysis should also be closely coordinated with the construction programme, to include the numbers and likely configurations of equipment that will be used. Not an easy task.</p>
<hr>
<p><strong>Schematic Of Suggested Analysis</strong></p>
<p>In the figure below, the suggested approach is shown. Of course all of the conditions are not known, for example the locations of sea/river bed, the hydrostatic conditions between the sheet piled retaining elements, etc.</p>
<p><a href="https://i.stack.imgur.com/pCsll.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pCsll.png" alt="combination boussinnesq and log spiral"></a></p>
<p>Construction loadings at the top of the section can be modelled using the track patterns/footprints and associated loadings. Boussinesq theory is used to compute the lateral stresses at the retaining structure as illustrated by the yellow and green stress envelopes, and these can be superimposed to accommodate any surface loading configuration that is desired.</p>
<p>The log spiral analysis, however, is an iterative process, where the origin of the curve, at point <strong>O</strong> must be perturbed such that the curve always intersects point <strong>A</strong> at right angles and also intersects point <strong>C</strong> at the base of the excavation. This yields a series of soil envelopes within <strong>ABC</strong> that reach a maximum value as illustrated by the curve and points above point <strong>A</strong>.</p>
<p>Note that this considers a curved failure surface. The assumption of passive conditions is difficult to assess, but near to corners of the cofferdam the <em>box</em> effect should provide sufficient rigidity. Towards the centre of the sides of the <em>box</em> this assumption needs further examination.</p>
<p>The traditional way to carry out the log spiral analysis is graphically. That is to construct a log spiral template to scale according to a scale drawing and shift it around the drawing under the constraints of points <strong>A</strong> and <strong>C</strong>. The area of <strong>ABC</strong> is calculated for each trial until a clear maximum is reached. However we have developed an algorithm that will carry this out computationally, so no graphical analysis is needed.</p>
<p>Depending upon your geometry, you may not encounter a maximum, instead you may be limited by point <strong>D</strong>. In this case the envelope defined by <strong>DBC</strong> would be the value of interest.</p>
<p>One of the most difficult aspects of such an analysis will be to establish the <em>worst case</em> base condition. Careful consideration will be needed to determine which events could coincide, in terms of equipment configurations, fluctuations in water levels and other issues, such as potential de-watering risks. A risk-based approach may be advised, that warrants more than the traditional factor of safety methods.</p>
| 2516 | How to determine the lateral earth pressure in a double-walled cofferdam? |
2015-04-17T18:14:56.800 | <p>In a typical finite elements analysis package, a modal analysis gives the N first natural modes, and it is possible to get the equivalent stress and total deformations for each of these frequencies. </p>
<p>But what amplitude did the solver use to get these results? </p>
<p>Why can't we obtain a gain instead, for displacements and stresses and for each frequency?</p>
| |mechanical-engineering|finite-element-method|simulation| | <p>The short answer is, there is no amplitude used. Even more important though, is the fact that <strong>the displacements and stresses shown in the results of a modal analysis cannot be used to say anything about the physical behavior of the part in absolute terms.</strong></p>
<p>The basic equation of motion is </p>
<p>$$[M][\ddot{U}]+[B][\dot{U}]+[K][U]=F(t)$$</p>
<p>$M$, $B$,and $K$ are matrices for mass, dampening, and stiffness, respectively, and $U$ is the displacement. The material properties are known, and we are solving for the displacement. These are specified for the individual nodes created when meshing an FEA model. For a modal analysis, we ignore the damping effects, and assume there are no loads present. This essentially poses the question "If we constrain the part in a certain manner but apply no load to it, what are the possible ways in which it will vibrate?" </p>
<p>Look at the equation of motion when we neglect the damping and apply no load (F=0)</p>
<p>$$[M][\ddot{U}]+[K][U]=0$$</p>
<p>We are trying to solve this equation for non-zero values of $U$; that is, points at which the inertial forces and spring forces of the material are equal. In this idealized case of no damping, the part could theoretically vibrate through these points forever after an initial displacement/force is applied. </p>
<p>In solving the idealized equation above, we also let </p>
<p>$$[\ddot{U}]=\lambda[U]$$</p>
<p>where $\lambda$ is an eigenvalue. This is a consequence of harmonic motion, which makes sense when you think about it a bit. If a part is oscillating between two positions and the points of the part take linear paths between their two positions, the acceleration vector will always be a linear multiple of the displacement vector. </p>
<p>In the final form of the equation, we see that we only have 4 distinct terms:</p>
<p>$$\lambda[M][U]+[K][U]=0$$</p>
<p>Since $M$ and $K$ are known, we're just looking for a displacement matrix for which a constant $\lambda$ exists. But given the nature of this equation, it's easy to see that for a matrix $[V]=A[U]$ where $A$ is an arbitrary constant, $\lambda_U=\lambda_V$. </p>
<p>Lastly, it's useful to note that the units of $\lambda$ are $s^{-2}$, meaning that $\lambda=\phi^2$ with $\phi$ being the resonant frequency of the vibration mode in question. </p>
<p>So as I said at the top, there is no amplitude used to the actual main calculations in a modal analysis. To display the results and give the "displacement" values in the results, there is some standard amplitude or normalization done, but those numbers are not what you're looking for in a modal analysis, and they shouldn't be used in absolute form; how you <em>can</em> use them is in determining the proportionality of displacements between nodes. If you see point A has a displacement of 2mm when point B has a displacement of 1mm, you know that 2:1 ratio will always exist for this vibration mode. </p>
<p>To determine how much a part will actually vibrate, you have to do a full vibrational analysis, defining the loads and their frequencies yourself, using the knowledge you've gained from the modal analysis. </p>
<p>Two good references:</p>
<ul>
<li><a href="http://en.wikipedia.org/wiki/Modal_analysis_using_FEM" rel="noreferrer">Wikipedia article on <em>Modal Analysis using FEM</em></a></li>
<li><a href="http://machinedesign.com/archive/vibration-analysis-designers" rel="noreferrer">A take on vibration and modal analysis as it applies to design</a></li>
</ul>
| 2520 | What amplitude is used for a finite element modal analysis excitation? |
2015-04-17T19:32:28.337 | <p>In my mechanical vibrations class we studied the method to orthonormalize a set of differential equations by the mass matrix (principle coordinates). This is where you take the matrix of eigenvectors from the un-damped system and normalize it by the mass matrix. </p>
<p>Multiplying the mass matrix by the modal matrix gives:</p>
<pre><code>X'MX = I
</code></pre>
<p>And multiplying the stiffness matrix by the modal matrix gives:</p>
<pre><code>X'KX = W^2(ii)
</code></pre>
<p>Where W^2(ii) is the matrix of squared natural frequencies.</p>
<p>My question is: is it appropriate to try and orthonormalize the damping matrix in the same fashion? </p>
<pre><code>X'CX = 2*zeta*omega(ii)
</code></pre>
<p>Should this transform the system's damping into principle coordinates?</p>
| |mechanical-engineering| | <p>The one-word answer is "maybe".</p>
<p>In real structures, damping is often nonlinear, and hard to model from first principles. Also, a physically realistic but "arbitrary" damping matrix does not have nice mathematical properties - for example the mode shapes will be complex (i.e. different parts of the structure will vibrate out of phase with each other).</p>
<p>As a consequence, in class (and in many real world situations) you will learn about models of damping that are mathematically convenient but which don't have much basis in physics. Your "modal damping" model based on $2\zeta\omega$ is a good choice for lightly damped structures. Instead of trying to model or measure the $N^2$ terms of the complete damping matrix, you only need to estimate $m$ damping coefficients for the $m$ modes in the frequency range of interest (and often $m \ll N$). You will probably also learn about the Rayleigh damping model, which arbitrarily assumes the damping matrix is of the form $a M^\alpha + b K^\beta$ where $a$, $b$, $\alpha$, $\beta$ are scalar parameters, and most often $\alpha = \beta = 1$.</p>
| 2521 | Orthonormalization of Damping Matrix -- modal analysis |
2015-04-18T15:17:38.743 | <p>We sell products that attach to a motor drive's DC bus. We also formerly sold diode kits that let you hook one product up to multiple drives. We stopped selling those diode kits because they were unreliable with modern hardware, and we had better solutions. My customer tells me he wants to keep using the old diode kits, because he can have his repair techs disconnect power to one drive and have it replaced, while the others are still powered on.</p>
<p>I maintain that this is an unsafe practice, because diodes are not safety-rated devices. (See question <a href="https://electronics.stackexchange.com/questions/165389/is-it-safe-to-treat-a-system-downstream-of-a-reverse-biased-diode-as-de-energize/165390#165390">here</a>.) But my customer is rather insistent. What is my ethical obligation in this case?</p>
| |electrical-engineering|ethics|sales|safety| | <p>Your absolute minimum responsibility is to inform your customer that you consider the intended usage to be unsafe and that you do not recommend or endorse using it in this way. This should be done in writing and be acknowledged by the customer. </p>
<p>The next level depends on context. It is not impossible that your customer has good reason to do this and has their own procedures to ensure that it is used in a safe way. However if you believe that they are just cutting corners then we all know that it would not be ethical to supply them with the means to do it. </p>
<p>If you want to do this properly then the only way is to have an in-depth discussion with your customer to ensure that they will use your product responsibly. </p>
<p>Note that the correct ethical approach goes somewhat beyond discharging your nominal legal responsibilities and in any case you may still end up being legally liable even if you tick the boxes on something you know to be dangerous. </p>
| 2534 | My customer wants to use my products to do something unsafe. What is my ethical obligation? |
2015-04-19T05:49:23.140 | <p>This is the top part of a basic sit/stand base I'm making. It is a monitor platform and sliding keyboard tray.</p>
<p><a href="https://i.stack.imgur.com/NPFC4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NPFC4t.jpg" alt="enter image description here"></a></p>
<p>Instead of the large base with 2 rods on each end <a href="http://www.codeovereasy.com/2013/09/diy-adjustable-desk-for-under-25/" rel="nofollow noreferrer">(like this design see video at bottom of page)</a> I'm looking for another way to raise and lower the desk.</p>
<p>The one issue with the base design, is that in the lowered position, the platform is already 4 inches up off the desk and will then have to sit on top of the base another 3.5 inches or so. The addition of a base anymore than an inch or so would be too high...(unless the platform does <em>not</em> have to sit on the base)</p>
<p>I'm looking for some sort of electronic or manual jack that could fit under the over-hanging portion of the platform. Perhaps a small one on each of the 4 corners.... or maybe 1 longer one for each side? (The platform is 33"X18")</p>
<p>I would like to raise the desk about 12 inches from where the top sits now. Also will be in an office setting so can't be too loud.</p>
<p>Are there any type of devices that could be engineered... or purchased... or somewhere in between, that can work for this application or have been use for something similar in the past?</p>
<p>Some other random ideas I have had: some sort of accordion or scissor lift that attaches across to the bottom of each vertical leg of the platform....or maybe air bags?</p>
<p>Here is a close up of the leg and edge hanging over.<br>
<a href="https://i.stack.imgur.com/LxgO4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LxgO4t.jpg" alt="enter image description here"></a></p>
<p><strong>Update1</strong> </p>
<p>or for a more manual design, I came up with scissor sliding rods with rails. On one side, it would have a sliding rail attached to the top of the over hang, and then a rail attached to some sort of base - that would fit beneath the over hang: </p>
<p><img src="https://i.stack.imgur.com/2WoGV.png" alt="enter image description here"></p>
<p>Has something like this ever been proven to work decent?</p>
<p>The only problem with the <a href="https://www.youtube.com/watch?v=XDU43UTL2Yg" rel="nofollow noreferrer">scissors as a whole unit</a> is that the top would need to be attached to the legs of the platform...and not sure how many inches it would add..in addition, the whole desk would move forward and back, along with up and down (and I have no idea what the kid is saying...probably some thing like I'm 8 and smarter than you).</p>
| |mechanical-engineering|power-electronics|hydraulics| | <p>Put the fixed point of the rail on the back on both the upper <em>and</em> lower rails in the video you linked they are on opposite sides leading to a side to lateral movement. </p>
| 2547 | Miniature electric or manual jacks? |
2015-04-19T16:52:08.270 | <p>How small/compact can a pump or other mechanism be, that would allow compressing gas (or liquid) to pressures of excess of 10GPa, continuously?</p>
<p>Throughput may be minuscule, of order of milligrams per hour; even lower is acceptable. Also, for "size of the device" let us count all the necessary infrastructure: if it requires energy of a nuclear power plant to operate, it's not compact. Let's say we have some 30 watt of energy "for free"; anything above that goes out of our "compactness budget".</p>
<p>Input gas is pressurized to "reasonable" levels - order of 30KPa; if more pressure is required, the container goes into "compactness budget" with at least 20kg of the gas inside.</p>
<p>There's abundant heat sinking capacity, but any extra heating goes out of our "budget". Operating (ambient) temperature - choose anything between 40K and 400K as you like. It should have operational lifetime of at least a year, preferably ~10 years.</p>
<p>(clarifying: "continuous" - we can assume a small "buffer" container on the output side, emitting the pressurized gas at rate that is the average of the intake speed, so short "discontinuity" like piston motion is fully acceptable - simply keep the "buffer" above 10GPa with losses of order of assumed average input (1mg/hour, or less if required, but fairly constant).)</p>
| |pressure|pumps|compressors| | <p>Such high pressures can be achieved via <a href="https://en.wikipedia.org/wiki/Diamond_anvil_cell" rel="nofollow">diamond anvil cells</a> (DAC). The heart of these cells are two diamonds mounted on tungsten seats symmetrically in front of each other which are separated by metallic plate. Metallic plate has an aperture with loaded sample (gas, crystal, powder, etc.). These diamonds are then brought closer to each other by tightening bolts or by extension of piezo actuators. Pressures in such DACs can reach up to 200GPa or higher depending on the size of the diamond, it's quality, temperature and type of material being compressed. Amount of sample which can be compressed usually ranges from micrograms to milligrams.</p>
<p>One would need a special chamber in order to load gas into DAC. In these chambers operator controls separation between diamonds in order to fill the aperture in metal gasket between diamonds with liquefied gas and enclose it safely for further compression</p>
| 2551 | Gigapascal pressures in a small package |
2015-04-19T19:12:42.510 | <p>Is it possible to convert the energy you would normally lose when braking your bicycle to rotational energy by attaching an extra wheel which would start to rotate once you use your brakes? E.g., when facing a red light after riding down a hill.</p>
<p>If you could restore that energy, you could regain part of your velocity without as much effort after stopping at the light.</p>
<p>I am quite sure I'm not the first one who had this idea (as I heard about some projects trying to achieve the same thing but then with cars) so I wonder what the technical difficulties are and why I have not seen bicycles like this on the road?</p>
| |mechanical-engineering|bicycles|regenerative-braking|flywheels| | <p>There are a few fundamental difficulties. </p>
<p>The energy stored in a flywheel is a combination of its moment of inertia and its speed. To get a higher moment of inertia you either need more mass or a larger diameter. The problem being that bicycles tend to be very sensitive to both weight and space. This leaves you with speed, firstly this will require quite a high gear ratio to get from the relatively low speeds of a bicycle wheel to the high speeds required for an effective flywheel as well as a mechanism to engage and disengage it smoothly and transfer between extracting and imputing energy, this adds weight and complexity. </p>
<p>Also a flywheel with a useful amount of energy will have a string gyroscopic effect which is a particular issue on a bicycle unless you have an even more complex system to mitigate it. </p>
<p>Equally high speed flywheels pose materials challenges for bearings and the flywheel itself which is trying to explosively pull itself apart and obviously even a light, low speed flywheel is going to need to be fully enclosed for safety. </p>
<p>The short answer is that with current technology an electrical system with batteries is always going to be more attractive, the technology is mature and well developed and for anything like reasonable cost is going to provide much better energy density and is much easier to package on a bicycle as well as allowing for a much greater degree of control. For example a digitally controlled electrical system can have many different modes of operation without any additional mechanical parts. </p>
| 2557 | What are the technical obstacles for using flywheel-based regenerative braking on bicycles? |
2015-04-21T18:29:41.173 | <p>Is there a specific chemical that is used frequently in the production of liquid crystal displays? During my internet research so far, it seems as if the specific chemical composition of the liquid crystal does not matter very much. <a href="http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=16378837">This</a> website gives a list of apparently commonly used LCs. Are there one or two chemicals that are used in practically all common LCDs?</p>
| |chemical-engineering|optics|liquid| | <p>The article <a href="http://www.ijestr.org/IJESTR_Vol.%201,%20No.%207,%20July%202013/Liquid%20Crystal%20Display.pdf" rel="nofollow">Liquid Crystal Display: Environment & Technology</a> (2013) provides a detailed list of chemicals often used in LCDs. There are a myriad of chemicals that are used, each serving a specific purpose for the functionality of LCDs.</p>
<p>However, for liquid crystals (LCs) themselves:</p>
<blockquote>
<p>The composition of
LC includes a bicyclohexyl compound (35-50% by weight), a cyclohexyl phenyl compound (15-25% by weight), a
bicyclohexyl phenyl compound (20-25% by weight), and a cyclohexyl biphenyl compound (15-20% by weight).</p>
</blockquote>
<p>The article states that there are typically 10-25 compounds used to manufacture LCs, with many mixtures used, depending on the application the LCD is used for.</p>
<p>The variations in mixtures result in differing chemical
properties of alkyl or alkoxy side chains.</p>
<p>The back light unit has an interesting component:</p>
<blockquote>
<p>The backlight unit (BLU) major constituent of LCD contains hazardous mercury to operate. </p>
</blockquote>
| 2571 | What liquid crystals are used most frequently in displays? |
2015-04-21T19:11:45.827 | <p>Oblique cutting generates transverse cutting force which reduces tool life. Oblique cutting involves principal and auxiliary cutting edges in the cutting action which increases friction and advances wear in the cutting tool. </p>
<p>Orthogonal cutting doesn't have those complexities, but oblique cutting is found to be used in more often in an industrial setting. </p>
<p>Why are oblique cuts made instead of orthogonal cuts when it is more efficient to make an orthogonal cut?</p>
<p><img src="https://i.stack.imgur.com/VTLB2.png" alt="Orthogonal and oblique cutting"></p>
| |mechanical-engineering|materials|mechanical-failure| | <h1>Short answer</h1>
<p>Orthogonal cutting increases pressure and decreases tool life. Thus isn't used if aim isn't cutting. Lathe operation is a shaping operation, not a cutting operation most of the time. </p>
<h1>Long answer</h1>
<p>Orthogonal cutting isn't suitable because</p>
<ol>
<li>Cutting forces act on a smaller area of cutting surface on tool if orthogonal cutting is employed so it increases wear on tool tip. However oblique cutting distribute cutting forces to greater area thus increases tool life. </li>
<li>You usually don't want corners on your part. Corners reduce strenght and you want chamfers. </li>
<li>Chip flow is another problem in orthogonal cutting. </li>
<li>It's not possible to obtain perfect tool tip for orthogonal cutting, so addition of ploughing force increases total cutting force.</li>
</ol>
<p>However, if you're trying to cut off a part you'll use orthogonal cutting. Below are some examples of cutting tool shapes
<img src="https://i.stack.imgur.com/okdvD.jpg" alt="Cutting tool shapes"></p>
<p>Check <a href="https://books.google.com.tr/books?id=r8paflRca90C&lpg=PP1&hl=tr&pg=PA76#v=onepage&q&f=false" rel="nofollow noreferrer">Fundamentals of Metal Cutting and Machine Tools</a> pages 117-139 which are accessible in Google Books.</p>
| 2572 | Oblique cutting implemented in Industrial applications |
2015-04-22T06:36:20.740 | <p>I'm building a press for cider. I'm going to be using a 24" x 24" press board. I'm looking to deliver 40 psi to the cider. How can I determine the amount of weight required, in tons? What sort of equation would I use?</p>
| |mechanical-engineering|pressure|hydraulics| | <p>The obvious place to start is "what does psi mean?" It's an abbreviation for "pounds per square inch".</p>
<p>If we have a 24" x 24" board, then we have </p>
<p>$$24~in\times24~in=576~in²$$</p>
<p>To get 40psi, i.e. 40 pounds for each square inch, we would need </p>
<p>$$40~\frac{lbs}{in²}*576~in² = 23040 ~lbs$$.</p>
<p>You've asked for an answer in tons, so we need to convert 23040 pounds to tons. Assuming you're using the US short ton, there are 2000 pounds in a ton. So we need </p>
<p>$$23040 ~lbs/2000 ~\frac{lbs}{ton} = 11.52~US~tons$$</p>
| 2577 | How do I calculate the weight needed on a press board to deliver a certain amount of pressure? |
2015-04-22T15:04:50.520 | <p>I'm dealing with some hydrocarbons (diesel, gas oil, kerosene, ethanol) where I'm putting them in a chamber and shining light through them to hit a sensor on the other side in order to measure their colour. Unfortunately what I'm finding is that the lenses and reflective surfaces in the chamber that I'm using for these optical measurements are being stained by the products and subsequently affecting the measurements. When the chamber is emptied, small amounts of product remain clinging to the internal surfaces. As this dries a residue is left which remains on the surface.</p>
<p>I've tried a couple of different materials: stainless steel reflectors, transparent Grilamid plastic lenses, a stain-resistant transparent potting and regular glass. All of them end up with a residue that clings to the surface that needs to be wiped clean. Unfortunately, for my application I'm not going to be able to get access to the internal surface to clean it so I need to find a way to prevent this residue from clinging to the internal surfaces. </p>
<p>The main restriction for me is that the system needs to basically be 100% maintenance free, and unfortunately I can't control the fluid coming in and out and therefore can't rinse the system out. </p>
<p>I was wondering if there are any products (both transparent and reflective) that are basically completely "hydrophobic" but for hydrocarbons and will prevent the product clinging at all to the surface and staining the lenses and reflectors?</p>
| |materials|optics| | <p>A super-hydrophobic coating would work well for you because super-hydrophobic materials are self-cleaning. <a href="http://www.sandia.gov/" rel="nofollow">Sandia National Labs</a> has a nice report on super-hydrophobic coatings which can be found <a href="http://www.sandia.gov/research/research_development_100_awards/_assets/documents/2008_winners/Superhydrophobic_SAND2008-2215W.pdf" rel="nofollow">here</a> (associated patent <a href="http://www.google.com/patents/US7485343" rel="nofollow">here</a>). In it they report the invention of a coating who's hydrophobicity can be varied by application of UV light. The <a href="http://en.wikipedia.org/wiki/Contact_angle" rel="nofollow">contact angle</a> of their coating is greater than 150$^\circ$, and they reference two other patents which reach contact angles of 169$^\circ$ and 172$^\circ$ which are both very high, the maximum is 180$^\circ$. </p>
<p>The problem is, all of this research is fairly recent and doesn't seem to have found its way to industry yet. I can't find many coating houses who sell super-hydrophobic optical coatings. <a href="http://www.aculon.com/super-hydrophobic-coatings.php?_kk=superhydrophobic%20coatings&_kt=18bf94de-5d58-4823-9fe3-53bac7ca1fbf&gclid=COu7_NqgisUCFXMF7AodmVAAbw" rel="nofollow">Aculon's website</a> says that they sell one, but they don't publish the contact angle, and the picture they show makes it appear to only be ~90$^\circ$. </p>
<p>Aculon also claims to sell an <a href="http://www.aculon.com/oleophobic-coatings.php?gclid=CKOdjMGhisUCFdcYgQodkhYAKg" rel="nofollow">oleophobic</a> (oil repelling) coating. This may work for you, but you would need to test a sample with the materials you plan to use. The video on their website shows someone trying to write on it with a Sharpie and it just won't take to the surface. </p>
<p>The problem with all of these coatings is longevity. Rain-X is an excellent example of a hydrophobic coating, but it must be reapplied once a month or so. My eyeglasses came with a hydrophobic coating which was very impressive when they were new. Now that they are 10 months old, everything sticks to them again. So, whatever you decide to build, you should be able to service it regularly.</p>
| 2583 | How to avoid dyed liquids staining reflective/transparent surfaces |
2015-04-23T09:56:08.000 | <p>I am designing a metal plate that will be laser-cut (or machine-cut) and then folded. I want to know how to size the pre-folded plate in order to get the right dimensions after folding.</p>
<p><img src="https://i.stack.imgur.com/Y5Pq3.png" alt="Diagram of a folded 2mm aluminium plate with holes and window"></p>
<p>My actual part is not exactly like this (I simplified it for ease of drawing) but it shows what I want to achieve. In this case it's a 2mm aluminium plate, the red arrows show the inner dimensions after folding that I want to specify and achieve. The holes must also line up and the window should be correctly placed.</p>
<p>Intuitively I would expect some compression along the inner part of the folds and stretching on the outer parts - ideally along the centre of the plate - but I don't know if this is what will happen.</p>
<p>Assuming the red arrows are 100mm each, should the plate be 300mm? I'm guessing not, so how do I calculate the radius of curvature that will be achieved and if I need to add (or remove) material at the folds in order to achieve my required dimensions?</p>
| |machining|metal-folding| | <p>Yes use Aluminum 5052 H32 for sheet metal parts with bends. r =or> T.
I do it like this, get the lengths of the straight lines and set aside.
The t/T is a little more than .50, say .53 until you get the real number.
K= t/T, radius of the neutral line is inside r + t = r + .53T
For a 90 degree bend, length of the bend is 2*pi*(r+t)/4 = pi*(r+t)/2 = pi*(r+.53T)/2, for any angle of bend, length of the bend is 2*pi*(r+t)*angle/360
length = staight lengths + length of bend, continue to add more bends if needed
eg: angle 1-1/4in x 2-1/4in outside dimensions, 90deg bend, 1/8in thk aluminum 1/8in inside radius, remove t and r to get straights 1in and 2in add 2*pi*(.125+.53(.125))in/4 = 3in + .3in = 3.3in </p>
| 2589 | How do I size metal plates to get the correct dimensions after folding? |
2015-04-23T21:00:11.883 | <p>I need to release pressure of around 50psi, after 3-4 seconds of initiating, from a vessel using a mechanical switch. I had considered a pneumatic solenoid, but I want to forgo the inconvenience of charging the device with electricity in addition to the air, as well as save the space that would be needed for the bulky solenoid, circuitry, and battery. </p>
<p>I therefore need what would essentially be a "mechanical-fuse" that releases pressure after a delay when triggered. I considered a screw-type piston that gets forced along a track by the air in the vessel itself, which at a set psi could guarantee a specific time of release. I also need the system to be no longer than 5 inches along the axis of released pressure. Does a "screw-fuse" sound like a viable idea? Does anyone have any other sources for a mechanical fuse of this type, or an idea of their own?</p>
| |mechanical-engineering|pressure|compressed-air| | <p>There are three possibilities here that come up offhand, between my own mind and ratchet freak's comment:</p>
<ul>
<li>The choke that ratchet freak mentioned -- this is the simplest approach if you are OK with a slow decay of pressure</li>
<li>The dampened piston that ratchet freak also mentioned, where the mechanical switch pulls out a lock from the piston to allow pressure to retract it against the calibrated force from the damper</li>
<li>A clockwork mechanism that is triggered by the switch and provides the three-to-four second delay by letting a spring unwind</li>
</ul>
<p>The main issue with the second and third mechanisms is that they would require some sort of manual or automatic reset functionality (a reset could be triggered by pressure building up to a certain point).</p>
| 2594 | Are there any precedent designs for a mechanically-delayed pressure release valve? |
2015-04-24T06:16:10.763 | <p>A quote from my textbook :</p>
<blockquote>
<p>Turbomachinery flows are steady only in a time-averaged sense; that is, the flow is periodic,
with a period equal to the time taken for a blade to move a distance equal to the spacing
between adjacent blades. Despite the unsteadiness, in elementary analysis all variables are
assumed to have steady values.</p>
</blockquote>
<p>Can you please explain this 'unsteadiness'? What effect does a blade moving a distance equal to the spacing between adjacent blades has on making the flow periodic?</p>
| |mechanical-engineering|fluid-dynamics| | <p>The words <strong>unsteady</strong> and <strong>periodic</strong> have different meaning depending on the context they are used in. My answer is mainly with regards to axial or radial turbomachines.</p>
<p>Lets say our machine is running at a specific operating point. The rotation of the blades (the shaft) does not change. The massflow-rate through the system is constant.
Even though this operating-point is <em>steady</em> the flow inside of the machine does <em>not</em> have to be steady.
For practical purposes one usually designs every blade row in its specific (relative) <em>frame of reference</em>. This means the designer 'sits' on the air-foil.
Now we assume further: steady flow- and boundary conditions. Doing this makes the aero and mechanical design feasible. When designing the blades and vanes for a turbomachine one will not design 'every' blade but assume that the flow around the blades will be the same for all blades of one blade-row. (Even though this assumption is usually not valid at the boundaries of the operating range it is a very practical approach to get the design-process started.) This means the flow is believed to be <strong>steady</strong> and <strong>periodic</strong>.</p>
<p>However! There is one thing called the 'Unsteadiness Paradox' <a href="https://i.stack.imgur.com/KOQnc.png" rel="nofollow noreferrer">1</a>. It basically boils down to the fact that a turbomachine <strong>has</strong> to be unsteady in order to work. But the <em>unsteadiness</em> in this case refers to the rotation of the blade-row.</p>
<p><img src="https://i.stack.imgur.com/KOQnc.png" alt="Turbomachine"></p>
<p>The figure shows a sketch from Greitzer. If one assumes the turbo-machine to be 'black box' which just adds or extracts work than it is not possible to describe the turbo-machines analytically. It only works when adding a stationary and rotating black-box.</p>
<p>Without any simplifications of the flow and the geometry it would not be possible to design a turbomachine and predict its operating characteristics. </p>
<p>Two things have to be simplified:</p>
<ol>
<li>Aerodynamics</li>
<li>Geometry</li>
</ol>
<p>In case of the aerodynamics the unsteadiness (eddies and turbulence) in the flow is usually averaged out. In case of the geometry one assumes <strong>periodicity</strong> (see the figure below from Oxford University):</p>
<p><img src="https://i.stack.imgur.com/E45q4.png" alt="multi-passage-simulation"></p>
<p>For the version (a) only one passage was simulated and simply copied (all passages look the same). In version (b) all passages were simulated showing slight differences under certain operating conditions.</p>
<p>The second simplification to the geometry is the introduction of <strong>mixing planes</strong> so the rotating and stationary part of the turbomachine can be calculated separately (see figure below from Cambridge University):</p>
<p><img src="https://i.stack.imgur.com/64vi5.png" alt="multi-row-simulation"></p>
<p>The unsteadiness which is introduced by the relative motion of blades and vanes (right) is averaged-(smeared)-out (left).</p>
<p>These simplifications are necessary in order to have fast optimisation and development of the turbomachine and are later corrected and checked by experiments or high fidelity simulations if necessary.</p>
<p><a href="https://i.stack.imgur.com/KOQnc.png" rel="nofollow noreferrer">1</a>
Dean, R.C., "On the Necessity of Unsteady Flow in Fluid Machines", ASME J. Basic Eng. March 1959, pp 24-28</p>
| 2595 | What makes a flow periodic in Turbomachines? |
2015-04-24T19:27:27.740 | <p>I want to embed a wire inside a plastic and heat it to specific temperatures (80$^\circ$C 100$^\circ$C 130$^\circ$C 180$^\circ$C). I want to find a material that is resistive enough for resistive heating, but also I want the wire's resistivity to vary fairly strongly with temperature - this way I can know the actual temperature that I'm getting by looking at the resistance during the heating process. </p>
<p>My problem is that I cannot seem to find a material with the properties I want. I tried nichrome wire and carbon fibers, and while they are resistive enough to give me heat, their resistance does not change enough (for my desired temperature ranges) to be measurable with my setup.</p>
| |electrical-engineering|materials|heat-transfer| | <p>Just about anything you can find will exhibit a temperature dependence on its resistivity. The first hit looking up nichrome says it's temperature dependence is 0.004/°C.</p>
<p>A better model for resistivity is the Steinhardt-Hart(sp?) equation, but you probably don't need something that elaborate. Since you only want to control the temperature to 4 discrete values, you can figure out ahead of time what the total resistance of a particular diameter and length of wire will be at each of those 4 temperatures.</p>
<p>To measure the resistance, you need to measure the voltage across the wire and the current thru it, then divide. With the right analog electronics, these values can be presented to A/D inputs of a microcontroller, which can compute the resistance, determine the temperature, then vary the current via a controller accordingly.</p>
<p>I'd probably implement the resistance to temperature function as a piecewise-linear table lookup. That's quick to compute and can model any non-linearities without lots of cycles.</p>
| 2605 | Material which can be used for resistive heating and simultaneous temperature measurement |
2015-04-25T08:40:28.170 | <p>Why are tremors of an earthquake felt most on the upper floors of a building in comparison to lower floors? Does this have something to do with a third class lever?</p>
| |mechanical-engineering|civil-engineering|applied-mechanics|seismic| | <p>Harmonic coupling with the building's natural vibration frequencies, dissipation of energy in a massive building, and the variety of vibration modes that can experience coupling could result in more or less displacement and more or less velocity at lower floors or upper floors. There's no universal relationship between either velocity or displacement between upper and lower floors.</p>
<p>Furthermore, earthquakes have two prominent types of waves, S waves and P waves. S waves move at a 45 degree angle to the propagation and will slowly fall behind P waves to the point that the earthquake actually separates into two phases. Having experienced this personally, it's actually easy to tell if you were at the epicenter or far away because the building motion changes and the quake has two distinct peaks. You don't need a seismometer for a large enough quake far enough away. The importance of the two wave types is that they can excite unique vibration modes, twisting vs bending, so there isn't even a universal answer for one building.</p>
<p>A tall building is the easiest way to visualize how the velocity and frequency at the ground and upper floors are not necessarily related. </p>
<p>The foundation is usually relatively fixed, though some designs attempt to de-couple ground movement from structure movement through the use of mechanical bearings of various sorts where the structure meets the foundation. At the fixed foundation, a tall building moves according to the earthquake. Heavy though the foundation may be, the earth is still heavier.</p>
<p>At the upper floors, if the building is in harmony with the motion, it will be as if pushing a light pole at its natural frequency. You will get it to break most likely and the motion at the top will be severe.</p>
<p>In tall buildings, typically this frequency is much larger than the earthquake frequency, so while some energy does accumulate, very little of it is conserved because of all the destructive interference and dissipation in the large number of structural connections. The dangerous buildings occur right around ten floors. Shorter than this, the building wants to vibrate faster than the quake. Taller than this, the building wants to vibrate more slowly than the quake. At extreme heights, energy dissipation becomes more pronounced as there is just too much mechanical loss through hysteresis for 2n or higher vibration modes to have much effect.</p>
<p>Here's really cool video of buildings that built up energy swaying after an earthquake. Note that the lateral movement at the top is large, but you can imagine that the velocities at the top related to the buildup are not extreme relative to the fixed foundations.</p>
<p><a href="https://www.youtube.com/watch?v=g0cz-oDfUg0" rel="noreferrer">https://www.youtube.com/watch?v=g0cz-oDfUg0</a></p>
| 2609 | Why are tremors of an earthquake felt more on upper floors? |
2015-04-25T12:17:45.800 | <p>I am making a device for measurements. I would like to measure the distance within accuracy of 1 mm. Range could be 2 cm to 15 cm . I looked at <a href="https://www.sparkfun.com/datasheets/Sensors/Infrared/GP2D120XJ00F_SS.pdf" rel="noreferrer">Proximity Sensors</a> but the readings displayed by these sensors are not steady.</p>
<p>I wish to measure the thickness of the plate (carbon steel). The two sensors will be mounted on a structure. The sensors will give me the distance of the surface from the sensor. Then I will calculate the thickness of the plate.</p>
<p>What are possible types of <strong>low cost sensors</strong> I can use?</p>
<p><img src="https://i.stack.imgur.com/eYc3e.png" alt="Sensor Arrangement"></p>
| |mechanical-engineering|measurements|sensors|tolerance| | <p>Was looking for other distance solutions, and came across this question.</p>
<p>A sensor that I've found works well is the Sharp GP2Y0E02B (digital version) (newer version is the GP2Y0E03). Configuring it as an I2C sensor, using an Arduino, I've been able to get <em>sub-millimeter</em> resolution (0.156 mm - Range is ~630 mm, with 12 bit resolution). </p>
<p>Almost didn't believe it when I read the spec, so I put the sensor on a Bridgeport with a test target, and used this as 'ground truth'. Out to 50 mm on a simple test, the error was less than 0.2 mm; out at 100 mm, I got about a 0.4 mm error.</p>
| 2612 | Distance sensors with accuracy of 1 mm? |
2015-04-26T04:07:32.957 | <p>I have a simulation software which shows energy consumption at 30 minutes timestep. If a system consumes 0.06 kWh at 6am, and 0.09 kWh at 630am, how to calculate the energy consumption between 6-7am? Can I assume it is the average of both timestep? Does finer timestep, eg. 10 minutes interval returns a different result?</p>
| |mechanical-engineering|electrical-engineering|hvac| | <p>In SI units the measure of "amount" of electrical energy is the Joule.<br>
Energy is a measure of "work done".<br>
Energy can be (and often is) expressed in terms of 'rate of doing work'.<br>
The rate at which work is done or at which energy is supplied or utilised is also named "Power".<br>
Work done = 'rate of doing work' x time period involved<br>
= Power x time (involved) </p>
<p>So 1 Joule of energy = 1 Watt of power for 1 Second = 1 W.s
1 Joule is a small amount of energy compared to typical usage so a larger multiple is used of 1 kWh<br>
= 1000 Watts work rate x 3600 seconds<br>
= 3,600,000 Joule = 1 MJ = 1 kWh </p>
<p>kWh (KiloWatt hours) is a typical unit of "amount" for electrical energy use for eg heaters, motors, lighting etc.</p>
<p>SO if you operate at a power level of 120 Watts from 6am to 6.30 am you will consume 120 Watts x 1/2 hour = 60 Watt hours = 0.060 kWh. </p>
<p>180 Watts from 6:30am to 7am = 180 x 1/2 Hour = 90 Wh = 0.090 kWh.</p>
<p>As kWh = amount of energy, successive figures can be added.<br>
So 0.060 kWh + 0.090 kWh = 0.150 kWh.</p>
<p><strong>Key summary:</strong></p>
<p>Energy = <strong>Joule</strong> or Watt-hour (W.h) or KiloWatt hour (<strong>KWh</strong>) </p>
<p>Energy = Watts of power x period used </p>
<p>Energy = W1.T1 + W2.T2 + .... WnTn</p>
| 2614 | Calculating kWh for different timestep |
2015-04-26T23:39:46.737 | <p>This may fall a little under chemistry processes but I felt it has enough pertaining to aerospace to put it here. Basically, I'm in the process of attempting to develop a way to derive $I_{sp}$ as pertaining to rocket engines rather than rely on charted information.</p>
<p>Since specific impulse is essentially the exhaust gas velocity working against gravitational force,</p>
<p>$$
I_{sp} = \frac{v_e}{g_0}
$$</p>
<p>...it stands to reason that the ideal exhaust gas velocity equation can be substituted in here, giving something like</p>
<p>$$
I_{sp} = \frac{\sqrt{\frac{TR}{M}\cdot\frac{2\gamma}{\gamma-1}\cdot(1-\frac{\rho_e}{\rho}^{\frac{\gamma - 1}{\gamma}})}}{g_0}
$$</p>
<p>The obvious problem here is that this is the <em>ideal</em> exhaust gas velocity, so this is a sort of "perfect-universe" $I_{sp}$. Because most rocket engines use either hydrocarbon or hydrogen/oxygen propellants, water vapor is a major fraction of this exhaust. And as is usually taught early on in chemistry courses, water vapor is a textbook failure of ideal gas behaviors because of its intermolecular forces. </p>
<p>So my question is - is there a "real" specific impulse formula; something like the van der Waals equation for this application? </p>
| |aerospace-engineering| | <p>There is no general formula for isp that would provide accurate values all the propellant combinations and nozzle expansion ratios one might use. The <a href="http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm" rel="nofollow">NASA Chemical Equilibrium (CEA) program</a> can provide the information you are looking for and would be an answer to your question. It is useful for any of the common propellant systems and many exotic propellant systems, e.g., fluorine, metals, etc.</p>
<p>Extracted from nasa CEA webpage: </p>
<blockquote>
<p>CEA is a program which calculates chemical equilibrium product concentrations from any set of reactants and determines thermodynamic and transport properties for the product mixture. Built-in applications include calculation of theoretical rocket performance (isp at any nozzle expansion ratio), Chapman-Jouguet detonation parameters, shock tube parameters, and combustion properties.</p>
</blockquote>
<p>The code in fortran is freely available (check web site). This code is the "bible" for the rocket industry. I used this 50 years ago while working on the Apollo program and working on other rocket systems.</p>
<p>The program is very fast. You could set CEA up to be a call from your application or you could run a bunch of cases with the propellants you are most interested in and call a curvefit to the data for the specific situations you are looking at.</p>
| 2622 | Calculating real or more accurate specific impulse |
2015-04-27T04:40:18.970 | <p>I have a structure like such:</p>
<p><img src="https://i.stack.imgur.com/CZvAm.png" alt="Structure"></p>
<p>Due to my lack of knowledge of ANSYS, I have made the singular distributed loading (GI) represented as two distributed loadings (GH and HI).
Would this be an accurate model?
<img src="https://i.stack.imgur.com/iFnkL.png" alt="Model"></p>
<p>I feel like there should be some bending in the middle (HE to EB).</p>
<p>When I model it is a singular 240kN force on the center points I get:
<img src="https://i.stack.imgur.com/WL5eo.png" alt="Model with Singular force"></p>
<p>Which of these would be more accurate?</p>
| |structural-engineering|structures|modeling|stresses| | <p>Adding just a slight notion:</p>
<p>Your experience would always expect the beam HE to buckle, because it is a Eulerian buckling beam (don't know if it's also called like this in English).
The numeric only sees an evenly distributed load, so there will be no buckling... Try to alter your forces to the left and right of H a bit and you will get a moment which will cause HE to bend (If its support is rigid).</p>
| 2623 | Is a distributed load in two parts equal to a full distributed load? |
2015-04-27T07:12:50.343 | <p>Some sources say that higher pressure in hydrocyclones implies better performance it leads to a finer filtration. But why?</p>
<p>Is it because water is almost incompressible and the particles can be pressured and their density increased? </p>
| |mechanical-engineering|fluid-dynamics|pressure|cyclones| | <p>The use of the word <strong>performance</strong> might be misleading since a connection to <em>pressure loss</em> is quite reasonable. However, in this context I guess <strong>performance</strong> is connected to the filtration. </p>
<p>Using higher pressure implies in deed a higher pressure loss but it also results in a higher circumferential velocity. Since the main working principle of a hydrocyclone is to divide a solid-liquid suspensions by density a higher circumferential velocity will increase the radial acceleration. Since small particles will encounter higher drag du to their smaller Reynoldsnumber it will take longer for them to travel to the outside wall (based on <a href="https://books.google.de/books?id=8dj5IJbEwu4C&lpg=PP2&ots=3jH8p-NrNE&lr&pg=PR4#v=onepage&q&f=false" rel="nofollow noreferrer">Svarovsky (1981)</a>, residence time theory ).</p>
<p><img src="https://i.stack.imgur.com/686U0.png" alt="hydrocyc"></p>
<p>In order to have a <em>better</em> filter a higher radial acceleration is needed which (all other things kept equal) asks for higher pressure to produce higher circumferential velocities.</p>
| 2626 | Hydrocyclone effectiveness vs. absolute pressure |
2015-04-27T18:16:46.357 | <p>If a permanent room or platform is suspended from a larger structure (say, on a cable that is attached to a hook on a track in a ceiling) and designed to be in constant but not necessarily repetitive motion, does the load that the room/platform places on the larger structure count as a dead load or a live load?</p>
<p>On the one hand, it is permanent and won't ever be removed (like a dead load,) but on the other hand, it moves (like a live load.) I'm not talking about the loads that would be incurred by its motion, only by its weight.</p>
| |structural-engineering|structures| | <p>A moving load should always be classified as a live load even when its weight is known and remains the same. A dynamic body of mass has an extra degree of freedom. This positional uncertainty calls for a higher factor of safety. Your "permanent" room is not much different from a "permanent" mobile trailer in that both are capable of movement and that the gravitational load do not vary much.</p>
<p>A dead load, on the other hand, is permanent in both magnitude and position. I.e. it is a static load. A static load, being predictable, requires a lower factor of safety.</p>
<p>I would further argue that the classification of load depends on what you are designing. For the hook itself, you may even go for a higher safety factor by classifying it under crane load by reason that failure can result in catastrophic consequences to anyone below or inside the suspended room.</p>
| 2633 | Is a permanent platform that is hanging and moving a dead load or a live load? |
2015-04-29T08:05:42.903 | <p>An idea that keeps popping up in my head, that I don't know enough structural/civil engineering to know if it's good from a technical point of view.</p>
<p>Take a piece of <a href="http://en.wikipedia.org/wiki/Polyethylene" rel="nofollow">polyethelene</a> (PE) piping, fill it with concrete, let the concrete bind. My understanding is that under compression, a concrete (or other) pillar will 'want' to shear apart in a plane 45° to the direction of the compression (assuming we don't bend our pillar). This is an outward movement. This shear force could be contained by the surrounding pipe, since PE is quite good in tension. Othe materials for the pipe might work the same, I say PE because it's quite corrosion resistant.</p>
<p>I know everyone is building concrete pillars with rebar inside, I don't claim my idea is superior. I'm not interested in understanding why everyone is building pillars the way everyone is building them, I want to understand the flaws and limitations in my idea. Some thoughts:</p>
<p>I think it's useful to think as pressure in the pipe, as we have that data easily available - if we use PN10, it can take 10 bar, etc., and we don't need to think about the thickness of the pipe and it's yield strength. But how would the compressive force translate into pressure on the pipe? The concrete is no liquid, so pressue will be less than compressive load / pipe area. How much? I think understanding this will tell us how much, or little, our pillar will carry. </p>
<p>Another possible issue is that PE is quite smooth. The pipe can't take forces via friction and in my mind that translates to point loads at the places where the concrete happens to deform most.</p>
| |structural-engineering|columns| | <p>This approach has been used in many, many scientific studies, and is an active area of research in the materials engineering community, but with a different material. FRP (<a href="https://en.wikipedia.org/wiki/Fibre-reinforced_plastic" rel="nofollow noreferrer">Fiber Reinforced Plastic</a>) is used to wrap around an existing concrete column (or using an FRP pipe filled with concrete), with very good results.</p>
<p>Sources: <a href="http://dx.doi.org/10.1016/j.conbuildmat.2008.07.008" rel="nofollow noreferrer">1</a> <a href="http://www.sciencedirect.com/science/article/pii/S0263822305000231" rel="nofollow noreferrer">2</a> <a href="http://www.mdpi.com/2073-4360/6/4/1040/pdf" rel="nofollow noreferrer">3</a> <a href="http://ieeexplore.ieee.org/document/4737811/?reload=true" rel="nofollow noreferrer">4</a></p>
| 2651 | Would a PE-pipe filled with concrete make a good pillar? |
2015-04-30T00:15:41.107 | <p>Here is the link to the current CA PE Exam application: </p>
<p><a href="http://www.bpelsg.ca.gov/pubs/forms/ceapp.pdf">http://www.bpelsg.ca.gov/pubs/forms/ceapp.pdf</a> </p>
<p>Page 2, which is the list of engagements/references, states: </p>
<blockquote>
<p>For each engagement claimed as qualifying experience, list the name of the person who will serve as your reference. Individuals serving as references must be licensed as Professional Engineers in California in the discipline of licensure for which you are applying or legally exempt from licensure. Individuals serving as references must have been appropriately licensed, or exempt from licensure, during the period of the engagement. YOU MUST LIST A MINIMUM OF FOUR REFERENCES. At least one of the applicants licensed references must be from someone who is or was in a supervisory capacity over the applicant, for each engagement for which the applicant desires credit. References must be from individual legally authorized to practice civil engineering in the state or country where the projects are located.</p>
</blockquote>
<p>This statement is very confusing. First it says: </p>
<blockquote>
<p>Individuals serving as references must be licensed as Professional Engineers in California in the discipline of licensure for which you are applying or legally exempt from licensure.</p>
</blockquote>
<p>Then later on it says: </p>
<blockquote>
<p>References must be from individual legally authorized to practice civil engineering in the state or country where the projects are located.</p>
</blockquote>
<p>So, which is it: do the references have to be CA PE's, or not? Can someone tell me for sure? </p>
| |licensure| | <p>As it turns out, I have been approved to sit for the Civil PE exam in California this fall even though none of my references were CA PEs. So I suppose that answers my question. </p>
| 2658 | Do references for the California PE application have to be California PE's? |
2015-04-30T06:57:52.027 | <p>Why is a Non-causal PD Controller:
$C(s) = K_ds + K_p$</p>
<p>When a causal PD Controller is:
$C(s) = (K_ds + K_p)/(\tau s + 1)$</p>
<p>The difference being the denominator. $\tau$ is the time constant and is "<em>sufficiently small and does not perturb the other poles substantially</em>".</p>
| |control-engineering|control-theory| | <p>Because with the form:</p>
<p>$C(s) = K_ds + K_p$</p>
<p>you have a non-proper transfer function. In process engineering, you always need to have <a href="http://en.wikipedia.org/wiki/Proper_transfer_function" rel="nofollow noreferrer">proper transfer functions</a>, where the degree of the numerator is less than or equal to the degree of the denominator.</p>
<p>A work around is to introduce an "approximate derivative", which is the second form:</p>
<p>$C(s) = (K_ds + K_p)/(\tau s + 1)$</p>
<p>if $\tau$ becomes too large, it changes the dynamics of the transfer function significantly, and therefore the overall closed-loop response. This also helps reduce the amount of noise generated by the derivative term (acts as a low-pass filter, see <a href="http://en.wikipedia.org/wiki/PID_controller#Noise_in_derivative" rel="nofollow noreferrer">PID controllers</a> on Wikipedia).</p>
<p>In practice, however, derivative gains are rarely used, PI controllers being the preferred implementation.</p>
<p>The causality aspect is because in the non-causal PD controller, you have more zeroes than poles, see <a href="https://dsp.stackexchange.com/questions/10204/why-more-poles-than-zeroes">this discussion</a>.</p>
| 2663 | Causal and Non-Causal PD Controller |
2015-04-30T10:44:18.687 | <p>I am currently doing a project which has led me to my first encounter with IR cameras, and I am therefore quite curious as to how they operate. Specifically, I would like to know the following</p>
<ol>
<li>How is heat converted to an electrical signal (current or voltage)?</li>
<li>How is the <em>spectral bandwidth</em> of an IR camera so well defined?</li>
<li>Why are IR cameras so much more expensive than color video cameras? (Color cameras have IR suppressors, right?)</li>
<li>How are 'regular' IR cameras different from radiometric cameras?</li>
<li>What is the difference between IR cameras which can detect temperatures up to, say 1000 °C, as compared to IR cameras which can detect temperatures up to 400 °C?</li>
</ol>
| |electrical-engineering|thermodynamics|heat-transfer|optics|temperature| | <p>There are a lot of questions in your question, and it should probably be broken up into multiple different questions. I don't want to wait until that happens though so I'll address the ones that I know the answers to. </p>
<blockquote>
<ol>
<li>How is heat converted to an electrical signal (current or voltage)?</li>
</ol>
</blockquote>
<p>A <a href="http://en.wikipedia.org/wiki/Microbolometer" rel="noreferrer">microbolometer</a> is just a special case of a <a href="http://en.wikipedia.org/wiki/Bolometer" rel="noreferrer">bolometer</a> which contains a material whose resistance is very sensitive to its temperature. The change in resistance caused by heating from incident electromagnetic (EM) radiation is read out by a circuit similar to what you will find in a voltmeter. These devices can be designed to be sensitive to incredibly low amounts of power, and generally also have a high dynamic range. The ones I've used in the laser industry are sensitive all the way from 10 mW to 100 W, a dynamic range of 10<sup>4</sup>. </p>
<blockquote>
<ol start="2">
<li>How is the spectral bandwidth of a camera so well defined? </li>
</ol>
</blockquote>
<p>Bolometers are known for having incredibly broad spectral bandwidths. Since the device actually measures the heat deposited by the EM radiation, the bandwidth of the detection material itself (usually either <a href="http://en.wikipedia.org/wiki/Amorphous_silicon" rel="noreferrer">amorphous silicon</a> or <a href="http://en.wikipedia.org/wiki/Vanadium_oxide" rel="noreferrer">vanadium oxide</a>) is defined by the wavelengths at which it absorbs. The bandwidth of the microbolometer detectors must therefore be defined with external optics which reject or absorb the other wavelengths. My guess would be that they use an absorbing <a href="http://www.edmundoptics.com/optics/optical-filters/bandpass-filters/infrared-ir-bandpass-filters/3387/" rel="noreferrer">IR bandpass filter</a> in front of the detector surface. </p>
<blockquote>
<ol start="3">
<li>Why are IR cameras so much more expensive than color video cameras? (Color cameras have IR suppressors, right?)</li>
</ol>
</blockquote>
<p>I don't know exactly, but the ability to manufacture these things <em>en masse</em> only became possible in the last few years while <a href="http://en.wikipedia.org/wiki/Charge-coupled_device" rel="noreferrer">charge-coupled device (CCD)</a> detectors have been in mass production since the 1980's. You are correct that CCD detectors incorporate an IR filter, but the underlying materials are only sensitive down to ~1-2 μm so they don't work in the deep IR like the microbolometers do. </p>
<blockquote>
<ol start="5">
<li>What is the difference between IR cameras which can detect temperatures up to, say 1000 °C, as compared to IR cameras which can detect temperatures up to 400 °C?</li>
</ol>
</blockquote>
<p>All warm bodies emit radiation across the full range of the EM spectrum. The spectral content of the radiation emitted at a given wavelength is given very nearly by <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/bbrc.html" rel="noreferrer">Plank's blackbody curve</a> (shown below). One of the key features of this curve is that the peak of the emitted radiation shifts towards longer wavelengths with lower temperatures. The peak of the emitted radiation is given by <a href="http://en.wikipedia.org/wiki/Wien%27s_displacement_law" rel="noreferrer">Wien's law</a> which is
$$
\lambda_{max}=\frac{b}{T}
$$
where $b$ is Wien's constant ($b=2.8977721\cdot10^{3}\:\mathrm{m*K}$) and the temperature $T$ is in units of Kelvin. From this you can calculate that the peak wavelengths of the temperatures you ask about are: $\lambda_{1000}=2.3\ \mu \text{m}$ and $\lambda_{400}=4.3\ \mu\text{m}$. So, detectors which are designed to be sensitive to those different temperatures are simply tuned (probably by tuning the bandpass filter in front) to be more sensitive at the different peak wavelengths. </p>
<p><a href="http://en.wikipedia.org/wiki/File:Wiens_law.svg" rel="noreferrer"><img src="https://i.stack.imgur.com/VKNeq.png" alt="Blackbody curve"></a></p>
| 2664 | How does a microbolometer (IR camera) work? |
2015-04-30T22:50:14.330 | <p>I'm trying to design a bioreactor to produce citric acid with whey and Aspergillus Niger. </p>
<p>The first step in the process would be to put whey into the reactor with some dextrose (approx. 10%). Then, this solution should be sterilized, and I was thinking of using batch steam sterilization by direct injection of the steam in the reactor. The sterilization should be done at 121°C.</p>
<p>I'm not sure how to calculate the sterilization time and the amount of steam required for a given reactor volume (approx. 600 m<sup>3</sup>). I was thinking of calculating it as if I had a 600 m<sup>3</sup> autoclave, but I doubt the approximation holds. In that case, the death kinetics would be given by the <a href="http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Projects00/sterilize/batch.html">Arrhenius equation</a>:</p>
<p>$$
k = Ae^{\frac{-E}{RT}}
$$</p>
<p>I'd need to determine the activation energy, $E$, the pre-exponential factor $A$, or some approximation for $k$. Then, I'd just have to integrate the equations for the heat and hold energy required:</p>
<p>$$
\begin{align}
\nabla heat &= ln(\frac{N_0 V_0}{N_1 V_1}) = \int_{t_0}^{t_1} k(t) dt \\
\nabla hold &= ln(\frac{N_1 V_1}{N_2 V_2}) = k(t2_-t_1) dt
\end{align}
$$</p>
<p>To figure out $t_1$ and $t_2$. </p>
<p>Am I approaching this problem correctly?</p>
| |chemical-engineering|process-engineering|steam| | <p>It all depends on the sterilization level you want to achieve. As you say, you would need to know the death kinetics parameters, or equivalently the log-decimal reduction time (D) and how it changes with temperature (Z). As sterilization is a problem that takes place almost everywhere in food industry, a standard sterilisation cycle lasts for 30 min. at 121 degC at the cold spot of your reactor. This ensures that the spores of many <em>Bacillus</em>, <em>Clostridium</em> and other thermophilic microorganisms cannot survive.<br>
For a 600 m3 reactor the easiest is to have a temperature probe inside of the reactor.
As how much steam? That depends on how well isolated your reactor is.
The best is to make a feedback loop with setpoint 121 degC that would send steam whenever temperature starts to drop. Otherwise, you can try to do the same manually. </p>
| 2673 | Steam sterilization of whey in a bioreactor |
2015-05-02T09:20:04.410 | <p>I'm trying to develop a thermoelectric (Peltier tile-based) beverage cooling system. Ideally, I'd also like a device to solve the warm beer problem if the thermodynamics give me any confidence.</p>
<p>First, I'll detail my thought flow in the theory sense and then I'll get to my specific setup that I had in mind.</p>
<p>Obviously, if one immerses a 100 W, 10% efficient Peltier tile in a glass of water at 298 K (let's say, containing 10 mol exactly), if the container system is assumed to be adiabatic in nature, to drop the temperature to 274 K (roughly 35 °F) requires:</p>
<p>$$
Q = m \, c_p \, \Delta T = 180.2 \times 4.1813 \times 24 = 18083.3 \:\mathrm{J}
$$</p>
<p>Then, since the efficiency gives the nominal "heat transfer" power to be 10 W:</p>
<p>$$
t = \frac{18083.3\:\mathrm{J}}{10\:\mathrm{J/s}} = 1808.33 \:\mathrm{s}
$$</p>
<p>So, the cooling would take about 30 minutes to chill the glass down, much more when you consider that a glass of water doesn't adhere closely to adiabatic behavior.</p>
<p>The actual setup (poorly-drawn hand sketch follows):</p>
<p><img src="https://i.stack.imgur.com/FNid4.jpg" alt="The experimental setup"></p>
<p>Now, obviously there are four sets of values to consider here:</p>
<ul>
<li>The specific heat and thermal conductivity of the beverage itself</li>
<li>The specific heat and thermal conductivity of the glass container holding the beverage, easily <em>approximated</em> by making the assumption that the glass can be characterized by fused silica</li>
<li>The specific heat and thermal conductivity of the water transfer fluid, well-established</li>
<li>The specific heat and thermal conductivity of the 6061 aluminum working chamber, which is again well-established</li>
</ul>
<p>My problem is how to model the system and get model calculations for the transfers between each step, so, for example, solving the equation so I know if I use a given cumulative wattage of Peltier tile, it'll reduce the temperature of the system by a given amount in a certain amount of time.</p>
<p>As before, I'm willing to concede, for the purposes of discussion, that this system can be assumed to be adiabatic with respect to the external environment; that is, no heat is absorbed from the environment during the cooling process.</p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <p>There are two wattage ratings for a peltier plates. One is the power consumed, the other is the heat transported across the chip <strong>when there is a temperature difference of zero</strong>. An important note is that the amount of heat transported across the plate is dependent on the temperature difference across the plate.</p>
<p><img src="https://i.stack.imgur.com/fIOiJ.gif" alt="Peltier graph"></p>
<p>Image courtesy of:
<a href="http://www.heatsink-guide.com/peltier.htm" rel="nofollow noreferrer">http://www.heatsink-guide.com/peltier.htm</a></p>
<p>Using the graph above as an example, if the hot side of the peltier plate is not attached to a heat sink then it will quickly heat up to 70 degrees hotter than the beverage unit and heat will stop flowing outward. At that point if this plate was 10% efficient it would still be producing out about 400W of heat. That heat would be produced on the hot side of the chip* but that means that without a heat sink of some sort the hot side would quickly become even hotter than 70 degrees above the beverage unit and heat would start to flow back into the beverage unit, heating it rather than cooling it.</p>
<p>Thus the heat sinks on your beverage "cooling" unit will be very, very important to how well your system cools (or heats) your beverage, and without any information on them my guess is that your system will heat your beverage.</p>
<p>*The heat is actually produced throughout the chip but it gets transferred out along with the other heat and the graph is compensated to account for this extra heat transfer, and only show the heat transferred away from the cool side.</p>
| 2676 | Determining behavior of heat exchange system |
2015-05-02T10:03:48.727 | <p>I have 3-axis accelerometer and 3-axis gyroscope sensor data coming from my bike trip. Until now I have extracted only basic information like overspeeding, hard acceleration and sudden turns.</p>
<p>What additional motion characteristics can I infer from this data? I also have x, y coordinates.</p>
| |mechanical-engineering|electrical-engineering|sensors| | <p>This information you have, in theory, allows complete reconstruction of location and orientation of the bike along the entire path. This is basically what is referred to as <i>inertial navigation</i>.</p>
<p>However, the problem in real life is that there will be some error on all the signals. Since determining location and orientation require integrating the signals you have, these errors accumulate over time. Note that position is the second integral of acceleration. After some time, the resulting location and orientation data will be so far off as to be useless.</p>
<p>For small and cheap sensors like you can put on a bicycles, the integrals may only be useful for a few seconds, tens of seconds at most. I once worked on a device that tracked head motion during a golf swing, and it likewise had 6-axis acceleration and rotation rate data. The useful integration time window was about 2 seconds.</p>
<p>So to really answer the question, you can only use this kind of data for characterizing short term conditions or events. If used in conjunction with something that is long-term stable, like GPS, then you can use it to fill in the high frequency information between GPS readings. However, for an ordinary bike ride, the extra position accuracy is of little use.</p>
<p>You can use this data on its own for its high frequency information, like smoothness/roughness of the road, number of larger pothole incidents, how wobbly/steady the bike is, how much it sways side to side when you are pedaling, when and how hard you braked, etc.</p>
| 2677 | What kind of motion information can be extracted from 3-axis accelerometer and 3-axis gyroscope sensor data |
2015-05-02T13:48:56.413 | <p>I need to connect two flat bars of metal with some kind of bearing that doesn't increase the joint thickness by more than 2-3mm (the less the better), as shown here:</p>
<p><a href="https://i.stack.imgur.com/9nH2w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9nH2wm.png" alt="two bars of 50x5 mm cross-section fastened near their ends on a 30- to 180-degree swivel with axis of rotation parallel to the 5 mm dimension"></a></p>
<p>There are almost no axial loads but the peak lateral (shearing) stress on the bearing would be of order of 2000 N with 800 N sustained being the norm and the directions of the stress varies (although axial respective to the lower beam would be most common and strongest). </p>
<p>There is a serious restriction on thickness of the joint - the present 10 mm is already pushing the envelope; another 2-4 mm would be acceptable though unwelcome.</p>
<p>The bars will be steel or duraluminum. There are extra braces, etc., which prevent buckling and limit stress in the perpendicular direction to negligible levels, but the two bars need to carry axial stress and lateral stress "in the thick direction".</p>
<p>Moderate friction is acceptable - 5 N·m friction torque is still a reasonable value. Rotation is moderately fast, at most ~1 rev/s, usually much less. Also the range of rotation is limited; only 30-180 degrees angle between the two bars is needed (and even a little less than that if necessary). Lifetime doesn't need to exceed 1000 hours; less for the prototype.</p>
<p>This will be prototyped using a common DYI workshop without specialized machinery like CNC devices, also preferably no very expensive solutions - of course purchase of a pre-made bearing fitting the criteria would be perfectly acceptable if the price is not excessive.</p>
<p>What kind of bearing or joint would fulfill these prerequisites?</p>
<p>EDIT: Let me add a picture explaining the general layout of forces:</p>
<p><img src="https://i.stack.imgur.com/BoZKh.png" alt="drawing explaining layout of the forces"></p>
<p>In general situation any common bolt, bushing (with flanged axis) etc. would be sufficient. Here the problem is getting it all to be narrow enough not to extend way out of the narrow plane occupied by the two bars, and not fall apart.</p>
| |mechanical-engineering|bearings| | <p>I recommend a sliding contact bearing. It's easy to manufacture and seems to do the job for a prototype, especially if it has to be cheap. Steel seems to be the better choice in this case.</p>
<p>Rather use two thinner bars for one of the links, so you get a double lap. This will reduce the moment on the bearing and yields a more even contact stress distribution. So if you can really spend 3 mm more, you can split into two 4 mm bars instead of one 5 mm.</p>
<p>Use a drill press and a reamer to achieve small clearence. All pieces of the bearing point can be drilled and reamed together. But for the middle bar, the inner one, you may use a press fitting. Thus the bolt will only move on the outer bars where a transition fitting shall be a good choice.</p>
<p>Perhaps you want to invest in bronze sleeves for the bearings. It will lower the friction and result in longer lifetime.</p>
<p>P. S.: If you have to chill something to a low temperature, it's often enough to use dry ice (CO2) and ethanol or aceton. This yields about -75 °C. It's not LN2 but often you don't need -195 °C.</p>
| 2680 | How to design a narrow bearing that supports shear direction load |
2015-05-04T04:18:36.493 | <p>I am rebuilding a mini soil compactor roller. </p>
<p>I replaced the old 8 hp Kubota diesel engine unit with a new Dongfeng unit. I want to weld the coupling shaft, which is built from 2" pipe with rubber damper at one end, to the hydraulic motor. </p>
<p>How can I ensure the shaft is balanced after we weld it to the flywheel? We would like to do this cheaply if possible.</p>
| |mechanical-engineering|flywheels| | <p>With some low friction bearings you can spin it on the shaft and wait for it to come to a rest and mark the low point, do this a few times and you may see them gathering at the bottom.</p>
<p>Then drill out a portion of the wheel where it is heaviest and repeat.</p>
| 2686 | Balancing flywheel and coupling shaft at home |
2015-05-05T05:09:47.133 | <p>I have been wondering about this question for quite some time. Assuming an ideal case, the energy from photons hitting solar cells is converted into electric energy as described by the equation:</p>
<p>$RI^2t=W\equiv E=\hbar\nu$</p>
<p>where $\nu$ is the frequency of photons. Using a lens won't increase the frequency of photons, thus no extra electricity is generated.</p>
<p>Am I correct in thinking that no extra electricity will be generated by solar cells when a lens is used to focus light onto them?</p>
| |electrical-engineering|optics|photovoltaics|solar-energy| | <p>A new way to convert solar power has been discovered at the University of Michigan. Please check out :
<a href="https://phys.org/news/2011-04-solar-power-cells-hidden-magnetic.html" rel="nofollow noreferrer">https://phys.org/news/2011-04-solar-power-cells-hidden-magnetic.html</a></p>
<p>It uses the magnetic component of light which manifested itself when high intensity light goes through a transparent but non-electrically conductive material, glass for example. The light must be focused to an intensity of 10 million watts per square centimeter. Sunlight isn’t this intense on its own, but new materials are being sought that would work at lower intensities.</p>
<p>Your concentration of light using lenses and mirrors has limited potential for extracting more energy from the conventional solar cell, but it would certainly increase electrical energy with this new method.</p>
| 2690 | Will more electricity be generated by using a lens to focus sunlight onto solar cells? |
2015-05-05T18:31:05.723 | <p>I have made a Python program in which I have computed the deflection of a beam.</p>
<p>I am plotting the deflected beam and am now asked to show the length of the beam.</p>
<p>If the beam is not deflected, it is easy to show the length of the beam but when the beam is deflected, I guess the length of the beam is longer than the undeflected beam. How can I compute the length of the deflected beam? Is it correct to distinguish between the length of the deflected and undeflected beam?</p>
| |mechanical-engineering|structural-engineering|structures|beam|statics| | <p>A subtle bit of wording - the exact length does not change, as pointed out by Air above in his answer. However, the horizontal projection of the beam does change, as the shortest path between two points will always be a straight line. Thus, by curving, the tip of the beam will have to move backwards a bit in the x direction, to account for the deflection in the y direction.</p>
<p>The exact change is minimal. For a beam whose deflection $y(x)$ has been defined in terms of x, the change in "length" can be determined by the slope function $\theta(x) = y'(x)$</p>
<p>$$\Delta L = -\frac{1}{2}\int^L_0(\theta(x))^2dx $$</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/0071742476" rel="nofollow">(Ref Roark's formulas for stresses and strains, 8th edition)</a>, Eq. 8.1-14. Note it is always a shrinking of the beam.</p>
| 2693 | Does the length of a beam change upon deflection? |
2015-05-05T20:20:10.200 | <p>If I use an ultrasonic sensor will it detect the water level? </p>
<p>I was thinking about a product to read water level on water boxes (common in Brazil). I researched about instrumentation for this measure, and I think that an ultrasonic sensor is the best option. Will the water correctly reflect ultrasound and not change the normal measurements against a solid obstacle? </p>
| |electrical-engineering|measurements|sensors|control-engineering| | <p>Or you could just buy one. They are used in oil industry to measure liquid level in wells. They are necessary to turn the pump jacks off /on. </p>
| 2695 | Can I use a ultrasound sensor to measure water level? |
2015-05-06T02:29:30.643 | <h3>Weight of Structural Steel</h3>
<p>When calculating the weight of a steel structure, I usually calculate the weight of all of the various members and plates first. I then add an additional percentage to account for the weight of nuts and bolts. This percentage is usually between 2% and 5% of the total weight.</p>
<p>It makes sense that bolts will add additional weight beyond just the weight of the members. It is also complicated and time consuming to do it the "proper" way, e.g. deduct the weight of the hole and add the weight of the correct length and size of bolt. This is done with a spreadsheet or by hand and without the aid of specialized steel detailing software. Another complication is that furnished bolt lengths are partially determined by the fabricator.</p>
<h3>The Problem</h3>
<p>My company has always done this. The one time that a third party reviewer asked why the steel weight was ~5% higher, we were unable to cite a code reference where this was called for.</p>
<p><strong>How is the weight of fasteners typically included in steel weight calculations?</strong> This can either be for a bid weight or a design dead load weight.</p>
<p>As additional information, in all welded construction, the weight of bolts might be so minor that it is negligible. Some of my projects use completely bolted connections including built-up members.</p>
| |structural-engineering|steel| | <p><a href="http://www.dot.ca.gov/hq/esc/techpubs/manual/bridgemanuals/bridge-design-aids/page/bda_11.pdf" rel="nofollow">Caltrans Bridge Design Aids, Section 11</a> adds 3% for welds and bolts:</p>
<blockquote>
<h3>Furnish Structural Steel (Bridge) and Erect Structural Steel (Bridge)</h3>
<ol>
<li>Segregate by type of steel and estimate in kilograms.</li>
<li>Include bearings (except PTFE Spherical Bearings), anchor bolts, shear connectors, and expansion dams (except where expansion dams are
galvanized or embedded in concrete).</li>
<li>Ignore small, non-repetitive cuts, copes, and bolt holes.</li>
<li>Add 3% for welds and bolts.</li>
</ol>
</blockquote>
<p>This is applicable to new bridges -- retrofits may have a much higher percentage of bolt weight.</p>
| 2699 | Is there a standard percentage to increase structural steel weight to account for bolts? |
2015-05-06T12:49:24.117 | <p>I'm looking for a power source that is able to deliver 12V at 15-30A (yes, amps). I was looking at high capacity batteries, but looking at the specs, they either say they have a lower discharge voltage than 12V, or they don't say anything at all. </p>
<p>Do such power sources exist?</p>
<p>I'm hoping to use this in a hydraulic system. The pumps I've seen require 12V/24V and consume around 10-15A minimum. I've been looking at Hydraulic Power Units (HPUs), and their specs say that power consumption is really high.</p>
<p>Reading around, it seems most, if not all, hydraulic systems are powered by small diesel engines. I would really like to avoid that, if it is possible.</p>
<p>If such high capacity power sources do not exist, what would be my alternative?</p>
<hr>
<p>Apologies for the above sounding like a "find things for me" question. This was not the intention. Maybe I haven't actually understood fully how hydraulic systems actually are designed. Also, I'm not very good with words.</p>
<p>I am new to hydraulic systems, and I am in the process of designing a portable hydraulic system, similar to ones used in vehicles, like construction vehicles. However, my hydraulic system will not be part of a vehicle. I realise vehicles use their engines and some form of petrol/diesel to electricity generator to obtain the power required, but I cannot see a way of obtaining that same power without using a vehicle's engine. </p>
<p>How would I go about getting that same power without the use of a vehicle? Ideally, the system can be moved around without it being attached to AC wall electricity, but that can be left until a later date.</p>
<hr>
<p>Details/Background of my project:</p>
<p>I would like to build a hydraulic exoskeleton that uses as little electrical components as possible. I have gone for hydraulic over pneumatic as hydraulic packs a lot more punch than pneumatic. Most of the sites online sell the valves, pumps, pistons etc, but I have previously found it difficult to find a way of powering these pumps, up until I asked on here <a href="https://engineering.stackexchange.com/questions/481/motor-for-a-hydraulic-pump-in-a-hydraulic-system">a while back</a>, where I was introduced to HPUs. From what I have understood (do correct me if I'm wrong), I need a way of powering these HPUs to generate the necessary RPM and flow rates.</p>
<p>I have found the pistons I would like to use, the hydraulic fluid, how much to use, the reservoir tanks, the filtering system, etc, but I am stuck on the power source. I just have no idea how to harness the power I need for this. This is where I started looking at ultra-high capacity batteries.</p>
<p>To be honest, I don't even know which would be best, a DC- or AC-input motor. My thought was that the constant forward drive of the motor required a DC source, hence the batteries. The link I provided in the comments below points to AC motors.</p>
<p>I'll keep adding details if and when they are required. As I said, I'm not good with words and I'm new to hydraulics.</p>
| |electrical-engineering|power-electronics|hydraulics| | <p>Pretty much all electric motors demand a lot more power for starting from a dead stop than for generating motion in a steady state. There are a whole lot of schemes for smoothing out this 'inrush current' demand, either from the electrical side or from the mechanical side (by decoupling the load and gradually re-connecting it once the motor spins up.) The problem with HPUs designed for use on vehicles is that they are designed with the assumption that the power source has a battery already there to provide the starting current for the engine starter. These starting batteries can typically provide hundreds of amps for a few seconds - much more than a modest HPU will require.</p>
<p>Because they usually have such a large power source available, and because all of the strategies for reducing inrush current have some cost and complexity associated, vehicle HPUs are generally designed with no intention to minimize the inrush current requirements. This is made further more complicated as solenoids and relays are often used for the control systems on mobile hydraulics, which have inrush currents of their own, often applied at the same time as the pump starting.</p>
<p>While automotive batteries can provide extremely high currents (for a short time) most AC to DC power supplies can only deliver maybe 2 times as much power momentarily as they can in the steady state. This means that if you connect a vehicle hydraulic system to a AC to DC power supply rated for the same current that the HPU needs in the steady state, you likely won't have enough current to start the system. As a result, the pump will be starved for power and you may hear a whine or just notice it getting very hot without turning. Additionally, any relays of solenoids will likely 'chatter.' This is because on their own, there is enough power form the power supply to actuate the solenoid, but as soon as the motor is connected, the voltage drops too low and the solenoid opens again, restarting the cycle. Both of these conditions will damage your HPU and should not be sustained for more than a second.</p>
<p>The simplest strategy is to provide an automotive battery (typically lead-acid,) like the HPUs are designed to run off of. Depending on the size of your HPU and the length of time you need it to run on a charge will dictate the size of the battery you need. Automotive batteries are produced in a range of sizes designer for everything from starting very small cars to powering entire RVs for days without running the engine. They can also easily be combined in parallel. You mention that voltage form these batteries can sometimes drop below their 12 volt nominal value. That is true as they loose charge, but most HPUs can handle a significant voltage drop below nominal (10-20%) before performance is seriously effected. If the use will be steady state, a battery can be provided in addition to an Ac to DC power supply, which will maintain the voltage and keep the battery charged, just like a battery in a car is fed by an alternator once the engine is running.</p>
<p>If you must power your HPU without a battery at all, you need to plan for your AC to DC power supplies to have 2 or 3 times as much available current as the nameplate current demand of the HPU, or implement your own inrush limiting power supply (typically some combination of chokes and capacitors, but I don't know much about how they work.) Power supplies in the range of 100-200A at 12VDC are commercially available. If you need more than this, you can choose power supplies designed to be put into a parallel master-slave configuration and connect two or three power supplies together. One manufacturer of such supplies is MeanWell.</p>
| 2705 | Hydraulic power source |
2015-05-06T15:19:00.057 | <p>In <a href="http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000036534.pdf">this NASA document</a>, it is mentioned on page 221 (239 of the pdf) that a "23 tol gear box" was used. What does that mean?</p>
<p>A google search for "23 tol gear box" (with quotes) comes up completely empty, and "tol gear box" shows only irrelevant results. A search for just tol doesn't come up with any convincing results either. I can't rule out that it is a typo of some sort. </p>
<p>From the context I presume it to mean that the gearbox output runs 23 times as fast as the powered side, but I have to be certain. </p>
| |mechanical-engineering|gears| | <p>Oh, I get it. The typo is a missing space. It's supposed to mean "23 to 1".</p>
| 2714 | What does "tol" mean in a gear box? |
2015-05-06T19:44:03.523 | <p>Note: This may be UK-specific, I'm not sure.</p>
<p>I overheard part of a conversation about 3-phase electricity, one guy was saying that mains generators use 3 coils (or something) and as it spins it produces 3 different waves of alternating current that are 120 degrees apart. That's why the power lines often have three (or six) sets of cables, one for each phase. He said that if different phases are electrically connected then bad things happen, and sometimes if only one phase goes down, every third house on a street could lose power.</p>
<p>This raised a few questions for me:</p>
<p>Does that mean all power stations have to be in sync with each other? If so how is this achieved?</p>
<p>Is it true about every third house, I've never seen this?</p>
<p>Does this mean we effectively have 3 grids overlaid on top of each other?</p>
| |electrical-engineering| | <p>In North America, you generally don't see 3 phase power being delivered to houses. The transformers (whether they're up in a pole, or in a green box on the ground) take apart the 3-phase 1100V power delivered to them and step it down to 110V. For 220V, two feeds are connected appropriately, and 220V is delivered to each breaker box.
In commercial and industrial settings, you can get three phase power delivered. We have big air compressors and hoists that run on 3 phase 220V. A motor has to be designed to run that kind of power. We have had outages on one and on two of the three phases. Bad things happen. Things draw WAY more amperage than they should from the remaining power and smoke and unhappiness results. In our old building, this happened it was because it was old and a circuit-breaker for that purpose wasn't installed. In our new building, the electrician was incompetent and didn't install it. Large installations of fluorescent lights will use 3 phase 367V power. The outage in the new building wrecked the lighting so that afterward, 2 of the 5 bulbs in each fixture wouldn't light. All the ballasts needed to be replaced.
We also have transformers in the building that step up the 3 phase 240V delivered to 3 phase 400V and 600V for some specialized equipment.</p>
<p>But in residential neighborhoods, what gets delivered to the house is one phase power 110V on two buses that, when wired correctly can give you 220V power. And if all three phases aren't being delivered correctly to the transformer, a circuit-breaker in that trips and nobody gets any power.</p>
<p>Burning down people's home because of power snafus isn't good karma. The power company ensures that it won't be their fault! </p>
| 2721 | Does a 3-phase electricity grid need to be synchronised and are the phases split between houses? |
2015-05-06T20:51:48.677 | <p>In order to make this a manageable question, let's add a few simplifications. </p>
<ol>
<li>The dust particles can be well described as uniform spheres of radius $R$ and density $\rho$. </li>
<li>The space is enclosed and there is no bulk flow, i.e the air is still in a macroscopic sense.</li>
<li>The air is at the <a href="http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure">standard temperature and pressure (STP)</a>; $T=20\ ^\circ\mathrm{C}$ and $P=1\ \mathrm{atm}$.</li>
</ol>
<p>Under these conditions, what is the settling time for dust particles? At what size/density does Brownian motion of the air become important? </p>
| |fluid-mechanics|air-quality| | <p>Solid particle settling time in air depends mainly on the size of the particle. Different forces become significant depending on what size range you're talking about, so it's hard to give an answer that's both concise and accurate.</p>
<p>I'll do my best to synthesize the important points rather than parrot a reference; that said, where practical applications in the field of air quality are concerned, the text I recommend is <a href="http://www.waveland.com/browse.php?t=380&pgtitle=Air%20Pollution%20Control%20by%20Cooper-Alley" rel="noreferrer">Air Pollution Control by Cooper & Alley</a>. In particular, I'm going to pull many of the details for this answer from Section 3.3: Particulate Behavior in Fluids.</p>
<h2>Gravitational Settling Overview</h2>
<p>Dust doesn't behave like <a href="http://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment" rel="noreferrer">Galileo's bocce balls</a>; small particles of different sizes fall at different rates. <strong>For solid particles, variation in settling velocity is due mainly to the influence of drag forces.</strong></p>
<p>You might expect that Brownian motion would "juggle" very small particles around, keeping them from settling. Sufficiently small dust particles <em>can</em> remain entrained indefinitely but, practically speaking, that has more to do with the air never being perfectly still than it does with Brownian motion. In the context of air quality, we care about Brownian motion mainly when considering impaction (e.g., on water droplets in a <a href="http://www.epa.gov/ttncatc1/dir1/cs6ch2.pdf" rel="noreferrer">PM wet scrubber</a>) or deposition (e.g., <a href="http://www.flow3d.com/wp-content/uploads/2014/08/Particulate-Matter-Modeling-in-Near-Road-Vegetation-Environments.pdf" rel="noreferrer">on foliage near roadways</a>). Neither of these mechanisms are relevant to the case of pure gravitational settling.</p>
<p>In fact, when a solid particle gets small enough to start considering the motion of discrete air molecules, we find that it actually settles a bit <em>more</em> quickly than <a href="http://en.wikipedia.org/wiki/Stokes%27_law" rel="noreferrer">Stokes' law</a> implies. This is when we apply the experimentally-determined <a href="http://en.wikipedia.org/wiki/Cunningham_correction_factor" rel="noreferrer">Cunningham slip correction factor</a> to reduce the Stokes drag coefficient. The correction factor in air is related to the particle diameter <span class="math-container">$d_p$</span> and the <a href="http://en.wikipedia.org/wiki/Mean_free_path#Mean_free_path_in_kinetic_theory" rel="noreferrer">mean free path</a> <span class="math-container">$\lambda$</span> by:</p>
<p><span class="math-container">$$C = 1 + 2.0 \frac{\lambda}{d_p} \left [ 1.257 + 0.40 \exp(-0.55 \frac{d_p}{\lambda}) \right ]$$</span></p>
<p>As for what "small enough" actually means, the Cooper & Alley text says:</p>
<blockquote>
<p>For particles smaller than 1 micron, the slip correction factor is
always significant, but rapidly approaches 1.0 as particle size
increases above 5 microns.</p>
</blockquote>
<p>That could be justification enough to spare yourself the time or processing cycles required to calculate the correction factor when all you're concerned with are relatively large particles.</p>
<h2>Equation of Motion</h2>
<p>We can derive an equation of motion in one dimension as follows.</p>
<ol>
<li>Apply Newton's second law to the particle in terms of its relative velocity in the fluid.*
<span class="math-container">$$m_p v_r' = F_g - F_B - F_D$$</span></li>
<li>Stokes' law gives the drag force in terms of the viscosity of the fluid and the velocity and diameter of the particle; the buoyant force is equal to the weight of the displaced fluid.
<span class="math-container">$$m_p v_r' = m_p g - m_{air} g - 3 \pi \mu d v_r$$</span></li>
<li>Divide by the mass of the particle.
<span class="math-container">$$v_r' = g - \frac{m_{air}}{m_p} g - \frac{3 \pi \mu d}{m_p} v_r$$</span></li>
<li>Express mass as the product of volume and density, where the volume of the particle and the volume of displaced air are the same.
<span class="math-container">$$v_r' = g - \frac{\rho_{air}}{\rho_p} g - \frac{3 \pi \mu d}{\rho_p V} v_r$$</span></li>
<li>Using <span class="math-container">$V_{sphere} = \frac{1}{6} \pi d^3$</span>, simplify the drag force term and move it to the left side.
<span class="math-container">$$v_r' + \frac{18 \mu}{\rho_p d^2} v_r = (1 - \frac{\rho_{air}}{\rho_p}) g$$</span></li>
</ol>
<p>This is a linear ODE with a known coefficient (at STP) representing the following <em>characteristic time</em> for settling particles:
<span class="math-container">$$\tau = \frac{\rho_p d^2}{18 \mu}$$</span></p>
<p>The characteristic time is a useful parameter for comparing the behavior of different systems of particles dispersed in fluids, similar to how the Reynolds number can be used to identify when different systems will have similar flow regimes. Applying the Cunningham slip correction factor gives the slip-corrected time <span class="math-container">$\tau' = C \tau$</span> and the equation of motion that I'll use in the next section:
<span class="math-container">$$v_r' + \frac{v_r}{\tau'} = (1 - \frac{\rho_{air}}{\rho_p}) g$$</span></p>
<p><sub>* The coordinate system for this example is defined such that the falling velocity is positive.</sub></p>
<h2>Terminal Velocity</h2>
<p>For a solid particle falling in air, <span class="math-container">$\dfrac{\rho_{air}}{\rho_p}$</span> is close to zero. Under that assumption, setting <span class="math-container">$v_r' = 0$</span> in the equation of motion gives the terminal settling velocity of the particle:
<span class="math-container">$$v_t = \tau' g$$</span></p>
<p>Using that terminal velocity, the solution of the equation of motion can be expressed as:
<span class="math-container">$$\frac{v_r}{v_t} = 1 - e^{-t \over \tau'}$$</span></p>
<p>By the time <span class="math-container">$t = 4 \tau'$</span>, the particle has already reached 98% of its terminal velocity. If you calculate the characteristic time for dust particles, you'll see that this takes only fractions of a second; dust particles spend most of their settling time falling at terminal velocity. The velocity itself varies significantly with particle diameter, but <strong>it can take anywhere from hours to days for <a href="http://www.epa.gov/pmdesignations/faq.htm#0" rel="noreferrer">fine particulates</a> to settle just a few meters</strong>.</p>
<h2>Larger Dust</h2>
<p>This is all well and good for smaller dust, but what about the bigger stuff that gets in your eyes and makes you cough? Well, bad news from Cooper & Alley:</p>
<blockquote>
<p>For a particle larger than 10–20 microns settling at its terminal
velocity, the Reynolds number is too high for the Stokes regime
analysis to be valid. For these larger particles, empirical means are
required to obtain the settling velocity...</p>
</blockquote>
<p>"Empirical means" is a nice way of saying <em>figure it out yourself</em> or else get used to reading charts that plot fitted curves with ugly decimal exponents to the results of previous experimentation.</p>
| 2725 | How long does it take for dust to settle out of the air? |
2015-05-07T08:23:58.640 | <p><strong>Background</strong>
I am working on a project that involves heating air inside a small box, then measuring the temperature over time. The purpose of the box is to be a basic test chamber for PID experiments.</p>
<p><strong>Initial Test</strong>
I made a 10 cm * 10 cm * 10 cm box from 3 mm plywood. I then placed a small ~1 W heater inside with a temperature probe and observed the temperature rise... very slowly. </p>
<p><strong>Proposed New Prototype</strong>
I am going to construct a new prototype once I decide how to calculate the approximate time to warm the air inside the box to a set-point. Some specifications and assumptions are laid out below:</p>
<ul>
<li>The ~1 W heater consists of 4 parallel 100 ohm resistors and a 5 V power supply (there is also a 330 ohm resistor and an LED for visual indication). I don't want to change this heater design, as the intention is that this is easy to replicate and uses readily accessible 5 V power sources and electronic components.</li>
<li>Currently no fan has been used but I have a small 20 mm, 5 V fan on order, so will integrate this into the design at some point.</li>
<li>The box will be used at room temperature between 21–23 °C</li>
<li>Ideally a temperature rise up to 30 °C (or more) would be possible over a 5-10 minute time interval</li>
<li>The new proposed box size is 5 cm * 5 cm * 5 cm, 8 times smaller than the previous prototype. This is open to change depending on the previous 2 requirements of maximum temperature rise and time taken to achieve that change.</li>
</ul>
<p>I am not after exact timings - I am aware that losses through the box will have an effect. But if this effect is minimal, then a simplified approximate solution is preferable as some design parameters may change slightly. I am looking for any guidance and calculations that will save me time instead of having to make different boxes and learn by trial and error.</p>
| |thermodynamics|heat-transfer|heat-exchange| | <p>A simple approach would be to just consider the heat added.</p>
<p>The specific heat equation gives the temperature change due to heating as</p>
<p>$$Q=mc\Delta T$$ where $Q$ is heat added, $m$ is mass. $c$ is the specific heat of the material in question (~1.005kJ/kgK for air at room temperature) and $\Delta T$ is the temperature change.</p>
<p>A 5x5x5 cm box will have an air mass of ~0.15 g (air density ~1.2 kg/m$^3$).</p>
<p>For a 1W (1 J/s) heater this gives a temperature change of
$$\Delta T = \frac{Q}{mc}= \frac{1}{0.00015*1005}= 6.6 K/s$$ </p>
<p>I will suggest this is much higher than you observed with the larger box (even accounting the increased volume) which suggests there is a factor not being considered.</p>
<p>Two factors leap to mind.</p>
<p>1) Heat losses from the box. Unless your box is well insulated it will lose heat via radiation and conduction. However for temperatures near room temp. I would expect these loses to be low.</p>
<p>2) Heating the box. Inevitably if you heat the air in the box this will transfer to the box itself. The box has much larger weight so will require more energy to heat (probably 100x or more). Intuitively I would think the transfer of heat from the air to the box would be slow compared to the heating of the air. But I am not an expert in this so may be wrong. </p>
<p>More likely I think the problem may be that most of the heat is going into the box initially if the heater is in contact with it, due to much much better thermal conductivity of wood than air. This is what I would look at to improve the rate of heating.</p>
| 2734 | Heating the air in a small box |
2015-05-07T17:39:16.427 | <p>I understand the difference between a #3 finish on a piece of stainless steel compared to a #4 finish being that the surface roughness is different. Specifically, a #4 finish is classified as having an Ra value of less than 25 micro-inches where as a #3 finish has an Ra value of less than 40 micro-inches. </p>
<p>The basis of my question is if I have two cabinets made from 304 SST, one has a #3 finish, one with a #4 finish in the same outdoor environment, which will rust faster? Are there any other concerns between the two cabinets as far as performance of the cabinet (welds, hinges, bends, etc.)? Have there been any studies conducted? </p>
| |materials|steel| | <p>In addition to hazzey's answer I'd just like to add that <strong>passivation</strong> of stainless is very important to reduce the risk of corrosion, especially in welded products.</p>
<p>Typically the steel is cleaned then dipped into various chemical baths following welding. This allows the steel to restore its chromium-oxide rich "passive" layer. </p>
<p>Depending on the application it may be possible to remove surface contaminants from welds with a stainless wire brush rather than using the chemicals but this will affect the surface finish.</p>
<p>Some of the pickling chemicals used in the passivation process can dull a very shiny finish. The #3 and #4 brushed finishes may be affected so it would be worth consulting the passivation company before going ahead with it. For high shine finishes it is worth considering electro-polishing.</p>
| 2742 | Will stainless steel with a #3 finish corrode faster than a #4 finish? |
2015-05-08T13:06:06.750 | <p>I am a high school student wondering if the pulley system described below will work, and if so, how to go about it.</p>
<p>Basically, I have a small system where as you pull back an object, it causes two lengths, labeled $d$ and $p$, to move forward. $d$ is dependent on many factors other than just the pulley system, but $p$ is just growing linearly as the object is pulled back. I've determined that the relationship between $d$ and $p$ is: $$p=\sqrt{A+Bd^2}-C$$
Where $A$, $B$, and $C$ are positive constants.</p>
<p>Here is a diagram to visualize how $p$ is connected to the system:</p>
<p><img src="https://i.stack.imgur.com/JhFO0.jpg" alt="enter image description here"></p>
<p>The length $p$ is the distance from the smaller block to the edge of the base. As you pull the object back, the string is pulled which pulls the spring and increases the length $p$. My question is, is there any sort of pulley system or just general construction I can insert between the object you pull and the pulley which is connected to the smaller block such that as you pull the object back, $p$ does not increase at a linear rate, but instead increases at the exact same rate of $d$.</p>
| |mechanical-engineering|pulleys| | <p>If I understand your question correctly, then you want to know if there is some mechanism which can convert a movement of distance $d$ to a movement of distance $p$, given that you want $p=\sqrt{A+Bd^2}-C$. If $d$ is caused by pulling on a rope, then the easiest way I can think to achieve your objective is a spring-loaded <a href="https://en.wikipedia.org/wiki/Cam" rel="nofollow">cam</a>. You would need to relate $d$ to the angle of turn of the cam and then make the radius of the cam the appropriate function of angle.</p>
<p>However, this depends on you having a low requirement for range of motion. You could of course, use a <a href="https://en.wikipedia.org/wiki/Block_and_tackle" rel="nofollow">block and tackle</a> to amplify the movement, but you will run up against size and frictional limitations.
Additionally, it may be difficult to fabricate your cam.</p>
<p>I don't know the details of your project, but it might be worthwhile to look at the range of $d$ and $p$ that you expect and see if you can come up with a linear relation that approximates your desired relation to the desired degree of accuracy. It is much easier to design linear mechanisms and you may find that your equation is needlessly complex.</p>
| 2752 | How to design and build a pulley system for a given purpose? |
2015-05-10T07:34:37.737 | <p>I have a Fisher and Paykel Washing Machine (model number WL80T65C, well past warranty) which got rained on (a lot) and one of the capacitors blew up. I've done an amateur replacement of the capacitor which might work but when taking the circuit apart I forgot to take a photo until half way through. I've managed to place everything except for two cables, one of which is Green/Yellow and coming from the mains. This is Australia so I presume it's a ground cable.</p>
<p>The only hint I can find as to which is the ground and which is the other is the sad face that's printed on the circuit board, as shown in the photo attached.</p>
<p><img src="https://i.stack.imgur.com/DQhWF.jpg" alt="Rude frowny face mocking me"></p>
<p>The cables connect to where the blue marks are, one on the left and one on the right.</p>
<p>So, is a frown face in a PCB a standard thing and does it tell me which is ground or are Fisher and Paykel mocking me? Is there a place where I can look for the schematics of this thing?</p>
<p>Edit: Here's a better photo of the sad face and its friend, taken in the day time. </p>
<p>NB: It's become apparent that the component was ground and it didn't matter which way around I fixed the cables but I'm still curious as to what the sad face means.</p>
<p><img src="https://i.stack.imgur.com/AjUHi.jpg" alt="enter image description here"></p>
| |electrical-engineering|circuits|embedded-systems| | <p>The sad faces are where the hand insert components go. If any sad faces are showing, the operator hasn't finished their job.</p>
| 2764 | What does a Sad Face on a circuit board mean? |