CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2015-06-27T13:31:11.017 | <p>Normally, in investment casting, the mould is preheated before pouring the liquid metal in to reduce thermal shock from rapid cooling. </p>
<p>In a vacuum, the heat transfer due to gas convection is reduced to zero, which will also reduce the thermal shock. Does that mean that, if performing investment casting in a vacuum, I do not have to preheat the mould?</p>
| |metallurgy|metals| | <h3>Convection, Conduction, Radiation</h3>
<p>Of the three modes of heat transfer, only one is affected by the vacuum. As you noted in your question, gaseous convection should be eliminated. That being said, the thermal shock is present as soon as the hot material hits the mold. As soon as the two materials come into contact, the heat transfer will be by conduction. This is the same whether in a vacuum or not.</p>
<p>The mold should still be heated.</p>
| 3353 | Is it necessary to preheat an investment casting mould when working in a vacuum? |
2015-06-27T23:22:44.917 | <p>I'm trying to understand the importance of protective coatings on steel structures like the Golden Gate Bridge. For example, if we stopped maintenance on the Golden Gate Bridge today and left it packed with empty cars, what would be its expected lifetime (until a car falls into the water) and the variance of that lifetime? If we additionally removed all of the paint/coatings on any steel parts, what would be the expected lifetime and variance? I'm hoping for a rough answer from an expert in the field.</p>
| |civil-engineering|bridges|corrosion| | <p>If you are really considering catastrophic failure, what you are asking actually has several parts. The first is the coatings aspect, the second is the metal structure, the third is the connections, the fourth is the cables, the fifth would be the substructure and the sixth is the foundation. A failure of any one of these systems (with the exception of coatings) could lead to the type of catastrophic failure you are considering.</p>
<p>The Golden Gate Bridge was constructed in the mid to late 1930s (around 1933 to 1938 I believe, see <a href="http://www.baybridgeinfo.org/timeline" rel="nofollow noreferrer">timeline</a>) and I will come back to this in later paragraphs.</p>
<p>It is important to note that the loading conditions currently on the bridge in terms of axle loads and traffic volumes are perhaps magnitudes higher than the original designer could have envisaged. These loads will have a significant impact upon the risk of catastrophic failure, and for this reason there has likely been significant structural strengthening carried out in the Golden Gate Bridge over the years. </p>
<p>One thing that I noted on the timeline was that there already has been a catastrophic failure of the upper and lower decks due to a 7.1 earthquake on October 17, 1989.</p>
<p><a href="https://i.stack.imgur.com/EboUt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EboUt.png" alt="Golden Gate Catastrophic Failure 1989"></a></p>
<p><strong>Coatings</strong></p>
<p>Generally speaking coatings are designed for a given service life. Provided that they are applied with adequate workmanship and quality, for a given material in a given environment the performance is fairly predictable; ie they will lose their protective capability over time due to oxidation, temperature, solar radiation and abrasion effects. Many designs are geared towards a certain loss in terms of mils/yr, so that for a 10 year service interval there will be a certain dry film thickness of coating specified. <a href="http://corrosion.ksc.nasa.gov/pubs/nasa-std-5008b.pdf" rel="nofollow noreferrer">NASA</a> Kennedy Space Center has detailed documentation on these issues. Once this protective measure is breached then corrosion of the underlying steel may take place. So it is important that coatings are renewed at regular intervals.</p>
<p><strong>Structural Steel</strong></p>
<p>Structural steel generally has a well defined atmospheric corrosion rate for a given environment and this is based upon it's proximity to the sea. Studies done by NASA in Florida on metal coupons from seafront to a distance inland about 30 km indicate that the danger zone is within roughly 1000 m of the sea. After that the atmospheric corrosion rates dropped off significantly. But the Golden Gate Bridge is clearly within a danger zone. From <a href="http://corrosion.ksc.nasa.gov/atmos.htm" rel="nofollow noreferrer">NASA</a> data, Pont Reyes, California would have an expected atmospheric corrosion rate of about 20 mils/yr. Certainly Caltrans will have similar research data. If left unprotected by coatings, and if this number were valid for your immediate environment, then you could take the thickness of the structural section and reduce exposed surfaces by 20 mils/yr to roughly estimate what would happen to an unprotected structure with time. Of course the NASA studies are using modern grades of steel, the Golden Gate steel was produced in the 1930s, so there could be a greater annual loss. This is atmospheric corrosion; the structural steel that is within range of the seawater spray and splash zone will actually experience a more severe corrosion condition, and protective measures there will likely include sacrificial anode systems in addition to coatings. That is, an anode material such as zinc is electrically connected to the structural steel in seawater spray exposure zones because it is more susceptible to corrosion. It is sacrificed in order to direct corrosion activity away from the structural steel. These also must be replaced at intervals. There are also impressed current systems too, these are called cathodic protection and they work in a similar fashion.</p>
<p>Now what does this mean for the structure? A structural analysis would be required to determine where the critical loading point occurs; ie where the force of a fully loaded bridge would exceed the ultimate loading capacity of the corroded, diminished sections. When structural elements oxidize, the oxidized portion of the section can no longer support load, and the remaining sound steel under the corrosion product is the issue of importance. But you can see that if coatings/protective measures are not reapplied when required then the corrosion rate of structural steel could endanger the structural capacity of the bridge very quickly, perhaps within a few years.</p>
<p><strong>Connections</strong></p>
<p>Structural connections are very sensitive to fatigue and this also exacerbates their sensitivity to corrosion. In other words elements of higher stress cycling are also more susceptible to corrosion. The two factors are quite significant in determining lifespan. For this reason, and I do not have any specific knowledge of this matter for the Golden Gate Bridge, I suspect that the connections in the bridge today have been essentially entirely replaced, perhaps several times at high stress locations, since initial construction. These are the most critical part of the structural steel framework for the bridge, from a load and lifespan perspective. If they fail, then catastrophic failure follows. It is also possible that additional protective measures are taken for the connections in terms of additional coating thickness or specialized treatments.</p>
<p><strong>Cables</strong></p>
<p>The Golden Gate Bridge is a suspension bridge so the cables are an integral part of the load capacity of the structure. These cables normally would have a protective sleeve around them to ensure that they are isolated from the environment, and cable sleeves today are typically filled with protective lubricants. I do not have any knowledge specific to the Golden Gate Bridge but I would expect that the design does include protective sleeves, at least for the main cables. Corrosion of cables can very quickly lead to very serious structural problems and for that reason they are inspected frequently. One of the problems with sleeved cables is that the cable itself is hidden from view, unlike other elements of the bridge, so the inspection would require non-destructive techniques that may include ultrasonic and radiographic testing. I am not aware of any loss per year allowances for cables, if they show any signs of corrosion then I would expect they will be scheduled for replacement or repair.</p>
<p><strong>Substructure</strong></p>
<p>The concrete substructure (foundation piers) elements would take the brunt of the most severe exposure condition, that is sea spray and splash zone. Wetting and drying conditions in the sea environment are the most corrosive. Here any exposed concrete may degrade at the surface due to salt scaling effects and this, along with carbonation (ie conversion of calcium hydroxide to calcium carbonate in the concrete) can reduce the effective protection of reinforcing steel in the concrete by reducing the thickness of the protective concrete cover.</p>
<p>There are generally well known formulas for concrete in sea splash and spray zones for time to corrosion of reinforcing steel. Pioneers in this field (among many others since) were Beaton and Stratfull (<a href="http://www.dot.ca.gov/research/researchreports/1961-1963/62-17.pdf" rel="nofollow noreferrer">1962</a>) and Stratfull (<a href="http://www.dot.ca.gov/research/researchreports/1964-1965/64-06.pdf" rel="nofollow noreferrer">1964</a> and <a href="http://www.dot.ca.gov/research/researchreports/1966-1967/67-09.pdf" rel="nofollow noreferrer">1967</a>) of Caltrans. More recently Collepardi and Clear have introduced improvements (Collepardi introduced diffusion theory in the early 1970s and Clear's work in the mid 1970s was a refinement of the Beaton and Stratfull analysis). These formulas are used at the time of design and can be used throughout the life of the structure for estimating remaining life. This has to do with the concentration of chlorides with depth in the concrete, that are carried into the concrete by seawater and increase in concentration with time.</p>
<p>For this reason protective coatings and other systems such as cathodic protection are used to prevent reinforcing steel corrosion. I would suspect that one or more of these treatments are in place at the Golden Gate Bridge, and I would also expect that portions of the concrete and reinforcement have been replaced in the past.</p>
<p>Nonetheless, if the corrosion were to be left untreated in the substructure then there could be a very serious catastrophic failure. If left completely untreated, the concrete of the type used on the Golden Gate Bridge in the 1930's would probably reach a critical stage within 25 to 30 years. The age of the bridge, being some 75 to 80 years of age means that this substructure concrete has likely had substantial repair, replacement and protective treatment by now.</p>
<p><strong>Foundations</strong></p>
<p>Concrete foundations, if permanently below the low tide line, generally are not at high risk, but can present problems. It is not commonly known, but the concrete of foundations of the Golden Gate Bridge contain Pozzolans, which was an advanced technique for its time. Pozzolans are usually added for improved durability, but they can play a major role in reducing temperature related distortions that can lead to cracking.</p>
<p><a href="https://i.stack.imgur.com/povJu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/povJu.png" alt="enter image description here"></a></p>
<p>In fact one of the more tragic bridge collapses in history occurred for this very reason during May 1980. The <a href="http://www.metrojacksonville.com/article/2013-sep-sunshine-skyway-bridge-disaster-may-9-1980" rel="nofollow noreferrer">Sunshine Skyway Bridge</a> in Tampa collapsed killing 35 persons. The engineer's investigation concluded that the cause of the collapse was directly attributable to cracks in the below water pier foundations due to the high temperature of the concrete at the time of construction.</p>
| 3356 | How do protective coatings affect the lifetime of bridges? |
2015-06-28T02:57:28.337 | <p>I recently discovered that one of my roommates had attempted to boost the WiFi signal by cutting up a soda can and placing it near the router, like this:</p>
<p><img src="https://i.stack.imgur.com/Xisb7.png" alt="Airport router with carved up soda can"></p>
<p>My roommate was probably following an internet guide like <a href="http://www.wikihow.com/Make-a-Wi-Fi-Booster-Using-Only-a-Can" rel="noreferrer">this one</a>, which advises shaping the modified can "like a radar dish" and placing it around the external antenna of the router. However, our router has no external antenna.</p>
<p>Will either configuration actually boost the router's reception? I don't know too much about antennae, but I suspect that the second configuration works more like a shield.</p>
| |electrical-engineering|signal|wifi| | <p>It is too small a surface area to be an effective microwave reflector. The shape is wrong too, it must be parabolic with respect to the transmitting fractal antenna of the router. Yours is perpendicular to it.<br>
<a href="http://www.instructables.com/id/Easy-to-Build-WIFI-24GHz-Yagi-Antenna/" rel="nofollow noreferrer">http://www.instructables.com/id/Easy-to-Build-WIFI-24GHz-Yagi-Antenna/</a></p>
<p>With the advent of embedded fractal antennas, we no longer need external antennas sized according to the wavelength of the carrier frequency. A good development or we would have to be carrying around antennas for our cell phones. I'm not saying you can't shape an aluminum can to enhance directional transmission and reception. It just takes some design and calculation.</p>
| 3358 | Is it possible to "boost" a WiFi signal using an aluminum soda can? |
2015-06-28T21:49:59.043 | <p>I'm interested in metalworking as a hobby, to make small parts for my numerous other hobbies. I'm looking at those cheap, typically Chinese import mini-lathes and mini-mills. I've read about their limitations as compared to larger machines; this question is not about that.</p>
<p>My question is, <strong>what are the pros and cons of a combo mini lathe/mill as opposed to separate machines?</strong></p>
| |mechanical-engineering|machining| | <p>If the cost and workspace are not the problems, you should take 2 separate machines. Otherwise, a combo machine will save the money and workspace</p>
| 3367 | What is the difference between using separate lathe and mill or a combination machine? |
2015-06-29T14:03:34.793 | <p>There is a room on the south side of the building, which doesn't have any ventilation vents, but does have a window.</p>
<p>Opening a window doesn't cause good air circulation (by which I mean that the air in the room is still hot, and the air on the street is still cold).</p>
<p>How does one create good air circulation in a room --with a single window- without using a fan or air conditioner?</p>
| |fluid-mechanics|hvac|airflow| | <p>The fundamental problem is that hot air must leave the room and cold air must enter it, both at the same time through the same single opening. The conflicting forces will mostly cancel each other out, resulting in very little actual exchange of air.</p>
<p>What you really need is old technology.</p>
<p><em>Double hung</em> windows are what our ancestors used. The bottom sash lifts, just as in most windows, but the top sash, rather than being fixed, can be lowered to create an opening at the top.</p>
<p>With both a high and a low opening in the window, hot air tends to be forced out the top as cooler air enters at the bottom.</p>
| 3374 | How does one create a draft in a room (with a single window)? |
2015-06-29T14:06:55.073 | <p>What type of a mechanical device can I use and program to apply a controllable force within the range of what a human can to an area about the size of a fingertip? It is desirable for the applied force to be precise and known.</p>
| |mechanical-engineering| | <p>I agree the compressed air is the way to go. There are many vendors that sell small cylinders and even have bumpers available for the end of the actuator rod that could approximate a fingertip.</p>
<p>For control, you don't mention any requirements, but if they're simply manually controlled on/off application of force, a regulator with some simple manual valving should do the trick, just make sure you have a way to vent the air in the cylinder when you're trying to release the force. Keep in mind that unless you control the flow into the cylinder, the force will be applied quite suddenly unless you use a flow control valve or orifice as well. </p>
<p>What I've done for applications like this is to use an I/P transducer, controlled by computer which would lets me precisely control the pressure applied to the cylinder by varying the current flowing in a loop, typically 4-20mA. (There are also voltage controlled versions, but I have no experience with them) This way, you could ramp the pressure at any rate you wish and the process is much more repeatable. </p>
| 3375 | Applying an adjustable force to a small area |
2015-06-29T15:28:41.657 | <p>I am interested in finding a way to calculate the force necessary to do a backwards extrusion. Typically these would be for forgings 0.5 to 3" in diameter (or square) of different kinds of steel. The diagram below shows a typical type of forging I might want to do:</p>
<p><img src="https://i.stack.imgur.com/FCVagm.png" alt="hex die"></p>
<p>This would be die formed in A2 steel by backwards extrusion using a hex-shaped punch. The forging would be done with the work piece at its forging temperature.</p>
<p>I want a way to calculate the force (forging load) necessary do such extrusions.</p>
| |mechanical-engineering|forging| | <p>You can find formulas for Backward extrusion in <a href="https://books.google.com.tr/books?id=_ZVqPgAACAAJ&printsec=frontcover&dq=editions:ISBN0470505923&hl=tr" rel="nofollow">Groover's Principles of Modern Manufacturing 4th edition</a> section 17.5.2. I summarized required formula but if you need more info ask or check the book. Also you can <a href="http://www.egr.msu.edu/~pkwon/me477/bulkforming2.pdf" rel="nofollow">check these lecture notes</a> I googled.</p>
<hr>
<p><strong>Reduction ratio</strong>: $r_x=\frac{A_0}{A_f}$</p>
<p>$A_0$ and $A_f$ are initial and final area in mm respectively. </p>
<p>In bacwards extrusion, effect of friction can be ignored so <strong>true strain</strong> is $\varepsilon = ln r_x$</p>
<p><strong>Average flow stress</strong> is $\overline{Y_f}=\frac{K\varepsilon^n}{1+n}$</p>
<p><strong>K</strong> is strength coefficient and <strong>n</strong> is strain hardening exponent. There are a few examples on <a href="https://en.wikipedia.org/wiki/Strain_hardening_exponent" rel="nofollow">wikipedia</a> but you'll probably get A2 steel's info from datasheet or materials books. (K is in MPa so $\overline{Y_f}$ is in MPa)</p>
<p><strong>Pressure required</strong> for indirect extrusion is $p=\overline{Y_f}\varepsilon_x$</p>
<p><strong>Force required</strong> is $F=pA_0$</p>
| 3381 | How can I calculate backward extrusion force |
2015-06-30T06:09:49.630 | <p>If I have a vehicle (V1) with one 100 kW engine and an identical vehicle (V2) with 2 of these 100 kW engines, will this mean that V2 will have double the torque of V1 and double the top speed of V1?</p>
<hr>
<h3>Suspected Answer</h3>
<p>It's not like in a series circuit in which, when you have two identical batteries in series, the the voltage doubles. </p>
<p>I'm guessing that there will be some increase (but not double) in torque, but the top speed would remain the same between V1 and V2 whilst pulling an identical load. I suspect that the difference between V1 and V2 will be that V2 may be quicker to achieve the top speed, but the top speed won't increase by much. I.E. V1's max speed pulling 30 kg will be 20 kph but might take 10 seconds, but V2s max speed pulling 30 kg will be 20 kph but might take 8 seconds.</p>
| |mechanical-engineering|automotive-engineering| | <p>There is no simple answer. What is the top speed limiting factor? For a truck, it's mostly handling so no amount of power can increase that. For a sports car, it's mostly aerodynamic resistance. While the aerodynamic resistance (more precisely it's called <a href="https://en.wikipedia.org/wiki/Drag_%28physics%29" rel="nofollow">drag</a>) increases proportionally to velocity squared, the power needed to overcome it increases proportionally to the velocity cubed. So 8x the power would yield 2x top speed and 2x more power means a 1.26x faster vehicle (when considering aerodynamics alone).</p>
<p>The torque will be definitely double, if the transmission and wheels can handle the doubled amount. At wost, you can add secondary set of entire propulsion set, just as they did with <a href="https://en.wikipedia.org/wiki/Citro%C3%ABn_2CV#.22Sahara.22_four-wheel_drive" rel="nofollow">Citroen 2CV Sahara</a>.</p>
<p>You should also research trainsets, as there are some real-life examples of almost identical vehicles differentiated only by amount of power plants.</p>
| 3392 | Does having 2 engines together increase anything? |
2015-06-30T13:02:46.953 | <p>In a Vacuum Metallurgy (Casting), a mold needs to be heated (500 to 700 degrees) to reduce thermal shock on the metal surface. Normally, a gas fired oven is employed for such job, but in a vacuum environment how do you achieve the same?</p>
| |metallurgy|metals| | <p>A second idea for added control, if needed:</p>
<p>If your molders can accomplish it, you could also bury a <a href="https://en.wikipedia.org/wiki/Heating_element" rel="nofollow">heating coil</a> in the mold before you harden it. The heating cable would need to be able to withstand the liquid metal temperatures, so something like <a href="https://en.wikipedia.org/wiki/Tungsten" rel="nofollow">tungsten</a> would be required. This would also allow some control of the process after pouring - if you needed to get to a different <a href="https://en.wikipedia.org/wiki/Steel#Material_properties" rel="nofollow">metallurgical state</a>, for example. The heater near the edges would keep the edges hot while the inside cools.</p>
| 3394 | Heating Sand/Ceramic/Refractory Material under vacuum |
2015-06-30T15:21:46.420 | <p>I am looking to complete a materials science/engineering-based Master's degree in the UK. Those which I have been looking at, particularly Leeds and Manchester, do not seem to be accredited by the <a href="http://www.iom3.org/" rel="nofollow">IOM3</a> or other institutions.</p>
<p>I am interested in becoming chartered/incorporated in the future. If I were to complete an unaccredited course, what impact would the lack of accreditation have on my career prospects in the field of materials science/engineering?</p>
| |materials|education|licensure| | <p>Try asking the <a href="http://www.iom3.org/" rel="nofollow">IOM3</a> & other accrediting institutions why the courses you are interested in are not accredited and if there is a possibility that they could be accredited in the near future.</p>
<p>The courses may not cover material that such institutions consider vital for professionals engaged in that field and which would be required for membership of the institutions.</p>
<p>If you were to study one of those courses you may need to study the missing subjects elsewhere prior to being eligible to for membership of the institutions.</p>
<p>Regarding employment prospects, it's safer to study an accredited course. In times of economic downturns or gluts of professionals in particular fields employers can be very discriminating. Anyone without an accredited qualification can more easily have their application disregarded. </p>
<p>It's not just job prospects in the UK that you need to consider; materials science/engineering is an international profession. Larger companies track the performance of universities and the courses they offer and they know which universities offer courses that are of most interest to them.</p>
<p>Companies also prefer to employ people with accredited qualifications so as to minimize any legal issues should there be a problem arising from their products or services and the personnel responsible for those products or services did not have an accredited qualification.</p>
| 3396 | What are the career implications of choosing a non-accredited postgraduate degree? |
2015-06-30T18:20:20.170 | <p>The rubber casing for my iPhone cable has frayed and split near the phone port, and I would like to seal it somehow to stop the split from spreading further. Would Krazy Glue be an effective sealant? I've heard that superglue can eat away at some materials.</p>
| |materials|adhesive| | <p>Where a cable enters a connector is a point strain, point strains are remarkably prone to failure. I would not use glue at all particularly a rigid glue like super glue. I would use dental floss or string, and caulk or a curable plastic putty.</p>
<p>To begin take a length of string about two foot long and fold it in half place the edge of your connector (the par directly opposite from where the cable enters the connector) in the fold of the string. Tye a couple knots around the connector ending with each end of the string next to the cord, but on opposites sides. A drop or two of glue to hold the string to the connector body may be needed. Push the cable gently into the connector to relieve strain. alternating the two string tails, tie a series of half hitches up the cable. Mask off the connector edge. Cover the connector body and the cord end that is tied with caulk. Let cure. Remove the mask, trim the excess caulk cut the string from the connector edge and trim the string ends. Will outlast your phone.</p>
| 3397 | Can Krazy Glue be used as a sealant? |
2015-06-30T20:29:08.570 | <p>How can I "tie" pressure change due to a mass leakage from the flow?
In a "closed" system I'd use $pV=nRT$ but this is an "open" system...</p>
<p>My open system is an horizontal (no gravity issue) pipe (constant sections) in which flows a gas. We may think pipe walls as permeable walls. Hence we have one incoming flow to the pipe inlet, and two outgoing flows: one from the pipe outlet and the other from the pipe wall</p>
<p>Assume the simplest case: Mach<0.30 and density is almost constant. To be complete, leakage is a known variable and I need to find outlet pressure</p>
| |fluid-mechanics|gas| | <p>You can't use the <a href="https://en.wikipedia.org/wiki/Ideal_gas_law" rel="nofollow">ideal gas law</a> in an open system, but you can use <a href="https://en.wikipedia.org/wiki/Bernoulli's_principle" rel="nofollow">Bernoulli's equation</a>. </p>
<p>Starting with the conservation of energy: </p>
<p>$$ E_{in} = E_{out} $$</p>
<p>We have one source in and two sources out:</p>
<p>$$ E_{in} = E_{out} + E_{leak} $$</p>
<p>The energy is the sum of the enthalpy and kinetic energy (since the pipe is horizontal):</p>
<p>$$ (h_1 + v_1^2/2)*m_1 = (h_2 + v_2^2/2)*m_2 + m_{leak} * (h_3+ v_3^2/2)$$</p>
<p>By definition, the enthalpy is the internal energy plus $P/\rho$.</p>
<p>Now for some important assumptions:</p>
<ol>
<li>The velocity of the leak far from the pipe would be 0 (i.e., it's reached the atmosphere and not causing tornadoes).</li>
<li>The leak's final pressure would be atmospheric pressure (via the same assumptions)</li>
<li>The entire system is under roughly constant temperature, so the internal energy, u of the gas at all three points is the same</li>
</ol>
<p>$$ (u + P_1/\rho + v_1^2/2)*m_1 = (u + P_2/\rho + v_2^2/2)*m_2 + m_{leak}*(u + P_{atm}) $$</p>
<ol start="4">
<li>Now we assume constant flow. The pressure, internal energy and velocity at the inlet should be the same with respect to time. Same with the outlet and leak. So, differentiating with respect to time, the only thing that changes at each point is the mass:</li>
</ol>
<p>$$ P_{in} = P_{out} + P_{leak} $$</p>
<p>$$ (u + P_1/\rho + v_1^2/2)*\dot{m}_1 = (u + P_2/\rho + v_2^2/2)*\dot{m}_2 + \dot{m}_{leak}*(u + P_{atm}) $$</p>
<p>Now we know by conservation of mass:</p>
<p>$$ m_{in} = m_{out} + m_{leak} $$</p>
<p>Which works when taking the derivative with respect to time. So, we can remove the u terms. Finally, it's easier to express mass flow as:</p>
<p>$$ \dot{m} = \rho A V $$</p>
<p>So, substituting in the mass flow, and multiplying the entire equation by the constant density to convert from head to pressure:</p>
<p>$$(P_1 +\rho\frac{V_1^2}{2})\rho AV_1 = (P_2 + \rho \frac{V_2^2}{2})\rho AV_2 + \dot{m}_{leak}(P_{atmosphere})$$</p>
<p>Note the units are Watts*density - to make your mass loss easier to solve. While you could solve for the leak in this equation, using your knowledge of the pressure outside and the pressure at two places, you would still need to know at least one of the velocities to solve for the leak. The other velocity, you could solve for because of conservation of mass again:</p>
<p>$$ \dot{m}_{in} = \dot{m}_{out} + \dot{m}_{leak} $$
$$ \rho V_1 A = \rho V_2 A + \dot{m}_{leak} $$</p>
<p>And of course, I did make these assumptions: <em>constant density</em>, and <em>no loss due to pipe friction</em> under the assumption the leaks are so light that the temperature didn't change, and so density didn't change - just decrease the velocity. Ultimately, to figure out the velocities, you need to know the flow rate - so the only way to tell is with flowmeters, not pressure gauges.</p>
| 3402 | Pressure drop due to mass loss |
2015-07-01T05:09:35.830 | <p>I've had trouble with leaky pipe flanges, since my flanges are made of plastic. The problem comes down to the fact that sealing a flange is based on torque. Torque relies on bolt friction, which can change even between different bolt diameters. Therefore, it's a terrible indicator of the clamping force. </p>
<p>Is there any other way to measure flange sealing force than by using a torque wrench? I'm interested in any method that works, whether it's a known system or something crazy.</p>
<p>Ideal solutions would take around the same time to measure as the torque wrench (maybe double), can be used to check flanges that are currently in operation as well as new installations, and can work with a wide variety of flange, bolt and gasket materials.</p>
| |mechanical-engineering|piping|bolting|threads| | <p>I'm sure there are specific strategies relating to pipe flanges, but for the broader issue of controlling that all of your bolts are tight enough, here is some background information.</p>
<p>When tightening bolts in any clamping application, what's really important is the total clamping force. In any simple joint, and specifically when dealing with pipe flanges, the clamping force in the joint is the same thing as the tension in the bolt (also known as preload.) Torque is sometimes used as a surrogate for preload since the two are correlated, but for a number of reasons, it is not a very good tool for measuring clamping force.</p>
<p>The relationship between torque and preload is based on some things that are fairly easy to quantify like the thread pitch of the thread, but also on things that are hard to quantify like the accuracy of the threadform, the quality of the coating, the cleanliness of the fasteners, etc. The amount of torque that is translated into tension is a relatively small percentage of the overall torque, with the remainder overcoming friction, often quoted as 10%. Because many of these factors are hard to quantify and hard to control without very detailed quality control, torque does not always offer an acceptably accurate way of predicting preload. (More on the torque-tension relationship: <a href="http://www.portlandbolt.com/technical/faqs/tension-vs-torque-explained-sort-of/" rel="nofollow">1</a> <a href="http://www.assemblymag.com/articles/83789-fastening-the-truth-about-torque-and-tension" rel="nofollow">2</a> <a href="http://www.smartbolts.com/wp-content/uploads/2012/05/TheTorque-TensionRelationship.pdf" rel="nofollow">3</a>)</p>
<p>To overcome this problem, when the amount of preload is critical, we measure the tension in the fastener directly to calibrate our tightening system. One way to do this is with a device colloquially known by it's brand name, <a href="http://www.skidmore-wilhelm.com" rel="nofollow">Skidmore-Wilhelm</a>. A bolt can be tightened around this device, and the tension is measured. In structural applications in the US, per the <a href="http://www.boltcouncil.org/files/2014RCSCSpecification-withErrata.pdf" rel="nofollow">RCSC (PDF Warning)</a> it would be typical to tighten 3 bolts out of each batch against the machine, using either a torque wrench, or by turning the nut a certain number of degrees past 'snug-tight.' If those 3 bolts fall into the acceptable tension range, then that torque or angle is used to tighten the actual production bolts until that batch of bolts runs out, or the next day, when the calibration procedure is performed again. Torque or angle are still being used as surrogates for pretension, but they are being calibrated for all of the unknown variables regularly.</p>
<p>More recently, some methods for measuring the tension directly have emerged that are affordable enough that they can be permanently used for each bolt (whereas the Skidmore-Wilhelm costs thousands of dollars.) These include bolts with heads that change color when preloaded, washers that squirt out paint when preloaded, and bolts with a splined end that breaks off when preloaded. These methods are generally considered to be more reliable than conventional bolting by calibrated torque or turn-of-the-nut (the method that uses angles) but do increase materials costs, so they may or may not be appropriate for your application.</p>
| 3409 | Measuring pipe bolt torque without a torque wrench |
2015-06-23T18:04:15.747 | <p>I have a small issue that is causing a lot of problems. I'm on a <a href="https://en.wikipedia.org/wiki/Formula_SAE" rel="nofollow noreferrer">Formula SAE</a> team and our battery is consistently shorting between our positive terminal and our chassis. Our chassis is the grounding point. </p>
<p>Our battery is underneath our seat, our seat is made of carbon and therefore when the driver moves shorts occur. I've tried using tape, rubber covers and plastic boxes. The box cracks under the weight of the driver and the tape and covers fall off during a long race.</p>
<p>How can I prevent shorts with a solution that wont break, nor fall off?</p>
<p>In case a short does occur, we use breakers.</p>
<p>Below is a sketch of the position of the battery under the seat. At point one, when we are static, the seat does not touch the battery, but under high acceleration, the driver will rub on to the battery terminals. This knocks off tape and covers that are on it. In the course of a long race, eventually shorts start to occur after the tape/covers have been rubbed off. We noticed that the issues happened on tight turns and moments of high acceleration.</p>
<p><img src="https://i.stack.imgur.com/GgLnr.jpg" alt=""></p>
| |electrical-engineering|automotive-engineering|battery| | <p>A piece of thick plastic or timber will make a lid that will solve this or use a gel battery and lay it on its side</p>
| 3415 | How to prevent shorts between exposed battery terminals and racing vehicle chassis |
2015-07-02T02:12:59.257 | <p><em>The question</em>: Consider $1\ kg$ of air at $32\ C$ that is expanded by a reversible polytropic process with $n=1.25$ until the pressure is halved. Determine the heat transfer.<br>
Specific heat constant volume for air is $0.1786\ kJ/kg.K$</p>
<p>The correct answer is: $17.02\ kJ$ heat added</p>
<hr>
<p>My work so far:</p>
<p>$$\frac{T_{1}}{T_{2}} = \frac{P_{1}}{P_{2}}^{\frac{n-1}{n}}$$
$$n = 1.25$$
$$\frac{P_{1}}{P_{2}} = 0.5$$
$$T_{1} = 32+273$$</p>
<p>You'd get $T_2 = 350.352$</p>
<p>$$ Q = m*Cv*(T_2-T_1)$$ </p>
<p>I get $8.00\ kJ$ something. </p>
<p><strong>What am I doing wrong?</strong></p>
| |mechanical-engineering|heat-transfer| | <p>You can't use the specific heat constant of constant volume - a polytropic process isn't constant volume or constant pressure.</p>
<p>Instead, what is derived is the specific heat of polytopic expansion:</p>
<p>$$c_n = c_v\frac{n-k}{n-1}$$</p>
<p>Where k is the <a href="https://en.wikipedia.org/wiki/Heat_capacity_ratio" rel="nofollow">heat capacity ratio</a>. Air is considered a diatomic ideal gas, so k = 1.4. When I lookup air's specific heat for volume via <a href="http://www.engineeringtoolbox.com/specific-heat-capacity-gases-d_159.html" rel="nofollow">engineering toolbox</a> is $c_v = 0.718 kJ/kg*K$, so I'm using that.</p>
<p>So, here's how I would approach it (note you did have p2/p1 backwards as well):</p>
<p>$$ T_2 = T_1(\frac{p_2}{p_1})^{1-1/n} = 305K*(0.5)^{0.2} = 265.52K $$</p>
<p>$$ c_n = 0.718\frac{-0.15}{.25} = -0.4308 \frac{kJ}{kg*K} $$</p>
<p>$$ Q = m c_n (T_2 - T_1) = 1kg * (-0.4308)*(265.5K - 305K) = 17.02 kJ $$.</p>
<p>Source for equations - the ever useful <a href="http://rads.stackoverflow.com/amzn/click/1591264146" rel="nofollow">MERM</a>. Don't take the PE exam without it.</p>
| 3420 | The Heat Added in a Reversible Polytropic Process |
2015-07-02T06:01:49.487 | <p>I major in microelectronics. I just want to apply a fixed and adjustable force to my flexible sample.</p>
<ul>
<li>'Fixed' means that the force could be hold for given time without change.</li>
<li>'Adjustable' means that the force would be changed after my setting or its value could be changed with time and rule.</li>
</ul>
<p>It would be better if the force was horizontal.</p>
<p>My sample can endure the force with value about 10 N. And I hope the force could be adjustable between 0 and 10 N.</p>
<p>I think a stepper motor with load cell may be the solution but I don't know how to realize it exactly. The only thing I can think of is to hang different balance weight to one side of my sample. I know some kind of universal testing machines could also get this done but they are too expensive.</p>
| |mechanical-engineering|mechanical-failure| | <p>The poor man's solution would be to simply stack weights onto your sample.</p>
<p>While that seems low tech, you need surprisingly little different weights, which you can usually buy in sets, to get a massive range of forces.</p>
<p>The applied force is guaranteed to be independent of deflection, and this solution is also rather cheap compared to building specialized rigs.</p>
| 3423 | How can I apply a fixed and adjustable force to my flexible sample? |
2015-07-02T12:19:03.220 | <p>The Bond number is a dimensionless number typically used to analyze cases where two fluids of different densities are in contact and subject to gravity only. I'm analyzing a system where gas and liquid are entrained in a pipe (think garden hose with trapped air pockets), and both are subject to high accelerations, much higher than gravity. I'd like to modify the equation for the Bond number to use my known value of acceleration in place of gravitational acceleration, but I can't find any published literature where someone has done this. </p>
<p>What I want to do is this: $$Bo=\frac{\Delta \rho a L^2}{\sigma}$$ where $a$ has replaced $g$ in the traditional equation: $$Bo=\frac{\Delta \rho g L^2}{\sigma}$$ Anyone have experience or at least comments on my approach?</p>
| |fluid-mechanics| | <p>After some more thought on this, I'm realizing that the Bond number is more of a measure of the ratio of <em>body</em> forces to surface tension. Since I am looking at acceleration by way of <em>surface</em> forces, I'm not sure the traditional Bond number is applicable here. </p>
<p>@Mark's answer regarding dimensional analysis still seems valid, so I think the best thing for me to do is keep my modified version of the equation, but not call it the Bond number anymore; just call it the ratio of surface forces to surface tension, if there isn't already a number with that definition.</p>
<p>After some searching, I have found the Euler number (ratio of pressure force to inertial force), and the Weber number (ratio of inertial force to surface tension). The product of these two numbers effectively gives me what I want (ratio of surface force due to pressure to surface tension):</p>
<p>$$
Eu = \frac{\Delta p}{\rho v^2}
$$
$$
We = \frac{\rho v^2 l}{\sigma}
$$
$$
Eu\cdot We=\frac{\Delta p\cdot l}{\sigma}
$$</p>
| 3431 | Bond number for accelerating flow |
2015-07-02T12:43:50.503 | <p>So one of the proposed suspension systems that will be used on the <a href="https://en.wikipedia.org/wiki/Hyperloop" rel="nofollow">hyperloop</a> include the externally pressurized air cushions. These cushions lift (or at least help lift) the capsule and reduce drag when the capsule is in motion. Also, the tube itself is supposed to have a very low pressure. But won't 100s of capsules moving in the tube releasing significant amounts of compressed air pressurize the tube and thereby significantly increase drag? I suppose this could be solved by perpetually depressurizing the tube, but I did not see any information on it. Does anyone know how this problem was addressed or am I just missing something here?</p>
| |fluid-mechanics|transportation|drag| | <p>The air used for the air cushions will come from the air still in the tube pressurized by the pod itself.</p>
<p>Thus the system remains closed (all air that is released is sucked out of the environment).</p>
| 3432 | How does the pressurized suspension of the Hyperloop not affect the tube pressure? |
2015-07-02T15:26:59.073 | <p>Due to the fact that pipe flanges rely on bolts, and bolts rely on friction, it seems obvious to me, as a manufacturer of piping, to recommend that bolt torques need to be checked every so often, to ensure piping doesn't leak. I've got a few questions about this, but here is the first one:</p>
<p>How often do you normally check torques on bolts at a plant? In other words, what is a good industry standard preventative maintenance schedule when it comes to threaded fasteners that are loaded below yield? I would prefer that answers provide a reputable source for reference.</p>
| |mechanical-engineering|piping|bolting|maintenance| | <p>There is no answer because there are too many variables, amount of vibration and thread geometry being just two of them. A joint could last for 10,000 years or work loose tomorrow. There is no way to know.</p>
| 3441 | How often do you need to check torques on bolted flanges? |
2015-07-02T18:31:29.740 | <p>I have been reading about embedded systems, and particularly the reset modes. As I understand a microcontroller can have several reset modes. </p>
<p>What is brownout reset (BOR) and power on reset (POR)? What is the difference between BOR and POR?</p>
| |electrical-engineering|embedded-systems| | <p>When the V<sub>dd</sub> drops below a brown out threshold voltage, BOR will hold the microcontroller in reset state. Not all devices have BOR detection, but most do, and some have multiple voltage thresholds to select from.<br>
Between a BOR and Power On Reset the whole range of startup voltages can be covered to protect for proper operation after a power drop at the V<sub>dd</sub> line.</p>
| 3443 | What is the difference between BOR and POR micro-controller reset modes? |
2015-07-02T20:42:34.323 | <p>I have seen a number of structures within rivers, which resemble steps, and which allow the water to cascade down them instead of flowing naturally down the course of the river. </p>
<p>Example 1: River Avon in Bath</p>
<p><a href="https://i.stack.imgur.com/rPrQj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rPrQjm.jpg" alt="enter image description here"></a></p>
<p>Example 2: River Seine in Paris (from <em>Les Miserables</em> (2012))</p>
<p><a href="https://i.stack.imgur.com/cOL4a.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/cOL4am.jpg" alt="enter image description here"></a></p>
<p>What is the purpose of these structures?</p>
| |civil-engineering|hydrology|flow-control| | <p>The two photos in the post show the same structure: <strong>Pulteney Weir</strong>, downstream of <a href="https://en.wikipedia.org/wiki/Pulteney_Bridge" rel="noreferrer">Pulteney Bridge</a> on the River Avon in Bath. The shot of the "Seine" in <em>Les Miserables</em> was <a href="http://www.imdb.com/title/tt1707386/locations" rel="noreferrer">filmed on location</a> in Bath.</p>
<p>Pulteney Weir was designed by architect <a href="http://www.theguardian.com/news/2003/jul/14/guardianobituaries.artsobituaries" rel="noreferrer">Neville Conder</a>, and built between 1968 and 1972. It's one component of the Bath Flood Prevention Scheme, which was carried out after the disastrous <a href="http://www.bathintime.co.uk/image/1116261/floods-on-the-lower-bristol-road-bath-december-1960" rel="noreferrer">flood of December 1960</a>. The Bath in Time website has photos of the <a href="http://www.bathintime.co.uk/image/1099732/fishing-on-the-pulteney-weir-c-1960s" rel="noreferrer">old weir</a> (which descended in a single step) and of the new weir under construction (<a href="http://www.bathintime.co.uk/image/317625/flood-prevention-barriers-being-constructed-at-pulteney-weir-c-1968" rel="noreferrer">1968</a>, <a href="http://www.bathintime.co.uk/image/818458/panorama-of-the-pulteney-weir-scheme-under-construction-21-july-1971" rel="noreferrer">1972</a>).</p>
<p>An upstream view of the weir shows that there's more to the structure than the horseshoe-shaped steps: these meet a mid-river artificial island, and a channel to the left of the island terminates in a <a href="https://en.wikipedia.org/wiki/Sluice" rel="noreferrer">sluice gate</a>.</p>
<p><img src="https://i.stack.imgur.com/Uewv2.jpg" alt="upstream view of the weir"></p>
<p>(<a href="http://wikimapia.org/3675688/Pulteney-Weir-Bath#/photo/965114" rel="noreferrer">Photo by GeodesyMike from Wikimapia</a>, licensed under CC-BY-SA.)</p>
<p>As with any <a href="https://en.wikipedia.org/wiki/Weir" rel="noreferrer">weir</a>, it has multiple purposes: to control the level of the river above the weir for navigation, fishing, and irrigation; to preserve water in long periods of drought; to hold flood waters back and release them gradually downstream; and to control the speed and force of the water, preventing the river from scouring the banks and damaging the foundations of the bridge and nearby buildings.</p>
<p><a href="https://www.youtube.com/watch?v=8k0o_nwjm5M" rel="noreferrer">This video</a> shows the weir in operation during flood. You can see that the shape of the weir directs the force of the water to the centre of the river, preventing it from damaging the banks.</p>
| 3450 | What is the purpose of these "steps" in rivers? |
2015-07-03T02:07:55.673 | <p>I am answering this question:</p>
<blockquote>
<p>Steam enters the superheaters of a boiler at a pressure of 25 bar and
dryness of 0.98% and leaves at the same pressure at a temperature of
370 °C. Calculate the heat energy supplied by the superheaters.</p>
</blockquote>
<p>The answer is supposed to be 405.51 J.</p>
<p>The Properties of Steam are:</p>
<p>At 25 bar and 370 °C: $h=3171.8\ \mathrm{kJ/kg}$</p>
<p>$h_{f}=962.11\ \mathrm{kJ/kg}$, $h_{fg}=1841.01\ \mathrm{kJ/kg}$</p>
<p>$h=h_f+xh_{fg}$</p>
<p>$h=962.11+0.98*1841.01=4070.474\ \mathrm{kJ/kg}$</p>
<p>$Q=m(h_2-h_1)=4070.474-3171.8=898.674$</p>
<p>What am I doing wrong?</p>
| |thermodynamics|steam| | <p>You just got the math wrong</p>
<p>$$h_1 = 962.11 + (0.98*1841.01) = 2766.2998 \ \mathrm{kJ/kg}$$
$$q = (h_2-h_1) = 3171.8 - 2766.2998 = 405.5 \ \mathrm{kJ/kg}$$</p>
<p>(I omitted your $m$ in final equation since we are calculating per kg.)</p>
| 3455 | Calculating the heat energy supplied by a superheater |
2015-07-04T05:40:35.293 | <p>I am trying to solve the problem:</p>
<blockquote>
<p>What is the lifting force of in kN for a 10 m diameter spherical balloon with helium inside at 101 kPa and 320 K surrounded by air at 101 kPa and 298.15 K?</p>
</blockquote>
<p>The given answer is 5.28 kN.</p>
<p>I start from the Buoyancy Formula:</p>
<p><span class="math-container">$$F_{b} = (p_{air} - p_{gas})*g*V$$</span></p>
<p>And Ideal Gas Law:
<span class="math-container">$$PV = mRT$$</span>
<span class="math-container">$$P/RT = m/V$$</span></p>
<p>Density of air: <span class="math-container">$101/0.287*298.15=1.18$</span></p>
<p>Density of He: <span class="math-container">$101/0.287*320=1.10$</span></p>
<p>Volume of helium balloon: <span class="math-container">$(4\pi/3)*(10/2)^3=(500/3\pi)$</span></p>
<p>Substitute:</p>
<p><span class="math-container">$$F_b = (1.18 - 1.10)*9.81*(500/3\pi)$$</span></p>
<p>I get 0.4101 N; am I missing something or is the given answer wrong?</p>
| |fluid-mechanics| | <p>Just for additional information, incase you didn't know how to get the specific gas constant of He.</p>
<p>The general formula is Universal gas constant over the molecular weight of the specific gas is equal to the specific gas constant:</p>
<p><strong>Ru/MW = Rgas</strong></p>
<p><strong>8.314/4.002 = 2.077 KJ/Kg . K</strong></p>
| 3468 | What is the lifting force of a helium balloon in air? |
2015-07-05T10:09:05.673 | <p>I am solving the problem:</p>
<blockquote>
<p>A gas bubble rising from the ocean floor is 1 inch in diameter at a depth of 50 feet. Given that specific gravity of seawater is 1.03, the buoyant force in lbs being exerted on the bubble at this instant is nearest to:</p>
</blockquote>
<p>The given answer is 0.020 lbs.</p>
<p>I start from the equation for buoyancy:</p>
<p>$$F_b=(P_{fluid}-P_{bubble})Vg$$</p>
<p>$g=32.2\:\mathrm{ft/s^2}$</p>
<p>$V=(4/3)\pi(1\:\mathrm{in}/2)^3 * (1 \:\mathrm{ft}/12 \:\mathrm{in})^3=3.03*10^{-4}\:\mathrm{ft^3}$</p>
<p>$P_{fluid}=sp.gr(saltwater)*densityH_2O=1.03*62.4=64.272\:\mathrm{lb_m/ft^3}$</p>
<p>P_gas:
Pressure, $P=62.4*50\:\mathrm{ft}=3120\:\mathrm{lb/ft^2}$</p>
<p>$PV=mRT$</p>
<p>$P_{bubble}=m/V=P/RT=3120/53.34*492R (STP)=0.11889\:\mathrm{lb_m/ft^3}$</p>
<p>When I substitute all of the values I only get 0.607 lbs. What am I doing wrong?</p>
| |fluid-mechanics|thermodynamics| | <p>the difference in densities is essential. A bubble of air will have much more of a buoyant force than a bubble of oil in the water.</p>
| 3473 | Calculating the buoyant force of a rising bubble |
2015-07-05T11:12:49.380 | <p>I am solving the following problem:</p>
<blockquote>
<p>Air having an initial pressure of 6516 kPa and an initial volume of 0.113 m<sup>3</sup> is compressed adiabatically to a final volume of 0.057 m<sup>3</sup>. Calculate the work done by the gas as it compresses to a final pressure of 17237 kPa.</p>
</blockquote>
<p>The given answer is -175.9 kJ.</p>
<p>I started from: $$PV^k = C$$</p>
<p>Work is therefore: $$W = \int\limits_1^2 P\,dV$$</p>
<p>$$\frac{(P_2V_2 - P_1V_1)} {(1-k)} =\frac{(17237*0.057)-(6516*0.113)} {(1-1.4)} = 615.025 \:\mathrm{kJ}$$</p>
<p>Is my relation correct? What am I missing and why is the answer negative?</p>
| |thermodynamics| | <blockquote>
<p>Is my relation correct?</p>
</blockquote>
<p>Yes, it's correct. And here is the derivation:</p>
<p>mechanical work is defined as:
$$W = \int_{V_i}^{V_f}P\,dV$$</p>
<p>multiplying by $\frac{V^{k}}{V^{k}}$:
$$W = \int_{V_i}^{V_f}\frac{PV^{k}}{V^{k}}\,dV$$</p>
<p>Since $PV^{k}$ is constant, we can safely put it out the integration, yielding:
$$W = PV^k\int V^{-k}\,dV = PV^k\left[\frac{V^{-k+1}}{1-k}\right]^{V_f}_{V_i}$$</p>
<p>Finally, putting $V_f = V_2$ and $V_i = V_1$:
$$W = \frac{(P_2V_2 - P_1V_1)} {(1-k)}$$</p>
<blockquote>
<p>why is the answer negative?</p>
</blockquote>
<p>It's just a sign convention, if work is done by a system work is positive, if work is done on a system work is negative (as in your case).</p>
<blockquote>
<p>what am I missing?</p>
</blockquote>
<p>When I tried to solve the problem I got the same result as yours. Are you sure there is nothing missing in the problem description?</p>
| 3475 | Why do I get negative work done for adiabatic compression of air? |
2015-07-05T13:50:10.037 | <p>In fluid kinematics I can't understand the meaning of these terms : vorticity and circulation.<br>
Can somebody give me a description of these terms so that a lay person can understand them easily?</p>
| |fluid-mechanics| | <p>There are fundamental types of motion (or deformation) for a fluid element: translation, rotation, linear strain and shear strain. Usually all these types of motion occur concurrently which makes the analysis of fluid dynamics somehow difficult.</p>
<p>One can express the rate of translation vector mathematically by the velocity vector <span class="math-container">$\vec{V}$</span>:
<span class="math-container">$$\vec{V} = u\vec{i} + v\vec{j} + w\vec{k}$$</span></p>
<p>When it comes to expressing the rate of rotation of a fluid element it becomes quite challenging, Why? because a fluid element translates and deforms as it rotates, imagine an initially rectangular fluid element that starts to rotate while each line of the rectangule having a different angular velocity than the other. You can check White's book for the complete derivation but we can express the rotation vector <span class="math-container">$\vec{\omega}$</span> for now as follows:</p>
<p><span class="math-container">$$\vec{\omega} = \frac{1}{2} [(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z})\vec{i}
+ \frac{1}{2} (\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x})\vec{j}
+ \frac{1}{2} (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y})\vec{k}]$$</span></p>
<p>Putting it simply as half the curl of the velocity vector (just a mathematical manipulation):
<span class="math-container">$$\vec{\omega} = \frac{1}{2}\vec{\nabla}\times \vec{V}$$</span></p>
<p>Now, let us define a vector which is called <strong>Vorticity vector</strong> which is twice the angular velocity (we just got rid of the silly <span class="math-container">$\frac{1}{2}$</span>) and we might call it <span class="math-container">$\xi $</span>:
<span class="math-container">$$\vec{\xi} = \vec{\nabla}\times \vec{V}$$</span></p>
<p>Okay, enough with the math. What does it mean?</p>
<p>For an arbitary point in a flow field:</p>
<ul>
<li>Any fluid element (particle) that occupy that point having a <strong>non-zero vorticity</strong>, that point is called <strong>rotational</strong>.</li>
<li>Vice versa, Any fluid element (particle) that occupy that point having a <strong>zero vorticity</strong>, that point is called <strong>irrotational</strong> which means particle is not rotating.</li>
</ul>
<p><img src="https://i.stack.imgur.com/J0awM.png" alt="enter image description here" />
Flow from A to B is rotational (has voriticity) while flow from A to C is irrotational (has no vorticity).</p>
<p>You can find many examples for rotational flows such as in wake regions behind blunt bodies and flow through turbomachines.</p>
<p>According to this <a href="http://www.ess.uci.edu/%7Eyu/class/ess228/lecture.4.vorticity.all.pdf" rel="nofollow noreferrer">lecture</a>:</p>
<blockquote>
<p>• Circulation and vorticity are the two primary measures of rotation
in a fluid.</p>
<p>• Circulation, which is a scalar integral quantity, is a macroscopic
measure of rotation for a finite area of the fluid.</p>
<p>• Vorticity, however, is a vector field that gives a microscopic
measure of the rotation at any point in the fluid.</p>
</blockquote>
| 3478 | What is the meaning of vorticity and circulation? |
2015-07-06T00:26:33.297 | <p>I am trying to find the enthalphy of Helium if its internal energy is 200 kJ/kg. The answer is given as 333.42 kJ/kg, but I keep getting the wrong answer.</p>
<p>I've tried the following:</p>
<p>$\begin {align}
H &= U + PV \\
H &= U + mRT \\
\end {align}$</p>
<p>Universal Gas Constant, $R_u = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}}$</p>
<p>Individual Gas Constant, $R_i = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}} \div 4 \:\mathrm{\dfrac{g}{mol}} = 2.0785 \:\mathrm{\dfrac{kJ}{kg \cdot K}}$</p>
<p>$\begin {align}
H &= 200 \:\mathrm{kJ/kg} + 1\:\mathrm{kg}*2.0785 \:\mathrm{kJ/kgK}*273.15 \:\mathrm{K} \\
H &= 767.74 \;...\\
\end {align}$</p>
<p>Why is my answer different?</p>
| |thermodynamics| | <p>Following your method: You can't assume that temperature of Helium at that particular state is $273.15\ K$, but we can calculate it as follows:</p>
<p>$$u = C_{v}\Delta T$$
$$\Delta T = \frac{u}{C_v} = \frac{200}{3} = 66.66667\ K$$
$$h = C_{p}\Delta T = 5*66.66667 = 333.3 \ kJ/kg$$</p>
<p>As John said, Values from your text book of $C_p$ and $C_v$ should yield a value closer to your expected answer.</p>
| 3484 | Finding the enthalphy of Helium given internal energy |
2015-07-06T15:40:46.590 | <p>While deciding what tool (library) to use for a simulation of a hybrid gas turbine plant, I found that different libraries/tools of thermodynamic properties yield different values than the other.</p>
<p>I made a simple test to calculate $c_p$ of air at $300\ K$(same goes for enthalpy by the way) only for simplification and comparison and here are the results.</p>
<pre><code>Source | Cp at 300 K (J/kg.K)
--------------------------------------------
Thermopy (python) | 1004.89
Cantera (python) | 1010.06
CoolProps (python) | 1006.627
EES Software | 1005
Engineering Toolbox | 1005
Incropera's book | 1007
Cengel's book | 1007
</code></pre>
<p><strong>What do you think is the reason for the difference in values?</strong> </p>
<p>Apparently the one closest to the properties sheets from Incropera and Cengel is <code>CoolProps</code> but <strong>is this level of accuracy sufficient?</strong></p>
<p>The case might not be convincing for $c_p$, but it did occur to me once while calculating adiabatic flame temperature of some fuels and the error were in hundred Celsius range! </p>
<p>Link : <a href="https://gist.github.com/anonymous/3ec84b03c80928a5b5c0" rel="nofollow">used Python code</a></p>
| |mechanical-engineering|thermodynamics|simulation|numerical-methods| | <p><strong>What do you think is the reason for the difference in values?</strong></p>
<p>It's likely that they have calculated the coefficients of whatever model they are using from different data sets (or sub-sets). It's also possible that they are using different models.</p>
<p><strong>is this level of accuracy sufficient?</strong></p>
<p>That depends on what you're using it for. In general you probably have an acceptable level of error in your output. To get an idea of whether your input accuracy is acceptable you can do a <a href="https://en.wikipedia.org/wiki/Sensitivity_analysis" rel="nofollow">sensitivity analysis</a> on your simulation to see what kind of accuracy you need for your inputs.</p>
<p>My guess however, is that a variation of less than 1% will probably not be a significant source of error in your output compared to the simplifications of your model, and thus the sensitivity analysis is unnecessary for the case of errors in $C_p$</p>
<p>However, I do not know the accuracy of the rest of the package, so I'd recommend running a sensitivity analysis on your model, and an error estimation of each of your inputs to give you an idea of how much accuracy you need and if the package you use can deliver that accuracy.</p>
| 3492 | Different calculations for values of thermodynamic properties |
2015-07-07T15:56:36.713 | <p>My Lego model has a winch on the front. It's a simple spindle with a reel on it with a rubberised cog for winding at one end and two cogs at the other (as shown below).</p>
<p>Is there a technical name for the cog/gear on the blue spindle that stops the cog on the grey spindle spinning freely? Is this a common mechanism in real machinery?</p>
<p>Also, assuming this relies on friction and that the effectiveness will be reduced with wear and tear, how would that be mitigated in real equipment?</p>
<p><img src="https://i.stack.imgur.com/VE37S.jpg" alt="The gears/cogs in question"></p>
| |gears|pulleys| | <p>I would say that you have a <a href="https://en.wikipedia.org/wiki/Ratchet_(device)" rel="nofollow">ratchet</a>. Ratchets do fail. The <a href="https://en.wikipedia.org/wiki/Brownian_ratchet" rel="nofollow">Brownian Ratchet</a> is a simple thought experiment done by thermodynamic physicists that shows that ratchets need to fail - otherwise you could extract energy from absolutely nothing - a contradiction of the <a href="https://en.wikipedia.org/wiki/Laws_of_thermodynamics" rel="nofollow">Laws of Thermodynamics</a>.</p>
<p>Wear and tear on ratchets are mitigated via lubrication of parts of course, but bigger systems rely on things like <a href="https://en.wikipedia.org/wiki/Governor_(device)" rel="nofollow">governers</a> for real mechanical control instead of a ratchet.</p>
| 3506 | Gear that prevents another gear spinning freely |
2015-07-08T13:31:04.777 | <p>Does anyone have a feel for the flatness of Portland Cement when poured in a standard/typical way? Say, for example, that I wanted to pour a cement "cube" (20"x 18"x 19.25") in a mold. How flat would the top surface be without any special procedures? </p>
<p>Background:
I am going to mount a forceplate on top of this cement block and I am trying to determine if I need to specify the flatness tolerance in the drawing. </p>
<p>Is there somewhere I can look for this information?</p>
| |concrete|tolerance| | <p>If you live in a country that uses ACI standards, you might consider specifying flatness using an F<sub>L</sub> or F<sub>F</sub> number. ACI 117, Commentary Table R4.8.4 has this information, with flatness ranging from "Conventional" to "Super Flat." For the dimensions you have listed, specifying anything beyond "Flat" is probably overkill since your forceplate will likely cause the underlaying slightly-unflat cement to consolidate to the face of the plate anyway.</p>
<p>However, what I said above may be overkill for what you're doing. I'd just tell them to overfill and then strake it off using the top of the form as a guide, just like what you'd do scooping flour for a cookie recipe. That should get you what you want.</p>
<p>And to clarify - if the cement is a dry mix, my comment about the consolidation of the cement is valid. However, if this is a cement mix with water that is allowed to cure, the consolidation comment is not strictly applicable.</p>
| 3517 | Typical flatness of Portland cement |
2015-07-08T20:47:36.880 | <p>I've been shopping around for gears, pulleys and other transmission components for building a milling machine over the past few weeks. Many of these parts are so rudimentary and ubiquitous, yet the prices are very high. A ribbed pulley, for instance, about 3 inches in diameter often sells for over $30 online. Many of these parts are either stamped or cast, making them cheap to manufacture. What factors into the cost of these components? </p>
| |mechanical-engineering|linkage| | <p>In such cases you don't pay for the rubber, but for the administrative and storing costs. Manufacture a such rubber and then try to sell it on the ebay. With taxes, all of the law regulations, with post costs, etc.</p>
<p>For example, if you only want a rare screw to repair a device, you can easily pay even tens of dollars for that. Why? Producing a kilo of such screws would only cost fewer as a single dollar. But storing a thousands of similar boxes full with rare screws, and then sent them on post to countries thousands of kilometers away, it costs much more.</p>
<p>And, there is another effect, which an already deleted answer said. They won't calculate the price on the costs plus a profit, but they will see how is it priced on the ebay.</p>
<p>It is a highly unfortunate feature of the global society.</p>
<p>You can make significantly cheaper your costs if you pay a lot of them, and know the market (so you can buy it from cheaper manufacturers), and pay from them directly. But you can only do that if you aren't a hobbyist, but also a large manufacturer.</p>
| 3524 | What factors into the high cost of transmission components? |
2015-07-09T21:26:55.567 | <p>Over in <a href="http://chat.stackexchange.com/transcript/message/22676589#22676589">Programmers chat</a> we were discussing whether a gold skeleton would be feasible. That got me thinking:</p>
<p>If I were to build a Terminator (T-800) using pure gold as the endoskeleton, would the skeleton be able to support the <a href="https://scifi.stackexchange.com/q/32239/31563">weight of the terminator</a>? If so, what type of force would cause it to deform under stress? Would it be able to handle normal human tasks such as running, walking, or jumping? What about the superhuman feats of strength the terminator performs in the movies?</p>
<p>For the sake of this calculation, assume a T-800 weighs an extra 30% more than a human its size. Based on what Google says Arnold's size is and a rounding slightly, that assumption is 2m tall and 150kg.</p>
<p>The question <a href="https://physics.stackexchange.com/q/39512/57109">Stacking gold Bars</a> indicates that gold has quite a bit of strength given its reputation as a soft metal. However, that question talks about the compressive strength: also, I am not an engineer and unable to calculate the types of stresses a skeleton would undergo and whether those would be greater than the tensile strength of gold.</p>
<p>Finally, assume the same or similar motors and supports: there is very little information out there, but clearly this machine is capable of supporting great loads compared to humans, throwing grown men across rooms like dolls, etc. Only the skeleton is gold: motors, electronics, etc. could remain as other metals (steel, titanium, copper).</p>
| |materials|stresses|metals| | <p>The following table gives some of the material properties of four metals: gold, aluminium, titanium and iron. </p>
<pre><code>Metal Tensile Shear Bulk Young's Brinell Density
Strength Modulus Modulus Modulus Hardness
MPa GPa GPa GPa GPa g/cm^3
Gold 120 27 180 79 188-216 19.300
Aluminium 125-300 26 76 70 160-550 2.700
Titanium 210 44 110 116 716-2770 4.506
Iron 150-430 82 170 211 200-1180 7.874
</code></pre>
<p>From the table, gold has tensile and shear strengths similar to that of aluminium but its bulk density is more than double that of aluminium and slightly stronger than that of iron.</p>
<p>The Young's modulus for gold is similar to that of aluminium, so both metals have similar level of stiffness and from the Brinell hardness numbers gold is a soft metal.</p>
<p>An endoskeleton made of gold could support itself but having shear and tensile properties similar to aluminium it would not be able to withstand large external forces that would try to destroy it.</p>
<p>The other issue with using gold as the main metal for the endoskeleton is moving the endoskeleton and other body parts would require a lot of energy and strong motors due to its weight as gold is more than twice as dense as iron.</p>
| 3534 | Would a T-800 with a gold skeleton deform under stress? |
2015-07-10T07:04:24.203 | <p>Would the body of a normal round pot shaped pressure cooker (built to cook foods at 15 psi) be able to withstand being evacuated down to high-vacuum pressures?</p>
<p>Apparently it has been done before: <a href="http://www.instructables.com/id/Pressure-Cooker-Vacuum-Chamber/" rel="nofollow">i.e. this instructable</a> but was this person just lucky it didn't implode?</p>
| |mechanical-engineering|structural-engineering|pressure| | <p>Round profiled vessels fail at vacuum collapse pressures far below those designed for internal pressures. If more vacuum resistance is required, corrugated designs are preferred. Thin bellows survive upto 30% of external pressure as vacuum.</p>
| 3538 | Would a structurally sound pressure cooker be able to withstand an internal vacuum? |
2015-07-10T16:30:23.910 | <p>I have a strip of stainless steel encastree'd at one end to which is applied a constant pressure on one side, and I need to know what the deflection equation y = f(x) is at equilibrium.
If the deflection was small I would be able to use one of the very well known deflection formulas which assume vertical loads, but it isn't. </p>
<p>I am trying to work out the bending moment M=f(x) using integrals to use:
$$y(x)=\int\int_{beam} \frac{M}{EI}dx^2$$
But I'm getting more and more confused. What is the right approach to solving this problem?</p>
<p>Here is a diagram:</p>
<p><a href="https://i.stack.imgur.com/ZdihL.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ZdihLm.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|beam| | <p>For large deformation, there are changes to the strain-displacement relation and vertical equilibrium, but horizontal equilibrium remains unchanged.</p>
<p>Let N be the axial force and M the bending moment, for finite rotations, these are defined as:
$$
N=EA\epsilon^o
\\
M=EI\kappa
$$ where
$$\epsilon^o = \frac{du}{dx} + \frac{1}{2}\left(\frac{dw}{dx}\right)^{2}
$$
and
$$
\kappa = -\frac{\frac{d^{2}w}{dx^{2}}}{\left[1 + \left(\frac{dw}{dx}\right)^2\right]^{3/2}}
$$
so vertical equilibrium is given by:
$$
\frac{d^2M}{dx^2} + N\frac{d^{2}w}{dx^2} + P = 0
$$
where $w$ is the vertical deflection
The solution is then the solution to this differential equation dependent upon your set boundary conditions</p>
| 3541 | Large deflection of a cantilever beam with distributed normal load |
2015-07-10T16:32:20.300 | <p>I am answering the following question:</p>
<blockquote>
<p>Gas initially at a pressure of 101.325 kPa and temperature of 60 °C
undergoes the following cycle:</p>
<ul>
<li>1-2: Adiabatic Compression through compression ratio of 4.5:1</li>
<li>2-3: Heating at constant volume through a pressure ratio of 1.35:1</li>
<li>3-4: Constant entropy expansion to initial pressure</li>
<li>4-1: Constant pressure cooling to initial volume</li>
</ul>
<p>If $Cp = 1\ \mathrm{\frac{kJ}{kg\ K}}$ and $Cv = 0.678\
> \mathrm{\frac{kJ}{kg\ K}}$ for the gas, determine the thermal
efficiency for the cycle.</p>
</blockquote>
<p>The given answer is 41.41%.</p>
<p>Since it sounds pretty much like the Otto cycle, I try using the formula with compression ratio $r_k = 4.5$:</p>
<p>$$k = Cp/Cv = 1.47$$</p>
<p>$$e = 1-(1/(r_k^{k-1})) = 50.68\% $$</p>
<p>What am I doing wrong? Is it not an Otto cycle?</p>
| |mechanical-engineering|thermodynamics| | <blockquote>
<p>What am I doing wrong? Is it not an Otto cycle?</p>
</blockquote>
<p>No, it's not an Otto cycle. It would be if it wasn't for the last process, heat rejection (cooling) in Otto cycle is performed at constant volume not constant pressure.</p>
<p>This should be the <span class="math-container">$P-v$</span> diagram of your engine (sorry for the lousy image):
<img src="https://i.stack.imgur.com/He5SS.png" alt="enter image description here" /></p>
<p>So, naturally you can't apply Otto's efficiency to this cycle.</p>
<p><strong>Side note:</strong> compression ratio in Otto cycle <span class="math-container">$r$</span> is not <span class="math-container">$\frac{P_2}{P_1}$</span> as in your calculations, it's defined as the ratio between maximum and minimum volume in the cycle <span class="math-container">$\frac{V_{max}}{V_{min}} = \frac{V_4}{V_3}$</span></p>
| 3542 | Thermal efficiency of a given complete cycle |
2015-07-11T02:12:58.180 | <p>I am solving the following problem:</p>
<blockquote>
<p>The tank of an air compressor has a volume of 3 cubic feet and is
filled with air at a temperature of 40 °F. If a gauge on the tank
reads 150 psig, what is the mass of the air in the tank?</p>
</blockquote>
<p>The given answer is 1.78 lbs.</p>
<p>This is just $PV = mRT$, so I try:</p>
<p>$$ T = 40+460 = 500\ \mathrm{°R}$$</p>
<p>$$ P = (150\ \mathrm{psi_g} + 14.7\ \mathrm{psi_{atm}})*144 = 23,716.8\ \mathrm{\frac{lbf}{ft^2}}$$</p>
<p>$$ R = 53.34\ \mathrm{\frac{ft*lbf}{lbm*°R}}$$</p>
<p>$$\frac{PV}{RT} = m = \frac{23,716.8\ \mathrm{\frac{lbf}{ft^2}}*3\ \mathrm{ft^3}}{53.34\ \mathrm{\frac{ft*lbf}{lbm*°R}}*500\ \mathrm{°R}} = 2.67\ \mathrm{lbm}$$</p>
<p>Is the problem set wrong or am I missing something?</p>
| |mechanical-engineering|compressors| | <p>You forgot to divide by 1.5 at the bottom of your equation.
Try it and see, but I suggest you use exact values to start good habits.</p>
| 3544 | Finding the mass in the tank of an air compressor from gage pressure |
2015-07-11T16:18:46.103 | <p>I am contemplating buying a new car. However, the approach to the underground garage in my apartment has a 90 degree frustrating turn. Given the dimensions of the approach and car, what is the max turn circle for the car to fit the garage and turn?</p>
<p><img src="https://i.stack.imgur.com/A7A8w.jpg" alt="garage and car dimensions"></p>
<p><img src="https://i.stack.imgur.com/qoVZt.jpg" alt="more legible dimensions"></p>
<p>given Ackerman steering and the overhanging front part of the car I believe you can use Pythagoreas' theorem to get R min and R max. delta R should be less than the shortest path in the pathway, ie 2.5m. unfortunately the result does not seem plausible. feedback would be greatly appreciated.<img src="https://i.stack.imgur.com/Gj2A4.jpg" alt="enter image description here"></p>
| |automotive-engineering| | <p>Something important to consider is that if the corridor taking you down to the underground is narrower then the width of the way turning in, then there are certain sizes of cars that are able to go in but are unable to go out from the underground garage. So these cars can only go out in reverse.</p>
| 3547 | How much clearance does a car need when turning a corner? |
2015-07-12T03:57:26.417 | <p>Air in the cylinder of a Diesel Engine is at $30^o\ $$C$ and $138\text{ kPa}$ in compression. If it is further compressed to 1/18th of its original volume. Which of the following most nearly equals the work done in compression of the displacement volume of the cylinder is $14.2$ Liters</p>
<p>Answer is $11\ \text{kJ}$</p>
<p>$$T_1 = 30^o\ C + 273.15 = 303.15\ K$$
$$P_1 = 138\ kPa$$
$$V_d = 14.2\ L*(1\ m^3/1000L) = 0.0142\ m^3$$</p>
<h2>Attempt 1: Using isentropic relation</h2>
<p>$$r_k = 18 = V_1/V_2$$
$$V_1 = \frac{mRT}{P_2} = \frac{1*0.287*303.15}{138} = 0.63046\ m^3$$
$$V2 = 0.03502578502\ m^3$$
$$P2 = \frac{mRT}{V2}=\frac{287*303.15}{0.0350257} = 2484\ Pa$$
for air:
$$k = 1.4$$
$$W = \frac{(P_2V_2 - P_1V_1)}{k - 1} = \text{(A very very small value)}$$</p>
<hr>
<h2>Attempt 2: Using $r_k$ and displacement volume:</h2>
<p>$$r_k = \frac{(V_d + V_c)}{V_c}$$
$$V_d = 14.2L = 0.0142\ m^3$$
$$V_c = 8.3529411*10^{-4} = V_2$$
$$V_1 = V_c + V_d = 0.01583529\ m^3$$
$$P_1V_1^k = P_2V_2^k $$
$$\frac{138*0.01583529^{1.4}}{(8.3529411*10^{-4})^{1.4}} = P2 = 8487.512\ kPa$$</p>
<p>Plug in everything into the work function
$$W = \frac{P_2V_2 - P_1V_1}{k-1} = 12.1 \ kJ$$</p>
<p>Ok, I'm getting close but I don't know what I'm doing wrong... Any hints? And why doesnt' method 1 work?</p>
| |mechanical-engineering|thermodynamics|engines| | <p>Your second attempt is correct, you just had two mistakes: </p>
<ol>
<li><p>If you re-calculated $V_1$ you'll find that it equals $0.01503$ not
$0.0158$ $m^3$.</p></li>
<li><p>same issue with $P_2$.
$$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$</p></li>
</ol>
<p>Substituting in isentropic work equation:
$$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$</p>
<p>Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer <em>nearly equals</em>.</p>
| 3550 | Methods for determining Work of Diesel Engine's Isentropic Process |
2015-07-13T16:43:00.907 | <p>Almost all of the new Snapdragon CPU series (610, 810) have an overheating problem. This overheating problem may occur in the first few hours of using mobile phones that are equipped with the CPU, as we can see from the various reviews from around the web on Xiaomi mi4i, LG G4, Sony Xperia Z3+, etc.</p>
<p>I assume Qualcomm, a billion dollar company, has a quality control department working on (or detecting) this issue before they release the CPU. How does such a significant issue get past QC without being detected way before the launch date?</p>
| |quality-engineering|computer-engineering| | <p>There is no good reason to think that Qualcomm's QC people did not detect the problem. It's what you do about it that is the problem.<br>
That there is or ever was a problem is denied by Qualcomm, some users and some independent testers. But at least one reasonably well documented test set indicates that when running benchmark programs the processors tend to become more heat affected sooner than previous Qualcomm processors and than competitive alternatives.</p>
<p>High performance ICs generate substantial thermal output and all things being equal , output increases approximately linearly with throughput (as a major factor in energy loss is related to capacitive node charge / discharge, with energy dissipation being proportional to switching cycles per second).</p>
<p>In new device aimed at being near the front in a competitive market, the manufacturer may strive for raw output first and efficiencies second. Temperature relates to both thermal output and the ability to handle the heat. A device that is pushed towards the upper limits of a technology may be less efficient and require the device user to take more extensive efforts to cool it. In the finite space available 'corners may be cut'. Some manufacturers (eg Samsung) have gone to newer lower lithographic width devices to achieve the needed efficiencies. Qualcomm chose not to and as the devices are made by a third party, the drive to be faster sooner apparently swamped the awareness of the need to be better cooled. Some prospective users have addressed the problem by "baling out". Others by denying the problem exists as long as possible and then "writing software" to fix it. This is no doubt a painful learning experience. The project manager is unlikely to have the same problem again as (one can guess) he probably now works somewhere else, and Qualcomm will no doubt take greater notice of their QC department.</p>
<hr>
<p>The "answer" very much deep-ends on who you believe.<br>
I'm not an expert at all on this. I asked Google - and found that a lot of people are not experts but there does seem to be at least one good looking answer available as to what is really happening. </p>
<p>It seems they pushed too far too fast, used a fab house who was good but not as good as their cooperative opposition (Samsung) and allowed themselves to be less 'hands on' over the results until too late. (Almost*) ALL multicore processors in top end phones run into thermal speed limiting with time. Some limit more than others. The 810 is worst than most. (The Snapdragon 801 in eg the HTC One M8 at 2.3 Ghz seems to be an exception to the general rule). </p>
<p>The following does not so much advise "how" but rather what did or didn't happen and how transparent people were about it and what is really happening and when it was known. The following are almost in 'order I found them'. I could have reordered by date but its interesting to see that the date and what is said largely do not correlate well. </p>
<p><a href="http://www.forbes.com/sites/jaymcgregor/2015/05/06/qualcomm-finally-speaks-out-about-samsung-and-snapdragon/" rel="nofollow">Here</a> on May 6th Qualcomm deny that there is a problem in commercial products, as opposed to preproduction prototypes.</p>
<p><a href="http://www.techradar.com/news/phone-and-communications/mobile-phones/snapdragon-10-overheating-lg-g-flex-2-htc-one-m9-1290204" rel="nofollow">Here</a> on April 4th various user and manufacturere's spkesmen are being less than traansparent.</p>
<p><a href="http://www.theinquirer.net/inquirer/news/2413064/sony-admits-snapdragon-810-is-causing-xperia-z3-overheating-issues" rel="nofollow">Here</a> on June 15th !!! Sony are claimed to have admitted to problems with the 810 in their Z3+ after users demonstrated the ability to crash it by recording video. Sony advice a software fix will be provided. A software fix is a rather suspect way of fixing thermal issues. </p>
<p><a href="http://www.theinquirer.net/inquirer/news/2391263/samsung-to-ditch-qualcomm-snapdragon-chip-in-galaxy-s6-smartphone" rel="nofollow">Here</a> on January 29th Qualcomm adumbrate that Samsung had decided not to use the 810 and to use an internal Samsung processor instead.</p>
<p><strong>BUT</strong> on April 3oth <a href="http://www.theinquirer.net/inquirer/news/2406556/qualcomm-theres-nothing-wrong-with-our-snapdragon-810-chip" rel="nofollow">the same source</a> Qualcomm indicate that samsung may manufacture the snapdragon for the in future, and they say a decisin by LG to use a qualcomm hex core processor instead of the octal core snapdragon was made a year ago and is not due to any overheating issues and probably says that they will limit power use by reducing processing speeds for selected applicaations (where selected might mean "all". . </p>
<p>And/but on Feb 5th it is advised <a href="http://www.technobuffalo.com/2015/02/05/qualcomm-reportedly-fixes-its-overheating-snapdragon-810-processor/" rel="nofollow">here</a> that there never was a problem and BESIDES it has already been fixed, maybe.</p>
<p>Which leads to the </p>
<p><strong>Best picture you'll get, probably:</strong></p>
<p>Fairly definitively, on Feb 24th <a href="http://arstechnica.com/gadgets/2015/04/in-depth-with-the-snapdragon-810s-heat-problems/" rel="nofollow">here</a> some actual thermal tests. Nice time versus processor speed tests for range of processors. What happened? Qualcomm pushed things a bit far. the processor was made for them by TSMC in Taiwan. The device uses a number of ARM Cortex cores. How these are implemented varies with manufacturer. Samsung are using a smaller transistor process than TSMC. The smaller lower power process is really needed. </p>
<p>They say:</p>
<ul>
<li><p>In short, chips throttle, but the 810 throttles more than most, and it's severe enough that the 810 is actually slower than the 801 or 805 in some CPU-bound tasks over the long haul. The Exynos 7 Octa, which has similar specs on paper, is much better in practice.</p>
<p>At this point, Qualcomm has implied to us several times that its use of ARM Cortex CPU cores was a stopgap measure—Apple got the 64-bit A7 chip to market around a year before anyone expected it to. Chips are designed over a period of two or three years, so using the ready-made Cortex cores were the quickest way to get a 64-bit response to market.</p>
<p>The results, unfortunately, don’t look great. It might be because the 810 is using a 20nm TSMC manufacturing process instead of the 14nm Samsung process used for the Exynos, or it might be that Samsung has more experience working Cortex CPU cores into its designs. Whatever the reason, our testing of real phones with these SoCs in them shows that 810-based phones are slower and have worse battery life.</p>
<p>Qualcomm's next major flagship is the Snapdragon 820, the first to use its custom-designed 64-bit “Kryo” architecture. Rumor has it that the chip will be made on the same 14nm Samsung process as the Exynos 7 Octa.</p>
<p>A return to its own CPU cores plus a newer manufacturing process should hopefully mean a return to the kind of performance and battery life we’ve gotten from Qualcomm-based phones in years past. All signs point to the 810 being a one-time slip-up and not the start of a trend—let’s hope that those signs are accurate.</p></li>
</ul>
<hr>
<p><strong>AND</strong> just to add balance, on March 2nd,
<a href="http://semiaccurate.com/2015/03/02/behind-fake-qualcomm-snapdragon-810-overheating-rumors/" rel="nofollow">these people</a> - "SemiAccurate" - say there is no problem never was a problem, it's all made up and here are the lab tests to prove it. They claim to have investigated in depth and it was essentially a FUD campaign by a Korean phone manufacturer whose name begins with Samsung. </p>
<hr>
<p>I liked the time/temperature graphs. </p>
| 3554 | How do products like the overheating Snapdragon CPU make it through quality control? |
2015-07-14T12:29:28.053 | <p>I am designing an NFC device but am still a little unsure about the principles behind NFC/RFID. From what I understand:</p>
<ol>
<li><p>A primary coil constantly emits a 'carrier frequency' magnetic
field.</p></li>
<li><p>This field induces an e.m.f. on a passive secondary coil which is in
a secondary circuit.</p></li>
<li><p>The secondary circuit comprises a system that converts data into a
modulation signal, which is realised by a modulating impedance in
the second circuit.</p></li>
<li><p>This impedance causes something in the primary circuit to change,
thus receiving data.</p></li>
</ol>
<p>It is step 4. that I am most confused by. Sources I've read simply say the load impedance is "felt" by the primary coil, whatever that means.</p>
<p>Also, what is the function of having a separate transmit and receive antenna on NFC transceivers?</p>
<p>As a physicist I have very little prior exposure to electrical engineering or system design!</p>
| |electrical-engineering|wireless-communication| | <p>The transmitter to receiver coupling is similar to how a transformer works. for a ideal transformer, whatever impedance is connected to the secondary appears as the impedance of the primary divided by the square of the turns ratio.</p>
<p>For example, imagine a 7 V AC source and a 1:2 stepup transformer. If you connect a 3 Ω resistor directly to the 7 V, it will draw 2.3 A. If you instead put the transformer between the source and the resistor, the resistor now sees 14 V applied to it. It will draw 4.7 a from the secondary, which means 9.3 A must be going into the primary, and the resistance seen by the 7 V source is 3/4 Ω.</p>
<p>Now to get back to your question, surely you can see that a circuit could detect this 750 mΩ on it's 7 V supply if designed to do so.</p>
<p>The coupling between transmitter and receiver is not like a ideal transformer, but changes in impedance presented to the secondary (the coil in the receiver) still cause apparent changes in the impedance of the primary (the coil in the transmitter). It's a small matter of engineering to properly detect the pulses of different impedance the receiver is putting on its receiving coil, turn them into 1s and 0s, and decode this digital stream to get the data payload encoded onto it by the receiver.</p>
| 3566 | How do NFC/RFID devices transmit data? |
2015-07-14T17:41:09.513 | <p>The common description for a continous PID-controller is written like this:
$$y(t)=K_p⋅e(t)+K_i\int_0^t e(τ)dτ+K_d\dfrac{de(t)}{dt}$$
The best value of the constants $K_p$, $K_i$ and $K_d$ for a given controlled system will depend on its time constant(s), be it a $\text{PT}_1$ system, or $\text{PT}_2$ system, etc. ...</p>
<p>What do you do if the time constant of such a system is variable. Lets say, it varies between $T_a$ and $T_b$ ($T_a < T_b$). How do you design the PID-constants?</p>
| |control-engineering|control-theory| | <p>One way would be to implement some form of adaptive control. If your range of time constants is small and known, you could use something called <a href="http://www.controleng.com/search/search-single-display/back-to-basics-how-gain-scheduling-works/52cd5aac6a456dfb6b1352494e4dd169.html" rel="nofollow">"gain scheduling"</a>
where you determine before hand all the time constants you'll be dealing with (hopefully it is finite) and use if/then logic to define P I and D. It can be challenging to make sure you have covered enough variability to ensure stability and performance through the range. A good success story for gain scheduling is the Chinook helicopter. It can be done.</p>
<p>If you don't have a feasible prediction for what the time constants will be, you could look into using Model Reference Adaptive Control(MRAC). In this control scheme you have a reference model (your ideal system) with your chosen PID controller. The MRAC minimizes the error between what the plant is actually doing and what your reference model is doing. In this way you force your changing plant to act like your LTI model.</p>
<p>Or you could try using Model Identification Adaptive Controller (MIAC). Here the control scheme does system identification in real time and uses an update law for your controller. This one requires the most advanced skill of the three ideas. </p>
<p>Since your system is changing time constants over time, it is no longer LTI. This means you need to either do gain scheduling (pretty easy if you know the range of time constants) or system identification with update law for your PID.</p>
| 3571 | How do you set up a PID-Control if the time constants of the controlled system are variable? |
2015-07-15T01:58:10.700 | <p>I am solving the following problem:</p>
<blockquote>
<p>A Pipe 200 mm outside diameter and 20 m length is covered with a layer
of 70 mm thick insulation having thermal conductivity of 0.05 W/m·K
and a thermal conductance of 10 W/m<sup>2</sup>·K at the outer surface. If the
temperature of the pipe is 350 °C and the ambient temperature is 15
°C, calculate the external surface temperature of the lagging.</p>
</blockquote>
<p>I have calculated the thermal resistances as:</p>
<p>$${ R_{pipe} = \frac{1}{A \times h_o} = \frac{1}{ (\pi \times 200^2 \times (1/10^3)^2 \times ()} = 3.183098862 }$$</p>
<p>$${ R_{insulation} = \frac{1}{2\pi KL} = \frac{1}{ (2\pi \times 0.05 \times 20)} = 0.04776312933 }$$</p>
<p>The heat transfer is supposed to be:</p>
<p>$${ Q = \frac{\triangle T}{ R_{total}} }$$</p>
<p>What does it mean by "temperature of the lagging"? The change in temperature from the outside and inside already given.</p>
| |thermodynamics|heat-transfer| | <blockquote>
<p>What does it mean by "temperature of the lagging" ?
</p>
</blockquote>
<p>Air's comment:</p>
<blockquote>
<p>The lagging is not really the insulation layer itself but the cladding around the insulation layer. The insulation has no real structural properties to speak of; lagging not only holds the insulation in place and protects it from the environment (including insects and rodents), it can also provide a lower-emissivity surface for the insulation layer, further reducing heat transfer.</p>
</blockquote>
<p>however, since there is no given $k$ (thermal conductivity) value for the lagging I'll assume that the required temperature is the outer surface temperature of the insulation layer.</p>
<p>
Your $R_{pipe}$ term should be the resistance due to convection heat transfer from insulation surface to the ambient air:
$$ R_{pipe} = \frac{1}{A_o h_o} = \frac{1}{\pi D_{ins} L h_o} = \frac{1}{\pi * 0.17 * 2 * 20 * 10} = 0.00468\ \text{°C/Watt}$$
and since the thermal insulation is a cylinder with inner and outer radius of $r_1$ and $r_2$ respectively, its thermal resistance should be calculated as follows:
$${ R_{ins} = \frac{ln(\frac{r_2}{r_1})}{2\pi kL} = \frac{ln(\frac{170\ mm}{100\ mm})}{ (2\pi \times 0.05 \times 20)} = 0.0844\ \text{°C/Watt} }$$</p>
<p>$$Q = \frac{\triangle T}{ R_{total}} = \frac{350 - 15}{ 0.00468+0.0844} = 3760.664 \text{ Watt}$$
finally temperature of outer surface of insulation:
$$ \triangle T = Q\times R_{pipe} = 3760.664* 0.00468=17.6\ \mathrm{°C} = T_{outer} - T_{ambient}$$
$$T_{outer} = T_{ambient} + \triangle T = 15 + 17.6 = 32.6\ \mathrm{°C}$$</p>
| 3574 | Calculating the temperature of the lagging around an insulated pipe |
2015-07-15T17:42:40.840 | <p><a href="https://i.stack.imgur.com/mOgz0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOgz0m.jpg" alt="enter image description here"></a></p>
<p>$\rho, p_o, \alpha, F_g, h, b$ are given. $F_T$ is wanted. There is a plate that can rotate around $A$. The plate measures b perpendicularly to the screen surface.</p>
<p><strong>Pressure gradient</strong></p>
<p>$p(z) = \rho g (2h - z)$</p>
<p>I neglect $p_0$ since it it's the same pressure on both sides of the plate.</p>
<p><strong>Pressure Force</strong></p>
<p>$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}l$$</p>
<p>With</p>
<p>$sin(\alpha)=\dfrac{dz}{dl}$</p>
<p>$dl=\dfrac{dz}{sin(\alpha)}$</p>
<p>$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$</p>
<p>Now I can integrate from $0$ to $h$ since I substituted $l$ with $z$ and I get</p>
<p>$$F_p = b \rho g \dfrac{3h^2}{2sin(\alpha)}$$</p>
<p>This is the pressure force that works on the plate. To calculate its torque it has to be multiplied with its lever $l_a$.</p>
<p>In order to find $l_a$ I have to regard the actual torque which is</p>
<p>$$F_{p} \cdot l_a = \int l \cdot p \, \mathrm{d}A$$ </p>
<p>$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}l$$</p>
<p>Again substitute $dl$ with $dz$ yields</p>
<p>$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$</p>
<p>I assume I can now insert insert $0$ for $l_0$ and $h$ for $l_1$ again. And at this point I'm unsure but I think since I integrate over $dz$ again I can insert $\frac{z}{sin(\alpha)}$ for $l$.</p>
<p>$$F_p \cdot l_a = \int_{0}^{h} \int_0^b \! z \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$</p>
<p>Inserting and integrating</p>
<p>$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = \int_{0}^{h} \int_0^b \! \frac{z}{sin(\alpha)} \cdot \rho g (2h-z) \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$</p>
<p>$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = b \rho g \frac{1}{sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$</p>
<p>$$\dfrac{3h^2}{2} \cdot l_a = \frac{1}{sin(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$</p>
<p>$$l_a = \frac{1}{sin(\alpha)} \left[\frac{2}{3}h-\frac{2}{9}h \right]$$</p>
<p>$$l_a = \frac{4h}{9sin(\alpha)}$$</p>
<p>I'm very unsure with how I substituted and inserted. I would appreciate someone to review if I did it correctly and maybe elaborate on what I got wrong if possible.</p>
| |mechanical-engineering|fluid-mechanics| | <p>It looks like, from your math, you attempt to find the total normal force due to the pressure on the incline $F_p$, and then try to find a distance $l_a$ from point A such that if $F_p$ was applied there the torque would be equivalent to the torque caused by the pressure.</p>
<p>I'm not sure what you would use $l_a$ for, but if you're trying to equlibrate torques, it's easier to just calculate the torque due to the pressure (which you do in order to find $l_a$ anyway)</p>
<p>So I'm going to start from your calculations and note the errors in bold.</p>
<p>Throughout your calculation you've used the center dot to indicate scalar multiplication, while also using implied multiplication. Generally if the center dot is not everywhere, then it implies the dot product of vectors, so I've removed all instances.</p>
<p>The actual torque is:</p>
<p>$$\tau = \int \vec{l} \times \vec{n} \, p \, \mathrm{d}A$$</p>
<p>Where $\vec{n}$ is the vector normal to the surface, and $\vec{l}$ is the vector to the integrating point from axis A. As these vectors will always be perpendicular we can simplify using $l$ as the distance to point A.</p>
<p>$$\tau = \int l \, p \, \mathrm{d}A$$</p>
<p>$$\tau = \int_{l_0}^{l_1} \int_0^b \! l \, p \, \mathrm{d}y\mathrm{d}l$$</p>
<p>At this point I'd integrate over y to get:</p>
<p>$$\tau = b\int_{l_0}^{l_1} \! l \, p \, \mathrm{d}l$$</p>
<p>But we can delay that as you did.</p>
<p>Now we'll change our integration variable from $l$ to $z$ This step requires substituting the differential <strong>and changing the bounds</strong></p>
<p>$$z=l\,\sin(\alpha)$$
$$\mathrm{d}z=\mathrm{d}l\,\sin(\alpha)$$
$$\mathrm{d}z \frac1{\sin(\alpha)}=\mathrm{d}l$$</p>
<p>$$\tau = \int_{z_0}^{z_1} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$</p>
<p>Now inserting $0$ for $z_0$ and $h$ for $z_1$ makes sense <strong>but we can't substitute $z$ for $l$</strong>.</p>
<p>$$\tau = \int_{0}^{h} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$</p>
<p>Now we need to substitute $l$ for $z\frac1{\sin(\alpha)}$</p>
<p>$$\tau = \int_{0}^{h} \int_0^b \! \frac{z}{\sin(\alpha)} \, \rho g (2h-z) \frac{1}{\sin(\alpha)}\, \mathrm{d}y\mathrm{d}z $$</p>
<p>$$\tau = b \rho g \frac{1}{\sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$</p>
<p>So while your substitutions weren't valid in some intermediate states the final torque was correct.</p>
| 3582 | Calculating pressure force on inclined area |
2015-07-16T02:41:15.393 | <p>Given the question:</p>
<blockquote>
<p>Calculate the energy transfer rate across a 6 in wall of firebrick with a temperature difference across the wall of 50 °C. The thermal conductivity of the firebrick is $0.65\ \frac{\text{Btu}}{\text{hr ft }^\circ\text{F}}$ at the temperature of interest.</p>
</blockquote>
<p>The correct answer is 369 W/m<sup>2</sup></p>
<hr>
<p>I used the following approach:</p>
<p>$${ x = 6in => 0.5ft }$$
$${ \triangle T = 50^oC => 122^oF }$$
$${ k = 0.65Btu/hr-ft-F }$$</p>
<p>$${ \frac{Q}{A} = \frac {k\triangle T}{x} = 158.6Btu/hr*ft^2 => 500W/m^2 }$$</p>
<p>But I did not calculate the correct answer. Is there a step that I am missing?</p>
| |thermodynamics|heat-transfer| | <p>L=6 inches(2.54cm/(1 inch))=15.24cm</p>
<p>∆T=50℃</p>
<p>k= 0.65 Btu/hr-ft℉((1W/cm℃)/(57.79Btu/hr-ft℉))= 65/5779 W/cm℃</p>
<p>q_k/A=k/L ∆T</p>
<p>q_k/A= ( 65/5779 W/cm℃)/15.24cm × 50℃</p>
<p>q_k/A=0.0369 W/〖cm〗^2 = 369 W/m^2</p>
| 3585 | Problem with Energy Transfer Rate Conversion |
2015-07-16T06:25:14.473 | <p>For an assignment, I was given:</p>
<blockquote>
<p>Calculate the heat loss per linear ft from $2$ $in$ nominal pipe. ($2.375$ $in$ outside diameter covered with $1$ $in$ of an insulating material having an average thermal conductivity of $0.0375$ $Btu/hrft^oF$. Assume that the inner and outer surface temperatures of the insulation are $380^oF$ and $80^oF$ respectively.</p>
</blockquote>
<p>The correct answer is $116$ $Btu/lb^oF$</p>
<hr>
<p>I used the formula from conduction through pipes:</p>
<p>$${ Q = \frac{\triangle t}{R_{total}} = \frac{\triangle t}{ \frac{ln \frac{r_2}{r_1} }{2 \times \pi \times k \times L} } = \frac{\triangle t}{ \frac{ln \frac{d_2}{d_1} }{2 \times \pi \times k \times L} } }$$</p>
<p>Where:</p>
<p>$${ \triangle t = 380-80= 300^oF }$$
$${ d_2 = 3.375 }$$
$${ d_1 = 2 }$$
$${ k = 0.0375 }$$</p>
<p>And I get close, but not correct. I calculated it to be: $135.090464$ $Btu$</p>
<p>What am I doing wrong?</p>
| |heat-transfer|pipelines| | <p>I get 115.706...<br />
= 116
ie their answer.</p>
<p>What you are doing wrong (in this problem and your other recent one) is going too fast and not visualising how the data given translates into the real world situation. You are applying formulae correctly and seem to have a good understanding of what is involved to turn given parameters into correctly formulated expressions. Now all you have to do is slow down a bit and think about what you are doing. Do that and you'll do well ! :-)</p>
<p>The pipe is described as</p>
<ul>
<li>2 inch nominal pipe.
(2.375 inch outside diameter covered with 1 inch of an insulating material).</li>
</ul>
<p>One interpretation of that would be a pipe of 2.375 inch finished OD and 1 inch thick insulation under the outer surface. That gives a 0.375 inch internal pipe ID. Doesn't sound likely.</p>
<p>A second interpretation is a 2.375 inch ID internal pipe with 1 inch of external insulation over it for an external OD of ???</p>
<p>A third interpretation is the one you used.</p>
<p>I used the 2nd one and get the correct answer.<br />
Look at the 2nd version above.<br />
What is the OD?<br />
plug in the results and see what you get.</p>
<p>Short answer: Doh!!! :-)<br />
Try to aim at no doh!s - we all manage them but they can be minimised with due care.</p>
| 3588 | Heat Loss per Linear Ft |
2015-07-16T09:24:31.383 | <p><a href="https://engineering.stackexchange.com/questions/370/how-do-i-calculate-the-forces-on-a-desk-and-its-legs">This question about calculating loads on parts of a desk</a> made me finally ask my first question on SE.</p>
<p>What generally applicable standards are there for office furniture (desks, let's say), that define the loads that they must be able to survive?</p>
<p>In aerospace, the regulations ask for all structural parts to comply with all foreseeable loads in the complete operational scenarios during the entire life of the aircraft (static loads, dynamic loads, fatigue loads, etc); and 1.25 or 1.5 times that load is the ultimate load that each structure must be able to bear (even if for a few seconds).</p>
<p>I'd like to know open source materials, and real life experiences if available.</p>
| |mechanical-engineering|standards| | <p>In the United States at least, the design of mass-produced furniture if covered under ANSI/BIFMA (Business and Institutional Furniture Manufacturer's Association) <a href="https://www.bifma.org/store/ViewProduct.aspx?id=1375047">X5.5</a>. I don't have a copy of the standard to give you specific values, but in broad strokes the requirements for consumer goods like this (where the consequences of failure are much less than in an airplane or a building) codes focus less on design and calculation and more on testing of a physical sample.</p>
<p>One important aspect for furniture (not unlike airplanes) is that in addition to handling a maximum load, it needs to be able to handle many cycles of loading and unloading. Because of this, the design may well be governed more by the number of cycles the desk needs to be tested to than the maximum weight it needs to support one time.</p>
<p>Another major distinction is that in the US, manufacturers are not required by law to conform to the standard. Unlike airplanes, where the FAA has the authority to require conformance with applicable standards, furniture is not centrally regulated. As a way of controlling quality though, many large purchasers like schools or corporations will require conformance on all products that they buy, causing some manufacturers to develop products that conform. The difference of course is that it is perfectly legal (and common) to sell a desk that has not been tested by the standard.</p>
| 3591 | What loads are office desks required to withstand? |
2015-07-17T21:56:42.583 | <p>If I have a flat span of steel material, 3.0m x 150mm wide and 5mm thick, and apply an increasing downward force to the middle, at some point it will fail.</p>
<p>But if the profile of that piece of steel was T shaped instead of a flat span, it is much stronger. </p>
<p>Is the Safe Working Load (SWL) for different profiles calculatable by a specific formula? </p>
<p>Assuming so, otherwise how could you design things like cranes, does the same formula apply to different materials e.g. aluminium v steel and different thicknesses of the same material?</p>
| |mechanical-engineering|materials|structures| | <p>There are better ways of <em>uniformly</em> stiffening a plate e.g., by the <em>Sandwich</em> principle. A somewhat compromised version is the <em>Isogrid</em>, these are the distributed I-beams.</p>
<p>In both methods, effective bending rigidity is increased by making plate central parts hollow and so lighter in weight. Stresses are reduced.</p>
<p>Please google under these names. In place of 5 mm is a thick plate, you can use 1/8 inch plate and tack weld on isogrid cells or at least orthogonal ribs/grills of same 1/8 inch thickness, say 3 inch cell sizes. The webs are like bottle crates. Stiffness could go up by 4/5 times. The extra fabrication is worth it. </p>
<p>If alternate material choice is ok, even a 1-inch plywood sheet can be bonded with 1/16 thick steel sheets usind a good adhesive on either side for reduced deformation. </p>
<p>When improving designs, detail is important.</p>
| 3605 | How much strength does adding a T piece add? |
2015-07-18T00:11:36.530 | <p>I am solving the following problem:</p>
<blockquote>
<p>A liquid-to-liquid counterflow heat exchanger is used to heat a cold fluid from 120 °F to 310 °F. Assuming that the hot fluid enters at 500 °F and leaves at 400 °F, calculate the log mean temperature difference for the heat exchanger.</p>
</blockquote>
<p>The given answer is 232 °F</p>
<hr>
<p>I used:</p>
<p>$${ LMTD = \frac{\triangle max - \triangle min }{ ln( \frac {\triangle max}{\triangle min} ) } }$$</p>
<p>$${ \triangle max = 500-120 = 380 }$$
$${ \triangle min = 400-310= 90}$$</p>
<p>I plug them all together and I got 201.3383331.</p>
<p>What am I doing wrong? </p>
| |heat-transfer|heat-exchanger| | <p>$\triangle max$ and $\triangle min$ are not defined as maximum and minimum temperature differences in a heat exchanger. Quoting from Wikipedia:</p>
<blockquote>
<p>The LMTD is a logarithmic average of the <strong><em>temperature difference between the hot and cold feeds at each end</em></strong> of the double pipe exchanger.</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/35kev.png" alt="Temperature distribution of HEX inlets and outlets"></p>
<p>$${ \triangle T_{max} = 400-120 = 280 \text{ F}}$$
$${ \triangle T_{min} = 500-310= 190\text{ F}}$$
$${ LMTD = \frac{\triangle T_{max} - \triangle T_{min} }{ ln( \frac {\triangle T_{max}}{\triangle T_{min}} ) } } = \frac{280-190}{ln( \frac {280}{190})}= 232\text{ F}$$</p>
| 3606 | Log Mean Temperature Difference of a Counterflow Heat Exchanger |
2015-07-18T14:44:59.970 | <p>Pressure drop across a porous medium can be described by the Forchheimer extension to Darcy’s law:
$$-\frac{dp}{dx} = \frac{\mu}{K_1}\cdot v + \frac{\rho}{K_2}\cdot v^2$$</p>
<p>where $K_1$ and $K_2$ are permeability coefficients that depends on medium geometry (Pitz-Paal
et al., 1996).</p>
<p>I just came across this paragraph in a paper studying the flow stability in porous media in solar volumetric receivers, which I couldn't digest at all:</p>
<blockquote>
<p>In the flow through a porous sample, the mass flow density is determined by the pressure difference between the two sides of the sample. <strong><em>The pressure drop is produced by a blower. Instability occurs when a pressure drop causes different mass flow densities</em></strong></p>
</blockquote>
<p>What blower? and what makes pressure drop cause different mass flow? (Keeping in mind previous equations)</p>
| |mechanical-engineering|fluid-mechanics|porous-medium| | <p>Expanding Dan's answer (after some research), The main reasons for the so-called instability in porous media that is subjected to a heat flux (e.g. Solar irradiance) with air as a working medium are the behavior of dynamic viscosity of air with temperature (Viscosity increases as temperature increases) and the bell-shaped distribution of concentrated incident solar flux on the porous receiver.</p>
<p>Many experiments mentioned in the same paper suggested a relationship between incident flux and velocity of air, If a local high flux spot was formed in a porous medium mass flow rate is decreased and temperature is increased (high local dynamic viscosity leading to a lower local flow rate) and vice versa for local low flux spots, which means that the porous medium temperature can exceed the design temperature leading to failure even if the outlet temperature is low.</p>
<p>This can be shown by integrating the Forchheimer extension of Darcy's law using a correlation for viscosity behavior with temperature and ideal gas law, the following curve relating the quadratic pressure difference and outlet temperature can be obtained:</p>
<p><a href="https://i.stack.imgur.com/GGdoJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GGdoJ.png" alt="enter image description here"></a></p>
<p>It can be shown that at specific incident fluxes at the same quadratic pressure difference there could exist two different outlet temperatures (hence the instability).</p>
| 3609 | What is "flow instability" in porous media? |
2015-07-18T18:23:49.867 | <p>I work for a small engineering and manufacturing company. To date our production process for each product has been defined in each critical aspect by the design engineer, then handed to the production department and rarely looked at again by engineering. Many noncritical aspects are undefined and left to the variable judgement of the assembly workers. I'm considering assigning an engineer to the task of analyzing and documenting the entire production chain of a product for consistency and quality purposes, and to look for cost saving improvements. What discipline of engineering would this be considered, if any?</p>
| |documentation|procedure| | <p>Most engineering discipline are capable of defining manufacturing and test process. It is the management function to ensure the processes are followed. Generally depending on the size and structure of the organization most defining manufacturing and testing process are mostly handled by manufacturing test and process engineering teams. If a <a href="http://asq.org/index.aspx" rel="nofollow noreferrer">Quality</a> organization exists this team gets heavily involved in ensuring the implementation and monitoring of processes. </p>
<p>In the event of the quality related issue the quality organization gets involved in root cause and analysis of the quality related issue. In event of issues with the manufacturing process the changes are made the to process through the use of a change management process. </p>
<p>Also periodic quality audit are established to ensure if the manufacturing and and test process are followed per the definition. This is an proactive measure.</p>
<p>This question and response is loosely tied to one of your previous questions <a href="https://engineering.stackexchange.com/questions/2264/how-can-i-arrange-my-eco-system-to-enforce-the-principle-of-least-privilege/2267#2267">How can I arrange my ECO system to enforce the principle of least privilege?</a></p>
<p>There are statically process control, six sigma - continuous improvement that can address some of the above mentioned issues.</p>
<hr>
<p><strong>Summary</strong></p>
<p>A Quality Engineer with direction from management as well as significant support from manufacturing and design engineering can address the this issue.</p>
<hr>
<p><strong>References:</strong> </p>
<ul>
<li><a href="http://asq.org/manufacturing/why-quality/overview.html" rel="nofollow noreferrer">Quality In Manufacturing</a></li>
<li><a href="https://engineering.stackexchange.com/questions/3315/how-to-identify-low-quality-electronic-components-early">How to identify low quality electronic components early?</a></li>
</ul>
| 3612 | What engineering discipline deals with ensuring that a manufacturing and testing process is clearly defined and followed? |
2015-07-19T11:51:35.880 | <p>I have some tomato plants on my balcony (4th floor in a 7 story building). The balcony is a recessed one.</p>
<p><img src="https://i.stack.imgur.com/Qamnt.jpg" alt="Photo of recessed balkonies"></p>
<p>The building has lightning protection on top.</p>
<p>To support them, I put a 2 m long aluminum rod into each pot. At the tip of the rods I put some isolating tape, so that it covers the tip of the rod.</p>
<p>Will these rods attract lightning?</p>
| |electrical-engineering|building-design|safety| | <p><a href="https://what-if.xkcd.com/16/" rel="nofollow noreferrer">xkcd</a> has a fantastic description of how lightning works. <a href="https://www.youtube.com/watch?v=6MUYsIjTKvk" rel="nofollow noreferrer">This video</a> is also quite good (especially as of 0:56, but the rest is worth watching as well) since it shows lightning in super slow motion, so you can see it "searching" for a path down to earth. The important thing here is to learn that lightning searches for a path in steps. Xkcd mentions 60-meter steps and the video mentions 50 yards (at 1:00), which for our purposes are close enough.</p>
<p>This means that for your rods to be struck by lightning, they need to be the path of least resistance within 60 m (or whatever) of the step leader (the searching tip of the lightning). For this to happen, it's pretty safe to assume that the step leader would basically have to be on (or very close to) the same horizontal plane as your rods and probably around 60 m away. This is because if it were coming from above, then the building's lightning rod would take care of it. If it were coming at something of an angle, then it would probably strike one of the floors above you (any exterior electrical sockets would do the trick or maybe even the railings as @Olin mentions in <a href="https://engineering.stackexchange.com/a/3619/1832">his answer</a>). And if it were on the same horizontal plane but very close, then once again, there's probably a better route on one of the floors (maybe yours) through electrical sockets or railings.</p>
<p>So, basically, out of an infinity of possible points where the step leader may be, only a very, very small subset of those points would really risk your rods getting struck by lightning. So you're probably safe. Or, as Olin points out, just use wooden rods.</p>
| 3617 | Will lightning strike an aluminum rod in a pot on my balcony? |
2015-07-19T07:33:37.257 | <p>Is it possible to use an IR sensor to differentiate accurately between carbon fiber and a human hand kept on the carbon fiber surface?</p>
<p>I would like to move the sensor from one end of the carbon fiber object (a rectangular surface) to the other and be able to detect a significant difference in reflected intensity when the sensor passes over the hand. Will this work? If not, what are alternative sensors that I could use? The sensor will be approximately 10 cm above the carbon fiber surface.</p>
| |materials|sensors| | <p>Some IR sensors just receive. In this case they distinguish only by difference in infrared emission from the environment. These sensors could not see a hand if it was inside an oven mitt for example or if the hand was the same temperature as the surface, but in most cases it would work.</p>
<p>In a photo gate setup you could have an IR led constantly on to illuminate the surface. It would work as long as the reflectiveity of your surface was always different from a hand (not sure if we are talking raw carbon fiber here or shiny epoxy coated carbon fiber). Dirt on a shiny surface would cause problems as would external heat sources like sunlight.</p>
<p>An infrared proximity sensor uses triangulation and will be able to distinguish your targets more reliably, but will still have sunlight issues. <a href="http://www.societyofrobots.com/member_tutorials/book/export/html/71" rel="nofollow">More info on IR sensors</a></p>
<p>You will want to look at an ultrasonic sensor. An ultrasonic sensor could easily handle the task and works better in dirty and sunny environments. A capacitive sensor would also work, but only at short range.</p>
<p>Sensor Examples:<br>
<a href="https://www.sparkfun.com/search/results?term=ultrasonic%20proximity%20sensor" rel="nofollow">Ultrasonic and infrared proximity sensors</a><br>
<a href="http://www.automationdirect.com/adc/Overview/Catalog/Sensors_-z-_Encoders/Ultrasonic_Proximity_Sensors/18mm_Round,_400mm_Maximum_Sensing_Distance_(UK_Series)" rel="nofollow">Ultrasonic proximity sensors</a><br>
<a href="http://www.omega.com/pptst/E2K-X.html?pn=E2K-X4ME1" rel="nofollow">Capacitive proximity sensor</a><br></p>
| 3623 | IR sensor with carbon fibre |
2015-07-19T20:37:25.677 | <p>Referring to von Guiricke's early experiment to hold vacuum using <a href="https://en.wikipedia.org/wiki/Magdeburg_hemispheres" rel="nofollow">Magdeburg hemispheres</a>, what is the best known shape to hold say 10 atmosphere vacuum (external pressure) using <em>any material or material combination</em> without buckling its walls? I.e., using the least amount of material so that the vacuumed vessel is as lightweight as possible.</p>
<p>EDIT1:</p>
<p>Material choice relaxed in question as better materials are now available compared to Magdeburg test time.</p>
| |mechanical-engineering| | <p>When I design pressure vessels to hold vacuum, here are the preferences I usually give to customers:</p>
<ol>
<li>Spherical is always the best in terms of minimizing the thickness, but not the most economical in therms of amount it can hold or for being friendly to the manufacturer</li>
<li>Ellipsoidal (2:1) heads on a shell whose length is equal to the diameter is second best. It comes out a bit oblate compared to the sphere (1.5:1), but per given packing area holds plenty of volume. The ellipsoidal heads work out to equal the thickness requirements of the shell, but is not structurally as efficient as the sphere.</li>
<li>Avoid Flat areas - especially flat bottoms. If you need it to be flat on the bottom, weld a skirt to the bottom. </li>
</ol>
| 3624 | What is the most lightweight geometry for a vacuum holding vessel? |
2015-07-20T04:43:09.243 | <p>Making extremely accurate machines seems like a chicken and egg problem. How do you make one if you don't already have one?</p>
<p>For example, how was the first <a href="https://en.wikipedia.org/wiki/Indexing_head">indexing head</a> made without an indexing head to make the indexing plate? How was the first lathe made without a lathe to make the spindle? Is it an iterative process where you start with something hand-made, and each machine can make one slightly better than the last? I suppose that <a href="https://www.mathsisfun.com/geometry/constructions.html">geometric constructions</a> are the source of accuracy of some primitive machines. But how do you get from there to something like the precision screw required for an accurate lathe or micrometer?</p>
| |mechanical-engineering|engineering-history| | <p>For plenty of detail on building accurate machine tools from scratch, read the excellent and detailed "Foundations of Mechanical Accuracy"
By Wayne R. Moore. It's an old book and there are plenty of copies around if you use your favourite web search engine.</p>
<p>The basic principles for increasing accuracy are lapping and comparison of surfaces.</p>
<p>From memory some of Moore's key steps are:</p>
<ul>
<li>Flat lapping 3 surface plates against each other to form a master surface plate</li>
<li>Copying straight edges from the master surface plate</li>
<li>Centreless grinding to make the first cylinders</li>
<li>Cylindrical lapping to refine cylinders</li>
<li>Lapping accurate screw threads from rough initial screws</li>
</ul>
| 3627 | How are increasingly accurate machines made? |
2015-07-21T01:47:14.253 | <p>I am doing B.E. in Mechatronics Engineering. I am due to take a subject called 'Sensors and instrumentation'. So far I only have the syllabus for the course. I decided to do some reading in advance of the course starting.</p>
<p>The syllabus refers to "low pass sensors" and "band pass sensors".<br>
When I web search for these terms I find many mentions of low pass <em>filters</em> and band pass <em>filters</em>, but nowhere on the Net are "low pass <em>sensors</em>" and "band pass <em>sensors</em>" mentioned.</p>
<p>Do the terms "low pass sensors" and "band pass sensors" have their own meanings and, if so, what are they?</p>
<hr>
<p>I have been googling for the past five days for these two words but none of the results satisfactorily explained the meaning of 'low pass sensor' or 'band pass sensor'.</p>
| |electrical-engineering|sensors|power-electronics| | <p>You are combining two concepts. </p>
<ol>
<li><p><strong>A sensor</strong> is any device that responds to a desired signal in some manner to produce an output that is used to detect the presence of the desired signal or to allow some desired attribute of the signal to be measured - such as its amplitude, frequency, rate of change, relative phase, etc. Even colour, temperature, dimensions, current/voltage/power ... . </p></li>
<li><p>The terms <strong>low-pass</strong> and <strong>band-pass</strong> relate to the processing of a sensed signal so that only a portion of the signal is used to provide the sensor output. It may seem counter intuitive to ignore part of a signal, but this is useful when the signal may be accompanied noise or by other interfering or unwanted signals. </p></li>
</ol>
<p>In the above context, a low-pass sensor is a sensor, intended to detect or measure a specific signal or condition, that accepts the lower frequency components of the signal but rejects input above a selected frequency. A band-pass sensor is one which processes only signals with frequencies within a given range of frequencies, and rejects signal higher or lower in frequency than the selected band of frequencies. </p>
<p><strong>As examples:</strong></p>
<p>A low pass sound sensor may be a microphone plus amplifier plus a filter set to exclude frequencies above say 4000 Hz. Such a sensor would respond to speech signals and the fundamental sounds of many musical instruments but would not be affected by high frequency noise signals. </p>
<p>A bandpass doppler radar may have the range of frequencies that it acts on limited to a range of Doppler frequencies corresponding to velocities from the speed of a person walking slowly up to about 100 mph. It would therefore tend to ignore signals from eg a door being very slowly closed (V << walking speed) or to signals from reflections from aircraft (V > 100 mph). </p>
| 3639 | What is meant by the terms "low pass sensor" and "band pass sensor" in a course syllabus? |
2015-07-21T09:48:56.713 | <p>What is this tiny rotating piece of hardware that I found inside my Samsung GE1202? It has E9C written in blue on it.</p>
<p><a href="https://i.stack.imgur.com/H4kSJm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/H4kSJm.png" alt=""></a></p>
| |electrical-engineering|motors|consumer-electronics| | <p>That is a micro vibrating motor to produce the "buzz" effect from your phone.
I used much <a href="http://www.chicagovibrator.com/Store/tabid/84/c/truck-vibrators/Default.aspx">much larger vibrating motors</a>, air driven, on the bottom of my Class 7 dump truck. Mounted under the bed, they would vibrate the bed, loosening the material so it would flow out cleanly. </p>
| 3644 | What is this small rotating part from a cell phone? |
2015-07-21T14:43:24.177 | <p>Would it be possible to power a small LED light using the headset jack of a regular smartphone, and if so, how would you connect the LED light with an audio jack. How would such an implement look like. I'm not very versed in electrical terminology so simplified explanations are appreciated. </p>
| |electrical-engineering|audio-engineering| | <p>Another option to consider is that many phones have a high intensity LED built in near the camera lens which works as a flash. It may be possible to use mirrors or a light pipe to make use of this. </p>
| 3648 | Powering LED light using smartphone headset jack |
2015-07-21T15:41:57.677 | <p>I need to design an automatic shifting system, for a <a href="http://students.sae.org/cds/formulaseries/about.htm" rel="nofollow">FSAE car</a>. Clutch and shifts are electronically engaged and disengaged using a <a href="http://pi-innovo.com/" rel="nofollow">Pi Innovo</a> engine control unit (ECU). </p>
<p>The issue I am worried about is downshifts, I would like to rev-match, for quick, clean downshifts and clutch life.
There is no possibility of my team using an electronic throttle, therefore I can not "blip" the throttle electronically. </p>
<p>My idea is when a downshift is needed, I send a signal to FULLY disengage the clutch electronically, then when the engine RPM reaches what is consider matched, I send a signal to shift.</p>
<p>Would allowing my engine to rev while the clutch is fully disengaged damage the clutch? Also, how is important is rev-matching? We already replace our clutch after a week of racing, would not rev-matching be that much of an issue?</p>
| |electrical-engineering|automotive-engineering|control-engineering| | <p><strong>Overview:</strong><br>
I wouldn't be worrying about clutch life. From my experience with FSAE events they're not long enough to actually wear down a clutch. </p>
<p><strong>However(!!):</strong><br>
One of the main reason racers actually blip the throttle/rev match is to avoid <strong>compression lockup</strong>. This is when the engine braking caused by downshifting (and resultantly having to bring the engine up to a higher RPM to match the wheel speed) causes your wheels start rotating slower which causes the car to oversteer. </p>
<p><strong>As per your other questions:</strong><br>
Allowing the engine to rev while the clutch is disengaged will not damage the clutch. Your proposed solution of allowing the engine to decelerate freely and only engaging the clutch at a preset level could very well work from a clutch-life point of view but it will probably result in slow downchanges with an unpredictable application of power/traction (suddenly the engine will be connected to the rear wheels when a split second before they had just been free-wheeling). This would be pretty disastrous/hard-to-compensate-for in corners. </p>
<p><strong>Alternative solutions:</strong><br>
Usually when these types of systems (sequential dog-boxes with some type of smart shifting mechanism but no throttle control) are employed the driver will manually downshift (such as in V8 race cars in Australia) allowing them to rev match and whatnot. </p>
<p>Or you could look at installing a <strong>slipper clutch</strong> on your engine/transmission which would then completely remove the need to rev-match at all and make the downshifting system just as easy as the upshifting one (without introducing additional clutch-wear or compression lockup). This would be by far the easiest solution (but unfortunately quite costly). </p>
| 3649 | Automatic Shifting without Electronic throttle |
2015-07-22T18:10:00.393 | <p>Searched, but didn't find this discussed here previously. I'm upgrading a hobby-level CNC router (a Shapeoko 3) to have wider belts (it came w/ 6 mm GT2 (2 mm)), I'm going up to 9 mm (otherwise same specs).</p>
<p>I ordered new pulleys and belting from SDP/SI, but when I went to install them was concerned that the end of the motor shaft (6.35 mm diameter NEMA23) is inset from the end of the pulley by almost 5 mm (4.88 mm). With the previous pulleys, they were flush with the motor end (but probably should have been slightly off to allow the belt to run true).</p>
<p>Is that too great a distance? What would be the maximum distance that the pulley could be off the shaft? What guidelines are there for this? How is it determined? (I've tried searching but haven't been able to find anything – am I not using the correct terminology?)</p>
<p>What is more important, that the belt run true or the pulley be on the shaft? There's only a couple of inches between the idlers (pairs of flanged bearings on M6 bolts) and a great deal of engagement.</p>
<p>The pulley in question is SDP/SI part # A 6A51-020DF0908.</p>
| |mechanical-engineering|pulleys| | <p>If I understand you correctly, you're worried about the radial load on the motor shaft due to belt tension when the pulley is located axially 5mm further from the motor. </p>
<p>Whether or not the loading scenario you've described is OK depends on many factors; the shaft dia, material properties of the shaft, the motor bearings, the speed of the shaft, tolerable shaft run-out, the belt tension as well as the axial location of the pulley.</p>
<p>The better motor suppliers will specify an allowable radial load at a specific axial location on the shaft, typically the center halfway between the motor bearing and the end of the shaft. That doesn't mean you have to load the shaft radially through that point, but the further away from the motor the radial load is applied, the less radial load is acceptable. I'm guessing that you don't have this information.</p>
<p>The most significant consequences of applying a radial load further away from the motor will be increased radial load on the motor bearings and increased shaft deflection. The increase in bearing load will reduce the life of the bearings and the deflection will cause greater shaft vibration. How significant these are really depends on your application. </p>
<p>I'm assuming that you're using a typical toothed drive belt, which shouldn't require a lot of tension to function properly, so the effect <em>should</em> be relatively small. If you're concerned, you could measure the effect with a dial indicator mounted to the motor, measuring the radial deflection of the shaft in line with the direction of the belt pull with no tension on the belt and compare that to the same measurement made with the belt properly tensioned. </p>
<p>Without more information, I would say to try it because the alternative of intentionally misaligning the belt in an attempt to keep the radial load closer to the motor is going to result in increased belt wear and unwanted friction. </p>
| 3657 | Guidelines for motor shaft length and pulley engagement |
2015-07-23T13:47:44.550 | <p>Recently, tube steel manufacturers in the US have begun using the new ASTM A1085 specification for forming tube steel (a.k.a. hollow structural section or HSS) shapes, as opposed to the existing A500 specification.</p>
<p>From what I've read (<a href="https://www.aisc.org/WorkArea/showcontent.aspx?id=33730">here</a> and <a href="http://www.atlastube.com/astm-a1085">here</a>), the A1085 material spec has the following benefits over the existing A500 spec:</p>
<ul>
<li><p>tighter tolerances (no need to take reduced wall thickness for design)</p></li>
<li><p>identical yield strengths for all types of tube members (as opposed to A500, in which the yield strength differs for round and rectangular shapes and depending on the grade)</p></li>
<li><p>set maximum yield stress of 70 ksi (useful for seismic applications, though I am not well versed in this area)</p></li>
<li><p>standard Charpy V-notch requirements corresponding AASHTO Zone 2 (I assume useful for the transportation industry for fatigue reasons)</p></li>
</ul>
<p>These benefits come with a small premium - I've read 7% to 10% increased cost depending on the mill.</p>
<p>I primarily work as a structural engineer in the nuclear industry. Most of the work I do is in services, where we perform mostly small modifications to plants (i.e., no design of large buildings for the most part, but smaller supports, platforms, and the like).</p>
<p>If I have no need to restrict the maximum yield strength of the material and I have no fatigue concerns, is there any benefit to specifying the newer A1085 tube steel material over the existing A500 specification? For practicing engineers in non- or light-seismic zones, have you realized any benefit using the new A1085 spec?</p>
| |structural-engineering|steel|specifications|columns|hss| | <p>AISC's magazine, Modern Steel Construction periodically publishes a guide to specifying grades of steel to help engineers stay aware of developments in the market. Their latest <a href="http://msc.aisc.org/globalassets/modern-steel/steelwise/2015/steelwise.pdf" rel="nofollow">edition</a>, from this February mentions A1085, but still suggests A500 Grade C as the standard. They recommend checking that A1085 is available and affordable in your area - it sounds like you already have. This may not be true in all regions.</p>
<p>Other than the maximum yield strength which isn't useful in your practice, there are a few other convenient features. Firstly, according to <a href="http://msc.aisc.org/globalassets/modern-steel/steelwise/steelwise-september-2013.pdf" rel="nofollow">this</a> article, it has the same mechanical properties specified regardless of product form. By contrast, A500 Gr C tube has a different Fy for square/rectangular and round tubes. Also, A500 allows actual wall thickness to be reduced by a large variance. This was intended to allow for cheaper manufacture by less accurate methods, but with modern steel mills, tubes are consistently produced undersized. Because of this, calculations require reducing the wall thickness by 7% from nominal. While these issues are not major considerations in specifying steel, they will have some appeal in simplifying the design process and reducing opportunities for error.</p>
<p>The standard also specifies minimum corner radii, which reduces the risk of corner cracks. Previously, AWS D1.1 and the AISC Manual have warned about the risk of corner cracks in square and rectangular A500 members subject to significant stress from welding or galvanizing. I'm not certain if this fear will go away, or just be easier to quantify if using the new standard since A500 is produced with fairly uniform radii already.</p>
<p>The main down side of specifying the A1085 tube is cost, as you point out. Additionally, you may not be able to count on availability right now if you are specifying projects outside of your specific area. One other down side if you do lighter work is that A1085 is not available with 1/8" wall thickness like A500 is. </p>
<p>In your situation, these considerations may be a wash, meaning you should continue to specify A500 to reduce costs as long as it is still available. It seems probable that the distinct advantages for seismic design will lead to widespread use of this new tube standard. If so, it will eventually become the default nationally, and you may have to switch simply because A500 becomes less readily available. In the mean time, it might be appropriate to allow tube steel in your designs to conform with either spec, since the design values are fairly similar.</p>
| 3661 | Tube Steel - ASTM A1085 vs. A500 |
2015-07-23T14:21:05.247 | <p>I'm running an FEA on a vessel that has a "splashing water" effect with vibrations. The problem I'm really coming up against is <em>what kind of Young's Modulus and Poisson's ratio to use for the fluids?</em> - air and water. I have been successful in using the bulk Modulus, and treat it as uncompressible - i.e. Poisson's ratio of 0.5, but I never used it for vibrational analysis, just static analysis. <strong>Are these still good assumptions in a dynamic analysis?</strong></p>
| |mechanical-engineering|modeling|vibration|finite-element-method| | <p>Section 2.1.14 of the Abaqus Example Problems Manual is 'Water sloshing in a baffled tank', which might be useful (even if you don't use Abaqus at your own institution, the manuals are hosted by many institutions that allow public access). The Abaqus water model uses an equation of state (EOS). If you download the input file from the manual, you can cut and paste the material section or convert it to suit the code that you're using.</p>
| 3664 | Modelling splashing water in FEA |
2015-07-23T19:02:23.740 | <p>Is there a calculation / rule of thumb that can determine a containers rough maximum negative pressure (vacuum) from it's rated internal pressure (assuming a single layer construction so there is no additional bracing preventing expansion).</p>
<p>As an example for a 50 litre stainless steel keg (cylindrical with convex end caps) with an integral 580 +/- 140 psi bursting disc. </p>
<p>Would it be valid to assume that the weakest point (excluding the bursting disc) is calculated to withstand 400 psi of internal pressure (minus approx 14 psi from external atmosphere), directed outward so placing the material under tension. By reversing pressure differential (My logic says the weakest point still remains the same) the weak spot would come under a compressive force and the maximum pressure under vacuum would be a function of the ratio of that particular stainless steal grades tensile:compressive strength.</p>
<p>EDIT
I just realized the bursting disc pressure is not relevant since that only indicates when a rupture would occur. In the reverse situation deformation into a flat can would occur well before then. However if instead of the bursting pressure you were to apply the 'Maximum working pressure' of 43.5 psi which presumably implies no deformation does the above apply?</p>
| |structural-engineering|pressure|stresses|mechanical-failure|pressure-vessel| | <p>Failure of cylindrical shells under external pressure depends on length in addition to all the factors for internal pressure. If you can get your hands on a copy of the ASME boiler and pressure vessel code Section VIII division 1, and section II part D, you can calculate it yourself using the formulae in UG-28. Alternatively, there are a few pressure vessel software suites that offer free trials.</p>
| 3668 | Calculating maximum internal pressure from maximum rated external pressure? |
2015-07-24T14:55:40.963 | <p>I know that when a motor runs it generates torque and that torque can be used to do useful work. On the other hand, the motor needs strong support that absorbs the reaction torque. In our case let us assume that that support is provided by a workshop floor on which the motor is firmly attached (the workshop floor is essentially the Earth). The Earth receives the reaction torque and being the massive object it is, it doesn't move.</p>
<p>Now let us imagine that we took our motor into space where there is no gravitational field. What would happen if we tried to run the motor? Assuming the motor is powered by a battery pack. The battery pack and its control electronics are neatly packed around the stator.</p>
<p>Would the motor rotate at all? Would the rotor and the stator rotate in opposite directions? Would there be a transfer of energy from the batteries to rotational mechanical energy?</p>
| |electrical-engineering|motors|torque| | <p>The stator and the rotor would rotate relative to each other in the same way as if one end were attached to the Earth. If your frame of reference was attached to either the stator or the rotor, you would see the things attached to the other part rotating in the same way as if you were standing on Earth. The only difference is that, in space, you would need to consider the conservation of angular momentum to determine how much the two portions rotate with respect to some external reference frame such as the stars. </p>
<p>Using a motor to spin a flywheel is one of the ways that spacecraft <a href="https://en.wikipedia.org/wiki/Attitude_control" rel="nofollow">rotate while in space</a>. Interestingly, the people who designed the Voyager spacecrafts neglected to account for this effect in the spinning tape recorders that were used to record data. They ended up having to engage external thrusters every time the recorders were in use in order to maintain the correct pointing of the spacecraft. </p>
<p>To answer your questions more directly:</p>
<ol>
<li><p><strong>Would the motor rotate at all? Would the rotor and the stator rotate in opposite directions?</strong> Yes and yes. The motor produces the same amount of torque between the rotor and stator (for a given load and speed) no matter what each is attached to.</p></li>
<li><p><strong>Would there be a transfer of energy from the batteries to rotational mechanical energy?</strong> Yes, rotational energy would be put into the object attached to the stator as well as the object attached to the rotor. </p></li>
</ol>
| 3674 | What would happen if we tried to run a motor in space when it is not attached to anything to provide support to it? |
2015-07-24T15:35:05.160 | <p>This is a schematic for a plastic piece that has an oval hole on it. I understand that the Ø12 part of the measurement is for a diameter, but I do not understand what the <strong>"X25L"</strong> at the end is indicating nor do I understand the <strong>"2-"</strong> at the beginning. What is the meaning of those numbers?</p>
<p><a href="https://i.stack.imgur.com/3MEBC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3MEBCm.png" alt=""></a></p>
| |mechanical-engineering|drafting| | <p>The "2-" refers to the fact there are two of these holes somewhere on the print. Rather than making that same note every time the hole appears, you can say "2-" or "6-" and leave it to be assumed that all similar holes are identical to the dimensioned one. (At my company, we use "2x" or "6x", but this person may not have wanted to do this considering there's another 'x' in the note, but some proper spacing should easily clarify that.)</p>
<p>The Ø12X25L is the full dimension of the hole size. I actually don't like the diameter symbol in front of the twelve, because it's not really a circle, but it's still relatively clear. In any case, the twelve is the diameter of the semicircles at each end, and thus also the width of the slot. The 25L is the length of the slot. Because it's not a circle, you do need two dimensions to determine its size. (Don't confuse this with bolt sizes, which I often see as M12x80L, where M12 is the bolt size and 80L is the actual length. Oval bolts are a bad idea generally.)</p>
| 3677 | How does a dimension like "2-Ø12X25L" describe an oval hole? |
2015-07-25T23:03:44.690 | <p>I need to select a solid body in ANSYS Design Modeler, with script, in a design with multiples bodies. With the following command I select all the solid bodies, but this is not what I need.</p>
<pre><code>ag.bodyPick;
ag.gui.SelectAll();
</code></pre>
<p>I'm creating a script to model two spur gears engaged, first creating one tooth then making circular pattern to create the entry gear, with the following script I create the pinion but when I'm going to make the circular pattern for the gear, the pattern command give me an error, because I select all the solid bodies including the pinion, and I need to pick only the solid of the gear tooth to make the circular pattern them.</p>
<p>This is my script:</p>
<pre><code>ag.m.ClearAllErrors();
ag.gui.setUnits(ag.c.UnitMillimeter, ag.c.UnitDegree, ag.c.No);
var p = new Object();
//Plane
p.Plane = agb.GetActivePlane();
p.Origin = p.Plane.GetOrigin();
p.XAxis = p.Plane.GetXAxis();
p.YAxis = p.Plane.GetYAxis();
p.Sk1 = p.Plane.NewSketch();
p.Sk1.Name = "Pinion";
p.Sk2 = p.Plane.NewSketch();
p.Sk2.Name = "Wheel";
//PINION
with (p.Sk1)
{
p.Sp1 = SplineBegin();
with(p.Sp1)
{
SplineFlexibility = agc.Yes;
SplineXY(-1.76564478418, 24.6869398366);
SplineXY(-1.51020858813, 24.7038816792);
//......... spline coordinates of the pinion profile
//......... removed because character limit
SplineXY(5.26099091644, 24.1843849328);
SplineFitPtEnd();
}
p.Ln2 = Line(-1.76564478418, 24.6869398366, -1.24843570599,17.4554120057);
p.Ln3 = Line(3.71989256718, 17.1000701545, 5.26099091644, 24.1843849328);
p.Cr4 = ArcCtrEdge(
0, 0,
3.71989256718, 17.1000701545,
-1.24843570599, 17.4554120057);
p.Ext1 = agb.Extrude(agc.Add, p.Sk1, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 0.0, agc.No, 0.0, 0.0);
}
agb.Regen();
var PF1 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "0";
agb.Regen();
var PF2 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "34.0";
agb.Regen();
var LF1 = agb.LinePt();
LF1.AddSegment(PF1.GetPoint(1, 0),PF2.GetPoint(1, 0), 0);
agb.Regen();
var Pat = ag.gui.CreatePattern();
ag.listview.ActivateItem("Pattern Type");
ag.listview.ItemValue = "Circular";
ag.listview.ActivateItem("Geometry");
ag.bodyPick;
ag.gui.PickFilter(5, true);
ag.gui.PickFilter(11, false);
ag.gui.PickFilter(11, false);
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("Axis");
ag.edgePick;
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("FD2, Angle");
ag.listview.ItemValue = "16.3636363636";
ag.listview.ActivateItem("FD3, Copies (>0)");
ag.listview.ItemValue = "21.0";
agb.Regen();
//GEAR
with (p.Sk2)
{
p.Sp1 = SplineBegin();
with(p.Sp1)
{
SplineFlexibility = agc.Yes;
SplineXY(1.9304744152, 30.2623156532);
//......... spline coordinates of the gear profile
//......... removed because character limit
SplineXY(-5.79031304774, 30.3548985987);
SplineFitPtEnd();
}
p.Ln2 = Line(1.9304744152, 30.2623156532, 0.209834175566,173.751999593);
p.Ln3 = Line(-0.629381853016, 173.762062957, -5.79031304774, 30.3548985987);
p.Cr4 = ArcCtrEdge(
0, 191.250741537,
-0.629381853016, 173.762062957,
0.209834175566, 173.751999593);
p.Ext2 = agb.Extrude(agc.Add, p.Sk2, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 34.0, agc.No, 0.0, 0.0);
}
var PF3 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "77.5";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "0";
agb.Regen();
var PF4 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "77.5";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "34.0";
agb.Regen();
var LF2 = agb.LinePt();
LF2.AddSegment(PF3.GetPoint(1, 0),PF4.GetPoint(1, 0), 0);
agb.Regen();
var Pat = ag.gui.CreatePattern();
ag.listview.ActivateItem("Pattern Type");
ag.listview.ItemValue = "Circular";
ag.listview.ActivateItem("Geometry");
ag.gui.PickFilter(5, true);
ag.gui.PickFilter(11, false);
ag.gui.PickFilter(11, false);
ag.bodyPick; //<- Here is where I need to select only the gear body
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("Axis");
ag.edgePick; //<- Here is where I need to select only the gear rotation axis for the circular pattern
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("FD2, Angle");
ag.listview.ItemValue = "2.74809160305";
ag.listview.ActivateItem("FD3, Copies (>0)");
ag.listview.ItemValue = "130.0";
agb.Regen();
</code></pre>
| |mechanical-engineering|design|computer-aided-design|ansys| | <p>You can select a single body by this code:</p>
<pre><code>agb.AddSelect(agc.TypeBody, bodyname);
</code></pre>
<p>Here is the script:</p>
<p>(Key ideas: First, assign name "GEAR" to the gear body <code>ag.fm.Body(2).Name = "GEAR";</code>. Second, creat gear object by <code>gear = selectNode("GEAR");</code> where the function <code>selectNode(target)</code> is at the end. Third, use <code>agb.AddSelect(agc.TypeBody, gear);</code> after <code>ag.bodyPick;</code>)</p>
<pre><code>ag.m.ClearAllErrors();
ag.gui.setUnits(ag.c.UnitMillimeter, ag.c.UnitDegree, ag.c.No);
var p = new Object();
//Plane
p.Plane = agb.GetActivePlane();
p.Origin = p.Plane.GetOrigin();
p.XAxis = p.Plane.GetXAxis();
p.YAxis = p.Plane.GetYAxis();
p.Sk1 = p.Plane.NewSketch();
p.Sk1.Name = "Pinion";
p.Sk2 = p.Plane.NewSketch();
p.Sk2.Name = "Wheel";
//PINION
with (p.Sk1)
{
p.Sp1 = SplineBegin();
with(p.Sp1)
{
SplineFlexibility = agc.Yes;
SplineXY(-1.76564478418, 24.6869398366);
SplineXY(-1.51020858813, 24.7038816792);
//......... spline coordinates of the pinion profile
//......... removed because character limit
SplineXY(5.26099091644, 24.1843849328);
SplineFitPtEnd();
}
p.Ln2 = Line(-1.76564478418, 24.6869398366, -1.24843570599,17.4554120057);
p.Ln3 = Line(3.71989256718, 17.1000701545, 5.26099091644, 24.1843849328);
p.Cr4 = ArcCtrEdge(
0, 0,
3.71989256718, 17.1000701545,
-1.24843570599, 17.4554120057);
p.Ext1 = agb.Extrude(agc.Add, p.Sk1, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 0.0, agc.No, 0.0, 0.0);
}
agb.Regen();
var PF1 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "0";
agb.Regen();
var PF2 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "34.0";
agb.Regen();
var LF1 = agb.LinePt();
LF1.AddSegment(PF1.GetPoint(1, 0),PF2.GetPoint(1, 0), 0);
agb.Regen();
var Pat = ag.gui.CreatePattern();
ag.listview.ActivateItem("Pattern Type");
ag.listview.ItemValue = "Circular";
ag.listview.ActivateItem("Geometry");
ag.bodyPick;
ag.gui.PickFilter(5, true);
ag.gui.PickFilter(11, false);
ag.gui.PickFilter(11, false);
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("Axis");
ag.edgePick;
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("FD2, Angle");
ag.listview.ItemValue = "16.3636363636";
ag.listview.ActivateItem("FD3, Copies (>=0)");
ag.listview.ItemValue = "21";
agb.Regen();
//GEAR
with (p.Sk2)
{
p.Sp1 = SplineBegin();
with(p.Sp1)
{
SplineFlexibility = agc.Yes;
SplineXY(1.9304744152, 30.2623156532);
//......... spline coordinates of the gear profile
//......... removed because character limit
SplineXY(-5.79031304774, 30.3548985987);
SplineFitPtEnd();
}
p.Ln2 = Line(1.9304744152, 30.2623156532, 0.209834175566,173.751999593);
p.Ln3 = Line(-0.629381853016, 173.762062957, -5.79031304774, 30.3548985987);
p.Cr4 = ArcCtrEdge(
0, 191.250741537,
-0.629381853016, 173.762062957,
0.209834175566, 173.751999593);
p.Ext2 = agb.Extrude(agc.Add, p.Sk2, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 34.0, agc.No, 0.0, 0.0);
}
var PF3 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "77.5";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "0";
agb.Regen();
ag.fm.Body(2).Name = "GEAR";
var PF4 = ag.gui.CreateSpot();
ag.listview.ActivateItem("Type");
ag.listview.ItemValue = "Construction Point";
ag.listview.ActivateItem("Definition");
ag.listview.ItemValue = "Manual Input";
ag.listview.ActivateItem("FD8, X Coordinate");
ag.listview.ItemValue = "0";
ag.listview.ActivateItem("FD9, Y Coordinate");
ag.listview.ItemValue = "77.5";
ag.listview.ActivateItem("FD10, Z Coordinate");
ag.listview.ItemValue = "34.0";
agb.Regen();
var LF2 = agb.LinePt();
LF2.AddSegment(PF3.GetPoint(1, 0),PF4.GetPoint(1, 0), 0);
agb.Regen();
var Pat = ag.gui.CreatePattern();
ag.listview.ActivateItem("Pattern Type");
ag.listview.ItemValue = "Circular";
ag.listview.ActivateItem("Geometry");
ag.gui.PickFilter(5, true);
ag.gui.PickFilter(11, false);
gear = selectNode("GEAR");
ag.bodyPick;
agb.AddSelect(agc.TypeBody, gear);
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("Axis");
ag.edgePick;
ag.gui.SelectAll();
ag.listview.ItemValue = "Apply";
ag.listview.ActivateItem("FD2, Angle");
ag.listview.ItemValue = "2.74809160305";
ag.listview.ActivateItem("FD3, Copies (>=0)");
ag.listview.ItemValue = "130";
agb.Regen();
function selectNode (target)
{
var DM = ag.wb.AppletList.Applet( "AGApplet" ).App;
var Nodes = DM.Script.ag.tree.Nodes;
var count = Nodes.Count;
var name, current;
for (var i =1; i <= count; i++)
{
current = Nodes(i);
name = current.Text.toLowerCase();
if (name == target.toLowerCase())
{
DM.Script.agTree_LeftClick(current, false);
var obj = ag.listviewSelectedObject;
return obj;
}
}
}
</code></pre>
| 3685 | What is the script function to select Solid Body in ANSYS Design Modeler? |
2015-07-27T14:34:30.440 | <p>How can I calculate the propulsive force produced by ejection of a gas? I suppose it depends on the gas, the velocity with which it ejects, the amount of gas, and some other parameters, like the geometry of the nozzle for example. I would like a calculator to play with the gas density, the gas velocity etc, and see the various propulsive forces generated.</p>
| |aerospace-engineering|gas|propulsion| | <p>It's pretty basic conservation of momentum kind of problem. A converging diverging nozzle is used to accelerate the gas such that it's supersonic leaving the nozzle. These problems are pretty well-defined. There are lots of calculators out there that let you monkey around with varying different parameters without getting into a full on mathematical understanding of compressible flow. Here's one: </p>
<p><a href="https://engineering.purdue.edu/~wassgren/applet/java/cdnozzle/" rel="nofollow">https://engineering.purdue.edu/~wassgren/applet/java/cdnozzle/</a></p>
| 3703 | How to calculate propulsive force of gas ejection |
2015-07-28T11:41:34.847 | <p>I wish to investigate an active vehicle braking method for use in extreme situations. Very high accelerations are acceptable in situations where this would prevent the much worse accelerations present in a collision. </p>
<p>I was thinking of a system that will use ejected gas/liquid to reposition the position of a car and/or assist in braking. We know that ejected mass (gas or other) create a propulsive force. I was thinking to utilize this mechanism to create a force that will decelerate the car. Or maybe change its position/trajectory (eg. when it moves out of the road in a steep turn). </p>
<p>Would such system be viable?<br>
What parameters would such system have?<br>
eg to decelerate a car travelling at 30m/s within 3 meters
(about 15g deceleration!).<br>
How much mass should we eject and at which speed?<br>
Have such systems being proposed/researched/built?</p>
| |automotive-engineering|projectiles| | <p>15 g might seem like a lot, and it is, if the 15g is sustained as it would be in something like an aircraft's aerobatic maneuver. 15g isn't unreasonable as a limit to what humans can take in a short duration event like a crash, Indy car drivers have sustained horizontal impact forces many times that without significant injury. However, consider how those drivers are restrained to avoid injury, 5 pt harness, helmet, arm restraints, head restraints and so on. The level of G-force that humans can tolerate depends on the individual, their training and also the direction in which the G-force is applied.</p>
<p>Here's an interesting read on human tolerance of impact forces</p>
<p><a href="http://ftp.rta.nato.int/publiC/PubFullText/RTO/EN/RTO-EN-HFM-113/EN-HFM-113-06.pdf" rel="nofollow">http://ftp.rta.nato.int/publiC/PubFullText/RTO/EN/RTO-EN-HFM-113/EN-HFM-113-06.pdf</a></p>
| 3713 | Vehicle emergency high-deceleration braking and steering system |
2015-07-29T14:47:09.493 | <p>I'm designing a compact mechanism for the deployment of a mast. The idea is that the linear actuator for the deployment is on the other side of the pivot, and a slider increases the lever arm to enable the actuator to deploy the mast. </p>
<p>I ran through the maths (shown in details below), and wrote the code for the simulation. I want to know how the actuator force and the lever length relate, and how the force varies with the angle.
<a href="https://i.stack.imgur.com/KLVsB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KLVsB.png" alt="enter image description here"></a></p>
<p>However, when I run the code I get huge spikes in the force (see below, I capped the force to 10^6 to better see the pattern), but for example I don't see why the mechanism would jam between 40 and 80°. For some reason the cross product of HP with OP (what I call the signed lever arm of F) goes through zero but I see no reason why it would given the current dimensions and angle range set. How can I fix it?</p>
<p>Original:</p>
<p><a href="https://i.stack.imgur.com/wBrvl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wBrvl.png" alt="enter image description here"></a></p>
<p>With the force capped at 200N:
<a href="https://i.stack.imgur.com/UVzIY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UVzIY.png" alt="enter image description here"></a></p>
<p>And the denominator of F:</p>
<p><a href="https://i.stack.imgur.com/3a7Rh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3a7Rh.png" alt="enter image description here"></a></p>
<h2>Annex</h2>
<p>Code:</p>
<pre><code>close all
clear all
%See diagram for description of variables, units are metres
a = (2+10)*10^-3;
b = a;
c = 165*10^-3;
d = 70*10^-3;
h = 180*10^-3;
m = 8; %Mass of the mast in kg
g = 9.81; %Acceleration of gravity in m/s²
%Range of the input variables
thetaMin = 0*pi/180;
thetaMax = 100*pi/180;
tMin = 10*10^-3;
tMax = 100*10^-3;
%Vectors
theta = thetaMin:(thetaMax-thetaMin)/100:thetaMax;
t = tMin:(tMax-tMin)/100:tMax;
%Solve for F
F = zeros(length(t), length(theta)); %Actuator force
L = F; %Length of linear actuator
denominator = F; %Denominator of F, or "signed lever arm"
for i = 1:length(t)
for j = 1:length(theta)
M = [cos(theta(j)) sin(theta(j)) 0;
-sin(theta(j)) cos(theta(j)) 0;
0 0 1];
HP = M*[-t(i);
a;
0];
OP = [-b;
h;
0]+HP;
L(j,i) = norm(OP);
HG = M*[c;
-d;
0];
W = [0;
-m*g;
0];
weightMoment = HG(1)*W(2)-HG(2)*W(1); %cross(HG,W)
forceLever = 1/norm(OP)*(HP(1)*OP(2)-HP(2)*OP(1)); %1/norm(OP)*cross(HP,OP)
denominator(j,i) = forceLever;
F(j,i) = -weightMoment/forceLever;
end
end
%Plot results
[T,THETA] = meshgrid(t,theta);
figure(1)
surface(T*1000,THETA*180/pi,F)
ylabel('Angle (deg)')
xlabel('Slide length (mm)')
zlabel('Force (N)')
title('Actuator force')
figure(2)
surface(T*1000,THETA*180/pi,L*1000)
ylabel('Angle (deg)')
xlabel('Slide length (mm)')
zlabel('Actuator length (mm)')
figure(3)
surface(T*1000,THETA*180/pi,denominator*1000)
ylabel('Angle (deg)')
xlabel('Slide length (mm)')
zlabel('Signed lever arm')
</code></pre>
<p>Maths: solve
$$\vec{moment}(H)=\vec{HP}\times \vec{F}+\vec{HG} \times \vec {W}=\vec{0}$$
With
$$\vec{F}=\frac{\vec{OP}}{OP}F$$
$$\vec{HP}=
\left[ {\begin{array}{cc}
cos(\theta) & sin(\theta) \\
-sin(\theta) & cos(\theta)
\end{array} } \right]
\left( {\begin{array}{cc}
-t \\
a
\end{array} } \right)
$$
$$\vec{HG}=
\left[ {\begin{array}{cc}
cos(\theta) & sin(\theta) \\
-sin(\theta) & cos(\theta)
\end{array} } \right]
\left( {\begin{array}{cc}
c \\
-d
\end{array} } \right)
$$
$$\vec{OP}=\vec{OH}+\vec{HP}$$
$$\vec{W}=\left( {\begin{array}{cc}
0 \\
-mg
\end{array} } \right)$$
$$\vec{OH}=\left( {\begin{array}{cc}
-b \\
h
\end{array} } \right)$$</p>
| |mechanical-engineering|modeling|mathematics| | <p>If you plot the x values of OP and HP, they cross zero, which seems unphysical.</p>
<p>Looking more closely, you are doing your rotations backwards, i.e. you are doing our rotations in +theta whereas you want to go in -theta.</p>
<p>This is something I often do too. To fix, just replace theta with -theta in M.</p>
<p>Here's a figure of F when you do this which looks much more reasonable.</p>
<p><a href="https://i.stack.imgur.com/wUPCu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wUPCu.png" alt="enter image description here"></a></p>
| 3729 | Discontinuity in mechanism simulation: is the mechanism jammed or is my analysis wrong? |
2015-07-29T19:27:19.243 | <p>I am an engineering student and I took my second dedicated controls class last semester. It was cool math and whatnot, but after all was said and done: I still have no idea how to make an actual working control system. </p>
<p>In other words, there was no tie-in to reality at all; it was all Laplace-domain mathematics that somehow affected our result.</p>
<p>What I want to know is: how do I go from real-space, to controls-space, and back to real-space? To help with understanding, I have prepared an example:</p>
<p>I have platform that is supported by a vertical rod. This rod can freely move up and down, or rather, it would be able to if it was not held in position by a motor (either a servo or a gear). The job of this motor is to keep the platform at the same height. E.g., if I place a mass on the platform, the motor must be able to adjust to it and respond accordingly such that after the response, the height of the platform is the same as before. (Of course, with a specified rise time, maximum overshoot, yada yada...).</p>
<p>This might sound trivial to you controls engineers out there, but I really don't know how to do this. I get that it is a closed loop system, but what I don't know is what all these laplace-space block diagrams correlate to in the real world. How do I get the block diagram for this system, and how do I build a controller inside of it?</p>
<p>The point of this example is to extend it to other problems, like airplane control, but I figure if I don't understand this I probably won't be able to do those other (cooler) things!</p>
| |control-engineering|control-theory| | <p>If you are in the unlikely position to have/know the transfer function of your plant/equipment then nothing beats a good predictor.</p>
| 3734 | How can I approach the application of control theory to real control systems? |
2015-07-29T22:55:47.283 | <p>Is there a term used in systems engineering that describes a situation in which the design of one subsystem (call it "A") affects the the requirements for another subsystem (call it "B") <em>while <strong>at the same time</strong> the design of B affects the requirements for A</em>? </p>
| |terminology|systems-engineering|systems-design| | <p>Another term that would be applicable is <strong><em>mutually reliant</em></strong>.</p>
| 3737 | Is there a term for a circular dependency in system design and engineering |
2015-07-30T16:14:07.703 | <p>Occasionally I have to design T-connections between rectangular tube steel (HSS) members where I have a flare-bevel weld between the corner of the tube and the connecting flat member. If the weld stress is high enough, a reinforcing fillet may be required.</p>
<p>Per AWS D1.1-2010, Section 2.4.2.7, </p>
<p><a href="https://i.stack.imgur.com/LBR1B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LBR1B.png" alt="Section 2.4.2.7"></a></p>
<p>The effective throat is illustrated by Figure 3.3 and Annex A (pasted below)</p>
<p><a href="https://i.stack.imgur.com/F5Tf1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F5Tf1.png" alt="Figure 3.3"></a></p>
<hr>
<p><a href="https://i.stack.imgur.com/vfo49.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vfo49.png" alt="Annex A"></a></p>
<p>Thus far, I have been unable to easily calculate the minimum effective throat of a flare-bevel with reinforcing fillet due to the complex geometry of the weld throat. The best method I have found so far is to draw it in AutoCAD to scale and figure it out from there, but this is relatively time-consuming. It is not immediately evident to me which plane of the weld is the minimum throat since the contour of the weld changes along its width.</p>
<p>Has anyone seen a formulaic (i.e., programmable) methodology for determining the effective throat of a flare-bevel with reinforcing fillet? I mostly work my calculations in <a href="http://www.ptc.com/engineering-math-software/mathcad" rel="nofollow noreferrer">Mathcad</a>, so having something I can program into steps would be highly beneficial to my workflow.</p>
| |structural-engineering|welds|hss| | <p>I've spent some time working out the geometry for this configuration and believe I've developed a set of cases that will work to determine the total effective throat of this connection.</p>
<p>The solution is divided into three cases.</p>
<h2>Case 1 - Undersized Fillet Weld $(L_2 < RO)$</h2>
<p>For this case, the fillet weld is sized such that the vertical leg is less than the root opening $RO$ of the weld (see the figure below).</p>
<p><a href="https://i.stack.imgur.com/SviQS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SviQS.png" alt="Case 1"></a><br></p>
<p>The values $\alpha_{RO}$ and $RO$ are found as</p>
<p>$$ \alpha_{RO} = \arcsin \left( \frac{R_1 - E}{R_1} \right) $$</p>
<p>and</p>
<p>$$ RO = G + R_1 \left[ 1 - \cos(\alpha_{RO}) \right]. $$</p>
<p>Knowing these, one can determine if $L_2 < RO$, in which case $E_T = E$.</p>
<h2>Case 2 - Medium Fillet Weld $ (RO < L_2 \leq R_1\tan (\alpha_{RO})) $</h2>
<p>In this case, the fillet weld is large enough that the toe of the vertical leg is past the height of the root opening, but the fillet isn't large enough that the total effective throat passes through part of the fillet.</p>
<p><a href="https://i.stack.imgur.com/NPPoD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NPPoD.png" alt="Case 2"></a><br></p>
<p>With $\alpha_{RO}$ and $RO$ already found, </p>
<p>$$ E_T = \sqrt{E^2 + (L_2 - RO)^2}. $$</p>
<h2>Case 3 - Large Fillet Weld $ (L_2 > R_1\tan (\alpha_{RO})) $</h2>
<p>For Case 3, the fillet weld is large enough that the total effective throat passes through the body of the fillet.</p>
<p><a href="https://i.stack.imgur.com/1wSuC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1wSuC.png" alt="Case 3"></a><br></p>
<p>The angle of the fillet, $\alpha_F$ is found as</p>
<p>$$ \alpha_F = \arctan \left( \frac{L_1}{L_2} \right). $$</p>
<p>With this, one can set up a system of equations to solve for $E_T$. I've skipped this step for brevity. It can be shown that, </p>
<p>$$ E_T = \frac{L_1 + E - RO \tan(\alpha_F)}{\sin(\alpha_{RO})\tan(\alpha_F) + \cos(\alpha_{RO})}. $$</p>
<p>This methodology is now formalized in the article "Measuring the Effective Throat of Groove Welds," which appears in the February 2017 edition of the AWS <em>Welding Journal</em>.</p>
| 3745 | Effective Throat of Flare-Bevel Weld with Reinforcing Fillet |
2015-07-31T08:31:21.547 | <p>I'm looking for a material with a Young's modulus of less than 100 Pa. I've looked at elastomers, but I haven't had any luck.</p>
| |stresses|biomechanics|elastic-modulus| | <p>You are asking for an unrealistic material. If you peruse the <a href="http://www.engineeringtoolbox.com/young-modulus-d_417.html" rel="nofollow">engineering toolbox</a> you'll find that the Young's modulus of typical materials is measured in 10<sup>9</sup> Pa so the value you are asking for is 7 orders of magnitude smaller than typical materials. Even the Young's modulus of the rubber in rubber bands is ~10<sup>7</sup> Pa.</p>
<p>Another simple thought experiment will convince you of how unreasonable such a material is. The change in altitude required to change the pressure by 100 Pa is 10 meters which is about 3 building stories. Since Young's modulus is defined by
$$
\frac{\Delta \ell}{\ell}=\frac{P}{E},
$$
where $E$ is Young's modulus, the strain on your material would be 1 if you carried it from the top of a 3 story building to the ground floor. This means that the material would compress to nothing from such a minor change in elevation. Real materials would, of course, be well into the non-linear regime at that point.</p>
| 3752 | Material with very small Young's modulus |
2015-07-31T14:14:40.423 | <p>I have a wheel I CNC milled <a href="http://www.lowes.com/pd_304090-210-304090.0___?productId=3122447&pl=1&Ntt=foam" rel="nofollow">out of foam</a> and I need to mount a motor to it. I designed and printed a part that would attach the motor to the foam, but I anticipated that the wheels would be wood back then. It connects to the wheel using screws, but it does not seem like screws would hold well in foam because the foam can't hold threads. Is there any other way to do this? I was thinking of using longer screws with nuts on them, but I am open to other ideas. </p>
| |mechanical-engineering|materials|motors| | <p>I would design a hub with a flange to fit against the side of the wheel and use an adhesive that's chemically compatible with the foam and has good bonding strength to metal. 3M has some solutions that we've used for bonding foam sheets. You'll want to take advantage of the fact that adhesives are strongest in shear rather than tension/peel. You could also have an end plate with adhesive that would bolt through the foam wheel creating a "sandwich". The torque should be transmitted by the adhesive surfaces since the bolts will tear up the foam.</p>
<p>If you want a mount that is removable, use an adhesive that's soluble.</p>
| 3755 | How do you mount a polystyrene foam wheel to a motor? |
2015-07-31T15:14:27.970 | <p>Temporary structures are usually constructed in a less precise manner than permanent structures. This is more so in a braced deep excavation where the walls are in constant motion as the excavation progresses downwards. During the assembly of walers and struts, there could be a small gap between the strut-waler connection due to fabrication imperfections. How does one decide whether the fillet weld connecting the strut to the waler need to be designed against compression due to possibility of such a small gap?</p>
| |structural-engineering|welds| | <blockquote>
<p>How does one decide whether the fillet weld connecting the strut to the waler need to be designed against compression <strong>due to possibility of such a small gap?</strong></p>
</blockquote>
<p>In our office, we generally design welds to take compression load due to the exact reason you're specifying. Unless you can ascertain that you have full bearing between the two pieces (i.e., both pieces are milled to mate perfectly), I would say you should check compression on the weld.</p>
<p>For a code reference, the closest I can find is from the AISC 360-10 specification, section J1.1:</p>
<blockquote>
<p>The <em>required strength</em> of the connection shall be determined by <em>structural analysis</em> for the specified <em>design loads</em>, consistent with the type of construction specified, or shall be a proportion of the required strength of the connected members when so specified herein.</p>
</blockquote>
| 3757 | Should a welded connection between strut and waler be designed to take compression load? |
2015-08-02T07:23:43.680 | <p>For a Board Exam Review: </p>
<blockquote>
<p>A triple thread worm has a diameter of 3 inches. The wheel has 25
teeth and a pitch diameter of 5 inches. Material for both the worm and
the wheel is of phosphor Bronze. Compute the Helix Angle</p>
</blockquote>
<p>Answer is 0.2</p>
<p>Helix Angle can be obtained by trigonometry:</p>
<p><a href="https://i.stack.imgur.com/SOxup.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SOxup.png" alt="enter image description here"></a></p>
<p>$${ \tan(H) = \frac{\pi \times D_{worm}}{L} }$$
$${ L = N_t * P_a }$$</p>
<p>Where L is the lead, N is the number of threads and P is the Axial Pitch.</p>
<p>$${ L = 3threads * 3in }$$</p>
<p>Well thats where I stopped. I don't know how to find the diameter of the worm gear without it being explicitly given to me. I tried other formulas but they were more of the diameter of the wheel instead of the gear. Is there a formula for solving the diameter of the gear?</p>
| |design|gears| | <p>If I get your descriptions right you have everything you need given. You can use the pitch diameter.</p>
<p>$$tan(H) = \frac{\pi \cdot d_p}{N_t \cdot P_a}$$</p>
<p>$$tan(H) = \frac{\pi \cdot 5\:\mathrm{in}}{25 \cdot 3\:\mathrm{in}}$$</p>
<p>$$tan(H) = \frac{5\pi}{75}$$</p>
<p>$$tan(H) = \frac{\pi}{15}$$</p>
<p>$$H = arctan \left( \frac{\pi}{15}\right)$$</p>
<p>$$H = ~ 0.20 \:\mathrm{rad}$$</p>
<p>There you go. Please remember to include the units if you post an answer. An angle of 0.2 could have been degree, too. At least I wasn't sure until after I calculated it.</p>
| 3777 | Finding the diameter of a Triple Thread Worm Gear |
2015-08-03T05:27:15.603 | <p>Reviewing for a National Board Exam: </p>
<blockquote>
<p>How many 5/16 inch holes can be punched in one motion in a steel plate
made of SAE 1010 Steel 7/16 inch thick using a force of 55 tons. The
ultimate strength for shear is 50 ksi and use 2 Factor of Safety</p>
</blockquote>
<p>Answer is 5.</p>
<p>This is a common strength of materials question given among problem sets in engineering. Here is my attempt:</p>
<p>$${ S_{plate} = \frac{F}{N\times d \times t} }$$</p>
<p>$${ 50ksi = \frac{ 55tons \times\frac{2204.622lb_f}{1ton} }{N \times \frac{5}{16}in \times \frac{7}{16}in} }$$</p>
<p>I get N = 17...</p>
<p>I think what I'm getting wrong is the conversion of tons; I've tried UK Tons, US Tons but nothing. Is there a correct conversion factor?</p>
| |steel|stresses| | <p>The shear force required should depend upon the shear area:
$$N \pi \biggl(\frac{5}{16}''\biggr) \biggl(\frac{7}{16}''\biggr)$$
or number times circumference times height, since the side-walls of $N$ cylinders are resisting.</p>
<p>That would drop your answer by almost exactly the right amount. Short tons could take it the rest of the way. Doing the math quickly with short tons gets me $N = 5.122$, which of course we can't have part of a same-sized hole, so we need the floor, or $5$, the desired solution.</p>
<p>With long tons we get $N=5.634$, which has the same floor.</p>
<p>As pointed out by idkfa, I did not consider the safety factor. However, as the question asks about a production process and not an application for the end product, it isn't clear that the safety factor is relevant, in the same way that the SAE designation of the material is (mostly) irrelevant.</p>
| 3780 | Number of Holes punched through a Plate of SAE 1010 Steel |
2015-08-03T05:49:38.307 | <p>We know that demand for electricity is not constant. Power plants are not operated under full load conditions. What if power plants were operated under full load condition and excess electricity was stored in capacitors? I have read that capacitors of 10 kF capacitance are available. Could the use of such capacitors make it reasonable to operate power plants under full load at all times?</p>
| |energy-storage|power-engineering| | <p>Capacitors are currently "rather too costly" for this purpose.<br>
I've revised my estimated cost after some more research but you appear to be in the 'well over one hundred thousand dollars" range! </p>
<p>The dear way: If you were to assemble a 10kF 150 volt capacitor from available smaller capacitors now it would cost around 1 million dollars and store about 30 kWh of electricity - worth maybe 5 to 15 dollars retail depending where you are and a lot less wholesale. </p>
<p>Using 160 VDC rated parts, 1700 of <a href="http://www.digikey.com/product-search/en?pv14=8&FV=fff40002%2Cfff8000c&k=supercapacitor&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25" rel="nofollow noreferrer"><strong>these</strong></a> would cost about $2,000,000 (really), weigh about 9 tones and occupy about 17 cubic metres (!). You'd get a discount for quantity but overall they are not viable
<a href="http://www.maxwell.com/images/documents/160vmodule_ds_3000246-5.pdf" rel="nofollow noreferrer"><strong>Datasheet here</strong></a></p>
<p><strong>One of 1700:</strong></p>
<p><a href="https://i.stack.imgur.com/n5ATZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n5ATZ.jpg" alt="enter image description here"></a></p>
<p>"Doing it yourself" is hardly cheaper..<br>
In 1000 quantity <a href="http://www.digikey.com/product-detail/en/BCAP3400%20P285%20K05/1182-1038-ND/4896903" rel="nofollow noreferrer">these</a> 3400 <span class="math-container">$\mu$</span>F, 2.8fV cost a mere 53 dollars each. Each series string = 150/2.85 = 53 capacitors.<br>
Capacitance in series divides by number of caps so C per spring = 3400/53 = 64F.<br>
Number of strings = 10,000/64 = 156.<br>
Total caps = caps/string x strings = 53 x 156 = 8268<br>
Cost = 8268 x 53 = 438,204 dollars
That's better than one million plus above - but that's a lot of mounting to do AND balancing circuitry will be essential.<br>
1 to 2 million starts to look almost good. </p>
<p>Using smaller voltage high capacity units "off the shelf
Alternatives include:</p>
<p>"Flow batteries" using eg Vanadium oxide liquid in various oxidation levels. "Tanks" of liquid are pumped through a cell to "charge" and stored in another tank for subsequent discharge. Large trial systems exist. Not yet mainstream. May or may not "make it" commercially. </p>
<p>Lithium Ion - as being used in Tesla cars and their Powerbank home storage technology. JUST becoming economic with careful timing of charge and discharge to buy cheap power and use it at peak cost periods. Large commercial installations exist in eg Germany - costs are about break even overall so far with advantages of continuity being a factor. </p>
<p>Lithium Titanium (a LiIon variant) - coming - used in eg Suzuki LEAF along with traditional LiIon cells. Dearerthan than standard LiIon but immensely fast charge rates and can have 5,000 - 10,000 usage cycles with due care.</p>
<p>Molten salt - used to store energy thermally and make power off peak - eg station at Guila-Bend near Phoenix. This is solar thermal heated but electric heating could be used.</p>
<p>Pumped storage as John mentioned. Some use but not very common. Efficiency overall about 60%. UK has a large system used for grid leveling applications.</p>
<p>Flywheels - investigated for many decades - good in theory but to get acceptable energy densities you need large masses )(tons) rotating at 10s of thousands of RPM. Mechanical failure is not pretty. So far many have tried but there are no significant working systems.<br>
Flywheel added comment - March 2020: There are some flywheel storage systems available used for peaking load control with about 10 kWh capacity per flywheel. </p>
<p>Others exist, but that gives a feel. </p>
<p>. </p>
| 3781 | Can large capacitors allow power plants to run at full load and store excess electricity? |
2015-08-03T14:54:44.720 | <p>I have seen several products on the internet marketed as 'ball screws' that have both the screw itself and the special ball bearing nut. They are intended to have a motor mounted on one end of the screw to rotates it, causing the nut to be moved.</p>
<p>However, I'm looking for the particular kind of ball screw where the nut is part of a gear, and mounted via axial bearings, so that as it is rotated it remains stationary and the screw is extended.</p>
<p>Are these things sold in complete units without any machining necessary? What are they called/where can I find one?</p>
| |mechanical-engineering| | <p>Festo makes complete assemblies that fit your description. You can also buy the individual components from them, or from a company like McMaster Carr.</p>
<p>www.festo.com</p>
<p>www.mcmaster.com</p>
| 3785 | Name of the ball screw nut used in pistons? |
2015-08-05T08:58:25.907 | <p>Steel seems to be a very complex material, not least because of the different phases and microstructures. At the moment, I'm especially interested in the bainite formation that happens when austenitic steel is cooled rapidly (but not rapidly enough to drive the martensite formation). </p>
<p>My question: What is the mass density (or range of values, or functions describing the densities in relation to specific conditions) of these two microstructures?</p>
| |materials|steel|metallurgy| | <p>There is a lot more going on in this question than appears at first glance. Density of austenite is fairly straightforward: it is approximately the atom-weighted sum of the face-centered cubic densities of the substitutional constituents as the microstructure consists of a single phase. In other words, Fe, Mo, V, etc. The interstitial constituents, i.e. C, N, add mass as they do not replace Fe lattice sites but instead fit between them, increasing mass without increasing volume, therefore increasing density. However in austenite there is at most 2% by weight of carbon in solution at $1130\:\textrm{C}$ as shown in the phase diagram at the bottom. The density of pure iron is $7870\:\textrm{kg}/\textrm{m}^3$ (from Google), and the density increase is probably close to about 2% of the density of graphite, $2250\:\textrm{kg}/\textrm{m}^3$ (from Google), or about $45\:\textrm{kg}/\textrm{m}^3$, for a total of about $7915\:\textrm{kg}/\textrm{m}^3$ for the density of fully saturated austenite at room temperature. It isn't clear how you would produce such an unstable material, but we can at least estimate in theory. There are slight volume changes in the FCC lattice caused by the substitutional and interstitial atoms, but without atomistic models these changes are challenging to predict. They may be determined experimentally, of course.</p>
<p>For bainite, there are two phases: ferrite and cementite. The ferrite phase consists of virtually all of the substitutional microconstituents in a body-centered cubic structure, so its density may be calculated by atom-weighted average. There is also a trivial quantity of carbon in solution at equilibrium, increasing density as before. The density of the microstructure is then the volume-weighted average of the densities of the two phases. Plain-carbon steel with no other alloying additions should have ferrite density very close to that of pure iron, or about $7870\:\textrm{kg}/\textrm{m}^3$ (from Google), and cementite has a theoretical density of approximately $7640\:\textrm{kg}/\textrm{m}^3$ (<a href="http://iicarbide.com/archives/IIC_iron_carbide_manufacturing_process_rev_01.pdf" rel="nofollow noreferrer">source</a>). For steels the maximum amount of cementite is approximately 32% by volume, from a tie-line construction on the phase diagram. The maximum carbon concentration in a steel, by definition, is close to 2.14 weight percent, whereas in cementite is 6.67 weight percent. So the volume-weighted density would be a minimum of approximately $7795\:\textrm{kg}/\textrm{m}^3$.</p>
<p>These numbers may be adjusted for temperature using coefficients of linear expansion cubed for the volume change (assuming linearity). They may also be adjusted by considering the densities of substitutional alloy additions. Finally I would caution that the numbers are theoretical for austenite, as plain-carbon steel, pure, retained austenite should be practically impossible to produce in bulk. I'd also caution that there are more complex phenomena to consider such as phase changes, diffusion, as well as other thermodynamic and kinetic phenomena that govern how steels form microstructures and phases, and which will definitely have an effect on the densities.</p>
<p><a href="https://i.stack.imgur.com/9MD97.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9MD97.jpg" alt="FeC Phase Diagram"></a></p>
| 3803 | Densities of different phases of steel (austenite, bainite) |
2015-08-05T10:41:00.317 | <p>I want to algorithmically process <strong>energy meter data</strong>. The energy meter measures a heat or power producer or a heat or power consumer (but not both, so the measured energy will always have a positive sign). No additional information is known about the energy system (like maximum load) neither about the type of energy meter - only data stored in a database can be accessed. Processing will be done by an algorithm looking at data for a given time interval (no live processing).</p>
<p>Usually, data are <strong>weakly monotonic</strong> of the form</p>
<p>2015-04-01 00:00 20.78 kWh</p>
<p>2015-04-01 00:05 30.80 kWh</p>
<p>2015-04-01 00:10 73.99 kWh</p>
<p>2015-04-01 00:20 82.30 kWh</p>
<p>2015-04-01 00:25 82.30 kWh</p>
<p>2015-04-01 00:30 83.44 kWh</p>
<p>...</p>
<p>The energy produced or consumed for a given period is simply the difference of the energy meter counts. So far, so good. However, the algorithm has to deal with the following three problems:</p>
<p><strong><em>1. Outliers "above" have to be detected as invalid data.</em></strong></p>
<p>2015-04-01 00:00 20.78 kWh</p>
<p>2015-04-01 00:05 30.80 kWh</p>
<p><strong>2015-04-01 00:10 500 kWh</strong></p>
<p>2015-04-01 00:20 82.30 kWh</p>
<p>2015-04-01 00:25 82.30 kWh</p>
<p>2015-04-01 00:30 83.44 kWh</p>
<p>....</p>
<p><strong><em>2. Outliers "below" have to be detected as invalid data.</em></strong></p>
<p>2015-04-01 00:00 20.78 kWh</p>
<p>2015-04-01 00:05 30.80 kWh</p>
<p><strong>2015-04-01 00:10 20 kWh</strong></p>
<p>2015-04-01 00:20 82.30 kWh</p>
<p>2015-04-01 00:25 82.30 kWh</p>
<p>2015-04-01 00:30 83.44 kWh</p>
<p>....</p>
<p>In unlikely cases, there might be several consecutive outliers above or below or a combination of both.</p>
<p><strong><em>3. A reset of the energy meter has to be detected automatically.</em></strong></p>
<p>2015-04-01 00:00 20.78 kWh</p>
<p>2015-04-01 00:05 30.80 kWh</p>
<p><strong>2015-04-01 00:10 3.99 kWh</strong></p>
<p><strong>2015-04-01 00:20 12.30 kWh</strong></p>
<p><strong>2015-04-01 00:25 12.30 kWh</strong></p>
<p><strong>2015-04-01 00:30 13.44 kWh</strong></p>
<p>...</p>
<p>After a reset, counting starts a again from another level (a reset is simply a level shift). The level the counting starts from after the reset is often zero, but can also be any other positive number. A reset can occur at an arbitrary point in time (usually not too often).</p>
<p>To my eyes, problems 1. - 3. seem ubiquitous in measurement engineering and must have been already addressed. Nevertheless, I couldn't find any literature on this topic. Does anybody know about existing solutions to this problem? All help will be highly appreciated.</p>
| |measurements|energy|metrology| | <p>There are two ways to do it.</p>
<h3>The old way</h3>
<p>The traditional way is to develop a set of somewhat arbitrary rules based on the errors you manually classify. You filter out non-monotonicity (easy), identify resets (easy), and try to spot other bad values (trickier). That gives you a set of values to mark as missing, and then you analyse the rest of the data. This method is not well grounded in theory, but you will have the (somewhat unsatisfactory) defence that: "it's how lots of other people do it".</p>
<h3>Best practice</h3>
<p>The best-practice way to do it is to write down the probability of everything relevant, and then apply Bayes' Theorem to work out what the most likely real time series was, given your recorded observations.</p>
<p>You start with a prior distribution for the rate of use of energy, based on preceding work.</p>
<p>And then create probability distributions for the ways that errors can happen: a meter reset, a dropped decimal point in recording; a dropped digit; a completely junk reading. Add in a distribution for the measurement error of the meter itself: they've usually got either a datasheet or an accredited standard which has an error range defined.</p>
<p>The statistics should account for things like a real usage spike and a reset coinciding. You might need to specify a joint distribution if they are linked: for example, a power cut could conceivably result in a meter reset <em>and</em> a power surge, as things like heaters, fridges and freezers would all come back on at full power when power is restored.</p>
<p>And then you calculate a posterior distribution for actual energy use, which is the thing you're interested in.</p>
<h3>Pros and cons</h3>
<p>The second method has the advantage of being rooted in rock-solid theory. It is, however, quite a lot of work to set up the distributions; and in pretty much every real-world case, there isn't an analytic solution, so you have to find for a numerical solution (e.g. using markov-chain monte-carlo). Software packages such as <a href="http://mc-stan.org/" rel="nofollow noreferrer">Andrew Gelman's STAN</a> will do that part of the work for you.</p>
<h3>Before you start, chart</h3>
<p>Either way, start by charting your raw data. The eye will pick up informative patterns.</p>
| 3807 | How to automatically process energy meter data considering outliers and meter reset? |
2015-08-05T17:02:16.623 | <p>My drafting program outputs a .dwf format file. In it, arcs are encoded as </p>
<pre><code>(Circle 602331619,844923003 537072960 32715,65588)
</code></pre>
<p>where my so-far affirmed interpretation is <code>(Circle, X-center,Y-center Angle Unknown1,Unknown2)</code>.</p>
<p>The image for the arc is shown below (highlighted in Gold):</p>
<p><a href="https://i.stack.imgur.com/7HBwn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7HBwn.png" alt="The arc in concern is the highlighted on."></a></p>
<p>Hence, each arc has its center and radius given. Are the pair of points at the end the starting and ending angles?</p>
<p>If you take <code>(32715,65588)</code> as <code>(3.2715 rad,6.5588 rad)</code>, the output starts making sense. Is this correct?</p>
| |civil-engineering|drafting|autocad| | <p>The file format for *.dwf files is defined. A reference for the various drawing commands can be found <a href="http://paulbourke.net/dataformats/whip/" rel="nofollow noreferrer">here</a>. Specifically, the definition of the <code>Draw Circle/Circular Arc/Circular Wedge function</code> is <a href="http://paulbourke.net/dataformats/whip/DrwCircArcWdg.html" rel="nofollow noreferrer">here</a>.</p>
<p>From that link an arc is described by:</p>
<blockquote>
<p>X,Y,R Start,End</p>
</blockquote>
<p>where:</p>
<blockquote>
<p>X,Y - Center point (in logical coordinates) of the circle to be drawn</p>
<p>R - Radius (in logical coordinates) of the circle to be drawn</p>
<p>Start, End - The angles (in 360/65,536ths of a degree) that describe a "pie-slice" of the full circle to be rendered; legal values range from 0 to 65,536</p>
</blockquote>
| 3815 | AutoCAD .DWF Circle Output Meaning |
2015-08-06T06:07:46.417 | <p>In an air-conditioned house equipped with an air extractor that removes for instance wet, smelly, etc. air from the bathrooms and kitchen, what are sensible cost-effective systems to prevent:</p>
<ul>
<li>Heat loss during the winter by bringing in cold air and extracting hot air from inside the house</li>
<li>Heat gain during the summer by bringing in hot air and removing the chilled air inside</li>
</ul>
<p>That is, I would like to minimize the energy cost of extracting chilled/heated air by somehow recycling that energy before the air is dumped outside.</p>
| |heat-transfer|hvac|airflow| | <p>What you want is called a <i>heat exchanger</i>. Imagine two long air tubes with a thin wall between them. The air exiting the house travels in one tube, and entering air in the other, but in opposite directions. Over the length of the tubes, heat transfers thru the thin wall. Ideally, by the time the house air gets to the far end where the outside air comes in, it's at the outside air temperature, and vice versa for the outside air at the other end.</p>
<p>In practice, there are various ways to construct heat exchangers, and these things are either far from ideal, or very bulky and expensive. The cheapest commercial ones are usually cross flow instead of opposite flow. This is much easier to construct, but the efficiency is lower. Cross flow units in series increases efficiency.</p>
<p>Most of the time, the payback for a heat exchanger for something like a bathroom vent is too long to make sense. Price out some units and do the math before jumping into this.</p>
| 3824 | How to minimize heat loss in an air extractor? |
2015-08-07T13:05:36.557 | <p>I am a beginner in Electronics. I am trying to make a small robot controlled by an Arduino. The problem is that I do not know how to calculate the power required to rotate a structure with a servo motor placed on its axis (see picture below).</p>
<p>My servo motor is a Dynamixel MX-28 T with an average stall torque of 2.5 Nm at 12 V. The product page for the servo can be found <a href="http://www.trossenrobotics.com/dynamixel-mx-28-robot-actuator.aspx" rel="nofollow noreferrer">here</a> and the specification sheet for it is <a href="http://support.robotis.com/en/product/dynamixel/mx_series/mx-28.htm" rel="nofollow noreferrer">here</a>.</p>
<p>Do you think it is possible to rotate it using this servo motor?</p>
<h2>Informations:</h2>
<p>Structure weight: 2,2 kg (2200 g) (include battery-arduino-servo)</p>
<p>Structure Dimension: 900mm/120mm/120mm (see picture below)</p>
<p>Structure type: Aluminum</p>
<p>Servo weight: 72 g</p>
<p><strong>total weight 2.2 kg will be distributed fairly. 1.1 kilo on the top and on the bottom 1.1 kilo</strong></p>
<p><a href="https://i.stack.imgur.com/jhJwzm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jhJwzm.png" alt=""></a>
<a href="https://i.stack.imgur.com/RQwWT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RQwWTm.jpg" alt=""></a></p>
<p>(click for full resolution)</p>
<p><a href="https://i.stack.imgur.com/7Eld1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Eld1.png" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering|stepper-motor|robotics| | <p>With the structure (and masses) being distributed perfectly even on both sides of the servo, the servo will not require any additional force to rotate the structure since the mass on one side is always balanced out or negated by the mass on the other side.</p>
<p>You will still have to overcome friction forces and depending on how you mount and use the servo (directly connected to the driving shaft of the servo or not), you may have to take into account whether your servo's motor shaft can handle such radial forces (and maybe axial forces too depending on your application).</p>
<p>So yes, it is possible to rotate the structure with the servo.<br>
But in order to know how fast you can accelerate or decelerate the structure, you will have to use its moment of inertia.</p>
| 3834 | The couple of my servo motor is it powerful enough? |
2015-08-07T22:25:28.387 | <p>In a sensor project I'm involved with, we have a requirement to bypass the airflow from one sensor while it does some further measurements. To do this, I implemented a flap driven by a <a href="https://www.pololu.com/product/1053" rel="nofollow noreferrer">standard servo</a> which rotates the flap through approximately 90 degrees, opening one port and closing another.</p>
<p>I'd like to try and improve this with a better, less leaky, mechanism. The first thing I've thought of is a cylinder rotating within another:</p>
<p><a href="https://i.stack.imgur.com/WIIaq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WIIaq.png" alt="enter image description here"></a></p>
<p>The green outer cylinder contains three ports to which the inlet pipe and two outlet pipes (~20mm dia) are connected. The inner cylinder (~30mm dia) rotates to select one or other of the outlet pipes - thus it has either a really elongated inlet port or four ports. I figure I can place the servo in the centre - this dictates the inner cylinder diameter.</p>
<p>Is there an existing off-the-shelf product that would do this for me or will I have to try and 3-D print it (or find two waste pipes that will fit appropriately)?</p>
<p>Is this a reasonably way to do airflow selection or will it turn out to be even more leaky than the airflap (which wasn't particularly well done as it happens - it can be improved)?</p>
<p><strong>Edit</strong>: As an addition in the OP rather than as an answer/comment: This is another way of doing this. Once you know "pneumatic" is the term you can add terms like "servo valve" to google image search and get all sorts of neat ideas (including little lego valves!):</p>
<p><a href="https://i.stack.imgur.com/pU2l0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pU2l0.png" alt="enter image description here"></a></p>
| |mechanical-engineering|design|airflow| | <p>Are you bound to the shape/configuration of part you constructed in your question? If not, a simple 3-way <strong>pneumatic solenoid valve</strong> would make this very simple and leak proof(relatively speaking). They are inexpensive too.(eBay) You would have an input and the valve will allow flow through one outlet and then energize the solenoid and you will get flow through the second outlet. </p>
| 3842 | Airflow selection mechanism |
2015-08-08T02:39:39.027 | <p>Power cords for laptop computers have a transformer built into them. The power supply has a transformer that steps down mains level power to a lower voltage and rectifier that converts the current from AC at the mains to DC.</p>
<p>Power surges con occur for a variety of reasons, such as a nearby lightning strike. If a damaging power surge were to occur, would the power supply prevent the power surge from damaging the laptop computer?</p>
| |power-electronics| | <p>An occupational hazard of mine is working with a lot of people who own laptop computers. I also happen to live in the central plains of the United States, and we get our fair share of thunderstorms. They're not as pretty as the lightning storms of Phoenix, AZ, but they're generally a lot more powerful.</p>
<p>Anecdotally, I've heard of multiple times where the power supply (aka "the brick") for the laptop computer becomes a sacrificial component and will be destroyed by a power surge whereas the laptop itself is just fine.</p>
<p>I have heard of similar stories with desktop computers, but on a much more rare basis. Generally, damaging surges will take out the power supply and the mainboard with a desktop.</p>
<p>Laptop computers have an additional block of circuitry that directs power to the battery for charging purposes and / or directing power to the rest of the laptop circuitry. I would hazard a guess that this additional charging circuitry provides an additional measure of shunting and helps protect laptop computers from damaging power surges.</p>
<p>Between laptops and desktops, I have probably heard of an equal number of instances of laptop power supplies and desktops being destroyed. I have heard of the actual laptop being destroyed less frequently.</p>
<p>Your most likely risk is losing the laptop power supply due to the damaging power surge. I'm not certain how to evaluate that overall risk as almost everyone I know connects their electronics through protective power strips.</p>
<p>But you're not completely risk-free of the laptop being damaged. Damaging power surges are notoriously fickle and can have incredible variance in intensity. Lightning strikes vary in intensity; the distance between the strike and the equipment will vary; and the number of connected circuits between the strike and your equipment will also vary.</p>
<p>From a protective standpoint, one of the best things you can do is put a surge protector block within the circuit panel supplying mains power to your dwelling. From there, make sure that any sensitive electronics are powered through surge protecting power strips. Those two measures will provide you a significant reduction of risk at a modest price. And if you're really concerned, power everything down and disconnect the power strip when a storm is rolling through.</p>
| 3846 | Evaluating the probability of damage to a laptop computer from power surges |
2015-08-08T13:50:08.957 | <p>Given a successful deuterium-tritium fusion reactor such as that conceived by ITER, could the world's energy needs theoretically be met without the input of any petroleum or natural gas products?</p>
<p>Deuterium is currently typically produced by the girdler sulfide process from seawater which requires a natural gas input, although methods involving electrolysis or distillation are known.</p>
<p>Lithium is currently mined, although there are methods known of extracting it from seawater.</p>
<p>Could the extractions of these elements be powered exclusively from the energy produced by fusion reactors in the future? This would leave seawater and the reserve of lithium from the earth's crust as the only fuel sources (or just seawater, if we are extracting the lithium from it). </p>
| |renewable-energy| | <p>The energy cost of obtaining Deuterium and Lithium is minimal compared to the energy released by the fusion process. There are many references that provide relevant information but a sensible one to refer to is the official ITER site "Fusion for energy". <a href="http://fusionforenergy.europa.eu/understandingfusion/merits.aspx" rel="noreferrer"><strong>Fusion for Energy</strong></a> (F4E) is the European Union’s Joint Undertaking for ITER and the Development of Fusion Energy.</p>
<p>On their <a href="http://fusionforenergy.europa.eu/understandingfusion/merits.aspx" rel="noreferrer">understanding fusion page</a> they say:</p>
<blockquote>
<p>A 1,000-megawatt electric fusion power plant would consume around 100 kg of deuterium and three tonnes of natural lithium in a year whilst generating 7 billion kilowatt-hour. To generate the same amount of electricity, a coal-fired power plant would need around 1.5 million tonnes of coal.</p>
</blockquote>
<p>and</p>
<blockquote>
<p>If used to fuel a fusion power station, the lithium in one laptop battery, complemented with half a bath of water, would produce the same amount of electricity as burning 40 tonnes of coal.<br />
(eg ~= 8000 dollars at 5c/kWh wholesale or 40,000 dollars at NZ 25c/kWh retail).</p>
</blockquote>
<p>However what is required is Tritium - Lithium is simply an easily obtained precursor. Obtaining Tritium from Lithium is the "hard part" and while it is proposed that this step be incorporated in the ITER system, and is liable to succeed if ITER succeeds, the process is at this stage "theoretical and experimental".</p>
<p>Tritium can be, and currently is, bred from Lithium using fast neutrons from existing reactors, the amount obtained in this manner is enough to initiate a national or international fusion energy system but nowhere near enough to sustain it. What is essential is the "breeding" of the essential Tritium from Lithium using a suitable source of fast neutrons from the ITER process itself</p>
<p>ITER's own site, in the section <a href="https://www.iter.org/mach/tritiumbreeding" rel="noreferrer"><strong>Tritium Breeding</strong></a> says</p>
<blockquote>
<p>... While deuterium can be extracted from seawater in virtually boundless quantities, the supply of tritium is limited, estimated currently at twenty kilos. .... tritium can be produced within the tokamak when neutrons escaping the plasma interact with a specific element—lithium—contained in the blanket. This concept ... is an important one for the future needs of a large-scale fusion power plant.</p>
<p><strong>ITER will procure the tritium "fuel" necessary for its expected 20-year lifetime from the global inventory</strong>. But for DEMO, the next step on the way to commercial fusion power, about 300g of tritium will be required per day to produce 800 MW of electrical power. No sufficient external source of tritium exists for fusion energy development beyond ITER, making the successful development of tritium breeding essential for the future of fusion energy.</p>
<p>ITER will provide a unique opportunity to test mockups of breeding blankets, called Test Blanket Modules (TBM), in a real fusion environment. Within these test blankets, viable techniques for ensuring tritium breeding self-sufficiency will be explored.</p>
</blockquote>
<p><strong>______________________</strong></p>
<p><strong>How Tritium is obtained at present:</strong></p>
<p>Construction & Operation of a Tritium Extraction Facility at the Savannah River Site<br />
Final Environmental Impact Statement
<a href="http://energy.gov/sites/prod/files/EIS-0271-FEIS-01-1999.pdf" rel="noreferrer"><strong>DOE/EIS-0271 1999</strong></a> - 140 page pdf</p>
<p><strong>__________________________</strong></p>
<p>With respect to generating hydrocarbons from "other sources of cheap energy"</p>
<p>Yes. The process could be totally "geologically stored hydrocarbon" free, which I think is the 'spirit' of what you are asking.</p>
<p>Once you achieve "energy too cheap to meter" [tm] (2nd time lucky perhaps) you can produce hydrocarbons from current biological waste - or from newly grown feedstock if you so desired. Various people have demonstrated 'just about anything organic' to petroleum products converters. Videos of some such are available on you-tube - caveat emptor as ever, and I knew a man in Louisiana who I have lost touch with who was doing just this several years ago, and having a lot of trouble interesting anyone in the process.</p>
| 3850 | Could future fusion energy exist without the input of any petroleum or natural gas products? |
2015-08-09T04:21:47.880 | <p>It does make sense centrifugal compressors exist, because gases can compress.</p>
<p>We all know that liquids don't compress like gas. Then why do centrifugal or reciprocating pumps exists? </p>
<p>See even a the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Centrifugal_pump" rel="nofollow">Centrifugal Pumps</a>.</p>
| |fluid-mechanics|pumps| | <p>All real materials are compressible. If you look at thermodynamics tables, even small change in pressure causes density change in fluid. But as an engineer, when working on pumps you can omit small and effectless changes in density of fluids. So pumps, compressors, and fans all do the same thing: they compress material and create flow.</p>
| 3854 | Why do pumps really exists? |
2015-08-10T01:58:17.750 | <p>Current gear manufacturing methods for gears usually depend on power tools, or computerised machines, or something like that.</p>
<p>However, watchmakers and clockmakers at one point in history needed to make gears by hand. So what tools did they use? How did they make gears (especially really tiny ones)?</p>
| |gears|engineering-history|manufacturing-engineering| | <p>Have look at these two:-</p>
<p><a href="https://i.stack.imgur.com/YcCuf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YcCuf.jpg" alt="clock1"></a></p>
<p><a href="https://i.stack.imgur.com/Sk6Ez.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sk6Ez.jpg" alt="clock2"></a></p>
<p>These are medieval clockwork mechanisms. You could make them with a smithy and hammer. Cast and bash was the principle technique. And then a bit of heavy bashing followed by more light bashing. To be honest, the greater skill lies in the mechanical arrangement of parts. They had torture racks and stuff and rudimentary locks including ridiculously large padlocks. But they had to invent and evolve these sophisticated mechanisms to keep some sort of decent time.</p>
<p>A great report of the restoration of the Salisbury cathedral clock (1386!) is <a href="https://en.wikipedia.org/wiki/Salisbury_cathedral_clock" rel="nofollow noreferrer">here</a>. Look at the <a href="https://en.wikipedia.org/wiki/Salisbury_cathedral_clock#1956_restoration" rel="nofollow noreferrer">1956 restoration</a> bit which details making parts by hand forging (bashing).</p>
| 3863 | How were gears manufactured before power tools, CNC, etc |
2015-08-10T12:25:25.637 | <h1>The Tibetan Bridge Type</h1>
<p>I have a question regarding the bridge type called "Tibetan Bridge".
Recently close by my home town a new tourist attraction was built, the <a href="http://www.highline179.com/en/guinness-world-record-for-the-highline179/" rel="nofollow noreferrer">Highline 179</a> bridge.
<a href="https://i.stack.imgur.com/BVm9t.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BVm9t.jpg" alt="Highline 179 from Wikipedia"></a></p>
<p>It was planned as the longest suspension bridge for pedestrians in the world, but before completion it was already outclassed by
the Skypark Bridge in Sochi.</p>
<p>Now the Bridge is denoted as the longest Tibet-style pedestrian suspension bridge and I could not find information what a Tibet-style or Tibetan bridge is.</p>
<h2>Other Tibetan Bridges</h2>
<p>I could find other Tibetian bridges (e.g. Tibetan Bridge Claviere), but I could not find any information what Tibet style could be from an engineering perspective. The Wikipedia article on bridge types does not list this particular type.</p>
| |civil-engineering|bridges|architecture| | <p><a href="https://en.wikipedia.org/wiki/Suspension_bridge#Precursor" rel="nofollow">The section on the precursor to modern suspension bridges, in the wikipedia article that you mentioned</a>, talks about bridges of the "Tibetan saint and bridge-builder Thangtong Gyalpo" and points out that they "did not include a suspended deck bridge which is the standard on all modern suspension bridges today".</p>
<p>I confess I am only guessing, and haven't found anything definitive to back this up, but I infer that a "Tibet-style" suspension bridge is hence one which doesn't have a <em>suspended bridge deck</em>. What does this mean? In simple terms it means that the vertical cables (hangars) in a Tibet-style bridge are all the same length, such that the bridge deck has a significant sag in it. Modern bridges, with their suspended decks, have main cables with a significant sag, but varying lengths hangars such that the bridge deck is made (relatively) flat.</p>
| 3865 | What is a Tibetan Style Bridge? |
2015-08-11T13:24:00.290 | <p>When an ideal gas with constant specific heat is throttled adiabatically, with negligible changes in kinetic and potential energies: </p>
<ul>
<li>(a): ∆h = 0, ∆T = 0 </li>
<li>(b): ∆h =0, ∆S > 0 </li>
</ul>
<p>The answer is supposed to be (b). </p>
<p>Please explain why answer (a) is wrong and (b) is correct.</p>
| |mechanical-engineering|thermodynamics| | <p>You answer this by applying first and second laws of thermodynamics.</p>
<p>First Law
$$Q - W = \triangle E_{sys}$$
since throttling process is adiabatic and there is no mechanical work applied on or extracted from the system then $Q$ and $W$ = 0.</p>
<p>that leaves us with $\triangle E_{sys} = 0$, and for steady state system we can express $E$ as the sum of enthalpy, kinetic energy and potential energy:</p>
<p>$$E = m(h + v^2/2 + gz)$$
and since we have negligible changes in kinetic and potential energies we finally get:</p>
<p>$$\triangle h = 0$$</p>
<p>And finally, as we know from second law of thermodynamics that for an adiabatic process <a href="https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Entropy" rel="nofollow">$\triangle S \geq 0$</a></p>
<p>(You should have a look on the throttling process on an h-s diagram of a reversed carnot cycle (refrigeration cycle).)</p>
<p>The temperature behavior in throttling process is governed by the Joule-Thomson coefficient $\mu$
$$\mu = \left ( \frac{\partial T}{\partial P} \right )_h $$</p>
<p>The Joule-Thomson coefficient is a measure of the change in temperature at constant-enthalpy process, if it equals zero then temperature remains constant, if greater than zero temperature decreases and if less than zero temperature increases.</p>
<p>However, for an ideal gas the enthalpy is a function of only the temperature $h = h(T)$, and since enthalpy remains constant in throttling processes and $\mu = 0$ (You can check Cengel's Thermodynamics book for the proof), <strong>the temperature also remains constant.</strong></p>
<p>And so, all the choices are actually correct.</p>
| 3881 | Adiabatic throttling of ideal gas |
2015-08-12T16:45:16.400 | <p>Is this how it works?</p>
<p>The fingers generate moment.<br>
The load generates moment.<br>
Finger moment > load moment, so the load is cut.</p>
<p>How do I use an equation to identify the most vulnerable point of the product?</p>
<p>More questions on this topic:</p>
<p>What is the typical accelaration of scissors? $F=ma$, so that I can assume $F$? Other method of giving a logical assumption of F is also acceptable.</p>
| |mechanical-engineering|statics| | <p>Your general idea is correct. Your fingers apply a force which can be translated to a moment at the rotation axis. The reaction forces applied by the object to be cut must generate an equal moment at the axis. If the necessary reaction forces are greater than the object's shear resistance forces, then the object is sheared through.</p>
<p>For this reason, it is best to place the object as close to the axis as possible, so that the forces required to balance the moment are maximized. Too close, however (at which point the angle between the blades becomes greater than 90 degrees), and you run into the problem that the forces start becoming primarily horizontal (pushing the object out of the scissors) as opposed to vertical (cutting through the object).</p>
<p>This is because the blade's force is applied perpendicular to the blade, so the greater the angle between the blades, the greater the horizontal component of the force becomes. This component is useless since it does not serve to cut through the object, but instead simply tries to force the object out of the scissors. Only the vertical component actually shears the object. This is because the horizontal components of both blades point in the same direction, while the vertical components point in opposite directions.</p>
| 3906 | How to analyze the free body diagram of a pair of scissors? |
2015-08-12T18:52:57.830 | <p>Looking for some hints on how to create a <strong>rough estimate</strong> for the following problem.</p>
<p>Given two steel gates with the same dimensions, same material - e.g. everything is the same. The only difference is that the middle parts have different structures.</p>
<p>When applying some force to the top, the gate will start to be more and more deformed, and at some force the gate will touch the ground at the place where the blue arrow points.</p>
<p><strong>I'm looking for a <em>rough estimate</em> of how much more</strong> force is needed for the second gate - i.e. how much more "sturdy" is the second gate.</p>
<p>I really don't need any exact calculation, but probably will need some material data, so:</p>
<ul>
<li>common steel <a href="https://commons.wikimedia.org/wiki/File:Closed_thin-walled_beam.png" rel="noreferrer">thin-walled beam</a> (25mm x 25mm x 2mm wall thick)</li>
<li>each joint point is welded, we can be simplify and assume that the welds are exactly as strong as the material itself</li>
<li>the suspension points can hold infinite force</li>
<li>and any other possible simplification - this problem isn't for any rocket-science but for solving an evening talk with a friend. </li>
</ul>
<p><a href="https://i.stack.imgur.com/f2njv.png" rel="noreferrer"><img src="https://i.stack.imgur.com/f2njv.png" alt="enter image description here"></a></p>
| |structural-engineering|beam|stresses|reinforcement| | <p>As grfrazee said, you won't know for sure until you do a finite element analysis. I was intrigued by this question as a colleague and I got into a discussion about this. While we both agreed the diagonal bracing would be better at resisting deflection, we wondered by what factor it would be better.</p>
<p>We were really curious so we settled the debate and did a quick structural analysis on <a href="http://skyciv.com/skyciv-structural-features" rel="noreferrer">SkyCiv Structural 3D</a> (can try for free for one month if anyone is wondering). It took around an hour to set up both gates and analyze them mainly because we had to generate the node positions from scratch. Anyways here are the results of the linear static analysis which take into account the assumptions and simplifications you made. We applied a 5 kN POINT LOAD at both F1 and F2 and made each support a pin support at the locations you specified. Note that in the 3D colored results the deflection is 12X greater than the actual deflection of the gate in both scenarios - it is exaggerated so you can see the deflected shape of the gates.</p>
<h2>Gate #1</h2>
<p>$\text{y-deflection at the bottom-left of the gate} = 31.74\text{ mm}$</p>
<p>$\text{Max total deflection} = 32.10\text{ mm}$</p>
<p><a href="https://i.stack.imgur.com/r8i2b.png" rel="noreferrer"><img src="https://i.stack.imgur.com/r8i2b.png" alt="SkyCiv Structural 3D Deflection Result for Gate 1"></a></p>
<hr>
<h2>Gate #2</h2>
<p>$\text{y-deflection at the bottom-left of the gate} = 7.84\text{ mm}$</p>
<p>$\text{Max total deflection} = 7.55\text{ mm}$</p>
<p><a href="https://i.stack.imgur.com/K6ASN.png" rel="noreferrer"><img src="https://i.stack.imgur.com/K6ASN.png" alt="SkyCiv Structural 3D Deflection Result for Gate 1"></a></p>
<p>Diagonal bracing (Gate #2) is clearly the winner. So when both gates are subjected to the same load it looks like Gate #2 resists deflection better (i.e. is more stiff) by a factor of <strong>4.25</strong>. </p>
<p>Some more interesting points:</p>
<ul>
<li>There's a pretty high bending stress at that top right support in both scenarios ~ 350 MPa.</li>
<li>The analysis didn't take into account self-weight of the gates.</li>
</ul>
<p>Also let me add that there looks to be a scaling issue with the diagonal grid you have drawn, because when I modeled it I found that there were far less points than what was suggested by your diagram. I ensured that the parallel spacing between each rhombus was 300mm. This means the diagonal of each rhombus is roughly 424mm. Your gate is 3300mm in length so that means around 8 rhombi should fit across your gate in the x-direction - but you've drawn around 12. Just thought I'd let you know.</p>
| 3909 | Strength of a welded steel gate with vertical bars vs. crossed diagonal bars |
2015-08-13T13:21:31.570 | <p>I'm trying to crimp some pins onto 20 AWG wire. </p>
<p>I can either use a tool like this:</p>
<p><a href="https://i.stack.imgur.com/ihyNm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ihyNmt.jpg" alt="specialty crimper"></a></p>
<p>Or I can just use pliers, which are obviously cheaper and more readily available. Why should I use the more expensive tool?</p>
| |tools| | <p>The proper crimper will indent the cylindrical body of the contact, leaving the outside diameter unchanged. </p>
<p>Pliers will simply squash the contact, making it oval, and larger than the original diameter in one direction. This may prevent the contact from seating correctly (or at all) in the connector housing. The distorted contact will probably make it impossible to use the contact removal tool to remove the contact from the connector.</p>
| 3926 | What is the advantage of using a four-sided crimper vs. pliers? |
2015-08-13T14:37:52.000 | <p>In my lab we use a number of materials which are <a href="https://en.wikipedia.org/wiki/Hygroscopy" rel="noreferrer">hygroscopic</a> and must therefore be stored in a dry environment. For that purpose we have numerous dry cabinets very similar to the one shown below (<a href="http://rads.stackoverflow.com/amzn/click/B00871AXHE" rel="noreferrer">Amazon link here</a>). </p>
<p>These cabinets run on electricity, do not require desiccant, and do not have a water outlet such as one finds with home dehumidifiers, yet they are able to maintain low relative humidity (RH) values of < 40%. The temperature inside the cabinet is also the same as (or very close to) the temperature outside of the cabinet, and there are no other connections aside from the electrical plug.</p>
<ul>
<li>How do these cabinets maintain such low humidity values?</li>
<li>Assuming they are removing water from the air, where does the water go?</li>
</ul>
<p><a href="https://i.stack.imgur.com/GK0BN.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/GK0BNm.jpg" alt="enter image description here"></a></p>
| |hvac|dehumidification| | <p>I believe warmer air has more capacity to hold vapour so instead of condensing onto the items inside the cabinet, the moisture stays in the air. I have 2 cabinets and neither have any outlet so the water vapour can't go anywhere. Also they are relatively cheap to buy as have a simple design.</p>
<p>They seem pretty effective though as mine run at the value I set (45% rh) with no noise and consume around 10 Watts.</p>
| 3928 | How do dry storage cabinets work? |
2015-08-13T16:13:38.813 | <p>We regularly bolt an angle iron against an existing building element (flange down), and use the flange as support for a new concrete slab. Is there any design approach to assist with the sizing / spacing of the chemical anchors as well as confirming the suitability of the angled section? Epoxy anchors are one form of chemical anchor we use. There are other types of (chemical) bonding agents available like polyesters, vinyl esters as well as hybrid systems.</p>
<p>We have tried various design software packages none of them is very helpful with this configuration. Therefore a manual approach is all that is left, it seems.</p>
| |structural-engineering| | <blockquote>
<p>...and use the flange as support for a new concrete slab</p>
</blockquote>
<p>By this, I assume you mean a shelf angle (or a connection angle) to support a concrete floor. The image below is an example from the <em>PCI Design Handbook</em>, 7th Edition.</p>
<p><a href="https://i.stack.imgur.com/fXPY5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fXPY5.png" alt="Shelf Angle"></a><br></p>
<p>Section 6.6.5 of the <em>PCI Design Handbook</em>, 7th Edition goes through a design example of an unstiffened connection angle. You will likely find similar design examples in previous editions of that <em>Handbook</em>, thought not necessarily with the same section number.</p>
<p><a href="http://www.structuremag.org/wp-content/uploads/C-StrucDesign-Mcginley-May131.pdf" rel="nofollow noreferrer">This article</a> from <a href="http://www.structuremag.org/?p=219" rel="nofollow noreferrer">STRUCTURE Magazine</a> also goes through some design methodology for shelf angles.</p>
| 3931 | Angle Iron Cleat at Slab-Slab connection |
2015-08-14T07:53:00.993 | <p>I'm hoping to rebuild a joystick (sidewinder strategic commander), though the parts inside it are a bit too wobbly because they have been worn beyond tolerances. Due to this wobbling on one (rotation) axis, the sliding action is rough on the x/y plane.</p>
<p>Any recommendation on how and what to use, if there is a compound I can apply to make up for what was lost through friction?</p>
<p>How can I bring the plastic mechanisms back to spec? Also what lubricant is suggested afterwards?</p>
| |plastic|tolerance| | <blockquote>
<p>Any recommendation on how and what to use, if there is a compound I can apply to make up for what was lost through friction?</p>
</blockquote>
<p>At first thought, I'm thinking that you scuff the surface of the plastic parts and try to rebuild the missing area with <a href="http://www.jbweld.com/collections/epoxy-adhesives/products/j-b-weld-twin-tube" rel="nofollow">JB Weld</a>. After it cures, you can file away the unneeded areas to get the part back to "original" shape.</p>
<p>You'll want as smooth of a contact surface as you can get to reduce wear going forward, so don't forget to sand with increasingly fine grits of sandpaper. I would probably start at 150 grit and work to at least 600 grit, maybe even 1000 grit.</p>
<blockquote>
<p>Also what lubricant is suggested afterwards?</p>
</blockquote>
<p>For lubricant, it really <a href="http://machinedesign.com/mechanical-drives/lubrication-tips-plastic-gears-and-more" rel="nofollow">depends on the type of plastic</a> your parts are made of. The site I linked recommended a thicker synthetic lubricant (silicone-based) or mineral oil.</p>
| 3945 | How to rebuild and lubricate worn plastic parts? |
2015-08-14T14:03:25.020 | <p>I'm using a OnePlusOne and while using my sunglasses I can see the phone screen in portrait mode, however if I rotate the screen to landscape I can't see anything at all.</p>
<p>My sunglasses are polarized.</p>
<p>What is the reason for this?</p>
| |electrical-engineering| | <p>The screen of your phone uses a polarizing screens as well. When you hold you phone in portrait mode, the screens are oriented in the same direction so you can see the image. When you rotate it 90deg, the screens are not aligned and light cannot pass. </p>
| 3949 | Why can't I see my phone screen in landscape with sunglasses? |
2015-08-04T21:20:15.227 | <p>The TI MSP430F20XX series has a 12-bit internal ADC output, which is right-justified.</p>
<p>What is the difference between a left-justified output and a right-justified output? What are their pros and cons?</p>
| |electrical-engineering|embedded-systems| | <p>On this processor, the register that holds the conversion result is 16 bits wide. </p>
<p>A right-justified result means that bits [(<em>N</em>-1):0] (where <em>N</em> is the number of bits of precision) of the register contain the ADC value and the most-significant bits of the register are set to zero.</p>
<p>A left-justified result means that bits [15:(16-<em>N</em>)] of the register hold the result, and bits [(15-<em>N</em>):0] are set to zero.</p>
<p>For example, if your actual conversion result is 0x123, it would be read as 0x0123 if the register was right-justified, and as 0x1230 if it was left-justified.</p>
<p>An advantage of left-justified results (on processors that support it) is that you can take just the most-significant byte of the register, giving you 8-bits of precision instead of the native precision. This can be useful if you don't need the extra precision, or have RAM constraints and want to store a large number of samples.</p>
<p>On the other hand, a right-justified value can be used directly without the scaling that a left-justified value would need. </p>
| 3959 | What are left justified and right justified ADC results? |
2015-08-05T18:12:00.707 | <p>I am writing a driver for the <code>Samsung K9WAG08U1D NAND flash</code> chip. The specification of the memory chip mentions it has a page size of 2048 bytes (2kB). I am using a <code>TI MSP430F2619</code> which has 4096 Bytes (4kB) of RAM. This means I need to allocate a 2k memory buffer just to write to flash. My application is a protocol converter and hence requires an additional buffer for handling to and fro transmission.
Please suggest me better approach to reduce the RAM requirement due to flash page size.</p>
| |electrical-engineering|embedded-systems| | <p>You don't need to fill the page register all in one go.</p>
<p>You begin a page write (i.e. the "Page Program" operation) by writing the Serial Data Input command (<code>0x80</code>), the column address, and the row address. Then you transfer the data to the page register (up to 2112 bytes). This transfer can be broken up into chunks, with any delay between chunks you need.</p>
<p>When you have filled the page register, you begin the transfer from the page register to the array with the Page Program Confirm command (<code>0x10</code>).</p>
| 3961 | Saving RAM Memory when writing to 2K page size NAND flash |
2015-07-30T10:33:42.867 | <p>The development board for the touchscreen chip we got is too good ;)</p>
<p>The chip provides the readouts over I2C, passes it to FPGA (I could provide the model if I find a magnifying glass... but I'd prefer a more generic answer anyway) and the FPGA passes the data over RS232 to an FDDI which connects over USB to the PC.</p>
<p>Except I don't want the signal over USB or RS232, I need it in I2C.</p>
<p>There's a jumper that allows me to remove power from the FPGA, and there are convenient test points on the I2C bus which I can connect to the bus with my device, but the FPGA still remains connected.</p>
<p>What kind of problems/interference can I expect from it - present on the bus, unpowered?</p>
| |electrical-engineering|embedded-systems| | <p>It does, unfortunately depend a little on the FPGA itself.</p>
<p>But most cases that will probably not work, unfortunately. In many logic chips with programmable I/O type pins, the final stage looks a bit like this:</p>
<p><img src="https://i.stack.imgur.com/x1GT6.png" alt="schematic"></p>
<p>This is both a side-effect of the cheapest to make final stage and for protection. </p>
<p>Now if you turn off the power to that stage, the Vdd/Vcc voltage goes to 0V, so the top diode, D2, will conduct towards that. Since I2C is powered with resistors from the power rail in an open-collector type scheme, the diode will easily conduct those few mili-amps away, so damage is very unlikely, but the bus will never see a logic one.</p>
<p>If the FPGA is 1.8V or 2.7V or 3.3V itself, but 5V tolerant on the pins in question, the top diode will not be there and the problem may very well disappear, but other characteristics may still have similar influences. (JFETs that turn on because their biasing disappears at power-off, though highly unlikely, technically possible).</p>
<hr>
<p>An option would be to break the traces and then (if you want the FPGA again later) to add decent analogue switches. Sometimes traces come along the board in such a way that a simple dual Analogue Switch can be dropped right across a gap you made with the scalpel. I've done that a couple of times. It's not something you'd design in, but for 400kHz I2C there are plenty chips that could do it and as an after-market mod, it's not even the ugliest you could think of.</p>
| 3965 | What is the behavior of unpowered FPGA pins on I2C bus? |