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2017-11-07T11:04:06.037
|
<p>I want to provide a gradual increase in force provided to a piston (perhaps 0-100N over the space of a couple of seconds maximum), up till the point it begins moving, and then drive it to a varying speed after that. </p>
<p>I was thinking of using a linear actuator and applying a variable voltage to control the speed and force, but apparently using such motors in stall like this is generally a bad idea due to burning out the coils. However stepper motors and solenoid actuators are both designed for holding loads in place while applying a certain torque - would a stepper motor be able to do this without damage? Or could a solenoid actuator provide the gradual force increase in stall, and then the rest of the movement be accomplished using a seperate motor? </p>
<p>Many thanks in advance </p>
|
|motors|pressure|torque|stepper-motor|linear-motors|
|
<p>There is nothing inherently wrong with running a motor in stall. The difficulty is that the power dissipation <i>may</i> be higher than the motor can handle long term.</p>
<p>Every motor can dissipate <i>some</i> amount of power safely indefinitely. In stall, the power being dissipated as heat is the full voltage times current. As long as that doesn't exceed what the motor can handle, you can run the motor as long as you want in stall.</p>
<p>Torque (or force for a linear motor) is proportional to current, and the power dissipation proportional to the square of the current. Both these are independent of the motor speed.</p>
<p>One way to run a motor safely at any speed is to regulate the current thru it. As long as that does not exceed the maximum current the motor can handle continuously, you can apply it continuously.</p>
<p>In your case, you want to control torque. That works out well with the above. Find the current to torque constant, and regulate the current to get the desired torque. As long as you limit this current to what the motor is rated for continuously, there is no special issue about the motor being stalled.</p>
<p>One possible problem for you is the "lumpy" nature of the torque over a magnetic cycle. This is also called "cogging". Stepper motors usually have high cogging, so may not be appropriate for what you want. A brushless DC motor being properly controlled can have quite low torque ripple. That is probably a better choice for you, since you are trying to control torque.</p>
|
17937
|
How to vary the force output of a motor while stalled?
|
2017-11-07T12:17:33.103
|
<p>I want to model a chemical process. I was searching in the literature to find similar works. In some of papers they say that "we proposed an unstructured model". What does it mean? What is the difference between a structured model and an unstructured model?</p>
|
|chemical-engineering|modeling|
|
<p>A model for a biological reaction, bioreactor, etc. is called unstructured when the biomass components are not diferentiated, i.e. the cells with their internal metabolites are all lumped into a single variable (commonly expressed as X). We normally use Monod kinetics to model growth in unstructured models. </p>
<p>In contrast, a structured model may model as different variables the cell count, their content in i.e. pyruvate, fats, ATP, etc. Using this information, they can also account for different growth phases for the cell</p>
|
17939
|
What is an unstructured mathematical model?
|
2017-11-07T19:10:58.120
|
<p>I am working on a project which involves applying electrical load across a metal contact filled with lubricants (bearing applications). However, I am missing one of the blocks as attached and I cannot figure it out what material this is. Looking at the posts here, I am presuming this must be some copper alloy such as brass, however, I know that the rotating disc that this block sits on is of brass material and that can raise issues such as wear due to the presence of similar materials?</p>
<p>Any suggestions what material I can use for this application to not damage the discs while providing good conductivity?</p>
<p><a href="https://i.stack.imgur.com/g9roUl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g9roUs.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/ENbOsl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ENbOss.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/gEhXjl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gEhXjs.jpg" alt="enter image description here"></a></p>
|
|electrical-engineering|materials|
|
<p>It looks like phosphor bronze to me which makes sense as it has good bearing properties and reasonable electrical conductivity. </p>
<p>It could also potentially be sintered (ie porous), oil filled bronze bushings are fairly common. </p>
|
17944
|
Selection of copper alloy for electrical and wear applications
|
2017-11-08T14:47:04.497
|
<p>I want to lift something in the air with a stepper motor. If the object is 1kg and I want to keep it in the air for 12 hours. Is it a good idea to keep the power on the motor all that time? Will it draw excessive amount of power? Ideally I would like to cut power to the motor when the movement is done, but then, will the motor be able to hold its position?</p>
|
|electrical-engineering|stepper-motor|
|
<p>Depends on how much weight and the motor.</p>
<p>Some stepper motors have detent torque. This is the torque they will hold without consuming any power. </p>
<p>They also have a holding torque for which you need to supply power.</p>
<p>The most fool-proof solution is to use a external brake.</p>
|
17950
|
How much power does a stepper motor draw when holding weight?
|
2017-11-08T16:03:00.023
|
<p>I'm making sort of a <a href="https://www.google.com/search?tbm=isch&q=desk%20cycle" rel="nofollow noreferrer">desk cycle</a>. What are the ways to make silent circular motion with resistance ? (I don't need to adjust resistance and resistance should be just light.)</p>
<p>If I didn't need resistance, I could use a bearing. Or maybe there are regular bearings that have some specific resistance ?
I see there are also magnetic bearings but I had really difficulties to find their prices and those ones that I found costs in hundreds. Are there some cheap ones, ideally ~ 15 or less bucks ?
Or do you have an idea how to achieve silent circular motion with resistance ? I was thinking of something like winding/unwinding - continual never ending loop against some weight but that would require changing of the motion from wind to unwind to make it never ending and honestly it seems it isn't very simple to design.</p>
<p>Many thanks</p>
|
|design|
|
<p>you might turn a generator and either dump the power into a resistor or use it to power something.</p>
<p>Of course silent is a bit of a subjective term.</p>
|
17952
|
Silent circular motion with resistance
|
2017-11-08T18:16:45.140
|
<p>I have several (at least six) piles of mineral adding up to 10 000 tons. I want to mix them in one homogeneous big pile. How can I do it?</p>
<p>Thank you very much!</p>
|
|manufacturing-engineering|
|
<p>This happens all the time in the minerals industry where people need to blend material from different stockpiles to produce a supposedly uniform product for a processing plant.</p>
<p>Depending on the size of each stockpile either a front end loader or a bucket wheel excavator are used to take some material from one pile and deposit it elsewhere. You can get fancy and have a number of loader feeding hoppers for conveyor belt and have a number of conveyor belt form one pile.</p>
<p>Depending on what is being optimized: grade of a metal, minimization of contaminant minerals/metals, hardness or softness of the material being fed to crushers/grinders the mix from each stockpile will be varied - such as one bucket from from Pile 1, three from Pile 2 & one from Pile 3. The blend mix will be up to you to decide.</p>
|
17954
|
How to obtain an homogeneous mixture of 10 000 tons of minerals, each one in a different pile?
|
2017-11-09T06:00:29.347
|
<p>I had a typical <a href="https://previews.123rf.com/images/mybaitshop/mybaitshop1005/mybaitshop100500092/6989210-A-pink-bendy-straw-on-a-white-background--Stock-Photo-bent.jpg" rel="nofollow noreferrer">bending drinking straw</a> and when I twisted the outside tip, I expected the straw to rotate in an L-shape, so that the other end of the straw would be rotating sideways. Instead, I found that the straw remained perfectly in place, and the other end twisted in place regardless of the bent angle.</p>
<p>So I was just curious: Do they ever use this particular mechanism in engineering, rather than, say, gears?</p>
|
|mechanical-engineering|
|
<p>There are a few types of joint which fall broadly into this category as well as the spring type arrangement used in flexible drives for pendant drills etc there is a joint called a <a href="https://en.wikipedia.org/wiki/Rag_joint" rel="nofollow noreferrer">rag joint</a> which uses reinforced rubber of leather to transmit rotation through an angle. </p>
<p>Equally any sort of flexible tubing like fuel hose can be an effective way to improvise a flexible shaft coupling for prototyping etc. </p>
<p>While things like paired universal joints and constant velocity joints are now more common for power applications an elastic joint has some advantages in that it is simple, easy to miniaturise and can help to isolate vibration and shock loading. </p>
<p>Equally CV joints are often protected by convoluted rubber gaiters (boots) which work on the same principal, although these are there to provide an external seal to contain grease and keep out contamination rather than transmit power. </p>
|
17961
|
Is a mechanism similar to a bendy-straw ever used to change the direction of a rotation in engineering?
|
2017-11-10T05:20:13.943
|
<p>I was at a well-known electric car company today and they claimed "instant torque." The quote was something like the following:</p>
<blockquote>
<p>Driving this car is very different from a gas-powered vehicle. For example, it has instant torque when you press the accelerator. Unlike a gas vehicle where there is a slight delay.</p>
</blockquote>
<p>My question is, where does this "instant torque" come from? Is the explanation as simple as the fact that electrons/electricity travel/s at the speed of light. But it takes time for an internal combustion engine to convert chemical potential energy into kinetic energy via thousands of tiny explosions? Or is there something else that could illuminate this explanation?</p>
<p>Edit: <a href="https://mechanics.stackexchange.com/questions/25944/why-do-electric-motors-produce-peak-torque-instantly-vs-gasoline-engines/25946">I just found this related question on a separate SE.</a> Can anyone add more of a "fundamental physics explanation" to the set of answers?</p>
|
|automotive-engineering|electrical|car|
|
<p>Instant torque means that you have have <100 milliseconds delay from pedal/lever/throttle movement to maximum corresponding acceleration of the vehicle, which is controlled by a phase controller that sends pulses to the motor. The controller and motor are at most 100 ms time response, at least 10 ms. </p>
<p>The clutch prevents their being a direct connection in between the motor and the wheels. It also prevents continuous acceleration from low to high gear. From standstill, the delay for clutch engagement is about 1-3 seconds, compared to 50-100 milliseconds on EV's.</p>
<p>An electric car has zero rotation difference in between the motor axle and the wheel rotation. </p>
<p>The motor-coil-pulses apply force to magnets, that's why you need a slightly slow controller to redistribute DC battery energy into multiple motor "phase" wires. </p>
|
17972
|
Why do electric cars claim "instant torque?"
|
2017-11-10T14:29:06.513
|
<p>I thought I understood what a proportionality was, but it was recently called into question. I want to make sure I fully understand how the term is generally used in the engineering world. The basic explanation of proportionality is that two variables are related by a constant multiplier such that a change in one of them results in a change in the other, scaled by this constant of proportionality. So we could say:</p>
<p>$$
A(x_1, x_2, ... x_n) \propto B(x_1, x_2, ... x_n)
$$</p>
<p>or</p>
<p>$$
A(x_1, x_2, ... x_n) = cB(x_1, x_2, ... x_n)
$$</p>
<p>where $A$ and $B$ are the proportional variables and $c$ is the constant of proportionality.</p>
<p>My question deals specifically with what restrictions are placed on this constant. In my previous description, $A$ and $B$ were functions of a number of variables, they are necessarily the same variables. Is this the case?</p>
<p>Additionally, in engineering, few things are every truly a constant, so our constant of proportionality could be a function of a number of parameters:</p>
<p>$$
A(x_1, x_2, ... x_n) = c(y_1, y_2, ... y_n)B(x_1, x_2, ... x_n)
$$</p>
<p>However, it is my understanding that when discussing proportionality it must be assumed that all of those $y_n$ parameters must be held constant within your system. This is where my understanding starts to come into question. It was suggested that $c$ need not be independent of $A$ or $B$ and this is still a proportionality:</p>
<p>$$
A = c(B)B
$$</p>
<p>Now, I had always assumed that this is a prime example of a situation where $A$ is not proportional to $B$. But upon thinking about it further, I came up with 3 limited domain cases given the following relationship:</p>
<p>$$
A = (e^B - 1)B
$$</p>
<p>case 1: $B < -10$</p>
<p>case 2: $B > 10$</p>
<p>case 3: $-\infty < B < \infty$</p>
<p>It seems to me that for case 1, we could perhaps say that $A \propto B$ given the small change in the coefficient. But for cases 2 and 3, would it ever be reasonable to consider $A \propto B$?</p>
<p>To clarify my questions: Is anything about my understanding as outlined above wrong? Are there any situations where it is correct to say $A \propto B$ and that the constant of proportionality is a strong function of $B$? Does this answer change if we can consider $B$, and therefore $A$ to be nearly a constant? Is that a useful situation to consider?</p>
|
|mathematics|
|
<p>Proportionality is a little more complex than that. In general, it only means they are correlated, and the correlation is monotonic. </p>
<p>Generally two variables are proportional, if $$A = f(B)$$</p>
<p>where f is a <em><strong>monotonic</strong></em> function. </p>
<ul>
<li>The basic variant is "directly proportional" - a linear relation:</li>
</ul>
<p>$$
A = cB + d
$$</p>
<p>with constant (or independently parametric) $c$ and $d$; $c > 0$. Usually, when you talk about proportionality without other qualifiers, it's what is meant. $d$ is called <em>offset</em> or <em>bias</em>, and you may talk about <em>biased proportionality</em> when it's non-zero.</p>
<ul>
<li>Then there's the very common "inversely proportional."</li>
</ul>
<p>$$
A = {c \over B} + d
$$</p>
<ul>
<li>$A = cB + d$ for negative $c$ is "linearly proportional, with negative coefficient" - sometimes mislabeled as "inversely proportional". General $A = cB + d$ for non-zero $c$ with sign not given is just "linearly proportional".</li>
</ul>
<p>Then there are others:</p>
<ul>
<li><p>"quadratically proportional", $A = cB^2 (+dB +e)$</p></li>
<li><p>"exponentially proportional", $A = e^{cB} (+ d)$</p></li>
<li><p>"logarithmically proportional", $A = c \ log B (+ d)$</p></li>
<li><p>In other cases, you'll see "proportional to <em>f</em> of B", e.g. the name of $A = cB^4$ is not called "tesseractically proportional", just "proportional to fourth power of B".</p></li>
</ul>
<p>But note all these functions are strongly monotonic. There's no such thing as "sinusoidally proportional". </p>
|
17980
|
What exactly is a proportionality?
|
2017-11-10T19:07:23.243
|
<p>I'm working on a design that uses fairly large plastic components (ABS, ~400mm longest dimensions). I'm looking at adding a steel component that encapsulates this plastic part. And, the steel part needs to be tight against the ABS part.</p>
<p>Because of the thermal properties of ABS, the plastic part will change size fairly drastically relative to the Steel part. Since the steel part needs to tightly encapsulate the ABS part, I've been considering cooling the ABS part to around -10 to -20 Celsius before attaching the steel which would be at around +20 Celsius. </p>
<p>My question is, with ABS cooled to that temperature would riveting the steel part to it still be an option, or should I expect fracturing from the rivet? I've tried looking into the hardness, brittleness, and Younges modulus of ABS as a function of temperature but I can't find any good data. </p>
<p>Rivet would be a 7/32 sized rivet through a 3mm thick piece of ABS.</p>
<p>Any help would be greatly appreciated. </p>
|
|mechanical-failure|thermal-expansion|abs|
|
<p>Cooling ABS to -20 C should be within its operating conditions. </p>
<p>But why are you concerned that your encapsulation does not sit tight? What is the temperature range in which this part is used? If its above room temperature, that ABS part will expand more and most likely „press-fit“ into the encapsulation.</p>
<p>Do you have a drawing of that ABS part and encapsulation?</p>
|
17983
|
Temperature vs Material Properties - ABS
|
2017-11-11T18:49:14.297
|
<p>We have a two year-old, two-story town house with:</p>
<ul>
<li>fiberglass insulation in the walls;</li>
<li>a Tyvek vapor barrier under foam board on the exterior walls;</li>
<li>blown cellulose insulation in the attic;</li>
<li>stucco outside;</li>
<li>and a refrigeration cycle air conditioning.</li>
</ul>
<p>We live in Phoenix, AZ (desert climate) and our indoor humidity levels are consistently 20% to 35% higher than outside. During the summer we consistently have humidity levels in excess of 80% (not 70%), even with the damper fully open. It has been so humid inside that the tile floors had sweat on them.</p>
<p>We have no indoor plants or water features. Our dryer vent is operational, and our exhaust fans from the bathrooms are vented externally through the roof. There are no unusual ceiling vaults and we do have ceiling fans to stir the air. Ours is an end unit, so we have one shared wall with a neighboring town house, which has no vapor barrier.</p>
<p>We have had the builder heavily involved for warranty work to determine the root cause of the humidity sources. We have tried:</p>
<ul>
<li>Slowing the AC blower motor down to allow the water in air to condense;</li>
<li>Looking for water leaks with an infrared camera (FLIR);</li>
<li>Looking for blockages in the vents in the eaves;</li>
<li>Calibrating the thermostat (Ecobee);</li>
<li>Balancing the air flow by adjusting the vents inside with flow meter hood;</li>
<li>Adjusting the air conditioner refrigerant cycle pressures;</li>
<li>Fully opening the damper to let in dry air;
<h2>- Increasing the exhaust fan to run for 30 min in a 60 min cycle.</h2></li>
</ul>
<p>The problem hasn't gone away and we're not sure what to do next. Is there anything else we can try before we resort to installing a whole house dehumidifier, or adding additional attic ventilation?</p>
<p>What are your thoughts on the RH trend over time?</p>
<p><a href="https://i.stack.imgur.com/trhP4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/trhP4.jpg" alt="Indoor Temp and Humidity over time"></a></p>
<p><a href="https://i.stack.imgur.com/sRZAG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sRZAG.jpg" alt="Aug 1, 2018 indoor temp and RH"></a></p>
<p>Also, I did a rough calculation that the home needs to express ~350 pints of water per day (ASHRAE) psychometric chart corrected for altitude, but the HVAC company measured the AC system is operating properly and measured the AC will remove 77 pints of water per day. I read that whole home dehumidifiers typically remove ~100 pints per day. What are your recommendations on how to best ventilate the attic while controlling cost?</p>
<blockquote>
<p><a href="https://github.com/derkooh/HumidityData" rel="nofollow noreferrer">Historical Indoor RH and Temp. Including limited outdoor data</a></p>
</blockquote>
|
|thermodynamics|hvac|building-design|environmental-engineering|
|
<p>If I understood correctly you have one year old house. Do your neighbours who don't have your problem also have one year old houses? Once I have heard of people having similar problems with humidity due to moving in before construction materials had time to dry so humidity was stored in concrete floors and it cused problems. This link could help explain better <a href="https://blog.yourpacesetter.com/blog/understanding-humidity-in-your-new-home" rel="nofollow noreferrer">humidity in new house</a>. If this is your case maybe it could be solved using industrial dehumidifiers for some period of time, maybe there are companies who deal with such requests and can give you better answers.</p>
|
17992
|
Humidity in home is much higher than outside
|
2017-11-12T19:21:30.203
|
<p>In the making of firework explosives, the explosive mixtures are handled with great care to avoid static electricity induced detonation of the explosives. Is it a good idea that they store the explosives in a plastic container - isn't there a risk of static buildup that could detonate the explosives?</p>
<ul>
<li><a href="https://www.youtube.com/watch?v=3mlrjrLUpyg" rel="nofollow noreferrer">How It's Made Fireworks</a></li>
</ul>
|
|electrical-engineering|materials|design|electromagnetism|electrical|
|
<p>Static electricity is a major issue when handling or dealing with explosives.</p>
<p>When blast holes are loaded with ANFO (ammonium nitrate fuel oil) as the main explosive, and a specific density of ANFO is required in the holes, the ANFO is blown into the holes via compressed air and a tube.</p>
<p>For short holes drilled with hand held machines, the tube is made of aluminum because of static electricity concerns. For large holes drilled with larger machines, the tube is made of plastic. To ensure that static electricity will not be an issue low static plastic tubing is used.</p>
<p>Generally black plastic tubing is used. To differentiate low static tubing for conventional tubing, low static tubing has a red or orange stripe that runs the full length of tubing.</p>
<p>Providing the plastic containers used to store high explosives are made of low static plastic, static electricity will not be an issue. However, the plastic containers need to be identifiable as being made from low static plastic.</p>
|
18007
|
Quantifying Static Electricity Hazard with High Explosives
|
2017-11-13T10:49:37.763
|
<p>Need some help designing a bolt pattern for a pressure tight enclosure.</p>
<p>How do i determine the force applied at a distance from a bolt location so that I meet the minimum force to compress the gasket along the entire length of the enclosure but while using the minimum number of bolts?</p>
<p>And if anyone know how to model this on solidworks that would be very useful as well!</p>
|
|mechanical-engineering|civil-engineering|solidworks|bolting|pressure-vessel|
|
<p>I found these documents relating to my question. However i suspect they're all for circular gasket designs with the bolt pattern being circular. </p>
<p>EngineersEdge:
<a href="https://www.engineersedge.com/general_engineering/gasket_cover_compression.htm" rel="nofollow noreferrer">https://www.engineersedge.com/general_engineering/gasket_cover_compression.htm</a></p>
<p>Journal Paper:
International Journal of Pressure Vessels and Piping 120-121 (2014) Comparative study of bolt spacing formulas used in bolted joint designs
Abdel-Hakim Bouzid</p>
|
18015
|
Determine force distribution from a bolt pattern
|
2017-11-14T16:36:59.887
|
<p>For Linear actuators, a common mechanism is to use ballscrews with ballnuts.</p>
<p>The ballscrew is supported at the two sides by a set of bearings which permit the rotation of the screw while keeping the axial reaction force to the ballnut movement.</p>
<p>Looking for those ballscrew supports, I found there are several categories:</p>
<ul>
<li>Floating (identified by <strong>F</strong>)</li>
<li>Fixed (Identified by <strong>K</strong>)</li>
</ul>
<p>An other letter indicate several types/shapes of supports: <strong>E, B, F,</strong> etc.</p>
<p>Those pieces are usually sold in a set of one fixed and one floating support:</p>
<blockquote>
<p>e.g. </p>
<p>BK + BF </p>
<p>EK + EF</p>
<p>FK + FF</p>
</blockquote>
<p>....</p>
<p>I could not find understandable information about this.</p>
<p>What types of ballscrew/leadscrew supports exits?
What is the difference between them?</p>
|
|mechanical-engineering|terminology|linear-motion|
|
<p>BK, BF refer to the mount format - in this case a rectangular face mounted block with one (BF) or two (BK) bearings and mount holes.</p>
<p>1st Letter == mount format;
2nd letter == bearing set ;
numeric == ID of the bearings in mm;</p>
<p>A quick internet search will identify all of these bearing types and specifications.
<a href="https://i.stack.imgur.com/cQh6U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQh6U.png" alt="F series face mounted"></a></p>
<p><a href="https://i.stack.imgur.com/HvJvC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HvJvC.png" alt="E series base mount"></a></p>
<p><a href="https://i.stack.imgur.com/HEZeH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HEZeH.jpg" alt="B series heavy duty"></a></p>
|
18027
|
What type of supports exists for ballscrews or leadscews?
|
2017-11-15T13:56:22.277
|
<p>I am interested in comparing the different materials that can be used to make castings of cast irons, and therefore was wondering which material is commonly used/which materials can be used to make permanent casting molds (to be used specifically for cast iron). Thank you.</p>
|
|materials|metallurgy|metals|casting|
|
<p>You can cast iron into cast iron moulds as long as the moulds have plenty of mass compared to the quantity of metal you are pouring into them. Small artisan foundries sometimes use cast iron moulds to make small ingots as a 'signature piece' to use up small quantities of surplus metal. Often these are open face moulds. </p>
<p>Obviously steel works as well but steel moulds tend to be significantly more difficult to make. </p>
<p>Typically in industry multiples of the same casting are made in green sand moulds using a pattern which is pressed into the mould which means that although the specific mould is broken up after each casting the sand can be reused and it is fairly straightforward to mass produce moulds. </p>
<p>With metal moulds you need to be quite careful about draft angles and avoiding undercuts. Shrinkage stresses can also be a problem with more complex and hollow shapes which can end up too constrained by the mould. </p>
<p>You may also need to allow additional venting compared to sand moulds as unlike sand gasses can't escape into the walls of the mould </p>
|
18037
|
Which materials can be used to make permanent casting molds for cast iron?
|
2017-11-16T12:43:15.897
|
<p>This comes from the composites store website:</p>
<blockquote>
<p>These high performance carbon shapes are produced by packing a high volume of carbon fiber into epoxy resin under tension resulting in an unmatched tensile and compression stength.</p>
<p>These shapes are twice the stiffness of Aluminum.</p>
<p>Compared to a wet lay-up of Carbon Tow the shapes are:</p>
<p>350% Stronger in Tension</p>
<p>550% Stronger in Compression</p>
</blockquote>
<p><a href="http://www.cstsales.com/graphlite_data.html" rel="nofollow noreferrer">Quote source</a></p>
<p>It doesn't explain why. Carbon Tow are long continuous strands of carbon fiber. So somehow putting it under tension first, then fixing it in position with the epoxy makes it stronger? How?</p>
|
|composite|carbon-fiber|
|
<p>Its not about the tension in the final product. [They are claiming] tension during fabrication buys them a much higher fiber loading. See 67 vol%.. a typical "wet layup" is maybe half that. Having the fibers very very straight probably contributes somewhat as well.</p>
<p>Note also the 3+GPa claimed strength is for some "micro" composite. If you look elsewhere on that site you will see practical size rods of "graphlite" are more like 2GPa. </p>
|
18051
|
Why is graphlite so strong in tension?
|
2017-11-17T08:46:22.640
|
<p>I have a temperature sensor and I want to normalize the sensor values. The measured room has a pretty constant temperature but the sensor shows slightly different values e.g. <code>30.20, 30.00, 30.30, 30.00</code>. For display reasons, I want to have it less variation. Also sometimes it deviates unreasonable e.g. <code>30.00, 30.00, 10.00, 30.00, 30.00</code>. But it can change from one moment to another also in a reasonable way e.g. opening a window: <code>30.00, 30.00, 30.00, 30.00, 25.00, 15.00,10.00</code></p>
<p>I do not have target curve otherwise I thought about a PID. </p>
<p>At the moment I sum the last values and build an average. What can I do to</p>
<ol>
<li>normalize the output values</li>
<li>eliminate runaway values </li>
</ol>
|
|control-theory|
|
<p>I simple <a href="https://en.wikipedia.org/wiki/Low-pass_filter" rel="nofollow noreferrer">lowpass</a> filter will probably solve the problem. It is easy to implement and you can adjust how much you want to filter with the time constant.</p>
<p>Here is an example of a low-pass filter in actual code. It's technically Python but reads like pseudo-code. It takes a list of temperature values as input and outputs the filtered values as a list.</p>
<pre><code>Input: input_series, time_constant
Output: output_series
output_series = []
output_series.append(input_series[0])
for current_unfiltered_value in input_series:
previous_filtered_value = output_series[-1]
B = 1.0 / time_constant
A = 1.0 - B
new_filtered_value = (A * previous_filtered_value
+ B * current_unfiltered_value)
output_series.append(new_filtered_value)
return output_series
</code></pre>
<p>The downside of using lowpass filters is that it introduces phase lag. The filtered value will lag slightly behind the measured signal, depending on the time constant. A highly filtered signal will lag more than a less filtered signal.</p>
<p>There are other filters you can use for better filtering result. For example the <a href="https://en.wikipedia.org/wiki/Kalman_filter" rel="nofollow noreferrer">Kalman filter</a> or the <a href="https://en.wikipedia.org/wiki/Alpha_beta_filter" rel="nofollow noreferrer">Alpha-beta filter</a>. They are both model-based, which means you have to make a model for your temperature changes for it to work. There is an excellent <a href="https://github.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python" rel="nofollow noreferrer">notebook</a> on Github that introduces these filters if you are unfamiliar with them.</p>
<p>I would personally go with a simple lowpass filter in the beginning. If you don't control a process based on the filtered values, but want to filter the values to make them prettier in a user interface, then that's usually more than good enough.</p>
<p>For eliminating runaway values I would calculate the <a href="https://en.wikipedia.org/wiki/Variance" rel="nofollow noreferrer">variance</a> of the measured signal at every timestep (you could, for example, calculate the variance for the last 50 samples). If the calculated variance passes a certain threshold then it can be assumed to be a spike, which can then be rejected.</p>
|
18061
|
Control Variations in a Measurement (Temperature Sensor)
|
2017-11-17T15:09:46.160
|
<p>I am currently trying to design a simple truss (pin-jointed) structure able to withstand a 1.2 kN load. </p>
<p>I carried all the analysis and found the forces, stresses and strains in the members. My question now is how good and accurate are those forces I calculated.</p>
<p>Say I found that in a particular member the force was 10000N (tensile) and the proof stress of the alloy I'm using = 100 MPa. Theoretically, using any bar with area > 100mm^2 should be fine, right? However, in the analysis I made several assumptions (massless bars, no firction, loaded through centroid...) and the calculated Force might not be exactly the real-life one. As a rule of thumb, what margins should I take to account for these assumptions? Should I systematically allow for 10% error for example? </p>
<p>Secondly, how safe is it (with regards to failure of the member) to go close to the proof stress? Should I there again allow for a margin for error?</p>
<p>Thank you very much for your answers!!</p>
<p>Best wishes,</p>
<p>NB1: I do not have access to a Finte Elements Sotware to run it computationally and check!</p>
<p>NB2: The structure is statically determinate.</p>
|
|structural-engineering|civil-engineering|structures|
|
<p>This is a question which cannot be answered conclusively, because the issues you are raising are matters of risk and probability. There is no such thing as a calculation which <em><strong>guarantees</strong></em> that the final structure will be secure. All you can do is state that the adopted calculations imply in an acceptable (and very, very low) level of risk.</p>
<p>After all, as you mention, there are a multitude of random variables with behaviors out of your control:</p>
<ul>
<li>Might the load be greater than what you actually expect?</li>
<li>Might the material be faulty, with a strength below what you expect?</li>
<li>Will the structure behave exactly like the structural model, or will imperfections modify the internal stresses?</li>
<li>And countless other potential risks...</li>
</ul>
<p>So, what is that acceptable level of risk? Who has the authority to determine that? Well, the answer is "not you." Instead, every country/state/whatever has a set of codes which determine the adequate methods and considerations to be taken into account when performing your calculations in order to reach the level of risk which the country/state/whatever believes is acceptable.</p>
<p>When dealing with structures, there tend to be two general methodologies: allowable strength design (ASD) and load and resistance factor design (LRFD).</p>
<p>In ASD, the required strength to resist a given load is calculated and compared to the "allowable strength" of the material, which is a fraction (50%, for example) of the material's nominal strength.</p>
<p>In LRFD, the expected loads are increased by a given factor and then the required strength for this increased load is compared to the allowable strength of the material, which is again a fraction of the nominal strength (but higher than in ASD).</p>
<p>For example, in Brazil, bridges have their dead loads increased by 35% and their live loads by 50% (and the dynamic factor). Meanwhile, concrete bridges have their strength reduced by ~30%, while steel bridges are reduced by only ~13%. This is an example of the LRFD method. (Brazil doesn't really use ASD nowadays).</p>
<p>Why are dead loads increased by 35% while live loads are only increased by 50%, though?</p>
<p>Well, it's because you can be pretty reasonably sure of what the final weight of your bridge will be. After all, you are in control: you selected the dimensions and the material. Something will need to go pretty seriously wrong in either of those variables (dimensions or concrete density, for example) for your bridge to be significantly off your expected weight.</p>
<p>However, you have no control over the live loads that drive over your bridge. Bridges are usually calculated with a codified live load, which describes a heavy truck or train. But, who knows what trucks will weigh in 20 years? And codes usually allow you to only consider one (or whatever) trucks driving by your bridge at a given moment. Sure, the odds of having two of these massive trucks driving by at once are astronomical, but what if you have one of them followed by a bunch of other, lighter trucks? Basically, you have no control over what happens on top of your bridge, so you need to put a much higher safety factor on live loads.</p>
<p>The same concept applies to why concrete strength is discounted so severely, compared to steel. Concrete is often done <em>in-situ</em> (on site), where quality control is more problematic. Even when it's mixed in a controlled environment and then trucked over to the site, there are a multitude of factors which govern concrete strength: was it mixed properly, are the aggregates adequate for the site's environment, was it poured correctly, was it hydrated adequately post-pour, what will its strength profile over time be, etc, etc, etc. Meanwhile, steel is manufactured in a factory, put in a truck, positioned on the site, and then welded/screwed in. There are basically no random variables as to the structural element's strength once it leaves the factory, which means that quality control at the factory door is essential and sufficient to be very sure of the material's actual strength. So the smaller safety factor is basically to cover the potential for screwups at the factory and for potential loss of strength over time due to the elements. The only other variable to be considered for steel is the linkages (welding/screwing), but guess what, there are codes for that as well.</p>
|
18068
|
Percentage error in calculated Forces during simple structural analysis for a truss pin-jointed strucures
|
2017-11-18T13:25:54.333
|
<p>I had the question myself it those high voltage transmission lines, disturb other electrical devices? Let's say I have an high voltage next to a mobile basestation, does it effekt the electrical components in the base station or the wireless communication of the antennas ?</p>
<p>Is there any way I could calculate that?</p>
|
|electrical-engineering|electrical-grid|
|
<p>Generally, no. Transmission lines can act like antennas, but they operate at around 50 Hz. The wavelength of a 50 Hz EM-wave is $6\times 10^6$ m. In order to effectively couple that energy, you'd have to have a receive antenna of $3\times 10^6 $ m length. That's not feasible. </p>
<p>Not only this, but of course the gain of the antenna scales with frequency and effective area. For low frequencies, even antennas of large effective area tend to be bad radiators. </p>
<p>Even if it COULD feasibly couple with basestation electronics, it just wouldn't happen. A base station's carrier frequency is more likely to interract with its own electronics than a power line's. And, because of this, basestation electronics is very well shielded from most frequencies of radiation, especially RF/HF radiation. </p>
<p>You can get calculations for all of this stuff in basic antenna design books i.e. Modern Antenna Design by Thomas Milligan.</p>
|
18072
|
Do high votlage transmission lines disturb other electrical devices?
|
2017-11-18T20:05:32.907
|
<p>I would like to learn to weld (stick welding) for my own purposes (mostly for small repairs and building simple metal constructions). So far I'm trying to educate myself by reading and watching instructional videos, but so far I haven't found an answer to one question which is bothering me (I know it may sound stupid):</p>
<p>Assuming that welding equipment is fully functional (free of any defects and everything is set up properly), is there a risk of electric shock if a welder inadvertently touch part of construction which he is welding?</p>
|
|electrical-engineering|welding|
|
<p>I am a welder, and here to tell you, you can definitely get shocked. True, if everything is in good condition, proper PPE you are golden. This means your ground has good connection (the bolt holding the wire to the clamp is tight, strong spring for clamp, clean surface on base metal in contact with the ground clamp, not going through any moving parts i.e. bearings, swivels), wearing gloves and good thick soled shoes, the more you sweat the higher chance you have, being engineers you understand salt water is a great conductor I am sure. High Frequency increases your chance to get a pants wetting experience (never happened to me but have seen guys pee themselves while TIGing). </p>
<p>Which is really why I am scouring the web to understand more about why HF will reach out and get ya. I get that's what it's for in welding, the electrical arc will reach out to the base metal instead of touching the electrode (tungsten) resulting in a contamination, but <em>how</em> does it?</p>
<p>Back to it though, some reasons for getting shocked. </p>
<ol>
<li>bad ground of course</li>
<li>filings inside the welder from grinding (make sure you blow it out every 6 months
or so) Best is to always unplug and take off the housing.</li>
<li><p>being wet (sweat, standing in water, early morning dew soaked gloves, ect.) making
yourself a better conductor than the copper wire ground. This goes for other people
around you. Early in my career I almost killed a guy (stick welding) on the other
end of a project (around 20' away) because he was standing in water with soaked
shoes touching it with bare hands.</p>
<p>My advice is to never count on perfect conditions and always act like the gun is loaded, so to speak.</p>
<p>There are many other hazards to welding but this should answer your question on getting shocked and how to best avoid it.</p>
<pre><code> (source for the lower portion)
</code></pre>
<p><a href="https://www.brighthubengineering.com/power-plants/89792-ac-and-dc-shock-comparison/" rel="nofollow noreferrer">https://www.brighthubengineering.com/power-plants/89792-ac-and-dc-shock-comparison/</a> </p>
<p>AC current is alternating in nature and follows a sine curve. It is continuously changing direction and passing through zero to a maximum positive value and then to a maximum negative value. The voltage of an AC current is a RMS or root mean square value, and the peak or maximum value is 1.4 times the RMS value. However DC current will make a single continuous contraction of the muscles compared to AC current, which will make a series of contractions depending on the frequency it is supplied at. In terms of fatalities, both kill but more milliamps are required of DC current than AC current at the same voltage.</p></li>
</ol>
|
18074
|
Welding and risk of electric shock
|
2017-11-20T00:27:55.087
|
<p>Given that hot air rises, it's easy to understand how attic insulation can help keep a building warm; say, in the winter, for example. But when winter changes to summer, does the attic insulation help keep the building cool? Or even worse, does attic/roof insulation make the building hotter by resisting the flow of heat up and out?</p>
<p>In the below diagram, my idea is that the R-factor of the attic insulation will resist the flow of heat up and out of the building by impeding flow from zone 1 to zone 2 in the below diagram.</p>
Figure 1. Premise: R-factor resists heat flow from zone 1 to zone 2 thus trapping warm air inside the house in the summertime?
<p><a href="https://i.stack.imgur.com/kYlgf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kYlgf.jpg" alt="enter image description here"></a></p>
<p>How would we quantify the overall effect on house temperature based on this and other competing effects? Both theoretically and empirically?</p>
|
|thermodynamics|building-design|building-physics|thermal-insulation|
|
<p>Attics are much hotter than the outside air when the sun is shining on them (say 120F in an attic). When the air conditioning system is running the internal air temperature is less than the outside (lets say 72F in the house and 85F outside). The ceiling temerature might be warmer than the floor, but both will be much cooler than the attic (say 73F for the ceiling and 71F for the floor). Heat will transfer from the hotter attic (120F) into the house (73F ceiling temp), so ceiling insulation is very important for keeping a house cool.</p>
|
18084
|
Does attic insulation help cool a building?
|
2017-11-20T03:49:20.673
|
<p>I am trying to understand how to turn a basic click button into something that behaves like a pressure sensitive button, meaning the harder you press it the greater the value.</p>
<p>There are PWM modules that you can use to get different current but I am thinking of something like a button.</p>
<p>Could I do this with a micro-controller, a click button and some software or are there special buttons that do the job and send signals in a range based on how hard/ fast it's pressed?</p>
|
|electrical-engineering|pwm|
|
<p>"<strong>Quantum-tunnelling composite</strong>" is a rubber material impregnated with metallic particles. Its resistance varies as a function of mechanical pressure.</p>
<p>Basically, put a piece of QTC membrane between a pair of metal plates and squeeze it, the resistance of the QTC will decrease with an increase in pressure. This allows more or less current to flow from one contact to the other.</p>
<p>This stuff works remarkably well, I've used it as the lower half of a resistor voltage divider and you can get very smooth and reproducible results over a wide range of pressure. You can even use it to form a crude but effective speed controller for (very small) motors.</p>
<p>In your case you'd use the ADC of your microcontroller to measure the voltage and convert that into an integer value that quantifies the pressure on the button.</p>
<p><a href="https://en.wikipedia.org/wiki/Quantum_tunnelling_composite" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Quantum_tunnelling_composite</a></p>
<p>It's widely available and not very expensive if you only need a couple of square inches of it.</p>
|
18087
|
Applying PWM to get analog button
|
2017-11-21T18:06:46.203
|
<p>This is a more technical rather than professional question. I have a hyper-threaded quad core CPU. While building models in SpaceClaim I noticed that it only uses one logical core. A single core will be on a very high usage percentage while the others are nearly idle. How can I set SpaceClaim to use more cores and make better benefit of the CPU? </p>
|
|cad|ansys|
|
<p>Software must be programmed specifically to use multiple cores. There is nothing restricting SpaceClaim or any other CAD software from being programmed this way, but it is currently more complex/expensive to program and there is no immediate incentive for companies to work this direction. Their software is already very expensive, so their customers don't mind buying a faster computer every year. Customers won't demand multi threading support until the price of software drops below the price of new computers.</p>
<p>Here is an (unfortunately old 2007) list I copied from <a href="http://www.tomshardware.co.uk/forum/243332-28-programs-cores-quad-core-chip" rel="nofollow noreferrer">toms hardware</a>. It helps paint the picture of how few programs are multi threaded. The main thing a multiple core processor buys you is the ability to run different programs at 100% simultaneously.</p>
<blockquote>
<ul>
<li>3D Studio MAX using Mental Ray Renderer (>99 % of 4 cores)</li>
<li>Adobe Premiere Elements v3.0.2 (52 - 85 % of 4 cores depending on source type, filters, etc.)</li>
<li>AutoGK v2.40 (30 - 53 % of 4 cores depending on source type, filters, etc.)</li>
<li>Cinema 4d Rendering (>99 % of 4 cores)</li>
<li>Dr. DivX v2.0.0 (47 - 65 % of 4 cores depending on source type, filters, etc.)</li>
<li>DVDShrink v3.2 (~90 % of 4 cores)</li>
<li>Lightwave 3D (>99 % of 4 cores)</li>
<li>Nero Suite 7.x (>90 % of 4 cores when encoding)</li>
<li>Noise Ninja v2.13 (~80 % of 4 cores when doing the noise reduction on an image)</li>
<li>Sony Vegas 7.0e (83 - 100 % of 4 cores depending on source type, filters, etc.)</li>
<li>TMPG XPress v4.2.3.193 (65 - 100 % of 4 cores depending on source type, filters, etc.)</li>
<li>Winrar v3.70 (~85 - 90 % of 4 cores on benchmark; ~75% in practice)</li>
<li>x264 v0.55.663 (>99 % of 4 cores when doing the 2nd pass of a 2 pass encode)</li>
</ul>
</blockquote>
|
18112
|
How to use more CPU cores in SpaceClaim?
|
2017-11-21T21:26:32.150
|
<p>I'm trying to design a hover car track. As in, you have a 'track/circuit' that have objects being slowly moved along by the force of the air from beneath and sides. The propulsion to move the objects forwards comes from the air from the sides that are angled in the direction of the track. Picture <a href="https://en.wikipedia.org/wiki/Scalextric" rel="nofollow noreferrer">Scalextric</a> cars combined with air hockey.</p>
<p>These objects that are like carriers that hold things and are under 750 g (in total) and regarding contact with track is a circle that is 135 mm in diameter. </p>
<p>Track length isn't known at the moment but I think 3–5 m.</p>
<p>How would I effectively 'power' and 'transport' the air supply? I've looked into centrifugal fans in series and parallel and their designs and outputs but I'm stuck at calculating what the requirements would be to do this and if it can be done effectively at low cost, like something I could do in a medium sized garage. </p>
<p>The track will split and merge too. And there would be multiple carriers on the track.</p>
<p>Carriage is cylindrical in shape, weight = 750g, diameter =135mm, height 110mm</p>
<p>track is flat with no change in elevation. width of track would be big enough to fit a single carriage approx 150mm
The track would be like a square trench for the carriage to fit in. so the walls would be 110mm too.
The walls would be a manifold that has jets angled in the direction of the track, say 60 degrees away from the normal of the wall and parallel with the floor of the track. </p>
|
|mechanical-engineering|airflow|compressed-air|
|
<p>Have a look at the <a href="https://en.wikipedia.org/wiki/Hovertrain" rel="nofollow noreferrer">hovertrain</a> wiki.</p>
<p>Also, <a href="https://en.wikipedia.org/wiki/Pneumatic_tube" rel="nofollow noreferrer">pneumatic tube transport</a> is often used at bank drive up windows.</p>
<p>A small demo could be constructed low cost in a garage. Do some research on those links above and if you have specific questions regarding fan selection or design; ask them at that time.</p>
|
18117
|
How to power and transport air to a small hover car track
|
2017-11-22T03:00:44.637
|
<p>I am working on a project for an Environmental Engineering class. Our concept is a system that collects rain water for use in toilet flushing, i.e. requires minimal filtration. I have no experience in engineering statics, so I am curious if it is possible, with calculations to back it up, to attach a 50 gallon drum for water collection to the side of a house for purposes of gravity feeding the water into a toilet on a lower level of the house. And if not, could this be accomplished at low cost with the rain barrel instead at ground level, collecting runoff from the rain gutters on the roof, using a water pump to displace the water to the bathroom? Any help is appreciated thank you :D
Additional details: we are students at the University of Washington, so our system is initially proposed for rainy cities such as Seattle where collecting large volumes of rainwater is fairly easy, and the availability of the rainwater isn't of much concern. This solution intends to solve problems off runoff, and to lower monthly water utility cost, as well as to save on water usage.</p>
|
|mechanical-engineering|structural-engineering|civil-engineering|mechanical|
|
<p>You should evaluate the source : ie amount of rainfall over the year, when it is available, how many days it rains etc.</p>
<p>Then evaluate the use over the year : flushes per day etc.</p>
<p>Then you can work out the storage volume necessary based on the number of days between rainfall filling the tank and the use per day.</p>
<p>Once you have the volume of tank required then you can work out the weight and start designing a support or decide where it can be located (my parents house had the water tank halfway up the hill to provide pressure with no pumps necessary...).</p>
|
18120
|
Collecting Rainwater off a Domestic Roof for Flushing Toilets
|
2017-11-23T00:12:12.960
|
<p>So recently I went to Supercomputing 17 conference and picked this from a booth. </p>
<p>I would like to know as to what this device is? What's its name? Where to find its documentations?</p>
<p>Thanks<a href="https://i.stack.imgur.com/t4pJp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t4pJp.jpg" alt="enter image description here"></a></p>
|
|electrical-engineering|sensors|embedded-systems|iot|
|
<p>It looks like the devkit associated with the NodeMCU community.
If you turn it around with the metal portion facing away from you it should say something like "devkit v1.x".</p>
<p>It's a developer board with an ESP8266 built into it. You can actually see the ESP8266 board sitting on top of the other board in the photo!</p>
<p>The ESP8266 specifies only the microcontroller (under the metallic cover) whereas the ESP-12N specifies the board on which the microcontroller is included (the board piggy backing on the bigger board). </p>
<p>The function of the underlying board is to simply provide you with a USB interface (UART - USB) and a voltage stabilization for convenience and standardization of the I/Os.</p>
<p>Check this wikipedia article for more information: <a href="https://en.wikipedia.org/wiki/NodeMCU" rel="nofollow noreferrer">NodeMCU</a></p>
|
18140
|
What is this device?
|
2017-11-23T09:38:19.090
|
<p>Does anybody know what the thread size is on a bicycle pump connector, such as this one?</p>
<p><a href="https://i.stack.imgur.com/2dh0w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2dh0w.jpg" alt="enter image description here"></a></p>
<p><strong>Note</strong>: I'm not talking about the female end that goes on the valve stem on the bike (that's a Schrader valve thread), I'm talking about the male end that connects with the pump. It looks almost like an M5, but doesn't quite fit in an M5 female connector. </p>
<p>I have spent quite a bit of time googling the question, but have drawn a blank so far.</p>
|
|mechanical-engineering|threads|
|
<p>3/16 BSF (3/16in dia 32 turns per inch)</p>
|
18149
|
Bicycle pump connector thread size
|
2017-11-23T19:39:41.560
|
<p>So I've noticed that computer science has actually been developed so that it has become more human and less technological in the sense that development has raised the abstraction level up to a "more human" level. A modern computer scientist is able to think about computing in very "humanly" ways, rather than as merely machine-like ways. That's because programming languages have become to be able to express concepts more naturally, rather than as some sort of memory address manipulation.</p>
<p>Is a similar kind of phenomenon observable in some other fields of engineering?</p>
|
|computer-engineering|
|
<p>An important factor is that computing inherently has a numbers of layers of abstraction and function between the end user and the actual functionality. To some extent this is a from of automation in that regularly used actions are packaged into a single simple set of commands. </p>
<p>There is also the fact that more computing power allows more visually orientated interfaces which, can to a degree hide what is going on not to far under the surface. Equally most people experience computing via software, which to a large extent represents picking form a menu of pre-defined options created by somebody else, even when it doesn't necessarily feel like it. </p>
<p>There are analogies to this in other forms of engineering. A trivial example is standardisation of something like screw threads. It is rare that anyone would now bother to design a thread from scratch, most of the time you are picking form a list of available options and the same goes for things like bearings, gears and motors. </p>
<p>There are also ways in which the same advances you are talking about in computing have had a direct effect on mechanical and manufacturing engineering. An obvious example is the proliferation of CAD/CAM software. You can now design a part on a computer, send the file to a manufacturer and have it delivered shortly afterwards and (at least in theory) it could be more or less untouched by human hands in between. </p>
<p>This process cuts out what would otherwise be a huge amount of labour in terms of producing accurate, dimensioned drawings again it is more a case of automating and streamlining a previously existing process than anything entirely radical. But it does strip away one layer of specialist skill/knowledge. </p>
<p>Where it does become more radical is when you st art to integrate physical modelling with automated drafting so you can test and optimise parts and assemblies in a way which isn't really possible with physical prototypes, especially when dealing with cost-prohibitive manufacturing processes. </p>
|
18157
|
Can you perceive similar kind of "humanizing" of technology in other fields of engineering than computer science?
|
2017-11-24T16:15:54.433
|
<p>I want to repair a dent in a piece of cast iron with brass. I want to fill the site with brass.</p>
<p>For some reason, I want to do it by myself. I can buy an instrument for this work in a limited budget(within $200)</p>
<p>I dont like to buy a gas welder to be used at my home since the storage is a risk.</p>
<p>Is there any other option for this work?</p>
|
|gas|welding|
|
<p>Brazing is generally a good way to repair cast iron but it tends not to be very easy to fill large gaps. You certainly can braze with a sufficiently powerful venturi burner using propane only but if you want to build up metal you will need the more precise flame that you get from oxy/gas. </p>
<p>Oxy/propane will certainly work and propane is much less dangerous to store than acetylene. </p>
<p>Stick welding can be successful for repairing cast iron, generally nickel electrodes are used, brass electrodes don't exist but you can get bronze ones which should work. However you will probably still need to pre-heat the work to reduce the risk of cracking. </p>
<p>Another alternative, which is quite attractive if you don't need high temperature resistance is to use a metal filled resin. You can make this up yourself from eg epoxy casting resin and fine brass powder. This has the advantage that you don't have the thermal stresses which are what cause the problems in welding or brazing cast iron. </p>
<p>There are also techniques which involve cutting keyed slots into cast iron and peening in cold metal slugs (usually nickel) but this is more for crack repair and may not be suitable for this application. </p>
|
18161
|
Is there any option other than a gas welder
|
2017-11-24T17:18:41.950
|
<p>There is a type of screw/ bolt that is flat on top so that it's almost flush when fully seated, but it is <em>not</em> counter-sunk, and it usually takes a hex driver (allen key). I see it used on Ikea furniture frequently, and the only pic I could find of it was from an Ikea manual:</p>
<p><a href="https://i.stack.imgur.com/6ir15.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ir15.png" alt="enter image description here"></a></p>
<p>Note that it is not counter-sunk so from the side it looks like a 'T'. </p>
<p>Does anyone know what the correct mech eng jargon term for this screw would be? E.g. if I'm telling someone to switch from a pan-head to this type of screw? </p>
<p>Thanks. </p>
|
|mechanical-engineering|bolting|
|
<p>"Low-profile cap head bolt" would describe it also. Googling that returns images like
<a href="https://i.stack.imgur.com/pDUE5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pDUE5.png" alt="enter image description here"></a></p>
<p>but the accepted answer is technically correct.</p>
|
18163
|
Correct term for non-countersunk flat hex screw/ bolt?
|
2017-11-25T00:33:15.857
|
<p>How is "16.7 million colors" in a display calculated?</p>
<p>Also, is it a software or a hardware thing (the 16.7 million colors)?</p>
|
|computer-hardware|
|
<p>Each color red, green, blue that makes a pixel has a range of 0 to 255. So each of the colors has 256 possible settings. 256*256*256=16.77 million possible colors.</p>
|
18169
|
How is "16.7 million colors" in a display calculated?
|
2017-11-25T12:47:54.067
|
<p>$$\propto \dfrac {\text {Cross-Sectional Area}}{\text {Volume}} $$</p>
<p>I've seen this equation on my textbook (physics). </p>
<p>Let us imagine that there are two pen with $7$ and $3$ lenght. Which is harder to break? According to the equation, $3$ is harder to break. Can you explain it? </p>
<p>Kindest Regards!</p>
|
|building-physics|geometry|
|
<p>I suppose your statement is correct. </p>
<p>If $$\text{Cross-Sectional Area}\times \text{Length} = \text{Volume}$$
then,
$$x\propto \frac{\text{Cross-Sectional Area}}{\text{Cross-Sectional Area}\times \text{Length} }$$
$$\therefore x\propto \text{Length}^{-1}$$</p>
<p>So, if $x$ is a measure of how hard something is to break, then, yes, something of three units length will be comparably harder to break than something of seven units length.</p>
|
18175
|
What does this equation mean?
|
2017-11-26T05:40:27.770
|
<p>I have to work out the air mass flow coming out each vent on a server and have proposed to use a hot wire anemometer to measure the air velocity at each vent by placing it slightly inside one outlets of the vent (it is a grid pattern), then multiply that by the vented area and air density to find mass flow.</p>
<p>My concern is that I am instructing technicians to do it on site, and I am worried that the measurements may vary significantly depending on where on the vent they are taken from i.e. if they take it from the corner or side of the vent it may be different from the centre.</p>
<p>This is my first time trying to measure something like this and was wondering how others in the industry do it. Is there a better way of doing this?</p>
|
|fluid-mechanics|airflow|
|
<p>If the size of your vents is consistent, you could construct a section of duct that hooks up to a vane anemometer. As long as your setup doesn't restrict flow, you can get fairly accurate, repeatable measurements. The duct needs to capture all of the flow, but not restrict it. This might not be possible in your setup, but probably worth looking at.</p>
<p>Typical measurements are in flow rate, not just velocity.</p>
<p><a href="https://i.stack.imgur.com/HjHk4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HjHk4.jpg" alt="enter image description here"></a></p>
|
18187
|
How to accurately measure mass flow through server vents using an anemometer
|
2017-11-26T11:17:39.917
|
<p>Most of the real life structures are statically indeterminate.</p>
<p>What is the benefit of designing a statically indeterminate structure instead of a statically determinate one?</p>
<p>I don't mean only beams and trusses.Per example, the shaft of a merchant vessel is considered as a typical statically indeterminate structure.</p>
|
|mechanical-engineering|structural-engineering|structural-analysis|structures|statics|
|
<p>For example, take a look at the following static systems.
Assume they have the same length and the same (constant) cross-section. Thus an equal allowed bending moment $M_u$.</p>
<p><img src="https://i.stack.imgur.com/S8qjr.png" width="300" ></p>
<p>The first system is statically determinate, as it is supported only by simple supports. The maximum moment developing within the beam is $M=\frac{QL}{4}$, thus the load under which the beam fails is $$Q_u=4\frac{M_u}{L}$$
This coincides with the formation of a plastic hinge at point B, which leaves the systems statically under-determinate, a mechanism.</p>
<p>In the second system, the beam is supported by two clamped supports, which both reduce the maximum moment at the point where the load is applied. If you determine the static forces within the beam you will find that, neglecting residual stresses, the maximum moment $M=\frac{QL}{8}$. Thus,
$$ Q_u=8\frac{M_u}{L} $$
To turn into a mechanism, three plastic hinges have to be formed, which requires more work.</p>
<p>Hence the second system theoretically can bear twice the load of the first system.</p>
<p>Statically indeterminate systems in general are more stable and more rigid, but harder to calculate. Another important factor are the above-mentioned residual stresses, resulting amongst others from thermal expansion or the manufacturing process (e.g. inhomogeneous cooling of steel).</p>
|
18190
|
What is the advantage of statically indeterminate structures?
|
2017-11-26T19:49:26.350
|
<p>I am getting confused with the maths in the max distortion strain energy part of the question. Plus there is a misprint as the formula for distortion should be a^2 +b^2-2ab=(yield stress)^2, here they have missed the 2 in 2ab. Or maybe I am wrong</p>
<p>Can someone please solve it correctly showing the working(maths). I just need a step by step working as I am making some error solving it myself. And the one solved here is missing steps.
*F.S=1 that's why not mentioned.</p>
<p>Thanks<img src="https://i.stack.imgur.com/D4soc.jpg" alt="enter image description here"></p>
|
|mechanical-engineering|design|mathematics|
|
<p>Maximum distortion strain energy formula from your example is reduced <a href="https://www.google.hr/url?sa=t&source=web&rct=j&url=https://nnf.mit.edu/sites/default/files/documents/sr-2009-2b.pdf&ved=0ahUKEwi_q6-wkt3XAhXLLFAKHZ1GDRkQFgiuATAT&usg=AOvVaw0Bt-cRVhhDf2ZPbKHfkIGt" rel="nofollow noreferrer">von Mises</a> formula,
so the calculation should be correct. It does not comply the binomial theorem because of specific physical properties of material and stress tensor components relations. </p>
<p>It's very easy ask google question. Though always check wikipedia's answer and try finding some lecture notes or other literature if possible peer reviewed.</p>
|
18197
|
Machine design problem
|
2017-11-27T03:05:02.320
|
<p>I am implementing a second order filter.
It simulates hydraulic servo on an airplane. Its about 35-40 years old.</p>
<p><a href="https://i.stack.imgur.com/Wlurl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wlurl.png" alt="Second order filter to simulate servo"></a></p>
<p>What confuses me is the notation. Like "2(.7)(52)S", what does it mean? If its just products, why then not write the resulting number?</p>
<p>I tried with doing 2704/(S^2+72.8S+2704), but it seem a bit slow.</p>
|
|control-engineering|aerospace-engineering|simulation|servo|
|
<p>It might be written that way because the <a href="http://www.kves.uniza.sk/kvesnew/dokumenty/DREP/Filters/2nd%20order%20transfer%20function.pdf" rel="noreferrer">transfer function</a> of a second order system is written as:</p>
<p>$$
h(s) = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}
$$</p>
<p>where $\zeta$ is the relative damping ratio and $\omega_n$ is the natural frequency.</p>
<p>So in your system</p>
<p>\begin{align}
\zeta &= 0.7 \\
\omega_n &= 52
\end{align}</p>
<p>It's easy to see what the parameters are when it's written with $\zeta$ and $\omega_n$ factored out.</p>
|
18202
|
What filter is this?
|
2017-11-27T06:11:20.867
|
<p>I'd noticed that for the small motors, the ones used in toy cars, there are rpm values and voltage values written on the specifications.</p>
<p>For example, I have a motor at a range of (12V - 24V) and rpm of (4600 - 9300) respectively.</p>
<p><strong>Does this mean that I need to provide 12V to the toy car so that the wheels will rotate at 4600 rpm and 24V to rotate the wheels at 9300 rpm?</strong></p>
<p>I would also like to know does the other way work? I mean if I were to use the motor as a generator.</p>
<p><strong>Does this mean that if I were to provide 4600 rpm, I can generate 12V of voltage?</strong> </p>
<p>Any response is appreciated.</p>
<p>Thank you</p>
|
|mechanical-engineering|electrical-engineering|motors|
|
<p>It means that if you were to apply 12V to the motor terminals, and the motor was not attached to the car in any way (i.e. no resistance to the shaft turning at all), then it would spin at approximately 4600 ±tolerance (and likewise ≈ 9300rpm @ 24v). As soon as you add in any resistance, then the speed will drop as the torque required to keep the motor moving increases.</p>
<p><a href="https://i.stack.imgur.com/QuZng.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QuZng.gif" alt="Typical PMDC Motor Performance"></a></p>
<p>Furthermore, it's likely that there's some sort of gear-train between the motor output shaft and the wheels, which will reduce the speed of the wheels compared to the numbers printed on the motor even further.</p>
<p>You won't get 12V back for 4600rpm input, due to the efficiency of the motor. Note also that peak efficiency is not measured at 'free speed'. Neglecting that for an example, however, and assuming the motor is 80% efficient in both directions, this means 4600rpm corresponds to 9.6V of ideal input, and converting back, 4600rpm could only create ~7.7V.</p>
|
18205
|
Motor for Toy Cars
|
2017-11-27T08:01:21.203
|
<p>Diesel engines are considered "steady torque", which means that torque is same no matter the revolutions whereas torque depends on revolutions for an engine that runs on Otto cycle.</p>
<p>What is the reason for this?</p>
|
|mechanical-engineering|automotive-engineering|torque|diesel|
|
<p>The diesel engines are working on diesel cycle. The heat addition(ingnition) is taking place during the constant pressure process. </p>
<p>After added energy is spent by the expansion of piston.
The amount of expansion is fixed in terms of stroke length. </p>
<p>So for given fixed pressure and fixed expansion we ought to get constant work output. That's why the torque is relatively fixed in diesel engines. </p>
<p>In case of otto cycle/petrol/gasoline engines the heat is added during constant volume process. So you can rise the pressure as much you can for the same expansion by adding more heat. Hence you can get quick pickup/high torque in petrol/gasoline engines by adding more fuel while starting the engine.</p>
<p>Hope this helps.</p>
|
18207
|
Why diesel engines are considered steady torque engines?
|
2017-11-28T18:43:28.757
|
<p>We define the durability with the equation as seen below. </p>
<p>$$x\propto \frac{\text{Cross-Sectional Area}}{\text{Cross-Sectional Area}\times \text{Length} }$$</p>
<p>Hence</p>
<p>$$\therefore x\propto \text{Length}^{-1}$$</p>
<p>This equation was only for matters which has same density. Let us imagine that there is a tree with $3\text h$ and an iron pole with $3\text h$. Now we see that our equation doesn't work in this condition. The iron pole will be more durable than the tree even if they have same lenght. Then, how do we define the durability as per density? </p>
<p>According to our equation, </p>
<p>$$\text {Tree} = \frac {1}{3h}$$</p>
<p>$$\text {Iron Pole} = \frac {1}{3h}$$</p>
<p>It seems like their durabilities are same. However, Iron pole has greater density than tree. By the way, Iron pole is more durable. How do you explain this? or Is there any equation that we can use when there are two matter which doesn't have same density?</p>
|
|building-physics|
|
<p>This is clearly building on <a href="https://engineering.stackexchange.com/q/18175/1832">this question</a>, and specifically its accepted answer.</p>
<p>Not to toot my own horn, but I recommend you also read my answer to that question. More specifically, the first paragraph.</p>
<p>What's important to notice here is that the equation given above isn't describing an equation, but a relationship of <em><strong>proportionality</strong></em>. When you say that "durability" (an unclear term) $x \propto 1/L$, you are stating that durability is inversely proportional to length. As I stated in my answer to your other question, that is correct for most common failure states.</p>
<p>However, this statement makes no attempt at being a complete description of the variables that determine an element's "durability". It merely states that one of the variables that goes into "durability" is the inverse of the element's length. There may be (and indeed are) other variables that also influence "durability".</p>
<p>Abandoning that useless term and taking the example of a column under uniaxial load, we can calculate that the buckling load is <strong><em>equal</em></strong> to</p>
<p>$$P_E = \frac{\pi^2 EI}{(KL)^2}$$</p>
<p>I won't bother describing what the variables are because that's immaterial to this question. What's important to note here is that this is an <strong><em>equation</em></strong>, as described by the use of the equality symbol ($=$) instead of the proportionality symbol ($\propto$). An equation attempts to be a complete description of the dependent variable (in this case $P_E$).</p>
<p>So long as you accept the assumptions that go into deriving this equation, then $P_E$ is exactly and exclusively equal to $\dfrac{\pi^2 EI}{(KL)^2}$. That is a true, undeniable fact.</p>
<p>However, here are a few other true, undeniable facts:</p>
<ul>
<li>$P_E \propto E$</li>
<li>$P_E \propto I$</li>
<li>$P_E \propto 1/K^2$</li>
<li>$P_E \propto 1/L^2$</li>
</ul>
<p>That is, $P_E$ is directly proportional to $E$. If you double $E$, you double $P_E$. There are other variables that are relevant to actually determine the value of $P_E$, sure, but there's no denying that a greater $E$ implies a greater $P_E$. The same applies to all of the other statements of proportionality above.</p>
<p>That's what's wrong with your case. You are looking at a statement of proportionality describing that "durability" is proportional to the inverse of the element's length and assuming this states that there are no other relevant variables. That is not true. All that statement is saying is that an increase in length reduces "durability". It makes no assertions as to the existence (or lack thereof) of other variables which may also influence the "durability".</p>
|
18232
|
How do we define the durability as per density?
|
2017-11-28T19:26:10.770
|
<p>I am having a problem with transferring the heat flux boundary conditions into a temperature to be able to put it into a matrix. I understand that
deltat = deltax*q''/k but I do not know how to code it so that I can loop it into the matrix in MATLAB. I have not had heat transfer and it is a steady state problem, so it should be relatively simple.</p>
<p>How could you index code to be able to loop through since deltat = T(i,j) - T(I+1,J)?</p>
<p><a href="https://i.stack.imgur.com/seZCR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/seZCR.png" alt="code"></a>
<a href="https://i.stack.imgur.com/E3KgP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E3KgP.png" alt="Schematic for Problem"></a></p>
<p>Thank you for any help!</p>
|
|heat-transfer|mathematics|matlab|
|
<p>You haven't specified which method you are using, so i am going to assume Finite Volume.
In that case you have a staggered grid like so:</p>
<p><a href="https://i.stack.imgur.com/u1GIe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u1GIe.png" alt="enter image description here"></a></p>
<p>The vertical lines are the faces of you cells and the circles are the centers of you cells. Your left boundary is located at the face at $x=0$ and your right boundary is located at the face at $x=L$. The domain of size $L$ is divided into $N$ cells. To simplify the boundary treatment, I have included two ghost / virtual nodes $i=0$ and $i=N+1$. I assume a uniform gridspacing $\Delta x=L/N.$</p>
<p>Now how do you impose a flux condition at $x=0$ in terms of local temperatures? Well from a Taylor expansion around $x=0$ we find:
$$
T_{i=0} = T_{x=0} - \left.\frac{dT}{dx}\right|_{x=0}\left(\frac{\Delta x}{2}\right) + \frac{1}{2}\left.\frac{d^2T}{dx^2}\right|_{x=0}\left(\frac{\Delta x}{2}\right)^2 + O\left(\Delta x^3\right)
$$
and:
$$
T_{i=1} = T_{x=0} + \left.\frac{dT}{dx}\right|_{x=0}\left(\frac{\Delta x}{2}\right) + \frac{1}{2}\left.\frac{d^2T}{dx^2}\right|_{x=0}\left(\frac{\Delta x}{2}\right)^2 + O\left(\Delta x^3\right)
$$
Subtracting:
$$T_{i=1} - T_{i=0} = \left.\frac{dT}{dx}\right|_{x=0}\Delta x + O\left(\Delta x^3\right)$$
If $\Delta x \ll 1$ we can neglect the $O\left(\Delta x^3\right)$ terms and we find the flux at $x=0$ in terms of the local temperatures at $i=0$ and $i=1$:
$$\left.q"\right|_{x=0}=-\lambda\left.\frac{dT}{dx}\right|_{x=0} = -\lambda\frac{T_{i=1} - T_{i=0}}{\Delta x}$$</p>
<p>So to impose the flux at $x=0$ you need to impose the local temperature of the ghost node at $i=0$ according to above equation.</p>
<p>A similar analysis can be done for the boundary at $x=L$, I'll leave that for OP.</p>
|
18233
|
Laplace Equation -- Heated Plate -- Heat Flux Boundary Condition
|
2017-11-29T01:29:45.420
|
<p>*Sorry everyone for the very wordy question.</p>
<p>I am trying to write up a method to accurately measure the airflow through the inlet and various outlets on one of our servers in order to measure the heat output from the server partitions/cards. The boss wants to know the rate of heat production from each card, the current airflow provided to each card, and what the airflow actually should be to keep the cards running at a certain temperature. The server is partitioned into three sections and the outlet vents look similar to these:
<a href="https://i.stack.imgur.com/uWPmc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uWPmc.png" alt="enter image description here"></a></p>
<p>I have purchased a TSI 9565 velocity meter that comes with a hot wire anemometer and a small pitot tube that measures down to 0.15m/s. I plan on using the pitot tube attachment to measure the air velocity, but my concern is that the velocity will vary depending on where on the vent I take it from, so I won't just be able to average it. The problem server is on site and the technician has limited time to access the server, which means he will only be able to take one measurement at each vent. We do have an exact copy of the server setup here in the main office, but the rack is quite different which means the airflow going into each server will be different. </p>
<p>However, I was thinking that because the server setups are the same, maybe the ratios between the velocities for each vent would also be the same. I don't have any justification for this, but if it is roughly true, I could record each individual velocity reading from each vent perforation on the server here in the office, where I have plenty of time to do it. Then I could specific a certain spot on each vent for the technician on site to take a single velocity reading from and find the factor of difference between that server and the office server. From there I could multiply all the velocity values for that vent that I recorded on the office server. It's a bit hard to explain so I've included a quick example.</p>
<p><a href="https://i.stack.imgur.com/9yr3r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9yr3r.png" alt="enter image description here"></a></p>
<p>My question is, is my assumption that the servers would have the same relative velocity profiles valid? If not, what would be an accurate way to measure the airflow through each vent?</p>
|
|fluid-mechanics|thermodynamics|airflow|
|
<p>"Is my assumption that the servers would have the same relative velocity profiles valid?"
Assuming your server's output air temperature is $343~K$ max. And from the experimental data, max velocity is about ~$3~m/s$ also assuming the individual port of ~$5~mm$ in size. </p>
<p>So in the field server, the Reynold's number ($Re$) is approximately ~$10^4$ . and in office (same temperature & size but the velocity is 1.5 m/s) ($Re$) < 10000. both are well within in the laminar region (it is only the exit condition). </p>
<p>Since the velocity distribution is caused by same geometry(in field & office) and Reynold's no indicates that the regime of flow are ~same (laminar regime), assuming linearity between two of your measurements might work. (for velocity profiles & mass flow rate only).</p>
<p>Linking from the previous question, you are intended to find mass flow rate & heat thrown out from the server. </p>
<p>When you come to heat transfer calculations, you have to measure the temperature profile for both the servers in the same way you did for velocity profiles. </p>
<p>Temperature profiles not necessarily similar because, total heat transfer is depends on local heat transfer coefficients (mainly convection by air) which are non linear function of local Nusselt number & local Re number.
So it may not bring the similar temperature profile at the outlet.</p>
<p>So the flows with similar velocity profiles may not have similar temperature profiles. </p>
<p>Hope this helps. </p>
|
18244
|
How to measure airflow going through individual server vents
|
2017-11-29T17:29:45.073
|
<p><a href="https://i.stack.imgur.com/94VLZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/94VLZ.jpg" alt="Venturi Tube Problem"></a></p>
<p>I'm trying to calculate the volume flow. I've tried setting up Bernoulli and using the hydrostatic pressure and the conservation of mass, as one does to solve a horizontal tube, but I keep hitting up against needing to know the heights z1 and z2, which are not given.</p>
<p>I can't figure out a way to remove it from my equations.</p>
<p>Density is constant. </p>
<p>The system is in equilibrium. </p>
<p>No turbulence. </p>
<p>No friction. </p>
<p>Pressures are constant across the diameters. </p>
|
|fluid-mechanics|
|
<p>If I understand correctly, the $\Delta p$ is the net pressure drop shown by the gauge. The pressure indicator will always show the 'net effect', i.e. it doesn't understand the physical reasons contributing towards the reading. Thus, it <em>includes</em> the hydrostatic pressure drop as well as dynamic pressure drop. Assuming there is the pressure drop along the direction of flow, and assuming that the pressure gauge shows absolute value of the pressure drop (i.e. magnitude), we can write the Bernoulli's equation (which we can safely do since the flow is irrotational, incompressible and inviscid) along the streamline as,</p>
<p>\begin{equation}
p_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g z_2
\end{equation}
i.e.
\begin{equation}
\frac{1}{2} \rho \left( v_1^2 - v_2^2 \right) = \underbrace{\overbrace{\left(p_2 - p_1 \right)}^{-ve} + \overbrace{\rho g \left( z_2 - z_1 \right)}^{+ve}}_{-\Delta p}
\end{equation}</p>
<p>We can see that, tilting of the Venturi exerts back-pressure at section $1$. i.e. it resists the flow. </p>
<p>According to the continuity equation, </p>
<p>\begin{equation}\label{cont}
v_1 = \frac{A_2}{A_1} v_2
\end{equation}</p>
<p>Substituting above equation in Bernoulli's equation, </p>
<p>\begin{equation}
\frac{1}{2} \rho \left( \left(\frac{A_2}{A_1}\right)^2 - 1\right)v_2^2 = {-\Delta p}
\end{equation}</p>
<p>Which gives, </p>
<p>\begin{equation}
v_2 = \sqrt{ \left(\frac{-2 \Delta p}{\rho}\right) \left( \frac{A_1^2}{A_2^2 - A_1^2}\right) } = 0.1326 \frac{m}{s}
\end{equation}</p>
<p>and hence, $v_1 = 0.0398 \frac{m}{s}$</p>
<p>The required flow rate is then; $v_1 A_1 = v_2 A_2 = 0.004 \frac{m^3}{s}$. </p>
|
18260
|
Volume flow in a non-horizontal Venturi tube
|
2017-11-30T22:17:58.583
|
<p>in stress concentrations, say in a hole made in a beam loaded axially, we need the value of K ( the stress concentration value) which equals max stress\ avg stress for the given body. </p>
<p>the avg stress is given by the apllied force over the net area, but im not sure what's the max stress that satisfies this relation. </p>
<p>sorry if the question is more of a defenition. </p>
|
|mechanical-engineering|materials|stresses|
|
<p>In manual calculations you will pick or calculate a value of K from a table or graph based on the geometry or calculate from a formula. Depending on the material there may also be additional factors to take into account relating to surface roughness etc. There are also various online <a href="http://www.amesweb.info/StressConcentrationFactor/CentralCircularHoleInFiniteWidthPlate.aspx" rel="nofollow noreferrer">calculators</a> which can work it out for you.</p>
<p>Essentially K is the ratio between average stress and max stress so you need to know at least two of the three values to calculate the third. </p>
<p>So in practice the average stress is easily calculable, multiply this by the appropriate K value and you get the max stress (assuming that the assumptions used to calculate K are valid). </p>
|
18278
|
what's the max stress value ?
|
2017-12-01T17:14:50.487
|
<p>I was told I should ask here. Please note I am not an engineer and I apologize if this isn't an appropriate place, as I suspect this is for engineers to exchange ideas (though I found no "rules" at all).</p>
<p>With no expectation of a high degree of precision, what should I do to mestimate (measure/estimate) the force required to completely close the lid of a large box? It is hinged and latches, much like the hood of a car, and has gas charged struts holding it open. It doesn't take much to get it down - I think the struts only slightly offset gravity - but from there it takes considerable force to fully close/engage the latch mechanism (significant friction plus a strong spring). It is too difficult for the local seniors to get closed, but it is important for theft prevention.</p>
<p>I'd like to put a small motor inside that grabs the latch and pulls it closed that last inch. I realize I'll have a lot to do to design the "grab" and also to consider things like pinch protection... but for now I just want to see if a suitable motor exists. What I don't know how to do is calculate the force.</p>
<p>Can I simply stand on a scale and take the delta when it latches, along with the distance the latch traveled? Intuitively - for my non-engineering brain anyway - I can't really see how the distance matters (if the force is enough to move it an inch isn't it enough to move it a mile?), but the definition for the term "foot pounds" suggests that shorter distance requires less force. So let's say my scale tells me I weigh 200 lbs and when I press on the hood at the point it latches it says I weigh 100 lbs, and it traveled an inch... does that mean I need a motor rated for 8.33 foot pounds? Or am I completely off base?</p>
<p>Thank you for your indulgence.</p>
|
|mechanical-engineering|
|
<p>I would suggest that a good way to estimate this sort of thing is by using sand bags (or similar). Just pile them on the lid untill it closes and then weight them. </p>
<p>Having said that it sounds like your real issue is a bit more nuanced and the real problem is getting the latch closed and there might be better ways to approach this than just adding a powered actuator. To me it sounds like a less stiff latch is an all round better solution. </p>
<p>Another option is to change the relationship between the rate of the gas struts and the weight of the lid. As you will be aware a car trunk usually closes itself with relatively little effort so it may be that using different struts (or changing their mounting position) or adding weight to the lid or just lubricating the hinges will solve the problem. </p>
<hr>
<p>In response to comments. </p>
<p>100lb seems like a lot to close a latch, but if it is designed to have someone lean their weight on it then maybe. The difficulty here though is that when you lean on something to close it, especially in this sort of situation is that dynamic forces and momentum come into play and the 'effort'required to operate a 'sticky'mechanism can be hard to quantify. </p>
<p>My initial instinct in approaching this problem would be to just add a bit more weight to the lid and see if that helps. </p>
|
18293
|
Practical low-tech real world method for measuring how much force to close a lid
|
2017-12-01T18:42:44.697
|
<p>For a low-carbon steel part that will be put into a long lifetime use (10 years plus) what are recommended load cases to consider creep? The part I'm specifically working with will have a relatively low cycle count (20k load cycles over ten years) at about 10% of it's yield strength Loads are compression and tension loads with no load being it's 'normal' position.</p>
<p>Is there a percentage of yield strength that calculating creep worth considering? Are there material properties that are more critical? Is cycle count more critical? </p>
<p>And, how do you go about calculating creep? I vaguely remember doing this in university but can't find any simple questions.</p>
|
|structural-engineering|strength|creep|
|
<p>creep is caused when applied stresses in a metal part can be relieved by movement of metal atoms via diffusion within the crystal lattice. diffusive transport only becomes important when the service temperature exceeds about 1/3 to 1/2 the melt temperature (rough rule of thumb, using degrees K)- meaning steel doesn't creep at room temperature but lead/tin solder does. the materials science text by Van Vlack covers this topic in more detail. </p>
|
18295
|
Calculating Creep
|
2017-12-01T20:29:54.020
|
<h2>Question in short</h2>
<p>The following drawing is a ball-screw with both ends mechanized and a ball-nut on it. </p>
<p><strong>What is the meaning of the indicated notations?</strong></p>
<p><a href="https://i.stack.imgur.com/3v730.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3v730.png" alt="enter image description here"></a></p>
<h2>Context</h2>
<p>What is clear to me are:</p>
<ul>
<li>Indications of sizes/distances</li>
<li>Angles</li>
<li>Parameters of precision/error</li>
<li>Diameters</li>
</ul>
<p>It seem that the little letter (e.g. <code>2</code>) are related with the values in the box, but I can't understand the meaning. The <code>1</code> is completely obscure to me.</p>
|
|technical-drawing|
|
<p>What you are asking is mainly in reference to the GD&T feature control frames on the drawings. Your drawing looks like it's metric, so it may refer to an ISO standard. </p>
<p>If your drawing references a specification, you should research that as this answer would essentially need to teach you GD&T and how to interpret it. Maybe start here:
<a href="https://en.m.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing</a></p>
<p>But there are several resources online.</p>
<p>That being said:</p>
<ol>
<li>Machined Surface Finish requirement(material removal required) as indicated by the integrated horizontal line below the value. </li>
<li>It looks like a datum callout</li>
<li>Concentricity or Coaxiality(depending on spec and interpretation)(not commonly specified) </li>
<li>Circular Runout(typically used over concentricity)</li>
<li>Perpendicularity</li>
</ol>
<p>What GD&T does is to clearly define what features are important and how they are to be inspected. They are used to control position and/or form of the part. In your case, that's a rotating shaft so typically the cylinders need to be concentric so they are balanced and aligned for assembly. </p>
|
18296
|
Understanding some drawing notations
|
2017-12-02T10:40:26.650
|
<p>I’m trying to understand the effects of a bike’s trail on its steering stability. There, however, is a concept of slippage velocity of the tires during steering of which I cannot make heads or tails of. </p>
<p>Book- Motorcycle Dynamics</p>
<p>Author- Vittore Cossalter</p>
<p>Page no. 13-14</p>
<p><a href="https://i.stack.imgur.com/0afjY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0afjY.jpg" alt="enter image description here"></a></p>
<p>1st Question: Why should the final resultant velocity of the front tire be in the same initial direction of motion ? Shouldn’t it be in the plane or rotation of the front tire, now that a steering input has been given?</p>
<p>2nd Question: What is this slippage velocity ? </p>
<p>I’ve tried Google as well. But there isn’t any information on this topic in simple words. </p>
|
|mechanical-engineering|automotive-engineering|dynamics|
|
<p>When a wheel is rolling in a straight line, the velocity of the wheel at the point of contact with ground is zero. That is, the backwards spinning speed of the wheel cancels the forward speed of the bike. In the frame of the wheel, everything is moving in the x-direction. Here $\bar{v}$ is the speed of the wheel at the contact point:</p>
<p>$$ \bar{v}_{straight,x} = v_{bike} - \omega r = 0 $$</p>
<p>$$ \bar{v}_{straight,z} = 0 $$</p>
<p>From the perspective of the wheel, the relative speed of the bike is different after the wheel is turned:</p>
<p>$$ \bar{v}_{turned,x} = v_{bike}\cos\theta - \omega r = -(1-\cos\theta)v_{bike} $$</p>
<p>$$ \bar{v}_{turned,z} = -v_{bike}\sin\theta $$</p>
<p>The terms $\bar{v}_{turned}$ terms define the sliding velocity of the wheel.</p>
|
18303
|
What is slippage velocity?
|
2017-12-02T20:15:23.167
|
<p>I would like a lock for a garage door (roll up style). I would like the lock to be something I can control electronically, and also I would like it to retain it's state when the power is removed. </p>
<p>The power requirement seems to rule out any type of electromagnetic lock, and I considered a solenoid, but I'm not sure if I can get one that stays in the current position when the power is removed. </p>
<p>I think there are two types that will work - one with a latch on a solenoid (<a href="https://www.adafruit.com/product/1512" rel="nofollow noreferrer">https://www.adafruit.com/product/1512</a>) or possibly a small linear actuator that could be extended through a plate to lock the door. Are there any other options I'm overlooking? </p>
|
|mechanical-engineering|
|
<p>You can get latching solenoids which fire back and forth given a current pulse. They stay in position when the current is removed.</p>
|
18309
|
Type of electronic lock for a garage door?
|
2017-12-03T14:05:27.950
|
<p>I'm looking for the best orientation (North-East-South-West) of a greenhouse (simplifiable as a rectangular solid).<br>
My question is more specifically about the <strong>sun position effects</strong> (over the times of the day and over the days of the year).<br>
With <em>sun effect</em> I mean both <strong>solar exposure</strong> for plants and <strong>temperature increase</strong> within the greenhouse. Anyway I think both aspects go hand-in-hand.</p>
<p>Thus, how can I calculate the position of the greenhouse?</p>
<p><strong>PS</strong>: if can help to better <em>focus</em> my question, the question could also be re-arranged e.g. for house's rooms: which is the coldest/hottest room in my house? </p>
<p><strong>EDIT:</strong> solar exposure's issue was solved. Now it remains how to deal with temperature gain. In short: with maximum solar exposure I will get automatically the higher temperature inside the greenhouse?</p>
|
|civil-engineering|building-design|building-physics|
|
<p>The angle will also depend on your objective. I use my green house to winter over various plants so my angle is set to maximize winter light , normal to the sun in mid November/January. If you are growing bedding plants from seed ,you probably want maximum sun in March / April ( northern hemisphere). I found trying to make a variable roof angle was too complicated for a home greenhouse. </p>
|
18315
|
Place a greenhouse: how to orient it for the best effect?
|
2017-12-04T05:08:05.777
|
<p>I have been assigned to work out how much air flow is needed through each partition of our servers in order to keep our processors at (or below) 60C (333K).</p>
<p>I have worked out the current mass flow rate going through each partition, and the heat flow rate, and I know the temperature of each processor from the system logs. </p>
<p>However I don't know how to calculate the required mass flow rate to bring down the temperature of the CPU's to a certain value.</p>
<p>For example, let's assume on one partition I have found:</p>
<pre><code>air mass flow rate = 0.01 kg/s
heat flow rate = 150 J/s
temperature of inlet air = 298K
temperature of outlet air = 308K
temperature of CPU = 353K
Goal CPU temperature = 333K.
</code></pre>
<p>So I need to reduce the CPU temperature by approximately 6% (1 - 333/353).</p>
<p>Does this mean I simply need to increase the airflow by 6%? I originally figured this could be true because a decrease in temperature of the CPU would be a linear decrease in the heat flow. So according to the heat capacity formula (q=mc$\Delta$T), where q is the heat flow, m is the mass flow, c is the heat capacity at constant pressure, and $\Delta$T is the change in temperature:</p>
<pre><code>1.06 * 150 J/s = 1.06 * 0.01kg/s * 1000J/kgK * (308K - 298K)
Therefore the mass flow rate needs to be 0.0106m/s.
</code></pre>
<p>The problem with this is I am assuming the outlet temperature remains the same, which I don't expect would happen. And the mass flow increases by such a marginal amount. I just don't feel like an increase in mass flow by 6% would decrease the CPU temperature by 20C. </p>
<p>Could someone please show me how I should be doing this? (Please note the goal is simply to find how much I need to increase the airflow)</p>
|
|fluid-mechanics|thermodynamics|heat-transfer|
|
<p>For anyone interested in how to solve this, I have reached out to my thermo professor who has come back with a very helpful way of working this out using the formula: </p>
<p>$$rate.of.cooling [W] = rate.of.heat.production = h A (T_{surface} - T_{air})$$</p>
<p>Where, $h$ is the convection coefficient, $A$ is the surface area of the processor, $T_{surface}$ is the surface temperature of the processor and $T_{air}$ is the average temperature of the inlet air. </p>
<p>The rate of cooling will remain the same no matter the airflow. The convection coefficient is proportional to the velocity of the air, so any changes to it's value will require an equivalent change in velocity. </p>
<p>The initial $T_{surface}$ value is $353K$, so the temperature difference that is driving the heat exchange is $353K - 298K = 50K$. The goal $T_{surface}$ value is $333K$, which has a temperature difference of only $30K$ driving the cooling process. </p>
<p>Since the rate of cooling remains the same at $150J/s$, the other side of the equation must also remain the same. So:</p>
<p>$$hA(50) = x\cdot h A (30)$$
$$x = 50/30 = 1.7$$</p>
<p>So if the convection coefficient is increased by a factor of 1.7, and velocity is proportional to the convection coefficient, the velocity must be increased a factor of 1.7 in order to maintain the same rate of cooling, but decrease the surface temperature of the processor to $333K$. This is assuming the vented area remains the same.</p>
|
18324
|
How to calculate required air mass flow to keep CPU below certain temperature?
|
2017-12-05T02:42:54.080
|
<p><a href="https://i.stack.imgur.com/EmGun.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EmGun.png" alt="enter image description here"></a>
I am having some difficulty determining the constants $a, b, k$ in the open circuit voltage transfer function $T(s) = v_{0}/v_{i}(s)$, which is given by $T(s) = \dfrac{k [s + a]}{[s + b]}$</p>
<p>I know that the constants must be in terms of the elements $\{C_{1}, C_{2}, gm1, gm2, GL[=1/RL]\}$ positive. I also know that the $gm$ can be negative when switching the input terminal.</p>
<p>Any help is much appreciated.</p>
|
|electrical-engineering|
|
<p>The op amp in the left of the circuit don't do anything;
The op amp in the right is connected like a comparator, with positive feedback. That means it's output will be VCC or GND depending on Vi, not intermediate values.</p>
|
18341
|
Small Signal OTA Circuit
|
2017-12-05T17:28:35.537
|
<p>Why don't we use fillets when making buildings and houses (i.e why do we leave the edges sharp?)? Does stress concentration factor not matter in making structures?</p>
<p>I am talking about the building like this--></p>
<p><a href="https://i.stack.imgur.com/RnQrR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RnQrR.jpg" alt="Building with sharp edges"></a></p>
<p>I know this edge is not <code>highly stress concentrated</code>(only the atmospheric pressure is acted on these edges, which is easily bearable by this building.) but assume the time of any natural disaster, at that time these type of edges would be more likely to fail.And making the edges round doesn't affect the cost much.I think.</p>
<p>NOW ANOTHER EXAMPLE.</p>
<p><a href="https://i.stack.imgur.com/ADZfW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ADZfW.jpg" alt="Stiars"></a></p>
<p>This is also stress concentrated.As we climb the stairs there will definitely some forces will act.</p>
|
|mechanical-engineering|civil-engineering|structures|
|
<p>In general architecture there is a cost pressure towards standardisation and it is often cheaper to use standard parts which are easy to assemble, specify and transport. </p>
<p>There is also the fact that in architecture the crucial loads are often supporting the span of the roof so the big loads tend to be compression and thrust on the walls with pinned joints where fillets don't really add much. </p>
<p>Equally in steel framed buildings it is often cheaper to over-specify the sections used rather than optimise them. </p>
<p>Having said that there are certainly plenty of example of architecture where material limited (eg stone) have produced extrmely elegant structurally optimised designs : </p>
<p><a href="https://i.stack.imgur.com/kOulb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kOulb.jpg" alt="Vaulted Ceiling"></a></p>
|
18349
|
Why we don't use fillets while making buildings?
|
2017-12-05T20:35:01.053
|
<p>I am dealing with a setup similar to the image below. The horizontal plate is bolted to a horizontal concrete surface (Bridge Abutment). And a horizontal force is being applied to the the vertical. My question is how do you determine how thick the stiffener plate needs to be. I am looking to understand the general process and then I will turn around and try to apply my code (CHBDC).</p>
<p><a href="https://i.stack.imgur.com/arjvr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/arjvr.png" alt="Steel stiffener"></a></p>
<p>My thoughts are you need to check to see if it will buckle. Treat it like a column and performa k*L/r check. k=2 for fixed fixed connection due to welds. not sure about section properties as they vary. Also not how to convert a horizontal point load in the middle of the plate into axial load? in the plate.</p>
<p>Was looking to stay out of FEA. Wondering if there was some basic first principles formula or something along those lines.</p>
<p>I am actually trying to size the green peices:</p>
<p><a href="https://i.stack.imgur.com/o3V5x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o3V5x.png" alt="double stiffeners"></a></p>
|
|structural-engineering|structural-analysis|
|
<p>This is pretty much the job of a structural engineer but... <a href="https://www.internationaljournalssrg.org/IJCE/2016/Volume3-Issue12/IJCE-V3I12P102.pdf" rel="nofollow noreferrer">https://www.internationaljournalssrg.org/IJCE/2016/Volume3-Issue12/IJCE-V3I12P102.pdf</a></p>
|
18351
|
How to calculate the required thickness of a steel stiffener plate in a corbel/bracket
|
2017-12-06T01:56:23.627
|
<p>I have been wondering about how a positively charged plasma might be used to generate electrical power. So, I was wondering if, similar to a moving negative charge, a moving (or oscillating) positive charge would generate a voltage across a coil.</p>
|
|electrical-engineering|
|
<blockquote>Does a moving positive charge generate a magnetic field?</blockquote>
<p>Yes. The real question is why would you think it wouldn't. Maxwell's equations are clear and have stood the test of time.</p>
<p>The polarity of the charge only effects the polarity of the resulting magnetic field.</p>
|
18355
|
Does a moving positive charge generate a magnetic field?
|
2017-12-06T03:54:01.437
|
<p>In order to conduct a study, I need to place a rectangular plate freely placed (1 cm by 2cm) inside water and control it from outside (better wireless). I need to give oscillatory motion with specific frequency/amplitude.
At present I used LEGO Next servo motors and I could not achieve it successfully due to mechanical instability and methods to control freq and amp. Also the gear assemblies used affected the results.
May I ask people here to suggest a way to produce vibration motion of this plate with minimal interfering mechanical set up?
I can choose any kind of plate (plastic or metal) I can employ magnetic or electric methods to produce vibration.
any direction way forward will be appreciated.</p>
|
|mechanical-engineering|fluid-mechanics|
|
<p>Professor Wiciak has done some similar researches using <a href="http://acoustics.ippt.gov.pl/index.php/aa/article/view/1419" rel="nofollow noreferrer">piezoelements</a> to induce vibrations.</p>
|
18357
|
producing vibration motion of a plate (1cmx2cm) placed in water
|
2017-12-06T18:34:48.007
|
<p>western United States.</p>
<h2>Problem / Question</h2>
<p>The bad power in my house is somehow producing EMI in my audio equipment. Will a simple Faraday cage help? (Using an <a href="https://www.homedepot.com/p/Andersen-20-11-16-in-x-55-13-32-in-Stone-Aluminum-Casement-Insect-Screen-C5-1344020/100655642" rel="nofollow noreferrer">aluminum metal screen</a>)
If not that, then what would be the best, most practical, cost-effective solution to blocking this interference.</p>
<ul>
<li><p>As is usual on S.E., if you think I'm wrong about something, please tell me why.</p></li>
<li><p>I'm not inexperienced with engineering / physics, although not a professional. Technical explanations are welcome but not required.</p></li>
</ul>
<h2>Story</h2>
<p>I recently moved into a new rental house. It has old wiring (pre-60s) which is mostly un-grounded and quite degraded. The landlord offered to ground a few of the outlets, but not replace all the wiring. For now, I'm going to have to live with the wiring as-is.</p>
<h2>Why I think this is EMI</h2>
<ul>
<li>Using an amp with a cord as an antenna, the noise behaves exactly like radio wave reception in a TV or a radio. It changes depending on angle, and dissipating somewhat when I block it with my body.</li>
<li>I'm using a new power conditioner with ground noise eliminating circutry.</li>
<li>There was no noise at all, even without the power conditioner, at my last place.</li>
<li>Several other non-audio electrical appliances around the house hum, and sound quite a bit like the noise I hear in the amps.</li>
<li>The produced audio frequencies from the noise are all above 500hz or so. It's my understanding that ground loops are usually low-frequency.</li>
<li>Using other outlets, grounded or not, all have the same issue.</li>
</ul>
<h2>Things I already tried that didn't work</h2>
<ul>
<li>Plugging in to the grounded outlets around the house using an extension cord.</li>
<li>Turning off as many things as possible to find the source of the noise.</li>
<li>Using different circuit breakers, turning off as many as possible.</li>
<li>Using different combination of audio equipment to reduce noise. Using shielded, balanced audio cables greatly reduces the hum. However, that's not possible with everything I own.</li>
<li>Standing in such a way as to reduce the noise. This works somewhat, but is untenable. (this lead me to the Faraday cage idea).</li>
</ul>
|
|electrical-engineering|rf-electronics|power-transmission|
|
<h2>After having the power company fix my mains connection, most of the noise dissapeared</h2>
<p>I finally got around to calling my power company, and had them come out and test my mains connection. They told me that the grounding on it was bad. After a repair team fixed it, I found about 75% of the noise was gone.</p>
<p>Still more noise than past places, (again, I suspect the old wiring) but at least now the noise is at a manageable level.</p>
|
18361
|
Faraday cage to reduce EMI in audio amplifier
|
2017-12-07T08:05:56.607
|
<p>On <a href="https://en.wikipedia.org/wiki/Water_jet_cutter" rel="nofollow noreferrer" title="Water jet cutter">Wikipedia</a> it states that the main benefit of a water jet cutter is that there is no heat-affected zone.</p>
<p>I am wondering about the physics behind this. I have found experiments supporting this, but I cannot produce a theoretical argument. Am I wrong in assuming that by the time it would take to heat up the material, a new influx of cold water keeps the temperature down?</p>
|
|thermodynamics|heat-transfer|cutting|
|
<p>The item to be cut is completely submerged in a large water filled tank. Some as large as 10x10 feet, or more. What little heat that is created by the abrasion, and the pump, is immediately carried away from the kerf. I've had slots cut in Kydex® plastic with no apparent heat increase in the material. </p>
|
18365
|
Why is there no heat-affected zone with water jet cutting?
|
2017-12-07T22:43:49.893
|
<p>Here is how this artist rendered <a href="https://en.wikipedia.org/wiki/Isambard_Kingdom_Brunel" rel="nofollow noreferrer">Brunel's</a> <a href="https://en.wikipedia.org/wiki/Royal_Albert_Bridge" rel="nofollow noreferrer">Saltash Bridge</a>:</p>
<p><a href="https://i.stack.imgur.com/axaDo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/axaDo.jpg" alt="enter image description here"></a></p>
<p>Here is how it appears today. </p>
<p><a href="https://i.stack.imgur.com/0P9km.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0P9km.jpg" alt="enter image description here"></a></p>
<p>Did the appearance change, or was some artistic licence taken in the painting?</p>
<p>My question is: <strong>Did this artist take some licence with the Royal Albert Bridge - or has its appearance changed?</strong></p>
|
|bridges|
|
<p><a href="https://commons.wikimedia.org/wiki/File:Royal_albert_bridge_hist.jpg" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:Royal_albert_bridge_hist.jpg</a></p>
<p>The picture from when the original was being jacked into place shows the lens closed at each end.</p>
<p>The engraving would have been from before the bridge was constructed -- perhaps while the railway corporation was trying to raise money.</p>
|
18371
|
Did this artist take some licence with the Royal Albert Bridge - or has its appearance changed?
|
2017-12-07T23:19:05.033
|
<p>Most air conditioners have three modes: Auto, Sun, Snowflake</p>
<p>What is the difference it terms how it works?</p>
|
|airflow|temperature|electrical|
|
<p><strong>The sun is heating mode</strong>. When the room temperature reaches the set temperature, the air conditioner stops operating until the temperature falls below the set temperature and the starts operating again. When in heating mode, the air conditioner does not cool. This setting is used during cold weather periods, such as in winter.</p>
<p><strong>The snowflake is cooling mode</strong>. When the room temperature reaches the set temperature, the air conditioner stops operating until the temperature rises above the set temperature and starts operating again. When in cooling mode, the air conditioner does not heat the room. This setting is used in hot weather periods, such as summer.</p>
<p><strong>Auto is automatic mode</strong> where the air conditioner can either heat or cool as required. It tries to achieve the set temperature by switching from heat mode to cooling mode automatically. If you don't want to change between heat and cool modes when the seasons change, just set the air conditioner to automatic mode.</p>
|
18372
|
What's the difference between aircon modes Auto, Sun, Snowflake?
|
2017-12-09T15:46:05.237
|
<p>I am researching about ways to continuously determine black spots (size 50+ micron) in our product stream. This stream consists of imperfect spherical polymer beads (size 3 mm) of different materials (ABS, SEBS, etc.) and colors (opaque and transparent). I want to be able to count them as well as characterize them (size). </p>
<p>I have been looking into a couple of options:</p>
<ol>
<li>Visual observation using computer vision</li>
<li>NIR spectroscopy:</li>
</ol>
<p><a href="https://i.stack.imgur.com/3PsWp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3PsWp.jpg" alt="enter image description here"></a></p>
<p>What other options do I have and which one is most effective (preferably also least expensive)?</p>
<p><strong>Update:</strong>
In our production plant we produce at 500 kg/hr and the pellets are transported along a conveyor as a monolayer of around 50 cm width at around 5 cm/s. However i can imagine that is simply too complex to begin with and i am not opposed to having a secondary stream with much lower massflow and then upscale in a later stage or just have labscale setup to check samples taken of the productstream. I am ok with an approximate count as long as it is reproducible. I would also be ok with initially only detecting 100+ micron black spots.</p>
|
|materials|measurements|sensors|optics|spectrometry|
|
<p>The simplest method will be a multi-step approach. The first step will be a defective product detection, that gathers general data about the product stream and quarantines defective products. On a parallel path, a small batch of samples will be also sequestered from the main line to be analyzed continuously for detectable defects. A laboratory analysis could then confirm size and count using digital tools.</p>
<p>Recognizing that these are, for the most part, polarized plastics, they should have a fairly consistent low to modest <a href="https://en.wikipedia.org/wiki/Relative_permittivity" rel="nofollow noreferrer">relative permittivity</a>, double to triple the value of air. Regardless of opacity and/or material, a product line with high amount of defects will be unpolymerized or coked plastics, consisting primarily of carbon, which has a detectable higher relative permittivity, nearly five times the value of even the most polar plastics. A static (ultra low frequency) dielectric detector can detect these defects, and quarantine the offending batch.</p>
<p><a href="https://i.stack.imgur.com/x60N1.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x60N1.gif" alt="enter image description here"></a></p>
<p>As the relative voltage across the plates changes, the increase in stored voltage can be detected as a defect for sequestering.</p>
<p>On a parallel path, a small portion can run through a dielectric spectrometer. These have had some uses before, (See <a href="https://www.osapublishing.org/as/abstract.cfm?uri=as-71-3-456" rel="nofollow noreferrer">here</a> for detection of different polymers based upon response spectra, and <a href="http://ieeexplore.ieee.org/document/4772907/?reload=true" rel="nofollow noreferrer">here</a> where it was used to detect areas of low polymerization in LDPE.) It works the same way as the dielectric detector, but uses an alternating current. This can get a full spectrum analysis for comparison between polymers:</p>
<p><a href="https://i.stack.imgur.com/Wj7W7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wj7W7.jpg" alt="enter image description here"></a></p>
<p>Since this detection method is slower, it will need to be able to analyze samples one at a time. However, this will ensure the main detector is not fluctuating substantially due to contaminants in the air or other problems. It will also review a history of defective compounds. As the library of spectrum is built, this could replace entirely the need to utilize laboratory detection, except for the cases of control chart defects (see below).</p>
<p>Some advantages of this is that it can detect defects inside the beads that would not be visible via normal spectroscopy alone.</p>
<p>Final laboratory counts to measure black spots could then be utilized to verify the automated machines are working within parameters. The 7 tools of quality, particularly the control chart, will be useful as a countermeasure to ensure these detectors are working properly.</p>
<p><a href="https://i.stack.imgur.com/9BfLY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9BfLY.jpg" alt="enter image description here"></a></p>
|
18392
|
How can I detect black spots?
|
2017-12-10T17:32:02.407
|
<blockquote>
<p>Calculate the radius of iridium atom, given that Ir has an FFC crystal
structure, a density of $22.4\text{g/cm}^3$. and atomic weight of
$192.2\text{g/mol}$.</p>
</blockquote>
<p>Everyone whom I asked this question is giving different answers. By the way, I was going to ask something about this. Why are they finding that the atomic radius of iridium atom is $1.34\text{pm}, 1.35\text{pm}, \text{1.36pm}$? According to textbook solution, it seems $1.36\text{pm}$. However, they said that all answers correct. Thereupon, I wanted to check their answer from wikipedia, that gives anohter answer too. What would you say? Are all answers correct? </p>
<p>Kindest Regards!</p>
|
|materials|chemical-engineering|chemistry|
|
<p>The radius $r$ can be calculated purely from geometry. Consider the FCC lattice. Its lattice parameter $a$ is related to radius through the arrangement of atoms. What is the close-packed direction in the FCC lattice? Along the face diagonal, because two corner atoms touch the face-centered atom. The length of the face diagonal is $\sqrt{a}$ by the pythagorean theorem. That length is also four $r$. You can draw the lattice and the atoms of a face to convince yourself this is true. So $4r=\sqrt{a}$.</p>
<p>Now it remains to calculate the lattice parameter. How many atoms per unit cell? Again you can draw the lattice and atoms of the unit cell to convince yourself there are 4 atoms per unit cell (remember corners count for 1/8 atoms and faces 1/2). How many unit cells per mole? Well there are $6.022\times 10^{23}$ atoms per mol, divided by atoms per unit cell gives unit cells per mole. How big is the unit cell? Of course $a^3$ cm cubed per unit cell. Multiply by unit cells per mole to get cm cubed per mole. Now multiply by density (grams per cm cubed) to get grams per mole. Since those are units of atomic weight we can set what we have equal to the value given. Solve for $a$, then substitute into the earlier expression to obtain $r$.</p>
<p>The probable reason there are different values is that different methods may have been used to calculate the atomic radius. What we just did is one method, and accurate enough for bulk calculation purposes in an engineering context. Another method is to use Density Functional Theory to model the electrostatic force given lattice parameters and electron orbitals and other properties. It is also possible that the values are for Ir present in various alloys or compounds, or at different temperatures. Atomic radii vary for the same atom when bonded to others, and for different temperatures due to thermal expansion.</p>
|
18401
|
Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of 22.4 g/cm3 and atomic weight of 192.2g/mol
|
2017-12-11T09:57:12.473
|
<p>There are cars with petrol engines having 1,0 l or even less engine displacement but I have never seen a car with diesel engine having less than 1,5 l. Are there technical reasons that diesel engine cannot have less than about 1,5 l? </p>
|
|diesel|
|
<p>Ignition lag in diesels makes it difficult to run them over a range of rpms that peak out above about 4500 rpm. You can tinker with them to run just at higher rpms, but not to run over the wide range of rpms required of a vehicle diesel. The little 1 liter gas jobs can be perfectly driveable at 7000 rpms or more. Hence they don't need to have as large a displacement.</p>
<p>Honda's new little <a href="https://www.topgear.com/car-reviews/honda/civic-type-r/16-i-dtec-sport-5dr/spec" rel="nofollow noreferrer">aluminum diesel</a> apparently tops out at 4000 rpm.</p>
|
18408
|
Why do diesel engines tend to have larger engine displacements?
|
2017-12-11T11:29:42.863
|
<p>The biggest diameter of my pressure sensor input port has a size of 1.93 mm. The silicon hose which I want to connect with the sensor's port has an inner diameter size of 2.0 mm.</p>
<p>My honeywell pressure sensor: <a href="https://i.stack.imgur.com/qPYAa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qPYAa.jpg" alt="Honeywell pressure sensor"></a></p>
<p>My silicon hose: <a href="https://i.stack.imgur.com/aPcwr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aPcwr.jpg" alt="enter image description here"></a></p>
<p>As you can imagine - it does not fit. The port is to small or the inner diameter of the silicon hose to large.</p>
<p>What is the best way to bring both together? Are there adapters availabe? What is the name therefore? What is a good way in engineering to fix such an issue?</p>
<p>Thanks in advance for your help!</p>
|
|mechanical-engineering|pressure|sensors|pneumatic|
|
<p>The fittings on your pressure sensor look like barb fittings. If that's the case, a 1.6mm ID silicone tubing will easily stretch over a 1.93mm barb fitting. 1.6mm (or 1/16") tubing is a fairly common size and you should be able to get it from standard laboratory suppliers.</p>
<p>You'll then need an adapter to go from your 2mm ID to 1.6mm ID silicone tubing. Again, 2.4mm (3/32") is a fairly common size and a 2mm ID silicone tubing will easily stretch over a 2.4mm barb fitting. So something like AD-6005 from <a href="http://www.nordsonmedical.com/" rel="nofollow noreferrer">Nordson Medical</a> or equivalent should do it:</p>
<p><a href="https://i.stack.imgur.com/St0S5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/St0S5.jpg" alt="enter image description here"></a></p>
|
18411
|
How can I connect a silicon hose with 2mm inner diameter to a smaller 1,9mm port
|
2017-12-11T18:20:54.707
|
<p>Define "production tooling"?
What are its advantages?
And why production tooling is essential for manufacturing?</p>
|
|mechanical-engineering|
|
<p>If you are to make one part or only samples, you would have prototype tooling. </p>
<p>If you are to mass produce a part(high volume), then you have production tooling. Depending on your budget and quantity to be mass produced, it will drive the design of the production tooling. </p>
<p>The considerations for production tooling are as follows but not limited to:</p>
<ol>
<li>Quantity to be made before service is required</li>
<li>Setup</li>
<li>Serviceability</li>
<li>Mistake proofing/human factors</li>
<li>Safety</li>
<li>List item</li>
<li>Quality</li>
<li>Process Monitoring/inspection system</li>
<li>Speed of production</li>
<li>Budget</li>
</ol>
<p>Your question is vague, so I don't know what you are to manufacture, but apply the above points. </p>
<p>Production Tooling is the tools used to mass
Produce a product. Production tooling considers the above points when being designed and manufactured. </p>
|
18415
|
What is production tooling?
|
2017-12-12T21:23:23.573
|
<p>I have seen this kind of mark on various cast metal parts. This is an image of the base casting of a hard disk drive made by HGST.</p>
<p><a href="https://i.stack.imgur.com/UzKGi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UzKGi.jpg" alt="Markings on an HGST HDD base casting"></a></p>
<ul>
<li>What are these marks for?</li>
<li>What are they called?</li>
</ul>
|
|manufacturing-engineering|metals|casting|molding|
|
<p>these are part-tracking marks. they indicate which cavity in the mold the part was cast from, in which mold set, in what molding machine, in what year, month, day and shift. this way, a part that fails its finished goods inspection or failed in the field can be tracked backwards through the factory to its source, and the reason for the failure fixed.</p>
|
18441
|
Purpose and name of fine pitch marks on a cast metal part
|
2017-12-13T00:48:03.733
|
<p>I'm relatively new to hydraulics, and I'm trying to understand this schematic. I've got the majority of the system figured out, but I cant for the life of me understand what the symbols at positions 18/20/23 represent.</p>
<p>The unit actively controls a constant pressure in the two cylinders attached using an electronic pressure switch. Its the control unit of a hydraulic tensioning system in a ski lift. My feeling is that 18/20 may have something to do with electrical pressure cells?</p>
<p><a href="https://i.stack.imgur.com/WNch1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WNch1.png" alt="enter image description here"></a></p>
<p>Thanks</p>
|
|mechanical-engineering|hydraulics|instrumentation|
|
<p>For future reference, 20 is a sharp edged orifice.</p>
|
18444
|
Need help identifying component on hydraulic schematic
|
2017-12-13T00:53:03.557
|
<p>This question got into my mind when I am cleaning my motorcycle.</p>
<p>If spring shock-absorbers like the one below absorbs shock from road imperfections, reducing strain in the motorcycle body as your drive, then why does car manufacturers use rigid bumper for cars instead of semi-rigid or flexible bumpers?</p>
<p>Isn't it good for the car that its bumper dissipates some energy in collision so that the time of impact increases, therefore reducing impact force? What are the pro's and con's of this idea?</p>
<p><a href="https://i.stack.imgur.com/21i8h.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/21i8h.jpg" alt="shock absorber"></a></p>
|
|mechanisms|
|
<p>The difference is the amount of crash energy dissipated and limited play course of a spring loaded dumper.
The energy absorbed by a spring is the area under force line:
1/2 k.x^2.m. - m is the bumper mass.
which is very small compared to inertia of the car. And at the end of it's course it delivers the huge leftovers of impact energy to the rest of the car while now adding it's own mass to attacking car's inertia.</p>
<p>The idea behind integrated crash structural design is a highly energy absorbing gradual and ductile crumpling of the entire structure surrounding the occupants as a scrifice zone while reducing the negative acceleration to tolerable levels. No loose part is going to be of any help.
If you look at the corrogated pressed patterns on the body frame of the new cars you see they have been designed to absorb much energy in a crash while crumpling away from the cab of the car. You will have the entire front of the car's mass and length to help dampen the impact.</p>
|
18445
|
Rigid bumpers vs shock absorber bumpers
|
2017-12-13T11:21:08.667
|
<p>I really know that the real calulation of the range of UMTS, GSM or LTE antennas is depending on a lot of issues, reflection, and other stuff which makes it really complex. </p>
<p>I need only a rough model which I can calculate the range of them, so I thought I use the free-space path loss formula and look for the "r" in the formula. Is there anything wrong with my idea? Despite the fact that of course in really you'll almost never have just a free space but rather buildings,hills and other stuff but lets say, for my model I say there is nothing that disturbs my signal.....</p>
<p>If I can use the formula, my question would be so the only difference of those systems would be the frequency and the transmitted power or received power of the mobile phones, is that correct?</p>
<p><a href="https://en.wikipedia.org/wiki/Free-space_path_loss" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Free-space_path_loss</a></p>
|
|electrical-engineering|radio|
|
<p>The usual $ r^{-4} $ round-trip link loss is based on the antenna being a point radiator, or at least so far from the target as to appear almost like a point radiator. The full formula has all sorts of correction factors (aka "fudge factors") for the actual beam profile, and targets are represented as a standard shape of equivalent total reflectance, e.g. a sphere or a flat plate mounted perpendicular to the propagation axis. </p>
<p>That should suffice for a rough approximation of the link loss.</p>
|
18448
|
Calculating the link budget of antenna and receiver with free space-path-loss formula
|
2017-12-13T15:56:32.410
|
<p>I'm in a bit of a pickle with one of the quality techns that works for us (the company) over a profile of surface tolerance. I initially submitted something resembling figure 1 (top capture) and quality came back to me saying figure 2 (bottom capture) would be better.</p>
<p>I've never seen ''auto-referencing'' of datum and I don't see the point because it doesn't add anything to the GD&T. I've never seen anything like this in any book, reference or drawing I stumbled upon over my (short) carreer. Are there any arguments to which practice would be best? I'm open to any source saying one way should be the preferred one.</p>
<p>Thanks!</p>
<p><a href="https://i.stack.imgur.com/4skyG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4skyG.png" alt="What I'm proposing"></a></p>
<p><a href="https://i.stack.imgur.com/zZtt3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zZtt3.png" alt="What quality guy suggests"></a></p>
|
|mechanical-engineering|
|
<p>Don't do it. It's probably not legal in any GD&T spec. </p>
<p>You can't reference a feature back to itself(think: chicken or egg?). </p>
<p>Your first version is more appropriate, but not ideal. How do you plan to inspect the position of the first hole(datum b)? I see what you're trying to accomplish(intent), but it should be clear from an inspection stand point. I think you need to remove the tolerance of position for that datum B hole and give datum C twice the tolerance...</p>
<p>Btw, do your holes sizes have a tolerance or fall back on a general note?</p>
|
18450
|
Drawing best practice and datum self-referencing
|
2017-12-14T21:39:56.063
|
<p>The water held underground in a geothermal system is under high temperature and pressure. Are pumps used to raise the very hot underground water to the surface or does the water rise under its own pressure? One source says that the water is pumped up: <a href="https://energy.gov/eere/geothermal/electricity-generation" rel="nofollow noreferrer">https://energy.gov/eere/geothermal/electricity-generation</a> while another says that it rises on its own: <a href="https://www.youtube.com/watch?v=kjpp2MQffnw" rel="nofollow noreferrer">https://www.youtube.com/watch?v=kjpp2MQffnw</a> I am not sure if I am just misunderstanding the wording. Does anyone know which is correct?</p>
|
|pressure|heat-transfer|pumps|geotechnical-engineering|
|
<p>They are both correct. There are actually a number of different types of geothermal power plant with the most common being, dry steam, flash steam, and binary cycle. </p>
<p>Dry steam plants require pumps to draw from underground resources of steam. This is piped directly from underground wells to the power plant. </p>
<p>Flash steam plants (most common) use geothermal reservoirs that have naturally high temperatures (180 Celsius +). The hot water from the reservoir flows up through a well under its own pressure. As it reaches the surface the pressure decreases and the water begins to boil and turn in to steam. Any water left over from this process is then pumped back in to the reservoir. </p>
<p>Binary cycle power plants are more complex but are capable of operating at much lower temperatures (100-180 Celsius). These use hot water pumped from an underground reservoir to heat an organic fluid with a low boiling point. This organic fluid is vaporised in a heat exchanger which turns a turbine. The leftover water is "injected" back in to the reservoir. </p>
<p>If you want to read up more on the specific types of geothermal plant, this website has more information: <a href="https://www.conserve-energy-future.com/geothermalpowerplanttypes.php" rel="noreferrer">https://www.conserve-energy-future.com/geothermalpowerplanttypes.php</a> </p>
<p>Hope that helps!</p>
|
18469
|
Are pumps used in geothermal systems
|
2017-12-15T13:07:23.373
|
<p>I'm trying to derive the first order response of a sensor with step input. However, I'm stuck at step just before conducting the Laplace transform. In the following image, how is the left side equal to the right side? I know the right side equation is necessary to obtain to perform Laplace transform. <a href="https://i.stack.imgur.com/faNZG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/faNZG.png" alt="enter image description here"></a></p>
|
|mechanical-engineering|sensors|
|
<p>It's simple maths. Starting from the RHS to get to the LHS:</p>
<p>$$
AK \left[ \frac{1}{s} - \frac{1}{s - \frac{-1}{\tau}} \right]
= AK \left[ \frac{1}{s} - \frac{1}{s + \frac{1}{\tau}} \right]
= AK \left[ \frac{1}{s} - \frac{\tau}{\tau s +1} \right]
= AK \left[ \frac{\tau s + 1 - \tau s}{s(\tau s +1)} \right]
= \frac{AK}{s(\tau s + 1)}
$$</p>
|
18474
|
Deriving the First order response of a sensor
|
2017-12-17T00:52:42.527
|
<p>Symmetrical truss is given:</p>
<p><a href="https://i.stack.imgur.com/7DVfu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7DVfu.jpg" alt="enter image description here"></a></p>
<p>I have issue with the sign convention for joint H.</p>
<p>See below the solution for joint I. As you can see, when calculating equilibrium in y (vertical) direction, forces that act upwards are written with "+" sign, and forces acting downwards with "-" negative sign.</p>
<p><img src="https://i.stack.imgur.com/tly7C.jpg" alt=""></p>
<p>Yet for joint H below, forces acting downwards are written as positive, which is why $F_{HJ}$ turns out to be a negative force (compression force). But it's not consistent with the solution for previous joint.</p>
<p><img src="https://i.stack.imgur.com/8glFL.jpg" alt=""></p>
<p>How would you go about finding force in member HJ?</p>
|
|civil-engineering|structural-analysis|
|
<p>The answer presented is not inconsistent, but it is badly laid out.</p>
<p>They have not visibly defined "upwards as positive" (they should have...). The equation that's confusing you is:</p>
<p>$$\sum F_y = 0; F_{HJ}\cos(45) + 100\sin(75) + 273.2\cos(45) = 0$$</p>
<p>In this instance, every single force on the Free Body Diagram in the 'y' direction is pointing downwards, so, to follow the "upwards is potitive" convention, we could write:</p>
<p>$$ -F_{HJ}\cos(45) - 100\sin(75) - 273.2\cos(45) = 0$$</p>
<p>Which, I'm sure you'll agree, is equivalent to the originally presented formula.</p>
<p>When answering questions like this, you should have always define the positive directions by an annotation with the FBD, and show any extra steps (such as multiplying an equation through by '-1'), to prevent confusion for casual readers.</p>
<p>Note further, that a 'trailing minus' at the start of an equation can get lost, and is often the root of a sign-error; what they did made sense, but they should have made it clearer.</p>
|
18489
|
Structural Analyses of Trusses, method of joints
|
2017-12-17T12:20:40.443
|
<p>A component of a control loop is approximated by the following relation between input $x$, and output, $y$:</p>
<p>$y = 5x^2$</p>
<p>During normal operation, the value of input to this component ranges between $1$ and $1.5$. In
stability analysis of the overall control loop, what is the gain of this component?</p>
<p>So, I'm not sure how to solve this problem. I don't know how to create a transfer function by performing a Laplace transform because this isn't a differential equation and there is only one component. Would really appreciate it if someone could point me in the right direction.</p>
|
|pid-control|
|
<p>The input and the output of the component need to have a linear relation (such as y = 5 * x), so you have to linearize that ecuation in order to aproximate it to a linear ecuation that you could use to obtain the gain of the component.</p>
<p>To linearize :</p>
<p>If x = a; </p>
<p>y = (f(a) + f’(a) * (x - a))</p>
<p>So if you take a = 1 (one of the values you gave)</p>
<p>y = 5 + 10 * (x - 1)
y = 10 * x - 5</p>
<p>So you could use a gain of 10 and add a sumator with a minus for substracting 5. </p>
<p>I hope this helps you.</p>
|
18490
|
PID Gain Calculation
|
2017-12-18T09:54:01.530
|
<p>I am searching for information on how finite element code (abaqus /ansys) could be used to model the contact stress behaviour between 2 bearing surfaces?</p>
<p>The key point is that there is an additional hard coating n both surfaces so as to reduce wear.</p>
<p>How would one model this additional hard layer?</p>
<p>Would this hard layer be represented by an additional layer of elements with a higher elastic modulus?</p>
|
|mechanical-engineering|structural-engineering|materials|gears|structural-analysis|
|
<p>I would suggest a two step process:</p>
<p>1) Do a high resolution simulation (with four layers) of a small region to estimate the effective contact stiffness of the system.</p>
<p>2) Do a two surface simulation with the contact stiffness from step 1.</p>
<p>Doing a high resolution simulation of the whole system will be computationally expensive, and not suggested unless you have access to parallel computing. </p>
<p>The reason is that the thin layer will have to be discretized using at least three elements through the thickness; leading to millions (if not billions) of elements for the entire assembly.</p>
|
18500
|
Contact stress of bearing surfaces
|
2017-12-19T10:00:46.237
|
<p>I want to calculate the deformation of the following structure:
<a href="https://i.stack.imgur.com/ynvss.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ynvss.png" alt="enter image description here"></a></p>
<p>I have numbered the vertical beam 1 and the horzontal beam 2. The hight of beam 2 is given by:
$h(x)=h\sqrt{1-x/L_2}$</p>
<p>So I thought I could solve this with Castigliano's method. Where beam 1 undergoes an axial load $F$ and a constant bending moment of $Fa$. Beam 2 undergoes a bending moment of $0$ on the left and $Fa$ on the right. The energy is given by: $U=\int_0^L\frac{F(x)^2}{2AE}dx+\int_0^L\frac{M(x)^2}{2EI}dx$
So for beam 1 and 2 the energy would be:</p>
<p>$$ U_1 = \int_0^{L_1}\frac{F^2}{2bhE}dx+\int_0^{L_1}\frac{F^2a^2}{2E\frac{1}{12}bh^3}dx = \frac{F^2L_1}{2bhE} + \frac{F^2a^2L_1}{2E\frac{1}{12}bh^3} $$</p>
<p>$$U_2 = \int_0^a\frac{F^2x^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3}dx =
\frac{4F^2L_2^2}{Ebh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right) $$</p>
<p>Where $b$ is the thickness of the part</p>
<p>Then I applied Castigliano's method, which gives the deformation of the shape in the same direction as $F$:</p>
<p>$$\delta=\frac{\partial U}{\partial F}= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2^2}{bh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right)\right) $$</p>
<p>But this equation gives me wrong results. For instance, with $F=1500, E=113.76*10^9, b=0.005, h=0.007, L_1=0.05, L_2=0.04$ and $a=0.032$
I get a deformation of 9.8 mm, where as a FEA simulation (Nastran In-Cad) gives a deformation of 6.2 mm (in the same direction as the force).</p>
<p>I tried this without the parabolic property and this does generate correct results. That equation is:</p>
<p>$$\frac{F}{E}\left(\frac{L_1}{A_1}+\frac{L_2^2L_1}{I_1}+\frac{L_2^3}{3I_3}\right)$$</p>
<p>So I think there something wrong with my parabolic equation, but I can't figure out what it is. </p>
<p>Edit: did some trial and error, and if I fill in $h(h/2)$ for $h$ in the last term I get pretty accurate results. To clarify, the height used (in this situation) in the last therm ($\frac{8L_2^2}{bh^3}$) is the hight of beam 2 where beam 1 ends, on the left. But I still have know idea why this is and if this works for every situation</p>
|
|mechanical-engineering|beam|
|
<p>Thanks to <a href="https://engineering.stackexchange.com/questions/18510/deformation-of-two-perpendicular-beams-with-parabolic-properties/18543#comment34150_18510">Biswajit</a> for his comment, which pointed out my mistake. My coordinates where wrong. The correct equation for $U_2$ is:</p>
<p>$$U_2 = \int_0^a\frac{F^2(x-a)^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3} dx$$</p>
<p>So the correct equation for the deformation is:
$$ \delta= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2}{bh^3}\left(-3a^2+12aL_2+8L_2 \sqrt{1-\frac{a}{L_2}}(L_2-a)-8L_2^2\right)\right) $$</p>
|
18510
|
Deformation of two perpendicular beams with parabolic properties
|
2017-12-20T05:57:19.473
|
<p>At my workplace I have been assigned to create an accurate 3D model of our Dell PowerEdge R820 servers so we can insert our custom cards and run a CFD and thermal analysis on them. The end goal is to have an exact understanding off all the airflow and heat transfer within the servers and server racks to help in designing future systems and fix overheating issues we are having with some of our current ones.</p>
<p>My problem is that the insides of these servers are quite complex and it is going to take an era for me to reverse engineer all the parts with a pair of verniers and find out how much heat is produced and where.So I am wondering if anyone else has come across something like this before and how they created an accurate CFD of their system. </p>
<p>Would 3D scanning be an option? I am concerned that 3D scanning will be too accurate, in the sense that what should be a straight edge might be turned into 200 different vertices etc. And I am not sure of how much cleanup it would require. I plan to do the CFD in Solidworks which requires solid modelling - it may have issues with the surfacing of the scans. Would like to hear from anyone with experience regarding this.</p>
<p>On the other hand, does anyone have experience with getting 3D CAD models from companies like Dell? I work at a large company and we buy a lot of their products for our projects but I wouldn't have a clue what department to call to see what needs to be organised to get CAD files. Would welcome any input here.</p>
<p>Lastly, for the thermal modelling, my plan of attack is to use some IR cameras to see where the heat is being produced. But I don't know how to work out HOW much heat is being produced at these spots so I can model the heat production from these processors in the CFD. What do people normally do for this?</p>
|
|cad|cfd|solidworks|
|
<p>Your options are a bit limited. Either you get the model from Dell, or you model it yourself. </p>
<p>Getting the model is a long shot, but if you are a really big customer that might happen. However this is not necceserily a godsend, since:</p>
<ul>
<li>The model may not be accurate enough for your need. They may have ripped some data out to protect their IP</li>
<li>It may be too accurate in places that are of no interest for you. And mostly the data is just in your way.</li>
<li>The model may be unsuitable for your task. So its geometry but your cfd needs the material info and heat generation info.</li>
</ul>
<p>That said modeling a suitably absteacted version that simulates faster than imported models is not necceserily a huge job. Its just very scary, as it is just raw work.</p>
<p>Would scanning help? In my experience not really, though it might speed up data gathering, as you now have a digital model to check against. So you do not need to go and measure that one dimension you realized you need. Thats a good thing if you can not have the object on the table next to you. However, scanning is another can of worms as a work process.</p>
<p>Concentrate on abstracting things as much as you can. Do you really need every detail?</p>
<p>Also note thermal camera does not directly neasure what you want it measures how much things radiate at a specific band which varies by material and its surface quality. worse some materials like the silicon in your chips are transparent at typical infrared camera ranges. You still need to know how much the materials transfer energy to air. That said the thermal camera is a good validator target for your model.</p>
<p>This is a lot of work no matter how you look at it.</p>
|
18517
|
How to model the inside of a server for CFD and thermal studies?
|
2017-12-21T16:12:23.943
|
<p>Are there some general guidelines for approximating the plastic modulus of steel under tension so that a bilinear stress strain curve can be defined for Loading beyond the elastic region?</p>
|
|structural-engineering|materials|structural-analysis|applied-mechanics|
|
<p><a href="https://archive.org/details/en.1993.1.5.2006" rel="nofollow noreferrer">Eurocode 1993-1-5 Design of Steel Structures</a> - Plated structural elements, <em>Annex C</em> Clause <em>C.6 Material properties</em> allows the following assumptions:</p>
<p>a) elastic-plastic without strain hardening <em>(i.e. elastic-perfectly plastic)</em></p>
<p>b) elastic-plastic with a nominal plateau slope <em>(this uses a very small young's modulus post-yield, E/10000, to help with convergence difficulties experienced in finite element analysis)</em></p>
<p>c) elastic-plastic with linear strain hardening <em>(this uses E/100 to model the post-yield young's modulus)</em></p>
<p>d) true stress-strain curve modified from test results</p>
<p><a href="https://i.stack.imgur.com/nRWy3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nRWy3.png" alt="Figure C.2 showing options a to d as indicative stress-strain curves"></a></p>
<p>Option c would be applicable for your situation: beyond the elastic region you can approximate the Young's Modulus as 1/100th of the elastic Young's Modulus.</p>
|
18536
|
Plastic modulus
|
2017-12-22T08:52:01.853
|
<p>The core challenge of building a tank is deciding where to focus its armor and how much speed / cost you're willing to trade for protection, so this got me thinking, with modern detection technology, like laser, could we potentially place a single very thick plate of armor on an axle and once a projectile is detected, rotate it (around the outside of the tank) in front of the projectile in time to stop it (assuming it isn't a close quarter engagement).</p>
<p>If so, tanks could stay much lighter and much cheaper and still have all the benefits of extremely tough armor, and more importantly, have that same protection on all sides of the tank rather than just the front.</p>
<p>Question is: Is that physically possible? Is it physically possible with known modern technology to detect an incoming projectile, then rotate a wheel with enough torque and speed to intercept the projectile's path before it passes?</p>
<p>Factors I know:</p>
<ul>
<li>Fastest projectile should be traveling at 2000 m/s. </li>
<li>Radius to be traveled around the sides of an average tank is 15 meters including the gun barrel.</li>
<li>A 0.5 x 0.5 meter steel plate with a thickness of 120 mm will weigh 246 kg.</li>
<li>Processing / detection time isn't relevant. The time between scanning the incoming projectile (twice, to calculate angle and speed) and sending the correct information to the armor acceleration mechanism is easily less than 1ms, during which the projectile has only traveled 2 meters since detection.</li>
</ul>
<p>What I don't know:</p>
<ul>
<li><p>Assuming we use the best viable method known, how fast can that 246kg plate be accelerated to the correct location? </p></li>
<li><p>Would it be better if the plate was kept in constant motion and simply adjusted once the projectile is detected so it doesn't need to be accelerated from a zero energy state.? </p></li>
</ul>
<p>Other factors to consider:</p>
<ul>
<li>The armor doesn't <em>have</em> to be stopped right away. It can be accelerated to exactly the correct speed to be in front of the projectile at the calculated time and place of collision, then freely continue swinging around the tank and slow down gradually. </li>
<li>There might need to be 2 equally sized plates on opposite ends of the tank to counter each other so the tank can remain stable - so potentially twice the weight. </li>
</ul>
|
|sensors|
|
<blockquote>
<p>Assuming we use the best viable method known, how fast can that 246kg
plate be accelerated?</p>
</blockquote>
<p>It's not just a plate to accelerate, it's a <a href="https://en.wikipedia.org/wiki/Moment_of_inertia" rel="nofollow noreferrer">moment of inertia</a> to fight and it depends on the position of the plate regarding the center of rotation, its geometry, thus depends on the design of the tank, then it depends on the acceleration curve of the motor and the initial position of the plate, etc.…</p>
<p>I think there is no engineer answer to your question, too many guesses and unsupported assumptions need to be made on too many variables without access to (classified) accurate data.</p>
<p>Like :</p>
<ul>
<li>"<strong>How far a laser can detect an incoming projectile accurately</strong>" : "accurately" needs to be defined, what is your maximum acceptable error of position ? Then, is it by clear weather or in the dust/storm/etc. ?</li>
<li>"<strong>What's the fastest scan rate that can be achieved by the laser detection system? We need to read 2 locations to determine the projectile's attack vector.</strong>" : you certainly need to use satellites here, since the tank might be blinded by mountains etc. Thus, you need to take into account the latency of the transmission, the risk of data loss, etc. And you need at least 3 readings to achieve proper accuracy in a short amount of time. After that, you need to achieve a third-order accuracy reading on the missile acceleration in order to get a first-order accuracy on the position of the impact (since it's a double integral). As you see, it's not just sensing and measuring, it's <a href="https://en.wikipedia.org/wiki/Numerical_methods_for_ordinary_differential_equations" rel="nofollow noreferrer">differential equations</a> to solve in real-time, thus the numeric overhead is significant.</li>
<li>"<strong>The armor doesn't have to be stopped right away. It can be accelerated to exactly the correct speed to be in front of the projectile at the calculated time and place of collision, then freely continue swinging around the tank and slow down gradually.</strong>" : actually, using a moving armor is quite clever and might make good use of the conservation of the <a href="https://en.wikipedia.org/wiki/Angular_momentum" rel="nofollow noreferrer">angular momentum</a>. However, the very aspect of the automation control is a challenge here. Assuming you have the perfect motor (actuator) to do so, ensuring a proper control to achieve precision on position, speed and acceleration based on an acceleration and position input needs at least a 3rd-order accurate controler with a very fast numeric solver. Since your oversized motor won't probably have a linear response, you will have to correct that as well. So it's a lot of math to process with a lot of measures to gather and a lot of errors to handle in a short amount of time, and by the time you get the parameters to input in your motor controler, you might just be dead already.</li>
</ul>
<p>Yes, physically, this is just basic newtonian mechanics and tennis playing with missiles, and it is possible in theory. </p>
<p>However, there is no definitive answer to the feasibility aspect until someone does it, and this is more sci-fi than engineering.</p>
|
18542
|
Would it be physically possible for rotating tank armor to move in front of an incoming projectile fast enough to intercept?
|
2017-12-22T23:09:52.120
|
<p>The part is a clevis and was insterted in a steering rack . Part was in Tension, torsion and bending. It failed at the end of the threaded bolt part. Is it a common problem for bolted joints ? How can it be resolved ?</p>
<p>At first a bend was observed in the threaded part in the direction in which it is kept in the picture before failure. </p>
<p><a href="https://i.stack.imgur.com/XduOE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XduOE.jpg" alt="enter image description here"></a><a href="https://i.stack.imgur.com/10OcP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/10OcP.jpg" alt="enter image description here"></a><a href="https://i.stack.imgur.com/mbb8E.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mbb8E.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/8UPOs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8UPOs.jpg" alt="enter image description here"></a></p>
|
|materials|failure-analysis|
|
<p>I agree , reversed bending fatigue, no torsion, likely sharp thread roots, maybe a loose nut. But the surface looks granular, not smooth as typical of fatigue. That suggests very low strength steel. So, radius the thread roots and be sure the nut is properly torqued and check if the hardness meets spec.</p>
|
18551
|
What type of failure is this?
|
2017-12-23T21:44:17.103
|
<p>This is a clevis made for linking the tie-rods to the steering rack. This part is under dynamic loading and is getting loosened. We tried using jam nuts but it did not work well. Due to loosening the thread came out of the rack and the part fractured maybe due to bending fatigue. </p>
<p><a href="https://i.stack.imgur.com/NXnQs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NXnQs.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/iZ8Sy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iZ8Sy.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/cdFGY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cdFGY.jpg" alt="enter image description here"></a></p>
|
|mechanical-engineering|mechanical-failure|
|
<p>to prevent the nut from unscrewing, you must use either 1) a thread-locking compound (Loctite or equivalent) on the threads during assembly, 2) split lock washers, 3) an aircraft-grade castellated nut-and-safety-wire assembly, or 4) a nut with a nylon insert to prevent unscrewing (Nylock or equivalent).</p>
|
18556
|
How to prevent screw from loosening?
|
2017-12-25T10:29:43.173
|
<p>i am designing a new car and planning to use an existing engine, my field is aerospace engineering so i have a good knowledge in machine design and thermodynamics, at first where can i find the technical specs of an engine other than disassembling and figuring out needed information? secondly, what are the good software that are suitable for individual?</p>
<p>edit: looks like some people don't understand what is written above, the question is:</p>
<p>where do i obtain specifications, data, blueprints, whatever of an engine, for example GM's LS3 or Honda K20?</p>
<p>know one of the users below told me to purchase a workshop manual, i will appreciate any additional info.</p>
<p>thank you for your time.</p>
|
|design|automotive-engineering|car|
|
<p>Design a stroker kit - first pick an engine where there is room to change the distance the piston travels - most engines are now designed with interference ie the piston and the valves occupy the same space but at different times.</p>
|
18568
|
how can i improve a current engine by redesigning internals such as pistons , ...etc?
|
2017-12-26T11:31:58.617
|
<p>I have a system that has the following transfer function:</p>
<p>$G(s) = \frac{1}{s^2(s^2+4+1)}$</p>
<p>As can be seen it is a 4th order system.
This is the bode plot of the system:
<a href="https://i.stack.imgur.com/HWMlb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HWMlb.png" alt="bode plot of plant"></a></p>
<p>I need to use a lead compensator in order to generate sufficient phase margin. If I use a tame PDD controller I am limiting my crossover frequency around the position of the green arrow, as the double derivative action can theoretically only provide 180 phase lead. However if I use a tame PDDD action I can place my crossover frequency in the vicinity of the orange arrow. This also means that I can make my system infinitely fast, if there is no limit on control action...</p>
<p>My question is: is it realistic to use a triple tame derivative action? I can't find a similar controller anywhere on the internet!</p>
<p>Thanks in advance!</p>
|
|control-engineering|control-theory|
|
<p>In the synthetic example you provide there is no real disadvantage to the triple derivative. In a practical application the story is different. Generally, the triple derivative will amplify (high-frequent) measurement noise and additional low-pass filtering is needed to guarantee performance. The phase lag associated with the low-pass filter will limit the achievable bandwidth.</p>
<p>It is generally (in any controller) a good idea to include a low-pass filter as this will limit control action at higher frequencies. This is desirable as generally measurement noise will increase at higher frequencies (which limits the effectiveness of the feedback loop) and it is generally necessary to limit forces generated by the actuators (e.g. to limit heat dissipation, to limit the maximum generated force or because the available actuator bandwidth is limited).</p>
<p>Alternatives to including a triple derivative with additional low-pass filters in the controller include lead-lag filters or band-limited differentiators.</p>
|
18583
|
What are the disadvantages of PDDD controller
|
2017-12-26T15:29:48.383
|
<p>I know little about internal combustion engines and I feel a little confused about brake and shaft powers. I read a book that only talks about shaft power, and other that only talks about brake power, without making a proper definition of the physical meaning of these quantities. I'm now thinking that they both mean the same thing: <strong>useful power on the crankshaft (the power produced by the engine cylinders less the losses and the power that is used in accessory components)</strong>. </p>
<p>For example, in <em>Mechanics and thermodynamics of propulsion (1992) (Hill Peterson)</em>, pag. 155, the the brake specific fuel consumption uses shaft power in its definition, and not the brake power (as the name suggests it).</p>
<p>I would really appreciate some clarification, or some book suggestions, since the ones that I read are too vague. </p>
|
|mechanical-engineering|aerospace-engineering|power|propulsion|
|
<p>The short answer is these terms are defined by standards organizations based on specific test methods. These methods specify the engine condition, environmental factors, fuel chemistry, inlet and exhaust systems configurations, and the correction factors applied to the measurements. The specific tests and standards used vary by country and by industry. So neither term has a specific meaning unless you call out the document that defines the terms.</p>
<p>In general, though, shaft horsepower measures power after the transmission, brake horsepower measures before the transmission. On a cruise ship, you might have many gensets feeding a couple of high voltage buses that power the prop shaft motors. You want to talk about shaft power here, not brake power. Similarly, turboshaft jet engines have builtin gearsets to drive, say, a helicopter rotor. Shaft power is the relevant term here as well. Shaft power usually makes more sense when comparing different types of power plants that use different transmission arrangements.</p>
|
18585
|
What is the difference between shaft power and brake power?
|
2017-12-27T08:05:18.800
|
<p><h1>Background</h1><br>
I am currently preparing for Entrance Examination for Engineering college , somehow I noticed that I don't even know which branch I want to take and what is the kind of Education I am opting for .I always wanted to make cool stuffs not like Robots in particular but like something worth usage like that, but I just know there are few branches for but which is the one which can lead me into getting the required knowledge needed to make devices maybe like TV or a phone things dealing with semiconductors , Transistors ?
<br></p>
<p><h1>Main</h1><br>
So what is the branch that can enable me to build gadgets like modern day use somehow maybe I want to relate with the creation of gadgets that can acts as an interface b/w user and the digital world
Thats all what I want to know.</p>
<p><h2>Note: </h2>I couldn't find the required tag for the need so maybe it wont be good enough to rely on that.</p>
|
|building-design|circuits|
|
<p>To manufacture something like an elctronic device means you should have an interdisciplinary knowldege,You will need sensors which requires you to know instrumentation,If the sensor is electronic in nature you should know how to interface it thus you need knowledge of embedded systems,for embedded systems to work you need to program the chip,To program a chip efficently with proper resource managment then you should understand the arhitecture of chip which involves embedded and vlsi the list will go on so what I meant to say is you should first decide whether you should get a degree in either a hardware oriented or software oriented field then go on to specialize usually people with passion in this field take up course in Electrical and elcectronics if more hardware oriented or electronics and communication then take a masters in one of the fields in computer science.</p>
|
18591
|
Which is the branch of engineering that deals with making gadgets?
|
2017-12-29T03:27:29.883
|
<p>In control theory, are manipulated variables and control variables the same concept or is there a subtle difference between them?</p>
|
|control-engineering|control-theory|
|
<p>Manipulated variable and controlled variable are not necessarily the same thing. For example, take a water tank level control system where make up water flow into the tank is the manipulated variable and tank level is the controlled variable. There are many other examples besides this one.</p>
|
18618
|
The differences between manipulated variables and control variables
|
2017-12-29T04:50:55.460
|
<h2>The Problem</h2>
<p>I'm proposing a 2D problem in the x-y plane. Say there is a free-floating disk of radius R and mass Md (no velocity), at which a projectile is shot. The projectile (velocity Vp, mass Mp) has a velocity which is parallel to the x axis. If taking the disk's center for the origin, the projectile contacts the disk at a distance Y above the x axis, and it sticks (no energy lost).</p>
<h2>The Question</h2>
<p>I'd like to know what the ending velocity and angular velocity of the disk/projectile system is. Using conservation of energy, I end up with two variables, and I can't figure out what else to use to create a system (I'm very rusty on conservation of momentum, but I figure that's where the solution lies?).</p>
<h3>Disclaimer</h3>
<p>I'm not a student, if you're worried about doing my homework. Back in those days, I wouldn't have blinked at this, but now... software engineering has done me wrong XD</p>
|
|mechanical-engineering|energy|
|
<p>When it comes to solving any collision problem, it is always important to apply <em>conservation of linear momentum</em> and <em>conservation of angular momentum</em>. This gives us the first two equations.</p>
<p><strong>Conservation of linear momentum</strong></p>
<p>For two rigid bodies 1 and 2, the conservation of linear momentum states the following:</p>
<p>$$m_1 \mathbf{u}_1+ m_2 \mathbf{u}_2 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 $$</p>
<p>$m_i$ is the mass of body $i$, $\mathbf{u}_i$ is the velocity vector of body $i$ just before the collision, and $\mathbf{v}_i$ just after the collision.</p>
<p>Let body 1 be the projectile, and body 2 be the disc. Since the disc is initially stationary, we can simplify this to:</p>
<p>$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$</p>
<p>This is the first equation we’ll need to solve the collision.</p>
<p><strong>Conservation of angular momentum</strong></p>
<p>For angular momentum about an arbitrary point $P$ to be conserved, the following mess of an equation should be (briefly) observed:</p>
<p>$$\left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{u}_1\right)+I_1\omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{u}_2\right)+ I_2\omega_2\mathbf{k} = \left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{v}_1\right)+I_1\Omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{v}_2\right)+ I_2\Omega_2\mathbf{k} $$</p>
<p>$\mathbf{r}_i$ is the position vector of the centre of mass of body $i$ , $\mathbf{r}_P$ is the position vector of point $P$, $I_i$ is the moment of inertia about the axis of rotation for body $i$, $\omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just before the collision, and $\Omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just after the collision.</p>
<p>By setting $\mathbf{r}_P=\mathbf{r}_2$, noting that body 2 is initially stationary, and noting that body 1 is a particle and thus has no moment of inertia, the equation simplifies to:</p>
<p>$$\mathbf{r}\times\left(m_1 \mathbf{u}_1\right)= \mathbf{r}\times\left(m_1 \mathbf{v}_1\right)+ I_2\Omega_2\mathbf{k}$$</p>
<p>where $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2=-\sqrt{R^2-Y^2}\mathbf{i} + Y\mathbf{j}$.</p>
<p>Note that all the terms are parallel to the (out-of-plane) $z$-axis, i.e. parallel to $\mathbf{k}$. Therefore we can use the dot product to multiply all of the terms by $\mathbf{k}$ without loss of information from the equation:</p>
<p>$$m_1\left(\mathbf{r}\times\mathbf{u}_1\right)\cdot\mathbf{k}= m_1\left(\mathbf{r}\times\mathbf{v}_1\right)\cdot\mathbf{k} + I_2 \Omega_2$$</p>
<p>By noting that:
$$\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}= \left(\mathbf{b}\times\mathbf{c}\right)\cdot\mathbf{a} = \left(\mathbf{c}\times\mathbf{a}\right)\cdot\mathbf{b} $$</p>
<p>we can rewrite the equation as:</p>
<p>$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$</p>
<p>where $\mathbf{r^*}=\mathbf{k}\times\mathbf{r}=-Y\mathbf{i}-\sqrt{R^2-Y^2}\mathbf{j}$</p>
<p>In this form, we now have the second of the equations we need to solve the collision problem.</p>
<p><strong>The third and final equation</strong></p>
<p>We need one more equation to solve this problem. This final equation will arise from the fact that the two bodies stick to one another.</p>
<p>((In your question, you specify that there is no energy loss as a result of the collision. However, it is not actually possible for the total kinetic energy of the system to be conserved if both bodies stick* after the collision. If the collision were perfectly elastic (no energy loss), both bodies must rebound from each other after colliding. For this reason, it is important that conservation of kinetic energy is not applied.</p>
<p>*If the bond that sticks one body to the other is not perfectly rigid (i.e. it is possible to pull apart the body apart as if there was a spring joining them), then it would be theoretically possible for mechanical energy to be conserved. However, this would result in complex oscillating behaviour between the two bodies, much like an undamped mass-spring system.))</p>
<p>If the two bodies are to stick post-collision, it important that velocities of the bodies at the point of contact are the same. This results in the following kinematic condition:</p>
<p>$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{k}\times\mathbf{r}$$</p>
<p>This can be simplified to give the final equation:</p>
<p>$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$</p>
<p><strong>In summary</strong></p>
<p>Three equations have been derived:</p>
<p>$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$</p>
<p>$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$</p>
<p>$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$</p>
<p>There are three unknowns:
$$\mathbf{v}_1 \quad \mathbf{v}_2 \quad \Omega_2$$</p>
<p>These are the velocity of the projectile, the velocity of the centre of the disc, and the angular speed of the disc, respectively, immediately after the collision.</p>
<p>By using vector algebra, it is possible to solve these equation for the three unknowns. Then it may be necessary to perform some substitutions, including:</p>
<p>$$m_1=M_P$$
$$m_2=M_D$$
$$I_2 = \frac{1}{2}M_D R^2$$
$$\mathbf{u}_1=V_P \mathbf{i}$$</p>
|
18619
|
Momentum of a free-floating body being hit by a projectile
|
2017-12-29T18:40:51.140
|
<p>For a single 4x4 wood column loaded with say 300lbs of weight is there any horizontal thrust being generated in any direction? Columns don't seem to be referenced in this manual.</p>
<p><a href="http://www.engineeringwiki.org/wiki/Main_Page" rel="nofollow noreferrer">Engineering Wiki - Civil Engineering</a> </p>
|
|structural-engineering|
|
<p>Columns can and do create horizontal reaction if that's what you mean.</p>
<p>Some examples are:</p>
<ul>
<li>A column loaded by a vertical force not passing through its axial center.</li>
<li>A column rigidly or semi-rigidly connected to a beam.</li>
<li>A column adjacent to a non-structural member such as a partition wall, incidental stacked material such as in a storage.</li>
<li>A column subject to lateral forces of a shear strap or lateral force of shear wall.</li>
<li>A column not loaded symmetrically.</li>
</ul>
|
18626
|
Does a column produce horizontal thrust?
|
2017-12-30T16:00:25.773
|
<p>I'm having a trouble drawing this isometric shape.
I hope to find the drawing for this shape. ( would be so grateful if it's drawn in that isometric cube where each square of the cube is one unit of the dimensions )
<a href="https://i.stack.imgur.com/nP2qa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nP2qa.jpg" alt="enter image description here"></a></p>
<p>Edit:
My try according to my understanding to the answer of @kamran
<a href="https://i.stack.imgur.com/lTQX0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lTQX0.jpg" alt="enter image description here"></a></p>
<p>Is it correct ?</p>
|
|drawings|
|
<p>Close!</p>
<p><a href="https://i.stack.imgur.com/AfVtt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AfVtt.png" alt="enter image description here"></a></p>
<p><em>Figure 1. The error.</em></p>
<p>Since there is no hidden line on the cut-away face then that edge must be in-line with the front edge. Project the bottom corner straight back as shown.</p>
|
18640
|
Drawing isometric from two orthographic views
|
2017-12-30T16:13:48.690
|
<p>There is a common problem with fruit trees in <strong>temperate climates</strong> where the trees <strong>flower before the last frost</strong>. Generally speaking, if the tree has flowered, and the temperature drops below freezing, the flowers die and do not produce fruit.</p>
<p>This is certainly not a new problem. It's my understanding that the problem affects almost the entire fruit farming industry in temperate climates. </p>
<p>I've heard that fruit farmers have <a href="http://www.omafra.gov.on.ca/english/crops/facts/85-116.htm#freeze" rel="nofollow noreferrer">large-scale commercial techniques</a> that they use to mitigate the issue, such as:</p>
<ul>
<li>Covering</li>
<li>Fog or smoke</li>
<li>Wind Machines</li>
<li>Sprinkling</li>
<li>Heating</li>
</ul>
<p>While these methods might be practical for commercial farmers, <strong>not all of them are practical for small-scale urban farmers like me</strong>.</p>
<hr>
<p>One technique that us gardeners often wonder about is: </p>
<p><strong>Instead of actively protecting the open flowers, would it be possible to trick fruit trees into not flowering as early?</strong> In other words, keep the ground frozen longer than it would be naturally, to delay flowering (I'm making an assumption that trees don't flower when the ground is frozen). </p>
<p>While such a technique would likely be impractical for large-scale farmers, it might be resonably practical for small-scale gardeners.</p>
<p><em>Note: I imagine that there are more factors at play than just ground temperature. For example, the amount of sun, angle of the sun, and the temperature of the air might also be factors when it comes to the flowering timing.</em></p>
<p>I suspect that prolonged frozen ground conditions could significantly delay the tree from flowering (maybe by a couple of weeks) which would make a huge difference when it comes to avoiding late frosts from killing flowers.</p>
<p><strong>Can I prolong frozen ground conditions to trick fruit trees into flowering later?</strong></p>
<hr>
<p><strong>Idea:</strong></p>
<p>An idea would be to insulate the ground around the tree with snow. The theory being that the insulation would prevent heat from the sun from warming up the ground where the roots are.</p>
<p><a href="https://i.stack.imgur.com/WfMSI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WfMSI.jpg" alt="enter image description here"></a></p>
<p>However, an obvious problem with snow is that it melts, and not always when you want it to. From past experience, I'm guessing that this pile of snow will mostly melt before it is helpful to me (before late spring).</p>
<p><strong>Edit:</strong></p>
<p>I'm dealing specifically with plants in the prunus genus:</p>
<p>Examples:</p>
<ul>
<li>Prunus tomentosa</li>
<li>Prunus besseyi x prunus salicina hybrids</li>
<li>Prunus avium x prunus salicina hybrids</li>
</ul>
|
|temperature|environmental-engineering|
|
<p>Options:</p>
<h3>Site location:</h3>
<p>NE slopes get spring sun at a lower angle of attack, and so warm up more slowly.</p>
<p>In addition NOT being at the bottom of the slope may save you from pools of cold air.</p>
<h3>Sprinklers.</h3>
<p>Set up sprinklers on tall poles. I suggest using Senniger wobble head sprinkers as they have very large droplets, which take longer to freeze. Turn them on when temps are about +4.</p>
<p>You need a water supply sufficient to run them for all the trees/shrubs at once.</p>
<h3>Deep freezing + insulation</h3>
<p>In the winter pack the snow down with skis or snowshoes. This will allow the ground to freeze deeper.</p>
<p>As soon as the snow is gone, spread insulated tarps (about $20 used for a 10x20 tarp). Try to get white ones.</p>
<p>In addition at the end of winter when the cold is nearly done, use your snowblower to bury the trees 2-3 feet.</p>
<p>Roots generally require a tempurature of 5-7 C before they get active.</p>
<h3>Shading.</h3>
<p>South and west of your plants erect poles that are tall enough to hang shade cloth. Shade cloth comes in widths up to 15 feet and densities up to blocking 70% of light. I would suggests stringing high tensile strength wire on the tops, and use common curtain rings to hang the cloth. Set a grommet about every 16" in the edge of the cloth. Available at anyplace that sells greenhouse supplies.</p>
<h3>Pruning</h3>
<p>One of the problems with prunus is that pollenizers may not bloom at the same time as the fruit. One way around this is to deliberately leave some branches on pollenizers low to the ground, so that you can bury them in snow to delay their bud break. Sandcherry is used this way for hybrid plums.</p>
|
18641
|
Prolong frozen ground conditions (to trick fruit trees into flowering later)
|
2017-12-30T19:29:44.530
|
<p>Basically, I'm thinking about a system, that filtrates, and makes a lake's water drinkable, drives water into my house's faucets without electricity, and all these should be able to work when it is -20 Celsius outside. The water heating is not a problem, it is going to happen with a good old wood consumer stove.</p>
<p>*I even thought about adding a microsystem, that makes energy, like this: <a href="https://youtu.be/rPdpnunr1k0" rel="nofollow noreferrer">https://youtu.be/rPdpnunr1k0</a></p>
<p>The ram pump I would like to build is something like this:</p>
<p><a href="https://youtu.be/CG0laNqJWY0" rel="nofollow noreferrer">https://youtu.be/CG0laNqJWY0</a></p>
<p><a href="https://youtu.be/nFZYD05I29s" rel="nofollow noreferrer">https://youtu.be/nFZYD05I29s</a></p>
<p>I draw a basic thing I don't even dare to call a plan:</p>
<p><a href="https://i.stack.imgur.com/pWcFd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pWcFd.png" alt="Water moving"></a></p>
<p>The lake water goes down, into a simple filtration system and makes the mentioned hydro energy tool work, then flows (underground) into the other filtrating tool (under the house), that makes drinkable water. The next stop is a huge container or a few smaller, then the ram pump makes the water go up about 3 meters into a cold water container, and from that to a heatable storage.</p>
<p><em>The ram pump makes "wastewater" but I could make that into a smaller pond or something.</em></p>
<p>I wish to know more about this topic so I appreciate every idea to make this notion better and make the pump, the hydropower tool and the whole system more efficient.</p>
<p>Thank you in advance.</p>
<p>Edit: The electricity made with the turbine is to store it in a battery or more batteries, so I could use it to make light or charge a laptop.</p>
<p>The problem I have is to make the system work in -20 Celsius and to use a turbine to make electricity <em>efficiently.</em></p>
|
|mechanical-engineering|pumps|heating-systems|
|
<p>My grandfather's house had a system like that. According to my dad, it used a tiny pump in a stream to supply the house. A plate about the size of a playing card flopped up and down in the tiny creek and drove a piston pump the size of my pinky. It fed a hose to a tank on the roof about 100 yards away. The tubing was maybe 1/4 inch ID. Assuming about 1/4 cubic inch per stroke and 60 strokes per min, that's 15 cubic inches/min or 3.9 gal per hour. Sounds reasonable. This was more than a century ago, though. You could carry the whole thing in your pocket.</p>
<p>Here's a vastly overcomplicated version called <em>The Plata Pump</em>: <a href="http://www.fao.org/docrep/010/ah810e/AH810E214.gif" rel="nofollow noreferrer">http://www.fao.org/docrep/010/ah810e/AH810E214.gif</a><a href="https://i.stack.imgur.com/brDvf.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brDvf.gif" alt="enter image description here"></a></p>
|
18643
|
How to make water go into my house, from a lake without energy?
|
2017-12-31T02:20:50.517
|
<p>Which part of the LED is affected when it is exposed to liquid nitrogen? I have seen experiments showing LED's brightness increasing and color shifting, when the LED is submerged in liquid nitrogen. I have also seen one where the color went DOWN in frequency, how can you explain this inconsistancy?</p>
|
|electrical-engineering|lighting|
|
<p>The <a href="https://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials" rel="nofollow noreferrer">LED</a> is a semiconductor that was designed for a certain voltage at a certain temperature. This certain voltage is called "<a href="https://en.wikipedia.org/wiki/Band_gap#List_of_band_gaps" rel="nofollow noreferrer">band gap</a>" and is set by properly <a href="http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/MetalBonding/MetalBonding.html" rel="nofollow noreferrer">doping</a> it during manufacturing. The "band gap" is the voltage (think pressure) required for the electrons to hop the "gap". The larger the band gap, the higher the voltage to cross, the higher the energy of light emitted, and the higher the frequency.</p>
<p><a href="https://i.stack.imgur.com/HdiGc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HdiGc.jpg" alt="enter image description here"></a></p>
<p>When a semiconductor is cooled its resistance increases as well its band gap. This larger band gap requires a larger corresponding voltage for electrons to hop. This increases the frequency (blue shift) of the light emitted, like the orange to green shift from liquid nitrogen demonstrated in this <a href="https://www.youtube.com/watch?v=4qgPhNHJPB8" rel="nofollow noreferrer">youtube video</a>. Notice that the measured voltage drop across the LED also increased.</p>
<p>Similarly, solar photo-voltaic cells are also semiconductors and increase output voltage as the temperature gets colder.</p>
<p>Note that the voltage to frequency relationship doesn't always make sense though. I am not in the LED manufacturing industry, but other things are going on physically inside the LED when the opposite occurs. For example, this <a href="https://youtu.be/hxBJ9ToVWmA?t=34" rel="nofollow noreferrer">youtube video</a> shows a yellow LED change to green and green to yellow when submersed in liquid nitrogen. Yellow to green makes sense as we previously discussed. Green to yellow doesn't though.</p>
<p>Perhaps the doping chemical in that specific green led becomes more conductive at low temperatures and actually reduces the band gap. I am not a material scientist, but something similar to that has to occur, because the band gap voltage sets the emitted light frequency.</p>
<p>You can also see color changes by increasing the voltage on an LED beyond its design voltage as shown in this <a href="https://www.youtube.com/watch?v=vCY_HS_RCqc" rel="nofollow noreferrer">youtube video</a>. The life of the LED is significantly reduced, but its a simple experiment. This color shift is most likely due to over heating the semiconductor inside its plastic housing. At higher temperatures, the band gap decreases and the corresponding frequency decreases (red shift).</p>
<p>Here is a good <a href="https://www.digikey.com/en/articles/techzone/2013/may/thermal-effects-on-white-led-chromaticity" rel="nofollow noreferrer">Digikey article on chromatic shift in LEDs due to temperature</a>. Mid way down the page it better explains the physical nature of the band gap:</p>
<blockquote>
<p>As the temperature of the LED rises, the atoms in the lattice vibrate more, which slightly increases the lattice constant. This in turn decreases the band gap which increases the wavelength of the emitted photon.</p>
</blockquote>
|
18647
|
LED color change in extreme cold
|
2017-12-31T07:28:20.850
|
<p>All I know is sand blasting is used for cleaning and etching purposes and abrasive jet machining (AJM) is used for making cavities and slots. But I think AJM can well also be used for cleaning and etching since it involves spraying abrasive particles just like sand blasting. If thats true isnt sand blasting a type of AJM process? How are they different? Are they different in machining procedure, or the equipment used?</p>
|
|mechanical-engineering|machining|
|
<p>Sand blasting is not a type of AJM because you would never use it to machine anything. Sand blasting would never be used to cut/mahine an intricate shape in a brittle material, the jet does not have enough velocity and has too big a diameter which means it would never be accurate enough for a machining task. It's mainly used just to clean say rust off a casting. And AJM would never be used to clean a thread in a casting because it could potentially damage it.</p>
|
18651
|
How is sand blasting different from abrasive jet machining
|
2018-01-02T22:02:56.327
|
<p>I have a heat pump which moves heat from outside to inside. I also have a solar heating array that collects radiant heat from the sun. Given that my goal is to heat the inside (with minimal energy consumption from the heat pump), my question is: <strong>Is it better to heat the air inside with the solar array or to heat the heat pump's evaporator outside?</strong></p>
<p>Breaking this down, it's a question of whether it's better to use these sources in parallel or in series. In parallel, the solar array heats the inside directly and the heat pump heats the inside directly. In series, the solar array heats the evaporator and the heat pump heats the inside directly.</p>
<p>Some other relevant details, the evaporator exchanges heat by convection and is limited to a minimum temperature of 0°C. This is because any colder and you risk freezing the convector, thus stopping convection. The environment can get as low as 3°C at night, meaning heat is exchanged across a minimum of a 3°C difference.</p>
<p>Assumptions for this, the heat pump can move 5kW of heat across a 12.8°C difference (8.3-21.1°C), consuming 2kW. The indoor heat loss is (let's say) 5kW for a 15°C temperature difference (5-20°C). The solar array acts as a heat reservoir with 10kWh stored at 40°C (5°C to 40°C). Finally, let's assume the heating time is 10 hours.</p>
|
|heat-transfer|heating-systems|solar-energy|
|
<p>Assuming you don't have solar panels on your heating array, it would be more efficient to heat the load directly as you're only moving the heat once, so you wouldn't lose as much in the process.</p>
<p>However if you have solar panels, these work more efficiently at lower temperatures. Depending on your climate, you might be able to move more heat from the solar array into the heat pump evaporator than you would be able to move indoors. This could mean you are able to cool the panels a little more, making them generate slightly more power. (You'd have to check this though, if you're getting down to 3°C you probably don't need any panel cooling!)</p>
<p>You don't say where the heat pump evaporator is located, but in some places it is located underground, below the frost line. If this is the case, then pumping heat from the solar array to the underground evaporator will also allow some level of heat storage by heating up the ground, which will then be accessible to the heat pump overnight, improving efficiency.</p>
<p>You will likely also be able to get away with a simpler system if you move heat from the solar array to the heat pump evaporator, as you won't have to worry about maintaining a comfortable temperature and can just pump everything. The heat pump won't complain if it gets too warm, unlike the indoor occupants.</p>
|
18677
|
Is it better to heat a heat pump evaporator or the load directly?
|
2018-01-03T17:56:33.140
|
<p>I've been watching drag races of semi trucks and found that, especially during the initial phase of acceleration, they all bend to their right-hand side. </p>
<p>I understand, that the huge torque in rear wheels would lift the whole front of the truck, but why does it slant to one side?
Can rotational inertia of the drive shaft be strong enough to cause such a phenomenon?</p>
<p><a href="https://i.stack.imgur.com/iSlRR.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/iSlRR.jpg" alt="enter image description here"></a></p>
|
|mechanical-engineering|automotive-engineering|torque|traction|
|
<p>Go outside start your car and rev it up you will feel it pull to the right and it's all about the load. Look around you'll see the same trucks racing without trailers and they spend the tires. Once you add that much weight all that spinning has to go somewhere I don't care how strong your frame is it's going twist watch any kind of high horsepower drag racing</p>
|
18680
|
Why do drag race semi trucks slant to one side?
|
2018-01-03T20:07:30.460
|
<p>I work with left ventricular assist devices in patients with heart failure. These devices operate by maintaining a fixed speed of a centrifugal or axial rotor. The native heart continues to contract creating pressure differences between the pump inflow (heart) and outflow (aorta).</p>
<p>The device and the heart are in parallel by my reckoning. Volume A enters the heart. Volume B exits the aortic valve. Volume C exits the pump. Flow in the body is A= B+C. If the pump is off, flow in the body is A=B. If the aortic valve is sewn shut, the flow in the body is A=C. It is considered dogma that when flow through the device increases, device power consumption increases. The heart and the pump share the pulsatility of the heart and are subject to the same afterload in most situations.</p>
<p>I want to understand why power increases when pressure across the pump decreases. I understand that there will be more mass-flow through the pump with a lower delta-P across the pump. What I don't understand is why this mass-flow slows the pump, ostensibly requiring more power to overcome apparent resistance. It seems more logical that increased flow decreases energy of rotor rotation as the reduction in pressure difference would seem to have the action of imparting energy to the rotor.</p>
<p>Why is more power required to turn the rotor with a lower delta P across the pump and what principles/equations explain this effect?</p>
|
|pumps|
|
<p>This is entirely a design choice. If you imagine a positive displacement pump such as a piston pump running at constant speed, then the work done is basically a function of pressure difference, since the flow rate is constant. The greater the pressure, the more work required. The opposite extreme would be a pump like the squirrel cage fan in your house's heating system. If you apply enough back pressure, the flow goes to zero. Inside the scroll, the air is just spinning around with the fan blades. There is no torque and no work required to just keep the system running, doing nothing. As you begin to reduce back pressure, the fan begins to move air. Now there is a torque on the blades because the incoming air mass has to be accelerated to scroll perimeter speed. The loaded motor will slow down a bit. </p>
<p>Axial pumps can be designed to be very stiff, having a big power rise in the face of backpressure rise, or soft, having a small power rise in the face of increasing back pressure. But if the flow really gets choked down, the power drops in all cases. These devices would be set up to top out below normal diastolic pressure, so they would be choked and just free spinning during the ventricular contraction. What the power response looks like as flow increases from stall is at the designers discretion.</p>
|
18683
|
Power in a continuous flow pump
|
2018-01-04T01:16:02.583
|
<p>Consider this triangle with its centroid at $C$:</p>
<p><a href="https://i.stack.imgur.com/GW2LT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GW2LT.png" alt="Text-book triangle"></a></p>
<p>Then this is how I believe we can calculate the <em>Second Moment of Area</em> along the $x_{C}$ and $y_{C}$ axis:</p>
<p>$$\begin{align}
I_{xc} &= \frac{bh^3}{36} \\
I_{yc} &= \frac{hb(b^2-st)}{36}
\end{align}$$</p>
<p>So far so good.</p>
<p>But then we have this triangle:</p>
<p><a href="https://i.stack.imgur.com/bTXV1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bTXV1.png" alt="Slanted triangle"></a></p>
<p><strong>Questions</strong></p>
<ul>
<li>Is there any similar way of calculating the second moment of area for a slanted triangle, like the one above?</li>
<li>Will even the same equations work?</li>
<li>How is $s$ and $t$ derived in that case? Can they be negative?</li>
</ul>
<p>Please assume we can <strong>NOT</strong> rebase the triangle — we must use $LR$ as the base.</p>
<p><a href="https://www.engineeringtoolbox.com/area-moment-inertia-d_521.html" rel="nofollow noreferrer">Source for equations</a></p>
|
|statics|moments|
|
<p>Slanted triangle can be obtained by cutting off piece of right triangle, and you can calculate moment ot inertia or second moment of area by subtracting smaller right triangle from the larger one using Steiner's rule. You can express all dimensions considering your base and edges using trigonometric functions.</p>
|
18695
|
Second Moment of Area for slanted triangle?
|
2018-01-04T16:52:38.737
|
<p>My office is afforded a prime view of a construction project where the workers are currently covering the leveled ground in a wire mesh, raking gravel across the entire mesh, and then covering that gravel layer in another layer of dirt. </p>
<p>What is the function of this, engineering-wise? I know gravel in a foundation typically is used to help with water runoff, but I can't guess why the mesh is being rolled out underneath, and this is going to become the foundation of a 3-story health pavilion, not some single-family house or storage shed (the entire footprint here with gravel and mesh will have a building on top of it, not a parking deck or a parking lot).</p>
<p>Is this simply a base layer of some kind, upon which concrete and rebar will be placed?</p>
<p><a href="https://i.stack.imgur.com/oxKX8.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/oxKX8.jpg" alt="enter image description here"></a></p>
<p>I am in the United States, in Georgia.</p>
|
|structural-engineering|geotechnical-engineering|foundations|soil|
|
<p>You are correct, this is the subbase that the foundations will be built on top of. </p>
<p>The gravel can be used as you say for water run off but it also allows for a level base on which to lay the concrete. In areas where the ground becomes highly saturated it can protect the concrete foundation from erosion. </p>
<p>The wire mesh is usually used for load-bearing concrete, simply to give it some extra strength and prevent cracking. </p>
<p>The dirt going on top of the subbase is the base material. This is used to help the workers get the correct grade and to keep the foundation flat. It also supports the workers and equipment during the concrete pouring.</p>
|
18721
|
Why is this construction site pouring gravel over a wire mesh, and then covering it with dirt?
|
2018-01-07T12:55:36.193
|
<p>I am currently reading “Fundamentals Of Vehicle Dynamics” by Thomas D. Gillespie. </p>
<p><a href="https://i.stack.imgur.com/PsY1r.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PsY1r.jpg" alt="enter image description here"></a></p>
<p>The author mentions how the asymmetric pressure distribution across the length of the contact patch of a tire while rolling results in Rolling resistance. This, then inevitably creates a moment about the axis of the wheel which resists its movement. </p>
<p>The author states, however, that this asymmetric pressure distribution only occurs when a tire is rolling. There isn’t any asymmetry when it’s in a dead-stop position.</p>
<p>What I want to know now is the reason for this asymmetric pressure distribution. What causes it? How does Rolling generate asymmetry?</p>
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|mechanical-engineering|automotive-engineering|
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<p>I came across <a href="http://the-contact-patch.com/book/road/c2020-the-contact-patch" rel="nofollow noreferrer">a site</a> that, among others, tries to explain exactly the phenomena in your question. It goes quite into depth about the different components of the tyre and how they interact with each other and the effects it has on the contact patch and rolling resistance. I also especially enjoyed the part speaking of the standing wave.</p>
<p>Just as a post script:</p>
<p>I really like @Mark's <a href="https://engineering.stackexchange.com/a/18787/14440">answer</a> and I've drawn a sketch showing how I understand it:
<a href="https://i.stack.imgur.com/DEtWF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DEtWF.png" alt="enter image description here"></a></p>
<p>Basically, the moment when an element of the tyre touches the ground it has to stop to a standstill (relative to the ground). In order to have this happen, a force in the opposite direction needs to be applied to decelerate it. fx isn't shown in the diagram but that's the required forward force to decelerate the element, also causing rolling resistance.</p>
<p>The opposite obviously applies to the trailing edge, needing an upward force to accelerate it again.</p>
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18759
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What causes Rolling Resistance?
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2018-01-07T15:02:01.907
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<p>I have been having a debate with a coworker regarding heat exchanger fluid temperatures and while I thought I firmly grasped the principles, I was not arguing my point very well. </p>
<p>I am looking at an air to water coil for an air handler. I was selecting the coil based on a leaving air temp of ~60F (it's an industrial space with a lot of ventilation). I am using condensing boilers so I sized for 180 leaving water temperature with 140 return on the coldest day which gives a 40 degree temperature delta.
On the air side, I am bringing in 5 degree air (EAT) at a rate of 16000 CFM on one of the air handlers. My coworker wants to lower the temperature to 140 supply and 100 degree return, his reason being that it would still be a 40 degree temperature delta on the water side so the thermal output would be the same. I believe the hotter supply temperature will result in more rapid heat transfer but I didn't really have any good argument against his logic. He was simply relating the two rules of thumb equations for air and water: [Btu/hr]=1.08*CFM*deltaT and [Btu/hr]=500*gpm*deltaT to prove his point.
My argument was that this does not account for the heat transfer taking place from one side of the coil to the other, but I was not able to adequately illustrate my side to the point that he would agree. </p>
<p>First of all, am I even right about operating the system at 180 degrees being a more efficient solution, and, if so, how do I fit the heat exchanger heat transfer into the system of equations?</p>
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|thermodynamics|heat-transfer|
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<p>Simple answer is this, the Heat exchange equation accounts for the amount of time the fluids are in contact, while the conservation of energy equation assumes the fluids are in equilibrium. So if the temperature differences between the 2 fluids is smaller you need the fluids to be in contact longer to get the amount of heat exchange you are calculating for the higher temperature difference. Since the conservation of energy equations have flow rates in the terms, this is the equivalent to saying the fluids are traveling across the heat exchanger surface at a constant rate, the only way to get the same amount of heat transfer with a smaller temperature differential is to increase the surface of the heat exchanger, so the fluids stay in contact for a longer period of time. </p>
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18763
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Heat Exchanger water to air temperatures
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2018-01-07T17:02:44.080
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<p>Most large cities these days have some sort of metropolitan commuter train system. What are the reasons for the commuter trains, why couldn't we instead of having train tracks why not clear a large distance of straight road and have a vehicle, such as a large bus riding on it?</p>
<p>Looking at the speed of trains (in my city they go at an average of around 70 km/h) surely the speed alone is not a factor since it would probably be feasible to have some sort of large vehicle achieve this speed. Trains also require large infrastructure such as tracks and power lines. On the other hand, such large vehicles would probably be safer on tracks instead of roads, and a train on tracks would probably be easier to automatize than a vehicle with wheels.</p>
<p>I can think of a lot of reasons for and against trains over wheeled vehicles, but what are the real reasons why commuter trains are so popular a choice for cities?</p>
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|transportation|
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<p>I don't know if its a driver, but an advantage to rail is that you can run parallel lines very close together and in close proximity to structures and such.</p>
<p>I don't think you'd want buses approaching each other at 100+mph on a paved road that would fit in the space of an existing rail corridor.</p>
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18764
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Advantages of a commuter train instead of a vehicle travelling in a straight road
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