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2023-04-07T21:02:51.717 | <p>Below is a problem I made up and did. I am hoping that you will tell me that my solution is right.</p>
<p>Problem:</p>
<p>A <span class="math-container">$5$</span>g piece of steel at <span class="math-container">$30$</span>°C is placed in <span class="math-container">$400$</span> ml of water at
<span class="math-container">$20$</span>°C. Assume that all the excess heat from the steel goes into the
water. What is the final temperature of the water?<br />
Answer:</p>
<p>Using the Internet we find that the heat capacity of steel is <span class="math-container">$0.420$</span> J per gram.
Let <span class="math-container">$t_f$</span> be the final temperature of the water. It is also the final
temperature of the steel.
<span class="math-container">\begin{align*}
5(0.42)(30-t_f) &= 400(1)(t_f - 20) \\
(0.42)(30-t_f) &= 80(t_f - 20) = 80t_f - 1600\\
0.42(30) - 0.42t_f &= 80t_f - 1600\\
12.6 - 0.42t_f &= 80t_f - 1600\\
12.6 &= 80.42t_f - 1600\\
t_f &= 20.05
\end{align*}</span>
Hence the final temperature is <span class="math-container">$20.05$</span>°C.</p>
<p>Is my solution right? It seems to low to me.</p>
| |steel|temperature|homework| | <p>You are almost right. One thing you have to take into account is <a href="https://en.wikipedia.org/wiki/Specific_heat_capacity" rel="nofollow noreferrer">heat capacity of the water</a> 4.184 J/(g K). With that, the equation will look like:</p>
<p><span class="math-container">$$5\cdot 0.42\cdot (30-t_f) = 400\cdot 4.184\cdot (t_f-20)$$</span></p>
<p>So the final temperature <span class="math-container">$t_f$</span> is:</p>
<p><span class="math-container">$$t_f = \frac{5\cdot 0.42\cdot 30 + 400\cdot 4.184\cdot 20}{5\cdot 0.42 + 400\cdot 4.184} \dot{=} 20.0125 °C$$</span></p>
<p>So as you can see, it is even lower due to much greater mass as well as thermal capacity of water.</p>
| 54810 | Cooling steel by putting it in water ; what is the final temperature |
2023-04-10T20:04:29.023 | <p>I have large areas to cover (84 cm x 70 cm, 150 cm x 84 cm, and 150 cm x 70 cm) with stretch film (22 µm thick, 45 cm wide).</p>
<p>The stretch film is not wide enough so I have to overlap several sheets to cover the whole area. Those areas represents the walls of a rectangle parallelepiped which will be filled in with water. I have enough stretch film rolls to wrap the frame with enough layers so that it can hold the water pressure.</p>
<p>This is what I want to achieve :
<a href="https://i.stack.imgur.com/z6hqn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z6hqn.jpg" alt="Frame covered with stretch film" /></a></p>
<p>The <a href="https://ca-ressemble-a-du-fake.github.io/recuperateur_d_eau_en_film_etirable/" rel="nofollow noreferrer">project goal is to create a custom size water tank</a>(the project page hasn't been updated since I haven't found the correct solution so far).</p>
<p>To avoid leaks I need to make the first sheet thoroughly waterproof. That's why I want to kind of weld two overlapping sheets together so that they act together like a larger sheet (like when I weld steel parts).</p>
<p>So far I tried to "weld" two overlapping sheets with a hair-dryer (1900 W - maybe 120°C) but it just seems to stick the sheets (that can then be separated). I also tried with a thermal cleaner (400 - 550°C) and with a heat gun (300 °C) but both seem to be to hot as they tend to burn the external sheet and create small holes that are undesirable.</p>
<p>Consequently I am asking if it is even possible to weld two sheets of stretch films so that they act like a single one (it seems to be possible eg <a href="https://www.tech-pol.com/en/plastic_welding.php" rel="nofollow noreferrer">that company</a>), and if so which tool I should use or which temperature I should reach and which one I should not exceed.</p>
<p>Thank you very much for advises !</p>
| |plastic|welding| | <p>It's not yet an answer but that's useful hints! So consider it as an interim solution.</p>
<p>I have just made some experiments and I can tell when holes appear in stretch film while heating it with heat gun and when they don't : actually it depends on whether the film is already stretched (even slightly) or not.</p>
<p>So one solution would be to make the walls aside from the walls (apart, eg on the floor) and then gently put it on the frame. That's not easy but as soon as the film is stretched it's very hard to avoid holes when "welding" two sheets together. Whereas without tension the sheets can melt and merge together before holes appear.</p>
| 54849 | Can I weld stretch film? |
2023-04-11T03:22:10.540 | <p>Hi and thank you in advance. I understand that one of Euler's formulas states:</p>
<pre><code>cos(x) + isin(x) = e^ix
</code></pre>
<p>I know that if you were to have the same coefficient attached to sin and cos the following would hold true:</p>
<pre><code> 4*cos(x) + 4*isin(x) = 4*e^ix or z*cos(x) + z*isin(x) = z*e^ix
</code></pre>
<p>Additionally a constant could be applied inside the trig term and this would hold true:</p>
<pre><code>cos(3x) + isin(3x) = e^3ix or cos(xt) + isin(xt) = e^ixt
</code></pre>
<p>I have trouble understanding how we can utilize this though say if any of the above coefficients were to not be uniform.
Say:</p>
<pre><code>4*cos(x) + 3*isin(x) = ?
</code></pre>
<p>If you were to say the above is equivalent to 7e^ix, this not true. How can you use eulers formula with different coefficients? Is it possible?</p>
<p>I ask this because a fellow student used the following:
<a href="https://i.stack.imgur.com/CQwFR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CQwFR.png" alt="Euler's method application" /></a></p>
<p>and somehow through Euler's converted it to the following:</p>
<p><a href="https://i.stack.imgur.com/wToAx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wToAx.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|systems-engineering|euler| | <p>For your example: 4<em>cos(x) + 3</em>isin(x) = ?
Plot a point at (4,3), find the distance from origin to point = <span class="math-container">$(x^2 + y^2)^½$</span></p>
<p>Then <span class="math-container">$4*cos(x) + 3*isin(x) = 5e^{i\theta}$</span></p>
<p><span class="math-container">$\theta$</span> is the angle from the x axis.</p>
| 54851 | Euler's formula with different coefficients and constants |
2023-04-12T21:44:15.877 | <p>How to Create a <a href="https://en.wikipedia.org/wiki/Lima%C3%A7on" rel="nofollow noreferrer">Limacon of Pascal</a> in a CAD drawing for extrusion etc? Need a .STEP file ultimately.</p>
| |cad| | <p>One possibility would be in OpenSCAD:</p>
<pre><code>module limacon(a, b) {
r = function(a, b, theta) a + b * sin(theta);
polygon([for (t=[0 : 1 : 359])
[r(a,b,t)*sin(t), r(a,b,t)*cos(t)]]);
};
// draw a limacon with the specified parameters:
limacon(3, 5);
</code></pre>
<p>With this, we can draw the 3D extruded polygon like this:</p>
<pre><code>linear_extrude(5)
limacon(3, 5);
</code></pre>
<p>Which results in a drawing like this:</p>
<p><a href="https://i.stack.imgur.com/27gXw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/27gXw.png" alt="enter image description here" /></a></p>
<p>From there, things get a little more tedious than you'd probably like. At least as far as I know, OpenSCAD can't export directly to a .STEP file, but you can import the OpenSCAD file into FreeCAD, then export the object to a .STEP file from there.</p>
| 54875 | How to Create a Limacon of Pascal in a CAD drawing? |
2023-04-13T08:27:41.593 | <p>In turbojet (gas turbine engines), we analyze fluid in terms of pressure. We say that when fluid/air hits stator of compressor, its velocity decreases but static pressure increases due to Bernoulli equation. It assumes that total energy of fluid is conserved.</p>
<p>But why are we assuming this? In my opinion, when fluid hits a surface such as stator(fixed blades), its energy should be down, because it loses energy by hitting something else. It has to automatically transfer some energy to stator.</p>
<p>Instead, we say that its kinetic energy is transformed to static pressure when it hits stator, which is a bit interesting to me. If its velocity decreases naturally in a tube for a normal flow without external affects, that's understandable. However here, there is another object to hit, and this is not a normal velocity decrease.</p>
<p>How do you explain this?</p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|energy| | <p><a href="https://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow noreferrer">Bernoulli's principle</a> is proven, and can be proven by anyone (including yourself) experimentally; it is not an "assumption". Generally it is better to experiment and learn to trust the physics on simpler systems before advancing to complex systems like turbines where there is a lot going on.</p>
<p>An <a href="https://en.wikipedia.org/wiki/Ideal_gas_law" rel="nofollow noreferrer">ideal gas</a> and the <a href="https://en.wikipedia.org/wiki/Kinetic_theory_of_gases" rel="nofollow noreferrer">Kinetic theory of gases</a> state that the particles only undergo <a href="https://en.wikipedia.org/wiki/Elastic_collision" rel="nofollow noreferrer">elastic collisions</a> which means <a href="https://en.wikipedia.org/wiki/Momentum" rel="nofollow noreferrer">momentum</a> is conserved. Think of them like rubber balls. You can think of a stator blade like wall. When a rubber ball comes and hits the wall, it bounces off with opposite momentum but with 100% of its energy. As more and more of these balls are thrown at the wall they not only bounce off the wall but off of each other. The concentration and speed of balls near the wall surface will increase, resulting in an increase in pressure and temperature in the area near the wall.</p>
| 54882 | When fluid hits a surface or object, why doesn't its energy decrease? |
2023-04-14T18:53:55.050 | <h5>Good Damping</h5>
<p>Carbon bicycle frames have the unusual property that they are superior to aluminum frames in high-frequency absorption. A cyclist on a carbon-frame bike will feel tarmac chatter far less than on an aluminum-frame bike. The frame yields at these high frequencies, isolating the rider from them.</p>
<h5>High Strength</h5>
<p>Yet carbon frames are also famously solid. A carbon-frame bike remains solid underneath a cyclist using arm muscles, efficiently transferring all their power to the drivetrain with little perceptible frame bending.</p>
<h5>Question</h5>
<p>How can a material be good at damping, yet also be strong?</p>
<p>Credit for this phrasing of the question goes to TimWescott.</p>
| |strength|carbon-fiber| | <p>The trick is in the angle of winding the layers of carbon fiber lamination, or layup.</p>
<p>A tube of the bike could possibly be wound up of layers of 0 , 20, 45 degrees, and even different materials having the optimal shear, tensile modulus appropriate for that angle.</p>
<p>So, for example, much of the stiffness of the tube against, say, torsion is due to the 45-degree layup.</p>
<p>Furthermore, because different materials strain differently, there is inherent energy absorption during deformation.</p>
<p>Some of the winding gets altered and improved after stress tests, even along the length of the tube, say when the tube is attached to the saddle.</p>
<p>A carbon fiber layup example showing the orientation of layers. The image shows a total of 8 layers of varying
carbon fiber directional layup. Image from Velocite (Velocite, 2015)</p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/up2O8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/up2O8.png" alt="tube" /></a></p>
| 54904 | How can a material be good at damping, yet also be strong? |
2023-04-17T02:02:51.250 | <p>I own an evaporative air cooler of width 28 inches and height 34 inches, I was wondering if a taller air cooler with lesser width (but same surface area as the one I own) would have been more efficient in cooling the room.</p>
<p>My logic is that a cooler with lesser height would throw the air at a lower height and as cool air tends to sink, taller coolers can be more efficient. But also my bed rests lower on the ground so a cooler with more width and lesser height would throw the cool air horizontally (which would be better) while lowering the direction of the taller cooler would result in throwing the air at a negative angle from the horizontal which intuitively feels to me that more cold air is lost as its basically cooling the ground more.</p>
<p>How do I calculate the difference in efficiency of both the coolers given their surface area and all other factors are same, at different levels of height in the room. Assuming the fan is not turned on i.e there is minimal vertical air flow. What all criteria should be taken into account? How would it affect the calculations if the fan is in fact turned on i.e more vertical air flow?</p>
| |heat-transfer|airflow|cooling|heat| | <p>Nope, the density difference between hot and cold air, and humid and dry air, is insufficient to make any difference in the efficiency of a swamp cooler. Ditto for the exit angle of the cooled air atream coming out of the cooler.</p>
<p>The reason all of them are more or less cubical in shape is that this minimizes the amount of steel needed to build the cooler housing relative to the surface area of the sides containing the evaporation pads, thereby minimizing the manufacturing cost for the unit.</p>
<p>The primary determinant of the power consumption of a swamp cooler is the size of the motor that drives the fan, and the fan design itself. This is why all but the cheapest coolers use <em>squirrel-cage fans</em> that are <em>direct-driven</em> (no belts, gears, etc. between the motor shaft and the fan assembly). These waste less energy than a propeller-type fan per cubic foot of air delivered. They also generate less noise than a propeller fan, which is an important consideration when the cooler is used in a home or office.</p>
| 54925 | Efficiency difference of an evaporative air cooler based on its dimensions (wider or taller) given that all the other aspects are same |
2023-04-17T14:00:03.193 | <p>I'm looking for a way to calculate possible balanced positions of 5 equal masses on the circumference of a spinning wheel such that the distance along the circumference between the masses are all different, and no distance is an integer multiple of any of other the other distances. This is like the balanced centrifuge problem, but without defined slots.</p>
<p>Masses arranged in a 5 point star would be balanced, but the distances would all be the same. 3 masses in an equilateral triangle with 2 additional masses places across from each other would also be balanced, but distances between the 3 masses would be the same, and the distances between the 2 are the same clockwise and counter clockwise. Is there a possible arrangement of 5 masses where all the distances are different?</p>
| |geometry|flywheels| | <p>Based on <a href="https://engineering.stackexchange.com/a/54938/41450">Greg Locock's answer</a> I made this diagram that starts off with a balanced arrangement and then I moved the vectors to create these pentagons. This particular balanced arrangement doesn't meet the criteria that the distances be non-integer multiples of other distances, but this is definitely a step closer to the solution I'm seeking.
<a href="https://i.stack.imgur.com/4Zeth.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Zeth.png" alt="5 balanced vectors translated into pentagons" /></a></p>
<p>Here is a diagram of fixing two of the vectors (black arrows) and moving the other vectors (red arrows) to find a new balanced arrangements.
<a href="https://i.stack.imgur.com/C2Pmh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C2Pmh.png" alt="moving foci on pentagon to find new balanced positions" /></a></p>
| 54930 | 5 balanced masses on wheel with different distances between them |
2023-04-17T21:50:18.067 | <p>I have a 2 mm thick high-quality carbon fiber composite sheet. I want to cut a shape in it with a 0.8 mm diamond flute end mill. I need a cut this narrow because of the shape's detail. The shape's total cutting length is about 30 cm, cut depth 2.1 mm (three passes of 0.7 cut depth). The plate is glued to a solid plastic base and pressed with several clamps, then submerged in water.</p>
<p>The mill end breaks after about 10 cm (HSS) to 40 cm (tungsten carbide). Current mill parameters are 40 mm/min, cut depth 0.7 mm, 10 000 rpm, but I tried anything starting at 10 mm/min. Does not seem to change a lot.</p>
<p>The CNC is 3018 upgraded with a 500 W 10 000 rpm brushless DC motor. When looking at a rotating end mill, no soft edges are seen in hard light (there is instead an illusion that the mill is not rotating at all) so I think that the spindle's vibration is low. But anyway, I changed the spindle to a small brushless DC 20 000 rpm and tried it at rpm from 8 000 to 20 000. Same thing, the end mill breaks after some dozen cm.</p>
<p>A 3018 CNC is not top quality for sure, but it seems to work smoothly. It is controlled with upgraded electronics - drivers are TMC2209, firmware is GRBL. Maybe there is a ~0.1 mm backlash when looking at a circular cutout, but I'd guess that a backlash is something which may pause the spindle's XY for a moment, not something which jerks it, at least not at only ~ 30 mm/min.</p>
<p>I had never problems with this machine when using end mills of diameters >= 2 mm (can't say about diameters > 0.8 mm, < 2.0 mm because I have never used them). Also, the said 0.8 mm end mills do not have any problems with a very hard maple wood.</p>
<p>This microscopic photo shows an average quality of the cut, pass depth 0.7 mm:<a href="https://i.stack.imgur.com/Ck8GB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ck8GB.jpg" alt="enter image description here" /></a></p>
<p>Spindle power consumption at 15 000 rpm, 0.4 mm cut depth as suggested by fred_dot_u:</p>
<ul>
<li>free: 17.3 W</li>
<li>isolated groove 0.4 mm deep: 18.2 W</li>
<li>isolated groove 1.6 mm deep: 18.7 W</li>
<li>adjacent groove 1.6 mm deep: 18.3 W</li>
</ul>
<p>Any ideas? Is it really that hard to cut a carbon fiber plate with a sub-mm end mill or something is wrong here?</p>
| |cnc|carbon-fiber| | <p>I've read your question twice, and may yet have missed one detail of value. How deep is your cut? I see "feed 0.7" but not "feed cut depth."</p>
<p>According to an <a href="https://inventables.zendesk.com/hc/en-us/articles/360015957214-Cut-Depth-and-Depth-Per-Pass" rel="nofollow noreferrer">Inventables web site</a>:</p>
<blockquote>
<p>As a general rule, your depth per pass should never exceed half the
diameter of your bit. For example, the depth per pass for a 1/4"
(.25") bit should not exceed .125" per pass, the depth per pass for a
1/8" bit (.125") should not exceed .0625" per pass, etc.</p>
</blockquote>
<p>This means that for your 0.8 mm bit, you would want to make five passes of 0.4 mm (or more passes, reduced depth) in order to meet the above referenced requirement. I will presume that the carbon fiber you are cutting is a composite, rather than raw fiber, which may complicate the circumstances.</p>
<p>Performing the cut in a water bath probably resolves any heating of the epoxy which would certainly complicate the project.</p>
<p>Ensure that your work piece is well secured to the sacrificial sub-board. You very much do not want vibration on the work while cutting. Double stick carpet tape is a common substance and may require heating to remove without breaking the part, but carbon fiber composite can handle temperatures well enough as long as the epoxy resin is not damaged.</p>
| 54936 | Cutting carbon fiber plate with a sub-mm end mill |
2023-04-18T17:18:18.243 | <p>As far as I understood from <a href="https://youtu.be/PxyjjgHeDAA" rel="nofollow noreferrer">this video</a>, there are torque converters and fluid couplers (a hydraulic clutch), but certain types of torque converters kind of work by performing the two functions at different speeds.</p>
<p>I just can't find hydraulic torque converters that serve as reduction boxes in bigger ratios, like gears, pulleys or transmission belts.</p>
<p>I guess the closest I could find was <a href="https://youtu.be/qxZFSNITK-c" rel="nofollow noreferrer">hydrostatic transmission</a>, but this kind of transmission is used with hydraulic pumps and hydraulic motors separated, but connected. Which I couldn't find a precise amount of efficiency, but it was said to be really low.</p>
| |torque|hydraulics|power-transmission|hydrostatics| | <p>Torque converters as used in passenger cars commonly operate over ranges of between 2:1 and 4:1 for output RPM. The maximum torque multiplication (and rpm "division") occurs essentially at the <em>stall condition</em> (output shaft prevented from rotating) but at stall, the efficiency is zero (all the power input is dissipated in fluid friction).</p>
<p>It is possible to construct torque converters with higher torque multiplication ratios than this but they are custom jobs used in things like drag racers and not commonly available.</p>
<p>It is also possible to place two torque converters in series to get higher overall ratios, as used in the 1963 Buick Riviera <em>Twin-Turbine Drive</em>, (sometimes known as the "Slush-O-Matic") but the power transmission efficiency of this arrangement is very poor.</p>
<p>A far more efficient arrangement is to combine a torque converter with a two-speed transmission, yielding the famous General Motors/Chevrolet <em>PowerGlide</em> of the mid-1960s.</p>
| 54949 | Are there hydraulic torque converters that are specifically built to work like reduction gearboxes? |
2023-04-18T19:06:14.893 | <p>In a <a href="https://youtu.be/MTeJWE_Ou0g" rel="nofollow noreferrer">hydrostatic transmission system</a>, hydraulic pumps drive hydraulic motors (basically a reverse pump).</p>
<p><a href="https://i.stack.imgur.com/rlCBu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rlCBu.png" alt="Illustration of a hydrostatic transmission system" /></a></p>
<p>If you were use a smaller hydraulic pump as an input and a bigger hydraulic motor as the output, would it work like a reduction gearbox? Where the force is multiplied and the speed divided?</p>
| |pumps|hydraulics|power-transmission|hydrostatics| | <p>The pump and motor shown are variable which means the flow and delivery pressure can be set or chosen or feedback controlled. However, a pump should be matched so that the max capability of the motor can be provided by the pump, even though the pump output can be varied.</p>
<p>You can check out more detail here:</p>
<p><a href="https://store.boschrexroth.com/Hydraulics/Pumps/Axial-piston-pumps/Variable-displacement/AXIAL-PISTON-PUMP_R902504834?cclcl=en_US" rel="nofollow noreferrer">https://store.boschrexroth.com/Hydraulics/Pumps/Axial-piston-pumps/Variable-displacement/AXIAL-PISTON-PUMP_R902504834?cclcl=en_US</a></p>
<p>And the Bosch Rexroth site has so much information - was known as Mannesman Rexroth..</p>
| 54953 | Would connecting a small hydraulic pump to a larger hydraulic motor result in force multiplication? |
2023-04-19T13:59:03.660 | <p>Let's say I have 1000 W worth of solar panels hooked to a 2 kWh battery system like the <a href="https://www.anker.com/a1780?utm_source=google&utm_medium=search&utm_campaign=us_anker_pps_conversion_search_A1780_purchase_ost&utm_content=A1780&utm_term=%7B19750921396%7D_%7B150388444470%7D_%7B650568854773%7D&gclid=CjwKCAjwov6hBhBsEiwAvrvN6PQmm_GA4A6IK0grGulrAUleAxqYm9b2bYnRzkw-9c0UPzpb3Zyl0RoC6LAQAvD_BwE" rel="nofollow noreferrer">Anker 767</a>. Let's also say that the panels are running at full clip and I have a 900 W load on the system.</p>
<p>Does the load degrade the batteries at all? Is the current actually going through the battery or is it bypassing the battery since it's less than the input voltage?</p>
| |photovoltaics|solar| | <p>It can do if the circumstances are right. It generally means you have an active BMS that is expecting a very low rate of change for current and voltage on the load side and will make adjustments based on a reasonable sampling rate of say 1 kHz. And you have a DC load that expects a steady voltage supply and runs a current chopper at say 1 kHz, but the two aren't perfectly synced. You can end up with the BMS operating in bang-bang mode and having more current flowing through the battery than the load is actually drawing. Modern, low impedance batteries are far more suseptable than wet lead, which acted like a giant capacitor and added had a lot of damping to any high-frequency dynamics.</p>
<p>The link below shows the filter elements typical of a grid connected PV system with batteries that can be islanded. It is drawn from a document that describes the design strategies for designing the filters so that they work properly in all modes of operation. The performance of grid-tied PV systems is tightly regulated and there are quite of lot of problems cause by the increasing numbers of non-compliant grid-tied PV sources. All the components labeled in red and blue are filter components. They are grouped into three filter assemblies, and they have to be designed carefully to work with each other. Since they may live on separate components, that can be tricky for the DIYer</p>
<p><a href="https://link.springer.com/article/10.1007/s42452-020-2747-7/figures/1" rel="nofollow noreferrer">https://link.springer.com/article/10.1007/s42452-020-2747-7/figures/1</a>
<a href="https://i.stack.imgur.com/lXULI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lXULI.png" alt="enter image description here" /></a></p>
| 54962 | Does a load on a charging solar system degrade the battery if the panels are producing more current than the load? |
2023-04-20T13:13:33.283 | <p>I tried to understand <strong>flow coefficent C<sub>v</sub></strong> of a valve. There are many similar discussions on the Internet, e.g.,
<a href="https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/" rel="nofollow noreferrer">https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/</a></p>
<p>There are two formulas for C<sub>v</sub>, one for <strong>incompressible fluids</strong>:
<span class="math-container">$$
C_v = Q \sqrt{\frac{SG}{\Delta P}} \tag{1}
$$</span></p>
<p>and another for <strong>compressible fluids</strong>:
<span class="math-container">$$
C_v = \frac{Q}{960\sqrt{\frac{\Delta P(P_1+P_2)}{SG * T}}}
= \frac{Q}{960}\sqrt{\frac{SG}{\Delta P}\frac{T}{(P_1+P_2)}} \tag{2}
$$</span></p>
<p>Obviously, the units of C<sub>v</sub> given by the above two formulas are completely different, while
(1) gives the units of the C<sub>v</sub>:
<span class="math-container">$$
\rm \frac{Gallon}{Min}
$$</span>
<strong>[Notes:</strong> As David Bailey pointed out my mistake, above units should be
<span class="math-container">$$
\rm \frac{Gallon}{Min} \frac{1} {\sqrt{PSI}} = \frac{GPM}{\sqrt{PSI}}
$$</span>
<strong>]</strong></p>
<p>(2) gives the units of the C<sub>v</sub>:
<span class="math-container">$$
\rm \frac{SCFH.\sqrt{^\circ R}}{PSI} = \frac{Ft^3.\sqrt{^\circ R}}{H.PSI} \sim \frac{Gallon}{Min} \frac{\sqrt{^\circ R}}{PSI} \qquad {\color{red} \neq} \qquad \frac{Gallon}{Min} \frac{1} {\sqrt{PSI}}
$$</span></p>
<p><strong>WHY different?</strong> Can someone give me some explanation?</p>
<p><strong>Furthermore, I never figured out how formula (2) was derived, any suggestion？</strong></p>
<p>Any reply would be greatly appreciated.</p>
| |hydraulics|pneumatic|compressible-flow| | <p>Under the hint of David Baily's answer of my question, we may have a simple relation between formula (2) and (1).</p>
<p>Let's derive from (2) to (1).</p>
<p>We rewrite (2)</p>
<p><span class="math-container">\begin{align}
C_v &= \frac{Q{(\rm SCFH)}}{964} \sqrt{\frac{SG_{\color{red}c}}{\Delta P}\frac{T}{(P_1+P_2)}} \tag{2} \\
&= \frac{Q{(\rm SCFH)}}{1364} \sqrt{\frac{1}{\Delta P} \frac{\rho}{\rho_{air}}\frac{T}{P}}
\end{align}</span></p>
<p>According to above accepted answer by David Baily,
<span class="math-container">$$\rho_{air} = \frac{\rho_{water}}{817.5},$$</span>
and according to <a href="https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/" rel="nofollow noreferrer">https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/</a> :
<span class="math-container">$$P = \frac{P_1+P_2}{2} =14.7 ~{\rm PSI}, \qquad T=520 ~{\rm ^\circ R}\sim 60 {\rm ^\circ F}.$$</span>
Then
<span class="math-container">\begin{align}
C_v &= \frac{Q{(\rm SCFH)}}{1364} \sqrt{\frac{1}{\Delta P} \frac{817.5\rho}{\rho_{water}}\frac {520~ {\rm ^\circ R}}{14.7 ~ {\rm PSI}}} \\
&= \frac{Q{(\rm SCFH)}}{8.02} \sqrt{\frac{1}{\Delta P} \frac{\rho}{\rho_{water}}\frac { {\rm ^\circ R}}{{\rm PSI}}} \\
&= \frac{Q{(\rm SCFH)}}{8.02} \sqrt{\frac{SG_{\color{red}i}}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\
&= Q{(\rm GPM)} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\
% &= \frac{Q{(\rm GPM)}}{8.02\times8.00} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\
% &{\color{red}\neq}~ Q{(\rm GPM)} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \tag{1}
\end{align}</span></p>
| 54973 | What are the Units of flow coefficient Cv? And relationship between flow coefficient Cv formulas of incompressible and compressible fluids |
2023-04-22T07:44:39.390 | <p>I tried solving the Navier Stokes equation for a Newtonian fluid in a pipe.</p>
<p><span class="math-container">$\frac{du}{dt}-\frac{\sigma}{\rho}\frac{du}{dx}=g$</span> with initial condition <span class="math-container">$u(0,t)=2$</span> which means that the flow velocity at <span class="math-container">$x=0$</span> is <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$\frac{dt}{dx} = -\frac{\rho}{\sigma}\rightarrow t =-\frac{\rho}{\sigma}x+c\rightarrow c = t+\frac{\rho}{\sigma}x $</span></p>
<p>We the introduce two ned variable <span class="math-container">$\xi = c , \eta=t$</span> to put the PDE into canonical form:</p>
<p><span class="math-container">$\xi_{t} = 1 ,\xi_{x} =\frac{\rho}{\sigma} $</span> and</p>
<p><span class="math-container">$\eta_{t}=1,\eta_{x}=0$</span></p>
<p><span class="math-container">$u_{x} = \xi_{x}u_{\xi}+\eta_{x}u_{\eta} = \frac{\rho}{\sigma}u_{\xi}$</span></p>
<p><span class="math-container">$u_{t} =\xi_{t}u_{\xi}+\eta_{t}u_{\eta} = u_{\xi}+u_{\eta}$</span></p>
<p>Substituting into the original equation we get:</p>
<p><span class="math-container">$u_{\eta}=g\rightarrow u(\xi,\eta) = \eta g+f(\xi)$</span> But <span class="math-container">$\eta = t$</span> and <span class="math-container">$\xi = t+\frac{\rho}{\sigma}x$</span> so we get <span class="math-container">$u(x,t) = gt+f(t+\frac{\rho}{\sigma}x)$</span>. By applying the condition <span class="math-container">$u(0,t)=2\rightarrow gt+f(t)=2\rightarrow f(t) = 2-gt \rightarrow f(t+\frac{\rho}{\sigma}x) = 2-gt-g\frac{\rho}{\sigma}x$</span> and by substituting <span class="math-container">$f(t+\frac{\rho}{\sigma}x) \rightarrow u(x,t) = 2-g\frac{\rho}{\sigma}x$</span> which doesnt make sense since it can take negative values and predict a negative flow velocity.What am I doing wrong?</p>
| |mechanical-engineering|fluid-mechanics| | <p>According to <a href="https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations" rel="nofollow noreferrer">Wikipedia in the Navier Stokes equation</a> the <a href="https://en.wikipedia.org/wiki/Cauchy_stress_tensor" rel="nofollow noreferrer">Cauchy stress tensor</a> <span class="math-container">$\sigma = \tau-p$</span> where <span class="math-container">$\tau$</span> describes viscocity and <span class="math-container">$p$</span> describes the volumetric pressure</p>
<p>Since we can make the meaningful assumption that when <span class="math-container">$|p|>0$</span> fluid will flow in the pipe <span class="math-container">$|p|>|\tau|\rightarrow \tau-p<0$</span> so in my equation the term <span class="math-container">$-g\frac{\rho}{\sigma}>0$</span> so <span class="math-container">$2-g\frac{\rho}{\sigma}x$</span> cannot be <span class="math-container">$<0$</span>(for <span class="math-container">$x>0$</span>)</p>
| 54993 | Navier Stokes solution doesnt make sense to me |
2023-04-22T09:33:09.280 | <p>I recently saw some DIY battery builds and something has been bothering me. So for example in one of the builds they needed 14V nominal so they put together 4x3.7v batteries in series so they achieved around 14V. But there exists a lot of voltage booster ICs that could boost the voltage from for example 2 batteries in series(7.4V) to 14V and the other two battery could then be connected in parallel and thus having a higher capacity. I know these ICs are around 85% or so efficient but wouldn't the double capacity of the battery outweigh this inefficiency? So my question is why doesn't anybody use voltage booster ICs instead of using a lot of series batteries to boost voltages? Or is there a rule that says that when the voltage gets doubled using a circuit the capacity is halved?</p>
| |electrical-engineering| | <p>Voltage booster circuits are inefficient. They spend energy in the conversions, effectively resulting in a lower capacity of the energy storage system overall. Worse, you get even less energy than a system that would have stored a higher voltage (Storage will inherently result in some losses so it's odd to say less than you put in)</p>
<p>Reducing (buck) DC voltage can be significantly more efficient compared to increasing (boost) if you can tolerate the waves introduced by a switching regulator.</p>
<p>If you already have to pay the storage penalty, you can theoretically get a boost cheaply(less additional cost) by charging in parallel, discharging in series setup, but changing hookups comes with its own costs.</p>
<p>Similarly, alternating current already pays penalties due to the nature of its electromagnetic emissions, so converting via coils of a transformer doesn't incur much additional penalty. DC boosting tends to convert to something closer to AC, incurring some pretty severe losses in the process.</p>
<p>In order for any change to occur, entropy must increase. That is the rule (2nd law of thermodynamics) and it prevents anything from truly being 100% efficient (There are things such as heat pumps that can get a coefficient of performance greater than 100% by claiming they didn't care about some of that entropy-penalty-related energy expenditure so it was free).</p>
| 54996 | Why use batteries in series if voltage booster circuits exist? |
2023-04-23T08:58:04.883 | <p>An alternator generates AC current, whereas batteries (as far as I know) provide/are charged with DC current. So I'm wondering, how is this apparent mismatch explained?
Do car manufacturers employ a rectifier downstream the alternator in order to convert AC to DC?
Wouldn't it be simpler to use a dynamo instead, to directly generate DC current?
Also considering that, when present, regenerative braking systems provide DC power (so are they connected to the battery pack via a different power transmission line?).</p>
<p>P.S. I apologize in advance if the question does not meet some community requirements, I am new to Stack Overflow, plus it regards a discipline I am still trying to learn about.</p>
| |electrical-engineering|automotive-engineering|car|generator|power-generation| | <p>The brush arrangement in a DC generator wears out with use and when it stops working, the driver is stranded with a car that can't start and won't run. There are no brushes in an alternator to wear out and it is therefore less expensive to build and more reliable in use.</p>
<p>The diodes that do the job of the commutator brushes do not wear out and are less expensive than the parts they replace.</p>
| 55004 | Why does a car generator employs an alternator (i.e. AC) when power supply is provided by a battery (i.e. DC)? |
2023-04-23T12:56:59.030 | <p>In a development by Festo (BionicCobot), a rotary pneumatic actuator was used.
<a href="https://i.stack.imgur.com/ecfUo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ecfUo.png" alt="enter image description here" /></a></p>
<p>It is also similar to Kinetrol vane pneumatic actuators.
<a href="https://i.stack.imgur.com/t0w2m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t0w2m.png" alt="enter image description here" /></a></p>
<p>I have a question what is the formula for the torque of such an actuator depending on the pressure.</p>
<p>And are there any restrictions on the angle of rotation of such a drive?</p>
| |mechanical-engineering|pneumatic| | <p>I can't really answer any questions on the Festo as they're not showing much in the videos. The yellow fellah is much simpler.</p>
<blockquote>
<p>I have a question what is the formula for the torque of such an actuator depending on the pressure.</p>
</blockquote>
<p>Torque is force by distance so in this case I suspect that you'll have to do an integral. Let's assume that the vane has constant width, <span class="math-container">$w$</span>, minimum radius, <span class="math-container">$R_1$</span> and maximum radius, <span class="math-container">$r_2$</span>. Pressure is <span class="math-container">$P$</span> (and is force per unit area in your chosen units).</p>
<p><span class="math-container">$$T = \int^{R_2}_{R_1}Pwr\ dr = Pw\int^{R_2}_{R_1}r\ dr$$</span></p>
<p>If my maths is correct so far this shoult work out as</p>
<p><span class="math-container">$$ T = \frac {Pw}{2}(R_2^2 - R_1^2) $$</span></p>
<blockquote>
<p>And are there any restrictions on the angle of rotation of such a drive?</p>
</blockquote>
<p>Yes. The yellow fellah is restricted to 90° as the vane will bump into the casting. The Festo unit will have similar restrictions.</p>
| 55006 | How to calculate the torque of pneumatic rotary vane actuators? |
2023-04-24T20:26:18.663 | <p>I feel like this should be simpler than I'm finding it, but here goes. I need a transfer function for the function <code>y(t)=u(t)H(u(t)-m)(1-H(u(t)-M))</code>. This is the simplest way I could think to represent a clamp function, with max value <code>M</code> and min value <code>m</code>. As per usual, <code>u(t)</code> is the input, <code>y(t)</code> is the output.
Thanks in advance for any help!</p>
<p>Edit: <code>H(x)</code> represents the Heaviside function (0 if <code>x<0</code>, 1 otherwise)</p>
| |transfer-function| | <p>As fibonatic pointed out, the clipping function is non-linear and therefore cannot be represented as a transfer function.</p>
| 55017 | Clamp function transfer function |
2023-04-25T10:27:54.987 | <p>I've been asked to peer review a paper where the authors claim to be presenting an <strong>open source hardware</strong> item that is to be manufactured of aluminium using "CNC milling machines" or "standard lathes and milling machines". The maximum size of a part is 150 mm cubed. There are many parts involved.</p>
<p>However they only supply a series of STEP CAD files for the parts and no instructions for manufacture whatsoever. They claim the parts can alternatively be purchased from a single company which is part of the authorship (and declared in CoI). I can't give more details due to pre-publication confidentiality.</p>
<p>Since I have little experience with metalworking manufacturing my question to you is:</p>
<p><strong>What minimum documentation of 'know how' should be provided for such a project, making complex mechanical parts out of aluminium?</strong></p>
<p>For example, should details of the model of suitable lathes/CNC machines be provided? Are there any special machine files or other documentation required? Or is it reasonable to expect that any specialist aluminium metalworker can just make parts from STP files without further manufacturing know how info?</p>
<p>Thanks.</p>
| |mechanical-engineering|manufacturing-engineering|cnc| | <p>An STP file is basically a drawing isn't it? People can make parts from drawings, but not blindly. That's one reason why 3D printing is so attractive: 3D printers approach all parts the same way so you can directly feed them the STP file. Not so with machine tools. There's thousands of machines, tools, and approaches that can be used to make the same part with various tradeoffs in precision, speed, set up time, and initial investment. You have to sit down and work out the processes, tooling, and fixturing you want, and the order. Translating a drawing to a production process is its own big thing.</p>
<blockquote>
<p>What minimum documentation of 'know how' should be provided for such a project, making complex mechanical parts out of aluminium?</p>
</blockquote>
<p>Depends on what you mean by "minimum".</p>
<p>You could argue the drawing is the minimum...a bit lacking though if it doesn't have tolerances specfied and such. Just add process design experience.</p>
<p>Then there is the process manual which details the step-by-step the tooling and operation to be performed on the part, intermediary drawings, and tolerancing. Not cheap to develop, and also specific to the equipment you have available, but also not necessarily the only way to make to a part and not something you leave laying around. Used in industries like aerospace. I've been told each part actually has an its own copy of the manual following it around where the measurements after each process are filled in and signed off on.</p>
| 55024 | 'Know how' requirements for CNC / lathe manufacturing for open hardware projects |
2023-04-26T01:24:51.257 | <p>I'm trying to design a system that will have glass bottles upside down filled with liquid. They will be pressurized via a bulkhead air input and will dispense through bulkhead liquid output.</p>
<p><a href="https://i.stack.imgur.com/NEfC3.png" rel="nofollow noreferrer">Here is a crude drawing of the system concept</a></p>
<p>The bulkheads are 1/4" Push-to-Connect fittings, I asked a mechE friend if i could simply use a <a href="https://a.co/d/2ySZMTk" rel="nofollow noreferrer">one-way check valve like these</a> in between the bulkhead and the air pump but he said there will likely still be backflow issues (he didn't elaborate why).</p>
<p>What is the standard method for preventing backflow for system like this?</p>
| |mechanical-engineering|fluid-mechanics|valves| | <p>One obvious/common method would be a rigid tube sticking up above the water line:</p>
<p><a href="https://i.stack.imgur.com/I5I8H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I5I8H.png" alt="enter image description here" /></a></p>
<p>Air inlet on the left, liquid outlet on the right. If (for whatever reason) you prefer not to use rigid tubing, a sufficient length of flexible tubing with a float on the end is also sometimes used.</p>
<p>Obviously neither of these works if the bottle will start out entirely full. For a case like that, another common possibility is to have the air fill a bladder that can expand to (at least nearly) fill the bottle. As the simplest version, consider simply a child's birthday balloon. When you blow air in, the balloon inflates, filling the bottle.</p>
<p>One more common possibility is to simply put the air pump well above the top of the bottle, or at least loop the hose outside the bottle up well above the water line (far enough to assure that if the pump stops, it won't form a siphon).</p>
| 55030 | How to prevent liquid backflow when pumping an upside-down liquid container with air to pressurize it? |
2023-04-26T02:39:33.333 | <p>I'm looking for a mechanism I can employ so that I can pull liquid containing bottles that are upside down out of a system, fill it, put the cap (with mechanism) back on. Then be able to turn the bottle upside down and push it back into the system to allow liquid flow. I want it to not leak anything until i push it into the main unit.</p>
<p>Basically, a combination bulkhead+valve that where the valve is activated by connecting the bulkhead on one side.</p>
<p>I realized I've seen a "push to connect" bulkhead thingie like this before. My water filter in my fridge pushes in to start the water flow. How does it do this? What kind of bulkhead fitting is on it. <a href="https://i.stack.imgur.com/HE6hB.jpg" rel="nofollow noreferrer">Some pictures of it here</a></p>
<p>How does it work, presumably the water is ready to go on the other side and doesn't flow until you push in this unit which is horizontal.</p>
| |mechanical-engineering|fluid-mechanics|flow-control| | <p>Often a spring loaded ball valve that is opened as the unit gets pushed into its final location.</p>
<p>The spring pressure may be light as the system pressure can also be part of holding the valve closed, however if the system pressure is significant there can be a pressure balance across the valve.</p>
| 55032 | How do these push-to-connect water filters activate water flow once pushed in? |
2023-04-29T12:04:44.503 | <p><a href="https://i.stack.imgur.com/l09Rf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l09Rf.png" alt="enter image description here" /></a></p>
<p>The rectangular gate will open authomatically, when the depth of water becomes
very large. What is the minimum value of d that will cause the gate to just open?
The width of the gate is 2 m.</p>
| |fluid-mechanics|civil-engineering|pressure|hydrostatics|water-pressure| | <p>The hydrostatic pressure can be divided into constant part causing a force per width <span class="math-container">$\rho\cdot g\cdot h\cdot d$</span> plus linear part with force per width <span class="math-container">$\rho\cdot g\cdot \frac{h^2}{2}$</span>, where <span class="math-container">$h=3m$</span> is the height of the gate. The constant part can be replaced by a force acting in the center of the gate, so with lever arm <span class="math-container">$L_1 = h/2-1.35$</span> above the hinge. The force from linear pressure part will act at the <span class="math-container">$h/3$</span> from the bottom, so with a lever arm <span class="math-container">$L_2 = 1.35-h/3$</span> below the hinge. When the resulting moment is zero, the moments on the hinge from both forces must be equal:</p>
<p><span class="math-container">$$h\cdot d\cdot L_1 = \frac{h^2}{2}\cdot L_2$$</span></p>
<p>From that:
<span class="math-container">$$d = \frac{h}{2}\frac{L_2}{L_1} = 3.5m$$</span></p>
| 55069 | Fluid Mechanics about Hydrostatic Pressure |
2023-04-29T21:58:37.077 | <p><a href="https://i.stack.imgur.com/yERhL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yERhL.jpg" alt="Water tank with faucet on the bottom side" /></a></p>
<p>Consider I have water tank as shown by the picture. By normal condition, the flow rate when the tank is full is higher than when the tank almost finish, and till it finally finish it will stop (in this case, what we consider is until the water level reaches the upper side of the faucet). But I want the water inside the tank flows from the faucet at constant flow rate without any additional electronic/microcontroller device. I want a pure mechanic, hopefully I may have a simple one. My question then, is any method to do it?</p>
| |flow-control|hydrostatics|water-pressure| | <p>Another type of 'constant flow' valve is a tornado valve, sometimes used to create stormwater overflow in sewerage systems. Incoming fluid at increasing pressure makes the tornado spin faster, and the held spinning volume provides back pressure, but the outflow is only from the free center -- where the height is only the height of the 'valve'.</p>
| 55080 | How to make a constant flow rate of water from a tank with a faucet at the bottom side? |
2023-05-01T11:09:00.943 | <p>I want to calculate the error of a control loop. Therefore, the system and the general processes and models are defined as follows:<a href="https://i.stack.imgur.com/Zdmlp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zdmlp.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/CiLhK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CiLhK.png" alt="enter image description here" /></a></p>
<p>The equation for error is described as follows:
<a href="https://i.stack.imgur.com/qT9gh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qT9gh.png" alt="enter image description here" /></a></p>
<p>I do not understand how they eliminate the denominator in the first term (1 + A1s + . . + Ans^n) and the denominator (B0 + B1s + . . + Bms^m) in the second term.</p>
<p>Thank you in advance!</p>
| |mechanical-engineering|electrical-engineering|control-engineering|control-theory|modeling| | <p>As is stated in your equation:
<span class="math-container">$M_1(s)\,M_2(s) = (bo + B_1s + \cdots + b_m\,s^m)\frac{1}{1+a_1\,s+\cdots+a_n\,s^n} $</span> from here we observe that</p>
<ul>
<li><span class="math-container">$M_1(s) = bo + B_1s + \cdots + b_m\,s^m$</span> , and</li>
<li><span class="math-container">$M_2^{-1}(s) = 1+a_1\,s+\cdots+a_n\,s^n$</span>.</li>
</ul>
<p>From your block diagram we can observe that <span class="math-container">$$e(s)= Y(s)\,M_2^{-1} - U(s)\, M_1(s)$$</span>
Replacing the above equations, gives us the result you were seeking, and as @Lidell says, nothing is eliminated.</p>
| 55096 | Error Equation of a Control Loop |
2023-05-02T14:02:53.407 | <p>Looking for a compliant mechanism that can witsdent sheat force but still move in another axis</p>
<p>Is there such compliant mechanism that can resist shear force but still bend like a hinge?</p>
<p>Is it even possible?</p>
| |mechanisms|shear|hinge| | <p>Yes, such a device is called a compliant hinge or a flexure hinge. Here is an example: <a href="https://www.youtube.com/watch?v=WceyxTb_xVc" rel="nofollow noreferrer">https://www.youtube.com/watch?v=WceyxTb_xVc</a></p>
| 55106 | Compliant mechanism that can resist shear force |
2023-05-09T17:31:22.280 | <p>I've made a plywood carbon sheet and tested its breaking force, for a 200mm long, 50mm wide, 16mm thick it was 100kg +-</p>
<p>I was wondering if it's enough to help me predict what will be the breaking force when its the same beam but 1000mm long and 2000mm?</p>
<p>Is it linear? am I'm missing some vars in order to calculate it?</p>
<p>It's harder and more costly for me to test longer beam, as I'm in the prototyping phase and trying to determine to right thickness and plywood carbon ratio of the beam</p>
| |structural-engineering|structural-analysis|beam|strength| | <p>As Tiger Guy says one test is absolutely not reliable. There are extensive tests under precisely controlled conditions by commercial structural lumber manufacturers. here is one. <a href="https://www.sciencedirect.com/science/article/pii/S0263822320313477" rel="nofollow noreferrer">source</a></p>
<p>But let's say you did a hundred tests and the average was 100kg.
A rough relation between rapture load and beam properties, when a load F is applied at mid-span with length L, is</p>
<p><span class="math-container">$$M_{rapture}= \frac{F_{rapture}L}{4}=Constant\quad F_{rapture}\propto \frac{1}{L}$$</span></p>
<p>So if we keep all dimensions the same except the length L, if you multiply the L=200mmby 5 to get L=1000mm you can expect it will barely support,
<span class="math-container">$$F_{1000mm}= 100/5=20kg$$</span></p>
<p>And for l=2000mm
<span class="math-container">$$F_{2000mm}= 100/10=10kg.$$</span></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/HUwCi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HUwCi.png" alt="beam with load F" /></a></p>
| 55174 | Predicting breaking force of beam per length based on one testing |
2023-05-10T11:45:37.713 | <p>I am designing a cylindrical shaft that has a round disc at its top. I am hoping to seat this shaft in 2 radial bearings and couple to a motor. The image below gives an overview of the loading diagram. There is an axial force applied when the system is static, so I am aiming to calculate the equivalent radial force in the system to specify the radial bearings adequately to handle both radial and axial forces.</p>
<p>Can someone point me toward the correct beam theory to use for calculating what radial force the applied downward force would create on the bearings?</p>
<p>I have checked overhanging beam theory and cantilever beam theory, which point to a moment like this creating zero shear force, which I believe would mean 0 radial force, which doesn't make a lot of sense to me when I consider how the shaft would react to this downward force applied at a distance from the spin axis.</p>
<p><a href="https://i.stack.imgur.com/81AiS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/81AiS.png" alt="Load Diagram. I am wondering how I should approach calculating R1 and R2" /></a></p>
<p>Let me know if you need more information, thank you</p>
| |beam|bearings|shear|bending| | <p>The sketch shows the distribution of static reactions (C & V). Hope this helps.</p>
<p><a href="https://i.stack.imgur.com/rvctJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rvctJ.png" alt="enter image description here" /></a></p>
| 55179 | Correct beam theory for calculating radial force in bearing due to a bending moment in beam? |
2023-05-10T15:41:53.060 | <p>Lets say I have a block of metal clamped in a vice on my mill, and that part has a hole in the face that's facing me (parallel to the Y axis).</p>
<p>What's the best way to zero the X axis on that hole? The hole diameter is about 6mm</p>
<p>The only way I can think of is to first locate the hole relative to a flat surface, and then zero the mill on that surface. I figure that's not ideal due to relying on the surface accuracy, and tolerance stack up. It would be really nice to zero on the hole directly. Is there a way to do that?</p>
| |machining|cnc| | <p>Honestly using another datum is probably the simplest way... (Assuming once you do the tolerance stack calcs, that the possible deviation is within acceptable limits...)</p>
<p>Since you are asking to use the hole as a datum, however, I'm assuming it has a precise fit/tolerance. If so you could consider a precision dowel fitted into the hole, and then touch off both outside faces of that dowel, before splitting the difference to find the centre.</p>
| 55182 | What's the best way to zero a mill using a hole on the SIDE of the part |
2023-05-12T19:11:30.173 | <p>So imagine this, an hourglass that is the size of a two storey house in height but is wide as an average sedan car that can trickle sand for 5 hours, and at the middle there's a gear which is turned as the sand trickles through it (thus creating electricity **just guessing here), on the outside of the hour glass in the middle, there is a turning point to allow the hour glass to be turned when there's no more sand in the upper part of the hour glass.
That turning point will be turned by a motor that gets power from an inverter, which is connected to a battery that is charged by solar.</p>
<ol>
<li><p>Will this be able to create electricity continuously or not (since there are two sources of energy, radiation from the sun and gravity from the earth)</p>
</li>
<li><p>If not please tell me why and be gentle about it, still young :).</p>
</li>
<li><p>I assumed that it would take less energy to turn the hourglass itself than to actually lift the sand inside upwards (am I wrong about this, if so, why? )</p>
</li>
<li><p>I also heard someone say you cannot get more energy that you put in (now to me this was out of context because I didn't really understand it ,please try to simplify this to me)</p>
</li>
</ol>
<p>TL,DR : I'm a teenager who wants to generate electricity by trickling the sand in an hour glass ,and turning the hour glass using a motor ,connected to a battery ,recharged by solar, is it possible or feasible for one household?</p>
| |renewable-energy|energy-storage|solar|electricity| | <p>The initial potential energy is given by
<span class="math-container">$$E = mgh$$</span>
where <span class="math-container">$m$</span> = mass of the sand, <span class="math-container">$g$</span> is the acceleration due to gravity and <span class="math-container">$h$</span> is the height above the turbine. All the potential energy between the turbine and the bottom of the hourglass will be lost as the sand free-falls there without energy conversion. This matters greatly because the obvious place to locate the turbine in your scheme is in the neck of the hourglass.</p>
<p>You'll also have to consider the turbine efficiency. Let's say it's 90%.</p>
<p>Then we have to rotate the hourglass and lift the sand from the bottom of the glass back up to its initial height. The energy required this time is <span class="math-container">$mgH$</span> where <span class="math-container">$H$</span> is the height from the bottom (not the turbine as before) back to the storage height. This again will have a motor and gearbox efficiency. Let's say that's 80%.</p>
<p>If we say, for example, that <span class="math-container">$H = 2h$</span> we can recalculate that energy as follows:
<span class="math-container">$$E_{out} = mgh\times 0.9 = 0.9mgh$$</span>
<span class="math-container">$$E_{in} = mgH \times 0.8 = 2mgh \times 0.8 = 1.6mgh$$</span>
That means that the system will consume <span class="math-container">$\frac {1.6}{0.9} = 1.77$</span> times as much energy as it generates.</p>
<p>In this situation it would be far more efficient to just use the solar. The only reason not to would be if your sand battery was a cheaper solution to energy storage than a chemical battery.</p>
<p>In practice there are better ways of doing this. Hydro pumped storage is the most common and water would be much easier to manage than sand. You could also have a look at "gravity battery" concepts that are being trialled but they seem a bit daft as they are replacing water with concrete.</p>
| 55205 | Can we create electricity through an hourglass? |
2023-05-16T02:22:06.860 | <p>Was reviewing some notes on fluid dynamics, and the notes go as follows (conservation of mass for a qubic CV),
<span class="math-container">$$\frac{dm_{out}}{dt} = \rho u (dydz)_{x+dx} + \rho v (dxdz)_{y+dy} + (similarly,forZ) = \rho u dydz +\frac{\partial \rho u dydz}{\partial x} dx + (similarly,forYandZ)$$</span>
I don’t understand what the “x + dx” subscript means. Is that a Taylor expansion? I’m having a hard time following that step</p>
| |mechanical-engineering|fluid-mechanics|mass-transfer| | <p>It just means "located at the plane <span class="math-container">$x+dx$</span>", and it really should be applied to the velocity, not the <span class="math-container">$dydz$</span>.</p>
<p>Your cube has 6 faces. Flow in the <span class="math-container">$x$</span> direction can enter or exit the Control Volume by crossing the plane perpendicular to <span class="math-container">$x$</span> direction at location <span class="math-container">$x$</span>, and also at the location <span class="math-container">$x+dx$</span> (<span class="math-container">$dx$</span> is the side length of your cube). Likewise for flow in the <span class="math-container">$y$</span> or <span class="math-container">$z$</span> directions.</p>
<p>To build the continuity equation, you are counting up all the mass flow rates into and out of the CV.</p>
<p>I think a better way to write your equation is:</p>
<p><span class="math-container">$$ \frac{dm_{out}}{dt} = \rho u(x+dx) dydz + \rho v(y+dy) dxdz + \rho w(z+dz) dxdy
$$</span></p>
<p>(With the <span class="math-container">$x+dx$</span> etc as the argument of the velocities)</p>
<p>Then you also have</p>
<p><span class="math-container">$$ \frac{dm_{in}}{dt} = \rho u(x) dydz + \rho v(y) dxdz + \rho w(z) dxdy
$$</span></p>
<p>Conservation of Mass requires:</p>
<p><span class="math-container">$$ \frac{dm_{in}}{dt} - \frac{dm_{out}}{dt} = \frac{dm}{dt}$$</span></p>
<p>And the mass <span class="math-container">$m$</span> inside the CV equals the density times volume:</p>
<p><span class="math-container">$$m=\rho \,dxdydz$$</span></p>
<p>Try to combine these results to get your continuity equation.</p>
| 55243 | Notation on Continuity Eqn Derivation |
2023-05-16T22:57:58.170 | <p>I'm trying to understand why the Mythbuster's <a href="https://www.youtube.com/watch?v=VUiGhyHC-1A" rel="nofollow noreferrer">dimple car</a> gets better fuel economy - if it actually would as I'm not</p>
<p>I understand that a golf ball functions by creating turbulent air flow and via the disruption of the boundary layer it reduces drag and increases lift.</p>
<p>The only thing I can think of is maybe there's an area of low pressure directly behind the car and by adding the dimples it is creating turbulent flow that maybe prevents this? That seems like a stretch though.</p>
| |automotive-engineering|aerospace-engineering|aerodynamics| | <p>When a round object travels thru air, one would expect aerodynamic smoothness, in contrast to flat, geometric angled objects. But the resulting air flow behind it creates a vortex with in turn draws more air against it creating drag.<br />
A dimpled ball breaks that up by creating a micro layer of air preventing the ball from "Attaching" to the surface. Nearly all sports balls use perforated or oddly grooved surfaces to deter aerodynamic catching.</p>
<p><a href="https://i.stack.imgur.com/yt6ZX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yt6ZX.jpg" alt="enter image description here" /></a></p>
<ol>
<li>Basketballs (albeit thats more for grip adherence)
<a href="https://i.stack.imgur.com/pq1Wp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pq1Wp.jpg" alt="enter image description here" /></a></li>
<li>Volleyballs
<a href="https://i.stack.imgur.com/0d2EA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0d2EA.jpg" alt="enter image description here" /></a></li>
</ol>
| 55251 | Why did the Mythbuster's dimple car get better fuel economy? |
2023-05-18T18:56:43.973 | <p>I was milling a pocket recently with a 7/16in endmill. It came out 20 thousandths undersized. Worn tool maybe? To try to fix this, I expanded the coordinates by 10 thou and milled the pocket again. Surprisingly, when I measured afterward it had barely changed.</p>
<p>My guess is that the cut was too light and the endmill was deflecting instead of cutting? Is that likely what happened? I didn't try to go farther out of fear that it would start cutting all of a sudden and oversize the pocket.</p>
<p>The material is 7075 aluminum, and I had to clamp only about 0.5in of the endmill in the collet, so that I could reach down far enough into the part. I know that's not ideal. The distance from the collet to the blades was about 1in, and I cut approximately 0.6in depth per pass (just on this operation).</p>
<p>What do you think happened, and how can I remedy this?</p>
<p><strong>Edit:</strong> I repeated the original cut using plunge milling followed by a finishing pass to cut off the bumps and that fixed it.</p>
| |machining| | <p>Yes, you can definitely take too small a cut. This becomes quite evident when working on a manual machine. The smallest cut you can effectively take depends on the material, the sharpness of the cutting tool, and to an extent what the cutting tool is made of.</p>
<p>Materials have spring to them so taking too light a can cause the material to deform (I'm talking about the workpiece here but I suppose it also holds for the cutting tool) under cutting pressure and not "bite" preventing a proper cut from happening. When this happens the cutting tool instead rubs producing excessive heat made worse by the absence of chips to carry the heat away which kills the cutting tool, or deforms the material instead which works harden the material.</p>
<p>Therefore a sharper tool lets you take smaller cutters since it reduces cutting pressure is required.</p>
<p>All that said, I don't think that should be happening at 10 thousandths in aluminum in a tool of good condition. But I am not familiar with 7075.</p>
<p>Take a cut on a test piece and see how much is actually cut versus the coordinates of the tool. Sometimes it's more and sometimes it's less.</p>
| 55277 | When milling, is there such a thing as a cut that is too small? |
2023-05-19T19:20:54.810 | <p>In some electric bikes, a motor is mounted to the hub of the front wheel. When turned on, this hub motor applies a torque, turning the wheel, reacting against the ground, with a resultant force acting against the inertia of bike / rider / air resistance etc.</p>
<p>How can I work out the force applied to the centre of the axle, given a motor acting with specific torque t
and wheel or radius r, assuming the tyre does not overcome the friction force of the ground?</p>
<p>Lets say we fix the wheel, axle, and motor to a fork, that is anchored (so nothing can move). Lets assume the fork cannot flex.</p>
<p>Forgive my crude drawing, but this might look something like this:</p>
<p><a href="https://i.stack.imgur.com/O7Uju.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7Uju.jpg" alt="enter image description here" /></a></p>
<p>The Force of the motor acts on the rim, then the reaction force from the road acts on the fork (I think). The net foce is to the right hand of the page (where the motor is at the centre of the wheel).</p>
<p>Assuming the motor has a Torque of 40Nm, what is the force on the fork centred at the axle?</p>
<p>(I'm trying to determine if my fork can support a front motor, so the next stage will be to look at whether this fork can support the force at that distance)</p>
| |mechanical-engineering|applied-mechanics|torque|forces| | <blockquote>
<p>Assuming the motor has a Torque of 40Nm, what is the force on the fork centred at the axle?</p>
</blockquote>
<p>Since the system is fixed in place, there is no acceleration or rotation. Therefore any torque developed by the motor is completely countered by friction with the ground, and that force is transmitted to the fork. So this is just <span class="math-container">$\frac{40\rm{Nm}} {311\rm{mm}} = 129\rm{N}$</span>. If the wheel could rotate and the frame could accelerate, that force would be reduced.</p>
<p>Presumably if instead of anchoring your fork you allowed it to move with minimal resistance, you could attach a scale to it and directly measure the force that the motor is creating.</p>
| 55295 | How to calculate force acting on axle from hub motor with known torque? |
2023-05-21T17:54:13.587 | <p>I've got a machine that sucks water. The inlet is half inch. I had been using braided pvc till now but it has started to breathe air which is causing problems with the machine. It also bends under stress or in hot weather. I'd like a half inch pipe that doesn't breath and doesn't bend and can be pulled over a barbed nipple. Any ideas what I should be looking for?</p>
| |pipelines|waterproofing|air|tubing| | <p>You need a pipe rated for suction.</p>
<p>Or use a corrugated steel inner which won’t collapse - did that for one application…</p>
| 55313 | Suitable pipe suggestion required |
2023-05-23T10:58:28.247 | <p>and thanks in advance for taking the time to help me.</p>
<p><strong>TL;DR: I work for a company where I need to make technical drawings—mostly using SolidWorks—of some mechanical components. Do I have to include the scale for the technical drawings even if 1) no one in the manufacturing team pays attention to it and 2) I provide all of the relevant dimensions on the drawing?</strong></p>
<p>I'm an engineer, and I currently work for a company as an R&D and product development engineer. A big part of my job is to design and create technical drawings for the manufacturing team to carry out the necessary processes to make what I designed a real product.</p>
<p>In our company, we manufacture parts for medical devices, which makes every aspect of our business quite complicated by having to adhere to a very strict set of rules and standards, which also applies to the format of our technical drawings. We are now updating our guidelines and standards for our technical drawings, and one of the topics that was brought up is the importance of providing scale information in the technical drawings (for example, 2:3 @ A4).</p>
<p>Now I know that adding the scale information is not that difficult or time-consuming, but I couldn't help but notice that apart from the discussions we had recently on this particular topic, no one, especially in the manufacturing team, has ever looked at or paid any sort of meaningful attention to the scale information despite us (the design and technical drawing team) always showing it in the drawings.</p>
<p>The thing is, we use mainly SolidWorks to do our job, and sometimes it is easy to change the scale of a drawing without updating the relevant field that automatically shows the universal scale of the drawing. This obviously causes the true scale of the drawing to conflict with the scale provided in the relevant field in the title blocks. That is why I'm wondering whether or not it is an acceptable practice to omit such a detail, especially when taking into consideration that 1) no one ever asked or paid attention to the scale information, and 2) we already provide all of the relevant dimensions clearly, which totally negates the need for a scale in our drawings in our company.</p>
<p>I'm almost sure that our management will accept whatever I deem "acceptable", but I couldn't find any conclusive answers in that regard, and most of the articles discussing this topic are targeting technical drawings created by architects and civil engineers, which is quite different from what we do here.</p>
<p>Thanks for taking the time, and I appreciate your help.</p>
| |mechanical-engineering|solidworks|technical-drawing|standards|drawings| | <p>You should not ommit the scale. Even if no one seems to read it, it is still important information. The task is quite trivial, I don't see any reason why to remove the scale because it can cause several issue later on.</p>
<ol>
<li><p>Since you are working for a medical device company and might be exposed to an audit from an quality organ, such as FDA or Tüv. They will look at everything and if it is not correctly documented the company might not pass the audit.</p>
</li>
<li><p>When a new employee has to look up old information, maybe you will no longer work there. It could become tedious to try and figure out what was done if data is missing from the drawing.</p>
</li>
</ol>
| 55333 | Can I omit the scale information in the technical drawing of a mildly complex mechanical part if I provide all of the necessary dimensions clearly? |
2023-05-23T18:27:04.157 | <p>I am designing a cylindrical container that will be filled with water. One end of the container will be solid with flow port and the other end will be a 3mm rubber sheet. I want to oscillate the volume of the container by pushing and pulling the rubber sheet. My main concerns are preventing leaks and making sure the rubber sheet does not tear over thousands of actuations. My current design is:</p>
<p><img src="https://i.stack.imgur.com/nm6rd.png" alt="Layout" /></p>
<p><img src="https://i.stack.imgur.com/kcBoW.png" alt="Dimensions" /></p>
<p>Here are the details of my rubber sheet:</p>
<p><a href="https://i.stack.imgur.com/t2lgf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t2lgf.png" alt="enter image description here" /></a></p>
<p>My machinist expressed concern that the current design would shear the rubber sheet too much and cause early failure. I am unsure how to even approach this design challenge. Current design was created with some trial and error and 3d printed physical testing. Sandwiching the rubber between flat surfaces did not hold the sheet in place well enough. I am unsure what to even look up to find the right calculations for this application. Any help is appreciated. Thanks!</p>
| |fluid|shear|elastic-modulus|seals| | <p>Consider some inspiration from diaphragm valves, which do something similar:
<a href="https://i.stack.imgur.com/VxJcG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VxJcG.png" alt="Diaphragm valve" /></a></p>
<p>[Image source: https://en.wikipedia.org/wiki/Diaphragm_valve]</p>
<p>Additionally, to prevent damaging the rubber sheet over many actuation cycles, I could imagine building a chamber on the other side (like where the yellow sort of plunger sits in the image) and working with pressured air there. This way you might be able to move the rubber sheet in a more gentle way than for example mechanically.</p>
| 55339 | How can I anchor a rubber sheet that will be stretched? |
2023-05-24T05:45:13.903 | <p>Surface roughness standards generally say that the thing you specify is the maximum your willing to accept. But manufacturer could make a finer job if they wished.</p>
<p>I however have a situation where the surface has to be quite rough and im having problems with my supplier naturally making better surface quality than I want.</p>
<p>So for now ive clarified this in text and that works fine for now. But just for clarity, how would i have actually supposed to document this in a standards compilant way in the drawing?*</p>
<p>* Not that i think it would have helped as im pretty sure the machine shop would not have understood it without clarification but just out of curiosity.</p>
| |mechanical-engineering|technical-drawing| | <p>I am not sure what the current recommendation in the international standards is, since in mechanics I realized that a lot of things vary from industries to industries, and in the same industry from companies to companies.</p>
<p>However, I have most of the time seen and used the recommendation below, taken from this <a href="https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/08/surface-roughness.pdf" rel="nofollow noreferrer">Cheat Sheet on Drawing Indictation of Surface Texture</a>.</p>
<p><a href="https://i.stack.imgur.com/VWVwE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VWVwE.png" alt="Indication of a range of surface roughness (RA) on a technical drawing" /></a></p>
| 55342 | surface roughness define how rough surface has to atleast be |
2023-05-25T15:42:28.577 | <p>I need to calculate the principal stresses for the given cylinder with thick walls inside filled with gas shown on this picture:<a href="https://i.stack.imgur.com/bcyNW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bcyNW.png" alt="enter image description here" /></a></p>
<p>I used the Lamé equations, namely:</p>
<p><span class="math-container">$$\sigma_{x} = p\cdot\frac{r_{1}^2}{r_{2}^2-r_{1}^2}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\sigma_{r,\phi} = \sigma_{x}\cdot[1\mp(\frac{r_{2}}{r})^2].$$</span></p>
<p>I got the 3 stresses, namely: the axial, the radial and the circumferential stress.</p>
<p>Also I have 2 moments acting on the cylinder.</p>
<p>I calculated the normal stress, which I will denote as <span class="math-container">$\sigma_{xn}$</span> as follows:</p>
<p><span class="math-container">$$\sigma_{xn} = \frac{M_{y}}{I_{y}}\cdot z$$</span></p>
<p>or actually:</p>
<p><span class="math-container">$$\sigma_{xn} = \frac{M_{y}}{I_{p}}\cdot z$$</span></p>
<p><span class="math-container">$$I_{p} = (d_{2}^4-d_{1}^4)\cdot\frac{\pi}{32}.$$</span></p>
<p>I calculated the normal stress and everything is fine until now. But now I have 4 stresses total, namely the 3 from the Lamé equations and 1 normal stress. I have to calculate the principal stress on the points A and B on the picture.</p>
<p>I know that I have to use the superposition principle to add the stresses together, but I don't know how.</p>
<p>Is there some sort of equation like:</p>
<p><span class="math-container">$$\sigma_{1,2} = \frac{\sigma_x + \sigma_r}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_r}{2}\right)^2 + \tau_{r\phi}^2}$$</span></p>
<p>for this task?</p>
<p>How to add the stresses when they are all of different nature?</p>
<p>Thank you.</p>
| |structural-engineering|structural-analysis|applied-mechanics|statics|strength| | <p>If you have multiple load cases of the same structure with the same supports, you can calculate elastic stress components for each load case separately and then simply add the resulting stresses to get stress for all the loads acting together. Adding the stresses is straight forward, you just have to be careful about coordinates or directions. Stress at a point is tensor with respect to specific orientation, so if you need to add 2 stress tensors (by components), they both need to be in the same coordinates.</p>
<p>In your case, principal Lamé stresses will be in cylindrical coordinates (axial, radial, circumferential) and the principal stress from bending moments would be only in axial direction (if the moments are torque, you get shear circumferential/axial shear stresses), so you would be adding only axial stress components.</p>
<p>When in doubt, just write down the whole stress tensors from both loads in the same coordinates and add their corresponding components.</p>
| 55350 | Superposition of cylinder stress and normal stress |
2023-05-26T17:58:12.547 | <p>I've been studying the contour method for measuring residual stress. See <a href="https://www.lanl.gov/contour/" rel="nofollow noreferrer">https://www.lanl.gov/contour/</a> if you are not familiar.</p>
<p>In the contour method, a single cut is made all the way through the workpiece.</p>
<p>What if the cut was made only part of the way through the workpiece? Would such a cut be sufficient to estimate the component of stress normal to the cutting plane?</p>
<p>Has anyone ever tried this or seen any work which tries to do this?</p>
| |mechanical-engineering|stresses|finite-element-method| | <p>It depends what you mean by normal stress. Mean value of normal stress across a full section should be 0 if the specimen is not loaded in any way. So I presume you are interested in actual distribution of normal stress across the section.</p>
<p>I don't think partial cut would "cut it" in general case. You need to make the cut all the way through, so the newly created surface can fully relax and take a new shape due to residual stress.</p>
<p>If you do just a partial cut, there are infinitely many possibilities for residual stress in the uncut part of the cutting plane and also the cut section may not have free space to relax if there were local compressive stresses.</p>
| 55359 | Possible Modification to Contour Method for Residual Stress Measurement |
2023-05-29T21:05:21.790 | <p>I am making a machine which, from the curve of the top of a violin, cuts the feet of a bridge.</p>
<p>I thought that a sensor attached to a motorized rail could take precise measurements of the distance from the top of the rail to the top of the violin, which would be translated to a curve (from the distance measurements in the X and Y directions) used to cut the feet of a bridge on a CNC machine.
This sensor wouldn't need to have a high range (the rail could be as low as needed for the sensor) but a high precision would be required. Since the tolerance of my CNC machine is +/- 0.005 in, I would like a sensor which could measure somewhere in the range of that precision.</p>
<p>However, I cannot seem to find a sensor that would do the job. I want to use a Raspberry Pi to control the motorized rail system which the sensor would be attached to, which means that the sensor would have to interface in some way with the Raspberry Pi.</p>
<p>After some research, it seems that a linear displacement sensor could work, as would a Triangulation Laser Distance Sensor. However, I searched for products and could not find any. I took a look at laser parts like the <a href="https://rads.stackoverflow.com/amzn/click/com/B089F5B9WP" rel="nofollow noreferrer" rel="nofollow noreferrer">Sharp GP2Y0E03</a> sensor, but the only information about the accuracy I could see was "High Precision Measurement", which isn't exactly helpful.</p>
<p>I would think that a part sensor similar digital caliper but for distance would exist?</p>
<p>Here is (an admittedly bad) drawing of what I am trying to achieve.</p>
<p><a href="https://i.stack.imgur.com/SHifI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SHifI.png" alt="Poorly drawn image of the curve and the measuring tool" /></a></p>
| |sensors|distance-measurement| | <p>If your CNC machine is a CNC router and not a CNC lathe nor a CNC Machine Center, you may have what you require, short of a dial indicator.</p>
<p>Consider to secure the curved portion to the bed of the machine. Attach a dial indicator to the gantry with the contact at the highest (closest) point of the curve. Your CNC readout should provide a reference point. Jog the gantry left and right to ensure you are at the closest point and note the y-axis coordinate.</p>
<p>As you jog in the x-direction a specific amount, say five millimeters, the dial indicator will change specific to the curved portion. This provides the y-direction/distance required to plot a curve to match.</p>
<p>You can reduce the x-axis distance as needed if you require greater than five millimeter resolution or increase it if you don't need such resolution.</p>
| 55384 | Sensor to measure distance with high precision |
2023-05-30T17:29:51.193 | <p>My apologies if this isn't the right Stack to ask about dielectric elastomers, but I couldn't find any other search result on other stacks about dielectric elastomers, except here.</p>
<hr />
<p>So, no matter how much I search, there isn't any kind of artificial muscle that even reaches the efficiency of electric motors (above 90%).</p>
<p>Even carbon nanotubes and graphene soft actuators can't even reach 2% of efficiency (as far the articles I read that actually tell its efficiency).</p>
<p>I just stumbled upon Dielectric Elastomers, which are considered an promising option, but I can't find an elastomer that can even reach such efficiencies. The most efficient that I heard about was an specific type of Dielectric Elastomer that the "ChatGPT" of Quora suggested that it has 42% of efficiency: "<a href="https://www.science.org/doi/10.1126/science.abn0099" rel="nofollow noreferrer">A processable, high-performance dielectric elastomer and multilayering process</a>".</p>
<hr />
<p>In anyway, I found this helical Dielectric Elastomer, which is basically a spiral tape with the positive side on the top part and the negative side on the bottom part, since it is an spiral, the top half will always be attracted to the bottom half: "<a href="https://www.researchgate.net/publication/331725568_Force_optimization_and_numerical_validation_of_helical_dielectric_elastomer_actuator" rel="nofollow noreferrer">Force optimization and numerical validation of helical dielectric elastomer actuator</a>".</p>
<p><a href="https://i.stack.imgur.com/0lcjM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0lcjM.png" alt="Illustration showing the dielectric elastomer" /></a></p>
<hr />
<p>By replacing the conductive coating for actual copper wiring, how much efficient would this system be?
Or it would only work with copper wiring if thes spirals where covered into small copper coils like electromagnets?</p>
| |mechanical-engineering|electrical-engineering|energy-efficiency|actuator| | <p>Copper works hardens, then fractures.
Don’t think this will last long.</p>
| 55391 | What would be the efficiency of dielectric elastomers if these were made with copper wire instead? |
2023-05-30T20:38:29.307 | <p>I'm interested in the residual stress profile induced by shot peening. Most of the papers I have read do some measurement and include a graphic which looks like: <a href="https://i.stack.imgur.com/r8liL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r8liL.png" alt="enter image description here" /></a></p>
<p>This graphic identifies how the residual stress varies with depth. Presumably, if the depth direction is Z and the shot peening balls strike a surface parallel to the X-Y plane then this graphic illustrates the residual stress in the directions X and Y. Is it correct to assume that the stresses are symmetric in the X and Y directions? Also, would shot peening have any impact on the component of stress in the Z direction, the direction of depth? Is it assumed that the component of stress in the Z direction is negligible compared to the components in the X and Y directions?</p>
| |mechanical-engineering|stresses|finite-element-method| | <p>Stress component in <span class="math-container">$Z$</span> direction <span class="math-container">$\sigma_z$</span> must be in equilibrium with normal pressure at the surface. If there is no pressure, <span class="math-container">$\sigma_z$</span> is zero (it can still be nonzero below the surface). However, this component is important during the peening, because during the impact, this component is definitely nonzero at the surface and the compression must be so intensive, that it induces some plastic deformation. After that, <span class="math-container">$\sigma_z$</span> drops to zero at the surface.</p>
<p>Assuming shot peening is done perpendicularly to the surface, symmetry in <span class="math-container">$X-Y$</span> could depend on the shape and size of specimen. The symmetry would probably be worst for small specimens of irregular shapes.</p>
| 55393 | Residual Stress Induced By Shot Peening |
2023-06-02T09:09:29.047 | <p>The technical field of the question is gantries in <a href="https://corexy.com/theory.html" rel="nofollow noreferrer">CoreXY</a> 3D printers, where a gantry moves in Y direction and it holds a carriage sliding in X direction. In the following image, X is horizontal and Y is vertical.</p>
<p><a href="https://i.stack.imgur.com/6f77j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6f77j.png" alt="enter image description here" /></a></p>
<p>When the gantry changes vertical (Y) direction suddenly, it acts as a beam supported at both ends and it deflects/resonates. This results in reduced quality (the 3D prints show waves on some surfaces) which forces more conservative settings for the acceleration and longer print times.</p>
<p>Different materials are commonly used for the gantry, but even if I know what moment of inertia is, I cannot grasp the (theoretical) connection between moment of inertia, density of the material, Young modulus.</p>
<p>A typical case is a carriage 300 g heavy mounted on one of the following profiles as stiffening support.</p>
<ul>
<li><p>aluminium 2020 extrusions (I=7.4x10^3 mm^4, 0.25 kg, Young modulus 70 GPa)</p>
</li>
<li><p>square hollow aluminium profile (Young modulus 70 GPa)</p>
<ul>
<li>1 mm wall (I=4.5x10^3 mm^4, 0.1 kg)</li>
<li>2 mm wall (I=7.8x10^3 mm^4, 0.2 kg)</li>
</ul>
</li>
<li><p>square steel hollow profile (Young modulus 200 GPa)</p>
<ul>
<li>1 mm wall (I=4.5x10^3 mm^4, 0.27 kg)</li>
<li>2 mm wall (I= 7.8x10^3 mm^4, 0.54 kg)</li>
</ul>
</li>
</ul>
<p>I ignored the steel rail on which the carriage actually slides, and the use of carbon fibers beams.</p>
<p>Which combination would result in:</p>
<ul>
<li>lowest amplitude of the resonance</li>
<li>highest resonance frequency</li>
</ul>
<p>I can see that a 2 mm alu profile has the same weight and Young's modulus as a 2020 profile, but a moment of inertia about 10 times as much: it will be better from any point of view, it's easy to understand because only one parameter changes.</p>
<p>However, how do I compare those performances of interest (amplitude and frequency of resonance) between steel and aluminium? for a given moment of inertia steel weight is almost 3 times as high, but so is Young's modulus.</p>
| |beam|stiffness| | <p>Not moment of inertia, but second moment of inertia.</p>
<p>With regards to your last sentence of aluminum vs steel: Steel might have a modulus 3x as high, but stiffness of the structure increases to the cubed of the distance away from the bending axis. That means that with aluminum being 1/3 the density of steel, you can use a lot more volume of it for the same mass, and with all that volume the end structure can be made much stiffer than steel because you can position so much more material farther away from the bending axis.</p>
<p>So for same volume (i.e. aka same geometry) steel will be stiffer. For same mass. aluminum can be designed to be stiffer.</p>
<p>Dynamic stiffness and resonances is beyond that which I do not know but I believe it involves other material parameters such as strain-stress curves (Poisson ratio) and lossiness (viscous damping ratio?)</p>
| 55410 | How to relate stiffness, weight, moment of inertia to obtain a stiffer system? |
2023-06-03T17:38:42.027 | <p>This question is based on the following <a href="https://www.thehindu.com/news/cities/Tiruchirapalli/express-train-hits-lorry-tyre-placed-on-track-near-tiruchi-suffers-detention/article66924462.ece" rel="nofollow noreferrer">news article</a>. I just want to know if rubber tyres are strong enough to cause rail derailment if placed on the track</p>
| |rail| | <p>The tires did not derail the train. More generally if they could then why do saboteurs waste time with explosives and so on?</p>
| 55427 | Can tyres placed on track derail trains? |
2023-06-05T13:55:13.263 | <p>So, a pulley drive with the drive pulley having 1cm of diameter and the driven pulley 10cm of diameter would have something around a 10:1 reduction ratio.</p>
<p><a href="https://i.stack.imgur.com/cdXE6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cdXE6.png" alt="Pulley drive illustration" /></a></p>
<p>However, I don't know if the same logic would apply to a compound pulley, assuming that the fixed pulley has 10cm of diameter and the movable pulley has 1cm of diameter, on top of that, the winch driving the system has 1cm of diameter.</p>
<p><a href="https://i.stack.imgur.com/h8SBX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h8SBX.png" alt="Illustration showing a compound pulley" /></a></p>
| |mechanical-engineering|mechanisms|pulleys| | <h3>The size of the sheaves doesn't matter to pulley systems.</h3>
<p>Pulleys gain advantage by multiplying the constant tension of the rope by the pulley advantage. Gears and belts give an advantage to the torque of the sheaves/gears, not the tension on the belt.</p>
<p>In your example, there is a ~2:1 advantage because there are two ropes pulling the weight up. Both ropes have the same tension, so there is twice the force on the weight, at the expense of half the motion. The belt example gives advantage because the sheaves turn at different rates.</p>
| 55438 | Would the reduction ratio of a pulley drive also be applied to a compound pulley? |
2023-06-05T16:17:41.960 | <p>If I have a thin-walled and not very stiff cylinder, such as a rubber-band, and I put it between two pins and pull it apart radially, is the force required considered the <strong>hoop force</strong> or <strong>radial force</strong>? What would the difference be between the two? I have read various derivations for hoop force/stress but I am still confused on if my above example is describing <strong>hoop</strong> or <strong>radial</strong> force.</p>
| |stresses|pressure-vessel|thin-films| | <p>In a thin walled vessel it is simple. A cylinder uses coordinates <span class="math-container">$(r, \theta, z)$</span>.</p>
<p>Hoop forces mean azimuthal forces (in the <span class="math-container">$\theta$</span> direction) that attempt to increase the circumference of the vessel. These are the highest-magnitude forces cause by internal pressure in a cylinder, and this is why tanks and pressure vessels often fail by the skin splitting, like so:</p>
<p><a href="https://i.stack.imgur.com/ljwSF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ljwSF.png" alt="" /></a>
Source: <a href="https://blog.iqsdirectory.com/you-can-stop-pressure-vessel-failure/" rel="nofollow noreferrer">https://blog.iqsdirectory.com/you-can-stop-pressure-vessel-failure/</a></p>
<p>Even overcooked hot dogs tend to split this way.</p>
<p>Radial forces are those trying to increase or decrease the wall thickness of the vessel (<span class="math-container">$r$</span> direction). These are usually very small in a thin-walled vessel, and are neglected in first order thin-wall analysis, because the wall is so thin there is no material there to become stressed, with the radial reaction forces essentially being distributed in the hoop direction.</p>
<p>Longitudinal or axial forces are those trying to increase (or decrease) the length, in the <span class="math-container">$z$</span> direction.</p>
<p>I will add: in your rubberband example, it's not quite the same as a pressure vessel, but the stress you are describing is primarily hoop (<span class="math-container">$\theta$</span> direction) stress.</p>
| 55440 | In a thin-walled cylinder, what exactly is Hoop Stress vs Radial Stress? |
2023-06-06T09:48:18.267 | <p>I have to find final temperature of water in an heat exchanger.<br />
the heat exchanger must cool down air using water, meaning initial temperature of water must be cold,
initial temperature of air is 298.15 Kelvin</p>
<p>I can use this equation to find initial temperature of water, <span class="math-container">$Q_{max} = C_{min}*\delta T$</span>
where <span class="math-container">$C_{min} = \frac{kg}{s} * \frac{J}{kg*K} = \frac{W}{K}$</span></p>
<p>and this is not a problem, but here comes my question: in order to find final temperature from this equation <span class="math-container">$T_{out, water} = T_{in, water} +/- \frac{Q}{C}$</span></p>
<p>how can I know which sign to plug into this equation?</p>
<p>this is my reasoning, let me know if it's right:
if the text says water is cooling down air, that means initial temperature of water is becoming warmer than before by taking heat from air (i.e the hot fluid), and therefore equation becomes: <span class="math-container">$T_{out, water} = T_{in_water} + \frac{Q}{C}$</span></p>
<p>if this is true, then minus sign means the contrary.</p>
<p>but sometimes I have an exercise that say the same thing, but it requires me to flip the signs, for example, "air is being extracted from a server room and it has to be cooled down by cold water", in this case, exercise requires to use this equation: <span class="math-container">$T_{out, water} = T_{in_water} - \frac{Q}{C}$</span> so I think my reasoning is wrong or at least it cannot be applied all the time for every situation.</p>
| |heat-exchanger|building-physics| | <p>I think your confusion comes from unclear definition of <span class="math-container">$Q_{max}$</span> and <span class="math-container">$\delta T$</span>.</p>
<p>Let's say you have 2 streams with possible heat exchange between them. Following stream parameters will be denoted by subscripts <span class="math-container">$1$</span> and <span class="math-container">$2$</span>:</p>
<ul>
<li>mass flow <span class="math-container">$\dot{m}_i$</span></li>
<li><a href="https://en.wikipedia.org/wiki/Specific_heat_capacity" rel="nofollow noreferrer">specific heat capacity</a> <span class="math-container">$c_{p,i}$</span></li>
<li>inlet temperature <span class="math-container">$T_{in,i}$</span></li>
<li>outlet temperature <span class="math-container">$T_{out,i}$</span></li>
</ul>
<p>Now it is useful to denote the heat flow between the stream in a way that makes the direction obvious, e.g. <span class="math-container">$\dot{Q}_{1\rightarrow2}$</span> for heat flow from stream <span class="math-container">$1$</span> to stream <span class="math-container">$2$</span>. For a heat exchanger in steady state with following parameters:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Heat_transfer_coefficient#Composition" rel="nofollow noreferrer">overall heat transfer coefficient</a> <span class="math-container">$U$</span></li>
<li><a href="https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference" rel="nofollow noreferrer">logarithmic mean temperature difference</a> <span class="math-container">$\Delta T_{1-2}$</span> between streams <span class="math-container">$1$</span> and <span class="math-container">$2$</span></li>
<li>heat exchange area <span class="math-container">$A$</span>,</li>
</ul>
<p>two things must be true in steady state:</p>
<ul>
<li>energy balance (heat leaving one stream end up in the other one):
<span class="math-container">$$\left(T_{in,1}-T_{out,1}\right)\cdot c_{p,1}\cdot \dot{m}_1 =
\left(T_{out,2}-T_{in,2}\right)\cdot c_{p,2}\cdot \dot{m}_2$$</span></li>
<li>heat rate proportional to negative temperature difference and overal heat transfer coefficient (Fourier law; heat always flows against temperature gradient):
<span class="math-container">$$\dot{Q}_{1\rightarrow2} = \Delta T_{1-2}\cdot U\cdot A$$</span></li>
</ul>
<p>I am not sure if this is obvious, but the equations can be connected in this way:
<span class="math-container">$$\dot{Q}_{1\rightarrow2} = \left(T_{in,1}-T_{out,1}\right)\cdot c_{p,1}\cdot \dot{m}_1$$</span>
<span class="math-container">$$\dot{Q}_{1\rightarrow2} = \left(T_{out,2}-T_{in,2}\right)\cdot c_{p,2}\cdot \dot{m}_2$$</span></p>
<p>As you can see, the assumption about which stream is cooler or hotter does not change these equations. It's just sometimes <span class="math-container">$\dot{Q}_{1\rightarrow2}$</span> will be positive and other times negative.</p>
| 55446 | how to find final temperature of a fluid in an heat exchanger? |
2023-06-06T19:21:16.457 | <p>In Short: I'm working on a project where we need to cool 50 ml of liquid from 60 °C to 30 °C or 20 °C. The liquid will be in a 150 ml copper container with an open top.</p>
<p>Due to the lack of skills in the project team (myself & my buddy), we've been looking at the three things we can think of.</p>
<ol>
<li><p>Peltier style cooling: we have mains power as an option but I'm not sure the time it will take to cool via conduction of roughly 7 cm contact area.</p>
</li>
<li><p>CO<sub>2</sub> discharge with some copper pipe submerged in the liquid and a few short discharges of liquid CO<sub>2</sub>. We have a few concerns here as to if this would work practically or efficacity with the amount of CO<sub>2</sub> at hand. A further thought on this was a wrapped coil around the container, same concerns as above.</p>
</li>
<li><p>Think low-tech, give up on a smart solution and dunk it in an ice bath.</p>
</li>
</ol>
<p>I know this is pretty light on content but I'm struggling to work out what way to focus and trying to be efficient in my use of time and not go down a rabbit hole only to work out the science behind what I was dreaming of would never work out anyway.</p>
| |electrical-engineering|thermodynamics|heat-transfer|heat-exchanger|building-physics| | <p>If you're wondering how long it would take in order to cool that liquid.</p>
<p><span class="math-container">$$mc_p\frac{dT}{dt} = \frac{1}{\frac{1}h + \frac{\Delta x}k}A(T_C-T) $$</span></p>
<p>Where:</p>
<p><span class="math-container">$m$</span> is the mass of liquid</p>
<p><span class="math-container">$c_p$</span> is the heat capacity of liquid</p>
<p><span class="math-container">$h$</span> is the convective coefficient of the fluid you're using for cooling (<a href="https://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm" rel="nofollow noreferrer">https://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm</a>)</p>
<p><span class="math-container">$\Delta x$</span> is the thickness of the copper wall</p>
<p><span class="math-container">$k$</span> is the thermal conductivity of copper, ~400 W/m/K</p>
<p><span class="math-container">$A$</span> is the contact area</p>
<p><span class="math-container">$T_C$</span> is the coolant temperature... which I'm assuming constant. It will be constant if you use a bath of ice or almost constant if you use a lot of coolant, as a thermal reservoir.</p>
<p>But we can pretty much ignore the resistance to mass transfer coming from conductivity, convection is the bottleneck here, so integrating <span class="math-container">$ mc_p \frac{dT}{dt} = hA(T_C-T) $</span>,</p>
<p><span class="math-container">$$
t = \frac{mc_p}{hA}\cdot\ln\left(\frac{T_0-T_C}{T_f-T_C}\right)
$$</span></p>
<p><span class="math-container">$T_0$</span> is the initial temperature, <span class="math-container">$T_f$</span> is the final.</p>
| 55451 | Advice needed on how to cool 50 ml liquid from 60 °C to 30 °C |
2023-06-07T20:37:00.730 | <p>I am designing a control system :</p>
<p><a href="https://i.stack.imgur.com/EXXRe.pngith" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EXXRe.pngith" alt="" /></a></p>
<p>with open loop gain equal to : <span class="math-container">$\frac{K}{s(s^{2}+4s+3)}$</span></p>
<p>The closed loop gain is obviously :<span class="math-container">$\frac{K}{s(s^{2}+4s+3)+K}$</span></p>
<p>I tried finding the critical frequency but found something else interesting while trying to find the critical frequency.</p>
<p>I applied Ruth's algorithm to the denominator of the closed loop gain and I get this result:</p>
<p><a href="https://i.stack.imgur.com/vtQqs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vtQqs.png" alt="" /></a></p>
<p>Obviously for a stable system k doesnt exist but this seemed super odd so I decided to set k = 2 and find all the roots <a href="https://www.symbolab.com/solver/roots-calculator/roots%20x%5E%7B3%7D%2B4x%5E%7B2%7D%2B3x%2B2?or=input" rel="nofollow noreferrer">using a calculator</a> and I found that the root is in the left half plane of the Re,Im plane so the system SHOULD be stable.So what am I doing wrong?</p>
| |control-engineering|control-theory| | <p>Your setup for the Routh-Hurwitz stability criterion isn't correct. Your setup should look like:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>var</th>
<th>c1</th>
<th>c2</th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="math-container">$s^3$</span></td>
<td><span class="math-container">$1$</span></td>
<td><span class="math-container">$3$</span></td>
</tr>
<tr>
<td><span class="math-container">$s^2$</span></td>
<td><span class="math-container">$4$</span></td>
<td><span class="math-container">$K$</span></td>
</tr>
<tr>
<td><span class="math-container">$s^1$</span></td>
<td><span class="math-container">$ \frac{12-K}{4} $</span></td>
<td><span class="math-container">$0$</span></td>
</tr>
<tr>
<td><span class="math-container">$s^0$</span></td>
<td><span class="math-container">$K$</span></td>
<td><span class="math-container">$0$</span></td>
</tr>
</tbody>
</table>
</div>
<p>So your system is stable when <span class="math-container">$0 \leq K \leq 12$</span>.</p>
| 55467 | Routh's algorithm and root analysis give contradictory results |
2023-06-11T12:54:24.130 | <p>Suppose that I'm boiling water in an open pot. As the pot is heated by the stove, it loses mass in the form of vapour.</p>
<p>The pot is my control volume. That system could be modeled by the following equations:</p>
<p><strong>Mass balance:</strong></p>
<p><span class="math-container">$$
\frac{dm}{dt} = -\dot m_{out}
$$</span></p>
<p><strong>Energy balance:</strong>
<span class="math-container">$$
\frac{dU}{dt} = -\dot H_{out} + \dot Q
$$</span></p>
<p>Considering <span class="math-container">$U = mc_v^{liq}T$</span>, if the boiling process occurs at constant temperature <span class="math-container">$T_b$</span>,</p>
<p><span class="math-container">$$
c_v^{liq} T_b\frac{dm}{dt} = -\dot m_{out} c_p^{vap} T_b + \dot Q
$$</span></p>
<p>Let's throw in some numbers:</p>
<ul>
<li><p><span class="math-container">$c_v^{liq} = 4.0$</span> kJ/kg°C, the specific heat capacity of liquid water at constant volume</p>
</li>
<li><p><span class="math-container">$c_p^{vap} = 2.0$</span>kJ/kg°C, the specific heat capacity of vapour at constant pressure</p>
</li>
<li><p><span class="math-container">$T_b = 100$</span> °C</p>
</li>
<li><p><span class="math-container">$\dot Q = 400$</span> kW (the stove supplies a positive heat inflow to the system, whereas it also loses heat to the surroundings by convection at a constant rate <span class="math-container">$q = hA(T_{surr}-T_b)$</span>, but I figure that the overall <span class="math-container">$\dot Q$</span> must be a positive number, in order to supply the latent heat necessary to vaporize the water.)</p>
</li>
</ul>
<p><span class="math-container">$$
-400\cdot\dot m_{out} = -200\cdot\dot m_{out} + 400
$$</span>
<span class="math-container">$$
(-200\text{ kJ/kg})\cdot\dot m_{out} = 400\text{ kW}
$$</span>
<span class="math-container">$$
\dot m_{out} = -2.0\text{ kg/s}
$$</span></p>
<p>So... I found a negative mass outflow... as if it was actually an inflow, with vapour condensing into liquid water. Moreover, 2.0 kilograms/second is <strong>a lot</strong>! Even though that's just a toy model, I believe the numbers I provided are at least in the ballpark of a realistic situation...</p>
<p>And is it even realistic that the rate of vapour outflow would be constant?</p>
<p>Where did I screw up my model? Can anyone show me a better, more realistic, one?</p>
<p><strong>EDIT 1:</strong> Thank you Transistor for reminding me in the comments to account for <em>heat of vaporization</em>, what a blunder! Heating rate should be in watts, rather than kilowatts as well.</p>
<p>Redoing the calculations</p>
<p><span class="math-container">$$
c_v^{liq} T_b\frac{dm}{dt} = -\dot m_{out} (c_p^{vap} T_b + \Delta h_{vap}) + \dot Q
$$</span></p>
<p><span class="math-container">$$
-400\cdot\dot m_{out} = -(200 + 2250)\cdot\dot m_{out} + 0.40
$$</span>
<span class="math-container">$$
(1650\text{ kJ/kg})\cdot\dot m_{out} = 0.40\text{ kW}
$$</span>
<span class="math-container">$$
\dot m_{out} = 2.4\times10^{-4}\text{ kg/s}
$$</span></p>
<p>That's more like it!</p>
<p><strong>EDIT 2:</strong> I thought a little bit further, and I believe there were more things wrong with my previous reasoning.</p>
<p>The internal energy of the system during the phase change is <span class="math-container">$U = m_Lu_L + m_Vu_V$</span> (liquid + vapour). If <span class="math-container">$m_L + m_V = m_0$</span>,</p>
<p><span class="math-container">$$
U = m_L u_L + (m_0 - m_L)\cdot (u_L + \Delta u_{vap})
$$</span>
<span class="math-container">$$
\frac{dU}{dt} = -\Delta u_{vap} \frac{dm_L}{dt}
$$</span></p>
<p>Moreover, I was messing up with the reference for calculating the enthalpies. If my reference is liquid water at 0°C, then the enthalpy of the vapour leaving the system is actually <span class="math-container">$-\dot m_{out} (c_p^{liq} T_b + \Delta h_{vap})$</span>, with <span class="math-container">$c_p^{liq}$</span> insetad of <span class="math-container">$c_p^{vap}$</span>.</p>
<p>Hence,</p>
<p><span class="math-container">$$
-\Delta u_{vap} \frac{dm_L}{dt} = -\dot m_{out} (c_p^{liq} T_b + \Delta h_{vap}) + \dot Q
$$</span></p>
<p>The mass balance would be:</p>
<p><span class="math-container">$$
\frac{d}{dt} (m_L + m_V) = 0
$$</span></p>
<p><span class="math-container">$$
\frac{dm_L}{dt} = -\frac{dm_V}{dt} = -\dot m_{out}
$$</span></p>
<p><span class="math-container">$\Delta u_{vap} \approx 2090$</span> kJ/kg is a little bit lower than <span class="math-container">$\Delta h_{vap} \approx 2250$</span> kJ/kg at 1 atm.</p>
<p><span class="math-container">$$
\dot m_{out} = \frac{\dot Q}{c_p^{liq}T_b+\Delta h_{vap}+\Delta u_{vap}}
$$</span></p>
<p>Using the values above, <span class="math-container">$\dot m_{out} \approx 8\times10^{-5}$</span> kg/s. Now it's quite a bit smaller than I would expect. Maybe I got my signs wrong and it's actually <span class="math-container">$\Delta h_{vap} - \Delta u_{vap}$</span>? That would make more sense, and <span class="math-container">$\dot m \approx 7\times10^{-4}$</span>kg/s, which seems more reasonable (around 23 minutes to boil 1 kg of water).</p>
<p><strong>EDIT 3:</strong> I've got the answer I was looking for in the Physics StackExchange. Thank you for all the help!</p>
<p><a href="https://physics.stackexchange.com/questions/767996/mass-flow-of-vapour-produced-while-boiling-water-in-an-open-system/768019#768019">https://physics.stackexchange.com/questions/767996/mass-flow-of-vapour-produced-while-boiling-water-in-an-open-system/768019#768019</a></p>
| |chemical-engineering|vaporization| | <p>The governing equations are mass and energy balances. We can use an enthalpy balance rather than an internal energy balance because the process is at constant temperature and pressure.</p>
<p><span class="math-container">$$\dot{m}_{vap} + \frac{dm_{liq}}{dt} = 0$$</span></p>
<p><span class="math-container">$$\dot{m}_{vap}\tilde{H}_{vap}(T_b) + \frac{d(m_{liq}\tilde{H}_{liq}(T_b))}{dt} = \dot{q}_{stove}$$</span></p>
<p>The enthalpy balance is derived for a system with no heat loss. All stove heat goes directly to vaporize the liquid. A simple adjustment in a non-ideal case is to decrease the stove heat value by the expected (constant) heat loss, for example to the surroundings.</p>
<p>The vapor and liquid are at their normal boiling point <span class="math-container">$T_b$</span>. The change is the vaporization enthalpy. Substitutions give a form as below.</p>
<p><span class="math-container">$$\dot{m}_{vap} \Delta_{vap}\tilde{H} = \dot{q}_{stove}$$</span></p>
<p>Suppose that you need to heat the liquid from a lower temperature and boil it. The simplest adjustment is to neglect vaporization until you reach the normal boiling point. Calculate an initial time needed to heat <span class="math-container">$\Delta t_o$</span>.</p>
<p><span class="math-container">$$m_{liq,o}\tilde{C}_{p,liq} \left(T_b - T_o\right) = \dot{q}_{stove} \Delta t_o$$</span></p>
| 55485 | Calculating the mass flow of water vapour in a boiling pot. A little help with mass and energy balances needed |
2023-06-14T18:24:02.213 | <p><strong>How can I calculate the stresses in corners where different geometries meet?</strong></p>
<p>This image is a quick example of the positions i am referring to.</p>
<p><a href="https://i.stack.imgur.com/p7yqt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p7yqt.png" alt="enter image description here" /></a></p>
<p><strong>Additional information</strong></p>
<p>In class we studied the membrane model for thin walled pressure vessels, and we used that to calculate <span class="math-container">$\sigma_m$</span> and <span class="math-container">$\sigma_t$</span> in a random location along the cylindrical and conical profile.</p>
<p>We didn't mention of a way to calculate stresses in proximity of changes in profile, and this annoys me a bit as I have no knowledge of the material behaviour there, but due to the abrupt change in geometry I guess there will be stress concentrations there.</p>
<p>First I attempted to model the transition zone as a corner with a small fillet, as to remove the discontinuity, and analyze that, but I got quickly lost in math as the volume of revolution was not so easy to calculate.</p>
<p>Then I assumed the pressure constant along the fillet, as it's small compared to the whole vessel, but I got constant stresses back, so I guess that's not the right answer.</p>
<p>I checked Peterson for stress concentrations and found nothing, same for Roark's. The only place where I found something is on Strength of materials by Timoshenko, but he explains the answer using a formula from plates in bending, a topic which I have yet to cover, so I am not comfortable with plate theory yet.</p>
<p>So the question is:</p>
<p><strong>How can I calculate the stresses in corners where different geometries meet?</strong></p>
<p>Eventually, are there simpler paths other than plate theory or am I forced to learn that before understanding how to calculate these stresses?</p>
| |mechanical-engineering|structural-analysis|stresses|solid-mechanics|pressure-vessel| | <h1>Junction between cylindrical and conical shell</h1>
<h2>Simple conservative solution</h2>
<p>You can use longitudinal membrane forces and put them into equilibrium in vertical direction. Cylindrical shell has the whole longitudinal force <span class="math-container">$N_{\vartheta V}$</span> in the vertical direction, but in case of conical shell, you have to split the longitudinal force <span class="math-container">$N_{\vartheta K}$</span> into vertical and radial components.</p>
<p><a href="https://i.stack.imgur.com/sPRaz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sPRaz.png" alt="enter image description here" /></a></p>
<p>As you can see, <span class="math-container">$N_{\vartheta K}\cdot \cos(\vartheta)$</span> has to have the same magnitude as <span class="math-container">$N_{\vartheta V}$</span>. From membrane theory perspective, the radial force <span class="math-container">$N_R = N_{\vartheta K}\cdot \sin(\vartheta)$</span> is a force with which the conical shell is trying to deform edge of the cylindrical shell and membrane shell has no local radial stiffness to resist this. But you can design a ring stiffener just for this purpose.</p>
<p>In case of internal pressure <span class="math-container">$P$</span> and using mean cylindrical shell diameter <span class="math-container">$D_m$</span>, the <span class="math-container">$N_{\vartheta V}$</span> force is:
<span class="math-container">$$N_{\vartheta V} = \frac{P\cdot D_m}{4}$$</span></p>
<p>It is also true, that:
<span class="math-container">$$N_{\vartheta V} = N_{\vartheta K}\cdot \cos(\vartheta) = N_R\cdot \frac{\cos(\vartheta)}{\sin(\vartheta)}$$</span></p>
<p>Therefore:
<span class="math-container">$$N_R = N_{\vartheta V}\cdot \tan(\vartheta) = \frac{P\cdot D_m}{4}\cdot \tan(\vartheta)$$</span></p>
<p>A ring stiffener with cross-section <span class="math-container">$A$</span> can be welded to the junction. In case of internal pressure, it will be compressed and the compressive stress <span class="math-container">$\sigma_c$</span> can be calculated from the radial force equilibrium (this is similar to calculating hoop stress in cylindrical shell):</p>
<p><span class="math-container">$$2\cdot \sigma_c \cdot A = N_R\cdot D_m$$</span></p>
<p>So the compressive stress will be:</p>
<p><span class="math-container">$$\sigma_c = N_R\cdot \frac{D_m}{2A} = \frac{P\cdot D_m^2\cdot \tan(\vartheta)}{8A}$$</span></p>
<p>If you put a transition zone defined by radius there, it is more complicated, but this zone will also have to be able to resist the radial force. And as you decrease the radius, the force does not change much, but the size of the zone does, so the membrane stress there rises up to infinity for zero radius.</p>
<h2>Reality</h2>
<p>In reality, the shell junction does have some resistance against the radial deformation, so you can take this into account. Derivation is quite complex, but important thing is, that the length of the shell you can include into the stiffening area is calculated using formula:</p>
<p><span class="math-container">$$L = k\cdot \sqrt{e_n\cdot D_m}$$</span></p>
<p>where <span class="math-container">$e_n$</span> is nominal thickness and <span class="math-container">$k$</span> is a factor, 1.4 in this case I think. So the required additional stiffening area is less than calculated in the previous solution and it can take a form of local thickness increase instead of ring stiffener. Transition zone defined by radius also helps.</p>
<p>This is used for example in vertical cylindrical storage tank design (API 620), where you have a junction between the conical roof and the cylindrical shell. Pressure vessels could also be based on this, but more advanced techniques are used in some standards.</p>
<h1>Point on the axis</h1>
<p>Regarding the point on the axis, the stresses there would be basically 0. The vessel would probably not end with a sharp spike like this.</p>
| 55515 | Thin wall pressure vessels: local stresses in bends |
2023-06-15T01:21:55.843 | <p>I have a discreate PI controller implemented in stm32 MCU, it has the following form:
<span class="math-container">$$
PI=K_p\cdot[r(t_i)-y(t_i)] + (K_I)\cdot\sum_{n=0}^{i}[r(t_n)-y(t_n)]\cdot\Delta t
$$</span>
where</p>
<ul>
<li><span class="math-container">$r(t_i)$</span> is the target (or input) value,</li>
<li><span class="math-container">$y(t_i)$</span> is the output value and</li>
<li><span class="math-container">$\Delta t $</span> is the sampling time.</li>
</ul>
<p>I use the <span class="math-container">$\sum$</span> function to implement the integral function while the error term is <span class="math-container">$[r(t_i)-y(t_i)]$</span>.</p>
<p>I need to simulate this PI controller in MATHLAB Simulink.</p>
<p>Besides, I have found another form of the PI controller: it doesn't has the sum equation, and but has the following structure:
<span class="math-container">$$
PI=K_p\cdot[r(t_i)-y(t_i)]+ (K_I)[r(t_i)-y(t_i)]\cdot\Delta t
$$</span>
For the integral term of this PI controller, the <span class="math-container">$Z$</span>-transform is
<span class="math-container">$$
K_I(z)=e(z)\cdot T\cdot\frac{ z}{z-1}.
$$</span></p>
<p>This form does not include the sum term.</p>
<p><strong>My question is</strong>: if I use the <code>sum equation</code> to sum the error term, how should I represent the PI controller correctly in the <span class="math-container">$Z$</span>-transform domain?</p>
| |control-theory|pid-control|matlab|simulink| | <p>Your model is already in discrete form but you need to introduce the <span class="math-container">$z$</span>-operator to your equation. I perceive a missconception about the z-Transform. I hope the following lines will clarify that...
consider the following</p>
<ul>
<li><span class="math-container">$e(t) := r(t)-y(t)$</span></li>
</ul>
<p>The PI-controller can be represented as</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;">Continous Time</th>
<th style="text-align: left;">Frequency Domain</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;"><span class="math-container">$$u(t)=K_p\,e(t)+K_I\int_0^{\tau}e(\tau)d\tau$$</span></td>
<td style="text-align: left;"><span class="math-container">$$U(s)=(K_p+\frac{K_I}{s})\,E(s)$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>As probably you notice the proportional part is not a challenge if with want to discretize it, but the integral is! This term could be approached through Euler's forward or backwards method. Using the fordward Euler method leads to</p>
<p><span class="math-container">$$u(k\,T) =\int_0^{k\,T}e(\tau)d\tau \approx \sum_{i=0}^{k-1}e(i\,T)\cdot T$$</span></p>
<p>where T is the sample period, i.e. the loop time in your microcontroller and the notation <span class="math-container">$u(k\,T)$</span> denote the value of the variable <span class="math-container">$u$</span> at time <span class="math-container">$k\,T$</span> with <span class="math-container">$k=0,1,2...$</span>.</p>
<p>In the Z-Transform domain the operator <span class="math-container">$z$</span> is used to denote a forward shift by one sampling interval, e.g. <span class="math-container">$z\cdot\,x(k)$</span> is equal to <span class="math-container">$x(k+1)$</span>. Analogously, the backward shift operator is denoted by <span class="math-container">$z^{-1}$</span>, e.g. <span class="math-container">$z^{-1}\cdot x(k) = x(k-1)$</span>.</p>
<p>In digital control systems one can approximate the <span class="math-container">$s$</span>-Domain in <span class="math-container">$z$</span>-Transform domain, e.g. using a euler backward integration we have</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;">Integation in Discrete Time</th>
<th style="text-align: left;">Integration in Z-Domain</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;"><span class="math-container">$$x(k+1)\approx x(k) + T\cdot u(k+1)$$</span></td>
<td style="text-align: left;"><span class="math-container">$$zX(z)=X(z)+T\,z\,U(z)\,\text{, reordering}\\ \frac{X(z)}{U(z)}=\frac{T\,z}{z-1}$$</span></td>
</tr>
</tbody>
</table>
</div>
<p>Which if you noticed, is what you found in the "another" PI-Controller.
Now how the approximation is made? easy, we can recall the integration as <span class="math-container">$\dot{x}(t)=u(t)$</span> which in laplace is <span class="math-container">$\frac{X(s)}{U(s)}=\frac{1}{s}$</span>. Then,</p>
<p><span class="math-container">$$\frac{1}{s}\approx\frac{T\,z}{z-1}\,\text{, which leads to}\, s\approx \frac{z-1}{T\,z}$$</span></p>
<p>The advantage of the last equation is that we can have our Transfer function in Freq Domain and then replace the approximation of <span class="math-container">$s$</span>, i.e. <span class="math-container">$C(z)\approx C(s)|_{s=\frac{z-1}{T\,z}}$</span></p>
| 55520 | What is the Z transform of the Discreate PI control model？ |
2023-06-15T04:54:52.197 | <p>I would like to seat some bearings into a piece of aluminum, and I only have access to a small CNC machine. The CNC has ball screws and is calibrated to maybe 50um error in the actual travel on x and y axes. I assume this hole is not as circular or straight as what could be produced on a lathe or boring machine. I can’t say much about the bearings as I haven’t selected them yet - they will be angular contact and the application is for a small desktop robotic arm. I expect to pay about $10-15 for each bearing and would ideally like to get my money’s worth by “doing it as right as possible”.</p>
<p>The outer diameter of the bearing might be on the order of 24mm. I am confident that I can cut a circular hole using the CNC and get some sort of a “press fit”, but I have the understanding that when people seat bearings they usually use a boring machine, lathe, and/or a reamer tool. Is using the small CNC machine to create an undersized hole and then hand-reaming this a feasible approach?</p>
<p>My understanding is that reamers will “follow” an existing hole and that this is why hand reaming (which lacks the rigidity of a machine) could work.</p>
<p>Finally, one of the framers I have is Morse taper so not sure if people hand ream with those or if it’s pretty much only used on a mill or lathe.</p>
| |bearings|cnc|holes| | <blockquote>
<p>can’t say much about the bearings as I haven’t selected them yet</p>
</blockquote>
<p>Well that's the problem right there. You need to figure out what precision level bearing you need first, and then decide if you can manufacture something to that spec. i.e. you need to consider things like the speed, load, life, reliability, etc. For some bearings, a hand reamed hole might be overkill, for others it might be totally unacceptable.</p>
<blockquote>
<p>I expect to pay about $10-15 for each bearing</p>
</blockquote>
<p>Without knowing anything else about the bearing but the cost, my initial reaction is that a hand reamed hole is probably fine. For $10 bearings I've used a drill press (without reaming) and gotten acceptable results.</p>
<p>If your application requires \$2000 bearings, then you need to pay very close attention to details. For \$10 maybe not so much.</p>
| 55522 | Is it acceptable to use a hand reamer on an X-Y CNC machined hole? |
2023-06-15T09:52:40.643 | <p>I have a stainless steel pot whose welds show the formation of an oxide layer. Colors around those welds range from gold to dark blue (see images below). The main question is: How do I remove those oxide layers with household items? I assume the standard process would be pickling those welds, is that right?</p>
<p>Am I too concerned? As far as google brought me, discoloration due to welding is acceptable up to shades of gold but dark blue should definitely not occur (e.g. in regarding corrosion resistance).</p>
<p><img src="https://i.stack.imgur.com/j6WRP.jpg" alt="Weld 1" />
<img src="https://i.stack.imgur.com/n4ggM.jpg" alt="Weld 2" /></p>
| |materials|steel|welding| | <p>The oxide color layers shown are of no concern. The colors are caused by thickness of oxide , however they are only a few to several layers thick. Household items that will remove them are sand paper and possibly steel wool.</p>
| 55526 | How to remove oxide layer around stainless steel welds with household items |
2023-06-15T20:14:27.320 | <p><a href="https://i.stack.imgur.com/yR7Xx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yR7Xx.jpg" alt="enter image description here" /></a>
I have some “hybrid” bearings that I purchased a long time ago on Amazon. They have silicon carbide balls and steel inner and outer races, as you can see in the photos. What I am wondering is what are the other pieces that seem to be holding the balls and covering the balls? Are they likely made of plastic?</p>
<p>I realize that this is hard to answer without any traceability or part number, but clearly these are mass- produced bearings common to many people so maybe someone knows.</p>
<p>The bearings seem to spin very fast without much resistance but also have a lot of play in their non-preloaded condition.</p>
<p>A second aspect to my question is based on this play observed and the fact that there looks like a plastic part in there - could these be some sort of gimmick for use in fidget spinners and not for any actual constrained or load bearing application?</p>
<p>My fault for buying them in the first place but curious if anyone knows more about these mysterious bearings from the cave of alibaba, and the design choices that led them to be what they are.</p>
<p>Edit:
I think this is called a bearing retainer based on the answer below. I found a high end fancy bearing from campagnolo bicycles that seems to have a retainer that looks similar .. so just because it’s plastic doesn’t mean it’s cheap or not good quality. <a href="https://i.stack.imgur.com/6Q2gs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Q2gs.jpg" alt="enter image description here" /></a></p>
| |bearings|ceramics| | <p>All ball bearings have what is called a ball retainer or ball cage that holds the balls in the raceways. It can be metal or plastic. You can just Google images "ball bearing anatomy" for exploded pictures of a variety of constructions.</p>
| 55533 | In this “hybrid” ceramic 608 bearing is there likely a plastic piece holding the bearings? |
2023-06-17T22:56:04.160 | <p>What is the ideal construction for an insulative box to keep my frosty beverages frosty? I am interested in finding the theoretical maximum as well as the practical maximum.</p>
<p>I was sitting on my back porch drinking a cold beverage and observing how quickly the ice in my cooler was melting. I began wondering what the ideal insulative container for my beverages might be. I know I could just go buy a Yeti but I was having fun with the thought experiment so I got on YouTube and began watching some lectures on heat transfer, fluid dynamics, etc.</p>
<p>I am NOT a physicist or mechanical engineer and so while the math all makes sense to me I don't have the background to fully flesh out the optimal design.</p>
<p>Here are some of the things I'm not sure about:</p>
<ul>
<li>Should the inside of the box be reflective (lower emissivity) or something else?</li>
<li>Should the walls be "solid" insultation or have narrow air gaps where each side is lined with reflective material.</li>
<li>If there are air gaps do both walls need to be reflective or does just the cold side need to be reflective to insulate from radiative heat transfer. What would the optimal surface be for the heat side)</li>
<li>Do reflective surfaces lose their radiation reflection properties if sandwiched against other materials.</li>
<li>Do the answers change as the size of the box changes (for instance is there a different answer if my box is the size of a boxcar or something)</li>
</ul>
<p>Here are some of the constraints:</p>
<ul>
<li>inside volume of box is one cubic meter</li>
<li>Wall thickness of no more than 6 cm</li>
<li>Direct sunlight exposure</li>
<li>Outside temperature 30°C</li>
<li>Container no more than 50% full of ice at -18°C</li>
<li>24 pack of standard sized aluminum canned beverages pre-chilled to 4°C</li>
</ul>
| |heat-transfer|thermal-conduction|thermal-insulation| | <p>There are two paths to the answer, as follows:</p>
<p>First of all, regarding emissivity and so on, radiative heat transfer scales as the 4th power of absolute temperature so for temperatures near room temp you needn't worry about it. All you want is a cooler box material that has the lowest possible thermal conductivity and the thickest possible walls.</p>
<p>What does this mean in <em>practical terms</em>? Well, the cheap answer is to build a box out of styrofoam with 3" thick walls and preload it with crushed ice to surround your refreshing beverages. This is how things that must be kept cold like special medicines are sent through the mail or via FedEx. Thicker than 3" is considered a waste. In fact, these can be bought for a few dollars each, so why not just buy one?</p>
<p>The physics answer is to build a cylindrical container with double walls and a few microns of sputtered aluminum on the inside surface of the outer cylinder and on the outside surface of the inner cylinder. Provide a narrowed-down neck through which individual cans of refreshing beverage can be loaded along with ice. Draw a hard vacuum on the inter-cylinder space, and plug the neck with a cork made of real cork. In fact, this is called a <em>dewar flask</em> and can be bought for several hundred dollars, so why not just buy one?</p>
| 55543 | What is the ideal theoretical construction for an insulative box to keep my frosty beverages frosty? |
2023-06-18T18:53:27.207 | <p>Originally I was wondering how topside edges/outside fillets were actually machined so I could understand what I am seeing in a drawing and how to properly model it.
I stumbled on <a href="https://engineering.stackexchange.com/questions/477/how-do-i-manufacture-a-fillet">this question</a> before asking which was very informative and confirm some of what I already expected. For wood work I would call it a quarter round router bit.
However with metal it appears to be call a radius cutter or a corner rounding end mill.</p>
<p>So I have been struggling with this part and parts like if for over a week now with various degrees of success. The blueprint is from a 1943 so standards may have changed a little since then.</p>
<p><a href="https://i.stack.imgur.com/ied95.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ied95.png" alt="enter image description here" /></a></p>
<p>In the front view we can see that the fillets on the part the sticks out to the front have a radius of 0.21. In the TRUE VIEW OF BASE ALONG LINE 'B', we can see the partial fillet along the length and the arc of the termination cut. With no radius given on the arc I am assuming it either covered in the spec which I do not have or there was a standard set of tools at the time.</p>
<p>At first I thought I would just make a section of the tool and revolve it at the point partially along the edge where I needed the fillet to terminate. Because of the acute angle of the faces, I would need to do this twice, once with the cutters axis parallel to each face. However in doing so I noted that there are to problems:</p>
<ol>
<li>The remaining material edge arc is convex instead of concave</li>
<li>Since the to outside faces are are an acute angel, the remaining material did not blend into the cut as anticipated.</li>
</ol>
<p><a href="https://i.stack.imgur.com/PkPMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkPMf.png" alt="Failed router cut" /></a></p>
<p>So that got me thinking and why I did a bit of research on how fillets are made. My next thought was to make a circle the diameter of an end mill bit and have the quadrant of the circle follow the arc of the fillet. The principal seemed sound and would give the right arc shape to match the drawing, but when it came tome to perform the sweep I kept getting errors.</p>
<p>The is the circle and path selection (note the arc is mainly blue instead of dashed yellow).<br />
<a href="https://i.stack.imgur.com/n8zYy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n8zYy.png" alt="End Mill Selection" /></a></p>
<p>And here is the error message I receive. I actually just discovered that when I click on the final error message it produces the green circular section that I had previously not seen and confuse me as I had the behaviour set to keep the circle section in the same orientation:</p>
<p><a href="https://i.stack.imgur.com/MLmv7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MLmv7.png" alt="End Mill failure" /></a></p>
<p><strong>QUESTION</strong></p>
<p><strong>How does one model this type of fillet termination in inventor?</strong></p>
<p>I am using Inventor LT 2021</p>
<p>On a side note, is that a normal way to dimension the length of the fillet?</p>
<p>I have also tried rails selecting the fillet radius as the inside path and an offset matching fillet arc for the rail guide. It did not work.</p>
<p>I tried creating a cylinder and sweeping the cylinder. It did not work.</p>
<p>I tried creating two circles on the different plane offset from each other and applied an edge rail. It failed.</p>
| |machining|technical-drawing|autodesk-inventor| | <p>I did this with solidworks, but I used the revolved cut feature which should be available in inventor too.</p>
<p>Basically the steps are:</p>
<p><strong>You make a fillet</strong></p>
<p><a href="https://i.stack.imgur.com/L82Jt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L82Jt.png" alt="enter image description here" /></a></p>
<p><strong>You revolve cut the end corners</strong></p>
<p><a href="https://i.stack.imgur.com/wxzBt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wxzBt.png" alt="enter image description here" /></a></p>
<p>In your case the corners are not exactly 90 degrees, in which case you can modify the revolving sketch as follows:</p>
<p><strong>For acute angles</strong></p>
<p><a href="https://i.stack.imgur.com/M3MJX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M3MJX.png" alt="enter image description here" /></a></p>
<p><strong>For obtuse angles</strong></p>
<p><a href="https://i.stack.imgur.com/uHZrI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uHZrI.png" alt="enter image description here" /></a></p>
<p>Remember to keep the revolution axis <strong>normal to the upper face</strong>.</p>
<p>In the case of very obtuse angles, like in the picture above, you won't exactly get the circular line perpendicular to the external profile, due to geometry:</p>
<p><a href="https://i.stack.imgur.com/brmMQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brmMQ.png" alt="enter image description here" /></a></p>
<p>but as long as acute angles are nearly 90° this isn't very noticeable.
I think you can see this on your blueprint too, albeit very slightly.</p>
| 55549 | Model partial length top side fillet |
2023-06-19T19:34:00.543 | <p>I am embarking on a project for building a low-noise air pump for use with things such as air mattresses or pool toys. I just find that the pumps sold at stores are way too noisy, plus it looked like a fun project.</p>
<p>As a first step, I am trying to first build an air pump that works. I have designed and 3d printed the air pump below. It is powered by a brushless dc motor. My idea was that by using a lower rpm, higher torque motor I could use a bigger fan to provide the same amount of air in a much quieter way. I opted for a radial design, rather than centrifugal, as from what I can understand, centrifugal pumps are significantly noisier.</p>
<p><a href="https://i.stack.imgur.com/P2LZa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P2LZa.png" alt="Complete air pump view" /></a></p>
<p><a href="https://i.stack.imgur.com/do0c5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/do0c5.png" alt="Cut view of the air pump" /></a></p>
<p><a href="https://i.stack.imgur.com/nXY09.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nXY09.png" alt="Air pump fan" /></a>
<a href="https://i.stack.imgur.com/O2Kji.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O2Kji.png" alt="Top view of air pump housing" /></a></p>
<p>The motor is rated for 7030 RPM in a no load setting. The air inlet is on top, and the outlet is on the bottom. The outlet is significantly smaller than the inlet. The fan rotates counter-clockwise.</p>
<p>When turning on the pump I find that barely any air comes out of the outlet, but instead most of the air actually comes out of the inlet. If I attach an inflation nozzle to the outlet, most of the air actually stops coming out from the outlet, and more air comes out from the inlet. Why does this happen and how can I avoid it?
From what I could figure out, the problem is probably related to the static pressure the fan is able to produce, since the outlet is smaller than the inlet, so the pump must be able to compress the air enough to push it through the outlet. How do I increase the static pressure produced by the fan (preferably without increasing rpm due to noise)?</p>
| |fan| | <p>If your fan is throwing back air out the inlet, you have <em>choked</em> the fan with an exit hole that is too small. Note that what is actually happening is that part of the fan disc is pulling air into the pump while another part of the fan disc is allowing that air to escape, at the same time.</p>
<p>Note that using a diffuser downstream of an axial-flow fan to convert kinetic energy in the airflow into static pressure is the wrong way to design an air pump for this application. The entire point of an axial fan is to slightly increase the pressure of a large volume of airflow, and <em>not</em> to develop high pressure in a small airflow.</p>
<p>This fact is chapter one, verse one of the art of <em>turbomachinery</em> and is the reason that high pressure is generated either by centrifugal pumps or piston/diaphragm pumps and not by axial fans.</p>
| 55559 | Why does my self-made air pump not blow air in the correct direction? |
2023-06-22T00:13:52.040 | <p>I need help creating a compound gear system that will result in the sought after drive ratio or very close to it. Naturally gear teeth need to be whole numbers. Websites like 'GearGenerator' (dot-com) help me solve my problem via guesswork. I am looking to solve my problem with the appropriate application of mathematics be it Algebra or Differential Equations. I'll give you an example of one of several gear ratios I am after.</p>
<p>Example A-
Constraints: gears cannot be less than 15 teeth(m1), gear cannot exceed 100 teeth(m1).</p>
<p>I need to use a compound stud gear system to result in a ratio of 7.81
Presently the solution that does not suit my constraints is a 20 tooth driving gear and a 156 tooth driven gear. I know I can guess around and then make minor adjustments to come up with a four to six gear system. I do have several more oddball ratios to solve and I would like to approach this with the aim of becoming a better engineer by applying physics and mathematics.</p>
<p>So, how do I go about this problem?</p>
<p>Sample A-
An example of a solution to an easier ratio of 2.55 is as follows:
<a href="https://geargenerator.com/#225,312.5,50,1,1,2,8288.700000012579,4,1,20,5,4,27,0,0,0,0,0,0,0,20,5,4,27,-86,0,0,0,0,1,0,51,12.75,4,27,0.3000000000000007,0,0,0,0,2,1,20,5,4,27,5.300000000000001,0,0,0,0,0,1,3,-229" rel="nofollow noreferrer">https://geargenerator.com/#225,312.5,50,1,1,2,8288.700000012579,4,1,20,5,4,27,0,0,0,0,0,0,0,20,5,4,27,-86,0,0,0,0,1,0,51,12.75,4,27,0.3000000000000007,0,0,0,0,2,1,20,5,4,27,5.300000000000001,0,0,0,0,0,1,3,-229</a>
<a href="https://i.stack.imgur.com/U4Usr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U4Usr.png" alt="enter image description here" /></a>
On the top left is a driving gear of 20 teeth, then another gear of the same size(functionally doing nothing but reversing direction), then the 51 tooth gear driving a gear on a shared axle. This results in a sought ratio of 2.55.</p>
<p>Please Note:
-Having a close answer to my question will help me have a more accurate celestial model. That is why 'meh, close enough' is not an ideal answer to give myself. Also, I want to learn.</p>
<p>-I'd appreciate your patience and understanding with my cavalier word choice and casual engineering background. Please don't hold back on presenting me solutions no matter the complexity. I have time.</p>
| |mechanical-engineering|gears|applied-mechanics|mechanical|experimental-physics| | <p>The solution that comes to mind for me would be to start with your target gear ratio and create a fractional form of the target gear ratio, in this case <span class="math-container">$\frac{781}{100}$</span>.</p>
<p>At this point, that numerator is far too big, so you would factor 781 and 100 and find a combination of the factors that does the job.</p>
<p>When you have gears in a train, the places where the gears mesh are where we need to consider the gear ratio, so in the case of 7.81 we can consider <span class="math-container">$\frac{781}{100} = \frac{11*71}{10*10} = \frac{11}{10}*\frac{71}{10}$</span> which means you need to have a place that the gears mesh for each of those reductions. This could mean a driving gear with 10 teeth driving a double gear with the driven gear having 11 teeth and the attached gear having 10 teeth which is then driving a gear with 71 teeth.</p>
<p>Thinking of it as <span class="math-container">$\frac{drivenA}{drivingA} * \frac{drivenB}{drivingB} = G.R.$</span> and <span class="math-container">$\frac{double}{driver}*\frac{driven}{double}$</span> where A and B are your needed gear ratios and a double gear has one driving gear and one driven gear from different ratios can be a useful way to make a gear train as you chain these together.</p>
| 55576 | gear size selection for a compound gear system to fulfill a specific output ratio |
2023-06-23T22:26:38.173 | <p><a href="https://i.stack.imgur.com/LII3e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LII3e.png" alt="enter image description here" /></a></p>
<p>Apparently the "CUP HOLDER" has a long history. The 1944 Spitfire cupholder above has me a tad confused about the manufacturing process. I am looking at the encased wired that is running the perimeter of the sloped top side edges.</p>
<p>I think this is simply a way of terminating the sheet metal so as not to leave an overly sharp edge or snag points. In order to create it, I originally thought it was just an edge tab rolled over to form a tube then the wire was inserted. Then I thought they might of used the wire are part of the forming process and rolled the edge around it. My final thought was that its actually a circular tube with 26 Gauge wall with a 14 Gauge wire insert as described at the top of the page item "C".</p>
<p><strong>So my question is how would one make this "tube" of a different gauge than the adjacent sheet metal which it would be attached to?</strong> I noted no welding symbols so I am curious as to the process.</p>
| |mechanical-engineering|cad|machining|technical-drawing| | <p>As comments mentioned, and your second guess surmised, the 24 Gauge sheet metal A is wrapped around the (solid) 14 Gauge wire.</p>
<p>The confusion appears to be the result of a mistake on the print. The callouts of A B and C refer to materials each part is made from, with designations as to the quantity required to make this part.</p>
<ul>
<li><p>A is cut and bent from 11.5"x13.5" of 24 Gauge sheet metal</p>
</li>
<li><p>B is 2 pieces, 4" each of 0.6"x0.5" 20 Gauge L angle</p>
</li>
<li><p>C is 6.1" of 0.8"x0.55" 20 Gauge L angle</p>
</li>
<li><p>26" of 14 Gauge wire is likely what was meant by the error.</p>
</li>
</ul>
<p>Material is Alclad so aluminum chart likely applies for Gauge to thickness.</p>
<p>The print is not without other errors/omissions - the Tenax fastener appears to be shown on only two of the three views.</p>
| 55593 | How is the edge piece made? |
2023-06-25T13:36:27.220 | <p>In fluid dynamics, when I have to compute Reynolds number, I was taught to differentiate between two cases: <span class="math-container">$$ 1)\space \frac{\rho\cdot v\cdot d}{\mu}$$</span> or <span class="math-container">$$2)\space \frac{v\cdot d}{\mu}$$</span></p>
<p>but, why do I have to do this? Why shouldn't I simply use the first formula? what is the criteria to use in order to include or omit the density (i.e <span class="math-container">$\rho$</span>)?</p>
| |fluid-mechanics|building-physics| | <p>I've never seen the second formula.
It's incorrect, unless it's a specific application like water where density is numerically 1 g/cc and they just don't write it.</p>
<p>The first formula</p>
<p><span class="math-container">$$\mathrm{Re} =\frac{\rho vd}{\mu}$$</span></p>
<p>is the definition of Reynolds number and is always true. But you may sometimes see it written in terms of kinematic viscosity <span class="math-container">$\nu$</span> which is equal to <span class="math-container">$\mu/\rho$</span>, and so you will see:</p>
<p><span class="math-container">$$\mathrm{Re} =\frac{vd}{\nu}$$</span></p>
<p>In other more specialized geometries like a porous bed or perforated plate, you may see the Reynolds number defined in terms of other parameters like the mass flow rate and some dimension. But for flow in a pipe, the first formula is always true.</p>
| 55601 | Doubt about Reynolds number formula |
2023-06-29T22:56:11.603 | <p>At first I thought it was a bad translation from Chinese to English, but it turns out that many data-sheets of kWh power meters mention a maximum altitude, usually below 2 or 3 km.</p>
<p>For example:</p>
<ul>
<li><a href="https://www.distribution.voltalia.com/media/spreadsheets/6230a7130640a_GW_Smart%20Meter_Datasheet-EN.pdf" rel="nofollow noreferrer">GoodWe smart meter</a></li>
<li><a href="https://www.se.com/ng/en/faqs/FA341626/" rel="nofollow noreferrer">Schneider power meter</a></li>
</ul>
<p>What exactly is the reason for the height (altitude) restriction?</p>
| |power|power-electronics|power-engineering| | <p>Here's a good answer on <a href="https://electronics.stackexchange.com/questions/366636/how-is-clearance-affected-by-altitude-air-pressure#:%7E:text=Here%20is%20a%20description%20of%20what%20happens%20with,curve%20on%20the%20chart%20above%20tells%20the%20story.">electronics.stackexchange</a></p>
<p>Briefly: air is ionized by being hit by fast electrons. When there is less air, the electrons move farther between collisions, and become faster and more energetic before hitting an air molecule.</p>
<p>This also creates the counter-intuitive effect that it requires more voltage to get an arc as items get very close: if the electrons don't have enough space to accelerate, they don't get enough energy to ionize air molecules.</p>
<p>At very high altitudes (the edge of space) there isn't any air at all, and arcing becomes less of a problem.</p>
| 55638 | Why do kWh (smart) meters have a maximum operating altitude? |
2023-06-30T16:35:42.113 | <p>I'm trying to determine electric motor power sizing with requirements given for covering a particular distance within a certain time. The knowns are the vehicle mass, the allotted time, the target distance to cover, and the vehicles initial velocity. The result would be the absolute minimum possible motor output power rating that would satisfy the requirement.</p>
| |mechanical-engineering|motors|power| | <p>We calculate the acceleration of the linear motion, <span class="math-container">$\alpha$</span>,then apply</p>
<p><span class="math-container">$$F=m\alpha$$</span>
To get F.
and power is
<span class="math-container">$$P=\frac{F*x}{t}$$</span>
So.
<span class="math-container">$$x=V_{0}t+\alpha t^2/2$$</span>
<span class="math-container">$$\alpha= \frac{2(x-v_0t)}{t^2}$$</span>
From this we get <span class="math-container">$$F=m* \frac{2(x-v_0t)}{t^2}$$</span>
anf required power is</p>
<p><span class="math-container">$$ P =\frac{x*m* \frac{2(x-v_0t)}{t^2}}{t}= x*m* \frac{2(x-v_0t)}{t^3}$$</span></p>
<p>x,m and t are given.</p>
| 55642 | Calculate constant power applied to displace a mass a given distance within a given time |
2023-07-02T12:24:24.500 | <p>I want to find the reaction force of support A and also find an equation for deflection using Castigliano's energy method for this beam.
I know that the reaction force of A is 65.625 N but my solution returns 62.5 N.
I don't know if there is something that I'm doing wrong.</p>
<p>Assuming L=10 m and omega = 50 N</p>
<p><a href="https://i.stack.imgur.com/Xhq1s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xhq1s.png" alt="enter image description here" /></a></p>
<p>This is my solution:
<a href="https://i.stack.imgur.com/BRIj2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BRIj2.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|deflection| | <p>Two aspects require additional attention when reviewing this solution: (1) the distributed load, which is itself changing spatially, and (2) the piecewise nature of this load.</p>
<p>The current solution gives "omega" as 50 N, but the units don't match, as we're talking about a distributed load that must have units of N/m.</p>
<p>This load increases linearly from 0 N/m to 50 N/m over 5 m, or a rate of increase of 10 N/m². It's easy to mix up these values if the units aren't correctly tracked.</p>
<p>You haven't defined any of the variables used in the solution, but since <span class="math-container">$w$</span> appears near to <span class="math-container">$x^2$</span> in a moment expression, let's take it to be the maximum distributed load (50 N/m in this problem).</p>
<p>At any point <span class="math-container">$x$</span> between <span class="math-container">$0$</span> and <span class="math-container">$L/2$</span>, the average distributed load to the left is linearly interpolated as <span class="math-container">$0+\frac{1}{2}\frac{w-0}{L/2-0}(x-0)=\frac{wx}{L}$</span>. The total shear to the left is therefore <span class="math-container">$Q(x)=R_A-\frac{wx^2}{L}$</span>, where <span class="math-container">$R_A$</span> is the reaction load at A. (You could also write this as <span class="math-container">$Q(x)=R_A-\frac{w^\prime x^2}{2}$</span>, where <span class="math-container">$w^\prime$</span> is the rate of distributed load increase.)</p>
<p>On the other half of the beam, the right side, at any point <span class="math-container">$x$</span> between <span class="math-container">$L/2$</span> and <span class="math-container">$L$</span>, the shear is <span class="math-container">$Q(x)=R_A-\frac{wL}{4}$</span>.</p>
<p>Integrating, we find the moment to be <span class="math-container">$M(x)=R_Ax-\frac{wx^3}{3L}$</span> on the left side and <span class="math-container">$M(x)=R_Ax-\frac{wLx}{4}+\frac{wL^2}{12}$</span> on the right side (the final term arises as a constant for continuity).</p>
<p>Working through the integration, we have</p>
<p><span class="math-container">$$y_A=0=\frac{1}{EI}\left[\int_0^{L/2}\left( R_Ax^2-\frac{wx^4}{3L}\right)\,dx+\int_{L/2}^{L} \left(R_Ax^2-\frac{wLx^2}{4}+\frac{wL^2x}{12} \right)\,dx\right] ;$$</span></p>
<p><span class="math-container">$$0=\left[\frac{R_Ax^3}{3}\right]^L_0+\left[-\frac{wx^5}{15L}\right]^{L/2}_0+\left[-\frac{wLx^3}{12}\right]_{L/2}^L+\left[\frac{wL^2x^2}{24}\right]_{L/2}^L;$$</span></p>
<p><span class="math-container">$$0=\frac{R_AL^3}{3}-\frac{7wL^4}{160};$$</span></p>
<p><span class="math-container">$$R_A=\frac{21wL}{160},$$</span></p>
<p>which gives the expected answer of <span class="math-container">$R_A=65.625\,\text{N}$</span>.</p>
| 55652 | Find deflection and reaction force of a beam using Castigliano's energy method |
2023-07-02T17:36:49.943 | <p>I've just read this article on a new - albeit very small scale ATM - way of generating electricity directly from water vapour:</p>
<p><a href="https://www.theguardian.com/science/2023/jul/02/it-was-an-accident-the-scientists-who-have-turned-humid-air-into-renewable-power" rel="nofollow noreferrer">https://www.theguardian.com/science/2023/jul/02/it-was-an-accident-the-scientists-who-have-turned-humid-air-into-renewable-power</a></p>
<p>There's also a link to the paper in the article.</p>
<p>To think of the evaporated water vapour in the atomsphere as a giant energy store driven by the sun smoothing out daily and yearly cycles is very compelling.</p>
<p>Somehow they've mananged to generate an electric field by differentially binding water vapour within the material and then have a current run along that field - again fed by water vapour as fuel. What I don't understand, however, is the conservation of energy side, here. We have water vapour going in and electricity and what going out? Liquid water?</p>
<p>Has anyone seen articles on that paper or the paper itself too and can elighten me?</p>
<p>Thanks,
Damian</p>
| |material-science|renewable-energy| | <blockquote>
<p>A distinct mechanism was revealed, in which a spontaneous water adsorption was found to build up across the film thickness... and induce differentiated charge interaction for current.</p>
</blockquote>
<p>It appears that the researchers don't really understand what is happening, though they posit possibilities. They state that there is a steady-state effect, so it's not just a dry material pulling in water. It seems possible to me that there is some type of condensation but the scale is so small as to not be measurable. The researchers state that they believe that the energy source is electrostatic movement of the water molecules, but it seems to me that they are providing a reason why it's not free energy. What the total-system effect is of drawing out electrostatic energy is beyond my understanding of molecular physics.</p>
| 55655 | Energy balance of "air-gen" direct water vapour to electricity generation |
2023-07-03T13:57:59.890 | <p>imagine the scenario where two pipe groups are connected in series. RG1 and RG2<br />
RG1 Consists of two pipes R1 and R2 in Series<br />
RG2 consists of two pipes R3 and R4 in Parallel<br />
the diameter and surface roughness of all four pipes are equal but the length of them varies</p>
<p>In this scenario, water as an incompressible fluid flows through R1 and exits the network at end at R3 and R4. The flow is subject to pressure loss in the form of friction and pressure loss at Parallel T-junction in RG2.</p>
<p>My question is does the Volume Flow Q hence flow speed V stay constant along RG1, and for RG2 since the diameters are the same then according to continuity equation the volume flow in R3 and R4 is half as much as in RG1 ? If so, then can we calculate pressure loss by calculating Reynolds number and then friction factor for each of the segments and then sum the pressure drop of the series pipes?</p>
<p>Edit:</p>
<p>the reason for my doubt is the bernouli equation. imagine E for entry and A for exit from a pipe.
<span class="math-container">$\frac{P_E}{\rho} + \frac{W_E^2}{2} = \frac{P_A}{\rho} + \frac{W_A^2}{2} + \Delta P$</span>
<br />
hence the flow speed W is dependant on the pressure difference across the pipes (also friction)</p>
<p>if possible please provide an example for demanded pressure at R1 for desired total volume flow at R4 and R3</p>
| |fluid-mechanics| | <p>Flow (speed or mass flow rate) is dependent on the TOTAL head loss.</p>
<p>Clearly flow can't change inside the system - the same fluid that goes into the series portion also goes into the parallel section. You can either calculate the flow from the total pressure drop or the pressure drop if you know the flow.</p>
<p>You can use equations to do this, but most engineers are going to use tables based on the type of pipe and add equivalent lengths of transitions, and elbows.</p>
| 55660 | Pressure Loss in a network calculation with Darcy Weisbach |
2023-07-05T18:08:06.763 | <p><strong>Background:</strong> Residential single story post-tension slab approximately 2300 sq. ft. new construction. Contractor has installed formwork and laid out post tension cables.</p>
<p>Contractor has "overstripped" the sheathing from the post tension cables approximately 3" on both ends. In other words there is 3" of exposed cable between the sheathing and the anchorage points on both ends of the cable. This is the part inside of the formwork.</p>
<p><strong>Question:</strong> Is it acceptable practice for the steel tension cables to be left exposed to the concrete in this manner when it is poured?</p>
<p>I've spent hours trying to find an answer but can't. Unfortunately it is 2 days before pour is scheduled and the job site isn't close and I don't have a good picture.</p>
| |concrete| | <p><a href="https://i.stack.imgur.com/JCMSk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JCMSk.png" alt="anchor live end" /></a></p>
<p>Yes, it's okay. The cable is covered with grease for protection against corrosion. That over-stripping is to make sure the anchorage engages the steel, not the plastic sleeve!</p>
<p>And the grease allows for the cable to be pulled by the hydraulic jack.
There are a variety of alternative anchors and details, grouted sleeves or greased sleeves, etc. But this is common for one or two-story light construction slabs on grade.</p>
<p>Of course, one should follow the approved plans and shop drawings.</p>
| 55680 | Residential Post Tension Slab Setup |
2023-07-06T17:21:03.773 | <p>Does creep exists for all materials at any stress at room temperature? I searched for this and could not find any answer other than creep isn't significant at low stresses and low temperatures. What I want to know is if the creep is present at low temperatures (way below the melting point) and even at the slightest of stress (way below yield strength) even if in the smallest negligible amount (creep) so that deformation or elongation will happen noticeably if we waited for a sufficient time period (even if very long). For example, a structural-grade steel cube suspended by a string from above will/will not elongate under its own weight and become a long and thin wire that ultimately touches down to the ground if we observe it for say 100 000 or more years? Or is there a creep threshold which determines that below that no irreversible creep is possible?</p>
| |structural-engineering|materials|stresses|deformation|creep| | <p>Yes, creep occurs at all temperatures, although the dropoff is exponential. The reason is that at any nonzero temperature (which is all temperatures), there’s a nonzero chance for a thermal defect such as a <a href="https://en.wikipedia.org/wiki/Vacancy_defect" rel="noreferrer">vacancy</a> to form, and to minimize strain energy, these defects preferentially resolve in a manner that relieves the existing stress state. This results in viscous flow.</p>
<p>The flow rate is marked on some deformation mechanism maps:</p>
<p><a href="https://i.stack.imgur.com/aVY4O.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/aVY4O.gif" alt="enter image description here" /></a></p>
<p>Observe the sagging of lead pipes decades after installation:</p>
<p><a href="https://i.stack.imgur.com/Sl38T.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/Sl38T.gif" alt="enter image description here" /></a></p>
<p>You can expect to wait the longest (or gain the longest safe operating time, however you look at it) for strong, refractory materials at low temperatures and load states.</p>
| 55684 | Creep of the materials |
2023-07-08T22:23:10.230 | <p>I'm using SolidWorks 2023 and I have a curve like this one:</p>
<p><a href="https://i.stack.imgur.com/3klKA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3klKA.jpg" alt="enter image description here" /></a></p>
<p>And I want to find it's full length. I've already tried to do it with Measure tool, but it doesn't show the length to me. How to do it?</p>
| |solidworks| | <p>Okay, I've found out that I can just Ctrl-Click every segment and the total length will appear in the bottom-right corner.</p>
| 55699 | How to measure length of multiple arcs and lines in SolidWorks? |
2023-07-09T22:28:29.020 | <p>This might be a stupid question, I'm not an engineer, and not really sure where to begin to research this. I tried Google but of course it's all biased toward news articles about Japan.</p>
<p>So let's start with what I know. Nuclear wastewater has tritium and other radioactive isotopes. Regular salt water can be boiled off leaving salt behind. Why can't we boil the wastewater off to leave behind the isotopes? I know they're atomic scale, so it's not quite the same but are there any publications where it has been tried and the results were documented?</p>
| |waste-water-treatment|waste-disposal| | <p>To use your example, salt can be removed from water by boiling the water and condensing the resultant steam. This produces two products, distilled water and a dry salt.</p>
<p>The reason why this is possible is because salt is dissolved in water. Simplifying things, lets only consider common table salt, sodium chloride (NaCl). This applies to all salt. When sodium chloride dissolves in water it <strong>does not</strong> chemically react with water (H<sub>2</sub>O) to form hydrogen chloride, HCl, which is hydrochloric acid, or oxidized sodium (NaO). The salt stays in solution and when water is removed salt remains.</p>
<p>You mention <a href="https://en.wikipedia.org/wiki/Tritium" rel="nofollow noreferrer">tritium</a>, which is one of the isotopes of hydrogen. What defines an element (hydrogen, oxygen, carbon, iron, etc.) is the number of protons in its nucleus. Hydrogen has one proton, carbon has six protons, oxygen has eight.</p>
<p>The other particle that an atomic nucleus can contain is a neutron. Protons have positive charge, neutrons have no electrical charge. Electrons have negative electrical charge, they orbit the nucleus.</p>
<p>Commonly, the nucleus of a hydrogen atom only contains one proton. However, sometimes it can contain one or two neutrons. This makes such atoms of hydrogen heavier than a typical atom with only one proton. A hydrogen atom with one proton is called <a href="https://en.wikipedia.org/wiki/Deuterium" rel="nofollow noreferrer">deuterium</a> and one with two protons is called <a href="https://en.wikipedia.org/wiki/Tritium" rel="nofollow noreferrer">tritium</a>. When such atoms combine with oxygen to produce a water molecule the atoms are locked together.</p>
<p>Boiling such water only produces steam. If you want to separate deuterium and tritium from water, the water molecule must be split to produce hydrogen gas and oxygen gas. This can be done using <a href="https://en.wikipedia.org/wiki/Electrolysis" rel="nofollow noreferrer">electrolysis</a>, where an electric current is passed through water.</p>
<p>Regarding the other radioactive wastes in the water, such a strontium and cesium etc. These elements can be removed from the water as you suggest, by boiling the water. This produces another problem, what to do with the radioactive power that remains when the water has been removed? How much energy must be used to do this and what does it cost? What form of energy will be used to do this: nuclear, renewables, coal, gas, solar evaporation? How much radioactive power would be produced and how is it then collected for "disposal" and how and where is it then disposed of?</p>
<p>Dumping radioactive water into the ocean is cheap and rids the utility of a inconvenient water storage problem. It gives the utility an opportunity to "move on" after so many long tortuous and expensive years.</p>
| 55709 | Question about nuclear wastewater treatment and disposal |
2023-07-12T19:43:00.747 | <p>I am a hobbyist and in the process of designing a human-powered hydrofoil (see Hydroped for reference) I'm kind of stuck in the calculation of the the maximum buckling load for a strut.</p>
<p>The strut is modeled as a vertical column with a wing at the lower end and the load at the upper end. My question is: how should I set the end conditions for this problem? I was thinking that the lower part could be considered fixed (the wing), because it's relatively still with respect to the water flow, and the upper part could be considered free. But the wing can actually move around in the water, so it should really be considered free as well?</p>
<p>A next step would be figuring out how to calculate the stress on a strut which has a sweep angle, but I guess the best approach is to take one step at a time...Any help is appreciated!</p>
| |structural-analysis|buckling| | <p>I think the most important force is the shock and torque and moments imparted to the post by the hydrofoil. or to the hydroped by the rider.</p>
<p>You would need to take into account that sometimes the thing flies off the crust of the big waves, sours up to 20 feet, and land or crash into choppy waters.</p>
<p>The dynamics of the lift and turning of the foil have significant stresses.</p>
<p>Buckling is the least of concerns when we have moments applied both on the top connection of the post to the board and the bottom connection of the post to the foil.
You design the post as a beam-column member.</p>
<p>Some manufacturers of gear use carbon fiber for extra strength!
<a href="https://i.stack.imgur.com/tjJNq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tjJNq.png" alt="hydrofoil" /></a></p>
| 55736 | How to calculate maximum buckling load for hydrofoil strut? |
2023-07-13T13:21:31.297 | <h2>How can I find the optimal shape of a cantilever beam subject only to its own weight, and having the following restrictions?</h2>
<ul>
<li>The cross section is a rectangle</li>
<li>Only the height of the rectangle can vary along the length of the beam</li>
<li>Euler-Bernoulli beam theory is used</li>
<li>Optimal in this case means that fibers at the extremes of <strong>any cross section at any position x</strong> will be used at 100%, so they will reach the yield strength of the material, ie <span class="math-container">$\sigma=\frac{M}{I}y_{max}=\sigma_y$</span></li>
<li>The only load acting on the beam is that originated due to the weight of the material itself</li>
<li>The beam is assumed to fail due to bending only, so vertical shear stresses will be ignored</li>
</ul>
<p>I am only interested in the analytical solution, without regards for production challenges, costs ecc.</p>
<h2>Context</h2>
<p>I am an engineering student and out of curiosity I am trying to find optimal shapes of beams subject to different load conditions, such that for each cross section the variable dimension cannot be any smaller or the material will yield. I managed to easily get the solutions for point load with variable base and variable height, but this one is giving me troubles.</p>
<h2>What I tried</h2>
<h3>Attempt 1</h3>
<p>I started with the following scheme:</p>
<img src="https://i.stack.imgur.com/kPbaf.png" width="600"/>
<p>I assumed the load q(x) to be unknown because it is given by the cross section. The second moment of area I(x) is also unknown.</p>
<p>By following the logic depicted in the image below</p>
<img src="https://i.stack.imgur.com/AT0Mt.png" width="600"/>
<p>I computed the load q(x) as:</p>
<p><span class="math-container">$$ q(x) = \frac{dF}{dx} = \rho g b h(x) g \frac{dx}{dx} = \rho g b h(x)$$</span></p>
<p>I computed the bending moment at a random point x due to the load on the right portion of the beam as:</p>
<p><span class="math-container">$$M(x) = \int_0^{M(x)}dM = \int_0^{L-x}q(t) \cdot t \, dt = \int_0^{L-x}\rho b g h(t)\cdot t \, dt$$</span></p>
<p>And the second moment of area I(x):</p>
<p><span class="math-container">$$I(x) = \frac{bh(x)^3}{12}$$</span></p>
<p>To finally get the stress at the extreme fibers that I then impose equal to the yield strength, <span class="math-container">$\sigma=\sigma_y$</span></p>
<p><span class="math-container">$$
\begin{aligned}
\sigma & = \frac{M(x)}{I(x)} \cdot y_{max} \\
& =\frac{\int_0^{L-x}\rho g b h(t)\cdot t \, dt}{\frac{bh(x)^3}{12}} \cdot \frac{h(x)}{2} \\
& =\frac{6 \rho g \int_0^{L-x} h(t)\cdot t \, dt}{h(x)^2} \\
& = \sigma_y
\end{aligned}
$$</span></p>
<p>At this point I try to differentiate the expression to eliminate the integral and get a differential equation which is usually easier to deal with than integral equations (I used Leibniz rule to differentiate under integral sign):</p>
<p><span class="math-container">$$
\begin{aligned}
\sigma_y h(x)^2 & = 6 \rho g \int_0^{L-x} h(t)\cdot t \, dt \\
\frac{d}{dx}\left[ \sigma_y h(x)^2 \right] & = \frac{d}{dx} \left[ 6 \rho g \int_0^{L-x} h(t)\cdot t \, dt \right] \\
2 \sigma_y h(x)h'(x) & = -6 \rho g \cdot h(L-x)\cdot (L-x)
\end{aligned}
$$</span></p>
<p>And the problem is that I don't know how to solve this differential equation where h has two different arguments, first x, then L-x. I tried thinking about substitutions but they just seem to move the problem to another place.</p>
<h3>Attempt 2</h3>
<p>Another thing I tried is choosing the origin at the tip of the beam, with the beam that goes towards the right this time.
I followed the same procedure shown above and I managed to get another equation, but this time I got the following:</p>
<p><span class="math-container">$$
\sigma_y h(x)^2 = 6 \rho g \int_0^x h(t) \cdot (x-t) \, dt
$$</span></p>
<p>I differentiated once both sides and got:</p>
<p><span class="math-container">$$
2 \sigma_y h(x) h'(x) = 6 \rho g \int_0^x h(t) \, dt
$$</span></p>
<p>Then differentiated again and got:</p>
<p><span class="math-container">$$
2 \sigma_y [h(x) h''(x) + h'(x)^2] = 6 \rho g h(x)
$$</span></p>
<p><span class="math-container">$$
h(x) h''(x) + h'(x)^2 - \frac{3 \rho g}{\sigma_y} h(x) = 0
$$</span></p>
<p>Which is a nonlinear 2dn order ODE, but again I am a bit stuck with the solution of it as I am a bit rusty with ODEs and I don't know if this is an equation that can be solved nicely or if it could keep me up all day.</p>
| |structural-analysis|beam|solid-mechanics|strength| | <p><a href="https://i.stack.imgur.com/2Q7zb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Q7zb.png" alt="Red is beam depth blue is stress" /></a></p>
<p>Red is beam depth blue is stress. That's a numerical solution that's why it goes silly at the tip</p>
| 55742 | Optimal shape for cantilever beam subject to body weight only |
2023-07-14T17:25:49.393 | <p>I purchased a cheap project box made of ABS plastic (so, relatively soft material). The box has pilot holes (with counterbore) on one half and matching smooth bosses/protrusion for driving the screws on the other, but the box came without any assembly screws.</p>
<p>After measuring the hole diameter, which was 2mm, I went ahead and ordered a bunch of M2 screws, and I'm pretty sure they'll do the job, but then I started thinking about and now I'm wondering if I actually ordered the best size for the job.</p>
<p>Digging around tells me that M2 specifies a screw with 2mm <em>Major Diameter</em>, i.e. 2mm is the diameter across the outer thread. In a perfect platonic world in which all dimensions are perfectly accurate, this screw would just graze the smooth sides of the shaft when inserted, wouldn't it?</p>
<p>With dimension tolerances, the screws I ordered will probably bite in sufficiently to do the job. But what size screws should I have ordered for the "best" results? should I choose the screw so the root diameter match the hole diameter, to get the thread to embed as much as possible into the sides? Should I aim for the thread to sink in about half way into the sides, so that the plastic can deform into the recesses of the thread? How much does it matter? are these considerations different for different kinds of materials? How?</p>
| |mechanical-engineering|fasteners| | <p>Thanks to @TigerGuy for reference.</p>
<h2>Vendor #1 - Stanley</h2>
<p><a href="https://www.stanleyengineeredfastening.com/-/media/web/sef/resources/docs/other/threaded_fasteners_for_plastics.ashx" rel="nofollow noreferrer">https://www.stanleyengineeredfastening.com/-/media/web/sef/resources/docs/other/threaded_fasteners_for_plastics.ashx</a>
suggests 75-85% thread engagement as starting point. This means that the root diameter matters also needs to be taken into account.</p>
<p>The document contains a handy table on p13 "Hole Sizes Per Percentage of Thread Engagement".
A few examples (80% TE):</p>
<pre><code>Major D/Hole D
===============
2.5mm/1.8mm
3.0mm/2.34mm
3.5mm/2.7mm
</code></pre>
<h2>Vendor #2 - Celo</h2>
<p>Similar guidance from another vendor, this time sorted by material.
Roughly, they recommend a hole diameter of 0.75d for softer plastics (like plain ABS) and 0.83d for harder plastics (like PC).</p>
<p><a href="https://www.celofasteners.com/en/content/189-boss-dimensions-for-celospark" rel="nofollow noreferrer">https://www.celofasteners.com/en/content/189-boss-dimensions-for-celospark</a></p>
<h2>Vendor #3 - Plastics/Chemical company</h2>
<p>See
<a href="https://techcenter.lanxess.com/scp/americas/en/docguard/Joining_Guide.pdf?docId=77016" rel="nofollow noreferrer">https://techcenter.lanxess.com/scp/americas/en/docguard/Joining_Guide.pdf?docId=77016</a></p>
<p>It suggests that several manufacturers recommend a hole size of 3.0mm for a major diameter screw of 3.5mm. which generally agrees with the previous doc.</p>
<p>So for screws around this size, a rule of thumb would be Screw Major diameter = Hole diameter + 0.5mm.</p>
<p><em>Damn, I should have ordered M2.5.</em></p>
| 55756 | How do I choose the optimal screw diameter for a given hole size? |
2023-07-14T19:05:29.153 | <p>Does this affect the gear's performance?
<a href="https://i.stack.imgur.com/9VqWh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9VqWh.png" alt="Source: McMaster-Carr" /></a></p>
| |mechanical-engineering|gears| | <p>the curve in the tooth is to clear the minor diameter of the worm gear so the worm and the tooth can engage fully.</p>
| 55759 | What are these flat surfaces on a worm gear's teeth for? |
2023-07-15T02:24:53.873 | <p>We are attempting to penetrate 1m of water featuring dissolved ions ( assume salinity of sea water) using radio waves. Aside for using wavelengths of very low frequency radio or below, what would be the best frequency for achieving this task?</p>
| |fluid|transmission|radio|radar| | <p>Underwater radio transmission at <em>VLF frequencies</em> has been used for many years by the US Navy to send signals to submerged submarines. Wikipedia has a good article on this which will answer all your questions. Here is a very brief summary:</p>
<p>VLF (10 to 30kHz) signals can penetrate ~tens of meters into seawater, but the wavelengths involved are in the range of 300 to 10 <em>kilometers</em> (!!!). Efficient radiation of RF power at those frequencies requires a vertical 1/4-wavelength antenna which is impossibly tall. For this reason, all practical VLF antennas are classed as <em>electrically short</em> which means extremely ineffective- which requires them to be driven with huge amounts of transmitter power (~megawatts) in order to produce useful RF signal strengths.</p>
<p>Note that 540 kilohertz is very close to the commercial AM broadcast band and to get seawater penetration will still require <em>many</em> megawatts of transmitter power and an antenna hundreds of feet tall.</p>
<p>Note also that you can shorten the required antenna by using the design rules for electrically short antennas but the transmitted radiation effectiveness (also called the <em>radiation resistance</em>) is extremely low so you will inevitably run right up against the severe limitations of short antennas.</p>
<p>The <em>ARRL Antenna Book</em> (used by ham radio operators) contains most of the design equations for VLF antennas.</p>
| 55763 | Optimal frequency for penetrating 1 meter of water using radio waves? |
2023-07-21T22:55:56.947 | <p>At a place I'm staying, there is a borewell whose design and construction I ignore and I have no means to figure out (total depth, cap etc.)</p>
<p>The entire house runs on a limited electrical supply from solar panels, therefore, the setup is NOT designed to run automatically (no floatswitch). Thus, every week or two, the pump is manually turned on in order to fill a couple of elevated tanks.</p>
<ul>
<li>The pump's inbuilt check-valve is not working properly.</li>
<li>The pump is 1 metre long.</li>
<li>The pump's outlet port is 2 metres below the water level.</li>
<li>The pump's outlet port is 9 metres from ground level, connected to a PVC pipe.</li>
</ul>
<p>Last few weeks when trying to perform the tank filling operation, the pump shows a near 80% failure to start pumping while the motor can clearly be heard rotating from above. One detail worthy of note is the owner of the house was <strong>always</strong> trying --futilely-- to "fill the water pipe" from above in the false belief the failing check-valve made it necessary to "prime" the pump.</p>
<p>After extracting the pump and testing it within a filled water tank, the pump was working fine, and majestically pumping water out of its 9m long pipe. It had a 100% success rate at starting and pumping.</p>
<p>However, in this "experimental" setup, it was easy to induce an airlock condition, where the pump would rotate but not move water at all by intentionally making sure the pump had air inside while a column of water was sitting above the unreliable but slow draining check valve.</p>
<p>Now, I utterly and thoroughly fail to imagine where air could come from while sitting 2m. below the water surface. But I'm not a physicist or engineer so I'm not sure if there could actually be a certain phenomenon, unknown to me, that could cause air entering the impeller section of the pump --except, perhaps, assuming the well's water level could drift below the pump's intake level--.</p>
<p>Assuming that I'm correct in assuming no air could enter the pump 2m. underwater, and once it is discarded or confirmed a possible water level drift, I suppose the well should be checked for dirt, leaves etc. that might be clogging the pump's intake. No?</p>
| |pumps| | <p>You have a submersible jet pump. In the bottom of the well, there will usually be a screen surrounding the pump to keep it from sucking up sand or pebbles out of the aquifer and getting them caught inside the pump. If the screen is plugged up, the pump will instead suck down all the "loose" water inside the well casing (where the pump is suspended), spit that water upwards into the well casing, suck it up again, etc. and churn it into a mix of water and water vapor and air. if the check valve is "above" the pump inside the well casing, then a leaky check valve will let the well pipe water level fall to the level of the aquifer, making it easy for the pump to suck the pipe dry and <em>cavitate</em> (go into endless churn without delivering any water).</p>
<p>So: start by replacing the check valve first, then test. if the system still pumps itself dry, unplug the screen.</p>
| 55812 | Can a fully submerged submersible pump become air locked? |
2023-07-22T15:22:50.907 | <p>Dimensionless numbers are often used when studying complex phenomena such as in fluid dynamics or thermodynamics. In my studies we often used them, especially in these two subjects, where they helped quite a lot.</p>
<h2>The question</h2>
<p>I need help in understanding if dimensionless numbers are always valid <strong>in reality</strong>. In other words, is it always true that: if a real system I am studying has the same dimensionless numbers as another system, the two are completely similar and all missing data of the studied system can be recovered from the dimensionless numbers?</p>
<p>(In the question I assume the two systems I am comparing are doing the same thing, eg two air flows out of two nozzles, not comparing a nozzle flow with a convective heat flow.)</p>
<h2>Some context</h2>
<p>I have this doubt for some reasons:</p>
<ul>
<li>I think there are some phenomena, like change of state of matter or magnetic saturation, that maybe aren't taken into account by dimensionless numbers, or at least I've never heard of such dimensionless numbers in my studies.</li>
<li>I once tried using dimensionless numbers in an electrodynamic simulation in Femm software. At first, changing some parameters, like width of the coil or current through the wire, while keeping the dimensionless numbers the same worked and I could calculate the missing parameters, like induced force on an iron piece, and indeed the hand calculations gave the same answer as the simulation. If a parameter became too different from the starting value however I would start getting different values by hand and from the simulation.</li>
</ul>
<h2>EDIT</h2>
<p>Reasons to why I think some dimensionless numbers are not always relevant in reality:</p>
<p>Sometimes the effect of the change of one or more dimensionless numbers is negligible on the behaviour of the system due to certain conditions that limit the physical mechanisms represented by such numbers to nearly zero.</p>
<p>I'll show an example:</p>
<p>Let's say we are analyzing a fluid flow in a tube and we have some simplifying conditions like the tube is insulated such that thermal exchanges with the outside are negligible (negligible also in reality, not only from a theoretical point of view) and we can consider the flow as adiabatic. In this case I wouldn't consider to check the Nusselt number because, while it can still be calculated, its value wouldn't influence the system by much, because the tube wall tends to block any heat exchange between the fluid and the ambient outside the tube. In this case I would be mainly interested in Reynolds number and Darcy friction factor, as these are the numbers that matter more in the system.</p>
<p>I would also be interested in Mach number, to be sure that the flow is still able to receive a pressure feedback from the opposite side of the flow and respond to it, and to also check for compressibility effects. <br/> Another number which would be of interest is one relating heat transfer through the tube by conduction and momentum of the fluid, to show that the second is more important than the first.</p>
| |modeling| | <p>What matters is the physical laws, not the dimensionless numbers themselves. If you write the Navier Stokes momentum and energy balances in non-dimensional form, (by suitable choices like <span class="math-container">$u'=u/U_0,\, x'=x/L,\, p'=p/(0.5\rho U_0^2)$</span> etc., the Reynolds number, Prandtl number and others will appear naturally as part of the equations. You can look this up in any common reference to see what I mean.</p>
<p>Some other dimensionless groups are more empirical, shown to predict behavior by experiment.</p>
<p>To your question on whether dimensionless numbers will predict a real life situation all the time: any physical model is just that, a model, with many assumptions and idealizations. It is meant to get us relatively close to an answer, which is often the best we can hope for with fluid dynamics. Also keep in mind that <em>all</em> dimensionless numbers must match for there to be a true similarity, e.g. Reynolds, Mach, geometric aspect ratios, which can be very difficult to achieve in practice. Often a lab like a wind tunnel testing a model aircraft must make do with an imperfect match, and get what data they can.</p>
| 55813 | Validity of dimensionless numbers |
2023-07-22T19:37:22.160 | <p>Which one is correct method to find work done by a isentropic compressor?</p>
<p><strong>Method-1:</strong><br />
Compressor power = mCpdT<br />
Where m = mass flow rate<br />
Cp = specific heat at constant pressure<br />
dT = Temperature difference at inlet and outlet</p>
<p><strong>Method-2:</strong><br />
Compressor work = (P2V2 - P1V1)/(1 - γ)<br />
Where γ = specific heat ratio<br />
As this formula represent work done by gas so inversely we can say work done by compressor.</p>
<p><strong>Method-3:</strong><br />
compressor power = mCvdT<br />
Where Cv = specific heat at constant volume<br />
We know dq = du + dw.
Here dq = 0 since it is isentropic process. So dw= -du and du = mCvdT.
It seems that if precise calculation is done method-2 and method-3 give same result.</p>
<p>I applied three method with a sample math. Three method gives three result but I am not sure which one is to be accepted. Please see the photos below for clear understanding.</p>
<p><a href="https://i.stack.imgur.com/tVSUM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tVSUM.jpg" alt="sample math of compressor" /></a>
<a href="https://i.stack.imgur.com/zGEBQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zGEBQ.jpg" alt="sample math of compressor" /></a></p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|compressors|turbomachinery| | <p>Method 1 is correct. The power equation has <span class="math-container">$c_p \Delta T$</span> which is enthalpy change.</p>
<p>Method 3 ignores that this is a flow process and therefore winds up with change in internal energy (hence use of <span class="math-container">$c_v$</span>). Multiply the answer by <span class="math-container">$\gamma = 1.4$</span>, so <span class="math-container">$c_p$</span> instead of <span class="math-container">$c_v$</span>, to get enthalpy change and you are back to Method 1 and its answer.</p>
| 55818 | How to find work done by compressors |
2023-07-25T18:49:55.690 | <p>I'm a final year mechanical eng student, I'm working on designing a prototype of a '<strong>Hybrid vertical axis wind turbine (VAWT)</strong>". This <strong>hybrid</strong> turbine consists of two types of VAWTs <strong>Savonius</strong> & <strong>Darrieus</strong>, one providing good 'starting torque' at 'low wind speed' another, needs 'high starting torque' and ' high wind speeds' but, can spin faster than <strong>Savonius turbine</strong> (no limiting drag). So I want to combine both of these to enjoy both of these turbines' advantages, thus trying to nullify each others' disadvantages. To do this I'll be needing a <strong>switching OR engage/disengage</strong> mechanism which can switch between both turbines' shafts... <strong>Savonius turbine shaft</strong> engaged with <strong>Darrieus's shaft</strong> until Savonius <strong>assists</strong> Darrieus in <strong>starting & speeding up</strong> and after <strong>reaching its drag-limiting speed</strong>, Savonius's shaft has to <strong>disengage from the Darrieus's</strong> thereby letting it <strong>spin independently</strong> with Savonius's shaft disengaged. Here the engaging & disengaging are going to happen at a <strong>wind speed of 7 m/s</strong>. Please suggest to me if there's a mechanism that can do this switching AND if there's a mechanism that <strong>doesn't use electrical power or servo motors</strong> for the switching actuation mechanism. Any suggestion is appreciated. Thanks in advance! :)
<a href="https://i.stack.imgur.com/neVz8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/neVz8.png" alt="Diagram of Savonius-Darrieus turbine, notice above's Savonius & below's Darrieus" /></a></p>
| |aerospace-engineering|aerodynamics|turbines|wind-power|shafts| | <p>You are looking for a one-way clutch or freewheel. You will be familiar with these from a bicycle wheel chain-drive cassette. The chain drives the wheel until the wheel's rotational speed exceeds that of the driving sprocket.</p>
| 55850 | Is there a way to 'switch OR engage/disengage' power between two different rotating shafts depending on a set RPM (wind speed)? |
2023-07-30T05:50:05.783 | <p>I'm confronted some aspects of Material Science I'm unfamiliar with.</p>
<ol>
<li><p>In general, if a material is subjected to an elastic deformation for
too long, does that become a plastic deformation?</p>
</li>
<li><p>Does this happens to metals, as well as polymers?</p>
</li>
<li><p>How long does it take for a material subjected to an elastic deformation to develop plastic deformation? [I just need to know an order of magnitude for the time].</p>
</li>
<li><p>If you repeatedly subject a material to elastic deformation, does
the material lose progressively its property to withstand elastic
deformations?</p>
</li>
<li><p>Do strained materials deteriorate more quickly than the ones that are not subject to strain?</p>
</li>
</ol>
<p>I realize some of this questions are rather broad, and I apologize for that. If it is more convenient, this question can be answered with a indication to a literature in the theory of elasticity that address such issues.</p>
<p>Any help is appreciated.</p>
| |materials|material-science|elasticity| | <p>In most common engineering materials, elastic deformation does not become plastic deformation except in those materials subject to what is called <em>creep</em>. Whole textbooks have been written on creep so this forum is not the right place to go into the details of the creep mechanism. BTW, plastics are particularly susceptible to creep. Google it and see what you get.</p>
<p>Cyclic loading within the elastic region does not degrade the strength of a machine element except in those materials susceptible to <em>fatigue</em> (same comment; do a Google search on "metal fatigue").</p>
<p>When under constant stress, metals like steel are much more easily corroded (search on "stress corrosion").</p>
<p>A piece of steel which has been strained by plastically deforming it will corrode more easily than when not deformed, because the microstructure contains residual stress (see above).</p>
| 55879 | Does elastic deformation becomes plastic deformation, given sufficient time? |
2023-07-30T16:06:17.537 | <p>Would it be safe to remove the built up rust on wheels' <a href="https://en.wikipedia.org/wiki/Control_arm" rel="nofollow noreferrer">control arms</a> by soaking them in vinegar?</p>
<p>Or is <a href="https://en.wikipedia.org/wiki/Hydrogen_embrittlement" rel="nofollow noreferrer">Hydrogen embrittlement</a> a concern for a run-of-the-mill, 10-years old control arm, which is (I think) made of cast iron? And one should better stick to a wire brush.</p>
| |car|wheels| | <p>You will get better results derusting with phosphoric acid instead. You can buy it in bottles ready-to-use as <em>concrete prep</em> or <em>concrete etch</em> in hardware supply stores, which is used to prepare broken concrete surfaces for patching.</p>
<p>I have derusted large thread-cutting taps this way, with no embrittlement. It also mildly undercuts the thread teeth, leaving them sharp enough to cut skin. In fact, you can resharpen dull files with an overnight soak in concrete prep.</p>
| 55882 | Derusting car wheel's control arm by soaking in vinegar? |
2023-07-30T16:07:15.020 | <p>In the 1890s, trains allowed access/egress at every cabin in a car: this obviated the need for a hallway and saved space.</p>
<p>Why was this eliminated as early as the 1920s?</p>
<p>1890's style car:</p>
<p><a href="https://i.stack.imgur.com/5Piqp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Piqp.jpg" alt="enter image description here" /></a></p>
| |rail| | <p>It was impossible to close the carriage on a cold morning. There were always some open doors somewhere after each station. And the doors were narrow, which made access difficult or impossible for prams, walkers, and wheelchairs.</p>
<p>But I think, like everything else, it comes down to cost: it was more expensive to put in more doors.</p>
<p>Those multiple doors made entrance and exit much faster for bulk masses of people at major stations. As the number of people taking commuter trains decreased towards the end of the last century, and handicapped access became more important, and automated doors were implemented for safety and control, the sheer speed of dumping hundreds of people all at once became relatively less important, and trains moved to fewer doors.</p>
<p>Something similar happened with trams and busses. Trams and ommni-buses like the San Francisco cable tram, which allowed fast exit, were replaced with single or double-door vehicles which forced the passenger past the operator for single-operator ticket sales and control. If you look at the SF cable tram now, you'll see that although they've kept the old vehicles, entrance and exit is strictly controlled.</p>
| 55883 | Train disembarking: why don't trains allow exit at any point1890 v 2023 |
2023-07-31T05:39:38.227 | <p>For example we are talking about special machine vision optics, cheaply manufactured by a chinese company, exclusively made for a well recognized western company with a brand name. The manufacturing process accords to strict standards assured by frequent audits by the said company, but the manufacturer is not withheld from selling the products elsewhere in general, but they must not either sell it with the western brand name, nor to any branded reseller. So one can actually directly buy from the manufacturer, without the "Western brand optics" label on it, but the product is exactly the same.</p>
<p>I would say it is "OEM", but in my perspective it would mean that the manufacturer does sell the products for other brands too, with different labels... but it does not: it either puts one specific label on it and ships to the western warehouse, or sells the products under no specific name "under the counter" on e.g. Aliexpress.</p>
<p>Is there a separate word for these products? I tried googling and inventing word combinations, but I did not find one suitable word/acronym. Should I just say OEM for this too?</p>
<p>Thank you for all ideas!</p>
| |product-engineering|project-management|engineering-managment| | <p>These are called "grey market goods". They're not fake and off the correct production line. However, they are lacking compared to the <em>real</em> products in some way. Usually, this is because they are parts recovered from the reject bin and sold.</p>
<blockquote>
<p>So one can actually directly buy from the manufacturer, without the "Western brand optics" label on it, but the product is exactly the same.</p>
</blockquote>
<p>This doesn't mean as much as you might think. The way factories work, especially in China, the same factory is contracted out to build products for different customers. Sometimes the same product is made for different customers. However, this does not mean the quality control are necessarily the same.</p>
<p>It is highly likely the original (branded) customer of the product demands higher quality control than the other customers. Indeed, if the other customers are selling the same product off the same manufacturing line there is very little incentive to maintain the same levels of quality control as the original customer. That's not to say the original customer can't mark up the price on their product due to branding. So you may be overpaying for the branded product, but the chances that the unbranded version doesn't have the same quality control is very high.</p>
<p>It is quite common for the off-brand parts to entirely consist of rejected parts from the original customer. Might not matter much for a simple consumer good, but matters very much if the product is something like an integrated circuit.</p>
| 55889 | Is there a word for "counterfeit" products that were in fact manufactured on the official assembly line, but sold cheap lacking brand label? |
2023-08-05T11:44:45.243 | <p>Every type of fuel cell has a catalyst and an electrolyte to carry electrons between the cathode and anode. What I don't understand is how the electrolyte is stored in a Alkaline Fuel Cell. In essence, is the electrolyte stored as a liquid in a tank with the electrodes at either end of the fuel cell, or in some sort of sponge that is soaked in the electrolyte and then periodically re-wet with additional electrolyte?</p>
<p><strong>How is the electrolyte stored in an Alkaline Fuel Cell?</strong></p>
| |chemical-engineering|energy-storage| | <p>After some research of my own, I've found that Alkaline fuel cells tend to store the electrolyte as a liquid in a very thin layer (around 1.2-1.7mm in thickness) which can have some sort of solid polymer matrix or other shape to maintain its shape and structure.</p>
<p>I also found that the electrolyte is normally a free-flowing liquid, so a pump and reservoir is often used to circulate the electrolyte.</p>
<p>Reference 1 - <a href="https://arrow.tudublin.ie/cgi/viewcontent.cgi?article=1027&context=engscheleart2" rel="nofollow noreferrer">https://arrow.tudublin.ie/cgi/viewcontent.cgi?article=1027&context=engscheleart2</a></p>
| 55939 | How is the electrolyte stored in an Alkaline Fuel Cell? |
2023-08-06T09:46:30.233 | <p>During my civil engineering studies, I've come across the notion of a "Polplan", a method for determining the possible infinitesimally flexible configuration of a statically indeterminate truss or more general for any multi-beam structure.</p>
<p><a href="https://i.stack.imgur.com/FBFY4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FBFY4.png" alt="An example of the Polplan" /></a></p>
<p>In the figure provided you see most of the parts that make up a Polplan: There are "main poles" (Hauptpole), which denote the center of a infinitesimal rotation of the corresponding rigid body. You can also see the "secondary poles" (Nebenpole), which denote the locations where any two rigid bodies exibit the same infinitesimal displacement.
In orange you can see the infinitesimal displaced figure.</p>
<p>There are certain rules (which can be found on the German Wikipedia <a href="https://de.wikipedia.org/wiki/Polplan#Regeln_f%C3%BCr_die_Erstellung_des_Polplanes" rel="nofollow noreferrer">here</a>), which allow to deduce the "Polplan" for any statically underdeterminant structure.</p>
<p>Is there an english term for this procedure?</p>
<hr />
<p><strong>EDIT: Not the original question, only the context:</strong></p>
<p>My goal with this question is to find out whether there is a nice mathematical framework with which one can deduce the polplan without solving the constraint satisfaction problem given by the rules. The english term would allow me to find english ressources on it.</p>
| |structural-engineering|mechanisms|terminology| | <p>I found this link that explains the rules of polplan clearly, and you can use the translate option of chrome to translate it into english.</p>
<p><a href="https://www.ingenieurkurse.de/baustatik-1/kurs-baustatik/statische-bestimmheit-polplan.html" rel="nofollow noreferrer">https://www.ingenieurkurse.de/baustatik-1/kurs-baustatik/statische-bestimmheit-polplan.html</a></p>
<p>Hope this helps you as it did for me.</p>
| 55948 | Is there an english equivalent of a "Polplan" in structural analysis? |
2023-08-06T20:51:00.303 | <p>The question is straight forward. What are the protrusions attached to the main cylinder of this electric motor?</p>
<p>This one has two protrusions that are the same size. Others, like the motor for my table saw have two of these protrusions, but one is larger than the other.<br />
<a href="https://i.stack.imgur.com/Ucc1i.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ucc1i.jpg" alt="enter image description here" /></a></p>
<p>This motor has only a single protrusion:<br />
<a href="https://i.stack.imgur.com/p9GHU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p9GHU.jpg" alt="enter image description here" /></a></p>
<p>And some motors don't appear to have any of these at all. So what gives? What function do these serve, and why does that function appear to be optional, or at least handled in different ways for different motors?</p>
| |electrical-engineering|motors| | <p>Those contain the <em>start capacitors</em> that are needed in <em>single-phase electric motors</em> to get the armature spinning when the motor is first turned on. The start capacitor is connected to the AC mains and a small set of windings that accompany the main <em>running</em> windings in the motor. The capacitor delays the phase of the current in the start winding so that when the field winding comes on and turns the armature a little bit, the start winding then adds a slightly delayed extra kick in the same direction- an action which gets the armature turning instead of just vibrating back and forth, which is what it would do in the absence of the start winding and start capacitor.</p>
<p>Once the motor comes up to full speed, a switch automatically cuts out the start circuit and the motor is left running on the run winding only.</p>
<p>This is a task that is handled differently in different kinds of AC motors. Three-phase AC motors do not require start caps at all, since the delayed phases are furnished by the mains. This means 3-phase motors are <em>self-starting</em>, as are all DC motors.</p>
| 55950 | What are the protrusions on the side of some electric motors? |
2023-08-07T02:39:19.037 | <p>I am trying to derive the differential equation for pulling vacuum on an air tank.</p>
<p>The solution to the ODE:</p>
<p><span class="math-container">$$P(t)=P_2 \cdot exp(\frac{dV}{dt} \cdot \frac{t}{V_s})$$</span>
Where <span class="math-container">$\frac{dV}{dt}$</span>=
flow rate out of tank in ACFM (assuming constant for all time t),
t = time in minutes,
<span class="math-container">$V_s$</span> = tank volume in cubic feet, <span class="math-container">$P_2$</span>
= final pressure, <span class="math-container">$P_1$</span>
= pressure at time t
<br>
Assumptions: ideal gas, constant gas flow rate</p>
<p>My derivation begins with writing the following differential equation:</p>
<p><span class="math-container">$$dP=P(1−\frac{dV}{V_s})$$</span></p>
<p>How do I get to the solution from this?</p>
| |fluid-mechanics|thermodynamics| | <p>I did get the same result but started from this:</p>
<p><span class="math-container">$$\frac{dP}{ P } = (\frac{dV}{dt} / Vs) * dt$$</span>
Then integrating both sides
<span class="math-container">$$ ∫\frac{1}{P} dP = ∫ (\frac{dV}{dt} / Vs) * dt$$</span></p>
<p>Integrating:</p>
<p><span class="math-container">$$ln|P| = (\frac{dV}{dt} / Vs) * t + C1$$</span>
Now, we'll exponentiate both sides:</p>
<p><span class="math-container">$$P = exp((\frac{dV}{dt} / Vs) * t + C1)$$</span> we drop the absolute value for P: it's always positive.</p>
<p>Combine the constants C1 and exp(C1) into one constant, let's call it C:</p>
<p><span class="math-container">$$P = C * exp((\frac{dV}{dt} / Vs) * t )$$</span></p>
<p>This is your answer by replacing C with its boundary value which is <span class="math-container">$P_2$</span>.</p>
<p><span class="math-container">$$P = P_2 * exp((\frac{dV}{dt} / Vs) * t )$$</span></p>
| 55955 | Vacuum Tank Emptying Transient Equation |
2023-08-08T16:12:48.307 | <p>I was performing a simple static structural analysis using Ansys to check the stress value for different materials. I have kept the below parameters constant:</p>
<ol>
<li>Geometry</li>
<li>Force (in N)</li>
<li>Area on which the force is applied</li>
</ol>
<p>I changed the materials from copper through PVC. However, the value of stress didn't changed. From what I have inferred about stress is: The stress remains constant in elastic regime, and once the material enters plastic regime, the value of stress changes. However, in Ansys (or any other Multiphysics software employing FEA, like Pro Mechanica of PTC Creo) the stress value is fairly constant, irrespective of materials. I am confused, for why this happens. Please see below stress values for Structural Steel, Aluminum & PVC respectively.<br />
Any help is greatly appreciated.</p>
<p>P.S. I have checked this question (<a href="https://engineering.stackexchange.com/questions/23876/stress-analysis-for-the-same-configuration-but-different-material">Stress analysis for the same configuration but different material</a>) as well, but this doesn't answers my questions
<a href="https://i.stack.imgur.com/7OnnZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7OnnZ.png" alt="Structural Steel" /></a>
<a href="https://i.stack.imgur.com/26RJ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/26RJ9.png" alt="Aluminum" /></a>
<a href="https://i.stack.imgur.com/uBbwm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uBbwm.png" alt="PVC" /></a></p>
| |mechanical-engineering|structural-engineering|stresses| | <p>Oh for goodness sake, this is statics 101. The stress in a beam in bending in the linear range is M/I=sigma/y</p>
<p>There are no material properties in that equation.</p>
| 55969 | Why is stress the same even if I change the material? |
2023-08-09T06:48:54.350 | <p>The Solar PV cell has an efficiency of 25-40%.</p>
<p>Now steam turbines run by coal or some other fuel have an efficiency of approximately 60-80%</p>
<p>What is the efficiency of steam turbines run by Solar energy? Are there any energy losses when using solar energy to heat up water instead of using something like coal?</p>
<p>For eg: See <a href="https://youtu.be/ccO0_hSXMyM?t=124s" rel="nofollow noreferrer">this</a> video. What is the efficiency of these solar powered turbines? Efficiency interms of total energy out put vs total solar energy incident. How does it compare to the efficiency of energy generated by Solar PV cells? Would PV panels installed in a similar volume generate more electric power?</p>
| |energy-efficiency|turbines|solar-energy|steam|solar| | <p>Less than 31.25% when going to electricity per wikipedia. (If you just needed heat for heating a room, it's a different story- 80% is not unheard of, but you'd be doing well to get half that into water to make PRESSURIZED steam because you can't really insulate on the solar collector itself and mechanisms such as reflectors and heat exchangers for collecting to an area with lower thermal losses will incur their own inefficiencies. Followed by your 70% turbine, that multiplies to ~30% at 1 significant digit )</p>
<p>Whether it's better than solar photovoltaic, depends on which panel, per your 25-40%.</p>
<blockquote>
<p>Of all of these technologies the solar dish/Stirling engine has the highest energy efficiency. A single solar dish-Stirling engine installed at Sandia National Laboratories National Solar Thermal Test Facility (NSTTF) produces as much as 25 kW of electricity, with a conversion efficiency of 31.25%.[66]</p>
<p>Solar parabolic trough plants have been built with efficiencies of about 20%.[citation needed] Fresnel reflectors have a slightly lower efficiency (but this is compensated by the denser packing).</p>
<p>The gross conversion efficiencies (taking into account that the solar dishes or troughs occupy only a fraction of the total area of the power plant) are determined by net generating capacity over the solar energy that falls on the total area of the solar plant. The 500-megawatt (MW) SCE/SES plant would extract about 2.75% of the radiation (1 kW/m²; see Solar power for a discussion) that falls on its 4,500 acres (18.2 km²).[67] For the 50 MW AndaSol Power Plant[68] that is being built in Spain (total area of 1,300×1,500 m = 1.95 km²) gross conversion efficiency comes out at 2.6%.</p>
<p>Efficiency does not directly relate to cost: total cost includes the cost of construction and maintenance.</p>
</blockquote>
<p>From the first web result after google's shenanigans of attempting to provide what people ask and guessed answers.</p>
<p><a href="https://en.m.wikipedia.org/wiki/Solar_thermal_energy#Electrical_conversion_efficiency" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Solar_thermal_energy#Electrical_conversion_efficiency</a></p>
<p>Efficiency comparisons only make sense when you have the same input and output. It wouldn't be good to have to burn coal to make light to feed a photovoltaic panel to generate electricity, just as it would be bad to have to grow plants, compress the hydrogen out them to make coal to then power something. Pay attention instead to what you need and what you have that you can exchange for it, seeking the most direct path.</p>
| 55976 | What is the efficiency of Solar energy run turbines? Are they better or worse than Solar PV cells? |
2023-08-10T10:53:48.810 | <p>For simplicity sake consider a solid metallic cylinder which is <span class="math-container">$2r$</span> in height with a radius of <span class="math-container">$r$</span>.</p>
<p>For a given metal and <span class="math-container">$r$</span>, how to calculate how fast can this spin in terms of rpm for a very long time without breaking or deforming?</p>
<p>Also which is the best metal to achieve huge rpm without breaking or deforming?</p>
<p>......</p>
| |mechanical-engineering|metals|strength|kinematics| | <p>For a thin spinning disk ignoring the frictions, such as air friction, there are two stresses, radial and hoop stress. Both must be below the accepted stress limit of the material. We assume no vibration and warping or fluttering.</p>
<h1>radial</h1>
<p>stress formula is</p>
<p><span class="math-container">$$ σ_r = \frac{\rho \omega^2 r}{t}$$</span></p>
<ul>
<li><span class="math-container">$\sigma_r$</span> is the radial stress (in Pascals or N/m²),</li>
<li><span class="math-container">$\rho$</span> is the material density, kg/m^3</li>
<li><span class="math-container">$\omega$</span> angular velocity, radians/s</li>
<li>t thickness, meters</li>
</ul>
<h1>hoop</h1>
<p>hoop stress is tangential to the circumference of the disk and acts along its thickness. It is caused by the resistance of the material to being stretched by the centrifugal force</p>
<p><span class="math-container">$$ σ_h = \frac{\rho^2 \omega^2 r^2}{2t} $$</span></p>
| 55989 | Relationship between how fast a metal cylinder can spin on its axis without breaking or deforming? |
2023-08-11T18:09:21.243 | <p>I know that for a first-order system, the bandwidth can be computed known the time constant, tau, where the bandwidth is equal to 1/τ.</p>
<p>Is there an equivalent formulation for a second-order system, particularly an overdamped system? I have done a search but nothing obvious popped up. I need a formula and derivation if such a thing exists rather than a call to a Matlab function for example. I did come across this on page 12: <a href="http://engineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002/Week2/Second-Order_System_2002.pdf" rel="nofollow noreferrer">http://engineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002/Week2/Second-Order_System_2002.pdf</a></p>
<p>but it doesn't say where the formulation comes from but suggests a good approximation is the natural frequency <span class="math-container">$\omega_n$</span>.</p>
| |frequency-response| | <p>This question is related to:</p>
<p><a href="https://engineering.stackexchange.com/questions/50712/contradiction-of-bandwidth-and-damping/51202#51202">Contradiction of bandwidth and damping</a></p>
<p>We know that the transfer function for a second-order system is:</p>
<p><span class="math-container">$$ H(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2} $$</span></p>
<p>Switching s to <span class="math-container">$j\omega$</span> and dividing top and bottom by <span class="math-container">$\omega_n^2$</span> and setting <span class="math-container">$u = \omega/\omega_n$</span> we obtain:</p>
<p><span class="math-container">$$ H(j\omega) = \frac{1}{1 + j 2 \zeta u - u^2} $$</span></p>
<p>From this, we can derive the amplitude A:</p>
<p><span class="math-container">$$ A = \frac{1}{\sqrt{(1-u^2)^2 + (2\zeta u)^2}} $$</span></p>
<p>Since the bandwidth is <span class="math-container">$1/\sqrt{2}$</span> of the value of <span class="math-container">$A$</span> at DC, we can find <span class="math-container">$u$</span> at this point which we call <span class="math-container">$u_c$</span>:</p>
<p><span class="math-container">$$ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{(1-u_c^2)^2 + (2\zeta u_c)^2}} $$</span></p>
<p>Solving for <span class="math-container">$u_c$</span> gives us:</p>
<p><span class="math-container">$$ u_c = \sqrt{1 - 2 \zeta^2 + \sqrt{(2 \zeta^2 - 1)^2 + 1}} $$</span></p>
<p>Since <span class="math-container">$u_c$</span> was defined as <span class="math-container">$\omega/\omega_n$</span> where <span class="math-container">$\omega$</span> is now equal to the bandwidth, <span class="math-container">$\omega_c$</span>, we can rewrite the above as:</p>
<p><span class="math-container">$$ \omega_c = \omega_n \sqrt{1 - 2 \zeta^2 + \sqrt{(2 \zeta^2 - 1)^2 + 1}} $$</span></p>
<p>This is the bandidth for as second-order system.</p>
| 56006 | What is the bandwidth for a 2nd order system? |
2023-08-13T03:15:38.467 | <p>I have a blueprint which states the following about two bolts which are used:<br />
<a href="https://i.stack.imgur.com/hrMsP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hrMsP.png" alt="enter image description here" /></a></p>
<p>Does that mean that the <em>radius</em> of the bolt decreases by 1/16" every foot, or the <em>diameter?</em></p>
| |bolting| | <p>That would be the diameter.</p>
<p>If you applied that to the radius, then the taper would not match the part it would be fitted to.</p>
| 56022 | How is the taper of a bolt correctly notated? (Imperial) |
2023-08-13T07:56:38.883 | <p>In an early paper on Simo, I came across an example titled "Cantilever 45-degree bend subject to fixed and follower end load". A force of <span class="math-container">$600N$</span> and <span class="math-container">$300N$</span> was applied at the end, resulting in a force-displacement curve. Given the physical parameters
<span class="math-container">$E=10^7$</span> and <span class="math-container">$G=0.5 \times 10^7$</span> , why isn't the material density required for the calculation? Subsequent studies have also used this example without mentioning the beam's density. How should this problem be approached dynamically? I'm interested in calculating the equilibrium state.</p>
<p>The description of this example<a href="https://i.stack.imgur.com/q0zeR.png" rel="nofollow noreferrer">1</a> as follows:</p>
<p>EXAMPLE 7.5. Cantilever 45-degree bend subject to fixed and follower end load. This example has been considered by Bathe and Bolourchi [10] under fixed end load. The bend has a radius of 100 with a unit square cross-section. The material properties are <span class="math-container">$E=10^7$</span> and <span class="math-container">$G=0.5 \times 10^7$</span>. These authors performed the analysis for conservative loading only using 8 three-dimensional degenerated beam elements. In the present calculation 8 linear elements are used. For comparison purposes with the results reported in [10] the bend is subject to a sequence of three load increments of magnitude 300,450 , and 600 .
<a href="https://i.stack.imgur.com/q0zeR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q0zeR.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/q0zeR.png" rel="nofollow noreferrer">1</a>: Simo, Juan C., and Loc Vu-Quoc. "A three-dimensional finite-strain rod model. Part II: Computational aspects." Computer methods in applied mechanics and engineering 58.1 (1986): 79-116.</p>
<p>Can this example be mathematically expressed? I'd like to compute it myself using Matlab or Python. Any comments or suggestions would be greatly appreciated.</p>
| |mechanical-engineering|structural-analysis|beam|finite-element-method| | <p>I'm not familiar with the finite strain approach to structural analysis, but I suspect they ignored density because density has no (direct) contribution to the strength or deflection of a structure.</p>
<p>There are some caveats:</p>
<ul>
<li>For some materials (e.g. concrete) density may have some correlation with stiffness, but density itself does not directly affect the displacement or strength of the structure.</li>
<li>When combined with an acceleration (e.g. gravity) density may add a <em>load</em> to the structure, but I suspect the authors of that paper are ignoring self weight loads as well.</li>
</ul>
| 56024 | Benchmarking Geometrically Exact Beams: Clarification Needed on Material Density? |
2023-08-16T14:27:07.210 | <p>This kind:</p>
<p><a href="https://i.stack.imgur.com/jZprM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jZprM.jpg" alt="rabbit with short fuzz coating" /></a></p>
<p>It's a very thin layer of very short (<1mm) hairs, coating a hard plastic underneath. I encounter this sort of lining on some mid-tier electronics that pretend to be more classy than they are, measurement tools boxes, higher (but not top) tier ballpoint boxes, and toys made of rigid plastic. In particular, I remember as a kid, the coating would rub off of more exposed parts, revealing bare plastic of a different color underneath, but there was no layer of cloth, glue or such visible - the "fuzz" seemed to be applied directly to the plastic surface.</p>
<p>What is this surface finish called, and how is it achieved? Where to look if I want to make my own?</p>
| |materials|material-science|plastic| | <p>It's called flocking and involves a way of mixing chopped fibers into a liquid which is then applied to the surface either by spray or by brush. when the solvents in the paint evaporate, the fibers are left glued to the surface.</p>
| 56051 | How is fuzzy velvet-like surface on plastic achieved? |
2023-08-18T20:25:39.247 | <p>I am installing an egress window in my basement to make it a legal bedroom. I am going to be doing most of the work myself (outside of cutting the concrete) but doing it by the book (permits, inspections, etc). I hired a structural engineer to do the drawings. I will try to follow his drawings to the T, but there are a few things I am unclear on, highlighted in the drawing.</p>
<ol>
<li>Did the engineer intentionally not include threaded rod for the header and I'm not seeing how it's secured? Or was that an oversight?</li>
<li>I don't fully understand the engineer's vision for securing the threaded rods to the vertical structural supports.</li>
<li>I have no idea what the symbol means where the vertical support and the header meet (maybe that will answer question #2?)</li>
</ol>
<p><a href="https://i.stack.imgur.com/QVgjo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QVgjo.jpg" alt="Structural Engineer's Drawings" /></a></p>
| |structural-engineering| | <p>As david mentioned, the symbol has to do with a weld. It means weld all around (the little circle) and groove should be square on the side pointed by the arrow. Where the arrow points could be a little clearer, but the intent there should basically be to preserve the surface of the side bars (rather than cutting into them for space for the weld material) to bear the load from the top.
There are various weld symbol tables online such as <a href="https://www.studiesaddict.com/2019/10/welding-symbols-their-description.html?m=1" rel="nofollow noreferrer">https://www.studiesaddict.com/2019/10/welding-symbols-their-description.html?m=1</a></p>
<p>Since the sides are welded to the top as one up-side down U, there is no need for rods in the top.</p>
<p>My concern would be heat from welding affecting the epoxy (I would have gone for inserts in the side bars making a threaded install of the rod into the side bars as they're stainless). At the same time, you have to consider accessibility for your welder since it is a retrofit. It appears the designer favored accessibility mentioning the adhesive prior to the weld.</p>
| 56079 | Questions on structural eng drawing for egress window |
2023-08-20T11:06:48.323 | <p>I was recently discussing with a relative about electric cars. It was claimed
that the motor of an electric car would age (wear and tear might be the english term?) and the efficiency would become
dramatically worse after a few years of usage.</p>
<p>While I can imagine that the motor just breaks after a while, I was always
under the assumption that it would perform until then just like on the first day.</p>
<p>Is it true that the motor of electric cars ages in a way that the energy
consumption increases / potential travel distance significantly drops, even if the battery is
ok?</p>
<p>What happens when it ages? Which parts break in which ways?</p>
| |electric-vehicles| | <p>Electric motors as used in cars do NOT experience serious wear and tear within a few years. The wear items in a DC electric motor are the bearings (which are chosen to furnish many years of operation without replacement) and the brushes (in motors that use them) which are cheap and easily replaceable.</p>
| 56088 | How do electric motors of cars age? |
2023-08-20T11:30:32.497 | <p>In <a href="https://speyer.technik-museum.de/" rel="nofollow noreferrer">Technik Museum Speyer</a> I have seen <strong>strange attachments to the keel</strong> of Rescue Ship John T. Essberger. What is their purpose? Are they anyhow related to hydrodynamics? They are quite small compared to the size of the ship.</p>
<p><a href="https://i.stack.imgur.com/3BdJx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3BdJx.jpg" alt="strange attachments to the keel" /></a></p>
| |ships| | <p>They are sacrificial zinc anodes to prevent corrosion. Every boat has them, big metal boats may have several tons of them. This type is a strap-on teardrop style.</p>
| 56089 | Keel Attachment |
2023-08-24T17:46:46.827 | <p>My question is rather specific and perhaps not of general interest. I'm working on a specific problem in physics. Let's suppose we have spherical vessel with a vacuum inside. The walls of the container are heated such that they have a fixed temperature, <em>T</em>. The walls radiate inwardly, but energy is steadily lost from inside the vessel through a nonstandard mechanism. My general understanding is that there should then be a gradient in the cavity temperature measured at each point, with the highest temperature at the walls and the minimum at the centre. Is this correct? How would one quantify this? I would be grateful for references on it also.</p>
| |thermal-radiation| | <h3>Speculative astrophysics questions cannot be answered by traditional heat transfer.</h3>
<p>A vacuum-filled sphere will only contain the UV radiation produced by the sphere. Since the sphere is a single temperature, all photons will be this temperature as well.</p>
<p>If you add speculative heat absorption and release mechanisms that involve neither traditional matter nor temperature, your solution is dependent on those (unknown) mechanisms. Your speculative physics will need to redefine temperature to fit your model.</p>
<p>Engineers use known science and math to solve problems. Scientists can exist on the fringe of observable or empirical information, but engineers stick to what is known.</p>
| 56119 | Blackbody radiation profile inside a heated, spherical conatiner with energy loss |
2023-08-24T20:47:11.023 | <p>I hate to ask a somewhat broad question, but here I am with no one else to turn to . Specifically, my question is: What should I do first?</p>
<p>Let me explain the context to make my question a little clearer: I absolutely love drinking tea and I want to be able to brew it on the go. After trying out different sorts of tumblers, I finally found one that I really like. Its only problem is that it doesn't have a removable strainer, which makes it difficult to clean, among other things. Here's the link to it: <a href="https://www.taotealeaf.com/tao-tea-tumbler/" rel="nofollow noreferrer">https://www.taotealeaf.com/tao-tea-tumbler/</a></p>
<p>The first thing I did was to design what I had in mind with a program called FreeCAD. After exporting it as an STL file and slicing it with Cura, I was able to print a prototype on my 3D printer and adjust my original design. While I am somewhat more confident in the design now, I'm still unsure about how big the gaps should be between the glass housing and the outer metal (currently 0.5mm), between the metal rod in the center and the inner metal (also 0.5mm), and between the threading for the top and bottom parts.</p>
<p>Since I never really had any training in this field, I don't know what steps I might have missed or what remains to be done before I find someone to etch my design out of a block of food-grade stainless steel. Again, the question is: What should I do first?</p>
<p><a href="https://i.stack.imgur.com/LAlwN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LAlwN.jpg" alt="rough sketch of the strainer" /></a></p>
| |steel|cad|3d-printing|product-design| | <p>First: Select an appropriate food-grade material.</p>
<p>Don't poison yourself or others with toxic materials that are in prolonged contact with hot water.</p>
<p>Some versions of brass have relatively high concentrations of lead. Materials in plumbing systems are now required to be "lead-free."</p>
| 56121 | Designing a tea strainer: What should I do first? |
2023-08-27T14:25:34.100 | <p>As in the title, what is the ideal oil for an electric razor, and why? I might add that the electric razor will only be used at room temperature.</p>
<p>What oil would an engineer select for their electric razor?</p>
<p>Throughout my life, I've always just used the cheapest "machine oil" or "sewing machine" oil (or similar) that I can get my hands on. I've often wondered, am I doing something wrong? Should I read up on this? It doesn't seem that easy to find information about, often I'd just end up watching a video with a random guy that makes no references to any papers, studies, or OEM specifications, nor any scientific terminology or concepts to substantiate why they use X or Y type of oil and it "works best for them".</p>
<p>I don't require an answer that refers to a randomized controlled trial or even a paper, but some reference to basic concepts involving lubrication in machine parts, would be nice. The more the merrier.</p>
<p>For anyone that understands engineering and oil, is it possible to say anything about what type of oil benefits a typical electric razor?</p>
<p>Should it have a certain viscosity, viscosity index? Heat resistance (maybe between the metal surfaces of the shaver head it gets quite hot?)? Oxidation properties? What about "additive package"? The "additive package" from what I understand can contain anti-wear agents, detergents, and a lot more. This seems interesting.</p>
<p>I don't claim to understand very well what any of these are (beyond a two sentence description), hence why I'm asking.</p>
<p>How about the fact that you'll be putting the shaver head against your skin afterwards and there might be residual oil even after wiping it away (?), are certain oils harmful for the skin?</p>
<p>Again, what would an engineer select for such an application?</p>
| |fluid-mechanics|metals|consumer-electronics| | <p>Lubricating an electric hair-cutting device is an easy job for just about any liquid oil on the market- it is a <em>noncritical application</em>.</p>
<p>All you really need is an oil with low enough viscosity to wick its way in between the stationary cutter and the oscillating cutter. You also do not want the oil to polymerize into wax when the razor is not in use. So-called "sewing machine oil" is perfect for this purpose.</p>
| 56139 | What is the ideal oil for an electric razor? |