# Tag Info

4

While there's a list of (rare) materials with negative thermal expansion, that 30% expansion ratio excludes most of them and narrows down your search to polymers or structures. Within the realm of polymers, two-way reversible shape memory polymers might be your option. They can have relatively high reversible strains and can be thermally activated: WM ...

4

The bolt tension could increase, decrease, or stay the same. It depends on the thermal stress distribution in the whole structure, which is likely to be eventually supported by something that remains at (approximately) room temperature. This is essentially the same situation as any bolted joint that will carry a load. The "pre-stress" put into the bolts by ...

3

I'm assuming that you want to know why we use the heat capacity at constant pressure, $c_p$, instead of at constant volume, $c_v$, in analyzing heat exchangers. The main reason is that the pressure drop through most heat exchangers will be minimal, so a constant-pressure assumption works well. This is especially true for liquids, whose thermodynamic ...

3

In my view, you are attacking this from the wrong angle. The pipe will expand by about 150 mm. You need to give your pipe run the required space to do so. Look if you can use existing bends (this is what we need an isometry of the pipe run for) to account for the expansion. There's a wall, there has to be a bend. If not possible, see if you can incorporate ...

2

You can take have to use temperature dependent material properties for the e-modulus and Poisson's ratio. This scan be achieved with ´MPTEMP' followed by ´MPDATA´. Linear interpolation is used by the software. Example with fictious values: MPTEMP,1, 20, 100, 150, 400 MPDATA,EX,1,, 204e9, 197e9, 193e9, 178e9 There is also the option to input a ...

2

I'll try to be a bit more verbose than the other answers. Essentially, I believe we should arrive at the same results (although I haven't checked). Like you have surmised, the temperature difference of $\Delta T=50ºC$ will initially cause the member to expand and bridge the gap -between the wall and the copper member- up to $0,08$ mm . Then $\Delta T$ ...

2

Imagin the block is heated on the bottom and has expanded to $l+\delta l$, but this expansion tries to expand the top part along with it. So the will be some strain between the hot lower part and the cold top part which will cause stress until the whole block is at temperature $T_i+\delta i$ Some years ago I had to supervise a steel frame structure. They had ...

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The meral bar is not likely to be "fixed" on a base that has less thermal expansion coefficient than it, when the heat reaches the support, it expands too. The source was mainly focused on the longitudinal displacement (1D), but you have a valid point for 3D volume expansion.

1

Assuming the cylinder wall is thin an approximate method to calculate the expansion of circumference is: Let's call the initial circumference, $C_0=2\pi r_0=2\pi* 230mm .$ The change in diameter due to temperature is $$\ d_c=c_1 - c_0= 2 π r_0 \Delta t *α \quad\\ \rightarrow dc = 2 π r_1 - 2 π r_0 = 2 π r_0 \Delta t* α$$. $\alpha=$ linear expansion ...

1

I guess nobody knows the analytical answer but I hazard to say the stresses if the support is really kept cool or has a largely smaller thermal index, will be very significant. A large bar with a sudden increase in temperture can even crack at the support or damage the support Roark's formulas for stress and strain had some empirical formulas. I am flying ...

1

Perhaps the support bracket accepts radial expansion. Or you could consider that the stress is only being evaluated over the free section and measurement starts a value of x distance from the support. At least that was how we set up our examination of the change in length of a copper bar. I think we had point zero about 2cm from the support and not only was ...

1

Why a bimetallic strip heating up is a bending problem When a bimetallic strip is heated by $\Delta T$, from its original temperature, what happens is that one strip tends to grow longer and the other less. because they are bonded together, they end up to bend and rotate. Figure 1: Diagram of a bimetallic strip showing how the difference in thermal ...

1

Plating will decrease the bending by a small nominal amount because the gold will expand equally on either side holding back the metal with highier thermal expansion index and encouraging the one with lower expansion index to expand. But at the typical thicknesses of 0.00001" to o.ooo1" it's effect is negligible.

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I made mistake on my earlier response (deleted). Allow me to try it again. Please let me know if there is mistake.

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We know that the spring and the copper bar reach an equilibrium at Fs =Fc. $Fc = KcXc$ where: $Kc =\frac{(\text{Young modulus of copper})A}{L}$ $Fs =KsXs$ $KcXc=KsXs$ $Xc/Xs= Ks/Kc$ and we already know $Xc-0.08-Xs= 1.325 -166.5*1.325/(Ks+Kc)$ I let you handle the rest.

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You are right that your numbers are over-estimates. They would be "correct" for a plate with no resistance to bending, which was only supported along two edges and heated across its full width between the unsupported edges. This is a hard problem to produce a simple "formula", because in the real situation the bending will be reduced by the stiffness of the ...

1

Found out about Response Surfaces and using all the points that I already created, created a Response Surface Optimization which let me easily find the answer without a significant amount more computation.

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No, the zip tie and conduit won't cause the pipe to develop an internal restriction. Even squeezing the outer cover will not compress the inner sleeve... There are metal pipe clamps that will compress the inner but they have serious leverage that you won't get from a nylon zip tie.

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The standard technique is to weld a thermocouple to the work. Then the time to temperature depends on the furnace, size , heat source, fan , etc. A serious furnace will have a survey ( like Mil H 6872 ) of temperature variations available from the manufacturer. If this is a typical small lab muffle , just weld the thermocouple to the valve you already ...

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You can use Newton's Law of Cooling to model both the heating and cooling of your valve. You'd use the data you already collected to establish the model, then use the model to see how long it takes the valve to approach equilibrium in the furnace. You can use the same model to estimate the temperature behavior between the time you removed the part from the ...

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