# Tag Info

1

You have made an error in assuming that R(factor,cream) will be the same in both cases. $R=L/(K.A)$ Here $K$ and $A$ are constant but $L$ in first case is twice of that in the second case. Thus if we assume the thermal resistance of the cream coloured layer to be $R$ in the first case, then in becomes $R/2$ in the second case. And as the dimensions of the ...

1

You start with the equation $$\dot Q = \frac{A \Delta T}{R} \,.$$ I write $\dot Q$ for $\mathrm d Q / \mathrm dt$, it is a common short notation. Since there is only one $R$ here, I do not need the subscript “factor”. Then the $R$ is given by $L/K$ where $L$ is the length of the material. The $K$ is a material constant. In your example you have one cream ...

1

I'll venture the guess, that the explanation you read was wrong. What I'd find plausible is: liquid mixed with gas has a higher effective thermal capacity - you don't need conductivity since the stuff is pumped through the cooling cycle, right? What I mean by this: the liquid can receive a lot of heat by boiling. It carries the heat away not only in the ...

4

It is all a matter of engineering balance, between the water circulating in the radiator circuit of the car, which enfolds the engine and with water-metal contact which takes heat away at a certain rate. In automobiles and motorcycles with a liquid-cooled internal combustion engine, a radiator is connected to channels running through the engine and ...

0

Combustion gas temperature can be as high as you quoted. But metal temperature is much lower than its melting temperature. At high temperature, metal will lose its strength. Engine design will not allow this to happen. During the cycle, combustion gas tries to heat up the metal. But there is coolant flow carrying away the heat. The net is that the metal ...

0

Heat transfer coefficients are a bit tricky, for good reason. You need to know something about the conditions that exist on both sides of the pipe, and you only stated that you have water in a stainless steel pipe. For heat transfer to occur, you need a second fluid on the outside of the pipe that you can transfer heat with. In addition to the above, ...

0

Have a look at Birds sitting on electric wires: potential difference between the legs this gives some very good answers, which suggest any pd across and current through is entirely negligible (assuming a standard bird as defined by European law). Essentially, the pd drop over significant distances is small by design so that it is correspondingly smaller over ...

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I think you might be missing information, as tom pointed out. For example you can look at this setup as a resistor (the bird) in parallel with the power line. You do need to specify something more about the bird, perhaps the power transferred...

1

The way you have asked the question there is not enough information... To work out R of the wire you have the conductivity and cross section of the wire..., but you also need the length of wire, which might be the unknown (the gap between the birds legs). If you have the voltage between the two legs you can go forward with this by working out the ...

1

There's not enough information to say. It looks like your model has spring-like (harmonic) potentials between neighboring atoms, but phonon scattering requires anharmonic potentials. Harmonic potentials mean that the superposition principle still holds, so the phonons just pass right through each other without scattering. Since you have harmonic ...

4

Why do these two methods give different results and which one is correct? Both are incorrect; but the second more, I'd say. As H.Tofaili pointed out, both often are very similar, since every of them makes some approximation which is only always allowed in some limit. In your second approach you assume, that there is no heat conducted across the ...

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