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Consider a double pipe parallel flow heat exchanger, the mean temperature difference in this case turns out to be logarithmic because the temperature profiles are exponential.

If we would have made use of Arithmetic temperature difference, when the profiles were exponential, we would have got erroneous results. However, if the temperature profiles were linear then, the mean temperature difference would've been equal to the arithmetic one.

I started off by considering linear temperature profiles and proceeded just in the fashion we do for calculating the LMTD, however the mean temperature still turns out to be logarithmic, it should come arithmetic isn't it? Can this logarithmic relation be further reduced to arithmetic one?enter image description here

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I don't think its possible for the parallel flow heat exchanger to have a linear profile.

The reasoning is the following. If the temperature profile is linear with respect to x (the length of the heat exchanger) then at each dx, will be dropping by the same dT . That means that the heat lost from the hot is constant (also that the heat gained from the cool), and let's denote it with $Q_{dx}$.

However, that $Q_{dx}$ need to be transmitted through the walls of the heat exchanger. However that means that independent of the position x, the same $Q_{dx}$ is always transmitted. In order for that to happen, if the other parameters of the heat exchanger (material and crosssection) remain the same, that means that the temperature difference should remain the same. However, that is not possible, since the temperature drop per x.

Therefore, because the temperature difference drops between the two flows, the temperature exchange will be reduced, therefore an exponential temperature profile is the expected outcome (unless very careful design imposes other constraints).

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  • $\begingroup$ Yes I do agree in real the temperature profile can't be linear. I read somewhere actually that had the temperature profile been linear (hypothetically) then AMTD would've been the true mean temperature difference. So I started off by considering a linear temperature profile and was expecting the mean temperature difference to turn out arithmetic. $\endgroup$
    – Harshit Rajput
    Aug 15 at 5:52
  • $\begingroup$ I am not sure I see the linear assumption anywhere apart your graph. Mathematically I don't see where you substitute the temperature difference with the constant slope. So in essence you are just doing the lmtd again. Please correct me if I am missing something. $\endgroup$
    – NMech
    Aug 15 at 5:57
  • $\begingroup$ I have also posted this question on physics stack exchange. Please also have a look there, may be that question was more articulate. Someone also gave an answer that might give a better insight. Add anything extra if you feel like. $\endgroup$
    – Harshit Rajput
    Aug 15 at 12:18
  • $\begingroup$ physics.stackexchange.com/questions/659280/… $\endgroup$
    – Harshit Rajput
    Aug 15 at 12:18
  • $\begingroup$ @HarshitRajput just like you were told in the comments in the other post, if the coefficient of transfer is constant that is not possible. Since you were talking about a hypothetical/non existent problem, I think the Physics SE was probably a more suitable forum to get an answer. $\endgroup$
    – NMech
    Aug 16 at 4:17

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