Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I have a fluid flowing through a pipe in the ocean and there is heat transfer from the ocean to the fluid in the pipe. I prepared a simulation and the results show that if I increase the mass flow rate of the fluid through the pipe then the heat transfer rate to the fluid increases and the outlet temperature increases.

These results don't seem right to me. I would think if the heat transfer rate increases, then the outlet temperature should also increase. Does anyone have any suggestions as to why my intuition might be wrong? It is also quite possible the simulation could be wrong.

share|improve this question
Eventhough the heat transfer rate is increased to heat is also distributed over a larger amount of fluid. –  user17615 Apr 19 '13 at 13:14

1 Answer 1

up vote 1 down vote accepted

Suppose you have a very slow flow rate. The outlet temperature will be the same temperature as the ocean because the water is in thermal contact with the ocean for a long time. However the heat flow rate will be very slow because there is only a small difference between the ocean temperature and the temperature of the water in the pipe, so the temperature gradient across the wall of the pipe is low.

Now suppose you have a very fast flow rate. The outlet temperature is reduced because any particular volume of water doesn't stay in contact with the ocean long enough to be significantly heated. However because the water in the pipe is now a lot cooler than the ocean there will be a large temperature gradient across the wall of the pipe and hence a high heat flow.

So as you increase the flow rate you would expect the outlet temperature to decrease but the overall rate of heat extraction from the ocean to increase.

share|improve this answer
Thank you very much. This makes perfect sense –  Greg Harrington Apr 19 '13 at 6:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.