# Heat exchange depending on coolant flow direction

Consider a simplest case of a heat exchanger - two parallel pipes of flowing liquids (say, hot and cold) that have physical contact along some part of their length. Hot water of a certain temperature goes from A to B. Cold water can go either from C to D or from D to C. Assume that heat exchange between liquids occurs only where pipes contact (at XY part). What is the favorable direction (meaning "the most heat is transferred) of a coolant flow relative to hot flow - in the same direction (C->D) or in the reverse direction (D->C)? How the coolant and hot flows' temperatures are distributed along the pipe contact?

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You want to make sure that the heat flow is always between fluids at the nearest possible temperature, to minimize the entropy production from the flow of heat. The best method is to flow the cold coolant in the opposite direction as the hot coolant, so that as the cold coolant gets hotter, it is transferring heat to hotter hot-coolant. If you adjust the pipes right, and make them long, you can do the entire circuit with as close to zero entropy generation as you like. This is the principle by which ducks can send blood to their feet and do so without losing any significant body heat in the feet, although they are immersed in cold water.

For the temperature profile, if you have the hot water be 100 degrees, and the cold water be 0 degrees, and you have a long linear pipe where they touch, then the profile can be exactly linear with a 1 degree difference in temperature at all points, assuming that that heat diffusion constant for the metal is constant over the range of temperature, and the specific heat of water is constant for the range of temperature, and both of these are close enough approximations for a practical heat exchanger.

If the pipe runs from 0 to L, the profile for hot water temperature is:

$T_H(x)= {100x\over L}$

The profile for the cold water

$T_C(x) = {100x\over L} - 1$

so that the difference between them is always 1 degree. You adjust the flow rate so that the heat transfer moves C units of heat energy in a length $L/100$, where C is the specific heat of water, and then the exchanger works with this profile. You can adjust the temperatures to be as close to each other as you like, and the entropy gain from the heat flow is:

$\Delta S = {Q\over T^2} \delta T \approx \Delta S_0 {1\over 230}$

Where you use the absolute temperature T in the denominator, so that in this system you only make about 1 percent of the entropy you would if you let the hot and cold water transfer heat by direct contact without an exchanger.

These profiles are universal attractors, so if you just set up the appropriate flow rate, you will approach the linear profile with time.

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My idea was that to maximize the heat exchange, one must maximize the temperature difference along the contact profile. Maximal difference of coolant temperatures is exhibited on their entry into the exchanger, thus the input flows must be in the same direction. As for the temperature profile in this case, I expected the temperature curve to decay exponentially, as the energy flow is proportional to the temperature difference, and this implies the exponential solution. Why our deductions are that very different? – mbaitoff Sep 10 '11 at 10:33
@mbaitoff: Your idea is exactly the opposite of the correct one. Your exchanger would be maximally inefficient (but have an exponential profile), and the exchanger described above is maximally efficient. Think about it this way: I can send water in at 100 degrees, have it cool down to 0 degrees, and come back to 99 degrees by doing one cycle around the long-linear exchanger. I lose next to none of the original heat by the cycling process. In your exchanger, you would lose everything. – Ron Maimon Sep 10 '11 at 16:48
I think we have a misunderstanding in exchanger schematics here. I assume a two distinct pipes, and never assume a closed cycle of a fluid. I cant imagine what you say "come back to 99 degrees". I'm going to post a descriptive picture of my problem. – mbaitoff Sep 10 '11 at 19:07
There, I updated the question with the picture. What's your idea now? – mbaitoff Sep 10 '11 at 19:24
@mbaitoff Suppose that you have (as in Ron's example) hot water at 100 degrees and cold water at 0 degrees. If the flow rates are equal, your idea would end up with the hot water being cooled to 50 degrees; Ron's idea would end up with the hot water being cooled to 1 degree, maximizing the heat transfer. – mmc Sep 10 '11 at 19:44