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I have been having a debate with a coworker regarding heat exchanger fluid temperatures and while I thought I firmly grasped the principles, I was not arguing my point very well.

I am looking at an air to water coil for an air handler. I was selecting the coil based on a leaving air temp of ~60F (it's an industrial space with a lot of ventilation). I am using condensing boilers so I sized for 180 leaving water temperature with 140 return on the coldest day which gives a 40 degree temperature delta. On the air side, I am bringing in 5 degree air (EAT) at a rate of 16000 CFM on one of the air handlers. My coworker wants to lower the temperature to 140 supply and 100 degree return, his reason being that it would still be a 40 degree temperature delta on the water side so the thermal output would be the same. I believe the hotter supply temperature will result in more rapid heat transfer but I didn't really have any good argument against his logic. He was simply relating the two rules of thumb equations for air and water: [Btu/hr]=1.08*CFM*deltaT and [Btu/hr]=500*gpm*deltaT to prove his point. My argument was that this does not account for the heat transfer taking place from one side of the coil to the other, but I was not able to adequately illustrate my side to the point that he would agree.

First of all, am I even right about operating the system at 180 degrees being a more efficient solution, and, if so, how do I fit the heat exchanger heat transfer into the system of equations?

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    $\begingroup$ Are you right about what? Please start a new paragraph and ask your question. The two situations lead to different coil selections. There is enough size here that you both could probably retire off the savings from a professionally engineered and optimized system. $\endgroup$ – Phil Sweet Jan 7 '18 at 20:06
  • $\begingroup$ I clarified what I was asking about a little bit. Hopefully it is more clear now. $\endgroup$ – Secundus Jan 8 '18 at 4:58
  • $\begingroup$ I am not familiar with those rules of thumb, what is the definition of delta T in those equations? $\endgroup$ – J. Ari Jan 8 '18 at 20:55
  • $\begingroup$ The delta T on the water side is the difference in the temperature of the water entering the heat exchanger and the temperature of the water leaving the heat exchanger. On the air side equation (1.08...) the delta T is the difference in entering air temperature and leaving air temperature. $\endgroup$ – Secundus Jan 9 '18 at 23:59
  • $\begingroup$ Yes a higher water temp will be more heat transfer at the air handler but it will and also be more expensive to heat water to a higher temp. It would be wasteful to have leave a 40 degree exit delta. Exchangers are relatively cheap. $\endgroup$ – paparazzo Jan 10 '18 at 15:19
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A proper analysis requires a log mean delta T calculation, and an understanding of the configuration of the heat exchanger (i.e parallel flow, cross flow, etc.). But at a very high level it comes down to a variation of the basic convective heat flow equation Q = UADT where the DT is the temperature difference between the water and the air. Assuming the heat transfer coefficient and area are the same, then a higher DT will provide a higher heat transfer rate.

The equations you cited apply to the required water flow to provide the heat, and the heat increase in the air. They don't have anything to say about how much heat will actually be transferred. Using those equations alone, you could bring in water at 73 degrees and send it out at 33 degrees, still getting the 40 degree drop, but obviously it won't work, because they don't account for the heat transfer rate from water to air.

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  • $\begingroup$ Thank you. I tried to explain that and LMTD calculation would be needed to actually determine the answer, but that didn't get me anywhere, and I couldn't quite remember to look up convective heat flow. I tried to use a similar example to your 73 and 33 degrees but didn't get that across either. Ultimately what you are saying is I need the Q on all three of those equations to hit the same value, would that be accurate? $\endgroup$ – Secundus Jan 10 '18 at 0:04
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    $\begingroup$ Yes. And ultimately the heat transfer equation will tell you how big the heat exchanger needs to be. Here’s a high level description of how to do it: che.ufl.edu/unit-ops-lab/experiments/he/he-theory.pdf $\endgroup$ – Mark Jan 10 '18 at 2:59
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Simple answer is this, the Heat exchange equation accounts for the amount of time the fluids are in contact, while the conservation of energy equation assumes the fluids are in equilibrium. So if the temperature differences between the 2 fluids is smaller you need the fluids to be in contact longer to get the amount of heat exchange you are calculating for the higher temperature difference. Since the conservation of energy equations have flow rates in the terms, this is the equivalent to saying the fluids are traveling across the heat exchanger surface at a constant rate, the only way to get the same amount of heat transfer with a smaller temperature differential is to increase the surface of the heat exchanger, so the fluids stay in contact for a longer period of time.

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