Copper tube HX capacity

Dan_9

How does one determine the maximum btu that a 60 ft./3/4" copper coil can extract from a nominal 160 degree H2O medium? It seems that the slower the flow, the greater the delta T, and the faster the flow, the lower the delta T. Does this mean that the potential to extract is a constant? I've thought myself in circles to the point of confusion. Can anyone straighten me out? Thanks in advance.

Tony Conner

The BTU/hr...

...delivered will likely be about the same, but you can either get more GPM of lower delta-T water, or take a lower GPM, and have a greater temp increase.

If you've got a water to water HX that will do 100,000 BTU/hr, you can get 20 GPM of water with a 10*F rise. Or 10 GPM of water with a 20*F rise. Or 5 GPM with a 40*F rise. You could say 200 GPM with a 1*F rise, but you'd never get 200 GPM through the 3/4" copper tube. Likewise for 1 GPM with a 200*F rise. You won't likely have the hot side inlet temp to make that happen.

Take the BTU/hr of any HX heating water. Divide it by 500. (This flips gallons per minute into lbs per hour - one BTU will heat 1 lb of water 1*F. It also takes "minutes" into "hours". 8.33 lbs of water per gallon X 60 minutes per hour. 8.33 X 60 = 499.8 Round it to 500.

After dividing the BTU/hr you have to work with by 500, the answer will be what you have if you mutiply GPM X delta-T. For the 100,000 BTU HX used for the example, and within the limitations of the pressure drops, and hot side inlet temp to the HX, GPM X delta-T must equal 200. When one factor gets larger, the other one must get smaller.