I am looking for a solution to this problem:
I need 15GPM with only a 12 Deg F rise (158F-170F)
I am looking for the cheapest solution using three phase electric.
Use is in a Heat Exchanger, Hot water is used to heat liquid then the cooler water needs to be reclaimed to pass through exchanger again. Heat loss in water is only 12F
Ok, thanks for the reply. But I kind of already knew the math to figure out the power involved. I guess more of what I was asking is the plumbing involved in hooking a system up in order to make sure I can mantain a steady flow ( at least 4hrs) of 170F water.
I was thinking maybe a single 80 Gal heater set as high as i could get it with a tankless water heater doing the rest. But I dont know if I could actually get my 15 GPM this way and I dont want to manifold a bunch of tankless together(too expensive). Please any other ideas or comments.
If you have 20 hours to recharge, your average power input can be much less than 26 kw, but the stored energy for ~30,000# of water raised 12F stays the same, at 3.6 therms.
Last edited by Thatguy; 11-03-2009 at 09:03 AM.
Thank you thatguy, I think I am following you correctly. But I dont think I explained my purpose correctly. I need to create a closed system that recirculates the same water, ie.
Heat water to 170 --> hot water is used at 15GPM and returned at 158F --> reheat water to 170F --> Reuse water at 15GPM. Cycle will be continued for 4 hours. Actual volume of water is only a couple of gallons (as much as needed to allow enough time for electric to reclaim heat lost).
Can this be done with just a single 80 gal tank and a tankless system, or maybe a few tanks linked together?
Thank you for your patience and help, I think I have stepped a little out of my element on this problem.
80 gals = 660# of water raised 12F = 7920 BTU of energy.
7920 BTU in 320 seconds = 25 BTU/sec of power = 89100 BTU/hr of power = 26.1 kw of power, which is 109A @ 240v.
With 5.5 kw, 240 v units from Home Depot you'd need 5 ea. of these.
5 x 5.5 = 27.5 kw at 240v and 20.7 kw at 208v.
Typical HD elements are 3 kw, 4.5 kw and 5.5 kw, all at 240v. Typical WHs have two elements.
At startup you need to heat the incoming water at 50F or whatever, to 170F. Then you run for 4 hrs, then you turn off the power.
Startup will take 660 x 120F = 79200 BTU of energy. Since you have 89100 BTU/hr of power this startup will take 60(79/89) = 53 minutes.
Running the numbers again for 3ea., 80 gal., 9kw WHs connected in parallel is your homework problem. . .
Last edited by Thatguy; 11-03-2009 at 04:03 PM.
Can you explain what you are trying to do? Or is it proprietary/sensitive? While the delta T is not great, it's a pretty large duty for electric (~90,000 Btu/hr or 26.4 KW...that's a lot of elements running at the same time.) You won't be able to build a "hot reservoir" above this temperature to reduce the instantaneous load.
Is the 15 gpm quantity flow controlled and the delta T a result of the heating demand?
One thing to factor in is pumping heat addition. I've had to design coolers for insulated storage tank recirculation on thermally sensitive materials, because of the pumping power input into the fluid. In this case, the pump will help rather than hinder.
I assume you don't have nat. gas available. But would a propane fired system make more sense for such a large duty?
Thank you thatguy. You don't just give me the food, you teach me to fish. I appreciate that.
It is to be used in a heat exchanger for a HTST Pasteurizer. The 15GPM and delta T are set by the heating and volume requirements of the HTST(controlled via pump). Nat. Gas is not available and propane would be nice but we are limited to Electric only.
I am not sure what you mean by pumping heat, but yes there will be a pump on this system.
Your best bet is probably to quiz the pasteurizer vendor about heating loop options.
Any time you have a pump, compressor, stirrer, agitator, etc. it puts the majority of its power into the working fluid. There is some loss to external bearings, cooling fan blades on the electric motor, gear box, etc. but most of the energy ends up in the circulated fluid. If you have a pump curve or load for the required flow (with suction and discharge pressures) then that will be approximately how much power the pump adds to the system. It will probably only be a small fraction of the rather large duty you need though so you can probably dismiss it.I am not sure what you mean by pumping heat, but yes there will be a pump on this system.