[Equest-users] ASHRAE baseline EF for service hot water heater
Bishop, Bill
bbishop at pathfinder-ea.com
Wed Dec 5 09:05:00 PST 2018
Hi Harshal,
Is your project's SHW heater ≤24 amps and ≤250 volts? Based on the storage capacity, I'm guessing it's much larger, in which case the formula you shared doesn't apply.
I'm also guessing that the 2000 L storage tank is not an integral part of the heat pump and that the 4 kW circulation pump serves a supply/return loop from the heat pump to the storage tank. I believe that would make it an "Unfired storage tank" per the referenced Table 7.8, with minimum performance of R12.5 tank insulation.
Lastly, I'm not aware of a direct formula for converting from energy factor (EF) to COP. However, I think EF and COP should be comparable (close to the same number), and not the inverse of each other. If you did in fact have a water heater with 0.233 EF, the COP would definitely be way less than 1.
I suggest you check the heat pump manufacturer's literature to see how it was rated. There may not be a formal test procedure applicable to the heater and it is probably not covered in the ASHRAE 90.1 tables. If that is the case, you will need to determine a non-ASHRAE method to determine savings.
Regards,
~Bill
William Bishop, PE, BEMP, BEAP, CEM, LEED AP
Senior Energy Engineer
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From: Equest-users <equest-users-bounces at lists.onebuilding.org> On Behalf Of Harshal Gupta via Equest-users
Sent: Tuesday, December 4, 2018 3:24 AM
To: equest-users at lists.onebuilding.org
Subject: [Equest-users] ASHRAE baseline EF for service hot water heater
Hello All,
Project Details
I have an issue for conversion of EF to COP of service hot water heater.
Their is a service hot water heater installed in my project of 2000 L or 528 US gallons capacity. Heater is equipped with a re circulation pump of 4kW.
Modeling Issue
As per ASHRAE 90.1 base case the minimum efficiency to be considered in base case as per table 7.8. Minimum efficiency formula as per table is 0.93-.00132 √V EF.
With the above formula considering V as 528 US gallons Energy Factor (EF) will be 0.233. In many articles it was marked COP is 1/EF, by this statement COP will be 4.299.
This is giving me the negative savings.
Can any one help me with correct process for conversion of EF to COP.
Secondly, what should be pump power in base case?
Thanks in Advance
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Warm Regards
Harshal Gupta
Energy Analyst
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