# ASHRAE baseline EF for service hot water heater

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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?

[cid:9b04a961-d79b-43b3-9d71-60b59107130d]

Warm Regards
Harshal Gupta
Energy Analyst

LEED AP || IGBC AP || GRIHA CP

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Best Regards

Harshal

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Joined: 2015-11-18
Reputation: 0

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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Thanks Bishop!!

Yes SHW heater ?24 amps but voltage range is 380 volts.
4kW is circulation pump.

Warm Regards
Harshal Gupta
Energy Analyst
LEED AP || IGBC AP || GRIHA CP
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Best Regards

Harshal

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Joined: 2015-11-18
Reputation: 0

As per ASHRAE 90.1 : 2010 appendix G electric resistant based water heater to be model and sized according to section 7.4.1.
In base case and efficiency of electric heater should be considered as per table 7.4.2.
In my project the capacity of heat pump based SWH is as follows:

1. Rated heating capacity : 19.8kW
2. Rated power input : 6.02 kW
3. Maximum power input : 6.8kW

As per table 7.8 performance required for resistance based electric should be 0.93-0.00132 V EF, which is again 0.233 EF.
In modeling tool eQUEST, EIR is required which is ratio of electric input power to nominal heater capacity.
EF is a combination of thermal efficiency and standby losses, in case if i modeled tank losses as 0 by putting Tank UA =0 in DWH in eQUEST. What should i model in EIR?

Please someone guide can guide on this.

Warm Regards
Harshal Gupta
Energy Analyst
LEED AP || IGBC AP || GRIHA CP