The DHW heater efficiency calculations for baseline are based on Table 7.8. For
Gas storage water heaters, the size >75,000 btu/h, and <4000 (Btu/h)/gal, the
baseline performance is:
80% Et(Q/800+110V^1/2) SL, Btu/h = 80%*(400000/800+100*130^1/2).
The input Q for the heater is 400,000 Btu/h, and V is 130 gal. So the SL
calculated is 3077 Btu/h.
Since the SL is defined as standby loss based on 70F temperature difference
between stored water and ambient requirements, is it right to calculate the
efficiency by Q-SL/Q? The result is 99.6%, which appears to be so high!
Could anyone show me if it is right or the right calculation for this?
Thanks in advance!
Ying
Ying,
I believe the correct interpretation is that there are two minimum
efficiency requirements implied by "80% Et(Q/800+110V^1/2) SL, Btu/h".
You don't multiply the two factors together.
The heater must have a minimum efficiency of 80%, and
the standby loss must be less than "(Q/800+110V^1/2)" in Btu/h.
You use these minimum values in your baseline model.
Using your heating and storage capacity numbers below,
SL = (400000/800)+(110*130^.5) = 1754 Btu/h
This is based on 70F temperature difference so
UA = (1754 Btu/h)/70F = 25.06 Btu/h-F
Use this value for your standby loss UA.
Regards,
William Bishop, EIT, LEED(r) AP