I had a question about calculating baseline 90.1 efficiency for electric
DWHs (tank type).
Based on Table 7.8, my efficiency needs to be SL = 20 + 35*sqrt(V) at a
70 degree delta T between water and ambient, where SL is the standby
loss in btuh, and V is the volume of the tank.
If, in my proposed model, I have a 240 gallon tank (meaning my baseline
has the same volume), it would appear that the maximum standby loss at a
70 degree delta should be 562 btuh [20 + 35 * sqrt(240)]
Since eQuest calculates the heat loss by multiplying the UA (thermal
conductivity of storage tank) by the tank, it would appear that my UA
for this particular example would be 562 / (70*240), or 0.0334.
That way, at a 70 degree delta, the heat loss (SL) would be 0.0334
Btu/hr-gal-F * 240 gallons * 70 F = 562 btuh
Does that sound about right?
Thanks to anyone that has a quick chance to review!
James Hansen, PE, LEED AP
Hmm… I read this quickly and it seemed off so I double checked:
TANK-UA
The overall heat loss coefficient (UA) of the heater. The heat loss is calculated as this value times the temperature difference between the water in the heater and the environmental temperature.
You got the 90.1 calc right: SL = 20 + 35*sqrt(240) = 562
Rephrasing the DOE2 help entry copied above: SL = UA * delta-T
Set: 562 = UA * 70
Then: UA = 562/70 = 8.03
As I understand it, TANK-UA rolls together the properties of both the insulation and surface area, so given TANK-UA you shouldn’t need to work in the tank volume or area to find the standby loss for a given delta-T. That’s my take – hope it helps =)!
~Nick
NICK CATON, P.E.
I typically use RMI’s EMIT tool for these calcs. http://www.rmi.org/rmi/ModelingTools