Steve,

To model a condensing boiler that is 96% efficient at 100% fire and 80?F return water temperature, enter "1.042" (=1/0.96) for the HIR if you use the default curve.

For condensing boilers, the HIR-fPLR curve modifies fuel consumption based on boiler part-load ratio and return water temperature (HWR). For boilers, I believe the hourly part-load ratio is simply the hourly load divided by the design boiler capacity.
The default performance curve for high efficiency condensing boilers is "CondBlr-HiEff-HIR-fPLR&HWR". The data points for this curve are shown below:

[cid:image002.jpg at 01CE91D0.4CAAF360]

The Indep-1 (X) values are part-load ratio, and the Indep-2 (Y) values are HWR (?F). Depend (Z) is the multiplier of HIR. Data point 1 is for PLR=1.0 (100% fire) and 80?F HWR, and the output for this point is "1.000", meaning no adjustment is made to the entered value for HIR. Therefore, the value you enter for HIR should be for 100% fire and 80?F. Your manufacturer data show 96% efficiency at this point, so you would enter 1.042 for your HIR.

The other manufacturer data point you mention is 86.5% efficiency at 100% fire and 160?F. That corresponds with Data Point 17 in the default curve, with an output (HIR multiplier) of 1.121. If you enter HIR=1.042, the hourly HIR will be 1.121x1.042=1.168. And 1/1.168=85.6%. This is close, though slightly less, than the manufacturer data. So the default HIR-fPLR curve does not perfectly match your boiler, but you may decide it is close enough. Alternatively, you could create your own curve, and share it with me. ;)

BTW, sorry for the delayed reply. Better late than never.

Regards,
Bill

William Bishop, PE, BEMP, BEAP, LEED AP