Can we calculate part-load heat loss if we know the design heat loss?
Hi,
Using Wrightsoft, our engineer calculated a design heat load of 9318 Btuh for our small, slab-on-grade house. The design temp. is -13F and indoor temperature is 70F. At 4 Pascals, our blower door air tightness is 14 CFM, and 111 CFM at 50 Pascals.
In addition to the design load, I would like to know the heat loss for warmer-than-design days/nights so that I can modulate the temperature of the hot water entering a fan coil we are about to install. (It’s designed for low-temperature water with a delta T across the 4-row coil of 20F). With a mixing valve, I will have some flexibility to set an appropriate water temperature to match the house’s heat loss as conditions vary.
I could just go with trial and error of course, but am curious whether there is an equation which can calculate this part-load? The fan coil (Jaga Briza 22) has a spreadsheet that will tell me the Btuh output based on, among other factors, supply water temperature.
Thank you in advance for any thoughts or recommendations.
Cheers
Richard (just west of Ottawa, ON.)
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Replies
The standard assumption is that heat loss is linear with temperature difference between indoors and outdoors.
So if your load is 9318 BTU/hr at -13F with an indoor temperature of 70F, that's 83F of difference between indoor and outdoor. So with a difference of half that -- 41.5F, or 28.5F -- the heating load would be half that, or 4659 BTU/hr.
It probably slightly overestimates the load but it's a close enough approximation.
Thank you! This is helpful indeed.
Our experience living in this house over the last 5 winters is that there's very little heat loss (if any) even when outside temps are in the low 30s. Of course, sunny days keeps us toasty in sub-zero weather (until sundown, that is..).
The fan coil installer will provide several temperature gauges (we're using a plate heat exchanger to separate potable from non-potable water) so I can monitor the supply and return temps as I adjust the mixing valve (Tekmar 710) as outside temps drop.
Thank you again for your answer.
A load of 9300 BTU/hr at -13F is an accomplishment, you should be proud.
Traditionally internal loads were basically ignored in heating calculations because they were so low compared to heating loads, but in well-built modern houses they become a factor. A hair dryer operating at 1500 W would provide more than half of the peak heating load in your house!