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Just Noobie Luck or a Valid Calculation Method?
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OttoPilot
Member Posts: 10
Heat loss calculations are essential when making system changes. But I considered them to be relatively inaccurate with old houses because the tightness of the house has much more importance for heat loss than (for example) the number of windows and wall area. And tightness is not easily measured.
When I put a new boiler into my 1925 gravity hot water home, I calculated actual heat loss—using the actual heat required at a specific outside temperature.
I timed the boiler on/off cycle to give me a heat load. Then I extrapolated to figure heat load on a design day.
For example:
• existing boiler nameplate output = 200,000 Btu/h
• average boiler on time = 8 minutes 40 seconds per hour
• indoor = 70 °F
• temperature gain in the house due to normal activity, sunshine, etc. = 5 °F
• actual outdoor = +10 °F
• design outdoor = −30 °F
Therefore, the boiler actual output was 200,000 times 8 minutes 40 seconds (8.67 minutes) divided by 60 min = 28,900 Btu/h
This value came about when the effective difference between indoor and outdoor temperature was = 70° indoor − 5° internal heat gain − 10 ° outdoor = 55 °F
At design the effective difference would be = 70° indoor − 5° internal heat gain − (−30) ° outdoor = 95 °F
Boiler output needed at design = 28,900 Btu/h times 95 °F / 55 °F = 49,900 Btu/h.
I bought a 50,000 Btu/h output boiler.
Six years later the overnight low was −32 °F. And my heating system performance?
When I awoke, the indoor temperature was 1° below setpoint and it took four hours for the temperature to regain setpoint. The boiler didn’t shut off during that time.
Did I take too big a risk on an untested method? In hindsight, a big YES.
But if a heating system is still operational, why wouldn’t this be a more accurate method to size a replacement versus a conventional heat loss calculation? (Subject to a greater than zero safety factor. ☺)
Gary
When I put a new boiler into my 1925 gravity hot water home, I calculated actual heat loss—using the actual heat required at a specific outside temperature.
I timed the boiler on/off cycle to give me a heat load. Then I extrapolated to figure heat load on a design day.
For example:
• existing boiler nameplate output = 200,000 Btu/h
• average boiler on time = 8 minutes 40 seconds per hour
• indoor = 70 °F
• temperature gain in the house due to normal activity, sunshine, etc. = 5 °F
• actual outdoor = +10 °F
• design outdoor = −30 °F
Therefore, the boiler actual output was 200,000 times 8 minutes 40 seconds (8.67 minutes) divided by 60 min = 28,900 Btu/h
This value came about when the effective difference between indoor and outdoor temperature was = 70° indoor − 5° internal heat gain − 10 ° outdoor = 55 °F
At design the effective difference would be = 70° indoor − 5° internal heat gain − (−30) ° outdoor = 95 °F
Boiler output needed at design = 28,900 Btu/h times 95 °F / 55 °F = 49,900 Btu/h.
I bought a 50,000 Btu/h output boiler.
Six years later the overnight low was −32 °F. And my heating system performance?
When I awoke, the indoor temperature was 1° below setpoint and it took four hours for the temperature to regain setpoint. The boiler didn’t shut off during that time.
Did I take too big a risk on an untested method? In hindsight, a big YES.
But if a heating system is still operational, why wouldn’t this be a more accurate method to size a replacement versus a conventional heat loss calculation? (Subject to a greater than zero safety factor. ☺)
Gary
1
Comments

As you have realized, you probably should have given a little more buffer. The 5degree internal gain is a serious WAG and you did not consider what happens when it is cloudy and windy. It is also possible that the old boiler was slightly overfired.
I do like using actual consumption to size new systems, I like to use data from a few different sources. In your case, clocking the meter, using a longterm degree day calc, and giving yourself a 1020% safety factor would have been a good idea.
Your design day temp drop of only 1degree tells you that your math was pretty darn close, you deserve an attaboy for that one"If you can't explain it simply, you don't understand it well enough"
Albert Einstein1 
I'm not a pro, but that's pretty interesting. How are you getting this value: "temperature gain in the house due to normal activity, sunshine, etc. = 5 °F". Seems to me this method relies on both the indoor and outdoor temperatures being accurately measured, and the system as a whole probably needs to be balanced before the new boiler gets installed. How did you extrapolate to get heat loss on a design day? My understanding is that's definitely not a linear calculation.0

"The 5degree internal gain is a serious WAG "
I used a spreadsheet and entered the supplier's figure for daily natural gas over one year. I took fuel use in summer as the base use all year. When I subtracted that from winter consumption, got what I assumed was home heating consumption.
I got daily average temperature from the weather bureau and entered that. Then I could calculate how much fuel was used per degree difference from inside to outside.
It didn't result in a linear correlation until I subtracted 5° from the indoor/outdoor difference. I assumed that was the internal heat gain of the house. So it wasn't a WAG. It was a SWAG (scientific wilda$$ed guess).
"How did you extrapolate to get heat loss on a design day? My understanding is that's definitely not a linear calculation."
My spreadsheet numbers showed a linear relationship.0 
Same, it’s linear enough for our purposes. Just add a little safety padding. The goal of a heat loss calculation is to tell you is how much fuel you use on the coldest day. Going straight to the source seems like an accurate method and it’s pretty damn quick using a meter or receipts.1

Thank you for that great article. The second I saw that scatter plot I realized I could have made the calculation much more simply than I did.0
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