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Heat Loss Calcs & Annual fuel useage
George Durisek
Member Posts: 3
I ran a heat loss calculation for a residence in Northern Ohio. I used a heating temperature diff. of 70 F. Here are my results:
House Sensible Total 1499 BTUH loss
Heat Loss 104,930 BTUH
Annual Gas Useage 134.86 MMBtu/yr.
The home owner asked me to run the calculation if he maintained the house at 66 F. The new heating temperature diff. is now 66 F. Here are my results:
House Sensible Total 1499 BTUH loss (does not change)
Heat loss 98,934 BTUH
Annual Gas Useage 134.86 MMBtu/yr.
The annual gas useage formula yielded the same result in both cases. I was expecting to see savings. The heat loss is of course less at the lower thermostat setting but the useage remined the same. It seems that the heat loss to temp. diff value are proportional thus yielding the same gas useage result. Am I off track here can someone point me in the right direction?
The formula I am using is as follows:
E(annual)=((Qdesign)*(degreee Day)*24*(Cd))/((1,000,000)*(Delta T design))
Thank You
George Durisek
House Sensible Total 1499 BTUH loss
Heat Loss 104,930 BTUH
Annual Gas Useage 134.86 MMBtu/yr.
The home owner asked me to run the calculation if he maintained the house at 66 F. The new heating temperature diff. is now 66 F. Here are my results:
House Sensible Total 1499 BTUH loss (does not change)
Heat loss 98,934 BTUH
Annual Gas Useage 134.86 MMBtu/yr.
The annual gas useage formula yielded the same result in both cases. I was expecting to see savings. The heat loss is of course less at the lower thermostat setting but the useage remined the same. It seems that the heat loss to temp. diff value are proportional thus yielding the same gas useage result. Am I off track here can someone point me in the right direction?
The formula I am using is as follows:
E(annual)=((Qdesign)*(degreee Day)*24*(Cd))/((1,000,000)*(Delta T design))
Thank You
George Durisek
0
Comments
-
i belive you should see a savings 1'/. every 3¡ãF is lowered
0 -
George-
A couple questions:
1. What's the "sensible total"? All heating is sensible, unless you're also humidifying.
2. I assume that you're using the second number as Qdesign. 104930/98934=70/66, as we would expect.
3. The formula looks close to what I'd use..
E(annual)=(24 [hrs/day] x degree days x Qdesign [Btu/hr] x Cd [correction factor for 65º]) / ( Boiler Efficiency x Delta T x Heating Value of Fuel [Btu/CF]
If you use 5500 DD, 84%, and Cd=1 you'd get:
E=(24*5500*104930*1)/(.84*70*100000)= 2355 CF of gas0 -
Annual Loads
Hello George,
In the equation that you posted, you are dividing Design Load in btuh by design delta t. That will yield the number UA, the units of which are btuh/deg dt.
Then you are multiplying UA x 24 x degree days to get annual degree hours.
The reason that the two answers come out the same is because the degree day tables are made up of measured values that ALL USE BASE 65 as the "balance point" of the building. The balance point is that outside air temperature at which the building first calls for heat. Different for different buildings and usually much less than 65.
You need to recalculate deg days using a different base appropriate for the building, which is a trick as the degree day data is compiled, historical data. There is an algorithm that will let you do this.
These days, with the computer power we have available, it's more accurate to actually do hourly heat loss calculations for every hour in the year (8760) using a TMY data. (typical meterological year). The TMY data is compiled for 239 locations in the US and is based on the historical data.
Anyway, if you do it this way, you can apply an appropriate dt for every hour of the year. Even lets you evaluate the effects of things like night set back or window insulation.
(what's Cd?)
Dale0
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