The mass flow of air would be 167ft^3/sec times the density of air 0.0739lb/ft³ which equal 12.3 lb/sec
Total surface area of the house is:
Floor = 1000 ft2
walls = 1138 ft2
roof = 1000ft2
Total=3138 ft2
The U factor that was calculated in a prior post will be .492
And the heat capacity of air will be as taken from Wikipedia 1.012 kJ/kg* K or .241874 Btu/lb*F
q required = [UA + mass flow rate(Cp)](degree days)
you have to do the above equation twice using both heating and cooling degree days (the numbers are found on a blog post). Then the q is in btu. When you use heating degree days, its the amount of btu to heat the house for a year and same for cooling.
IN IOWA CITY
Case 1, Heating:
q=[.492*3138+12.3*.241874] *6162= 9.5e6
Case 2, Cooling:
q=[.492*3138+12.3*.241874]*1060=1.6e6
Total: 11.1e6 Btu to 3253 Kilowatt Hours
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