The figure (Figure 1) shows a house whose walls consist of plaster (R 0.17), R-1

Question

The figure (Figure 1) shows a house whose walls consist of plaster (R 0.17), R-11 fiberglass insulation, plywood (R - 0.65), and cedar shingles (R 0.55). The roof has the same construction except it uses R-30 fiberglass insulation. The average outdoor temperature in winter is 22°F , and the house is maintained at 70°F . The house's oil furnace produces 100,000 Btu for every gallon of oil, and oil costs $2.20 per gallon. Now assuming that the house has ten single-glazed windows, each measuring 3.0 ft by 5.3 ft assuming that the house has ten single-glazed windows, each measuring 3.0 ft by 5.3 ft Four of the windows are on the south and admit solar energy at the average rate of 30 Btu/h ft2 All the windows lose heat, their R-factor is 0.90

Explanation / Answer

surface area of walls of the house, Sw = 36*10*2 + 28*10*2 + 0.5*14*tan(30)*28*2 ft^2 = 1506.321 ft^2
surface area of the roof, Sr = 2*36*14/cos(30) = 1163.9381 ft^2
surface area of windows, S = 3*5.3*10 = 159 ft^2

Energy input from four windows, per unit time, Pi = 30*3*5.3*4 = 1908 Btu/h
Heat lost due to the temperature difference = Po = (Ti - To)[(Sw - S)/R1 + (Sr/R2) + S/R3]
here, R1 is R value of the walls = 0.17 + 11 + 0.65 + 0.55 = 12.37
R2 is R value of the roof combination = 0.17 + 30 + 0.65 + 0.55 = 31.37
R3 is R value of the material of the windows = R3 = 0.9

Ti = 70 F = 21.111 C
To = 22 F = -5.5556 C

Po = (70 - 22)*[(1506.321 - 159)/12.37 + 1163.9381/31.37 + 159/9.9] = 15489.0547 Btu/ hr
Pb ( POwer from furnance) = x

for steady state
Po = Pi + Pb
Pb = 13581.054 btu per hour
Cost of Pb = 0.1358 gallons of oil per hour = 0.298 $ per hour = 215.123 $ per month

solar gain = 1908 Btu/h = 0.01908 gal/h = 0.041976 $ per hour = 30.22272 $ a month

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