The figure (Figure 1) shows a house whose walls consist of plaster (R-017), R-11

Question

The figure (Figure 1) shows a house whose walls consist of plaster (R-017), 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 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

Given,

Average outdoor winter temperature, t = 22° F

surface area of walls of the house, Sw = (36 x 10 x 2) + (28 x 10 x2) + (0.5 x 14 xtan(30) x 28 x 2 = 1506.321 ft2

surface area of the roof, Sr = 2 x 36 x 14/cos(30) = 1163.9381 ft2

surface area of windows, S = 3 x 5.3 x 10 = 159 ft2

Energy input from four windows, per unit time, Pi = 30 x 3 x 5.3 x 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

As, 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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