Farnsworth In Class Problem: Solution

T_outside = 40^oF = 40 + 451 = 491^oR

T_inside = 72^oF = 72 + 451 = 523^oR

T_sky = T_outside - 7^oF = 491 - 7 = 484^oR

A_facade = (9.5)(29)(2) + (2)(9.5)(77) = 2014ft²

A_roof = (29)(77) = 2233ft²

A_i = 4247 ft²

R_i = 11.0 + .025 = 11.025 ^oF (ft²)(hr)/BTU

Q_c = A_i (T_inside⁴ = T_outside⁴)/R_i = (4247)(523-491)/11.025 = 12,326.9 BTU/hr

Q_R = Sigma * (T_sky - T_outside) * (E_facade*A_facade + E_roof*A_roof)

Q_R = (.17 x 10^-8) ((491)⁴ - (484)⁴) [(.05)(2014)+(.80)(2233)]

Q_R = 10,407.5 BTU/hr

Q = Q_C + Q_R = 22,733.5 BTU/hr

The Farnsworth House loses almost twice as much heat as a space heater whose maximum capacity of 12,000 BTU/hr can provide.