In this problem you will estimate the heat lost by a typical house, assuming tha

Question

In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is T_{ m in} = 20^ C} and the temperature outside is T_{ m out} = 0^ C}. The walls and uppermost ceiling of a typical house are supported by 2 imes 6-inch wooden beams (k_{ m wood} = 0.12 W/(m K)) with fiberglass insulation (k_ins} = 0.04 {W}/(m K)}) in between. The true depth of the beams is actually 5.625 inches, but we will take the thickness of the walls and ceiling to be L_{ m wall} = 18 cm} to allow for the interior and exterior covering. Assume that the house is a cube of length L = 9.0;m} on a side. Assume that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air. Ignore heat loss through the ground. The effective thermal conductivity of the wall (or ceiling) k_eff, is the area-weighted average of the thermal conductivities of the wooden beams and the fiberglass insulation that make up each of them. Allowing for the fact that the 2 imes 6 beams are actually only 1 .625 inches wide and are spaced 16 inches center to center, a calculation of this conductivity for the walls yields k_ eff} = 0.048 ; {W}/(m K). For simplicity, assume that the ceiling also has the same value of k_eff. Part A What is H, the total rate of energy loss due to heat conduction for this house? Round your answer to the nearest 10 { m W}

Explanation / Answer

a)The effective thermal conductivity of the wall (or ceiling)is keff = (?Q/?t) * (1/A) *(x/?T) Here,(?Q/?t) = 0.12 W/m/K,A = 2 * 6,x = 16 inchesand ?T = (20 - 0) = 20oC b)We define the heat conduction by the formula H = k * A * (?T/x) Here,k = 0.12 W/m/k,A = 2 * 6 = 12 in2,?T =(20 - 0) = 20oC and x = 9.0 m c)The oil needed to supply the heat lost byconduction is Q = m * s * ?t or m = (Q/s * ?t) or m = [(1.4 * 108)/(4186 * 273)] or m = 122.5 kg

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