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_in = 20 degree C and the temperature outside is T_out = 0 degree C. The walls and uppermost ceiling of a typical house are supported by 2 times 6-inch wooden beams (k_wood = 0.12 W/(mK) with fiberglass insulation (k_ins = 0.04 W/(mK)) 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_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 times 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/(mK). For simplicity, assume that the ceiling also has the same value of k_eff. What is H, the total rate of energy loss due to heat conduction for this house? Round your answer to the nearest 10 W. Let us assume that the winter consists of 150 days in which the outside temperature is 0 degree C. This will give the typical number of "heating degree days" observed in a winter along the northeastern US seaboard. (The cumulative number of heating degree days is given daily by the National Weather Service and is used by oil companies to determine when they should fill the tanks of their customers.) Given that a gallon (3.4 kg) of oil liberates Q_g = 1.4 times 10^8 J when burned, how much oil will be needed to supply the heat lost by conduction from this house over a winter? Assume that the heating system is 75% efficient. Give your answer numerically in gallons to two significant figures.

Explanation / Answer

According to the given problem,

Using the given statement,

The part of this that bothers ME is that we're not told the thickness of the insulation. Are we to take it to be the whole 18 cm? Or only 5 5/8 inches? I'm going to assume the latter. The effective thermal conductivity will be just the weighted average of the k_wood and the k_insulation, based on their relative widths: k_eff = [ (13/8)(0.12 W/(m K) + (16 - (13/8)) (0.04 W/(m K)) ] / 16
= [ 0.195 W/(m K) + (0.565 W/(m K) ] / 16 = [0.76 W/(m K)] / 16 = 0.0475 W/ (m K)

(A) Perhaps the reason why they gave the 18-cm total wall thickness is that the area of the interior walls will be somewhat less than 9 m x 9 m. Each of the 4 interior side walls will have an area of
8.82 m x 8.64 m = 76.2 m^2; the interior ceiling area might be (8.64 m)^2 = 74.65 m^2.
Yes, H = -kA(delta-T) / (thickness) = -(0.0475 W/ (m K)) (379.45 m^2) (20 K) / (45/8 of an inch)
= - (-.95 W/m ) (379.45 m^2) / 0.1429 m = 2.52 kW

(B) I'm going to use my answer to (B) even though I have doubts about the thickness.
2.52 kW times 86400 s/day times 150 days = 32.7 gigajoules
If a gallon of oil produces 0.14 gigajoules but its use is only 75% efficient,
then apparently a gallon of oil delivers only 0.105 gigajoules to heating the house...
so that 311 gallons are needed to supply the heat lost to conduction.

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