Question

In this problem, you will estimate the heat lost by a typical house, assuming that the temperature inside is Tin=20∘C and the temperature outside is Tout=0∘C. The walls and uppermost ceiling of a typical house are supported by 2×6 inch wooden beams (kwood=0.12W/(mK)) with fiberglass insulation (kins=0.04W/(mK)) in between. The thickness of the wall is Lwall=18cm allowing for interior and exterior covering. Assume that the house is a cube of length L=9.0m on a side. Approximate that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air and you may ignore heat loss through the ground. The effective thermal conductivity of the wall (or ceiling) keff, is the area-weighted average of the thermal conductivities of the wooden beams and the fiberglass insulation. For a typical wall, this value is keff=0.048W/(mK). You may also assume that the ceiling also has the same value of keff.

1.What is the total rate of energy loss due to heat conduction for this house? Round your answer to the nearest 10 W.

2. Let us assume that the winter consists of 150 days in which the outside temperature is 0∘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 Qg=1.4×108J 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.

Answer 1

SOLUTION

a)

Starting of the following equation:

$\frac{dQ}{dt}=k_{eff}.A.\left [ \frac{T_{H}-T_{C}}{L} \right ]\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (1)$

or

$\frac{dQ}{dt}=\frac{k_{eff}.5L^{2}\left ( T_{in}-T_{out} \right )}{L_{wall}}$

Replacing values of data given:

$\frac{dQ}{dt}=\frac{(0.048\,W/mk).5(9.0\,m)^{2}\left ( 20\,C^{\circ}-0\,C^{\circ} \right )}{18\times 10^{-2}\,m}$

Solving we obtain:

$\frac{dQ}{dt}=2160\,W$

b)

Therefore, the needed galons per winter can be obtained :

$N=\frac{Q_{t}}{\epsilon .Q_{galon}}=\frac{H_{t}.\Delta t}{\epsilon H}$

Replacing values:

$N=\frac{\left (2160\,W \right ).\left (150\,days \right ).(86400\,s)}{0.75.(1.4\times 10^{8}\,J)}$

Solving we obtain:

${\color{Blue} N=266.6\, \,gallons\approx 2.7\times 10^{2}\,gallons}$