(a) If the walls, windows, and roof have an average thickness of 22.0 cm and the

Question

(a) If the walls, windows, and roof have an average thickness of 22.0 cm and the thermal conductivity (including the windows) is 0.48 W/(m · °C), determine how many cubic meters of gas is used up during the 120-day winter season? Assume air infiltration and heat loss through the ground is negligible.
m3

(b) Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install an extra 8 cm of fiberglass to the roof. If energy costs $0.90 per one hundred million joules and the insulation was $4.00 per square meter, then calculate the simple payback time taking the 120-day heating season to be equivalent to one year.
yr

Explanation / Answer

energy required=thermal conductivity*area*temperature difference*time/thickness

area calculation:

total area=area of walls +area of roof

=2*(W*h1+L*h1+0.5*W*h2)+2*L*sqrt(h2^2+0.25*W^2)

=304 m^2

then energy required=0.48*304*16*120*24*3600/0.22=1.1*10^11 J

then natural gas required=1.1*10^11/(3.89*10^7)=2.8278*10^3 cubic meter

b)for second question, 8 cm is thickness or length?

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.