A wall in a house contains a single window. The window consistsof a single pane
ID: 1726205 • Letter: A
Question
A wall in a house contains a single window. The window consistsof a single pane of glass whose area is 0.15 m2 andwhose thickness is 5 mm. Treat the wall as a slab of the insulatingmaterial Styrofoam whose area and thickness are 17 m2and 0.10 m, respectively. Heat is lost via conduction through thewall and the window. The temperature difference between the insideand outside is the same for the wall and the window. Of the totalheat lost by the wall and the window, what is the percentage lostby the window?
Explanation / Answer
Surface area of glass, A1 = 0.14 m^2 Thickness of glass, d1 = 8 X 10 -3m Case I : ( Ordinary glass window) Thermal conductivity of glass window, K1 = 0.96W / mK Surface area of Styrofoam wall, A2 = 12 m^2 Thickness Styrofoam wall, d2 = 0.10 m Thermal conductivity of styrofoam, K2 = 0.033 W/ mK Heat lost by conduction = K A ( 1 -1 ) t / d ( 1 - 2 ) and t are same forboth. Hence, Heat lost is proportional to K A / d Heat lost by window / Total heat lost by wall +window = ( K1 A1/ d1 ) / [ ( K1 A1 / d1 ) + K2 A2 / d2 ) ] = ( 0.96 * 0.14 / 8 X 10 -3 ) / [ ( 0.96 * 0.14 / 8 X 10-3 ) + ( 0.033 * 12 / 0.1 ) ] = 16.8 / ( 16.8 + 3.96 ) = 0.8092 Percentage of heat lost by window = 80.92 % Case II : ( Fiber glass window) By Using fiber glass window: Thermal conductivity of fiber glasswindow, K1 = 0.048 W / mK Heat lost by window / Total heat lost by wall +window = ( K1 A1/ d1 ) / [ ( K1 A1 / d1 ) + K2 A2 / d2 ) ] = ( 0.048 * 0.14 / 8 X 10 -3 ) / [ ( 0.048 * 0.14 / 8 X10 -3 ) + ( 0.033 * 12 / 0.1 ) ] = 0.84 / ( 0.84 + 3.96 ) = 0.175 Percentage of heat lost by window = 17.5 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.