The average thermal conductivity of the walls (including windows) and roof of a
ID: 1699019 • Letter: T
Question
The average thermal conductivity of the walls (including windows) and roof of a house in the figure is 4.8 10-4 kW/m·°C, and their average thickness is 19.5 cm. The house is heated with natural gas, with a heat of combustion (energy given off per cubic meter of gas burned) of 9300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 26.0°C if the outside temperature is 0.0°C? Disregard radiation and energy loss by heat through the ground.
The average thermal conductivity of the walls (including windows) and roof of a house in the figure is 4.8 10-4 kW/mC, and their average thickness is 19.5 cm. The house is heated with natural gas, with a heat of combustion (energy given off per cubic meter of gas burned) of 9300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 26.0C if the outside temperature is 0.0C? Disregard radiation and energy loss by heat through the ground.Explanation / Answer
CONCEPT: HEAT .........HEAT TRANSFER....THERMAL CONDUCTIVITY ....... ******************************************************** LET, We Use Fourier's Law q''=-k(?T/l) Where q'' is the heat per unit time per unit area, k is the thermalconductivity, l is the thickness and ?T is the temperaturedifference q''=-(4.8 10-4kW/m·°C)[(0-26)/(0.195m)] q''=0.064 kW/m2 Now we calculate the heat q per unit area, multiply q'' by time q=q''t=(0.064 kW/m2)(86400s)=5529.6 kJ/m2 since time t = 1 day = 24*60*60 Now to find the number of cubic meters (volume V) of gas to burneach day, you divide the heat q by the heat of combustion V=q/?H Where ?H is the heat of combustion by solving, V=5529.6 kJ/(9300kcal/m3) V=142.104 m3 [as 1 J = .239 cal]
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