Suppose a house has 20 cm of blown-in cellulose insulation above the ceiling and

Question

Suppose a house has 20 cm of blown-in cellulose insulation above the ceiling and inside all exterior walls. The thermal conductivity of the insulation is 0.044 W/(K·m). The interior of the house is a box 17 m long by 11 m wide by 3.2 m high. When the outside air temperature is 5.4°C, and the interior temperature of the air in the house and the ground under the house is 21.9°C, how much heat is lost through the walls and ceiling per hour?

______________ J/hour

The house has a furnace that runs on natural gas. If the natural gas provides heating of 1.081 megajoules per standard cubic foot (MJ/cf), how much natural gas needs to be burned per hour to keep the air temperature steady in the house?

_______________ cf/hour

If an additional 11 cm of insulation is added above the ceiling, by what percentage does the natural gas usage drop?

________________ %

Explanation / Answer

Given that

The thickness of the insulation is =20cm =0.2m

The thermal conductivity of the insulation is = 0.044 W/(K·m).

When the outside air temperature is = 5.4°C,

and the interior temperature of the air in the house and the ground under the house is = 21.9°C,

Now by using the fourier law of heat conduction , we have

Heat lost =KAdT/dx

Q =(0.044)(17*11)*(21.9-5.4)/0.2 =678.81W

Then 678.81*3600J/h =2.443*106J/h =2.443MJ/h

b)

The amount of natural gas needed to be burnt is =2.443/1.081 =2.26cf/h

c)

If there is a addition of 11cm of insulation then

Heat lost =KAdT/dx

Q =(0.044)(17*11)*(21.9-5.4)/0.31 =437.941W

Then 437.941*3600J/h =1.576*106J/h =1.576MJ/h

Now the natural usafe of gas is 1.576/1.081 =1.458cf/h

Now the % drop in usage of natural gas is ((2.26-1.458)/2.26)*100=35.486%

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