Suppose a house has 10 cm of blown-in cellulose insulation above the ceiling and

Question

Suppose a house has 10 cm of blown-in cellulose insulation above the ceiling and inside all exterior walls. The thermal conductivity of the insulation is 0.044 W/(Km). The interior of the house is a box 16 m long by 10 m wide by 3.0 m high. When the outside air temperature is 4.7°C, and the interior temperature of the air in the house and the ground under the house is 20.0°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 1.071 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 14 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 =10cm =0.1m

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

When the outside air temperature is = 4.7°C,

and the interior temperature of the air in the house and the ground under the house is = 20.0°C,

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

Heat lost =KAdT/dx

Q =(0.044)(16*10)*(20-4.7)/0.1 =1077.12W

Then 678.81*3600J/h =3.877*106J/h =3.877MJ/h

b)

The amount of natural gas needed to be burnt is =3.877/1.071 =3.620cf/h

c)

If there is a addition of 14cm of insulation then

Heat lost =KAdT/dx

Q =(0.044)(16*10)*(20-4.7)/0.24 =448.8W

Then 437.941*3600J/h =1.615*106J/h =1.615MJ/h

Now the natural usafe of gas is 1.615/1.071 =1.5085cf/h

Now the % drop in usage of natural gas is ((3.877-1.5085)/3.877)*100=61.091%

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