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11 October, 11:28

The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k = 0.029 W / m K. The measured temperature difference across a 25-mm-thick sheet of the material is T1 - T2 = 12°C. a. What is the heat flux through a 3 mx 3 m sheet of the insulation? b. What is the rate of heat transfer through the sheet of insulation? c. What is the thermal resistance of the sheet due to conduction?

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  1. 11 October, 11:48
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    a. The heat flux through the sheet of insulation is 19.92 W/m^2

    b. The rate of heat transfer through the sheet of insulation is 125.28 W

    c. The thermal resistance of the sheet due to the conduction is 0.86 Km^2/W.

    Explanation:

    From the heat conduction Fourier's law it can be state for a wall of width e and area A:

    q = Q/ΔT = k*A * (T2-T1) / e

    Where q is the rate of heat transfer, k the conductivity constant, and T2 and T1 the temperatures on the sides of the wall. Replacing the values in the correct units, we obtained the rate of heat transfer:

    q = 0.029 W/*mK * (3m*3m) * (12°K) / (0.025m)

    (The difference in temperatures in Kelvin is the same than in Celcius degres).

    q = 0.029 W/*mK * (9 m^2) * (12°K) / (0.025m) = 125.28 W

    The heat flux is calculated by dividing q by the area of the wall:

    q/A = k * (T2-T1) / e = 0.029 W/*mK * (12°K) / (0.025m) = 19.92 W/m^2

    The thermal resistance of the sheet is defined as:

    R = e / k

    Replacing the values in the proper units:

    R = 0.025 m / 0.029 W/*mK = 0.86 Km^2/W
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