Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C,

Question

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C, whereas the right face is exposed to a flow of 22°C air with a convection heat transfer coefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

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Thông Đạt 5 years 2021-08-14T01:17:36+00:00 1 Answers 718 views 1

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  1. Cherry
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    2021-08-14T01:19:24+00:00
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    Answer:

    The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

    Explanation:

    Thickness of the wall is  L=  20cm = 0.2m

    Thermal conductivity of the wall is  K = 2.79 W/m·K

    Temperature at the left side surface is T₁ =  50°C

    Temperature of the air is T = 22°C

    Convection heat transfer coefficient is  h = 15 W/m2·K

    Heat conduction process through wall is equal to the heat convection process so

    Q_{conduction} = Q_{convection}

    Expression for the heat conduction process is

    Q_{conduction} = \frac{K(T_1 - T)}{L}

    Expression for the heat convection process is

    Q_{convection} = h(T_2 - T)

    Substitute the expressions of conduction and convection in equation above

    Q_{conduction} = Q_{convection}

    \frac{K(T_1 - T_2)}{L} = h(T_2 - T)

    Substitute the values in above equation

    \frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

    Now heat flux through the wall can be calculated as

    q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

    Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

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