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The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at temperature of 20 0 c and 50 c, respectively. determine the rate of heat
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Feb 23, 2018
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The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at temperature of 20 0 c and 50 c, respectively. determine the rate of heat transfer through the wall, in w.
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The working equation to be used here is written below:
Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall
The solution is as follows:
Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m
Cybersam
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Mar 1, 2018
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