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The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and 10 C at the
Question
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and 10 C at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of 1.4 W/m K, what is the rate of heat loss through the slab
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2021-09-02T14:24:26+00:00
2021-09-02T14:24:26+00:00 1 Answers
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Answers ( )
Answer:
Q = – 4312 W = – 4.312 KW
Explanation:
The rate of heat of the concrete slab can be calculated through Fourier’s Law of heat conduction. The formula of the Fourier’s Law of heat conduction is as follows:
Q = – kA dt/dx
Integrating from one side of the slab to other along the thickness dimension, we get:
Q = – kA(T₂ – T₁)/L
Q = kA(T₁ – T₂)/t
where,
Q = Rate of Heat Loss = ?
k = thermal conductivity = 1.4 W/m.k
A = Surface Area = (11 m)(8 m) = 88 m²
T₁ = Temperature of Bottom Surface = 10°C
T₂ = Temperature of Top Surface = 17° C
t = Thickness of Slab = 0.2 m
Therefore,
Q = (1.4 W/m.k)(88 m²)(10°C – 17°C)/0.2 m
Q = – 4312 W = – 4.312 KW
Here, negative sign shows the loss of heat.