Question

An enclosure has dimensions 1.2 m x 2 m x 2.5 m as length x width x height. Walls and ceiling are maintained at 600 k and floor at 450 k. Emissivity of ceiling, wall and floor are 0.5, 0.6 and 0.7 respectively. The net radiation to floor is _________w

Answer 1

Answer:

Net rate of heat radiation to floor is given as

$\frac{dQ}{dt} = -75455.7 Watt$

Explanation:

Net radiation of the floor is given as

$\frac{dQ}{dt} = \sigma e_1 A T^4 - \sigma e_2 (2A_2 + 2A_1) T_s^4 - \sigma e_3 A T_s^4$

here we know that

$A = 1.2 \times 2 = 2.4 m^2$

$A_1 = 1.2 \times 2.5 = 3 m^2$

$A_2 = 2 \times 2.5 = 5 m^2$

now we have

$\frac{dQ}{dt} = (5.67 \times 10^{-8})0.7 (2.4)(450^4) - (5.67 \times 10^{-8})0.6 (6 + 10)(600^4) - (5.67 \times 10^{-8})0.5 (2.4)(600^4)$

$\frac{dQ}{dt} = 3906.1 - 70543.87 - 8817.9$

$\frac{dQ}{dt} = -75455.7 Watt$

#Learn

Topic : Radiation heat transfer

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