Question

Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity K = 0.149 J/ (m-°C-sec) , thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at Tl = 127°C. If the ambient temperature is Ta = 27°C, calculate the rate of heat loss per unit area (in watt/m2) due to radiation from the lid.

Answer 1

Answer:595W/m^2

Explanation:

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Answer 2

The rate of heat loss per unit area due to radiation from the lid is 595 watt/m².

Explanation:

Given that,

Conductivity $k= 0.149 J/m^{\circ}C sec$

Thickness t = 5 mm

Emissivity = 0.6

Temperature of lid = 127°C

Temperature of ambient = 27°C

We need to calculate the rate of heat loss per unit area due to radiation from the lid

Using Stefan's-Boltzmann law

$E=\epsilon\sigma(T_{l}^4-T_{a}^4)$

Put the value into the formula

$E=0.6\times\dfrac{17}{3}\times10^{-8}\times((127+273)^4-(27+273)^4)$

$E=595\ watt/m^2$

Hence, The rate of heat loss per unit area due to radiation from the lid is 595 watt/m².

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Topic : Rate of heat loss per unit area

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