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Mashcka [7]
2 years ago
14

A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-4

00K. If this radiation transfer process is characterized by a radiation heat transfer coefficient h, calculate the value of h (a) 14.4 W/m2.C (b) 114.4 W/m2C (c) 314.4 W/m2.C ( 514.4 W/m2.c
Engineering
1 answer:
Nat2105 [25]2 years ago
5 0

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

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a) of the attached figure we have:

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Solution

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ΔW = 3035 J

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