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The net radiation from the surfaces of two parallel plates maintained at is to be reduced by at least 99%. Number of shields which should be placed between surfaces to achieve this reduction is (εshield = 0.05), (εsurface = 0.8)
Option~
(A) 3
(B) 4
(C) 5
(D) 6
Correct answer is 'B'. Can you explain this answer?
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The net radiation from the surfaces of two parallel plates maintained...
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Q ∝ 1/R
where R is the resistance to radiation heat exchange between the plates. When there is no shield
R1 = 1-ε1/Aε1 + 1/A + 1-ε2/Aε2
R1 = 1-0.8/A x 0.8 + 1/A + 1-0.8//A x 0.8
R1 = 1.5/A
When n shields are placed in between the plates
= 1.5 + 39 n/A
Based on given condition
Q2/Q1 = 0.01
∴ 1.5/1.5 + 39 n = 0.01
∴ n = 3.8
i.e.n ≃ 4
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The net radiation from the surfaces of two parallel plates maintained...
Problem Statement: To reduce the net radiation from the surfaces of two parallel plates maintained at to be reduced by at least 99%. Number of shields which should be placed between surfaces to achieve this reduction is (εshield = 0.05), (εsurface = 0.8)
Solution:
Net radiation between two parallel plates can be calculated using the following equation:
q = σε(A1A2(T1^4 - T2^4))
where,
q = net radiation between two plates
σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m^2K^4
ε = emissivity of the surface
A1, A2 = area of the surfaces
T1, T2 = temperatures of the surfaces
Given:
εsurface = 0.8
εshield = 0.05
Reduction in net radiation = 99%
To achieve a 99% reduction in net radiation, we need to reduce the net radiation to 1% of the original value.
Let the number of shields required be n.
The net radiation after n shields can be calculated using the following equation:
q'n = σε(A1A2(T1^4 - T2^4))(εsurface + (1-εsurface)εshield)^n
q'n = 0.01q
Substituting the given values in the above equation, we get:
0.01q = σ(0.8)(1)(T1^4 - T2^4)(0.05 + 0.95(0.05))^n
0.01 = (0.8)(1)(0.05 + 0.95(0.05))^n
Taking log on both sides, we get:
n = ln(0.01/0.8)/ln(0.05 + 0.95(0.05))
n ≈ 4
Therefore, the number of shields required to achieve a 99% reduction in net radiation is 4.
Answer: Option B. 4
Solution:
Net radiation between two parallel plates can be calculated using the following equation:
q = σε(A1A2(T1^4 - T2^4))
where,
q = net radiation between two plates
σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m^2K^4
ε = emissivity of the surface
A1, A2 = area of the surfaces
T1, T2 = temperatures of the surfaces
Given:
εsurface = 0.8
εshield = 0.05
Reduction in net radiation = 99%
To achieve a 99% reduction in net radiation, we need to reduce the net radiation to 1% of the original value.
Let the number of shields required be n.
The net radiation after n shields can be calculated using the following equation:
q'n = σε(A1A2(T1^4 - T2^4))(εsurface + (1-εsurface)εshield)^n
q'n = 0.01q
Substituting the given values in the above equation, we get:
0.01q = σ(0.8)(1)(T1^4 - T2^4)(0.05 + 0.95(0.05))^n
0.01 = (0.8)(1)(0.05 + 0.95(0.05))^n
Taking log on both sides, we get:
n = ln(0.01/0.8)/ln(0.05 + 0.95(0.05))
n ≈ 4
Therefore, the number of shields required to achieve a 99% reduction in net radiation is 4.
Answer: Option B. 4
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The net radiation from the surfaces of two parallel plates maintained at is to be reduced by at least 99%. Number of shields which should be placed between surfaces to achieve this reduction is (εshield = 0.05), (εsurface = 0.8)Option~(A) 3(B) 4(C) 5(D) 6Correct answer is 'B'. Can you explain this answer?
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