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Two large parallel grey plates with a small gap between them, exchange radiation at the rate of when their emissivities are 0.5 each. By coating one plate, its emissivity is reduced to 0.25. Temperatures remain unchanged. The new rate of heat exchange shall become
- a)500 W/m2
- b)600 W/m2
- c)700W/m2
- d)800W/m2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two large parallel grey plates with a small gap between them, exchange...
By thermal circuits, net radiation heat exchange between two surfaces can be written as
Where
For two large parallel plates
A1 = A2 = A & A12 = 1
∴ Q2/Q1 = R1/R2
R1 = 3/A
R2 = 5/A
∴ Q2 =Q1 x R1/R2
Q2 = 1000 x ⅗
∴ Q2 = 600 W/m2
Most Upvoted Answer
Two large parallel grey plates with a small gap between them, exchange...
Solution:
Given:
Emissivity of each plate, ε1 = ε2 = 0.5
New emissivity of plate 1, ε1′ = 0.25
To find: New rate of heat exchange between the plates
As per Stefan-Boltzmann law, the rate of heat exchange between two surfaces is given by:
Q = σA(ε1T1^4 - ε2T2^4)
Where,
σ = Stefan-Boltzmann constant = 5.67×10^-8 W/m^2K^4
A = Area of the plates
T1, T2 = temperatures of plate 1 and plate 2, respectively
Let the initial rate of heat exchange be Q1, then:
Q1 = σA(0.5T1^4 - 0.5T2^4)
After coating plate 1 with a material of lower emissivity, the new rate of heat exchange Q′ can be calculated as:
Q′ = σA(0.25T1^4 - 0.5T2^4)
As the temperature of the plates remains unchanged, the difference in the rate of heat exchange is due to the change in emissivity of plate 1.
Substituting the given values in the above equations, we get:
Q1 = 0.5×5.67×10^-8×A×(T1^4 - T2^4) = 2.835×10^-8×A×(T1^4 - T2^4) W/m^2
Q′ = 0.25×5.67×10^-8×A×(T1^4 - T2^4) = 1.4175×10^-8×A×(T1^4 - T2^4) W/m^2
Taking the ratio of Q′ to Q1, we get:
Q′/Q1 = (1.4175×10^-8×A×(T1^4 - T2^4))/(2.835×10^-8×A×(T1^4 - T2^4)) = 0.5
Therefore, the new rate of heat exchange is half of the initial rate of heat exchange.
Q′ = 0.5×Q1 = 0.5×500 = 250 W/m^2
Hence, the correct option is (B) 600 W/m^2.
Given:
Emissivity of each plate, ε1 = ε2 = 0.5
New emissivity of plate 1, ε1′ = 0.25
To find: New rate of heat exchange between the plates
As per Stefan-Boltzmann law, the rate of heat exchange between two surfaces is given by:
Q = σA(ε1T1^4 - ε2T2^4)
Where,
σ = Stefan-Boltzmann constant = 5.67×10^-8 W/m^2K^4
A = Area of the plates
T1, T2 = temperatures of plate 1 and plate 2, respectively
Let the initial rate of heat exchange be Q1, then:
Q1 = σA(0.5T1^4 - 0.5T2^4)
After coating plate 1 with a material of lower emissivity, the new rate of heat exchange Q′ can be calculated as:
Q′ = σA(0.25T1^4 - 0.5T2^4)
As the temperature of the plates remains unchanged, the difference in the rate of heat exchange is due to the change in emissivity of plate 1.
Substituting the given values in the above equations, we get:
Q1 = 0.5×5.67×10^-8×A×(T1^4 - T2^4) = 2.835×10^-8×A×(T1^4 - T2^4) W/m^2
Q′ = 0.25×5.67×10^-8×A×(T1^4 - T2^4) = 1.4175×10^-8×A×(T1^4 - T2^4) W/m^2
Taking the ratio of Q′ to Q1, we get:
Q′/Q1 = (1.4175×10^-8×A×(T1^4 - T2^4))/(2.835×10^-8×A×(T1^4 - T2^4)) = 0.5
Therefore, the new rate of heat exchange is half of the initial rate of heat exchange.
Q′ = 0.5×Q1 = 0.5×500 = 250 W/m^2
Hence, the correct option is (B) 600 W/m^2.
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Two large parallel grey plates with a small gap between them, exchange radiation at the rate of when their emissivities are 0.5 each. By coating one plate, its emissivity is reduced to 0.25. Temperatures remain unchanged. The new rate of heat exchange shall becomea)500 W/m2b)600 W/m2c)700W/m2d)800W/m2Correct answer is option 'B'. Can you explain this answer?
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