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Two parallel square plates, each 4 m2 area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the plates
- a)61.176 k W
- b)51.176 k W
- c)41.176 k W
- d)31.176 k W
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two parallel square plates, each 4 m2area, are large compared to a gap...
Q 12 = (F g) 12 A 1 σ b (T 14 – T 24).
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Two parallel square plates, each 4 m2area, are large compared to a gap...
Solution:
Given data:
Area of each plate, A = 4 m2
Gap between plates, d = 5 mm = 0.005 m
Temperature of plate 1, T1 = 800 K
Temperature of plate 2, T2 = 300 K
Emissivity of plate 1, ε1 = 0.6
Emissivity of plate 2, ε2 = 0.9
Stefan-Boltzmann constant, σ = 5.67 x 10-8 W/m2K4
The net energy exchange by radiation between two parallel square plates can be calculated using the following formula:
Q = σε1ε2A(T14 - T24)/(1/d + 1/d)
Where Q is the net energy exchange by radiation between the plates.
Calculating the value of Q using the given data:
Q = (5.67 x 10-8)(0.6)(0.9)(4)(8004 - 3004)/(1/0.005 + 1/0.005)
Q = 51.176 kW
Therefore, the net energy exchange by radiation between the plates is 51.176 kW, which is closest to option B.
Given data:
Area of each plate, A = 4 m2
Gap between plates, d = 5 mm = 0.005 m
Temperature of plate 1, T1 = 800 K
Temperature of plate 2, T2 = 300 K
Emissivity of plate 1, ε1 = 0.6
Emissivity of plate 2, ε2 = 0.9
Stefan-Boltzmann constant, σ = 5.67 x 10-8 W/m2K4
The net energy exchange by radiation between two parallel square plates can be calculated using the following formula:
Q = σε1ε2A(T14 - T24)/(1/d + 1/d)
Where Q is the net energy exchange by radiation between the plates.
Calculating the value of Q using the given data:
Q = (5.67 x 10-8)(0.6)(0.9)(4)(8004 - 3004)/(1/0.005 + 1/0.005)
Q = 51.176 kW
Therefore, the net energy exchange by radiation between the plates is 51.176 kW, which is closest to option B.
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Two parallel square plates, each 4 m2area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the platesa)61.176 k Wb)51.176 k Wc)41.176 k Wd)31.176 k WCorrect answer is option 'B'. Can you explain this answer?
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Two parallel square plates, each 4 m2area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the platesa)61.176 k Wb)51.176 k Wc)41.176 k Wd)31.176 k WCorrect answer is option 'B'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about Two parallel square plates, each 4 m2area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the platesa)61.176 k Wb)51.176 k Wc)41.176 k Wd)31.176 k WCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two parallel square plates, each 4 m2area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the platesa)61.176 k Wb)51.176 k Wc)41.176 k Wd)31.176 k WCorrect answer is option 'B'. Can you explain this answer?.
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