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A black body of total area 0.045 m2 is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m 2 and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black body
- a)547.3 K
- b)287.4 K
- c)955.9 K
- d)222.2 K
Correct answer is option 'C'. Can you explain this answer?
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A black body of total area 0.045 m2is completely enclosed in a space b...
Qr = σ A (T b4 – Tw4), Q c = k A d t/δ = 1979.5 W. So temperature of black body is 955.9 K.
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A black body of total area 0.045 m2is completely enclosed in a space b...
Given data:
Total area of black body = 0.045 m²
Surface area of walls = 0.5 m²
Thickness of walls = 5 cm = 0.05 m
Thermal conductivity of walls = 1.07 W/mK
Temperature of inner surface of wall = 215°C
Temperature of outer surface of wall = 30°C
To find: Temperature of black body
Solution:
1. Calculation of heat transfer rate:
Heat transfer rate through the walls = kA (ΔT/Δx)
Here, ΔT = (215 - 30) = 185°C
Δx = thickness of walls = 0.05 m
A = surface area of walls = 0.5 m²
k = thermal conductivity of walls = 1.07 W/mK
Heat transfer rate through the walls = 1.07 × 0.5 × (185/0.05) = 1983 W
2. Calculation of emissive power of black body:
Emissive power of black body = σT⁴A
Here, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴
T = temperature of black body (unknown)
A = total area of black body = 0.045 m²
3. Calculation of absorbed radiation by black body:
Absorbed radiation by black body = emissive power of black body × absorptivity of walls
Here, absorptivity of walls = 1 (since black body is completely enclosed by the walls)
Absorbed radiation by black body = emissive power of black body × 1
4. Calculation of net heat transfer rate:
Net heat transfer rate = absorbed radiation by black body - heat transfer rate through the walls
5. Equating the net heat transfer rate to zero and solving for T:
Net heat transfer rate = 0
i.e., absorbed radiation by black body = heat transfer rate through the walls
σT⁴A = 1983 W
T⁴ = (1983 / (0.045 × 5.67 × 10⁻⁸))
T⁴ = 7.032 × 10⁸
T = (7.032 × 10⁸)^(1/4) = 955.9 K
Therefore, the temperature of the black body is 955.9 K.
Hence, option (c) is the correct answer.
Total area of black body = 0.045 m²
Surface area of walls = 0.5 m²
Thickness of walls = 5 cm = 0.05 m
Thermal conductivity of walls = 1.07 W/mK
Temperature of inner surface of wall = 215°C
Temperature of outer surface of wall = 30°C
To find: Temperature of black body
Solution:
1. Calculation of heat transfer rate:
Heat transfer rate through the walls = kA (ΔT/Δx)
Here, ΔT = (215 - 30) = 185°C
Δx = thickness of walls = 0.05 m
A = surface area of walls = 0.5 m²
k = thermal conductivity of walls = 1.07 W/mK
Heat transfer rate through the walls = 1.07 × 0.5 × (185/0.05) = 1983 W
2. Calculation of emissive power of black body:
Emissive power of black body = σT⁴A
Here, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴
T = temperature of black body (unknown)
A = total area of black body = 0.045 m²
3. Calculation of absorbed radiation by black body:
Absorbed radiation by black body = emissive power of black body × absorptivity of walls
Here, absorptivity of walls = 1 (since black body is completely enclosed by the walls)
Absorbed radiation by black body = emissive power of black body × 1
4. Calculation of net heat transfer rate:
Net heat transfer rate = absorbed radiation by black body - heat transfer rate through the walls
5. Equating the net heat transfer rate to zero and solving for T:
Net heat transfer rate = 0
i.e., absorbed radiation by black body = heat transfer rate through the walls
σT⁴A = 1983 W
T⁴ = (1983 / (0.045 × 5.67 × 10⁻⁸))
T⁴ = 7.032 × 10⁸
T = (7.032 × 10⁸)^(1/4) = 955.9 K
Therefore, the temperature of the black body is 955.9 K.
Hence, option (c) is the correct answer.
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A black body of total area 0.045 m2is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m2and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black bodya)547.3 Kb)287.4 Kc)955.9 Kd)222.2 KCorrect answer is option 'C'. Can you explain this answer?
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A black body of total area 0.045 m2is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m2and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black bodya)547.3 Kb)287.4 Kc)955.9 Kd)222.2 KCorrect answer is option 'C'. 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 A black body of total area 0.045 m2is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m2and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black bodya)547.3 Kb)287.4 Kc)955.9 Kd)222.2 KCorrect answer is option 'C'. 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 A black body of total area 0.045 m2is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m2and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black bodya)547.3 Kb)287.4 Kc)955.9 Kd)222.2 KCorrect answer is option 'C'. Can you explain this answer?.
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