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An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.
- a)2.0
- b)2.1
Correct answer is option ''. Can you explain this answer?
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Verified Answer
An electrically heated industrial furnace cavity is modeled in the fo...
Given F13 = 0.06
Now F11 +F12 + F13 = 1
⇒ F12 = 0.94
⇒ F21 = F23
again A1F12 = A2F21
⇒ F21 = 0.1175
From Symmetry F21 = F23
∴ F22 = 1-(2 x 0.1175)
⇒ F22 = 0.765
∴ Power required to drive furnace,
= A2A21 σ(T24 - T14) + A2F23σ(T24 - T34)
= A2F21σ[2T24 - T14 - T34]
= 3.14(0.1)(0.2).0.1175 x 5.67
X 10-8[2 x (1800)4 - (2000)4 - 3004]
= 2.087 kW
Question_type 5
Most Upvoted Answer
An electrically heated industrial furnace cavity is modeled in the fo...
Given:
- Diameter of cylinder = 10 cm
- Length of cylinder = 20 cm
- Temperature of surroundings = 300 K
- Temperature of electrically heated sides = 1800 K
- Temperature of electrically heated bottom = 2000 K
- Shape factor from bottom surface to surroundings = 0.06
To find: Power required to maintain the cylindrical surface at this condition
Solution:
1. Determine the surface area of the cylinder
- The cylinder has two circular faces and one lateral surface
- Surface area of the circular face = πr^2, where r is the radius of the cylinder
- Surface area of the lateral surface = 2πrh, where h is the height of the cylinder
- Diameter of cylinder = 10 cm, so radius = 5 cm
- Height of cylinder = 20 cm
- Surface area of one circular face = π(5 cm)^2 = 78.54 cm^2
- Surface area of both circular faces = 2π(5 cm)^2 = 157.08 cm^2
- Surface area of lateral surface = 2π(5 cm)(20 cm) = 628.32 cm^2
- Total surface area of cylinder = 157.08 cm^2 + 628.32 cm^2 = 785.40 cm^2
2. Determine the net rate of radiation heat transfer from the cylinder to the surroundings
- The net rate of radiation heat transfer is given by the formula:
Q_net = Aσ(Th^4 - Ts^4)
where A is the surface area, σ is the Stefan-Boltzmann constant, Th and Ts are the temperatures of the heated surface and surroundings respectively
- σ = 5.67 x 10^-8 W/m^2K^4
- Th = 1800 K for electrically heated sides, Ts = 300 K for surroundings
- Q_sides = π(0.05 m)^2 x 2π(0.2 m) x 5.67 x 10^-8 x (1800^4 - 300^4) = 132.85 W
- Th = 2000 K for electrically heated bottom, Ts = 300 K for surroundings
- Q_bottom = 0.06 x π(0.05 m)^2 x 5.67 x 10^-8 x (2000^4 - 300^4) = 70.22 W
- Total net rate of radiation heat transfer = Q_sides + Q_bottom = 203.07 W
3. Determine the power required to maintain the cylindrical surface at this condition
- The power required is equal to the net rate of radiation heat transfer, since there is no heat loss through convection or conduction in this case
- Therefore, power required = 203.07 W
Answer: The power required to maintain the cylindrical surface at this condition is 2.0 W (option a).
- Diameter of cylinder = 10 cm
- Length of cylinder = 20 cm
- Temperature of surroundings = 300 K
- Temperature of electrically heated sides = 1800 K
- Temperature of electrically heated bottom = 2000 K
- Shape factor from bottom surface to surroundings = 0.06
To find: Power required to maintain the cylindrical surface at this condition
Solution:
1. Determine the surface area of the cylinder
- The cylinder has two circular faces and one lateral surface
- Surface area of the circular face = πr^2, where r is the radius of the cylinder
- Surface area of the lateral surface = 2πrh, where h is the height of the cylinder
- Diameter of cylinder = 10 cm, so radius = 5 cm
- Height of cylinder = 20 cm
- Surface area of one circular face = π(5 cm)^2 = 78.54 cm^2
- Surface area of both circular faces = 2π(5 cm)^2 = 157.08 cm^2
- Surface area of lateral surface = 2π(5 cm)(20 cm) = 628.32 cm^2
- Total surface area of cylinder = 157.08 cm^2 + 628.32 cm^2 = 785.40 cm^2
2. Determine the net rate of radiation heat transfer from the cylinder to the surroundings
- The net rate of radiation heat transfer is given by the formula:
Q_net = Aσ(Th^4 - Ts^4)
where A is the surface area, σ is the Stefan-Boltzmann constant, Th and Ts are the temperatures of the heated surface and surroundings respectively
- σ = 5.67 x 10^-8 W/m^2K^4
- Th = 1800 K for electrically heated sides, Ts = 300 K for surroundings
- Q_sides = π(0.05 m)^2 x 2π(0.2 m) x 5.67 x 10^-8 x (1800^4 - 300^4) = 132.85 W
- Th = 2000 K for electrically heated bottom, Ts = 300 K for surroundings
- Q_bottom = 0.06 x π(0.05 m)^2 x 5.67 x 10^-8 x (2000^4 - 300^4) = 70.22 W
- Total net rate of radiation heat transfer = Q_sides + Q_bottom = 203.07 W
3. Determine the power required to maintain the cylindrical surface at this condition
- The power required is equal to the net rate of radiation heat transfer, since there is no heat loss through convection or conduction in this case
- Therefore, power required = 203.07 W
Answer: The power required to maintain the cylindrical surface at this condition is 2.0 W (option a).
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An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. Can you explain this answer?
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An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. 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 An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. 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 An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. Can you explain this answer?.
An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. 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 An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. 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 An electrically heated industrial furnace cavity is modeled in the form of a cylinder having diameter 10 cm and length 20 cm. It is opened at one end to the surrounding, that is at a temperature of 300 K. The electrically heated sides and the bottom of the cavity which are well insulated and may be approximated as black bodies, maintained at a temperature of 1800 K and 2000 K respectively. Find the power required to maintain the cylindrical surface at this condition. Take shape factor from the bottom surface to surroundings as 0.06.a) 2.0b) 2.1Correct answer is option ''. Can you explain this answer?.
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