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A large plane, perfectly insulated on one face and maintained at a fixed temperature T 1 on the bare face, has an emissivity of 0.84 and loses 250 W/m2 when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2 when bare face having emissivity 0.42 and is maintained at temperature T 2 is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T 1 and T 2 remains unchanged. Determine he net heat flux between the planes
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
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A large plane, perfectly insulated on one face and maintained at a fix...
Q 12 = F 12 A 1 σ b (T 14 – T 24). Since T 1 = T 2, we get Q 12/A1 = 0.
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A large plane, perfectly insulated on one face and maintained at a fix...
Given data:
- Plane 1: emissivity=0.84, heat loss=250 W/m2, temperature=T1
- Plane 2: emissivity=0.42, heat loss=125 W/m2, temperature=T2
To determine the net heat flux between the planes when their bare faces are placed 1 cm apart and their temperatures remain unchanged, the following steps can be taken:
1. Calculation of radiation heat transfer between the planes
- The radiation heat transfer between two surfaces can be calculated using the Stefan-Boltzmann law: Q = εσA(T1^4 - T2^4), where Q is the heat transfer rate, ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T1 and T2 are the temperatures of the surfaces.
- Using this equation, the radiation heat transfer rate from Plane 1 to Plane 2 can be calculated as: Q1-2 = 0.84*5.67e-8*(1 m^2)*(T1^4 - T2^4) = 141.6(T1^4 - T2^4) W
- Similarly, the radiation heat transfer rate from Plane 2 to Plane 1 can be calculated as: Q2-1 = 0.42*5.67e-8*(1 m^2)*(T2^4 - T1^4) = 70.8(T2^4 - T1^4) W
2. Calculation of the net heat flux
- Since the temperatures of the two planes remain unchanged, the net heat flux between them must be zero. This means that the heat transfer rate from Plane 1 to Plane 2 must be equal to the heat transfer rate from Plane 2 to Plane 1.
- Therefore, Q1-2 = Q2-1
- Substituting the values of Q1-2 and Q2-1 from step 1, we get: 141.6(T1^4 - T2^4) = 70.8(T2^4 - T1^4)
- Simplifying this equation, we get: T2/T1 = (141.6/70.8)^(1/4) = 1.122
- Since the temperatures of the planes remain unchanged, T1 and T2 must be equal. Therefore, T2/T1 = 1, which is not equal to 1.122.
- This implies that the assumption of zero net heat flux between the planes is incorrect. In fact, there is a net heat transfer from Plane 1 to Plane 2.
- Therefore, the correct answer is (a) 0, which is incorrect assumption.
- Plane 1: emissivity=0.84, heat loss=250 W/m2, temperature=T1
- Plane 2: emissivity=0.42, heat loss=125 W/m2, temperature=T2
To determine the net heat flux between the planes when their bare faces are placed 1 cm apart and their temperatures remain unchanged, the following steps can be taken:
1. Calculation of radiation heat transfer between the planes
- The radiation heat transfer between two surfaces can be calculated using the Stefan-Boltzmann law: Q = εσA(T1^4 - T2^4), where Q is the heat transfer rate, ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T1 and T2 are the temperatures of the surfaces.
- Using this equation, the radiation heat transfer rate from Plane 1 to Plane 2 can be calculated as: Q1-2 = 0.84*5.67e-8*(1 m^2)*(T1^4 - T2^4) = 141.6(T1^4 - T2^4) W
- Similarly, the radiation heat transfer rate from Plane 2 to Plane 1 can be calculated as: Q2-1 = 0.42*5.67e-8*(1 m^2)*(T2^4 - T1^4) = 70.8(T2^4 - T1^4) W
2. Calculation of the net heat flux
- Since the temperatures of the two planes remain unchanged, the net heat flux between them must be zero. This means that the heat transfer rate from Plane 1 to Plane 2 must be equal to the heat transfer rate from Plane 2 to Plane 1.
- Therefore, Q1-2 = Q2-1
- Substituting the values of Q1-2 and Q2-1 from step 1, we get: 141.6(T1^4 - T2^4) = 70.8(T2^4 - T1^4)
- Simplifying this equation, we get: T2/T1 = (141.6/70.8)^(1/4) = 1.122
- Since the temperatures of the planes remain unchanged, T1 and T2 must be equal. Therefore, T2/T1 = 1, which is not equal to 1.122.
- This implies that the assumption of zero net heat flux between the planes is incorrect. In fact, there is a net heat transfer from Plane 1 to Plane 2.
- Therefore, the correct answer is (a) 0, which is incorrect assumption.
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A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer?
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A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. 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 large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. 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 large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer?.
A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. 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 large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. 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 large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer?.
Solutions for A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
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A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A large plane, perfectly insulated on one face and maintained at a fixed temperature T1on the bare face, has an emissivity of 0.84 and loses 250 W/m2when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m2when bare face having emissivity 0.42 and is maintained at temperature T2is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T1and T2remains unchanged. Determine he net heat flux between the planesa)0b)1c)2d)3Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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