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Consider sun (radius R) to be a black body of temperature T. Find out the view factor of earth ( with respect to sun. The center to center distance between sun and earth is (l >>r,R).
Option~
(A)
(B)
(C)
(D)
Correct answer is 'A'. Can you explain this answer?
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Consider sun (radius R) to be a black body of temperature T. Find out...
By definition of view factor
Fse = Qes➝ne/Qes
Where Qes = σT4 x 4πR2
Consider an imaginary sphere of radius l with its center at center of sun. As sun is a symmetric body it emits radiation uniformly in all direction which is received uniformly at surface of imaginary sphere. The energy received by earth is actually the energy to be received by AB portion of the surface of imaginary sphere, where AB is interaction of imaginary sphere and earth. l >> r As therefore AB is nearly a circular surface of area πR2.
Total energy emitted by sum = σ(4πR2)T4
∴ Energy received per unit area of imaginary sphere
= σ(4πR2)T4/4πl2
= σR2T4/l2
∴ Energy received by earth (i.e. surface AB)
= σR2T4/l2 x πr2
Hence Fse =
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Most Upvoted Answer
Consider sun (radius R) to be a black body of temperature T. Find out...
By definition of view factor
Fse = Qes➝ne/Qes
Where Qes = σT4 x 4πR2
Consider an imaginary sphere of radius l with its center at center of sun. As sun is a symmetric body it emits radiation uniformly in all direction which is received uniformly at surface of imaginary sphere. The energy received by earth is actually the energy to be received by AB portion of the surface of imaginary sphere, where AB is interaction of imaginary sphere and earth. l >> r As therefore AB is nearly a circular surface of area πR2.
Total energy emitted by sum = σ(4πR2)T4
∴ Energy received per unit area of imaginary sphere
= σ(4πR2)T4/4πl2
= σR2T4/l2
∴ Energy received by earth (i.e. surface AB)
= σR2T4/l2 x πr2
Hence Fse =
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Community Answer
Consider sun (radius R) to be a black body of temperature T. Find out...
By definition of view factor
Fse = Qes➝ne/Qes
Where Qes = σT4 x 4πR2
Consider an imaginary sphere of radius l with its center at center of sun. As sun is a symmetric body it emits radiation uniformly in all direction which is received uniformly at surface of imaginary sphere. The energy received by earth is actually the energy to be received by AB portion of the surface of imaginary sphere, where AB is interaction of imaginary sphere and earth. l >> r As therefore AB is nearly a circular surface of area πR2.
Total energy emitted by sum = σ(4πR2)T4
∴ Energy received per unit area of imaginary sphere
= σ(4πR2)T4/4πl2
= σR2T4/l2
∴ Energy received by earth (i.e. surface AB)
= σR2T4/l2 x πr2
Hence Fse =
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Consider sun (radius R) to be a black body of temperature T. Find out the view factor of earth ( with respect to sun. The center to center distance between sun and earth is (l >>r,R).Option~(A) (B) (C) (D) Correct answer is 'A'. Can you explain this answer? for Chemical Engineering 2025 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about Consider sun (radius R) to be a black body of temperature T. Find out the view factor of earth ( with respect to sun. The center to center distance between sun and earth is (l >>r,R).Option~(A) (B) (C) (D) Correct answer is 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider sun (radius R) to be a black body of temperature T. Find out the view factor of earth ( with respect to sun. The center to center distance between sun and earth is (l >>r,R).Option~(A) (B) (C) (D) Correct answer is 'A'. Can you explain this answer?.
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