Chemical Engineering Exam > Chemical Engineering Questions > A thin shield of emissivity E3on both sides i...
Start Learning for Free
A thin shield of emissivity E 3 on both sides is placed between two infinite parallel plates of emissivities E 1 and E 2 and temperatures T 1 and T 2. If E 1 = E 2 = E 3, then the fraction radiant energy transfer without shield takes the value
- a)0.25
- b)0.50
- c)0.75
- d)1.25
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A thin shield of emissivity E3on both sides is placed between two infi...
The ratio of radiant energy transfer without and with shield is given by
(1/E 1 + 1/E 2 – 1)/ [(1/E 1 + 1/E 3 – 1) + (1/E 3 + 1/E 2 – 1)].
(1/E 1 + 1/E 2 – 1)/ [(1/E 1 + 1/E 3 – 1) + (1/E 3 + 1/E 2 – 1)].
Most Upvoted Answer
A thin shield of emissivity E3on both sides is placed between two infi...
The fraction of radiant energy transfer without the shield can be calculated using the Stefan-Boltzmann Law. According to this law, the rate of radiant energy transfer between two bodies is proportional to the fourth power of the temperature difference between them.
Let's consider two infinite parallel plates with temperatures T1 and T2, where T1 is greater than T2. The emissivities of the plates are denoted as E1 and E2, respectively. Without the shield, the radiant energy transfer between the plates can be calculated as:
Q1 = σ * E1 * A * (T1^4 - T2^4)
Where Q1 is the radiant energy transfer without the shield, σ is the Stefan-Boltzmann constant, and A is the surface area of the plates.
Now, let's introduce a thin shield with emissivity E3 on both sides between the plates. Since E1 = E2 = E3, the radiant energy transfer through the shield can be calculated as:
Q2 = σ * E3 * A * (T1^4 - T2^4)
To find the fraction of radiant energy transfer without the shield, we can divide Q1 by the total radiant energy transfer (Q1 + Q2):
Fraction without shield = Q1 / (Q1 + Q2)
Substituting the values of Q1 and Q2, we get:
Fraction without shield = (σ * E1 * A * (T1^4 - T2^4)) / (σ * E1 * A * (T1^4 - T2^4) + σ * E3 * A * (T1^4 - T2^4))
Simplifying the equation, we find:
Fraction without shield = (T1^4 - T2^4) / ((T1^4 - T2^4) + (E3/E1) * (T1^4 - T2^4))
Since E1 = E2 = E3, we can simplify the equation further:
Fraction without shield = (T1^4 - T2^4) / ((T1^4 - T2^4) + (T1^4 - T2^4))
Fraction without shield = (T1^4 - T2^4) / (2 * (T1^4 - T2^4))
The temperature difference (T1^4 - T2^4) cancels out, and we are left with:
Fraction without shield = 1 / 2
Therefore, the fraction of radiant energy transfer without the shield is 0.5 or 50%. Hence, the correct answer is option B.
Let's consider two infinite parallel plates with temperatures T1 and T2, where T1 is greater than T2. The emissivities of the plates are denoted as E1 and E2, respectively. Without the shield, the radiant energy transfer between the plates can be calculated as:
Q1 = σ * E1 * A * (T1^4 - T2^4)
Where Q1 is the radiant energy transfer without the shield, σ is the Stefan-Boltzmann constant, and A is the surface area of the plates.
Now, let's introduce a thin shield with emissivity E3 on both sides between the plates. Since E1 = E2 = E3, the radiant energy transfer through the shield can be calculated as:
Q2 = σ * E3 * A * (T1^4 - T2^4)
To find the fraction of radiant energy transfer without the shield, we can divide Q1 by the total radiant energy transfer (Q1 + Q2):
Fraction without shield = Q1 / (Q1 + Q2)
Substituting the values of Q1 and Q2, we get:
Fraction without shield = (σ * E1 * A * (T1^4 - T2^4)) / (σ * E1 * A * (T1^4 - T2^4) + σ * E3 * A * (T1^4 - T2^4))
Simplifying the equation, we find:
Fraction without shield = (T1^4 - T2^4) / ((T1^4 - T2^4) + (E3/E1) * (T1^4 - T2^4))
Since E1 = E2 = E3, we can simplify the equation further:
Fraction without shield = (T1^4 - T2^4) / ((T1^4 - T2^4) + (T1^4 - T2^4))
Fraction without shield = (T1^4 - T2^4) / (2 * (T1^4 - T2^4))
The temperature difference (T1^4 - T2^4) cancels out, and we are left with:
Fraction without shield = 1 / 2
Therefore, the fraction of radiant energy transfer without the shield is 0.5 or 50%. Hence, the correct answer is option B.
|
Explore Courses for Chemical Engineering exam
|
|
Similar Chemical Engineering Doubts
A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer?
Question Description
A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct 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 A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct 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 A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer?.
A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct 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 A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct 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 A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer?.
Solutions for A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
Download more important topics, notes, lectures and mock test series for Chemical Engineering Exam by signing up for free.
Here you can find the meaning of A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A thin shield of emissivity E3on both sides is placed between two infinite parallel plates of emissivities E1and E2and temperatures T1and T2. If E1= E2= E3, then the fraction radiant energy transfer without shield takes the valuea)0.25b)0.50c)0.75d)1.25Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
|
|
Explore Courses for Chemical Engineering exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup