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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Heat Transfer  >  Test: Network Approach For Heat Exchange - Mechanical Engineering MCQ

Test: Network Approach For Heat Exchange - Mechanical Engineering MCQ


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10 Questions MCQ Test Heat Transfer - Test: Network Approach For Heat Exchange

Test: Network Approach For Heat Exchange for Mechanical Engineering 2024 is part of Heat Transfer preparation. The Test: Network Approach For Heat Exchange questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Network Approach For Heat Exchange MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Network Approach For Heat Exchange below.
Solutions of Test: Network Approach For Heat Exchange questions in English are available as part of our Heat Transfer for Mechanical Engineering & Test: Network Approach For Heat Exchange solutions in Hindi for Heat Transfer course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Network Approach For Heat Exchange | 10 questions in 10 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study Heat Transfer for Mechanical Engineering Exam | Download free PDF with solutions
Test: Network Approach For Heat Exchange - Question 1
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The total radiant energy leaving a surface per unit time per unit surface area is represented by

Detailed Solution for Test: Network Approach For Heat Exchange - Question 1

It comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.

Test: Network Approach For Heat Exchange - Question 2
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Determine the radiant heat flux between two closely spaced, black parallel plates radiating only to each other if their temperatures are 850 K and 425 K. The plates have an area of 4 m2

Detailed Solution for Test: Network Approach For Heat Exchange - Question 2

Q 12 = F 12 A 1 σ b (T 14 – T 24) = .010.

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Test: Network Approach For Heat Exchange - Question 3
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What is the value of grey body factor for concentric cylinders?

Detailed Solution for Test: Network Approach For Heat Exchange - Question 3

Here, F 12 = 1.

Test: Network Approach For Heat Exchange - Question 4
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The net heat exchange between the two grey surfaces may be written as
Detailed Solution for Test: Network Approach For Heat Exchange - Question 4

This equation gives the electrical network corresponding to surface resistances of two radiating bodies.

Test: Network Approach For Heat Exchange - Question 5
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The net rate at which the radiation leaves the surface is given by
Detailed Solution for Test: Network Approach For Heat Exchange - Question 5

The net rate at which the radiation leaves the surface is given by the difference between its radiosity and the incoming irradiation.

Test: Network Approach For Heat Exchange - Question 6
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A ring (E = 0.85) of 8 cm inner and 16 cm outer diameter is placed in a horizontal plane. A small element (E = 0.7) of 1 cm2 is placed concentrically 8 cm vertically below the center of the ring. The temperature of the ring is 800 K and that of small area is 400 K. Find the radiant heat gain by the small ring
Detailed Solution for Test: Network Approach For Heat Exchange - Question 6

Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) = A 1 σ b (T 14 – T 24)/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2/A 1.

Test: Network Approach For Heat Exchange - Question 7
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Two opposed, parallel, infinite planes are maintained at 420 K and 480 K. Calculate the net heat flux between these planes if one has an emissivity of 0.8 and other an emissivity of 0.7
Detailed Solution for Test: Network Approach For Heat Exchange - Question 7

Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) and (F g) 12 = 1/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2/A 1.

Test: Network Approach For Heat Exchange - Question 8
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Consider the above problem, if temperature difference is doubled by raising the temperature 480 K to 540 K, then how this heat flux will be affected?
Detailed Solution for Test: Network Approach For Heat Exchange - Question 8

Q 2 = 0.59 (5.67 * 10 -8) (540 4 – 420 4).

Test: Network Approach For Heat Exchange - Question 9
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The total radiant energy incident upon a surface per unit time per unit area is known as
Detailed Solution for Test: Network Approach For Heat Exchange - Question 9

Some of it may be reflected to become a part of the radiosity of the surface.

Test: Network Approach For Heat Exchange - Question 10
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Which one of the following is true for opaque non-black surface?
Detailed Solution for Test: Network Approach For Heat Exchange - Question 10

For an opaque non-black surface of constant radiation characteristics, the total radiant energy leaving the surface is the sum of its original emittance and the energy reflected by it out of the irradiation impinging on it.

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