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Chemical Engineering Exam  >  Chemical Engineering Tests  >  Test: Adiabatic & Reradiating Surfaces - Chemical Engineering MCQ

Test: Adiabatic & Reradiating Surfaces - Chemical Engineering MCQ


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10 Questions MCQ Test - Test: Adiabatic & Reradiating Surfaces

Test: Adiabatic & Reradiating Surfaces for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Test: Adiabatic & Reradiating Surfaces questions and answers have been prepared according to the Chemical Engineering exam syllabus.The Test: Adiabatic & Reradiating Surfaces MCQs are made for Chemical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Adiabatic & Reradiating Surfaces below.
Solutions of Test: Adiabatic & Reradiating Surfaces questions in English are available as part of our course for Chemical Engineering & Test: Adiabatic & Reradiating Surfaces solutions in Hindi for Chemical Engineering course. Download more important topics, notes, lectures and mock test series for Chemical Engineering Exam by signing up for free. Attempt Test: Adiabatic & Reradiating Surfaces | 10 questions in 10 minutes | Mock test for Chemical Engineering preparation | Free important questions MCQ to study for Chemical Engineering Exam | Download free PDF with solutions
Test: Adiabatic & Reradiating Surfaces - Question 1
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Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when no other surface is present

Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 1

Q = F 12 A 1 σ B (T 1 4 – T 2 4).

Test: Adiabatic & Reradiating Surfaces - Question 2
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Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when the disks are connected by a cylindrical black no-flux surface

Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 2

Q = F 12 A 1 σ B (T 1 4 – T 2 4).

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Test: Adiabatic & Reradiating Surfaces - Question 3
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Heat exchange between two black surfaces enclosed by an insulated surface is given by

Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 3

This is the net heat exchange between two black surfaces enclosed by an insulated surface.

Test: Adiabatic & Reradiating Surfaces - Question 4
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Heat exchange between two gray surfaces enclosed by an adiabatic surface is given by
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 4

This is the net heat exchange between two gray surfaces enclosed by an adiabatic surface.

Test: Adiabatic & Reradiating Surfaces - Question 5
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A blind cylindrical hole of 2 cm diameter and 3 cm length is drilled into a metal slab having emissivity 0.7. If the metal slab is maintained at 650 K, make calculations for the radiation heat escape from the hole
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 5

Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11].

Test: Adiabatic & Reradiating Surfaces - Question 6
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A cavity in the shape of a frustum of a cone has diameter 30 cm and 60 cm and the height is 80 cm. If the cavity is maintained at temperature of 800 K, determine the heat loss from the cavity when the smaller diameter is at the bottom
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 6

Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11].

Test: Adiabatic & Reradiating Surfaces - Question 7
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Consider the above problem, find how this heat loss would be affected if the cavity is positioned with bigger diameter at the base
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 7

Percentage change in heat flow = 6577 – 1640/6577 = 0.7506.

Test: Adiabatic & Reradiating Surfaces - Question 8
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A conical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 8

Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11]. Here, F 11 = 0.649 and A 1 = 0.0503 m2.

Test: Adiabatic & Reradiating Surfaces - Question 9
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A cylindrical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 9

Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11]. Here, F 11 = 0.842 and A 1 = 0.11186 m2.

Test: Adiabatic & Reradiating Surfaces - Question 10
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What is the unit of coefficient of radiant heat transfer?
Detailed Solution for Test: Adiabatic & Reradiating Surfaces - Question 10

Its value can be calculated from the heat flux equation for any configuration.

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