Test: Network Approach For Heat Exchange


10 Questions MCQ Test Heat Transfer for Engg. | Test: Network Approach For Heat Exchange


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This mock test of Test: Network Approach For Heat Exchange for Chemical Engineering helps you for every Chemical Engineering entrance exam. This contains 10 Multiple Choice Questions for Chemical Engineering Test: Network Approach For Heat Exchange (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Network Approach For Heat Exchange quiz give you a good mix of easy questions and tough questions. Chemical Engineering students definitely take this Test: Network Approach For Heat Exchange exercise for a better result in the exam. You can find other Test: Network Approach For Heat Exchange extra questions, long questions & short questions for Chemical Engineering on EduRev as well by searching above.
QUESTION: 1

The total radiant energy leaving a surface per unit time per unit surface area is represented by

Solution:

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

QUESTION: 2

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

Solution:

12 = F 12 1 σ b (T 14 – T 24) = .010.

QUESTION: 3

What is the value of grey body factor for concentric cylinders?

Solution:

Here, F 12 = 1.

QUESTION: 4

The net heat exchange between the two grey surfaces may be written as

Solution:

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

QUESTION: 5

The net rate at which the radiation leaves the surface is given by

Solution:

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

QUESTION: 6

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

Solution:

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

QUESTION: 7

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

Solution:

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

QUESTION: 8

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?

Solution:

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

QUESTION: 9

The total radiant energy incident upon a surface per unit time per unit area is known as

Solution:

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

QUESTION: 10

Which one of the following is true for opaque non-black surface?

Solution:

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.