Engineering Questions with Answers - Multiple Choice Questions

# MCQs on Electrical Network Approach for Radiation Heat Exchange

1 - Question

The total radiant energy leaving a surface per unit time per unit surface area is represented by
d) Interchange factor

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

2 - Question

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
a) .040
b) .030
c) .020
d) .010

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

3 - Question

What is the value of grey body factor for concentric cylinders?
a) 3/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
b) 4/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
c) 1/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
d) 2/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].

Explanation: Here, F 12 = 1.

4 - Question

The net heat exchange between the two grey surfaces may be written as
a) (Q 12NET = E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 12 + 1 – e 2/A 2 e 2)
b) (Q 12NET = 2 E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 12 + 1 – e 2/A 2 e 2)
c) (Q 12NET = E b 1 – 2 E b 2/ (1 – e 1/A 1 e 1 + 1/A 12 + 1 – e 2/A 2 e 2)
d) (Q 12NET = 2 E b 1 – 3 E b 2/ (1 – e 1/A 1 e 1 + 1/A 12 + 1 – e 2/A 2 e 2)

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

5 - Question

The net rate at which the radiation leaves the surface is given by
a) e (E  J)/1 – 4 e
b) e (E  J)/1 – 3 e
c) e (E  J)/1 – 2 e
d) e (E  J)/1 – e

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

6 - Question

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
a) – 10.59 J/hour
b) – 11.59 J/hour
c) – 12.59 J/hour
d) – 13.59 J/hour

Explanation: Q 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.

7 - Question

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
a) 534.86 W/m2
b) 634.86 W/m2
c) 734.86 W/m2
d) 834.86 W/m2

Explanation: Q 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.

8 - Question

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?
a) 1803.55 W/m2
b) 1703.55 W/m2
c) 1603.55 W/m2
d) 1503.55 W/m2

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

9 - Question

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

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

10 - Question

Which one of the following is true for the opaque non-black surface?
a) J = E +2 p G
b) J = E + p G
c) J = 2 E + p G
d) J = ½ E + p G