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
a) Radiation
b) Radiosity
c) Irradiation
d) Interchange factor
View Answer
Answer: b
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
View Answer
Answer: d
Explanation: Q 12 = F 12 A 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)].
View Answer
Answer: c
Explanation: Here, F 12 = 1.
Explanation: Here, F 12 = 1.
4 - Question
The net heat exchange between the two grey surfaces may be written as
a) (Q 12) NET = E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
b) (Q 12) NET = 2 E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
c) (Q 12) NET = E b 1 – 2 E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
d) (Q 12) NET = 2 E b 1 – 3 E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
View Answer
Answer: a
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 b – J)/1 – 4 e
b) e (E b – J)/1 – 3 e
c) e (E b – J)/1 – 2 e
d) e (E b – J)/1 – e
View Answer
Answer: d
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
View Answer
Answer: b
Explanation: 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.
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
View Answer
Answer: c
Explanation: 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.
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
View Answer
Answer: a
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
b) Radiosity
c) Radiation
d) Irradiation
View Answer
Answer: d
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
View Answer
Answer: b
Explanation: 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.