Heat Transfer Questions and Answers – Radiations Shields

This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Radiations Shields”.

1. A radiation shield should
a) Have high transmissivity
b) Absorb all the radiations
c) Have high reflexive power
d) Partly absorb and partly transmit the incident radiation
View Answer

Answer: c
Explanation: Reflexive power is much high for radiation shield.

2. Radiation shield is used between the emitting surfaces such that
a) To reduce overall heat transfer
b) To increase overall heat transfer
c) To increase density
d) To reduce density
View Answer

Answer: a
Explanation: Many situations are encountered where it is desired to reduce the overall heat transfer between two radiating surfaces.

3. Which of the following can be used as a radiating shield?
a) Carbon
b) Thin sheets of aluminum
c) Iron
d) Gold
View Answer

Answer: b
Explanation: The shields are thin opaque partitions arranged in the direction perpendicular to the propagation of radiant heat.

4. Two large parallel planes with emissivity 0.4 are maintained at different temperatures and exchange heat only by radiation. What percentage change in net radiative heat transfer would occur if two equally large radiation shields with surface emissivity 0.04 are introduced in parallel to the plates?
Find percentage change in net radiative heat transfer occur if large radiation shields
a) 65.1%
b) 75.1%
c) 85.1%
d) 95.1%
View Answer

Answer: d
Explanation: When shields are not used, Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) = 0.2 A 1 σ b (T 14 – T 24) and when shields are used Q 12 = 0.0098 A 1 σ b (T 14 – T 24).

5. Determine the net radiant heat exchange per m 2 area for two infinite parallel plates held at temperature of 800 K and 500 K. Take emissivity as 0.6 for the hot plate and 0.4 for the cold plate
a) 6200 W/m2
b) 7200 W/m2
c) 8200 W/m2
d) 9200 W/m2
View Answer

Answer: a
Explanation: Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) and (F g) 12 = 0.135. Therefore, Q 12 = 6200 W/m2.
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6. Consider the above problem, what should be the emissivity of a polished aluminum shield placed between them if heat flow is to be reduced to 40 percent of its original value?
a) 0.337
b) 0.347
c) 0.357
d) 0.367
View Answer

Answer: b
Explanation: (F g) 12 = 1/E 1 +1/E 2 +2/E 3 – 2 = 7.936.

7. Consider radiative heat transfer between two large parallel planes of surface emissivities 0.8. How many thin radiation shields of emissivity 0.05 be placed between the surfaces is to reduce the radiation heat transfer by a factor of 75?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: (Q 12) ONE SHIELD = A σ b (T 14 – T 24)/ 1/E 1 +1/E 2 +2/E 3 – 2 and 75 = (Q 12) NO SHIELD / (Q 12) N SHIELD.

8. Two parallel square plates, each 4 m2 area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the plates
a) 61.176 k W
b) 51.176 k W
c) 41.176 k W
d) 31.176 k W
View Answer

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

9. The furnace of a boiler is laid from fire clay brick with outside lagging from the plate steel, the distance between the two is quite small compared with the size of the furnace. The brick setting is at an average temperature of 365 K whilst the steel lagging is at 290 K. Calculate the radiant heat flux. Assume the following emissivity values
For brick = 0.85
For steel = 0.65
a) 352.9 W/m2
b) 452.9 W/m2
c) 552.9 W/m2
d) 652.9 W/m2
View Answer

Answer: a
Explanation: Q 12 = (F g) 12 A 1 σ b (T 14 – T 24).

10. Consider the above problem, find the reduction in heat loss if a steel screen having an emissivity value of 0.6 on both sides is placed between the brick and steel setting
a) 5.56
b) 4.46
c) 3.36
d) 2.36
View Answer

Answer: d
Explanation: (F g) 12 = 0.247 and Q = 149.51 W/m2.

Sanfoundry Global Education & Learning Series – Heat Transfer.

To practice all areas of Heat Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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