# Heat Transfer Questions and Answers – Electrical Network Approach for Radiation Heat Exchange

This set of Heat Transfer Questions for entrance exams focuses on “Electrical Network Approach For Radiation Heat Exchange”.

1. 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. 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 A 1 σ b (T 14 – T 24) = .010.

3. 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. 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)

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

5. 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

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

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
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 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. 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 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. 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. 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. 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

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.

Sanfoundry Global Education & Learning Series – Heat Transfer.

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

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