This set of Heat Transfer Question Bank focuses on “Heat Exchange Between Non Black Bodies”.

1. Consider radiant heat exchange between two non-black parallel surfaces. The surface 1 emits radiant energy E _{1} which strikes the surface 2. Identify the correct option

a) The value of B is α E_{1}

b) The value of C is (1 – α_{1}) E _{1}

c) The value of D is (1 – α_{1}) (1 – α_{2}) ^{2 }E _{1}

d) The value of E is (1 – α_{1}) ^{2} (1 – α_{2})^{ }E _{1}

View Answer

Explanation: The surface 1 emits radiant energy E

_{1}which strikes the surface 2. From it a part α

_{2}

^{ }E

_{1 }is absorbed by the surface 2 and the remainder (1 – α

_{2}) E

_{1}is reflected back to surface 1 and so on.

2. A large plane, perfectly insulated on one face and maintained at a fixed temperature T _{1} on the bare face, has an emissivity of 0.84 and loses 250 W/m^{2} when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m^{2} when bare face having emissivity 0.42 and is maintained at temperature T _{2} is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T _{1 }and T _{2} remains unchanged. Determine he net heat flux between the planes

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: Q

_{12 }= F

_{12 }A

_{1 }σ

_{ b }(T

_{1}

^{4}– T

_{2}

^{4}). Since T

_{1}= T

_{2}, we get Q

_{12}/A

_{1}= 0.

3. Interchange factor for body 1 completely enclosed by body 2 (body 1 is large) is given by

a) 2/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

b) 1/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

c) 4/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

d) 3/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

View Answer

Explanation: This is the interchange factor for the radiation from surface 1 to surface 2.

4. A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K, find the rate of heat loss from the thermos bottle to the ambient air at 300 K

a) 5.38 W

b) 6.38 W

c) 7.38 W

d) 8.38 W

View Answer

Explanation: Q

_{12 }= F

_{12 }A

_{1 }σ

_{ b }(T

_{1}

^{4}– T

_{2}

^{4}). F

_{12}= 1/ (1/E

_{1}+ 1/E

_{2 }– 1) = 0.01266.

5. A 250 mm by 250 mm ingot casting, 1.5 m high and at 1225 K temperature, is stripped from its mold. The casting is made to stand on end on the floor of a large foundry whose wall, floor and roof can be assumed to be at 300 K temperature. Make calculation for the rate of radiant heat interchange between the casting and the room. The casting material has an emissivity of 0.85

a) 161120 W

b) 171120 W_{}

c) 181120 W

d) 191120 W

View Answer

Explanation: Q

_{12 }= F

_{12 }A

_{1 }σ

_{ b }(T

_{1}

^{4}– T

_{2}

^{4}). F

_{12}= 0.85 and A

_{1}= (0.25)

^{2}+ 4(1.5) (0.25) = 1.5625 m

^{2}.

6. Interchange factor for infinitely long concentric cylinders is given by

a) 1/ [A_{1}/A_{2 }(1/E_{ 2} – 1)].

b) [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

c) 2/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

d) 1/ [1/E_{ 1} + A_{1}/A_{2 }(1/E_{ 2} – 1)].

View Answer

Explanation: This is the interchange factor for the radiation from surface 1 to surface 2.

7. What is the geometric factor for infinitely long concentric cylinders?

a) 1

b) 0.5

c) 0.33

d) 0.75

View Answer

Explanation: The inner cylinder is completely enclosed by the outer cylinder and as such the entire heat radiations emitted by the emitted by the inner cylinder are intercepted by the outer cylinder.

8. What is the geometric factor for concentric spheres?

a) 0.85

b) 0.33

c) 1

d) 0.95

View Answer

Explanation: The inner sphere is completely enclosed by the outer sphere and as such the entire heat radiations emitted by the emitted by the inner sphere are intercepted by the outer cylinder.

9. The net heat interchange between non-black bodies at temperature T _{1} and T_{ 2 }is given by

a) f _{12 }F_{ 12} σ (T_{ 1}^{4} – T _{2}^{4})

b) f _{12 }F_{ 12} A _{1 }σ (T_{ 1}^{4} – T _{2}^{4})

c) f _{12} A _{1 }σ (T_{ 1}^{4} – T _{2}^{4})

d) _{ }F_{ 12} A _{1 }σ (T_{ 1}^{4} – T _{2}^{4})

View Answer

Explanation: The factor f

_{12}is called the interchanging factor from surface 1 to surface 2.

10. A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork (k = 0.03 W/m degree) would be required if the same insulating effect is to be achieved by the use of cork?

a) 26.8 cm

b) 25.8 cm

c) 24.8 cm

d) 23.8 cm

View Answer

Explanation: Q = k A (t

_{1}– t

_{2})/δ. So, δ = 0.268 m = 26.8 cm.

**Sanfoundry Global Education & Learning Series – Heat Transfer.**

To practice Heat Transfer Question Bank, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**

- Practice Chemical Engineering MCQs
- Practice Mechanical Engineering MCQs
- Apply for Chemical Engineering Internship
- Apply for Mechanical Engineering Internship
- Check Chemical Engineering Books