Heat Exchange Between Non-Black Bodies MCQ Quiz in தமிழ் - Objective Question with Answer for Heat Exchange Between Non-Black Bodies - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 18, 2025
Latest Heat Exchange Between Non-Black Bodies MCQ Objective Questions
Top Heat Exchange Between Non-Black Bodies MCQ Objective Questions
Heat Exchange Between Non-Black Bodies Question 1:
Two infinite parallel plates are placed at a certain distance apart. An infinite radiation shield is inserted between the plates without touching any of them to reduce heat exchange between the plates. Assume that the emissivities of plates and radiation shield are equal. The ratio of the net heat exchange between the plates with and without the shield is
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 1 Detailed Solution
Concept:
Heat transfer relation when emissivities of plates and shield is same:
where N is the number of shields.
Calculation:
Given:
N = 1
Heat Exchange Between Non-Black Bodies Question 2:
A 1m diameter spherical cavity is maintained at a uniform temperature of 500 K. The emissivity of the material of the sphere is 0.5; One 10 mm diameter hole is drilled. The maximum rate of radiant energy streaming through the hole will be
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 2 Detailed Solution
Concept:
Qradiation = σϵAT4
Calculation:
Given T = 500 k, d = 10 mm, ϵ = 0.5;
∴ Qr = 0.139 W
Heat Exchange Between Non-Black Bodies Question 3:
The increase in number of surface and space resistance induced between two radiation surfaces if 4 shields are introduced between them?
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 3 Detailed Solution
Explanation:
Consider a system
Without radiation shield:
Surface resistance = 2
Space resistance = 1
Now,
With 1 radiation shield
Surface resistance = 4
Space resistance = 2
∴ the general formula for 'n' shields is:
Surface resistance = 2n + 2
Space resistance = n + 1
Since here increase in resistance is asked not total resistance, therefore:
Increase in surface resistance = 2n + 2 - 2 = 2n
Increase in space resistance = n + 1 - 1 = n
For 4 shield placed in between:
Increase in surface resistance = 2n = 2 × 4 = 8
Increase in space resistance = n = 4
Heat Exchange Between Non-Black Bodies Question 4:
Surface A in the figure is coated with white paint and maintained at temperature of 27°C. It is located directly opposite to surface B which can be considered a black body and maintained at temperature of 727°C. Calculate the amount of heat that needs to be removed from surface A per unit area to maintain its constant temperature. (Consider emissivity of the white paint = 1 and view factor = 1, σ = Stefan-Boltzmann constant)
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 4 Detailed Solution
Concept:
The amount of heat removed from surface A per unit area to maintain its constant temperature,
where, F12 = view factor, σ = Stefan-Boltzmann constant
Calculation:
Given:
ϵ1 = ϵ2 = 1, T1 = 300 K, T2 = 1000 K, F12 = 1
Net Heat Exchange
= σ × 1 × 1 × (10004 – 3004)
Q = 9919 × 108 σ W/m2
Heat Exchange Between Non-Black Bodies Question 5:
What is the next radiant interchange per square meter for two very large plates at temperatures 800 K and 500 K respectively ? (The emissivity of the hot and cold plates is 0.8 and 0.6 respectively. Stefan Boltzmann constant is 5.67 × 10-8 w/m2 K4)
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 5 Detailed Solution
Concept:
Radiation heat exchange (q) between two infinitely long parallel plates is given by:
Calculation:
Given T1 = 800 K, T2 = 500 K; ϵ1 = 0.8, ϵ2 = 0.6;
Now the net heat exchange will be
⇒ q = 10268.123 W/m2;
⇒ q = 10.26 kW/m2;
Heat Exchange Between Non-Black Bodies Question 6:
A hemispherical furnace has diameter of 2 m. The floor of furnace is maintained at 1000 K and the emissivity is 0.67. The temperature of roof is 1300 K and it has an emissivity of 0.45. What is the net radiation heat transfer between the roof and the floor?
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 6 Detailed Solution
Concept:
Heat transfer is given by,
Where, T = temperature
A = area, ϵ = emissivity
Calculation:
T1 = 1000K, T2 = 1300K
E1 = 0.67, E2 = 0.45
F12 = 1
A1/A2 = ½
∴ Q21 = 157.167 kW
Heat Exchange Between Non-Black Bodies Question 7:
Two infinite large plates are exchanging radiation energy with emissivity of 0.3 and 0.4 respectively. If 3 radiation shields are introduced with emissivity 0.02 between these two plates, find the percentage increase in thermal resistance.
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 7 Detailed Solution
Concept:
General formula for heat-transfer between two infinite parallel plates:
where
If 'n' number of the shield is placed between two infinite parallel plates:
where
[Note: Applicable only if emissivities of all shield is same.]
Calculation:
Given:
ϵ1 = 0.3, ϵ2 = 0.4, ϵs = 0.02, n = 3
Increase in resistance:
Rwith shield - Rwithout shield
Percentage increase in resistance percentage:
6144.82 ≈ 6145 %
Heat Exchange Between Non-Black Bodies Question 8:
An enclosure measures 1.5 m × 1.7 m with a height of 2 m. The walls and ceiling are maintained at 250°C and the floor at 130°C. The walls and celling have an emissivity of 0.82 and the floor 0.7. Determine the net radiation to the floor (in kW).
Answer (Detailed Solution Below) 4.6 - 4.8
Heat Exchange Between Non-Black Bodies Question 8 Detailed Solution
Let, A1 = Total area of walls and ceiling
= 2 (1.5 × 2) + 2 (1.7 × 2) + 1.5 × 1.7 = 15.35 m2
A2 = Floor area = 1.5 × 1.7 = 2.55 m2
The floor is completely enclosed by the area A1
F21 + F22 = 1
F21 = 1
A1F12 = A2F21
F12 = A2/A1 = 2.55/15.35 = 0.166
Rate of heat transfer by radiation from the walls and ceiling to the floor
Q1 - 2 = 5.67 × 10-8 × 15.35 × 0.113 (5234 – 4034) = 4764.1 W = 4.76 kW
Heat Exchange Between Non-Black Bodies Question 9:
What is the net radiant interchange per square meter for two very large plates at temperatures 800 K and 500 K respectively. (The emissivity of the hot and cold plates are 0.8 and 0.6 respectively. Stefan Boltzmann constant is 5.67 × 10-8 W/m2 K4).
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 9 Detailed Solution
Concept:
Radiation heat exchange between two infinite parallel plates:
Calculation:
Given:
T1 = 800 K, T2 = 500 K, ϵ1 = 0.8, ϵ2 = 0.6
Q = 10.268 kW/m2
Heat Exchange Between Non-Black Bodies Question 10:
A radiation shield should have
Answer (Detailed Solution Below)
Heat Exchange Between Non-Black Bodies Question 10 Detailed Solution
Concept:
Radiation shield
- Radiation heat transfer between two surfaces is greatly reduced by inserting radiation shields between them.
- The radiation shields reduce the heat transfer by placing additional resistances in the path of heat flow.
- The radiation shields should have high reflectivity and low emissivity.
- The high reflectivity makes the radiation shield to reflect most of the radiation heat and thus decreases the heat transfer to the other surface.
- In the same way, the lower the emissivity, the higher the resistance and the lesser the heat transfer.