This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Intensity Of Radiations”.
1. The solid angle is defined by a region by the rays of a sphere, and is measured as
a) An/r 2
c) An/r 3
d) An/r 4
Explanation: Solid angle is represented by α. Where, An is projection of incident surface normal to line of propagation.
2. The plane angle is defined by a region by the rays of a circle, and is measured as
a) 3 L/ r
b) 2 L/ r
c) L/ r
d) 4 L / r
Explanation: It is the ratio of arc of length on the circle to the radius of the circle. Where L is the arc of length and r is the radius of the circle.
3. When the incident surface is a sphere, the projection of surface normal to the line of propagation is the silhouette disk of the sphere which is a circle of the diameter of
Explanation: It must be a circle of a diameter of a sphere.
4. If I n denotes the normal intensity and I α represents the intensity at angle α, then
a) I α = 2 I n cos α
b) I α = 3 I n cos α
c) I α = 4 I n cos α
d) I α = I n cos α
Explanation: The intensity of radiation in a direction from the normal is proportional to cosine of the angle.
5. The intensity of normal radiation I n is how much times the emissive power?
b) 2/ π
c) 3/ π
d) 4/ π
Explanation: I n = σ T 4/ π and E = σ T 4.
6. A small surface emits diffusively, and measurements indicate that the total intensity associated with emission in the normal direction I n = 6500 W/square m sr. The emitted radiation is intercepted by three surfaces. Mark calculations for intensity associated with emission
a) 3500 W/m2 sr
b) 4500 W/m2 sr
c) 5500 W/m2 sr
d) 6500 W/m2 sr
Explanation: For a diffusion emitter, the intensity of the emitted radiation is independent of direction.
7. Consider a deep-space probe constructed as 1 m diameter polished aluminum sphere. Estimate the equilibrium temperature that the probe reaches if the solar energy received is 300 W/m2. For solar radiation, absorptivity of aluminum is 0.3 and the average emissivity appropriate for aluminum at low temperature is 0.04
a) 415.67 K
b) 315.67 K
c) 215.67 K
d) 115.67 K
Explanation: Q in = α q A P = 70.7 W. Q out = E σ b A T 4.
8. The total emissive power of the emitter with area d A and temperature T is given by
a) E = 2 σ T 4 d A
b) E = 3 σ T 4 d A
c) E = σ T 4 d A
d) E = ½ σ T 4 d A
Explanation: E = I n π d A.
9. A black body of 0.2 m2 area has an effective temperature of 800 K. Calculate the intensity of normal radiations
a) 1234.65 W/m2 sr
b) 7396.28 W/m2 sr
c) 3476.74 W/m2 sr
d) 8739.43 W/m2 sr
Explanation: In = α T 4/π = 7396.28 W/m2 sr.
10. The energy radiated out decreases with increases in α and becomes zero at an angle of
Explanation: I α = I n cos α. So at 90 degree it becomes zero.
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.