This set of Heat Transfer online quiz focuses on “Spectral And Spatial Energy Distribution”.
1. The distribution of radiant energy is non uniform with respect to
c) Both wavelength and direction
Explanation: It should be non-uniform with respect to both wavelength and direction.
2. Spatial distribution is also known as directional distribution.
Explanation: Spatial or directional both are same.
3. Of the radiant energy 350W/m2 incident upon a surface 250W/m2 is absorbed, 60W/m2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material
Explanation: Absorptivity = 250/350 = 0.714.
4. Thermal radiation strikes a surface which has a reflectivity of 0.55 and transmissivity of 0.032. A quantity known as flux is found out to be 95 W/m 2. Determine the rate of incident flux.
a) 123.34 W/m2
b) 333.37 W/m2
c) 122.27 W/m2
d) 227.27 W/m2
Explanation: Incident flux = 95/1 – 0.032 – 0.55 = 227.27 W/m2.
5. What is the wavelength for visible light?
a) 3.9 * 10 -1 to 7.8 * 10 -1 micron meter
b) 4.9 * 10 -1 to 7.8 * 10 -1 micron meter
c) 5.9 * 10 -1 to 7.8 * 10 -1 micron meter
d) 6.9 * 10 -1 to 7.8 * 10 -1 micron meter
Explanation: This is the maximum and minimum wavelength for visible light.
6. Radiant energy with an intensity of 800 W/m2 strikes a flat plate normally. The absorptivity is thrice the reflectivity and twice the transmissivity. Determine the rate of absorption
a) 236.40 W/m2
b) 336.40 W/m2
c) 436.40 W/m2
d) 536.40 W/m2
Explanation: Rate of absorption = 0.5455 * 800 = 436.40 W/m2.
7. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. This plate attains a temperature of 295 K when one of its face receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 K, workout the reflectivity of the plate
Explanation: Heat loss by convection from both sides of the plates = 2 h A d t = 1.1 W. Energy lost by reflection = 2.0 – 1.1 = 0.9 W.
8. On a clear night there is radiation from earth’s surface to space. On such a night, the water particles on the plant leaves radiate to the sky whose temperature may be taken as 200 K. The water particles receive heat by convection from the surrounding air, the convection heat transfer coefficient has a value of 30 W/m2 K. If the water should not freeze, make calculations for the air temperature
a) 280.474 K
b) 345.645 K
c) 123.456 K
d) 874.387 K
Explanation: Heat radiated to sky = Heat received by convection. So, temperature of air = 224.22/30 + 273 = 280.474 K.
9. Isothermal furnaces, with small apertures, approximate a black body and are frequently used to calibrate heat flux gauges.
Explanation: A small hole leading into a cavity thus acts very nearly as a black body.
10. A surface element emits radiation in all directions, the intensity of radiation is however same in different directions.
Explanation: It must be different in different directions.
Sanfoundry Global Education & Learning Series – Heat Transfer.
To practice all areas of Heat Transfer for online quiz, here is complete set of 1000+ Multiple Choice Questions and Answers.