This set of Heat Transfer Questions & Answers for freshers focuses on “Critical Thickness Of Insulation”.
1. A cable of 10 mm outside is to be laid in an atmosphere of 25 degree Celsius (h = 12.5 W/m2 degree) and its surface temperature is likely to be 75 degree Celsius due to heat generated within it. How would the heat flow from the cable be affected if it is insulated with rubber having thermal conductivity k = 0.15 W/m degree?
a) 43.80 W per meter length
b) 53.80 W per meter length
c) 63.80 W per meter length
d) 73.80 W per meter length
Explanation: Q = 2 π d t/ (1/k) log e(r c/r 0) = 53.80 W per meter length.
2. Chose the correct one with respect to critical radius of insulation
a) There is more heat loss i.e. conductive
b) There occurs a decrease in heat flux
c) Heat loss increases with addition of insulation
d) Heat loss decreases with addition of insulation
Explanation: For a pipe heat loss is more at the critical radius.
3. A heat exchanger shell of outside radius 15 cm is to be insulated with glass wool of thermal conductivity 0.0825 W/m degree. The temperature at the surface is 280 degree Celsius and it can be assumed to remain constant after the layer of insulation has been applied to the shell. The convective film coefficient between the outside surface of glass wool and the surrounding air is estimated to be 8 W/m2 degree. What is the value of critical radius?
a) 9.31 mm
b) 10.31 mm
c) 11.31 mm
d) 12.31 mm
Explanation: Critical radius of insulation = k/h = 0.0825/8 = 0.01031 m = 10.31 mm.
4. For an object i.e. spherical the value of critical radius would be
Explanation: It depends on variation of angle with layers of insulation.
5. Maximum value of critical radius is
a) 0.01 m
b) 0.04 m
c) 0.06 m
d) 0.0001 m
Explanation: K for common insulating material is 0.05 W/ m degree.
6. An electric cable of aluminum (k = 240 W/ m degree) is to be insulated with rubber (k = 6 W/ square meter degree). If the cable is in air (h = 6 W/square meter degree). Find the critical radius?
a) 80 mm
b) 160 mm
c) 40 mm
d) 25 mm
Explanation: Critical radius = 0.15/6 = 0.025 m = 25 mm.
7. The value of critical radius in case of cylindrical hollow object is
Explanation: Unit is meter.
8. A wire of radius 3 mm and 1.25 m length is to be maintained at 60 degree Celsius by insulating it by a material of thermal conductivity 0.175 W/m K. The temperature of surrounding is 20 degree Celsius with heat transfer coefficient 8.5 W/ m2 K. Find percentage increase in heat loss due to insulation?
a) 134.46 %
b) 124.23 %
c) 100.00 %
d) 12.55 %
Explanation: Q = 8.5 (2 π 0.003 1.25) (60 – 20) = 8.01 W. % increase = (18.78 – 8.01/8.01) (100) = 134.46 %.
9. A pipe of outside diameter 20 mm is to be insulated with asbestos which has a mean thermal conductivity of 0.1 W/m degree. The local coefficient of convective heat to the surroundings is 5 W/square meter degree. Find the critical radius of insulation for optimum heat transfer from pipe?
a) 10 mm
b) 20 mm
c) 30 mm
d) 40 mm
Explanation: Critical radius of insulation = k/h0 = 0.1/5 = 0.02 m = 20 mm.
10. For insulation to be properly effective in restricting heat transmission, the pipe radius r0 will be
a) Greater than critical radius
b) Less than critical radius
c) Equal to critical radius
d) Greater than or equal to critical radius
Explanation: Addition of insulating material doesn’t always decrease in the heat transfer rate.
Sanfoundry Global Education & Learning Series – Heat Transfer.
To practice all areas of Heat Transfer for freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.