250+ TOP MCQs on HT Equipment – Combined Heat Transfer by Conduction and Convection

Heat Transfer Operations Interview Questions and Answers on “HT Equipment – Combined Heat Transfer by Conduction and Convection”.

1. In a Heat Exchanger, heat cannot be transferred by all the three means, that is Conduction, Convection and Radiation simultaneously.
a) True
b) False

Answer: a
Clarification: In a Heat Exchanger there can be maximum two possible means of heat transfer in practical use, they are Conduction and Convection.

2. In which one of the following, Heat transfer takes place without any Conduction or Convection?
a) Condenser
b) Recuperators
c) Cooling tower
d) Shell and Tube HE

Answer: c
Clarification: In a cooling tower, heat transfer takes place by evaporation which carries away the thermal energy from the liquid in the form of vapour hence cooling it, no instance of conduction or convection is observed.

3. How many Heat Exchangers can be classified on the basis of phase change of the fluids?
a) 1
b) 2
c) 3
d) 4

Answer: b
Clarification: For a heat exchanger to be classified on the basis of phase change of participating fluids, they can be classified into Condensers and Evaporators.

4. When we want to heat a stream of liquid by Steam, and the steam may condense into liquid. Then which of the following is best suited?
a) Steam on the side with the fins
b) Steam on the side with the fins and condensate drainage
c) Steam on the side with the fins and a brush to remove the condensate film from surface
d) Steam on the tube with the fins on inner surface of the tube passing through a pool of cooling liquid

Answer: c
Clarification: As we know that film condensing is less effective than drop-wise condensing, hence a brush continuously removing the film is the best suit for this purpose.

5. When we want to heat a stream of liquid by Steam, we usually keep removing the condensate from the surface of the tube.
a) True
b) False

Answer: a
Clarification: It is best suited when we keep removing condensate from the tube surface as film condensation decreases the heat transfer coefficient and removing it supports drop-wise condensation.

6. What is the critical radius of insulation for a cylindrical wall?
a) (frac{K}{H})
b) (frac{4K}{H})
c) (frac{K}{4H})
d) (frac{H}{K})

Answer: a
Clarification: The critical thickness of insulation is that radius of the insulating wall (with conductive heat transfer coefficient as K and the surrounding fluid convective heat transfer coefficient as H) that gives maximum heat transfer rate Q. Hence for a cylindrical insulating wall it is (frac{K}{H}).

7. What is Critical radius of insulation?
a) The maximum radius that can allow heat transfer
b) The maximum heat transfer coefficient at a maximum possible radius
c) The radius at which maximum heat transfer rate is observed
d) The minimum radius at which maximum heat transfer rate is observed

Answer: c
Clarification: The critical thickness of insulation is that radius (neither maximum nor minimum) of the insulating wall with conductive heat transfer coefficient as K and the surrounding fluid convective heat transfer coefficient as H, that gives maximum heat transfer rate Q.

8. What is critical thickness of insulation for a spherical insulator?
a) KH
b) (frac{2K}{H})
c) (frac{4K}{H})
d) HK

Answer: b
Clarification: The critical thickness of insulation is that radius of the insulating wall (with conductive heat transfer coefficient as K and the surrounding fluid convective heat transfer coefficient as H) that gives maximum heat transfer Q. Hence for a spherical insulating wall it is (frac{2K}{H}).

9. For the calculation of critical radius of insulation for a cylindrical tube, which of the following differentiation is correct?
a) (frac{d(frac{K}{HR_2}+ln(frac{R_2}{R_1}))}{dR_1}) = 0
b) (frac{d(frac{K}{HR_1}+ln(frac{R_2}{R_1}))}{dR_2}) = 0
c) (frac{d(frac{K}{HR_2}+ln(frac{R_1}{R_2}))}{dR_1}) = 0
d) (frac{d(frac{K}{HR_2}+ln(frac{R_2}{R_1}))}{dR_2}) = 0

Answer: d
Clarification: The total heat transfer rate can be defined as the net Temperature gradient to the total resistance of the medium. Hence the heat transfer rate can be represented as:
Q = (frac{2pi(T-T_infty)K}{frac{K}{HR_2}+ln(frac{R_2}{R_1})})
thus to calculate the maximum heat transfer rate at a given thickness, we differentiate the denominator to zero for the minimum of the denominator component which will fetch us the maximum heat transfer rate Q. We differentiate it with respect to R2 as it is the only variable with R1 a fixed constant.

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