This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Double Pipe Heat Exchangers – Construction and Operation”.
1. Gases cannot be used in Double pipe heat exchangers because of their high flow rates.
a) True
b) False
View Answer
Explanation: In a double pipe heat exchanger we can use any type of fluid for the purpose, be it a gas or a liquid, their flow rate does not matter until it crosses the pressure limit.
2. If we know the output and input temperatures of the heat exchanging fluids, then which one of the following calculation is not required to determine the number of bends in tube for the equipment?
a) Pressure drop in the equipment
b) Overall heat transfer coefficients
c) Total heat transfer area required
d) Pipe length
View Answer
Explanation: The pressure drop is calculated only to determine whether the equipment would work for a given flow rate and not the temperature requirements.
3. Which of the following has the maximum Log mean temperature difference for a Double Pipe Heat Exchanger?
a) Counter-flow
b) Parallel Flow
c) Cross Flow
d) Split Flow
View Answer
Explanation: The mean temperature difference for a counter-flow operation is uniform throughout the length of the tube and hence gives a higher value to the LMTD.
4. How many types of flow operations exist for a double pipe heat exchanger?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: There exist only two possible flow operations for a double pipe heat exchanger which are Counter-flow and Parallel flow operations.
5. Which one of the following is the determining reason for heat transfer in double pipe HE?
a) Conduction
b) Natural Convection
c) Forced Convection
d) Radiation
View Answer
Explanation: Forced convection is the major means of heat transfer taking place in the HE. In a double pipe HE, the fluid is in a constant uniform motion inside the pipes, it is the phenomena of forced convection that removes the fluid from the wall by dynamic mixing which increases the overall bulk temperature of the fluid.
6. Which one of these is not true when the steady state is reached by the heat exchanging fluids in a double pipe HE?
a) When the two liquids have same temperature
b) When their temperatures become stable
c) Wall temperature becomes constant
d) Rate of heat transfer becomes constant
View Answer
Explanation: When the steady state is reached by the fluids in the HE, their temperatures become stable but not necessarily equal, as the heat transfer rate becomes constant, heat is still entering the tubes.
7. We can use mean temperature difference too in place of LMTD if we can calculate it.
a) True
b) False
View Answer
Explanation: We can use mean temperature difference in place of LMTD but it is impossible to continuously measure the temperature difference along the whole length of the tube of the HE. Hence we use LMTD for a better approximation.
8. We can apply LMTD only when?
(i) There is no change in Specific heats
(ii) Overall heat transfer coefficient is constant
(iii) No heat loss
(iv) No pressure drop
a) (ii)(iii)(iv)
b) (i)(ii)(iii)
c) (ii)(iii)
d) (i)(ii)(iv)
View Answer
Explanation: The LMTD has limitations on its applicability which usually are no change in specific heats overall heat transfer coefficient is constant and no heat loss, but it is totally independent on the pressure drop that takes place in the equipment.
9. The density of the fluids, its viscosity and the thermal conductivity (K) is measured at ______
a) LMTD
b) Mean temperature
c) Median of the temperature
d) Square mean of the temperature
View Answer
Explanation: The LMTD has limitations on its applicability such as there should be no change in specific heats, the overall heat transfer coefficient should be constant and negligible heat losses. Hence owing to these limitations, the density and other physical properties are measured at the Mean temperature of the fluids.
10. Which one of the following is true about a hairpin used in double pipe HE?
a) It can handle high pressure drops
b) It cannot handle high pressure drops
c) It is very resistant to fouling
d) It is very expensive
View Answer
Explanation: One of the major advantages of hairpins is its U-shaped structure which makes it capable to handle very high pressure drops in the tube side, which is one of the contributing factors which make its use very common in industries. It is comparatively cheaper with respect to other designs.
11. To calculate the temperature difference in a double pipe heat exchanger, we use _____
a) LMTD
b) Mean temperature difference
c) Median of the temperature difference
d) Square mean of the temperature difference
View Answer
Explanation: We can use mean temperature difference in place of LMTD but it is impossible to continuously measure the temperature difference along the whole length of the tube of the HE. Hence we use LMTD for a better approximation.
12. In a double pipe heat exchanger, in the inner side fluid enters at 15℃ and leaves at 65℃. The annulus has steam at 1atm. What is the value of LMTD?
a) 39℃
b) 66.7℃
c) 70℃
d) 56.35℃
View Answer
Explanation: As steam is condensing at 100℃(Tvap at 1atm), we have LMTD = \(\frac{(100-15)-(100-65)}{Ln(\frac{100-15}{100-65})}\) = 56.35℃.
13. What do you understand by Rfo and Rfi?
a) Dirt factor and Fouling factor
b) Inner Tube outer surface fouling factor and inner surface fouling factor
c) Inner Tube outer surface fouling factor and inner surface fouling factor
d) Inner Tube outer surface fouling factor and Annulus tube inner surface fouling factor
View Answer
Explanation: As the heat transfer takes place by conduction through only the inner tube of the HE, we are concerned about fouling only on this tube. Thus we calculate inner tube outer surface fouling factor and inner tube inner surface fouling factor respectively.
14. What is the unit of Fouling factor (Rf) in S.I. system?
a) m2K/Pa
b) mK/Pa
c) m2K / W
d) mK/W
View Answer
Explanation: When a heat exchanger is used with a fouling liquid, it leaves traces/deposits on the surface of the separating wall which reduces the overall heat transfer coefficient of the HE. This extra factor which reduces it is specified as the dirt factor (RD) or the fouling factor (Rf), which is represented as \(\frac{1}{U_D} = \frac{1}{U_D} + R_D\), as U has dimension W/m2K, hence we have dimension of R as 1/U, i.e. m2K/W.
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