250+ TOP MCQs on Shell and Tube Heat Exchangers – Log-Mean Temperature Difference and Correction Factor

Heat Transfer Operations Interview Questions and Answers on “Shell and Tube Heat Exchangers – Log-Mean Temperature Difference and Correction Factor”.

1. When using ΔTLM for a shell and tube heat exchanger, we cannot use the conventional formula for LMTD.
a) True
b) False

Answer: a
Clarification: In a shell and tube heat exchanger we observe both counter current and concurrent flow at different regions of the shell.

2. What should be the minimum value of the Correction Factor to get multiplied to LMTD?
a) >1
b) <2
c) <1
d) >2

Answer: c
Clarification: The value of corrected LMTD should be less than the theoretical LMTD considering countercurrent flow. Thus the LMTD correction factor is less than 1.

3. Which of the following has the Log mean temperature difference is used for corrected LMTD for a Shell and tube Heat Exchanger?
a) Counter-flow
b) Parallel Flow
c) Cross Flow
d) Split Flow

Answer: a
Clarification: The mean temperature difference for a counter-flow operation is uniform throughout the length of the tube and hence gives the maximum value to the LMTD. Thus as Ft < 1, we get correct LMTD from this.

4. What should be the minimum value of LMTD Correction Factor F for steady state operation of the equipment?
a) <0.5
b) <1
c) 1 > F > 0.75
d) 1 > F > 0.5

Answer: c
Clarification: The value of corrected LMTD should be less than the theoretical LMTD considering countercurrent flow. Thus the LMTD correction factor is less than 1. Also for steady state operation of the HE, its value should be more than 0.75.

5. Why do we use LMTD Correction factor?
a) Incomplete Countercurrent flow
b) Leakage leading to heat loss
c) Counter current and Co-current at parts
d) Due to fouling

Answer: c
Clarification: In a shell and tube heat exchanger we observe both counter current and concurrent flow at different regions of the shell. This phenomenon makes the use of complete countercurrent LMTD unsuitable for calculations.

6. If we are given F value and R value, which is the only one variable we cannot determine by this information?
a) Value of P
b) Temperature of inlet and outlet of shell
c) Temperature of inlet and outlet of tube
d) Temperature of inlet and outlet of shell and tube

Answer: d
Clarification: Given R and F we can find P easily from F-curve, then we will be having 2 equations hence we can have maximum two variable to be able to do the calculations, hence, the incorrect option is temperature of inlet and outlet of shell and tube which has 4 variables which cannot be determined.

7. The value of correction factor can be equal to 1 in most cases of operation.
a) True
b) False

Answer: b
Clarification: In a shell and tube heat exchanger for most industrial purposes, it never achieves complete counter current flow regime for which the value of F is always < 1.

8. In a shell and tube heat exchanger, in the inner side fluid enters at 15℃ and leaves at 65℃. The shell has steam at 1atm. What is the value of corrected LMTD and correction factor?
a) 39℃, F = 0
b) 56.35℃, F = 0
c) 70℃, F = 1
d) 56.35℃, F = 1

Answer: d
Clarification: As steam is condensing at 100℃ (Tvap at 1atm) hence we have T1=T2, thus R = 0 and F = 1; we have
LMTD = (frac{(TH1-TC2)-(TH2-TC1)}{Ln(frac{TH1-TC2}{TH2-TC1})})
LMTD = (frac{(100-15)-(100-65)}{Ln(frac{100-15}{100-65})}) = 56.35℃.

9. Which one of the following are correct relation of R and P?
(T/t =shell/tube, 12 = inlet/outlet)
a) R = (frac{T_2-T_1}{t_2-t_1}), P = (frac{t_2-t_1}{T_2-T_2})
b) R = (frac{T_2-T_1}{t_2- t_1}), P = (frac{t_2- t_1}{T_1- t_1})
c) R = (frac{T_2-T_1}{t_2- t_1}), P = (frac{t_2- t_1}{T_2- T_1})
d) R = (frac{T_1-T_2}{t_2- t_1}), P = (frac{t_2- t_1}{T_1- t_1})

Answer: d
Clarification: R and P are temperature factors which are required to calculate the Correction factor from the F-curve chart. The correct formula for them is R = (frac{T_1-T_2}{t_2- t_1}), P = (frac{t_2- t_1}{T_1- t_1}).

10. In a shell and tube heat exchanger, in the inner side fluid enters at 15℃ and leaves at 65℃. The shell has oil entering at 105℃ and leaving at 85℃. What is the value of correction factor coefficients R and P?
a) R = 0.4, P = 0.55
b) R = 0.3, P = 0.55
c) R = 0.4, P = 0.66
d) R = 0.3, P = 0.66

Answer: a
Clarification: R and P are temperature factors which are required to calculate the Correction factor from the F-curve chart. The correct formula for them is R = (frac{T_1-T_2}{t_2- t_1}), P = (frac{t_2- t_1}{T_1- t_1}) hence putting the values we get R = (frac{105-85}{65-15}) = 0.4 and P = (frac{65 – 15}{105-15}) = 0.55.

12. What are the axes of the F-curve?
a) X – Correction Factor, Y – R
b) X – Correction Factor, Y – mean of P and R
c) Y – Correction Factor, X – R
d) Y – Correction Factor, X – P

Answer: d
Clarification: The F curve is a plot between the correction factor F and the coefficient P with different concave downward curves for different values of R ranging anywhere between the 0.2 to 5.

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