# Heat Transfer Operations Questions and Answers – Shell and Tube Heat Exchangers – Pressure Drop Calculations

This set of Heat Transfer Operations Interview Questions and Answers for Experienced people focuses on “Shell and Tube Heat Exchangers – Pressure Drop Calculations”.

1. If UD = Overall Dirt Heat Transfer coefficient and Uc = Overall Clean Heat Transfer coefficient, then which of the following relation is correct?
a) UD > Uc
b) UD < Uc
c) UD = Uc
d) UD >> Uc

Explanation: In a shell and tube HE, fouling decreases the heat transfer rate hence the heat transfer coefficient decreases, thus the expression UD < Uc is correct.

2. Our design dirt factor should be _____ the provided threshold dirt factor.
a) Less than
b) More than
c) Equal to
d) Much less than

Explanation: The dirt factor calculated considering the area required and flow parameters, should come more than the dirt factor of the fluid else there may be unwanted deposits of sediments on the tubes.

3. We have a fluid of dirt factor 0.0035 which we want to use in our operation. Then during design considerations, we find that for our design the dirt factor is not applicable, this value of the dirt factor can be ________
a) 0.00035
b) 0.0035
c) 0.0045
d) 0.0055

Explanation: The dirt factor calculated considering the area required and flow parameters, should come more than the dirt factor of the fluid else there may be unwanted deposits of sediments on the tubes. As the value 0.00035 is very less than the given dirt factor of the fluid, our setup would not have a long life as there would be lot of sediments deposited to its tubes.

4. We have a fluid of dirt factor 0.00035 which we want to use in our operation. During design calculations we find the dirt factor to be less than that of the fluid, we pass the setup for industrial use.
a) True
b) False

Explanation: The dirt factor calculated considering the area required and flow parameters, should come more than the dirt factor of the fluid else there may be unwanted deposits of sediments on the tubes.

5. If we desire to use a fluid with high pressure operation, we prefer to keep it in the tube side.
a) True
b) False

Explanation: When we are required to use high pressure flow, we prefer to keep it at the tube side as keeping it at the shell side can damage the tubes, baffles, fins etc.

6. For the calculation of Friction Factor on the Shell side, which one of the following is the correct formula?
a) F = 0.0035+ $$\frac{0.264}{Re^{0.42}}$$
b) F = 0.0027+ $$\frac{0.264}{Re^{0.42}}$$
c) F = 0.0035+ $$\frac{0.42}{Re^{0.264}}$$
d) F = 0.0027+ $$\frac{0.42}{Re^{0.264}}$$

Explanation: Friction factor is a dimensionless quantity which signifies the frictional losses due to the wall in the flow of fluid, which contributes to pressure drop. The correct formula is F = 0.0035+ $$\frac{0.264}{Re^{0.042}}$$

7. Consider we have a 2-4 Shell and Tube Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and shell of inner diameter 100mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15m/s(tube) and 21 m/s(shell) respectively, friction factor of 0.00096 and length of the tube as 1m. What is the value of Frictional Head Loss for the inner tube?
a) 24.6 m
b) 27.6 m
c) 2.76 m
d) 2.46 m

Explanation: The Frictional head loss hfs can be calculated by the formula, hfs = $$\frac{FL}{D} \frac{v^2}{2g}$$ = 2.76 m, where we have F = 0.00096, L = 1×4m, D = 0.016m, v = 15m/s, g = 9.81m/s2.

8. Consider we have a 2-4 Shell and Tube Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and Shell of inner diameter 100mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15 m/s(tube) and 21 m/s(shell) respectively, friction factor of 0.00096 and 2 tubes of length 0.4 m. What is the value of Pressure drop for the inner tube due to frictional losses?
a) 10.67KPa
b) 20KPa
c) 25KPa
d) 24KPa

Explanation: The frictional head loss hfs can be calculated by the formula, hfs = $$(\frac{FL}{D}) \frac{v^2}{2g}$$ = 1.1 m, where we have F = 0.00096, L = 0.4×4 = 1.6m, D = 0.016m, v = 15m/s, g = 9.81m/s2. Hence Pressure Drop = hfs×density×g = 1.1×990×9.8 = 10.67 KPa.

9. What is the Pressure drop equivalent for 7 tube passes?
a) 50ρ $$\frac{v^2}{2}$$
b) 28ρ $$\frac{v^2}{2}$$
c) 20$$\frac{v^2}{2}$$
d) 7$$\frac{v^2}{2}$$

Explanation: The Pressure drop equivalent for a single tube pass is $$\frac{4ρv^2}{2}$$, hence for seven tube passes it would be 28ρ $$\frac{v^2}{2}$$.

10. Consider we have a 2-4Shell and Tube Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and shell of inner diameter 100mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15 m/s and 21 m/s respectively, friction factor of 0.00096 and 4 tube of length 0.4 m and 2shell passes. If the maximum pressure drop allowed on shell side is 10atm, is the setup suitable for industrial use?
a) No, since pressure drop is less than specified
b) Yes, since pressure drop is less than specified
c) No, since pressure drop is more than specified
d) Yes, since pressure drop is more than specified

Explanation: The hfs can be calculated by the formula, hfs=$$(\frac{FL}{D}+4×2)\frac{v^2}{2g}$$ = 180 m, where we have F = 0.00096, L = 0.4 m, D = 0.100 m, v = 21m/s, g = 9.81m/s2. Hence Pressure Drop = hfs×density×g = 180×990×9.8 = 1747198 Pa = 17atm >> 10atm hence it is not suitable.

11. In the calculation of Reynold’s Number for the Friction factor of the shell, which is the correct formula for equivalent diameter De?
a) $$\frac{4×(D_{oi}^2 – D_{io}^2)×\frac{π}{4}}{π(D_{io})}$$
b) $$\frac{4×(D_{io}^2 – D_{oi}^2)×\frac{π}{4}}{π(D_{io})}$$
c) Depends on the pitch
d) $$\frac{4×(D_{io}^2 – D_{oi}^2)×\frac{π}{4}}{π(D_{io}+D_{oi})}$$

Explanation: While friction factor is calculated for the shell, the wall friction is provided by both the sides of the tube. Hence when we are calculating the Hydraulic diameter for the shell we consider the pitch, i.e triangular pitch or square pitch.

12. Consider we have a shell and tube Heat Exchanger, with the inner tube of diameter 30mm (neglect thickness) and shell of diameter 100mm. We have two fluids A & B (both with viscosity 2.5×10-5Pa-s), we desire to have their flow rates as 4 Kg/s(tube) and 5.3 Kg/s(shell) respectively. What is the Friction Factor (F) for this setup for the shell if it has triangular pitch with Pitch = 40mm?
a) 0.0086
b) 0.0103
c) 0.086
d) 0.00103

Explanation: The Equivalent diameter for a triangular pitch is correctly given by the formula,
D=$$\frac{4(\frac{0.86}{2}Pt^2-\frac{\frac{π}{2}OD^2}{4})}{\frac{π}{2}OD} = \frac{4(\frac{0.86}{2}40^2-\frac{\frac{1}{2}30^2}{4})}{\frac{1}{2}π30}$$
We have Re = GD/µ, where D = 28.43 mm, hence
Re = 5.3×(28.43)×10-3/(2.5×10-5) = 6027.
Now the friction factor F can be calculated by F = 0.0035+ $$\frac{0.264}{Re^{0.42}}$$ = 0.0103.

13. If there are N shell passes in a shell and tube HE, then which one of the following relation is correct foe the pressure drop a shell side?
a) $$(\frac{FL}{D}+4×N)\frac{v^2ρ}{2}$$
b) $$(\frac{FL}{D}+N)\frac{v^2ρ}{2}$$
c) $$(\frac{4FL}{D}+N)\frac{v^2ρ}{2}$$
d) $$(\frac{4FL}{D}+4×N)\frac{v^2ρ}{2}$$

Explanation: For the shell side pressure drop of N shell passes we have the correct formula as
ΔP = $$(\frac{FL}{D}+4×N)\frac{v^2ρ}{2}$$, which is different from that double pipe heat exchanger when it comes to return losses.

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