# Heat Transfer Operations Questions and Answers – Double Pipe Heat Exchangers – Pressure Drop Calculations

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This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Double Pipe Heat Exchangers – Pressure Drop Calculations”.

1. If the pressure drop for an operation is more than the pressure assigned to the equipment, then the equipment is perfect to be used for that operation.
a) True
b) False

Explanation: In a double pipe heat exchanger we can use the pressure drop less than or equal to the maximum pressure assigned to the equipment, else it may result in high maintenance cost for the equipment and early wear and tear of the HE.

2. Consider we have a Double Pipe Heat Exchanger, with the inner tube of diameter 20mm (neglect thickness) and outer tube of diameter 30mm. We have two fluids A & B (both with viscosity 2.5×10-5Pa-s), we desire to have their flow rates as 15Kg/s and 21Kg/s respectively. What is the Friction Factor (F) for this setup for the inner pipe?
a) 0.0086
b) 0.086
c) 0.0068
d) 0.068

Explanation: We have Re = GD/μ hence Re= 15×20×103/(2.5×10-5) = 12000.
Now the friction factor F can be calculated by F = 0.0035+$$\frac{0.264}{Re^{0.042}}$$ = 0.0086.

3. Which of the following has the maximum Pressure Drop for a Heat Exchanger?
a) Counter-flow
b) Parallel Flow
c) Hairpins arrangement
d) Split Flow

Explanation: The arrangement with hairpins in the heat exchangers has higher pressure drop as the 180 degree turn increases the pressure drop.
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4. How many times do we calculate pressure drop for a double pipe heat exchanger?
a) 1
b) 2
c) 3
d) 4

Explanation: We calculate the Pressure drop for both the annulus as well as the inner tube and verify it with the critical values whether operation at that pressure is possible or not.

5. Consider we have a Double Pipe Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and outer tube of inner diameter 30mm. We have two fluids A & B, we desire to have their flow rates as 15 m/s and 21 m/s respectively, friction factor of 0.0096 and length of the tube as 1metre. What is the value of hfs for the inner tube?
a) 27.6m
b) 15m
c) 21.12m
d) 22m

Explanation: The hfs can be calculated by the formula, hfs = $$\frac{4FL}{D} \frac{Lv^2}{2g}$$ = 27.6 m, where we have F = 0.0096, L = 1m, D = 0.016m, v = 15m/s, g = 9.81m/s2.
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6. In the calculation of Friction Factor on the annulus/tube side for heat transferred, which is the correct formula for Friction Factor?
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+ $$\frac{0.264}{Re^{0.42}}$$.

7. Consider we have a Double Pipe Heat Exchanger, with the inner tube of Outer diameter 20mm(thickness 2mm) and outer tube of inner diameter 30mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15m/s and 21 m/s respectively, friction factor of 0.00096 and length of the tube as 1metre. What is the value of Pressure drop for the inner tube?
a) 25 KPa
b) 20 KPa
c) 27 KPa
d) 35 KPa

Explanation: The hfs can be calculated by the formula, head loss hfs = $$\frac{4FL}{D} \frac{v^2}{2g}$$ = 2.76 m, where we have F = 0.00096, L = 1m, D = 0.016m, v = 15m/s, g = 9.81m/s2. Hence Pressure Drop = hfs×density×g = 2.76×990×9.8 = 27 KPa.

8. For calculating the pressure drop in a double pipe heat exchanger,
(i) We calculate Reynolds number
(ii) We calculate the length of the tube required
(iii) We calculate heat transfer coefficient for both sides
(iv) We calculate the Friction factor for both sides
a) (ii)(iii)
b) (i)(ii)(iv)
c) (i)(ii)
d) (i)(ii)(iii)

Explanation: The steps to calculate pressure drop for a HE are-
i. Reynold’s number for all sides.
ii. Friction Factors.
iii. Pressure Drop. By the formula = hfs×density×g, where hfs = $$\frac{4FL}{D} \frac{v^2}{2g}$$.

9. When we calculate the pressure drop ΔP = $$\frac{XFL}{D} \frac{v^2}{2g}$$×ϸ, what is the value X?
a) 2
b) 4
c) 5
d) 8

Explanation: When we are required to calculate the Pressure Drop for a double pipe setup, the correct formula is ΔP = $$\frac{4FL}{D} \frac{v^2}{2g}$$×ϸ, where F = friction factor and ϸ is the density.

10. Consider we have a Double Pipe Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and outer tube of inner diameter 30mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15m/s and 21m/s respectively, friction factor of 0.00096 and 5 hairpins of length 0.4 metre. What is the value of Pressure drop for the inner tube?
a) 6atm
b) 2atm
c) 5atm
d) 4atm

Explanation: The hfs can be calculated by the formula, hfs = $$(\frac{4FL}{D}+5) \frac{v^2}{2g}$$ = 62.8 m, where we have F = 0.00096, L = 0.4×5 = 2m, D = 0.016m, v = 15m/s, g = 9.81m/s2. Hence Pressure Drop = hfs×density×g = 62.8×990×9.8 = 609 KPa = 6atm.

11. What is the Pressure drop equivalent for 7 Hairpins?
a) 5ρ $$\frac{v^2}{2}$$
b) 7ρ $$\frac{v^2}{2}$$
c) 5$$\frac{v^2}{2}$$
d) 7$$\frac{v^2}{2}$$

Explanation: The Pressure drop equivalent for a single Hairpin is $$\frac{v^2}{2}$$, hence for seven hairpins it would be 7ρ $$\frac{v^2}{2}$$.

12. Consider we have a Double Pipe Heat Exchanger, with the inner tube of Outer diameter 20mm (thickness 2mm) and outer tube of inner diameter 30mm. We have two fluids A & B, both of density 990 Kg/m3, we desire to have their flow rates as 15m/s and 21m/s respectively, friction factor of 0.0096 and 5 hairpins of length 0.4 metre. If the maximum pressure drop allowed is 10psi, 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 = $$(\frac{4FL}{D}+5) \frac{v^2}{2g}$$ = 112.5 m, where we have F = 0.0096, L = 0.4×5 = 2m, D = 0.016m, v = 15m/s, g = 9.81m/s2. Hence Pressure Drop = hfs×density×g = 112.5×990×9.8 = 1091.475 KPa = 158 PSI >> 10PSI hence it is not suitable.

13. In the calculation of Reynold’s Number for the Friction factor of the annulus tube, 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) $$\frac{4×(D_{oi}^2 – D_{io}^2)×\frac{π}{4}}{π(D_{io}+D_{oi})}$$
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 annulus pipe, the wall friction is provided by both the sides of the tube. Hence when we are calculating the Hydraulic diameter for the annulus, we consider the complete wetted perimeter, that is, P = π(Dio + Doi).

14. Consider we have a Double Pipe Heat Exchanger, with the inner tube of diameter 20mm (neglect thickness) and outer tube of diameter 30mm. We have two fluids A & B (both with viscosity 2.5×10-5Pa-s), we desire to have their flow rates as 15 Kg/s and 21 Kg/s respectively. What is the Friction Factor (F) for this setup for the annulus pipe?
a) 0.0086
b) 0.0103
c) 0.086
d) 0.00103

Explanation: We have Re = GD/µ, where D = $$\frac{4×(D_{io}^2 – D_{oi}^2)×\frac{π}{4}}{π(D_{io}+D_{oi})}$$ = Doi – Dio, hence
Re = 15×(30 – 20)×10-3/(2.5×10-5) = 6000.
Now the friction factor F can be calculated by F = $$0.0035+\frac{0.264}{Re^{0.42}}$$ = 0.0103.

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