1. 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.
2. 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+ \(\frac{0.264}{Re^{0.42}}\).
3. 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.
4. 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 = \(\frac{4FL}{D} \frac{v^2}{2g}\).
5. 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.
6. 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.
7. 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 \(\frac{v^2}{2}\).
8. 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.
9. 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).
10. 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.