1. 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.
2. 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.
3. What are the three parts of Extended Surface heat exchangers when it is classified based on flow pattern?
a) Compound flow, Plate Fin and Cross Counter Flow
b) Compound flow, Tube Fin and Cross Counter Flow
c) Compound flow, Tube Fin and Plate Fin
d) Compound flow, Cross Parallel and Cross Counter Flow
Explanation: When classified based on flow pattern, they can be divided to 3 parts, they are – Compound flow, Cross Parallel and Cross Counter Flow and when classified based on construction can be divided to 2 parts, they are – Plate Fin and Tube Fin.
4. Extended Surface heat exchangers are a subset of the broader category of _____ flow type Heat Exchangers.
a) Multi Pass
b) Single Pass
c) Compound
d) Split
Explanation: An extended surface HE is a subset of the multipass flow arrangement. When classified based on flow pattern under multipass flow, they can be divided to 3 parts, they are – Compound flow, Cross Parallel and Cross Counter Flow.
5. Compound flow Extended surface heat exchangers are superior to Counter flow operation.
a) True
b) False
Explanation: An extended surface HE is always superior to a normal HE. When classified based on flow pattern can be divided to 3 parts, they are – Compound flow, Cross Parallel and Cross Counter Flow.
6. In case of Classification based on construction, extended surface heat exchangers can be classified into _________
a) Plate Fin and Tube Fin
b) Cross Parallel and Plate Fin
c) Tube Fin and Cross Parallel
d) Compound flow and Tube Fin
Explanation: An extended surface HE when classified based on construction can be divided to 2 parts, they are – Plate Fin and Tube Fin.
7. Plate type Extended Fins HE is a classification based on the _____ of the heat exchanger.
a) Construction
b) Flow pattern
c) Phase pattern
d) Heat transfer mechanism
Explanation: An extended surface HE when classified based on construction can be divided to 2 parts, they are – Plate Fin and Tube Fin. Plate type Extended Fins HE is a classification based on the construction of the heat exchanger
8. Compound flow Fins extended surface heat exchangers are classification based on______
a) Construction
b) Flow pattern
c) Phase pattern
d) Heat transfer mechanism
Explanation: An extended surface HE when classified based on flow pattern can be divided to 3 parts, they are – Compound flow, Cross Parallel and Cross Counter Flow.
9. Fins primarily increase the ______
a) Turbulence
b) Heat transfer area
c) Heat transfer rate
d) Heat transfer coefficient
Explanation: Fins are primarily used to increase the heat transfer area, the rest of the factors given are consequences of this change.
10. Which of the following are not affected by adding Fins?
a) Net Heat Transfer
b) Heat transfer coefficient
c) Heat transfer rate
d) Turbulence
Explanation: The net heat transfer in a particular operation should remain the same, fins are used to decrease the area and time for that process.