Heat Transfer Operations Questions and Answers – Double Pipe Heat Exchangers – Heat Transfer Coefficients Calculations

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This set of Heat Transfer Operations test focuses on “Double Pipe Heat Exchangers – Heat Transfer Coefficients Calculations”.

1. For the calculation of Reynolds number in annulus side, we simply take the difference of the radius of the two tubes.
a) True
b) False

Explanation: In a double pipe heat exchanger we can use the difference of diameter for Reynolds number calculation as we find out that De = 4A/P = Do – Di.

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 15 Kg/s and 21 Kg/s respectively. What is the Reynolds number for this setup for the inner pipe?
a) 12000
b) 13000
c) 15000
d) 14000

Explanation: We have Re = GD/μ hence Re = 15×20×103/(2.5×10-5) = 12000.

3. Which of the following has the maximum Heat transfer rate for a Double Pipe Heat Exchanger?
a) Counter-flow
b) Parallel Flow
c) Cross Flow
d) Split Flow

Explanation: The mean temperature difference for a counter-flow operation is uniform throughout the length of the tube and hence gives a higher value to the LMTD. Hence as we have constant heat transfer coefficient for given equipment, more LMTD signifies higher transfer rate.

4. How many times do we calculate effective diameter/hydraulic diameter for a double pipe heat exchanger?
a) 1
b) 2
c) 3
d) 4

Explanation: We calculate the hydraulic diameter only for the outer annulus tube as its flow area is different than that of the inner tube.

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 15Kg/s and 21Kg/s respectively. What is the value of Hydraulic diameter for the annulus tube?
a) 20mm
b) 15mm
c) 25mm
d) 10mm

Explanation: The Hydraulic diameter for the annulus tube is De = Doi – Dio, which is De = 30-20 = 10mm.

6. In the calculation of Heat Transfer Coefficient on the annulus side for heat transferred to the inner 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: When the calculation is for Heat Transfer Coefficient on the annulus side to the inner tube, the wetted perimeter becomes the perimeter of the outer surface of the inner tube, so we use only πDio as the wetted perimeter.

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, we desire to have their flow rates as 15Kg/s and 21Kg/s respectively. Calculate the equivalent diameter dor the calculation of convective heat transfer coefficient for the ANNULUS side
a) 10mm
b) 20mm
c) 25mm
d) 35mm

Explanation: When the calculation is for Heat Transfer Coefficient on the annulus side to the inner tube, the wetted perimeter becomes the perimeter of the outer surface of the inner tube, so we use only πDio as the wetted perimeter. Hence the formula, De = $$\frac{4×(D_{oi}^2 – D_{io}^2)×\frac{π}{4}}{π(D_{io})}$$ = (302 – 202)/20 = 25mm.

8. We can apply the formula for Hydraulic diameter De = $$\frac{4×(D_{io}^2 – D_{oi}^2)×\frac{π}{4}}{π(D_{io}+D_{oi})}$$ only when _________
(i) We are calculating Reynolds number for the annulus tube
(ii) We are calculating Reynolds number for the inner tube
(iii) We are calculating heat transfer coefficient for the annulus tube
a) (ii)(iii)
b) (i)
c) (i)(ii)
d) (i)(ii)(iii)

Explanation: When we are calculating the Hydraulic diameter for the annulus, we consider the complete wetted perimeter, that is, P = π(Dio + Doi). Whereas when we are calculating the same for the Heat Transfer Coefficient on the annulus side to the inner tube, the wetted perimeter becomes the perimeter of the outer surface of the inner tube, so we use only πDio as P.

9. When we calculate the overall heat transfer coefficient U= $$\frac{X×ho}{X+ho}$$, what is the value X?
a) hi, heat transfer coefficient for inner tube for inner side
b) hio, heat transfer coefficient for inner tube for outer side
c) hoi, heat transfer coefficient for annulus for inner side
d) ho, heat transfer coefficient for annulus side

Explanation: When we are required to calculate the overall heat transfer coefficient U, we calculate it for the direction of interaction of the fluid’s heat energy, hence we calculate it with hio and ho, and not with hi which is tube’s inner side heat transfer coefficient.

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, we desire to have their flow rates as 15 Kg/s and 21 Kg/s respectively. Calculate the convective heat transfer coefficient from tube to annulus side. (Given Nusselt number on tube side = 496 and K = 15W/mK)
a) 2067 W/m2K
b) 1653 W/m2K
c) 1623 W/m2K
d) 1673 W/m2K

Explanation: We have been given Nu = 496 = hD/K or h = 496×15/0.016 = 2067 W/m2K.
But we are required to find tube side for annulus HT coefficient which is hiDi = hoDo or ho = hiDi/Do = 2067×16/20 = 1653 W/m2K.

11. To calculate the efficiency of a double pipe heat exchanger, the formula is
€ = $$\frac{X}{Maximum \, possible \, rate \, of \, Heat \, Transfer}$$, where X stands for _________
a) Clean tube rate of heat transfer
b) Experimental rate of heat transfer
c) Actual rate of heat transfer
d) Minimum rate of heat transfer

Explanation: The efficiency of a double pipe heat exchanger which is calculated by NTU or number of transfer units is given by € = $$\frac{Actual \,Heat\, transfer\, rate}{Maximum \, possible \, rate \, of \,Heat\, Transfer}$$. Here x is Actual rate of heat transfer.

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. What is the overall heat transfer coefficient if the convective heat transfer coefficient for annulus side is 496 W/m2K and for the inner tube is 530 W/m2K?
a) 228.6 W/m2K
b) 220 W/m2K
c) 256.2 W/m2K
d) 212 W/m2K

Explanation: Given hi and ho, to calculate U, we need to have hio and ho. Hence for hio we use hiDi = hioDo or hio = hiDi/Do = 530×16/20 = 424 W/m2K. Hence, U = $$\frac{hio×ho}{hio+ho}$$ = 228.6 W/m2K.

13. What is the value of Prandtl number if the value of Nusselt number is 496 and Reynolds number is 120000?
a) 4.67
b) 5.67
c) 4
d) 7.3

Explanation: As we are aware of the equation, Nu = 0.027×(Re)0.8×(Pr)0.33. Substituting the values we get the value of Pr = 4.67.

14. What is the unit of Fouling factor (Rf) in F.P.S system?
a) Btu/hr ft2
b) Btu/hr ℉
c) Btu/hr P℉
d) Btu/hr P2
Explanation: The fouling factor (Rf), which is resupsented as $$\frac{1}{U_D} = \frac{1}{U_C} + R_D$$, hence we have dimension of R as 1/U, i.e. Btu/hr ft2℉.