pump sizing
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Essential formulas and examples for pump sizingTRANSCRIPT
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Method 2, Equivalent Pipe Diameters
Extra length of pipe to allow for miscellaneous losses
735 25 103 18:4mSo, total length for DP calculation 120 18.4 138.4 m
DPf 8 0:0032(138:4)
(25 103) 9981:982
2 277,247N=m2
277 kN=m2(5:3a)
Note: The twomethods will not give exactly the same result. The method using velocity heads isthe more fundamentally correct approach, but the use of equivalent diameters is easier toapply and sufficiently accurate for use in preliminary design calculations.
5.4.3. Power Requirements for Pumping Liquids
To transport a liquid from one vessel to another through a pipeline, energy has to besupplied to
1. Overcome the friction losses in the pipes;2. Overcome the miscellaneous losses in the pipe fittings (e.g., bends), valves,
instruments etc.;3. Overcome the losses in process equipment (e.g., heat exchangers, packed beds);4. Overcome any difference in elevation from end to end of the pipe;5. Overcome any difference in pressure between the vessels at each end of the
pipeline.
The total energy required can be calculated from the energy equation:
gDz DP=r DPf=r W 0where
W work done by the fluid, J/kg;Dz difference in elevations (z1 z2), m;DP difference in system pressures (P1 P2), N=m2;DPf pressure drop due to friction, including miscellaneous losses, and equipment
losses (see section 5.4.2), N=m2;r liquid density kg=m3;g acceleration due to gravity, m=s2.
If W is negative, a pump is required; if it is positive, a turbine could be installed toextract energy from the system.
The head required from the pump DPf=rg DP=rg Dz (5:5a)The power is given by
Power (W m)=h, for a pump (5:6a)and (W m) h, for a turbine (5:6b)
5.4. PUMPS AND COMPRESSORS 251
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EBSCO Publishing - NetLibrary; printed on 5/9/2011 11:25:40 PM via Royal Melbourne Institute of TechnologyeISBN:9780750684231; Towler, Gavin P.; Sinnott, R. K. : Chemical Engineering Design Account: -277897657
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where
m mass flow rate, kg/s;h efficiency power out/power in.The efficiency will depend on the type of pump used and the operating conditions.
For preliminary design calculations, the efficiency of centrifugal pumps can be estimatedusing Figure 5.13.
P1
P2
Z1
Z2
LiquidLevel
Vessel 1Vessel 2
Datum
Pump
Figure 5.12. Piping system.
75
70
65
60
55
50
4510 20 30 40 50 60 70 80 90
200125
100
75
50
25
Head, m
Effic
ienc
y, %
Capa
city,
m3 /h
Figure 5.13. Centrifugal pump efficiency.
252 CHAPTER 5 PIPING AND INSTRUMENTATION
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Example 5.2
A tanker carrying toluene is unloaded, using the ships pumps, to an on-shore storagetank. The pipeline is 225mm internal diameter and 900m long. Miscellaneous lossesdue to fittings, valves, etc., amount to 600 equivalent pipe diameters. The maximumliquid level in the storage tank is 30m above the lowest level in the ships tanks. Theships tanks are nitrogen blanketed and maintained at a pressure of 1.05 bar.The storage tank has a floating roof, which exerts a pressure of 1.1 bar on the liquid.
The ship must unload 1000 metric tons (tonnes) within 5 hours to avoid demurragecharges. Estimate the power required by the pump. Take the pump efficiency as 70%.
Physical properties of toluene: density 874 kg=m3, viscosity 0:62mNm2 s.
Solution
Cross-sectional area of pipe p4(225 103)2 0:0398m2
Minimum fluid velocity 1000 103
5 3600 1
0:0398 1874
1:6m=s
Reynolds number (874 1:6 225 103)=0:62 103
507,484 5:1 105 (5:4a)
Absolute roughness commercial steel pipe, Table 5.2 0.046 mmRelative roughness 0.046/225 0.0002Friction factor from Figure 5.11, f 0.0019Total length of pipeline, including miscellaneous losses,
900 600 225 103 1035m
Friction loss in pipeline, DPf 8 0:00191035
225 103
874 1:622
2
78,221N=m2(5:3a)
Maximum difference in elevation, (z1 z2) (0 30) 30mPressure difference, (P1 P2) (1:05 1:1)105 5 103 N=m2
Energy balance
9:8(30) (5 103)=874 (78,221)=874W 0 (5:5)W 389:2 J=kg
Power (389:2 55:56)=0:7 30,981W, say 31 kW: (5:6a)
Note that this is the maximum power required by the pump at the end of theunloading when the ships tank is nearly empty and the storage tank is nearly full.Initially, the difference in elevation is lower and the power required is reduced. Fordesign purposes the maximum power case would be the governing case and would beused to size the pump and motor.
5.4. PUMPS AND COMPRESSORS 253
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EBSCO Publishing - NetLibrary; printed on 5/9/2011 11:29:17 PM via Royal Melbourne Institute of TechnologyeISBN:9780750684231; Towler, Gavin P.; Sinnott, R. K. : Chemical Engineering Design Account: -277897657
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