2 Pump-pipeline Sys

7/30/2019 2 Pump-pipeline Sys

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Water Supply and WastewaterRemoval

Pump-Pipeline

Systems

Instructor: Parjang Monajemi


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Introduction

Simply stated, a pump is a machine used tomove liquid through a piping system and toraise the pressure of the liquid. A pump can befurther defined as a machine that uses severalenergy transformations to increase the pressure

of a liquid.

There are actually three distinct reasons forraising the pressure of a liquid with a pump:

1. Static elevation A liquids pressure must be

increased to raise the liquid from one elevationto a higher elevation. This might be necessary,for example, to move liquid from one floor of abuilding to a higher floor.

Water Supply and Wastewater Removal2 Spring 1390


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2. Friction. It is necessary to increase thepressure of a liquid to move the liquid througha piping system and overcome frictional losses.Liquid moving through a system of pipes,valves, and fittings experiences frictional losses

along the way.

3. Pressure. In some systems it is necessary toincrease the pressure of the liquid for processreasons. In addition to moving the liquid over

changes in elevation and through a pipingsystem, the pressure of a liquid must often beincreased to move the liquid into a pressurizedvessel, such as a boiler or fractionating tower,

or into a pressurized pipeline.Water Supply and Wastewater Removal3 Spring 1390


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Pressure and Head

It is important to understand the relationshipbetween pressure and head. Pressure ismeasured in psi (pounds per square inch) orkilopascal (kPa), bar, or kilograms per squarecentimeter (kg/cm2), while the equivalent units

for head are meters (m) or feet(ft).



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Classification of Pumps

There are many ways to classify pumps:according to their function, their conditions ofservice, materials of construction, etc. Thepump industry trade association, the HydraulicInstitute, has classified pumps as follows:

1. Kinetic: In a kinetic pump, energy iscontinuously added to the liquid to increase itsvelocity. When the liquid velocity issubsequently reduced, this produces a pressure

increase. Although there are several specialtypes of pumps that fall into this classification,for the most part this classification consists ofcentrifugal pumps.



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Centrifugal pumps, involve a collection ofblades, buckets, flow channels, or passagesarranged around an axis of rotation to form arotor. Rotation of the rotor produces dynamiceffects that either add energy to the fluid or

remove energy from the fluid. centrifugalpumps are classified as axial-flow, mixed-flow,or radial-flow machines depending on thepredominant direction of the fluid motion

relative to the rotors axis as the fluid passesthe blades



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Positive Displacement In a positivedisplacement pump, energy is periodicallyadded to the liquid by the direct application of aforce to one or more movable volumes of liquid.This causes an increase in pressure up to the

value required to move the liquid through portsin the discharge line. The important points hereare that the energy addition is periodic (i.e., notcontinuous) and that there is a direct

application of force to the liquid. Typicalexamples shown include the common tire pumpused to fill bicycle tires, the human heart, andthe gear pump.



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The following are some key application criteriathat would lead to the selection of a P.D. pumpover a centrifugal pump:a) High viscosity

b) Self-priming

c) High pressure

d) Low flow

e) High efficiency

f) Low velocity

g) Low shearh) Fragile solids handling capability

i) Accurate, repeatable flow measurement

j) Constant flow/variable system pressure

k) Two-phase flow



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(Volk, 2005)



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Cavitation in Turbomachines

Cavitation refers to conditions at certainlocations within the turbomachine where thelocal pressure drops to the vapor pressure ofthe liquid, and as a result, vapor filled cavitiesare formed. As the cavities are transported

through the turbomachine into regions ofgreater pressure,

they will collapse rapidly,generating extremely highlocalized pressures. Signs of

cavitation in turbopumpsinclude noise, vibration, andlowering of the head-discharge and efficiency

curves.Water Supply and Wastewater Removal10 Spring 1390


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On the suction side of a pump, low pressuresare commonly encountered, with the possibilityof cavitation occurring within the pump. Therequiered head at the pump inlet to keep theliquid from cavitating or boiling is called Net

Possitive Suction Head (NPSH).

Consider the shown operating pump. Location 1is on the liquid surface on the suction side, andlocation 2 is the point of minimum pressure

within the pump.

2

2

s s vrequired

p V pNPSH

g



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The design requirement for a pump is thusestablished as follows

22

1 1

1

2

1

2

2 2

2

2

s s

s loss

atm s ss loss

atm s sloss

atm vloss

p Vp V

z z hg g

p p Vz z h

g

p p Vz h

g

p pz h NPSH

atm vavailable loss

p pNPSH z h

atm vloss

p pNPSH z h



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A centrifugal pump is to be placed above alarge, open water tank, as shown, and is topump water at a rate of 0.5 cfs. At this flowratethe required net positive suction head, is 15 ft,as specified by the pump manufacturer. If the

water temperature is and atmospheric pressureis 14.7 psi, determine the maximum height,that the pump can be located above the watersurface without cavitation. Assume that the

major head loss between the tank and thepump inlet is due to filter at the pipe inlethaving a minor loss coefficient k=20. Otherlosses can be neglected. The pipe on the suctionside of the pump has a diameter of 4 in.

(Munson, 2009)Water Supply and Wastewater Removal13 Spring 1390


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23

14.7 0.506915 20

62.2 / 2

atm vloss

p pNPSH z h

psi Vz

lb ft g

2

3

0.5069 5.73 /15 20 7.65

62.2 / 2

atmp psi ft sz z ft

lb ft g



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Calculate NPSHa for this system and verify theadequacy of the selected pump. (Volk, 2005)

Suction lift=12ftDesign capacity Q=2000 gpm

Design pump total head=175 ft

Liquid=water at 80F (s.g.=1.0)

Hf=3ft

P=14.2psia



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Solve the previous problem, except using waterat 160F (Volk, 2005).

14.2 2.31

32.81.0

1.2

32.8 12 3 1.2 16.6

v

a

ft

H ft

NPSH ft

From previous figure, at 2000 gpm, NPSH = 11.2 ft,

16.6 11.2a rNPSH ft NPSH ft

14.2 2.3133.5

0.98

11.2

33.5 12 3 11.2 7.3

v

a

ft

H ft

NPSH ft

7.3 11.2a rNPSH ft NPSH ft



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Performance Characteristics

The flow system used to test a centrifugal pumpat a nominal speed of 1750 rpm is shown. Datameasured during the test are given in the table.Calculate the net head delivered and the pumpefficiency at a volume flow rate of 1000 gpm.

Plot the pump head, power input, and efficiencyas functions of volume flow rate. (Pritchard,2011)



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2 1p p ph

1 2 3

2 2 3

3

0.92 62.4 1 0.49

36.9 62.4 3 38.2

38.2 0.4989.2

62.4

s

d

p

lb lbp p z ft psi

in ft

lb lbp p z ft psi

in ft

h ftlb

ft



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Basic Output Parameters

Usually V2 and V1 are about the same, z2 z1 isno more than a meter or so, and the net pumphead is essentially equal to the change inpressure head

The power delivered to the fluid simply equalsthe specific weight times the discharge timesthe net head change This is traditionallycalled the water horsepower. The power

required to drive the pump is the brakehorsepower where is the shaftangular velocity and T the shaft torque.

2 1

p

p ph

P Qh

bhp T



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Water at 10 C (nu=1.3x10

-6

m

2

/s) is to flowfrom reservoir A to reservoir B through a cast-iron pipe of length 20 m at a rate of Q=0.015m3/s as is shown. The system contains a sharp-edged entrance and six regular threaded 90

elbows. Determine the the required head of thepump that must be used if the pipe diameter is5 cm. (Munson, 2009)



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3

2

0.015 /

7.65 /0.05

4

m sQ

V m sA

7.65 0.05Re 294231

0.0000013

VD VD

0.260.005

5

mm

D cm

,Re 294000 0.03f

2 2 2 2 2

2 0.5 6 1.5 1 12 0.5 9 1 65.8 63.82 2 2 2 2p p

L V V V V Vh f m h m

D g g g g g

2 2

1 1 1 11 2 1 1

2 2p p

P V P VE h E headlosses z h z headlosses

g g



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The given pump adds 25 kW to the water andcauses a flowrate of 0.04 m3/s. Determine theflowrate expected if the pump is removed fromthe system. Assume f=0.016 for either case andneglect minor losses. (Munson, 2009)

2 2 2 2

1 1 2 21 2 1 1

31.8 30 14.262.5 0.16 68.7

2 2 20 0.06 20

P v P vZ hp Z head loss Z Z m

g g

3 310,000 / 0.04 /

25000100%

62.5

N m m s HQHP W

H m

3 3

2 2 2 2

0.04 / 0.04 /14.2 / 31.8 /

0.06 0.044 4

pipe exitm s m sv m s v m s

m m



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222 21 1 2 21 2

22

30

68.7 0.162 2 20 0.06 20

3068.7 0.16

20 0.06 20

pipeexit

pipeexit

vvP v P v

Z Z head lossg g

vv

2 2 2 20.06 0.04 0.444 4

pipe exit pipe exitv m v m v v

22

2 2 3

0.443068.7 0.16 23 /

20 0.06 20

23 / 0.04 0.03 /4

exitexitexit

vvv m s

Q m s m m s



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The efficiency is basically composed of threeparts: volumetric, hydraulic, and mechanical.

1. The volumetric efficiency is where QLis the loss of fluid due to leakage in theimpeller-casing clearances

2. The hydraulic efficiency where hf hasthree parts: (1) shock loss at the eye due toimperfect match between inlet flow and theblade entrances, (2) friction losses in the blade

passages, and (3) circulation loss due toimperfect match at the exit side of the blades.

3. the mechanical efficiency is

v

L

Q

Q Q

1f

h

s

h

h

1f

m

p

php



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where Pf

is the power loss due to mechanicalfriction in the bearings, packing glands, andother contact points in the machine.

By definition, the total efficiency is simply theproduct of its three parts v h m

Performance characteristicsfor a given pump gometryand operating speed areusually given in the form of

plots of and bhp versus Qcommonly referred to ascapacity as is llustrated



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Water is to be pumped from one large, opentank to a second large, open tank as shown.The pipe diameter throughout is 6 in. and thetotal length of the pipe between the pipeentrance and exit is 200 ft. Minor loss

coefficients for the entrance, exit, and the elboware shown on the figure, and the friction factorfor the pipe can be assumed constant and equalto 0.02. A certain centrifugal pump having the

given performance characteristics is suggestedas a good pump for this flow system. With thispump, what would be the flowrate between thetanks? Do you think this pump would be a goodchoice?(Munson, 2009)



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Q H

0 89 0

400 86 30

800 81 53

1200 75 73

1600 65 86

2000 53 88

2400 28 60



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As can be seen, although the

operating efficiency is not the peak

efficiency, which is about 86%, it is

close (about 84%). Thus, this pump

would be a satisfactory choice



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Solve the previous problem if two pumps wereused (a) in series, (b) in parallel

Q H

0 89 0

400 86 30

800 81 53

1200 75 73

1600 65 86

2000 53 88

2400 28 60

Q H

0 178

400 172

800 162

1200 150

1600 130

2000 106

2400 56

Q H

0 89

800 86

1600 81

2400 75

3600 65

4000 53

4800 28

Single pump Two pumps in parallel Two pumps in series



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0

20

40

60

80

100

120

0 400 800 1200 1600 2000 2400

Head,

ft

Flowrate, gal/min

2 Parallel pumps

0

20

40

60

80

100

120

140

160

180

200

0 400 800 1200 1600 2000 2400

Head,

ft

Flowrate, gal/min

2 pumps in series

1780 / min

88%

Q gal

2100 / min

85%

Q gal


Dimensionless Parameters and


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Dimensionless Parameters andSimilarity Laws

As we studied earlier, we know that theprincipal, dependent pump variables are theactual head rise ha, shaft power shaft, andefficiency . We expect that these variables willdepend on the geometrical configuration, which

can be represented by some characteristicdiameter D, other pertinent lengths L, andsurface roughness , In addition, the otherimportant variables are flowrate Q, the pump

shaft rotational speed , fluid viscosity , andfluid density .



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When the pump flow involves high Reynoldsnumbers experience has shown that the effectof the Reynolds number can be neglected. For

simplicity, the relative roughness, can also beneglected in pumps since the highly irregularshape of the pump chamber is usually thedominant geometric factor rather than thesurface roughness. Thus, with these

2

ReD



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simplification and for geometrically similar pumps(all pertinent dimensions, scaled by a commonlength scale), the dependent pi terms arefunctions of only Q/D3 so that:

3 5

2 5

3

Power Coefficient

Head Coefficient

Flowrate Coefficient

W

H

Q

WCD

gHC

D

QC

D



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An 8-in diameter centrifugal pump operating at1200 rpm is geometrically similar to the 12-indiameter pump having the shown performancecharacteristics while operating at 1000 rpm. Forpeak efficiency, predict the discharge, actual

head rise, and shaft horsepower for this smallerpump. The working fluid is water at 60 F(Munson, 2009).



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For a given efficiency the flow coefficient has the same value for a given familyof pumps.



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A centrifugal pump provides a flowrate of 500gpm when operating at 1750 rpm against a200-ft head. Determine the pumps flowrate anddeveloped head if the pump speed is increasedto 3500 rpm. (Munson, 2009)

3

1 2 223 3 3 3

1 1 2 2 1 2

Flowrate Coefficient

5001000

1750 3500

Q

QC

D

Q Q QQ gpm

D D D D

2 5

1 2 222 5 2 5 2 2

1 1 2 2

Head Coefficient

200800

1750 3500

H

gHC

DgH gH H

H ftD D


Specific Speed, Suction Specific


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Specific Speed, Suction SpecificSpeed

A useful pi term named specific speed can beobtained by eliminating diameter D between theflow coefficient and the head rise coefficient.

With an analysis similar to that used to obtainthe specific speed pi term, the suction specificSpeed Ss, can be expressed as



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Select a pump to deliver 500 gal/min of waterwith a pressure rise of 65 psi. Assume arotational speed not to exceed 3600 rpm.

(Potter, 2012)

(Munson, 2009)

3

3600 377 /30

65 144150

1.94 32.2

5001.11 / sec

7.48 60

rad s

pHp ft

g

Q ft

3/ 4 3/ 4

377 1.110.69

32.2 150s

p

QN

gH



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References:1. Munson B.R., Young D. F., Okishi T. H., and Huebsch W. W.,

Fundamentals of Fluid Mechanics, John Wiley and Sons inc,Sixth Edition, 2009.

2. Potter M. C., Wiggert D. C., and Ramadan B. H., Mechanics ofFluids, Cengage Learning, Fourth Edition, 2012.

3. Pritchard P. J., Fox and McDonalds Introduction to FluidMechanics, John Wiley and Sons inc, Eighth Edition, 2011.

4. Volk M., Pump Characteristics and Applications, Taylor &Francis Group, Second Edition, 2005.

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2 Pump-pipeline Sys