pipe system design

7/29/2019 Pipe System Design

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Fall 2009 CE154 1

Pipe System DesignCE154 - Hydraulic Design

Lecture 7


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Pump System

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Fall 2009 CE154 3

Pump Terminology

Pump head (dynamic head) H

Pump discharge Q

Pump speed n Pump power P


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Centrifugal Pump

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Rotary Pump

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Reciprocating Pump

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Fall 2009 CE154 7

Pump Terminology

Power P

Motor

efficiency m

PumpEfficiency p

Q, H


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Fall 2009 CE154 8

Pump Terminology

Pump Output (Water) Power (Q in gpm, H inft, s is specific gravity and dimensionless, andP in horsepower)

Pump Input (Brake) Power

3960QHsPw

p

QHsbhp3960


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Fall 2009 CE154 9

Pump Terminology

Electric Motor Power

Typically motor efficiency is approximately

98%

mp

QHsmhp

3960


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Fall 2009 CE154 10

Pump Terminology

Single or Double suction pump Single or multiple stage pump


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Fall 2009 CE154 11

Pump Specific Speed

Q is gpm per suction, n is rpm, H is ft per stage

Pump Type Ns (US unit)

Radial Pump 500-4200

Mixed flow Pump 4200-9000Axial flow Pump 9000-15000

HQN

ns 75.0

5.0


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Fall 2009 CE154 12

Pump Performance

Variable-Speed pumps may bedesirable when different operatingmodes require different pump head orflow

Homologous lawsQ1/Q2 = n1/n2

H1/H2 = (n1/n2)2P1/P2 = (n1/n2)3


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Pump Performance Curves

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Fall 2009 CE154 14

Pump Terminology Static Lift elevation difference

between pump centerline and thesuction water surface. If the pump ishigher, static lift is positive. If pump is

lower, static lift is negative. Static Discharge elevation differencebetween the pump centerline and theend discharge point. If pump is higher,static discharge is negative.

Total Static Head sum of static liftand static discharge.


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Fall 2009 CE154 15

Pump Terminology

Shutoff Head head at 0 flow

Operating point the point where thepump curve and the system curveintersect. A system curveis a curvedescribing the head-flow relationshipof the pipeline system.

QbaH 2


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Fall 2009 CE154 16

System Curve

friction losses


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Fall 2009 CE154 17

Operating Point


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Fall 2009 CE154 18

Pump Terminology

Net positive suction head (NPSH)- to ensure that water does not vaporize atthe pump impeller tip

- NPSH available = available suction head +atmospheric pressure vapor pressure suction head loss = determined by localcondition

- NPSH required = characteristic of andprovided by pump curve

hhhhNPSH svpsatmA


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Fall 2009 CE154 19

Example (p. 10.3 Mays HDH)


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Fall 2009 CE154 20

Atmospheric Pressure

At mean sea level,1 atomospheric pressure = 14.69 psia

= 1.03 kg/cm2

= 760 mm HgEl.(ft) 0 1000 2000 4000 6000 8000

Patm(psia) 14.69 14.17 13.66 12.69 11.8 10.9


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Fall 2009 CE154 21

Vapor pressure

Vapor pressure of water at atmosphericpressure

T

(F)

32 40 50 60 68 80 90

Pv(psia)

0.089 0.112 0.178 0.256 0.339 0.507 0.698

T(F) 100 120 140 160 180 200 212

Pv(psia)

0.949 1.69 2.89 4.74 7.51 11.53 14.70


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Fall 2009 CE154 22

Pipe System Design

Determine system curve- identify all loss-incurring elementsincluding friction, transitions, fittings,

valves and other special equipment inthe system (use Darcy-Weisbachequation to compute friction loss)

- identify suction and dischargeconditions


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Fall 2009 CE154 23

Design Scenario Ex. 8-1

Need to deliver water from SantaTeresa Water Treatment Plant atelevation 300 ft to Cisco Power Plant atelevation 330 ft. The maximum flow

demand is 200 cfs. Design pipe size andregulating devices for operation. Basic design approach

- consider steady-state, governing case- design pipe size and control equipment- check transientcondition to verifypipe pressure class


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Fall 2009 CE154 24

Example 8-1

1. Prepare a list of available design data:- Design discharge 200 cfs- Suction water elevation El. 300 ft

North Am. Vertical Datum (NAVD88 )- Discharge water elevation El. 330 ftNAVD88

2. Prepare a list of data requirements:- Topographic maps to route pipeline- Right of way maps- Utility and road crossing maps


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Fall 2009 CE154 25

Example 8-1

- Select a pump station site to setpreliminary pump elevation:

-- assume 2 Pumps set at El. 290 ft

-- station design includes 40 ft ofpipeline, 4 isolation valves and 2 pumpdischarge valves

- No other information on discharge siteis needed (for hydraulic design), otherthan the maximum tank level at 330 ft.


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Fall 2009 CE154 26

Example 8-1

3. Assume that we have selected a routethat results in a pipeline 8.0 miles long,and the fittings include- 80 bends of 90,- 25 bends of 45,- 10 butterfly valves for isolation

4. Determine pipe diameter- rule of thumb maintain operatingvelocity at 6-8 ft/sec (h V2,consider economic analysis later)


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Fall 2009 CE154 27

Example 8-1 Pipe Sizing

Design maximum discharge Q = 200 cfsassume V = 8 ft/secA = 200/8 = 25 ft2

D = 5.64 ft = 67.7 inchesSay D = 72 inch why?A = 28.27 ft2V = 7.1 ft/sec

Next, determine pipe material. For thissize, steel and reinforced concretepipes are available. Say RCP.


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Fall 2009 CE154 28

Example 8-1


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Fall 2009 CE154 29

Example 8-1 Friction loss

From roughness chart, for 72 concretepipes, in the mid roughness range, i.e., notnew but not seriously corroded,

e/D = 0.0005 f=0.017 From equation for turbulent rough flows,

f= 0.017 O.K.

D

e

f

log214.11


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Fall 2009 CE154 30


When the pipe is significantly corroded,the chart shows that

e/D = 0.0017 and f=0.023

When new pipe, the chart showse/D = 0.000165 and f=0.0135

Use the mid roughness values fordesign, and use the new pipe and roughpipe values to check the system designto ensure operability.


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Fall 2009 CE154 31


Compute friction loss:h = f L/D V2/2g

= 0.017 x 8 x 5280 / 6 x V2/2g

= 119.7 V2/2g= 119.7 x (7.1)2/2/32.2= 93.7 ft

Compute minor losses:Size the butterfly valves:

Use Val Matic valve data


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Fall 2009 CE154 32

Example 8-1 Valve data

Val Matic butterfly valve data


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Fall 2009 CE154 33

Example 8-1 Valve sizing

Q=Cv(P)1/2

Q = 200 cfs x 448.8 gpm/cfs = 89760 gpm

Considerations:- use the smallest size of valve that can

pass the design flow with acceptable headlosses

Valve size (in) Fully-open Q (gpm) P (psi)

72 266500 142 87100 1

54 144000 1


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Fall 2009 CE154 34

Example 8-1 Valve loss

For this example, lets try 72 valves. At 266500 gpm, the valve incurs 1 psiof loss.In terms of

Q = 266500 gpm = 593.8 cfsA = 28.27 ft2, V = 21.0 fps,

h = 1 psi = 2.31 ft, kv = 0.337 At 200 cfs, V=7.1 fps, h = 0.26 ft pervalve

gkh V2

2


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Fall 2009 CE154 35

Example 8-1(Val Matic BFV)


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Fall 2009 CE154 36

Example 8-1 Minor loss

Compute 90 bend losses:Assume r/D=2, from Slide 34 of lastlecture, kb90 = 0.19

There are 80 bends at 90.

Compute 45 bend losses (25 of them):Assume r/D=2, k

b45

= 0.1 (Crane TP410).There are 25 bends at 45


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Fall 2009 CE154 37

minor loss Reference

This web link provides a few pages fromthe Crane Co. technical paper 410(selling for $35 at Crane Co.) for

calculating minor losseshttp://www.lightmypump.com/help16.html
http://www.lightmypump.com/help16.htmlhttp://www.lightmypump.com/help16.htmlhttp://www.lightmypump.com/help16.htmlhttp://www.lightmypump.com/help16.html


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Fall 2009 CE154 38

Example 8-1 System head loss

Compute total head loss:

Since we have suction piping as well, computeloss in the suction pipe

g

g

gD

Lf

ggD

Lf

Vh

Vh

Vknknknh

Vknknkn

Vh

l

l

bbbbvvl

bbbbvvl

28.140

2)1.02519.080337.0107.119(

2)(

2)(

2

2

2

2

45459090

2

45459090

2


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Fall 2009 CE154 39

Example 8-1 Pipe sizing

Assume the pumpstation has 2 pumps.To have the samevelocity in the pipes,the pump dischargeand suction pipediameter is 54.

Additionally, 1 pipebifurcation and 1combining tocontribute to minor

losses

549.50

2

1

72

1

1

2

1

2

1

D

DA

A

D

D

E l 8 1 P S i


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Fall 2009 CE154 40

Example 8-1 Pump StationLoss

Losses at pump station:

f = 0.0175 from Slide 32, Lecture 8

hf = (0.0175 x 40 / 4.5)V542

/2g= 0.16V542/2g

Bifurcation loss: kbi = 0.3 V542/2g

Combining loss: kcm = 0.5 V542

/2g Valve loss: kv = (0.337 x 2 +3.0) V542/2g

Total loss = 4.63 V542/2g

E l 8 1 S t


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Fall 2009 CE154 41

Example 8-1 SystemSchematics

El. 300 ftEl. 290 ft

El. 330 ft

Hp


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Fall 2009 CE154 42

Example 8-1 System Curve

Total static headHst = 330-300 = 30 ft

Losses

pump station loss + pipeline loss= 4.63 V542/2g + 140.8 V722/2g= 4.63 Q542/(2gA542) + 140.8 Q722/(2gA722)= 4.63 Q

72

2/(8gA54

2) + 140.8Q72

2/51483.8)= 4.63Q2/65159.2 + 140.8Q2/51483.8= (0.000071 + 0.002734)Q2= 0.002805Q2


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Fall 2009 CE154 43


System CurveH = 30 + 0.002805 Q2

H

(ft)

30. 31.8 37.0 45.8 58.1 73.8 93.1 115.9 142.2 172.0

Q(cfs

)

0 25 50 75 100 125 150 175 200 225


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Fall 2009 CE154 44


The Q in system curve is the total flow

Since we have 2 pumps, each pump putsout half of the total Q

In this case, need to construct the 2-parallel-operating pump curve bydoubling the flows

Plot the system curve over the pumpcurve to determine the pump operationpoint


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Fall 2009 CE154 45

Example 8-1 Pump Selection


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Fall 2009 CE154 46

8-1 Pump Operating Point

Select 17.57 impeller


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Fall 2009 CE154 47

Ex. 8-1 Pump operating point

For new and old pipe conditions, revisesystem curves and determine if the pumpcan operate at these limits. In new pipes,pump may run-out (fall off the far end of

the pump curve), or not meeting the higherNPSH requirement. In really old pipes, pump may not be able

to deliver the required flow rate.


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Fall 2009 CE154 48

Ex. 8-1 New Pipe Operation

In this case, new pipe f=0.0135 insteadof 0.017. Friction loss coefficientbecomes 116.2 instead of 140.8. System

curves becomesH = 30 + 0.002326Q2

E l 8 1 N i


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Fall 2009 CE154 49

Example 8-1 New pipecondition


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Fall 2009 CE154 50

Ex. 8-1 Pump Suction Design Verify NPSHA meets NPSHR

Assume average temperature of 68F,

hatm = 14.53 psia, interpolated from Table on

Slide 18 hv = 0.339 psia

hs = 300-290 = 10 ft

hs = 4.63 V542/2g = 4.63 (6.29)2/64.4 = 2.84 ft NPSHA = (14.53-0.339)x2.31 + 10 2.84 = 39.9 ft

NPSHR = 7 ft from pump curve OK

hhhhNPSH svpsatmA


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Fall 2009 CE154 51

Example 8-1 System HGL

El. 300 ftEl. 290 ft

El. 330 ft

Hp

Normal operation HGL

PipePressure


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Fall 2009 CE154 52

Example 8-1 Steel Pipe design

Internal pressure p

Cutting the steel pipe in half, integratethe pressure over the top inside surface,the resulting force is supported by the

wall thickness at the two ends.

p

t

pipe diameter D


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Fall 2009 CE154 53

Example 8-1 Pipe design

Total force on the top side

Stress on the wall

pDprprdpr 2cossin 00

t

pDs2


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Fall 2009 CE154 54

Steel Pipe Specifications


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Fall 2009 CE154 55

Steel Pipe Specifications

Concrete cylinder pipe


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Fall 2009 CE154 56

Concrete cylinder pipe(Ameron)


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Fall 2009 CE154 57

Example 8-1 Pipe design

Separate the pipeline into differentsections of similar design pressure leveland select pipe classes to match thepressure requirements.

This pressure is the normal designpressure.

Consider transient conditions to ensuresafe operation, e.g., pump start, pumpshutdown, valve closure, & loss of power

Concepts of hydraulic


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Fall 2009 CE154 58

Concepts of hydraulictransients

Bulk modulus of elasticity

Change in density is accompanied by change

in pressure. This change is transmittedthrough the system at the speed of theelastic wave (sonic wave)

d

dpE


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Fall 2009 CE154 59

Wave speed

E = elastic modulus of pipe wall

K = bulk modulus of watere = pipe wall thicknessD = pipe diameter = water density

a = wave speed

E

K

e

D

K

a

1


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Fall 2009 CE154 60

Transient pressure

Joukowsky equation

V = instantaneous change in velocity

a = wave speed in pipeg = gravitational accelerationH = rise in head

g

VaH


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Fall 2009 CE154 61

Transient considerations

Consider normal (pump start, valveclosure, etc.) and abnormal (powerfailure, valve malfunction, etc.)

operations to determine pressurefluctuations

Vacuum pressure and subsequent vapor

pocket collapse Remedial actions


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Fall 2009 CE154 62

Example 8-1 Discussion

Pump suction design requirement ofsuction tank, suction pipe velocity,function of isolation valves

Pipe system design proportions ofhead loss elements, cost consideration,right of way consideration

Pump selection multiple pumps vs.single pump, power consumption


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Fall 2009 CE154 63

Homework #8

Design a pipeline system to deliver a maximumof 60 cfs of cooling water from AndersonReservoir with minimum water El. 300 ft to a

Calpine power plant 15 miles away at El. 250ft. The minimum water pressure at the endof the line should be 10 ft above atmospheric.Ground elevation at the reservoir is El. 220

ft. Draw the pipeline profile with normaloperation HGL. Try the attached pump curveto see if it is appropriate to use. If not, websearch for a pump curve or design your own.


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Homework #8

pipe system design

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