water hammer and anchor design

LECTURE NOTES XII

HYDROELECTRIC POWER PLANTS

Prof. Dr. Atl BULU

Istanbul Technical UniversityCollege of Civil EngineeringCivil Engineering Department

Hydraulics Division

Prof. Dr. Atl BULU1

Chapter 12

PENSTOCKPenstock TypesIn determining the number of penstocks for any particular installation various factors have tobe considered. Let us compare by a single penstock and by a system of n penstocks. Thefundamental condition of identical discharge can be realized by selecting diameters either,

a) For identical flow velocities,b) For identical friction losses.

Q = Discharge conveyed in a single penstock,D = Diameter of the penstock,V = Flow velocity,hL = Headloss,e = Wall thickness,G = Weight of the penstock.

a) Identical flow velocitiesDividing the discharge Q among n conduits, the diameter of each pipe should be determinedto ensure an identical flow velocity V. With each penstock discharging,

nQQn

The condition of identical velocity is expressed by the relation,

44 22 nn

DQ

DQV

Where Dn is the diameter of any penstock.

nD

QQDD nn

The head loss due to friction in case of single penstock installation,

52

2

22

216

422

DQ

gfLh

DQ

gDfL

gV

DLfh

L

L

Prof. Dr. Atl BULU2

52

1

1 26.0216

DQah

fLgfLa

L

And for n penstocks,

52

152252

15

2

1 DnQaDn

nQanDnQ

ah nL

nhh LLnThe wall thickness in case of single penstock arrangement,

steelsteel

papDe 22 2

Dae 2p = Static + water hammer pressuresteel = Tensile stress of the steel

For n penstocks,

nDaDae nn 22

neen

The total penstock weight in case of single penstock installation,

DeaGaDeG steelsteel

3

3

For one penstock of the n-number system,

ne

nDaeDaG nnn 33

nGGn

The total weight of the system of n penstocks,GnGn

Prof. Dr. Atl BULU3

b) Identical friction (head) lossesFor determining the diameter Dn ensuring a head loss identical with that in the singlepenstock,

5 2

5

2

152

1

nDD

DnQ

aDQah

n

nL

The flow velocity in each of the penstocks,

52

4nV

DnQ

Vn

n

The wall thickness of each of the n penstocks,

5 22 neDae nn

The weight of the each of the n penstocks,

5 43 nGeDaG nnn

The total weight of the penstock system is,,

GnnGn 5

The above results are compiled in the Table.The alternative based on identical head loss should be considered in the economical analysis,since; energetically this is equivalent to the single penstock arrangement. As can be seen, thetheoretical weight increases for several penstocks with n1/5 fold. Actually the difference isgreater since the weight of couplings and joints does not decrease in proportion with thediameter. The amount of steel required for solutions involving several penstocks may besignificantly higher than the amount required for a single penstock arrangement. Owing to theincreased number supporting piers, the costs of civil engineering construction will alsobecome higher.On the other hand, the use of two or more penstocks means added safety of operation and nocomplete shutdown will become necessary in case of repair. The number of penstocks shouldbe decided on the basis of thorough economical analysis of different alternatives.

Prof. Dr. Atl BULU4

Table. Comparison of single penstock and ofmulti penstock arrangements

n penstocks foridenticalOne

penstockVelocity Head loss

Discharge Q QnQn Qn

Qn

Diameter D 21nD 52nD

Velocity V V 51nVHead loss hL 21nhL hL

WallThickness

e 21ne 52ne

Total weight G G 51Gn

The penstock is made of steel. As regards the location of the penstock, two different solutionsmay be discerned which are characteristics of the method of support as well.

1. Buried penstocks are supported continuously on the soil at the bottom of a trenchbackfilled after placing the pipe. The thickness of the cover over the pipe should beabout 1.o to 1.2 m.

The advantages of buried pipes are the following:a) The soil cover protects the penstock against effect of temperature variations,b) It protects the conveyed water against freezing,c) Buried pipes do not spoil the landscape,d) They are safer against rock slides, avalanches and falling trees.

Disadvantages are:a) Such pipes are less accessible for inspection, faults cannot be determined easily,b) For large diameters and rocky soils their installation is expensive,c) On steep hillsides, especially if the friction coefficient of the soil is low, such pipes

may slide,d) Maintenance and repair of the pipe is difficult.

Prof. Dr. Atl BULU5

2. Exposed penstocks are installed above the terrain surface and supported on piers(briefly called supports or saddles). Consequently, there is no contact between theterrain and the pipe itself, and the support is not continuous but confined piers.

The advantages of exposed pipes are the following;a) The possibility of continuous and adequate inspection during operation,b) Its installation is less expensive in case of large diameters of rocky terrain,c) Safety against sliding may be ensured by properly designed anchorages,d) Such pipes are readily accessible and maintenance and repair operations can be carried

out easily.The disadvantages are;

a) Full exposure to external variations in temperature,b) The water conveyed may freeze,c) Owing to the spacing of supports and anchorages significant longitudinal stresses may

develop especially in pipes of large diameters designed for low internal pressures.As a general rule, buried pipes are applied only on mildly sloping terrain where the top layersdo not consist of rock. The exposed arrangement is more frequently applied. The mainadvantage of exposed penstocks is the possibility of continuous inspection during operation.Concrete blocks holding the pipeline may be simple supporting piers permitting slightlongitudinal movement of the pipe, or anchor blocks which do not permit movement of thepipe. Anchorages are usually installed at angle joints, while supporting piers are spaced ratherclosely (6 to 12 m) depending on the beam action of the pipe and the supporting capacity ofthe soil.In order to reduce the longitudinal stresses due to the temperature variations and other causes,rigid joints between pipe sections should in some places be substituted by elastic ones.Large power penstocks subject to heads of several hundred meters may be constructed ofbanded steel pipes.Simple steel pipes are used for,

cmkgpD 10000Banded steel pipes for,

cmkgpD 10000Where p (kg/cm2) internal pressure, and D (cm) pipe diameter.

Prof. Dr. Atl BULU6

Penstock Hydraulic CalculationsPractical empirical equations used to find out the diameter of a penstock will be given.

Maximum velocity in the penstocks may be taken as Vmax = 6 m/sec. Using the head losscondition,

grossL HRnLVh 05.03422

Ludin Bundschu has given empirical equations to compute the economical pipe innerdiameter by depending on the head shown in the Figure,

7 305.0100 QDmH gross (m)

732.5100

grossgross H

QDmH (m)

Example: Calculate the inner diameter of the penstock for a hydroelectric power plant for Q= 15 m3/sec discharge, and H = 120 m head. Water surface oscillations in the surge tank willnot be taken into account.Solution:

a) Choosing the velocity in the penstock as Vmax = 6 m/sec,

ADDA

mVQA

24

5.2615

2

2

Prof. Dr. Atl BULU7

mD 80.15.22

Hydraulic radius = mDR 45.0480.1

4The slope angle of the penstock will be assumed as = 45 0 and the length of the penstockwill be,

mL 170120120 22

Manning coefficient = n = 0.014The head loss,

mRnLVhL 48.345.0

014.0170634

2234

22

mmH 48.3612005.005.0Vmax = 6 m/sec velocity may be accepted.

b) Using empirical diameter equations,

732.5100120 H

QDmmH

mh

mDQV

mDR

mD

L 72.151.0014.017059.4

sec59.404.2154451.04

04.24

04.2120152.5

3422

22

73

The head gained by increasing the pipe inner diameter,mH 76.172.148.3

If the plant runs 180 days by 24 hours daily, the gained energy will be,

kwhEQHTE

9123842418076.11588

Prof. Dr. Atl BULU8

Forces Acting on PipesPipes must be designed to withstand stresses created by internal and external pressures,changes in momentum of the flowing liquid, external loads, and temperature changes, and tosatisfy the hydraulic requirements of the project.

1. Internal ForcesThe internal pressure within a conduit is caused by static pressure and water hammer. Internalpressure causes circumferential tension in the pipe walls which is given approximately by,

tpr (1)

Where, = Tensile stress,

p = Static + water hammer pressurer = Internal radius of the pipe,e = Wall thickness.Pipes are chosen to supply this condition. If the steel pipe is chosen, the thickness of its wallmay be calculated by,

steel

pre (2)

2. Water HammerWhen a liquid flowing in a pipeline is abruptly stopped by the closing a valve, dynamicenergy is converted to elastic energy and a series of positive and negative pressure wavestravel back and forth in the pipe until they are damped our by friction. This phenomenon isknown as water hammer.

FigureStatic + water hammer pressure

p/

0.2p/

ValveA

Prof. Dr. Atl BULU9

This results in a pressure rise which causes a portion of the pipe surrounding the element tostretch. A pressure in excess of hydrostatic cannot be maintained at the junction of pipe andreservoir, and the pressure at A drops to normal as some of the water in the pipe flows backinto the reservoir.The velocity c (celerity) of a pressure wave in any medium is the same as the velocity ofsound in that medium and is given by,

21

wwEc (3)

Ew = the modulus of elasticity of the water,w = the specific mass of water.c is about 1440 m/sec for water under ordinary conditions. The velocity of a pressure wavecreated by water hammer is less than 1440 m/sec because of the elasticity of pipe. Thevelocity of a pressure wave in a water pipe usually ranges from 600 to 1200 m/sec for normalpipe dimensions and materials. If longitudal extension of the pipe is prevented whilecircumferential stretching takes place freely, the velocity of a pressure wave cp is given by,

5.05.0

11

eEED

Ecp

ww

p (4)

Where,Ep = the modulus of elasticity of the pipe walls,D = the pipe diameter,e = the wall thickness.If the valve is closed instantaneously, a pressure wave travels up the pipe with the velocity cp.In a short interval of time dt, an element of water of length cpdt is brought to rest. ApplyingNewtons second law and neglecting friction,

dtdVAcAdpdtMdVFdtdtdVmF

p

Since velocity is reduced to zero, dV = -V and dp equals the pressure ph caused by waterhammer. Hence,

Vcp ph (5)

The total pressure at the valve immediately after closure is,

Prof. Dr. Atl BULU10

ppp htotal (6)

If the length of the pipe is L, the wave travels from valve to reservoir and back in time,

pcLt 2 (7)

This is the time that a positive pressure will be maintained at the valve. If the valve is closedgradually, a series of small pressure waves is transmitted up the pipe. These waves arereflected at the reservoir and return down the pipe as waves of normal pressure. If the valve iscompletely closed before the reflected wave returns from the reservoir, the pressure increaseis, Vcp ph . If the closure time tc > t, negative pressure waves will be superimposed on thepositive waves and the full pressure ph developed by gradual closure of the valve is givenapproximately by,

cp

cph

ch t

VLVctcLpt

tp 22 (8)

Example: Water flows at 2 m/sec from a reservoir into a 100 cm diameter steel pipe which is2500 m long and has a wall thickness e = 2.5 cm. Find the water hammer pressure developedby closure of a valve at the end of the line if the closure time is a) 1 sec , b) 8 sec.

Ew = 2109 (N/m2), Esteel = 21011 (N/m2), water = 1000 kg/m3

Solution:

sec1010714.01414025.010211021

11000102

11

5.0

119

9

5.05.0

mc

c

eEDE

Ec

p

p

steelw

wp

sec95.41010225002

pcLt

a) If tc = 1 sec < 4.95 sec,kPaVcp ph 61002.2210101000

b) If tc = 8 sec > 4.95 sec,

kPapttp hc

h66 1025.11002.28

95.4


Water hammer pressures can be greatly reduced by use of slow-closing valves, automaticrelief valves, air chambers, and surge tanks. If the velocity of flow is increased suddenly bythe opening of a valve or starting a pump, a situation opposite to water hammer develops.For practical purposes,

pph 20.0 (9)

may be taken.3) Forces at Bends and Changes in Cross-Section

A change in the direction or magnitude of flow velocity is accompanied by a change inmomentum comes from the pressure variation within the fluid and from forces transmitted tothe fluids from the pipe walls.

Figure . Forces at a horizontal pipe bendWhere p1 and p2 and V1 and V2 represent the pressure and average velocity in the pipe atsections 1 and 2, respectively. For a horizontal pipe bend of uniform section, V1 = V2 and p1p2.Example: A 1 m diameter pipe has a 300 horizontal bend in it, and carries water at a rate of 3m3/sec. If we assume the pressure in the bend is uniform at 75 kPa gauge pressure, the volumeof the bend is 1.8 m3, and the metal in the bend weighs 4 kN, what forces must be applied tothe bend by the anchor to hold the bend in place? Assume expansion joints prevent any forcetransmittal through pipe walls of the pipes entering and leaving the bend.


Solution: Consider the control volume shown in figure, and first solve for the x component offorce,

10

2,0

2211

12

30cos3100030cos VVFApApVVQF

xanchor

x

Where,

sec82.3785.03785.04

75000

21

22

21

21

mAQVV

mDAAPapp

NFF

xanchor

xanchor9423

130cos785.075000130cos82.31000,

00,

Solve for Fy:

NFF

QFApVVQF

yanchor

yanchor

yanchor

y

3516830sin785.07500030sin82.331000

30sin82.330sin30sin

,

00,

0,

022

10

2

Solve for Fz:

NFFWWVVQF

zanchor

zanchorwaterbend

zzz

2168098108.140000

,

,

12

Then the total force that the anchor will have to exert on the bend will be,

kjiFanchor 21658351689420


Example: This 130-cm overflow pipe from a small hydroelectric plant conveys water fromthe 70-m elevation to the 40-m elevation. The pressures in the water at the bend entrance andexit are 20 kPa and 25 kPa, respectively. The bend interior volume is 3 m3, and the bend itselfweighs 10 kN. Determine the force that a thrust block must exert on the bend to secure it ifthe discharge is 15 m3/sec.

Solution:

The geometric location of the bend in space, (x,y,z)= (0, 13, 60) mVelocity and pressure vectors at cross-sections 1 and 2 respectively,

kzjyixAQV

kjApFkjA

QV

kjiAQV

ml

p 61.0793.0)(61.0793.0

40.1600.10

40.1600.130

40.1610130

11

1

1

2221

1

kjiApFkjiA

QV

kjiAQV

ml

p 656.0623.0426.0656.0623.0426.05.300.20

5.300.19

5.300.13

50.30201913

22

2

2

2222

2

Weight k98103Using momentum equation;

p1A1+ QV1

p2A2+ QV2

1

2

Control Volume


12 VVQF

0426.0426.0 2,,1,2

AQQApF

VVQF

xblock

xxx

sec30.113.11544

33183430.125000

22

22

mDQ

AQV

NAp

NFF

xblock

xblock86343

30.11426.015100033182426.0,

,

NFF

NApQVApApF

yblock

yblock

yblock

2919333183623.026547793.0793.0623.030.11151000

26547430.120000

793.0623.0623.0793.0

,

,

21

21,

NFApApF

QVWApApF

zblock

zblock

zblock

260591000098103656.0610.0656.0610.030.11151000

61.0656.656.061.0

,

21,

21,

Then the total force vector which the thrust block exerts on the bend to hold it in place is,

kjiF 260592919386343Example: The sluiceway is steel lined and has nozzle at its downstream end. What dischargemay be expected under the given conditions? What force will be exerted on the joint that joinsthe nozzle and sluiceway lining? f = 0.01.Solution: First write energy equation from water surface in reservoir to outlet of sluices;

54.254.2480.1

2920300022

22

22

23

22

2222

1211

QVmA

LgV

df

gV

hzgVpzg

VpL


sec5000051.00079.021

6062.191

91.450.201.00.962.19

154.230

91.491.4450.2

3

222

22

2

32

23

mQQQ

QQ

QVmA

sec18.1091.40.50

sec69.1954.20.50

3

22

mV

mAQV

Now consider the force at the joint,

32 VVQFx

47550018.1069.19501000

33int

int33ApF

FApjo

jo

Writing energy equation at the joint,

gV

gVp

2222

233

30m

12m 3L=60mD=250cm

9m

OutletDiameter=180cm

1

2

W

p3A3Fjoint xQV3

QV2


NFF

Papmp

jo

jo222891

47550091.4142035142035981048.14

48.1462.1918.1069.19

int

int

3

223

In addition a force will have to be applied at the joint to resist the weight of the nozzle andweight of water in the nozzle. Depending upon the length of the nozzle this may be as muchas 30% or 40% of the force calculated above and it will act upward.Example: A pipe 40 cm in diameter has a 1350 horizontal bend in it. The pipe carries waterunder a pressure of kPa gage at a rate of 0.40 m3/sec.What is the magnitude and direction ofhorizontal external force necessary to hold the bend in place under the action of water?Solution:

sec17.3126.040.0

126.0440.0

42

22

mAQV

mDA

NFF

pAQVFQVpAFpAQV

F

x

x

x

x

x

369445cos1126.09000017.340.01000

45cos1045cos

0

0

0

0

045sin0

QVpAFF

y

y

45pA

QV

pAQV

y

x

Fx

Fy FHorizontal plane


NFFy

y8914

45sin126811340 0

383.089143694cos

964989143694 2222

FF

NFFFx

yx

Example: A 900 horizontal bend narrows from a 60 cm diameter upstream to a 30 cmdownstream. If the bend is discharging water into the atmosphere and pressure upstream is170 kPa gage, what is the magnitude and direction of the resultant horizontal force to hold thebend in place?Solution:

12

22

21

12

222

211

2211

22

22

2211

1

4430.0

60.044

22

VVDD

VV

DVDVAVAV

gV

Apzg

Vpz

gV

gV

gV

233.172

1600298101700000

21

21

21

p1A1QV1

patm=0

QV2

y

x

Fx

Fy F

Horizontal plane

ControlVolume


NFQVApF

mQmVmV

x

x

5446876.435.11000460.0170000

sec35.176.4460.0

sec04.1976.44sec76.4

2111

32

21

2570404.1935.110002QVFy

NF 602282570454468 22

904.06022854468cos F

Fx

4) TransitionsThe fitting between two pipes of different size is a transition. Because of the change in flowarea and change in pressure, a longitudal force will act on the transition. To determine theforce required to hold the transition in place, the energy, momentum, and continuity equationswill be applied.Example: Water flows through the contraction at a rate of 0.75 m3/sec. The head losscoefficient for this particular contraction is 0.20 based on the velocity head in the smallerpipe. What longitudal force (such as from an anchor) must be applied to the contraction tohold it in place? We assume the upstream pipe pressure is 150 kPa, and expansion jointsprevent force transmittal between the pipe and the contraction.

Solution: Let the x direction be in the direction of flow, and let the control surface surroundthe transition as shown in the figure.


First solve for p2 with the energy equation,

LhzgVpzg

Vp2

222

1211

22Where,

sec65.260.075.0430.159810

150000

21

1

1

mAQV

mp

sec72.445.075.04

22

2 mAQV

kPapmp

mgVh

zzL

14029.14981029.1423.062.19

72.462.1965.230.15

23.062.1972.420.0220.0

2

222

222

21

The anchor force,

NFF

FApApVVQF

anchor

anchor

anchor

x

1863715534

45.0150000460.0140000

65.272.475.0100022

2211

12

Then anchor must exert a force of 18637 N in the negative x direction on the transition.Example: A 50 cm diameter pipe expands to a 60 cm diameter pipe. These pipes arehorizontal, and the discharge of water from the smaller size to the larger is 0.80 m3/sec. Whathorizontal force is required to hold the transition in place if the pressure in the 50 cm pipe is70 kPa? Also, what is the head loss?Solution:


sec08.41963.080.01963.04

50.01

22

1 mVmA

mVVh

mVmA

L 08.062.1983.208.4

62.19

sec83.22826.080.02826.04

60.022

21

22

22

Writing energy equation between sections 1 and 2,

Papmp

phg

Vpg

VpL

7357550.7981050.7

08.062.1983.2

62.1908.4

981070000

22

22

222

222

211

Writing the momentum equation for the transition,

NFF

FApApVVQF

x

x

x

x

369425.180.010001963.0700002826.073575

08.483.280.01000221112

5) Temperature Stress and StrainTemperature stresses develop when temperature changes occur after the pipe is installed andrigidly held in a place. For example, if a pipe is strained from expanding when thetemperature changes T0, the pipe would be subjected to a compressive longitudal reflectionof,

TLL (10) = Coefficient of thermal expansion

Then the resulting effective longitudal strain would be,

Controlvolume

p1A1QV1

p2A2 QV2

Fx

x


TLL (11)

And the resulting temperature stress would be,TEE (12)

E = Bulk modulus of elasticity of the pipe material.Example: Find the longitudal stress in a steel pipe caused by temperature increase of 300C.Assume that longitudal expansion is prevented. For steel, E = 210106 (kN/m2, kPa), =11.710-6.Solution:

kPamkN ,7371030107.1110210 266To eliminate the temperature stress, expansion joints are used. These joints can be placed atregular intervals and must allow the pipe to expand a distance L, where L is the spacingbetween expansion joints.

6) Steel Pipe Weight

The area of the material of a steel pipe for a 1 m length with a wall thickness e,

DeeDeDeA

DeeDDA

DeDA

1

444

24222

22

If the ration 0De is taken as zero,

eDA (13)The weight of the L length pipe with specific mass (density) of steel steel = 7800 (kg/m3) is,

81.97800eDLGgeDLG steel

eDLG 240388 (Newton) (14)Pipe wall thickness e due to the static and water hammer pressure is,


ststatichammerstatic ppppp 12.012.0 (15)

22Dpee

Dp ss

The working stress of the steel is, st= 11000 (kN/m2),

23

3

101100002981012.0240388

10110002981012.0

DpG

Dpe

st

st

LDpG st 229.1 (16)

Where, pst = Static pressure (N/m2), D = Internal pipe diameter (m), L = Pipe length (m).7) External Pressure

When the pipe is empty, it must have a wall thickness enough to resist atmospheric pressure.This minimum wall thickness is calculated by Allievi equation as,

De 008.0min (17)In which emin is the minimum thickness of the pipe wall in m and D is the pipe inner diameterin m as well.

8) Longitudal BendingPipes should normally be designed to resist some bending in the longitudal direction eventhey are to be buried or they are laid on saddles.The maximum span which a simply supported pipe could accommodate is calculated below.The maximum stress is,

WM

Where M is the bending moment and W is the section modulus. For a pipe whose wallthickness e, is small in comparison with the internal diameter D,

eDW 42

If bending moment is,


82pLM

Where p is the steel pipe weight with water in it for unit length and L is the span,

peDL

eDpL

eDpL

st

st

22

22

2

2

2

24

8

If the pipe of specific weight st is conveying water with specific weight w,

DeDp stw42

eDDeL

stwst

48 (18)

st = Working stress of the steel.

Example: Calculate the maximum permissible simply supported span L for the 1 m diametersteel pipe with a wall thickness e = 0.012 m. st = 110000 kN/m2, w = 9810 N/m3, st = 80000N/m3.Solution: The weight of the 1 m length of pipe with the water in it is,

mNpp

DeDp stw

10721012.00.1800004

0.198104

2

2

Maximum permissible span,

012.00.1210721

22L

st

mLL

2810721

012.00.1210110 262


The angle of bottom support, 2 , is normally 1200 for concrete saddles. In order to prevent thedeflection during the filling of the pipe, it is recommended to take the bottom support angle,2 , as given below.

D 3 m 2 = 12003 < D 4 m 2 = 18004 < D 5 m 2 = 2100D > 5 m 2 = 2400

9) Freezing EffectsThe water in the pipe can freeze whenever there is no flow in it and the outside temperaturedrops to the low values. These precautions are recommended for the side effects,

1. The water velocity in the pipe should be kept grater than 0.50 m/sec.2. A discharge of 1 m3/hour for 1 m2 of pipe perimeter is to be supplied,3. A minimum required discharge may be calculated by,

10min

434.0LnLnDLkQ

w (19)

Where, 0 = t0 T0 , 1 = t1 T1 . t0 is the water temperature in 0C at the inlet of the pipewhich can be taken as +40C for the reservoirs, T0 is the outside temperature at the inlet, t1 isthe temperature of the water at the outlet (0C) and T1 is the outside temperature at the outlet.Generally, T0 = T1. The coefficient k = 1/200 for water. w 10000 N/m3.


Example: Calculate the minimum required discharge for a pipe of 1 m diameter and 1000 mlength if the temperature of air drops to -400C.Solution:

CTtCTt

0111

0000

4040044404

sec007.04044100010000.1200

1434.03

min mLnLnQ

Using the perimeter related solution,

sec00087.014.3114.31

33

2

mhourmQmDP

These equations are empirical equations and have to be used cautiously. The experience ofdesign engineers is very important when using empirical equations.

water hammer and anchor design

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