HYDROELECTRIC POWER PLANTS - İTÜweb.itu.edu.tr/~bulu/hyroelectic_power_files/lecture_notes_12.pdf · « HYDROELECTRIC POWER PLANTS » ... the diameter of each pipe should be determined

LECTURE NOTES – XII

« HYDROELECTRIC POWER PLANTS »

Prof. Dr. Atıl BULU

Istanbul Technical University College of Civil Engineering

Civil Engineering Department Hydraulics Division

Prof. Dr. Atıl BULU 1

Chapter 12

PENSTOCK

Penstock Types In determining the number of penstocks for any particular installation various factors have to be considered. Let us compare by a single penstock and by a system of n penstocks. The fundamental 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 velocities Dividing the discharge Q among n conduits, the diameter of each pipe should be determined to ensure an identical flow velocity V. With each penstock discharging,

nQQn =

The condition of identical velocity is expressed by the relation,

44 22n

n

DQ

DQV

ππ==

Where Dn is the diameter of any penstock.

nD

QQDD n

n ==

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

5

2

2

2

2

216

422

DQ

gfLh

DQ

gDfL

gV

DLfh

L

L

⋅=

⎟⎠⎞

⎜⎝⎛⋅=⋅⋅=

π

π


5

2

1

1 26.0216

DQah

fLgfLa

L =

==π

And for n penstocks,

5

2

152

252

15

2

1 DnQa

DnnQa

nDnQ

ahnL ==

⎟⎠

⎞⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=

nhh LLn

=

The wall thickness in case of single penstock arrangement,

steelsteel

papDeσσ 22 2 =→=

Dae 2=

p = Static + water hammer pressure σsteel = 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

==→= πγπγ

For one penstock of the n-number system,

ne

nDaeDaG nnn ⋅⋅== 33

nGGn =

The total weight of the system of n penstocks,

GnGn =


b) Identical friction (head) losses For determining the diameter Dn ensuring a head loss identical with that in the single penstock,

5 2

5

2

15

2

1

nDD

DnQ

aDQah

n

nL

=

⎟⎠⎞

⎜⎝⎛

==

The flow velocity in each of the penstocks,

52

4n

VDnQ

Vn

n ==π

The wall thickness of each of the n penstocks,

5 22neDae nn ==

The weight of the each of the n penstocks,

5 43nGeDaG 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, the theoretical weight increases for several penstocks with n1/5 fold. Actually the difference is greater since the weight of couplings and joints does not decrease in proportion with the diameter. The amount of steel required for solutions involving several penstocks may be significantly higher than the amount required for a single penstock arrangement. Owing to the increased number supporting piers, the costs of civil engineering construction will also become higher. On the other hand, the use of two or more penstocks means added safety of operation and no complete shutdown will become necessary in case of repair. The number of penstocks should be decided on the basis of thorough economical analysis of different alternatives.


Table. Comparison of single penstock and of multi penstock arrangements

n penstocks for

identical

One

penstock

Velocity Head loss

Discharge

Q

QnQn =

QnQn =

Diameter

D

21nD

52nD

Velocity

V

V

51nV

Head loss

hL

21nhL

hL

Wall

Thickness

e

21ne

52ne

Total weight

G

G

51Gn

The penstock is made of steel. As regards the location of the penstock, two different solutions may 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 trench backfilled after placing the pipe. The thickness of the cover over the pipe should be about 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.


2. Exposed penstocks are installed above the terrain surface and supported on piers (briefly called supports or saddles). Consequently, there is no contact between the terrain 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 layers do not consist of rock. The exposed arrangement is more frequently applied. The main advantage of exposed penstocks is the possibility of continuous inspection during operation. Concrete blocks holding the pipeline may be simple supporting piers permitting slight longitudinal movement of the pipe, or anchor blocks which do not permit movement of the pipe. Anchorages are usually installed at angle joints, while supporting piers are spaced rather closely (6 to 12 m) depending on the beam action of the pipe and the supporting capacity of the 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 of banded steel pipes. Simple steel pipes are used for,

( )cmkgpD 10000<

Banded steel pipes for,

( )cmkgpD 10000>

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


Penstock Hydraulic Calculations Practical 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 loss condition,

grossL HR

nLVh 05.034

22

≤′

=

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

7 305.0100 QDmH gross =→< (m)

7

32.5100gross

gross HQDmH =→> (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 will not be taken into account. Solution:

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

ππ ADDA

mVQA

24

5.26

15

2

2

=→=

===


mD 80.15.22 ≅=π

Hydraulic radius = mDR 45.0480.1

4===

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

mL 170120120 22 ≅+=′

Manning coefficient = n = 0.014

The head loss,

mR

nLVhL 48.345.0

014.0170634

22

34

22

=××

=′

=

mmH 48.3612005.005.0 >=×=

Vmax = 6 m/sec velocity may be accepted.

b) Using empirical diameter equations,

732.5100120

HQDmmH =→>=

mh

mDQV

mDR

mD

L 72.151.0

014.017059.4

sec59.404.2

1544

51.0404.2

4

04.2120

152.5

34

22

22

73

=××

=

=××

==

===

=×

=

ππ

The head gained by increasing the pipe inner diameter,

mH 76.172.148.3 =−=Δ

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

kwhEQHTE

9123842418076.11588

=××××=Δ=Δ


Forces Acting on Pipes Pipes 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 to satisfy the hydraulic requirements of the project.

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

tpr

=σ (1)

Where, σ = Tensile stress, p = Static + water hammer pressure r = 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 wall may be calculated by,

steel

preσ

= (2)

2. Water Hammer

When a liquid flowing in a pipeline is abruptly stopped by the closing a valve, dynamic energy is converted to elastic energy and a series of positive and negative pressure waves travel back and forth in the pipe until they are damped our by friction. This phenomenon is known as water hammer.

Figure…Static + water hammer pressure

p/γ

0.2p/γ

Valve

A


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

21

⎟⎟⎠

⎞⎜⎜⎝

⎛=

w

wEcρ

(3)

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

5.0

5.0

1

1

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

eEED

Ec

p

w

wp ρ

(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. Applying Newton’s second law and neglecting friction,

dtdVAcAdpdtMdVFdtdtdVmF

pρ=−=

=

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

Vcp ph ρ= (5) The total pressure at the valve immediately after closure is,


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 closed gradually, a series of small pressure waves is transmitted up the pipe. These waves are reflected at the reservoir and return down the pipe as waves of normal pressure. If the valve is completely closed before the reflected wave returns from the reservoir, the pressure increase is, Vcp ph ρ= . If the closure time tc > t, negative pressure waves will be superimposed on the positive waves and the full pressure ph’ developed by gradual closure of the valve is given approximately by,

cp

cph

ch t

VLVctcLp

ttp ρρ 22

==≈′ (8)

Example: Water flows at 2 m/sec from a reservoir into a 100 cm diameter steel pipe which is 2500 m long and has a wall thickness e = 2.5 cm. Find the water hammer pressure developed by closure of a valve at the end of the line if the closure time is a) 1 sec , b) 8 sec.

Ew = 2×109 (N/m2), Esteel = 2×1011 (N/m2), ρwater = 1000 kg/m3

Solution:

sec1010714.01414025.010211021

11000

102

1

1

5.0

11

9

9

5.0

5.0

mc

c

eEDE

Ec

p

p

steel

w

wp

=×=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

××××

+×⎟⎟⎠

⎞⎜⎜⎝

⎛ ×=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

ρ

sec95.41010

225002=

×==

pcLt

a) If tc = 1 sec < 4.95 sec,

kPaVcp ph

61002.2210101000 ×=××== ρ

b) If tc = 8 sec > 4.95 sec,

kPapttp hc

h66 1025.11002.2

895.4

×=××=≅′


Water hammer pressures can be greatly reduced by use of slow-closing valves, automatic relief valves, air chambers, and surge tanks. If the velocity of flow is increased suddenly by the 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 in momentum comes from the pressure variation within the fluid and from forces transmitted to the fluids from the pipe walls.

Figure . Forces at a horizontal pipe bend

Where p1 and p2 and V1 and V2 represent the pressure and average velocity in the pipe at sections 1 and 2, respectively. For a horizontal pipe bend of uniform section, V1 = V2 and p1 ≈ p2. Example: A 1 m diameter pipe has a 300 horizontal bend in it, and carries water at a rate of 3 m3/sec. If we assume the pressure in the bend is uniform at 75 kPa gauge pressure, the volume of the bend is 1.8 m3, and the metal in the bend weighs 4 kN, what forces must be applied to the bend by the anchor to hold the bend in place? Assume expansion joints prevent any force transmittal 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 of force,

( )

( )10

2,0

2211

12

30cos3100030cos VVFApAp

VVQF

xanchor

x

−××=+−

−=∑ ρ

Where,

sec82.3785.03

785.04

75000

21

22

21

21

mAQVV

mDAA

Papp

====

===

==

π

( ) ( )

NFF

xanchor

xanchor

9423130cos785.075000130cos82.31000

,

00,

−=

−××+−××=

Solve for Fy:

( )( )

NF

F

QFAp

VVQF

yanchor

yanchor

yanchor

y

35168

30sin785.07500030sin82.331000

30sin82.330sin

30sin

,

00,

0,

022

10

2

−=

××−×××−=

−=+

−=∑ρ

ρ

Solve for Fz:

( )

NFFWW

VVQF

zanchor

zanchorwaterbend

zzz

2168098108.140000

,

,

12

=×+=

=++

−=∑ ρ

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

kjiFanchor

rrr21658351689420 +−−=


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

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

( )kzjyixAQV

rrrr++⎟

⎠⎞

⎜⎝⎛=

( )( )kjApF

kjAQV

kjiAQV

ml

p

rrr

rrr

rrrr

61.0793.0)(

61.0793.0

40.1600.10

40.1600.130

40.1610130

11

1

1

2221

1−=

−⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ −+⎟⎠⎞

⎜⎝⎛=

=++=

( )( )( )kjiApF

kjiAQV

kjiAQV

ml

p

rrrr

rrrr

rrrr

656.0623.0426.0

656.0623.0426.0

5.300.20

5.300.19

5.300.13

50.30201913

22

2

2

2222

2+−−=

−+⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛ −+⎟⎠⎞

⎜⎝⎛=

=++=

Weight k

r98103×−=

Using momentum equation;

p1A1+ρQV1

p2A2+ρQV2

1

2

Control Volume


( )∑ −= 12 VVQF

rrrρ

( )

⎟⎠⎞

⎜⎝⎛ −=−

−=∑0426.0426.0 2,

,1,2

AQQApF

VVQF

xblock

xxx

ρ

ρ

sec30.113.1

1544

331834

30.125000

22

2

2

mDQ

AQV

NAp

=××

===

=×

×=

ππ

π

NFF

xblock

xblock

8634330.11426.015100033182426.0

,

,

=

×××+×=

( )

( )NF

F

NAp

QVApApF

yblock

yblock

yblock

29193

33183623.026547793.0793.0623.030.11151000

265474

30.120000

793.0623.0623.0793.0

,

,

2

1

21,

−=

×+×−−×××=

=×

×=

−=×−×+

π

ρ

( )

( )NF

ApApFQVWApApF

zblock

zblock

zblock

260591000098103656.0610.0656.0610.030.11151000

61.0656.656.061.0

,

21,

21,

=

+×+×−×+−×××=

+−=−×+×− ρ

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

kjiFrrrr

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

54.254.2

480.1

29

203000

22

22

2

2

23

22

2

222

1

211

QVmA

Lg

Vdf

gV

hzg

Vpzg

VpL

=→=×

=

∗+++=++

+++=++

π

γγ


sec5000051.00079.021

6062.19

191.450.2

01.00.962.19

154.2

30

91.491.4

450.2

3

22

2

2

2

2

32

2

3

mQQQ

QQ

QVmA

≅

+=

×××++×=

=→=×

=π

sec18.1091.40.50

sec69.1954.20.50

3

22

mV

mAQV

==

===

Now consider the force at the joint,

( )∑ −= 32 VVQFx ρ

( )

475500

18.1069.19501000

33int

int33

+−=

−××=+

ApF

FAp

jo

jo

Writing energy equation at the joint,

gV

gVp

22

22

233 =+

γ

30m

12m 3 L=60m D=250cm

9m

Outlet Diameter=180cm

1

2

W

p3A3 Fjoint xρQV3

ρQV2


( )

NF

FPap

mp

jo

jo

222891

47550091.4142035142035981048.14

48.1462.19

18.1069.19

int

int

3

223

−=

+×−==×=

=−

=γ

In addition a force will have to be applied at the joint to resist the weight of the nozzle and weight of water in the nozzle. Depending upon the length of the nozzle this may be as much as 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 water under a pressure of kPa gage at a rate of 0.40 m3/sec.What is the magnitude and direction of horizontal external force necessary to hold the bend in place under the action of water? Solution:

sec17.3126.040.0

126.04

40.04

222

mAQV

mDA

===

=×

==ππ

( )( )( )( ) ( )

NFF

pAQVF

QVpAFpAQV

F

x

x

x

x

x

369445cos1126.09000017.340.01000

45cos1

045cos

0

0

0

0

=−××+××=

−+=

=+−−+

=∑

ρ

ρρ

( ) 045sin

0

QVpAF

F

y

y

ρ+=

=∑

45

pA

ρQV

pA ρQV

y

x

Fx

Fy F

α

Horizontal plane


( )NF

F

y

y

8914

45sin126811340 0

=

+=

=→===

=+=+=

αα 383.089143694cos

964989143694 2222

FF

NFFF

x

yx

Example: A 900 horizontal bend narrows from a 60 cm diameter upstream to a 30 cm downstream. If the bend is discharging water into the atmosphere and pressure upstream is 170 kPa gage, what is the magnitude and direction of the resultant horizontal force to hold the bend in place? Solution:

12

22

2

1

1

2

222

211

2211

22

2

22

211

1

4

430.060.0

44

22

VVDD

VV

DVDV

AVAVg

VApz

gVpz

=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=

=

++=++

ππ

γ

gV

gV

gV

233.17

21600

298101700000

21

21

21

=

++=++

p1A1 ρQV1

patm=0

ρQV2

y

x

Fx

Fy

α

F

Horizontal plane

Control Volume


NF

QVApF

mQ

mVmV

x

x

5446876.435.110004

60.0170000

sec35.176.44

60.0

sec04.1976.44sec76.4

2111

32

21

=××+×

×=

+=

=××

=

=×=→=

π

ρ

π

2570404.1935.110002 =××== QVFy ρ

NF 602282570454468 22 =+=

=→=== αα 904.06022854468cos

FFx

4) Transitions

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

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


First solve for p2 with the energy equation,

Lhzg

Vpzg

Vp+++=++ 2

222

1

211

22 γγγ

Where,

sec65.260.075.04

30.159810

150000

21

1

1

mAQV

mp

=××

==

==

π

γ

sec72.445.075.04

22

2 mAQV =

××

==π

kPap

mp

mg

Vh

zz

L

14029.149810

29.1423.062.19

72.462.19

65.230.15

23.062.19

72.420.02

20.0

2

222

222

21

=×=

=−−+=

=×==

=

γ

The anchor force,

( )( )

NF

F

FApApVVQF

anchor

anchor

anchor

x

18637

15534

45.01500004

60.0140000

65.272.475.0100022

2211

12

−=

+×

×−×

×=

−××=+−

−=∑

ππ

ρ

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 are horizontal, and the discharge of water from the smaller size to the larger is 0.80 m3/sec. What horizontal force is required to hold the transition in place if the pressure in the 50 cm pipe is 70 kPa? Also, what is the head loss? Solution:


sec08.41963.080.01963.0

450.0

12

2

1 mVmA ==→=×

=π

( ) ( ) mVVh

mVmA

L 08.062.19

83.208.462.19

sec83.22826.0

80.02826.04

60.0

2221

22

2

2

=−

=−

=

==→=×

=π

Writing energy equation between sections 1 and 2,

Papmp

p

hg

Vpg

VpL

7357550.7981050.7

08.062.19

83.262.19

08.4981070000

22

22

222

222

211

=×=→=

−−+=

++=+

γ

γ

γγ

Writing the momentum equation for the transition,

( )( )

NFF

FApAp

VVQF

x

x

x

x

369425.180.010001963.0700002826.073575

08.483.280.010002211

12

=××−×−×=

−××=+−

−=∑ ρ

5) Temperature Stress and Strain

Temperature stresses develop when temperature changes occur after the pipe is installed and rigidly held in a place. For example, if a pipe is strained from expanding when the temperature changes ΔT0, the pipe would be subjected to a compressive longitudal reflection of,

TLL Δ=Δ α (10) α = Coefficient of thermal expansion Then the resulting effective longitudal strain would be,

Control volume

p1A1 ρQV1

p2A2 ρQV2

Fx

x


TLL

Δ=Δ

= αε (11)

And the resulting temperature stress would be,

TEE Δ== αεσ (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 = 210×106 (kN/m2, kPa), α = 11.7×10-6. Solution:

( )kPamkN ,7371030107.1110210 266 =××××= −σ To eliminate the temperature stress, expansion joints are used. These joints can be placed at regular intervals and must allow the pipe to expand a distance ΔL, where L is the spacing between 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

24

222

22

ππ

π

π

If the ration ⎟⎠⎞

⎜⎝⎛ ≅ 0

De is taken as zero,

eDA π= (13)

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

81.97800××==

eDLGgeDLG steel

πρπ

eDLG 240388= (Newton) (14)

Pipe wall thickness e due to the static and water hammer pressure is,


ststatichammerstatic ppppp 12.012.0 =≅+= (15)

σσ

22Dpe

eDp ss =→=

The working stress of the steel is, σst = 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 diameter in m as well.

8) Longitudal Bending

Pipes should normally be designed to resist some bending in the longitudal direction even they 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

=σ

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

eDW4

2π=

If bending moment is,


8

2pLM =

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

peDL

eDpL

eD

pL

st

st

σπ

ππσ

22

2

2

2

2

2

24

8

=

==

If the pipe of specific weight γst is conveying water with specific weight γw,

DeDp stw πγγπ+=

4

2

eDDeL

stw

st

γγσ4

8+

= (18)

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

( )mNp

p

DeDp stw

10721

012.00.1800004

0.19810

42

2

=

×××+×

×=

+=

ππ

πγπγ

Maximum permissible span,

012.00.1210721

2

2

××××

=π

σ Lst

mL

L

2810721

012.00.1210110 262

≅

×××××=

π


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

D ≤ 3 m → 2θ = 1200 3 < D ≤ 4 m → 2θ = 1800 4 < D ≤ 5 m → 2θ= 2100

D > 5 m → 2θ = 2400

9) Freezing Effects The water in the pipe can freeze whenever there is no flow in it and the outside temperature drops 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

434.0θθγ

πLnLnDLkQ

w −= (19)

Where, θ0 = t0 – T0 , θ1 = t1 – T1 . t0 is the water temperature in 0C at the inlet of the pipe which can be taken as +40C for the reservoirs, T0 is the outside temperature at the inlet, t1 is the 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. γw ≈ 10000 N/m3.


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

CTt

CTt0

111

0000

40400

44404

=+=−=

=+=−=

θ

θ

( ) sec007.040441000

10000.12001434.0

3min m

LnLnQ =

−×

××××=

π

Using the perimeter related solution,

sec00087.014.3114.31

33

2

mhourmQmDP

=×=

=×= π

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

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