LECTURE NOTES – XII
« HYDROELECTRIC POWER PLANTS »
Prof. Dr. Atıl BULU
Istanbul Technical University
College of Civil Engineering
Civil Engineering Department
Hydraulics Division
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,
Qn =
Q
n
The condition of identical velocity is expressed by the relation,
V=
Q
Qn
=
2
πD 4 πDn2 4
Where Dn is the diameter of any penstock.
Dn = D
Qn
D
=
Q
n
The head loss due to friction in case of single penstock installation,
hL = f ⋅
hL =
L V2
fL ⎛ 4Q ⎞
⋅
=
⋅⎜
⎟
D 2 g 2 gD ⎝ πD 2 ⎠
2
16 fL Q 2
⋅
2 gπ D 5
1
Prof. Dr. Atıl BULU
a1 =
16 fL
= 0.26 fL
2 gπ
hL = a1
Q2
D5
And for n penstocks,
2
⎛Q⎞
⎜ ⎟
Q 2 n5 2
Q2 n
n
hLn = a1 ⎝ ⎠ 5 = a1 2 5 = a1
n D
D5
⎛ D ⎞
⎜
⎟
⎝ n⎠
hLn = hL n
The wall thickness in case of single penstock arrangement,
e=
pD
2σ steel
→ a2 =
p
2σ steel
e = a2 D
p = Static + water hammer pressure
σsteel = Tensile stress of the steel
For n penstocks,
en = a2 Dn = a2
en =
D
n
e
n
The total penstock weight in case of single penstock installation,
G = γ steelπDe → a3 = γ steelπ
G = a3 De
For one penstock of the n-number system,
Gn = a3 Dn en = a3 ⋅
Gn =
D e
⋅
n n
G
n
The total weight of the system of n penstocks,
nGn = G
2
Prof. Dr. Atıl BULU
b) Identical friction (head) losses
For determining the diameter Dn ensuring a head loss identical with that in the single
penstock,
2
⎛Q⎞
⎜ ⎟
Q2
n
hL = a1 5 = a1 ⎝ 5⎠
D
Dn
Dn =
D
5
n2
The flow velocity in each of the penstocks,
Q
V
Vn = n 2 = 5
πDn
n
4
The wall thickness of each of the n penstocks,
en = a2 Dn =
e
5
n2
The weight of the each of the n penstocks,
Gn = a3 Dn en =
G
5
n4
The total weight of the penstock system is,,
nGn = 5 n ⋅ G
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.
3
Prof. Dr. Atıl BULU
Table. Comparison of single penstock and of
multi penstock arrangements
n penstocks for
identical
One
penstock
Velocity Head loss
Q
=Q
n
n
n
Q
=Q
n
Discharge
Q
Diameter
D
D n1 2
D n2 5
Velocity
V
V
V n1 5
Head loss
hL
hL n1 2
hL
Wall
Thickness
e
e n1 2
e n2 5
Total weight
G
G
Gn1 5
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)
b)
c)
d)
The soil cover protects the penstock against effect of temperature variations,
It protects the conveyed water against freezing,
Buried pipes do not spoil the landscape,
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.
4
Prof. Dr. Atıl BULU
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)
b)
c)
d)
The possibility of continuous and adequate inspection during operation,
Its installation is less expensive in case of large diameters of rocky terrain,
Safety against sliding may be ensured by properly designed anchorages,
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,
pD < 10000(kg cm )
Banded steel pipes for,
pD > 10000(kg cm )
Where p (kg/cm2) internal pressure, and D (cm) pipe diameter.
5
Prof. Dr. Atıl BULU
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,
hL =
V 2 L′n 2
≤ 0.05H gross
R4 3
Ludin – Bundschu has given empirical equations to compute the economical pipe inner
diameter by depending on the head shown in the Figure,
H gross < 100m → D = 7 0.05Q 3
(m)
5.2Q 3
H gross
(m)
H gross > 100m → D = 7
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,
Q 15
= = 2.5m 2
V
6
2
πD
A
→D=2
A=
π
4
A=
6
Prof. Dr. Atıl BULU
D=2
2.5
π
≅ 1.80m
Hydraulic radius = R =
D 1.80
=
= 0.45m
4
4
The slope angle of the penstock will be assumed as α = 45
will be,
0
and the length of the penstock
L′ = 120 2 + 120 2 ≅ 170m
Manning coefficient = n = 0.014
The head loss,
hL =
V 2 L′n 2 6 2 ×170 × 0.014 2
=
= 3.48m
0.45 4 3
R4 3
0.05 H = 0.05 × 120 = 6m > 3.48m
Vmax = 6 m/sec velocity may be accepted.
b) Using empirical diameter equations,
H = 120m > 100m → D = 7
5.2Q 3
H
5.2 ×153
= 2.04m
120
D 2.04
R= =
= 0.51m
4
4
4Q
4 × 15
V=
=
= 4.59 m sec
2
πD
π × 2.04 2
4.59 2 × 170 × 0.014 2
hL =
= 1.72m
0.514 3
D=7
The head gained by increasing the pipe inner diameter,
ΔH = 3.48 − 1.72 = 1.76m
If the plant runs 180 days by 24 hours daily, the gained energy will be,
ΔE = 8QHT
ΔE = 8 × 15 × 1.76 ×180 × 24 = 912384kwh
7
Prof. Dr. Atıl BULU
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,
σ=
pr
t
(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,
e=
pr
σ steel
(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.
0.2p/γ
p/γ
A
Valve
Figure…Static + water hammer pressure
8
Prof. Dr. Atıl BULU
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,
12
⎛E ⎞
c = ⎜⎜ w ⎟⎟
⎝ ρw ⎠
(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,
⎞
⎛
⎟
0.5 ⎜
⎛ Ew ⎞ ⎜ 1 ⎟
⎟⎟
c p = ⎜⎜
⎟
⎜
ρ
⎝ w ⎠ ⎜ 1 + ED ⎟
⎜
E p e ⎟⎠
⎝
0.5
(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,
dV
dt
Fdt = MdV
− Adpdt = ρAc p dtdV
F =m
Since velocity is reduced to zero, dV = -V and dp equals the pressure ph caused by water
hammer. Hence,
ph = ρc pV
(5)
The total pressure at the valve immediately after closure is,
9
Prof. Dr. Atıl BULU
ptotal = ph + p
(6)
If the length of the pipe is L, the wave travels from valve to reservoir and back in time,
2L
(7)
t=
cp
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, ph = ρc pV . 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,
p h′ ≈
2L
2 LρV
t
ρc pV =
ph =
tc
c ptc
tc
(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:
⎞
⎛
⎟
0.5 ⎜
⎛ Ew ⎞ ⎜
1
⎟
c p = ⎜⎜
⎟⎟
⎜
⎝ ρ ⎠ 1 + Ew D ⎟
⎟
⎜
E steel e ⎠
⎝
0.5
⎞
⎛
⎟
⎜
⎛ 2 × 10 9 ⎞ ⎜
1
⎟
⎟⎟ ×
c p = ⎜⎜
9
⎝ 1000 ⎠ ⎜ 1 + 2 × 10 ×1 ⎟
⎜
⎟
⎝ 2 × 1011 × 0.025 ⎠
c p = 1414 × 0.714 = 1010 m sec
t=
0.5
2 L 2500 × 2
=
= 4.95 sec
1010
cp
a) If tc = 1 sec < 4.95 sec,
ph = ρc pV = 1000 ×1010 × 2 = 2.02 × 10 6 kPa
b) If tc = 8 sec > 4.95 sec,
ph′ ≅
4.95
t
ph =
× 2.02 × 10 6 = 1.25 ×10 6 kPa
8
tc
10
Prof. Dr. Atıl BULU
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,
ph = 0.20 p
(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.
11
Prof. Dr. Atıl BULU
Solution: Consider the control volume shown in figure, and first solve for the x component of
force,
∑F
x
= ρQ(V2 − V1 )
(
p1 A1 − p2 A2 cos 30 0 + Fanchor , x = 1000 × 3 × V2 cos 30 0 − V1
)
Where,
p1 = p 2 = 75000 Pa
A1 = A2 =
V1 = V2 =
πD 2
4
= 0.785m 2
Q
3
=
= 3.82 m sec
A 0.785
(
)
(
)
Fanchor , x = 1000 × 3.82 × cos 30 0 − 1 + 75000 × 0.785 × cos 30 0 − 1
Fanchor , x = −9423N
Solve for Fy:
∑F
y
(
= ρQ V2 sin 30 0 − V1
)
(
p 2 A2 sin 30 0 + Fanchor , y = ρQ − 3.82 sin 30 0
)
Fanchor , y = −1000 × 3 × 3.82 × sin 30 0 − 75000 × 0.785 × sin 30 0
Fanchor , y = −35168 N
Solve for Fz:
∑F
z
= ρQ(V2 z − V1z )
Wbend + Wwater + Fanchor , z = 0
Fanchor , z = 4000 + 1.8 × 9810 = 21680 N
Then the total force that the anchor will have to exert on the bend will be,
r
r
r
Fanchor = −9420i − 35168 j + 21658k
12
Prof. Dr. Atıl BULU
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:
1
p1A1+ρQV1
Control Volume
2
p2A2+ρQV2
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,
(
r
r ⎛Q⎞ r r
V = ⎜ ⎟ xi + yj + zk
⎝ A⎠
)
l1 = 0 2 + 132 + 10 2 = 16.40m
r ⎛ Q ⎞⎛ r 13.00 r 10.00 r ⎞
V1 = ⎜ ⎟⎜ 0i +
j−
k⎟
16.40
16.40 ⎠
⎝ A ⎠⎝
r
r ⎛Q⎞
r
V1 = ⎜ ⎟ 0.793 j − 0.61k
⎝ A⎠
r
r
r
Fp1 = ( p1 A1 ) 0.793 j − 0.61k
(
)
(
)
l2 = 132 + 19 2 + 20 2 = 30.50m
r ⎛ Q ⎞⎛ 13.0 r 19.0 r 20.0 r ⎞
V2 = ⎜ ⎟⎜
i+
j−
k⎟
30.5
30.5 ⎠
⎝ A ⎠⎝ 30.5
r
r ⎛Q⎞
r
r
V2 = ⎜ ⎟ 0.426i + 0.623 j − 0.656k
⎝ A⎠
r
r
r
r
Fp2 = ( p2 A2 ) − 0.426i − 0.623 j + 0.656k
(
)
(
)
r
Weight = −3× 9810k
Using momentum equation;
13
Prof. Dr. Atıl BULU
∑ F = ρQ(V
r
r
2
r
− V1
)
= ρQ(V2, x − V1, x )
∑F
x
Q
⎛
⎞
Fblock , x − 0.426 p2 A = ρQ⎜ 0.426 − 0 ⎟
A
⎝
⎠
2
π ×1.30
= 33183 N
p2 A = 25000 ×
4
Q 4Q
4 × 15
=
V = =
= 11.30 m sec
2
A πD
π ×1.32
Fblock , x = 0.426 × 33182 + 1000 × 15 × 0.426 × 11.30
Fblock , x = 86343 N
Fblock , y + 0.793 × p1 A − 0.623 × p2 A = ρQV (0.623 − 0.793)
p1 A = 20000 ×
Fblock , y
π × 1.30 2
= 26547 N
4
= 1000 × 15 × 11.30 × (0.623 − 0.793) − 0.793 × 26547 + 0.623 × 33183
Fblock , y = −29193 N
Fblock , z − 0.61× p1 A + 0.656 × p2 A − W = ρQV (− .656 + 0.61)
Fblock , z = 1000 ×15 ×11.30 × (0.610 − 0.656 ) + 0.610 × p1 A − 0.656 × p2 A + 3 × 9810 + 10000
Fblock , z = 26059 N
Then the total force vector which the thrust block exerts on the bend to hold it in place is,
r
r
r
r
F = 86343i − 29193 j + 26059k
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;
p1
γ
+
V12
p V2
+ z1 = 2 + 2 + z 2 + hL
γ 2g
2g
0 + 0 + 30 = 0 +
A2 =
π × 1.80 2
4
V22
f V2
+9+ ∗ 3 L
d 2g
2g
= 2.54m 2 → V2 =
Q
2.54
14
Prof. Dr. Atıl BULU
30m
1
3
L=60m
D=250cm
12m
2
9m
Outlet
Diameter=180cm
π × 2.50 2
Q
4
4.91
2
1
Q
0.01 Q 2
1
30 =
9
.
0
×
+
+
×
×
× 60
2
2
2.54 19.62
2.50 4.91 19.62
21 = 0.0079Q 2 + 0.00051Q 2
A3 =
= 4.91m 2 → V3 =
Q ≅ 50 m3 sec
V2 =
Q 50.0
=
= 19.69 m sec
A2 2.54
V3 =
50.0
= 10.18 m sec
4.91
Now consider the force at the joint,
ρQV3
Fjoint
x
p3A3
ρQV2
W
∑F
x
= ρQ(V2 − V3 )
p3 A3 + F jo int = 1000 × 50 × (19.69 − 10.18)
F jo int = − p3 A3 + 475500
Writing energy equation at the joint,
p3
γ
+
V32 V22
=
2g 2g
15
Prof. Dr. Atıl BULU
(19.69
)
− 10.182
= 14.48m
γ
19.62
p3 = 14.48 × 9810 = 142035 Pa
p3
=
2
F jo int = −142035 × 4.91 + 475500
F jo int = −222891N
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:
ρQV
y
pA
45
x
ρQV
pA
Fx
α
Horizontal plane
A=
πD 2
=
π × 0.40 2
Fy
F
= 0.126m 2
4
4
Q 0.40
V = =
= 3.17 m sec
A 0.126
∑F
x
=0
ρQV + pA − Fx − ( pA + ρQV ) cos 450 = 0
(
Fx = (ρQV + pA) 1 − cos 450
)
(
Fx = (1000 × 0.40 × 3.17 + 90000 × 0.126 )× 1 − cos 450
)
Fx = 3694 N
∑F
y
=0
Fy = ( pA + ρQV )sin 450
16
Prof. Dr. Atıl BULU
Fy = (11340 + 1268)sin 450
Fy = 8914 N
F = Fx2 + Fy2 = 3694 2 + 8914 2 = 9649 N
cos α =
Fx 3694
=
= 0.383 → α =
F 8914
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:
F
Fy
Horizontal plane
ρQV1
α
p1A1
Fx
y
Control
Volume
x
patm=0
ρQV2
z1 +
p1
γ
+
V12
p V2
= z2 + 2 + 2
A2 2 g
2g
V1 A1 = V2 A2
V1
π
4
D12 = V2
2
π
4
D22
2
V2 ⎛ D1 ⎞ ⎛ 0.60 ⎞
=⎜ ⎟ =⎜
⎟ =4
V1 ⎜⎝ D2 ⎟⎠ ⎝ 0.30 ⎠
V2 = 4V1
0+
170000 V12
16V12
+
= 0+0+
9810
2g
2g
17.33 =
V12
2g
17
Prof. Dr. Atıl BULU
V1 = 4.76 m sec → V2 = 4 × 4.76 = 19.04 m sec
Q=
π × 0.602
× 4.76 = 1.35 m3 sec
4
Fx = p1 A1 + ρQV1
Fx = 170000 ×
π × 0.60 2
4
+ 1000 × 1.35 × 4.76 = 54468 N
Fy = ρQV2 = 1000 × 1.35 × 19.04 = 25704
F = 544682 + 25704 2 = 60228 N
cos α =
Fx 54468
=
= 0.904 → α =
F 60228
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.
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Prof. Dr. Atıl BULU
First solve for p2 with the energy equation,
p1
γ
+
p V2
V12
+ z1 = 2 + 2 + z 2 + hL
2 gγ
γ 2g
Where,
150000
= 15.30m
γ
9810
Q
4 × 0.75
V1 =
=
= 2.65 m sec
A1 π × 0.60 2
p1
=
Q
4 × 0.75
=
= 4.72 m sec
A2 π × 0.45 2
V2 =
z1 = z 2
hL = 0.20
V22
4.72 2
= 0.20 ×
= 0.23m
2g
19.62
2.65 2 4.72 2
−
− 0.23 = 14.29m
19.62 19.62
γ
p 2 = 9810 ×14.29 = 140kPa
p2
= 15.30 +
The anchor force,
∑F
x
= ρQ(V2 − V1 )
p1 A1 − p2 A2 + Fanchor = 1000 × 0.75 × (4.72 − 2.65)
Fanchor = 140000 ×
π × 0.60 2
4
− 150000 ×
π × 0.45 2
4
+ 1553
Fanchor = −18637 N
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:
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Prof. Dr. Atıl BULU
Control
volume
ρQV1
p1A1
x
p2A2
ρQV2
Fx
A1 =
π × 0.60
A2 =
2
π × 0.50
4
2
= 0.1963m 2 → V1 =
= 0.2826m 2 → V2 =
0.80
= 4.08 m sec
0.1963
0.80
= 2.83 m sec
0.2826
4
(V − V )2 (4.08 − 2.83)2 = 0.08m
hL = 1 2 =
19.62
19.62
Writing energy equation between sections 1 and 2,
V12 p2 V22
=
+
+h
2g
γ 2g L
p1
+
p2
=
p2
= 7.50m → p2 = 9810 × 7.50 = 73575 Pa
γ
γ
γ
70000 4.082 2.832
+
−
− 0.08
9810 19.62 19.62
Writing the momentum equation for the transition,
∑F
x
= ρQ(V2 − V1 )
p1 A1 − p2 A2 + Fx = 1000 × 0.80 × (2.83 − 4.08)
Fx = 73575 × 0.2826 − 70000 × 0.1963 − 1000 × 0.80 × 1.25
Fx = 3694 N
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,
ΔL = α L Δ T
(10)
α = Coefficient of thermal expansion
Then the resulting effective longitudal strain would be,
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Prof. Dr. Atıl BULU
ΔL
= αΔT (11)
L
And the resulting temperature stress would be,
ε=
σ = Eε = EαΔT (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:
σ = 210 ×106 ×11.7 ×10 −6 × 30 = 73710(kN m 2 , kPa )
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,
A=
A=
[(D + 2e) − D ]
4
π
π
(D
4
2
2
2
+ 4eD + 4e 2 − D 2
)
e⎞
⎛
A = πe(D + e ) = πeD⎜1 + ⎟
⎝ D⎠
⎛e
⎞
If the ration ⎜ ≅ 0 ⎟ is taken as zero,
⎝D
⎠
A = πeD
(13)
The weight of the L length pipe with specific mass (density) of steel ρsteel = 7800 (kg/m3) is,
G = πeDLρ steel g
G = πeDL × 7800 × 9.81
G = 240388eDL (Newton)
(14)
Pipe wall thickness e due to the static and water hammer pressure is,
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Prof. Dr. Atıl BULU
p = pstatic + phammer ≅ 0.12 pstatic = 0.12 pst
σ=
(15)
ps D
pD
→e= s
2e
2σ
The working stress of the steel is, σst = 11000 (kN/m2),
0.12 pst D × 9810
2 ×11000 ×103
0.12 × 9810
G = 240388 ×
× pst D 2
3
2 × 110000 × 10
e=
G = 1.29 pst D 2 L
(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,
e min = 0.008 D
(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,
σ=
M
W
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,
W=
πD 2
4
e
If bending moment is,
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Prof. Dr. Atıl BULU
M =
pL2
8
Where p is the steel pipe weight with water in it for unit length and L is the span,
pL2
pL2
σ st = 82 =
πD
2πD 2e
e
4
2πD 2eσ st
L2 =
p
If the pipe of specific weight γst is conveying water with specific weight γw,
p=
πD 2
L=
4
γ w + γ stπDe
8Deσ st
γ w D + 4γ st e
(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,
p = γw
πD 2
4
p = 9810 ×
+ γ stπDe
π ×1.0 2
4
p = 10721( N m )
+ 80000 × π × 1.0 × 0.012
Maximum permissible span,
σ st =
10721× L2
2 × π × 1.0 2 × 0.012
110 × 106 × 2 × π ×1.0 2 × 0.012
10721
L ≅ 28m
L2 =
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Prof. Dr. Atıl BULU
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,
Qmin =
0.434k (πDL )
γ w (Lnθ 0 − Lnθ1 )
(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.
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Prof. Dr. Atıl BULU
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:
θ 0 = t0 − T0 = 4 + 40 = 440 C
θ1 = t1 − T1 = 0 + 40 = 400 C
Qmin =
1
× π × 1.0 × 1000
200
= 0.007 m3 sec
1000 × (Ln 44 − Ln 40 )
0.434 ×
Using the perimeter related solution,
P = πD × 1 = 3.14m 2
Q = 1× 3.14 m3 hour = 0.00087 m3 sec
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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Prof. Dr. Atıl BULU