Engineering Hydraulics II slides 36-51.pdf

Hydraulics

1. BACHELOR OF SCIENCE IN CIVIL
ENGINEERING
HYDRAULICS II –CCE 2322
Dr. George Okwadha

2. Water Hammer
•Water flowing in a pipe possesses some momentum due to its motion.
•When the flowing water is suddenly brought to rest by closing the valve, its
velocity is brought to a halt, and its momentum is destroyed.
•This causes a very high pressure on the valve
•The high pressure is followed by a wave of pressure vibrations along the
pipe which sets up or creates noise known as knocking.
•Such a knocking noiseis often heard in water pipes if a tap is suddenly
turned off.
•This phenomenon of sudden rise in pressure in the pipe accompanied with
knocking noise is referred to as water hammer or hammer blow.
•This rise in pressure may be so large enough to even cause a pipe burst, and
its essential to take into account this pressure rise in the design of pipes.
•The magnitude of the pressure rise depends on :
–The speed at which the valve is closed
–The velocity of flow
–The length of the pipe
–The elastic properties of the pipe material and the flowing fluid

3. Water Hammer Cont’
•To reduce the occurrence of water hammer, the valves in a pipeline
should always be closed gradually
•The valve may be closed either gradually or instantaneously, and
according to the nature of the closure of the valve, expressions for
calculating the pressure head due to water hammer may be developed.
(a)Gradual closure of the valve
•Consider a uniform pipeline through which some liquid is flowing with
uniform velocity, and whose valve is closed gradually
•Let L = Length of the pipe
•Let a = cross sectional area of the pipe
•Let v = velocity of water in the pipe
•Let t = Time, in seconds, in which the water is brought to rest by the closure of the
valve
•Let p = increase in intensity of pressure
•Total mass of fluid in the pipe, �=
ω�??????
�
, where ωis spec. wt. of water and
ω
�
= ρ...
1

4. Water Hammer Cont’
•Rate of retardation of the water =
??????����??????��
????????????��
=
�
�
……………………
•Force available for causing retardation, F = mass x acceleration
•�=
ω�??????
�
x
�
�
=
ω�??????�
��
………………………………..
•Intensity of pressure rise in the valve, �=
�����
??????���
=
ω??????????????????
????????????
�
=
ω??????�
��
•Pressure (or inertia) head due to increase in intensity of pressure is
given by: H =
�
ω
=
ω??????�
ω��
=
??????�
��
……………………………….
(b)Instantaneous closure of the valve
•From the previous section, increase in intensity of pressure is given
by: �=
ω??????�
��
•If the valve is closed instantly, t = 0, and therefore the increase in
intensity of pressure will be infinite.
2
3
4

5. Water Hammer Cont’
•But, it is not possible because its not
possible to close the valve
instantaneously, and therefore the
infinite intensity of pressure is
hypothetical
•Consider a uniform pipeline shown
through which some liquid (say water)
is flowing with a uniform velocity and
whose valve is closed suddenly.
•When the valve Is closed suddenly, rise
in intensity of pressure takes place due
to which circumferential stress, σ
cand
longitudinal stress, σ
Lare produced in
the pipe wall.
t
p
σ
c
σ
c
D
σ
c
σ
c
σ
L
σ
L
L

6. Water Hammer Cont’
•(a) Circumferential (Hoop) Stress, σ
c
•This is the stress which is set up to resist the bursting effect of the applied
pressure.
•Let L = Length of the pipe
•p= internal intensity of pressure
•D= Inside diameter of the pipe
•t = thickness of the pipe
•Force tending to push the two halves of the cylinder apart due to internal
intensity of pressure, F = pxDxL-------------------------(1)
•Total resisting force owing to circumferential stress on the cylinder walls,
F = 2x σ
cx L x t (sinceσ
cLtis the force on one wall of the half cylinder) ----(2)
•Equating eqns(1) and (2) givespxDxL= 2x σ
cx L x t
??????
�=
��??????
2??????�
=
��
2�

7. Water Hammer Cont’
(b) Longitudinal Stress, σ
L
•Consider the figure shown.
•Total force due to internal intensity of pressure, p:
F = intensity of pressure x area = px
??????�
2
4
•Area of the metal resisting this force, A = π.D.t
??????
??????=
�����
??????���
=
�??????�
2
4π��
=
��
4�
•Strain Energy stored in the pipe material per unit
volume (From Strength of Materials)
•�
�=
1
2�
(??????
�
2
+??????
??????
2
−
2??????
????????????
??????
�
) where
1
�
and E are
poisons ratio and Modulus of elasticity for the pipe
material respectively
•Substituting for ??????
�??????��??????
??????gives
•�
�=
1
2�
(
��
2�
2
+
��
4�
2
−
2
��
2??????
)(
��
4??????
)
�
)
•�
�=
1
2�
(
�
2
�
2
4�
2
+
�
2
�
2
16�
2
−
�
2
�
2
4��
2
) and if
1
�
=
1
4
then Strain Energy per unit volume is given by
�
�=
1
2�
(
�
2
�
2
4�
2
+
�
2
�
2
16�
2
−
�
2
�
2
16�
2
)
Therefore, �
�=
�
2
�
2
8��
2
.
σ
L
σ
L
p p
D
t
D
σ
L
acts over the
thickness of the wall, t
Area resisting the force
= π.D.t

8. Water Hammer Cont’
•Total strain energy stored in the pipe
material is given by
•�=
�
2
�
2
8��
2
x total volume of the pipe material
•�=
�
2
�
2
8��
2
x πDtx L =
�
2
�
3
π??????
8��
•�=
�
2
π�
2
�??????
2�4��
=
�
2
??????�??????
2��
since ??????=
π�
2
4
•But loss of kinetic energy of water is given by
KE =
��
2
2
=
ρ????????????�
2
2
……………………….
•Since m = ρV and V = AL
•Gain in strain energy in water is given by
•�=
�
2
2�
x Volume (From Strength of
Materials) where k is the bulk modulus of
water Thus �=
�
2
????????????
2�
……………………. .
•For a rigid pipeline, Loss of KE of water =
Gain in strain energy in water (ignoring
elasticity of the pipe material)
•Equating eqns1 and 2 gives
•
ρ????????????�
2
2
=
�
2
????????????
2�
⇒�
2
=kρ�
2
•�=�??????ρ=�
�ρ
2
ρ
ρ�??????�ℎ���??????=
�
ρ
velocity of pressure wave
•Also, if the elasticity of the pipe material
is considered, then
•Loss of KE of water = Gain in strain
energy in water + strain energy stored in
water.
•Therefore
ρ????????????�
2
2
=
�
2
????????????
2�
+
�
2
??????�??????
2��
•Dividing both sides by
????????????
2
gives
•ρ�
2
=
�
2
�
+
�
2
�
��
= �
2
1
�
+
�
��
•�=
ρ�
2
1
??????
+
�
�??????
=�
ρ
1
??????
+
�
�??????
•NOTE: If the pipeline is considered rigid,
E is infinite, then
�
��
≈ 0 and the above
equation becomes �=�??????ρ
1
2

9. Example 1
•Water is supplied to a turbine through a pipe 3 km long. The water flows in the pipe with a
velocity of 2 m/s. A valve near the turbine end is closed in 30 seconds. Find the rise in
intensity of pressure behind the valve in kg/cm
2
.
•Solution
•Given: Length of pipe, L = 3 km = 3000m
•Velocity of water, v = 3 m/s
•Time of valve closure, t = 30 sec
•Let p = rise in intensity of pressure. But �=
ω??????�
��
=
1000�3000�2
9.81�30
= 2.04 kg/cm
2
•
Example 2
•A valve at the outlet of a rigid pipe is suddenly closed to bring the water to rest. If the
velocity of the water was 3 m/s, find increase in intensity of pressure due to sudden closure
of the valve.
•Solution
•Given: velocity of water, v = 3 m/s; Bulk modulus of water , k = 2.1x10
4
kg/cm
2
= 2.1x10
8
kg/m
2
•Let p = rise in intensity of pressure ⇒�=�??????ρ=�
�ω
�
= 3x
2.1x10
8
x1000
9.81
=
43.89x10
4
kg/m
2
or 43.89 kg/cm
2

10. Example 3
•Water flows through a pipeline 1.5 km long, 15 cm diameter and 5 mm thick
with a velocity of 3m/s. Determine the increase in intensity of pressure when
the valve at the outlet of the pipeline is closed suddenly. Take E, k and m for
the pipe as 2.1x10
6
kg/cm
2
, 21x10
3
kg/cm
2
and 4 respectively.
•Solution
•Given: Length of pipe, L = 1.5 km
•Diameter of pipe, D = 15 cm
•Thickness of pipe, t = 5 mm = 0.5 cm
•Velocity of water, v = 3 m/s = 300 cm/s
•Modulus of elasticity for the pipe, E = 2.1x10
6
kg/cm
2
,
•Bulk modulus of water, k = 21x10
3
kg/cm
2
and
•Poissonsratio,
1
�
= 0.25
•Let p = increase in intensity of pressure
•But �=�
ρ
1
??????
+
�
�??????
⇒�=300
1
1
21x103
+
15
2.1x106??????0.5
= 38129 g/cm
2
= 38.129kg/cm
2
,

11. Water Hammer Cont’
•Time required by pressure wave to travel from the valve to the tank
and from the tank to the valve
•Time taken, �=
�??????�������������������������������������
??????����??????�����ℎ�������������
•�=
??????+??????
�
=
2??????
�
;i.e,�=
2??????
�
•Where L = Length of the pipe and C = velocity of the pressure wave
•NOTE: 1. If �>
2??????
�
, the closure of the valve is said to be gradual
•NOTE: 2.If &#3627408481;<
2??????
&#3627408438;
, the closure of the valve is said to be sudden

12. Water Hammer Cont’
T =

13. Surge Tanks
•A Surge tank is a reservoir fitted at some
opening made on a pipeline (called
Penstock) between the reservoir and the
hydraulic machine (e.g. a Turbine) to store
water when the demand is low (valve
suddenly closed) or to discharge water
when increased discharge is required.
•Surge tanks are installed close to hydraulic
machines to protect them from pressure
fluctuations
•The water surface in the surge tank will be
lower than the reservoir surface by an
amount equal to the friction head loss in the
pipe connecting the reservoir and the surge
tank
•When the generator load is reduced, turbine
gates are closed and the water moving
towards the turbine move backward
•The rejected water is then stored in the
surge tank
Reservoir
Turbine
Surge Tank
Penstock
•The retarding head so built up
In the surge tank reduces the velocity of flow in
the pipeline corresponding to the reduced
discharge required by the turbine.
•When the generator load is increased, the
turbine gates open to increase discharge
•The increased demand is met by water
stored in the surge tank and the reservoir.

14. Surge Tanks
•Determination of Fluctuations Depth
in the Surge tank during low and
high load in the turbine
•In the figure shown, a pipeline of area a
conveys water at a velocity, v to the
hydraulic machine (Turbine)
•A Surge tank is located at a distance L from
the reservoir
•The Surge tank’s cross sectional area is A
and
??????
&#3627408462;
> 1
•At steady operation condition, the water
level in the tank would be equal to that of
the reservoir if frictional losses are
neglected.
•Let Q
1= Discharge into the Surge tank due
to transient
•Let z = Distance from the steady state water
level to a level caused by a transient
•Let v = Water velocity
•The transient would cause a change in
water level such that ??????
1=??????
&#3627408465;&#3627408487;
&#3627408465;&#3627408481;
•But by continuity eqn, av= Q
1+Q
•Differentiating with respect to time gives
•??????
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=
&#3627408465;??????1
&#3627408465;&#3627408481;
+
&#3627408465;??????
&#3627408465;&#3627408481;
……………………….
•But the purpose of Surge tank is to reduce
downstream fluctuations. Therefore,
&#3627408465;??????
&#3627408465;&#3627408481;
is
negligible compared to other two terms in
eqn. 1. Therefore
•??????
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=
&#3627408465;??????
1
&#3627408465;&#3627408481;
=??????
&#3627408465;
2
&#3627408487;
&#3627408465;&#3627408481;
2
……………………
•Acceleration of mass of water due to
transient is given by
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
Surge Tank
Area, A, Q
1
To hydraulic
Machine (Turbine)
Qv
Area, a
Reservoir
Z(t)
L
1
2

15. Surge Tanks
•Mass of water, m = ρaL
•Note: The resulting inertia is opposed by a
rise z in the liquid level in the Surge tank
•Extra pressure on the Surge tank caused by
the rise z is given by, &#3627408477;=ρ????????????
•Corresponding opposing force, &#3627408441;=????????????????????????.
•Force caused by the transient, &#3627408441;=&#3627408474;
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
…
•Equating eqns3 and 4 gives
•&#3627408474;
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=-????????????????????????……………………………
•ρaL
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=-????????????????????????⇒
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=-
&#3627408468;
??????
??????……………
•Substituting for
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
from the continuity eqn.
consideration in eqn. 2 gives
•??????
&#3627408465;&#3627408483;
&#3627408465;&#3627408481;
=??????
&#3627408465;
2
&#3627408487;
&#3627408465;&#3627408481;
2
⇒ -
&#3627408468;
??????
????????????=??????
&#3627408465;
2
&#3627408487;
&#3627408465;&#3627408481;
2
•⇒
&#3627408465;
2
&#3627408487;
&#3627408465;&#3627408481;
2
+
&#3627408468;
????????????
????????????=0………………………..
•Note: Eqn. 7 is the differential equation
controlling the water level in the Surge
tank, and solving it gives
•??????=??????
1sin
&#3627408462;&#3627408468;
????????????
&#3627408481;+??????
2??????&#3627408476;&#3627408480;
&#3627408462;&#3627408468;
????????????
t…………
•Since z is measured from the steady water
level in the tank, z = 0 at t = 0 ⇒??????
2=0.
Hence ??????=??????
1sin
&#3627408462;&#3627408468;
????????????
&#3627408481;…………
•The constant ??????
1is evaluated by considering
the situation immediately after a complete
shut down (rejection) to the turbine i.e. at t
= 0 with Q = 0. The entire pipe flow then
enters the Surge tank. Therefore,
•??????
&#3627408465;&#3627408487;
&#3627408465;&#3627408481;
&#3627408481;=0
=??????
&#3627408476;and Differentiating eqn. 9
gives
&#3627408465;&#3627408487;
&#3627408465;&#3627408481;
=??????
1
&#3627408462;&#3627408468;
????????????
??????&#3627408476;&#3627408480;
&#3627408462;&#3627408468;
????????????
t
3
4
5
6
7
8
9

16. Surge Tanks
•Or
&#3627408465;&#3627408487;
&#3627408465;&#3627408481;
&#3627408481;=0
=??????
1
&#3627408462;&#3627408468;
????????????
. Therefore
•
??????&#3627408476;
??????
=??????
1
&#3627408462;&#3627408468;
????????????
…………………………....
•??????
1=
??????&#3627408476;
??????
????????????
&#3627408462;&#3627408468;
•Substituting for ??????
1in eqn. 9 gives
•??????=
??????&#3627408476;
??????
????????????
&#3627408462;&#3627408468;
sin
&#3627408462;&#3627408468;
????????????
&#3627408481;…………………
•NOTE: The water level in the Surge tank is
highest when
&#3627408462;&#3627408468;
????????????
&#3627408481;=
π
2
•This yields ??????
&#3627408474;&#3627408462;&#3627408485;=
??????&#3627408476;
??????
????????????
&#3627408462;&#3627408468;
•The time period of oscillation is given by
τ=2π
????????????
&#3627408462;&#3627408468;
•Since the water level can fluctuate between
−??????
??????????????????and +??????
??????????????????measured from the
steady water level, the height of the Surge
tank should be greater than 2??????
??????????????????.
•Similarly, the minimum water level should
be greater than ??????
??????????????????so that the water under
oscillation does not drop to the opening of
the pipe which would then cause an airlock.
10
11
π
2
1
-1
The Sine wave
π 3π
2
2π

17. Surge tank
•Example
•Water to a Peltonwheel (A peltonwheel is the rotating part of an impulse turbine) is supplied
through a 1.5km long pipeline 1m in dia. at the rate of 3m
3
/s. A 2.5m dia. Surge tank is
installed close to the turbine for protection against transients caused by a nozzle movement.
Consider a condition of full rejection i.ewhen the pipe discharge end is fully closed suddenly.
Estimate the consequent time period of oscillation of the water column in the Surge tank, and
the time required to reach the maximum fluctuation from the instant the discharge is stopped.
•Solution
•Pipe area, ??????=
??????
4
1
2
= 0.7854m
2
•Surge tank area, ??????=
??????
4
2.5
2
= 4.9087m
2
•&#3627408483;=
??????&#3627408476;
??????
=
3
4.9087
= 0.611 m/s
•
????????????
&#3627408462;&#3627408468;
=
4.9087&#3627408485;1.5&#3627408485;10
3
0.7854&#3627408485;9.81
= 30.91s
•??????
&#3627408474;&#3627408462;&#3627408485;=
??????
&#3627408476;
??????
????????????
&#3627408462;&#3627408468;
= 0.611x 30.91 = 18.89m
•Time period, τ=2π
????????????
&#3627408462;&#3627408468;
= 2π&#3627408485;30.91=194.21&#3627408480;
•The maximum amplitude occurs at
τ
4
=
194.21
4
= 48.55s from the onset of rejection