Beirut Arab University Faculty of Engineering
Mechanical Engineering - Fluid Department
Pipe Line Design
Fall 2014-2015
WATER HAMMER
Presenters: - E. AbdAllah El-Masri Supervisors: - Prof.Ali Hammoud
Water hammer phenomena
General Introduction to WATER HAMMER
Water Hammer Causes and Effects
Calculation of Water Hammer
How to Avoid Water Hammer
Introduction to SURGE TANK
SURGE TANK Calculation
Summary
Solved Examples
General Introduction to Water Hammer
Water hammer (or, more generally, fluid hammer) is a pressure surge or wave resulting when a fluid (usually a liquid but sometimes also a gas) in motion is forced to stop or change direction suddenly (momentum change). Water hammer commonly occurs when a valve is closed suddenly at an end of a pipeline system, power failure, main breaks, pump start-up and shut-down operations, check-valve slam, rapid demand variation, opening and closing of fire hydrants and a pressure wave propagates in the pipe etc.
Transient generating events are capable of producing both positive and negative pressure waves which travel at approximately the speed of sound in water.
This pressure wave can cause major problems, from noise and vibration to pipe collapse. It is possible to reduce the effects of the water hammer pulses with accumulators and other features.
The excessive pressure may fracture the pipe walls or cause other damage to the pipeline system.
Rough calculations can be made either using the Joukowsky equation, or more accurate ones using the method of characteristics.
Water Hammer Causes and Effects
If the pipe is suddenly closed or opened at the outlet (downstream), the mass of water before the closure is still moving forward with some velocity, building up a high pressure and shock waves and hammering noise. Water hammer can cause pipelines to break if the pressure is high enough. Air traps or stand pipes (open at the top) are sometimes added as dampers to water systems to provide a cushion to absorb the force of moving water in order to prevent damage to the system. (At some hydroelectric generating stations what appears to be a water tower is actually one of these devices, known as a surge drum)
On the other hand, when a valve in a pipe is closed, the water downstream of the valve will attempt to continue flowing, creating a vacuum that may cause the pipe to collapse or implode. This problem can be particularly acute if the pipe is on a downhill slope. To prevent this, air and vacuum relief valves, or air vents, are installed just downstream of the valve to allow air to enter the line and prevent this vacuum from occurring.
Other causes of water hammer are Pump failure, and Check valve slam (due to sudden deceleration, a check valve may slam shut rapidly, depending on the dynamic characteristic of the check valve and the mass of the water between a check valve and tank).
Sudden opening or closing of valves in a pipeline.
Starting or stopping the pumps in a pumping system.
Operating errors or malfunctioning of equipment. Electricity shut off.
Improper operation of surge protection devices can do more harm than good. An example is oversizing the surge relief valve or improperly selecting the vacuum breaker-air relief valve
Water Hammer Causes and Effects
The sudden change of pressure due to a valve closure may be viewed as the
result of the force developed in the pipe necessary to stop the flowing water
column. The column has a total mass M and is changing its velocity at the rate
of dV/dt.According to Newton’s second law of motion,
F = m𝑑𝑉
𝑑𝑡
If the velocity of the entire water column could be reduced to zero instantly
F = 𝑚 𝑣0;0
0 =
𝑚𝑣0
0 = ∞
The resulting force (hence, pressure) would be infinite. Fortunately, such an
instantaneous change is almost impossible because a mechanical valve
requires a certain amount of time to complete a closure operation. In
addition, neither the pipe walls nor the water column involved are perfectly
rigid under large pressure. The elasticity of both the pipe walls and the water
column play very important roles in the water hammer phenomenon.
Water Hammer Causes and Effects
Calculation of Water Hammer
How to calculate the time of closing the valve to avoid water hammer:
1- Time of closing the valve , suddenly (high pressure ) or gradually ( pressure is less)
2- Velocity of flow in the which always should be less than 3 m/s , optimum 1.5 m/s
3- Length of the pipe, the shorter the pipe the higher possibility of water hammer
4- Elastic property of the pipe material and fluid elasticity
5- The speed of pressure wave
The time depends on :
A- The pipe length
B- The speed of pressure wave
Positive and negative pressure waves :
When the fluid flows in the pipe, if the valve close suddenly and stop the flow, the kinetic energy will be changed into elastic resilience and create a serial positive and negative pressure wave vibrating back and forth in the pipe until the energy lost by friction.
Liquid at valve stops, the kinetic energy of water it converts into potential, pressure here increases in ∆p or ∆h
Propagation of water hammer pressure wave 1- For t = 0, the pressure profile is steady, which is shown by the pressure head
curve running horizontally because of the assumed lack of friction. Under steady-
state conditions, the flow velocity is v0.
2- The sudden closure of the gate valve at the downstream end of the pipeline
causes a pulse of high pressure ∆h; and the pipe wall is stretched. The pressure
wave generated runs in the opposite direction to the steady-state direction of
the flow at the speed of sound and is accompanied by a reduction of the flow
velocity to v = 0 in the high pressure zone. The process takes place in a period of
time 0 < t <1/2 Tr, where Tr is the amount of time needed by the pressure wave
to travel up and down the entire length of the pipeline. The important
parameter Tr is the reflection time of the pipe. It has a value of 2L/a
Propagation of water hammer pressure wave 3- At t = 1/2Tr the pressure wave has arrived at the reservoir. As the reservoir
pressure p = constant, there is an unbalanced condition at this point. With a
change of sign, the pressure wave is reflected in the opposite direction. The flow
velocity changes sign and is now headed in the direction of the reservoir.
4- A relief wave with a head of ∆h travels downstream towards the gate valve
and reaches it at a time t = Tr. It is accompanied by a change of velocity to the
value ∆v0.
Propagation of water hammer pressure wave 5- Upon arrival at the closed gate valve, the velocity changes from -v0 to v = 0.
This causes a sudden negative change in pressure of ∆h.
6- The low pressure wave ∆h travels upstream to the reservoir in a time
Tr< t <3/2Tr, and at the same time, v adopts the value v = 0.
Propagation of water hammer pressure wave
Analysis of Water Hammer Phenomenon
The magnitude of water hammer depend on :
1- Time of closing the valve , suddenly (high pressure ) or gradually ( pressure is less)
2- Velocity of flow in the which always should be less than 3 m/s , optimum 1.5 m/s
3- Length of the pipe, the shorter the pipe the higher possibility of water hammer
4- Elastic property of the pipe material and fluid elasticity
5- The speed of pressure wave
Positive and negative pressure waves :
When the fluid flows in the pipe, if the valve close suddenly and stop the flow, the
kinetic energy will be changed into elastic resilience and create a serial positive and
negative pressure wave vibrating back and forth in the pipe until the energy lost by
friction.
Liquid at valve stops, the kinetic energy of water it converts into potential, pressure
here increases in ∆p or ∆h
Analysis of Water Hammer Phenomenon
The following cases of water hammer will be considered:
• Gradual closure of valve,
• Sudden closure of valve and pipe is rigid, and
• Sudden closure of valve and pipe is elastic.
The time required for the pressure wave to travel from the valve to the reservoir and back to the valve is:
Where:
L = length of the pipe (m)
C = speed of pressure wave, celerity (m/sec)
If the valve time of closure tc :
If the closure is considered gradual
If the closure is considered sudden C
Ltc
2
Analysis of Water Hammer Phenomenon
The speed of pressure wave “C” depends on :
1- The pipe wall material.
2- The properties of the fluid.
3- The anchorage method of the pipe.
; if the pipe is rigid
; if the pipe is elastic
And
bE
C
cE
C
eE
KD
EE pbc
11
Analysis of Water Hammer Phenomenon Where:
C = velocity (celerity) of pressure wave due to water hammer.
= water density ( 1000 kg/m3 ).
Eb = bulk modulus of water ( 2.1 x 109 N/m2 ).
Ec = effective bulk modulus of water in elastic pipe.
Ep = Modulus of elasticity of the pipe material.
e = thickness of pipe wall.
D = diameter of pipe.
K = factor depends on the connection and fixation of the pipe anchorage method:
K = for pipes free to move longitudinally,
K = for pipes anchored ( connection) at both ends against longitudinal movement
K = for pipes with expansion joints.
K= 1.0 (pipe is supported at one end)
where = poisson’s ratio of the pipe material (0.25 - 0.35).
It may take the value = 0.25 for common pipe materials
Analysis of Water Hammer Phenomenon If the longitudinal stress in a pipe can be neglected, k = 1.0, and equation can be
simplified
Ep = Modulus of elasticity of the pipe material
eE
D
EE pbc
11
Analysis of Water Hammer Phenomenon If the longitudinal stress in a pipe can be neglected, k = 1.0, and equation can be
simplified
Ep = Modulus of elasticity of the pipe material
eE
D
EE pbc
11
The effect of elasticity in the pipe
Pressure surge in an elastic pipe will cause the pipe to swell and some of the
energy will be absorbed by straining the pipe wall. This reduces the rise in
pressure. The more elastic the wall is, the less the pressure rise will be. Consider
the case shown.
Water Hammer pressure calculation
The Maximum pressure created by the water hammer
The total pressure experienced by the pipe is P = ∆P + 𝑃0
Water Hammer pressure calculation
There are 3 types for water hammer pressure calculation:
1- Gradual Closure of Valve
2- Sudden Closure of Valve (Pipe is Rigid)
3- Sudden Closure of Valve and Pipe is Elastic
Water Hammer pressure calculation 1- Gradual Closure of Valve:
Water is considered as a incompressible flow
Pipe is considered rigid
Liquid is brought to rest by uniform deceleration
In this case:
Consider a pipe line with a fluid flowing at a steady velocity V0 m/s . A valve is gradually closed thus decelerating the fluid uniformly from V0 to zero in a time "t “second. The rise of pressure is calculated as follows:
Volume of Fluid = A x L
Mass of fluid flow = AL
Deceleration of flow = 𝑉;0
𝑡 =
𝑉
𝑡
Mass flow rate experience the deceleration force F = mass x deceleration =AL (V/t) = Ax P
P = L
𝑡 v0
The pressure Head
P LV LVH
gt gt
Water Hammer pressure calculation
2- Sudden Closure of Valve (Pipe is Rigid) T<2L/C:
If the valve is closed suddenly then as “t” is very small the pressure rise is very
large. In reality, a valve cannot close instantly but very rapid closure results in
very large pressure hammer. When this happen the compressibility of the fluid
and the elasticity of the pipe is an important factor in reducing the rise of
pressure.
First we considered the pipe is a rigid & water is elastic .
Water traveling distance is “L ” in time “ dt” , the traveling wave speed C =L/dt
that will be = dt= L/C
The Deceleration of flow = 𝑉;0
𝑑𝑡 =
𝑉
𝑑𝑡=
𝑉
𝐿/𝐶=
𝑉𝐶
𝐿
Force F = mass x deceleration =AL (VC/L)
Now the pressure =F/A = AL (VC/L)/A=VC
H𝑀𝑎𝑥 = C𝑔
v0 ; Joukowski's Law
Water Hammer pressure calculation
2- Sudden Closure of Valve (Pipe is Rigid):
If the time of closure T<2L/C, then the closure is said to be Sudden.
The pressure head due caused by the water hammer is P = Cv0
But for rigid pipe ; so
Note: Eb= bulk modulus of water = ( 2.1 x 109 N/m2)
g
VCH 0
bE
C
bE
g
VH 0
bEVP 0
Water Hammer pressure calculation
3- Sudden Closure of Valve and Pipe is Elastic:
If the time of closure T<2L/C, then the closure is said to be Sudden.
The pressure head due caused by the water hammer is P = Cv0
But for rigid pipe ; so
g
VCH 0
cE
C)
1(
10
eE
KD
E
g
VH
pb
)1
(0
eE
KD
E
VP
pb
Stresses in the Pipe Wall
After calculating the pressure increase due to the water hammer, we can find the stresses in the pipe wall:
• Circumferential (hoop) stress ”fc “:
• Longitudinal stress ”fL “:
where:
D = pipe inside diameter
tp = pipe wall thickness
P = P0 + P = total pressure= initial pressure (before valve closure) + pressure increase due water hammer.
tp
PD
2fc
tp
PD
4fL
Time History of Pressure Wave (Water Hammer)
The time history of the pressure wave for a specific point on the pipe is a
graph that simply shows the relation between the pressure increase ( P ) and
time during the propagation of the water hammer pressure waves.
Applying the water hammer formulas we can determine the energy gradient
line and the hydraulic gradient line for the pipe system under steady flow
condition.
So the total pressure at any point M after closure (water hammer) is
PM = PM,before closure + P
Or HM = HM,before closure + H
Time History of Pressure Wave (Water Hammer)
For example, considering point “A” just to the left of the valve.
Time history for pressure at point “A” (after valve closure)
Note: friction (viscosity) is neglected.
Time History of Pressure Wave (Water Hammer)
The time history for point ”M“ (at midpoint of the pipe)
Note: friction (viscosity) is neglected.
Time History of Pressure Wave (Water Hammer)
The time history for point B (at a distance x from the reservoir )
Note: friction (viscosity) is neglected.
This is a general graph where we can substitute any value for x (within the pipe length) to obtain the time history for that point.
How to Avoid Water Hammer The following characteristics may reduce or eliminate water hammer:
Low fluid velocities: To keep water hammer low, pipe-sizing charts for some applications recommend flow velocity at or below 5 ft/s (1.5 m/s).
Slowly closing valves
High pipeline pressure rating (expensive)
Good pipeline control (start-up and shut-down procedures)
Air vessels work in much the same way as water towers, but are pressurized. They typically have an air cushion above the fluid level in the vessel, which may be regulated or separated by a bladder. Sizes of air vessels may be up to hundreds of cubic meters on large pipelines. They come in many shapes, sizes and configurations. Such vessels often are called accumulators or expansion tanks.
A hydropneumatic device similar in principle to a shock absorber called a 'Water Hammer Arrestor' can be installed between the water pipe and the machine which will absorb the shock and stop the banging
Air valves are often used to remediate low pressures at high points in the pipeline. Though effective, sometimes large numbers of air valves need be installed. These valves also allow air into the system, which is often unwanted
Shorter branch pipe lengths
Shorter lengths of straight pipe, i.e. add elbows, expansion loops. Water hammer is related to the speed of sound in the fluid, and elbows reduce the influences of pressure waves.
With looped piping; lower velocity flows from both sides of a loop can serve a branch.
Flywheel on pump
Pumping station bypass
Hydroelectric power plants must be carefully designed and maintained because the water hammer can cause water pipes to fail catastrophically
Introduction to SURGE TANK
Air Vessels for Transient Protection of Large Pipe Networks
Transient protection of water distribution systems may require use of devices
such as open surge tanks, air vessels, air/vacuum valves, pressure relief valves
etc. Selection and design of suitable transient protection devices is dictated by
the severity of transient causing events, distribution system characteristics,
system operational procedures etc. Bong and Karney (2006) observes that
“designing these critical transient protection systems is a challenging problem…”
and “selection, installation, and operation of these hydraulic devices strongly
depend on the specifics of the particular pipe system as well as
experience/comfort of the designer/operator.” If not properly designed, these
devices can worsen the transient response of the system (Bong andKarney 2006).
Introduction to SURGE TANK
Air vessels, also known as closed surge tanks, are effective in protecting the
distribution system against negative as well as positive pressures and are widely
used in water distribution systems. Air vessels are expensive and their size used
in practice varies from a few tens of cubic feet to few thousand cubic feet. A
1500ft3 bladder surge tank, which is a more advanced form of a general air
vessel, could cost nearly $50,000. Optimal sizing of air vessels thus becomes an
important task of transient modeling and
protection design studies.
Typical arrangement of an air vessel, shown
in Figure 1.1, consists of three components
(1) the vessel (2) the connector pipe and
(3) inlet and outlet orifices controlling flow to
and from air vessel.
Introduction to SURGE TANK
One-way surge tanks Standpipes can only be installed at points of a piping
system characterized by low-pressure heads. As a rule, a standpipe cannot
replace a downstream air vessel. Fitted with a swing check valve in the
direction of the flow and a filling mechanism (one-way surge tank), it is used
to stop the pressure falling below atmospheric at the high points of long
clean-water pipelines. Because of the possibility of malodorous fumes,
standpipes are rarely found in waste water installations. Standpipes and one-
way surge tanks are highly reliable pieces of equipment provided the
following points are observed:
• Continuous or regular changes of water (problem of hygiene).
• Filtering of air flow.
• Functional tests of the check valve on one-way surge tank arrangements.
• Monitoring of water level or filling device on one-way surge tank arrangement.
Where use Surge tank
If the pumping system is not controlled or protected, contamination and damage to equipment and the pipeline itself can be serious. The effects of surges can be as minor as loosening of pipe joints to as severe as damage to pumps, valves, and concrete structures. Damaged pipe joints and vacuum conditions can cause contamination to the system from ground water and backflow situations. Uncontrolled surges can be catastrophic as well. Line breaks can cause flooding and line shifting can cause damage to supports and even concrete piers and vaults. Losses can be in the millions of dollars so it is essential that surges be understood and controlled with the proper equipment.
Surge Tank connection detail
Surge Tank Sizing Surge tanks Sizing can done by three method:
1- Computer programs which are very accurate and very expensive
2- By using the equations for incompressible flow
3- By using a quick calculation
There are computer programs available for analyzing pipeline systems with
and without water hammer protection. These programs are almost a
prerequisite nowadays to obtain fast and accurate answers and to optimize
the sizing of water hammer protection systems. However, engineering
intuition and simple design guides are still of great practical use in selecting
the correct form of water hammer protection, if required, as well as for the
initial sizing and planning of the throttling system, if installed. simple form of
analysis is available in the form of the differential. A equations of motion for
incompressible flow. Whereas pump trip in an unprotected pipeline results in
a rapid drop in pressure and an elastic wave traveling up and down the
pipeline, in the case of a protected pipeline, such as with an air vessel, the
decelerations and accelerations are slower. Therefore incompressible flow or
surge theory is sometimes sufficient for the analysis. The number of variables
is considerably reduced and a more general analysis may be done in
dimensionless form in order to provide quick estimate nomographs.
Surge Tank Sizing using the equation for
incompressible flow The incompressible flow differential equations of motion were analyzed for a number of
cases in order to obtain a generalized air vessel volume as a function of the minimum
relative head at the pumping station.
Incompressible flow theory suggests the following relationship between decelerating head
on a water column and the rate of deceleration:
which may be integrated to obtain the maximum cavity volume remaining upstream
before the water column reverses:
By entering this equation to the computer program, the computer plot the nomenclature
and minimum and maximum head envelope for a generalized pipeline, and the initial air
volume, incompressible liquid!. That is, the line gives dimensionless gas volume as a
function of relative minimum head at the pump station. Also plotted are points from
various full elastic water hammer analyses. The upper line is a plot of dimensionless total
vessel volume
Surge Tank Sizing using the equation for
incompressible flow
Fig.1 Maximum and minimum head envelopes using incompressible flow theory
Surge Tank Sizing using the equation for
incompressible flow
Fig.2 Air and vessel volumes
Surge Tank Sizing using the equation for
incompressible flow Outlet and Inlet Pipe Sizes:
By applying the continuity equation to get a relation between diameter
entering the vessel and the head .
Fig.3 Air vessel inlet diameter as function of head rise to initial flow
The following symbols are used in this paper:
A = cross-sectional area of pipe;
De = diameter of outlet pipe ~internal diameter or bore!;
Di = diameter of inlet pipe;
Dp = main pipeline diameter;
g = gravitational acceleration;
H = head above pump level plus atmospheric head in meters of liquid ~absolute
head!;
Hmax = maximum head above pump level plus atmospheric head;
Hmin = minimum head above pump level plus atmospheric head;
H0 = static head above pump level plus atmospheric head;
h = head difference along pipeline at given point in time;
hmax = maximum head above H0 (Hmax5H01hmax);
hmin = minimum head below H0 (Hmin5H02hmin);
K = head loss coefficient;
k = gas expansion coefficient;
L = length of pipeline;
p = pressure;
Q0 = initial flow rate;
S = air vessel volume5S01Sw ;
S8 = dimensionless parameter S0gH0 /ALVo 2 ;
Sw = initial water volume in vessel;
S0 = initial gas volume in vessel;
T = time to decelerate water column;
t = time;Ve 5 velocity of liquid ~water! in outlet pipe from vessel;
Vr = return velocity in pipeline;
V0 = initial pipeline water velocity; and
x = distance along pipeline from pump.
Surge Tank Sizing using the equation for
incompressible flow Example:
A pumping pipeline, 900 mm in diameter, 18,000 m long, with a static head of 410 m, conveys water at an initial velocity of 1.4 m/s. The air vessel characteristics are calculated below to limitminimum head to 40% of the static head, and maximum to 40% above static, neglecting friction.
From Fig. 1, S’=1.0=S0gH0 /AL𝑉02. Therefore, air volume
S0=1x0.785x0.92x18,000x1.42/9.8x410=5.6 m3 of air.
From Fig. 2, SgH0 /AL𝑉02 =3. Therefore, vessel volume S=16.8 m3.
Outlet pipe diameter from
De=0.15 m
This is rather small and would result in a theoretical water velocity of nearly 40 m/s, so a compromise large diameter, e.g., 250 mm may be used, which would increase maximum heads, however.
Inlet pipe diameter from Fig. 3 for ghmax/𝑉02 =816
Di=0.1x0.9=0.09 m
A diameter of 100 mm would probably be selected.
Surge Tank Sizing using simple equation Solved example:
In a single pipeline (or main path through looped system), run a steady state
analysis and note the total flow from pump or pumps. Use the Inventory
Calculator to find the total length of pipe (note you can define a subset in group
mode for looped systems). Calculate the approximate time it takes for a wave to
travel down the pipeline and back. Take the average wave speed and multiply by
the pipe length. For example, if the wave speed is 3000 ft/s and the pipe length
is 10000 ft, the time is 10000x2/3000 = 6.67 seconds. If the initial flow is 15
cubic feet per second, then the approximate volume for the surge tank is 15 x
6.67 = 100 cubic feet.
Summary Water hammer occurs when the kinetic energy of a fluid is converted into
elastic energy. But only rapid changes of the flow velocity will produce this
effect, for example the sudden closure of a gate valve or the sudden failure
or tripping of a pump. Due to the inertia of the fluid, the flow velocity of the
liquid column as a whole is no longer capable of adjusting to the new
situation.
The fluid is deformed, with pressure transients accompanying the
deformation process. The reason why surge pressure is so dangerous is that it
travels at the almost undiminished speed of sound (roughly 1000 m/s for a
large number of pipe materials) and causes destruction in every part of the
piping system it reaches.
To avoid water hummer can use surge tank and the total volume of the air
vessel consists of the volume of fluid in the vessel and the gas volume. The
usual minimum and maximum values used for initial gas volume are 25% and
75% of total tank volume.
Solved example Water Hammer
Solved Examples A- Consider a long pipe AB as shown in Fig.1.1 connected at one end to a tank
containing water at a height of H from the center of the pipe. At the other end
of the pipe, a valve to regulate the flow of water is provided. When the valve is
completely open, the water is flowing with a velocity, V in the pipe. If now the
valve is suddenly closed, the momentum of the flowing water will be destroyed
and consequently a wave of high pressure will be set up. This wave of high
pressure will be transmitted along the pipe with a velocity equal to the velocity
of sound wave and may create noise called knocking. Also the wave of high
pressure has the effect of hammering action on the walls of the pipe and hence
it is also known as water hammer.
Fig. 1.1 The pressure rise due to water hammer depends upon :
(i) the velocity of flow of water in pipe, (ii) the length of
pipe, (iii) time taken to close the valve, (iv) elastic
properties of the material of the pipe. The following
cases of water hammer in pipes will be considered:
Gradual closure of valve,
Sudden closure of valve and considering pipe rigid, and
Solved Examples A-
Gradual Closure of Valve. Let the water is flowing through the pipe AB shown in
Fig. 11.32, and the valve provided at the end of the pipe is closed gradually.
Let A= area of cross-section of the pipe AB
L= length of pipe
V= velocity of flow of water through pipe
T= time in second required to close the valve, and
P= intensity of pressure wave produced.
Mass of water in pipe AB= ƿ x volume of water = ƿ x A x L
The valve is closed gradually in time 'T' seconds and hence the water is brought
from initial velocity V to zero velocity in time seconds.
Retardation of water = change of vilocity
𝑡𝑖𝑚𝑒 =
V;0
𝑇 =
V
𝑇
Retardation force = Mass x Retardation =ƿAL x V
𝑇
…(i)
If ƿ is the intensity of pressure wave produced due to closure of the valve, the
force due to pressure wave,
= ƿ x area of pipe = ƿ x A …(ii)
Solved Examples B-
Sudden Closure of Valve and Pipe is Rigid
Equation gives the relation between increase of pressure due to water hammer in
pipe and the time required to close the valve. If t = 0, the increase in pressure
will be infinite. But from experiments, it is observed that the increase in
pressure due to water hammer is finite, even for a very rapid closure of valve.
Thus equation is valid only for (i) incompressible fluids and (ii) when pipe is rigid.
But when a wave of high pressure is created, the liquids get compressed to some
extent and also pope material gets stretched. For a sudden closure of valve [the
valve of t is small and hence a wave of high pressure is created] the following
two cases will be considered:
Sudden closure of valve and pipe is rigid, and
Sudden closure of valve and pipe is elastic
Solved Examples B-
Consider a pipe AB in which water is flowing as shown in Fig. 1. Let the pipe is rigid and valve
fitted at the end B is closed suddenly.
Let A = Area of cross-section of pipe AB,
L = Length of pipe
V= Velocity of flow of water through pipe,
p = intensity of pressure wave produced,
K=Bulk modulus of water
When the valve is closed suddenly, the kinetic energy of the flowing water is converted into
strain energy of water if the effect of friction is neglected and pipe wall is assumed perfectly
rigid.
Loss of kinetic energy = 1
2 x mass of water in pipe x 𝑉2
= 1
2 x ƿAL x 𝑉2
Gain of strain energy = 1
2
𝑝2
𝐾x volume =
1
2
𝑝2
𝐾 x AL
Equating loss of kinetic energy to gain of strain energy 1
2ƿAL x 𝑉2=
1
2
𝑝2
𝐾 x AL
Solved Examples B-
Or 𝑝2= 1
2ƿAL x 𝑉2 x
2K
𝐴𝐿 = ƿ K𝑉2
p= ƿ K𝑉2 = V 𝑘ƿ = V √kƿ2
ƿ
= ƿV x C (* K / ƿ = C)
Where C = velocity* of pressure of pressure wave
Solved Examples C-
Sudden Closure of Valve and Pipe is Elastic
Consider the pipe AB in which water is flowing as shown in Fig. 1.1. Let the
thickness 't' of the pipe wall is small compared to the diameter D of the pipe and
also let the pipe is elastic.
Let E = Modulus of Elasticity of the pipe material, 1
𝑚= Poisson's ratio for pipe material,
p = Increase of pressure due to water hammer,
t= Thickness of the pipe wall,
D = Diameter of the pipe.
When the valve is closed suddenly, a wave of high pressure of intensity p will be
produced in the water. Due to this high pressure p, circumferential and
longitudinal stresses in the pipe wall will be produced.
Let ƒ𝑡 = Longitudinal stress in pipe
ƒ𝑐 = Circumferential stress in pipe,
The magnitude of these stresses are given as ƒ𝑡 = 𝑝𝐷
4𝑡 and ƒ𝑐 = =
𝑝𝐷
2𝑡
Solved Examples C-
Now from the knowledge of strength of material we know strain energy stored in
pipe material per unit volume
= 1
2𝐸 = ƒ²𝑡 + ƒ²𝑐 −
2 ƒ𝑡.ƒ𝑐
𝑚
= 1
2𝐸
𝑝𝐷
4𝑡
2+
𝑝𝐷
2𝑡
2−
2 .𝑝𝐷
4𝑡.𝑝𝐷
2𝑡
𝑚
= 1
2𝐸
𝑝2𝐷2
16𝑡2 + 𝑝2𝐷2
4𝑡2 − 𝑝2𝐷2
4𝑚𝑡2
Taking 1
𝑚=
1
4 (i.e Poisson ratio =
1
4 )
Strain energy stored in pipe material per unit volume
= 1
2𝐸=
𝑝2𝐷2
16𝑡2 + 𝑝2𝐷2
4𝑡2 − 𝑝2𝐷2
4𝑡2𝑥 4 =
1
2𝐸 x
𝑝2𝐷2
4𝑡2 = 𝑝2𝐷2
8𝐸𝑡2
Solved Examples C-
Total volume of pipe material = 𝜋D x t x L
Total strain energy stored in pipe material
= Strain energy per unit volume x total
volume
= 𝑝2𝐷2
8𝐸𝑡2 x 𝜋D x t x L = 𝑝2𝜋D3L
8𝐸𝑡
= 𝑝2 𝑥 𝜋𝐷2𝑥 𝐷𝐿
8𝐸𝑡 =
𝑝2 𝐴 . 𝐷𝐿
2𝐸𝑡
𝑛𝐷2
4= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑖𝑝𝑒 = 𝐴
Now loss of kinetic energy of water = 1
2m V² =
1
2ƿAL x V²
Gain of strain energy of water 1
2
𝑝2
𝐾 x volume =
1
2
𝑝2
𝐾x AL
Then, loss of kinetic energy of water = Gain of strain energy in water + Strain
energy stored in pipe material 1
2ƿAL x V² =
1
2
𝑝2
𝐾 x AL +
𝑝2 𝐴 . 𝐷𝐿
2𝐸𝑡
Divided by AL, ƿ𝑉2
2 =
1
2
𝑝2
𝐾 +
𝑝2𝐷
2𝐸𝑡 =
𝑝2
2
1
𝐾+
𝐷
𝐸𝑡 or pV² = p²
1
𝑘+
𝐷
𝐸𝑡
p² = pV²
1
𝑘:
𝐷
𝐸𝑡
or p = pV²
1
𝑘:
𝐷
𝐸𝑡
= V x 𝑝
1
𝑘:
𝐷
𝐸𝑡
Solved Examples D-
Time Taken by Pressure Wave to Travel from the Valve to the Tank and from
Tank to the Value
Let T = The required time taken by pressure wave
L = Length of the pipe
C = Velocity of pressure wave
Then total distance = L + L =2L
Time, T = 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒘𝒂𝒗𝒆 =
𝟐𝑳
𝑪
Problem 11.52 the water is flowing with a velocity of 1.5 m/s in a pipe of length
2500 m and of diameter 500 mm. At the end of the pipe, a valve is provide. Find
the rise in pressure if the valve is closed in 25 seconds. Take the value of C= 1460
m/s
Solution Given:
Velocity of water, V = 1.5 m/s
Length of pipe, L = 2500 m
Diameter of pipe, D = 500 mm = 0.5 m
Time to close the valve, T= 25 seconds
Value of, C = 1460 m/s
Let the rise in pressure = p
Solved Examples D-
The ration, 2L/C = (2.2500 )/1460=3.42
From equation (11.33), we have if T >2L/C , the closure of valve is said to be
gradual.
Here T = 25 sec and 2L/C = 3.42
T >2L/C and hence valve is closed gradually
For graduall closure of valve, the rise in pressure is givenn by equation (11.31) as
p = (ƿVL )/(T ) = 1000 x 2500 x 1.5/2.5 = 150000 N/m²
= 150000/〖10〗^4 N/(cm^2 ) = 15.0 N/(cm^2 ). Ans.
Solved Examples E-
if in problem D, the valve is closed in 2 sec, find the rise in pressure behind the
valve. Assume the pipe to the rigid one and take Bulk modulus of water.
i.e., k = 19,62 x104 N/𝑐𝑚2
Solution. Given :
V = 1.5 m/s. L = 2500 m
D = 500 mm = 0.5 m
Time to close the valve, T = 2 sec
Bulk modulus of water, K = 19.62 x 104 N/𝑐𝑚2
= 19.62 x 104 x 104 N/𝑚2
Velocity of pressure wave is given by,
C= 𝐾
ƿ =
19.62 .10 3
1000 = 1400 m/s (ƿ = 1000)
The ratio, 2𝐿
𝐶 =
2 .2500
1400 = 3.57 T <
2𝐿
𝐶
From equation (11.34), if T < 2𝐿
𝐶, valve is closed suddenly, For sudden closure of
valve, when pipe is rigid, the rise in pressure is given by equation (11.35) or
(11.36) as
p = V = 𝐾ƿ = 1.5 19.62 . 103. 1000 (ƿ = 1000)
= 210.1 x 104 N/𝑐𝑚2.Ans.
Solved Examples F-
if in Problem D, the thickness of the pipe is 10 mm and the valve is suddenly
closed at the end of the pipe, find the rise in pressure if the pipe is considered to
be elastic. Take E = 19.62 x 1010 N/𝑚2 for pipe material and K = 19.62 x 104
N/𝑐𝑚2 for water. Calculate the circumferential stress and longitudinal stress
developed in the pipe wall.
Solution. Given :
V = 1.5 m/s, L = 2500 m, D= 0.5 m
Thickness of pipe, t = 10 mm = 0.01 m
Modulus of elasticity , E = 19.62 x 1010 N/𝑚2
Bulk modulus, K = 19.62 x 104 N/𝑐𝑚2 = 19.62 x 10108 N/𝑚2 For
sudden closure of the valve for an elastic pipe, the rise in pressure is given by
equation (11.37) as
p = V x 𝑝
1
𝐾:
𝐷
𝐸𝑡
= 1.5 x 1000
1
19.62 .108 : 0.5
19.62 .1010.0.1
= 1.5 x 1000
5.09 . 10−10: 2.54 .10−10
= 1715510 N/𝑚2 = 171.55 N/𝑐𝑚2. Ans.
Circumferential stress ( 𝑐) is given by
= 𝑝 .𝐷
2𝑡 =
171.55 . 0.5
2 . 0.1 = 4286.9 N/𝑚2
Longitudinal stress is given by, 𝑡 = 𝑝 .𝐷
4𝑡 =
171.55 . 0.5
4 . 0.1 = 2143.45 N/𝑐𝑚2 . Ans.
Solved Examples G-
A valve is provided at the end of a cast iron pipe of diameter 150 mm and of
thickness 10 mm. the water is flowing through the pipe, which is suddenly stopped by
closing the valve. Find the maximum velocity of water, when the rise of pressure due
to sudden closure of valve is 19.62 x 104 N/𝑐𝑚2and E for cast iron pipe as 11.772 x
106 N/𝑐𝑚2 .
Solution. Given
Diameter of pipe, D = 150 mm = 0.15 m
Thickness of pipe, t = 10 mm = 0.01 m
Rise of pressure, p = 196.2 N/𝑐𝑚2 = 196.2 x 104 N/𝑚2 Bulk modulus, K = 19.62 x 104 N/c𝑚2 = 19.62 x 103 N/𝑚2 Modulus of elasticity, E = 11.772 x 106 N/𝑐𝑚2 = 11.772 x 1010 N/𝑚2
For sudden closure of valve and when pipe is elastic, the pressure rise is given by
equation (11.37) as
p = V x 𝑝
1
𝐾:
𝐷
𝐸𝑡
= V x 1000
1
19.62 .108 : 0.15
11.72 . 1010 . 0.1
or 196.2 x 104 = V x 1000
5.09 .10−10 :1.274 .10−10
= V x 1000
6364 . 10−10 = V x 125.27 x 104
V = 196.2 . 104
125.27 . 104 = 1.566 m/s
Maximum velocity = 1.566 m/s. Ans.
Solved Examples H-
A valve is provided at the end of a cast iron pipe of diameter 150 mm and of
thickness 10 mm. the water is flowing through the pipe, which is suddenly stopped by
closing the valve. Find the maximum velocity of water, when the rise of pressure due
to sudden closure of valve is 19.62 x 104 N/𝑐𝑚2and E for cast iron pipe as 11.772 x
106 N/𝑐𝑚2 .
Solution. Given
Diameter of pipe, D = 150 mm = 0.15 m
Thickness of pipe, t = 10 mm = 0.01 m
Rise of pressure, p = 196.2 N/𝑐𝑚2 = 196.2 x 104 N/𝑚2 Bulk modulus, K = 19.62 x 104 N/c𝑚2 = 19.62 x 103 N/𝑚2 Modulus of elasticity, E = 11.772 x 106 N/𝑐𝑚2 = 11.772 x 1010 N/𝑚2
For sudden closure of valve and when pipe is elastic, the pressure rise is given by
equation (11.37) as
p = V x 𝑝
1
𝐾:
𝐷
𝐸𝑡
= V x 1000
1
19.62 .108 : 0.15
11.72 . 1010 . 0.1
or 196.2 x 104 = V x 1000
5.09 .10−10 :1.274 .10−10
= V x 1000
6364 . 10−10 = V x 125.27 x 104
V = 196.2 . 104
125.27 . 104 = 1.566 m/s
Maximum velocity = 1.566 m/s. Ans.
Solved Examples
I-
A steel pipe 5000 ft long laid on a uniform slope has an 18-in. diameter and a
2-in. wall thickness. The pipe carries water from a reservoir and discharges it
into the air at an elevation 150 ft below the reservoir free surface. A valve
installed at the downstream end of the pipe allows a flow rate of 25 cfs. If the
valve is completely closed in 1.4 sec, calculate the maximum water hammer
pressure at the valve. Neglect longitudinal stresses.
Solved Examples
I-
1
𝐸𝑐 =
1
𝐸𝑏 +
𝐷
𝐸𝑝𝑒
Where 𝐸𝑏 = 3.0 .105 psi, and 𝐸𝑝 = 2.8 . 107 psi, the above equation may thus be
written as 1
𝐸𝑐 =
1
3.0 .105 + 18
2.8 . 107 . 2.0
Hence, 𝐸𝑐 = 2.74 . 105 psi
C = 𝐸𝑐
ƿ=
2.74 . 105(144)
1.94= 4510 𝑓𝑡/𝑠𝑒𝑐
The time required for the wave to return to the valve is
t = 2𝐿
𝐶 =
2 .5000
4510 = 2.22 sec
because the water velocity in the pipe before valve closure is
𝑣0 = 25
𝜋
4 . (1.5)2
= 14.1 ft/sec
The maximum water hammer pressure at the valve can be calculated.
∆P = 𝜌𝑉0𝐶 = 1.94 . 14.1 .4510 = 1.23 . 105 lb/f𝑡2 (854 psi)
References
http://en.wikipedia.org/wiki/Water_hammer
http://en.wikipedia.org/wiki/Water_Hammer_Arrestor
Engineer's Handbook of Water Hammer Arresters Jay R. Smith Mfg. Co.P.O.
Box 3237 Montgomery, www.jrsmith.com
Accident at Russia’s Biggest Hydroelectric Sayano-Shushenskaya -2009 August
17-by Euler Cruz Consulting Engineer – Turbines /Rafael Cesário Mechanical
Engineer Brasil – 2009 Aug 24.
Water Hammer Practical Solutions by B.B Sharp & D.B Sharp.
WATER HAMMER ARRESTORS FOR HEAVY EQUIPMENT USA: 1-800-465-2736
www.mifab.com CAN: 1-800-387-3880
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