presentation water hammer 24 12-2014

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.


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


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.

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


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


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

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


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


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


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


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


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.


The time history for point ”M“ (at midpoint of the pipe)



The time history for point B (at a distance x from the reservoir )


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


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).


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.


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


incompressible flow

Fig.1 Maximum and minimum head envelopes using incompressible flow theory


incompressible flow

Fig.2 Air and vessel volumes


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.


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



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



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