Hydraulic Engineering Lec # 7:

Unsteady Flow through pipe lines, Water hammer.

Instantaneous and slow closure of valves

Prof. Dr.-Ing. Saqib Ehsan

Unsteady Flow

 Discharge through orifices and over weirs

under varying heads.

 Unsteady Flow through pipe lines, Water

hammer. Instantaneous and slow closure
of valves

 Surge wave in open channels.

 Unsteady Flow through pipe lines
 Let there be an incompressible
1 2
accelerating flow and dV is the change in
velocity in time dt. The energy equation
for unsteady flow (accelerating flow) can
be written as L
2 2
P1 V P V
  Z1  2 
1
 Z2  H L  Hi
2
 2g  2g Density=mass/volume
Volume=AL
Where : H i  Head consumed in accelerating the flow
b/w section 1 &2 γ=ρg

Mass of fluid = AL
g The last term in eq. 1 will be
 dV –ve if the flow is retarding
Force required to accelerate the flow =F=ma= AL
g dt flow and vice versa.

dV If the valve at section 2 is
Pressure required for accelerating =P=F/A= L
g dt suddenly/very rapidly closed,
P L dV the pressure head at this
Hence Head required to accelerate flow=hi= =
 g dt section will suddenly rise up
V12
P1 P2 V22 L dV by an amount of LdV/gdt and
   Z1    Z2  H L   Eq.1
 2g  2g g dt vice versa
 Establishment of Steady Flow:
1
 Determining the time for the flow to
become steady in a pipeline when
a valve is suddenly opened at end H
L
of the pipe can be accomplished
through application of Eq. 1.
 Immediate after valve is opened, V2 V 2 L dV L
2
H K  ; KK f
the head H is available to 2g 2 g g dt D
accelerate the flow. Thus flow Let us define the steady flow velocity by Vo .
commences, but as the velocity Noting that for steady flow dV/dt=0, we get
increases the accelerating head is
2 gH
reduced by fluid friction and minor Vo 
(1  K )
losses.
substituting H from this eq. into above eq.
 Let us assume the total head hL
 2 L  dV
can be expressed as KV2/2g. dt    2
 1  k  Vo  V
2

 Applying eq. 1 between section 1 & Integrating and noting that constant of integration
2.
is zero, since V=0 at t=0 and lnVo/Vo=0,
L V V
t ln o  Eq.2
1  K Vo Vo  V
 Establishment of Steady Flow:
 Eq. 2 indicated that V approaches
Vo asymptotically and that the 1
equilibrium will be attained only
after an infinite time, but it must be
H
remembered that this is an L
idealized case. In reality there will
be elastic wave and damping, so 2
that true equilibrium will be
reached in a finite time.
1.0

V
Vo

t
 Water Hammer (Hammer Blow)

 Whenever velocity in a pipe line is reduced instantaneously or in a

very short time a sudden increase in pressure takes place. This
sudden rise in pressure in a pipe line due to stoppage of flow is known
as water hammer or hammer blow.
 The increase in pressure is so high that the pipe may burst if it is not
strong enough to withstand that pressure.
 Water hammer is used for all type of fluids, its not specific for water.
Water Hammer (Hammer Blow)
 (a). Normal steady flow  (c). Valve closed
taking place: (Pressure head instantaneously : (Pressure
throughout is P/γ) wave travels further and pipe
wall stretches)
P/γ V
Pi/γ
P/γ V
L
Cp

 (b). Valve closed  (c). After Time L/Cp: (The

instantaneously: (Pressure entire pipe is under very high
wave travels u/s with a pressure)
velocity Cp. The pressure
Pi/γ V
rises to Pi/γ and the wall
stretch)
Pi/γ L
P/γ V
Cp=Celerity=Velocity of
Cp pressure wave
Movement of Pressure Wave
 1. Initial Condition

Po
V V
 2. Transmission of original
wave (Valve closed at t=0) L

Po Cp V=0 Pi  5. Reflected wave reaches

the valve
V Po
 3. Wave reaches reservoir V
Pi V=0  6. Reflection of negative
wave from closed end
 4. Reflected wave canceling
original pressure moving V
Po Pi
back towards the valve Cp V=0

Po Cp V=0 Pi
V
Movement of Pressure Wave
 7. Negative wave reaches
the reservoir end.
Pi V
V=0
L
 8. Reflection of negative
wave from reservoir as +ve  The Movement of wave is
wave canceling the pressure summarized below:
drop
V Pressure Wave at t=0
Po Pi
Cp V=0
Wave of unloading

Wave of reflection
 9. Initial Condition again
Wave of unloading
Pi
V Pressure wave
Movement of Pressure Wave
 Ideally, there would be a series of pressure waves
traveling back and forth over the length of pipe b/w high
and low pressures about the valve for zero flow.
 In actual practice, as a result of friction of the pipe and
the different values of elasticity for pipe material and
liquid, the amplitude of each reflected wave would be
less than the previous wave. This reduction in the
pressure of wave is know as damping of waves and the
result is that after sometimes the situation becomes
normal.
 However, the most critical condition in pipeline would be
when valve is instantaneously closed and critical section
is the one very close to the valve.
 Due to movement of pressure wave u/s and d/s of
pipeline, a sound is sometimes heard which is known a
knocking.
Pressure Variation with time at different
sections.

x
C B A

L
2L/Cp

+ +
Point A Time
- -

2x/Cp 2L/Cp

+ +
Point B Time
- -
2(L-x)/Cp
Equation for Water Hammer Pressure

 Types of Valve Closures

 Instantaneous Valve Closure: t=0
 Rapid Valve Closure: t<2L/Cp
 Slow Valve Closure: t>2L/Cp
 Where Cp= Celerity or velocity of pressure wave
 Equation for Water Hammer Pressure (Rigid Pipe
Theory)

 Let the valve is closed  For rapid valve closure

instantaneously. Let the
pressure wave cover a distance  Pressure=Ph = ρ.Cp.dv
Cp.dt in time period dt and  For slow valve closure
change in velocity is dv. Volume  Pressure=Ph =2ρL(V1-V2’)/t
of water the is compressed  For a rigid pipe
against valve and comes to rest  Cp=(Ev/ ρ)1/2
in time dt is
 For an elastic pipe
 Volume =Cp.dt.A
 Mass = ρ.Cp.dt.A
 Cp=(Ev/ (ρ(1+D/t+Ev/E))1/2
 Force = ρ.Cp.dt.A. dv/dt  Where
 Pressure=Ph = ρ.Cp.dv  Ev=Vol. Modulus of Elasticity
 Where h stands for water hammer  E=Young’s Modulus of Elasticity
pressure  D/t= Diameter to thickness ratio
 ρ = mass density of pipe
Methods of eliminating/controlling water hammer
pressure in pipelines

 1. Slow Valve Closure

 2. Pressure Relief Valve
 3. Surge Tanks
 Questions
 P1. water from a reservoir flowing through a rigid pipe 15cm diameter with a
velocity 2.4m/sec is completely stopped by a valve situated at 1000m from
the reservoir. Determine the maximum rise of pressure when valve closure
takes place in
 A) 1 sec B) 5 sec
Assume that pressure increase at uniform rate & there is no damping of
pressure wave. Take Cp=1433 m/sec

A): 2 L / Cp  2 x1000 /1433  1.4sec B): 2 L / Cp  2 x1000 /1433  1.4sec

 tc <1.4 sec  Rapid Valve Closure  tc>1.4  Slow Valve Closure
Ph   Cpdv   Cp V  0   V V '  V 0 
Ph  2  L  1 2   2  L  1 
 t   t 
 1000 x1433 x 2.4
 2 x1000 x1000  2.5 / 5 
 3439200 N / m2
 1000000 N / m 2
Questions
 P2. A cast iron pipe 15cm diameter and 1.5 cm thick is conveying water
when outlet valve is suddenly closed. Calculate the maximum permissible
discharge if pressure rise is not to exceed 1700KN/m 2. Take Ev for water as
2.06 E6 KN/m2 and E for cast iron as 117 E6 KN/m 2

Cp 
Ev Ph   Cpdv
 D Ev 
 1    1700 x103  1000 x1323.48 xV
 t E 
V
2.06 x109
Cp  
 15 2.06 x109  Q D 2 xV
1000 1   9  4
 1.5 117 x10 
 1323.48m / sec
Q
Questions
 P3. A pipeline 60cm diameter is 1500m long & is carrying a discharge of 0.5
m3/sec. The valve at the outlet end is closed in 10 secs in such a way that
the time of retardation at any instant is proportional to the time elapsed since
the commencement of closure.
 A: calculate the rise in pressure
 B: Calculate the rise of pressure if the closure were practically instantaneous
 Take Ev=1895 E6 N/m2

dv dv
  t    ct Po Vo2 Po Vo2 L dv
dt dt   Zo  hf    Zo  hf 
 dv  ctdt  2g  2g g dt
0 10 Po  P1 Vo2 L dv
 
 dv    ctdt
Vo 0
 2 g g dt
Po  P1 Vo2 L
ct 2 10
   0.3536t 
V Vo  
0
 2g g
2
Po  P1
1

C  0.03536        m

dv
  0.3536t
dt
Practice Problems

 Ex. 13.4
 13.29, 13.30, 13.32