1 9.14 WATER HAMMER IN PIPES Introduction Consider the flow of water between two points A and B in a long pipe as shown in Fig 9.12. When the valve is suddenly closed, the momentum of flowing water is destroyed and a wave of high pressure is set up which is transmitted along the pipeline. This creates noise called knocking and the wave has the effect of hammering action on the walls of the pipe, hence known as water hammer or or hammer blow.
H
Valve A
B V
Fig 9.12
The pressure rise due to water hammer depends upon; (i) The velocity of flow of water in the pipe (ii) The speed at which the valve is closed (iii) The length of the pipe (iv) The elastic properties of pipe material as well as that of flowing fluid. The rise in pressure in some cases may be so large that the pipe may even burst. Hence it is essential to take this pressure into account in the design of pipes. Time Taken by Pressure Wave to Travel From Valve to Tank and Back Let: t = time taken by pressure wave L = length of pipe C = velocity of pressure wave Total distance= L+L=2L ∴
t =
Or
t =
dis tan ce travelled from valve to tan k and back velocity of pressure wave 2L C
…(9.1)
Valve closure is said to be; (i)
Gradual (slow) if
(ii)
Sudden if
t <
t >
2L C
2L C
Cases of water hammer in Pipes: The following cases of water hammer in pipes are to be considered; (i) Gradual closure of valve (ii) Instantaneous closure of valve in rigid pipes (iii) Instantaneous closure of valve in elastic pipes
h.m.m.
2 (i)
Pressure Due to Gradual Closure of Valve Consider a long pipe conveying a liquid ( Fig 9.12) and provided with a valve which is closed gradually. Let A = area of cross-section of the pipe L = length of pipe V= velocity of flow through pipe t = time in seconds required to close valve p= intensity of pressure wave produced Due to gradual closure, water is brought from initial velocity V to zero velocity in time t seconds. ∴Retardation of water = change of velocity/time
=
V−0 t
=
V t
Axial force producing retardation = mass of water in pipe AB × retardation = ρAL ×
V t
…(i)
Force due to pressure wave, = pressure × area of pipe = p ×A Equating equations (i) and (ii); ρAL
×
V t
…(ii)
= p ×A ρLV
∴ p=
…(9.2)
t
Pressure head, H=
or
H =
p
=
w
LV gt
ρLV
t
×
1 w
=
ρLV
t
×
1 ρg
=
ρLV ρgt
…(9.3)
Equation (1) gives the relation between increase of pressure due to water hammer in the pipe and time required to close valve. If t = 0, increase in pressure seems infinite, but from experiments, this is finite. The equation is thus valid for incompressible fluids, and rigid pipes. In the case of very high pressure, the liquid is compressed to some extent, and the pipe material is also stretched. (ii) Pressure Due to Instantaneous Closure in Rigid Pipes Consider pipe AB shown in Fig 9.12 when the valve is closed suddenly. Kinetic energy (K.E.) of the flowing water is converted to strain energy (S.E.) of water (if friction is neglected). Loss of K.E.,
=
1 2
mV 2 =
1 2
× ρAL × V
2
Gain of S.E. of water, = mean rise in pressure × volumetric strain of water in pipe × volume
h.m.m.
3
p
=
p
×
2
K
× AL =
1 p2 2 K
AL , where K= bulk modulus of water
It may be noted that K, like other moduli of elasticity may also be thought of as stress/strain i.e. p/(dV/V), and is also called the coefficient of compressibility. Equating K.E. to S.E., 1 2
ρALV
or ∴ p=
or
2
1 p2
AL 2 K 1 2K 2 = ρKV p 2 = ρALV 2 × 2 AL
ρKV
=
2
=
2
ρKV ×
ρ
= ρV
ρ
C=
p ρV
K
K ∵ ρ
= ρVC
ρ
= C
…(9.4)
Where C is the propagation velocity (or celerity of a pressure wave); and is the same as velocity of sound in a liquid of density ρ. The above theory is closely related to Mach’s number which is given as, M=
V K
ρ
(iii) Pressure Due to Instantaneous Closure in Elastic Pipes Consider pipe AB shown in Fig 9.12 when the valve is closed suddenly. Let: E = modulus of elasticity of the pipe material µ = Poison’s ratio for the pipe material ( µ =1/4, and varies from 0.25 to 0.34 for mild steel) p = increase in pressure due to water hammer d = internal diameter of the pipe t = thickness of pipe wall (small compared to outside diameter D of the pipe) σC= circumferential stress in pipe σL= longitudinal stress in pipe Sudden closure of valve produces a wave of high pressure of intensity p, which causes circumferential and longitudinal stresses in the pipe wall. Original K.E. of fluid will be converted partly into S.E. of the fluid, and partly into S.E. of pipe. Consider a length of pipe, x ;
Volume of fluid in length x = Original K.E. of fluid,
1 2
π
4
mV 2 =
d2x
1 π 2 2 ⋅ρ× d x⋅ V 2 4
S.E. of fluid = mean rise in pressure ×volumetric strain× volume p p π = ⋅ ⋅ d2 x 2 K 4
=
1 p2 ⋅ 2 K
⋅
π
4
d2 x
From knowledge of strength of materials, Longitudinal strain of pipe =
σL
E
−µ
σC
E
h.m.m.
4
σC
Circumferential strain in pipe = ∴ S.E.
−µ
E
σL
E
stored in the pipe material per unit volume of pipe wall,
∑ average stress ×
=
=
=
1 2
σ L
σL
1 2E
(σ
2
+ σC
L
1 σ + σC C E 2 E
2
− 2µσ C σ L
1 stress × strain 2
∑
σ C
−µ
E
strain =
−µ
σ L
E
)
For a thin walled pipe, it can be shown that: σL =
pd 4t
and
pd
σC =
2t
∴ Total
S.E. stored in pipe material, = S.E. per unit volume of pipe material × total volume of pipe material 2 2 1 pd pd pd pd = + − 2µ ⋅ × πdtx 2E 4 t 2 t 2t 4t =
=
=
=
p2 d2 1 ⋅ πdx 2E 16t πdx
2E
p2 d2 1 4t 4
p2 d2 4t
+ 1−
−
2µ ⋅
p2 d2 8t
2µ 2
p2 d (1 + 4 − 4µ ) 4 2 × 4E 2 p2 d πd x ⋅ ( 5 − 4µ ) 4 2 × 4E πd
2
⋅
+
x
⋅
Original K.E. of fluid = S.E. of fluid + S.E. of pipe 1
⋅
π
2 4
d2x × ρ ⋅ V2 = 1
or
∴
2
p
=
V
1 2K
1
Since
µ =
∴
p=V
4
+
ρV
2
1 p2 2
⋅
p2
K
⋅
π
4
d2x +
πd
2
x
4
⋅
p 2d 2 × 4Et
(5 − 4µ )
p 2d
(5 − 4µ ) 2 × 4Et d 1 (5 − 4µ ) + = p2 2K 2 × 4Et =
2K
+
1 2 ρV 2 d ( 5 − 4µ ) 2 × 4Et
=
V
ρ
1 2K
2
+
d ( 5 − 4µ ) 2 × 4Et
=
V
ρ
1 K
+
d ( 5 − 4µ ) 4Et
(for mild steel pipes) ρ
1 K
+
d
…(9.5)
tE
h.m.m.
5 Derivation of equations for longitudinal stress (σL), and hoop stress (σC): Longitudinal stress, σL: Consider a thin walled pipe acted upon by pressure due to water hammer as shown in Fig 9.13 (a).
p⋅
π
4
d
2
x σH
d
d σL
t (a)
(b)
Fig 9.13
Due to pressure on the ends of the cylinder, there is a longitudinal stress in the cylinder wall. Total force on the end of the cylinder wall due to fluid pressure p, = pressure × area π
= p × d2
…(i)
4
Force due to longitudinal stress σL in the pipe wall to counteract fluid pressure, = stress × mean circumferential thickness = σ L × πd × t (approximately) …(ii) (If D is taken as mean diameter, then area of wall tube is πDt exactly) For equilibrium, equate forces in equations (i) and (ii); σ L × πd × t
∴ ∴Longitudinal
stress, σ L
π
= p × d2 4
=
pd 4t
…(9.6)
Hoop stress, σC: Now consider equilibrium of wall and contents of a length x of half the cylinder as shown in Fig 9.13 (b); Due to fluid pressure, there is a downward force, = pressure × area = p×xd …(iii) This force must be balanced by upward force due to circumferential σC (or hoop stress σH), = stress × area = σH×2(x×t) …(iv) Equating equations (iii) and (iv); σH×2(x×t) = p×xd
∴Hoop
stress, σC = σH =
pd 2t
…(9.7)
In the above equations, p is the gauge pressure.
h.m.m.
6 Velocity of Pressure Wave in a Fluid: The velocity of pressure wave (same as velocity of sound) transmitted through a fluid depends on the density and the bulk modulus of the fluid. Consider a tube of fluid of unit cross-sectional area, through which a pressure wave is being transmitted from right to left as shown in Fig 9.14, having velocity V. Now bring the wave to rest by imagining the fluid to have a velocity V in the opposite direction. At section A, let p, v, and V be the intensity of pressure, velocity and the volume respectively. Let the corresponding values at section B be (p+dp), (v+dv) and (V+dV), with the maximum pressure occurring at B. Between sections A and B, pressure increases by dp, and the volume compressed per second increases by dV. At section A, V= a×v = v, (∵ area 'a' is unity ) …(i)
At section B, volume compressed/s, V+dV = a(v+dv) But a =1 and V = v Hence, dV = dv (comparing the two equations above) A
B V
…(ii)
unit area, a V+dV p +dp
p
p +dp
Fig 9.14
Force on fluid between sections A and B, = change of momentum per second = mass/s ×change of velocity ∴
[p − (p + dp)]a = ρaV × dv
or But Hence or
dp = -ρVdv dv = dV dp = -ρVdv dp dV
= −ρV
…(iii)
The bulk modulus of a fluid is given by; Bulk mod ulus, K =
∴
or
increase in pressure decrease in volume / original volume *
[* (decrease in volume/ original volume) = volumetric strain] dp dp K= = -V -dV/V dV dp K =- …(iv) dV V
h.m.m.
7 From equations (iii) and (iv), ρV =
K V
But V = v, then K
ρV =
v
, or v =
K ρ
The pressure wave is often denoted by C, ∴
K
C=
…(9.8)
ρ
Example 9.1: Water flows at a velocity of 1.5m/s in a pipe of length 2500m and diameter 500mm Determine the rise in pressure if a valve at the end of the pipe is closed in 25 seconds. Take C= 1460m/s Solution Given: V = 1.5m/s, L =2500m, Diameter =0.5m, t=25s, C=1460m/s 2L
The ratio t = Since t >
2L
C
=
2 × 2500 1460
= 3.42s
, closure of valve is gradual
C
For gradual closure, p=
ρLV
=
t
10 3 × 2500 × 1.5 25
= 150kN/m
2
Example 9.2: If the valve in Example 9.1 is closed in 2 seconds, determine the pressure 9 2 exerted at the valve, assuming the pipe to be rigid. Take K water = 2×10 N/m . Solution C=
K
=
2 × 10 9
ρ
The ratio t = ∴ t <
2L C
10 3 2L C
=
= 1414m / s
2 × 2500 1414
= 3.54s
, hence closure of valve is rapid
For rapid closure of valve in rigid pipe, p = ρ V
K
== ρVC = 103 × 1.5 × 1414 = 2101kN/m 2
ρ
Example 9.3: If in Example 9.1 , the thickness of the pipe is 10mm and the valve is suddenly closed, determine: (a) the rise in pressure (b) the circumferential and longitudinal stress developed in the pipe wall. Take E =200×10 9 N/m2,and K water = 9 2 2×10 N/m .
h.m.m.
8 Solution
(a)
p=V
ρ
1 K
+
d tE 10 3
= 1.5
1 2 × 10
+
9
6
2
=1.732×10 N/m
0.5 0.01 × 200 × 10 9
2
= 1732 kN/m
(b) (i) Circumferential stress: σC =
pd 2t
=
1.732 × 10 6 × 0.5 2 × 0.01
= 43.3 × 10
6
2
N / m 2 = 43300kN/m .
(ii) Longitudinal stress: σL =
pd 4t
=
1.732 × 10 6 × 0.5 4 × 0.01
=
2
21.65 × 10 6 N / m 2 = 21650kN/m .
h.m.m.
