Page 189 - Water Engineering Hydraulics, Distribution and Treatment

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5.10 Additional Hydraulics Topics
The tensile stress, S , developed in a moving fluid can
Increase in water hammer pressure is caused by the sud-
t
den closing of a valve or a power outage. The change in
pressure is calculated by
S = (ΔP) r∕t
t
k
(ΔP) = c (Δv)
(5.53)
where
where
2
2
2
S = tensile stress of pipe, lb/ft (N/m )
ΔP = change in pressure, lb/ft (N/m )
t
2
3
3
ΔP = change in pressure, lb/ft (N/m )
= fluid density, slug/ft (kg/m )
c = celerity, ft/s (m/s)
r = pipe radius, ft (m)
t = pipe wall thickness, ft (m)
Δv = change in fluid velocity, ft/s (m/s)
k
For rigid pipes, such as cast iron pipes, the celerity of be calculated by the following equation: 2 2 (5.56)
Equation (5.56) can be used for moving fluid as well as
the pressure wave is obtained by non-moving fluid in a pipe or cylindrical liquid storage tank.
When the equation is used for a tank, ΔP is the internal–
0.5
c = (E ∕ ) (5.54) external pressure difference, or the measured gauge pressure
B
where (actual pressure =−14.7 psi).
c = celerity, ft/s (m/s) A spherical liquid storage tank is twice as strong as a
2
2
E = bulk modulus of the fluid, lb/ft (N/m ) cylindrical tank with hemispherical ends. In such a case:
B
3
3
= density, slug/ft (kg/m )
1
S = ∕ 2(ΔP) r∕t k (5.57)
t
For non-rigid pipes, such as PVC pipes, the expression
for celerity is where
2
2
S = tensile stress in tank wall, lb/ft (N/m )
t
0.5
c = {E ∕ [1 + (E ∕E)(D∕t )]} (5.55) ΔP = internal–external pressure difference, that is, gauge
B
w
B
2
2
where pressure, lb/ft (N/m )
2
2
E = modulus of elasticity of pipe wall, lb/ft (N/m ) r = tank radius, in. (mm)
D = inner diameter of pipe, ft (m) t = tank wall thickness, in. (mm)
k
t = pipe wall thickness, ft (m)
w
EXAMPLE 5.57 VELOCITY OF PRESSURE WAVES
◦
◦
Determine the velocity of pressure waves (celerity) traveling along a rigid pipe containing water at 60 F (15.6 C).
Solution 1 (US Customary System):
◦
From Appendix 3 at 60 F,
3 2 7 2
E = 311 × 10 lb∕in. = 4.478 × 10 lb∕ft .
B
3 2
= 1.938 slug∕ft (1 slug = 1lb-s ∕ft).
c = (E ∕ ) 0.5 (5.54)
B
7 0.5
c = (4.478 × 10 ∕1.938) .
c = 4,807 ft∕s.
Solution 2 (SI System):
◦
From Appendix 3 at 15.6 C,
7 2
E = 214 × 10 N∕m .
B
3
= 999.1kg∕m .
c = (E ∕ ) 0.5 (5.54)
B
7 0.5
c = (214 × 10 ∕999.1) .
c = 1,463 m∕s.