Page 192 - Water Engineering Hydraulics, Distribution and Treatment

170
Chapter 5
Water Hydraulics, Transmission, and Appurtenances
where
In the horizontal x-direction:
M = mass whose momentum changed in time t,slug
)
( ∑
(Mv ) ±
F
(5.60)
t = (Mv )
(kg)
y
y 1
y 2
t = time, s
F = sum of linear impulse forces, lb (N)
where
∑
(Mv ) = initial linear momentum in x-direction, slug-ft/s
v = exit velocity, ft/s (m/s)
2
x 1
(kg-m/s)
v = entrance velocity, ft/s (m/s)
1
3
3
(Mv ) = initial linear momentum in y-direction, slug-ft/s
Q = flow, ft /s (m /s)
y 1
2
2
(kg-m/s)
g = acceleration due to gravity, 32.2 ft/s (9.81 m/s )
3
3
(Mv ) = final linear momentum in x-direction, slug-ft/s
= specific weight of fluid, 62.4 lb/ft (9.8 kN/m )
x 2
3
3
(kg-m/s)
= density, slug/ft (kg/m )
(Mv ) = final linear momentum in y-direction, slug-ft/s
y 2
In the horizontal x-direction: (kg-m/s)
∑
F = sum of linear impulse forces in x-direction, lb
x
(N)
∑
F = sum of linear impulse forces in y-direction, lb
) y
( ∑
(Mv ) ± F t = (Mv ) (5.59)
x 1 x x 2 (N)
EXAMPLE 5.60 IMPULSE–MOMENTUM ANALYSIS
3
A 24 in. (610 mm) pipe is connected to a 12 in. (305 mm) pipe by a standard pipe reducer fitting. The water flow rate is 31.4 ft /s
2
3
2
(0.90 m /s) and the water pressure in the 24 in. pipe is 40 lb/in. (278 kN/m ). Determine the force that is exerted by water on the
reducer neglecting the minor head loss.
Solution 1 (US Customary System):
Q = A v = A v .
2 2
1 1
2
2
2
2
3
31.4 ft /s = [( ∕4)(24∕12) ft ]v = [( ∕4)(12∕12) ft ]v .
2
1
v = 10 ft/s.
1
v = 40 ft/s.
2
2
2
P / + (v ) ∕2g + z = P / + (v ) ∕2g + z + h .
1
2
f
2
2
1
1
2
2
40 × 144∕62.4 + (10) /(2 × 32.2) + 0 = P / + (40) /(2 × 32.2) + 0 + 0.
2
2
2
P / = 40 × 144∕62.4 + (10) /(2 × 32.2) − (40) /(2 × 32.2) = 69 ft of water.
2
3
2
P = 69 ft × 62.4 lb/ft = 4,306 lb/ft = 30 psi.
2
2
2
F = P A = (40 × 144 lb/ft )( ∕4)(24∕12) = 18,086 lb toward the right.
1
1
1
2
2
F = P A = (4,306 lb/ft )( ∕4)(12∕12) = 3,380 lb toward the left.
2
2
2
( )
∑
F = ( ∕g) Q (v − v ) (5.58)
2
1
(18,086 lb − 3,380 lb − F ) = (62.4∕32.2)(31.4)(40 − 10).
x
F = 14,706 − 1,825 = 12,881 lb to the left acting on the water.
x
The force that is exerted by water on the reducer = 12,881 lb to the right.
The vertical component F is 0.
y
Solution 2 (SI System):
Q = A v = A v .
1
1
2
2
3
2
2
2
2
0.90 m /s = [( ∕4)(0.610) m ]v = [( ∕4)(0.305) m ]v .
1
2
v = 3.05 m/s.
1