Page 193 - Water Engineering Hydraulics, Distribution and Treatment

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v = 12.2 m/s.
2
2
2
P / + (v ) ∕2g + z = P / + (v ) ∕2g + z + h .
2
f
1
1
2
1
2
2
2
278∕9.8+ (3.05) /(2 × 9.8) + 0 = P / + (12.2) /(2 × 9.8) + 0 + 0.
2
2
3
Here = 9.8 kN/m ,and g = 9.8 m/s
2
2
P / = 278∕9.8 + (3.05) /(2 × 9.8) – (12.2) /(2 × 9.8) = 21 m of water.
2
3
2
P = 21 m × 9.8 kN/m = 195 kN/m .
2
2
2
2
F = P A = (278 kN/m )( ∕4)(0.61) m = 81 kN toward the right
1
1
1
2
2
2
F = P A = (195 kN/m )( ∕4)(0.305) m = 14 kN toward the left.
2
2
2
(
∑
F ) = ( ∕g) Q (v − v ) 5.10 Additional Hydraulics Topics (5.58)
2
1
(81 − 14 − F ) = (9.8∕9.8)(0.90)(12.2 − 3.05).
x
F = 67 − 8 = 59 kN to the left acting on the water.
x
The force that is exerted by water on the reducer = 59 kN to the right.
The vertical component F is 0.
y
5.10.4 Drag and Lift Forces where
F = drag force, lb (N)
D
Drag is the component of the resultant hydraulic force exerted
F = lift force, lb (N)
L
by a fluid on a subject parallel to the relative motion of the
C = drag coefficient, dimensionless
fluid. Lift is the component of the resultant hydraulic force D
C = lift coefficient, dimensionless
exerted by the fluid on a subject perpendicular to the relative L
A = area projected on a plane perpendicular to the rela-
motion of the fluid. 2 2
tive motion of the fluid, ft (m )
The drag and lift forces are given by the following equa-
v = relative velocity of the fluid with respect to the sub-
tions:
ject, ft/s (m/s)
2
2
g = acceleration due to gravity, 32.2 ft/s (9.81 m/s )
3
3
2
F = C A (v ∕2 g) (5.61) = specific weight of fluid, 62.4 lb/ft (9.8 kN/m )
D
D
2
F = C A(v ∕2g) (5.62)
L
L
EXAMPLE 5.61 DETERMINATION OF DRAG FORCE
A flat plate 4 ft by 4 ft (1.22 m by 1.22 m) moves at 23 ft/s (7 m/s) normal to its plane. Determine the resistance to the plate moving
◦
◦
through water at 60 F (15.6 C). Assume a drag coefficient = 1.16 for length/width = 1.
Solution 1 (US Customary System):
◦
3
At 60 F, = 62.37 lb/ft (Appendix 3)
2
F = C A (v ∕2 g) (5.61)
D
D
2
2
3
2
= 1.16 (62.37 lb∕ft )(4 × 4ft )(23 ft∕s) ∕(2 × 32.2ft∕s )
= 9,513 lb.
Solution 2 (SI System):
◦
3
At 15.6 C, = 9.798 kN/m (Appendix 3)
2
F = C A (v ∕2g) (5.61)
D
D
2
2
2
3
= 1.16 (9.798 kN∕m )(1.22 × 1.22 m )(7 m∕s) ∕(2 × 9.8m∕s )
= 42.4kN.