Page 136 - Water Engineering Hydraulics, Distribution and Treatment

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Water Hydraulics, Transmission, and Appurtenances
Chapter 5
Solution 2 (SI System):
Method A. Pressure balance method:
2
3
138.8 kN/m + (9.8 kN/m )(0.84 m + h) = (13.57 × 9.8 kN/m )h + P .
3
P = 0kN/m assumed.
A
138.8 + 9.8 × 0.84 + 9.8h = 13.57 × 9.8h.
h = 1.19 m.
Method B. Water column length method:
1psi = 2.307 ft of water = 0.703 m of water = 6.94 kN/m .
1mHg = 13.57 m of water. 2 3 A
2
1kN/m = (0.703/6.94) m H O = 0.1013 m H O
2
2
138.8 × 0.1013 m + 0.84 m + h = 13.57h.
h = 1.1 m.
EXAMPLE 5.7 DETERMINATION OF WATER FORCE ON RISER PIPE
Referring to Fig. 5.2, determine (a) the force exerted by water on the bottom plate AB of the 2-ft (0.61-m)-diameter riser
pipe for water storage and (b) the total force on plane C. The dimensions given are a = 2 ft (0.61 m), b = 12 ft (3.66 m),
c = 6 ft (1.83 m), and d = 10 ft (3.05 m).
Water surface
b c
a
d

A B
C
Figure 5.2 Riser pipe analysis.
Solution 1 (US Customary System):
2
3
3
a. P fAB = (16 × 2 × 0.785 ft )(62.4 lb/ft ) = 3,135 lb.
2
2
2
2
p AB = P fAB /A = 3,135 lb/[(2 × 0.785 ft )(144 in. /ft ) = 6.93 psi.
3
3
2
3
2
b. P = [(12 × 0.785 × 6ft ) + (2 × 0.785 × 10 ft )] (62.4 lb/ft ) = 44,282 lb.
fC
2
2
2
2
p = 44,282 lb/[(2 × 0.785 ft )(144 in. /ft )] = 97.93 psi.
C
Solution 2 (SI System):
2
3
3
a. P fAB = (1.83 + 3.05) × 0.61 × 0.785 m (1,000 kg/m ) = 1,425.4 kg.
2
2
2
p AB = P fAB /A = 1,425.4 kg/(0.61 × 0.785 m ) = 4,880 kg/m .
2
3
3
2
3
b. P = [(3.66 × 0.785 × 1.83 m ) + (0.61 × 0.785 × 3.05 m )](1,000 kg/m ) = 20,131 kg.
fC
2
2
2
p = 20,131 kg/(0.61 × 0.785 m ) = 68,919 kg/m .
C
EXAMPLE 5.8 HYDRAULIC FORCE EQUILIBRIUM
2
2
2
2
In Fig. 5.3, the areas of the hydraulic plunger A and cylinder B are 6.5 in. (42.52 cm ) and 650 in. (4,193.15 cm ), respectively.
The weight of B is 9,500 lb (4,313 kg). The vessel and the connecting passages are filled with water of specific gravity 1.0. What
force P is required for hydraulic force equilibrium, neglecting the weight of A? Distance d = 16 ft = 4.88 m.
f