Page 141 - Water Engineering Hydraulics, Distribution and Treatment

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5.1 Fluid Mechanics, Hydraulics, and Water Transmission
◦
y = (1.68 m/Cos 60 ) + (1.37 m∕2) = 4.04 m.
cg
3
4
3
4
I = bh ∕12 = (1.98 × 1.37 m )∕12 = 0.4243 m .
cg
3
y = [I /y A] + y ,where A = b × h,and I = bh ∕12.
cp
cg
cg
cg
cg
4
y = [I /y A] + y = [0.4243 m /(4.04 m × 1.37 m × 1.98 m)] + 4.04 m = 4.08 m.
cg
cp
cg
cg
◦
h = y Sin 30 = 4.08 m × 0.5 = 2.04 m.
cp
cp
EXAMPLE 5.14 WATER, SPECIFIC WEIGHT, FORCE, AND MOMENT OF INERTIA
Water rises to level E in the pipe attached to water storage tank ABCD in Fig. 5.8. Neglecting the weight of the tank and riser pipe,
determine (a) the resultant force acting on area AB, which is 8 ft wide (2.4 m); (b) the location of the resultant force acting on area
AB; (c) the pressure on the bottom BC; and (d) the total weight of the water.
E
A = 1 ft 2 12 ft (3.7 m)
2
(0. 093 m )
6 ft (1.8 m) A D
B C
20 ft (6.1 m)
Figure 5.8 Water tank for Example 5.14.
Solution 1 (US Customary System):
a. The resultant force, P ,actingonareaAB:
f
P = ( h )A
f cg
3
= (62.4lb∕ft )(12 ft + 3 ft)(6 ft × 8ft)
= 45,000 lb.
b. The location of the resultant force acting on area AB:
y cp = [I ∕y A] + y cg
cg
cg
I cg = (1∕12)bh 3
3
y cp = [8(6) ∕12]∕[15(6 × 8)] + 15
= 0.20 + 15
= 15.2ft.
c. The pressure, p, on the bottom BC:
p = h
3
2
= (62.4lb∕ft )(6 ft + 12 ft)∕144 in. ∕ft 2
= 7.8psi.
d. The total weight, W,ofthe water:
W = V
3
= 62.4lb∕ft (20 × 6 × 8 + 12 × 1)ft 3
= 60,700 lb.