Page 152 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
Solution 1 (US Customary System):
P = ( h )A = P .
f
cg
EB
◦
EB = 8.5 ft/Sin (60 ) = 9.81 ft.
h = 8.5∕2 = 4.25 ft.
cg
2
A = 9.81 ft × 4.5 ft = 44.2 ft .
3
2
= (62.4 lb/ft )(4.25 ft)(44.2 ft ) = 11,722 lb.
P
EB
= 11,722 lb acting at (2∕3) × 9.81 ft = 6.54 ft from E, and perpendicular to gate EB.
Force P
EB
3
P
= (62.4 lb/ft )(8.5 ft)(3.5 ft × 4.5 ft) = 8,354 lb, acting at the center of gravity of BC since the pressure on BC is uniform.
BC
Taking moment about B (clockwise plus)
Unbalanced moment M = 11,722 lb × (1∕3) × 9.81 ft − 8,354 lb × (1∕2) × 3.5ft
= 38,331 lb-ft − 14,620 lb-ft
= 23,711 lb-ft clockwise.
Solution 2 (SI System):
P = ( h )A = P .
EB
cg
f
◦
EB = 2.59 m /Sin (60 ) = 2.99 m.
h = 2.59 m∕2 = 1.3 m.
cg
2
A = 2.99 m × 1.37 m = 4.1 m .
2
3
P EB = (9.81 kN/m )(1.3 m)(4.1 m ) = 52.23 kN.
Force P EB = 52.23 kN acting at (2∕3) × 2.99 m = 1.99 m from E, and perpendicular to gate EB.
3
P BC = (9.81 kN/m )(2.59 m)(1.07 m × 1.37 m) = 37.25 kN, acting at the center of gravity of BC since the pressure on BC is
uniform. Taking moment about B (clockwise plus)
Unbalanced moment M = 52.23 kN × (1∕3) × 2.99 m − 37.25 kN × (1∕2) × 1.07 m
= (52.05 − 19.93) kN-m
= 32.12 kN-m clockwise.
Note: 1 kN-m = 737.6 lb-ft.
EXAMPLE 5.26 HYDRAULIC GRADE LINE AND ENERGY GRADE LINE
3
3
In Fig. 5.14, water flows from A to B at a flow rate of 15 ft /s (0.4245 m /s) and the pressure head at A is 25 ft (7.62 m). Determine
the pressure head at B assuming no loss of energy from A to B, d = 12 in. (30.48 cm), d = 24 in. (60.96 cm), Z = 10 ft
A
A
B
(3.048 m), Z = 25 ft (7.62 m).
B
Energy grade line (EGL)
2 2
v A v B
2g 2g
Hydraulic grade line
P B
(HGL) γ
24″
P A B
γ
A Z B
12″
Z A
Datum plane
D D
′′
Figure 5.14 Hydraulic and energy grade lines. Conversion factor: 1 = iin. = 2.54 cm.