Page 187 - Water Engineering Hydraulics, Distribution and Treatment

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EXAMPLE 5.54 DETERMINATION OF DISCHARGE COEFFICIENT
Determine the discharge confident, C , for a hydraulic system when water flows through a standard 1-in. (25.4-mm)-diameter orifice
d
3
3
under an 18 ft (5.4864 m) head at a rate of 0.111 ft /s (0.0031435 m /s).
Solution 1 (US Customary System):
Q = C A (2gh)
d
2
0.111 = C ( ∕4)(1∕12) (2 × 32.2 × 18) .
d
C = 0.60.
d
Solution 2 (SI System):
0.5
Q = C A (2gh) 0.5 0.5 5.10 Additional Hydraulics Topics (5.45)
(5.45)
d
2 0.5
0.0031435 = C ( ∕4)(0.0254) (2 × 9.81 × 5.4864) .
d
C = 0.60.
d
EXAMPLE 5.55 DETERMINATION OF CONTRACTION COEFFICIENT
Water flows through 1-in. (25.4-mm)-diameter orifice under an 18 ft (5.4864 m) head. The jet strikes a wall 5 ft (1.524 m) away and
0.4 ft (0.1219 m) vertically below the center line of the contracted section of the water jet. Determine the coefficient of contraction,
C , if the coefficient of discharge, C , is known to be 0.60 from Example 5.54.
c d
Solution 1 (US Customary System):
x = vt (5.50)
1
y = ∕ gt 2 (5.51)
2
Eliminating t from the two equations one gets
2 2
x = (2 v ∕g)y.
2 2
(5) = (2 × v ∕32.2)(0.4).
v = 32.1ft∕s.
0.5
v = C (2gh) . (5.47b)
v
0.5
32.1 = C (2 × 32.2 × 18) .
v
C = 0.95.
v
C = C ∕C . (5.48a)
c
v
d
C = 0.60∕0.95 = 0.63.
c
Solution 2 (SI System):
x = vt (5.50)
1
y = ∕ gt 2 (5.51)
2
Eliminating t from the two equations one gets
2 2
x = (2 v ∕g)y.
2 2
(1.524) = (2v ∕9.81) (0.1219).
v = 9.67 m∕s.