Page 153 - Water Engineering Hydraulics, Distribution and Treatment

Solution 1 (US Customary System):
Q = (A )(v ) = (A )(v ).
A
B
B
A
3
2
2
2
15 ft /s = [(12∕12) × 0.785 ft ](v ft/s) = [(24∕12) × 0.785 ft ](v ft/s).
A
v = 19.1 ft/s.
A
v = 4.78 ft/s.
B
2
2
2
(v ) ∕2g = (19.1 ft/s) ∕2 × 32.2 ft/s = 5.6 ft.
A
2
2
2
(v ) ∕2g = (4.78 ft/s) ∕2 × 32.2 ft/s = 0.35 ft.
B
Apply the Bernoulli equation from A to B.
Use Eq. (5.5a), in which H , H ,and H are all zero. 2 B 5.2 Fluid Transport 131
a
l
e
2
2
[P / + (v ) ∕2g + Z ] + 0 − 0 − 0 = P / + (v ) ∕2g + Z .
A
B
A
A
B
B
[25 + 5.6 + 10] = P / + 0.35 + 25 assuming datum D–D.
B
[25 + 5.6 + 0] = P / + 0.35 + 15 assuming datum A.
B
For either datum, we obtain P / = 15.25 ft of water.
B
The total energy at any section can be plotted based on a chosen datum plane. Using D–D in this case,
2
Energy at A = P / + (v ) ∕2g + Z = 25 + 5.6 + 10 = 40.6 ft.
A A A
2
Energy at B = P / + (v ) ∕2g + Z = 15.25 + 0.35 + 25 = 40.6 ft.
B B B
It is noted that transformation from one form of energy to another occurs during flow. In this case a portion of both the pressure
energy and the kinetic energy at A is transformed to potential energy at B.
Solution 2 (SI System):
Q = (A )(v ) = (A )(v ).
B
B
A
A
2
3
2
2
2
0.4245 m /s = [(0.3048) × 0.785 m ](v m/s) = [(0.6096) × 0.785 m ](v m/s).
A
B
v = 5.82 m/s.
A
v = 1.46 m/s.
B
2
2
(v ) ∕2g = (5.82) ∕2 × 9.8 = 1.72 m.
A
2
2
(v ) ∕2g = (1.46) ∕2 × 9.8 = 0.109 m.
B
Apply the Bernoulli equation from A to B.
Use Eq. (5.5a), in which H , H ,and H are all zero.
a
e
l
2
2
[P / + (v ) ∕2g + Z ] + 0 − 0 − 0 = P / + (v ) ∕2g + Z .
B
B
B
A
A
A
[7.62 + 1.72 + 3.048] = P / + 0.109 + 7.62 assuming datum D–D.
B
[7.62 + 1.72 + 0] = P / + 0.109 + 4.57 assuming datum A.
B
For either datum, we obtain P / = 4.66 m of water.
B
The total energy at any section can be plotted based on a chosen datum plane. Using D–D in this case,
2
Energy at A = P / + (v ) ∕2g + Z = 7.62 + 1.72 + 3.048 = 12.39 m.
A A A
2
Energy at B = P / + (v ) ∕2g + Z = 4.66 + 0.109 + 7.62 = 12.39 m.
B
B
B
It is noted that transformation from one form of energy to another occurs during flow. In this case a portion of both the pressure
energy and the kinetic energy at A is transformed to potential energy at B.