Page 170 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
Solution 1 (US Customary System):
2
3
= ( ∕4)(12∕12) (8.025) = 6.3 ft /s.
Q
= A v
AB AB
AB
2
= Q
Q
= ( ∕4)(6∕12) (v ).
= 6.3 = A v
CD
CD
CD CD
AB
2
2
(v ) ∕2g = (32.1) /(2 × 32.2) = 16 ft.
= 32.1 ft/s
v
CD
CD
2
2
= 8.025 ft/s
v
(v ) ∕2g = (8.025) /(2 × 32.2) = 1ft.
AB
AB
2
2
v
= 8.025 ft/s
(v ) ∕2g = (8.025) /(2 × 32.2) = 1ft.
EF
EF
The energy grade line drops in the direction of water flow by the amount of the lost head. The HGL is below the energy line by
the amount of the velocity head at any cross-section. It is important to note that the HGL can rise where a change (an enlargement in
pipe size) occurs. This is so because the enlargement decreases the velocity, hence the velocity head decreases and the pressure head
increases (conservation of total energy).
2
(h ) = f(L/D)(v ∕2g) = 0.02(200∕1)(1) = 4ft.
f AB
2
(h ) = K(v ∕2g) = 0.37 × 16 = 5.9 ft. (Note: K = 0.37 from Table 5.2)
f BC
2
(h ) = f(L/D)(v ∕2g) = 0.015(100∕0.5)(16) = 48 ft.
f CD
2
2
(h ) = (v – v ) ∕2g = (32.1 – 8.085) /(2 × 32.2) = 9 ft (see Section 5.2.3).
f DE D E
2
(h ) = f(L/D)(v ∕2g) = 0.020(100∕1)(1) = 2ft.
f EF
2
At From Head loss, h (ft) EGL elevation (ft) v ∕2g (ft) HGL elevation (ft)
f
A Level = 0 E 1 301.0 1.0 H 1 300.0
B A to B 4 E 2 297.0 1.0 H 2 296.0
C B to C 5.9 E 3 291.1 16 H 3 275.1
D C to D 48 E 4 243.1 16 H 4 227.1
E D to E 9 E 5 234.1 1.0 H 5 233.1
F E to F 2 E 6 232.1 1.0 H 6 231.1
Plotting and connecting the dots of H to H will form the HGL. Plotting and connecting the dots of E to E will form
1
6
6
1
the energy grade line as shown in Fig. 5.19.
Solution 2 (SI System):
2
3
Q AB = A v = ( ∕4)(0.3048) (2.446) = 0.1784 m /s.
AB AB
2
Q = Q = 0.1784 = A v = ( ∕4)(0.1524) (v ).
CD AB CD CD CD
2
2
v = 9.784 m/s (v ) ∕2g = (9.784) /(2 × 9.80) = 4.884 m.
CD CD
2
2
v = 2.446 m/s (v ) ∕2g = (2.446) /(2 × 9.80) = 0.305 m.
AB AB
2
2
v = 2.446 m/s (v ) ∕2g = (2.446) /(2 × 9.80) = 0.305 m.
EF EF
Read the explanation in Solution 1.
2
(h ) = f(L/D)(v ∕2g) = 0.02(61∕0.3048)(0.305) = 1.22 m.
f AB
2
(h ) = K(v ∕2g) = 0.37 × 4.884 = 1.807.
f BC
2
(h ) = f(L/D)(v ∕2g) = 0.015(30.5∕0.1524)(4.884) = 14.66 m.
f CD
2
2
(h ) = (v − v ) ∕2g = (9.784 − 2.446) /(2 × 9.8) = 2.747 m (see Section 5.2.3).
D
E
f DE
2
(h ) = f(L/D)(v /2g) = 0.020(30.5∕0.3048)(0.305) = 0.61 m.
f EF