Question 10.6: Compute the lost head for a 150 ft run of standard pipe, hav......
Compute the lost head for a 150 ft run of standard pipe, having a diameter of 3 in. The pipe run has three standard 90-degree elbows, a globe valve, and a gate valve. One hundred gpm of water flows in the pipe.
The equivalent length of the various fittings will first be determined by using Figs. 10-22a and 10-22b.
Globe valve: K_{1} = 340 f_{t}, f_{t} = 0.018 (Fig. 10-22a and Table 10-2)
K_{1} = 340 (0.018) = 6.1
L = 86 ft (Fig. 10-22b)
Elbow: K = 30f_{t}, f_{t} = 0.018
K = 30(0.018) = 0.54
L = 8 ft
Gate valve:K_{1} = 8 f_{t}, f_{t} = 0.018
K_{1} = 8 (0.018) = 0.14
L = 2 ft
The total equivalent length is then
Actual length of pipe 150 ft
One globe valve 86 ft
Three elbows 24 ft
\underline{One gate valve } \underline{ 2 ft }Total 262 ft
From Fig. 10-20 the lost head l^{′}_{f} is 2.3 ft per 100 ft of length, or
l^{′}_{f} = 2.3 × 10^{−2}, ft/ft of length
The lost head for the complete pipe run is then given by
l^{′}_{f} = L_{e} l^{′}_{f} = (262)2.3 × 10^{−2} = 6.0 ft
| Table 10-2 Formulas, Definition of Terms, and Values of ft for Fig. 10-22 | |||
| Formula 1: K_{2} = \frac{K_{1} + (\sin \frac{θ}{2})0.8 (1 – β²) + 2.6(1 + β²)²}{β^{4}} | |||
| Formula 2: K_{2} = \frac{K_{1} + 0.5(\sin \frac{θ}{2}) (1 – β²) + (1 + β²)²}{β^{4}} | |||
|
β = \frac{D_{1}}{D_{2}}; β² = \left\lgroup \frac{D_{1}}{D_{2}}\right\rgroup^{2}; |
|||
| Nominal
Size, in. |
Friction
Factor f_{t} |
Nominal
Size, in. |
Friction
Factor f_{t} |
| \frac{1}{2} | 0.027 | 4 | 0.017 |
| \frac{3}{4} | 0.025 | 5 | 0.016 |
| 1 | 0.023 | 6 | 0.015 |
| 1 \frac{1}{4} | 0.022 | 8-10 | 0.014 |
| 1 \frac{1}{2} | 0.021 | 12-16 | 0.013 |
| 2 | 0.019 | 18-24 | 0.012 |
| 2 \frac{1}{2}, 3 | 0.018 | ||
