Example 4.18: A piping network is given below. Find the volumetric flow rate of water through different pipes. The following data is applicable.

Pipe	L/ft	d/ft	RR
1	2500	0.2557	0.00058
2	1000	0.3355	0.00045
3	2000	0.17225	0.00087

All pipes are at same elevation. Pressure drop (between points A and B) for pipe #1 is 40 psi. Viscosity of the fluid is 6.72 ´ 10^-4 lb/(ft·s).

Solution:

C As elevation of the points A, and B are same, frictional loss in pipe #1 is

C Using the technique of earlier examples, the group ReÖf for pipe #1 can be found as

C Figure 4.4 gives us friction factor of 0.005

C Reynolds number can be found to be 1.307 ´ 10⁵ (Figure 4.3).

C We can now find the velocity of water in the pipe as 5.507 ft/s.

C The volumetric flow rate can be calculated to be 0.283 ft³/s.

C We can find the value of the group ReÖf for pipe #2.

C Using the previous methods we can find

f₂ (0.0059), Re₂ (3.8 ´ 10⁴), V₂ (1.22 ft/s),

V₃ (4.62 ft/s), Re₃ (7.39 ´ 10⁴), and f₃ (0.0056).

C A corrected value for V₂ is

C We can repeat these calculations till we achieve convergence.

C The final values are V₂ = 1.25 ft/s, f₂ = 0.0059, V₃ = 4.73 ft/s, and f₃ = 0.0056.