The equivalent length of a fitting or an obstruction is the length of the pipe that offers the same pressure drop due to friction. It is usually expressed in terms of the equivalent pipe diameters L_e/d.

Example 4.13: Water flows through a smooth pipe having inside diameter 0.5 in at a velocity of 2 ft/s. The pipe is 1000 ft in length. Piping arrangement contains a fully-open, a 3/4-open, a 1/2-open, and a 1/4-open gate valves. Use equivalent length of valves, L_e, to estimate the total pressure drop. Use the following information:

Density = 62.37 lb/ft³; viscosity = 7.53 ´ 10^-4 lb/(ft·s)

Solution:

C Calculate Re:

C Friction factor for smooth pipe f = 0.0086.

C Obtain equivalent lengths of the valves using Figure 4.5

Total equivalent length L_e = 52.35 ft.

Valve	Lei/ft	ni	ni Lei /ft
Gate valve, full-open	0.35	1	0.35
Gate valve, 3/4-open	2	1	2
Gate valve, 1/2-open	10	1	10
Gate valve,1/4-open	40	1	40
å			52.35

C Calculate the effective length of the pipe as the sum of the pipe length and the equivalent lengths of fittings and valves.

C Calculate frictional loss h_f

C If the points in consideration are at the same elevation, then pressure drop is