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 fullyopen, a 3/4open, a 1/2open, and a
1/4open gate valves. Use equivalent
length of valves, L_{e}, to estimate the total pressure
drop. Use the following information:
Density = 62.37 lb/ft^{3};
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 
L_{ei}/ft 
n_{i} 
n_{i} Le_{i} /ft 
Gate valve, fullopen 
0.35 
1 
0.35 
Gate valve, 3/4open 
2 
1 
2 
Gate valve, 1/2open 
10 
1 
10 
Gate valve,1/4open 
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
_{} 