One can express the friction loss as the equivalent length of a straight pipe L
e, or by a factor "n" (number of velocity heads, Newtons) by which [ρ]V
2/2 has to be multiplied to give the same pressure drop as the particular fitting (valve) in question.
Published data for globe valves:
Opening L
e/D n
Full 330 6
Half 470 8.5
For regular gate valves:
Full 7 0.13
3/4 40 0.8
1/2 200 3.8
1/4 800 15
For a preliminary design basis one generally assumes the valve is one size smaller than the pipe. Factors for both reducers should then be added to those of the valves.
An example taken from the literature:
Line size: 8 inches. A butterfly valve is contemplated with K=3. The reducing fittings account for K=0.29. Thus the total would be K=3.29.
To put the resistance coefficient in terms of the main pipe size, 3.29 is multiplied by (D/d)
4. In this example:
3.29(7.981/6.065)4= 9.87
K is the resistance factor when the velocity head is expressed in ft, as in u
2/2g.
For more information I suggest reading the pertinent chapters on
Control-Valve Selection and Sizing by Les Driskell published by the ISA.