Question

A fluid with density of 61.8 lbm/ft³ is pumped at a rate of 345 gal/min to a tank vented to the atmosphere. The fluid is pumped at 77° F through a 2” schedule-40 steel pipe with an equivalent length of 500 ft. The ID of the pipe is 2.067” and the friction factor is 0.029213. The line has a globe control valve in the full-open position with an equivalent length of 110 ft. Calculate the pressure drop across the control valve in lbf/in². How was the friction factor calculated to 5 places?

L/D ratio is not given for globe valve

Answer 1

ANSWER :

Aabove equation to be used for the Calculation of pressure drop across Valve.

f = friction factor = 0.029213

L = Valve equivalent Lenght = 110 ft

D = Diameter of Pipe or Valve = 2 inch = 0.166 ft

p = density of fluid = 61.8 lbm/ft3

Flow in pipe = 345 gal/min

= 0.7685 ft3/sec

Area of Pipe = 22/7 × (Dia2 / 4) =3.142*(0.166*0.166/4)

= 0.0216 ft​​​​​​2

V = Velocity of Fluid

= Flow / Area of Pipe

= 0.7685 / 0.0216

= 35.57 ft/s

V2/2 = 632.76 ft2/s2

p x (V2/2) = 61.8 x 632.76

= 39,105 lbm/ft s2

L/D = 110 / 0.166

= 662.65

f X (L/D) = 0.029213 X 662.65

= 19.35

Pressure Drop ΔP = f x (L/D) x (pV2/2) = 19.35 x 39,105 = 756,995 lbm/ft s2 = 23,528 lbf