Question

A double-acting duplex pump with 6.5-in. liners, 2.5-in. rods, and 18-in. strokes was operated at 3,000 psig and 20 cycles/min. for 10 minutes with the suction pit isolated from the return mud flow. The mud level in the suction pit, which is 7 ft wide and 20 ft long, was observed to fall 18 in. during this period. Compute the pump factor, volumetric pump efficiency, and hydraulic horsepower developed by the pump

Answer 1

Answer:

Pump factor = Fp =  7.854 gal/cycle

Ev = 82.00 %

P_H = 183.29 hp

Explanation:

Given data:

Dimension of duplex pump

6.5 inch liner  

2.5 inch rod

18 inch strokes

Pressure 3000 psig

Pit dimension

7 ft wide

20 ft long

Ls = 18 inch

Velocity = (18)/10

volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100

we know that flow rate is given as = Area * velocity

Theoritical flow rate $= \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times N$

$Ev = \frac{7\times 12 \times 20\times 12\times 12 \times \frac{18}{10} inch^3/min}{\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 20}$

Ev = 82.00 %

Pump factor $Fp = = \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times Ev$

$Fp =\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 0.82$

Fp = 1814.22 in^3/cyl

Fp =  7.854 gal/cycle

Flow rate $q = NFp = 20 \times 7.854 = 157.08 gal/min$

Power $Ph = \frac{\DeltaP q}{1714} = \frac{3000 \times 157.08}{1714} = 274.93 hp$