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eimsori [14]
2 years ago
11

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
Engineering
1 answer:
Stella [2.4K]2 years ago
6 0

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

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2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam
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\eta = 91.7%

Explanation:

Determine the initial velocity

v_1 = \frac{\dot v}{A_1}

    = \frac{0.1}{\pi}{4} 0.08^2

     = 19.89 m/s

final velocity

v_2 =\frac{\dot v}{A_2}

      = \frac{0.1}{\frac{\pi}{4} 0.12^2}

      =8.84 m/s

total mechanical energy is given as

E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}

\dot v = \dot m v                       ( v =v_1 =v_2)

E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}

                = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}

                 = \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}

              = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}

E_{mech} = 36.34 W

Shaft power

W = \eta_[motar} W_{elec}

    =0.9\times 44 =39.6

mechanical efficiency

\eta{pump} =\frac{ E_{mech}}{W}

=\frac{36.34}{39.6} = 0.917  = 91.7%

8 0
2 years ago
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