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Question

5.3 The plungers of a three -cylinder pump have diameters of 12 cm and a stroke length of 50 cm. The pressure during the delivery stroke is 1 000 kPa. Calculate the following: 5.3.1 The power required to drive the pump at 350 r/min if the efficiency of the motor is 90% 5.3.2 The volume of the water delivered per minute in l/min if there is a slip of 13%

Answer 1

(5.3.2)

Consider Single - acting pump

Number of Cylinders n = 3

Piston diameter d = 12 cm

d = 12/100 m = 0.12 m

Stroke length s = 50 m

s = 50/100 = 0.5 m

Slip = 13% = 0.13

Speed N = 350 RPM

Cross sectional area of piston A = (¼)πd²

A = (¼)π(0.12)²

A = 0.0113 m²

Volume of Water discharge Q = (A×s×n×N)/231

Q = (0.0113×0.5×3×350)/231

Q = 0.0257 Gallons per minute

But due to slip 13% is lost.So

Q = 0.0257× 0.13 Gallons per minute

Q = 0.003341 Gallons per minute

1 Gallons per minute = 3.78541 litres/minute.

Q = 3.78541×0.003341

Q = 0.01264 litres/minute.

The volume of water delivered is 0.01264 litres/minute.

Answer 2

(5.3.1)

Consider Single - acting pump

Number of Cylinders n = 3

Piston diameter d = 12 cm

d = 12/100 m = 0.12 m

Stroke length s = 50 m

s = 50/100 = 0.5 m

Slip = 13% = 0.13

Speed N = 350 RPM

Prassure P = 1000 kPa = 1 MPa

1 MPa = 145.0377 psi

P = 145.0377 Psi

Efficiency η = 90 % = 0.9

Cross sectional area of piston A = (¼)πd²

A = (¼)π(0.12)²

A = 0.0113 m²

Volume of Water discharge Q = (A×s×n×N)/231

Q = (0.0113×0.5×3×350)/231

Q = 0.0257 Gallons per minute

Fluid Hourse Power FHP = (P x Q) / (1714 x η)

P = Pressure in psi ( per square inch )

Q = Flowrate in GPM ( Gallons per minute )

FHP = (145.0377 x 0.0257) / (1714 x 0.9)

FHP = 0.0024 HP

1 Hp = 745.699872 Watts

Power = 0.0024 x 745.699872

Power = 1.7896 Watts

The power required to drive the pump is 1.7896 Watts.