power

Question

5.3 The plungers of a three -cylinder pump have diameters of 75 mm and stroke lengths of 250 mm. The pressure during a delivery stroke is 750 kPa Calculate the power required to drive the pump at 150 r/min if the efficiency of the motor is 75%.

Answer 1

The Diameter of the 75 mm and Stroke length 250 mm.

The pressure of the delivery stroke is 750 KPa.

Efficiency = 75% = 0.75

Pressure = Force / area.

Force = Pressure * area.

Area of a cylinder is πr²

Diameter = 75mm  then radius r = 37.5mm

Area = π(37.5)²

Area = 4417.8 mm².

Then Force = Pressure * Area

Force = 750,000 N-m² * 4417.8 mm²

Force = 3313.35 N

Output Power = Force * Velocity

Velocity = 150 r/min.

We need to convert this into m-sec

Velocity(m-sec) = ((3.141*d)/60)* Velocity in r/min

V = 150* ((3.141 * 0.075 )/60) = 0.588 m/sec

P(out) = 3313.5 * 0.589

P(out) = 1951.65 Nm/Sec

Now Efficiency = (P(out) / P(in)) * 100

P(in) = (P(out)/Efficiency)*100

P(in) = (3313.5/0.75)*100

P(in) = 441800 Nm/Sec.

Therefore Power delivered to drive the pump = 441800 Nm/Sec.