A double-acting reciprocating pump, running at 50 r.p.m. is discharging 900 litres of water per minute. The pump has stroke of 400 mm. The diameter of piston is 250 mm. The delivery and suction heads are 25 m and 4 m respectively. Find the slip of the pump and power required to drive the pump.

Leave a Comment / Fluid Mechanics, Numerical (Water) / By
Double-Acting Reciprocating Pump Analysis

Problem Statement

A double-acting reciprocating pump, running at 50 r.p.m. is discharging 900 litres of water per minute. The pump has stroke of 400 mm. The diameter of piston is 250 mm. The delivery and suction heads are 25 m and 4 m respectively. Find the slip of the pump and power required to drive the pump.

Given Data & Constants

  • Speed, \(N = 50 \, \text{r.p.m.}\)
  • Actual discharge, \(Q_{\text{act}} = 900 \, \text{L/min} = \frac{900}{1000 \times 60} = 0.015 \, \text{m}^3/\text{s}\)
  • Stroke length, \(L = 400 \, \text{mm} = 0.4 \, \text{m}\)
  • Piston diameter, \(D = 250 \, \text{mm} = 0.25 \, \text{m}\)
  • Delivery head, \(h_d = 25 \, \text{m}\)
  • Suction head, \(h_s = 4 \, \text{m}\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Theoretical Discharge (\(Q_{\text{th}}\))

For a double-acting pump, discharge occurs on both the forward and backward strokes. We neglect the volume of the piston rod for this calculation.

$$ \text{Area of piston, } A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.25)^2 \approx 0.049087 \, \text{m}^2 $$ $$ Q_{\text{th}} = \frac{2 \cdot A \cdot L \cdot N}{60} $$ $$ Q_{\text{th}} = \frac{2 \times 0.049087 \, \text{m}^2 \times 0.4 \, \text{m} \times 50 \, \text{r.p.m.}}{60} $$ $$ Q_{\text{th}} \approx 0.03272 \, \text{m}^3/\text{s} $$

2. Find the Slip of the Pump

Slip is the difference between the theoretical and actual discharge.

$$ \text{Slip} = Q_{\text{th}} - Q_{\text{act}} $$ $$ \text{Slip} = 0.03272 - 0.015 = 0.01772 \, \text{m}^3/\text{s} $$

3. Calculate the Power Required to Drive the Pump

The power required (indicated power) is based on the theoretical discharge and total head. This represents the power needed to move the theoretical volume of water against the head, before accounting for slip or mechanical friction.

$$ \text{Total Head, } H = h_s + h_d = 4 \, \text{m} + 25 \, \text{m} = 29 \, \text{m} $$ $$ \text{Power, } P = \rho \cdot g \cdot Q_{\text{th}} \cdot H $$ $$ P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.03272 \, \text{m}^3/\text{s} \times 29 \, \text{m} $$ $$ P \approx 9308.6 \, \text{W} $$
Final Results:

Slip of the pump: \( \approx 0.0177 \, \text{m}^3/\text{s} \) (or 17.7 L/s)

Power required to drive the pump: \( \approx 9.31 \, \text{kW} \)

Explanation of Key Terms

Double-Acting Pump: Unlike a single-acting pump which discharges water only on the forward stroke, a double-acting pump discharges water on both the forward and backward strokes of the piston. This results in a more continuous flow and a theoretical discharge that is approximately twice that of a single-acting pump with the same dimensions.

Slip: This is the leakage within the pump. The theoretical discharge of 32.7 L/s is what the pump *should* deliver based on its size and speed, but it only delivers an actual 15 L/s. The difference of 17.7 L/s is the "slip". A high slip, as seen here, can indicate significant wear on the piston seals or valves.

Power Required: The calculated power of 9.31 kW is the "indicated power". It represents the power required to move the theoretical volume of fluid against the total head. The actual power delivered to the water ("water power") would be lower (calculated using \(Q_{act}\)), and the power drawn by the motor ("shaft power") would be even higher to account for pump inefficiencies.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top