Reciprocating pump working, slip and air vessel

Working of a single-acting pump

   Delivery valve (opens on delivery stroke)
        ___|___
       |  ___  |  Cylinder
 ======|=|___|=|=====> piston / plunger (crank + connecting rod)
       |_______|
   Suction valve (opens on suction stroke)
        |
   Suction pipe -> sump

• Suction stroke: the piston moves away from the cylinder cover, creating a vacuum. The delivery valve is held closed, the suction valve opens, and water is pushed into the cylinder by atmospheric pressure.
• Delivery stroke: the piston moves forward and compresses the water; the suction valve closes, the delivery valve opens, and water is forced into the delivery pipe.
• One revolution of the crank gives one suction and one delivery stroke, so discharge is intermittent (only for half the revolution).

Ideal indicator diagram: a plot of pressure head in the cylinder against piston displacement (stroke) is a rectangle: a horizontal line at $H_s$ below the atmosphere for suction, and a horizontal line at $H_d$ above the atmosphere for delivery. The height of the diagram is $(H_s+H_d)$ and its area represents work per stroke.

 H
 Hd |-----------------------|   delivery
    |                       |
 Hatm ---------------------------
    |                       |
 Hs |-----------------------|   suction
    0                       L  (stroke)

Air vessels: closed chambers with air at the top, fixed on the suction and delivery pipes close to the cylinder. The air is compressed when the flow is more than the mean and expands when the flow is less than the mean. Thus they:

1. Make the flow in the pipes nearly uniform, avoiding acceleration and friction head loss in the long pipes;
2. Reduce the pressure fluctuation and shock, so the pump can run at higher speed and with less risk of separation;
3. Save about 80 to 85 % of the work lost in friction in a delivery pipe for a single-acting pump (power saved).

Numerical

Given: $D=0.15$ m, $L=0.3$ m, $N=60$ rpm, $Q_{act}=5.0$ L/s, $H_s=3.5$ m, $H_d=22$ m, $\eta_o=0.80$.

Plunger area: $A=\frac{\pi}{4}(0.15)^2=0.01767$ m².

Theoretical discharge (single acting):

$$Q_{th}=\frac{ALN}{60}=\frac{0.01767\times0.3\times60}{60}=0.00530\ \text{m}^3/\text{s}=5.30\ \text{L/s}$$

Slip:

$$\text{Slip}=Q_{th}-Q_{act}=0.30\ \text{L/s}\ \ \Big(\text{percentage slip}=\frac{0.30}{5.30}\times100=5.7\%\Big)$$

Coefficient of discharge:

$$C_d=\frac{Q_{act}}{Q_{th}}=\frac{5.0}{5.30}=0.943$$

Power: total head $H=H_s+H_d=25.5$ m.

$$\begin{aligned} P_{water}&=\rho gQ_{act}(H_s+H_d)=1000\times9.81\times0.005\times25.5=1250.8\ \text{W}\\ P_{shaft}&=\frac{P_{water}}{\eta_o}=\frac{1250.8}{0.80}=1563\ \text{W}=1.56\ \text{kW} \end{aligned}$$

Answer: $Q_{th}=5.30$ L/s; slip $=0.30$ L/s ($5.7\%$); $C_d=0.943$; power $=1.56$ kW.