Pump

A centrifugal pump raises the pressure of a liquid by the centrifugal action of a rotating impeller: the impeller adds kinetic energy and pressure energy to the liquid, and the casing converts a part of the kinetic energy into pressure.

   Delivery pipe
        ^
        |      Casing (volute)
   +----+-------+
   |   .---.    |
   |  / ( ) \ <-+-- impeller with vanes
   | |  eye  |  |
   +-+---+---+--+
       |   |
 Suction pipe  <- foot valve with strainer
       |
   ~~~~~~~~~~~~~ sump

Main parts

1. Impeller: a wheel with backward-curved vanes (shrouded, semi-open or open) fixed on the shaft; it imparts energy to the liquid.
2. Casing: an airtight passage enclosing the impeller. Types: volute (spiral, with increasing area), vortex (a circular chamber between impeller and volute) and diffuser (guide vanes around the impeller, as in turbine pumps).
3. Suction pipe with foot valve and strainer: the foot valve is a non-return valve which keeps the pump primed; the strainer prevents debris entry.
4. Delivery pipe with delivery valve: the valve controls discharge, and is closed at starting to reduce the starting power.
5. Shaft, bearings, stuffing box/gland to seal the shaft.

Working

When the impeller rotates in a casing full of liquid, the liquid in the vanes is thrown outward by centrifugal force. A low pressure forms at the eye, so liquid from the sump (acted on by atmospheric pressure) flows continuously into the eye. As the liquid leaves the impeller at high velocity, the volute casing expands the flow area and the kinetic energy turns into pressure energy before the liquid enters the delivery pipe.

Priming

If air is inside the casing, the pump produces only a small pressure rise because the density of air is very small (air binding), so it cannot lift liquid. Priming is filling the suction pipe and casing with the liquid before starting, so the pump can create enough suction. It is done by the foot valve, a filling funnel, a vacuum pump or by an ejector.

The delivery valve is kept closed at starting (for the radial pump the power is minimum at zero discharge) and opened slowly after the pump has reached full speed.

Given: $D_2=0.4$ m, $B_2=0.03$ m, $N=1450$ rpm, $\beta_2=30^\circ$, $V_{f2}=2.8$ m/s, $V_{w1}=0$, $\eta_{man}=0.78$, $\eta_o=0.70$.

Outlet velocity triangle

Blade speed:

$$u_2=\frac{\pi D_2N}{60}=\frac{\pi\times0.4\times1450}{60}=30.37\ \text{m/s}$$

        V2   Vf2
         \   |
          \  |   Vr2 (at beta2 = 30 deg to u2 direction)
           \a|  /
     --------*-----> u2 (blade direction)
      |<-Vw2->|<-Vf2 cot b2->|

Whirl velocity:

$$V_{w2}=u_2-\frac{V_{f2}}{\tan\beta_2}=30.37-\frac{2.8}{\tan30^\circ}=30.37-4.85=25.52\ \text{m/s}$$

(a) Absolute velocity and direction

$$V_2=\sqrt{V_{w2}^2+V_{f2}^2}=\sqrt{25.52^2+2.8^2}=25.67\ \text{m/s},\qquad \tan\alpha_2=\frac{V_{f2}}{V_{w2}}\Rightarrow\alpha_2=6.3^\circ$$

(b) Heads

Euler head (no whirl at inlet):

$$H_e=\frac{V_{w2}u_2}{g}=\frac{25.52\times30.37}{9.81}=79.00\ \text{m}$$ $$H_m=\eta_{man}H_e=0.78\times79.00=61.62\ \text{m}$$

(c) Discharge

$$Q=\pi D_2B_2V_{f2}=\pi\times0.4\times0.03\times2.8=0.1056\ \text{m}^3/\text{s}=105.6\ \text{L/s}$$

(d) Power

Water power:

$$P_w=\rho gQH_m=1000\times9.81\times0.1056\times61.62=63.81\ \text{kW}$$ $$P_{shaft}=\frac{P_w}{\eta_o}=\frac{63.81}{0.70}=91.2\ \text{kW}$$

Answer: (a) $V_2=25.67$ m/s at $6.3^\circ$; (b) $H_e=79.00$ m, $H_m=61.62$ m; (c) $Q=0.1056$ m³/s; (d) 91.2 kW.

(a) Cavitation

Cavitation is the formation of vapour bubbles in a flowing liquid where the local absolute pressure falls to the vapour pressure, and the sudden collapse of these bubbles when they move into a higher-pressure region.

Causes: high suction lift, low atmospheric pressure (high altitude), high liquid temperature (high vapour pressure), high pump speed, large losses in the suction pipe and an insufficient NPSH. The pressure is lowest at the impeller eye, on the suction side of the vane.

Effects:

• Noise and vibration.
• Pitting and erosion of the impeller and casing by the shock of collapsing bubbles.
• Fall in head, discharge and efficiency.

Prevention: lower the pump (reduce suction lift) or use a flooded suction, use a larger suction pipe with fewer bends and valves, reduce the speed, cool the liquid, use an inducer or a pump with a lower required NPSH, use cavitation-resistant materials (stainless steel, bronze).

(b) Numerical

Available NPSH (net head above vapour pressure at the pump suction):

$$\text{NPSH}_a=\frac{p_a-p_v}{\rho g}-H_s-h_{fs}$$ $$\frac{p_a-p_v}{\rho g}=\frac{(101.3-4.25)\times10^3}{996\times9.81}=9.933\ \text{m}$$ $$\text{NPSH}_a=9.933-4.0-0.8=5.13\ \text{m}$$

Since $\text{NPSH}_a=5.13\ \text{m}>\text{NPSH}_r=4.2$ m, there is a margin of $5.13-4.2$ m and the pump will not cavitate.

Maximum suction lift (taking $\text{NPSH}_a=\text{NPSH}_r$):

$$H_{s,max}=9.933-0.8-4.2=4.93\ \text{m}$$

(A safety margin of at least 0.5 m is kept in practice.)

At 79.5 kPa:

$$\frac{p_a-p_v}{\rho g}=\frac{(79.5-4.25)\times10^3}{996\times9.81}=7.702\ \text{m}$$ $$\text{NPSH}_a=7.702-4.0-0.8=2.90\ \text{m}<4.2\ \text{m}$$

The available NPSH is now short by about 1.30 m, so the pump will cavitate. The suction lift must be reduced (or a flooded suction used) to get the margin back.

Answer: $\text{NPSH}_a=5.13$ m (no cavitation); $H_{s,max}=4.93$ m; at 79.5 kPa $\text{NPSH}_a=2.90$ m, so cavitation occurs.

Specific speed

The specific speed of a centrifugal pump is the speed of a geometrically similar pump that delivers unit discharge (1 m³/s) against unit head (1 m).

$$N_s=\frac{N\sqrt{Q}}{H_m^{3/4}}$$

($N$ in rpm, $Q$ in m³/s, $H_m$ in m.) It is the type number of the pump:

It helps in selecting the pump type for a duty and in predicting performance of similar pumps. (The head $H_m$ is for a single stage in a multistage pump, and $Q$ is for one side for a double suction impeller.)

(a) Specific speed and shaft power

$$N_s=\frac{1450\times\sqrt{0.05}}{25^{0.75}}=\frac{1450\times0.2236}{11.180}=29.0$$

This lies in the radial-flow (centrifugal) range.

$$P_1=\frac{\rho gQH}{\eta_o}=\frac{1000\times9.81\times0.05\times25}{0.72}=17.03\ \text{kW}$$

(b) Speed changed to 1750 rpm

Speed ratio $r=\dfrac{1750}{1450}=1.2069$. For the same pump:

$$Q\propto N,\qquad H\propto N^2,\qquad P\propto N^3$$ $$\begin{aligned} Q_2&=0.05\times1.2069=0.0603\ \text{m}^3/\text{s}\ (60.3\ \text{L/s})\\ H_2&=25\times1.2069^2=36.4\ \text{m}\\ P_2&=17.03\times1.2069^3=29.94\ \text{kW} \end{aligned}$$

Answer: (a) $N_s=29.0$, $P_1=17.03$ kW; (b) $Q_2=0.0603$ m³/s, $H_2=36.4$ m, $P_2=29.94$ kW.

Method: the operating point is the intersection of the pump curve and the system curve $H_{sys}=12+1500Q^2$.

(a) Single pump

$$40-2500Q^2=12+1500Q^2\ \Rightarrow\ 4000Q^2=28\ \Rightarrow\ Q=\sqrt{\frac{28}{4000}}=0.0837\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.0837^2=22.50\ \text{m}$$

(b) Two pumps in series

For series operation the same discharge flows through both pumps and the heads add:

$$H_{ser}=2(40-2500Q^2)=80-5000Q^2$$ $$80-5000Q^2=12+1500Q^2\ \Rightarrow\ 6500Q^2=68\ \Rightarrow\ Q=0.1023\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.1023^2=27.69\ \text{m}\ \ (13.85\ \text{m per pump})$$

(c) Two pumps in parallel

For parallel operation the heads are equal and the discharges add. Each pump delivers $Q/2$:

$$H_{par}=40-2500\left(\frac{Q}{2}\right)^2=40-625Q^2$$ $$40-625Q^2=12+1500Q^2\ \Rightarrow\ 2125Q^2=28\ \Rightarrow\ Q=0.1148\ \text{m}^3/\text{s}$$ $$H=12+1500\times0.1148^2=31.76\ \text{m}\ \ (\text{each pump delivers }0.0574\ \text{m}^3/\text{s})$$

Comparison

For this system the parallel arrangement gives the larger discharge, because the system curve is flat (small friction constant 1500 compared with the pump curve constant 2500). Note that two pumps never give double the discharge, since the system head rises with the flow.

When to use:

• Series: when the system needs a high head (steep system curve, large static lift); the pumps share the head, as in multistage pumps.
• Parallel: when a large discharge at nearly the same head is needed or the demand changes widely (one pump can be switched off); it also gives standby capacity.

Answer: Single: $Q=0.0837$ m³/s, $H=22.50$ m; series: $Q=0.1023$ m³/s, $H=27.69$ m; parallel: $Q=0.1148$ m³/s, $H=31.76$ m.

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.

(a) Losses and efficiencies

Energy losses in a centrifugal pump:

1. Hydraulic losses: shock at impeller entry (off-design flow), friction in impeller passages, casing and volute, and eddy losses due to sudden changes of area.
2. Leakage (volumetric) losses: liquid leaking back from the delivery side to the suction side through the clearance between the impeller and the casing.
3. Mechanical losses: friction in bearings and stuffing box, disc friction on the outer faces of the impeller.

(b) Characteristic curves

Curves are drawn for constant speed, with discharge on the x-axis.

 H, P,       H-Q (falls with Q)
 eta   \  .-------eta
        \/   ..  
        /\       P rises with Q
       /   \
      0---------------------- Q
 (shut-off head at Q = 0)

• Head-discharge (H-Q): head is highest at shut-off ($Q=0$) and falls as discharge increases (for backward curved vanes).
• Power-discharge (P-Q): power rises almost linearly with discharge; power at zero discharge is small, hence the pump is started with the delivery valve closed.
• Efficiency-discharge ($\eta$-Q): zero at $Q=0$, rises to a maximum at the best efficiency point (BEP) and falls again.
• Main (constant-speed) and operating curves show the performance at different speeds; constant efficiency (Muschel) curves show the lines of equal efficiency in the H-Q plane. The system curve cuts the H-Q curve at the operating point.

(c) Selection of a pump

• Required discharge and head (duty point), and the nature of the system curve; the duty point should be near the BEP.
• Type of liquid: clean water, sewage, slurry, corrosive or viscous liquid (viscosity reduces head and efficiency and raises the power).
• Suction conditions: NPSH available, to avoid cavitation.
• Specific speed to choose between radial, mixed and axial types; choose multistage if the head is high.
• Efficiency and power cost; the motor and speed; variation in demand (parallel units or variable speed).
• First cost, maintenance, spare parts availability, space, and the reliability required.
• For small discharge at very high head, or for viscous liquids and metering duties, a reciprocating or other positive displacement pump is selected.