centrifugal pump

• A machine which imparts energy to a liquid

causing the liquid to flow or rise to a higher

level or both.

• Demour’s centrifugal pump – 1730

• Theory – conservation of angular momentum

– conversion of kinetic energy to potential

energy

• Pump components Rotating element – impeller - takes the

power (mechanical) from the rotating shaft

and accelerates the fluid.

Enclosing the rotating element and sealing

the pressurized liquid inside – casing

or housing – acts as a diffuser and transforms

high fluid velocity (kinetic energy) into pressure.

Centrifugal Pump

Schematic diagram of basic elements of

centrifugal pump

Centrifugal Pump

which is the rotating part of the centrifugal

pump.

It consists of a series of backwards curved

vanes (blades).

The impeller is driven by a shaft which is

connected to the shaft of a prime mover, e.g.,

an electric motor, an IC engine, etc.

1. Impeller

Main Parts of Centrifugal Pumps

2. Casing

Which is an air-tight passage surrounding

the impeller

designed to direct the liquid to the

impeller and lead it away

Volute casing. It is of spiral type in which

the area of the flow increases gradually.

Main Parts of Centrifugal Pumps

The shaft which carries the mechanical power

from the engine/motor drive to the impeller.

3. Suction Pipe

4. Delivery Pipe

The conduit connecting the sump

(reservoir) up to the centerline of the

pump impeller.

The conduit connecting the pump

impeller centerline up to the delivery

point.

5. Impeller Shaft

6. Engine/Motor Driver

Supplies mechanical power to the rotating

impeller shaft. It can be mounted directly

on the pump, above it, or adjacent to it.

Broad range of applicable flows and heads. Higher discharges and heads

can be achieved by increasing the rotational speed or the diameter of the

impeller although with corresponding change in power input.

𝑄 ∝ 𝐷3 𝐻 ∝ 𝐷2 𝑃 ∝ 𝐷5

𝑄 ∝ 𝑁 𝐻 ∝ 𝑁2 𝑃 ∝ 𝑁3

Effect of Impeller Diameter and Speed

Discharge varies directly with the change in speed or as cube of the

change in impeller diameter

Head varies as square of the change in speed or impeller diameter.

Power input varies as the cube of the change in speed or fifth power to

the impeller diameter.

For instance, the pump speed is doubled:

Discharge will double.

Head will increase by a factor of 22 = 4.

Power input will increase by a factor of 23 = 8.

Centrifugal pumps (radial-flow pumps) are the most used pumps for hydraulic

purposes where high discharge with relatively low head is required such as

irrigation, cooling tower of central A/C system/power plants, etc.

Application of Centrifugal Pumps

Types of Centrifugal Pump Impellers

(b) Double-volute (casing)

pump (a) Volute (casing)

pump

(c) Turbine (casing with

guide vanes) pump

Types of Centrifugal Pumps (casing)

𝑣𝑤2 (𝑢2−𝑣𝑤2)

𝑣𝑤1

Inlet Velocity Triangle

Euler Equation for Centrifugal Pumps

Outlet Velocity Triangle Using subscripts 1 for inlet and 2 for outlet quantities:

𝑉1 and 𝑉2 =Absolute velocities at inlet and outlet

𝑢1 =𝜋𝐷1𝑁

60 and 𝑢2 =

𝜋𝐷2𝑁

60=Tangential velocities at

inner and outer periphery of the impeller

𝑣𝑟1 and 𝑣𝑟2 =Relative velocities at inlet and outlet

𝛼1 and 𝛼2 =Direction of absolute velocities at inlet

and outlet. 𝛼 is the angle made by the

absolute velocity vector 𝑉 with the positive

direction of the peripheral velocity 𝑢.

𝛽1 and 𝛽2 =Vane (blade) angle at inlet and outlet made by the

relative velocity vector 𝑣𝑟 with the negative direction of

the peripheral velocity 𝑢.

𝑣𝑤1and 𝑣𝑤2 =Whirl velocities at inlet and outlet (Tangential

components of 𝑉1 and 𝑉2 respectively).

𝑣𝑓1and 𝑣𝑓2=Flow velocities at inlet and outlet (Radial components of

𝑉1 and 𝑉2 respectively) .

𝐷1 and 𝐷2 =Diameters at inlet of outlet of the impeller

𝑁 =Rotational Speed of the impeller in rpm

𝑣𝑤2 (𝑢2−𝑣𝑤2)

𝑣𝑤1


𝑯𝒆 =𝑽𝟐𝟐 − 𝑽𝟏

𝟐

𝟐𝒈+𝒗𝒓𝟏𝟐 − 𝒗𝒓𝟐

𝟐

𝟐𝒈+𝒖𝟐𝟐 − 𝒖𝟏

𝟐

𝟐𝒈

𝑯𝒆 =𝟏

𝒈 𝒖𝟐𝒗𝒘𝟐 − 𝒖𝟏𝒗𝒘𝟏


Euler Head, 𝑯𝒆: [Theoretical head transferred to liquid by the

impeller or work done by the impeller per kg of

liquid]

Outlet Velocity Triangle

𝑣𝑟12 = 𝑢1

2 + 𝑉12 − 2𝑢1𝑉1𝑐𝑜𝑠𝛼1

𝑣𝑟22 = 𝑢2

2 + 𝑉22 − 2𝑢2𝑉2𝑐𝑜𝑠𝛼2

⇒ 𝐻𝑒=1

𝑔 𝑢2𝑉2𝑐𝑜𝑠𝛼2 − 𝑢1𝑉1𝑐𝑜𝑠𝛼1

From the inlet and outlet velocity triangles, we have:

Using these relations in 𝐻𝑒 equation and rearranging, we get:

𝑯𝒆 =𝟏

𝒈 𝒖𝟐𝒗𝒘𝟐

Assuming Radial Entry at inlet,

Euler Head,

Actual Head Developed by the impeller,

Manometric Head (Measured), 𝑯𝒎

Manometric (Hydraulic) Efficiency,

𝜼𝒎 =𝑯𝒎𝑯𝒆

𝑉22−𝑉1

2

2𝑔 represents the increase in absolute

kinetic energy of fluid; 𝑢22−𝑢1

2

2𝑔 represents the increase in static

pressure due to centrifugal action;

𝑣𝑟12 −𝑣𝑟2

2

2𝑔 represents the change in the kinetic

energy due to retardation of flow.


𝑉1 = 𝑉𝑓1

𝑢1

𝑣𝑟1

𝛼1 𝛽1


(radial entry, 𝜶𝟏 = 𝟗𝟎°)

Here:

At the time of start, the fluid velocities are zero and the only head that is operating

is the centrifugal head 𝑢22−𝑢1

2

2𝑔 . This centrifugal force must overcome the

manometric head for the fluid to move, i.e.,

𝑢22 − 𝑢1

2

2𝑔≥ 𝐻𝑚

𝑵𝒎𝒊𝒏 = 𝟖𝟒. 𝟓𝟗𝟔 𝑯𝒎

𝑫𝟐𝟐 −𝑫𝟏

𝟐

𝑢22 − 𝑢1

2

2𝑔≥ 𝜂𝑚𝐻𝑒

𝑢22 − 𝑢1

2

2𝑔= 𝜂𝑚

𝑢2𝑣𝑤2𝑔

Now, 𝑢1 =𝜋𝐷1𝑁

60 𝑎𝑛𝑑 𝑢2 =

𝜋𝐷2𝑁

60

So that, equating

𝑢22 − 𝑢1

2

2𝑔= 𝐻𝑚

𝑵𝒎𝒊𝒏 =𝟏𝟐𝟎𝜼𝒎𝝅

𝒗𝒘𝟐𝑫𝟐

𝑫𝟐𝟐 −𝑫𝟏

𝟐

Again,

Equating

Minimum Starting Speed of a Centrifugal Pump

Usually the impeller outer diameter is designed as twice the inner (inlet) diameter,

i.e., 𝑫𝟐 = 𝟐𝑫𝟏 . Using this relationship, the minimum diameter of the impeller for

fluid to move,

𝑢22 − 𝑢1

2

2𝑔≥ 𝐻𝑚

𝑫 𝟐 𝒎𝒊𝒏 = 𝟗𝟕. 𝟔𝟖𝑯𝒎𝑵

Here, 𝑢1 =𝜋𝐷1𝑁

60=𝜋𝐷2𝑁

120 𝑎𝑛𝑑 𝑢2 =

𝜋𝐷2𝑁

60

So that, equating

𝑢22 − 𝑢1

2

2𝑔= 𝐻𝑚

Minimum Impeller Diameter at a Given Speed

𝜋𝐷2𝑁

60

2

−𝜋𝐷2𝑁

120

2

= 2𝑔𝐻𝑚

This equation is used in practical situations to

design impeller for liquid pumping at a given

speed.

When the pump casing and the suction conduit are completely filled with water, as the

impeller rotates, the pressure at the pump suction side becomes lower than the

atmospheric pressure. Due to this difference in pressure head between the water surface

of the sump and the inlet of the pump, the atmospheric pressure pushes the water from

the sump to the pump casing. However, an impeller running in air would produce only a

small head. This cannot create the necessary differential head of water between the sump

and the pump inlet as the density of air is much less than that of water. Consequently, the

pump does not do its work of pumping of water.

Further, dry running of the pump may damage several parts of the pump. This is,

therefore, necessary to ensure that the pump casing, impeller, suction pipe and the portion

of the pump delivery pipe up to the delivery valve are always filled with water before the

start of the pump. Filling is done by pouring water into the funnel or priming-cup

provided for this purpose. An air vent in the casing is provided for the air to escape. This

air vent must be closed after filling. This filling process is called the “priming” of the

pump. Most centrifugal pumps are not self-priming, so they always need priming.

However, a one-way valve, called foot valve is used at the entrance of the suction pipe,

which keeps the suction conduit filled-up even when the pump is stopped. Thus when the

pump is restarted, it does not need priming.

Priming of Centrifugal Pump

A pump throws liquid in the opposite direction of a free falling body from a certain height. A

free falling body starts from zero velocity at a certain height and then reached maximum

velocity at the ground. On the other hand, a pump throws liquid at a maximum velocity at its

impeller exit to reach zero velocity at a certain height.

The height of liquid travel from impeller exit of a pump with impeller diameter of 500 mm

running at 1800 rpm, can be determined as follows:

Peripheral velocity of the Impeller at exit point u2 =𝜋𝐷2𝑁

60 =

𝜋×0.5×1800

60= 47.12 𝑚/s, at this

velocity the liquid will be thrown by the impeller at the impeller exit.

Therefore, the height of liquid travel is given by: 0 = 𝑢22 − 2𝑔ℎ ⇒ ℎ = 𝑢2

2/2𝑔 ⇒ ℎ = 113 𝑚

It can be observed that no liquid property enters the equation, maximum velocity of the liquid

depends only on the impeller size and speed.

Thus a pump always produce the same head regardless of the type of liquid being pumped.

However, pressure will increase or decrease in direct proportion to a liquid’s specific gravity

(𝑝 = 𝜌𝑔𝐻) and power input to the pump will also vary directly with liquid’s specific gravity.

Thus a centrifugal pump can develop the same 100 m of head when pumping water, brine or

kerosene. The resulting pressures, however, will vary. Therefore, the pump pressures are

expressed in liquid height in meters or feet (liquid head).

Why pump pressure is expressed in liquid head (liquid height in meters or feet)?

Problem–1: A centrifugal pump impeller runs at 950 rpm (𝑁). Its external and internal

diameters are 500 mm (𝐷2)and 250 mm (𝐷1) respectively. The vanes are set back at an

angle of 35o (𝛽2) to the outer rim. If the radial velocity of water through the impeller is

maintained constant at 2 m/s (𝑉𝑓1 = 𝑉𝑓2), find the angle of the vanes at inlet (𝜷𝟏), the

velocity and direction of water at outlet (𝑽𝟐, 𝜶𝟐) and the work done by the impeller per kg of

water (𝑯𝒆).

Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1

Example Problems


60=𝜋×0.25×950

60= 12.44 𝑚/𝑠

𝑡𝑎𝑛𝜷𝟏 =𝑉𝑓1

𝑢1=

2

12.44= 0.16 ⇒ 𝜷𝟏 = 𝟗. 𝟏𝟑°


60=𝜋×0.5×950

60= 24.87 𝑚/𝑠

𝑡𝑎𝑛𝛽2 =𝑉𝑓2

𝑢2−𝑣𝑤2⇒ tan35° =

2

24.87−𝑣𝑤2

⇒ 𝑣𝑤2 = 22.1 𝑚/𝑠

𝑽𝟐 = 𝑉𝑓22 + 𝑣𝑤2

2 = 22 + 22.12 = 𝟐𝟐. 𝟐 𝒎/𝒔

𝜶𝟐 = tan−1 𝑉𝑓2

𝑣𝑤2= tan−1

2

22.1= 𝟓. 𝟏𝟕°

𝑯𝒆 =𝑢2𝑣𝑤2

𝑔=24.87×22.1

9.81= 𝟓𝟔 𝒌𝒈 𝒎/𝒔

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Problem–2: A centrifugal pump is used to lift water at a rate of 0.085 m3/s (𝑄). The outer

diameter of the impeller is 325 mm (𝐷2) and the breadth of the wheel at outlet is 15 mm (𝑏2). The manometric head of the pump is 38 m (𝐻𝑚) and manometric efficiency is 85% (𝜂𝑚). If

the pump runs at 1500 rpm (𝑁), determine the blade angle at exit (𝜷𝟐).



60=𝜋×0.325×1500

60= 25.52 𝑚/𝑠

𝜂𝑚 =𝐻𝑚

𝐻𝑒=

𝐻𝑚𝑢2𝑣𝑤2𝑔

⇒ 0.85 =38

25.52×𝑣𝑤29.81

⇒ 𝑣𝑤2 = 17.19 𝑚/𝑠

𝑄 = 𝜋𝐷2𝑏2𝑉𝑓2 ⇒ 0.085 = 𝜋 × 0.325 × 0.015 × 𝑉𝑓2 ⇒ 𝑉𝑓2= 5.55 𝑚/𝑠

𝑡𝑎𝑛𝜷𝟐 =𝑉𝑓2

𝑢2−𝑣𝑤2

⇒ 𝜷𝟐 = tan−1 5.55

25.52−17.19= 𝟑𝟑. 𝟔𝟔°

Example Problems

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Problem–3: The inner and outer diameters of the impeller of a centrifugal pump are 300

mm (𝐷1) and 600 mm (𝐷2) respectively. The constant velocity of flow is 2.2 m/s (𝑉𝑓1 = 𝑉𝑓2)

and the vanes are curved backward at an angle of 45o at the exit (𝛽2) . If the manometric

efficiency is 75% (𝜂𝑚), find the minimum starting speed (𝑵𝒎𝒊𝒏) of the pump.


Example Problems


60=𝜋×0.6×𝑁𝑚𝑖𝑛

60= 0.0314159𝑁𝑚𝑖𝑛


𝑢2−𝑣𝑤2⇒ tan45° =

2.2

0.0314159𝑁𝑚𝑖𝑛−𝑣𝑤2

⇒ 𝑣𝑤2 = (0.031416𝑁𝑚𝑖𝑛 − 2.2) 𝑚/𝑠

=120 × 0.75

𝜋 0.031416𝑵𝒎𝒊𝒏 − 2.2 × 0.6

0.62 − 0.32

⇒ 𝑵𝒎𝒊𝒏= 63.662 0.031416𝑵𝒎𝒊𝒏 − 2.2

⇒ 𝑵𝒎𝒊𝒏= 𝟏𝟒𝟎 𝒓𝒑𝒎

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

𝑵𝒎𝒊𝒏 =120𝜂𝑚

𝜋 𝑣𝑤2𝐷2

𝐷22−𝐷1

2

Problem–4: A centrifugal pump delivers 0.20 m3/s water against a head of 26 m while

running at 950 rpm. The constant velocity of flow is 2.9 m/s and the vanes are curved

backward at an angle of 30o. If the manometric efficiency is 77%, find the diameter and the

width (or breadth) of the impeller at outlet.

Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1 𝑄 = 𝜋𝑫𝟐𝒃𝟐𝑉𝑓2

⇒ 0.20 = 𝜋 × 𝐷2 × 𝑏2 × 2.9 ⇒ 𝑫𝟐 𝒃𝟐 = 0.02195

Example Problems


60=𝜋×𝐷2×950

60= 49.74𝑫𝟐


𝑢2−𝑣𝑤2⇒ tan30° =

2.9

49.74𝑫𝟐−𝑣𝑤2

⇒ 𝑣𝑤2 = 49.74𝑫𝟐 − 5.023

𝜂𝑚 =𝐻𝑚

𝐻𝑒=

𝑔𝐻𝑚

𝑢2𝑣𝑤2

⇒ 0.77 =9.81 × 26

49.74𝑫𝟐 49.74𝑫𝟐 − 5.023

⇒ 𝑫𝟐2 − 0.134𝑫𝟐 − 0.101 = 0

∴ 𝑫𝟐 = 0.3918 𝑚 = 𝟑𝟗𝟐 𝒎𝒎

And 𝒃𝟐=0.02195

𝐷2= 0.056 𝑚 = 𝟓𝟔 𝒎𝒎

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Problem–5: A centrifugal pump impeller has an outside diameter of 200 mm (𝐷2) and it

rotates at 2900 rpm (𝑁). Determine the head generated (𝑯𝒎) if the vanes are curved

backward at an angle of 25o to the outer rim (𝛽2) and the velocity of flow throughout the

wheel is constant at 3 m/s (𝑉𝑓1 = 𝑉𝑓2). Assume hydraulic efficiency of 75% (𝜂𝑚). Determine

also the power in kW (𝑷) required to run the impeller if the breadth of the wheel at outlet is

15 mm (𝑏2). Neglect the effect of vane thickness and mechanical friction and leakage

losses.


Example Problems


60=𝜋×0.2×2900

60= 30.37 𝑚/𝑠


𝑢2−𝑣𝑤2⇒ tan25° =

3

30.37−𝑣𝑤2

⇒ 𝑣𝑤2 = 23.94 𝑚/𝑠

𝑯𝒎 = 𝜂𝑚𝐻𝑒 = 𝜂𝑚𝑢2𝑣𝑤2

𝑔

⇒ 𝑯𝒎 = 0.75 ×30.37 × 23.94

9.81= 𝟓𝟓. 𝟔 𝒎

𝑷 = 𝛾𝑄𝐻𝑚/𝜂𝑚 = 𝛾(𝜋𝐷2𝑏2𝑉𝑓2)𝐻𝑚/𝜂𝑚

⇒ 𝑷 = 9810 × 𝜋 × 0.2 × 0.015 × 3 × 55.6/0.75 ⇒ 𝑷 = 𝟐𝟎. 𝟓𝟔 𝐤𝐖

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Expressions for Manometric Head

Typical Pump Setup

Applying Bernoulli’s Equation between points 1 and 4

𝑝𝑎𝛾+ 0 + 0 + 𝐻𝑚 =

𝑝𝑎𝛾+ 𝐻𝑠 + 𝐻𝑑 + (𝐻𝑓𝑠+𝐻𝑓𝑑) +

𝑉𝑑2

2𝑔

⇒ 𝐻𝑚= 𝐻𝑠 + 𝐻𝑑 + (𝐻𝑓𝑠+𝐻𝑓𝑑) +𝑉𝑑2

2𝑔

⇒ 𝐻𝑚= 𝐻𝑠𝑡𝑎𝑡 + Σ𝐿𝑜𝑠𝑠𝑒𝑠 +𝑉𝑑2

2𝑔

⇒ 𝐻𝑚= 𝐻𝑠𝑡𝑎𝑡 + Σ𝐿𝑜𝑠𝑠𝑒𝑠

Manometric head of a pump is the gross head that

must be provided by the impeller for the liquid to

flow from the sump to the delivery point.

Applying Bernoulli’s Equation between points 2 and 3

𝐻𝑚 =𝑝3𝛾+𝑉𝑑2

2𝑔+ 𝑍3 −

𝑝2𝛾+𝑉𝑠2

2𝑔+ 𝑍2

𝐻𝑚 =𝑝3𝛾−𝑝2𝛾+𝑉𝑑2

2𝑔−𝑉𝑠2

2𝑔

𝐻𝑚 =𝑝3𝛾−𝑝2𝛾

Ideal Increase in Pressure Head in the Impeller

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Energy of a liquid at inlet + Energy added by the

impeller = Energy of liquid at outlet

𝑝1𝛾+𝑉12

2𝑔+ 𝑍1 +

𝑢2𝑣𝑤2𝑔

=𝑝2𝛾+𝑉22

2𝑔+ 𝑍2

Increase in piezometric head at the pump

𝑝2𝛾+ 𝑍2 −

𝑝1𝛾+ 𝑍1 = Δ𝐻𝑝 =

𝑉12

2𝑔−𝑉22

2𝑔+𝑢2𝑣𝑤2𝑔

𝑝2𝛾−𝑝1𝛾= 𝐻𝑚 = Δ𝐻𝑝 =

𝑉12

2𝑔−𝑉22

2𝑔+𝑢2𝑣𝑤2𝑔

Δ𝑝𝑖𝛾= 𝐻𝑚 & 𝐻𝑚 = Δ𝐻𝑝 = 𝐻𝑒 − Δ𝐾𝐸

Δ𝐻𝑝 =Δ𝑝𝑖𝛾=1

2𝑔𝑉12 − 𝑉2

2 + 2𝑢2𝑣𝑤2

Ideal Increase in Pressure Head in the Impeller


𝑢2−𝑣𝑤2

⇒ 𝑣𝑤2 = 𝑢2 −𝑉𝑓2

tan𝛽2= 𝑢2 − 𝑉𝑓2𝑐𝑜𝑡𝛽2

𝑉22 = 𝑣𝑤2

2 + 𝑉𝑓22

= 𝑢22 + 𝑉𝑓2

2 cot2 𝛽2 − 2𝑢2𝑉𝑓2𝑐𝑜𝑡𝛽2 + 𝑉𝑓22

= 𝑉𝑓22 (1 + cot2 𝛽2) + 𝑢2

2 − 2𝑢2𝑉𝑓2𝑐𝑜𝑡𝛽2

= 𝑉𝑓22 cosec2 𝛽2 + 𝑢2

2 − 2𝑢2𝑉𝑓2𝑐𝑜𝑡𝛽2

Δ𝐻𝑝 =Δ𝑝𝑖

𝛾=

1

2𝑔𝑉12 − 𝑉2

2 + 2𝑢2𝑣𝑤2

Δ𝑝𝑖𝛾=1

2𝑔𝑉𝑓12 − 𝑉𝑓2

2 cosec2 𝛽2 − 𝑢22 + 2𝑢2𝑉𝑓2𝑐𝑜𝑡𝛽2 + 2𝑢2(𝑢2−𝑉𝑓2𝑐𝑜𝑡𝛽2)

∴Δ𝑝𝑖𝛾=1


2 cosec2 𝛽2 + 𝑢22

Outlet

𝛽1

𝛼2

𝑢1

𝑢2

𝑉1 = 𝑉𝑓1

Inlet

𝑣𝑟1

𝑉2 𝑣𝑟2

𝛽2

𝑣𝑤2

Head Losses in Centrifugal Pumps

Mechanical losses are the frictional losses

in bearings, glands, packages, etc. and the

disc friction between the impeller and the

liquid which fills the clearance space

between the impeller and the casing.

Some leakage loss also take place

between impeller and casing, at

mechanical seals, glands, etc.

Hydraulic losses are due to:

Circulatory flow at the passages of the impeller and independent of the discharge.

Fluid friction at the flow passage: this loss depends on the fluid contact area and the

roughness of the surface and hence equal to 𝐾1𝑄2 where 𝐾1 is a coefficient.

Shock losses at the entrance to impeller: this loss occurs due to improper entry angle of the

flow with respect to the blade angle. At design condition, this loss is practically zero and

increases at reduced or increased flow from normal values.

Head Losses in Centrifugal Pumps

To account for various losses, several efficiencies are defined.

Volumetric efficiency, 𝜼𝒗 =𝑸

𝑸+𝑸𝑳

where 𝑄 is Discharge reaching the pump outlet, 𝑄𝐿 is the leakage flow which does not

reach the pump outlet and 𝑄 + 𝑄𝐿 is Discharge entering the eye of the impeller.

Manometric (hydraulic) efficiency, 𝜼𝒎 =𝑯𝒎

𝑯𝒆=

𝒈𝑯𝒎

𝒖𝟐𝒗𝒘𝟐=

𝑯𝒎

𝑯𝒎+𝑯𝒇𝒍

where 𝐻𝑓𝑙 is hydraulic losses in the impeller and the casing.

Mechanical efficiency, 𝜼𝒎𝒆𝒄𝒉 =𝜸 𝑸+𝑸𝑳 𝑯𝒆

𝑷 or 𝜼𝒎𝒆𝒄𝒉 =

𝑯𝒆

𝑯𝒆+𝑯𝒎𝒆𝒄𝒉

where 𝑃 is the mechanical power input to the impeller shaft by the prime mover and

𝐻𝑚𝑒𝑐ℎ is the mechanical head losses.

Overall efficiency, 𝜼𝒐 =𝜸𝑸𝑯𝒎

𝑷 or 𝜼𝒐 = 𝜼𝒗𝜼𝒎𝜼𝒎𝒆𝒄𝒉 =

𝑸

𝑸+𝑸𝑳×𝑯𝒎

𝑯𝒆×𝜸 𝑸+𝑸𝑳 𝑯𝒆

𝑷

Effect of Outlet Blade Angle

Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐

Theoretical Head-Discharge Relationship of a Pump

Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐

Theoretical Head-Discharge Relationship of a Pump

Head-Discharge (He-Q)

relationship for ideal pumps

Head-Discharge (He-Q)

relationship for actual pumps

From the ideal 𝐻 − 𝑄 curve on the left, it can be inferred that forward curved vanes produce higher head at higher discharge. However, both forward-curved and radial vane pumps result in poor efficiency. Forward-curved vane produce larger absolute velocities that require very efficient diffusers to convert the exit kinetic energy into pressure energy. So the energy losses are high. Further, an instability called pump surging occurs in forward-curved vanes.

Therefore, in actual practice, backward-curved vanes in the range of 20° – 40° are of common use. In actual 𝐻 − 𝑄 curve shown on the right the head decreases with increase in discharge for all types of vanes due to hydraulic losses present in real-world applications.

Typical Characteristics of Actual Centrifugal Pumps

Head-Discharge (𝑯−𝑸) characteristics

Variation of Efficiency with discharge

Variation of Power with discharge

Diameter = Constant

Diameter = Constant

Constant

From the 𝐻 − 𝑄 curve, it can be observed that for the same pump (same impeller/diameter) as the speed increases, the head also increases (𝐻 ∝ 𝑁2) at the same discharge, or the discharge increases (𝑄 ∝ 𝑁) at the same head. The higher the pump speed, the higher will be the head or discharge.

From the 𝜂 − 𝑄 curve, it can be observed that for the same pump (same impeller/diameter) as the speed increases, both the maximum efficiency and the maximum discharge increase. The higher the speed, the higher will be the efficiency and discharge.

However, the 𝑃 − 𝑄 curve shows that for a given discharge, higher speed needs higher power input.

Main Characteristics Curve of Actual Centrifugal Pumps

𝜼𝒎𝒂𝒙 (𝑩𝑬𝑷)

Head (Duty Point)

Discharge (Duty Point)

Pump manufacturers provide information on the performance of their pumps in the form of curves, commonly called pump characteristic curve (or simply pump curve). In pump curves the following information may be given: (i) the discharge on the x-axis, (ii) the head on the left y-axis, (iii) the pump efficiency as a percentage on the right (or left) y-axis, (iv) the pump power input on the left (or right) y-axis, (v) the NPSH of the pump on the y-axis (vi) the speed of the pump one the y-axis.

The discharge-head (𝑸,𝑯) values corresponding to BEP (𝜼𝒎𝒂𝒙) is called the ‘Duty Point’ of the pump.

Similarity Ratios (Affinity Laws) for Centrifugal Pumps

Similarity law: 𝑷

𝝆𝑵𝟑𝑫𝟓= 𝒇

𝒈𝑯

𝑵𝟐𝑫𝟐,𝑸

𝑵𝑫𝟑,𝝆𝑵𝑫𝟐

𝝁

⇒ = 𝑓( ,2 , )431

=𝟏 𝑃

𝜌𝑁3𝐷5 𝐴=

𝑃

𝜌𝑁3𝐷5 𝐵= 𝐶𝑝 = 𝑃𝑜𝑤𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

=𝟑 𝑄

𝑁𝐷3 𝐴=

𝑄

𝑁𝐷3 𝐵= 𝐶𝑄 = 𝐹𝑙𝑜𝑤 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

=𝟐 𝐻

𝑁2𝐷2 𝐴=

𝐻

𝑁2𝐷2 𝐵= 𝐶𝐻 = 𝐻𝑒𝑎𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (dropping ‘𝑔’ term)

𝑭𝒐𝒓 𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒂𝒔 𝒘𝒆𝒍𝒍 𝒂𝒔 𝒅𝒚𝒏𝒂𝒎𝒊𝒄 𝒔𝒊𝒎𝒊𝒍𝒂𝒓𝒊𝒕𝒚 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒑𝒖𝒎𝒑𝒔 𝑨 𝒂𝒏𝒅 𝑩, 𝒂𝒍𝒍 𝒕𝒉𝒓𝒆𝒆 ′𝒔 𝒎𝒖𝒔𝒕 𝒃𝒆 𝒆𝒒𝒖𝒂𝒍

=𝟒 𝜌𝑁𝐷2

𝜇= 𝑅𝑒 = 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑐𝑎𝑛 𝑏𝑒 𝑖𝑔𝑛𝑜𝑟𝑒𝑑)

Again, 3

1/2

23/4 =

𝑁 𝑄

𝐻3/4 = 𝑁𝑠 is called specific speed . For dynamic similarity

between two pumps or model and its prototype, the value of Ns should be same.

Specific Speed

Specific Speed, 𝑵𝒔 =𝑵 𝑸

𝑯𝒎𝟑/𝟒

Note that The values of 𝑵,𝑸 𝒂𝒏𝒅 𝑯𝒎 are taken at Best Efficiency Point (BEP)

There is another expression for non-dimensional specific speed (also called shape factor or shape number)

𝑺𝒒 =𝑵 𝑸

𝒈𝑯𝒎𝟑/𝟒

Please note the usage of units while calculating Specific Speed. Different units will result in different values for the same pump.

Ns = Specific Speed 𝑁𝑠 =𝑁 𝑄

𝐻3/4 𝑁𝑠 = 𝑐

𝑁 𝑄

𝐻3/4

N = Speed in rpm N = Speed in rpm

Q = Discharge in US gallons per minute Q = Discharge in m3/hr

H = Head in ft H = Head in m

[1 US gallon = 3.785 Liters] 𝒄 = 𝟎. 𝟖𝟔𝟏 𝑖𝑓 𝑄 𝑖𝑛 𝑚3/ℎ𝑟, 𝒄 = 𝟔. 𝟔𝟕 𝑖𝑓 𝑄 𝑖𝑛 𝑚3/𝑚𝑖𝑛 and

𝒄 = 𝟓𝟏. 𝟔𝟔 𝑖𝑓 𝑄 𝑖𝑛 𝑚3/𝑠

In US Customary Unit (FPS)

If the factor 𝑐 is omitted, the calculated specific speeds will be

different than the US customary values, for example, specific

Reciprocating Pump : 50 to 500 speed of Centrifugal Pump: 10 to 220 if Q is in m3/s

Centrifugal Pump: 500 to 10000 The factor 𝑐 (might be omitted) is used to compare the value

Radial Flow Pump : 500 to 4000 obtained with that obtained from US customary unit which are

Mixed Flow Pump : 2000 to 8000 prevalent in usage among practicing professionals.

Axial Flow/Propeller Pump: 7000 to 20000

This does not indicate the size, rather indicated the shape or type of the fluid machinery.

Specific Speed of a pump is defined as the speed of an imaginary pump

which will produce the same amount of discharge under unit head.

This is a numerical engineering tool for the selection of the type of the

pump for installation*.

*For instance, in an application, if the required discharge and head are known, the prime mover (motor or engine) rpm

is also known, then using these values 𝑁𝑠 can be calculated and a particular type of pump suitable in this 𝑁𝑠 range can

be selected for installation.

• In general, cavitation occurs when the liquid pressure at a given location is reduced to the

vapor pressure of the liquid.

• For a piping system that includes a pump, cavitation occurs when the absolute pressure at

the inlet falls below the vapor pressure of the water.

• This phenomenon may occur at the inlet to a pump and on the impeller blades, particularly

if the pump is mounted above the level in the suction reservoir.

• Under this condition, vapor bubbles form (water starts to boil) at the impeller inlet and

when these bubbles are carried into a zone of higher pressure, they collapse abruptly and

hit the vanes of the impeller (near the tips of the impeller vanes). causing:

Cavitation of Pumps and NPSH

Damage to the pump (pump impeller)

Violet vibrations (and noise).

Reduce pump capacity.

Reduce pump efficiency

• To avoid cavitation, the pressure head at the inlet should not fall below a certain minimum

which is influenced by the further reduction in pressure within the pump impeller. The

parameter used for the determination of cavitation is called ‘Net Positive Suction Head

(NPSH)’.

Net Positive Suction Head (NPSH)

NPSH is the difference between the total head at the pump inlet and the water vapor

pressure head 𝐻𝑣 = 2.5 𝑚 𝑤𝑎𝑡𝑒𝑟 , i.e., 𝑵𝑷𝑺𝑯 = 𝑯𝒑𝒊 +𝑽𝒔𝟐

𝟐𝒈−𝑯𝒗

the datum is taken through the centerline of the pump impeller inlet (eye).

There are two values of NPSH of interest. The first is the required NPSH, denoted

(NPSH)R , that must be maintained or exceeded so that cavitation will not occur and

usually determined experimentally and provided by the manufacturer.

The second value for NPSH of concern is the available NPSH, denoted (NPSH)A ,

which represents the head that actually occurs for the particular piping system. This

value can be determined experimentally, or calculated if the system parameters are

known.

For proper pump operation (no cavitation) : (NPSH)A > (NPSH)R

NPSHavailable (at the installation site) > NPSHrequired (for pump)

As stated above, NPSHrequired is usually given for a particular pump by the manufacturer

for its installation without cavitation. NPSHavailable is calculated at the installation site.

Calculation of NPSH

𝑵𝑷𝑺𝑯 = 𝑯𝒑𝒊 +𝑽𝒔𝟐

𝟐𝒈−𝑯𝒗

A cavitation parameter called “Thoma’s cavitation number” is defined as:

𝜎 =𝑁𝑃𝑆𝐻𝐴𝐻𝑚

=𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 − 𝐻𝑣

𝐻𝑚

datum

hs 𝐻𝑠 𝐻𝑎𝑡𝑚 Applying the Bernoulli’s equation between

point (1) and (2), datum at pump center line

𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 = 𝐻𝑝𝑖 +𝑉𝑠2

2𝑔

∴ 𝑵𝑷𝑺𝑯𝑨 = 𝑯𝒂𝒕𝒎 −𝑯𝒔 −𝑯𝑳 −𝑯𝒗

A critical value 𝜎𝑐 (for no cavitation, 𝜎 ≥ 𝜎𝑐)defined as 3% criteria, corresponds to

critical value of NPSH. This critical value of NPSH is known as NPSHR.

𝜎𝑐 =𝑁𝑃𝑆𝐻𝑅𝐻𝑚

⇒ 𝑁𝑃𝑆𝐻𝑅 = 𝜎𝑐𝐻𝑚

Thus, NPSHR is defined as the excess absolute head above 𝐻𝑣 , required by the pump

to obtain satisfactory pumping head (i.e., no more than 3% reduction in head or

efficiency at constant flow) and to prevent cavitation. It is determined by the pump

manufacturer through tests.

𝑯𝑳 = 𝑯𝒇 + 𝜮𝑯𝒎𝒊𝒏𝒐𝒓

=𝒇𝒍𝒗𝟐

𝟐𝒈𝒅+ 𝜮𝒌

𝒗𝟐

𝟐𝒈

To install a pumping station that can be effectively operated over a large range of

fluctuations in both discharge and pressure head, it may be advantageous to install

several identical pumps at the station in parallel or in series operation.

Pumps in Parallel: Pumping stations frequently contain several (two or more)

pumps in a parallel arrangement.

Any number of the pumps can be operated simultaneously.

The discharge is increased but the pressure head remains the same as with a

single pump.

A common feature of sewage pumping stations where the inflow rate varies

during the day.

Multiple-Pump Operation

Multiple-Pump Operation

Pumps in Series:

Increases the pressure head keeping the discharge approximately the same as that of a

single pump.

Basis of multistage pumps; the discharge from the first pump (or 1st stage) is delivered to

the inlet of the second pump (or 2nd stage), and so on.

The same discharge passes through each pump receiving a pressure boost in doing so..

Note that, however, all pumps in a series system must be operated simultaneously

Multi-stage Pump

Similar to series arrangement of identical pumps;

series of impellers mounted on a single compact

shaft instead to save space. the discharge from the

1st stage impeller is delivered to the inlet of the 2nd

stage impeller and so on.

Usually even numbers of impellers are used, the

inlets of one-half of the impellers are facing

opposite to the inlets of the other half of the

impellers to produce zero axial thrust on the shaft

which is known as ‘opposed mounting’.

Used when high heads are required but both

impeller speed and size limitations are prohibitive.

The performance analysis is done per stage basis as for a single-stage pump.

The specific speed is calculated based on manometric head per stage.

The total manometric head is calculated by multiplying manometric head per stage

with the number of stages.

Applications include Boiler feed, deep-well pumping, water supply to very high-rise buildings, etc.

Submersible pumps, Deep-well turbine pumps are some of the examples.

Multi-stage pump with four stages

Multi-stage Centrifugal Pumps

Submersible

Pump

Deep-well

Turbine Pump

Other Types of Pumps

Propeller

pump

(axial-

flow/roto-

dynamic

type)

arranged in

vertical

operation

Jet Pump - uses a jet, often of steam, to create a

low pressure. This low pressure sucks in fluid and

propels it into a higher pressure region.

Screw pump (positive displacement

type) - a screw pumps works across

one pair of mating threads as shown in

the figure. The liquid is similarly

carried across all the pairs of mating

threads.

Liquid is trapped

between threads at

the suction end…

Selection of A Pump

The approximate ranges of application of each type of pumps are shown in the following Figure.

Selection of A Pump

The centrifugal

pumps (radial flow)

occupy a substantial

part of the space.

The regions of

appropriate choice of

jet pumps (mixed

flow) and propeller

pumps (axial flow)

are also indicated by

the thick border.

The positive

displacement pumps

have a niche space in

the high-head-low-

discharge category.

Problem–6: A centrifugal pump delivers 0.055 m3/s (𝑄) of water to a total height of 16 m

(𝐻𝑠𝑡𝑎𝑡). The diameter of the pipe is 150 mm (𝑑) and it is 22 m (𝑙) long. If the overall

efficiency is 75% (𝜂𝑜) , calculate the power (𝑷) required to drive the pump. Take 𝑓 = 0.05 for

the pipe.

Solution: Water velocity in the pipe, 𝑉 =4𝑄

𝜋𝐷2=4×0.055

𝜋×0.152= 3.11 𝑚/𝑠

𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 + 𝐻𝑓 +𝑉2

2𝑔

⇒ 𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 +𝑓𝑙𝑉2

2𝑔𝑑+𝑉2

2𝑔

⇒ 𝐻𝑚 = 16 +0.05×22×3.112

2×9.81×0.15+

3.112

2×9.81= 20.11 𝑚

Example Problems

𝜂𝑜 =𝛾𝑄𝐻𝑚

𝑃

⇒ 𝑷 =𝛾𝑄𝐻𝑚𝜂𝑜

=9810 × 0.055 × 20.11

0.75= 𝟏𝟒. 𝟒𝟕 𝒌𝑾

Problem–7: A centrifugal pump delivers 10 l/s (𝑄) of water at 1500 rpm (𝑁). The internal

and external diameters of the impeller are 125 mm (𝐷1) and 250 mm (𝐷2) respectively.

Width of the impeller at the inlet is 13 mm(𝑏1) and at the outlet 7 mm (𝑏2). Vanes are curved

backward at an angle of 30o at the outlet (𝛽2). If water enters radially at the inlet (𝑉1 = 𝑉𝑓1),

find the pressure rise in the impeller (𝚫𝒑/𝜸).

Solution: 𝑄 = 𝜋𝐷1𝑏1𝑉𝑓1

⇒ 10 × 10−3 = 𝜋 × 0.125 × 0.013 × 𝑉𝑓1 ⇒ 𝑉𝑓1= 1.958 𝑚/𝑠

Example Problems

𝑄 = 𝜋𝐷2𝑏2𝑉𝑓2

⇒ 10 × 10−3 = 𝜋 × 0.25 × 0.007 × 𝑉𝑓2 ⇒ 𝑉𝑓2= 1.819 𝑚/𝑠


60=𝜋×0.25×1500

60= 19.63 𝑚/𝑠

𝜟𝒑

𝜸=

1


2 𝑐𝑜𝑠𝑒𝑐2𝛽2 + 𝑢22

∴𝜟𝒑

𝜸=

1

2 × 9.81 1.9582 −

1.8192

𝑠𝑖𝑛2 30° + 19.632 = 𝟏𝟗. 𝟏𝟔 𝒎

Example Problem

Problem–8:

In US customary units, 𝑵𝒔 = 51.66 × 42.4 = 𝟐𝟏𝟗𝟎

Note: For radial flow pumps, 𝑁𝑠 = 500 𝑡𝑜 4000

Example Problem

Problem–9: A centrifugal pump delivers 10 liter/s at 900 rpm against a head of 20 m. What

head will be developed when the pump runs at 600 rpm? What will be the quantity of water

delivered at that head?

𝑁√𝑄

𝐻3/4 1=

𝑁√𝑄

𝐻3/4 2= 𝑁𝑠 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑝𝑒𝑒𝑑

For the same pump, 𝑫𝟏 = 𝑫𝟐 = 𝑫

⇒ 900×√10

203/4=600×√(𝑄2)

8.893/4 ⇒ 𝑄2 =

900

600

2×

8.89

20

3/2= 6.67 𝑙𝑖𝑡𝑒𝑟/𝑠

=𝟐 𝐻

𝑁2𝐷2 1=

𝐻

𝑁2𝐷2 2= 𝐶𝐻 = 𝐻𝑒𝑎𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

=𝟐 𝐻

𝑁2 1=

𝐻

𝑁2 2 ⇒

20

9002=

𝐻2

6002 ⇒ 𝑯𝟐 = 𝟖. 𝟖𝟗 𝒎

Solution:

=𝟑 𝑄

𝑁𝐷3 1=

𝑄

𝑁𝐷3 2= 𝐶𝑄 = 𝐹𝑙𝑜𝑤 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

Again for the same pump, 𝑫𝟏 = 𝑫𝟐 = 𝑫

=𝟑 𝑄

𝑁 1=

𝑄

𝑁 2 ⇒

10

900=

𝑄2

600 ⇒ 𝑸𝟐 = 𝟔. 𝟔𝟕 𝒍𝒊𝒕𝒆𝒓/𝒔

𝐀𝐥𝐭𝐞𝐫𝐧𝐚𝐭𝐢𝐯𝐞: 𝒖𝒔𝒊𝒏𝒈 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅

Problem–10: The speed of two geometrically similar centrifugal pumps is 1000 rpm. The

outside diameter of the impeller of the first pump is 360 mm. It delivers 27 l/s of water

against a head of 17 m. If the flow rate of second pump is half of the first pump, find the

diameter of the impeller and head for the second pump.

Solution:

Example Problems

𝑁√𝑄

𝐻3/4 1=

𝑁√𝑄

𝐻3/4 2= 𝑁𝑠 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑝𝑒𝑒𝑑

=𝟑 𝑄

𝑁𝐷3 1=

𝑄

𝑁𝐷3 2= 𝐶𝑄 = 𝐹𝑙𝑜𝑤 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

For the first pump, 𝑁𝑠 = 51.66 ×1000 27×10−3

173/4 = 1014

For the second pump, 1014 = 51.66 ×1000 27×10−3/2

𝐻3/4 ⇒ 𝐻2 = 10.7 𝑚

For the two pumps, 𝑵𝟏 = 𝑵𝟐 = 𝟏𝟎𝟎𝟎 rpm

=𝟑 𝑄

𝐷3 1=

𝑄

𝐷3 2 ⇒

𝑄

0.363=𝑄/2

𝐷3 ⇒ 𝑫𝟐 = 𝟎. 𝟐𝟖𝟓𝟕 𝒎 = 𝟐𝟖𝟔 𝒎𝒎

=𝟐 𝐻

𝑁2𝐷2 1=

𝐻

𝑁2𝐷2 2= 𝐶𝐻 = 𝐻𝑒𝑎𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

=𝟐 𝐻

𝐷2 1=

𝐻

𝐷2 2 ⇒

17

.362=

𝐻2

.28572 ⇒ 𝑯𝟐 = 𝟏𝟎. 𝟕 𝒎

Again, from

𝐀𝐥𝐭𝐞𝐫𝐧𝐚𝐭𝐢𝐯𝐞: 𝒖𝒔𝒊𝒏𝒈 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅

Problem–11: The NPSHR of a centrifugal pump is given by the manufacturer as 7.5 m abs.

The pump is employed to pump water at 0.3 m3/s from a sump whose water level is 2.05 m

below the pump inlet. The atmospheric pressure at the site is 97 kPa abs and the vapor

pressure at the relevant temperature is 2.35 kPa abs. Total head loss in the suction pipe is

estimated to be 0.95 m. Determine the NPSHA and comment on the suitability of the

installation against the cavitation.

Solution:

Example Problems

𝑁𝑃𝑆𝐻𝐴 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 − 𝐻𝑣

⇒ 𝑁𝑃𝑆𝐻𝐴 =97 × 103

9810− 2.05 − 0.95 −

2.35 × 103

9810= 6.65 𝑚 𝑎𝑏𝑠

Since 𝑵𝑷𝑺𝑯𝑨(𝟔. 𝟔𝟓 𝒎) < 𝑵𝑷𝑺𝑯𝑹 (7.5 m), the pump will have cavitation problem.

Problem–12: A pump has a critical cavitation constant = 0.12. This pump has to be installed

in a well with a pipe of 10 m length and 200 mm diameter. There are an elbow (ke=1) and a

valve (kv= 4.5) in the system. The flow is 0.035 m3/s and the total dynamic head Hm = 25 m.

The atmospheric pressure is 9.7 m water abs and vapor pressure = 2.0 m water abs.

Calculate the maximum suction height for the pump to run without cavitation. [f = 0.02]

Solution:

𝑁𝑃𝑆𝐻𝑅 = 𝜎𝑐𝐻𝑚 = 0.12 × 25 = 3 𝑚 = 𝑁𝑃𝑆𝐻𝐴|𝑚𝑖𝑛

⇒ 𝑉𝑠 =4𝑄

𝜋𝑑2=4 × 0.035

𝜋 × 0.22= 1.11 𝑚/𝑠

𝐻𝐿 = 𝐻𝑓 + 𝐻𝑚𝑖𝑛𝑜𝑟 =𝑓𝑙𝑉𝑠

2

2𝑔𝑑+ 𝑘𝑒

𝑉𝑠2

2𝑔+ 𝑘𝑣

𝑉𝑠2

2𝑔= 0.063 + 0.063 + 0.283 = 0.409 𝑚

𝑁𝑃𝑆𝐻𝐴|𝑚𝑖𝑛 = 𝐻𝑎𝑡𝑚 −𝑯𝒔|𝒎𝒂𝒙 − 𝐻𝐿 − 𝐻𝑣 ⇒ 3 = 9.7 − 𝑯𝒔|𝒎𝒂𝒙 − 0.409 − 2.0

∴ 𝑯𝒔|𝒎𝒂𝒙 = 𝟒. 𝟑 𝒎

Exercise Problems Problem–1: A centrifugal pump delivers 120 l/s of water against a head of 25 m while running at 1500

rpm. The outside diameter of impeller is 300 mm and the breadth of the impeller at exit is 50 mm. If the

manometric efficiency is 80%, find the blade angle at outlet. Water enters the impeller radially at the

inlet. [Ans. 13.6o]

Problem–2: The outer and inner diameters of the impeller of a centrifugal pump are 500 mm and 245

mm respectively. The vanes are curved backward at an angle of 40o. The constant velocity of flow is

2.2 m/s at both inlet and outlet. The manometric head of the pump is 9.5 m. If the pump speed is 500

rpm, find its manometric efficiency and the vane angle at inlet. [Ans. 68%, 18.9o]

Problem–3: A centrifugal pump delivers 0.20 m3/s of water to a head of 35 m at a speed of 1500 rpm.

The outer diameter and width of the impeller at the outlet are 300 mm and 50 mm respectively. (i) If the

manometric efficiency is 0.75, calculate the blade angle at the outlet, (ii) If the impeller diameter at the

inlet is 150 mm, calculate the blade angle at inlet. [Ans. (i) 45.75o, (ii) 19.8o]

Problem–4: A two-stage centrifugal pump delivers 110 l/s of water at 1200 rpm. The outer diameter and

width of the impeller at the outlet are 450 mm and 25 mm respectively. The blades are curved backward

at an angle of 30o at the outlet. Due to the thickness of the blades, 10% of the exit area is blocked. If the

manometric efficiency is 86% and the overall efficiency is 80%, find the head developed by the pump

and power required to drive the pump. [Ans. 55.22 m, 148.97 kW] [Hint: 𝑄 = 𝜋𝐷2𝑏2 1 − 0.10 𝑉𝑓2]

Problem–5: The outside and inside diameters of the impeller of a centrifugal pump are 550 mm and 275

mm respectively. The inlet vane angle is 30o and the outlet vane angle is 45o. The constant velocity of

flow is 3 m/s. Find the speed of the pump, work done per unit of water and the pressure rise in the

impeller. [Ans. 361.14 rpm, 7.85 Nm, 5.06 m water]

Problem–6: Two geometrically similar pumps A and B have the same speed of 1500 rpm. Pump A has a

diameter of 0.35 m and discharge of 36 L/s against a head of 25 m. Pump B gives a discharge of 18 L/s.

Estimate the total head and impeller diameter of pump B.

[Ans. 15.86 m, 278 mm]

Exercise Problems Problem–7: A centrifugal pump delivers 60 l/s of salt water (sp. gr. 1.2) or petrol (sp. gr. 0.71) against

a pressure of 410 kPa. The overall efficiency of the pump is 66%. Prove that the same power is

required to drive the pump for both the liquids. [Ans. 37.27 kW]

Problem–9: This is required to pump 1.3 m3/s of water to a total head of 45 m. How many pumps of

specific speed 2070 (US Customary unit) and running at 1450 rpm would be needed when connected in

parallel? The dynamic head in the system can be neglected.

[Ans. 6] [Hints: 𝑁𝑠 = 2070 = 51.661450× 𝑄

450.75, ⇒ 𝑄 = 0.23 𝑚3/𝑠(for one pump),

So Number of pumps required = 1.3 / 0.23 ≈ 6 pumps]

Problem–10: A discharge of 0.4 m3/s of water is needed to be pumped to a total head of 240 m. How

many pumps connected in series and each having a specific speed of 1810 (US Customary unit) and

running at a speed of 1500 rpm would be needed for the job? The dynamic head in the system can be

neglected. [ Ans. 3]

[Hints: 𝑁𝑠 = 1810 = 51.661500× 0.4

𝐻0.75, ⇒ 𝐻 = 81.3 𝑚 (for one pump),

So Number of pumps required = 240 / 81.3 ≈ 3 pumps]

Problem–8: The scale ratio of the model and prototype of a centrifugal pump is 1/4. The prototype

delivers 1550 l/s of water at 550 mm against a head of 31 m and absorbs 750 kW. If the model works

against a head of 11 m, find the speed, discharge and power required by the model. [Ans. 1310.5 rpm,

61.55 l/s, 9.91 kW]

Problem–11: A four-stage centrifugal pump delivering 0.76 m3/s of water at 1000 rpm against a

manometric head of 66 m. Vanes are curved backward at an angle of 60o at the outlet. The ratio of the

velocity of flow and the peripheral velocity at outlet is 0.25. If the hydraulic loss is 1/3 of the exit kinetic

energy per unit weight of water, find the outside diameter of each impeller and the manometric efficiency.

[Ans. 286 mm, 84.4%]

Hints: Manometric head per stage, 𝐻𝑚 =66

4, Hydraulic loss, 𝐻ℎ𝑦𝑑𝑟𝑜_𝑙𝑜𝑠𝑠 =

1

3

𝑉22

2𝑔= 36.99𝐷2

2

Exercise Problems

Problem–14: A single stage centrifugal pump delivers 0.5 m3/s of water at 2000 rpm against a head of

32 m. The outside diameter of impeller is 250 mm. A geometrically similar multistage pump is required

to deliver 0.75 m3/s of water at 1500 rpm against a head of 220 m. Find out the number of impellers of

the multistage pump and the outside diameter of each of the impellers of the same pump. [Ans. 8, 700

mm]

Problem–16: The scale ratio of the model and prototype of a centrifugal pump is 0.5. The outside

diameter of the impeller of the model is 150 mm. The model supplies 0.045 m3/s of water at 7000 rpm

against a head of 42 m. If the efficiency of the model and prototype is same, find the discharge, head

and speed of the model. Find also the specific speed of the pump. [Ans. 0.09 m3/s, 10.5 m, 1750 rpm,

88.74 rpm]

Problem–15: A centrifugal pump delivers 130 l/s of water at 1050 rpm. The outside diameter of impeller

is 300 mm and it is 65 mm wide at exit. The blade angle at outlet is 30o. If the manometric efficiency is

86%, find the specific speed of the pump. [Ans. 60.39]

Problem–13: The external and internal diameters of an impeller of a centrifugal pump are 450 mm and

225 mm respectively. The pump delivers 200 l/s water at 1250 rpm. The outside and inside widths of the

impeller are 70 mm and 150 mm. The vanes are curved backward at an angle of 30o at exit. If the

manometric efficiency is 82%, find the Euler head and manometric head. [Ans. 77.90 m, 63.88 m]

Problem–12: A three-stage centrifugal pump delivers 65 l/s at 950 rpm. The outside diameter and

outside width of each impeller are 380 mm and 26 mm respectively. The vanes are curved backward at

an angle of 45o at the exit. If the manometric efficiency is 86%, find the manometric head developed by

the pump. [Ans. 83.55 m water]

[Hint: 𝐻𝑚 = 3𝐻𝑚│𝑝𝑒𝑟 𝑠𝑡𝑎𝑔𝑒]

centrifugal pump

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