SME1202 Fluid mechanics and Machinery Unit -4 PUMPS Prepared by: Mr.Kumaravel

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UNIT - 4 PUMPS

Impact of jets - Theory of roto-dynamic machines - various efficiencies-velocity components at entry and exit

of the rotor- velocity triangles. Centrifugal Pumps: Definition-Operations- Velocity Triangles Performance

curves - Cavitations-Multi-staging.

Reciprocating Pumps: Operation - Slip - indicator Diagram - Separation - Air vessels.

HYDRAULIC PUMP

A hydraulic pump is a mechanical source of power that converts mechanical power into hydraulic energy .

It generates flow with enough power to overcome pressure induced by the load at the pump outlet. When a

hydraulic pump operates, it creates a vacuum at the pump inlet, which forces liquid from the reservoir into the

inlet line to the pump and by mechanical action delivers this liquid to the pump outlet and forces it into the

hydraulic system.

Centrifugal Pump :The main components of a centrifugal pump are: i) Impeller ii) Casing iii) Suction pipe

iv) Foot valve with strainer, v) Delivery pipe vi) Delivery valve.

Impeller is the rotating component of the pump. It is made up of a series of curved vanes. The impeller is

mounted on the shaft connecting an electric motor.

Casing is an air tight chamber surrounding the impeller. The shape of the casing is designed in such a way

that the kinetic energy of the impeller is gradually changed to potential energy. This is achieved by gradually

increasing the area of cross section in the direction of flow

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Suction pipe: It is the pipe connecting the pump to the sump, from where the liquid has to be lifted up.

Foot valve with strainer: The foot valve is a non-return valve which permits the flow of the liquid from the sump towards the pump. In

other words the foot valve opens only in the upward direction. The strainer is a mesh surrounding the valve, it prevents the entry of

debris and silt into the pump.

Delivery pipe is a pipe connected to the pump to the overhead tank. Delivery valve is a valve which can regulate the flow of liquid from

the pump.

Working: A centrifugal pump works on the principle that when a certain mass of fluid is rotated by an

external source, it is thrown away from the central axis of rotation and a centrifugal head is

impressed which enables it to rise to a higher level.

Working operation of a centrifugal pump is explained in the following steps. 1) Close the

delivery valve and prime the pump. 2) Start the motor connected to the pump shaft, this causes

an increase in the impeller pressure. 3) Open the delivery valve gradually, so that the liquid

starts flowing into the deliver pipe. 4) A partial vacuum is created at the eye of the centrifugal

action, the liquid rushed from the sump to the pump due to pressure difference at the two ends

of the suction pipe. 5) As the impeller continues to run, move & more liquid is made available

to the pump at its eye. Therefore impeller increases the energy of the liquid and delivers it to

the reservoir. 6) While stopping the pump, the delivery valve should be closed first, otherwise

there may be back flow from the reservoir.

It may be noted that a uniform velocity of flow is maintained in the delivery pipe. This is due to

the special design of the casing. As the flow proceeds from the tongue of the casing to the

delivery pipe, the area of the casing increases. There is a corresponding change in the quantity

of the liquid from the impeller. Thus a uniform flow occurs in the delivery pipe.

Centrifugal pump converts rotational energy, often from a motor, to energy in a moving fluid.

A portion of the energy goes into kinetic energy of the fluid. Fluid enters axially through eye of

the casing, is caught up in the impeller blades, and is whirled tangentially and radially outward

until it leaves through all circumferential parts of the impeller into the diffuser part of the

casing. The fluid gains both velocity and pressure while passing through the impeller. The

doughnut-shaped diffuser, or scroll, section of the casing decelerates the flow and further

increases the pressure. The negative pressure at the eye of the impeller helps to maintain the

flow in the system. If no water is present initially, the negative pressure developed by the

rotating air, at the eye will be negligibly small to suck fresh stream of water. As a result the

impeller will rotate without sucking and discharging any water content. So the pump should be

initially filled with water before starting it. This process is known as priming.

Use of the Casing

From the illustrations of the pump so far, one speciality of the casing is clear. It has an

increasing area along the flow direction. Such increasing area will help to accommodate newly

added water stream, and will also help to reduce the exit flow velocity. Reduction in the flow

velocity will result in increase in the static pressure, which is required to overcome the

resistance of pumping system.

NPSH - Overcoming the problem of Cavitation

If pressure at the suction side of impeller goes below vapor pressure of the water, a dangerous

phenomenon could happen. Water will start to boil forming vapor bubbles. These bubbles will

move along with the flow and will break in a high pressure region. Upon breaking the bubbles

will send high impulsive shock waves and spoil impeller material overtime. This phenomenon

is known as cavitation. More the suction head, lesser should be the pressure at suction side to

lift the water. This fact puts a limit to the maximum suction head a pump can have.

However Cavitation can be completely avoided by careful pump selection. The term NPSH

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(Net Positive Suction Head) helps the designer to choose the right pump which will completely

avoid Cavitation. NPSH is defined as follows.

Where Pv is vapor pressure of water

V is speed of water at suction side

For a given pumping system it will have an NPSH called 'Available NPSH'. Pump anufacturer

will specify the minimum NPSH required for each pump for its safe operation, known

as 'Required NPSH'. If the pump needs to work without Cavitation the 'Available

NPSH' should be greater than

'Required NPSH'.

FLOW AT IMPELLER INLET AND EXIT

The rotating impeller of a centrifugal pump imparts energy to the fluid. As mentioned in earlier

lesson, the impeller contains radial flow passages formed by rotating blades (vanes) arranged in

a circle. A disk in the back (base plate) connects the impeller assembly to the shaft and another

disk (crown plate) covers the blades on the front. The flow enters axially near the center of

rotation and turns in the radial direction inside the impeller as shown in Fig. Thus, the liquid

enters the impeller at its center and leaves at its outer periphery.

The flow follows certain streamlines inside the rotating impeller, approximately parallel to the

blade surfaces. The shape of the blades and the resulting flow pattern in the impeller determine

how much energy is transferred by a given size of the impeller and how efficiently it operates.

The theoretical energy increase [i.e., theoretical head rise (Hmth) through the impeller] can be

found by applying the principle of conservation of angular momentum.

The components of flow through an impeller can be best studied by means of velocity vectors

as illustrated in Figure. In this figure, the inlet and outlet velocity diagrams of an impeller with

backward curved vanes are shown. Note that this figure shows a portion of the impeller of a

centrifugal pump with one blade only. The velocity vector diagram is triangular and hence, it is

known as a ‘velocity triangle’. It can be drawn for any point the flow path through the

impeller.

However, velocity triangles are usually drawn at the impeller inlet and outlet (exit) and are

called ‘inlet or entrance velocity triangle’ and ‘outlet or discharge velocity triangle’,

respectively.

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Let V = absolute velocity of the liquid (measured in the stationary frame of reference),

u = peripheral (tangential) velocity of the impeller (i.e., rotational velocity),

Vr = relative velocity of the liquid (i.e., velocity of the liquid relative to the impeller),

Vf = velocity of flow of the liquid, and

Vw = velocity of whirl of the liquid.

Further, let 𝜃 = impeller vane angle at the inlet, and ϕ = impeller vane angle at the outlet.

Similarly, α = angle between the absolute velocity of entering liquid and the peripheral velocity

of the impeller at the inlet point, and β = angle between the absolute velocity of leaving liquid

and the peripheral velocity of the impeller at the outlet (exit) point.

Since there are no guide vanes at the entrance to the impeller, the direction of absolute velocity

of liquid at this point is not directly known. However, for the best efficiency of the centrifugal

pump, it usually assumed that the liquid enters the impeller radially. Thus, a = 90º and the

velocity of whirl (Vw) at the inlet is zero, and hence V and Vf at the inlet are the same.

Moreover, it is desired that the liquid enters and leaves the vane without shocks. This can be

ensured if the inlet and outlet tips of the vane are parallel to the direction of the relative

velocities at the two tips. As such it is assumed that the peripheral velocities u and u1 are

parallel to the tangents to the impeller at the inlet and outlet vane tips, respectively . The

relative velocity of the liquid (Vr) is obtained by the vector sum of the absolute velocity (V)

and the peripheral velocity of the impeller (u).

Problem 1

The internal and external diameter of the impeller of a centrifugal pump are 200 mm and 400

mm respectively. The pump is running at 1200 rpm. The vane angles of the impeller at inlet

and outlet are 20° and 30° respectively. The water enters the impeller radially and velocity of

flow is constant. Determine the work done by the impeller per unit weight of water.

Given,

D1=0.2 M ; D2 = 0.4 M ; N=1200 RPM ; θ= 20 o ; ϕ = 30 o ;

Wkt , By Condition ,α=90o ;vw1 = 0 ; vf1=vf2

u1 = Π D1N / 60

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=( 3.14 X 0.20X1200 ) / 60 = 12.56 m/s

u2 = ΠD2N / 60

=( 3.14 X 0.40X1200 ) / 60 = 25.13 m/s

Wkt , tan θ = vf1 / u1

vf1 = 4.57 m/s

tan ϕ = vf1 / (u2 –vw2)

vw2 = 17.215 m/s

work done = vw2 u2 / g

17.215x25.13 / 9.81 = 44.1 Nm

MULTISTAGE CENTRIFUGAL PUMP

A simple centrifugal pump use a single impeller mounted on a shaft to produce a specific head

and a specific discharge rate but what if you need n times greater head or discharge rate. This

problem can be solved by using the multistage centrifugal pump

There are two possible reason why multistage centrifugal pumps are used.

1. Need high head at constant discharge rate

2. Need more discharge rate at constant head

Mounting more than one impeller on a same shaft and closing them in same casing will

produce higher head than single impeller pump but the discharge rate will be same as single

impeller.

If two pumps are installed parallel to each other at same sump then the discharge rate will be

increase but head will be same as that of single pump

PUMP IS SERIES

In series arrangement of pumps, more that one impellers are mounted on the single shaft of a

centrifugal pump and closed under a same casing. This arrangement can increase the head of

the pump by keeping the discharge rate constant.This type is used to deliver small quantity of

liquid at high head.

In this type of arrangement impeller one take is input from the sump and discharge is at a

specific head and discharge rate. This out put if impeller one is the input for the impeller

number two. Output of impeller number two will be same discharge and twice the head.

H total = n H

n = number of impeller

H = head made by single impeller

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Fig: Two stage pump-Impellers in series

Advantages

Less frictional losses

Reduce stresses

Less slip leakage

Thrust can be eliminated

High suction lift at relatively low impeller speed

PUMP IN PARALLEL

For a single centrifugal pump it is impossible to deliver a huge quantity of liquid at small head

but it is possible with parallel arrangement of pump. More that one pumps are install at a

single source and both of them work separately to produce a specific discharge rate then their

output is merged in a single delivery pie to get a greater discharge rate than single pump.

Q total = n Q

n = number of pump installed

Q = Discharge of single pump

Fig: Pumps in Parallel

Problem 2

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A three stage centrifugal pump has impeller 40 cm in diameter and 2 cm wide at outlet.the vane are

curved back at the outlet at 45 o and reduce the circumference area by 10 %.the manometric efficiency

is 90 % and the overall efficiency is 80 % .determine the head generated by the pump when running at

1000 rpm .delivering 50 litre per second.what should be the shaft horse power?

Given ,

n = 3 ; D2= 0.4 m ; B2 = 0.02 m ; ϕ = 45 o reduction in area = 0.1 ; area at outlet = 0.9 x π x 0.4 x0.02

= 0.02262 m2

manometric efficiency = 0.90 ; overall efficiency = 0.80 ; N= 1000 rpm ; Q = 50 lit/s

vf2 = Q/A = 0.05/0.02262 = 2.21 m/s

u2 = ΠD2N / 60 = 20.94 m/s

tan ϕ = vf2 / (u2-vw2)

vw2 =18.73 m/s

manometric efficiency = gHm / vw2 u2

Hm=35.98 m

Total head generated = n x Hm = 3 x 35.98 = 107.94 m

Power output of the pump = (weight of water lifted x total head ) / 1000

=( 1000 x 9.81 x 0.05 x 107.94 ) / 1000 = 52.94 kw

Overall efficiency = power output / power input = 52.94/ shaft power

Shaft power = 66.175 kw

RECIPROCATING PUMP

If the mechanical energy is converted into hydraulic energy by sucking the liquid into a

cylinder in which a piston is reciprocating, which exerts the thrust on the liquid and increases

its hydraulic energy is know as reciprocating pump. A reciprocating pump is a positive

displacement pump. It is often used where relatively small quantity of liquid is to be handled

and where delivery pressure is quite large.

Reciprocating pump consists of following parts.

1. A cylinder with a piston 5. suction pipe

2. piston rod 6. delivery pipe

3. connecting rod 7. suction valve

4. crank 8. delivery valve

WORKING OF A SINGLE-ACTING RECIPROCATING PUMP

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Single acting reciprocating pump:- A single acting reciprocating pump, which consists of a

piston which moves forwards and backwards in a close fitting cylinder. The movement of the

piston is obtained by connecting the piston rod to crank by means of a connecting rod. The

crank is rotated by means of an electric motor. Suction and delivery pipes with suction valve

and delivery valve are connected to the cylinder. The suction and delivery valves are one way

valves or non-return valves, which allow the water to flow in one direction only. Suction valve

allows water from suction pipe to the cylinder which delivery valve allows water from cylinder

to delivery pipe.

The rotation of the crank brings about an outward and inward movement of the piston in the

cylinder . During the suction stroke the piston is moving towards right in the cylinder, this

movement of piston causes vacuum in the cylinder. The pressure of the atmosphere acting on

the sump water surface forces the water up in the suction pipe . The forced water opens the

suction valve and the water enters the cylinder. The piston from its extreme right position

starts moving towards left in the cylinder. The movement of the piston towards left increases

the pressure of the liquid inside the cylinder more than atmospheric pressure. Hence suction

valve closes and delivery valve opens. The liquid is forced into the delivery pipe and is raised

to a required height.

For one revolution of the crank, the quantity of water raised up in the delivery pipe is equal to

the stroke volume in the cylinder in the single acting pump and twice this volume in the double

acting pump.

Discharge through a single acting reciprocating pump.

D = diameter of the cylinder

A = cross section are of the piston or cylinder

r = radius of crank

N = r.p.m of the crank

L = Length of the stroke = 2 x r

hs = Suction head or height of the axis of the cylinder from water surface in sump.

hd = Delivery head or height of the delivery outlet above the cylinder axis.

Discharge of water in one revolution = Area x Length of stroke

= A x L

Number of revolution per second = N/60

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Discharge of the pump per second

Q = Discharge in one revolution x No.of revolution per second

Problem 3

A single acting reciprocating pump ,running at 50 rpm , delivers 0.01 m3/s of water.the diameter of

The piston is 200 mm and stoke length 400 mm. Determine i)theoretical discharge ; ii)co-efficient of

discharge ; iii)slip and percentage of slip

Given ,N= 50 RPM ; Q = 0.01 m3 /s ; D= 0.20 m; L=040 m

A= π X 0.02 / 4 = 0.0314 m2

i)Qth =ALN / 60 = (0.0314 X .40 X 50 )/ 60

= 0.01047 m3/s

ii)Cd =Q ACT / Q THE = 0.01 /0.01047 = 0.955

iii) slip = Q THE - Q ACT =0.01047-0.01=0.00047 m3/s

percentage of slip = (Q THE - Q ACT ) / Q THE =4.489 %

WORKING OF A DOUBLE-ACTING RECIPROCATING PUMP

A double acting reciprocating pump is shown in Fig. It consists of two suction and two delivery

pipes connected to one cylinder. Each suction and delivery pipe is provided with a one-way

valve. The piston has a connecting rod-crank mechanism to drive it. It is connected to a prime

mover. When the prime mover is started the piston moves forward and backward. During each

forward and backward stroke suction and delivery takes place simultaneously.

Suction valve remains closed when delivery takes place either on the front side or rear side of

the piston.

Similarly, delivery valve remains closed when suction takes place either on the front side or

rear side of the piston. The discharge is twice that of a single acting pump.

Discharge through a double-acting reciprocating pump.

Discharge , Qth = 2 LAN

60 m

3

s⁄

The above equation gives the discharge of a double-acting reciprocating pump. This discharge is two

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times the discharge of a single-acting pump.

SLIP OF A RECIPROCATING PUMP

Slip is defined as the difference between theoretical discharge and actual discharge. If actual

discharge is greater than theoretical discharge negative value is found this negative value is

called negative slip.

When Qa > Qth then the slip of the pump becomes negative. Where Qa is the actual discharged

Qth is the theoretical discharge. This is possible when the length of suction pipe is

considerably long, delivery head low and pump is running at high speed.

% of slip = (Qth - Qa ) / Qth

INDICATOR DIAGRAM:

A pressure volume diagram (or PV diagram) is used to describe corresponding changes in

volume and pressure in a system.

PV diagrams, originally called indicator diagrams, were developed for understanding the

efficiency of steam engines. A PV diagram plots the change in pressure P with respect to

volume V for some process or processes. Typically in thermodynamics, the set of processes

forms a cycle, so that upon completion of the cycle there has been no net change in state of the

system; i.e. the device returns to the starting pressure and volume.

The figure shows the features of a typical PV diagram. A series of numbered states (1 through

4) are noted. The path between each state consists of some process (A through D) which alters

the pressure or volume of the system (or both).

A key feature of the diagram is that the amount of energy expended or received by the system

as work can be estimated as the area under the curve on the chart. For a cyclic diagram, the net

work is that enclosed by the curve. In the example given in the figure, the processes 1-2-3

produce a work output, but processes from 3-4-1 require a smaller energy input to return to the

starting position / state; thus the net work is the difference between the two.

AIR VESSEL IN RECIPROCATING PUMP

Air vessels are a closed container, in which the half part is filled with water & upper half part is

filled with compressed air. These air vessels installed very near to the suction valve & delivery

valve to avoid the separation.

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An air vessel usually fitted in the discharge pipe work to dampen out thepressure variations

during discharge. As the discharge pressure rises the airis compressed in the vessel, and as the

pressure falls the air expands. The peak pressure energy is thus stored in the air and returned to

the system when pressure falls. Air vessels are not fitted on the reciprocating boiler feed pumps

since they may introduce air into the de-aerated water.

The top half contains compressed air and lower half contains fluid being pumped. Air and

water are separated by a flexible diaphragm which is movable as per difference of pressure

between two fluids.

Air vessel is connected very near to the pump at nearly pump level. Without air vessel

frictional head increases and reaches a maximum value at mid stroke and decreases to zero.

With air vessel frictional head is constant throughout the stroke.

Impact of Jets

The liquid comes out in the form of a jet from the outlet of a nozzle which is fitted to a pipe

through which the liquid is flowing under pressure.

The following cases of the impact of jet, i.e. the force exerted by the jet on a plate will be

considered:‐

1. Force exerted by the jet on a stationary plate

Plate is vertical to the jet

Plate is inclined to the jet

Plate is curved

2. Force exerted by the jet on a moving plate

Plate is vertical to the jet

Plate is inclined to the jet

Plate is curved

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Force exerted by the jet on a stationary vertical plate

Consider a jet of water coming out from the nozzle strikes the vertical plate

V = velocity of jet, d = diameter of the jet, a = area of x – section of the jet The force exerted by

the jet on the plate in the direction of jet.

Fx = Rate of change of momentum in the direction of force

= initial momentum – final momentum / time

= mass x initial velocity – mass x final velocity / time

= mass/time (initial velocity – final velocity)

= mass/ sec x (velocity of jet before striking – final velocity of jet after striking)

= ρaV (V -0)

Fx= ρaV2

Force exerted by the jet on the moving plate

1st Case: Force on flat moving plate in the direction of jet

Consider, a jet of water strikes the flat moving plate moving with a uniform velocity away from the jet.

V = Velocity of jet

a = area of x-section of jet;

u = velocity of flat plate

Relative velocity of jet w.r.to plate = V – u

Mass of water striking/ sec on the plate = ρa(V - u)

Force exerted by jet on the moving plate in the direction of jet

Fx = Mass of water striking/ sec x [Initial velocity – Final velocity]

= ρa(V - u) [(V - u) – 0]

In this case, work is done by the jet on the plate as the plate is moving, for stationary plate the w.d is zero.

Work done by the jet on the flat moving plate

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= Force x Distance in the direction of force/Time

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2nd Case: Force on inclined plate moving in the direction of jet

Consider, a jet of water strikes the inclined plate moving in the direction of jet with a relative

velocity.

V = Velocity of jet

a = area of x-section of jet

u = velocity of flat plate

Relative velocity of jet w.r.t plate = V – u

If the plate is smooth, it is assumed that the loss of energy due to impact of jet is zero, then

the jet of water leaves the inclined plate with a velocity (V – u ).

Force exerted by jet on the inclined plate in the direction normal to the jet

Fn = Mass of water striking/ sec x [Initial velocity – Final velocity]

= ρa(V - u) [(V - u) sinθ – 0]

This normal force can be resolved into two components one in the direction of jet and other

perpendicular to the direction of jet.

Component of Fn in the direction of jet. Component of Fn in the direction perpendicular to

the direction

of jet

Work done by the jet on the flat moving plate

= Force x Distance in the direction of force/ Time

Force exerted by the jet of water on series of vanes

Force exerted by jet of water on single moving plate (Flat or curved) is not feasible one,

it is only theoretical one.

Let,

V = Velocity of jet

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a = area of x-section of jet. u = velocity of vane

In this, mass of water coming out from the nozzle/s is always in constant with plate.

When all plates are considered.

Mass of water striking/s w.r.t plate = ρaV

Jet strikes the plate with a velocity = V – u

Force exerted by the jet on the plate in the direction of motion of plate

= Mass/sec x (Initial velocity – Final velocity)