Reciprocating pump

• Pumps are used to increase the energy level of water by virtue of

which it can be raised to a higher level.

• Reciprocating pumps are positive displacement pump, i.e. initially, a

small quantity of liquid is taken into a chamber and is physically

displaced and forced out with pressure by a moving mechanical

elements.

• The reciprocating pumps are used where relatively small quantity

(fixed) of liquid per unit time is to be transferred but at relaively

higher pressure.

• For industrial purposes, they have become obsolete due to their

high initial and maintenance costs as compared to centrifugal

pumps.

• Small hand operated pumps are still in use that include well pumps,

etc.

• These are also useful where high heads are required with small

discharge, as oil drilling operations.

Main components

• A reciprocation pumps consists of a plunger or a piston that moves

forward and backward inside a cylinder with the help of a connecting

rod and a crank. The crank is rotated by an external source of power.

• The cylinder is connected to the sump by a suction pipe and to the

delivery tank by a delivery pipe.

• At the cylinder ends of these pipes, non-return valves are provided. A

non-return valve allows the liquid to pass in only one direction.

• Through suction valve, liquid can only be admitted into the cylinder

and through the delivery valve, liquid can only be discharged into the

delivery pipe.

[The terms ‘piston’ and ‘plunger’ are used interchangeably in

reciprocating pump. Although they perform the same action, there

is a structural difference between a piston and a plunger. A plunger is a

smooth cylindrical rod attached to a slider or rod mechanism and a

stationery seal is used around the plunger. Plungers are used for high

discharges. The piston is attached to a rod called piston (connecting)

rod that imparts reciprocating motion. The seal moves with the piston.

Pistons are used for low pressures.]

Main components

Working of Reciprocating Pump

• When the piston moves from the left to the right, a suction pressure

is produced in the cylinder. If the pump is started for the first time or

after a long period, air from the suction pipe is sucked during the

suction stroke, while the delivery valve is closed. Liquid rises into

the suction pipe by a small height due to atmospheric pressure on

the sump liquid.

• During the delivery stroke, air in the cylinder is pushed out into the

delivery pipe by the thrust of the piston, while the suction valve is

closed. When all the air from the suction pipe has been exhausted,

the liquid from the sump is able to rise and enter the cylinder.

• During the delivery stroke it is displaced into the delivery pipe. Thus

the liquid is delivered into the delivery tank intermittently, i.e. during

the delivery stroke only.

Classification of Reciprocating pumps

Following are the main types of reciprocating pumps:

• According to use of piston sides

– Single acting Reciprocating Pump:

If there is only one suction and one delivery pipe and the liquid is

filled only on one side of the piston, it is called a single-acting

reciprocating pump.

– Double acting Reciprocating Pump:

A double-acting reciprocating pump has two suction and two

delivery pipes, Liquid is receiving on both sides of the piston in

the cylinder and is delivered into the respective delivery pipes.

Classification of Reciprocating pumps

• According to number of cylinder

Reciprocating pumps having more than one cylinder are called multi-

cylinder reciprocating pumps.

– Single cylinder pump

A single-cylinder pump can be either single or double acting

– Double cylinder pump (or two throw pump)

A double cylinder or two throw pump consist of two cylinders

connected to the same shaft.

– Triple cylinder pump (three throw pump)

A triple-cylinder pump or three throw pump has three cylinders,

the cranks of which are set at 1200 to one another. Each cylinder

is provided with its own suction pipe delivery pipe and piston.

– There can be four-cylinder and five cylinder pumps also, the

cranks of which are arranged accordingly.

Classification of Reciprocating pumps

Double Acting Reciprocating Pump

Triple Cylinder Reciprocating Pump

Double Cylinder Reciprocating Pump

Discharge through a Reciprocating Pump

Let

A = cross sectional area of cylinder

r = crank radius

N = rpm of the crank

L = stroke length (2r)

Discharge through pump per second= Area x stroke length x rpm/60

This will be the discharge when the pump is single acting.

60

NLAQth

Discharge through a Reciprocating Pump

Discharge in case of double acting pump

Discharge/Second,

where, Ap = Area of cross-section of piston rod

However, if area of the piston rod is neglected

Discharge/Second =

• Thus discharge of a double-acting reciprocating pump is twice than that of a

single-acting pump.

• Owing to leakage losses and time delay in closing the valves, actual discharge

Qa usually lesser than the theoretical discharge Qth.

60

)(

60

LNAAALNQ P

th

60

)2( LNAAQ P

th

60

2ALN

Slip Slip of a reciprocating pump is defined as the difference between the

theoretical and the actual discharge.

i.e. Slip = Theoretical discharge – Actual discharge

= Qth – Qa

Slip can also be expressed in terms of %age and given by

10011001

100%

d

th

act

th

actth

CQ

Q

Q

QQslip

Here Cd is known as co-efficient of discharge and is defined as the ratio of the actual

discharge to the theoretical discharge.

Cd = Qa / Qth.

Value of Cd when expressed in percentage is known as volumetric efficiency of the

pump. Its value ranges between 95---98 %. Percentage slip is of the order of 2% for

pumps in good conditions.

Negative slip

• It is not always that the actual discharge is lesser than the theoretical

discharge. In case of a reciprocating pump with long suction pipe, short

delivery pipe and running at high speed, inertia force in the suction pipe

becomes large as compared to the pressure force on the outside of delivery

valve. This opens the delivery valve even before the piston has completed

its suction stroke. Thus some of the water is pushed into the delivery pipe

before the delivery stroke is actually commenced. This way the actual

discharge becomes more than the theoretical discharge.

• Thus co-efficient of discharge increases from one and the slip becomes

negative.

Power Input / Output

Consider a single acting reciprocating pump.

Let

hs = Suction head or difference in level between centre line of cylinder and the sump.

hd = Delivery head or difference in between centre line of cylinder and the outlet of

delivery pipe.

Hst = Total static head = hs + hd

Theoretical work done by the pump per second (Hydraulic power output)

= ρ Qth g Hst

= 𝜌𝐴𝐿𝑁

60𝑔 ℎ𝑠 + ℎ𝑑

Theoretical Power input to the pump = Theoretical work done by the pump per second

= 𝜌𝐴𝐿𝑁

60𝑔 ℎ𝑠 + ℎ𝑑

However, due to the leakage and frictional losses, actual power input will be more

than the theoretical power.

Let η = Efficiency of the pump.

Then actual power input to the pump ds hhgALN

60

1

Problem-1: A single-acting reciprocating pump discharge 0.018 m3 /s of water

per second when running at 60 rpm. Stroke length is 500 mm and the diameter

of the piston is 220 mm. If the total lift is 15 m, determine:

a) Theoretical discharge of the pump

b) Slip and percentage slip of the pump

c) Co-efficient of discharge

d) Theoretical Power required for running the pump

Solution:

Example Problems

(a) 𝑄𝑡ℎ = 𝐴 × 𝐿 ×𝑁

60=

𝜋

4 𝐷2 𝐿𝑁

60

𝑄𝑡ℎ =𝜋

4× 0.222 ×

0.5 × 60

60

𝑸𝒕𝒉 = 𝟎. 𝟎𝟏𝟗 𝒎𝟑/𝒔

(b) Slip = Qth – Qa = 0.019 – 0.018

= 0.001 m3 /s

Percentage slip = (Qth - Qa)/ Qth

= (0.019 – 0.018)/0.019

= 0.0526 or 5.26%

(c) Cd = Qa / Qth = 0.018/0.019

= 0.947

(d) Theoretical Power Input

= ρ Qth g Hst

(Neglecting Losses)

= 1000 x 0.019 x 9.81x 15

= 2796 w or 2.796 kW

Problem-2: A three-throw reciprocating pump is delivering 0.1 m3 /s of

water against a head of 100 m. Diameter and stroke length of the

cylinder are 250 mm and 500 mm respectively. Friction losses amount to

1 m in the suction pipe and 16 m in the delivery pipe. If the velocity of

water in the delivery pipe is 1.4 m/s, pump efficiency 90% and slip 2%,

determine the pump speed and the input power supplied.

Solution:

Slip, s = (Qth - Qa)/ Qth

0.02 = 1 – Qa / Qth

Qa / 0.98 = Qth = 3/60xπ/4 D2xLxN

0.1/ 0.98 = 3/60xπ/4 (0.25)2x0.5xN

N = 83.15 rpm

Total head generated

H = Hst + hfs + hfd + Vd2/(2g)

H = 100+1+16+ (1.4)2/(2x9.81)

H = 117.1 m

60

3ALNQth

Example Problems

Power required = 1/ ηh ( ρ Qth g H)

= 1/0.9 (1000 x 0.1/0.98 x 9.81 x 117.1)

= 130.21 x 103 W

= 130.21 KW

Comparison of Centrifugal and Reciprocating Pumps

Installation of a Reciprocating Pump

𝑻𝒐𝒕𝒂𝒍 𝑺𝒕𝒂𝒕𝒊𝒄 𝑯𝒆𝒂𝒅, 𝑯𝑻

𝑯𝒅

𝑯𝒔 𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒍𝒔

𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒍𝒅

𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒅𝒊𝒂, 𝒅𝒔

𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒅𝒊𝒂, 𝒅𝒅

𝑺𝒕𝒂𝒕𝒊𝒄 𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝑯𝒆𝒂𝒅, 𝑯𝒔

𝑺𝒕𝒂𝒕𝒊𝒄 𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝑯𝒆𝒂𝒅, 𝑯𝒅

𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝑯𝒆𝒂𝒅, 𝒉𝒎𝒔

𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝑯𝒆𝒂𝒅, 𝒉𝒎𝒅

𝑻𝒐𝒕𝒂𝒍 𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑯𝒆𝒂𝒅, 𝑯𝒎

𝒓

𝑳𝒄

𝑺𝒕𝒓𝒐𝒌𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝑳

𝑷𝒊𝒔𝒕𝒐𝒏 𝑪𝒚𝒍𝒊𝒏𝒅𝒆𝒓 𝒅𝒊𝒂. , 𝑫

𝑳 = 𝟐𝒓

𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝑽𝒔

𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝑽𝒅

𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒗𝒂𝒍𝒗𝒆

𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒗𝒂𝒍𝒗𝒆

Effect of Acceleration of Piston on Velocity and Pressure in

the Suction and Delivery pipes

Assuming simple harmonic motion of the piston (𝐿𝑐 >> 𝑟), its velocity may be

obtained in terms of the crank radius 𝑟 and crank angle 𝜃 and ultimately in terms of

the angular velocity 𝜔 and time 𝑡 as follows:

𝑝𝑖𝑠𝑡𝑜𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡, 𝑥 = 𝑟 − 𝑟𝑐𝑜𝑠𝜃 = 𝑟 − 𝑟𝑐𝑜𝑠(𝜔𝑡)

𝑝𝑖𝑠𝑡𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 =𝑑𝑥

𝑑𝑡= 𝜔𝑟𝑠𝑖𝑛 𝜔𝑡 = 𝜔𝑟𝑠𝑖𝑛𝜃

𝑝𝑖𝑠𝑡𝑜𝑛 𝑎𝑐𝑐𝑙𝑛, 𝑓 =𝑑𝑣

𝑑𝑡= 𝜔2𝑟𝑐𝑜𝑠𝜃

𝐿𝑒𝑡 𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑖𝑠𝑡𝑜𝑛, 𝑎 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒, 𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒

𝑎𝑉 = 𝐴𝑣 ⇒ 𝑉 =𝐴

𝑎 𝑣 ⇒ 𝑉 =

𝐴

𝑎 𝜔𝑟𝑠𝑖𝑛𝜃

𝐴𝑐𝑐𝑙𝑛 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 =𝑑𝑉

𝑑𝑡=

𝐴

𝑎𝜔2𝑟𝑐𝑜𝑠𝜃

𝜃 = 𝜔𝑡 =2𝜋𝑁

60𝑡

Effect of Acceleration of Piston on Velocity and

Pressure in the Suction and Delivery pipes

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 =𝛾𝑎𝑙

𝑔

𝐿𝑒𝑡, 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑐𝑐𝑙𝑛 = 𝑝𝑎

𝐹𝑟𝑜𝑚 𝑁𝑒𝑤𝑡𝑜𝑛′𝑠 2𝑛𝑑 𝑙𝑎𝑤 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑝𝑎 × 𝑎 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑙𝑛

⇒ 𝑝𝑎 × 𝑎 =𝛾𝑎𝑙

𝑔×

𝐴

𝑎× 𝜔2𝑟𝑐𝑜𝑠𝜃

𝑃𝑟𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑐𝑐𝑙𝑛, 𝐻𝑎 =𝑃𝑎

𝛾=

𝑙

𝑔

𝐴

𝑎 𝜔2𝑟𝑐𝑜𝑠𝜃

𝑆𝑜 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒, 𝜃 = 0° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 1, 𝑖. 𝑒. , 𝐻𝑎 =𝑙

𝑔

𝐴

𝑎 𝜔2𝑟

𝐴𝑡 𝑡ℎ𝑒 𝑚𝑖𝑑𝑑𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒, 𝜃 = 90° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 0, 𝑖. 𝑒. , 𝐻𝑎 = 0

𝐴𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒, 𝜃 = 180° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = −1, 𝑖. 𝑒. , 𝐻𝑎 = −𝑙

𝑔

𝐴

𝑎 𝜔2𝑟

Note: If the assumption of SHM of the piston is not valid due to shorter length of the connecting rod,

then acceleration head in the pipes is given by: 𝐻𝑎 =𝑙

𝑔

𝐴

𝑎 𝜔2𝑟𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃 +

cos 2𝜃

(𝐿𝑐/𝑟)

Effect of Friction in Suction and Delivery pipes

This indicates that the friction head is parabolic.

At the middle of the stroke, 𝜃 = 90° 𝑎𝑛𝑑 𝑠𝑖𝑛𝜃 = 1, ℎ𝑓𝑚 = 𝑓𝑙

𝑑

1

2𝑔

𝐴

𝑎 𝜔𝑟

2.

At the end of the stroke, 𝜃 = 180° 𝑎𝑛𝑑 𝑠𝑖𝑛𝜃 = 0, ℎ𝑓 = 0.

So, average friction loss in the pipe, ℎ𝑓𝑎𝑣=

2

3 ℎ𝑓𝑚

=2

3𝑓

𝑙

𝑑

1

2𝑔

𝐴

𝑎 𝜔𝑟

2

Indicator Diagram of a Reciprocating Pump

(Theoretical)

Indicator Diagram of a Reciprocating Pump (Theoretical)

An Indicator Diagram of a reciprocating pump is a plot showing the variation of

pressure in the cylinder with the displacement of the piston at various stages of

the piston strokes.

The area of the indicator diagram represents the work done by the pump per unit

weight of liquid in one complete revolution of the crank.

Total work done in one complete revolution of the crank per unit weight of liquid

(or Total Head) = 𝐻𝑆 + 𝐻𝑑

The unit weight of liquid pumped in one complete revolution of the crank is

𝜸𝑨𝑳.

Total work done in one complete revolution of the crank= 𝛾𝐴𝐿(𝐻𝑠 + 𝐻𝑑)

Theoretical work done per second (Hydraulic Power) = 𝛾𝐴𝐿𝑁

60 (𝐻𝑠 + 𝐻𝑑)

Indicator Diagram of a Reciprocating Pump

(with Acceleration Head)

Indicator Diagram of a Reciprocating Pump

(Actual)

Indicator Diagram of a Reciprocating Pump (Actual)

Total work done in one complete revolution of the crank per unit weight of liquid

= 𝐻𝑆 +2

3𝐻𝑓𝑠𝑚 + 𝐻𝑑 +

2

3 𝐻𝑓𝑑𝑚

Total work done in one complete revolution of the crank

= 𝛾𝐴𝐿(𝐻𝑆 +2

3𝐻𝑓𝑠𝑚 + 𝐻𝑑 +

2

3 𝐻𝑓𝑑𝑚)

Theoretical work done per second (Hydraulic Power)-single acting

= 𝛾𝐴𝐿𝑁

60(𝐻𝑆+

2

3𝐻𝑓𝑠𝑚 + 𝐻𝑑 +

2

3 𝐻𝑓𝑑𝑚)

=𝛾𝐴𝐿𝑁

60𝐻𝑆 +

2

3𝑓

𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟

2+ 𝐻𝑑 +

2

3𝑓

𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑 𝜔𝑟

2

For double acting pump,

Theoretical work done per second (Hydraulic Power)

=2𝛾𝐴𝐿𝑁

60𝐻𝑆 +

2

3𝑓

𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟

2

+ 𝐻𝑑 +2

3𝑓

𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑 𝜔𝑟

2

Indicator Diagram of a Reciprocating Pump

(Actual-including velocity head)

Separation in a Reciprocating Pump

If the absolute pressure inside the cylinder (i.e., absolute pressure on the

piston or plunger) is less than or equal to vapor pressure of the liquid, then

separation (cavitation) will occur.

i. e. , 𝐭𝐨 𝐚𝐯𝐨𝐢𝐝 𝐬𝐞𝐩𝐚𝐫𝐚𝐭𝐢𝐨𝐧, 𝑯𝒄𝒚𝒍 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆 ≥ 𝑯𝒗𝒂𝒑𝒐𝒓

There are two situations when separation can happen. One is at the beginning

of the suction stroke and another is at the end of the delivery stroke.

The absolute pressure at the cylinder or on the piston at any instant during

the suction stroke,

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠

= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠𝜔2𝑟𝑐𝑜𝑠𝜃 − 𝑓

𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠𝜔𝑟𝑠𝑖𝑛𝜃

2

It can be proved that the 𝐻𝑐𝑦𝑙 is minimum when 𝜃 = 0, i.e., at the beginning of the

suction stroke. So cavitation can occur at the beginning of the suction stroke.

Taking water vapor pressure at standard atmospheric condition as 2.5 m of water

absolute, it can be observed that to avoid separation (cavitation) during suction,

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 ≥ 2.5, 𝑖. 𝑒., 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝑙𝑠

𝑔

𝐴

𝑎𝑠𝜔2𝑟 ≥ 2.5 𝑎𝑏𝑠

This equation shows that as 𝜔 increases the absolute cylinder pressure decreases,

so, 𝜔 cannot be increased more than a maximum value. Thus, equating the above

equation and substituting 𝜔 =2𝜋𝑁

60 the maximum speed of the pump can be

determined.

𝐻𝑎𝑡𝑚 −𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2

𝑟 = 2.5 (𝑎𝑏𝑠), OR, − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2

𝑟 = −7.8 (𝑔𝑎𝑢𝑔𝑒)

Maximum Possible Speed of the Pump

The absolute pressure at the cylinder or on the piston at any instant

during the delivery stroke,

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑

= 𝐻𝑎𝑡𝑚 + 𝐻𝑑 +𝑙𝑑

𝑔

𝐴

𝑎𝑑𝜔2𝑟𝑐𝑜𝑠𝜃 + 𝑓

𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑𝜔𝑟𝑠𝑖𝑛𝜃

2

It can be proved that the 𝐻𝑐𝑦𝑙 is minimum when 𝜃 = 180, i.e., at the end of the

delivery stroke.

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 −𝑙𝑑

𝑔

𝐴

𝑎𝑑𝜔2𝑟

Note:

o The friction in pipes does not affect the minimum absolute pressure.

o If velocity head in the pipes is appreciable, it can be included as follows:

Minimum Absolute Pressure in the delivery stroke

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠 −Vs

2

2g

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑 +𝑉𝑑

2

2𝑔

An air vessel is a closed chamber having an opening at its base. The top half

contains compressed air and lower half contains liquid being pumped. Air and

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

pressure between two fluids.

Usually one air vessel is connected to suction side and one to the delivery pipe.

The vessels are connected as close to the pump as possible.

Effect of Air Vessels

Air vessel acts like a flywheel.

The compressed air at the top

contracts or expands to

absorb most of the pressure

fluctuations. The friction head

loss becomes rectangular

instead of parabolic. There is

no acceleration head in the

pipes beyond the air vessels.

The acceleration head

remains confined into a

shorter length between the

pump and the air vessels i.e.,

between 𝑙𝑠2 and 𝑙𝑑2.

𝑙𝑑2

𝑙𝑑1

𝑙𝑠2

𝑙𝑠1

Effect of Air Vessels:

The air vessels smooth out the flow in the suction and delivery pipes and the

flow is continuous beyond the air vessels. Fluctuations remain confined between

the air vessels and the pump.

By fitting air vessel as close to the pump as possible, the length of the pipe in

which acceleration head occurs is reduced. This reduces acceleration head and

the pump can be run at a much higher speed without any danger of separation.

As the acceleration head and frictional head are considerably reduced, the work

done is also reduced, hence, the power input is also reduced.

Effect of Air Vessels

Friction head

without air vessel Friction head

with air vessel

Delivery Pipe:

Let 𝑙𝑑1 be the length of the delivery pipe beyond the air vessel and 𝑙𝑑2

be the length between the cylinder and the air vessel.

𝐴𝑐𝑐𝑙𝑛𝐻𝑒𝑎𝑑, 𝐻𝑎𝑑2 =𝑙𝑑2

𝑔

𝐴

𝑎𝑑(𝜔2𝑟𝑐𝑜𝑠𝜃)

𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑, 𝐻𝑓𝑑2= 𝑓

𝑙𝑑2

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑 𝜔𝑟𝑠𝑖𝑛𝜃

2

𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑡 𝑏𝑒𝑦𝑜𝑛𝑑 𝑎𝑖𝑟 𝑣𝑒𝑠𝑠𝑒𝑙, 𝐻𝑓𝑑1= 𝑓

𝑙𝑑1

𝑑𝑑

1

2𝑔 𝑉𝑑

2

Average velocity 𝑉𝑑 in the delivery pipe: 𝑉𝑑 =𝑄

𝑎𝑑=

𝐴𝐿𝑁

60×

1

𝑎𝑑

(single acting pump)

Total pressure head (abs):

𝐻│𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +𝑉𝑑

2

2𝑔

= 𝐻𝑎𝑡𝑚 + 𝐻𝑑 +𝑙𝑑2

𝑔

𝐴

𝑎𝑑𝜔2𝑟𝑐𝑜𝑠𝜃 + 𝑓

𝑙𝑑2

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑 𝜔𝑟𝑠𝑖𝑛𝜃

2

+ 𝑓𝑙𝑑1

𝑑𝑑

1

2𝑔 𝑉𝑑

2 +𝑉𝑑

2

2𝑔

Effect of Air Vessels

Suction Pipe:

Let 𝑙𝑠1 be the length of the suction pipe beyond the air vessel and 𝑙𝑠2

be the length between the cylinder and the air vessel.

𝐴𝑐𝑐𝑙𝑛𝐻𝑒𝑎𝑑, 𝐻𝑎𝑠2 =𝑙𝑠2

𝑔

𝐴

𝑎𝑠(𝜔2𝑟𝑐𝑜𝑠𝜃)

𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑, 𝐻𝑓𝑠2= 𝑓

𝑙𝑠2

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟𝑠𝑖𝑛𝜃

2

𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑡 𝑏𝑒𝑦𝑜𝑛𝑑 𝑎𝑖𝑟 𝑣𝑒𝑠𝑠𝑒𝑙, 𝐻𝑓𝑠1= 𝑓

𝑙𝑠1

𝑑𝑠

1

2𝑔 𝑉𝑠

2

Average velocity 𝑉𝑠 in the suction pipe: 𝑉𝑠 =𝑄

𝑎𝑠=

𝐴𝐿𝑁

60×

1

𝑎𝑠

(single acting pump)

Total pressure head (abs):

H│𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠2 − 𝐻𝑓𝑠2 − 𝐻𝑓𝑠1 −𝑉𝑠

2

2𝑔

= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠2

𝑔

𝐴

𝑎𝑠𝜔2𝑟𝑐𝑜𝑠𝜃 − 𝑓

𝑙𝑠2

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟𝑠𝑖𝑛𝜃

2

− 𝑓𝑙𝑠1

𝑑𝑠

1

2𝑔 𝑉𝑠

2 −𝑉𝑠

2

2𝑔

Effect of Air Vessels

Work done and Power required for Pumps fitted with Air Vessels:

Work done in one revolution of the crank (single acting),

= 𝛾𝑄 𝐻𝑠 + 𝐻𝑎𝑠2 +2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

= 𝛾𝐴𝐿 𝐻𝑠 + 𝐻𝑎𝑠2 +2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

= 𝛾𝐴(2𝑟) 𝐻𝑠 + 𝐻𝑎𝑠2 +2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

= 2𝛾𝐴𝑟 𝐻𝑠 + 𝐻𝑎𝑠2 +2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

Work done per second (single acting), i.e., Power (hydraulic) developed 𝛾𝐴𝐿𝑁

60𝐻𝑠 + 𝐻𝑎𝑠2 +

2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

=2𝛾𝐴𝑟𝑁

60𝐻𝑠 + 𝐻𝑎𝑠2 +

2

3𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

Effect of Air Vessels

Work done and Power required for Pumps fitted with Air Vessels:

Neglecting the velocity head and the friction head in the smaller pipe

sections (𝑙𝑠2 𝑎𝑛𝑑 𝑙𝑑2), we have,

Work done in one revolution of the crank (single acting),

= 𝛾𝑄 𝐻𝑠 + 𝐻𝑓𝑠1 + 𝐻𝑑 + 𝐻𝑓𝑑1

Work done per second (single acting), i.e., Power (hydraulic) developed

=2𝛾𝐴𝑟𝑁

60𝐻𝑠 + 𝐻𝑓𝑠1 + 𝐻𝑑 + 𝐻𝑓𝑑1

Effect of Air Vessels here

𝐻𝑓𝑠1 = 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠

𝜔𝑟

𝜋

2

𝐻𝑓𝑑1 = 𝑓𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑

𝜔𝑟

𝜋

2

and

𝐻𝑓𝑠2 = 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠𝜔𝑟

2

𝐻𝑓𝑑2 = 𝑓

𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑𝜔𝑟

2

and

Work done when air vessel is NOT fitted:

Work done against friction during the suction stroke,

𝑃1 = 2𝛾𝐴𝑟2

3 𝐻𝑓𝑠𝑚

⇒ 𝑃1 = 2𝛾𝐴𝑟2

3 𝑓

𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟

2

⇒ 𝑃1 =4𝛾𝐴𝑟

3 𝑓

𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠 𝜔𝑟

2

Work done when air vessel is fitted:

𝑃2 = 2𝛾𝐴𝑟 𝐻𝑓𝑠1

= 2𝛾𝐴𝑟 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔 𝑉𝑠

2

𝑉𝑠 =𝑄

𝑎𝑠=

𝐴𝐿𝑁

60×

1

𝑎𝑠=

𝐴

𝑎𝑠

𝜔

2𝜋(2𝑟) =

𝐴

𝑎𝑠

𝜔𝑟

𝜋

𝑃2 = 2𝛾𝐴𝑟 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠

𝜔𝑟

𝜋

2

Effect of Air Vessels

Work saved in percentage for single acting pump:

𝑃1 − 𝑃2

𝑃1× 100%

=

2𝛾𝐴𝑟 𝑓𝑙𝑠𝑑𝑠

12𝑔

𝐴𝑎𝑠

𝜔𝑟2 2

3 −

1𝜋2

4𝛾𝐴𝑟3

𝑓𝑙𝑠𝑑𝑠

12𝑔

𝐴𝑎𝑠

𝜔𝑟2

× 100%

=3

2×

2

3−

1

𝜋2× 100% = 𝟖𝟒. 𝟖%

Work saved in percentage For double acting pump:

𝑄 =2𝐴𝐿𝑁

60= 2

𝐴

𝑎𝑠

𝜔𝑟

𝜋

𝑃1 − 𝑃2

𝑃1× 100%

=3

2×

2

3−

4

𝜋2× 100% = 𝟑𝟗. 𝟐%

Effect of Air Vessels

Problem–3: A single acting reciprocating pump has a plunger diameter of 125 mm

and stroke length of 300 mm. The length of the suction pipe is 10 m and diameter

75 mm. (i) Find acceleration head at the beginning, middle and end of suction

stroke. (ii) If the suction head is 3 m, determine the pressure head in the cylinder

at the beginning of stroke when the pump runs at 30 rpm. (iii) Under the

circumstance, what can be the maximum running speed of the pump without

separation (cavitation). Take atmospheric pressure as 10.3 m of water, and the

water vapor pressure as 2.6 m of water abs.

Solution: (i) 𝐻𝑎𝑠 =𝑙𝑠

𝑔

𝐴

𝑎𝑠𝜔2𝑟𝑐𝑜𝑠𝜃 =

10

9.81×

0.1252

0.0752 ×2𝜋×30

60

2× 0.15 × 𝑐𝑜𝑠𝜃 = 4.192𝑐𝑜𝑠𝜃

(ii) 𝐻𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑎𝑏𝑠 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠 = 10.3 − 3 − 4.192 − 0 = 3.108 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 (𝑎𝑏𝑠)

(iii)𝐻𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑏𝑠 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠 = 2.6 (𝑎𝑏𝑠)

𝐻𝑎𝑡𝑚−𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2𝑟 = 2.6 (𝑎𝑏𝑠)

⇒ 10.3 − 3 −10

9.81×

0.1252

0.0752 ×2𝜋𝑁𝑚𝑎𝑥

60

2× 0.15 = 2.6 ∴ 𝑁𝑚𝑎𝑥 = 31 𝑟𝑝𝑚

Example Problems

Condition 𝜽 𝑯𝒂𝒔

Beginning of stroke 0° 4.192 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

Middle of stroke 90° 0

End of stroke 180° −4.192 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

Problem–4: A single acting reciprocating pump has a piston of diameter 75 mm

and a stroke of 150 mm. It draws water from a sump 3.5 m below the pump

through a pipe of 5 m long. If separation occurs at 78.46 kPa below atmospheric

pressure when the pump runs at 45 rpm, find the diameter of suction pipe for no

separation. Assume simple harmonic motion of the piston.

Solution:

−𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2𝑟 ≥

−78.46×103

9810

−3.5 −5

9.81×

0.0752

𝑑𝑠2 ×

2𝜋 × 45

60

2

× 0.075 ≥ −8

𝑑𝑠 ≥ 32.6 𝑚𝑚

Example Problems

Problem–5: A single acting reciprocating pump has the following characteristics:

Cylinder diameter = 225 mm; Stroke length = 450 mm.

Suction head = 4.5 m; Diameter of suction pipe = 225 mm;

Suction pipe length = 20 m;

Atmospheric pressure = 10 m water (abs)

Cavitation pressure = 2 m water (abs)

Determine the maximum speed at which the pump can be run without cavitation.

Solution:

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2

𝑟 = 2 (abs)

10 − 4.5 −20

9.81×

0.2252

0.2252×

2𝜋 × 𝑁𝑚𝑎𝑥

60

2

× 0.225 = 2

𝑁𝑚𝑎𝑥 = 26 𝑟𝑝𝑚

Example Problems

Problem–6: A single acting reciprocating pump has the following characteristics:

Piston diameter = 100 mm; Stroke length = 300 mm.

Suction head = 4 m; Diameter of suction pipe = 75 mm;

Suction pipe length = 4 m;

Atmospheric pressure = 10 m water (abs)

Cavitation pressure = 2.5 m water (abs)

Determine the maximum speed at which the pump can be run without cavitation.

Assume Frictional losses = 1 m

Solution:

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2

𝑟 − 𝐻𝑓𝑠 = 2.5 (abs)

10 − 4 −4

9.81×

0.12

0.0752×

2𝜋 × 𝑁𝑚𝑎𝑥

60

2

× 0.15 − 1 = 2.5

𝑁𝑚𝑎𝑥 = 45 𝑟𝑝𝑚

Example Problems

Problem–7: A double acting reciprocating pump has the following characteristics:

Cylinder diameter = 200 mm; Stroke length = 200 mm.

Diameter of suction pipe = 200 mm;

Suction pipe length = 15 m;

Atmospheric pressure = 10 m water (abs)

Cavitation pressure = 2.5 m water (abs)

Speed of the pump = 60 rpm

Determine the maximum suction lift without separation.

Neglect frictional losses.

Solution:

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁

60

2

𝑟 − 𝐻𝑓𝑠 = 2.5 (abs)

10 − 𝐻𝑠 −15

9.81×

0.22

0.22×

2𝜋 × 60

60

2

× 0.1 − 0 = 2.5

𝐻𝑠 = 1.46 𝑚

Example Problems

Example Problems

𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑 ≥ 2.75 (𝑎𝑏𝑠)

𝐻𝑎𝑡𝑚 + 𝐻𝑑 −𝑙𝑑

𝑔

𝐴

𝑎𝑑𝜔2𝑟 + 0 ≥ 2.75

9.75 + 40 −45

9.81×

0.22

0.12×

2𝜋𝑁𝑚𝑎𝑥

60

2

× 0.4 = 2.75

𝑁𝑚𝑎𝑥 = 24 𝑟𝑝𝑚

Problem–8: A single acting reciprocating pump has a piston of diameter 200

mm with a crank of radius 400 mm. It delivers water to a tank with 100 mm

diameter and 45 m long delivery pipe. Water is lifted to a height of 40 m above

the axis of the cylinder. Find the maximum speed at which the pump can be run

without cavitation. Assume atmospheric pressure 9.75 m water abs, and

cavitation occurs at 2.75 m water abs.

Solution:

Example Problems

Solution:

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁𝑚𝑎𝑥

60

2𝑟 = 2.5

10 − 3.5 −7.5

9.81×

0.32

𝑑𝑠2 ×

2𝜋 × 40

60

2

× 0.1875 = 2.5

𝑑𝑠 = 237.9 𝑚𝑚

Problem–9: A double acting single cylinder reciprocating pump has a piston of

diameter 300 mm and a stroke of 375 mm. The pump axis is 3.5 m above the

water level in the sump. The suction pipe is 7.5 m long. Find the diameter of the

suction pipe so that minimum stipulated pressure head from cavitation

considerations is not violated. Assume atmospheric pressure 10 m water abs,

and cavitation occurs at 2.5 m water abs.

Problem–10: A single acting reciprocating pump has a piston of diameter 200

mm and a stroke of 300 mm. It draws water from a sump 3.5 m below the pump

through a pipe of 5.5 m long. The separation pressure head is 2.5 m water abs,

and atmospheric pressure is 10.3 m abs. When the pump runs at 60 rpm, find

the minimum diameter of suction pipe for no separation. Assume simple

harmonic motion of the piston.

Solution:

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁

60

2𝑟 ≥ 2.5

10.3 − 3.5 −5.5

9.81×

0.22

𝑑𝑠│𝑚𝑖𝑛2 ×

2𝜋 × 60

60

2

× 0.15 = 2.5

𝑑𝑠│𝑚𝑖𝑛 = 0.1757 𝑚 = 176 𝑚𝑚

Example Problems

Problem–11: A double acting reciprocating pump has a 250 mm cylinder with a

stroke of 400 mm. The suction pipe is 5 m long and the suction lift is 3 m. If the

speed of the crank is 25 rpm, determine the minimum diameter of the suction

pipe to prevent occurrence of cavitation. The minimum pressure from cavitation

consideration is limited to 2.5 m of water abs. Assume atmospheric pressure as

10 m of water abs.

Solution:

𝐻𝑎𝑡𝑚 − 𝐻𝑠 −𝑙𝑠

𝑔

𝐴

𝑎𝑠

2𝜋𝑁

60

2𝑟 ≥ 2.5

10 − 3 −5

9.81×

0.252

𝑑𝑠│𝑚𝑖𝑛2 ×

2𝜋 × 25

60

2

× 0.2 = 2.5

𝑑𝑠│𝑚𝑖𝑛 = 0.0985 𝑚 = 98.5 𝑚𝑚

Example Problems

Problem–12: A single acting reciprocating pump has 125 mm diameter cylinder

with a stroke of 500 mm. The length and diameter of the suction pipe are 5.2 m

and 100 mm respectively. The suction lift is 3.25 m and the delivery lift is 12 m.

The pump speed is 45 rpm. If an air vessel is fitted very close to the cylinder in

the delivery side, calculate power required to pump water. Assume the frictional

head in the delivery pipe to be 0.15 m and the velocity heads in the pipes can

be neglected. Take, 𝜂𝑝𝑢𝑚𝑝 = 0.9 𝑎𝑛𝑑 𝑓 = 0.02.

Example Problems

𝑃 = 50.16 × [(3.25 + 4.6 +2

3× 0.18 + 0) + (12 + 0.15 + 0)]

= 1009.22 𝑊𝑎𝑡𝑡 = 𝟏. 𝟎𝟏 𝒌𝑾 𝑨𝒏𝒔.

Solution: Power required,

2𝛾𝐴𝑟𝑁

60𝜂𝑝𝑢𝑚𝑝=

2 × 9810 ×𝜋4

× 0.1252 × 0.25 × 45

60 × 0.9= 50.16 𝑁/𝑠

Here,

𝑃 = 2𝛾𝐴𝑟𝑁

60𝜂𝑝𝑢𝑚𝑝𝐻𝑠 + 𝐻𝑎𝑠 +

2

3𝐻𝑓𝑠 +

𝑉𝑠2

2𝑔+ 𝐻𝑑 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +

𝑉𝑑2

2𝑔

𝐻𝑓𝑠 = 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠𝜔𝑟

2

= 0.02 ×5.2

0.1×

1

2 × 9.81×

0.1252

0.12×

2𝜋 × 45

60× 0.25

2

= 0.18 𝑚

𝐻𝑎𝑠 =𝑙𝑠

𝑔

𝐴

𝑎𝑠𝜔2 𝑟 =

5.2

9.81×

0.1252

0.12×

2𝜋 × 45

60

2

× 0.25 = 4.6 𝑚

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 = 0.15 𝑚

Therefore,

Problem–13: A single acting reciprocating pump has 100 mm diameter piston with

a stroke of 200 mm. The static suction head is 4 m, the diameter of the suction pipe

is 75 mm and the length of the suction pipe is 8 m. The diameter and length of the

delivery pipe are 100 mm and 30 m respectively. The static delivery head is 25 m.

An air vessel is fitted very near to the cylinder on the suction side and another at a

distance of 3 m from the cylinder on the delivery side. If the pump speed is 60 rpm,

estimate the power required to drive the pump. Take, 𝜂𝑝𝑢𝑚𝑝 = 0.9 𝑎𝑛𝑑 𝑓 = 0.025.

Example Problems

𝑃 = 17.12 × 4 + 0.017 + 25 + 1.207 +2

3× 0.015 + 0.014 = 517.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟏𝟖 𝒌𝑾 𝑨𝒏𝒔

Solution: Power required,

2𝛾𝐴𝑟𝑁

60𝜂𝑝𝑢𝑚𝑝=

2 × 9810 ×𝜋4

× 0.12 × 0.1 × 60

60 × 0.9= 17.12 𝑁/𝑠

Here

𝑃 = 2𝛾𝐴𝑟𝑁

60𝜂𝑝𝑢𝑚𝑝𝐻𝑠 + 𝐻𝑓𝑠1 + 𝐻𝑑 + 𝐻𝑎𝑑2 +

2

3𝐻𝑓𝑑2 + 𝐻𝑓𝑑1

𝐻𝑓𝑠1 = 𝑓𝑙𝑠

𝑑𝑠

1

2𝑔

𝐴

𝑎𝑠

𝜔𝑟

𝜋

2

= 0.025 ×8

0.075×

1

2 × 9.81×

0.12

0.0752×

2𝜋 × 60

60 × 𝜋× 0.1

2

= 0.017 𝑚

𝐻𝑎𝑑2 =𝑙𝑑2

𝑔

𝐴

𝑎𝑑𝜔2 𝑟 =

3

9.81×

0.12

0.12×

2𝜋 × 60

60

2

× 0.1 = 1.207 𝑚

𝐻𝑓𝑑2 = 𝑓𝑙𝑑2

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑𝜔𝑟

2

= 0.025 ×3

0.1×

1

2 × 9.81×

0.12

0.12×

2𝜋 × 60

60× 0.1

2

= 0.015 𝑚

𝐻𝑓𝑑1 = 𝑓𝑙𝑑1

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑

𝜔𝑟

𝜋

2

= 0.025 ×27

0.1×

1

2 × 9.81×

0.12

0.12×

2𝜋 × 60

60 × 𝜋× 0.1

2

= 0.014 𝑚

(Neglecting velocity heads in the pipes)

Example Problems

= 2 × 9810 ×𝜋

4× 0.32 × 0.225 0.03 ×

40

0.2×

1

2 × 9.81×

0.32

0.22 ×

2𝜋 × 60

60× 0.225

2

×2

3−

1

𝜋2

= 545.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟒𝟓 𝒌𝑾

𝑃1 − 𝑃2 = 2𝛾𝐴𝑟 𝑓𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑𝜔𝑟

2

2

3 −

1

𝜋2

𝑃2 = 2𝛾𝐴𝑟 𝐻𝑓𝑑 = 2𝛾𝐴𝑟 𝑓𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑

𝜔𝑟

𝜋

2

Work (Power) saved in overcoming friction,

𝑃1= 2𝛾𝐴𝑟2

3𝐻𝑓𝑑𝑚 =

2𝛾𝐴𝑟

3

2

3𝑓

𝑙𝑑

𝑑𝑑

1

2𝑔

𝐴

𝑎𝑑 𝜔𝑟

2

Problem–14: A single acting reciprocating pump has an air vessel in the delivery

side fitted very close to the cylinder. The cylinder has a diameter of 300 mm and a

stroke length of 450 mm. The delivery pipe is 40 m long and has a diameter of 200

mm. The speed of the pump is 60 rpm. Determine the power saved by the air

vessel in overcoming friction in the delivery pipe. Take friction factor, f = 0.03.

Solution: Friction Work without air vessel,

Friction Work with air vessel,

Exercises Problem-1: A single acting reciprocating pump has the piston diameter of 300 mm, stroke length of 500

mm, rotational speed of 40 rpm and the total lift (of water) of 25 m. If the actual discharge delivered at the

pump outlet is 1380 liters/minute, calculate the slip, coefficient of discharge and theoretical power in kW

required to drive the pump. [Ans. Slip = 2.34%, Coefficient of discharge = 0.9766, Power = 5.78 kW]

Problem-2: A double acting reciprocating pump running at 40 rpm is discharging 1 m3/min. The pump

has a stroke of 400 mm and the diameter of the piston is 200 mm. The delivery and suction heads are 20

m and 5 m respectively. Find the slip of the pump and the power required to drive the pump.

[Ans. Slip = 0.53%, Power = 4.08 kW]

Problem-3: For a single acting reciprocating pump, piston diameter is 150 mm, stroke length is 300 mm,

rotational speed is 50 rpm and the water is to be raised through 18 m. Determine (i) theoretical

discharge, (ii) if the actual discharge is 4 liters/s, determine volumetric efficiency, slip and actual power

required. Take the mechanical efficiency as 80%. [Ans. 4.4185 l/s, 90.53%, 94.72%, 882.9 watt]

Problem-4: A double acting reciprocating pump has a cylinder of diameter 200 mm and stroke of 300 mm.

The piston makes 30 strokes/minute. Estimate the maximum velocity and acceleration in the suction pipe

of diameter 200 mm and delivery pipe of diameter 250 mm.

[Ans. Suction = 0.471 m/s and 1.480 m/s2 ; Delivery = 0.301 m/s and 0.947 m/s2]

Problem-5: A Plunger is fitted to a vertical pipe filled with water. The lower end of the pipe is submerged in

a sump. If the plunger is drawn up with an acceleration of 5 m/s2, find the maximum height above the sump

level at which the plunger will work without separation. Assume atmospheric pressure = 10 m water abs,

and separation occurs at 2 m water abs. Take acceleration due to gravity as 10 m/s2. [Ans. Hs = 5.33 m]

Problem-6: A single acting reciprocating pump has plunger diameter 200 mm and stroke 300 mm. The

suction pipe is 100 mm in diameter and 8 m long. The water surface from which the pump draws water is 5

m below the pump cylinder axis. If the pump is working at 30 rpm, find the pressure in the cylinder at the

beginning, middle and end of suction stroke. Take the friction factor, f = 0.04.

[Ans. 0.4708 (beginning), 4.7205 (middle) and 10.129 (end) m water abs]

Exercises Problem-7: A single acting reciprocating pump has a stroke length of 150 mm. The suction pipe is 7 m

long. The water level in the sump is 2.5 m below the cylinder. The diameters of suction pipe and the

plunger are 75 mm and 100 mm respectively. If the speed of the pump is 75 rpm, determine the pressure

head on the piston at the beginning, middle and end of the suction stroke. Take Darcy-Weisbach friction

factor, f = 0.02. [Ans. -8.369 m water (gauge), -2.604 m water (gauge) and +3.868 m water (gauge)]

Problem-8: A single acting reciprocating pump has cylinder diameter of 350 mm and a stroke length of

350 mm. The static suction head is 3 m. The diameter of suction pipe is 200 mm and length 6 m. The

diameter of delivery pipe is 200 mm and length 25 m. The static delivery head is 20 m. If the speed of the

pump is 20 rpm, estimate the power required to drive the pump. Take Darcy-Weisbach friction factor, f =

0.02 and pump efficiency = 0.9. [Ans. 1.7 kW]

Problem-9: A single acting reciprocating pump has a stroke length of 375 mm and a cylinder of

diameter 225 mm. The suction pipe is 12 m long and has a diameter of 150 mm. The water level in the

sump is 3 m below the level of the cylinder. If the speed of the pump is 20 rpm, determine the pressure

head on the piston at the beginning of the suction stroke (i) when no air vessel is fitted and (ii) when an

air vessel is fitted to the suction pipe at the level of the cylinder and at a distance of 1.5 m from the

cylinder. Take f = 0.02. [Ans. (i) 5.263 m water (vacuum), (ii) 3.289 m water (vacuum)]

Problem-10: A single acting reciprocating pump has a stroke length of 400 mm and a cylinder of

diameter 250 mm. The delivery pipe is 20 m long and has a diameter of 150 mm. A large diameter air

vessel is fitted to the delivery pipe. For a crank speed of 40 rpm, determine the quantity of water going in

or coming out of the air vessel when the crank angle is (i) 15o (ii) 90o and (iii) 120o. Also, determine the

crank angle at which there is no flow into or out of the air vessel.

[Ans. (i) 0.0025 m3/s into the air vessel (ii) 0.028 m3/s goes out of the air vessel (iii) 0.016 m3/s goes out

of the air vessel; θ = 18, 58° or 161.41°]