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 multicylinder 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 Cylinder Reciprocating Pump
Double Acting Reciprocating Pump
Triple 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
Qth  A  L 
N
60
This will be the discharge when the pump is single acting.
Discharge through a Reciprocating Pump
Discharge in case of double acting pump
Discharge/Second,
 ALN ( A  AP ) LN 
Qth  


60
60


Qth 
(2 A  AP ) LN
60
where, Ap = Area of cross-section of piston rod
However, if area of the piston rod is neglected
Discharge/Second =
•
•
2ALN
60
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.
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
% slip 
Qth  Qact
100
Qth
 Q
 1  act
Qth


 100  1  Cd 100

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
=𝜌
𝐴𝐿𝑁
60
𝑔 ℎ𝑠 + ℎ𝑑
Theoretical Power input to the pump = Theoretical work done by the pump per second
𝐴𝐿𝑁
=𝜌
𝑔 ℎ𝑠 + ℎ𝑑
60
However, due to the leakage and frictional losses, actual power input will be more
than the theoretical power.
Let η = Efficiency of the pump.
ALN 
Then actual power input to the pump  1  

 g hs  hd 

 60 
Example Problems
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:
(a)
(b)
𝑁
𝜋
𝐿𝑁
𝑄𝑡ℎ = 𝐴 × 𝐿 × 60 = 4 𝐷2 60
𝜋
0.5 × 60
2
𝑄𝑡ℎ =
× 0.22 ×
4
60
𝟑
𝑸𝒕𝒉 = 𝟎. 𝟎𝟏𝟗 𝒎 /𝒔
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
Example Problems
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
Qth 
3ALN
60
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
𝑺𝒕𝒂𝒕𝒊𝒄 𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝑯𝒆𝒂𝒅, 𝑯𝒔
𝑺𝒕𝒂𝒕𝒊𝒄 𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝑯𝒆𝒂𝒅, 𝑯𝒅
𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒍𝒅
𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝑯𝒆𝒂𝒅, 𝒉𝒎𝒔
𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝑽𝒅
𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝑯𝒆𝒂𝒅, 𝒉𝒎𝒅
𝑻𝒐𝒕𝒂𝒍 𝑴𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝑯𝒆𝒂𝒅, 𝑯𝒎
𝑯𝒅
𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚 𝒑𝒊𝒑𝒆 𝒅𝒊𝒂, 𝒅𝒅
𝑻𝒐𝒕𝒂𝒍 𝑺𝒕𝒂𝒕𝒊𝒄 𝑯𝒆𝒂𝒅, 𝑯𝑻
𝑫𝒆𝒍𝒊𝒗𝒆𝒓𝒚
𝒗𝒂𝒍𝒗𝒆
𝑳𝒄
𝑪𝒚𝒍𝒊𝒏𝒅𝒆𝒓
𝒅𝒊𝒂. , 𝑫
𝑺𝒖𝒄𝒕𝒊𝒐𝒏
𝒗𝒂𝒍𝒗𝒆
𝑯𝒔
𝒓
𝑷𝒊𝒔𝒕𝒐𝒏
𝑺𝒕𝒓𝒐𝒌𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝑳
𝑳 = 𝟐𝒓
𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒍𝒆𝒏𝒈𝒕𝒉, 𝒍𝒔
𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚, 𝑽𝒔
𝑺𝒖𝒄𝒕𝒊𝒐𝒏 𝒑𝒊𝒑𝒆 𝒅𝒊𝒂, 𝒅𝒔
Effect of Acceleration of Piston on Velocity and Pressure in
the Suction and Delivery pipes
2𝜋𝑁
𝜃 = 𝜔𝑡 =
𝑡
60
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)
𝑑𝑥
𝑝𝑖𝑠𝑡𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 =
= 𝜔𝑟𝑠𝑖𝑛 𝜔𝑡 = 𝜔𝑟𝑠𝑖𝑛𝜃
𝑑𝑡
𝑑𝑣
𝑛
𝑝𝑖𝑠𝑡𝑜𝑛 𝑎𝑐𝑐𝑙 , 𝑓 =
= 𝜔2 𝑟𝑐𝑜𝑠𝜃
𝑑𝑡
𝐿𝑒𝑡 𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑖𝑠𝑡𝑜𝑛, 𝑎 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒,
𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒
𝐴
𝐴
𝑎𝑉 = 𝐴𝑣 ⇒ 𝑉 = 𝑣 ⇒ 𝑉 = 𝜔𝑟𝑠𝑖𝑛𝜃
𝑎
𝑎
𝑑𝑉
𝐴 2
𝑛
𝐴𝑐𝑐𝑙 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 =
= 𝜔 𝑟𝑐𝑜𝑠𝜃
𝑑𝑡 𝑎
Effect of Acceleration of Piston on Velocity and
Pressure in the Suction and Delivery pipes
𝛾𝑎𝑙
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 =
𝑔
𝐿𝑒𝑡, 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑐𝑐𝑙 𝑛 = 𝑝𝑎
𝐹𝑟𝑜𝑚 𝑁𝑒𝑤𝑡𝑜𝑛′ 𝑠 2𝑛𝑑 𝑙𝑎𝑤 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛,
𝑝𝑎 × 𝑎 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑙 𝑛
𝛾𝑎𝑙 𝐴
⇒ 𝑝𝑎 × 𝑎 =
× × 𝜔2 𝑟𝑐𝑜𝑠𝜃
𝑔
𝑎
𝑃𝑎 𝑙 𝐴 2
𝑃𝑟𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑐𝑐𝑙 𝑛 , 𝐻𝑎 = =
𝜔 𝑟𝑐𝑜𝑠𝜃
𝛾
𝑔𝑎
𝑆𝑜 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒,
𝑙𝐴 2
𝜃 = 0° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 1, 𝑖. 𝑒. , 𝐻𝑎 =
𝜔 𝑟
𝑔𝑎
𝐴𝑡 𝑡ℎ𝑒 𝑚𝑖𝑑𝑑𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒,
𝐴𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑜𝑘𝑒,
𝜃 = 90° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = 0, 𝑖. 𝑒. , 𝐻𝑎 = 0
𝑙𝐴 2
𝜃 = 180° 𝑎𝑛𝑑 𝑐𝑜𝑠𝜃 = −1, 𝑖. 𝑒. , 𝐻𝑎 = −
𝜔 𝑟
𝑔𝑎
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 𝐴
𝑔𝑎
cos 2𝜃
𝑐 /𝑟)
𝜔2 𝑟𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃 + (𝐿
Effect of Friction in Suction and Delivery pipes
At the middle of the stroke, 𝜃 = 90° 𝑎𝑛𝑑 𝑠𝑖𝑛𝜃 = 1, ℎ𝑓𝑚 =
𝑙 1
𝑓
𝑑 2𝑔
At the end of the stroke, 𝜃 = 180° 𝑎𝑛𝑑 𝑠𝑖𝑛𝜃 = 0, ℎ𝑓 = 0.
This indicates that the friction head is parabolic.
So, average friction loss in the pipe, ℎ𝑓𝑎𝑣 =
2
3
ℎ𝑓𝑚 =
2 𝑙 1
𝑓
3 𝑑 2𝑔
𝐴
𝑎
𝜔𝑟
2
𝐴
𝑎
2
𝜔𝑟 .
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
2
2
= 𝐻𝑆 + 𝐻𝑓𝑠𝑚 + 𝐻𝑑 + 𝐻𝑓𝑑𝑚
3
3
Total work done in one complete revolution of the crank
2
2
= 𝛾𝐴𝐿(𝐻𝑆 + 𝐻𝑓𝑠𝑚 + 𝐻𝑑 + 𝐻𝑓𝑑𝑚 )
3
3
Theoretical work done per second (Hydraulic Power)-single acting
𝛾𝐴𝐿𝑁
2
2
=
(𝐻𝑆 + 𝐻𝑓𝑠𝑚 + 𝐻𝑑 + 𝐻𝑓𝑑𝑚 )
=
𝛾𝐴𝐿𝑁
60
𝐻𝑆 +
60
2 𝑙𝑠 1
𝑓
3 𝑑𝑠 2𝑔
3
2
𝐴
𝜔𝑟
𝑎𝑠
+
3
2 𝑙 1
𝐻𝑑 + 𝑓 𝑑
3 𝑑𝑑 2𝑔
𝐴
𝑎𝑑
𝜔𝑟
2
For double acting pump,
Theoretical work done per second (Hydraulic Power)
2
2𝛾𝐴𝐿𝑁
2 𝑙𝑠 1 𝐴
2 𝑙𝑑 1 𝐴
=
𝐻𝑆 + 𝑓
𝜔𝑟 + 𝐻𝑑 + 𝑓
𝜔𝑟
60
3 𝑑𝑠 2𝑔 𝑎𝑠
3 𝑑𝑑 2𝑔 𝑎𝑑
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.
Maximum Possible Speed of the Pump
The absolute pressure at the cylinder or on the piston at any instant during
the suction stroke,
𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠
𝑙𝑠 𝐴 2
𝑙𝑠 1 𝐴
= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝜔 𝑟𝑐𝑜𝑠𝜃 − 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
𝑔 𝑎𝑠
𝑑𝑠 2𝑔 𝑎𝑠
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 𝐴 2
𝜔 𝑟
𝑔 𝑎𝑠
≥ 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
2𝜋𝑁
equation and substituting 𝜔 =
the maximum speed of the pump can be
60
determined.
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥
𝐻𝑎𝑡𝑚 −𝐻𝑠 −
𝑔 𝑎𝑠
60
2
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥
𝑟 = 2.5 (𝑎𝑏𝑠), OR, − 𝐻𝑠 −
𝑔 𝑎𝑠
60
2
𝑟 = −7.8 (𝑔𝑎𝑢𝑔𝑒)
Minimum Absolute Pressure in the delivery stroke
The absolute pressure at the cylinder or on the piston at any instant
during the delivery stroke,
𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑
𝑙𝑑 𝐴 2
𝑙𝑑 1 𝐴
= 𝐻𝑎𝑡𝑚 + 𝐻𝑑 +
𝜔 𝑟𝑐𝑜𝑠𝜃 + 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
𝑔 𝑎𝑑
𝑑𝑑 2𝑔 𝑎𝑑
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 𝐴 2
𝜔 𝑟
𝑔 𝑎𝑑
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:
𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑠𝑢𝑐𝑡𝑖𝑜𝑛
Vs2
= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠 − 𝐻𝑓𝑠 −
2g
𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑉𝑑2
= 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑 +
2𝑔
Effect of Air Vessels
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.
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 .
𝑙𝑑1
𝑙𝑑2
𝑙𝑠2
𝑙𝑠1
Effect of Air Vessels
Friction head
without air vessel
Friction head
with air vessel
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
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.
𝑙𝑑2 𝐴
𝑛
𝐴𝑐𝑐𝑙 𝐻𝑒𝑎𝑑, 𝐻𝑎𝑑 2 =
(𝜔2 𝑟𝑐𝑜𝑠𝜃)
𝑔 𝑎𝑑
2
𝑙𝑑2 1 𝐴
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑, 𝐻𝑓𝑑 = 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
2
𝑑𝑑 2𝑔 𝑎𝑑
𝑙𝑑1 1 2
𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑡 𝑏𝑒𝑦𝑜𝑛𝑑 𝑎𝑖𝑟 𝑣𝑒𝑠𝑠𝑒𝑙, 𝐻𝑓𝑑 = 𝑓
𝑉
1
𝑑𝑑 2𝑔 𝑑
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 +
𝑙𝑑2 𝐴
= 𝐻𝑎𝑡𝑚 + 𝐻𝑑 +
𝑔 𝑎𝑑
𝑉𝑑2
2𝑔
𝑙𝑑2 1 𝐴
𝜔2 𝑟𝑐𝑜𝑠𝜃 + 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
𝑑𝑑 2𝑔 𝑎𝑑
2
𝑙𝑑1 1 2 𝑉𝑑2
+𝑓
𝑉 +
𝑑𝑑 2𝑔 𝑑 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.
𝑙𝑠2 𝐴
𝑛
𝐴𝑐𝑐𝑙 𝐻𝑒𝑎𝑑, 𝐻𝑎𝑠 2 =
(𝜔2 𝑟𝑐𝑜𝑠𝜃)
𝑔 𝑎𝑠
2
𝑙𝑠2 1 𝐴
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑, 𝐻𝑓𝑠 = 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
2
𝑑𝑠 2𝑔 𝑎𝑠
𝑙𝑠1 1 2
𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑡 𝑏𝑒𝑦𝑜𝑛𝑑 𝑎𝑖𝑟 𝑣𝑒𝑠𝑠𝑒𝑙, 𝐻𝑓𝑠 = 𝑓
𝑉
1
𝑑𝑠 2𝑔 𝑠
Average velocity 𝑉𝑠 in the suction pipe: 𝑉𝑠 =
𝑄
𝑎𝑠
=
𝐴𝐿𝑁
60
×
1
𝑎𝑠
(single acting pump)
Total pressure head (abs):
H│𝑠𝑢𝑐𝑡𝑖𝑜𝑛
𝑙𝑠2 𝐴
= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝑔 𝑎𝑠
𝑉𝑠2
= 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝑎𝑠2 − 𝐻𝑓𝑠2 − 𝐻𝑓𝑠1 −
2𝑔
𝑙𝑠2 1 𝐴
2
𝜔 𝑟𝑐𝑜𝑠𝜃 − 𝑓
𝜔𝑟𝑠𝑖𝑛𝜃
𝑑𝑠 2𝑔 𝑎𝑠
2
𝑙𝑠1 1 2 𝑉𝑠2
−𝑓
𝑉 −
𝑑𝑠 2𝑔 𝑠
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),
2
𝑉𝑠2
2
𝑉𝑑2
= 𝛾𝑄 𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
2
𝑉𝑠2
2
𝑉𝑑2
= 𝛾𝐴𝐿 𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
2
𝑉𝑠2
2
𝑉𝑑2
= 𝛾𝐴(2𝑟) 𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
2
𝑉𝑠2
2
𝑉𝑑2
= 2𝛾𝐴𝑟 𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
Work done per second (single acting), i.e., Power (hydraulic) developed
𝛾𝐴𝐿𝑁
2
𝑉𝑠2
2
𝑉𝑑2
𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
60
3
2𝑔
3
2𝑔
2𝛾𝐴𝑟𝑁
=
60
2
𝑉𝑠2
2
𝑉𝑑2
𝐻𝑠 + 𝐻𝑎𝑠2 + 𝐻𝑓𝑠2 + 𝐻𝑓𝑠1 +
+ 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
Effect of Air Vessels
here
𝐻𝑓𝑠2
𝑙𝑠 1 𝐴
=𝑓
𝜔𝑟
𝑑𝑠 2𝑔 𝑎𝑠
𝐻𝑓𝑠1
𝑙𝑠 1 𝐴 𝜔𝑟
=𝑓
𝑑𝑠 2𝑔 𝑎𝑠 𝜋
2
and
2
𝐻𝑓𝑑2
𝑙𝑑 1 𝐴
=𝑓
𝜔𝑟
𝑑𝑑 2𝑔 𝑎𝑑
2
𝐻𝑓𝑑1
𝑙𝑑 1 𝐴 𝜔𝑟
=𝑓
𝑑𝑑 2𝑔 𝑎𝑑 𝜋
2
and
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
=
𝐻𝑠 + 𝐻𝑓𝑠1 + 𝐻𝑑 + 𝐻𝑓𝑑1
60
Effect of Air Vessels
Work done when air vessel is NOT fitted:
Work done against friction during the suction stroke,
2
𝑃1 = 2𝛾𝐴𝑟
𝐻𝑓𝑠𝑚
3
2
2 𝑙𝑠 1 𝐴
⇒ 𝑃1 = 2𝛾𝐴𝑟
𝑓
𝜔𝑟
3 𝑑𝑠 2𝑔 𝑎𝑠
4𝛾𝐴𝑟
𝑙𝑠 1 𝐴
⇒ 𝑃1 =
𝑓
𝜔𝑟
3
𝑑𝑠 2𝑔 𝑎𝑠
2
Work done when air vessel is fitted:
𝑃2 = 2𝛾𝐴𝑟 𝐻𝑓𝑠1
𝑙𝑠 1 2
= 2𝛾𝐴𝑟 𝑓
𝑉
𝑑𝑠 2𝑔 𝑠
𝑄
𝐴𝐿𝑁
1
𝐴 𝜔
𝐴 𝜔𝑟
𝑉𝑠 =
=
× =
(2𝑟) =
𝑎𝑠
60
𝑎𝑠 𝑎𝑠 2𝜋
𝑎𝑠 𝜋
2
𝑙𝑠 1 𝐴 𝜔𝑟
𝑃2 = 2𝛾𝐴𝑟 𝑓
𝑑𝑠 2𝑔 𝑎𝑠 𝜋
Effect of Air Vessels
Work saved in percentage for single acting pump:
=
𝑃1 − 𝑃2
× 100%
𝑃1
2
𝑙𝑠 1 𝐴
2
1
2𝛾𝐴𝑟 𝑓
𝜔𝑟
− 2
3 𝜋
𝑑𝑠 2𝑔 𝑎𝑠
4𝛾𝐴𝑟 𝑙𝑠 1 𝐴
𝑓
𝜔𝑟
3
𝑑𝑠 2𝑔 𝑎𝑠
2
× 100%
3
2 1
= ×
− 2 × 100% = 𝟖𝟒. 𝟖%
2
3 𝜋
Work saved in percentage For double acting pump:
2𝐴𝐿𝑁
𝐴 𝜔𝑟
𝑄=
=2
60
𝑎𝑠 𝜋
𝑃1 − 𝑃2
× 100%
𝑃1
3
2 4
= ×
− 2 × 100% = 𝟑𝟗. 𝟐%
2
3 𝜋
Example Problems
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.
𝑙 𝐴
0.1252
10
Solution: (i) 𝐻𝑎𝑠 = 𝑔𝑠 𝑎 𝜔2 𝑟𝑐𝑜𝑠𝜃 = 9.81 × 0.0752 ×
𝑠
2𝜋×30 2
60
× 0.15 × 𝑐𝑜𝑠𝜃 = 4.192𝑐𝑜𝑠𝜃
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
(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 −
⇒ 10.3 − 3 −
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥 2
𝑟
𝑔 𝑎𝑠
60
10
0.1252
× 0.0752
9.81
×
= 2.6 (𝑎𝑏𝑠)
2𝜋𝑁𝑚𝑎𝑥 2
60
× 0.15 = 2.6 ∴ 𝑁𝑚𝑎𝑥 = 31 𝑟𝑝𝑚
Example Problems
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 𝐴 2𝜋𝑁𝑚𝑎𝑥 2
𝑟
𝑔 𝑎𝑠
60
≥
−78.46×103
9810
5
0.0752
2𝜋 × 45
−3.5 −
×
×
9.81
60
𝑑𝑠2
𝑑𝑠 ≥ 32.6 𝑚𝑚
2
× 0.075 ≥ −8
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:
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥
𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝑔 𝑎𝑠
60
2
𝑟 = 2 (abs)
20
0.2252
2𝜋 × 𝑁𝑚𝑎𝑥
10 − 4.5 −
×
×
2
9.81 0.225
60
𝑁𝑚𝑎𝑥 = 26 𝑟𝑝𝑚
2
× 0.225 = 2
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
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥
𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝑔 𝑎𝑠
60
2
𝑟 − 𝐻𝑓𝑠 = 2.5 (abs)
4
0.12
2𝜋 × 𝑁𝑚𝑎𝑥
10 − 4 −
×
×
9.81 0.0752
60
𝑁𝑚𝑎𝑥 = 45 𝑟𝑝𝑚
2
× 0.15 − 1 = 2.5
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
𝑙𝑠 𝐴 2𝜋𝑁
𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝑔 𝑎𝑠 60
2
𝑟 − 𝐻𝑓𝑠 = 2.5 (abs)
15
0.22
2𝜋 × 60
10 − 𝐻𝑠 −
×
×
9.81 0.22
60
𝐻𝑠 = 1.46 𝑚
2
× 0.1 − 0 = 2.5
Example Problems
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:
𝐻𝑐𝑦𝑙 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 = 𝐻𝑎𝑡𝑚 + 𝐻𝑑 + 𝐻𝑎𝑑 + 𝐻𝑓𝑑 ≥ 2.75 (𝑎𝑏𝑠)
𝐻𝑎𝑡𝑚 + 𝐻𝑑 −
𝑙𝑑 𝐴 2
𝜔 𝑟 + 0 ≥ 2.75
𝑔 𝑎𝑑
45
0.22
2𝜋𝑁𝑚𝑎𝑥
9.75 + 40 −
×
×
9.81 0.12
60
𝑁𝑚𝑎𝑥 = 24 𝑟𝑝𝑚
2
× 0.4 = 2.75
Example Problems
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.
Solution:
𝐻𝑎𝑡𝑚 − 𝐻𝑠 −
𝑙𝑠 𝐴 2𝜋𝑁𝑚𝑎𝑥 2
𝑟
𝑔 𝑎𝑠
60
2
= 2.5
7.5 0.3
2𝜋 × 40
10 − 3.5 −
× 2 ×
9.81
60
𝑑𝑠
𝑑𝑠 = 237.9 𝑚𝑚
2
× 0.1875 = 2.5
Example Problems
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 𝐴 2𝜋𝑁 2
𝑟
𝑔 𝑎𝑠 60
≥ 2.5
5.5
0.22
2𝜋 × 60
10.3 − 3.5 −
×
×
9.81 𝑑 2
60
2
× 0.15 = 2.5
𝑠│𝑚𝑖𝑛
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 𝐴 2𝜋𝑁 2
𝑟
𝑔 𝑎𝑠 60
≥ 2.5
5
0.252
2𝜋 × 25
10 − 3 −
× 2
×
9.81 𝑑
60
2
× 0.2 = 2.5
𝑠│𝑚𝑖𝑛
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.
Solution: Power required,
Here, 2𝛾𝐴𝑟𝑁
60𝜂𝑝𝑢𝑚𝑝
=
2 × 9810 ×
2𝛾𝐴𝑟𝑁
𝑃=
60𝜂𝑝𝑢𝑚𝑝
2
𝑉𝑠2
2
𝑉𝑑2
𝐻𝑠 + 𝐻𝑎𝑠 + 𝐻𝑓𝑠 +
+ 𝐻𝑑 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1 +
3
2𝑔
3
2𝑔
𝜋
× 0.1252 × 0.25 × 45
4
= 50.16 𝑁/𝑠
60 × 0.9
𝑙𝑠 𝐴 2
5.2
0.1252
2𝜋 × 45
𝐻𝑎𝑠 =
𝜔 𝑟=
×
×
𝑔 𝑎𝑠
9.81
0.12
60
𝑙𝑠 1 𝐴
𝐻𝑓𝑠 = 𝑓
𝜔𝑟
𝑑𝑠 2𝑔 𝑎𝑠
2
2
× 0.25 = 4.6 𝑚
5.2
1
0.1252 2𝜋 × 45
= 0.02 ×
×
×
×
× 0.25
0.1
2 × 9.81
0.12
60
2
𝐻 + 𝐻𝑓𝑑1 = 0.15 𝑚
3 𝑓𝑑2
2
Therefore, 𝑃 = 50.16 × [(3.25 + 4.6 + 2 × 0.18 + 0) + (12 + 0.15 + 0)]
3
= 1009.22 𝑊𝑎𝑡𝑡 = 𝟏. 𝟎𝟏 𝒌𝑾 𝑨𝒏𝒔.
= 0.18 𝑚
Example Problems
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.
Solution: Power required, 𝑃 =
2𝛾𝐴𝑟𝑁
60𝜂𝑝𝑢𝑚𝑝
2
𝐻𝑠 + 𝐻𝑓𝑠1 + 𝐻𝑑 + 𝐻𝑎𝑑2 + 𝐻𝑓𝑑2 + 𝐻𝑓𝑑1
3
𝜋
(Neglecting velocity heads in the pipes)
2
2
×
9810
×
×
0.1
×
0.1
×
60
2𝛾𝐴𝑟𝑁
Here
4
=
= 17.12 𝑁/𝑠
60𝜂𝑝𝑢𝑚𝑝
60 × 0.9
𝑙𝑠 1 𝐴 𝜔𝑟
𝐻𝑓𝑠1 = 𝑓
𝑑𝑠 2𝑔 𝑎𝑠 𝜋
2
8
1
0.12
2𝜋 × 60
= 0.025 ×
×
×
×
× 0.1
0.075
2 × 9.81
0.0752 60 × 𝜋
𝑙𝑑2 𝐴 2
3
0.12
2𝜋 × 60
𝐻𝑎𝑑2 =
𝜔 𝑟=
×
×
𝑔 𝑎𝑑
9.81
0.12
60
𝑙𝑑2 1 𝐴
𝐻𝑓𝑑2 = 𝑓
𝜔𝑟
𝑑𝑑 2𝑔 𝑎𝑑
𝑙𝑑1 1 𝐴 𝜔𝑟
𝐻𝑓𝑑1 = 𝑓
𝑑𝑑 2𝑔 𝑎𝑑 𝜋
𝑃 = 17.12 ×
2
2
= 0.017 𝑚
2
× 0.1 = 1.207 𝑚
3
1
0.12 2𝜋 × 60
= 0.025 ×
×
×
×
× 0.1
0.1
2 × 9.81
0.12
60
2
27
1
0.12 2𝜋 × 60
= 0.025 ×
×
×
×
× 0.1
0.1
2 × 9.81
0.12 60 × 𝜋
2
4 + 0.017 + 25 + 1.207 +
2
2
× 0.015 + 0.014
3
= 0.015 𝑚
= 0.014 𝑚
= 517.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟏𝟖 𝒌𝑾 𝑨𝒏𝒔
Example Problems
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,
2
2𝛾𝐴𝑟 2 𝑙𝑑 1 𝐴
𝑃1 = 2𝛾𝐴𝑟 𝐻𝑓𝑑𝑚 =
𝑓
𝜔𝑟
3
3 3 𝑑𝑑 2𝑔 𝑎𝑑
2
Friction Work with air vessel,
𝑙𝑑 1 𝐴 𝜔𝑟
𝑃2 = 2𝛾𝐴𝑟 𝐻𝑓𝑑 = 2𝛾𝐴𝑟 𝑓
𝑑𝑑 2𝑔 𝑎𝑑 𝜋
Work (Power) saved in overcoming friction,
𝑃1 − 𝑃2 = 2𝛾𝐴𝑟
𝑙 1
𝑓 𝑑
𝑑𝑑 2𝑔
2
𝐴
𝜔𝑟
𝑎𝑑
2
3
−
2
1
𝜋2
𝜋
40
1
0.32 2𝜋 × 60
2
= 2 × 9810 × × 0.3 × 0.225 0.03 ×
×
×
×
× 0.225
4
0.2 2 × 9.81
0.22
60
= 545.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟒𝟓 𝒌𝑾
2
×
2 1
−
3 𝜋2
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°]