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Module 1. Hydraulic Basics
Module 2. Hydraulic Systems
MODULE 3.
MODULE 4.
MODULE 5.
MODULE 6.
MODULE 7.
MODULE 8.
LESSON 21. Design of Reciprocating type of Pump
21.1 Introduction
Hydraulic systems are generally provided with positive displacement pumps. The piston sucks and pushes the same quantity (volume) of fluid every rotation of shaft .In these pumps there is hardly any slippage and the quantity of fluid pumped is nearly same for each rotation of shaft. Specially piston pumps exhibit this quality..
21.2 Design of Reciprocating type of Pump
In the reciprocating pump as the crank rotates the reciprocating motion transmit a liquid from a pump inlet to its outlet. When a piston moves to the, a partial vacuum is created which draws a liquid through an inlet valve into a cylinder. When a piston moves other way an inlet valve is closed and the fluid is forced out of a cylinder through an outlet valve.
Let us consider single acting reciprocating pump
Let D = diameter of cylinder
A = crosectional area
L = length of stroke
R = radius of crank
N = speed of crank
Qth = theoretical discharge
Qth = Ax LxN discharge per min
= Ax LxN/ 60 discharge per second
The ratio of actual discharge (Qac) to the theoretical discharge (Qth )is called discharge coefficient (cd). But when discharge coefficient is expressed in percentage it is called as volumetric efficiency.
cd = \[\frac{{{\text{Qac}}}}{{{\text{Qth}}}}\]
Slip
It can be defined as the difference between the theoretical discharge and the actual discharge. If S' is the slip,
then
S' = (Qth - Qac)
Percentage slip can be given as-
S = [ 1 - \[\frac{{{\text{Qac}}}}{{{\text{Qth}}}}\] ] x 100
Work done and Power Requirement of Reciprocating Pump
It can be defined as the product of of weight fluid lifted and the total height through which it has been lifted .
Let W = work done / s
w = weight of fluid lifted
H = Total height of fluid lifted
Q = discharge of pump
W = w x Q x H
H = (hs + hd)
hs = suction head
hd = delivery head
Example
A reciprocating pump is having diameter of piston 260 mm with stroke length of 510 mm operating at a speed of 32 rpm delivers 0.0125 m3/ s of fluid.Calculate the followings-
a) Theoretical discharge of pump
b) Coefficient of discharge
c) Slip
d) Percentage slip
Soln
Given
diameter of piston D = 260 mm = 0.26 m
stroke length of L = 510 mm =0.51 m
actual discharge Qac = 0.0125 m3/ s
Speed of pump N = 32 rpm
Qth = Ax LxN/ 60
A = \[\frac{{\pi {D^2}}}{4}\]
= \[\frac{{\pi {{\left( {0.26} \right)}^2}}}{4}\]
= 0.053 m2
Now Qth = 0.053x 0.51x32 / 60
= 0.0144 m3/ s
Coefficient of discharge cd = \[\frac{{{\text{Qac}}}}{{{\text{Qth}}}}\]
= \[\frac{{0.0125}}{{0.0144}}\]
= 0.868
= 86.8%
Slip S' = (Qth - Qac)
= (().0144 – 0.0125)
= 0.0019 m3/ s
Slip % S = [ 1 - \[\frac{{{\text{Qac}}}}{{{\text{Qth}}}}\] ] x 100
= [ 1 - \[\frac{{0.0125}}{{0.0144}}\] ] x 100
= 13.1 %