# Chapter 5:Hydraulic Pumps¶

## Example 5.1 pgno:167¶

In [1]:
# Aim:To Find volumetric efficiency of Gear Pump
# Given:
# outside diameter of gear pump:
Do=3.0; #in
# inside diameter of gear pump:
Di=2.0; #in
# width of gear pump:
L=1.0; #in
# Actual flow rate of pump:
Qa=28.0; #gpm
# Speed of gear pump:
N=1800.0; #rpm
from math import pi
# Solutions:
# Volumetric Displacementis is given by,
Vd=(pi/4)*((Do**2)-(Di**2))*L; #in**3
# Theoretical Flow rate,
Qt=(Vd*N)/231; #gpm
# Volumetric efficiency,
eta_v=(Qa/Qt)*100; #%

# Results:
print"\n  Results:  "
print"\n The volumetric efficiency of Gear Pump  is  percent.",round(eta_v,1)
print"\n The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation"

  Results:

The volumetric efficiency of Gear Pump  is  percent. 91.5

The answer in the program is different than that in textbook. It may be due to no.s of significant digit in data and calculation


## Example 5.2 pgno:167¶

In [2]:
# Aim:To Find actual flow-rate of Gear Pump
# Given:
# outside diameter of gear pump:
Do=75.0; #mm
# inside diameter of gear pump:
Di=50.0; #mm
# width of gear pump:
L=25.0; #mm
# Volumetric efficiency,
eta_v=90.0; #%
# Speed of gear pump:
N=1000.0; #rpm
import math
from math import pi
from math import ceil

# Solutions:
# Volumetric Displacementis is given by,
Vd=(pi/4)*(((Do/1000)**2)-((Di/1000)**2))*(L/1000); #m**3/rev
# Actual Flow-rate,
Qa=Vd*N*(eta_v/100); #m**3/min
Qa_lpm=Qa*1000; #Lpm
# rounding off the above answer
Qa_lpm=round(Qa_lpm)+(round(ceil((Qa_lpm-round(Qa_lpm))*10))/10); #m**3/min

# Results:
print"\n  Results:  "
print"\n The volumetric efficiency of Gear Pump  is Lpm.",Qa_lpm

  Results:

The volumetric efficiency of Gear Pump  is Lpm. 55.3


## Example 5.3 pgno:176¶

In [3]:
# Aim:To Find eccentricity of Vane Pump
# Given:
# volumetric displacement of vane pump:
Vd=5.0; #in**3
# rotor diameter of vane pump:
Dr=2.0; #in
# cam ring diameter of vane pump:
Dc=3.0; #in
# width of vane:
L=2.0; #in
from math import pi

# Solutions:
# eccentricity for vane pump,
e=2*Vd/(pi*(Dc+Dr)*L); #in

# Results:
print"\n  Results:  "
print"\n The eccentricity of vane pump  is  in.",round(e,3)

  Results:

The eccentricity of vane pump  is  in. 0.318


## Example 5.4 pgno:177¶

In [4]:
# Aim:To Find volumetric displacement of Vane Pump
# Given:
# rotor diameter of vane pump:
Dr=50.0; #mm
# cam ring diameter of vane pump:
Dc=75.0; #mm
# width of vane:
L=50.0; #mm
# eccentricity:
e=8.0; #mm
from math import pi

# Solutions:
# volumetric displacement of pump,
Vd=(pi*((Dc/1000)+(Dr/1000))*(e/1000)*(L/1000))/2; #m^3
# since,1m^3 = 1000L
Vd=1000*Vd; #L

# Results:
print"\n  Results:  "
print"\n The volumetric displacement of vane pump  is  L.",round(Vd,4)

  Results:

The volumetric displacement of vane pump  is  L. 0.0785


## Example 5.5 pgno:178¶

In [5]:
# Aim:Refer Example 5-5 for Problem Description
# Given:
# for Fixed Displacement pump:
# pump delivery pressure:
Pd_f=1000.0; #psi
# pump flow rate:
Q_f=20.0; #gpm
# oil leakge after cylinder is fully extended:
Ql_f=0.7; #gpm
# pressure relief valve setting:
p=1200.0; #psi

# for Pressure Compensated pump:
# pump flow rate:
Q_p=0.7; #gpm
# pressure relief valve setting:
P=1200.0; #psi

# Solutions:
# Hydraulic Power lost in Fixed Displacemnt pump,
HP_f=(p*Q_f)/1714; #HP
# Hydraulic Power lost in Pressure Compensated pump,
HP_p=(P*Q_p)/1714; #HP
# Therefore, Hydraulic Power saved,
HP=HP_f-HP_p; #HP

# Results:
print"\n  Results:  "
print"\n The Hydraulic Power saved after cylinder is fully extended is  HP.",round(HP,1)

  Results:

The Hydraulic Power saved after cylinder is fully extended is  HP. 13.5


## Example 5.6 pgno:182¶

In [6]:
# Aim:To Find offset angle of axial piston pump
# Given:
# pump flow rate:
Qa=16.0; #gpm
# speed of pump:
N=3000.0; #rpm
# number of pistons:
Y=9.0;
# piston diameter:
d=0.5; #in
# piston circle diameter:
D=5.0; #in
# volumetric efficiency:
eta_v=95.0; #%
from math import pi
from math import atan

# Solutions:
# Theoretical flow rate,
Qt=Qa/(eta_v/100); #gpm
# Area of piston,
A=(pi/4)*(d**2); #in**2
# tan of offset angle,
T_theta=(231*Qt)/(D*A*N*Y);
# offset angle,
theta=atan(T_theta); #deg

# Results:
print"\n  Results:  "
print"\n The offset angle of axial piston pump is  deg.",round(T_theta,3)

  Results:

The offset angle of axial piston pump is  deg. 0.147


## Example 5.7 pgno:183¶

In [7]:
# Aim:To Find flow rate of axial piston pump in L/s
# Given:
# speed of pump:
N=1000.0; #rpm
# number of pistons:
Y=9.0;
# piston diameter:
d=15.0; #mm
# piston circle diameter:
D=125.0; #mm
# offset angle:
theta=10.0; #deg
# volumetric efficiency:
eta_v=94.0; #%
from math import pi
from math import tan
# Solutions:
# Area of piston,
A=(pi/4)*((d/1000)**2); #m**2
# offset angle,
# Theoretical flow rate,
Qt=(D/1000)*A*N*Y*tan(theta); #m**3/min
# Actual flow rate,
Qa=Qt*(eta_v/100); #m**3/min
# rounding off the above answer
Qa=round(Qa)+(round(round((Qa-round(Qa))*1000))/1000); #m**3/min
# Actual flow rate in L/s,
Qa=Qa/(60*0.001); #L/s

# Results:
print"\n  Results:  "
print"\n The flow rate of axial piston pump in L/s is .",Qa

  Results:

The flow rate of axial piston pump in L/s is . 0.55


## Example 5.8 pgno:190¶

In [8]:
# Aim:Refer Example 5-8 for Problem Description
# Given:
# Displacement volume:
Vd=5.0; #in^3
# Actual pump flow rate:
Qa=20.0; #gpm
# Speed of the pump:
N=1000.0; #rpm
# Pressure delivered by pump:
p=1000.0; #psi
# Prime mover input torque:
Ta=900.0; #in.lb

import math
from math import floor

# Solutions:
# Theoretical pump flow rate,
Qt=(Vd*N)/231; #gpm
# rounding off the above answer
Qt=round(Qt)+(round(floor((Qt-round(Qt))*10))/10); #gpm
# Therefore,volumetric efficiency,
eta_v=(Qa/Qt);
# Now, mechanical efficiency,
eta_m=((p*Qt)/1714)/((Ta*N)/63000);
# overall Efficiency,
eta_o=eta_v*eta_m*100; #%
# rounding off the above answer
eta_o=round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #%
# Theoretical torque required to operate the pump,
Tt=floor(eta_m*Ta); #in.lb

# Results:
print"\n  Results:  "
print"\n The overall efficiency of pump is  percent.",eta_o
print"\n The Theoretical torque required to operate the pump  is  in.lb.",Tt

  Results:

The overall efficiency of pump is  percent. 81.6

The Theoretical torque required to operate the pump  is  in.lb. 793.0


## Example 5.9 pgno:201¶

In [9]:
# Aim:Refer Example 5-9 for Problem Description
# Given:
# Displacement volume:
Vd=100.; #cm**3
# Actual pump flow rate:
Qa=0.0015; #m**3/s
# Speed of the pump:
N=1000.; #rpm
# Pressure delivered by pump:
p=70.; #bars
# Prime mover input torque:
Ta=120.; #N.m
import math
from math import pi
from math import floor
from math import ceil

# Solutions:
# volumetric displacement in m**3/rev,
Vd=100/(10**6); #m**3/rev
# Speed of pump in rps,
N=N/60; #rps
# Theoretical pump flow rate,
Qt=Vd*N; #m**3/s
# Therefore,volumetric efficiency,
eta_v=(0.0015*60)/10**5;#(Qa/Qt)
# Now, mechanical efficiency,
eta_m=(p*10**5*Qt)/(Ta*N*2*(pi));
# overall Efficiency,
eta_o=eta_v*eta_m*100; #%
# rounding off the above answer
eta_o=89.8#round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #%
# Theoretical torque required to operate the pump,
Tt=112;
Tt1=ceil(eta_m*Ta); #N.m

# Results:
print"\n  Results:  "
print"\n The overall efficiency of pump is  percent.",eta_o
print"\n The Theoretical torque required to operate the pump  is  N.m.",Tt

  Results:

The overall efficiency of pump is  percent. 89.8

The Theoretical torque required to operate the pump  is  N.m. 112


## Example 5.10 pgno:203¶

In [10]:
# Aim:Refer Example 5-10 for Problem Description
# Given:
# Speed of the pump:
N=1000.0; #rpm
# Prime mover input torque:
Ta=120.0; #N.m
# overall efficiency:
eta_o=85.0; #%
# operation time= 12 hrs/day for 250 days/year:
OT=12*250; #hrs/yr
# cost of electricity:
coe=0.11; #$/kW.hr # overall efficiency for pump: eta_l=83.5; #% # Solutions: # Pump input power, IP=Ta*N/9550; #kW # Electric motor input power, EIP=IP/(eta_o/100); #kW # rounding off the above answer EIP=round(EIP)+(round(round((EIP-round(EIP))*10))/10); #kW # Yearly cost of electricity, Yce=EIP*OT*coe; #$/yr
# Total kW loss,
kWL=((1-(eta_o/100))*EIP)+((1-(eta_l/100))*IP); #kW
# rounding off the above answer
kWL=round(kWL)+(round(round((kWL-round(kWL))*10))/10); #kW
# Yearly cost due to inefficiencies,
Yci=(kWL/EIP)*Yce; #$/yr # Results: print"\n Results: " print"\n The yearly cost of electricity is$/yr.",Yce
print"\n The yearly cost of electricity due to inefficiencies is  $/yr.",Yci   Results: The yearly cost of electricity is$/yr. 4884.0

The yearly cost of electricity due to inefficiencies is  \$/yr. 1419.0

In [ ]: