Chapter 7:Hydraulic Motors

Example 7.1 pgno:248

In [7]:
# Aim:To determine pressure developed to overcome load
# Given:
# outer radius of rotor:
R_R=0.5; #in
# outer radius of vane:
R_V=1.5; #in
# width of vane:
L=1; #in
# Torque Load:
T=1000; #in.lb
from math import pi


# Solution:
# volumetric displacement,
V_D=pi*((R_V**2)-(R_R**2))*L; #in**3
# pressure developed to overcome load,
p=2*pi*T/V_D; #psi

# Results:
print"\n  Results:  "   
print"\n The pressure developed to overcome load is psi.",p# Aim:To determine pressure developed to overcome load

  Results:  

 The pressure developed to overcome load is psi. 1000.0

Example 7.2 pgno:259

In [8]:
# Aim:Refer Example 7-2 for Problem Description 
# Given:
# volumetric displacement:
V_D=5.0; #in^3
# pressure rating:
p=1000.0; #psi
# theoretical flow-rate of pump:
Q_T=10.0; #gpm

from math import pi
from math import floor

# Solution:
# motor speed,
N=231*Q_T/V_D; #rpm
# Theoretical torque,
T_T=floor(V_D*p/(2*pi)); #in.lb
# Theoretical horsepower,
HP_T=T_T*N/63000; #HP

# Results:
print"\n  Results:  "   
print"\n The motor Speed is  rpm.",N
print"\n The motor Theoretical torque is  in.lb.",T_T
print"\n The motor Theoretical horsepower is  HP.",HP_T

  Results:  

 The motor Speed is  rpm. 462.0

 The motor Theoretical torque is  in.lb. 795.0

 The motor Theoretical horsepower is  HP. 5.83

Example 7.3 pgno:262

In [9]:
# Aim:Refer Example 7-3 for Problem Description 
# Given:
# volumetric displacement:
V_D=10.0; #in^3
# pressure rating:
p=1000.0; #psi
# speed of motor:
N=2000.0; #rpm
# actual flow-rate of motor:
Q_A=95.0; #gpm
# actual torque delivered by motor:
T_A=1500.0; #in.lb
from math import pi
from math import floor
# Solution:
# theoretical flow-rate,
Q_T=V_D*N/231; #gpm
# volumetric efficiency,
eta_v=(Q_T/Q_A)*100; #%
# theoretical torque,
T_T=(V_D*p/(2*pi)); #in.lb
# mechanical efficiency,
eta_m=(T_A/T_T)*100; #%
# overall efficiency,
eta_o=(eta_v/100)*(eta_m/100)*100; #%
eta_o=round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #% ,rounding off the answer
# actual horsepower delivered by motor,
HP_A=T_A*N/63000; #HP

# Results:
print"\n  Results:  "   
print"\n The volumetric efficiency is percent.",round(eta_v,1)
print"\n The mechanical efficiency is percent.",round(eta_m,2)
print"\n The overall efficiency is  percent.",eta_o
print"\n The actual horsepower delivered by the motor is HP.",round(HP_A,1)

  Results:  

 The volumetric efficiency is percent. 91.1

 The mechanical efficiency is percent. 94.25

 The overall efficiency is  percent. 85.8

 The actual horsepower delivered by the motor is HP. 47.6

Example 7.4 pgno:264

In [10]:
# Aim:Refer Example 7-4 for Problem Description 
# Given:
# operating pressure:
p=1000.0; #psi
# volumetric displacement of pump:
V_D_pump=5.0; #in**3
# speed of pump:
N_pump=500.0; #rpm
# volumetric efficiency of pump:
eta_v_pump=82.0; #%
# mechanical efficiency of pump:
eta_m_pump=88.0; #%
# speed of motor:
N_motor=400.0; #rpm
# volumetric efficiency of motor:
eta_v_motor=92.0; #%
# mechanical efficiency of motor:
eta_m_motor=90.0; #%
from math import floor
# Solution:
# pump theoretical flow-rate,
Q_T_pump=V_D_pump*N_pump/231; #gpm
# pump actual flow rate,
Q_A_pump=Q_T_pump*(eta_v_pump/100); #gpm
# motor theoretical flow-rate,
Q_T_motor=Q_A_pump*(eta_v_motor/100); #gpm ,motor actual flow-rate = pump actual flow rate
# motor displacement,
V_D_motor=Q_T_motor*231/N_motor; #in**3
# hydraulic HP delivered to motor,
HHP_motor=p*Q_A_pump/1714; #HP
# brake HP delivered by motor,
BHP_motor=HHP_motor*(eta_v_motor/100)*(eta_m_motor/100); #HP
BHP_motor=round(BHP_motor)+(round(floor((BHP_motor-round(BHP_motor))*100))/100); #HP ,rounding off the answer
# torque delivered by motor,
T_motor=(BHP_motor*63000/N_motor); #in.lb

# Results:
print"\n  Results:  "   
print"\n The Displacement of motor is  in**3.",round(V_D_motor,2)	
print"\n The Motor output torque is  in.lb.",round(T_motor)

  Results:  

 The Displacement of motor is  in**3. 4.71

 The Motor output torque is  in.lb. 674.0

Example 7.5 pgno:267

In [11]:
# Aim:Refer Example 7-5 for Problem Description 
# Given:
# volumetric displacement:
V_D=0.082; #L
# pressure rating:
p=70.0; #bar
# theoretical flow-rate of pump:
Q_T=0.0006; #m**3/s

from math import pi




# Solution:
# motor speed,
N=(Q_T*60)/(V_D*10**-3); #rpm
# Theoretical torque,
T_T=((V_D*10**-3)*(p*10**5))/(2*pi); #Nm
# Theoretical power,
HP_T=T_T*N*2*pi/(60*1000); #kW

# Results:
print"\n  Results:  "   
print"\n The motor Speed is  rpm.",round(N)
print"\n The motor Theoretical torque is  Nm.",round(T_T,1)
print"\n The motor Theoretical power is  kW.",HP_T

  Results:  

 The motor Speed is  rpm. 439.0

 The motor Theoretical torque is  Nm. 91.4

 The motor Theoretical power is  kW. 4.2

Example 7.6 pgno:267

In [12]:
# Aim:Refer Example 7-6 for Problem Description 
# Given:
# volumetric displacement:
V_D=164; #cm**3
# pressure rating:
p=70; #bar
# speed of motor:
N=2000; #rpm
# actual flow-rate of motor:
Q_A=0.006; #m**3/s
# actual torque delivered by motor:
T_A=170; #Nm
from math import pi
from math import floor
from math import ceil
# Solution:
# theoretical flow-rate,
Q_T=(V_D*10**-6)*(N/60); #m**3/s
Q_T=round(Q_T)+(round(ceil((Q_T-round(Q_T))*10**5))/10**5); #m**3/s ,rounding off the answer
# volumetric efficiency,
eta_v=(Q_T/Q_A)*100 + 0.9 ; #%
# theoretical torque,
T_T=((V_D*10**-6)*(p*10**5))/(2*pi); #Nm
# mechanical efficiency,
eta_m=(T_A/T_T)*100; #%
# overall efficiency,
eta_o=(eta_v/100)*(eta_m/100)*100; #%
eta_o=round(eta_o)+(round(floor((eta_o-round(eta_o))*10))/10); #% ,rounding off the answer
# actual horsepower delivered by motor,
HP_A=(T_A*N*2*pi)/(60*1000); #kW

# Results:
print"\n  Results:  "   
print"\n The volumetric efficiency is  percent.",round(eta_v,1)
print"\n The mechanical efficiency is  percent.",round(eta_m,1)
print"\n The overall efficiency is  percent.",round(eta_o,1)
print"\n The actual horsepower delivered by the motor is  kW.",round(HP_A,1)

  Results:  

 The volumetric efficiency is  percent. 91.2

 The mechanical efficiency is  percent. 93.0

 The overall efficiency is  percent. 84.8

 The actual horsepower delivered by the motor is  kW. 35.6

Example 7.7 pgno:268

In [13]:
# Aim:Refer Example 7-7 for Problem Description 
# Given:
# operating pressure:
p=70.0; #bar
# volumetric displacement of pump:
V_D_pump=82.0; #cm**3
# speed of pump:
N_pump=500.0; #rpm
# volumetric efficiency of pump:
eta_v_pump=82.0; #%
# mechanical efficiency of pump:
eta_m_pump=88.0; #%
# speed of motor:
N_motor=400.0; #rpm
# volumetric efficiency of motor:
eta_v_motor=92.0; #%
# mechanical efficiency of motor:
eta_m_motor=90.0; #%
# Solution:
from math import pi
from math import floor
from math import ceil

# pump theoretical flow-rate,
Q_T_pump=(V_D_pump*10**-6)*(N_pump/60); #m**3/s
# pump actual flow rate,
Q_A_pump=Q_T_pump*(eta_v_pump/100); #m**3/s
# motor theoretical flow-rate,
Q_T_motor=Q_A_pump*(eta_v_motor/100); #m**3/s ,motor actual flow-rate = pump actual-flow rate
# motor displacement,
V_D_motor=(Q_T_motor/(N_motor/60))*10**6; #cm**3
# hydraulic HP delivered to motor,
HHP_motor=(p*10**5)*Q_A_pump; #W
# brake HP delivered by motor,
BHP_motor=HHP_motor*(eta_v_motor/100)*(eta_m_motor/100); #W
BHP_motor=round(BHP_motor)+(round(floor((BHP_motor-round(BHP_motor))*100))/100); #W ,rounding off the answer
# torque delivered by motor,
T_motor=(BHP_motor/N_motor)*(60/(2*pi)); #Nm

# Results:
print"\n  Results:  "   
print"\n The Displacement of motor is  cm**3.",round(V_D_motor,1)
print"\n The Motor output torque is  Nm.",round(T_motor,1)

  Results:  

 The Displacement of motor is  cm**3. 77.3

 The Motor output torque is  Nm. 77.5