# Chapter 7:Hydraulic Motors¶

## Example 7.1 pgno:248¶

In :
# 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
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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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 :
# 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