# Chapter 2:Dynamics of Electric Drives¶

## Example no:2.1,Page no:16¶

In [4]:
import math

#variable declaration
Jo=0.2     # inertia of the motor in kg-m2
a1=0.1     # reduction gear
J1=10      # inertia of the load in kg-m2
v=1.5      # speed of the translational load
M1=1000    # mass of the translational load
N=1420     # speed of the motor
n1=.9      # efficiency of the reduction gear
n1_=0.85   # efficiency of the translational load and the motor
F1=M1*9.81 # force of the translational load

#Calculation
Wm=N*math.pi/30 #angular speed
J=Jo+a1**2*J1+ M1*(v/Wm)**2 # total equivalent moment of inertia
Tl= a1*Tl1/n1+F1/n1_*(v/Wm) # total equivalent torque

#Result
print"\nEquivalent moment of inertia is :",round(J,1),"kg-m2"

Equivalent moment of inertia is: 0.4 kg-m2

Equivalent load torque : 117.53 N-m


## Example no:2.2,Page no:22¶

In [7]:
import scipy
from scipy import integrate
import math

# variable declaration
J=10           #moment of inertia of the drive in kg-m2
print("And load torque : T=100-0.1*N in N-m ")
print("load torque  when the direction is reversed T=-100-0.1*N  in N-m")

#Calculation
print("T-Tl=0")
print("100-0.1*N-0.05*N=0")
N=100/0.15     #Required speed of the motor in rpm during steady state
N2=-100/0.15   #During reversal speed is in opposite direction
print("\nJdWm/dt=-100-0.1*N-0.05*N during reversing")
print("dN/dt=30/(J*pi)*(-100-0.15*N)")
print("dN/dt=(-95.49-0.143*N)")
N1=N
N2=N2*0.95 #for speed reversal
x2 = lambda N: 1/(-95.49-0.143*N)

#result
print"\nHence Time of reversal is :",round(t[0],2),"s"

Passive load torque during steady state is :Tl=0.05*N in N-m
And load torque : T=100-0.1*N in N-m
load torque  when the direction is reversed T=-100-0.1*N  in N-m
T-Tl=0
100-0.1*N-0.05*N=0

JdWm/dt=-100-0.1*N-0.05*N during reversing
dN/dt=30/(J*pi)*(-100-0.15*N)
dN/dt=(-95.49-0.143*N)

Hence Time of reversal is : 25.51 s


## Example no:2.3,Page no:27¶

In [12]:
import math
from __future__ import division

#variable declaration
Tlh=1000     # load torque in N-m
Tmax=700     # maximum motor torque
Tll=200      # light load for the motor to regain its steady state
Tmin=Tll     # minimum torque
t_h=10       # period for which a load torque of 1000 N-m is apllied in sec
Jm=10        # moment of inertia of the motor in Kg-m2
No=500       # no load speed in rpm
Tr=500       # torque at a given no load speed in N-m

#Calculation
s=0.05              # slip at a torque of 500 N-m
Wmr=(1-s)*Wmo       # angular speed at a torque of 500 N-m in rad/sec

y=math.log((Tlh-Tmin)/(Tlh-Tmax))
x=Tr/(Wmo-Wmr)

J=x*t_h/y
Jf=J-Jm

#Result
#answer in the book is wrong
print"\n\nMoment of inertia of the flywheel : ", round(Jf,1),"Kg-m2"


Moment of inertia of the flywheel :  1937.2 Kg-m2