# Chapter 4: Selection of Motor Power Rating¶

## Example No:4.1, Page No:47¶

In [1]:
import math
from __future__ import division

#variable declaration
T_min=40        # minimum temperature rise in degree Celsius
T_r=15          # temperature rise when the load is declutched continously in degree Celsius
t_c=10          # time for which the motor is clutched to its load in sec
t_d=20          # time for which the motor is declutched to run on no-load in sec
C= 60           # time constants for both heating and cooling

#calculation
x=math.exp(-t_d/C)
y=math.exp(-t_c/C)

theta2= (T_min-T_r*(1-x))/x       #since T_min=T_r(1-x)+theta2*x
theta_ss=(theta2-T_min*y)/(1-y)   #since theta2=theta_ss(1-y)+T_min*y

#results
print"\n maximum temperature during the duty cycle :",round(theta2,1),"°C"
print"\n temperature when the load is clutched continously :",round(theta_ss,1),"°C"

 maximum temperature during the duty cycle : 49.9 °C

temperature when the load is clutched continously : 104.4 °C


## Example No:4.2, Page No:52¶

In [3]:
import math
from __future__ import division

#variable declaration
N=200         #full speed of the motor in rpm
Tc=25000      #constant torque in N-m
J=10000       #moment of inertia referred to te motor shaft in Kg-m2

#duty cycles
t1=10         #rolling at full speed and at constant torque
t2=1          #no load operation at full speed
t3=5          #speed reversal from N to -N
t4=1          #no load operation at full speed

T5=20000      #torque in N-m
t5=15         #rolling at full speed and at a torque T1

t6=1          #no operation at full speed
t7=5          #speed reversal from -N to N

#calculation
Tr=J*(N-(-N))*2*math.pi/60/5   #torque during reversal
x=Tc**2*t1+Tr**2*t3+T5**2*t5+Tr**2*t7
t=t1+t2+t3+t4+t5+t6+t7+t8      #total time
Trms=math.sqrt(x/t)            #rms torque

Trated=Trms                   #rated torque is equal to the rms torque
Pr=Trated*2*math.pi*200/60    #power rating
ratio=Tr/Trms                 #ratio of reversal torque to the rms torque

#results
#answer in the book is wrong
print"\n motor torque is :Trms=",round(Trms),"N-m"
if ratio<2:
print" motor can be rated as equal to Trms"
print" Power rating : P=",round(Pr*1e-3,3),"kW"

 motor torque is :Trms= 45975.0 N-m
motor can be rated as equal to Trms
Power rating : P= 962.895 kW


## Example No:4.3, Page No:53¶

In [4]:
import math
import scipy
from scipy import integrate

#variable declaration
Pmax=P2
#duty cycles in minutes
t1=5       #load rising from 0 to P1
t3=4       #regenerative power equal to P1
t4=2       #motor remains idle

#calculation
a = lambda x: (P1/5*x)**2
P11=math.sqrt(t[0]/t1)
x=P11**2*t1+P2**2*t2+P1**2*t3
t=t1+t2+t3+t4    #total time
Prms=math.sqrt(x/t)

#results
y=2*Prms
if P2<y:
print " Hence Pmax:",Pmax,"kW is less than twice Prms:",2*round(Prms,1),"kW"
print"\n Hence Motor rating is: ",round(Prms),"kW"

 Hence Pmax: 500 kW is less than twice Prms: 734.2 kW

Hence Motor rating is:  367.0 kW


## Example No:4.4, Page No:55¶

In [5]:
import math
from __future__ import division

#variable declaration
Cr=60      #heating time constant in minutes
Cs=90      #cooling time constant in minutes

#calculation

#part(i)
alpha=0  #constant copper losses are assumed to be proportional to Power**2 which is zero
tr=10    #time for the load motor to deliver in minutes
x=math.exp(-tr/Cr)
K=math.sqrt(1/(1-x))

#part(ii)
alpha=0 #constant copper losses are assumed to be proportional to Power**2 which is zero
tr=10   #intermittent load period allowed in minutes
ts=10   #shutdown period in minutes
x=math.exp(-(tr/Cr+ts/Cs))
y=math.exp(-tr/Cr)
K=math.sqrt((1-x)/(1-y))

#results

i)required permitted load: 51.0 kW



## Example No:4.5, Page No:56¶

In [12]:
import math
from sympy import Symbol
from __future__ import division

#variable declaration
P=100     #Half hour rating of the motor
Cr=80     #heating time constant in minutes
n=0.7     #maximum efficiency at full load

#calculation
Pc = Symbol('Pc')   #constant loss
Pcu=Pc/n**2   #coppper loss
alpha=Pc/Pcu
K=math.sqrt((1+alpha)/(1-math.e**(-30/Cr))-alpha)
Pco=P/K
print"Therefore continous rating is:",round(Pco,2),"kW"

Therefore continous rating is: 48.37 kW


## Example No:4.6, Page No:57¶

In [5]:
import math
from __future__ import division

#variable declaration
I=500       #rated armature current in A
Ra=0.01     #armature resistance in ohm
P=1000      #core loss in W
B=0.5

#duty cycles
tst=10       #interval for accelaration at twice the rated current
tr=10        #interval for running at full load
tb=10        #inteval fordecelaration at twice the rated armature current

#calculations
Es=tst*(2*I)**2*Ra+P
Eb=Es
p1s_tr=(I**2*Ra+P)*tr
p1r=I**2*Ra+P
gamma=(1+B)/2
x=(Es+p1s_tr+Eb)/p1r
y=gamma*tst+tr+gamma*tb
ts=(x-y)/B               #idling interval

fmax=3600/(tst+tr+tb+ts) #maximum frequency of drive operation

#results
#answer in the book is wrong
print"\nmaximum frequency of drive operation: fmax = ",round(fmax,2),"per hour"

maximum frequency of drive operation: fmax =  31.19 per hour