# CHAPTER 45 : RATING AND SERVICE CAPACITY¶

## EXAMPLE 45.1 , PAGE NO :- 1796¶

In :
'''An electric motor operates at full-load of 100KW for 10 minutes,at 3/4 full load for the next 10 minutes and at 1/2 load for
next 20 minutes,no-load for the next 20 minutes and this cycle repeats continously.Find the continous rating of the suitable
motor.'''

import math as m
l1 = 100.0        #kW        (load 1)
l2 = 0.75*l1      #kW        (load 2)
l3 = 0.5*l1       #kW        (load 3)

#coresponding time
t1 = 10.0         #minutes
t2 = 10.0         #minutes
t3 = 20.0         #minutes
t4 = 20.0         #minutes

#size of motor required
size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3 + l4*l4*t4)/(t1+t2+t3+t4/3))     #kW
print "Size of motor =",round(size),"KW."

Size of motor = 66.0 KW.


## EXAMPLE 45.2 , PAGE NO :- 1797¶

In :
'''An electric motor has to be selected for a load which rises uniformly from zero to 200KW in 10 minutes after which it remains
constant at 200KW for the next 10 minutes,followed by a no-load period of 15 minutes before the cycle repeats itself.Estimate a
suitable size of continuosly rated motor.'''

import math as m
l1 = 200.0/2        #kW        (load 1)
l2 = 200.0          #kW        (load 2)

#coresponding time
t1 = 10.0         #minutes
t2 = 10.0         #minutes
t3 = 15.0         #minutes

#size of motor required
size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3)/(t1+t2+t3/3))     #kW
print "Size of motor =",round(size),"KW."

Size of motor = 141.0 KW.


## EXAMPLE 45.3 , PAGE NO :- 1797¶

In :
'''A certain motor has to perform the following duty cycle:
(a) 100KW for 10 minutes               (No-load for 5 minutes)
(b) 50KW for 8 minutes                 (No-load for 4 minutes)
The duty cycle is repeated indefinitely.Draw the curve for the load cycle.Assuming that the heating is propotional to the square of
the load,determine suitable size of a continuosly-rated motor.'''

import math as m

l1 = 100.0      #KW      (load 1)
l3 = 50.0       #KW      (load 2)

#Time
t1 = 10.0       #minutes
t2 = 5.0        #minutes
t3 = 8.0        #minutes
t4 = 4.0        #minutes

#Size of the motor is
size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4))      #KW
print "Size of motor = ",round(size,-1),"KW."

Size of motor =  70.0 KW.


## EXAMPLE 45.4 , PAGE NO :- 1799¶

In :
'''A motor has to perform the following duty cycle :-
(i) 100 H.P (10 mins)            (ii) No Load (5 mins)
(iii)60 H.P (8 mins)             (iv) No Load (4 mins)
which is repeated infinitely.Determine the suitable size of continuosly rated motor.'''

import math as m

l1 = 100.0      #H.P      (load 1)
l3 = 60.0       #H.P     (load 2)

#Time
t1 = 10.0       #minutes
t2 = 5.0        #minutes
t3 = 8.0        #minutes
t4 = 4.0        #minutes

#Size of the motor is
size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4))      #H.P
print "Size of motor = ",round(size,2),"H.P."

Size of motor =  69.07 H.P.


## EXAMPLE 45.5 , PAGE nO :- 1800¶

In :
'''A motor working in a coal mine has to exert power starting from zero and rising uniformly to 100 H.P in 5 min
after which it works at a constant rate of 50 H.P for 10 min.Then, a no load period of 3 min.The cycle is repeated
indefinitely,estimate suitable size of motor.'''

import math as m

l1 = 100.0     #H.P    (load 1)
l2 = 50.0      #H.P    (load 2)

#Time
t1 = 5.0          #min    (time 1)
t2 = 10.0         #min    (time 2)
t3 = 3.0          #min    (time 3)

#Using Simpson's one-third rule of Integration
rating = m.sqrt((1.0/3*l1*l1*t1 + l2*l2*t2)/(t1 + t2 + t3) )     #H.P

print "Size of motor =",round(rating,2),"H.P ."

Size of motor = 48.11 H.P .


## EXAMPLE 45.6 , PAGE NO :- 1800¶

In :
'''A motor has following duty cycle
Load rising from 200 to 400 H.P              -              4 min.
Uniform load 300 H.P                         -              2 min.
Regenerative braking (50 H.P to 0)           -              1 min.
Idle                                         -              1 min.
Estimate suitable H.P rating of the motor that can be used.'''

import math as m

l1 = 200.0       #H.P      (load 1)
l2 = 400.0       #H.P      (load 2)
l3 = 300.0       #H.P      (load 3)
l4 = 50.0        #H.P      (load 4)

#Time
t1 = 4.0        #min       (time 1)
t2 = 2.0        #min       (time 2)
t3 = 1.0        #min       (time 3)
t4 = 1.0        #min       (idle time)

rating = m.sqrt((1.0/3*(l1*l1 + l1*l2 + l2*l2)*t1 + l3*l3*t2 + 1.0/3*l4*l4*t3)/(t1 + t2 + t3 + t4))   #H.P

print "Size of motor = ",round(rating,2),"H.P ."
print "Therefore, suitable size of motor is",round(rating,-2),"H.P"

Size of motor =  263.19 H.P .
Therefore, suitable size of motor is 300.0 H.P


## EXAMPLE 45.7 , PAGE NO :- 1802¶

In :
'''The load cycle of a motor for 15 min in driving some equipment is as follows :
0 - 5 min     -     30 H.P
5 - 9 min     -     No load
9 - 12 min    -     45 H.P
12 - 15 min   -     No load
The load cycle is repeated indefinitely.Suggest a suitable size of continuosly rated motor.'''

import math as m

l1 = 30.0       #H.P
l3 = 45.0       #H.P

#Time
t1 = 5.0        #min
t2 = 4.0        #min
t3 = 3.0        #min
t4 = 3.0        #min

#Size of motor is
Size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4))     #H.P
print "Suitable motor size =",round(Size,-1),"H.P ."

Suitable motor size = 30.0 H.P .


## EXAMPLE 45.8 , PAGE NO :- 1802¶

In :
'''A motor driving a colliery winder has the following acceleration period :
load cycle 0-15 sec    :   Load rises uniformly from 0-1000 H.P .
Full speed period      :   15-85 sec. Load constant at 600 H.P .
Decceleration period   :   85-95 sec. Regenerative braking the H.P returned uniformly from 200 to 0 H.P.
95 - 120 sec           :   Motor stationary.
Estimate the size of continuosly rated motor.'''

import math as m

l1 = 1000.0        #H.P         (load 1)
l2 = 600.0         #H.P         (load 2)
l3 = 200.0         #H.P         (load 3)

#Time
t1 = 15.0            #s
t2 = 70.0            #s
t3 = 10.0            #s
t4 = 25.0            #s

#Size of motor is

size = m.sqrt((l1*l1*t1/3 + l2*l2*t2 + l3*l3*t3/3)/(t1+t2+t3+t4))      #H.P

while(round(size)%5!=0):
size = size + 1

print "Size of continously rated motor = ",round(size),"H.P ."

Size of continously rated motor =  505.0 H.P .


## EXAMPLE 45.9 , PAGE NO :- 1807¶

In :
'''A 40KW motor when run continuosly on full load,attains a temperature of 35C , above the surrounding air.Its heating time
constant is 90 min.What would be the 1/2 hour rating of the motor for this temperature rise?Assume that the machine cools down
completely between each load period and that the losses are propotional to square of the load.'''

from sympy import Symbol,Eq,solve
import math as m
# Let 'P' KW be 1/2 hour rating of the motor
# theta1 - final temp rise at P KW
# theta2 - final temp rise at 40 KW
#Losses at P KW is directlt propotional to P^2

theta2 = 35.0      # *C
tau =  1.5         #hr        (time constant)
t = 0.5            #hr        (motor running time)

#Now,  (theta1/theta2) = loss at P KW/loss at 40KW = (P/40)^2
P = Symbol('P')
theta1 = theta2*(P/40)*(P/40)            #*C

#Now, theta2 = theta1*(1 - exp(-t/tau))

theta2a =  theta1*(1-m.exp(-t/tau))      #*C
eq = Eq(theta2,theta2a)
P = solve(eq)
P1 = P                #KW

print "1/2 hour rating = ",round(P1,2),"KW."

 1/2 hour rating =  75.13 KW.


## EXAMPLE 45.10 , PAGE NO :- 1807¶

In :
'''Determine the one-hour rating of a 15 H.P motor having heating time contant of 2 hours.The motor attains the temperature of
40*C on continuos run at full load.Assume that the losses are propotional to square of the load and the motor is allowed to cool
down to the ambient temperature before being loaded again.'''

from sympy import Symbol,Eq,solve
import math as m
# Let 'P' H.P be 1 hour rating of the motor
# theta2 - final temp rise at P H.P
# theta1 - final temp rise at 15 H.P
#Losses at P H.P is directlt propotional to P^2

theta1 = 40.0      # *C
tau =  2.0         #hr        (time constant)
t = 1.0            #hr        (motor running time)

#Now,  (theta2/theta1) = loss at P H.P/loss at 15 H.P = (P/15)^2
P = Symbol('P')
theta2 = theta1*(P/15)*(P/15)            #*C

#Now, theta1 = theta2*(1 - exp(-t/tau))

theta1a =  theta2*(1-m.exp(-t/tau))      #*C
eq = Eq(theta1,theta1a)
P = solve(eq)
P1 = P                #H.P

print "1 hour rating = ",round(P1),"H.P."

1 hour rating =  24.0 H.P.


## EXAMPLE 45.11 , PAGE NO :- 1808¶

In :
'''The heating and cooling time constants of a motor are 1 hour and 2 hours respectively.Final temperature rise attained is
100*C.This motor runs at full load for 30 minutes and then kept idle for 12 min and the cycle is repeated indefinitely.Determine
the temperature rise of motor after one cycle.'''

import math as m

theta2 = 100.0         #*C    (Final temperature rise)
tau_h = 1.0            #hr    (heating time constant)
tau_c = 2.0            #hr    (cooling time constant)
t1 = 30.0/60           #hr    (motor running time)
t2 = 12.0/60           #hr    (motor idle time)

#Heating cycle
theta1 = theta2*(1 - m.exp(-t1/tau_h))

#Cooling cycle
thetac = theta1*m.exp(-t2/tau_c)

print "Temperature rise of motor = ",round(thetac,2),"*C ."

Temperature rise of motor =  35.6 *C .


## EXAMPLE 45.12 , PAGE NO :- 1808¶

In :
'''Calculate the maximum overload that can be carried by a 20KW output motor,if the temperature rise is not to exceed 50*C after
one hour on overload .The temperature rise on full load,after 1 hour is 30*C and after 2 hour is 40*C . Assume losses propotional

from sympy import solve,Symbol,Eq
import math as m

# As theta = thetaf*(1 - exp(-t/T))

theta1 = 30.0      #*C        (temperature rise in time1)
t1 = 1              #hr        (time 1)
theta2 = 40.0      #*C        (temperature rise in time2)
t2 = 2              #hr        (time 2)
#Now   theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))
#Let us assume that x = exp(-1/T).Therefore
x = Symbol('x')
ratio1 = (1 - x**t1)/(1-x**t2)       #(theta1/theta2)
ratio2 = theta1/theta2

#As ratio1 = ratio2
eq = Eq(ratio1,ratio2)
x1 = solve(eq)
x = x1             #variable

#x = exp(-1/T) . Therefore,
T = -1/m.log(x)

#Now, theta1 = thetaf1*(1 - exp(-t1/T))

thetaf1 = theta1/(1-x**t1)           #*C

#Also theta3 = thetaf3*(1 - exp(-t3/T))
theta3 = 50.0        #*C         (max temp)
t3 = 1               #hr         (time 3)
thetaf3 = theta3/(1-x**t3)            #*C

#Given that temp is directly propotional to square of power output i.e   thetaf1/thetaf3 = (Power1/Power3)^2
P1 = 20.0    #KW
P3 = m.sqrt(thetaf3/thetaf1)*P1         #KW

print "Maximum overload that can be carried by motor = ",round(P3,2),"KW."

Maximum overload that can be carried by motor =  25.82 KW.


## EXAMPLE 45.13 , PAGE NO :- 1809¶

In :
'''In a transformer the temperature rise is 25*C after 1 hour and 37.5*C after 2 hours,starting from cold conditions.Calculate
its final steady temperature rise and the heating time constant.If the transformer temerature falls from the final steady state
value to 40*C in 1.5 hours when disconnected,calculate its cooling time constant.Ambient temperature is 30*C.'''

from sympy import solve,Symbol,Eq
import math as m

# As theta = thetaf*(1 - exp(-t/T))

theta1 = 25.0      #*C        (temperature rise in time1)
t1 = 1              #hr        (time 1)
theta2 = 37.5      #*C        (temperature rise in time2)
t2 = 2              #hr        (time 2)
#Now   theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))
#Let us assume that x = exp(-1/T).Therefore
x = Symbol('x')
ratio1 = (1 - x**t1)/(1-x**t2)       #(theta1/theta2)
ratio2 = theta1/theta2

#As ratio1 = ratio2
eq = Eq(ratio1,ratio2)
x1 = solve(eq)
x = x1             #variable

#x = exp(-1/T) . Therefore,
T = -1/m.log(x)

#As  theta1 = thetaf1*(1 - exp(-t1/T))
thetaf1 = theta1/(1-x**t1)              #*C
print "Final steady temperature rise =",round(thetaf1,2),"*C."

#Cooling conditions
theta_rise = 40.0 - 30.0           #*C      (temp rise above ambient conditions)
t3 = 1.5                     #hr      (time taken)

#Now,  theta_rise = thetaf1*exp(-t3/T)
T = -t3/m.log(theta_rise/thetaf1)             #hr
print "Cooling time constant =",round(T,2),"hr"

Final steady temperature rise = 50.0 *C.
Cooling time constant = 0.93 hr


## EXAMPLE 45.14 , PAGE NO :- 1809¶

In :
'''The initial temperature of machine is 20*C.Calculate the temperature of machine after 1.2 hours,if its final steady
temperature rise is 85*C and the heating time constant is 2.4 hours.Ambient temperature is 25*C.'''

import math as m
thetaf = 85.0         #*C      (final temp. rise)
theta1 = 20.0         #*C      (initial temp)
t1 = 1.2              #hr      (time taken)
T = 2.4               #hr      (heat time constant)
#Now, Temperature rise above coling medium is
theta = thetaf - (thetaf - theta1)*m.exp(-t1/T)    #*C

#Therefore, temp. of machine after t1 time is
temp = theta + 25.0

print "Temperature of machine =",round(temp,2),"*C."

Temperature of machine = 70.58 *C.


## EXAMPLE 45.15 , PAGE NO :- 1809¶

In :
'''The following rises were observed in a teperature rise test on a D.C machine at full loads. :-
After  1 hour  -     15*C
After  2 hours -     25*C
Find out    (i) Final steady temperature rise and time constant.
Assume that the final temperature rise on 50% overload is 90*C.'''

from sympy import solve,Symbol,Eq
import math as m

# As theta = thetaf*(1 - exp(-t/T))

theta1 = 15.0      #*C        (temperature rise in time1)
t1 = 1              #hr        (time 1)
theta2 = 25.0      #*C        (temperature rise in time2)
t2 = 2              #hr        (time 2)
#Now   theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))
#Let us assume that x = exp(-1/T).Therefore
x = Symbol('x')
ratio1 = (1 - x**t1)/(1-x**t2)       #(theta1/theta2)
ratio2 = theta1/theta2

#As ratio1 = ratio2
eq = Eq(ratio1,ratio2)
x1 = solve(eq)
x = x1             #variable

#x = exp(-1/T) . Therefore,
T = -1/m.log(x)                     #hr     (time constant)

#As  theta1 = thetaf1*(1 - exp(-t/T))
thetaf1 = theta1/(1-x**t1)          #*C     (Final steady temp. rise)
print "Final steady temperature rise =",round(thetaf1,2),"*C ."
print "Time constant =",round(T,2),"hr."

#(ii) Now at 50% overload .Final temp rise is
thetaf3 = 90.0              #*C
t3 = 1                      #hr     (time taken)
#As ,  theta = thetaf*(1 - exp(-t/T))
theta3 = thetaf3*(1 - m.exp(-t3/T))            #*C

print "The steady temperature rise =",round(theta3,2),"*C ."

Final steady temperature rise = 45.0 *C .
Time constant = 2.47 hr.
The steady temperature rise = 30.0 *C .


## EXAMPLE 45.16 , PAGE NO :- 1813¶

In :
'''The following data refers to a 500 H.P rolling mill,induction motor equipped with a flywheel.
No load speed                      ->   40 rpm
Slip at full load(torque)          ->   12%
Load torque during actual rolling  ->  41500 kg-m
Duration of each rolling period    ->   10 sec.
Determine inertia of flywheel required in the above case to limit motor torque to twice its full load value.Neglect no-load
losses and assume that the rolling mill torque falls to zero between each rolling period.Assume motor slip propotional to

import math as m
N = 40.0                    #rpm           (No load speed)
P = 500.0*(735.5)           #W             (Power)
w = 2*(3.14)*N/60           #rad/sec       (angular speed)
T0 = 0                      #kg-m          (initial torque)
Tl = 41500.0                #kg-m          (Torque load)
t = 10.0                    #sec           (time taken)
s = 0.12                    #              (slip)
g = 9.81                    #m/s^2
Tfull = P/(w*(1-s))         #N-m           (full load torque)
Tfull = Tfull/g             #kg-m
Tm = 2*Tfull                #kg-m          (Max torque)
S = 2*3.14*(0.12*40)/60
#Now,   S = K*Tfl
K = S/Tfull                   #constant
#Also,  Tm = Tl - (Tl-T0)*exp(-tg/IK) .Therefore I is
I =(-t*g)/(K*m.log((Tl-Tm)/(Tl-T0)))       #kg-m^2

print "Moment of Inertia = %e kg-m^2" %round(I,2)

Moment of Inertia = 2.947543e+06 kg-m^2


## EXAMPLE 45.17 , PAGE NO :- 1814¶

In :
'''A 6 pole,50 Hz Induction Motor has a flywheel of 1200 kg-m^2 as moment of inertia.Load torque is 100 kg-m for 10 sec.No load
period is long enough for the flywheel,to regain its full speed.Motor has a slip of 6% at a torque of 50 kg-m.Calculate
(i)Maximum torque exerted by motor
(ii)Speed at the end of deacceleration period.'''

import math as m

Tl = 100.0      #kg-m          (load torque)
t = 10.0        #s             (time taken)
g = 9.81        #m/s^2         (gravitational acceleration)
I = 1200.0      #kg-m^2        (moment of inertia)
p = 6           #              (poles)
f = 50.0        #Hz            (frequency)
s = 0.06        #              (slip)
Tfull = 50.0    #kg-m          (full load torque)
Ns = 120*f/p
Nr = (1-s)*Ns

#Now, S = K*T
S = 2*3.14*(Ns - Nr)/60    #rad/sec
K = S/Tfull         #constant

#As Tm = Tl*(1-exp(-t*g/I*K))
Tm = Tl*(1 - m.exp(-t*g/(I*K)))      #kg-m
print "Tm = ",round(Tm,2),"kg-m."

#(ii)Slip speed
S = K*Tm            #rad/sec      (slip speed)
S = S*(60/(2*3.14))   #rpm
N = Ns - S          #rpm           (actual speed)
print "Actual Speed =",round(N,2),"rpm."

Tm =  47.84 kg-m.
Actual Speed = 942.59 rpm.


## EXAMPLE 45.18 , PAGE NO :- 1815¶

In :
'''An Induction Motor equipped with a flywheel is driving a rolling mill which requires a load torque of 1900 N-m for 10 sec
followed by 250 N-m for 30 sec.This cycle being repeated indefinitely.The synchronus speed of motor is 750 rpm and it has slip of
10% when delivering 1400 N-m torque.The total Moment of Inertia of the flywheel and other rotating parts is 2100 kg-m^2.Draw the
curves showing the torque exerted by the motor and the speed for five complete cycles,assuming the initial torque is zero.'''

import math as m

Tl1 = 1900.0      #N-m         (load torque 1)
t1 = 10.0         #s           (time 1)
Tl2 = 280.0       #N-m         (load torque 2)
t2 = 30.0         #s           (time 2)
s = 0.1           #            (slip)
Ns = 750.0        #rpm         (synchronus speed)
I = 2100.0        #kg-m^2      (moment of inertia)
Tm = 1400.0       #N-m
S = Ns*s          #rpm         (slip speed)

K = S/Tm          #constant
T0 = 0            #N-m
#(i) During  First Cycle
Tm = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K))       #N-m
s2 = K*Tm*(60/(2*3.14))    #rpm
Speed1 = Ns - s2   #rpm
print "Speed 1 = ",round(Speed1,2),"rpm."

# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))
Tmb = Tm    #N-m
T0 = Tl2    #N-m         (No Load Torque)

Tm = T0 + (Tmb - T0)*m.exp(-t2/(I*K))
S2 = K*Tm*(60/(2*3.14))    #rpm
Speed2 = Ns - S2     #rpm
print "Speed 2 = ",round(Speed2,2),"rpm."
#################################################################

#(ii) During Second cycle
T0 = Tm
Tm2 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K))       #N-m
s2 = K*Tm2*(60/(2*3.14))    #rpm
Speed1 = Ns - s2   #rpm
print "Speed 1 = ",round(Speed1,2),"rpm."

# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))
Tm2b = Tm2    #N-m
T0 = Tl2    #N-m         (No Load Torque)

Tm = T0 + (Tm2b - T0)*m.exp(-t2/(I*K))
S2 = K*Tm*(60/(2*3.14))    #rpm
Speed2 = Ns - S2     #rpm
print "Speed 2 = ",round(Speed2,2),"rpm."
###################################################################

#(iii) During Third cycle
T0 = Tm
Tm3 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K))       #N-m
s2 = K*Tm3*(60/(2*3.14))    #rpm
Speed1 = Ns - s2   #rpm
print "Speed 1 = ",round(Speed1,2),"rpm."

# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))
Tm3b = Tm3    #N-m
T0 = Tl2    #N-m         (No Load Torque)

Tm = T0 + (Tm3b - T0)*m.exp(-t2/(I*K))
S2 = K*Tm*(60/(2*3.14))    #rpm
Speed2 = Ns - S2     #rpm
print "Speed 2 = ",round(Speed2,2),"rpm."
####################################################################

Speed 1 =  691.75 rpm.
Speed 2 =  731.62 rpm.
Speed 1 =  683.89 rpm.
Speed 2 =  731.0 rpm.
Speed 1 =  683.62 rpm.
Speed 2 =  730.98 rpm.


## EXAMPLE 45.19 , PAGE NO :- 1817¶

In :
'''A motor fitted with a flywheel supplies a load torque of 150 kg-m for 15 sec.During the no-load period,the flywheel regains
its original speed.The motor torque is required to be limited to 85 kg-m.Determine moment of inertia of flywheel.
The no-load speed of motor is 500 rpm and it has slip of 10% on full load.'''

from sympy import Symbol,solve,Eq,exp
import math as m

Tm = 85.0        #kg-m     (Max torque)
Tl = 150.0       #kg-m     (load torque with flywheel)
T0 = 0           #kg-m     (constant load torque)
t = 15.0         #s        (time)
N = 500.0        #rpm      (no load speed)
s = 0.1          #         (slip)
g = 9.82        #m/s^2
# s = K*T
K = 2*(3.14)*N*s/(60*Tm)      #constant

# As Tm = Tl*(1 - exp(-t*g/(I*K)))

I =(-t*g)/(K*m.log(1 - Tm/Tl))      #kg-m^2           (Moment of inertia)

print "Moment of inertia =",round(I,2),"kg-m^2."

Moment of inertia = 2860.94 kg-m^2.


## EXAMPLE 45.20 , PAGE NO :- 1817¶

In :
'''A 3-phase ,50 KW,6 pole,960 rpm induction motor has a constant load torque of 300 N-m and at wide intervals additional
torque of 1500 N-m for 10 sec.Calculate
(a)The moment of inertia of the flywheel used for load equalization,if the motor torque is not to exceed twice the rated torque.
(b)Time taken after removal of additional load,before the motor torque becomes 700 N-m.'''

from sympy import Symbol,Eq,solve
import math as m

P = 50.0e+3      #W          (output power)
Nr = 960.0       #rpm        (rotational speed)
p = 6.0          #           (no. of poles)
t = 10.0         #s          (time)
T0 = 300.0       #N-m        (constant load torque)
Tl = T0 + 1500.0 #N-m        (total load torque)
f = 50.0         #Hz         (frequency)

#   Power = T*w (torque*ang_speed)
T = P/(2*3.14*Nr/60)     #N-m        (Full-load torque)
Tm = 2*T                 #N-m        (Max torque)

Ns = 120*f/p      #rpm        (synchronus speed)

#Slip speed
sl = Ns-Nr        #rpm

#Now,  s = K*T
K = 2*3.14*sl/(60*T)      #constant

#As Tm = Tl - (Tl - T0)*exp(-t/I*K)

I = (-t)/(K*m.log((Tl - Tm)/(Tl - T0)))         #kg-m^2     (moment of inertia)

print "Moment of inertia of flywheel =",round(I,2),"kg-m^2."

#(b) Tm2 = T0 + (Tm-T0)*exp(-t/I*K)
Tm2 = 700.0            #N-m                         (Max torque - case 2)

t1= (-I*K)*m.log((Tm2 - T0)/(Tm - T0))         #s               (time after removal of load)

 Moment of inertia of flywheel = 1908.86 kg-m^2.


## EXAMPLE 45.21 , PAGE NO :- 1818¶

In :
'''A 3-phase,8 pole,50 cps.Induction Motor equipped with a flywheel supplies a constant load torque of 100 N-m and at wide
intervals an additional load torque of 300 N-m for 6 sec.The motor runs at 735 rpm at 100 N-m torque.Find moment of inertia of
the flywheel,if the motor torque is not to exceed 250 N-m.'''

from sympy import Symbol,Eq,solve
import math as m

T0 = 100.0           #N-m   (constant load torque)
Tl = T0 + 300.0      #N-m   (Total load torque)
f = 50.0             #Hz    (frequency)
P = 8.0              #      (poles)
Tm = 250.0           #N-m   (Max torque)
Ns = 120*f/P         #rpm   (Synchronus speed)
sl =  Ns - 735.0     #rpm   (Slip speed)
t = 6.0              #s     (time)
#Now, s = K*T0
K = 2*3.14*sl/(60*T0)    #constant

#Also,   Tm = Tl - (Tl-T0)*exp(-t/I*K)

I = -t/(K*m.log((Tl - Tm)/(Tl-T0)))           #kg-m^2       (moment of inertia)

print "Moment of Inertia =",round(I,2),"kg-m^2."

Moment of Inertia = 551.35 kg-m^2.


## EXAMPLE 45.22 , PAGE NO :- 1818¶

In :
'''A 6 pole,50 Hz,3-phase wound rotor Induction Motor has a flywheel coupled to its shaft.The total moment of inertia is
1000kg-m^2.Load torque is 1000 N-m for 10 sec followed by a no-load period which is long enough for the motor to reach its
no-load speed.Motor has a slip of 5% at a torque of 500 N-m.Find
(a)Maximum torque developed by motor
(b)Speed at the end of deacceleration period.'''

import math as m

P = 6.0                #           (No of poles)
I = 1000.0             #kg-m^2     (Moment of Inertia)
Tl = 1000.0            #N-m        (Load torque with  flywheel)
t = 10.0               #s          (time)
s = 0.05               #           (slip)
Tfl = 500.0            #N-m           (full load Torque)
f = 50.0               #Hz         (frequency)

Ns = 120*f/P           #rpm        (Synchronus speed)

#Now, s = K*Tfl
K = 2*3.14*(Ns*s)/(60*Tfl)        #constant

#K = 6.2e-3          #(considered value)

#Also Tm = Tl*(1-exp(-t/I*K)
Tm = Tl*(1 - m.exp(-t/(I*K)))              #N-m
print "Maximum torque developed by motor = ",round(Tm,2),"N-m."

#(b)    s = K*Tfl  where s = 2*3.14*(Ns - N)/60
N = Ns - (60/(2*3.14))*K*Tm            #rpm
print "Speed at the end of deacceleration period =",round(N,2),"rpm."

Maximum torque developed by motor =  615.35 N-m.
Speed at the end of deacceleration period = 938.47 rpm.


## EXAMPLE 45.23 , PAGE NO :- 1819¶

In :
'''A motor fitted with a flywheel supplies a load torque of 1000 N-m for 2 seconds.During no load period,the flywheel regains its
original speed.The motor torque is to be limited to 500 N-m.Find moment of inertia of the flywheel.No load speed of the motor is
500 rpm and its full load slip is 10%.'''

from  sympy import solve,Eq,Symbol
import math as m
N = 500.0             #rpm       (No load speed)
s = 0.1               #          (slip)
Tfl = 500.0           #N-m       (full load torque)
Tl = 1000.0           #N-m       (load torque with flywheel)
t = 2.0               #s         (time)
#Now, s = K*Tfl
K = (2*3.14*(N*s))/(Tfl*60)      #constant

#Also,  Tm = Tl*(1 - exp(-t/I*K))
I =-t/(K*m.log(1 - Tfl/Tl))       #(moment of inertia)

print "Moment of inertia of flywheel =",round(I,2),"kg-m^2."

Moment of inertia of flywheel = 275.67 kg-m^2.

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