# CHAPTER 6 - Electric Braking¶

## EXAMPLE 6.1 - PG NO.114¶

In [1]:
#Example 6.1, page 114
import math
T=172.#in N-m
w=(2.*math.pi*960.)/60.
E=215.#in V
Ia=(T*w)/E
Ra=.062#in ohm
v=220.#in v
Eg=v+(Ia*Ra)
N=960.#in Rpm
S=(N*round(Eg))/E
print'%s %.2f %s' %("The speed of shunt machine is",S,"rpm ")

The speed of shunt machine is 1004.65 rpm


## EXAMPLE 6.2 - PG NO.114¶

In [2]:
#Example 6.2, page 114
import math
N=480.#in rpm
T=318.3#in N-m
P=(2.*math.pi*N*T)/60.
#From graph
E=333.3#in V
Ia=48.#Amp
R=E/Ia
#disp(R)
print'%s %.2f %s' %("The total resistance of circuit is =",R,"ohm")

The total resistance of circuit is = 6.94 ohm


## EXAMPLE 6.3 - PG NO.116¶

In [3]:
#Example 6.3, Page 116
import math
per=.88
v=220.#in v
p=20.#in kw
I=(p*1000.)/(per*v)
T=(p*1000.*60.)/(2*math.pi*1200.)
#Part a
E_motor=v-(I*.1)
v_arm=v+E_motor
Ir=2.*I#Rated current
R=v_arm/Ir
print'%s %.2f %s' %('resistance at armature current=',R,'ohm')
R_Extra=R-.1
print'%s %.2f %s' %("\n Extra resistance added to motor armature is =",R_Extra," ohm ")
#Part b
#Part c
E=(E_motor*400.)/1200.
print'%s %.2f %s' %('E=',E,'v')
I_braking=(v+E)/R
T_braking=(T/103.3)*I_braking
print'%s %.3f %s' %("\n Braking  current is =",I_braking," A")
print'%s %.3f %s' %("\n Braking Torque is =",T_braking," N-m ")

resistance at armature current= 2.08 ohm

Extra resistance added to motor armature is = 1.98  ohm

Full load torque is 318.31  N-m

E= 69.89 v

Braking  current is = 139.397  A

Braking Torque is = 214.770  N-m


## EXAMPLE 6.4 - PG NO.124¶

In [3]:
#Example 6.4, page 124
import math
R1=.15#in ohm
Rs=.45#in ohm
x1=.6#in ohm
xz=1.8#in ohm
sf=0.05
Turn=1./math.sqrt(3.)
R_rotor=Rs*Turn#in ohm
X_rotor=xz*Turn**2.#in ohm

#Part 1
#BY FIGURE
E1=math.sqrt((3.**2.+.6**2)/(3.15**2+1.2**2))*440./math.sqrt(3.)
s=1.-sf
I2=E1/math.sqrt(x1**2+2**2)
R2=2.*60.#ohm
w=2.*math.pi*600.
T=(R2*3.*(I2**2.))/(s*w)
print'%s %.3f %s' %("Initial braking torque of rheostatic is =",T,"N-m")

#Part 2
s1=2.-sf
a=.15+(1.9/1.95)**2.
b=1.2**2.
I2=(440./math.sqrt(3.))*(1./math.sqrt(a+b))
T=(60.*1.9*3.*(I2**2.))/(1.95*w)
print('\n')
print'%s %.3f %s' %("Initial braking torque during reverse is =",T," N-m")
#the answers in textbook are less accurate due to approximations.

Initial braking torque of rheostatic is = 1225.596 N-m

Initial braking torque during reverse is = 1182.272  N-m


## EXAMPLE 6.5 - PG NO.128¶

In [4]:
#Example 6.5, page no 128
import math
import cmath
Kva=3000./3.#kva per phase
v=2300./math.sqrt(3.)#voltage per phase
#disp(v)
i=(1000.*1000.)/1330.#current per phase
#disp(i)
s=i*.2
x=math.sqrt((v+s)**2.+(s**2.))
temp=((x/i)**2.)-2.**2.#temp=(.2+R)**2
#print(temp)
t=cmath.sqrt(temp)
temp1=t-(.2)
#disp(temp1)
#Answer difference is because of round off value of x
r=1.97#in ohms
T_br=(3.*i*i*r*60.)/(2*math.pi*200.)
print'%s %.2f %s' %("Initial braking torque is =",T_br, "N-m")
#the answers in textbook are less accurate due to approximations.

Initial braking torque is = 159523.84 N-m


## EXAMPLE 6.6 - PG NO.131¶

In [5]:
#Example 6.6, Page 131
import math
pf=.85
N1=1500.#in rpm
N=1440.#in rpm
P=pf*230.*10.*math.sqrt(3.)
p_stator_loss=86.16#in w
p_rotor=P-p_stator_loss
rotor_copper_loss=((N1-N)/N1)*p_rotor
print'%s %.f %s' %("The rotor copper loss is =",round(rotor_copper_loss),"watt")
inertia=.0486#in kg-m2
E=2*.96*inertia*((2*math.pi*50)/2)**2
print'%s %.f %s' %('total energy dissipiated in rotar during starting and braking',E,'J')
stops_starts=7920/E
print'%s %.2f %s' %("Total number of starts and stops is =",stops_starts,"say 3")

The rotor copper loss is = 132 watt
total energy dissipiated in rotar during starting and braking 2302 J
Total number of starts and stops is = 3.44 say 3


## EXAMPLE 6.7 - PG NO.131¶

In [2]:
#Example 6.7, page 131
import math
v1=400/math.sqrt(3)#in v
ws=(2*math.pi*1000)/60#angular f
x=1#resistance in ohm
T_max=(3*v1**2)/(ws*2*x)
#print(T_max)
j=10#in kg-m2
s1=.05
s_maxT=0.2
a=(1.95**2-1)/(2*s_maxT)
temp=a+(.2*math.log(1.95))
r=((10*ws)/(2*T_max))*(temp)
#print(r)
Extra_R=r-(.2)
a=(1.95**2-1)/(2*1.45)
temp=a+(1.45*math.log(1.95))
t=((10*ws)/(2*T_max))*(temp)
#print(t)
print'%s %.3f %s' %("Minimum time to bring rotor to rest is= ",t,"sec")

Minimum time to bring rotor to rest is=  1.326 sec


## EXAMPLE 6.8 - PG NO.134¶

In [7]:
#Example 6.8, page 134
import math
#part a
k=6000./w
kw=6000.#n-m, initial brakin torque
Tf=300.#n-m, fictional torque
j=540.#kg-m2
tr=(j/k)*math.log((kw+Tf)/Tf)
#disp(tr)
s=math.e**((-k*tr)/j)
#disp(s)
temp=((j/k)*(kw+Tf)*(1.-s))-((Tf*tr))
Nr=(1./(2.*math.pi*k))*temp
#disp(Nr)
print'%s %.2f %s' %("Time taken for rheostatic braking is =",Nr,"s")
#part b
beta=3600./j
motor_rest_time=w/6.67
#disp(motor_rest_time)
rev=(1000./60.)*.5*(motor_rest_time)
print'%s %.1f' %("Number of revolutions made is =",rev)
#the answers in textbook are less accurate due to approximations.

Time taken for rheostatic braking is = 133.17 s
Number of revolutions made is = 130.8