Chapter 10: Braking: Mechanical Considerations and Control Equipment

Example 10.1, Page 300

In [8]:
#Variable declaration
I=[50,100,150,200,250];#
sp=[73.6,48,41.1,37.3,35.2];
T=[150,525,930,1335,1750];
v=600;#in volts
rm=0.6;#

#Calculations
eb=v-(I[1]*rm);#in volts
rh=3;#in ohms
tr=rh+rm;#in ohms
i=eb/tr;#in amperes
tr=T[2];#

#Result
print "braking torque is %.f (N-m)"%tr

braking torque is 930 (N-m)

Example 10.2, Page 300

In [13]:
import math

#Variable declaration
I=[20,40,60,80];#
emf=[215,381,485,550];#in volts
emf2=[202,357,455,516];#
T=40*9.81;# load torque in N-m
N=600.;#rpm
ia=56;#in amperes from curve
va=440.;#in volts from graph

#Calculations
il=T*(2*math.pi*(N/60));#input to load in W
tr=va/ia;# in ohms
tm=0.8;#in ohms
er=tr-tm;#in ohms

#Result
print "external resistance to be connected across the motor during break is %.3f ohm"%er

external resistance to be connected across the motor during break is 7.057 ohm

Example 10.3, Page 301

In [14]:
#Variable declaration
W=400.;#in tonne
We=1.1*W;# in tonne
S=2.;# distance in km
G=2.;# gradient in %
eta=75./100;# efficiency
D=2.;# distance in km
V1=40.;# in km
V2=20.;# in km
r=40.;#N/tonne

#Calculations&Results
Ea=(0.01072*We*(V1**2-V2**2))*10**-3;# in kWh
Ft=(W*r)-(98.1*W*G);
M=(-Ft*S*1000)/(1000*3600);
Et=Ea+M;# total energy available
Ee=eta*Et;
print "Electrical energy,Ee(kWh) = %.1f"%Ee  #answer varies due to roundinf-off errors
As=(V1+V2)/2;# average speed
At=D/As;# Average time taken
P=(Ee/At);
print "Average power,P(kW) = %.f"%P

Electrical energy,Ee(kWh) = 30.3
Average power,P(kW) = 454

Example 10.4, Page 301

In [16]:
#Variable declaration
W=2340;#in tonne
We=1.1*W;# in tonne
G=100./80;# gradient in %
eta=70./100;# efficiency
V1=60.;# in km
V2=36.;# in km
r=5*9.81;#N/tonne
t=5*60.;# in sec

#Calculations
Ea=(0.01072*We*(V1**2-V2**2))*10**-3;# in kWh
Ft=(W*r)-(98.1*W*G);#tractive effort in N
D=((V1+V2)/2)*(1000./3600)*t;# distance moved in m
M=(-Ft*D)/(1000*3600);
Et=Ea+M;
El=eta*Et;

#Result
print "Energy returned to the line,El(kWh) = %.1f"%El

Energy returned to the line,El(kWh) = 178.4

Example 10.5, Page 302

In [17]:
#Variable declaration
W=500;#in tonne
G=(20.*100.)/1000;# gradient in %
eta=75./100;# efficiency
V=40;# in kmph
r=40;#N/tonne

#Calculations
Ft=(W*r)-(98.1*W*G);#tractive effort in N
P=(-Ft*V)/3600;# Power available in kW
Pf=round(P*eta);

#Result
print "power fed into the line,Pf(kW) = %.f"%Pf

power fed into the line,Pf(kW) = 651

Example 10.6, Page 302

In [19]:
import math

#Variable declaration
OD=640.;# voltage represent by phasor OD
R=0.5;# reactor in ohm
Ia=OD/R;
V=400;# in volts
alfa=38.66;#Phase angle in degree

#Calculations
P=(V*Ia*math.cos(alfa*math.pi/180))*10**-3;

#Result
print "Power generated,P(kW) = %.3f"%P

Power generated,P(kW) = 399.804