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//Chapter 5:Dc Motor Drives
//Example 16
clc;
//Variable Initialization
//Ratings of the separately excited motor
V=220 // rated voltage in v
N=960 // rated speed in rpm
Ia=12.8 // rated current in A
Ra=2 // armature resistance in ohms
Vs=230 // source voltage in v
f=50 //frequency of the source voltage in hz
La=150e-3 // armature curcuit inductance in H
//Solution
E=V-Ia*Ra //back emf
Vm=sqrt(2)*Vs //peak voltage
Wm=2*%pi*N/60 //angular speed
W=2*%pi*f
//(i)When speed is 600rpm and the firing angle is 60 degrees
alpha=60 //firing angle in degrees
N1=600 //motor speed in rpm
x=W*La/Ra
phi=atan(x)
cot_phi=1/tan(phi)
Z=sqrt(Ra**2+(W*La)**2)
K=E/Wm
y=Ra*Vm/Z/K
alpha=alpha*%pi/180
b=sin(phi)*exp(-(alpha*cot_phi))
c=sin(alpha-phi)*exp(-(%pi*cot_phi))
a=1-exp(-(%pi*cot_phi))
Wmc=y*(b-c)/a //required angular speed in rps
Nmc=Wmc*60/2/%pi //required angular speed in rpm
Va=Vm/%pi*(1+cos(alpha))
E1=N1/N*E //value of back emf at the speed of N1
Ia=(Va-E1)/Ra
T=K*Ia
//(ii)When the torque is 20 N-m and firing angle is 60 degrees
T1=20 //required torque in N-m
alpha=60 //required firing angle in degrees
Ec=Nmc/N*E //motor back emf at critical speed of Nmc
Tc=K*(Va-Ec)/Ra //torque at the critical speed
Ia=T1/K
E1=Va-Ia*Ra
N1=E1/E*N //required speed
//Results
//if N1<Nmc then
mprintf("(i)The motor is operating under continuous condition")
mprintf("\nHence the required torque is :%.2f N-m",T)
//end
//if Tc<T1 then
mprintf("\n(ii)The motor is operating under continuous condition")
mprintf("\nHence the required speed is :%.1f rpm",N1)
//end
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