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//Chapter 5:Dc Motor Drives
//Example 19
clc;
//Variable Initialization
//Ratings of the separately excited motor
V=230 // rated voltage in V
N=960 // rated speed in rpm
Ia=200 // rated current in A
Ra=0.02 // armature resistance in ohms
Vs=230 // source voltage in V
//Solution
E=V-Ia*Ra //back emf
//(i) When the speed of motor is 350 rpm with the rated torque during motoring operation
N1=350 //given speed in rpm
E1=N1/N*E //given back emf at N1
Va=E1+Ia*Ra //motor terminal voltage
delta=Va/V //duty ratio
//(ii) When the speed of motor is 350 rpm with the rated torque during braking operation
Va=E1-Ia*Ra //motor terminal voltage
delta1=Va/V //duty ratio
//(iii)Maximum duty ratio is 0.95
delta2=0.95 //maximum duty ratio
Va=delta2*V //terminal voltage
Ia1=2*Ia //maximum permissable current
E1=Va+Ia1*Ra //back emf
N1=E1/E*N //maximum permissible speed
Pa=Va*Ia1 //power fed to the source
//(iv) If the speed of the motor is 1200 rpm and the field of the motor is also controlled
N2=1200 //given speed in rpm
//Now the field current is directly proportional to the speed of the motor
If=N/N2 //field current as a ratio of the rated current
//Results
mprintf("(i) Duty ratio is :%.3f",delta)
mprintf("\n(ii)Duty ratio is :%.2f",delta1)
mprintf("\n(iii)Maximum permissible speed is :%d rpm",N1)
mprintf("\nPower fed to the source is :%.1f kW",Pa/1000)
mprintf("\n(iv)Field current as a ratio of the rated current is :%.1f",If)
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