Working Principle: A variable resistance is connected in series with armature and the field winding is connected directly with separate source. The parameters of DC separately exited motors are:- Parameters values R 0.642 L L 0.012H RE 24062 LE 120H LZ 1.8H J 1 Kg m B. 0 Nms 1 0 Nm Initial speed 1 rad sec Rated current 11.78A Rated Power 5 HP DC Separately Exited Motor without Starter Initially, the back emf produced at the armature is zero. After full voltage is supplied across the armature, a very high current is obtained (la= 400A) since the value of armature resistance is low. The fuses will blow out due to this excess current and the brushes and commutator will also get damaged. To avoid damaging effect of high initial current, we will insert resistances in series with the armature (for certain time interval) which brings the initial current to safe value. The starter resistances are gradually shorted by the circuit breaker. The motor develops speed and back emf till it allains its rated values. DC Separately Exited Motor with Starter Initially, now starting armature current la = V/RI; which comes out to be very less Let the values of each resistance of the starter be Ra. Rb and Rc ohms. Therefore RI=Ra +Rb +Rc+RA R2 - Rb +RC+RA R3 = R; +RA only solve the report Q1 Q2 Q3 The values of Ra, Rb and Rc can be found out from the relation: Il max/12 min=R1/R2=R2/R3=R3 RA=K, Suppose that K-2.33 R3 RA-K-1.34,0.62.33, R3-2.33 0.6=1.34 22 R2 R3=K, R2-32 RI/R2- KRI-6,672 Hence we get the values of RI=6.67 ohm, R2-3 ohm, R3= 1.34 ohm. We can find the steps Ra. Rb and Rc Rc-0.74 12. Rb=1.66 22 and Ra=3.66 12 Hence initial current comes out to be Imax=V/RI=240/3.66+1.66+0.74+0,6)=36.6A Procedures: 1. Open File -> New---> Model 2- Open Simulink Library and browse the components. 3- Select DC Machine ratings 5hp, 240v, 16.2A, and 1220 rpm. 4- Connect the components as shown in fig. (1). 5- Plot the armature current, armature voltage, speed and the torque. Figure (1) Starting a DC Motor 00_ Bw.A12 . 1 1 * - Man 11 . 21 URRIA Corra solve Report: 1. What is the benefit of starter? 2. If k-1.5 find the step resistance. 3. Discuses plots of the armature current, armature voltage, speed and the torque. Working Principle: A variable resistance is connected in series with armature and the field winding is connected directly with separate source. The parameters of DC separately exited motors are:- Parameters values R 0.642 L L 0.012H RE 24062 LE 120H LZ 1.8H J 1 Kg m B. 0 Nms 1 0 Nm Initial speed 1 rad sec Rated current 11.78A Rated Power 5 HP DC Separately Exited Motor without Starter Initially, the back emf produced at the armature is zero. After full voltage is supplied across the armature, a very high current is obtained (la= 400A) since the value of armature resistance is low. The fuses will blow out due to this excess current and the brushes and commutator will also get damaged. To avoid damaging effect of high initial current, we will insert resistances in series with the armature (for certain time interval) which brings the initial current to safe value. The starter resistances are gradually shorted by the circuit breaker. The motor develops speed and back emf till it allains its rated values. DC Separately Exited Motor with Starter Initially, now starting armature current la = V/RI; which comes out to be very less Let the values of each resistance of the starter be Ra. Rb and Rc ohms. Therefore RI=Ra +Rb +Rc+RA R2 - Rb +RC+RA R3 = R; +RA only solve the report Q1 Q2 Q3 The values of Ra, Rb and Rc can be found out from the relation: Il max/12 min=R1/R2=R2/R3=R3 RA=K, Suppose that K-2.33 R3 RA-K-1.34,0.62.33, R3-2.33 0.6=1.34 22 R2 R3=K, R2-32 RI/R2- KRI-6,672 Hence we get the values of RI=6.67 ohm, R2-3 ohm, R3= 1.34 ohm. We can find the steps Ra. Rb and Rc Rc-0.74 12. Rb=1.66 22 and Ra=3.66 12 Hence initial current comes out to be Imax=V/RI=240/3.66+1.66+0.74+0,6)=36.6A Procedures: 1. Open File -> New---> Model 2- Open Simulink Library and browse the components. 3- Select DC Machine ratings 5hp, 240v, 16.2A, and 1220 rpm. 4- Connect the components as shown in fig. (1). 5- Plot the armature current, armature voltage, speed and the torque. Figure (1) Starting a DC Motor 00_ Bw.A12 . 1 1 * - Man 11 . 21 URRIA Corra solve Report: 1. What is the benefit of starter? 2. If k-1.5 find the step resistance. 3. Discuses plots of the armature current, armature voltage, speed and the torque