Complete Equivalent Circuit for the Shunt Motor

1. Complete Equivalent Circuit for the Shunt Motor Racir = resistance of the armature circuit Interpoles and Compensating Windings ECE 441

2. General Speed Equation For a DC Motor ECE 441

3. Example 10.8 • A 25-hp, 240-V shunt motor operating at 850 r/min draws a line current of 91 A when operating at rated conditions. A 2.14-Ω resistor inserted in series with the armature causes the speed to drop to 634 r/min. The respective armature-circuit resistance and field-circuit resistance are 0.221-Ω and 120-Ω. Determine the new armature current. ECE 441

4. Solution ECE 441

5. ECE 441

6. Example 10.9 • A shunt motor rated at 10hp, 240-V, 2500 r/min, draws 37.5 A when operating at rated conditions. • Ra=0.213Ω, RCW=0.065Ω, RIP=0.092Ω, Rf=160Ω ECE 441

7. Example 10.9 (continued) • Determine the steady-state armature current if a rheostat in the shunt field reduces the flux in the air gap to 75% of its rated value, a 1.0-Ω resistor is placed in series with the armature, and the load torque on the shaft is reduced to 50% rated ECE 441

8. At rated conditions ECE 441

9. At the new conditions ECE 441

10. ECE 441

11. Determine the new steady-state speed Check page 421 for = sign ECE 441

12. Speed Control of DC Motors • Armature Control • Insert a resistor or rheostat in series with the armature • Reduce speed below the base speed • Shunt Field Control • Insert a resistor or rheostat in series with the shunt field • Increase speed above the base speed ECE 441

13. For speed reduction Armature current decreases Armature current increases Torque decreases Torque recovers (increases) Speed decreases Speed settles to new value Counter-emf decreases ECE 441

14. For speed increase Decreasing field current reduces the flux Reduction in flux causes cemf to decrease Motor accelerates, cemf increases Armature current decreases Armature current increases Torque increases, machine accelerates Torque settles down Motor runs at new speed Speed increases ECE 441

15. Mechanical Power and Developed Torque Pmech = Total Power Input to the Armature – Copper Losses in the Armature Pmech = VTIa – Ia2Racir Racir = Ra + RIP + RCW Pmech = EaIa ECE 441

16. Power-flow Diagram – DC Motor Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray Pb = VbIa ECE 441

17. Power-flow Diagram – DC Generator Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray Pb = VbIa ECE 441

18. Starting a DC Motor • At “locked-rotor”, or “blocked-rotor” ECE 441

19. Manually-Operated DC Motor Starter “Start” with all of the rheostat resistance in the circuit All resistance is cut out when motor reaches full-speed “Off” position “Run” position “Holds” the lever in “Run” position (A “break” in the field circuit de-energizes the coil, shutting the motor down) ECE 441

20. Example 10.12Motor Starting • A 15-hp, 230-V, 1750 r/min shunt motor with a compensating winding draws 56.2A when operating at rated conditions. The motor parameters are Racir = 0.280 Ω and Rf = 137 Ω. ECE 441

21. Example 10.12 (continued) • Determine • (a) the rated torque ECE 441

22. Example 10.12 (continued) • (b) the armature current at locked-rotor if no starting resistance is used ECE 441

23. Example 10.12 (continued) • (c) the external resistance required in the armature circuit that would limit the current and develop 200% rated torque when starting ECE 441

24. ECE 441

25. ECE 441

26. Example 10.12 (continued) • (d) Assuming that the system voltage drops to 215 V, determine the locked-rotor torque using the external resistor in (c). • Assume that the flux density is proportional to the field current ECE 441

27. Example 10.12 (continued) ECE 441