1. Complete Equivalent Circuit for the Shunt Motor
Racir = resistance of the armature circuit
Interpoles and Compensating Windings
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2. General Speed Equation For a DC Motor
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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 Ω and 120-Ω. Determine the new armature current.
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4. Solution
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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
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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
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8. At rated conditions
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9. At the new conditions
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11. Determine the new steady-state speed
Check page 421 for = sign
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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
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13. Armature current decreases Armature current increases
For speed reduction
Armature current decreases
Armature current increases
Torque decreases
Torque recovers (increases)
Speed decreases
Speed settles to new value
Counter-emf decreases
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14. Decreasing field current reduces the flux
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
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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
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16. Power-flow Diagram – DC Motor
Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray
Pb = VbIa
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17. Power-flow Diagram – DC Generator
Ploss = Pacir + Pb + Pcore + Pfcl + Pf,w + Pstray
Pb = VbIa
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18. Starting a DC Motor
At “locked-rotor”, or “blocked-rotor”
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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)
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20. Example 10.12 Motor 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 = Ω and Rf = 137 Ω.
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21. Example (continued)
Determine
(a) the rated torque
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22. Example (continued)
(b) the armature current at locked-rotor if no starting resistance is used
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23. Example (continued)
(c) the external resistance required in the armature circuit that would limit the current and develop 200% rated torque when starting
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26. Example (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
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27. Example (continued)
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