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ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical.

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Presentation on theme: "ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical."— Presentation transcript:

1 ELEC 3105 Basic EM and Power Engineering Rotating DC Motor PART 2 Electrical

2 Motor / Generator Action Equivalent circuit Expression of V emf Loop Slide extracted from linear motor and modified for loop motor.

3 Motor / Generator Action Linear relation between speed and torque Current flows in a direction to charge the battery. Motor Slide extracted from linear motor and modified for loop motor. Stall torque Generator Link

4 Electrical Equivalent  E = motor voltage R a = armature resistance L a = armature inductance V = Applied motor voltage I a = armature current  = magnetic flux R f = field resistance L f = field inductance V f = Field voltage I f = field current

5 Electrical Equivalent  In steady state operation: FIELD SIDE reluctance Magnetic circuit

6 Electrical Equivalent  In steady state operation: ARMATURE SIDE Motor constant Back emf

7 Motor / Generator Action Equivalent circuit Expression of V emf Loop Slide extracted from linear motor and modified for loop motor.

8 Electrical Equivalent  In steady state operation: ARMATURE SIDE Motor constant Back emf

9 Electrical Equivalent  In steady state operation: ARMATURE SIDE Motor constant Developed torque Same motor constant in emf and developed torque

10 Motor / Generator Action Equivalent circuit Expression of V emf Loop Slide extracted from linear motor and modified for loop motor.

11 Electrical Equivalent  In steady state operation: ARMATURE SIDE Motor constant Developed torque Same motor constant in emf and developed torque Motor constant Back emf

12 Electrical Equivalent  Power flow: ARMATURE SIDE; Conservation of energy KVL POWER Power in Armature copper loss Power developed

13 Motor / Generator Action Equivalent circuit Expression of V emf Loop Slide extracted from linear motor and modified for loop motor.

14 Electrical Equivalent  Power flow: ARMATURE SIDE; Conservation of energy Power developed Electrical Mechanical copper

15 Electrical Equivalent  Power flow: ARMATURE SIDE Electrical Mechanical Rotational loss Copper

16 Electrical Equivalent  Motor sequence Speed of rotation limiting loop

17 Shunt Connected Field  R Developed torque Rotation rate

18 Shunt Connected Field  R Similar type of graph

19 Shunt Connected Field  R MotorGenerator Force motor to spin backwards Generator Force motor to spin to fast

20 Series Connected Field  Universal motor design: works for D.C. and for A.C. Since

21 Series Connected Field  Universal motor design: works for D.C. and for A.C.

22 Maximum Power Transfer  Power developed in the motor Find maximum with respect to the motor voltage

23 Maximum Power Transfer  For extremes of a function, take derivatives and set to zero

24 Calculation example  A 120 volt dc motor has an armature resistance of 0.70 Ω. At no- load, it requires 1.1 A armature current and runs at 1000 rpm. Find the output power and torque at 952 rpm output speed. Assume constant flux. Solution provided in class

25 Calculation example  A permanent magnet dc motor has the following information: 50 hp, 200 V, 200 A, 1200 rpm and armature resistance of 0.05 Ω. Determine the output power if the voltage is lowered to 150 V and the current is 200 A. Assume rotational losses are proportional to speed. Determine the rotational loss, armature resistance, no-load rpm, machine constant, efficiency? Solution provided in class

26 Calculation example  An 80 V dc motor has constant field flux, separately excited, and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 A and the no-load current is 0.5 A. Assume constant rotational losses. Solution provided in class


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