LEC12

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Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP Page 1 of 6 X-Tics & Speed Control of DC Series Motor Lecture Objectives X-Tics & Speed Control of Compound Motor Dynamic & Plugging Electrical Braking Numerical Problems Applying KVL: V DC Series Motor T = E A + I A (R A + R S Connected in series with the armature. ) Field Windings Field & armature carry the same current. As the high current has to pass through filed, so the cross section of the field must be greater than that of the field winding of shunt motor Small number of turns of thick wire with low resistance. . Applications: Electric traction, Cranes, Elevators, Air-compressors, Vacuum cleaners, Hair dryers, Sewing machines, etc. K is machine constant that depends on poles, number of conductors and parallel paths. As flux in this machine is directly proportional to its armature current (at least until metal saturates), therefore ф = KcI Induced Torque A where c is a constant of proportionality.

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Transcript of LEC12

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 1 of 6

    X-Tics & Speed Control of DC Series Motor Lecture Objectives

    X-Tics & Speed Control of Compound Motor Dynamic & Plugging Electrical Braking Numerical Problems

    Applying KVL: V

    DC Series Motor

    T = EA + IA (RA + RS

    Connected in series with the armature.

    ) Field Windings

    Field & armature carry the same current. As the high current has to pass through filed, so

    the cross section of the field must be greater than that of the field winding of shunt motor

    Small number of turns of thick wire with low resistance.

    .

    Applications: Electric traction, Cranes, Elevators, Air-compressors, Vacuum cleaners, Hair dryers, Sewing machines, etc.

    K is machine constant that depends on poles, number of conductors and parallel paths. As flux in this machine is directly proportional to its armature current (at least until metal saturates), therefore = KcI

    Induced Torque

    A

    where c is a constant of proportionality.

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 2 of 6

    Load-Torque (Ta / Ia) X-Tics of DC Series Motor

    Ta Ia

    Flux is proportional to Ia Ta I2

    a

    Load-Speed (Ia/N) Characteristics

    N Eb / N [V Ia (Ra + R se)] / Ia, [V Ia (Ra + R se) , N 1 / Ia Flux is proportional to armature current,

    N 1 / I2

    a

    Torque-Speed (Ta/N) Characteristics

    N Eb / N [V Ia (Ra + R se)] /

    If Ia, [V Ia (Ra + R se

    ) , N 1/Ia As Ta Ia

    N 1 / Ta

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 3 of 6

    Speed Control of DC Series Motor

    N E

    By Field Diverter

    b/ N = K (V IaRa - IaRse)/

    A variable resistor connected in parallel with field to control field current and flux

    If FDR, IF , , N N

    Provides 1/FDR

    speed above the normal speed

    . Lowest speed

    is obtained when the diverter is open (FDR = infinity).

    Maximum speed

    Used in electric traction.

    is obtained when diverter is short circuited (FDR = 0).

    Speed Control of DC Series Motor

    By Armature Resistance

    N E b / If R , V , E b , N

    N Most commonly used method for speed control

    of series motors.

    1 / Ra

    If terminal voltage is increased, the speed will increase for any given torque.

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 4 of 6

    Series field aids the shunt field

    Load-Torque (Ta/Ia) X-Tics. of Cumulative Compound Motor

    Ta Ia If Ia , then series field flux but shunt filed flux remains constant. Hence overall flux increases.

    Ta Ia The torque of cumulative compound motor is greater than that of shunt motor due to series field.

    Load-Speed (N/Ia) Characteristics of Cumulative Compound Motor

    N Eb

    / As Ia , then

    N 1/Ia

    N / Ta Characteristics Ta Ia N Eb

    / If Ia , then series field flux but shunt filed flux remains constant. Hence overall flux increases.

    N /Ta The torque of a cumulative compound motor is more than that of shunt motor but less than that of series motor. Due to presence of series field, the starting torque of this motor is increased.

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 5 of 6

    Speed Control of Cumulative Compound Motor A resistance is inserted in the armature circuit to reduce the speed.

    N E b / If R, [V Ia (Ra + Rse + R)] , Eb , N

    N 1 / R Electric Dynamic Braking Electric braking is used to stop the motor and its load quickly and smoothly. In Electric Dynamic Braking,

    armature of motor is disconnected from supply and is connected across variable resistor and field is still connected to supply. Armature, while slowing down, rotates in strong magnetic field and operates as a generator. It sends a large current to resistance and finally motor comes to rest.

    Electric Plugging / Reverse Braking

    In Reverse Braking , Armature connections are reversed to rotate the motor in reverse direction giving the braking effect. Back emf (Eb

    ) and applied voltage will act in the same direction. In order to limit the current, a variable resistance R is connected in series with the armature during the changing of armature connections.

  • Lec-12 Electrical Machines, B.Sc. Electrical Engineering Engr. Riaz Ahmad Rana, Assistant Professor, FOE, UCP

    Page 6 of 6

    Problem # 1 A 4-pole, 230 V series motor has a wave-connected armature with 1260 conductors. The flux per pole is 20 m Web when the motor is taking 50 A. Iron and friction losses amount to 1 KW. Armature resistance is 0.2 and series field resistance is 0.2 . Calculate: 1) Speed? 2) B.H.P? 3) Shaft torque? 4) Efficiency? Solution: 1) Speed? E b = NPZ / 60 A N = 60 A Eb / PZ E b = V Ia (Ra + RSE) = 230 50 (0.2 + 0.2) = 210 V N = 60 A Eb / PZ = 60 x 2 x 210 / 0.02 x 4x 1260 = 250 RPM {A = 2, wave-winding} 2) B.H.P? B.H.P = Motor output / 746 Motor O/P = Pm P f & W = Eb I a 2 KW = 210 x 50 1000 = 9500 W B.H.P = 9500 / 746 = 12.73 B.H.P 3) Shaft torque? Shaft torque = 9.55 x Pout / N =9.55 x 9500 / 250= 362.9 N-m 4) Efficiency? Efficiency = 9500 x 100 / 230 x 50 = 82.6 %