machine Lecture 6

IE243 ELECTRICAL MACHINES–I [Cr. Hrs = 2+1] [Marks: 100+50]

By

Asif Ahmed Memon

Introduction to DC motors

Speed Regulation

• THE EQUIVALENT CIRCUIT OFADC MOTOR

• THE MAGNETIZATION CURVE OF A DC MACHINE

• Type of dc motors

– SEPARATELY EXCITED AND SHUNT DC MOTORS

– THE PERMANENT-MAGNET DC MOTOR

– THE SERIES DC MOTOR

– THE COMPOUNDED DC MOTOR

THE EQUIVALENT CIRCUIT OF A DC MOTOR

• THE EQUIVALENT CIRCUIT OFADC MOTOR

• THE MAGNETIZATION CURVE OF A DC MACHINE

• Type of dc motors

– SEPARATELY EXCITED AND SHUNT DC MOTORS

– THE PERMANENT-MAGNET DC MOTOR

– THE SERIES DC MOTOR

– THE COMPOUNDED DC MOTOR

THE MAGNETIZATION CURVE OF A DC MACHINE

• THE EQUIVALENT CIRCUIT OFADC MOTOR

• THE MAGNETIZATION CURVE OF A DC MACHINE

• Type of dc motors

– SEPARATELY EXCITED AND SHUNT DC MOTORS

– THE PERMANENT-MAGNET DC MOTOR

– THE SERIES DC MOTOR

– THE COMPOUNDED DC MOTOR

SEPARATELY EXCITED AND SHUNT DC MOTORS

The Terminal Characteristic of a Shunt DC Motor

• A terminal characteristic of a machine is a plot of the machine's output quantities versus each other.

• For a motor, the output quantities are shaft torque and speed,

• so the terminal characteristic of a motor is a plot of its output torque versus speed.

The Terminal Characteristic of a Shunt DC Motor• How does a shunt dc motor respond to a load?

the load on the shaft of a shunt motor is increased

At steady state 𝜏𝑙𝑜𝑎𝑑 = 𝜏𝑖𝑛𝑑

𝜏𝑙𝑜𝑎𝑑 > 𝜏𝑖𝑛𝑑the motor will start

to slow down

When the motor slows down, its internal generated voltage drops

𝐸𝐴 = 𝐾∅𝜔 ↓

so the armature current in the motor increases 𝐼𝐴 = (𝑉𝑇 − 𝐸𝐴 ↓ ) 𝑅𝐴

As the armature current rises, the induced torque in the motor increases

𝜏𝑖𝑛𝑑 = 𝐾∅𝐼𝐴 ↑

and finally the induced torque will equal the load torque at a lower mechanical speed of rotation ω

The Terminal Characteristic of a Shunt DC MotorThe output characteristic of a shunt dc motor can be derived from the induced voltage and torque equations of the motor plus Kirchhoff's voltage law. (KVL)

The KVL equation for a shunt motor is

𝑉𝑇 = 𝐸𝐴 + 𝐼𝐴𝑅𝐴

The induced voltage 𝐸𝐴 = 𝐾∅𝜔

𝑉𝑇 = 𝐾∅𝜔 + 𝐼𝐴𝑅𝐴

Since 𝜏𝑖𝑛𝑑 = 𝐾∅𝐼𝐴

𝐼𝐴 =𝜏𝑖𝑛𝑑𝐾∅

Solve for IA

𝑉𝑇 = 𝐾∅𝜔 +𝜏𝑖𝑛𝑑𝐾∅

𝑅𝐴Therefore

Finally, solving for the motor's speed yields

𝜔 =𝑉𝑇𝐾∅

−𝑅𝐴𝐾∅ 2 𝜏𝑖𝑛𝑑

The Terminal Characteristic of a Shunt DC Motor

𝜔 =𝑉𝑇𝐾∅

−𝑅𝐴𝐾∅ 2

𝜏𝑖𝑛𝑑

• This equation is just a straight line with a negative slope. The resulting torque- speed characteristic of a shunt dc motor is shown in Figure

• If a motor has armature reaction, then as its load increases, the flux-weakening effects reduce its flux.

• As Equation shows, the effect of a reduction in flux is to increase the motor's speed at any given load over the speed it would run at without armature reaction.

• The torque-speed characteristic of a shunt motor with armature reaction is shown in Figure.

• If a motor has compensating windings, of course there will be no flux-weakening problems in the machine, and the flux in the machine will be constant.

The Terminal Characteristic of a Shunt DC MotorExample 9-1. A 50-hp, 250-V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding (see Figure ).

(a) Find the speed of this motor when its input current is 100 A. (b) Find the speed of this motor when its input current is 200 A. (c) Find the speed of this motor when its input current is 300 A. (d) Plot the torque-speed characteristic of this motor.

𝐸𝐴 = 𝐾∅𝜔

𝐸𝐴 = 𝐾′∅𝑛

𝐸𝐴2𝐸𝐴1

=𝐾′∅𝑛2𝐾′∅𝑛1

𝑛2 = 𝑛1𝐸𝐴2𝐸𝐴1

At no load, the armature current is zero

𝐸𝐴1 = 𝑉𝑇 = 250𝑉

𝑛1 = 1200 𝑟𝑝𝑚

The Terminal Characteristic of a Shunt DC MotorExample 9-1. A 50-hp, 250-V, 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding (see Figure ).

(a) Find the speed of this motor when its input current is 100 A.

𝐸𝐴 = 𝐾∅𝜔

𝐸𝐴 = 𝐾′∅𝑛

𝐸𝐴2𝐸𝐴1

=𝐾′∅𝑛2𝐾′∅𝑛1

𝑛2 = 𝑛1𝐸𝐴2𝐸𝐴1

At no load, the armature current is zero

𝐸𝐴1 = 𝑉𝑇 = 250𝑉

𝑛1 = 1200 𝑟𝑝𝑚

(a) 𝐼𝐿 = 100𝐴

𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 𝐼𝐿 −𝑉𝑇𝑅𝐹

𝐼𝐴 = 100𝐴 −250𝑉

50Ω= 95𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴

𝐸𝐴 = 250𝑉 − 95𝐴 0.06Ω = 244.3𝑉

𝑛2 = 𝑛1𝐸𝐴2𝐸𝐴1

= 1173 𝑟𝑝𝑚

The Terminal Characteristic of a Shunt DC MotorNonlinear Analysis of a Shunt DC Motor

Example 9-2. A 50-hp, 250-V, 1200 r/min dc shunt motor without compensating windingshas an armature resistance (including the brushes and interpoles) of 0.06 Ω. Its field circuithas a total resistance RF + Radj of 50 Ω, which produces a no-load speed of 1200 r/min. Thereare 1200 turns per pole on the shunt field winding, and the armature reaction produces ademagnetizing magnetomotive force of 840 A • turns at a load current of 200 A. Themagnetization curve of this machine is shown in Figure .

(a) Find the speed of this motor when its inputcurrent is 200 A.(b) This motor is essentially identical to theone in Example 9- 1 except for the absence ofcompensating windings. How does its speedcompare to that of the previous motor at aload current of 200 A?(c) Calculate and plot the torque-speedcharacteristic for this motor.

The Terminal Characteristic of a Shunt DC MotorNonlinear Analysis of a Shunt DC Motor

(a) Find the speed of this motor when its inputcurrent is 200 A.

If IL = 200 A, then the armature current of the motor is

𝐼𝐴 = 𝐼𝐿 − 𝐼𝐹 = 𝐼𝐿 −𝑉𝑇𝑅𝐹

= 195𝐴

𝐸𝐴 = 𝑉𝑇 − 𝐼𝐴𝑅𝐴 = 238.3 V

At IL = 200 A, the demagnetizing magneto motive force due to armature reaction is 840 A · turns, so the effective shunt field current of the motor is

The Terminal Characteristic of a Shunt DC MotorNonlinear Analysis of a Shunt DC Motor

Example 9-2. A 50-hp, 250-V, 1200 r/min dc shunt motor without compensating windingshas an armature resistance (including the brushes and interpoles) of 0.06 Ω. Its field circuithas a total resistance RF + Radj of 50 Ω, which produces a no-load speed of 1200 r/min. Thereare 1200 turns per pole on the shunt field winding, and the armature reaction produces ademagnetizing magnetomotive force of 840 A • turns at a load current of 200 A. Themagnetization curve of this machine is shown in Figure .

(a) Find the speed of this motor when its inputcurrent is 200 A.

𝐼𝐹∗ = 𝐼𝐹 −

𝐹𝐴𝑅𝑁𝐹

= 5.0𝐴 −840 𝐴 𝑇𝑢𝑟𝑛𝑠

1200 𝑡𝑢𝑟𝑛𝑠= 4.3𝐴

The Terminal Characteristic of a Shunt DC Motor

• THE EQUIVALENT CIRCUIT OFADC MOTOR

• THE MAGNETIZATION CURVE OF A DC MACHINE

• Type of dc motors

– SEPARATELY EXCITED AND SHUNT DC MOTORS

– THE PERMANENT-MAGNET DC MOTOR

– THE SERIES DC MOTOR

– THE COMPOUNDED DC MOTOR

THE SERIES DC MOTOR

)𝑉𝑇 = 𝐼𝐴(𝑅𝐴 + 𝑅𝑆

Induced Torque in a Series DC Motor