Motor Load Interaction

7/28/2019 Motor Load Interaction

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Motor-load Interaction

Drive Characteristics

Different types of load torque-speed characteristics

Different types of motor torque-speed characteristics

Motor-load steady state stability conditions

Static stability

Dynamic stability

Numerical example


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Drives Characteristics

m

For stable speed, motor torque (Tm) = load torque (TL)

L


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Load Torque-speed characteristics

1) Fan or pump

2) Compressor

3) Hoist or Elevator


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Motor Torque Speed Characteristics

Separately excited

d.c. motor

Tm

N

1. D.C. motor

(a) Separately excited dc motor


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Motor Torque Speed Characteristics (2)

Tm

N

Series excited

dc motor

(b) Series excited dc motor

N


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Motor Torque Speed Characteristics (3)2. Induction motor:(Note: Though the operating point on the left of the peak torque

in the bottom plot is stable, it is not permissible as the current will be very high dueto high slip.

Not permissible


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Steady-state stability (Static stability)

TmTL = J. dN/dt + BN

Where

Jmoment of inertia of rotating part

Bdamping coefficient (due to friction and windage)

Under steady state (stable) condition

dN/dt = 0

Therefore, Tm = TL + BN which is basically static stability.


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Steady-state stability (Dynamic stability)

At both I1 and I2, Tm = TL. However, which one of I1 and I2 is stable?I1 is unstable and I2 is stable.

I1 is unstable because if due to some perturbation

(i) Speed increases then Tm > TL causing the system to speed up.

(ii) Speed decreases, then Tm < TL causing the system to slow down.

I2 is stable because

(i) Any speed increment at I2 causes Tm < TL and the system will slow

down.

(ii) Any speed decrement at I2 causes Tm > TL and the system will speed up.

I2I1


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Steady-state stability (Dynamic stability) (2)

It can be shown (assignment problem 4.1) that to achieve dynamic stability at a

particular operating speed:

Note: There is a slight difference between this condition and equation (4.5)in the text , but both carries the same essence.


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Steady -state Stability (example)

A motor has a torque speed characteristics given by Tm=30-0.053N (torque in N-

m, speed in rad/sec). The motor is running a constant load torque of 25 N-m.

The system (motor and load) was initially running at steady state.

(i) Find the steady state speed.

(ii) Due to a single impulse type disturbance the speed decreases temporarily by

50 rpm. Find the speed of the system in rpm 0.6s after the disturbance has

ceased. J = 0.05 kg-m2.


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Steady-state Stability (example

solution)(i) Under steady-state, T

L= T

m.

25=30-0.053N.

Thus the steadystate speed is N=N1=5/0.053=94.34 rad/sec. (60/2p) = 900.88 rpm.

(ii) One can show using assignment problem 4.1 that

= 1

= ; =

1

= 0.053.

=

= 0.943.

=50.

0.6

0.943= 26.46 rpm.

Hence the final speed after 0.6s will be

N1-N= 900.88-26.46=874.42 rpm.


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Transient response to disturbance

0 2 4 6 8 10

840

850

860

870

880

890

900

910

X: 2.6

Y: 874.4

Time (s)

Speed(RPM)


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Four quadrant operation of a drive

Defining a quadrant using a simple hoist or winch drive as an example


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Four quadrant operation of a drive(2)

Steady state four quadrant boundary of a machine

Braking/

Generating

Reverse

Motoring

Motoring

Braking/

Generating

Documents

Motor Load Interaction