MEBS 6000 2010 Utilities services

MEBS 6000 2010 Utilities services M.Sc.(Eng) in building services

Department of Electrical & Electronic Engineering University of Hong Kong

K.F. Chan (Mr.) of 1818 July 2010

Joint speed torque characteristics of electric motors and mechanical loads

Electric motors exhibit a variety of speed-torque characteristics that are suitable for a

wide range of load demands. Mechanical loads also have a wide range of

speed-torque characteristics depending on their mechanical properties. For example,

the load torque of a hoist or a conveyor is largely independent of it speed. Load torque

of a fan or a pump is proportional to the square of its speed. When an electric motor is

connected to a mechanical load, the system operates at a speed-torque status that

matches the characteristics of the motor as well as the mechanical load.

Braking

If no braking is applied to a motor, after electric power is removed, it only stops when

all the kinetic energy is dissipated. Braking is a generic term used to describe a set of

operating conditions for electric drive systems. It includes rapid stopping of the

electric motor, holding the motor shaft, thus the driven equipment, to a specific

position, etc. Besides mechanical braking, all these aspects of braking can also be

done electrically. During braking, the energy flow changes its direction and utilization

of the braking energy enhances system efficiency.

Consider a PWM frequency inverter, in terms of power, the power input to the motor

is:

φcos3VIP =

where V is the phase to neutral voltage, I the phase current and φ the power factor.

In motoring operation, φ <90o, power is then positive and flows from the inverter to

the motor. A reduction in frequency of the supply reduces the synchronous speed,

before the rotor speed is reduced to the desired value, the rotor speed is higher than

the synchronous speed. This reversal of relative speed reverses the rotor induced emf,

rotor current and component of stator current which balances the rotor ampere turns.

Consequently, the power factor φ >90o, power becomes negative and flows from

motor to the source.

There are several forms of electric braking methods, collectively known as:

1. regenerative – feeding energy back to the power source,

2. dynamic – by dissipating the energy in an electrical resistance,

3. countercurrent – braking by reversing the direction of the stator field.

We will discuss, briefly, dynamic braking and, more in detail, regenerative braking of

ac induction motors.




Bi-directional drive systems

Consider the case of an electric motor driving an elevator.

Let us assume that the lift car is fully loaded with weight larger than the

counterweight. When the lift car is moving upwards, the motor torque is in the same

direction as the motor rotation. When the lift car is moving downwards, the motor

torque must be in opposite direction to the rotation to keep the lift car from falling

overspeed.

Four quadrant electric drive systems

The following conventions govern the power flow analysis of electric drive systems:

When the torque of the electric motor is in the same direction as the system

rotation, the motor consumes electric power from the power source and delivers

mechanical energy to the load

When the motor torque and the system rotation are in opposite directions, the

motor is in fact driven by the mechanical power and delivering electric power to

the electric power source.




(Adopted from EL-SHARKAWI, M.A., Fundamentals of Electric Drives)

(Adopted from DUBEY, G.K. Fundamentals of Electrical Drives)

The above shows the four quadrants of speed-torque characteristics to cover all

possible combinations of any electric drive systems.




Quad

rant

Direction of

rotation

Load

torque and

rotation

Power flow Operating

as

Example

1st Same as 2nd

quadrant but

opposite to 3rd

Opposite

direction

From motor

to load

Motor Fully loaded

elevator moving

upwards

2nd Same as 1st

quadrant

Same

direction

From load

to motor

Generator Empty lift car

moving

upwards

3rd Opposite to 1st

quadrant

Opposite

direction

From motor

to load

Motor Empty lift car

moving

downwards

4th Opposite to 1st

quadrant but

same as 3rd

Same

direction

From load

to motor

Generator Fully loaded

elevator moving

downwards

A bi-directional grinding machine, a horizontal conveyor, a bi-directional travellator,

are all examples of machines required to operate in the 1st and 3rd quadrant.

Example of drive system operating in all four quadrants (Adopted from DUBEY, G.K. Fundamentals of Electrical Drives)




A versatile motor drive system operates in all 4 quadrants. An elevator carrying

passengers or a hoist with counterweight are typical examples. For maximum

efficiency, the power electronic circuit must then be designed to allow the electric

power to flow in both directions.

Four quadrant operation can be obtained by any drive system with braking capability.

A reduction of the inverter frequency, to make synchronous speed less than the

spinning rotor speed, transfers the operation from quadrant 1 (forward motoring) to

quadrant 2 (forward braking). The inverter frequency and voltage are progressively

reduced as speed falls to brake the machine to desired speed.

Next phase sequence of the inverter output voltage is reversed by interchanging the

firing pulses between any 2 legs of the inverter. This transfer the operation to quadrant

3 (reverse motoring). The inverter frequency and voltage are increased to get the

required speed in the reverse direction. Then a reduction of inverter frequency will

make the synchronous speed less than the spinning rotor speed, transferring operation

from quadrant 3 to 4 (reverse braking).

Dynamic braking

(Adopted from Dubey, G.K., Fundamentals of Electrical Drives)

The above shows a simplified circuit for dynamic braking. A switch, SW, a braking

resistor R, and a self-commutated switch S, a transistor shown here, is added to the dc

link of a PWM frequency inverter circuit. During motor operation, the switch SW is

closed and the transistor S is opened, so the circuit acts as a normal PWM frequency

inverter.

During braking operation, the switch SW is opened, generated energy from the motor

flowing into the dc link charges the capacitor and so its voltage rises. When the

voltage rises above a set value, the transistor is closed, allowing the energy to be

dissipated in the resistor.




Regenerative braking

Regenerative braking occurs when the motor speed exceeds the synchronous speed.

This may happen when the load torque drives the motor beyond its synchronous speed.

In this case, the load is the source of energy and the induction motor is converting the

mechanical power into electrical power, which is delivered back to the electrical

system.

If the power source of a frequency inverter comes from a rechargeable dc battery like

the one shown below, the circuit by its nature has the capability of regenerative

braking.


However, if the power source comes from a ac/dc diode bridge like the one below,

some means shall be built into the frequency inverter drive to allow the regenerated

power during braking to return to the ac source.





One method is to use dual converter in the power source section, one converting the

source ac to the dc link, the other one inverting the regenerated power in the dc link

back into the ac power source.

(Adopted from MURPHY, JMD, & TURNBULL, FG. Power electronic control of AC motors)




Another method is to use a synchronous link converter.


Recall that diodes are typically installed across the transistors in the dc/ac inverter

section of a frequency drive to allow energy to flow back to the dc link. So one can

imagine that PWM inverter I, shown above in the power supply section, is an inverter

installed as a mirror image to PWM inverter II. PWM I and the inductor Ls constitute

a SLC.

PWM I is operated to produce voltage VI of required magnitude and phase – in phase

with Vs for motoring, and 180o out of phase with Vs for braking.

A SLI inverter schematics




Wind turbine using induction machines are good example of regenerative braking.

Induction machines are popular in wind applications because they are ideally suited

for variable power profile applications. Unlike the synchronous or dc machines,

induction machines become automatically synchronized with the external power

system.

The above figure shows the basic component of a wind turbine. At moderate wind

speed (called cutoff speed) the gearbox is designed to rotate the high speed shaft near

the synchronous speed of the induction machine. If the wind increases beyond the

cutoff speed, the induction machine rotates at a speed higher than its synchronous

speed. This is regenerative braking operation. The housing box can swivel at the top

of the tower to point the blades at the direction of maximum wind effects. When the

wind speed becomes excessive, the blades can be locked to prevent any mechanical

damage to the system.

Regenerative braking can be explained by the torque equation:

+

+

=2

2'2

1

'2

23

eqs

d

Xs

RRs

RVT

ω

where V is the voltage across each stator winding.




The negative slip occurs when the speed of the machine exceeds the synchronous

speed:

s

s

n

nns

−=

When the turbine blades drives the electric machine to a speed faster than the

synchronous speed, the slip becomes negative. Since the load torque is in the same

direction as the machine, it operates in the 2nd quadrant as a generator. In fact,

induction machine of the wind system is designed to operate at regenerative braking

in the 2nd quadrant only. When the wind speed is low so that the rotor speed is near or

below the synchronous speed, the blades are locked and the motor is disconnected

from the electrical supply to prevent the machine from running as a motor.

When used in wind applications, induction machines demand a significant amount of

reactive power from the utility system, mainly because they do not have their own

field circuit. When used in regenerative mode, the induction machine consumes

reactive power from the system while delivering real power. The inductive reactive

power Q is consumed in the magnetizing inductive reactance Xm and the equivalent

winding reactance Xeq.




eqm

XIX

VQ 2

2

2

'33 +=

+

++=

22

21

2

'

13

eq

eq

m Xs

RR

X

XVQ

where V is the phase to neutral voltage.

The reactive power Q is plotted in the following diagram:




Q is at its minimum at synchronous speed but is then almost linearly proportional to

speed when n>ns. To alleviate, a reactive power controller can be installed at the wind

farm.

In more general drive systems, the induction machine may operate in the first or

second quadrant only.

The above figure shows a constant but reversible load torque. When the load torque

changes from T1 to T2, the motor changes from 1st to 2nd quadrant operation, bear in

the mind that the motor is still rotating in the same direction.




The following figure shows another example. The figure shows 2 characteristics for

two different values of v/f control. The load torque is assumed to be constant. When a

speed reduction is desired from point 1, before the speed is changed, the motor

temporarily moves to operation point 2 in the 2nd quadrant under regenerative braking

then settles at point 3, the desired lower speed.




Example

A 400V, star connected, 3-phase, 6-pole, 50Hz induction motor has following

parameters referred to the stator:

R2’=R1=1Ω , X1=X2’=22Ω

For regenerative braking operation of this motor:

a) Determine maximum overhauling torque it can hold and range of speed for safe

operation.

b) Determine speed at which it will hold an overhauling load with a torque of

100Nm.

c) If a capacitive reactance of 2Ω is inserted in each phase of the stator, calculate

maximum overhauling torque the motor can hold as a ratio of maximum

overhauling torque without capacitor.

Answer

a) Synchronous speed = RPM10003

5060 =×=

3

502π =104.72rad/s

21

2

'2

maxRX

Rs

eq +±=

243.014

122max ±=

+±=s

For regenerative braking operation, the slip is negative, thus -0.243.

For generator action

++−

−=22

11

2

max

2

3

eqs XRR

VT

ω

[ ]22

2

max

411350

22

400

++−

−=

πT = -244.5Nm

Maximum torque occurs at smax of -0.243, i.e. at

(1+0.243)1000 = 1243RPM

Stable operation during regenerative braking occurs from synchronous speed to this

speed at which the torque is maximum. Thus range of speed for safe operation will

be from 1000 to 1243RPM.




b)

+

+

=2

2'2

1

'2

23

eqs

d

Xs

RRs

RVT

ω

Re-arranging, the above equation becomes

( ) 0''3

'2 22

22

21222

1 =+

−++ Rs

T

RVRRsXR

sdeq ω

Substituting the numerical values, the equation becomes

0126.1717 2 =++⇒ ss

Note that for a quadratic equation:

02 =++ cbxax

the solution can be written as:

a

acbbx

2

42 −±−=

So s = -0.957 or -0.063

However, -0.957 will give unstable operation, therefore s=-0.063.

Motor speed is thus

(1+0.063)1000 = 1063RPM

c)

If a capacitor is added per phase, the net effect is reducing the inductive reactance.

As 21

2

'2

maxRX

Rs

eq +±=

so ( )

447.0124

122max ±=

+−±=s

For generator action




++−

−=22

11

2

max

2

3

eqs XRR

VT

ω

( )

−++−

−=

22

2

max

2411350

22

400

πT =-618Nm

This is the maximum torque if a capacitor of 2Ω is added to each phase in series

with the stator winding. The ratio is

5.244

618

−−

=2.53

[This example adopted from Dubey, Gopal.K., Fundamentals of Electrical Drives]




Example

A 380V, 50Hz, six pole, star connected induction motor has the following parameters :

Ω=Ω=Ω= 5,4.0,6.0 '21 eqXRR

The motor is loaded by a 30 Nm bi-directional constant torque. If the load torque is

reversed, neglecting mechanical loss calculate

a) Motor speed

b) Power delivered to the electrical supply system

Answer

a) Phase to neutral voltage is V2203380 =

Synchronous speed is RPM10006

26050 =××

+

+

=2

2'2

1

'2

23

eqs

d

Xs

RRs

RVT

ω

Re-arranging, the above equation becomes

( ) 0''3

'2 22

22

21222

1 =+

−++ Rs

T

RVRRsXR

sdeq ω

Substituting the numerical values, the equation becomes

( ) ( )

+

+

=−

22

2

54.0

6.060

10002

4.0220330

ss π

( ) ( ) 016.097.1836.25 2 =++ ss

Note that for a quadratic equation:

02 =++ cbxax

The solution can be written as:

a

acbbx

2

42 −±−=

So s can be expressed as




( )( )( )22

1

22

221

2

22

212

2

21

2

'4'3

'2'3

'2

eq

eqsdsd

XR

RXRT

RVRR

T

RVRR

s+

+−

−±

−−

=ωω

Thus s = -0.0085 or -0.74

However, 21

2

'2

maxRX

Rs

eq +±= = 079.0±

Obviously, if the slip is -0.74, motor will run in unstable braking condition.

The regenerative speed is

( ) RPMsnn s 1009)0085.01(10001 =+=−=

b) ωTPd =

= Watt3170)60

10092)(30( =π

To calculate the stator and rotor copper loss we must first calculate the current '2I

Recall

gd PsP )1( −=

= )1(3'22'

2 ss

RI −

thus )0085.01()0085.0(

4.033170

2'2 +

−=− I

AmpI 72.4'2 =

So rotor copper loss is

WRI 7.26)4.0()72.4(33 2'2

2'2 ==

Stator copper loss is

WRI 1.40)6.0()72.4(33 21

2'2 ==

Therefore power delivered to the electrical system is

= 3170 – 26.7 – 40.1

=3103Watt

[Unless otherwise stated, text and figures adopted from EL-SHARKAWI, Mohamed

A., Fundamentals of Electric Drives]

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