AC Drives MSc2016

8/18/2019 AC Drives MSc2016

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Industrial AC Drives

3. Selection and Applications

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AC Drive selection

Pumps

Wind energy

Multiple Centrifugal and energy management

Others


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3. . Selection of AC Drive

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The selection of ASD and MOTOR for any applicationrequires considerable experience to get the selection

right.

In design phase two main items are important

SAFETY: System needed to safety margins when it is build

COST: cost to be minimized ,by selecting the optimum size of

MOTOR & CONVERTER for each application

Design process start with

Clear and accurate study for system requirements


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i. The following checklist covers the factors to be considered:

The nature of the application

Maximum torque and power requirements and how they change with speed

Starting torque requirements

The speed range - minimum and maximum speed

Acceleration & deceleration requirements (Is braking necessary?)

Compatibility with the mains supply voltage

Environmental conditions where the converter and motor will operate, ambient temperature,

altitude, humidity, water, chemicals, dust, etc

Ventilation and cooling for the converter and motor

Direction (uni- or bi-directional)

Accuracy of the speed control

Dynamic response (speed and torque response requirements)

Speed regulation requirements with changes in load, temperature and supply voltage

The duty cycle, including the number of starts and stops per hour

Overall power factor of the drive system and its effect on the mains supply


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EMI and harmonics in the mains power supply and in the motor and motor cable

Are EMI filters required?

Earthing, shielding and surge protection requirements

Torque pulsations in the rotor shaft

Control method - manual, automatic, analog, digital, communications

Control and communications interfaces required for the plant control system

Indications required

Reliability requirements, is a dedicated standby unit required

Protection features, in-built and external features required

Power and control cable requirements

Parameter settings, local or remote programming

Maintenance spares and repair considerations

Cost of the alternative systems, taking into consideration the capital cost, performanceadvantages, energy savings, and efficiency or process improvements.

Noise due to the harmonics in the motor

Mechanical resonance at certain motor speeds


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ii. The basic selection procedure

Most problems result due to less experience

incorrect selection and rating of AC Induction Motor

incorrect selection and rating of AC Converter

incorrect parameter settings installed in VSD control system

The AC drive system is correctly selected and rated when: Motor specification is correct The correct type and size of electric motor has been selected, whose output

torque, speed and accuracy are adequate for all load and environmentalconditions.

AC converter specification is correct The correct type and size of AC converter has been selected, whose output

(voltage, current, frequency) meets the motor requirements for all load and

environmental conditions.


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The correct procedure is:

The first step is to select a correctly rated electric motor

Only when this is completed, a suitable AC converter is chosen to

match the requirements of the motor

From the motor point of view, the main factors which need to

be considered are

1. the motor power rating (kW),

2. the number of poles (speed) and

3. the frame size so that the load torque on the motor shaft

remains within the continuous torque capability of the

motor at all speeds within the speed range.


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iii. load ability of converter fed squirrel cage motorsLoad ability: Is the continuous load torque capacity of a standard TEFC

squirrel cage induction motor used with VVVF converters.

• When selecting an AC motor for any drive application , the most important

requirement is to ensure that the motor does not become overloaded or stall under

all circumstances of speed and load.

• For AC motors connected to the power supply direct-on-line (DOL) called fixed

speed : It is usually sufficient to ensure that load torque is sufficiently below motor

torque at the rated speed of the motor.

• In the case of a variable speed drive. The load torque usually changes with speed, so

it is essential to check that the motor torque exceeds the load torque at all

speeds in the speed range.


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The AC VSD load ability curve:

• Motors fed from VVVF converters can be loaded continuously at torques below the

load ability limit line for the speed range.

• Many AC converters have an over-current capacity of up to 150% for 60 sec to cover

starting and transient operation.

The speed range and load torque capacity (load ability) of a TEFC squirrel cage motor when controlled by a

PWM-type VVVF converter

solid line marks the maximum

limits of continuous load torque


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Load power capacity of a TEFC squirrel cage motor

In the region below base speed,

known as the constant torque

region, the power capability

increases linearly from zero at zero

speed to full power at the base

speed.

Above base speed, the power output

capability cannot increase furtherand remains constant for further

increases in speed with frequency

and torque is reduced. This is

known as the constant power

region.


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iv. The Nature of the Machine load

For fixed speed drives: the power requirement in kW at the rated speed.

On larger drives: motor manufacturers usually ask for more information about the load

1-the moment of inertia

2- the acceleration requirements.

3- more details about the load characteristics are always necessary.

The output torque of an AC VSD is considered to be adequate when it:

• Exceeds the breakaway torque of the machine load.

• Can accelerate the load from standstill to its preset speed within the acceleration time

required by the process.

• Exceeds the load torque by an adequate margin during continuous operation at any speed inthe speed range and under all conditions.

• Motor current does not exceed the thermal ratings of all electrical components and remains

below the load ability curve during continuous operation.


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iv. The Nature of the Machine load

what needs to be known about the machine load can be

covered by the following:

1. The load torque, the type, magnitude and characteristics of the load

torque connected to the output shaft of the motor

2. The speed range, the minimum and maximum speed of the variable speed

drive

3. The inertia of the motor and mechanical load connected to shaft of the

motor


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The load torque

The torque required by the driven machine determines the

size of the motor because the continuous rated torque of the

motor must always be larger than the torque required by the

driven machine.

The load torque determines the cost of the motor because

the cost of an electric motor is Approximately proportional

to its rated output torque (not its rated power!).


14Torque characteristics of typical types of machine loads as a function o f speed, angle and time


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The load torque

Another important aspect of the load torque is that the figure

should apply at the shaft of the motor. When gearboxes, conveyers or

hoists are involved, the actual torque at the machine must be

converted to torque at the motor shaft.

The conversion formula to convert the load torque, speed and moment

of inertia to motor shaft values depends on the transformation ratio.

The load requirements are often given as the mechanical absorbed

power (Pm kW) at a particular speed (n rpm).

The mechanical load torque may then be calculated from the following

formula:


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The load torque


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Variable torque machine loads

Machines with variable torque over

their entire speed range, such as

centrifugal pumps and fans.

The following are some of the important factors associated with this type

of load:

The starting torque for normal centrifugal pumps and fans is very low and below the

loadability curve of the AC motor for all speeds. Slurry pumps can some times be a

problem, as they can have a high breakaway torque.

The required starting current is low, so the overload capacity of converters is seldom

required during acceleration.

Running for long periods at low speeds is seldom a problem.

However, running at speeds above the motor base speed could be a problem because

the power requirement of this drive increases as the cube of the speed. This is

incompatible with the capabilities of the constant power region


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Constant torque machine loads

Constant torque machine loads are those which

exhibit a constant torque over their entire speed

range, such as conveyors, positive displacement

pumps.

The following are the potential problems when driving constant torque

loads from a converter fed electric motor:

The starting torque is theoretically equal to the full speed load torque but, in practice, the real

starting torque can be much higher due to the additional requirements of:

Breakaway torque

Acceleration torque (dynamic torque)

Running for long periods at low speeds can result in motor thermal over-load, if the load

torque is above the motor loadability curve.

Separate forced cooling may be necessary in some cases.

Running at speeds above the motor base speed could also be a problem, with increased motor

slip and a higher possibili ty of stalling the motor.


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The speed range

The selection of the correct size of electric motor for a VSD is affected

by the speed range within which it is expected to run continuously.

The important factor is that the motor should be able to drive the load

continuously at any speed within the speed range without stalling or

overheating the motor

Running the motor at below base speeds (f < 50 Hz ) with a standard TEFC

cage motor has the following effects on the motor:

Reduces the motor cooling because the cooling fan, which is

attached to the motor shaft, runs at reduced speed. Therefore,

the temperature rise in the motor will tend to be much higher

than expected.


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Figure shows an example of the torque – speed curve for a variable speed pump drive, operating in

the range from 10 Hz to 50 Hz Some comments are:

1. The load torque is well within the load ability limits at all speeds.

2. The maximum speed is below the base speed of 50 Hz. The speed range should NOT be increased above

50 Hz because the load torque will exceed the load ability limit of the drive. (Load torque increases as the

square of the speed.)

3. Starting torque is low, so there should be no problems with breakaway.

4. The acceleration torque is high, so the drive can be expected to quickly reach its maximum speed, if fast

acceleration is required. However, with pumps, a long acceleration time is normally desirable to prevent

water hammer .

Load curve

Example of speed range and torque curve of a v ariable

speed pump drive when controlled by a PWM-type

VVVF converter


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Running the motor at above base speeds (f > 50 Hz ) with a standard TEFC

cage motor has the following effect on the motor The air-gap flux is reduced because the V/f ratio is reduced. Consequently, there is a

reduction in the output torque capability of the motor. The torque is reduced in

proportion to the frequency. The load torque is not permitted to exceed the pullout

torque of the motor, even for a short period, otherwise the motor will stall.

The maximum torque allowed at above-synchronous speeds depends on the motor

characteristics and frequency as follows:


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Some comments on the conveyer application are:

1. The load torque falls outside the load ability limits at low speeds below 28 Hz there could be problems

running the motor continuously at speeds below 28 Hz.

2. Although the maximum speed is below the base speed of 50 Hz, but the speed range could be increased

above 50 Hz to take advantage of the load ability characteristic above 50 Hz. (Load torque remains constant

with increases in speed.)

3. Starting torque is high, with a high breakaway, so there may be some problems with breakaway.

4. Acceleration torque is small, so the drive ramp-up time may have to take place over a long period to avoid

exceeding the VSD current limit.

Torque-speed curve for a variable speed conveyor

drive, operating in a range from 10 Hz to 50 Hz.

Load curve


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The Inertia of machine loads

During acceleration and deceleration, the moment of inertia of the load imposes an

additional dynamic acceleration torque on the motor.

The moment of inertia and the required acceleration time together affect the motor

torque and consequently the size and cost of the motor.

The dynamic acceleration torque TA is calculated as follows:

This can be rewritten as follows, with the speed in rev/min:


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Example:

A conveyer drive is to be accelerated from zero to a speed of 1500 rpm in 10 seconds.

The moment of inertia of the load J L = 4.0 kgm2. The torque of the conveyer load,

referred to the motor shaft, is a constant at 520 Nm. The motor being considered is a

110 kW, 1480 rpm with J M = 1.3 kgm2. Is this motor adequate for this duty?

The moment of inertia of the drive system is

J Tot = 4.0 + 1.3 = 5.3 kgm2

During acceleration, the dynamic torque required is

T A = 5.3 (2π/60)((1500-0)/10)=83.25 Nm

The machine load is a constant torque type with a value given above asT L = 520 Nm

During acceleration, the motor supply a total torque of

T Tot = T L + T A

T Tot = 520 +83.25 = 603.25 Nm


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The rated motor torque may be obtained from the manufacturer’s tables or calculated

from the rated power as follows:T N = (9550*110/1480)= 709.8 Nm

Because T N ≥T Tot , the motor is evidently suited for the drive requirements.

When the motor drives the mechanical load through a gearbox or pulleys, the inertia of

the load must be ‘referred’ to the motor shaft using the formula given below:

J M = J L ( )

( ) kgm2

Where J M = Inertia of the motor shaft

J L = Inertia of the load shaft


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The selection procedure may now be summarized as follows:STEP 1: Specify the initial data for the drive application

To select the correct motor/converter combination, the following information must be available:

Voltage and frequency of the power supply

The breakdown or starting torque

The load torque and its dependence on speed

Speed range of the variable speed drive

Acceleration requirements or ‘ramp times’

The moment of inertia of the motor and load

STEP 2: Specify the number of poles of the motor

The number of poles determines the synchronous speed of the motor and its usually selected

according to the maximum speed required by the application. Modern VVVF converter are

available with output frequencies of up to 400 Hz.

Above-synchronous speeds are of particular advantage for constant torque loads, where the

maximum speed should ideally, be in the range of 50-100 Hz.


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STEP 3: Select the motor power rating

Using the load torque requirements, the power rating of the motor can be selected from a motormanufacturer’s catalogue using the formula

STEP 4: Select a suitable frequency converter

A converter with a rating suitable for the motor selected should then be selected from the

manufacturer’s catalogue. Converters are usually manufactured for power ratings that match the

standard sizes of squirrel cage motors. Catalogues usually give the current rating as well as a check

to ensure that the motor current is below that of the converter.

The following factors must be considered:

Supply voltage and frequency

Rated current of the motor Duty type (variable torque or constant torque)

A converter is selected as that the rated current of the converter is higher than the rated current of

the motor. Also, the type of converter should be suitable for the duty required. Some

manufacturers have different converters for the two duty types.


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STEP 5: Final Checks

The following final checks should be made:

Is the continuous power rating of the motor (de-rated fro altitude, temperature,

harmonics, etc) greater than the continuous power requirements of the load?

Is the starting torque capability of the variable speed drive high enough to exceed the

breakaway torque of the load?

If the VSD is operating in the over-synchronous speed area, is the motor torque

capability at maximum speed adequate for the load torque?

Is the speed accuracy adequate for the application?


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Example Conveyer Application

Problem description

A variable speed drive application has been proposed for a conveyer designed to

transport a package of product from arriving point (initial distribution point ) to another

point .The required speed range is 600 rpm to 1400 rpm . The calculated power

requirement of the load, reduced to the motor shaft , is 66 kw at 1400 rpm The

breakaway torque is expected to be 110% of rated torque. The supply voltage is 415

volts, 50 Hz. Select the optimum size and rating of squirrel cage motor and converter

for the most cost effective solution


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According to the system description

1- Motor torque required calculation :

The load is a typical constant torque load type. From the pervious equations, the

constant load torque requirement across the speed range is

MOTOR selected must be with Toque TM > TB


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1- Selecting number of poles:

According to speed range the number of poles selected that determine base speed, so

two options available from catalogs :

4 – Pole Motor

Speed Range 1480 rpm

110 kw,415 V,188 A

Frame Size 280MRated Torque = 710 Nm

Needed =495 Nm

6 – Pole Motor

Speed Range 985 rpm

75 kw,415 V,135 A

Frame Size 280M

Rated Torque = 727 Nm

Needed =495 Nm


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Which is the Motor from the two available is the best choice?

Two motors are adequate but the converter choice will give some other clear

view For choice

Motor current is less than choice one, so the converter current required

is less in this case, and the initial cost is low.

Converter suitable for this motor:

110 Kw,415 V,220 A

Converter suitable for this motor:

75 Kw,415 V,140 A

This is the best choice because:


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3.2. Application of ASD in pumps

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Power electrical = I ×Vx pfWhere: Power electrical = [W], V = voltage [V], I = Current [A]

Power mechanical=2 π* T* N/60Where: N = Speed [rpm], T = Torque [Nm],

Power fluid = g Q H ρwaterWhere: Power fluid = [W], g = 9.81 m/s2 , Q = Flow quantity [m3 /s],

H=Pressure head [m], ρwater = density of water


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If you have a system using a motor that serves a varying load,

an Adjustable Speed Drive (ASD) may benefit you. Examplesof such systems are fans that reduce flow for non-designconditions or pumps that use control valves to vary flow.ASDs tailor motor performance to match present conditions.

Advantages include: Energy savings Improved process flow Softer/easier motor starting Faster response than valves or dampers


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Throttling verses Speed change for pump system

H-Q curves


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Throttling verses Speed change for pump system

H-Q curves

Operating

point

Reduced speedoperating point

flow

head

Throttled Operating

Point

q

H

q2

Hs

H2

3

2

2

1

2

2

1

2

2

1

Q2

Q

N

N

P

P

N

N

H

H

N

N

P

P2

Ps


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The following table shows the theoretical power required as

the motor speed & system flow decrease

P1/P2= (n1/n2)3

Power RequiredFlowSpeed

100%100%100%

72.9%90%90%

51.2%80%80%

34.3%70%70%21.6%60%60%

12.5%50%50%

6.4%40%40%

2.7%30%30%


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The table shows that reducing motor speed results in a

reduction in the power required. Significant energy savingsare the result. For example, to reduce the flow from 100percent to 50 percent cuts the required power to 12.5percent. Eight times less power is used (8 = 100%power/12.5% power).

The power required drops exponentially when motor speedis reduced, providing you with substantial energy savings.

ASD energy savings are estimated using computer analysis.

The overall ASD energy savings is calculated as thetheoretical energy savings minus the losses due to runningthe ASD and the drop in motor efficiency at lower speeds.


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Example:

The following example show the energy saving when operate apump (rated flow=0.25m3/sec, at speed 1500 rpm) withconstant speed to fill a canal with dimension (500 m length,5m width, 3m height).

The volume of water required = 5 x 3 x 500=7500 m3

The number of hours required =7500/(.25 x 60 x 60)=8.33 hr

Assume the input power to the pump = P kW

The energy consumes 8.33P kWh


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If operate the pump with VSD to fill the same volume

Operate 2 hours at rated speedq=0.25m3/sec, V1=0.25 x 2 x 60 x 60=1800 m

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Operate 4 hours at 0.75 x 0.25 m3/sec, V2=2700 m3

V3=7500-2700-1800=3000 m3

Operate at 0.5 x 0.25 m3/sec, it’s operating for 6.67 hr

From affinity laws P proportional to cubic speed

P (at 0.25)=P kW, P (at 0.75 x .25)=0.42P kW, P (0.5 x .25)=0.125PkW

Energy required 2 x P+4 x 0.42 x P+0.125 x 6.67 x P=4.5P kW

Percentage of energy saving = 8.33P-4.5P/8.33P=45.8%

AC Drives MSc2016

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