AC Drives MSc2016
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Transcript of AC Drives MSc2016
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1
Industrial AC Drives
3. Selection and Applications
2
AC Drive selection
Pumps
Wind energy
Multiple Centrifugal and energy management
Others
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3. . Selection of AC Drive
3
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
3. . Selection of AC Drive
4
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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3. . Selection of AC Drive
5
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
3. . Selection of AC Drive
6
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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3. . Selection of AC Drive
7
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.
3. . Selection of AC Drive
8
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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3. . Selection of AC Drive
9
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
3. . Selection of AC Drive
10
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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3. . Selection of AC Drive
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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.
3. . Selection of AC Drive
12
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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3. . Selection of AC Drive
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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!).
3. . Selection of AC Drive
14Torque characteristics of typical types of machine loads as a function o f speed, angle and time
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3. . Selection of AC Drive
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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:
3. . Selection of AC Drive
16
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
3. . Selection of AC Drive
17
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.
3. . Selection of AC Drive
18
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3. . Selection of AC Drive
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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.
3. . Selection of AC Drive
20
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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3. . Selection of AC Drive
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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:
3. . Selection of AC Drive
22
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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3. . Selection of AC Drive
23
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:
3. . Selection of AC Drive
24
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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3. . Selection of AC Drive
25
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
3. . Selection of AC Drive
26
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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3. . Selection of AC Drive
27
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.
3. . Selection of AC Drive
28
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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3. . Selection of AC Drive
29
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
3. . Selection of AC Drive
30
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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3. . Selection of AC Drive
31
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
3. . Selection of AC Drive
32
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
33
3.2. Application of ASD in pumps
34
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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3.2. Application of ASD in pumps
35
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
3.2. Application of ASD in pumps
36
Throttling verses Speed change for pump system
H-Q curves
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3.2. Application of ASD in pumps
37
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
3.2. Application of ASD in pumps
38
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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3.2. Application of ASD in pumps
39
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.
3.2. Application of ASD in pumps
40
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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3.2. Application of ASD in pumps
41
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
3
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%