Transcript The Basics of AC Drive Applications

Yaskawa Electric
America
The Basics of
AC Drive Applications
6/6/2002
PP.AFD.01.AfdDriveApplBasics
1 of 123
Basic Definitions
“Definitions”





To get a good understanding of applications, it is
advantageous to understand:
Power - the rate of doing work
Torque - twisting power
Load Torque - torque required to perform an application
Motor Torque - torque available from the prime mover
2
“Power and Torque”


Power is the rate
of doing work.
Power is a
product of torque
and speed.
P =T x n
Where:
T - torque
n - speed
3
“Power and Torque”
P =T x n
High
Med
Low
P T n
High
P T n
Med
Low
Power
increases
if torque
increases
4
“Power and Torque”
P =T x n
High
P T n
High
Med
Med
Low
Low
P T n
Power
increases
if speed
increases
5
“Power and Torque”

Power increases as speed increases
125%
High
P T n
100%
Med
X
75%
Torque
Power
50%
Low
25%
Power and speed
are low and torque
is constant.
0%
X
0%
50%
Speed (%)
100%
“Power and Torque”

Power increases as speed increases
125%
High
P T n
X
100%
Med
75%
Torque
Power
50%
Low
X
25%
Power increases
proportionally with
speed.
0%
0%
50%
Speed (%)
100%
“Power and Torque”

Power increases as speed increases
125%
High
P T n
X
100%
Med
75%
Torque
Power
X
50%
Low
25%
Power increases
proportionally with
speed.
0%
0%
50%
Speed (%)
100%
“Power and Torque”

Power increases as speed increases
125%
High
P T n
X
100%
Med
75%
X
Torque
Power
50%
Low
25%
Power increases
proportionally with
speed.
0%
0%
50%
Speed (%)
100%
“Power and Torque”

Power increases as speed increases
125%
High
P T n
X
X
100%
Med
75%
Torque
Power
50%
Low
25%
Power increases
proportionally with
speed.
0%
0%
50%
Speed (%)
100%
6
Types of Loads
“Types of Loads”

Loads can be grouped into four categories:

Variable Torque Loads


Constant Torque Loads


Load does not change with speed
Constant Horsepower Loads


Load changes with speed
Load considerations above motor base speed
Impact Loads

Load is intermittent, not connected to speed
7
“Variable Torque Loads”


With a variable torque
load, torque loading is
a function of the speed.
The torque will change
with the square of the
speed and the
horsepower will change
with the cube of the
speed.
This characteristic is
typical of centrifugal
type fans and pumps.
100%
80%
60%
Torque
Horsepower
40%
20%
0%
0%
50%
100%
Speed (%)
8
“Variable Torque Loads”


With a variable torque
load, torque loading is
a function of the speed.
The torque will change
with the square of the
speed and the
horsepower will change
with the cube of the
speed.
This characteristic is
typical of centrifugal
type fans and pumps.
100%
80%
60%
Torque
Horsepower
40%
20%
0%
0%
50%
Speed (%)
100%
“Variable Torque Loads”

These variable torque characteristics are a result of the
affinity laws of centrifugal machinery.

The ratio of torque 1 to
torque 2 is equal to the ratio
of speed 1 squared to speed
2 squared.

The ratio of horsepower 1 to
horsepower 2 is equal to the
ratio of speed 1 cubed to
speed 2 cubed.
T1
T2
=
(n1 )2
(n2 )2
(n1 )3
HP1
=
HP2
(n2 )3
9
“Variable Torque Loads”

The basic affinity laws can be converted for use with
centrifugal fans and pumps.

Flow is directly proportional
to speed.
F1
F2
Pressure is directly
proportional to the square of
the speed.
P1
P2
Horsepower is directly
proportional to the cube of
the speed.
(n1 )3
HP1
=
HP2
(n2 )3


=
=
n1
n2
(n1 )2
(n2 )2
10
“Variable Torque Loads”
Not all fans and pumps are variable torque loads:

Fans are grouped in 2 major categories:
 Fans and Blowers (variable torque)


Compressors (constant torque)
Pumps are grouped in 2 major categories:
 Centrifugal (variable torque)
 Positive Displacement (constant torque)
11
“Fan Loads”
Fan/Blowers
Compressors
Variable Torque
Constant Torque
Centrifugal
Axial
Reciprocating
Rotary
Radial Blade
Propeller
Piston
Rotary Lobe
Forward Curve
Tubeaxial
Diaphragm
Screw
Backward Incline
Vaneaxial
12
“Constant Torque Loads”

With a constant torque
load, torque loading is
not a function of the
speed. As the speed
changes the load
torque remains
constant. Horsepower
will change with speed.
125%
100%
75%
Torque
Horsepower
50%
25%

This characteristic is
typical of positive
displacement pumps
and conveyors.
0%
0%
50%
100%
Speed (%)
13
“Constant Torque Loads”

With a constant torque
load, torque loading is
not a function of the
speed. As the speed
changes the load
torque remains
constant. Horsepower
will change with speed.
125%
100%
75%
Torque
Horsepower
50%
25%

This characteristic is
typical of positive
displacement pumps
and conveyors.
0%
0%
50%
Speed (%)
100%
“Constant Torque Loads”

The torque and horsepower formulas for rotating
machinery can better illustrate the graph on the
previous slide.

The horsepower changes
linearly with speed when the
torque stays constant.
HP =
The torque remains constant
when the horsepower
changes linearly with speed.
HP x 5250
T =
rpm

T x rpm
5250
Where:
Torque is represented in ft-lbs
14
“Constant Horsepower Loads”


With a constant
horsepower load,
125%
torque loading is a
function of the speed in
the constant
100%
horsepower range. As
the speed changes, the 75%
load torque will
decrease at a rate
inversely proportional 50%
with speed (1/n).
25%
Horsepower
generally remains
constant.
0%
This characteristic is
typical of machine tool
spindle drives.
base speed
Torque
Horsepower
0%
50%
100%
150%
200%
Speed (%)
15
“Constant Horsepower Loads”


With a constant
horsepower load,
torque loading is a
125%
function of the speed in
the constant
100%
horsepower range. As
the speed changes, the
75%
load torque will
decrease at a rate
inversely proportional 50%
with speed (1/n).
Horsepower
25%
generally remains
constant.
0%
This characteristic is
typical of machine tool
spindle drives.
base speed
Torque
Horsepower
0%
50%
100%
Speed (%)
150%
200%
“Constant Horsepower Loads”

What happens to torque in the constant horsepower
area can be found below.

Example: A 5 HP, 1750 rpm, 60Hz
motor will be run to 3500 rpm (120Hz).
100% base speed and below
A
0.5 HP x 5250
= 15 ft-lbs
175 rpm
2.5 HP x 5250
= 15 ft-lbs
875 rpm
5 HP x 5250
C
= 15 ft-lbs
1750 rpm
B
HP x 5250
= T
rpm
above 100% base speed
D
5 HP x 5250
= 12 ft-lbs
2187 rpm
E
5 HP x 5250
= 10 ft-lbs
2625 rpm
F
5 HP x 5250
= 8.6 ft-lbs
3062 rpm
G
5 HP x 5250
= 7.5 ft-lbs
3500 rpm
16
“Constant Horsepower Loads”
Points A through G can be found on the speed torque
curve below.
16
15
A
B
C
14
D
12
E
10
10
F
8
G
6
5
4
Horsepower
Torque (ft-lbs)

Torque
Horsepower
2
0
1750
3500 0
Speed (rpm)
17
“Impact Loads”
This type of load
characteristic is
typical of a punch
press or any
equipment that uses
a flywheel.
MotorTorque
Load Torque
175%
Motor Speed
1800
150%
1500
125%
1200
100%
75%
900
50%
600
25%
300
0%
Motor Speed (rpm)

Impact loads
have intermittent
torque requirements,
which are not a
function of speed.
Torque (%)

0
Time
18
“Impact Loads”
This type of load
characteristic is
typical of a punch
press or any
equipment that uses
a flywheel.
MotorTorque
Load Torque
Motor Speed
175%
1800
150%
1500
125%
1200
100%
75%
900
50%
600
25%
300
0%
0
Time
Motor Speed (rpm)

Impact loads
have intermittent
torque requirements,
which are not a
function of speed.
Torque (%)

Calculation of
Torque
“Calculations”

The next section will explain how to calculate
the following:



Acceleration Torque
Deceleration Torque
Torque and Speed Through
a Gearbox
19
“Calculating Acceleration Torque”
Acceleration torque is the difference between the available
motor torque and the torque required to drive the load.
TORQUE

(200%)
Breakdown
Torque
(200-250%)
Pull Up
Torque
(125%)
(200%)
Motor Full
Load Torque
(150%)
(100%)
Torque available for
accel and decel.
(100%)
Torque required to
drive the load.
SPEED
20
“Calculating Acceleration Torque”

To calculate the
acceleration torque (TA), the
following items must be
known:


The total inertia
in lb-ft2. (WK2)

The change in motor
speed. (rpm)

The required
acceleration time in
seconds. (ta)
WK2 x  rpm
TA =
308 x ta
Where:
Torque is represented in ft-lbs
 Total inertia includes the inertia
of the load, inertia of the gearbox
and the inertia of the motor’s rotor.
21
“Calculating Acceleration Torque”

A typical example of calculating acceleration torque:

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired acceleration
time, between 0 and 1750 rpm, is 15
seconds.
TA =
WK2 x  rpm
308 x ta
10 lb-ft2 x 1750 rpm
3.8 ft-lbs =
308 x 15 seconds

The above information indicates the amount of
acceleration torque required, but does not tell us if
the 5HP motor and inverter are capable of achieving
the specified performance.
22
“Calculating Acceleration Torque”

Determine the motor rated torque and the torque
required to drive the load.

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired acceleration time,
between 0 and 1750 rpm, is 15 seconds.
80% of the motor rated torque is
required to drive the load.
Motor Rated Torque
5HP x 5250
15 ft-lbs =
1750
Note-In this example the motor rated current
and the inverter rated current are equal.
HP x 5250
Tr =
rpm
Load  Torque
12 ft-lbs 
torque required to move the load at constant speed
12 ft-lbs is 80% of 15 ft-lbs
23
“Calculating Acceleration Torque”

The maximum available motor torque will be
determined by the inverter ampacity overload
rating. Torque is proportional to current through a major
portion of the speed range. A typical inverter overload
rating is 150% for 1 minute.
(200%)
Maximum
available motor
torque.
(30 ft-lbs)
(22.5 ft-lbs)
Maximum Available Motor Torque
15 ft-lbs x 150% = 22.5 ft-lbs
Note-In this example the motor rated current
and the inverter rated current are equal.
(100%)
(15 ft-lbs)
(12 ft-lbs)
Torque required
to drive the
load.
24
Speed - Torque Curve
(300%)
(600%)
Slip
TORQUE
CURRENT
(300%)
(200%)
Locked Rotor
Torque (150%)
Breakdown
Torque
(200-250%)
Pull Up Torque
(125%)
Full Load
Torque (100%)
No Load
Current
(30%)
SPEED
Rated Speed Synch Speed
25
Speed - Torque Curve
(300%)
(600%)
Slip
TORQUE
CURRENT
(300%)
(200%)
Locked Rotor
Torque (150%)
Breakdown
Torque
(200-250%)
Pull Up Torque
(125%)
Full Load
Torque (100%)
Current
Torque
No Load
Current
(30%)
SPEED
Rated Speed Synch Speed
“Calculating Acceleration Torque”

The torque available for acceleration is the difference
between the available motor torque and the torque
required to drive the load.
Torque Available for Acceleration
22.5 ft-lbs - 12 ft-lbs = 10.5 ft-lbs
(200%)
Maximum
available motor
torque.
(30 ft-lbs)

(22.5 ft-lbs)
(100%)
Torque required to
drive the load.
(15 ft-lbs)
(12 ft-lbs)
Torque available
for acceleration.
(10.5 ft-lbs)
26
“Calculating Acceleration Torque”


Determine if the 5HP motor is capable of achieving the
specified performance.
Example: A 5HP, 1750 rpm, 60Hz motor and a 5HP inverter are used
to run a conveyor. The total inertia is 10 lb-ft2. The desired
acceleration time, between 0 and 1750 rpm, is 15 seconds. 80% of
the motor’s rated torque is required to drive the load.
Motor rated torque = 15 ft-lbs
Tr =
HP x 5250
rpm
Load torque = 12 ft-lbs
Maximum available motor torque = 22.5 ft-lbs
(Tr x 150%)
Torque available for acceleration = 10.5 ft-lbs
(Max. available - load torque)
Torque required for acceleration = 3.8 ft-lbs
TA =
WK2 x  rpm
308 x ta
Torque available for acceleration > Torque required for acceleration
27
Time to do an exercise
“Calculating Acceleration Torque”

Using the formulas below, determine if the performance
specification can be achieved for the the following application.

Example: A 10HP, 1750 rpm, 60Hz motor and a 10HP inverter are
used to run a boring machine. The total inertia is 19 lb-ft2. The
desired acceleration time, between 0 and 1750 rpm, is 4 seconds.
90% of the motor rated torque is required to drive the load.
Motor rated torque = _____ ft-lbs
Tr =
HP x 5250
rpm
Load torque = _____ ft-lbs
Maximum available motor torque = _____ ft-lbs
= (Tr x 150%)
Torque available for acceleration = _____ ft-lbs
= (Max. available - load torque)
Torque required for acceleration = _____ ft-lbs
TA =
WK2 x  rpm
308 x ta
Torque available for acceleration ____ Torque required for acceleration
Note-In this example the motor rated current
and the inverter rated current are equal.
28
“Calculating Acceleration Torque”

Using the formulas below, determine if the performance
specification can be achieved for the the following application.

Example: A 10HP, 1750 rpm, 60Hz motor and a 10HP inverter are
used to run a boring machine. The total inertia is 19 lb-ft2. The
desired acceleration time, between 0 and 1750 rpm, is 4 seconds.
90% of the motor rated torque is required to drive the load.
30
Tr =
Motor rated torque = _____ ft-lbs
HP x 5250
rpm
27
Load torque = _____ ft-lbs
45
Torque available for acceleration = _____
18 ft-lbs
Maximum available motor torque = _____ ft-lbs
27
Torque required for acceleration = _____ ft-lbs
= (Tr x 150%)
= (Max. available - load torque)
TA =
WK2 x  rpm
308 x ta
Torque available for acceleration ____
< Torque required for acceleration
The specified performance is not achieved!
Now what do
we do?
“Calculating Acceleration Torque”

When more acceleration torque is required than is
available, the inverter current rating may be increased
to handle the increase in current.
(200%)
Maximum
available motor
torque.
(100%)
(60 ft.lbs)
(45 ft.lbs)
(30 ft.lbs)
(27 ft.lbs)
Torque
required for
acceleration.
(27 ft.lbs)
Torque
available for
acceleration.
(18 ft.lbs)
Torque
required to
drive the load.
Torque available for acceleration
18 ft.lbs
Torque required for acceleration
27 ft.lbs
29
“Calculating Acceleration Torque”

Use the following formula to determine the necessary
current rating increase to achieve the required
acceleration torque.
Inverter
Output Rating
Increase
=
Torque Required for Acceleration + Load Torque
Torque Available for Acceleration + Load Torque
120%

=
27 ft-lbs + 27 ft-lbs
18 ft-lbs + 27 ft-lbs
Choose an inverter with a continuos output current of
120% of the motor full load current.
30
“Calculating Deceleration Torque”

Deceleration torque is the sum of the windage and friction
losses and either the motor breakdown torque or the inverter
braking torque (which ever is smaller).
Breakdown Torque
(approx 250% )
(approx 10%)
Motor
Braking Torque
(approx 10-15% )
Inverter
Smallest
+
Windage &
Friction
Losses
Available
Decel
Torque
31
“Calculating Deceleration Torque”

To calculate the
deceleration torque (TD) the
following items must be
known:


The total inertia
in lb-ft2. (WK2)

The change in motor
speed. (rpm)

The required deceleration
time in seconds. (td)
WK2 x  rpm
TD =
308 x td
Where:
Torque is represented in ft-lbs
 Total inertia includes the inertia
of the load, inertia of the gearbox
and the inertia of the motor’s rotor.
32
“Calculating Deceleration Torque”

When running a motor with an inverter in the motoring
mode, the power flows from the inverter to the motor.
(+)
L1
L2
L3
M
1FU
(-)
33
“Calculating Deceleration Torque”

In the regenerative mode, the power flows from the motor
to the inverter.
(+)
L1
L2
L3
M
1FU
(-)
34
“Calculating Deceleration Torque”

When in the regenerative mode, power cannot go through
the input diodes, back to the ac power line. This may cause
the DC bus voltage inside the inverter to climb to excessive
levels. If the DC bus voltage goes too high, the inverter will
protect itself by tripping on an overvoltage fault.
(+)
L1
L2
L3
M
1FU
(-)
35
“Calculating Deceleration Torque”

Since there is nowhere to dissipate the regenerative
energy, the amount of deceleration or braking torque
available from the motor and inverter combination is limited.
Braking torque for a motor and inverter combination is
typically in the range of 10 - 15%.
(+)
L1
L2
L3
M
1FU
(-)
36
“Calculating Deceleration Torque”
Energy to be
dissipated by
braking
resistors
Mechanical
energy in load
machine
Inverter Losses
IGBT Switching

Motor Losses
I2 R
Core


5% large inverters/motors
10-15% small inverters/motors
37
“Calculating Deceleration Torque”

A typical example of calculating deceleration torque:

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired deceleration
time, between 1750 and 0 rpm is 20
seconds
TD =
WK2 x  rpm
308 x td
10 lb-ft2 x 1750 rpm
2.8 ft.lbs =
308 x 20 seconds

The above information indicates the amount of
deceleration torque required, but does not tell us if the
5HP motor and inverter are capable of achieving the
specified performance.
38
“Calculating Deceleration Torque”

Determine the windage and friction losses of the
application. If this data is not known assume 0.

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired deceleration time,
between 1750 and 0 rpm is 20 seconds
Windage and friction losses are 7% of
the motor rated torque.
HP x 5250
Tr =
rpm
L = Tr x L%
Where L% - Percentage of windage and friction
losses
L - Windage and friction losses in ft-lbs
Motor Rated Torque
5HP x 5250
15 ft-lbs =
1750
Losses
1.05 ft-lbs
39
“Calculating Deceleration Torque”

Friction losses can be put into two categories:
Mechanical Friction

Bearing Friction - friction between front and rear rotor shaft
bearings
Air Friction

Rotor Fan Windage - friction of air on rotor fan blade surface

External Fan Windage - friction of air on external fan blade surface
Note: Windage and friction losses are basically constant throughout the
speed range.
40
“Calculating Deceleration Torque”

Determine the amount of braking torque required,
by subtracting the windage and friction losses from
deceleration torque required .

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired deceleration time,
between 1750 and 0 rpm, is 20 seconds.
Windage and friction losses are 7% of
the motor rated torque.
TB = TD - L
Where TB - Braking torque required in ft-lbs
1.75 ft-lbs = 2.8 ft-lbs - 1.05 ft-lbs
41
“Calculating Deceleration Torque”

Determine the percentage of braking torque required.

Example: A 5HP, 1750 rpm, 60Hz
motor and a 5HP inverter are used to run
a conveyor. The total inertia is
10 lb-ft2. The desired deceleration time,
between 1750 and 0 rpm, is 20 seconds.
Windage and friction losses are 7% of
the motor rated torque.
1.75 ft-lbs
11.6% =
15 ft-lbs
TB % =
TB
Tr
Where TB % - Percentage of braking torque
Torque required for deceleration < 15%
The specified performance
is achieved!
42
Time to do an exercise
“Calculating Deceleration Torque”

Using the formulas below, determine if the performance
specification can be achieved for the following application.

Example: A 10HP, 1750 rpm, 60Hz motor and 10HP inverter are
used to run a lathe. The total inertia is 15 lb-ft2. The desired
deceleration time, from 1750 to 0 rpm, is 2 seconds. Windage and
friction losses are 5%.
WK2 x  rpm
Torque required for deceleration = _____ ft-lbs
TD =
Motor rated torque = _____ ft-lbs
Tr =
Windage /Friction losses= _____ ft-lbs
L = Tr x L%
Braking Torque = _____ ft-lbs
TB = TD - L
Braking Torque = _____ %
TB % =
308 x td
HP x 5250
rpm
TB
Tr
Torque required for acceleration ______ 15%
43
“Calculating Deceleration Torque”

Using the formulas below, determine if the performance
specification can be achieved for the following application.

Example: A 10HP, 1750 rpm, 60Hz motor and 10HP inverter are
used to run a lathe. The total inertia is 15 lb-ft2. The desired
deceleration time, from 1750 to 0 rpm, is 2 seconds. Windage and
friction losses are 5%.
43
Torque required for deceleration = _____ ft-lbs
30
Motor rated torque = _____ ft-lbs
TD =
Tr =
1.5
WK2 x  rpm
308 x td
HP x 5250
rpm
L = Tr x L%
Windage /Friction losses= _____ ft-lbs
41.5
TB = TD - L
Braking Torque = _____ ft-lbs
138
TB % =
Braking Torque = _____ %
TB
Tr
Torque required for acceleration ______ 15% The specified performance
is not achieved!
>
Now what do
we do?
“Types of Braking”

If the braking torque required is more than 15%,
one of 3 braking methods must be used to
achieve the specified performance.



DC Injection Braking
Dynamic Braking
Regenerative Braking
44
“DC Injection Braking”

DC injection braking
utilizes the DC
current from the DC
bus of the inverter to
produce a stationary
magnetic field in the
motor. The rotor
passes through the
stationary field and a
braking torque is
produced.
N
S
Braking Torque - approx 50%
Duty Cycle - Typically 3-5% (motor dependent)
45
“Dynamic Braking”

Dynamic braking
is a process in
which regenerative
energy from the
load is dissipated
as heat across a
bank of resistors.
(+)
L1
L2
L3
M
1FU
(-)
Resistor
BTR
HEAT
Braking Torque - See chart on next page
Duty Cycle - See chart on next page
46
“Dynamic Braking”
47
“Regenerative Braking”

Regenerative braking requires the use of a DC to
AC converter to direct the regenerative energy
from the load, back onto the AC power line.
(+)
L1
L2
L3
(+)
Regenerative
Converter
M
1FU
(-)
(-)
Braking Torque - 100% continuos, 150% peak
Duty Cycle - Typically 100%
48
“Gearboxes”

A gearbox is
used to transmit
power from a
motor to the
driven machine.
It can change
the amount of
torque and
speed delivered
to the load.
Output
Input
Motor
Gearbox
49
“Gearboxes”

To determine the
output torque of a
gearbox, utilize
the formula below:
TO = TI x RG x EG
Output
Input
Motor
Gearbox
Where:
TO = Torque on the output shaft in ft-lbs
TI = Torque on the input shaft in ft-lbs
RG = The ratio of the gearbox
EG = Efficiency of the gearbox
50
“Gearboxes”

A typical example of calculating the torque available
from the output of a gearbox can be found below.

Example: A 5HP, 1750 rpm, 60Hz
motor is connected to a 30:1 gearbox.
The gearbox efficiency is 90%. What
is the expected torque on the output
shaft of the gearbox ?
HP x 5250
Tr =
rpm
TO = TI x RG x EG
5HP x 5250
15 ft.lbs =
1750 rpm
405 ft-lbs= 15 x 30 x 0.90
The expected torque on
the output shaft of the
gearbox is 405 ft-lbs
51
“Gearboxes”

To determine the
output speed of a
gearbox, utilize
the formula below:
nO =
nI
RG
Output
Input
Motor
Gearbox
Where:
nO = Speed on the output shaft in rpm
nI = Speed on the input shaft in rpm
RG = The ratio of the gearbox
52
“Gearboxes”

A typical example of calculating the speed available
from the output of a gearbox can be found below.

Example: A 5HP, 1750 rpm, 60Hz
motor is connected to a 30:1 gearbox.
The gearbox efficiency is 90%. What
is the expected speed on the output
shaft of the gearbox ?
58.3 rpm =
nO =
nI
RG
1750
30
The expected speed on
the output shaft of the
gearbox is 58.3 rpm.
53
Time to do an exercise
“Gearboxes”

Using the formulas below, determine if the performance
specification can be achieved for the the following application.

Example: A 5HP, 1750 rpm, 60Hz motor and 5HP inverter are
used to run a conveyor. The speed range of the motor/inverter is
45 - 1750 rpm. 200 ft-lbs is required to adequately run the
load. The customer wants to be able to adjust the speed of the
gearbox output shaft between 50 and 60 rpm. The gearbox is rated
at 20:1 and is 90% efficient.
HP x 5250
=
Tr
Motor rated torque = _____ ft-lbs
rpm
Gearbox Output Torque = _____ ft-lbs
Gearbox Output Speed Range = _____ rpm
TO = TI x RG x EG
SO =
SI
RG
Can the application be performed ?______
54
“Gearboxes”

Using the formulas below, determine if the performance
specification can be achieved for the the following application.

Example: A 5HP, 1750 rpm, 60Hz motor and 5HP inverter are
used to run a conveyor. The speed range of the motor/inverter is
45 - 1750 rpm. 200 ft-lbs is required to adequately run the
load. The customer wants to be able to adjust the speed of the
gearbox output shaft between 50 and 60 rpm. The gearbox is rated
at 20:1 and is 90% efficient.
HP x 5250
=
Tr
Motor rated torque = _____
15 ft-lbs
rpm
270
Gearbox output torque = _____ ft-lbs
TO = TI x RG x EG
2-88
Gearbox Output Speed Range = _________ rpm
YES
Can the application be performed ? ______
SO =
SI
RG
Dynamic Braking
“Dynamic Braking”

In most applications a standard dynamic braking package
will achieve the desired performance. In some applications
the standard dynamic braking package will not work. In
these cases a special dynamic braking package must
be designed to achieve the desired performance.
The next section will explain the following:


How to determine if a standard dynamic
braking package will achieve the desired
performance.
How to determine the specifications of a
non-standard dynamic braking package.
55
“Braking Packages”

A braking package consists of a braking transistor unit and
a resistor box.

A standard braking package has already been designed by
the inverter drive manufacturer to produce a specific
amount of braking torque at a specific duty cycle.

A non-standard braking package has been designed to
work with a specific application. The ohmic and wattage
values of the resistors and the number of braking transistor
units vary with application requirements.
Standard
Non-standard
Inverter
Braking
Transistor
50 
9kW
Inverter
Braking
Transistor
12 
14.5kW
Braking
Transistor
12 
14.5kW
56
“Standard Braking Package”

To determine if a standard braking package can be used,
follow the steps listed below:
 Determine the amount of braking torque required.
 Determine if the standard braking package can produce
more braking torque than is required.
 Determine the duty cycle required.
 Determine if the standard braking package duty cycle is
higher than the duty cycle required.
57
“Standard Braking Package”

Example: A 20HP, 1750 rpm, 230Vac, 60Hz motor and 20HP inverter are
used to run a conveyor. A 10:1 gearbox rated 90% efficient is
connected between the motor and the load. The load inertia on the output
of the gearbox is 3600 lb-ft2, the gearbox inertia is 2 lb-ft2 and the rotor
inertia is 2 lb-ft2. The desired deceleration time, between 175 and 0
gearbox output rpm, is 5 seconds. Windage and friction losses are 7% of
the motor rated torque. The conveyor will be started and stopped once
each 10 minutes, with 1 minute of off time. Can the desired performance
be achieved with the standard dynamic braking package?
Output
Input
Motor
Gearbox
58
“Determine Braking Torque”

Determine the amount of torque and speed on the input
and output of the gearbox.

Example: A 20HP, 1750 rpm, 230Vac,
60Hz motor and 20HP inverter
are used to run a conveyor. A 10:1
gear box rated 90% efficient is
connected between the motor and the
load.
60 ft-lbs
1750 rpm
Motor
HP x 5250
rpm
= TI x RG x EG
Tr =
TO
SO =
SI
RG
Output
Input
Gearbox
540 ft-lbs
175 rpm
10:1
90%
59
“Determine Braking Torque”

Determine the amount of inertia reflected back to the motor.

Example: The load inertia is 3600 lb-ft2.
Gearbox and rotor inertia are 2 lb-ft2 respectively.
WK2O
The desired deceleration time is 5
WK2I =
seconds between 1750 and 0 motor rpm.
RG 2
Windage and friction losses are 7% of
the motor rated torque.
36 + 2 + 2 = 40 lb-ft2 Input
Motor
Input
60 ft-lbs
1750 rpm
Output
Gearbox
3600 lb-ft2
Output
540 ft-lbs
175 rpm
60
“Determine Braking Torque”

Determine the amount of braking torque required to
achieve the desired performance.
WK2I x  rpm

Example: The load inertia is 3600 lb-ft2.
Gearbox and rotor inertia are 2 lb-ft2 respectively.
The desired deceleration time is 5
seconds between 1750 and 0 motor rpm.
Windage and friction losses are 7% of
the motor rated torque.
60 ft-lbs
1750 rpm
40-lb-ft2
Motor
308 x td
L = Tr x L%
TB = TD - L
TB % =
TB = 41.3 ft-lbs
Input
TB% = 69%
Input
TD =
Output
TB
Tr
Gearbox
Output
540 ft-lbs
175 rpm
3600-lb-ft2
61
“Determine Braking Torque”

Compare the amount of braking torque required, to
the amount of braking torque available from the
standard braking package.
Required
 Example: Refer to page 37 for a complete
list of the standard braking package
specifications.
TB% = 69%
HP
kW
Braking Torque
20
15
120%
The specified torque performance is achieved!
62
“Determine Duty Cycle”

Determine the duty cycle of the braking circuit.

Example: The conveyor will
be started and stopped once
each 10 minutes, with 1
minute of off time.
Where :
D =
td
D - Duty Cycle
t
t - Total cycle time
td - Decel time
t
5s
11m
0.75% =
5s
660s
63
“Determine Duty Cycle”

Compare the duty cycle required to the duty cycle of the
standard braking package. Refer to page 37 for
standard braking package duty cycle specification.

Example: The conveyor will
be started and stopped once
each 10 minutes, with 1
minute of off time.
HP
kW
Braking Torque
20
15
120%
Required
D = 0.75%
Duty Cycle
10%
The specified performance is achieved!
64
“Non-standard Braking Package”

Example: A 20HP, 1750 rpm, 230Vac, 60Hz motor and 20HP inverter are
used to run an automated lathe. A 2:1 gearbox rated 90% efficient is
connected between the motor and the load. The load inertia on the output
shaft of the gearbox is 234 lb-ft2, the gearbox inertia is 2 lb-ft2 and the
rotor inertia is 2 lb-ft2. The desired deceleration time, between
875 and 0 gearbox output rpm, is 4 seconds. Windage and friction
losses are 3% of the motor rated torque. The lathe will be started and
stopped once each 10 seconds, with 2 seconds of off time. Can the
desired performance be achieved with the standard dynamic braking
package?
Output
Input
Motor
Gearbox
65
“Non-standard Braking Package”

Determine the braking torque and duty cycle required to
achieve the desired performance.
60
Tr =
Motor rated torque = _____ ft-lbs
108
HP x 5250
rpm
TO = TI x RG x EG
Output torque = _____ ft-lbs
875
SO =
Output speed = _______ rpm
62.5
Total inertia = _______ lb-ft2
SI
RG
WK2I =
88.7
Torque required for deceleration = ________ ft-lbs
TD =
WK2O
RG 2
WK2I x  rpm
308 x td
66
“Non-standard Braking Package”

Determine the braking torque and duty cycle required to
achieve the desired performance.
1.8
Windage /Friction losses= _____ ft-lbs
86.9
Braking Torque = ________ ft-lbs
144 %
Braking Torque = ________
33
Braking Torque
20
15
120%
D =
TB
Tr
td
t
Required
Standard
kW
TB = TD - L
TB % =
Required Duty Cycle = ________ %
HP
L = Tr x L%
Duty Cycle
10%
144% Torque
33% Duty Cycle
67
“Non-standard Braking Package”

Use the following formula to determine
the ohmic value of the braking resistor.
VDC2
RB =
1.027 x (TB x 0.14) x rpm
Where:
VDC2 - Braking transistor turn on voltage squared
TB - Braking torque in ft-lbs
0.14 - ft-lbs to kg-m conversion
1.027 - Mechanical to electrical conversion factor
385 2
6.8  =
1.027 x ( 12.2 ) x 1750
The ohmic value required to produce 144% braking
torque is 6.8 
68
“Non-standard Braking Package”

Use the following formulas to determine the wattage value of the
braking resistor.
Pave = 1.027 x (TB x 0.14) x rpm x 10 -3
WB =
Pave
2
m
11kW = 1.027 x ( 12.2 ) x 875 x 10 -3
11kW
29kW =
0.38
Refer to the next page for a typical resistor m value.
The wattage value required to withstand a 33% duty
cycle, at 144% braking torque is 29kW.
69
“Non-standard Braking Package”
m value for an application with
repeating deceleration
10
1%
DUTY
10%
CYCLE
33%
100%
m value for an application with
non-repeating deceleration
1s DECEL TIME 10s
100s
3
20
m
1
0.5
0.38
10
m
x
2
0.1
These charts are for example only, actual resistor data
must be obtained from the resistor manufacturer.
1
70
“Non-standard Braking Package”

Use the following formula to determine
the number of braking transistor units
required based on the current requirements.
385
57A =
6.8 
IB =
N =
Where :
VDC
RB
IB
IP
IB - braking current
IP - balancing factor
57A
1.425 =
40A
N - number of braking transistor units required
VDC - Braking transistor turn on voltage
The braking current is 57A and the number of braking
transistor units is 1.425, or 2.
The number of braking transistor units calculated must be
verified with the overcurrent protection curve.
71
“Non-standard Braking Package”

The proposed braking package will be assembled as shown below.
Each braking unit will carry 28.5 amps.
Inverter
P
Braking
Transistor
12 
14.5kW 28.5 Amps
Braking
Transistor
12 
28.5 Amps
14.5kW
N

The following information will be used to determine if this configuration
exceeds the braking transistor unit overcurrent protection curve.

Braking Current ( IB ) divided by the proposed number of braking units ( N )

Deceleration Time ( td )

Duty Cycle ( D )
72
“Non-standard Braking Package”

The specifications for the application do not exceed the braking
transistor unit overcurrent protection curve.
IB  N = 28.5A
Decel time = 4 sec 400
5%
Deceleration Time ( td )
Duty Cycle = 33%
10%
100
5% Duty Cycle
20%
40
10% Duty Cycle
20% Duty Cycle
40% Duty Cycle
10
60% Duty Cycle
X
4
40%
60%
1
10
20
30
IB  N
40
50
73
“Non-standard Braking Package”

To achieve the desired performance the braking package will be
assembled as shown below.
Inverter
P
N
Braking
Transistor
12 
14.5kW
Braking
Transistor
12 
14.5kW

Never connect a braking resistor with a lower ohmic value than
specified by the manufacturer.

When using more than one braking transistor some manufacturers have a
master - slave configuration so both braking transistors will work as one.
74
“100% Duty Cycle”

Use the following formula to determine
the number of braking transistor units
required for 100% duty cycle.
IB =
VDC
RB
IB - Y + ( Y x 0.8 )
N =
Y
385
57 A =
6.8 
Where :
Y - braking transistor unit RMS current rating
0.8 - balancing factor
N - number of braking transistor units required
VDC - Braking transistor turn on voltage
57 - 15 + ( 15 x 0.8 )
3.6 =
15
The formula above must be used to determine the number of braking
transistor units when 100% duty cycle is required. If a 100% duty cycle was
required for this example, 4 braking transistor units would be required.
75
The effects of the
speed range on
motor heating
“Speed Control Range and Motor Heating”

There are 4 basic categories of speed ranges
for a general purpose induction motor.




More than 100% (above base speed)
50 - 100% (2:1)
16 - 50% (6:1)
Less than 16% (more than 6:1)
76
“Overall Consideration”



Whenever running a general purpose induction motor on
an inverter, the motor must be derated as shown below.
It is required to derate an
induction motor with a 1.0
service factor to 85% of its
rated power output.
The derating can be
eliminated if a motor with a
1.15 service factor is used,
and the load does not
exceed the 1.0 service factor
rating.
85%
1.15 SF but only
use 1.0 SF
77
“Overall Consideration”

There are 2 main reasons for the derating;
voltage


The rating of an induction motor is
based on a sinusoidal voltage and
current waveform. The output of an
inverter is not a pure sine wave and
includes extra harmonic currents.
More heat will be generated in the
motor, with no additional capability to
dissipate the heat.
current
When the motor speed is reduced
the motor’s cooling fan losses its
capacity to move air inside and/or
across the surface of the motor. This
reduction in cooling leads to a higher
running temperature.
78
“More Than 100% (above base speed)”
Variable Torque Load

Running a variable torque
application above base
speed is normally not
recommended. Typically the
maximum speed of the fan
or pump matches the
required flow when the
motor is running at rated
speed. Increasing the pump
speed will increase the flow
as well as the load, and
could possibly lead to an
overload condition.
Constant HP Load

Typically this situation does
not cause a problem. The
motor torque drops off
reducing the load and the
motor’s cooling fan
continues to cool the motor.
Consult the motor
manufacturer for the motor’s
mechanical maximum
speed. Only the overall
consideration should be
adhered to in this situation.
79
“50 - 100% (2:1)”
Overall Consideration
When running an induction
motor on an inverter inside
the 2:1 speed range, only
the overall considerations
previously discussed should
be adhered to.

Derate a motor with a 1.0 service
factor to 85% of its rated power.

The derating can be
eliminated if a motor with a
1.15 service factor is used,
and the load does not
exceed the 1.0 service factor
rating.
100/85 %
De-Rating

30Hz
50%
45Hz
75%
Motor Speed
60Hz
100%
1.0 SF
1.15 SF
80
“ 16 - 50% (6:1)”
Typically no special derating
other than the overall
consideration is required
when running a variable
torque load in the 6:1 speed
range. In some cases, such
as a centrifugal pump, the
load takes on the
appearance of a constant
torque load at low speeds. If
this is the case, the constant
torque specifications should
be followed.

Motor derating when running
a constant torque application
in the 6:1 range is shown
below.
100/85 %
De-Rating

Constant Torque Load
87/72 %
De-Rating
Variable Torque Load
75/60 %
10Hz
16%
1.0 SF
1.15 SF
20Hz
33%
Motor Speed
30Hz
50%
81
“Less than 16% (more than 6:1)”
Constant Torque Load

When running an induction motor
with more than a 6:1 speed range,
additional cooling considerations
are required. Separate motor
cooling must be provided for the
motor. Typically a separately
powered blower attached to the
motor is used to provide airflow
over the surface of the motor.
Since the blower is separately
powered, there is no reduction in
motor cooling at low motor speeds
which is inherent to a TEFC type
motor.
82
Motor Overload
Types
“Motor Overloads”

There are 6 basic categories of motor overload
protection devices.






Melting Alloy type
Magnetic
Bi-Metallic
Bi-Metallic Thermo Disc
Thermistor
Electronic Thermal Overload
83
“Melting Alloy Type”

Melting alloy type overload
relays are the most
common type overload
used in industry today.
Motor current flows through
an alloy heater coil and
produces a proportional
amount of heat. At a
specific temperature (thus
a specific current) the alloy
changes from a solid to a
liquid. The melting alloy
causes a mechanism to
release and open a set of
contacts. A cool off time is
required for the alloy to
return to a solid.
Reset
Button
L1
Spring
Heater
Coil
T1
84
“Melting Alloy Type”
Melting alloy type overload
relays are the most common type
overload used in industry today.
Motor current flows through
an alloy heater coil and produces a
proportional amount of heat. At a
specific temperature (thus a
specific current) the alloy changes
from a solid to a liquid. The melting
alloy causes a mechanism to
release and open a set of
contacts. A cool off time is
required for the alloy to
return to a solid.
Reset
Button
L1
Spring
Heater
Coil
T1
“Melting Alloy Type”
10.00
9.00

Melting alloy type overload
relays are normally identified
by class designations

Class 10 - designed to trip in
10 seconds at 600% of its
normal current rating.

Class 20 - designed to trip in
20 seconds at 600% of its
normal current rating.

Class 30 - designed to trip in
30 seconds at 600% of its
normal current rating.
Trip/minutes Class 10
8.00
Trip/minutes Class 20
Time/Minutes
7.00
6.00
Trip/minutes Class 30
5.00
4.00
3.00
2.00
1.00
0.00
100%
200%
300%
400%
Percent Load
500%
600%
85
“Melting Alloy Type”
10.00
9.00

Melting alloy type overload
relays are normally identified
by class designations

Class 10 - designed to trip in
10 seconds at 600% of its
normal current rating.

Class 20 - designed to trip in
20 seconds at 600% of its
normal current rating.

Class 30 - designed to trip in
30 seconds at 600% of its
normal current rating.
Trip/minutes Class 10
8.00
Trip/minutes Class 20
Time/Minutes
7.00
6.00
Trip/minutes Class 30
5.00
4.00
3.00
2.00
1.00
0.00
100%
200%
300%
400%
Percent Load
500%
600%
“Magnetic Type”

A magnetic type overload
has a current coil that
functions as a solenoid and
“pulls in” at a specific motor
current. When the coil pulls
in it opens the motor circuit
and stops the motor.
Normally, the trip time and
trip current can be adjusted
from potentiometers on the
overload. Magnetic type
overloads are typically found
in high temperature
environments or on motors
with high inrush currents.
There is no cool off time
required for this type of overload.
L1
L2
Trip Time
L3
NO
NO
Trip Current
NC
NC
T1
T2
T3
86
“Bi-Metallic Type”

A bi-metallic type overload works
on the same principle as an old
fashioned home thermostat. Two
dissimilar metals are bonded
together. As the motor
temperature increases the metals
start to bend. At a specified
temperature the bending metal
trips a mechanism and opens the
motor circuit. Bi-metallic type
Spring
overloads are normally found in
critical applications that require the
Bimotor to run very close to failure,
metallic
such as grocery store freezers,
Strip
unmanned pumping stations or
chemical process pumps. Bi-metallic
type overload normally reset
automatically.
T1
L1
87
“Bi-Metallic Type”

A bi-metallic type overload works
on the same principle as an old
fashioned home thermostat. Two
dissimilar metals are bonded
together. As the motor
temperature increases the metals
start to bend. At a specified
temperature the bending metal
trips a mechanism and opens the
motor circuit. Bi-metallic type
Spring
overloads are normally found in
critical applications that require the
Bimotor to run very close to failure,
metallic
such as grocery store freezers,
Strip
unmanned pumping stations or
chemical process pumps. Bi-metallic
type overload normally reset
automatically.
T1
L1
“Bi-metallic Thermo Disc”

Bi-metallic thermo discs function
the same as bi-metallic
overloads. The major difference is
that the thermo disc is actually
inserted into the motor frame.
When the motor temperature
increases to an unacceptable
level, the disc opens a normally
closed contact that turns off the
motor circuit.
88
“Thermistor”

Another form of motor
overload protection is a
thermistor. A thermistor
is a small resistive
device that is imbedded
in the stator windings of
the motor. The thermistor
sends a signal to a
controller that has been
calibrated to protect the
motor. At a specific
temperature the controller
activates a normally
closed contact and opens
the motor circuit.
NC
NC
Thermistor
89
“Electronic Thermal”
80
Trip/minutes at 0Hz
Trip/minutes at 6.7Hz
Trip/minutes at 13.4Hz
Trip/minutes at 20Hz
Trip/minutes at 33.3Hz
Trip/minutes at 46.6Hz
Trip/minutes at 60Hz
70
Time/Minutes
60
50
40

Another form of motor
overload protection
is an inverse time
electronic thermal overload.

Typically found in the
software of variable
frequency drives, this
function derives a set of
curves based on output
frequency and load to
determine a trip time.

A typical set of curves is
shown here.
30
20
10
0
50%
75%
100%
Percent Load
125%
150%
90
“Electronic Thermal”
80
Trip/minutes at 0Hz
Trip/minutes at 6.7Hz
Trip/minutes at 13.4Hz
Trip/minutes at 20Hz
Trip/minutes at 33.3Hz
Trip/minutes at 46.6Hz
Trip/minutes at 60Hz
70
Time/Minutes
60
50
40

Another form of motor
overload protection
is an inverse time
electronic thermal overload.

Typically found in the
software of variable
frequency drives, this
function derives a set of
curves based on output
frequency and load to
determine a trip time.

A typical set of curves is
shown here.
30
20
10
0
50%
75%
100%
Percent Load
125%
150%
Information and enrollment for Yaskawa Training
Classes can be accessed via:
 1-800-Yaskawa (927-5292); press 6 for Training Coordinator
assistance with class enrollment.
email to [email protected]. The Training Coordinator will
respond with answers and class enrollment assistance.
 Visit www.yaskawa.com - Choose “Training” on the
navigation bar to access class descriptions and schedules.
6/6/2002
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