Motor/Drive Sizing Exercise

Drives Business Unit General Motion Control (June 1999)

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Drives and Standard Productsfrom Siemens

Motor/Drive Sizing Exercise


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Y

X

Z

3 - Axis Conveyor


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Calculation of the Travel gear (X-Axis)

The Following Data is Given:¨ Mass to be transported m= 400 kg¨ Diameter of drive wheel D= 0.14 m¨ Max. speed V max= 1.6 m/s¨ Max. acceleration and deceleration a max= 6.4 m/s2

¨ Distance traveled s= 2 m¨ Cycle time T= 7 s¨ Mech.. efficiency mech.= 0.9¨ Specific travelling resistance w f= 0.1

¨ Mech.. accuracy s mech.= ±0.1 mm

¨ Overall accuracy required s tot= ±0.2 mm


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Travel Curve (X-Axis)

t t

t

T´

b v

tot

v

vmax Area corresponds to travel distance

Forwards Reverse

t

vmax

t kt p


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Calculation of the Time Segments (X-Axis)Since the travel is symmetrical, we only have to consider the forward movement!

So the new cycle time is:

sec5.32

7

2

TT

Now determine the remaining time values of the curve.

max

max

a

vtt vb s25.0

4.6

6.1

max

maxmax 22v

tv

tvs

t

vb

k s16.1

225.0

6.1225.0

6.12

s5.125.0125.0

totp tTt

vkbtot tttt

s25.15.3


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Calculation of the Maximum Speed (X-Axis)

Now we must calculate the Maximum speed under load at the drive wheel.

D

vnLoad

60maxmax rpm27.218

14.0

606.1

Due to low rpm, a gearbox should be used to better match the motor speed to the load speed. In this case a gearbox with a transmission ratio of i=10 is chosen, giving a resulting rpm at the motor of:

rpmnin LoadMot 7.218227.21810maxmax

Why is a gearbox necessary? Why not just select a motor with a lower rpm?

Answer: A smaller motor can be used - larger motor mean larger motor inertiaMotor/ Gear box is more economical


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Calculation of the Maximum Torque (X-Axis)Now we must calculate the Maximum Resistance Torque of the load at the drive wheel.

We also need to know the Maximum Acceleration and Deceleration Torque for the Load.

2

Dwgm fW Nm47.27

2

14.01.081.9400

D

aload

2max

2

2

DmJ load

loadloadloadbra J ,

24.9114.0

24.6 s

22

96.12

14.0400 kgm

Nm2.1794.9196.1

Maximum Acceleration and Deceleration Torque for a Rotational motion is given by:

loadloadloadbra J ,

Therefore:


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Calculation of the Maximum Torque (X-Axis) Cont.

Now we can get the Maximum Torque on the output side of the gear.

We also need to know the Maximum Acceleration and Deceleration Torque of the gear unit itself. The Technical Data of the gear unit is as follows:

¨ Gear Ratio i = 10¨ Max, Torque G = 400 Nm¨ Torsional Play G = 3’¨ Gear Unit Efficiency G.= 0.95- Inertia J G.= 0.001 Kgm2

mech

Wloadaload 1)(max Nm6.229

9.0

1)47.272.179(

Therefore:

iJ loadGGbra *, Nm914.0104.91001.0


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Calculation of the Position Accuracy (X-Axis)Before we choose a motor, we should determine whether or not the position accuracy required is met.

For the Encoder:

Total:

60360G

gear

Ds

mm061.0

60

3

360

14.0

I.e. + 0.0305 mm

For the Gear Unit:

zi

Dsencoder

mm04.0

102410

14.0

With a 2-pole resolver

encodergearmechtot ssss mm2.01705.004.00305.01.0

Hence, the required accuracy it complied with.


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Calculation of the Position Accuracy of other Feedback Devices

What would be the accuracy if a different feedback device had been chosen?

For a Pulse encoder with 2048 PPR?

For a Sin/Cos encoder or Absolute value (ERN/EQN)?

zi

Dsencoder

mm021.0

204810

14.0

zi

Dsencoder

mm00044.0

1010

14.05


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Calculation of the Maximum Torque reflected to the Motor (X-Axis)

Maximum Torque reflected to the motor is determined by adding all the individual torque values.

Therefore:

Hence, before we can figure out the Complete Torque a motor must be chosen. Since we know some of the torque required, an educated guess must be made in choosing an appropriate motor.

GmechWloadaGbraMotaMot i

1

)(max

95.09.010

1)47.272.179(914.0

Mota NmMota 08.25

2914104.91 sJJiJ MotMotloadMotMota Where:


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Calculation of the Maximum Torque reflected to the Motor (X-Axis)

Choose a Synchronous Servo motor (1FT6) based on the information known and the Dynamic Limit curves.

The first 1FT6 motor with nn=3000 rpm, which satisfies the condition of the Dynamic Limit curve is 1FT6084-8AF7 with the following characteristics:- Pn = 4.6 kW- n = 14.7 Nm- Max = 65 Nm- Jmot = 0.0065 kgm2- kTn = 1.34 Nm/A- mot = 0.92- 0 = 20 Nm

Explain why this motor was chosen over any other motor.


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Dynamic Limit curve of the 1FT6084-8AF7 Motor (X-Axis)

Mot max

1FT6084-8AF7with SIMOVERT MASTERDRIVE MC 3 AC 480V

0

100

200

300

400

500

600

700

0 1000 2000 3000

Speed [rpm]

Torq

ue

[lb

-in

]

0

10

20

30

40

50

60

70

Torq

ue

[Nm

]

S1 (100K)

Intermittent Operating Region

Continous Operating Region


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Dynamic Limit curve of the 1FT6082-8AF7 Motor (X-Axis)

1FT6082-8AF7with SIMOVERT MASTERDRIVE MC 3 AC 480V

0

50

100

150

200

250

300

350

400

0 1000 2000 3000

Speed [rpm]

Torq

ue

[lb

-in

]

0

5

10

15

20

25

30

35

40

45

Torq

ue

[Nm

]

S1 (100K)

Intermittent Operating Region

Continous Operating Region

Mot max


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Calculation of the Maximum Torque of the Motor (X-Axis)

The Acceleration Torque of the motor itself is thus:

NmMotbra 94.59140065.0,

NmaMotMot 03.3108.2594.5max

Now the Maximum motor torque can be determined.

Next a check must be done to ensure that the thermal limits of the motor are not exceeded. This is accomplished by determining the motor torques at every point of the travel curve.

We already have determined the acceleration torque. 31.03 Nm


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Calculation of the Travel Curve Torques (X-Axis)

We already have determined the acceleration torque.

NmaMotMot 03.3108.2594.5max

The Torque during constant travel now needs to be calculated.

Finally the torque during deceleration is required.

Gmech

WkMot i 1

Nm21.30959.010

147.27

)()(

1)(

WloadbrsignGmech

WLoadbrGaMotbrbrMoti

Nm83.1910

95.09.0)47.272.179(914.094.5


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Torque Characteristic Curve (X-Axis)

MMot

31.03 Nm

3.21 Nm

-19.83 Nm

0.25 s 1 s 0.25 s

3.5 s

t


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Checking the Validity of the selected Motor (X-Axis)

To verify it the motor we selected from the Dynamic Curve Profile is valid, the effective(rms) torque and speed must be determined.

By using the Travel Curve, the Effective Speed is calculated as follows:

The Effective Torque is calculated as follows:

T

tiiMoteff

2 Nm10

5.3

25.083.19121.325.003.31 222

T

tnn

ni

EB

mean2 rpm5.779

5.3

25.027.2182

17.218225.027.2182

By using the S1 Curve for the given motor, a determination can be made whether or not the motor selected is satisfactory.


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Checking the Validity of the selected Motor (X-Axis)

M/Nm

Meff

n/min-1nmean

12

10

8

6

4

2

01000 2000 30001500 2500500

14

16

18

20

Since the operating pointis well below the S1 Curve,

the motor selected is suitable.


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Selecting the Inverter (X-Axis)The Inverter is selected according to the maximum motor current and mean motor current.

The Mean motor current is calculated as follows:

Therefore the Max current is given by:

100

maxmax

Tn

MotMot k

I

A16.2334.1

03.31

Tk

tI

Tn

iiMot

meanMot100

A4.3

5.334.1

25.083.19121.325.003.31

For overload calculations, the motor current during constant travel is calculated:

100Tn

kMotkMot k

I

A4.234.1

21.3


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Selecting the Inverter (X-Axis)

Thus, checking:

Since the Accelerating and Decelerating times are < 250 ms and the time between is > 750 ms, a check should be made to see if the 300% overload capability of the Compact Plus unit can be utilized.

AIAI UnMot 30316.23max

AIAI UnmeanMot 2.104.3

AIAI UnkMot 3.991.04.2

From the data calculated select the drive that best fits.

Drive Selected: 6SE7021-0TP50 with an Iun = 10.2 A

Since the above criteria are met, a correct selection has been made.


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Determination of DC link Currents (X-Axis)

Power output during constant travel, calculation:

The Max DC link current and the mean DC link current must be determined for later rating of the rectifier unit. This is done by first determining all of the motor power levels within the travel curve.

Maximum power output during acceleration, calculation:

Maximum power output during deceleration, calculation:

9550

maxmax

MotaMotaMot

nP

kW09.7

9550

7.218203.31

9550

maxMotkMotkMot

nP

kW734.0

9550

7.218221.3

9550

maxmax

MotbrMotbrMot

nP

kW53.4

9550

7.218283.19


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Power Characteristic Curve (X-Axis)

0.25 s

1 s

0.25 s

PMot

t

7.09 kW

0.734 kW

-4.53 kW

Negative area corresponds to regenerativeoperation

3.5 s


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The Mean Power output during motor operation is given by:

Now that we have the Power Characteristic curve, the maximum power can be determined and the Maximum DC Link current can be calculated.

Maximum DC Link current during acceleration is given by:

lineInvMot

MotInvLink V

PI

35.1max

max A66.12

46035.198.092.0

7090

T

tPP

Pi

EMotBMot

meanMot2 kW463.0

5.3

1734.025.009.721

Now we need to determine the Mean value of DC current, but before we can do this we need the Mean Power.


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The next step involves determining the Braking Power.

Now that we have the Mean Power the mean DC Link currents can be calculated.

The Mean DC Link current is given by:

lineInvMot

meanMotmeanInvLink V

PI

35.1A83.0

46035.198.092.0

463


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Determination of Braking Power (X-Axis)

The Mean Braking Power is obtained from the negative characteristic of the motor output. The calculation is as follows:

The Braking Power and the Mean Braking Power are calculated for later sizing of the braking resistor.

The Maximum Braking Power is given by:

InvMotbrMotbr PP maxmax kW08.498.092.053.4

InvMot

iEbrMotBbrMot

meanbr T

tPP

P

2

kW146.098.092.05.3

25.0)53.4(21


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Calculation of the Travel Gear (Y-Axis)

Since the Y-Axis is the same as the X-Axis, you will now proceed with all of the same calculations for the Y-Axis on your own and present the results.

Ha Ha - Just Kidding!


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Calculation of the Lifting Drive (Z-Axis)

Since the Z-Axis is slightly different than the X-Axis and Y-Axis, we will now go over the major differences and dispense with the entire calculation.

What are the differences?

Travel Distance is represented in Height.The entire Travel Curve must be taken into consideration because of the different torque values for Lifting and Lowering even if it is Symmetrical.Rotational Force is translated to Linear Force via a Rack and Pinion.


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Calculation of the Torque of the Lifting Drive (Z-Axis)

Since the Z-Axis is a lifting drive the, the Resistance Torque Calculation is now called the Lifting Torque.

NmD

gmH 1.982

1.081.9200

2


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Torque Characteristic Curve of the Lifting Drive (Z-Axis)

MMot 16.58 Nm

11.47 Nm

6.37 Nm4.08 Nm

8.39 Nm

12.7 Nm

0.6 s 0.3 s 0.6 s 0.6 s 0.3 s 0.6 s7 s

Lifting Lowering

t


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Selection of the Inverter for the Lifting Drive (Z-Axis)


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Selection of the Inverter for the Lifting Drive (Z-Axis)


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Power output Characteristic Curve for the Lifting Drive (Z-Axis)

PMot

0.6 s 0.3 s 0.6 s

0.6 s 0.3 s 0.6 s

7 s

Lifting Lowering

t

4.97 kW

3.44 kW

1.91 kW

-1.22 kW

-2.52 kW

-3.81 kW

Negative area corresponds to regenerativeoperation


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Selection of the Rectifier Unit for the 3-Axis Conveyer

Now that all of the Inverters have been chosen, a rectifier needs to be selected.

Recall that the X and Z -Axis operate simultaneously.Also, recall that the Y-Axis is much smaller than any of the other axis’ and, thus, it does not need to be considered in the calculation of the rectifier.

First calculate the maximum DC Link current as follows:

maxmaxRe InvLinkctLink II AAA 84.2118.966.12


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Next, calculate the Mean value of the DC Link current as follows:

meanInvLinkmeanctLink II Re AAA 65.182.083.0

Now base on this calculated information, choose a Rectifier.


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Confirmation of this choice is given by:

AI ctLink 84.21maxRe AI nLink 6.656.1

AI meanctLink 89.1Re AI nLink 41


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Selection of the Braking Resistor for the 3-Axis Conveyer

What components are needed?Just the Braking Resistor! Why?The Chopper is integrated into the Compact Plus Rectifier!

What assumptions can be made?Since the X & Z-Axis travel simultaneously, both may brake at the same time.

So, first we need to calculate the maximum Braking Power Level as follows:

Invbrbr PP max kWkWkW 4.732.308.4


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Next, we need to determine the Mean Braking Power as follows:

meanInvbrmeanbr PP kWkWkW 426.028.0146.0

Now that all the calculation have been made a suitable resistor needs to be chosen.


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kWPbr 4.7max kWP 5.75.1 20

kWP meanbr 426.0 kWP 11.15.4/20

Checking our selection:

Since the above conditions have been met, our choice is valid.


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Other requirements for the 3-Axis Conveyer


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Other requirements for the 3-Axis Conveyer


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Other requirements for the 3-Axis Conveyer Cont.


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Ability of Drive to actually do the 3-Axis Conveyer

Now that everything has been calculated and all the options and accessories have been chosen, can the drive do the application????

Lets Look at the Function Diagram for the Operating Mode MDI [823]


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U953.32=___ (20)

1 0

0 1 10

U550.2 U559.2

0 1 10

U532 (0)

KKU550.3 U559.3

0 1 10

U533 (0)

KK

U534 (0)

KK

Position (X) Speed (F)

X

n540.12

F

MDI block number 0...10;Following is displayed:- at standstill: selected MDI block- during travel: currently traversed MDI block - no MDI mode ==> display "0"

U559.1 U550.1

<2>Recommended: U953.32=4

<4>

n540.13

-999 999 999... 999 999 999 LU 1... 100 000 000 [x 10 LU/min]90 30... 91 39

(90 30)

G

at MD1=3


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T2 = 4 x T0T3 = 8 x T0T4 = 16 x T0T5 = 32 x T0T6 = 64 x T0T7 = 128 x T0T8 = 256 x T0T9 = 512 x T0T10 = 1024 x T0

23456789

1011 ... 19

20

Parameter for setting the sampling timeValue range: 2 ... 20Factory setting: 20 (block is not calculated)Parameter value Sampling time

(T0 = 1/pulse frequency = 1/P340)

Sampling time at 10 kHzpulse frequency (T0 = 100 µs)

Reserved for future applicationsBlock is not calculated

0.4 ms0.8 ms1.6 ms3.2 ms6.4 ms

12.8 ms25.6 ms51.2 ms

102.4 ms

Sampling time at 5 kHzpulse frequency ( T0 = 200 micro sec)

0.8 ms01.6 ms

3.2 ms6.4 ms

12.8 ms25.6 ms51.2 ms

102.4 ms204.8 ms

Typically the Position Loop should be calculated at least 10 times!


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This gives you a value of 32 ms for accurate positioning.

Lets check this value with the Acceleration and Deceleration time required by the application.

msTimePosition 32 ms250

Since the Positioning Time of the Position Loop is less than the required Acceleration and Deceleration time, “WE CAN DO THAT”.


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Motor/Drive Sizing Exercise

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