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Motor/Drive Sizing ExercisePowerPoint Presentation

Motor/Drive Sizing Exercise

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Motor/Drive Sizing Exercise

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Siemens Motion Control

3 - Axis Conveyor

The Following Data is Given:

¨ Mass to be transportedm= 400 kg

¨ Diameter of drive wheel D= 0.14 m

¨ Max. speedV max= 1.6 m/s

¨ Max. acceleration and deceleration a max= 6.4 m/s2

¨ Distance traveleds= 2 m

¨ Cycle time T= 7 s

¨ Mech.. efficiency mech.= 0.9

¨ Specific travelling resistance w f= 0.1

¨ Mech.. accuracys mech.= ±0.1 mm

¨ Overall accuracy requireds tot= ±0.2 mm

- Since the travel is symmetrical, we only have to consider the forward movement!
- So the new cycle time is:

Now determine the remaining time values of the curve.

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

- 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:

- 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 inertia
- Motor/ Gear box is more economical

- 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.

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

- Therefore:

- 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 PlayG = 3’
- ¨ Gear Unit Efficiency G.= 0.95
- - Inertia J G.= 0.001 Kgm2

- Therefore:

- Before we choose a motor, we should determine whether or not the position accuracy required is met.

- For the Gear Unit:

I.e. + 0.0305 mm

- For the Encoder:

With a 2-pole resolver

- Total:

Hence, the required accuracy it complied with.

- 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)?

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

- Therefore:

Where:

- 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.

- 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.

Mot max

Mot max

- The Acceleration Torque of the motor itself is thus:

- 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

- We already have determined the acceleration torque.

- The Torque during constant travel now needs to be calculated.

- Finally the torque during deceleration is required.

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

- The Effective Torque is calculated as follows:

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

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

Since the operating point

is well below the S1 Curve,

the motor selected is suitable.

- The Inverter is selected according to the maximum motor current and mean motor current.

- Therefore the Max current is given by:

- The Mean motor current is calculated as follows:

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

- 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.

- From the data calculated select the drive that best fits.

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

- Thus, checking:

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

- 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:

- Power output during constant travel, calculation:

- Maximum power output during deceleration, calculation:

- 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:

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

- The Mean Power output during motor operation is given by:

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

- The Mean DC Link current is given by:

- The next step involves determining the Braking Power.

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

- The Maximum Braking Power is given by:

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

- 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!

- 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.

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

- Since the Z-Axis is a lifting drive the, ALL Torque cases must be evaluated.
- Lifting the load, motor torque during constant travel
- Lowering the load, motor torque during constant travel
- Lifting the load, motor torque during acceleration
- Lifting the load, motor torque during deceleration
- Lowering the load, motor torque during acceleration
- Lowering the load, motor torque during deceleration

- The same calculations used for the X & Y - Axis are followed here, too, with one major consideration change.
- What is it?
- The acceleration and Deceleration times are > 250ms, so only the 160% overload rating can be applied!

- Once again, since the Z-Axis is a lifting drive the, ALL Power cases must be evaluated for future determination of DC Link current needed for rectifier selection.
- Lifting the load, Power output during constant travel
- Lowering the load, Power output during constant travel
- Lifting the load, Power output during acceleration
- Lifting the load, Power output during deceleration
- Lowering the load, Power output during acceleration
- Lowering the load, Power output during deceleration

- 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:

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

- Now base on this calculated information, choose a Rectifier.

- To maintain a uniform design, a Compact Plus Rectifier should be chosen.
- Therefore we should choose:
- 6SE7024 - 1EP85-0AA0 with IDCn = 41 A

- Confirmation of this choice is given by:

- 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:

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

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

- Therefore, we should choose:
- 6SE7018 - 0ES87 - 2DC0, with P20 = 5 kW, & R = 80

- Checking our selection:

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

- What other components/options are needed?
- Given:
- Positioning to be carried out non-centrally.
- Need Technology Option, Order # 6SW1700-5AD00-1XX0
- All three motions are synchronized.
- Need SIMOLINK board, Order # 6SX7010-0FJ00
- PROFIBUS communication to PLC.
- Need CBP board, Order # 6SX7010-0FF00
- A Resolver is used.
- Need SBR1/2 board, Order # 6SX7010-0FC00 or 6SX7010-0FC00

- What other components/options are needed that are not mentioned?
- Motor Power Cables - assembled.
- Order # 6FX2002-5DA31-???0, ??? = Cable Length
- Feed Back Cables - assembled.
- Order # 6FX2002-2cf01-???0, ??? = Cable Length
- Other Miscellaneous options?
- I/O Expansion Boards
- RFI Filter
- Line Reactor
- Capacitor Module
- OP1S

- Other Miscellaneous options?
- -Z +K80, Safety Off - Special Order with extended Lead Time
- -Z +C91, 24 VDC Power Supply for Rect. - Special Order with extended Lead Time
- There will be some other Option Codes Available, but they all bring extended Lead times with them.

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

- What is the limiting factor?
- The Acceleration and Deceleration times and how they relate to the Drive Process time for the Position Loop.

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

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

- 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.

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

END