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Sähkömoottorin valinta

Mottorijärjestelmän malli

We see in the figure below a typical system, which uses an electric motor.

 

                           

 

The mathematical model of this system is as follows:

                      TTotal = J∙α+B∙ω+TFriction+TWork, where

  • TTotal is the total motor torque
  • J is the moment of inertia of the system
  • α is the angular acceleration
  • B is the viscous factor
  • TFriction is the total torque of friction in the system
  • TWork is the torque needed by the load

The moment of inertia is defined in the following equation as the sum of mass points m multiplied by their squared distance r from the axis:

                  

A mass m hanging from a drum of radius r causes the torque T:

                      T = m∙g∙r, where g is the acceleration of gravitation.

The centrifugal force F is calculated as:

                      F = m∙r∙ω2, where m is the mass, where the force is generated and ω is the angular velocity.

Gears

When we deal with gears, we are interested in the following things:

  1. Translation N
  2. Mechanical strength
  3. Efficiency η
  4. Life expectancy
  5. The moment of inertia J

The translation (N) is defined as the quotient of the rotational speeds of the input (ninput) and output axis (noutput) as follows

                      N = ninput/noutput.

The output torque (Toutput) of a gear is got from the input torque (Tinput) as shown here:

                      Toutput = N∙Tinput.

The moment of inertia of the load (JLoadOut) is seen to the motor as (JLoadIn)

                      JLoadIn = JLoadOut/N2.

The efficiency varies from gear to gear as listed here:

  • Planetary gear:    0.6-0.9
  • Cog gear              0.4-0.85
  • Screw gear           0.1-0.25-0.5
  • Precision gear      0.9-0.95

 

Force F by a belt causes a torque T in a cylinder of radius r as explained in the figure below.

            

                      T = F∙r/η.

Mass m on a belt causes the moment of inertia J in a cylinder of radius r as explained in the figure below.

                        

In these figures the belt cylinder system is seen as a gear with efficiency η.

 

For a screw transmission we get torque T from a longitudinal force F as shown in the figure below.

                 

 

The inertial moment caused by a mass m to the axis of a screw transmission can be calculated as explained in the below. Here q is the rise of the screw.

                                

In the figure below we see an example of dividing a motor load in small parts.

                           

 

Selecting the right gear for a servo motor

Katso: Yaskawa: How to select a servo motor

According to a video of Yaskawa, you should choose a suitable servo motor with the following 7 step procedure:

  1. Check what is the available voltage to your application.
  2. Determine the needed motion profile:

          a)     Repetitive operations, plot out motor speed throughout cycle, including acceleration/deceleration times.

          b)     For non-repetitive (like milling) operations calculate peak speed and acceleration

     3. Define the amount of torque “muscle” needed (difficult to calculate accurately but most forgiving):

          a)     Inertia J (influences to the acceleration)

          b)     Friction

          c)      External forces and torques.

     4. Calculate inertial ratio, IR, (often overlooked but important for determining servo performance).

               , where a is the gear reduction.

  • Basic servo drives may require inertia ratios of 3:1 or smaller
  • Advanced features (auto-tuning, vibration suppression, resonance filters, disturbance compensation functions) allow up to 30:1
  • A ratio of 1:1 gives excellent performance (but usually oversized motor).
  • Ratios less than 1:1 waste power with no performance advantage.

     5. Find a motor and drive that:

  • Matches supply voltage
  • Has rated speed, continuous torque, peak torque, that exceed calculated values and
  • Satisfies inertia ratio requirement in Step 4

     6. If there is a motor which is a close match you are finished. Otherwise continue to the step 7 to add gearing.

     7. Matching the motor to load with a gear. Gear matches the servo to the load by reducing speed, increasing torque and lowering inertia ratio:

  • Gear box manufactures list the inertia of the gear box to be added to the load.
  • Divide the motor speeds by required speed to get a starting gear ratio.
  • Divide the required torque by the gear ratio to find the new required torque.
  • This will narrow choices to a couple of motors. Find a motor with an acceptable inertia ratio. If two motors look equal choose the one with smaller inertia ratio.
  • Repeat this step a couple of times.

 

The different loads are reduced to the motor using the following equations, where the notations are obvious. The reduction is based to the equality of powers.

 

 

 

 

The inertial quantities are reduced to the motor with the equations. Here the reduction is based to the equality of the kinetic energies.

 

 

 

Exercise 1. Find a servomotor of NX type of Oriental Motor with the following caracteristics.

  1. Available power supply: 3 phase wye connected 230 V per phase.
  2. Rotational speed n = 300 rpm
  3. Load torque TL = 10 Nm
  4. Inertial moment JL = 4000 kg(cm)2
  5. Non-backlash

Solution:

  • Non-backlash means we must use PJ geared type.
  • Power is 2π(300/60)·10 Nm = 314 W. NX1040(640)AS/MS gives power 400 W, which is enough.
  • The load torque is (reduced to motor TL/a):
    - TL = 10 Nm > 1.27 Nm (NX640AS/MS): Not good
    - TL = 10 Nm > 5.08 Nm (NX1040AS/MS-J5): Not good
    - TL = 10 Nm < 10.2 Nm (NX1040AS/MS-J10): Fine, Also rotational speed 300 rpm is just enough
    - TL = 10 Nm < 25.4 (NX1040AS/MS-J25): Spins too slowly.
  • The inertial moment is (reduced to the motor is JL/a2):
    - JL = 4000 kg(cm)2 = 0.4 kgm2 > 0.2680 (NX1040AS/MS-J10): The 400 W-series is not good enough

 

  • We must now seek a motor-gear pair, which has large enough JL capability: The 750 W series is enough:
    - JL = 4000 kg(cm)2 = 0.4 kgm2 < 0.4710 (NX1075AS/MS-J10):

Thus NX1075AS/MS-J10 fulfils all the requirements; and it is the only motor-gear pair in the series which does this.

 

Exercise 2. Find a servomotor of NX type of Oriental Motor with the following caracteristics.

  1. Available power supply: 3 phase wye connected 230 V per phase.
  2. Rotational speed: Pulling wheel has radius rL = 20 cm.
  3. Load force FL = 10 N, load speed is vL = 10 m/s.
  4. Inertial mass mL = 1 kg,
  5.  Small or no backlash.

Solution:

  • Small or no backlash means standard or PS/PJ-geared solutions are all acceptable.
  • Rotational speed is got from the equation vL = rL · ωm. => ωm = vL/rL = (10m/s)/(0.2m) = 50 rad/s = 478 rpm. So only the standard or gear-ratio 5 is suitable.
  • Power is PL = vL · FL = 10 m/s · 10 N = 100 W.
  • The load torque is (reduced to motor TL/a):
    - TL = rL · FL = 0.2 m · 10 N = 2 Nm.
    - So Standard (50…400W) and NX610A/M-PS5 (TL = 1.43 NM) or NX810A/M-J5 (TL = 1.27 Nm) are not acceptable.
    - But Standard with power of 750 W (2.39 Nm) or NX920A/M-PS5 (200 W and 2.87 Nm) and NX820A/M-J5 (200W and 2.54 Nm) are good.
  • The inertial moment is (reduced to the motor JL/a2):
    - JL = mL · rL2 = 1 kg · (.2m)2 = 0.04 kgm2.
    - Standard is not accessible, only NX1040AS/MS-J5 (400W and 5.08 NM and 0.0669 kgm2) and NX1075AS/MS-J5 (750W and 9.56 Nm and 0.1180 kgm2) are possible solutions.

 

Thus the smaller one or NX1040AS/MS-J5 is the best choice.

 

 

 

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