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Induction Motor Data

Hi All

I am try to produce a torque curve using the standard torque equation, using the parameters given by the manufacturer as in the attached picture.

$$Torque =\dfrac{1}{\omega_s}\cdot\dfrac{V_1^2\cdot\dfrac{R_2}{s}}{(R_1+\dfrac{R_2}{s})^2+(X_1+X_2)^2}$$

At slip=1

Torque = (6350.85^2*(1.264/1))/(104.71*((0.535+1.264/1)^2+(6.086+3.309)^2))

Starting Torque = 5320.96Nm at slip = 1

However the company has given the starting torque as 8420Nm this is a massive 63% difference. Can anyone explain why such a big difference.

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  • \$\begingroup\$ The standard torque equation does not honor the squirrel cage configuration. The term R2/s needs to be modified so R2 becomes bigger at higher slip. \$\endgroup\$
    – Janka
    Commented May 10, 2018 at 7:00
  • \$\begingroup\$ Not sure if the equation is correct. It also may describe the quasi linear part of the IM characteristics - low slip, beyond max torque point. Check the equation, add link where you found it. \$\endgroup\$ Commented May 10, 2018 at 8:53
  • \$\begingroup\$ Hi do you or does anyone know an equation that works. Many papers I found online on this issue use the same equation or the Thevenin model. I think theses approaches are probably simplistic for the scale of the problem. The motor manufacturer may be using a more advanced induction motor model to generate their torque curve. \$\endgroup\$
    – ekobani
    Commented May 18, 2018 at 12:59

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You have neglected to multiply the results by the number of phases, but that results in calculated locked rotor torque much higher than the published value. The calculated value should be higher because the calculation neglects core losses and stray load losses, but it shouldn't be that much higher. Perhaps the published value is the minimum point of the torque vs. speed curve which doesn't necessarily occur at locked rotor. They may also be allowing for voltage drop in the power distribution system during starting.

Except for the missing multiplication by the number of phases, equation is as given by Fitzgerald, Kingsley, Umans, Electric Machinery 4th ed. It is derived from developed torque x speed = power dissipated in R2(1-s)/s. The power dissipated in R2(1-s)/s is calculated from the equivalent circuit shown in the question without the magnetizing branch (Xm & Rfe). Delivered torque is less than developed torque because friction and windage are subtracted.

The image below shows the torque vs. slip curves for both the rated and the locked-rotor values of the equivalent circuit elements. The values change as the rotor current frequency changes. The effect is a transition from the curve that is based on the locked rotor values of the circuit elements to the curve based on rated values. Sometime this transition results of a minimum value of resultant the torque vs. slip curve occurring between the locked rotor and break down points on the curve. See transition sketched in green.

enter image description here

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  • \$\begingroup\$ I now see that I need to multiply the results by 3 which will go well beyond the given tested result from manufacturer. The equation I am using may not be suitable, also from curve above, the locked rotor and rated parameters need to be used Thanks. \$\endgroup\$
    – ekobani
    Commented May 11, 2018 at 10:09

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