Electrical Machines and Power Electronics - 3 phase, 6 pole induction motor

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SUMMARY

The discussion focuses on the calculations for a three-phase, 6-pole, \Delta connected induction motor rated at 208V and 60Hz. The starting torque calculated for the motor in Y connection is 11.68 Nm, with both phase and line currents equal to 10 A. In the \Delta connection, the torque increases to 35 Nm, while the phase current is 17.34 A and the line current is 30 A. These calculations utilize specific equations related to induction machines and confirm the accuracy of the results presented.

PREREQUISITES
  • Understanding of three-phase induction motors
  • Familiarity with electrical circuit analysis
  • Knowledge of torque calculations in electrical machines
  • Proficiency in using complex numbers for impedance calculations
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Homework Statement



A three phase, 6-pole, \Delta connected induction motor is rated 208V, 60Hz. It has R_R\,=\,5.2\,\Omega, R_S\,=\,2.2\,\Omega, X_m\,=\,220\,\Omega, X_{lr}\,=\,7.5\,\Omega, X_{ls}\,=\,2.3\,\Omega.

Calculate the starting torque and both phase and line currents when the motor is started in \Delta and Y.

Homework Equations



Various equations from the notes about induction machines and multiple pole pairs.

I_{S,\,Y}\,=\,\frac{\frac{V_{rated}}{\sqrt{3}}}{(R_S\,+\,j\,X_{ls})+(j\,X_m\,||\,R_R\,+\,j\,X_{lr})}

T_Y\,=\,3\,\cdot\,\frac{P_{gap}}{\omega_s}\,\cdot\,\frac{p}{2}

I_{R,\,Y}\,=\,I_{S,\,Y}\,\cdot\,\frac{j\,X_m}{R_R\,+\,j\,X_{lr}\,+\,j\,X_m}

P_{gap}\,=\,|I_{R,\,Y}|\,\cdot\,R_R

The Attempt at a Solution



I_{S,\,Y}\,=\,\frac{\frac{208}{\sqrt{3}}}{(2.2\,+\,j\,2.3)+(j\,220\,||\,5.2\,+\,j\,7.5)}\,=\,10\,\angle\,-53.8^{\circ}

In Y, the line current is same as phase current above, right?

I get the torque in Y as T_Y\,=\,11.68\,Nm.

Does that look right?
 
Last edited:
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This is correct. The solutions also state...

I_{ph,\,Y}\,=\,|I_{S,\,Y}|\,=\,10\,A

I_{l,\,Y}\,=\,I_{ph,\,Y}\,=\,10\,A

And for \Delta:

T_{\Delta}\,=\,3\,\cdot\,T_Y\,=\,35\,A

I_{ph,\,\Delta}\,=\,\frac{208}{120}\,\cdot\,10\,=\,17.34\,A

I_{l,\,\Delta}\,=\,\sqrt{3}\,\cdot\,I_{ph,\,\Delta}\,=\,30\,A
 
Last edited:

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