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3-phase Delta Induction machine problem for Electric Machines and Power Electronics

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VinnyCee
#1
May1-11, 05:15 AM
P: 492
1. The problem statement, all variables and given/known data

A [tex]\Delta[/tex] connected induction machine is to start from a source of 480V, 60 Hz.

The motor ratings are :

4 poles, [tex]\Delta[/tex] connected, 480V, 60 Hz, [tex]X_{ls}\,=\,X_{lr}\,=\,4\,\Omega[/tex], [tex]X_m\,=\,50\,\Omega[/tex], [tex]R_s\,=\,0.25\,\Omega[/tex], [tex]R_r\,=\,0.4\,\Omega[/tex].

Find the current in the windings and the torque of the motor and the current at the line side.


2. Relevant equations

[tex]I_S\,=\,\frac{V_{l-l}}{\left(R_s\,+\,j\,X_{ls}\right)\,+\,\left(j\,X_m\,||\,R_r\,+\,j\,X_{ lr}\right)}[/tex]

[tex]I_R\,=\,I_S\,\frac{j\,X_m}{R_r\,+j\,X_{lr}\,+\,j\,X_m}[/tex]

[tex]P_{gap}\,=\,n_{ph}\,|I_R|^2\,\left(\frac{R_r}{s}\right)[/tex]

Where [tex]n_{ph}[/tex] is number of stator phases.

[tex]T\,=\,3\,\frac{P_{gap}}{\omega_s}\,\frac{p}{2}[/tex]

Where p is number of poles.

The above equations are not given explicitly anywhere, but an example uses them so I think they are right.



3. The attempt at a solution

At starting, s = 1.

[tex]I_S\,=\,\frac{480}{\left(0.25\,+\,j\,4\right)\,+\,\left(j\,50\,||\,0.4\ ,+\,j\,4\right)}\,=\,62.1^{\angle\,-85.6^{\circ}}[/tex]

[tex]I_R\,=\,\left(62.1^{\angle\,-85.6^{\circ}}\right)\,\frac{j\,50}{0.4\,+j\,4\,+\,j\,50}\,=\,57.5^{\ang le\,-85.2^{\circ}}[/tex]

[tex]P_{gap}\,=\,(3)\,(57.5)^2\,(0.4)\,=\,3967.5[/tex]

[tex]T\,=\,3\,\left(\frac{3967.5}{377}\right)\,\left(\frac{0.4}{1}\right)\,= \,63.14\,Nm[/tex]

Does that look right? How do I get the current at the line side?
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VinnyCee
#2
May1-11, 04:42 PM
P: 492
I found an equation that will possibly give line side current...

[tex]I_{line}\,=\,\sqrt{3}\,I_{ph}[/tex]

But what is [tex]I_{ph}[/tex]? Is it [tex]I_S[/tex] or [tex]I_R[/tex]?


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