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3phase Delta Induction machine problem for Electric Machines and Power Electronicsby VinnyCee
Tags: 3 phase, electric machines, induction machine, power electronics, power engineering 
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#1
May111, 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_{ll}}{\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? 


#2
May111, 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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