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

by VinnyCee
Tags: 3 phase, electric machines, induction machine, power electronics, power engineering
 P: 492 1. The problem statement, all variables and given/known data A $$\Delta$$ connected induction machine is to start from a source of 480V, 60 Hz. The motor ratings are : 4 poles, $$\Delta$$ connected, 480V, 60 Hz, $$X_{ls}\,=\,X_{lr}\,=\,4\,\Omega$$, $$X_m\,=\,50\,\Omega$$, $$R_s\,=\,0.25\,\Omega$$, $$R_r\,=\,0.4\,\Omega$$. Find the current in the windings and the torque of the motor and the current at the line side. 2. Relevant equations $$I_S\,=\,\frac{V_{l-l}}{\left(R_s\,+\,j\,X_{ls}\right)\,+\,\left(j\,X_m\,||\,R_r\,+\,j\,X_{ lr}\right)}$$ $$I_R\,=\,I_S\,\frac{j\,X_m}{R_r\,+j\,X_{lr}\,+\,j\,X_m}$$ $$P_{gap}\,=\,n_{ph}\,|I_R|^2\,\left(\frac{R_r}{s}\right)$$ Where $$n_{ph}$$ is number of stator phases. $$T\,=\,3\,\frac{P_{gap}}{\omega_s}\,\frac{p}{2}$$ 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. $$I_S\,=\,\frac{480}{\left(0.25\,+\,j\,4\right)\,+\,\left(j\,50\,||\,0.4\ ,+\,j\,4\right)}\,=\,62.1^{\angle\,-85.6^{\circ}}$$ $$I_R\,=\,\left(62.1^{\angle\,-85.6^{\circ}}\right)\,\frac{j\,50}{0.4\,+j\,4\,+\,j\,50}\,=\,57.5^{\ang le\,-85.2^{\circ}}$$ $$P_{gap}\,=\,(3)\,(57.5)^2\,(0.4)\,=\,3967.5$$ $$T\,=\,3\,\left(\frac{3967.5}{377}\right)\,\left(\frac{0.4}{1}\right)\,= \,63.14\,Nm$$ Does that look right? How do I get the current at the line side?
 P: 492 I found an equation that will possibly give line side current... $$I_{line}\,=\,\sqrt{3}\,I_{ph}$$ But what is $$I_{ph}$$? Is it $$I_S$$ or $$I_R$$?

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