Question 16.2: Starting Current and Torque Calculate the starting line curr...

For starting from a standstill, we have s = 1. The equivalent circuit is shown in Figure 16.15(a). Combining the impedances to the right of the dashed line, we have

Z_{\text{eq}}=R_{\text{eq}}+jX_{\text{eq}}=\frac{j50(0.6+j0.8)}{j50+0.6+j0.8}=0.5812+j0.7943 \ \Omega

The circuit with the combined impedances is shown in Figure 16.15(b).

The impedance seen by the source is

\begin{matrix} Z_s&=&1.2+j2+Z_{\text{eq}} \\ &=&1.2+j2+0.5812+j0.7943 \\ &=&1.7812+j2.7943 \\ &=&3.314 \underline{/57.48^\circ} \ \Omega \end{matrix}

Thus, the starting phase current is

\begin{matrix} \pmb{\text{I}}_{s,starting}&=&\frac{\pmb{\text{V}}_s}{Z_s}=\frac{440 \underline{/0^\circ}}{3.314 \underline{/57.48^\circ}} \\ &=& 132.8 \underline{/-57.48^\circ} \text{ A rms} \end{matrix}

and, because the motor is delta connected, the starting-line-current magnitude is

I_{\text{line,starting}}=\sqrt{3}\pmb{I}_{s,starting}=230.0 \text{ A rms}

In Example 16.1, with the motor running under nearly a full load, the line current is I_{\text{line}} = 37.59 \text{ A}. Thus, the starting current is approximately six times larger than the full-load running current. This is typical of induction motors.
The power crossing the air gap is three times the power delivered to the right of the dashed line in Figure 16.15, given by

\begin{matrix} P_{\text{ag}}&=&3R_{eq}(I_\text{}{\text{s,starting}})^2 \\ &=&30.75 \text{ kW} \end{matrix}

Finally, Equation 16.34 gives us the starting torque:

\begin{matrix} T_{\text{dev,starting}}&=&\frac{P_{\text{ag}}}{\omega_s} \\ &=&\frac{30,750}{2 \pi (60)/2} \\ &=&163.1 \text{ Nm} \end{matrix}

Notice that the starting torque is larger than the torque while running under full-load conditions. This is also typical of induction motors.