Question 17.2: Calculate the starting line current and torque for the motor...

For starting from a standstill, we have s = 1. The equivalent circuit is shown in Figure 17.15(a). Combining the impedances to the right of the dashed line, we have
Z_{eq} = R_{eq} + jX_{eq} = \frac{j50(0.6 + j0.8)}{ j50 + 0.6 + j0.8} = 0.5812 + j0.7943 Ω
The circuit with the combined impedances is shown in Figure 17.15(b).
The impedance seen by the source is
Z_s = 1.2 + j2 + Z_{eq}
= 1.2 + j2 + 0.5812 + j0.7943
= 1.7812 + j2.7943
= 3.314∠57.48° Ω

Thus, the starting phase current is
\textbf{I}_{s,  starting} = \frac{\textbf{V}_s} {Z_s} =\frac{ 440∠0°} {3.314∠57.48°}
= 132.8∠−57.48° A rms
and, because the motor is delta connected, the starting-line-current magnitude is
I_{line,  starting} = \sqrt{3}I_{s,  starting} = 230.0 A rms
In Example 17.1, with the motor running under nearly a full load, the line current is I_{line} = 37.59 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 17.15, given by
P_{ag} = 3R_{eq}(I_{s,  starting})^2
= 30.75 kW
Finally, Equation 17.34 gives us the starting torque:

T_{dev} = \frac{P_{ag}}{ω_s}            (17.34)
T_{dev,  starting} = \frac{P_{ag}}{ ω_s}
= \frac{30,750} {2π(60)/2}
= 163.1 Nm
Notice that the starting torque is larger than the torque while running under full-load conditions. This is also typical of induction motors.