Question 17.2: 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 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  \Omega

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°

Thus, the starting phase current is
I_{s,  starting}= \frac{V_s}{Z_s} = \frac{440 \angle 0°}{3.314 \angle 57.48°} = 132.8 \angle -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}}{\omega _s}           (17.34)

=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.