In [1]:
%pylab inline
In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works, an autotransformer is relatively expensive. A much less expensive way to reduce the starting current is to use a device called $Y - \Delta$ starter (described earlier in this chapter). If an induction motor is normally $\Delta$-connected, it is possible to reduce its phase voltage $V_\phi$ (and hence its starting current) by simply re-connecting the stator windings in Y during starting, and then restoring the connections to $\Delta$ when the motor comes up to speed. Answer the following questions about this type of starter.
In [2]:
Vstart_Vrun = 1/sqrt(3)
print('''
Vstart_Vrun = {:.1f} %
===================='''.format(Vstart_Vrun*100))
I.e., $V_{\phi,Y\text{-start}} = V_{\phi,\Delta\text{-run}}/\sqrt{3}$
The starting phase current will also decrease to same percentage of the normal starting current as the phase voltage does.
For the $\Delta$-connection: $I_{L,\Delta} = \sqrt{3}I_{\phi,\Delta}$
For the Y-connection: $I_{L,Y} = I_{\phi,Y}$
But, $I_{\phi,\Delta} = \sqrt{3}I_{\phi,Y}$ because of $V_{L_Y} = \sqrt{3}V_{\phi_Y}$ gives:
$$I_{L,\Delta} = \sqrt{3}I_{\phi,\Delta} = \sqrt{3} \cdot \sqrt{3}I_{L,Y} = \mathbf{3} \cdot I_{L,Y} $$
In [3]:
Ild_Iphid = sqrt(3)
Ily_Iphiy = 1
Iphid_Iphiy = sqrt(3)
Ild_Ily = Ild_Iphid * Iphid_Iphiy / Ily_Iphiy
print('''
Ild = {:.0f} Ily
==========='''.format(Ild_Ily))