```
In [1]:
```%pylab inline

```
```

A 460-V 50-hp six-pole $\Delta$ -connected 60-Hz three-phase induction motor has a full-load slip of 4 percent, an efficiency of 91 percent, and a power factor of 0.87 lagging. At start-up, the motor develops 1.75 times the full-load torque but draws 7 times the rated current at the rated voltage. This motor is to be started with an autotransformer reduced voltage starter.

- What should the output voltage of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor?

- What will the motor starting current and the current drawn from the supply be at this voltage?

```
In [2]:
```Vt = 460 # [V]
Wperhp = 746 # official conversion rate of "electrical horsepowers"
Pout = 50 * Wperhp # [W]
PF = 0.87
eta = 0.91
times_tor = 1.75
times_cur = 7

If a torque of 1.75 $\tau_{rated}$ is produced by a voltage of 460 V, then a torque of 1.00 $\tau_\text{rated}$ would be produced by a voltage of:

$$\frac{1.00\tau_\text{rated}}{1.75\tau_\text{rated}} = \left(\frac{V_{T2}}{460V}\right)^2$$```
In [3]:
```Vt2 = sqrt(1.00/times_tor * Vt**2)
print('''
Vt2 = {:.0f} V
==========='''.format(Vt2))

```
```

```
In [4]:
```Il2_Il1 = Vt2/Vt
Il1_Irated = times_cur
Il2_Irated = Il2_Il1 * Il1_Irated
print('''
Il2 = {:.2f} Irated
================='''.format(Il2_Irated))

```
```

The input power to this motor is: $$P_\text{in} = \frac{P_\text{out}}{\eta}$$

```
In [5]:
```Pin = Pout / eta
print('Pin = {:.1f} kW'.format(Pin/1000))

```
```

The rated current is equal to: $$I_\text{rated} = \frac{P_\text{in}}{\sqrt{3}V_TPF}$$

```
In [6]:
```Irated = Pin / (sqrt(3)*Vt*PF)
print('Irated = {:.2f} A'.format(Irated))

```
```

Therefore, the motor starting current is

```
In [7]:
```Il2 = Il2_Irated * Irated
print('''
Il2 = {:.1f} A
============='''.format(Il2))

```
```

The turns ratio of the autotransformer that produces this starting voltage is:

$$\frac{N_{SE}+N_C}{N_C} = \frac{V_T}{V_{T2}} = a$$```
In [8]:
```a = Vt/Vt2
print('a = {:.3f}'.format(a))

```
```

so the current drawn from the supply will be: $$I_\text{line} = \frac{I_\text{start}}{a}$$

```
In [9]:
```Iline = Il2 / a
print('''
Iline = {:.0f} A
============='''.format(Iline))

```
```