# Excercises Electric Machinery Fundamentals

## Problem 6-26



In [1]:

%pylab inline




Populating the interactive namespace from numpy and matplotlib



### Description

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.

#### (a)

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

#### (b)

• 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



### SOLUTION

#### (a)

The starting torque of an induction motor is proportional to the square of $V_{TH}$ ,

$$\frac{\tau_\text{start2}}{\tau_\text{start1}} = \left(\frac{V_\text{TH2}}{V_\text{TH1}}\right)^2 = \left(\frac{V_\text{T2}}{V_\text{T2}}\right)^2$$

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))




Vt2 = 348 V
===========



#### (b)

The motor starting current is directly proportional to the starting voltage, so

$$I_{L2} = \left(\frac{V_{T2}}{V_T}\right)I_{L1}$$


In [4]:

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




Il2 = 5.29 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))




Pin = 41.0 kW



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))




Irated = 59.13 A



Therefore, the motor starting current is



In [7]:

Il2 = Il2_Irated * Irated
print('''
Il2 = {:.1f} A
============='''.format(Il2))




Il2 = 312.9 A
=============



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))




a = 1.323



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))




Iline = 237 A
=============