Excercises Electric Machinery Fundamentals

Chapter 6 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))

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
=============