# Excercises Electric Machinery Fundamentals

## Problem 6-20

Note: You should first click on "CellRun All" in order that the plots get generated.



In [1]:

%pylab notebook
%precision 3




Populating the interactive namespace from numpy and matplotlib

Out[1]:

'%.3f'



### Description

A 208-V six-pole Y-connected 25-hp design class B induction motor is tested in the laboratory, with the following results:

• No load: 208 V, 24.0 A, 1400 W, 60 Hz
• Locked rotor: 24.6 V, 64.5 A, 2200 W, 15 Hz
• Dc test: 13.5 V, 64 A

Based on those measurements:

• Find the equivalent circuit of this motor, and plot its torque-speed characteristic curve.


In [2]:

Vnl =  208.0 # [V]
Inl =   24.0 # [A]
Pnl = 1400.0 # [W]
fnl =   60.0 # [Hz]
Vlr =   24.6 # [V]
Ilr =   64.5 # [A]
Plr = 2200.0 # [W]
flr =   15.0 # [Hz]
Vdc =   13.5 # [V]
Idc =   64.0 # [A]



### SOLUTION

From the DC test,



In [3]:

R1 = Vdc/Idc  /  2
print('''
R1 = {:.3f} Ω
============'''.format(R1))




R1 = 0.105 Ω
============



In the no-load test, the line voltage is 208 V, so the phase voltage is:



In [4]:

Vnl_phase = Vnl / sqrt(3)
print('Vnl_phase = {:.0f} V'.format(Vnl_phase))




Vnl_phase = 120 V



Therefore,

$$X_1 + X_M = \frac{V_\phi}{I_\text{A,nl}}$$


In [5]:

X1_Xm = Vnl_phase / Inl
print('X1 + Xm = {:.2f} Ω @ {:.0f} Hz'.format(X1_Xm, fnl))




X1 + Xm = 5.00 Ω @ 60 Hz



In the locked-rotor test, the line voltage is 24.6 V, so the phase voltage is:



In [6]:

Vlr_phase = Vlr / sqrt(3)
print('Vlr_phase = {:.1f} V'.format(Vlr_phase))




Vlr_phase = 14.2 V



From the locked-rotor test at 15 Hz , $$|Z'_{LR}| = |R_{LR} + jX'_{LR}| = \frac{V_\phi}{I_{A,LR}}$$



In [7]:

z_lr = Vlr_phase / Ilr
Slr = sqrt(3) * Vlr * Ilr
theta_lr = arccos(Plr / Slr)
print('''
z_lr     = {:.3f} Ω
theta_lr = {:.2f}°'''.format(z_lr, theta_lr))




z_lr     = 0.220 Ω
theta_lr = 0.64°



Therefore, $$R_{LR} = |Z'_{LR}| \cos{\theta_{LR}}$$



In [8]:

Rlr = z_lr * cos(theta_lr)
print('Rlr = {:.3f} Ω'.format(Rlr))




Rlr = 0.176 Ω




In [9]:

R2 = Rlr - R1
print('''
R2 = {:.3f} Ω
============'''.format(R2))




R2 = 0.071 Ω
============


$$X'_{LR} = |Z'_{LR}| \sin{\theta_{LR}}$$


In [10]:

X_lr = z_lr * sin(theta_lr)
print('X_lr = {:.3f} Ω @ {:.0f} Hz'.format(X_lr, flr))




X_lr = 0.132 Ω @ 15 Hz



At a frequency of 60 Hz,



In [11]:

Xlr = (60 / flr) *X_lr
print('Xlr = {:.3f} Ω @ 60 Hz'.format(Xlr))




Xlr = 0.528 Ω @ 60 Hz



For a Design Class B motor, the split is $X_1 = 0.4 \cdot X_{LR}$ and $X_2 = 0.6 \cdot X_{LR}$ .



In [12]:

X1 = 0.4 * Xlr; X1




Out[12]:

0.211




In [13]:

X2 = 0.6 * Xlr; X2




Out[13]:

0.317



Therefore,



In [14]:

Xm = X1_Xm - X1
print('''
Xm = {:.3f} Ω
============'''.format(Xm))




Xm = 4.793 Ω
============



The resulting equivalent circuit is shown below:

Plotting the torque-speed characteristic:

Calculate the Thevenin voltage and impedance:



In [15]:

V_th = Vnl_phase * ( Xm / sqrt(R1**2 + (X1 + Xm)**2) )
Z_th = ((Xm*1j) * (R1 + X1*1j)) / (R1 + (X1 + Xm)*1j)
R_th = real(Z_th)
X_th = imag(Z_th)



Now calculate the torque-speed characteristic for many slips between 0 and 1. Note that the first slip value is set to 0.001 instead of exactly 0 to avoid divide-by-zero problems.



In [16]:

s = linspace(0,50,51) / 50 # generate an array with 51 values between 0 and 50
s[0] = 0.001               # avoid division by zero
p = 6                      # number of poles
n_sync = 60/(p/2) * fnl    # [r/min]
w_sync = 2*pi/60 *n_sync   # [rad/s]
nm = (1-s) * n_sync



Calculate torque versus speed



In [17]:

tau_ind = (3 * V_th**2 * R2 / s)  /  (w_sync * ((R_th + R2/s)**2 + (X_th + X2)**2) )



Plot the torque-speed curve:



In [18]:

rc('text', usetex=True)   # enable LaTeX commands for plot
title(r'\bf Induction Motor Torque-Speed Characteristic')
xlabel(r'$n_m$ [r/min]')
ylabel(r'$\tau_{ind}$ [Nm]')
plot(nm, tau_ind,  linewidth = 2)
grid()




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