Excercises Electric Machinery Fundamentals

Chapter 6 Problem 6-21



In [1]:

    
%pylab notebook









    



Populating the interactive namespace from numpy and matplotlib

Description

A 460-V, 10 hp, four-pole, Y-connected, Insulation class F, Service Factor 1.15 induction motor has the following parameters:

Stator	Rotor	Power
$R_1 = 0.540\,\Omega$	$R_2 = 0.488\,\Omega$	$P_\text{core} = 150\,W$
$X_1 = 2.093\,\Omega$	$X_2 = 3.209\,\Omega$	$P_\text{f\&w} = 150\,W$
$X_M = 51.12\,\Omega$		$P_\text{misc} = 50\,W$



In [2]:

    
R1 =  0.54  # [Ohm]
R2 =  0.488 # [Ohm]
Xm = 51.12  # [Ohm]
X1 =  2.093 # [Ohm]
X2 =  3.209 # [Ohm]
Pcore = 150 # [W]
Pf_w  = 150 # [W]
Pmisc =  50 # [W]

V = 460  # [V]
p = 4
fse = 60 # [Hz]

For a slip of 0.02, find

(a)

The line current $I_L$

(b)

The stator power factor

(c)

The rotor power factor

(d)

The rotor frequency

(e)

The stator copper losses $P_{SCL}$

(f)

The air-gap power $P_{AG}$

(g)

The power converted from electrical to mechanical form $P_{conv}$

(h)

The induced torque $\tau_{ind}$

(i)

The load torque $\tau_{load}$

(j)

The overall machine efficiency $\eta$

(k)

The motor speed in revolutions per minute and radians per second

(l)

Sketch the power flow diagram for this motor.

SOLUTION

The equivalent circuit of this induction motor is shown below:

(a)

The easiest way to find the line current (or armature current) is to get the equivalent impedance $Z_F$ of the rotor circuit in parallel with $jX_M$, and then calculate the current as the phase voltage divided by the sum of the series impedances, as shown below.

Note that in this case we are ignoring the $P_\text{core}$ losses in the electrical calculations but we will consider them as part of the rotational losses later on. This is quite common practice as it is hard to determine the $P_\text{core}$ separate from the $P_\text{f\&w}$ losses in practice.

The equivalent impedance of the rotor circuit in parallel with $jX_M$ is:

$$Z_F = \frac{1}{\frac{1}{jX_M} + \frac{1}{Z_2}}$$



In [3]:

    
s = 0.02

Z2 = R2 + R2*(1-s)/s + X2*1j
Zf = 1 / (1/(Xm*1j) + 1/Z2)
Zf_angle = arctan(Zf.imag/Zf.real)
print('Zf = ({:.2f})Ω = {:.2f} Ω ∠{:.1f}°'.format(Zf, abs(Zf), Zf_angle/pi *180))









    



Zf = (17.98+11.09j)Ω = 21.12 Ω ∠31.7°

The phase voltage is:



In [4]:

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









    



V_phase = 266 V

so the line current $I_L$ is:

$$I_L = I_A = \frac{V_\phi}{R_1 + jX_1 +R_F + jX_F}$$



In [5]:

    
Il = V_phase / (R1 + X1*1j + Zf)
Ia = Il
Il_angle = arctan(Il.imag/Il.real)
Ia_angle = Il_angle
print('''
Il = {:.2f} A ∠{:.1f}°
===================='''.format(abs(Il), Il_angle/pi *180))









    



Il = 11.68 A ∠-35.5°
====================

(b)

The stator power factor is:



In [6]:

    
PF = cos(Il_angle)
print('''
PF = {:.3f} lagging
=================='''.format(PF))









    



PF = 0.815 lagging
==================

(c)

To find the rotor power factor, we must find the impedance angle of the rotor:

$$\theta_R = \arctan{\left(\frac{X_2}{R_2/s}\right)}$$



In [7]:

    
theta_R = arctan(X2 / (R2/s))
print('''
theta_R = {:.2f}°'''.format(theta_R/pi *180))









    



theta_R = 7.49°

Therefore the rotor power factor is:



In [8]:

    
PF_R = cos(theta_R)
print('''PF_R = {:.3f} lagging
===================='''.format(PF_R))









    



PF_R = 0.991 lagging
====================

(d)

The rotor frequency is:

$$f_r = sf_s$$



In [9]:

    
fr = s*fse
print('''
fr = {:.1f} Hz
==========='''.format(fr))









    



fr = 1.2 Hz
===========

(e)

The stator copper losses are: $$P_{SCL} = 3I^2_AR_1$$



In [10]:

    
Pscl = 3 * abs(Ia)**2 * R1
print('''
Pscl = {:.0f} W
============'''.format(Pscl))









    



Pscl = 221 W
============

(f)

The air gap power is:

$$P_{AG} = P_\text{in} - P_{SCL} = 3I^2_2\frac{R_2}{s} = 3I^2_A R_F$$

As noted earlier $P_\text{core}$ losses are lumbed into the rotational losses on the rotor side.

Note that $3I_A^2R_F$ is equal to $3I_2^2\frac{R_2}{s}$, since the only resistance in the original rotor circuit was $\frac{R_2}{s}$ , and the resistance in the Thevenin equivalent circuit is $R_F$. The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.



In [11]:

    
Pag = 3 * abs(Ia)**2 * Zf.real
print('''
Pag = {:.2f} kW
============='''.format(Pag/1000))









    



Pag = 7.36 kW
=============

(g)

The power converted from electrical to mechanical form is:

$$P_\text{conv} = (1-s)P_{AG}$$



In [12]:

    
Pconv = (1-s) * Pag
print('''
Pconv = {:.2f} kW
==============='''.format(Pconv/1000))









    



Pconv = 7.21 kW
===============

(h)

The synchronous speed of this motor is:

$$n_\text{sync} = \frac{120f_{se}}{P}$$



In [13]:

    
n_sync = 120*fse / p
w_sync = n_sync * (2*pi/60.0)
print('''
n_sync = {:.0f} r/min
w_sync = {:.1f} rad/s'''.format(n_sync, w_sync))









    



n_sync = 1800 r/min
w_sync = 188.5 rad/s

Therefore the induced torque in the motor is:

$$\tau_\text{ind} = \frac{P_{AG}}{\omega_\text{sync}}$$



In [14]:

    
tau_ind = Pag / w_sync
print('''
tau_ind = {:.1f} Nm
================='''.format(tau_ind))









    



tau_ind = 39.1 Nm
=================

(i)

The output power of this motor is:

$$P_\text{out} = P_\text{conv} - P_\text{f\&w} - P_\text{core} - P_\text{misc}$$

As you can see we are now taking into account the $P_\text{core}$ losses again which we ignored earlier on the electrical side.



In [15]:

    
Pout = Pconv - Pf_w - Pcore - Pmisc
print('Pout = {:.2f} kW'.format(Pout/1000))









    



Pout = 6.86 kW

The output speed is:

$$n_m = (1-s)n_\text{sync}$$



In [16]:

    
n_m = (1-s) * n_sync
print('n_m = {:.0f} r/min'.format(n_m))









    



n_m = 1764 r/min

Therefore the load torque is:

$$\tau_\text{load} = \frac{P_\text{out}}{\omega_m}$$



In [17]:

    
w_m = n_m * (2*pi/60.0)
tau_load = Pout / w_m
print('''
tau_load = {:.1f} Nm
=================='''.format(tau_load))









    



tau_load = 37.2 Nm
==================

(j)

The overall efficiency is:

$$\eta = \frac{P_\text{out}}{P_\text{in}} = \frac{P_\text{out}}{3V_\phi I_A\cos\theta}$$



In [18]:

    
eta = Pout / (3*V_phase*abs(Ia)*cos(Ia_angle))
print('''
η = {:.1f} %
=========='''.format(eta*100))









    



η = 90.5 %
==========

(k)

The motor speed in revolutions per minute is (these values were already calculated in part (i)):



In [19]:

    
print('n_m = {:.0f} r/min'.format(n_m))









    



n_m = 1764 r/min

The motor speed in radians per second is:



In [20]:

    
print('''
w_m = {:.1f} rad/s
================='''.format(w_m))









    



w_m = 184.7 rad/s
=================

(l)

The power flow diagram can be calculated from:

$$P_\text{out} = P_\text{in} - P_\text{SCL} - P_\text{RCL} - P_\text{f\&w} - P_\text{core} - P_\text{misc}$$

This means we still need to calulate $P_\text{in}$ and $P_\text{RCL}$:



In [21]:

    
Pin = sqrt(3)* V * abs(Il) * PF
Prcl = 3 * abs(Ia)**2 * R2

The diagram can then be created:



In [22]:

    
# This demonstrates how to create a simple diagram by implicitly calling the
# Sankey.add() method and by appending finish() to the call to the class.
from matplotlib.sankey import Sankey
sankey = Sankey(unit='W', scale= 1/Pin, format='%.0f', margin=0.25)
sankey.add(flows=[Pin, -Pag, -Pscl],
           labels=['$P_{in}=\sqrt{3}V_TI_LPF$', '$P_{AG}$', '$P_{scl}$'],
           orientations=[ 0, 0, -1],
           pathlengths=[0, 0.1, 0.6])
sankey.add(flows=[Pag, -Pconv, -Prcl],
           labels=[None, '$P_{conv}$',  '$P_{rcl}$'],
           orientations=[ 0, 0, -1],
           pathlengths=[0, 0.1, 0.5],
           prior=0, connect=(1, 0))
sankey.add(flows=[Pconv, -Pout, -Pmisc, -Pf_w, -Pcore],
           labels=[None,'$P_{out}$', '$P_{misc}$', '$P_{f&w}$', '$P_{core}$'],
           orientations=[ 0, 0, -1, -1, -1,],
           pathlengths=[0, 0.1, 0.1, 0.3, 0.45],
           prior=1, connect=(1, 0))
sankey.finish()
title("Sankey diagram of the power-flow");