# Excercises Electric Machinery Fundamentals

## Problem 5-1

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



In [1]:

%pylab notebook
%precision %.4g




Populating the interactive namespace from numpy and matplotlib

Out[1]:

'%.4g'



### Description

A 480-V, 60 Hz, 400-hp 0.8-PF-leading eight-pole $\Delta$-connected synchronous motor has a synchronous reactance of $0.6\,\Omega$ and negligible armature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. Assume that $|\vec{E}_A|$ is directly proportional to the field current $I_F$ (in other words, assume that the motor operates in the linear part of the magnetization curve), and that $|\vec{E}_A| = 480\,V$ when $I_F = 4\,A$.



In [2]:

Vt   = 480         # [V]
PF   =   0.8
fse  =  60         # [Hz]
p    =   8
Pout = 400 * 746   # [W] using the official "electrical horsepower" conversion
Xs   =   0.6       # [Ohm]



#### (a)

• What is the speed of this motor?

#### (b)

If this motor is initially supplying 400 hp at 0.8 PF lagging.

• What are the magnitudes and angles of $\vec{E}_A$ and $\vec{I}_A$ ?

#### (c)

• How much torque is this motor producing?
• What is the torque angle $\delta$ ?
• How near is this value to the maximum possible induced torque of the motor for this field current setting?

#### (d)

If $|\vec{E}_A|$ is increased by 30 percent.

• What is the new magnitude of the armature current?
• What is the motor’s new power factor?

#### (e)

• Calculate and plot the motor’s V-curve for this load condition.

### SOLUTION

#### (a)

The speed of this motor is given by:

$$n_m = \frac{120f_{se}}{P}$$


In [3]:

n_m = 120 * fse / p
print('''
n_m = {:.0f} r/min
==============='''.format(n_m))




n_m = 900 r/min
===============



#### (b)

If losses are being ignored, the output power is equal to the input power. This situation is shown in the phasor diagram below:

The line current flow under these circumstances is:

$$I_L = \frac{P}{\sqrt{3}V_T PF}$$


In [4]:

Pin = Pout
il = Pin / (sqrt(3) * Vt * PF)
il # [A]




Out[4]:

448.6



Because the motor is $\Delta$-connected, the corresponding phase current is:



In [5]:

ia = il / sqrt(3)
ia # [A]




Out[5]:

259



The angle of the current is:



In [6]:

Ia_angle = -arccos(PF)
Ia_angle /pi *180 # [degrees]




Out[6]:

-36.87




In [7]:

Ia = ia * (cos(Ia_angle) + sin(Ia_angle)*1j)
print('Ia = {:.0f} A ∠{:.2f}°'.format(abs(Ia), Ia_angle / pi *180))




Ia = 259 A ∠-36.87°



The internal generated voltage $\vec{E}_A$ is: $$\vec{E}_A = \vec{V}_\phi - jX_S\vec{I}_A$$



In [8]:

EA = Vt - Xs * 1j * Ia
EA_angle = arctan(EA.imag/EA.real)
print('''
EA = {:.0f} V ∠{:.1f}°
=================='''.format(abs(EA), EA_angle/pi*180))




EA = 406 V ∠-17.8°
==================



#### (c)

The induced torque is:

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


In [9]:

w_m = n_m * (1.0 / 60.0) * (2.0*pi/1.0)
tau_ind = Pout / w_m
print('''
tau_ind = {:.0f} Nm
================='''.format(tau_ind))




tau_ind = 3166 Nm
=================



The maximum possible induced torque for the motor at this field setting is the maximum possible power divided by $\omega_m$:

$$\tau_\text{ind,max} = \frac{3V_\phi E_A}{\omega_mX_S}$$


In [10]:

tau_ind_max = (3*Vt*abs(EA)) / (w_m * Xs)
print('''
tau_ind_max = {:.0f} Nm
======================'''.format(tau_ind_max))




tau_ind_max = 10345 Nm
======================



The current operating torque is about



In [11]:

tau_ind/tau_ind_max




Out[11]:

0.3061



times the maximum possible torque.

#### (d)

If the magnitude of the internal generated voltage $E_A$ is increased by 30%, the new torque angle can be found from the fact that $E_A\sin{\delta} \propto P =$ constant.



In [12]:

Ea_2 = 1.30 * abs(EA)
Ea_2 # [V]




Out[12]:

528.1


$$\delta_2 = \arcsin\left(\frac{E_{A1}}{E_{A2}}\sin{\delta_1}\right)$$


In [13]:

delta_1 = EA_angle
delta_2 = arcsin(abs(EA) / Ea_2 * sin(delta_1))
delta_2/pi *180 # [degrees]




Out[13]:

-13.62



The new armature current is: $$\vec{I}_{A2} = \frac{\vec{V}_\phi - \vec{E}_{A2}}{jX_S}$$



In [14]:

EA2 = Ea_2 * (cos(delta_2)+sin(delta_2)*1j)
Ia_2 = (Vt - EA2) / (Xs*1j)
Ia_2_angle = arctan(Ia_2.imag/Ia_2.real)
PF2 = cos(Ia_2_angle)
print('''
Ia_2 = {:.0f} A ∠{:.2f}°
====================
PF2  = {:.3f}
============'''.format(abs(Ia_2), Ia_2_angle/pi*180, PF2))




Ia_2 = 215 A ∠14.98°
====================
PF2  = 0.966
============



#### (e)

A Python program to calculate and plot the motor’s V-curve is shown below:

Initialize values:



In [15]:

Ea_plot = linspace(0.90, 1.70, 81)  * EA



Calculate $\delta_2$



In [16]:

delta_plot = arcsin(abs(EA) / Ea_plot * sin(delta_1))



Calculate the phasor $E_A$



In [17]:

EA_plot = Ea_plot * (cos(delta_plot)+sin(delta_plot)*1j)



Calculate $I_A$



In [18]:

Ia_plot = (Vt - Ea_plot) / (Xs*1j)



Plot the v-curve



In [19]:

title(r'Armature current versus $E_A$')
xlabel(r'$E_A$ [kV]')
ylabel(r'$I_A$ [A]')
plot(abs(EA_plot)/1000,abs(Ia_plot),  linewidth = 2)
grid()




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