<jupyter_start><jupyter_text>Excercises Electric Machinery Fundamentals Chapter 5  Problem 5-7<jupyter_code>%pylab inline<jupyter_output>Populating the interactive namespace from numpy and matplotlib<jupyter_text>Description A 208-V Y-connected synchronous motor is drawing 50 A at unity power factor from a 208-V powersystem. The field current flowing under these conditions is 2.7 A. Its synchronous reactance is $1.6\,\Omega$.Assume a linear open-circuit characteristic.<jupyter_code>Vt = 208   # [V]
Ia =  50   # [A]
Xs =   1.6 # [Ohm]
Ra =   0   # [Ohm]
PF2 =  0.8
If_1 = 2.7 # [A]<jupyter_output><empty_output><jupyter_text>(a) * Find $\vec{V}_\phi$ and $\vec{E}_A$ for these conditions.  (b) * Find the torque angle $\delta$ .  (c) * What is the static stability power limit under these conditions?  (d) * How much field current would be required to make the motor operate at 0.80 PF leading?  (e) * What is the new torque angle in part (d)?  SOLUTION  (a)The phase voltage of this motor is $V_\phi = 120 V$, and the armature current is $\vec{I}_A = 50\,A \angle 0°$ .Therefore, the internal generated voltage is:$$\vec{E}_A = \vec{V}_\phi - R_A \vec{I}_A - jX_S \vec{I}_A$$<jupyter_code>Vphi = Vt / sqrt(3)
EA = Vphi - Ra*Ia - Xs*1j*Ia
EA_angle = arctan(EA.imag/EA.real)
print('''
EA = {:.0f} V ∠{:.1f}°
=================='''.format(abs(EA), EA_angle/pi*180))<jupyter_output>EA = 144 V ∠-33.7°
==================<jupyter_text>(b)The tor que angle $\delta$ of this machine is<jupyter_code>delta = EA_angle
print('''
δ = {:.1f}°
=========='''.format(delta/pi*180))<jupyter_output>δ = -33.7°
==========<jupyter_text>(c)The static stability power limit is given by$$P_\text{max} = \frac{3V_\phi E_A}{X_S}$$<jupyter_code>Pmax = (3*Vphi*abs(EA)) / Xs
print('''
Pmax = {:.1f} kW
=============='''.format(Pmax/1000))<jupyter_output>Pmax = 32.5 kW
==============<jupyter_text>(d)A phasor diagram of the motor operating at a power factor of 0.78 leading is shown below. Since the power supplied by the motor is constant, the quantity $I_A cos \theta$ , which is directly proportionalto power, must be constant. Therefore,<jupyter_code>theta1 = 0 # [rad]
theta2 = arccos(PF2)
Ia2 = Ia*cos(theta1) / cos(theta2)
IA2 = Ia2 * (cos(theta2)+sin(theta2)*1j)
IA2_angle = theta2
print('IA2 = {:.1f} A ∠{:.2f}°'.format(abs(IA2), IA2_angle/pi*180))<jupyter_output>IA2 = 62.5 A ∠36.87°<jupyter_text>The internal generated voltage required to produce this current would be:$$\vec{E}_{A2} = \vec{V}_\phi - R_A \vec{I}_{A2} - jX_S \vec{I}_{A2}$$<jupyter_code>EA2 = Vphi - Ra*IA2 - Xs*1j*IA2
EA2_angle = arctan(EA2.imag/EA2.real)
print('EA2 = {:.0f} V ∠{:.1f}°'.format(abs(EA2), EA2_angle/pi*180))<jupyter_output>EA2 = 197 V ∠-24.0°<jupyter_text>The internal generated voltage $E_A$ is directly proportional to the field flux, and we have assumed in thisproblem that the flux is directly proportional to the fiel d current. Therefore, the required field current is:$$I_{F2} = \frac{E_{A2}}{E_{A1}}I_{F1}$$<jupyter_code>If_2 = abs(EA2)/abs(EA) * If_1
print('''
If_2 = {:.2f} A
============='''.format(If_2))<jupyter_output>If_2 = 3.69 A
=============<jupyter_text>(e)The new torque angle $\delta$ of this machine is<jupyter_code>delta2 = EA2_angle
print('''
δ_2 = {:.1f}°
============'''.format(delta2/pi*180))<jupyter_output>δ_2 = -24.0°
============
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