Excercises Electric Machinery Fundamentals

Chapter 4 Problem 4-2



In [1]:

    
%pylab notebook
%precision 0









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.0f'

Description

Given a 13.8-kV, 50-MVA, 0.9-power-factor-lagging, 60-Hz, four-pole, Y-connected synchronous machine with:

a synchronous reactance of $2.5\,\Omega$
an armature resistance of $0.2\,\Omega$.
at 60 Hz, its friction and windage losses are 1 MW
its core losses are 1.5 MW.
The field circuit has a dc voltage of 120 V,
the maximum $I_F$ is 10 A. The current of the field circuit is adjustable over the range from 0 to 10 A.

The OCC of this generator is shown in Figure P4-1 below



In [2]:

    
Vl = 13.8e3    # [V] 
PF = 0.9
Xs = 2.5       # [Ohm]
Ra = 0.2       # [Ohm]
P      =  50e6 # [W]
Pf_w   = 1.0e6 # [W]
Pcore  = 1.5e6 # [W]
Pstray =     0 # [W]
n_m = 1800     # [r/min]

(a)

How much field current is required to make the terminal voltage $V_T$ (or line voltage $V_L$ ) equal to 13.8 kV when the generator is running at no load?

(b)

What is the internal generated voltage $E_A$ of this machine at rated conditions?

(c)

What is the phase voltage $V_\phi$ of this generator at rated conditions?

(d)

How much field current is required to make the terminal voltage $V_T$ equal to 13.8 kV when the generator is running at rated conditions?

(e)

Suppose that this generator is running at rated conditions, and then the load is removed without changing the field current.

What would the terminal voltage of the generator be?

(f)

How much steady-state power and torque must the generator’s prime mover be capable of supplying to handle the rated conditions?

(g)

Construct a capability curve for this generator.

SOLUTION

(a)

If the no-load terminal voltage is 13.8 kV, the required field current can be read directly from the open-circuit characteristic. It is $\underline{\underline{I_F = 3.50\,A}}$.

(b)

This generator is Y-connected, so $I_L = I_A$ . At rated conditions, the line and phase current in this generator is:

$$I_A = I_L = \frac{P}{\sqrt{3}V_L}$$



In [3]:

    
ia = P / (sqrt(3) * Vl)
Ia_angle = -arccos(PF)
Ia = ia * (cos(Ia_angle) + sin(Ia_angle)*1j)
print('Ia = {:.0f} A ∠{:.1f}°'.format(abs(Ia), Ia_angle/pi *180))









    



Ia = 2092 A ∠-25.8°

The phase voltage of this machine is: $$V_\phi = V_T / \sqrt{3}$$



In [4]:

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









    



V_phase = 7967 V

The internal generated voltage of the machine is: $$\vec{E}_A = \vec{V}_\phi + R_A\vec{I}_A + jX_S\vec{I}_A$$



In [5]:

    
Ea = V_phase + Ra*Ia + Xs*1j*Ia
Ea_angle = arctan(Ea.imag/Ea.real)
print('''
Ea = {:.0f} V ∠{:.1f}°
=================='''.format(abs(Ea), Ea_angle/pi*180))









    



Ea = 11547 V ∠23.1°
==================

(c)

The phase voltage of the machine at rated conditions is:



In [6]:

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









    



V_phase = 7967 V
================

(d)

The equivalent open-circuit terminal voltage corresponding to an $E_A$ of the value calculated in (b) is:



In [7]:

    
Vt_oc = sqrt(3) * abs(Ea)
print('Vt_oc = {:.0f} kV'.format(Vt_oc/1000))









    



Vt_oc = 20 kV

From the OCC, the required field current is $\underline{\underline{I_F = 10\,A}}$.

(e)

If the load is removed without changing the field current then $V_\phi = E_A$:



In [8]:

    
abs(Ea)









    Out[8]:





11547

The corresponding terminal voltage would be $\underline{\underline{V_T = 20\,kV}}$.

(f)

The input power to this generator is equal to the output power plus losses. The rated output power is:



In [9]:

    
Pout = P*PF
print('Pout = {:.0f} MW'.format(Pout/1e6))









    



Pout = 45 MW

$$P_{CU} = 3I^2_AR_A$$



In [10]:

    
Pcu = 3 * abs(Ia)**2 * Ra
print('Pcu = {:.1f} MW'.format(Pcu/1e6))









    



Pcu = 2.6 MW



In [11]:

    
Pin = Pout +Pcu + Pf_w + Pcore + Pstray
print('Pin = {:.1f} MW'.format(Pin/1e6))









    



Pin = 50.1 MW

Therefore the prime mover must be capable of supplying $P_{in}$. Since the generator is a four-pole 60 Hz machine, to must be turning at 1800 r/min. The required torque is:

$$\tau_{app} = \frac{P_{in}}{\omega_m}$$



In [12]:

    
w_m = n_m * (2*pi/60.0)
tau_app = Pin / w_m
print('''
tau_app = {:.0f} Nm
==================='''.format(tau_app))









    



tau_app = 265924 Nm
===================

(e)

The rotor current limit of the capability curve would be drawn from an origin of:

$$Q = -\frac{3V^2_\phi}{X_S}$$



In [13]:

    
Q = - (3 * V_phase**2) / Xs
print('Q = {:.2f} Mvar'.format(Q/1e6))









    



Q = -76.18 Mvar

The radius of the rotor current limit is:

$$D_E = \frac{3V_\phi E_A}{X_S}$$



In [14]:

    
De = (3 * V_phase * abs(Ea)) / Xs
print('De = {:.0f} Mvar'.format(De/1e6))









    



De = 110 Mvar

The stator current limit is a circle at the origin of radius:

$$S = 3V_\phi I_A$$



In [15]:

    
S = 3 * V_phase * abs(Ia)
print('S = {:.0f} Mvar'.format(S/1e6))









    



S = 50 Mvar

Get points for stator current limit:



In [16]:

    
theta = arange(-95,95) # angle in degrees
rad = theta * pi/180   # angle in radians
s_curve = S * ( cos(rad) + sin(rad)*1j)

Get points for rotor current limit:



In [17]:

    
orig = Q*1j
theta = arange(65,115)  # angle in degrees
rad = theta * pi / 180  # angle in radians
r_curve = orig + De * ( cos(rad) + sin(rad)*1j )

Plot the capability diagram:



In [18]:

    
fig= figure()
ax=fig.add_subplot(1, 1, 1)
ax.plot(real(s_curve/1e6),imag(s_curve/1e6),'b')
ax.plot(real(r_curve/1e6),imag(r_curve/1e6),'r--')
ax.set_title('Synchronous Generator Capability Diagram')
ax.set_xlabel('Power (MW)')
ax.set_ylabel('Reactive Power (Mvar)')
ax.set_aspect('equal', 'datalim')
ax.legend(('stator current limit', 'rotor current limit'), loc=3);
ax.grid()