# Excercises Electric Machinery Fundamentals

## 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



Get points for rotor current limit:



In [17]:

orig = Q*1j
theta = arange(65,115)  # angle in degrees



Plot the capability diagram:



In [18]:

fig= figure()
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()




*{stroke-linecap:butt;stroke-linejoin:round;stroke-miterlimit:100000;}