Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-2



In [1]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[1]:





'%.4f'

Description

A single-phase power system is shown in Figure P2-1.

The power source feeds a 100-kVA 14/2.4-kV transformer through a feeder impedance of $38.2 + j140 \Omega$. The transformer’s equivalent series impedance referred to its low-voltage side is $0.10 + j0.4 \Omega$ . The load on the transformer is 90 kW at 0.8 PF lagging at $V_\text{load} = 2300\,V$.



In [2]:

    
Zline = 38.2  + 140.0j # [Ohm]
Zeq   =  0.10 +   0.4j # [Ohm]

V_high =  14e3 # [V]
V_low  = 2.4e3 # [V]

Pout   =  90e3 # [W] load
PF     =  0.8  # lagging

VS     = 2.3e3 # [V] secondary voltage

(a)

What is the voltage at the power source of the system?

(b)

What is the voltage regulation of the transformer?

(c)

How efficient is the overall power system?

SOLUTION

To solve this problem, we will refer the circuit to the secondary (low-voltage) side. The turns ratio of this transformer $a$ is:



In [3]:

    
a = V_high / V_low
a









    Out[3]:





5.8333

The feeder’s impedance referred to the secondary side is: $$Z'_\text{line} = \left(\frac{1}{a}\right)^2Z_\text{line}$$



In [4]:

    
Z_line = (1/a)**2 * Zline
print('Z_line = {:.2f} Ω'.format(Z_line))









    



Z_line = 1.12+4.11j Ω

The secondary current $I_S$ is given by : $$I_S = \frac{P_{OUT}}{V_SPF}$$



In [5]:

    
Is = Pout / (VS*PF)
print('Is = {:.2f} A'.format(Is))









    



Is = 48.91 A

The power factor is 0.80 lagging, so the impedance angel $\theta = \arccos(PF)$ is:



In [6]:

    
IS_angle = - arccos(PF)  # negative because lagging PF
print('θ = {:.2f}°'.format(degrees(IS_angle)))









    



θ = -36.87°

The phasor current is:



In [7]:

    
IS = Is * (cos(IS_angle)  + sin(IS_angle)*1j)
print('IS = {:.2f} A ∠{:.2f}°'.format(abs(IS), degrees(IS_angle)))









    



IS = 48.91 A ∠-36.87°

(a)

The voltage at the power source of this system (referred to the secondary side) is: $$\vec{V}'_\text{source}= \vec{V}_S + \vec{I}_SZ'_\text{line} + \vec{I}_SZ_\text{EQ}$$



In [8]:

    
V_source = VS + IS*Z_line + IS*Zeq
V_source_angle = arctan(V_source.imag/V_source.real) # angle of V_source [rad]
print('V_source = {:.1f} V ∠{:.1f}°'.format(
        abs(V_source), degrees(V_source_angle)))









    



V_source = 2484.3 V ∠3.2°

Therefore, the voltage at the power source is:

$$\vec{V}_\text{source} = \vec{V}'_\text{source} \cdot a$$



In [9]:

    
Vsource = V_source * a                             # [V]
Vsource_angle = arctan(Vsource.imag/Vsource.real)  # angle of Vsource [rad]
print('Vsource = {:.1f} kV ∠{:.1f}°'.format(
        abs(Vsource)/1000,   # display in kV
        degrees(Vsource_angle)))









    



Vsource = 14.5 kV ∠3.2°

(b)

To find the voltage regulation of the transformer, we must find the voltage at the primary side of the transformer (referred to the secondary side) under these conditions:

$$\vec{V}_P' = \vec{V}_S + \vec{I}_SZ_\text{EQ}$$



In [10]:

    
VP = VS + IS*Zeq
VP_angle = arctan(VP.imag/VP.real)   # angle of VP [rad]
print('VP = {:.1f} V ∠{:.1f}°'.format(abs(VP), degrees(VP_angle)))









    



VP = 2315.7 V ∠0.3°

There is a voltage drop of 15 V under these load conditions.

Therefore the voltage regulation of the transformer is:

$$VR = \frac{V_P' - V_S}{V_S} \cdot 100\%$$



In [11]:

    
VR = (abs(VP)-VS) / VS * 100  # [%]
print('VR = {:.2f} %'.format(VR))









    



VR = 0.68 %

(c)

The overall efficiency of the power system will be the ratio of the output power to the input power. The output power supplied to the load is $P_\text{OUT} = 90\,kW$. The input power supplied by the source is: $$P_\text{IN} = P_\text{OUT} + P_\text{LOSS} = P_\text{OUT} + I^2R$$



In [12]:

    
R = Z_line.real + Zeq.real
Pin = Pout + abs(IS)**2 * R               # [W]
print('Pin = {:.2f} kW'.format(Pin/1000)) # [kW]









    



Pin = 92.93 kW

Therefore, the efficiency of the power system is:



In [13]:

    
eta = Pout/Pin * 100  # [%]
print('η = {:.1f} %'.format(eta))









    



η = 96.9 %