Excercises Electric Machinery Fundamentals

Chapter 2 Problem 2-1

Preface:

The solution given here is based on the incorrect voltage ratio as given in text in the book. The mistake is that the author of the book was basing the turns-ratio on the voltage levels under load. This means that the internal losses (i.e., voltage drops) are included in the turns-ratio $a$ which normally should represent the physical turns-ratio $n$, i.e., at no-load.

I've decided to provide the solution which is based on the wrong 8000/277 turns ratio anyway so that you can compare this with the calculations you have done so far. In addition I've created a corrected version Problem_2-01_corrected.ipynb which uses the correct turns-ratio $n$ of 8000/283 (the correct no-load voltage is actually calculated in part (c) here).



In [1]:

    
%pylab inline









    



Populating the interactive namespace from numpy and matplotlib

Description

A 100-kVA 8000/277-V distribution transformer has the following resistances and reactances:

$R_P = 5\,\Omega \qquad R_S = 0.005\,\Omega$

$X_P = 6\,\Omega \qquad X_S = 0.006\,\Omega$

$R_C = 50\,k\Omega \qquad X_M = 10\,k\Omega$



In [2]:

    
RP = 5.0      #[Ohm] 
RS = 0.005    #[Ohm]
XP = 6.0j     #[Ohm]
XS = 0.006j   #[Ohm] 
RC = 50e3     #[Ohm]
XM = 10e3j    #[Ohm]

V_high = 8000 #[V]
V_low  = 277  #[V]
S = 100e3     #[VA]

The excitation branch impedances are given referred to the high-voltage side of the transformer.

(a)

Find the equivalent circuit of this transformer referred to the low-voltage side.

(b)

Find the per-unit equivalent circuit of this transformer.

(c)

Assume that this transformer is supplying rated load at 277 V and 0.85 PF lagging.

What is this transformer’s input voltage?
What is its voltage regulation?

(d)

What are the copper losses and core losses in this transformer under the conditions of part (c)?

(e)

What is the transformer’s efficiency under the conditions of part (c)?

SOLUTION

(a)

The turns ratio $a$ of this transformer is



In [3]:

    
a = V_high/V_low
print('a = {:.2f}'.format(a))

Therefore, the primary impedances referred to the low voltage (secondary) side are:

$$R_P' = \frac{R_P}{a^2}$$$$X_P' = \frac{X_P}{a^2}$$



In [4]:

    
R_P = RP / a**2
X_P = XP / a**2
print('R_P = {:.3f} Ω'.format(R_P))
print('X_P = {:.4f} Ω'.format(abs(X_P)))









    



R_P = 0.006 Ω
X_P = 0.0072 Ω

and the excitation branch elements referred to the secondary side are:

$$R_C' = \frac{R_C}{a^2}$$$$X_M' = \frac{X_M}{a^2}$$



In [5]:

    
R_C = RC / a**2
X_M = XM / a**2
print('R_C = {:.0f} Ω'.format(R_C))
print('X_M = {:.0f} Ω'.format(abs(X_M)))









    



R_C = 60 Ω
X_M = 12 Ω

The resulting equivalent circuit is:

(b)

The rated kVA of the transformer is 100 kVA, and the rated voltage on the secondary side is 277 V



In [6]:

    
S_base = 100e3  #[VA]
V_base = 277.0  #[V]

So the rated current $I_\text{base}$ in the secondary side is:



In [7]:

    
I_base = S_base / V_base
print('I_base = {:.0f} A'.format(I_base))









    



I_base = 361 A

Therefore, the base impedance on the primary side is: $$Z_{base} = \frac{V_{base}}{I_{base}}$$



In [8]:

    
Z_base = V_base / I_base
print('Z_base = {:.3f} Ω'.format(Z_base))









    



Z_base = 0.767 Ω

Since $Z_{pu} = Z_\text{actual} / Z_\text{base}$

The resulting per-unit equivalent circuit is as shown below:

(c)

To simplify the calculations, use the simplified equivalent circuit referred to the secondary side of the transformer:



In [9]:

    
Req = RP/a**2 + RS          #[Ohm]
Xeq = XP/a**2 + XS          #[Ohm]
print('Req = {:.3f} Ω'.format(Req))
print('Xeq = {:.4f} Ω'.format(Xeq))









    



Req = 0.011 Ω
Xeq = 0.0000+0.0132j Ω

And the transformer is supplying rated load at 277V and 0.85PF lagging.



In [10]:

    
VS = 277  # [V]
PF = 0.85

The secondary current $I_S$ in this transformer is:



In [11]:

    
Is = S / abs(VS)                            # absolute value of IS [A]
IS_angle = -arccos(PF)                      # angle of IS [rad]
IS = Is*cos(IS_angle) + Is*sin(IS_angle)*1j # value of IS [A]
print('IS = {:.0f} A ∠{:.1f}°'.format(*(abs(IS), degrees(IS_angle))))









    



IS = 361 A ∠-31.8°

Therefore, the primary voltage on this transformer (referred to the secondary side) is:

$$V_P' = V_S + (R_{EQ}+jX_{EQ})I_S$$



In [12]:

    
V_P = VS + (Req + Xeq)*IS
print('V_P = {:.0f} V ∠{:.1f}°'.format(*(abs(V_P), angle(V_P, deg=True))))









    



V_P = 283 V ∠0.4°

The voltage regulation $VR$ of the transformer under these conditions is:



In [13]:

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









    



VR = 2.13 %

(d)

Under the conditions of part (c) the transformer’s output power copper losses $P_{OUT}$ and core losses $P_{CU}$ are:

$$P_{OUT} = S \cos\theta$$$$P_{CU} = (I_S)^2R_{EQ}$$$$P_{core} = \frac{V_P'^2}{R_C}$$



In [14]:

    
P_out = S * PF                              
P_cu = abs(IS)**2 * Req
P_core = abs(V_P)**2 / R_C
print('P_OUT  = {:>6.1f} kW'.format(P_out/1000))
print('P_CU   = {:>6.1f} W'.format(P_cu))
print('P_core = {:>6.1f} W'.format(P_core))









    



P_OUT  =   85.0 kW
P_CU   = 1432.9 W
P_core = 1335.0 W

(e)

The efficiency of this transformer is:

$$\eta = \frac{P_{OUT}}{P_{OUT}+P_{CU}+P_\text{core}} \cdot 100\,\%$$



In [15]:

    
eta = P_out/ (P_out + P_cu + P_core) * 100
print('η = {:.1f} %'.format(eta))









    



η = 96.8 %