Excercises Electric Machinery Fundamentals

Chapter 2

Problem 2-1 (corrected)


Preface:

The solution given here is the corrected version of Problem_2-01.ipynb

The mistake was that the turns-ratio that was given in the book was based on the voltage levels under load.



In [1]:
%pylab inline


Populating the interactive namespace from numpy and matplotlib

Description

A 100-kVA 8000/283-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 = 8000.0 #[V]
V_low  = 283.0  #[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))


a = 28.27

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 = 0.0075 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 = {:.3f} Ω'.format(R_C))
print('X_M = {:.3f} Ω'.format(abs(X_M)))


R_C = 62.570 Ω
X_M = 12.514 Ω

The resulting equivalent circuit is (!!!Values in the figure are not corrected!!!):

(b)

The rated kVA of the transformer is 100 kVA, and the rated voltage on the secondary side is 283 V


In [6]:
S_base = 100e3  #[VA]
V_base = 283.0  #[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 = 353 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 = 0.801 Ω

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

The resulting per-unit equivalent circuit is as shown below (!!!Values in the figure are not corrected!!!):

(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+0.0135j Ω

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


In [10]:
VS = 277.0      #[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 = 2.18 %

(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   = 1467.1  W
P_core = 1280.3  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 = 96.9 %