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
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.
Assume that this transformer is supplying rated load at 277 V and 0.85 PF lagging.
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)))
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)))
The resulting equivalent circuit is (!!!Values in the figure are not corrected!!!):
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))
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))
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!!!):
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))
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)))
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)))
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))
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))
In [15]:
eta = P_out/ (P_out + P_cu + P_core) * 100
print('η = {:.1f} %'.format(eta))