HOME WORK # 3 – SOLUTIONS
1.
An ideal transformer has an input voltage of 480 V. The output current and
voltage are 10 A and 120 V. Determine the value of input current.
Solution:
I1 =
V2 I 2 (120 )(10 )
=
= 2.5 A
V1
480
2.
A 15-kVA, 2400:240-V, 60 Hz transformer has the following equivalent circuit
parameters:
R1 = 2.5Ω
Rc = 32k Ω
R2 = 0.025Ω
X m = 11.5k Ω
X 1 = 7.0Ω
X 2 = 0.070Ω
If the transformer is supplying a 10-kW, 0.8 PF lagging load at rated voltage, calculate:
(a)
(b)
(c)
Input voltage
Input current, and
Input power factor
Solution:
Reflect R2 and X 2 to the primary side:.
⎛N ⎞
⎛ 2400 ⎞
R2′ = ⎜ 1 ⎟ R2 = ⎜
⎟ ( 0.025 ) = 2.5 Ω
⎝ 240 ⎠
⎝ N2 ⎠
2
2
⎛N ⎞
⎛ 2400 ⎞
X 2′ = ⎜ 1 ⎟ X 2 = ⎜
⎟ ( 0.070 ) = 7.0 Ω
N
⎝ 240 ⎠
⎝ 2⎠
2
2
aV2 = (10 )( 240∠0° ) = 2400∠0° V
(a,b) Assume V2 on reference; then,
The load current is found by
I2 =
P2
10,000
=
= 52.08 A
V2 ( PF ) 240 ( 0.8 )
I 2 = I 2 ∠ − cos −1 ( 0.8 ) = 52.08∠ − 36.87° A
The reflected load current is
I 2′ =
1
I 2 = 5.208∠ − 36.87° A
10
E = I 2′ ( R2′ + jX 2′ ) + aV2 = ( 5.208∠ − 36.87° )( 7.433∠70.35° ) + 2400∠0°
The excitation branch voltage is found by use of KVL.
E = 2432.381∠0.50° V
The excitation branch current follows as
Io = Ic + Im =
2432.381∠0.50° 2432.381∠0.50°
E
E
+
=
+
32,000
11,500∠90°
Rc jX m
I o = 0.224∠ − 69.70° A
The input current can now be determined by KCL.
I1 = I 2′ + I o = 5.208∠ − 36.87° + 0.224∠ − 69.70°
I1 = 5.398∠ − 38.16° A
Application of KVL yields the input voltage.
V1 = I1 ( R1 + jX 1 ) + E = 5.398∠ − 38.16° ( 7.433∠70.35° ) + 2432.381∠0.50°
V1 = 2466.61∠0.99° V
V1 = 2466.61 V
PFin = cos ( ∠V1 − ∠I1 ) = cos ( 0.99° + 38.16° ) = 0.775 lagging
(c) With V1 and I1 known, the input power factor is given by
3.
A 20 kVA, 2500/250 V, 50Hz, single-phase transformer gave the following test
result:
Open-circuit test: 250V, 1.5A, and 100 watts measured on the low voltage side
Short-circuit test: 100V, 8A, and 320 watts measured on the on high voltage side
a)
Compute the parameters of the equivalent circuits and draw the equivalent circuit
of the transformer showing all the values.
b)
Calculate the equivalent circuit of the transformer in per unit.
Solution:
From the given test results data, we have
P
Rwdg = sc2 = 5Ω
I sc
⎛V ⎞
X l = ⎜ sc ⎟ − Rwdg 2 = 12Ω
⎝ I sc ⎠
2
and
Rcore
(V )
= oc
Poc
2
= 625Ω
⎛ V ⎞ ⎛V ⎞
I oc = ⎜ oc ⎟ + ⎜ oc ⎟
⎝ Rcore ⎠ ⎝ X m ⎠
Therefore, X m = 173Ω . Rcore and Xm are low-voltage side values, which can be
converted to the high voltage side: R_core = Rcore *(N1/N2)2 = 62.5 kΩ, and X_m = Xm
*(N1/N2)2 = 17.3 kΩ.
2
2
We can find the inductances:
Xl
Ll =
= 38.2mH
100π
X
Lm = m = 550.68mH
100π
(b) Equivalent circuit in per unit:
Z base1 =
25002
= 312.5Ω
20k
2502
Z base 2 =
= 3.125Ω
20k
625
Rcore , pu =
= 200, pu
Z base 2
173
X m , pu =
= 55.36, pu
Z base 2
12
X l , pu =
= 0.0384, pu
Z base1
5
Rwdg , pu =
= 0.016, pu
Z base1