Transformer sheet
1
A single-phase 100 kVA, 1000/ 100 V transformer gave the following test results:
open-circuit test 100 V, 6.0 A, 400 W short-circuit test 50 V, 100 A, 1800 W
(a) Determine the rated voltage and rated current for the HV and LV sides.
(b) Derive an approximate equivalent circuit referred to the HV side.
(c) Determine the voltage regulation at full load, 0.6 PF leading.
(d) Draw the phasor diagram for condition (c).
2
A 1 single-phase, 25 kVA, 220/440 V, 60 Hz transformer gave the following test results.
Open circuit test
: 220 V, 9.5 A, 650 W
Short-circuit test
: 37.5 V, 55 A, 950 W
(a) Derive the approximate equivalent circuit in per-unit values.
(b) Determine the voltage regulation at full load, 0.8 PF lagging.
(c) Draw the phasor diagram for condition (b).
3
A single-phase 10 kVA, 2400/ 120 V, 60 Hz transformer has the following equivalent circuit
parameters: Zeqp = 2 + j5 Ω, Rcp = 6.4 kΩ and Xmp = 2.6 kΩ Standard no-load and short-circuit
tests are performed on this transformer. Determine the following:
No-load test results: Voc , I oc ,
Poc
Short-circuit test results:
Vsc , I sc , Psc
For open circuit test on the primary side:
Voc = 2400V
Voltameter eadings
Ic =
Voc 2400
=
= 0.375 A
Rc 6400
Im =
Voc 2400
=
= 0.923 A
X m 2600
I oc = I c2 + I m2 = 0.375 2 + 0.923 2 = 0.997 A Ameter readings
The wattmeter readings is Poc = I c2 Rc = 0.375 2 * 6400 = 900W
For short circuit test on the primary side:
VA 10 * 10 3
I r1 =
=
= 4.167 A
Vr1
2400
(Ameter readings)
The voltameter readins is I sc * (Reqp + jX eqp ) = 4.167 * (2 + j5) = 22.441V
The wattmeter readings is
4-
( )
2
I sc
* Reqp = 4.167 2 * (2) = 34.73W
A single-phase, 250 kVA, 11 kV/2.2 kV, 60 Hz transformer has the following parameters.
RHV= 1.3 Ω XHV=4.5Ω, RLV = 0.05 Ω, XLV = 0.16, Rcs= 2.4 kΩ Xms = 0.8 kΩ.
Standard no-load and short-circuit tests are performed on high voltage of this transformer.
Determine the following:
No-load test results: Voc , I oc ,
(b)
Short-circuit test results:
Poc
Vsc , I sc , Psc
The HV winding of the transformer is connected to the 11 kV supply and a load,
Z L = 15∠ − 90 o Ω is connected to the low voltage winding. Determine:
(i)
Load voltage. (ii) Voltage regulation.
Solution:
a=
11000
=5
2200
Reqp = R p + a 2 R s = 1.3 + 5 2 * 0.05 = 2.55Ω
X eqp = X p + a 2 X s = 4.5 + 5 2 * 0.16 = 8.5Ω
Rcp = a 2 Rcs = 5 2 * 2.4 = 60kΩ
X mp = a 2 X ms = 5 2 * 0.8 = 20kΩ
For open circuit test on the primary side:
Voc = 11 kV
Ic =
Voltameter eadings
Voc 11000
=
= 0.183 A
Rc 60000
Im =
Voc 11000
=
= 0.55 A
X m 20000
I oc = I c2 + I m2 = 0.183 2 + 0.55 2 = 0.58 A Ameter readings
The wattmeter readings is Poc = I c2 Rc = 0.183 2 * 60000 = 2009W
For short circuit test on the primary side:
I r1
VA 250 * 10 3
=
=
= 22.73 A
Vr1 11 * 10 3
(Ameter readings)
The voltameter readins is I sc * (Reqp + jX eqp ) = 22.73 * (2.55 + j8.5) = 201.71V
( )
The wattmeter readings is I sc * Reqp = 22.73 * (2.55) = 1317.51W
2
(b)
2
Z L = 15∠ − 90 o Ω
Z L′ = a 2 * Z L = 5 2 * 15∠ − 90 o = 375∠ − 90 o Ω
Z L′
375∠ − 90 o
o
= 11000∠0 *
V2′ = V1
= 11270∠ − 0.4 o
o
Z L′ + Z eqp
375∠ − 90 + (2.55 + j8.5)
Then load voltage = V2 = V2′ / a = 11270 / 5 = 2254V
VR =
5
V1 − V2′ 11000 − 11270
=
= −2.396%
11270
V2′
A 1 φ , 10 kVA, 460/ 120 V, 60 Hz transformer has an efficiency of 96% when
it delivers 9 kW at 0.9 power factor. This transformer is connected as an
autotransformer to supply load to a 460 V circuit from a 580 V source.
(a)
Show the autotransformer connection.
(b)
Determine the maximum kVA the autotransformer can supply to the 460 V circuit.
(c)
Determine the efficiency of the autotransformer for full load at 0.9 power factor.
6
Reconnect the windings of a 1 φ , 3 kVA, 240/120 V, 60 Hz transformer so that it can supply
a load at 330 V from a 110 V supply. (a) Show the connection.
(b) Determine the maximum kVA the reconnected transformer can deliver.
7
Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a 3 φ
460/208 V transformer bank. The equivalent impedance of each transformer referred to the
high-voltage side is 1.0 + j2.0 Ω. The transformer delivers 20 kW at 0.8 power factor (leading).
(a) Draw a schematic diagram showing the transformer connection. (b) Determine the transformer
winding current. (c) Determine the primary voltage. (d) Determine the voltage regulation.
8
A 1 φ 200 kVA, 2100/210 V, 60 Hz transformer has the following characteristics. The
impedance of the high-voltage winding is 0.25 + j 1.5 Ω with the lowvoltage winding
short-circuited. The admittance (i.e., inverse of impedance) of the low-voltage winding is 0.025 - j
O.075 mhos with the high-voltage winding open-circuited.
(a)
Taking the transformer rating as base, determine the base values of power, voltage, current,
and impedance for both the high-voltage and low-voltage sides of the transformer.
(b)
Determine the per-unit value of the equivalent resistance and leakage reactance of the
transformer. (c)
(d)
Determine the per-unit value of the excitation current at rated voltage.
Determine the per-unit value of the total power loss in the transformer at full-load output
condition.
9- A 460 kVA, (4600/460) V, 60 Hz single phase transformer has the following parameters:
RP = 2.6 Ω, RS = 0.024 52, Re+h = 4 kΩ, XS = 0.06 Ω,
XP = 6.0Ω, Xm = 2.4 kΩ. The HV winding
of this transformer is connected to a 4.6 kV, 60 Hz supply while the LV winding is connected to a load of a
resistance of 6.0 Ω, an inductor of 14.0 mH and a capacitor of 502 uF all connected in shunt. Calculate the
voltage regulation, efficiency, input current and supply power factor.
Sol:
ZL =
1
1
1
+
+ j 2 * π * 60 * 0.000502
6 j 2 * π * 60 * 0.014
= 6∠0.076Ω
Z L′ = a 2 * Z L = 100 * 6∠0.076 = 600∠0.076Ω
600∠0.076Ω
600 ∠0.076 + 2.6 + j 6 = 602∠0.076Ω
602.638∠0.65Ω
1
1
1
1
+
+
4000 J 2400 600∠0.65
= 510.653∠12.85Ω
510.653∠12.85Ω
Z in = 510.653∠12.85Ω + 2.6 + J 6 = 514.55∠13.43Ω
V
4600
I1 = 1 =
= 8.94∠ − 13.43 A
Z in 514.55∠13.43Ω
So supply power factor is cos (13 .43) = 0.973 Lagging
E = V1 − I 1 * ( R1 + jX 1 ) = 4600 ∠0 − 8.94 ∠ − 13 .43 * (2.6 + j 6 ) = 4565 .17 ∠ − 0.59V
4565.17∠ − 0.59
= 7.58∠ − 1.233 A
602∠0.076
602∠0.65
V 2′ = E − I 2′ * (R 2′ + JX 2′ ) = 4565 .17 ∠ − 0.59V − 7.58∠ − 1.233 * (2.6 + j 6 ) = 4545 .19 ∠ − 1.16V
V ′ * I ′ cos(φ L ) 4545.19 * 7.58 * cos(0.076 )
η= 2 2
=
* 100 = 86.133%
V1 I 1 cos(φ inp )
4600 * 8.94 * cos(13.43)
I 2′ =
E
=
At no load the equivalent circuit becomes:
Noload equivalent circuit
1
Z excitation =
= 2057.98∠59.04Ω
1
1
+
4000 J 2400
4600
* 2057.98 ∠59.04 = 4585.56∠ − 0.024V
Then V NL =
2057.98∠59.04 + 2.6 + j 6
Then voltage regulation can be obtained as following:
V ′ − V2′ 4585.56 − 4545.19
VR = 2 NL
=
* 100 = 0.89%
V2′
4545.19
10- A single-phase, 10 kVA, 2000/200 V, 60 Hz distribution transformer has the following
characteristics:
Core loss at full voltage =120 W
Copper loss at half load =80 W
i. Determine the efficiency of the transformer when it delivers full load at 0.8 power factor
lagging.
ii. Determine the rating at which the transformer efficiency is a maximum. Determine the
efficiency if the load power factor is 0.9
iii. Determine the rating of transformer at 92% efficiency and 0.8 power factor.
iv. The transformer has the following load cycle:
No load for 6 hours
66% full load for 10 hours at 0.8 PF
85% full load for 8 hours at 0.9 PF
Determine the all day efficiency of the transformer
2- (i) Pout = 10 * 0.8 = 8kW
Pcore = 120W
Pcu
80
= x 2 Pcu , FL =
= 320W
2
Pcu , FL
⎛1⎞
⎜ ⎟
⎝2⎠
η=
8000
* 100 = 94.74%
8000 + 120 + 320
(ii) Maximum efficiency occurs at Pcore = Pcu
2
Then Pcu = 120 = xmax
Pcu , FL
xmax =
η=
Pcore
=
Pcu , FL
120
= 61.23%
320
0.6123 * 10000 * 0.9
* 100 = 95.83%
0.6123 * 10000 * 0.9 + 120 + 120
8000 * x
(iii) η =
2
= 92%
8000 * x + 120 + 320 * x
8000 * x = 0.92 8000 * x + 120 + 320 * x 2
294.4 x 2 − 640 x + 110.4 = 0
(
x=
)
640 ± 6402 − 4 * 294.4 * 110.4
= 1.087 ± 0.898
2 * 294.4
Then
x = 1.985 refused
or
x = 0.189
E24 = 0 + 10 * 0.66 * 0.8 *10 + 8 * 0.9 * 0.85 *10 = 114kWh
Ecore = 120 * 24 *10 −3 = 2.88kWh
(iv)
Ecu = 320 * 0.66 2 *10 + 320 * 0.85 2 * 8 = 3.244 kWh
E24
Then η all _ day =
*100
E24 + Ecore + Ecu
114
η all _ day =
*100 = 94.9%
114 + 2.88 + 3.244
11- A 6kVA, 250/500 V, transformer gave the following test results
short-circuite 20 V ; 12 A, 100 W and Open-circuit test : 250 V, 1 A, 80 W
i.
Determine the transformer equivalent circuit.
ii.
calculate applied voltage, voltage regulation and efficiency when the output is 10 A at 500
volt and 0.8 power factor lagging.
iii.
Maximum efficiency, at what percent of full load does this maximum efficiency occur? (At
0.8 power factor lagging).
iv.
At what percent of full load does the effeciency is 95% at 0.8 power factor lagging.
3- (I)
From O.C. Test
Po = Vo I o * cos ϕ o
∴ cos ϕ o =
Po
80
=
= 0.32
Vo I o 250 *1.0
= cos −1 0.32 = 71.3371o
Then I c1 = I o cos ϕ o = 1.0 * 0.32 = 0.32 A
I m1 = I o sin ϕ o = 1.0 * 0.7953 = 0.7953 A
V
250
Then Rc1 = o1 =
= 781.25 Ω
I c1 0.32
V
250
And X m1 = o1 =
= 314.35 Ω
I m1 0.7953
Then ϕ o
As shown in Fig.3.16, these values refer to primary i.e. low-voltage side
From Short Circuit test:
The rated current of the secondary side is:
I2 =
6000
= 12 A
500
It is clear that in this test instruments have been placed in the secondary i.e. highvoltage winding
and the low-voltage winding i.e. primary has been short-circuited.
Now,
Z eq 2 =
V2 sc 20
=
= 1.667Ω
I 2 sc 12
2
Z eq1 = a * Z eq 2
2
1
= ⎛⎜ ⎞⎟ *1.667 = 0.4167Ω
⎝ 2⎠
Psc = I 22sc Req 2
100
Then, Req 2 =
= 0.694 Ω
2
12
2
1⎞
⎛
2
Then, Req1 = a * Req 2 = ⎜ ⎟ * 0.694 = 0.174 Ω
⎝2⎠
Also,
Then,
2
2
2
2
X eq1 = Z eq
1 − Req1 = 0.4167 − 0.174 = 0.3786 Ω
As shown in the following figure, these values refer to primary i.e. low-voltage side
j0.3786 0.174
I0
V2′
V1
314.35
781.25
The parameters of series branch can be obtained directly by modifying the short circuit test
data to be referred to the primary side as following:
SC test 20 V ; 12 A, 100 W (refered to secondery)
SC test 20*a=10V ; 12/a=24A, 100 W (refered to Primary)
So, Z eq1 =
V1sc 10
=
= 0.4167Ω
I1sc 24
Also, Psc = I12sc Req1
Then, Req1 =
100
24 2
= 0.174 Ω
2
2
2
2
Then, X eq1 = Z eq
1 − Req1 = 0.4167 − 0.174 = 0.3786 Ω
It is clear the second method gives the same results easly.
(II) Output KVA = 10 * 500 * 0.8 = 4 kVA
Now, from the aproximate equivalent circuit refeared to secondery :
V1 ∠δ o = V2′ ∠ 0 o + I 2′ ∠ϕ o * Z eq1
Then,
V1 ∠δ o = 250 ∠ 0 o + 20 ∠ − 36.87 o * (0.174 + j 0.3786)
= 257.358 ∠0.89 o
V − V ′ 257.358 − 250
*100 = 2.943%
VR = 1 2 =
V2′
250
Pout = 10 * 500 * 0.8 = 4kW ,
Pi = Poc = 80W , and ,
Pcu = 10 2 * Req 2 = 100 * 0.694 = 69.4W or
2
2
⎛ I ⎞
⎛ 10 ⎞
Pcu = Psc * ⎜⎜ 2 ⎟⎟ = 100 * ⎜ ⎟ = 69.4 W
⎝ 12 ⎠
⎝ I 2 SC ⎠
Pout
4000
η=
*100 = 96.4%
=
Pout + Pi + Pcu 4000 + 80 + 69.4
(III) maximum effeciency ocures when Pc = Pcu = 80W
the
The percent of the full load at which maximum efficiency occurs is :
⎛ P ⎞
80
X= ⎜ c ⎟=
= 0.8945%
⎜P
⎟
100
cu
,
FL
⎝
⎠
Then, the maximum efficiency is :
η=
6000 * 0.8945 * 0.8
* 100 = 96.41%
6000 * 0.8945 * 0.8 + 80 + 80
(IV)
η=
=
Pout
Pout
= 0.95
+ Pi + Pcu
6000 * 0.8 * x
6000 * 0.8 * x + 80 + 100 * x 2
= 0.95
Then,
95 x 2 − 240 x + 76 = 0
Then, x = 2.155 (Unacceptable)
Or x = 0.3712
Then to get 95% efficiency at 0.8 power factor the transformer must work at 37.12% of full load.
12- A single phase, 50 kVA, 2400/460 V, 50 Hz two-winding transformer has an efficiency of 0.95%
when it delivers 45kW at 0.9 power factor. This transformer is connected as an auto-transformer to supply
load to a 2400 V circuit from 2860 V source.
(a) Show the transformer connection.
(b) Determine the maximum kVA that autotransformer can supply to 2400 V circuit.
(c) Determine the efficiency of the autotransformer for full load at 0.9 power factor.
Solution:
(a)
460
2860
2400
(b) I s , 2 w
50 *103
=
= 108.7 A
2460
Then, kVA ) Auto = 108.782860 = 310.87 kW
(c) η2 w
Then,
50 *103 * 0.9
=
= 0.95
50 *103 * 0.9 + Pi + Pcu , FL
Pi + Pcu , FL = 2368.42 W
η Auto =
310870 * 0.9
= 99.61 %
310870 * 0.9 + 2368.42
13- Three single phase, 30 kVA, 2400/240 V, 50 Hz transformers are connected to form 3 ϕ, 4160/240 V
transformer bank. The equivalent impedance of each transformer referred to the high voltage side is 1.5+j2Ω.
The transformer delivers 60 kW at 0.75 power factor (leading). (a) Draw schematic diagram showing the
transformer connection. (b) Determine the transformer winding current (c) Determine the primary voltage.
(d) Determine the voltage regulation. (e) determine the maximum efficiency if the maximum effeciency
occurred at 90% of full load at 0.9 power factor.
Solution:(a)
(b) kVA =
I 1 ph =
60
= 80kVA
0.75
80 *10 3
3 * 4160
= 11.1 A
I 1L = I 1 ph = 11.1 A
a=
2400
= 10
240
I 2 ph = a * I 1 ph = 10 * 11.1 = 111 A
I 2 L = 3 * I 2 ph = 3 *111 = 192.3 A
V2′ = 2400∠0 o V , I 2′ =
(
V1 = V2′ + I 2′ * Z eq1
)
I 2 ph
a
=
111
= 11.1∠41.41o A
10
= 2400∠0 + 11.103∠41.41o * (1.5 + j 2) = 2397.96∠0.66o V
Then the line primary voltage is
V1L = 3 * V1 = 3 * 2397.96 = 4153.39V
VR =
=
V1 − V2′
*100
V2′
2397.96 − 2400
*100 = −0.0875%
2400
I 2′ rated =
η max
S1 ph
V1 ph
=
30 *10 3
= 12.5 A
2400
Pout
0.9 * 90 * 10 3 * 0.9
=
=
* 100 = 98.46%
Pout + 2 Pcu 0.9 * 90 * 10 3 * 0.9 + 2 * 0.9 2 * 12.5 2 * 1.5 * 3
14- Three single-phase, 50 kVA, 2300/230 V, 60 Hz transformers are connected to form a three-phase,
4000/230 V transformer bank. The equivalent impedance of each transformer referred to low voltage is
0.012 + j0.016 Ω. The three-phase transformer supplies a three-phase 120 kVA, 230 V, 0.85 PF (lag) load.
(a) Draw a schematic diagram showing the transformer connection.
(b) Determine the transformer winding currents.
(c) Determine the primary voltage (line-to-line) required.
(d) Determine the voltage regulation.