1- A single-phase transformer is to be designed to have primary voltage of 33 KV and secondary voltage of 6.6 KV . If maximum density permissible is 1.2 Wb / m 2 and number of primary turns 1250 T , calculate the number of secondary turns and cross-sectional area of the core when the operating frequency is 50 Hz . 2- A single-phase 50 Hz transformer is required to step-down from 2200V to 250V . The cross sectional area of the core is 36cm 36cm 2 and the maximum value of the flux density is 6 Wb / m 2 . Determine the number of turns of the primary and secondary windings and also the turn ratio. 3- A single-phase 50 KVA transformer transformer has primary voltage of 6600V and secondary voltage of 254 V and has 32 secondary turns. Calculate the number of primary turns and primary and secondary currents. 4- A single-phase 50 Hz transformer transformer has primary voltage of 6000V and secondary voltage of 250 V . Find the number of turns in each winding if the flux is to be 80 mWb . 5- A 3000 / 200 V - 50 Hz single-phase transformer is built on a core having an effective cross-sectional area of 150 cm 2 and has also 80 turns in the low voltage winding. Calculate: a) The value of the maximum flux density in the core. b) The number of turns in the high voltage winding. 6- A 40 KVA 3300 / 240 V − 50 Hz single phase transformer has 660 turns on the primary. Determine: a) The number of turns on the secondary. b) The maximum value of flux in the core. c) The approximate values of primary and secondary full-load currents.
7- A 50 Hz single-phase transformer transformer has a turn ratio of 6 . The resistances are 0.9 Ω , 0.03 Ω and reactances are 5 Ω and 0.13 Ω for high voltage and low voltage windings respectively. Find: a) The voltage to be applied to the H.V. side to obtain full load current of 200 A in the L.V. winding on short- circuit. b) The power factor on short circuit. Solution: Turn ratio, k = N 1 / N 2 2
=
6
0.9 + 0.03 × 6 2
Re1
= R1 + R 2 .k =
X e1
= X 1 + X 2 .k = 5 + 0.13 × 6
Z e1
=
V 1
2
=
V 2 a) V sc
1.98 2 I 2
=
I 1
+ 9.68
k ⇒ I 1
2
=
=
I 2
=
Re1 Z e1
=
=
k
1.98 9.88
=
9.68Ω
9.88Ω 200 6
= 100 / 3 A
= I 1 . Z e1 = 100 / 3 × 9.88 =
b) cos ϕ sc
2
= 1.98Ω
=
329.3 V
0 .2
8-The equivalent circuit impedances of a 20 KVA 8000 / 400 V transformer are to be determined. The open-circuit test and the short-circuit test were performed performed on the primary side of the transformer and the following data were taken. Open-circuit test (on primary) Voc = 8000 V Ioc = 0.214 A Poc = 400 W
short-circuit test (on primary) Vsc = 489 V Isc = 2.5 A Psc = 240 W
Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit. Solution: - The power factor during the open-circuit is: i s:
Poc
cos ϕ o =
400
=
= 0.234 Voc. Ioc 8000 × 0.214 Ic = Io. cos ϕ o = 0.214 × 0.234 = 0.05 A
Im = Io. sin ϕ o = 0.214 × 0.972 = 0.208 Rc
=
Xm
Voc
=
=
Ic Voc
8000
= 160000 Ω = 160 K Ω
0.05 8000
=
Im
=
0.208
38460 Ω = 38.46 K Ω
- The power factor during the short-circuit test is: Psc 240 cos ϕ sc = = = 0.1963 Vsc. Isc 489 × 2.5
ϕ sc
78.7 o
=
Ze = Re =
Vsc
489
=
Isc Psc
= 195.6∠78.7Ω
2.5 240
=
I 2 sc
2 .5 2
Xe = 195.6 2
=
38. 4Ω
− 38.4
2
= 191.8Ω
or : Re = Ze. cos ϕ sc = 195.6 cos 78.7 = 38.4 Ω Xe = Ze. sin ϕ sc
= 195.6 sin 78.7 = 191.8 Ω
9 - Obtain the equivalent circuit of a 200/400 V, 50Hz single phase transformer referred to primary i.e. low voltage side, from the following test results: O.C. test: 200V , 0.7 A , 70W on L.V. side S.C. test: 15V , 10 A , 85W on H.V. side Solution: From O.C. it may be noted that in this test instruments have been placed in the primary i.e. low voltage winding and the high voltage winding i.e. secondary is left open. Now: P o 70 = = 0.5 and sin ϕ o = 0.866 cos ϕ o = U 1o I 1o 200 × 0.7
I c
= I o
cos ϕ o
=
0.7 × 0.5 = 0.35 A
I m
= I o
sin ϕ o
=
0.7 × 0.866 = 0.606 A
Rc
=
X m
V 1o I c
=
V 1o I m
=
200 0.35
=
=
200 0.606
571.4Ω =
330Ω
From S.C. it may be noted that in this test instruments have been placed in the secondary i.e. high voltage winding and the low voltage winding i.e. primary has been short circuited. Now: V 15 Z e = sc = = 1.5Ω I sc 10
P sc
Re
=
X e
=
I sc2
=
Z e2
85
=
100
0.85Ω
2
1.5 2
− Re =
− 0.85
2
= 1.236Ω
These values refer to secondary i.e. high voltage side. They must be referred to primary i.e. low voltage side by computing the transformation ratio as follows: 200 K = = 0.5 400 Now the values of Re and X e referred to primary i.e. low voltage side:
0.85 × 0.5 2
Re
=
X e
= 1.236 × 0.5
= 2
0.2125Ω =
0.309Ω
10 - A single phase 20 KVA , 50 Hz , 2200 / 220V transformer is tested as follows: O.C. test: 220V , 4.2 A , 148W on L.V. side S.C. test: 86V , 9.1 A , 360W on H.V. side On load test when the normal applied primary voltage is 2200V , the secondary voltage drops to 200V and the wattmeter in the primary circuit reads 16 KW . Determine: a- nominal currents of the transformer b- Percentage core losses c- Percentage copper losses d- Core loss resistance Rc and magnetizing reactance X m referred to primary e- Equivalent resistance Re and reactance X e referred to primary f - The percentage no-load current g- The percentage short-circuit voltage h- Regulation of the transformer i- The efficiency of the transformer Solution: a- Primary (H.V.) nominal current: I nH
=
Secondary (L.V.) nominal current: I nL 148
20 × 1000
=
220
× 100 = 0.74% 20.000 360 × 100 = 1.8% c- Percentage copper losses: 20.000
b- Percentage core losses:
=
2200 20 × 1000
9.1 A
=
91 A
d- From O.C. it may be noted that in this test instruments have been placed in the secondary i.e. low voltage winding and the high voltage winding i.e. primary is left open. Now: P o 148 = = 0.16 and sin ϕ o = 0.987 cos ϕ o = U 1o I 1o 220 × 4.2
I c
= I o
cos ϕ o
=
4.2 × 0.16 = 0.672 A
I m
= I o
sin ϕ o
=
4.2 × 0.987 = 4.15 A
Rc
=
X m
V 1o I c
=
=
V 1o
220
=
I m
=
327.4Ω
=
53Ω
0.672 220 4.15
These values refer to secondary i.e. low voltage side. They must be referred to primary i.e. high voltage side by computing the transformation ratio as follows: 2200 = 10 K = 220 Now the values of Rc and X m referred to primary i.e. high voltage side:
Rc X m
327.4 × 10 2
=
=
53 × 10 2
=
=
32740Ω
5300Ω
e- From S.C. it may be noted that in this test instruments have been placed in the primary i.e. high voltage winding and the low voltage winding i.e. secondary has been short circuited. Now: V 86 = 9.45Ω Z e = sc = I sc 9.1
P sc
Re
=
X e
=
2 sc
I
=
Z e2
360 9.12 2
=
− Re =
4.35Ω 9.45 2
− 4.35
2
=
8.39Ω
These values refer to primary i.e. high voltage side.
f- Percentage no-load current:
I o I n
=
g- Percentage short-circuit voltage:
4.2 91 V sc V n
h- Regulation of the transformer: ε % =
ε % =
220 − 200 220
× 100 =
9.1%
× 100 =
=
86 2200
V 2o
4.6% × 100 =
− V 2
V 2 o
3.9%
× 100 =
i- The efficiency of the transformer: P oc + P sc Input Power − Losses Losses η % = = 1− = 1− Input Power Input Power P 1
148 + 360
=1-
16.000
=
=
0.96825 = 96.825%
11 - A single phase 50 KVA transformer has primary voltage of 20000 V and secondary voltage of 400 V . The number of primary turns is N p = 4000 T and the cross sectional area of the core is S = 150 cm 2 and the length of the magnetic path in the iron core is 1.6 m . If the no load current drawn by the primary is 0.5 A , calculate: a) - The magnetic flux φ in the core b) - The flux density B in the core c) - The magnetizing force H (magnetic field intensity) d) - The permeability µ e) - The magnetic reluctance Rm f) - The number of turns of the secondary N s Solution: E = 4.44 N p f φ
φ = B
E φ
H = µ = Rm a
=
N 2
0.023
=
−4
= 1.53 Wb / m
S 150 × 10 N p I o 4000 × 0.5 =
l av B
1.6 1.53
=
H 1250 1 l av
=
1
µ S 1.224 × 10 N p V 1n 20000 =
=
N 1 50
=
V 2 n =
−3
=
400
40000 50
=
2
= 1250 AT / m
= 1.224 × 10
=
N s
0.023 Wb
=
4.44 × 4000 × 50
4.44 N p f
=
20000
=
.
−3
H / m 1.6
150 × 10
−4
= 8.7 × 10
4
50
80 T
12- A 50 KVA , 2200 / 220V , 50 Hz single phase transformer has a core loss, determined by the open-circuit test of 350W and a cupper loss at rated current of 630W determined by the short-circuit test, find the efficiency: a) At full load unity power factor b) At three fourths of full load unity power factor c) At full KVA rating 80 percent power factor d) At three fourths of rated KVA 80% power factor
Solution:
a)
η =
b)
η =
c)
η =
d)
η =
50000 × 1
= 98.1% 50000 × 1 + 350 + 630 3 / 4 × 50000 × 1
3 / 4 × 50000 × 1 + 350 + (3 / 4) 2 630
=
98.2%
50000 × 0.8
= 97.6% 50000 × 0.8 + 350 + 630 3 / 4 × 50000 × 0.8
3 / 4 × 50000 × 0.8 + 350 + (3 / 4) 2 630
=
97.7%
13 - The following readings were obtained in the open circuit and short circuit tests on 10 KVA , 450 / 120V transformer. O.C. Test V 1o = 450V
I o P 1o
= 1.12 A =
80W
S.C. test V sc = 9.65Ω
I sc
=
P sc
= 120W
22.2 A
Calculate: a) The equivalent circuit constants b) The efficiency and voltage regulation when delivering full load 0.8 lagging p.f. (power factor) c) The efficiency at half full-load and 0.8 lagging p.f. Solution: a): 10000 I 1n = = 22.2 A 450 10000 = 83.3 A I 2 n = 120 450 = 3.75 a= 120
80
cos ϕ o =
=
450 × 1.12 sin ϕ o = 0.987
0.159
Ic = Io. cos ϕ o = 1.12 × 0.159 = 0.178 A Im = Io. sin ϕ o = 1.12 × 0.987 = 1.1 A Rc
450
=
Xm
=
Ze = Re =
=
0.178 450
1.1 9.65
253Ω
=
409Ω
=
0.435Ω
22.2 120
(22.2) 2
=
0.243Ω
Xe = 0.361Ω b): 10000 × 0.8 η = = 0.9756 = 97.56% 10000 × 0.8 + 80 + 120 I 83.3 I 2' = 2 = = 22.2 A a 3.75 ε % = 22.2(0.243 × 0.8 + 0.361 × 0.6) / 450 = 2.04% c): At half full-load Iron losses = 80W 2 Cu losses = (1 / 2) × 120 = 30W 1 / 2 × 10000 × 0.8 η = = 0.9732 = 97.32% 1 / 2 × 10000 × 0.8 + 80 + 30
14 - A 15 KVA , 2300 / 230V 1-phase transformer is to be tested to determine its excitation branch components and its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer. O.C .Test
Vo = 2300V Io = 0.21 A Po = 50W S .C .Test Vsc = 47V Isc
=
6.0 A
Psc = 160W a) Find the equivalent circuit of the transformer referred to the high voltage side. b) Find the equivalent circuit of the transformer referred to the low voltage side. c) Calculate the full load voltage regulation at 0.8 lagging power factor, 1.0 power factor and at 0.8 leading power factor.
d) Calculate the efficiency of the transformer at full load with a power factor of 0.8 lagging and at half full load with a power factor of o.8 lagging. Solution: 15000 15000 = 6.52 A = 65.2 A , I 2 n = a) I 1n = 2300 230 The equivalent circuit referred to high voltage: 50 o cos ϕ o = = 0.1035 ⇒ ϕ o = 84 2300 × 0.21 I c = 0.21 × 0.1035 = 0.02174 A
I m
=
Rc
=
0.21 × 0.9945 = 0.21 A 2300 0.02174 2300
X m
=
X e
=
= 105795 Ω = 105.8 K Ω
= 10952 .4Ω = 10.95 K Ω 0.21 47 Z e = = 7.83Ω 6 160 Re = 2 = 4.45Ω 6
7.83 2
− 4.45
2
=
6.45Ω
The equivalent circuit referred to low voltage: 230 a= = 10 2300 Rc = 105.8 × 0.12 = 1.058 K Ω = 1058Ω
X m
2
= 10.952 × 0.1 = 109.52Ω
Re
=
4.45 × 0.12
=
0.0445Ω
X e
=
6.45 × 0.12
=
0.0645Ω
c) The regulation: The secondary current (load current) referred to the primary 2300 = 10 a= 230 I 65.2 I 2' = 2 = = 6.52 A a 10 6.52 ε %( L) = ( 4.45 × 0.8 + 6.45 × 0.6) = 0.021 = 2.1% 2300 6.52 ε %( R ) = (4.45 × 1 + 0) = 0.0126 = 1.26% 2300 6.52 ε %(C ) = ( 4.45 × 0.8 − 6.45 × 0.6) = −0.00088 = −0.088% 2300 d) The efficiency:
iS n cos ϕ 2
η =
iS n cos ϕ 2
+ p iron + i
2
=
p cu
1 × 15000 × 0.8 1 × 15000 × 0.8 + 50 + 12 × 160
0.5 × 15000 × 0.8
η =
0.5 × 15000 × 0.8 + 50 + 0.5 2 × 160
=
=
0.9828 = 98.28%
0.985 = 98.5%
15 - The efficiency of a 400 KVA single phase transformer is 98.77% when delivering full load at 0.8 p.f. and 99.13% at half load and unity power factor. Calculate: a) The iron losses b) The full load cupper losses Solution: P P 2 η = 2 = P 1 P 2 + Losses
0.9877 P iron
400 × 0.8
=
0.9913
=
=
losses
0.5 × 400 × 1 0.5 × 400 × 1 + P i + (0.5) 2 P cu
P i + 0.25 P cu = 1.76 By solving (I) and P i = 1 KW P cu
⇒
400 × 0.8 + losses (I) + P cu = 4
=
4 KW
=
200
⇒
200 + P i + 0.25 P c
(II) (II)
3 KW
16 - A single phase transformer is rated at 100 KVA , 2300 / 230V , 50 Hz . The maximum flux density B in the core is 1.2Wb / m 2 and the net cross sectional area of the core is 0.04m 2 . Determine: a) The number of primary and secondary turns needed b) If the mean length l of the magnetic circuit is 2.5m and the relative permeability µ r is 1200, determine the magnetizing current c) On short-circuit with full load current flowing, the power input is 1200W and on open-circuit with rated voltage, the power input was 400W . Determine the efficiency of the transformer at 75% of full load with 0.8 p.f. lagging. Solution: a): The number of primary and secondary turns E 1 = V 1 = 4.44 × f × N 1 × φ = 4.44 × f × N 1 × B × S
4.44 × 50 × N 1 × 1.2 × 0.04 ⇒
2300
=
N 1
216 turns
=
a
N 1
=
N 2
216 N 2
=
V 1n V 2 n
=
2300 230
= 10 ⇒ N 2 =
216 10
= 10
=
21.6 = 22 turns
b): The magnetizing current B B F = H .l = N . I ⇒ .l = N . I ⇒ .l = N . I ⇒ µ µ o µ r
1.2 × 2.5
= 216. I ⇒ 4π × 10 − 7 × 1200 I = 9.21 A c): The efficiency 0.75 × 100.000 × 0.8 η = 0.75 × 100.000 × 0.8 + 400 + (0.75) 2 × 1200
=
98.26%
17- A 120 KVA, 6000/400V, 3-phase transformer is star/delta connected. Calculate: - The phase voltages on each side. - The phase currents and the line currents on each side. - The transformation ratio. 18- A 100KVA 3-phase 50Hz 3300/400V transformer is Delta connected on the H.V. side and Star connected on the L.V. side. Calculate: a - The transformation ratio. b - The nominal line and phase currents in each winding. Solution: a- The transformation ratio is: 3300 K = = 14.29 400 / 3 b- The primary (H.V. side) nominal current: S n 100 × 1000 = = 17.5 A = I nLine ( H .V .) which is the line current. I n ( H .V .) = 3V n ( H .V .) 3.3300 Since the primary is delta connected so the phase current equals: I nLine( H .V .) 17.5 = = 10.1 A I nphase( H .V .) = 3 3 The secondary (L.V. side) nominal current: S n 100 × 1000 = = 144.3 A = I nLine ( L.V .) which is the line current. I n ( L.V .) = 3V n ( L.V .) 3.400 Since the secondary is star connected so the phase current is the same as line current: I nphase ( L.V .) = I nLine ( L.V .) = 144.3 A
R 1
M
jX1
R 2
jX2
I1
I2
V1
R c
V2
jXm
E1
ZL
E2
The Magnetic Circuit of the Transformer
R 1
jX1
ja²X2=jX2 Io
I1
V1
Ic
R c
Im
a²R 2= R 2
I2/a=I2
jXm
The Exact Electrical Equivalent Circuit of the Transformer
aV2=V2
ZL
R 1 Io Ic
V1
R c
Im
jXm
jX1 I1
ja²X2= jX2 a²R 2= R 2 I2/a=I2
aV2= V2
The Approximate Electrical Equivalent Circuit of the Transformer
ZL
Open-circuit Test: The purpose of this test is to:
1-
Calculate the phase angle φ between V 1 and I o , the core loss
current I c and the magnetizing current I m from the test results and construct the phasor diagram. 2Calculate the core loss resistance Rc and the magnetizing reactance
X m . 3Determine the core losses of the transformer. 4Determine the percentage of the no-load current. 5Deduce the transformation ratio. 6Draw the equivalent circuit for the transformer on no-load and insert these values into it.
A W AC
V 1
V 2
Connection Diagram for no-load Test
Equivalent circuit of the transformer on no-load
Phasor Diagram of currents in Exciting Impedance
Results: V 1o
V 1n
I o
P o
V 2o
(2 − 6)% I 1n
P iron
V 2 n
Calculations: P o cos ϕ o = V 1o . I o
I c
= I o
cos ϕ o
I m
= I o
sin ϕ o
Rc
=
X m
V 1o I c
=
V 1o I m
The ratio of transformation is: V V N a = 1 = 1o = 1n N 2 V 2 o V 2 n The percentage no-load current is: I i o = o × 100 I 1n
Short-circuit Test: The purpose of this test is to: 1- Calculate the equivalent winding resistance Re and leakage reactance X e
.
2- Determine the rated cupper losses P cu . 3- Calculate cos ϕ sc . 4- Calculate the percentage of the short circuit voltage v sc . 5- Draw the complete equivalent circuit for the transformer and insert the values of its parameters into it. .
A W
AC
V
Connection Diagram for Short-circuit Test jXe
Re ISC VSC
Equivalent circuit of the transformer on short circuit
Ze
jXe
X’2
X1
R 1
R’2
Results: V sc
I sc
P sc
(5 − 10)% V 1n
I n
P cu ( n )
ϕ sc
Calculations: V Z e = sc I sc
P sc
Re
=
X e
=
I sc2
cos ϕ sc
Z e2
− Re
2
=
P sc V sc . I sc
Re
sc = Z e . cos ϕ
X e
sc = Z e . sin ϕ
The percentage short circuit voltage is: V v sc % = sc × 100 V 1n
Re
Load Test: The purpose of this test is to determine the following: 1- The efficiency of the transformer η % . 2- The percentage voltage regulation ε %. 3- The load characteristics V 2 = f ( I 2 ) of the transformer
A1
A2 W V 1
AC
V 2
ZL
Connection Diagram for Short-circuit Test
1- The efficiency: Output power P 2 η = = Input Power P 1
P 2
=
P 2
+ iron losses ( P i ) + cupper losses ( P cu )
The percentage load of the transformer is: I i = 2 ⇒ I 2 = i. I 2 n I 2 n = V 2 I 2
The output power P 2
cos ϕ 2
= V 2 .i. I 2 n . cos ϕ 2 =
i.P 2 n
The iron losses P i are constant and don't depend on the load (they depend on the inserted voltage P i ∝ V
2 2
The cupper losses P cu
= Re I = Re .i
2
I n2
=
i 2 .P cu ( n )
Thus:
η =
i. P 2 n i. P 2 n
+ P i + i
2
i.S n cos ϕ 2
=
. P cun
i.S n cos ϕ 2
+ P i + i
2
. P cun
2- The percentage voltage regulation: ε % = ε % =
V 2o
− V 2
=
V 2o aV 2 n
− aV 2
aV 2 n
V 2 n
− V 2
V 2 n =
V 1n
'
− V 2
V 1n
From the equivalent circuit of the transformer: V 1n = V 2' + Re I 2' + jX e I 2'
V 1n
'
'
= V 2 + Re . I 2
cos ϕ 2
The voltage drop
V 1n
'
'
− V 2 = I 2 ( Re
'
+ X e I 2
sin ϕ 2 '
∆V = V 1n − V 2 =
cos ϕ 2
'
+ X e I 2 )
The percentage voltage regulation ε % =
ε % =
V 1n
'
− V 2
V 1n
=
I 2' V 1n
( Re cos ϕ 2
+ X e
V 1n
'
− V 2
=
V 1n
sin ϕ 2 )
Where: I I 2' = 2 a If the load is only resistive (R): I 2' I 2' ε % = ( Re cos 0 + X e sin 0) = . Re V 1n V 1n If the load is only capacitive (C): I 2' I 2' ε % = [ Re cos( −π / 2) + X e sin( −π / 2)] = − . X e V 1n V 1n
3- Load characteristics V 2
=
f ( I 2 )
C
V2 [V
V2O=V2n R L I2=IL [A In Load characteristics V 2
=
f ( I 2 )
:Three Phase Transformer Connections The two usual ways of connecting the three phase transformers are known as star Y and delta ∆ . .F.1. shows the two connections
F.1 Delta (a) and Star (b) connections The term V line refers to the line voltage, which is the voltage between any two lines of a three phase system. The term V phase refers to the phase voltage, which is the voltage between a line and a common reference potential (generally neutral).
:The voltages in the two types of connection Delta ( ∆ ) connected windings: It can clearly be seen from Fig.(2-a) that for the delta connected system the .phase voltage is the same as the line voltage V phase = V line Hence:
F.2.: Delta Connection Phasor Diagram This is demonstrated in F2. Where F2.a. shows how the voltages in the windings sum to zero. F2.b. shows the three voltages as separate phasors , symmetrically spaced at 120 o to each other. Note that there is no neutral point. Star (Y) connected windings: Consider the voltages in the system. From F.1.b, it is apparent that each line voltage is the phasor difference of two of the phase voltages. That is the line voltage V AB is obtained by subtracting V B from V A .
.F.3. shows the phasor diagram of the voltages in the star connected system
F.3. Star Connection Phasor Diagram From this, the relationship between the phase and the line voltages can be :calculated
V line
=
2 × V phase cos 30 = 3V phase
:The currents in the two types of connection .F.4. demonstrates the current flow in the windings for both connections
F.4. The Current Flow in Windings The term I line refers to the line current, which is the current through each connection line to the three phase system. The term I phase refers to the phase current, which is the current through each individual phase winding. Consider the currents in the two types of connection. Star (Y) connected windings: It can be clearly seen from F.4.b. that for the star connected system the phase current is the same as line current. I phase = I line Hence:
Delta ( ∆ ) connected windings: F.5. shows the phasor diagram for the currents in the delta connected system.
Fig. (5) The Phase Diagram for the Currents From this, the relationship between phase and line currents can be calculated:
I line
=
2 × I phase cos 30 = 3I phase
Power considerations in three phase transformers: - The operating conditions for the star connected system shown in F2.4 are represented in the phasor diagram of F.6. , in which ϕ represents the phase difference between the current and the voltage.
F.6. The Phase Diagram for the Star Connected System The conditions are summarized as follows: V V phase = line 3
I phase
= I line
Power per phase = V phase I phase
=
V line
I line cos ϕ 3 Hence since total power in the system is the sum of power in all three phases:
V line
I line cos ϕ = 3V line I line cos ϕ 3 - The voltages in the delta connection are represented by the triangle around the outside of F.6. This is reproduced in F.7. , so that the direction of the phasors is clearly shown. Total power = 3 × V phase I phase
=
3×
F.7. The Phase Diagram for Delta Connected System The conditions for delta connections are summarized as follows: V phase = V line
I phase
=
I line 3
Power in each phase = V phase I phase
= V line
I line
cos ϕ 3 Hence since total power in the system is the sum of power in all three phases: I line cos ϕ = 3V line I line cos ϕ Total power = 3 × V phase I phase = 3 × V line 3 Hence for both star and delta connections in three phase systems , the total power is given by: Total power =
3V line I line
This Wattmeter is suitable in ac/dc circuits when measuring power up to 1Kw. The instrument is a direct type requiring no external supply connections other than those to appropriate voltage and current terminals.
EXERCISES 1- Discuss the principle of working of a transformer. 2- Derive an expression for induced e.m.f. (E) in a transformer in terms of frequency, the maximum value of flux and the number of turns on the windings. 3- Draw the approximate equivalent circuit of a transformer referred to the primary side. 4- Draw the phasor diagram for a transformer when the load power factor (p.f.) is lagging. 5- Draw the connection diagrams for the open circuit and short circuit tests of a single phase transformer, showing all the necessary instruments. 6- State the various losses which take place in a transformer. On what factors do they depend? How can these losses be determined experimentally? 7- What is meant by magnetizing current I m ? 8- What is the effect of leakage flux in a transformer? 9- Why core loss ( P i ) is neglected in short circuit test and copper loss ( P c ) is neglected in open circuit test of a transformer? 10- How can copper loss P c be measured? 11- How can iron loss P i be measured? 12- Define the voltage regulation of a transformer. Deduce the expression for the voltage regulation. 13- Why does voltage drop in a transformer? 14- What type of load should be connected to a transformer for getting negative voltage regulation? (Highly capacitive load). 15- What is a function of transformer? (The basic function of a transformer is to transform energy from one voltage level to another, may be from low voltage to high voltage or vice-versa). 16- Differentiate between step-up and step-down transformers. (When transformer raises the voltage, it is called the step-up transformer, when it lowers the voltage; it is called the step- down transformer. 17- Is there a definite relationship between the number of turns and voltages in transformers? (The voltage varies in exact proportion to the number of turns in each winding). 18- What is the order of magnitude of no-load current of a transformer? (No-load current in transformers ranges from 2 to 5 per cent of full load primary current). 19- What is the order of magnitude of short circuit voltage? (Short circuit voltage in transformers ranges from 5 to 8 per cent of the rated voltage). 20- Does iron losses in a transformer increase with load? (Iron losses are constant from no load to full load). 21- Why does a transformer have an iron core? (The use of iron core ensures a high permeability of the magnetic circuit and because of high permeability the magnitude of exciting current necessary to create the required flux in the core is small. 22- Write short notes on 3-phase transformer connections and the voltage and current relationships.
