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Transformer
The Principle of Operation of a Transformer.
Transformer is an electrostatic device which transforms energy from a circuit to another by the Principle of Mutual Induction. The energy is transmit without change in frequencyby increasing or decreasing the voltage. The transformer is electrically isolated but magnetically coupled device.
Figure (i) shows a Single Phase Transformer with
= Primary voltage = Primary e.m.f. = Secondary voltage = Secondary e.m.f. = Primary Current = Primary Turns = Secondary Current = Secondary Turns = Flux in core
Types of transformer
- Core Type Transformer
In this transformer windings surround considerable (same size of) part of core.
- Shell Type Transformer
In this transformer core surrounds considerable portion of winding.
EMF Equation of a transformer
Average emf perturn = 4f volts
R.M.S value of emf per turn = form factor X Average emf = 1.11 x 4 f (^) m = 4.44 f (^) m
Thus,
A = 0.03 m^2. The number of primary turn is 31 and that of secondary is 375. The area of the core is 0.03 m 2
Example 2 A 25 kVA transformers have 500 turns on primary and 50 turns on secondary windings. Primary winding is connected to 300 volts, 50 HZ supply. Find the full load primaryand secondary currents, the secondary e. m. f. and maximum flux in the core. Solution Transformer capacity = 25 kVA = 25x 10³ VA, N 1 = 500, N 2 =50, E1=3000 V, Frequency = 50 Hz
From transformation ratio, K= EE^21 = NN^2 1 = I^1 I 2 Hence, E 2 = N 2 x E 1 N 1 = 50 x 3000 E 2 = 300 V^500 E 1 I 1 = E 2 I 1 = 25000 VA So, EI 1 1 I= 25000^1 = 25000 and^ E^2 II^22 ==^2500025000 = 8.3 A^300 = 83.3 A^300 Since we know that, E 1 = 3000 V So, 4.44 x N1 x f x Ø (^) m = 3000
Øm = 3000 4.44 x 500 x 50 = 0. = 27.03 mWb
Secondary current = 83.3A, primary current = 8.3A, Secondary e.m.f. = 300 V and maximum flux in core = 27.03 mWb
Transformer on no-load
When transformer is on no-load, the secondary current IPrimary turns draw a small current namely no-load current, I 2 = 0.0.
No-load current, I^0 is consisting of two components: a)b) IIμ = I w = I oo cos ø (working component or loss component)sin ø (magnetizing component)
Vector Diagran Power drawn by Primary is given by W 0 = V 1 I 0 Cos
Power factor , cos = 0. = cos-1^ (0.2) = 78.46 0 I = I 0 sin = (15) sin 78.46= (15) (0.9798) = 14.7 A. Example 5 The no-load current of transformer is 4.0 A at 0.25 power factor, when supplied at 250V, 50Hz. Primary turns are 200.Calculate i) R.M.S value of flux ii) Magnetizing current, andiii) Loss current. Solution I 0 = 4.0 A, cos =0.25, E 1 =250 V, f = 50 Hz, N 1 =200. E 1 = 4.44 N 1 f (^) m 250 = 4.44 x 200 x 50 x (^) m m = 0.00563 Wb. i) R.M.S value of flux = m = 0. = 0.00398 Wb = 3.98 mWb ii) cos = 0.25= cos-1 (^) (0.25) = 75.25 0 I (^) W =^ I 0 cos (^0) = 4 x 0.25= 1 A iii) I = I 0 sin = 4 sin (75.25 0 ) = 4 x 0. = 3.87 A
Transformer under load Primary winding secondary winding
When transformer is loaded , it draw secondary current I2. Primary current is given by I 1 = I 0 + I (^) 2’ Where I ‘ 2 = k I 2 ( k is transformation ratio ) I 0 is no-load current I (^) 2’ = k I 2 load component of primary current
Load can be either resistive ,inductive or capacitive. Note a ) ln pure a.c resistive circuit the current is in phase with voltage. b) ln pure a.c inductive circuit the current lags the voltage by 90 0. c) ln pure a.c capacitive the current leads the voltage by 90 0. Example 6 A single phase transformer with a ratio of 440/ 110 V takes a no-load5 A at 0.2 power factor lagging .lf the secondary supplies a current of 120 A at 0.8 current of power factor lagging , determine the primary current. Solution E 1 = 440 V , E 2 =110 V , | Io | = 5 A , cos (^) 0= 0.2 lag , I 2 = 120 A., cos 2 = 0.8 lag Transformation ratio , k = = = 0. Cos 0= 0.2^ lag
cos 2 = 0.8 lag secondary current , I2 = 0 2 = cos^ -1^ (0.8)^ =50 A = 36.86 0
Load component of primary current I (^) 2 = (^) = 0.5 x 50k I^2 = 25 A primary current, I (^) 1 = I^0 + I^2 = (0.4 – j0.92) +=^1 + 25 (20 – j15) = 20.4 - j 15.92 A= 25.9A
Example 8 The number of turns on primary and secondary of a transformer are 350 and 38respectively. If primary is connected to 2.2 kV, 50 Hz supply determine i) Secondary voltage on no- load.ii) Primary current when secondary current is 200 A at 0.8 power factor lagging. If he no-load current is 5 A at 0.2 power factor lag, find the power factor of current. Solution N 1 = 350 , N 2 = 38 , E 1 = 2.2 kV =2200 V , f = 50 Hz , I 0 = 5 A. i) Transformation ratio k == = = 0. E 2 = k E 1 = 0.11 x 2200 = 239.8 V ii) = 200 A and cos 2 = 0.8 lag 2 = cos^ -1^ (0.8) I = 36.86^0 2 == 200^0 A
cos 0 = 0.2 I 0 = 0 = cos= 78.46^ -1^ (0.2) 0 = 5A
I 2 = k I 2 = 0.11 x 200 A = 22. A I (^) 1 = I 0 + I0 01 F 2 1 = 44.22 0 = 5A + 22. cos 1 = cos 44.22 0 = 0.717 lag = (1- j4.9) + (17.6- j13.2) power factor = 0.717 lag. # = 18.6 - j18.1= 25.95 A = (^1)
Equivalent circuit of a transformer.
Secondary resistance and reactance are connected to primary.
here k is transformation ratio.
Primary resistance and reactance are connected to secondary
Where k is transformation ratio.
Approximate Voltage drop of a Transformer
Vector diagram:
In this diagram, The load inductive which has lagging power factor. When loaded,
For lagging power factor,
For leading power factor,
For unity (one) power factor,
Where,
Example 11
F 0^ A 230 V/460 V transformer has primary resistance of 0.2^ F 05 7^ and primary resistance of 0. 5 7secondary terminal voltage when supplying 10 A at0.8 power factor lagging.. Corresponding values for secondary are 0.75^ F 05 7 and 1.8^ F 05 7 , respectively. Determine
Solution
Hence,
Example 12 Calculate percentage of voltage regulation of transformer in which percentage resistancedrop is 1% and percentage drop is 5% when the power factor is,
i) 0.8 lagging ii) unity iii) 0.8 leading Solution
Losses in Transformer
i) Copper losses:a) Primary copper loss = I 12 R^1 W^ b) Secondary copper loss = I^ 2²R^2 W Total copper losses = ( I= l1²R1²R 01 1+l w2²R 2 ) wattswhere, R02=equivalent secondary resistance = I2²R 02 w R 01=equivalent primary resistance Copper losses is a variable loss ii) Core losses:a) Hysterisis loss b) Eddy current loss 1-
V1cosø 1 V 1I 12 cosø 1 I 1²Ro1 = W I
Or
Example 13
A 25 KVA, 2200 V/220 V, single phase transformer has primary resistance of 1 secondary resistance of 0.01 Ω. Find the equivalent secondary resistance and full load Ω efficiency at 0.8 power factor, if iron losses of the transformer is 80% of full load copper loss. Solution
Transformer ratings = 25KVA, RIron losses = 0.8 of full load copper loss. 1 = 1Ω, R 2 = 0.01Ω, cos ø = 0.8,
Transformer ratio, k = EE 12 = 2200220 = 0.
Equivalent secondary resistance, R (^02) = R= R (^02) + K²R+ R1’ 1 = 0.01 + (0.1) ²(1) = 0.02 Ω Apparent power = 25 x 10³ VA Real power = Apparent power x cosø
(Output) = 25 x10³ x 0. = 20 x 10³ w E^2 I^2 = 25000 VA^ Full load copper loss = I^ 2² R^02 I 2 = 25000 220 = (113.6) ²(0.02)= 258.1 W = 113.6 A Iron losses = 0.8 x full load copper loss = 0.8 x 258.1 = 206.5 W Full load efficiency = Output at 0.8 power factor (^) = output + full load copper loss + Iron loss 20000 20000 + 258.1 + 206. = 0. = 97.73 %
Example 14 A 20KVA, 440V/220V single-phase transformer has iron loss of 324 w. Half full loadCopper loss is found to be 100 w when delivering full load current. Determine
i) Efficiency when delivering full load current at 0.8 power factor lagging.ii) Percentage of full load when the efficiency is maximum.
Solution i) Transformer ratio, K = E1 = 440 = 2 E2 220 Half full load copper loss = 100 W Full load copper loss = 2² x 100 = 400 W Iron loss = 324 W Full load output = 20 x 10³ x 0.8 = 16000 W Full load efficiency = Output at 0.8 power factor Full load output + Full load copper loss + iron loss = 16000 16000 + 400 + 324
EXACT EQUIVALENT CIRCUIT REFERRED TO SECONDARY
APPROXIMATE EQUIVALENT CIRCUIT REFERRED TO SECONDARY
Testing of transformers
Open-circuit test or no-load test.
From open- circuit test we obtained iron loss i.e. W (^) o Besides,
Wo=Vo I o cos ø o
Cos øo = W o ; øo=Cosˉ¹ ( Wo )
Vo I o V 0 lo
For Low Voltage side (LV),
For High Voltage side (HV),
Short-circuit Test or Impedance Test
From short-circuit test, we obtained full load copper loss i.e Wsc. For High Voltage Side,
For Low Voltage Side,
Where k =
