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EET 306 ELECTRICAL MACHINES

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Presentation on theme: "EET 306 ELECTRICAL MACHINES"— Presentation transcript:

1 EET 306 ELECTRICAL MACHINES
TRANSFORMER Syafruddin Hasan Syafruddin Hasan

2 INTRODUCTION Transformer Fundamentals Loaded Transformers
Transformer Efficiency Parallel Operation Three-phase Transformers Impedance Matching Syafruddin Hasan

3 Introduction Transformer is an ac static machine that
(i) transfers electrical energy from one electric circuit to another (ii) does so without a change of frequency (iii) does so by the principle of electromagnetic induction (iv) has electric circuits that are linked by a common magnetic circuit. (v) the energy transfer usually take place with a change of voltage level, althought this is not alaway necessary Caution Transformer must not be connected to a dc source. If the primary winding of a transformer is connected to a dc supply, the flux produced will not vary but remain constant in magnitude and therefore no emf will be induced in the secondary winding, except at the moment of switching on. There will be no back induced emf in the primary winding and therefore a heavy current will be drawn from the supply which may result in the burning out of the winding. Syafruddin Hasan

4 FACTS ABOUT TRANSFORMERS
Transformers operate on mutual inductance. A transformer has a primary winding and a secondary winding. The coefficient of coupling is the portion of primary flux that links the secondary. With 100% coupling, the turns-per-volt ratio is the same for all windings. Transformers can have hysteresis, eddy current, and copper (I2R) losses. Syafruddin Hasan

5 MORE FACTS ABOUT TRANSFORMERS
Transformer losses can be reduced by using silicon steel cores, laminated cores, and small gage wires. As the load on a transformer increases, angle theta decreases. Three-phase transformers use a three-legged core. Transformer windings can be connected in series or parallel. Syafruddin Hasan

6 Transformer Fundamentals
Types of transformer Single phase transformer construction Syafruddin Hasan

7 Ideal transformer Assumptions:
Winding resistances are zero (perfect conductivity) No leakage flux (inductance) No iron loss Core material’s permeability is constant Magnetization current generates a flux that induces voltage in both windings Syafruddin Hasan

8 TRANSFORMER FUNDAMENTALS
Primary Secondary Load A transformer has a primary coil and a secondary coil. The primary is connected to a source. The secondary is connected to a load. During the first half-cycle, the flux builds up and collapses. This creates a half-cycle of induced voltage in the secondary. Syafruddin Hasan

9 TRANSFORMER FUNDAMENTALS
Primary Secondary Load During the next half-cycle, the flux again builds up and collapses. This creates another half-cycle of induced voltage in the secondary. Notice that the primary and secondary voltages are out-of-phase. Syafruddin Hasan

10 PRIMARY-SECONDARY TERMINOLOGY
Transformers are bidirectional devices. 120 V Primary Source 90 V Secondary Load This transformer was designed to step 120 V down to 90 V. 120 V Secondary Load Source 90 V Primary However, either winding can be used as the primary. Syafruddin Hasan

11 Mutual Inductance Transformers are constructed of two coils placed so that the charging flux developed by one will link the other. The coil to which the source is applied is called the primary coil. The coil to which the load is applied is called the secondary coil. Syafruddin Hasan

12 TURNS-PER-VOLT RATIO + _ 1.0 V + _ 0.25 V _ + 0.75 V
The 4-turn primary with a 1 volt source provides 4 turns-per-volt; therefore, each turn has 0.25 V across it. Each turn will produces a specific amount of flux in the core. This same flux will, in turn, produce 0.25 V in each secondary turn. and a 3 turn secondary provides 0.75 V. Thus, a 1 turn secondary provides 0.25 V, Syafruddin Hasan

13 Un-loadedTransformers
Current, voltages and flux in an unloaded ideal transformer Ep ImNp Vp Im Φ Closed loop Es Syafruddin Hasan

14 PRIMARY CURRENT Load Iref Ipri With no load, Ipri is the
Vpri Vpri Ipri Ien Ipri With no load, Ipri is the energizing current (Ien). With a load, Ipri is composed of Ien and Iref . The transformer acts like an inductor. Iref is the resistive load current reflected from the secondary to the primary. Theta is large; power is low. Syafruddin Hasan

15 Loaded Transformers Ep ImNp Vp Im Φm Es Loaded Es Is IsNs Φs Φp
Closed loop Currents and fluxes in a loaded ideal transformer Ep ImNp Vp Im Φm Es Loaded Es Is IsNs Φs Φp Φs = Φp I1 = Im + Ip Ip Syafruddin Hasan

16 Transformers Ideal transformer Turn ratio is:
The input and output complex powers are equal or Syafruddin Hasan

17 Real Transformer Syafruddin Hasan

18 EDDY CURRENT ILLUSTRATION
If an eddy current is induced in the core, one will also be induced in the aluminum disk placed on top of the core. The eddy current in the aluminum disk will produce a magnetic field that opposes the field that created the eddy current. Eddy current Aluminum disk Coil connected to a 60-Hz supply Syafruddin Hasan

19 TRANSFORMER LOSSES Copper (I2R) loss B H Eddy current loss (I2R in the core) Hysteresis loss (green area of the hysteresis loop) These losses are minimized by a narrow hystersis loop, thin laminations, and large diameter wire. Syafruddin Hasan

20 Transformers Real transformer
Figure Leakage flux in the real transformer Syafruddin Hasan

21 No-load Transformer Ic = Io cos φo Im = Io sin φo
No-load circuit equivalent Ic = Io cos φo Im = Io sin φo The exciting current (Io) magnitude about 1% - 5% of the rated current of the primary for power transformer, but may be much higher for small transformers Syafruddin Hasan

22 primary current Io has two components:
Active or working or iron (core) loss component Ic (in phase with V1), because it mainly supplies the iron loss plus small quantity of primary copper loss Ic = Io cos φo Magnetizing component Im (in quadrature with V1), because its function to sustain the alternating flux in the core Im = Io sin φo Syafruddin Hasan

23 Loaded Transformers Equivalent circuit of a real transformer
Syafruddin Hasan

24 Primary side I1 = Io + I2’ with, Secondary side with, and
Syafruddin Hasan

25 (c) Capacitive load (a) Resistive load (b) Inductive load
Syafruddin Hasan

26 Equivalent circuit refer to primary side
R2’ = a2 R2 = equivalent secondary resistance as referred to primary X2’ = a2 X2 = equivalent secondary reactance as referred to primary ZL’ = a2 ZL = equivalent load impedance as referred to primary E2’ = a E V2’ = aV I2’ = I2 / a Syafruddin Hasan

27 R01 = R1 + R2’ = R1 + a2 R2 = equivalent or effective resistance of transformer as referred to primary X01 = X1 + X2’= X1 + a2X2 = equivalent or effective reactance of transformer as referred to primary Z01 = √(R012 + X012) = equivalent or effective impedance of transformer as referred to primary Syafruddin Hasan

28 Equivalent circuit refer to secondary side
R02 = R2 + R1’ = R02 = R2 + R1/a2 = equivalent or effective resistance of transformer as referred to seconadry X02 = X2 + X1’= X2 + X1/a2 = equivalent or effective reactance of transformer as referred to secondary Z02 = √(R022 + X022) = equivalent or effective impedance of transformer as referred to secondary Syafruddin Hasan

29 Total Approximate Voltage Drop in a Transformer
= AN = AD + DN = I2R02 cosφ + I2X02 sinφ In general, approximate voltage drop is The approximate voltage drop as referred to primary is Note: (+) sign for inductive load (-) sign for capacitive load % voltage drop in secondary is Syafruddin Hasan

30 Determination of transformer parameters by measurement
Syafruddin Hasan

31 Transformer Test No-Load Test Objectives: or Syafruddin Hasan

32 Short Circuit Test Syafruddin Hasan

33 Regulation of a Transformer
Regulation of a transformer is the change in secondary terminal voltage from no-load to full-load and divided it by no-load terminal voltage at a primary input voltage and power factor are constant. and In further treatment, unless stated otherwise, regulation is to be taken as regulation ‘down’. The total voltage drop that was explained above can be used as % voltage regulation. The regulation may also be expressed in terms of primary values. Syafruddin Hasan

34 Losses in Transformer Pcu = I12 R1 + I22R2 = I12 R01 = I22 R02
Core or Iron loss. It includes both hysteresis and eddy current. Because the flux in a transformer remains practically constant for all loads ( its variation being 1 to 3% from no load to full load), the core loss is practically the same at all loads. Hysteresis loss, Eddy current loss, Copper Loss. This loss is due to the ohmic resistance of the transformer windings. Total Cu loss Pcu = I12 R1 + I22R2 = I12 R01 = I22 R02 Syafruddin Hasan

35 Efficiency Iron loss = Cu loss Condition for Maximum Efficiency
Variation of Efficiency with Power Factor Syafruddin Hasan

36 Differentiating both sides with respect to I1, we get
Primary input = V1 I1 cos φ1 Differentiating both sides with respect to I1, we get For efficiency (η) to be maximum, dη / dI1 = 0 Hence, the above equation becomes or Wi = I12 R01 or = I22 R02 Iron loss = Cu loss Syafruddin Hasan

37 TRANSFORMER EFFICIENCY
Source provides 1640 W Transformer 90 W loss (heat loss) Load 1550 W consumed The transformer core and copper losses cause the transformer to heat up as electric energy is converted to heat energy. Efficiency = 1550 W / 1640 W = Syafruddin Hasan

38 to increase power capability of transformer for maintenance
Parallel Operation of Single-phase Transformer Objectives: to increase power capability of transformer for maintenance The conditions that must be satisfied Primary windings of the transformers should suitable for the supply system voltage and frequency. The transformers should be properly connected with regard to polarity. The voltage rating of both primaries and secondaries should be identical. In other words, the transformers should have the same transformation ratio. The percentage impedances should be equal. Syafruddin Hasan

39 Parallel Operation of Single-phase Transformer
The current carries by each transformer The VA carries by each transformer and Syafruddin Hasan

40 Polarity Test 1. additive polarity 2. subtractive polarity
Connect the high voltage winding to a low voltage ac source Eg (say 120V) 2. Connect a jumper J between any two adjacent HV and LV terminals 3. Connect a voltmeter Ex between the other two adjacent HV and LV terminals 4. Connect another voltmeter Ep across the HV winding If Ex >> Ep, the polarity is additive. H1 and X1 are diagonally opposite. If Ex << Ep, the polarity is subtractive. H1 and X1 are adjacent. Syafruddin Hasan

41 SERIES-OPPOSING WINDINGS
120 V + _ 8 V 2 A 6 V 3 A 2 V _ + 2 A Connect two terminals with the same instantaneous polarities. Take the output from the other two terminals. The output voltage equals the difference between the two voltages. The current is limited to the lesser of the two winding currents. Syafruddin Hasan

42 SERIES-AIDING WINDINGS
120 V + _ 8 V 2 A 6 V 3 A 14 V _ + 2 A Connect two terminals that have opposite instantaneous polarities. Take the output from the other two terminals. The output voltage equals the sum of the two winding voltages. The current is limited to the lesser of the two winding currents. Syafruddin Hasan

43 PARALLEL SECONDARY WINDINGS
120 V + _ 9 V 4 A 9 V _ + 8 A The two windings must have equal voltage ratings. The two windings should have equal current ratings. Connect the negative terminals and the positive terminals. Take the output from the negative and positive terminals. Voutput = Vwinding and Ioutput = 2 x Iwinding Syafruddin Hasan

44 IMPEDANCE MATCHING 10:1 ratio 100 W 1 W 0.1 A 1 A 10 V 1 V 20 V 100 W
Notice the source in both circuits provides 0.1 A at 10 V. Thus, the transformer makes the 1-W resistor act like a 100-W resistor in terms of the load on the source. Syafruddin Hasan


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