Transformer 72j5c

AC Machines Transformer Lecture Notes By: Engr. Faheemullah Shaikh

Transformer Definition A transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. A varying current in the first or primary winding creates a varying magnetic flux in the transformer's core, and thus a varying magnetic field through the secondary winding. This varying magnetic field induces a varying electromotive force (EMF) or voltage in the secondary winding. This effect is called mutual induction

Transformer History † The phenomenon of electromagnetic induction was discovered independently by Michael Faraday and Joseph Henry in 1831. However, Faraday was the first to publish the results of his experiments and thus receive credit for the discovery. † Faraday's experiments including winding a pair of coils around an iron ring, thus creating the first toroidal closed-core transformer. † The first type of transformer to see wide use was the induction coil, invented by Rev. Nicholas Callan of Maynooth College, Ireland in 1836. He was one of the first researchers to realize that the more turns the secondary winding has in relation to the primary winding, the larger the increase in EMF. † In 1876, Russian engineer Pavel Yablochkov invented a lighting system based on a set of induction coils where the primary windings were connected to a source of alternating current and the secondary windings could be connected to several "electric candles" (arc lamps) of his own design. The coils Yablochkov employed functioned essentially as transformers

Transformer (History Continued) † In 1878, the Ganz Company in Hungary began manufacturing equipment for electric lighting, and by 1883 had installed over fifty systems in AustriaHungary. Their systems used alternating current exclusively, and included those comprising both arc and incandescent lamps, along with generators and other equipment. † Lucien Gaulard and John Dixon Gibbs first exhibited a device with an open iron core called a "secondary generator" in London in 1882, then sold the idea to the Westinghouse company in the United States. They also exhibited the invention in Turin, Italy in 1884, where it was adopted for an electric lighting system. Efficient, practical transformer designs did not appear until the 1880s. †

Russian engineer Mikhail Dolivo-Dobrovolsky developed the first threephase transformer in 1889.

Transformer The Ideal Transformer A lossless device with an input winding and an output winding.

Transformer The transformer has Np turns of wire on its primary side and Ns turns of wire on its secondary sides. The relationship between the primary and secondary voltage is as follows:

v p (t ) vs (t )

=

Np Ns

=a

where a is the turns ratio of the transformer.

The relationship between primary and secondary current is: Np ip (t) = Ns is (t) i p (t ) 1 = i s (t ) a

Note that since both type of relations gives a constant ratio, hence the transformer only changes value of the magnitude of current and voltage. Phase angles are not affected.

Transformer The dot convention in schematic diagram for transformers has the following relationship: • If the primary voltage is +ve at the dotted end of the winding wrt the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same wrt the dots on each side of the core. • If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.

Transformer Power in an Ideal Transformer The power supplied to the transformer by the primary circuit: Pin = Vp Ip cos θp Where θp = the angle between the primary voltage and the primary current. The power supplied by the transformer secondary circuit to its loads is given by: Pout = Vs Is cos θs Where θs = the angle between the secondary voltage and the secondary current. The primary and secondary windings of an ideal transformer have the SAME power factor – because voltage and current angles are unaffected θp - θs = θ

Pout =

Vp a

(aI )cosθ p

Pout = Vp Ip cos θ = Pin The same idea can be applied for reactive power Q and apparent power S. Output power = Input power

Transformer Theory of Operation of Real Single-Phase Transformers † Ideal transformers may never exist due to the fact that there are losses associated to the operation of transformers. Hence there is a need to actually look into losses and calculation of real single phase transformers. † Assume that there is a transformer with its primary windings connected to a varying single phase voltage supply, and the output is open circuit. † Right after we activate the power supply, flux will be generated in the primary coils, based upon Faraday’s law,

e ind =

dλ dt

where λ is the flux linkage in the coil across which the voltage is being induced. The flux linkage λ is the sum of the flux ing through each turn in the coil added over all the turns of the coil. N λ = ∑ φi i =1

Transformer †

This relation is true provided on the assumption that the flux induced at each turn is at the same magnitude and direction. But in reality, the flux value at each turn may vary due to the position of the coil it self, at certain positions, there may be a higher flux level due to combination of other flux from other turns of the primary winding. Hence the most suitable approach is to actually average the flux level as:

φ =

λ

N

Hence Faraday’s law may be rewritten as:

eind = N

dφ dt

Transformer The voltage ratio across Transformer Based upon Faraday’s Law, looking at the primary side of the transformer, we can determine the average flux level based upon the number of turns; where, φ =

1 NP

∫v

P

( t ) dt

This relation means that the average flux at the primary winding is proportional to the voltage level at the primary side divided by the number of turns at the primary winding. This generated flux will travel to the secondary side hence inducing potential across the secondary terminal.

Transformer †

† †

For an ideal transformer, we assume that 100% of flux would travel to the secondary windings. However, in reality, there are flux which does not reach the secondary coil, in this case the flux leaks out of the transformer core into the surrounding. This leak is termed as flux leakage. Taking into the leakage flux, the flux that reaches the secondary side is termed as mutual flux. Looking at the secondary side, there are similar division of flux; hence the overall picture of flux flow may be seen as below: Primary Side:

φ P = φ M + φ LP

φ P = total average primary flux φM = flux component linking both primary and secondary coils φLP = primary leakage flux

Transformer For the secondary side, similar division applies. Hence, looking back at Faraday’s Law, dφP dφM dφLP vP (t ) = N P

dt

= NP

dt

+ NP

dt

Or this equation may be rewritten into: vP (t ) = eP (t ) + eLP (t )

The same may be written for the secondary voltage. The primary voltage due to the mutual flux is given by

dφ M eP (t ) = N P dt And the same goes for the secondary (just replace ‘P’ with ‘S’)

Transformer From these two relationships (primary and secondary voltage), we have

eP (t ) dφM eS (t ) = = NP dt NS Therefore,

eP (t ) N P = =a eS (t ) N S

Transformer Magnetization Current in a Real transformer Although the output of the transformer is open circuit, there will still be current flow in the primary windings. The current components may be divided into 2 components: 1. Magnetization current, iM – current required to produce flux in the core. 2. Core-loss current, ih+e – current required to compensate hysteresis and eddy current losses. We know that the relation between current and flux is proportional since, F = Ni = φ R φR ∴i = N

Transformer Therefore, in theory, if the flux produce in core is sinusoidal, therefore the current should also be a perfect sinusoidal. Unfortunately, this is not true since the transformer will reach to a state of near saturation at the top of the flux cycle. Hence at this point, more current is required to produce a certain amount of flux. If the values of current required to produce a given flux are compared to the flux in the core at different times, it is possible to construct a sketch of the magnetization current in the winding on the core. This is shown below:

Transformer Hence we can say that current in a transformer has the following characteristics: † It is not sinusoidal but a combination of high frequency oscillation on top of the fundamental frequency due to magnetic saturation. † The current lags the voltage at 90o † At saturation, the high frequency components will be extreme as such that harmonic problems will occur. Looking at the core-loss current, it again is dependent upon hysteresis and eddy current flow. Since Eddy current is dependent upon the rate of change of flux, hence we can also say that the core-loss current is greater as the alternating flux goes past the 0 Wb. Therefore the core-loss current has the following characteristics: † †

When flux is at 0Wb, core-loss current is at a maximum hence it is in phase with the voltage applied at the primary windings. Core-loss current is non-linear due to the non-linearity effects of hysteresis.

Transformer

Now since that the transformer is not connected to any load, we can say that the total current flow into the primary windings is known as the excitation iex = im + ih + e current.

Transformer Current Ratio on a Transformer and the Dot Convention Now, a load is connected to the secondary of the transformer. The dots help determine the polarity of the voltages and currents in the core without having to examine physically the windings. A current flowing into the dotted end of a winding produces a positive magneto motive force, while a current flowing into the undotted end of a winding produces a negative magneto motive force.

Transformer In the figure above, the net magneto motive force is Fnet = NPiP - NSiS This net magneto motive force must produce the net flux in the core, so Fnet = NPiP - NSiS = R Where R is the reluctance of the core. secondary current is approx Fnet = NPiP - NSiS ≈ 0 Thus, NPiP ≈ NSiS

The relationship between primary and

as long as the core is unsaturated. iP N S 1 = = iS N P a

In order for the magneto motive force to be nearly zero, current must flow into one dotted end and out of the other dotted end.

Transformer As a conclusion, the major differences between an ideal and real transformer are as follows: † An ideal transformer’s core does not have any hysteresis and eddy current losses. † The magnetization curve of an ideal transformer is similar to a step function and the net mmf is zero. † Flux in an ideal transformer stays in the core and hence leakage flux is zero. † The resistance of windings in an ideal transformer is zero.

Transformer The equivalent circuit of a transformer Taking into real transformer, there are several losses that has to be taken into in order to accurately model the transformer, namely: † Copper (I2R) Losses – Resistive heating losses in the primary and secondary windings of the transformer. † Eddy current Losses – resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer. † Hysteresis Losses – these are associated with the rearrangement of the magnetic domains in the core during each half-cycle. They are complex, non-linear function of the voltage applied to the transformer. † Leakage flux – The fluxes which escape core and through only one of the transformer windings are leakage fluxes. They then produced self-inductance in the primary and secondary coils.

Transformer The exact equivalent circuit of a real transformer The Exact equivalent circuit will take into all the major imperfections in real transformer. Copper loss They are modelled by placing a resistor RP in the primary circuit and a resistor RS in the secondary circuit. Leakage flux As explained before, the leakage flux in the primary and secondary windings produces a voltage given by

dφLP eLP (t ) = N P dt

dφ LS eLS (t ) = N S dt

Transformer Since flux is directly proportional to current flow, therefore we can assume that leakage flux is also proportional to current flow in the primary and secondary windings. The following may represent this proportionality:

φ LS = ( PN S ) i S φ LS = ( PN S )iS Where

Thus

P = permeance of flux path NP = number of turns on primary coils NS = number of turns on secondary coils

di S d 2 e LS ( t ) = N S ( PN S ) i S = N S P dt dt di S d 2 e LS ( t ) = N S ( PN S ) i S = N S P dt dt

Transformer The constants in these equations can be lumped together. Then

diS eLS (t ) = LS dt diS eLS (t ) = LS dt Where LP = NP2 P is the self-inductance of the primary coil and LS = NS2 P is the self-inductance of the secondary coil. Therefore the leakage element may be modelled as an inductance connected together in series with the primary and secondary circuit respectively.

Transformer Core excitation effects magnetization current and hysteresis & eddy current losses The magnetization current im is a current proportional (in the unsaturated region) to the voltage applied to the core and lagging the applied voltage by 90° - modeled as reactance Xm across the primary voltage source. The core loss current ih+e is a current proportional to the voltage applied to the core that is in phase with the applied voltage – modeled as a resistance RC across the primary voltage source.

Transformer

Transformer Based upon the equivalent circuit, in order for mathematical calculation, this transformer equivalent has to be simplified by referring the impedances in the secondary back to the primary or vice versa.

Transformer Excitation Current Exciting current is the amount of amperage a transformer draws under a no load condition. Another way to look at it is that exciting current is the transformer's "idling" current. Exciting current is actually made up of two components: no load losses (normally expressed in watts) and reactive power (normally expresses in KVAR). Exciting current varies as a percent of the transformer's nameplate rating depending upon the transformer size. It is not unusual to have an exciting current of approximately 10% on very small transformers (under 1 KVA). On larger transformers, exciting current could be as low as a half of one percent.

Transformer Inrush Current When a transformer is initially connected to a source of AC voltage, there may be a substantial surge of current through the primary winding called inrush current. This is analogous to the inrush current exhibited by an electric motor that is started up by sudden connection to a power source, although transformer inrush is caused by a different phenomenon. (Discuss Later). We know that the rate of change of instantaneous flux in a transformer core is proportional to the instantaneous voltage drop across the primary winding. Or, as stated before, the voltage waveform is the derivative of the flux waveform, and the flux waveform is the integral of the voltage waveform. In a continuously-operating transformer, these two waveforms are phase-shifted by 90o. Since flux (Φ) is proportional to the magnetomotive force (mmf) in the core, and the mmf is proportional to winding current, the current waveform will be in-phase with the flux waveform, and both will be lagging the voltage waveform by 90o:

Transformer

Let us suppose that the primary winding of a transformer is suddenly connected to an AC voltage source at the exact moment in time when the instantaneous voltage is at its positive peak value. In order for the transformer to create an opposing voltage drop to balance against this applied source voltage, a magnetic flux of rapidly increasing value must be generated.

Transformer The result is that winding current increases rapidly, but actually no more rapidly than under normal conditions: Both core flux and coil current start from zero and build up to the same peak values experienced during continuous operation. Thus, there is no "surge" or "inrush" or current in this scenario.

Transformer Alternatively, let us consider what happens if the transformer's connection to the AC voltage source occurs at the exact moment in time when the instantaneous voltage is at zero. During continuous operation (when the transformer has been powered for quite some time), this is the point in time where both flux and winding current are at their negative peaks, experiencing zero rate-of-change (dΦ/dt = 0 and di/dt = 0). As the voltage builds to its positive peak, the flux and current waveforms build to their maximum positive rates-of-change, and on upward to their positive peaks as the voltage descends to a level of zero:

Transformer A significant difference exists, however, between continuous-mode operation and the sudden starting condition assumed in this scenario: during continuous operation, the flux and current levels were at their negative peaks when voltage was at its zero point; in a transformer that has been sitting idle, however, both magnetic flux and winding current should start at zero. When the magnetic flux increases in response to a rising voltage, it will increase from zero upwards, not from a previously negative (magnetized) condition as we would normally have in a transformer that's been powered for awhile. Thus, in a transformer that's just "starting," the flux will reach approximately twice its normal peak magnitude as it "integrates" the area under the voltage waveform's first half-cycle:

Transformer In an ideal transformer, the magnetizing current would rise to approximately twice its normal peak value as well, generating the necessary mmf to create this higherthan-normal flux. However, most transformers aren't designed with enough of a margin between normal flux peaks and the saturation limits to avoid saturating in a condition like this, and so the core will almost certainly saturate during this first halfcycle of voltage. During saturation, disproportionate amounts of mmf are needed to generate magnetic flux. This means that winding current, which creates the mmf to cause flux in the core, will disproportionately rise to a value easily exceeding twice its normal peak:

Transformer This is the mechanism causing inrush current in a transformer's primary winding when connected to an AC voltage source. As you can see, the magnitude of the inrush current strongly depends on the exact time that electrical connection to the source is made. If the transformer happens to have some residual magnetism in its core at the moment of connection to the source, the inrush could be even more severe. Because of this, transformer overcurrent protection devices are usually of the "slow-acting" variety, so as to tolerate current surges such as this without opening the circuit.

Transformer Autotransformer Auto-Transformer is a transformer with only one winding. That means, the primary and secondary windings are not electrically isolated from each other. other In other words, the primary and the secondary of autotransformer are physically connected. The theory is almost the same as two winding transformer

Transformer

Multiple Tap Autotransformer

Multiple Tap Autotransformer with load

Transformer

Step-down Autotransformer

Step-down Autotransformer

Transformer a

IH V2

IL

b I2 I1

VL

VH

V1 c

Two – winding Transformer

Connection as an Auto-Transformer

Transformer 1. Voltage Rating

a

V L = V1

V2

V H = V1 + V 2

IL

2. Current Rating

∴

I Rated I Rated

VH

V1 c

VA TW V2 I Rated

b I2 I1

VL

VA TW I1 = V1 I2 =

IH

3. Autotrans Turns-Ratio

autotrans

;

at high voltage side at low voltage side

, IH = I2

, I L = I H + I1

VH a = VL

Transformer 4. kVA Rating

5. Efficiency

VH I H kVA 1000 V I VAL = L L kVA 1000

Given ,

VAH =

Plosses and power factor , ∴ Pout = p. f × kVAautotrans Know that , Pin = Plosses + Pout ⎛ Pin − Plosses ⎞ Pout ⎟⎟ × 100% η= × 100% = ⎜⎜ Pin Pin ⎝ ⎠ ⎛ Plosses ⎞ ⎟⎟ × 100% η = ⎜⎜1 − Pin ⎠ ⎝

Transformer Example: A single phase, 100kVA, 2000/200V two-winding transformer is connected as an autotransformer as shown in Figure such that 2000V is obtained at secondary. The portion ab is the 200V winding, and the portion bc is the 2000V winding. Compute the kVA rating as an autotransformer. Also calculate the efficiency for a given losses of 2800W at 0.866 lagging power factor.

a

IH V2

IL

b I2 I1

VL

V1 c Figure

VH

Transformer VA TW 100 × 10 3 = = 50 A I1 = 2000 V1

Given , Plosses = 2800 W and p . f = 0 . 866 ,

VA TW 100 × 10 3 I2 = = = 500 A V2 200

∴ Pout = 0 . 866 × (1100 × 10 3 ) = 952 . 6 kW

∴

⎟ × 100 % η = ⎜⎜ 3 ⎟ × 955 . 4 10 ⎝ ⎠

I Rated

autotrans

;

I H = I 2 = 500 A I L = I H + I 1 = 500 + 50 = 550 A

V L = V1 = 2000 V V H = V1 + V 2 = 2000 + 200 = 2200 V VA

H

=

( 2200 )( 500 ) VH IH = = 1100 kVA 1000 1000

=

VLIL ( 2000 )( 550 ) = = 1100 kVA 1000 1000

OR VA

L

Pin = Plosses + Pout = 952 . 6 × 10 3 + 2800 = 955 . 4 kW ⎛ 952 . 6 × 10 3 ⎞

= 99 . 71 %

Transformer Transformer Taps & Voltage Regulation Distribution Transformers have a series taps in windings which permit small changes in turn ratio of transformer after leaving factory A typical distribution transformer has four taps in addition to nominal setting, each has a 2.5% of full load voltage with the adjacent tap This provides possibility for voltage adjustment below or above nominal setting by 5% Example: A 500 kVA, 13200/480 V distribution transformer has 4, 2.5 % taps on primary winding. What are voltage ratios? Five possible voltage ratings are: +5% tap 13860/480 V +2.5% tap 13530/480 V Nominal rating 13200/480 V -2.5% tap 12870/480 V -5% tap 12540/480 V

Transformer Taps on transformer permit transformer to be adjusted in field to accommodate variations in tap voltages While this tap can not be changed when power is applied to transformer Some times voltage varies widely with load, i.e. when high line impedance exist between generators & particular load; while normal loads should be supplied by an essentially constant voltage One solution is using special transformer called: “tap changing under load transformer” A voltage regulator is a tap changing under load transformer with built-in voltage sensing circuitry that automatically changes taps to preserve system voltage constant

Transformer some occasions it is desirable to change voltage level only by a small amount i.e. may need to increase voltage from 110 to 120 V or from 13.2 to 13.8 kV. This may be due to small increase in voltage drop that occur in a power system with long lines. In such cases it is very expensive to hire a two full winding transformer, however a special transformer called: ”auto-transformer” can be used. Diagram of a step-up auto-transformer shown in figure below: C: common, SE: series

Transformer A step-down auto-transformer : IH=ISE IL=ISE+IC In step-up autotransformer: VC / VSE = NC / NSE (1) NC IC = NSE ISE (2) voltages in coils are related to terminal voltages as follows: VL=VC (3) VH=VC+VSE (4) current in coils are related to terminal currents: IL=IC+ISE (5) IH=ISE (6)

Transformer Voltage & Current Relations in Autotransformer VH=VC+VSE Since VC/VSE=NC/NSE Î VH=VC+ NSE/NC . VC Noting that: VL=VC Î VH=VL+ NSE/NC . VL= (NSE+NC)/NC . VL VL / VH = NC / (NSE+NC)

(7)

Current relations: IL=IC+ISE employing Eq.(2) Î IC=(NSE / NC)ISE IL= (NSE / NC)ISE + ISE, since ISE=IHÎ IL= (NSE / NC)IH +IH = (NSE + NC)/NC . IH ÎIL / IH = (NSE + NC)/NC

(8)

Transformer Apparent Power Rating Advantage of Autotransformer Note : not all power transferring from primary to secondary in autotransformer through windings Therefore if a conventional transformer be reconnected as an autotransformer, it can handle much more power than its original rating . The input apparent power to the step-up autotransformer is : And the output apparent power is: And : Apparent power of transformer windings:

Sin=VLIL Sout=VH IH Sin=Sout=SIO

SW= VCIC=VSE ISE

Transformer This apparent power can be reformulated: SW= VCIC=VL(IL-IH) =VLIL-VLIH employing Eq.(8) Î SW= VLIL-VLIL NC/(NSE+NC) =VLIL [(NSE+NC)-NC /(NSE+NC)=SIO NSE /(NSE+NC)

SIO / SW = (NSE+NC) / NSE

(9)

This equation describes the apparent power rating advantage of an autotransformer over a conventional transformer. SIO is the apparent power entering the primary and leaving the secondary of the transformer, while SW is the apparent power actually traveling through the transformer's windings (the rest es from primary to secondary without being coupled through windings) Note that smaller the series winding , the greater advantage.

Transformer Limitations A failure of the insulation or the windings of an autotransformer can result in full input voltage applied to the output. This is an important safety consideration when deciding to use an autotransformer in a given application. Furthermore the input and output are not isolated; thus, if the "neutral" side of the input is not at ground voltage, the "neutral" side of the output will also be not grounded. it requires both fewer windings and a smaller core, an autotransformer for power applications is typically lighter and less costly than a two-winding transformer, up to a voltage ratio of about 3:1 - beyond that range a two-winding transformer is usually more economical.

Transformer In three phase power transmission applications, autotransformers have the limitations of not suppressing harmonics currents and as acting as another source of ground fault currents. A large three-phase autotransformer may have a "buried" delta winding, not connected to the outside of the tank, to absorb some harmonic currents. A special form of autotransformer called a "zig zag" is used to provide grounding (earthing) on three-phase systems that otherwise have no connection to ground (earth). A zig-zig transformer provides a path for current that is common to all three phases (so-called "zero sequence" current). Like multiple-winding transformer, autotransformers operate varying magnetic fields and so cannot be used directly on DC.

on

time-

Transformer Applications Autotransformers are frequently used in power applications to interconnect systems operating at different voltage classes, for example 138 kV to 66 kV for transmission. Another application is in industry to adapt machinery built (for example) for 480 V supplies to operate on a 600 V supply. On long rural power distribution lines, special autotransformers with automatic tapchanging equipment are inserted as voltage regulators, so that customers at the far end of the line receive the same average voltage as those closer to the source. The variable ratio of the autotransformer compensates for the voltage drop along the line. In audio applications, tapped autotransformers are used to adapt speakers to constant-voltage audio distribution systems, and for impedance matching such as between a low-impedance microphone and a high-impedance amplifier input.

Transformer Advantages of Auto-transformer • It effects a saving in winding material (copper or aluminum), since the secondary winding is part of the primary current. • Lower copper loss, therefore efficiency is higher than in the two winding transformer. • Lower leakage reactances, lower exciting current. • Variable output voltage can be obtained.

Transformer Problem: The 2400:240-V 50-kVA transformer is connected as an autotransformer, as shown in Fig. in which ab is the 240-V winding and bc is the 2400-V winding. (It is assumed that the 240-V winding has enough insulation to withstand a voltage of 2640 V to ground.) a. b.

Compute the voltage ratings of the high- and low-voltage sides, respectively, for this autotransformer connection. Compute the KVA rating as an autotransformer.

Transformer

a. Since the 2400-V winding bc is connected to the low-voltage circuit, VL = 2400 V. When Vbc = 2400 V, a voltage Vab = 240 V in phase with Vbc will be induced in winding ab (leakage-impedance voltage drops being neglected). The voltage of the high-voltage side therefore is VH = Vab + Vbc = 2640 V

Transformer b. From the rating of 50 kVA as a normal twowinding transformer, the rated current of the 240-V winding is 50,000/240 = 208 A. Since the highvoltage lead of the autotransformer is connected to the 240-V winding, the rated current at the high-voltage side of the autotransformer is equal to the rated current of the 240-V winding or 208 A. The kVA rating as an autotransformer therefore is

Note that, in this connection, the autotransformer has an equivalent turns ratio of 2640/2400. Thus the rated current at the low-voltage winding (the 2400-V winding in this connection) must be

Transformer At first, this seems rather unsettling since the 2400-V winding of the transformer has a rated current of 50 kVA/2400 V = 20.8 A. Further puzzling is that fact that this transformer, whose rating as a normal two-winding transformer is 50 kVA, is capable of handling 550 kVA as an autotransformer. The higher rating as an autotransformer is a consequence of the fact that not all the 550 kVA has to be transformed by electromagnetic induction. In fact, all that the transformer has to do is to boost a current of 208 A through a potential rise of 240 V, corresponding to a power transformation capacity of 50 kVA. This fact is perhaps best illustrated by Fig. b which shows the currents in the autotransformer under rated conditions. Note that the windings carry only their rated currents in spite of higher rating of the transformer.

Transformer Problem: An autotransformer is used to connect a 13.2-kV distribution line to a 13.8-kV distribution line. It must be capable of handling 2000 kVA. There are three phases, connected Y-Y with their neutrals solidly grounded. (a) What must the NSE / Nc turns ratio be to accomplish this connection? (b) How much apparent power must the windings of each autotransformer handle? (c) If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings be? (a) The transformer is connected Y-Y, so the primary and secondary phase voltages are the line voltages divided by 3 . The turns ratio of each autotransformer is given by:

Transformer (b) The power advantage of this autotransformer is

So 1/22 of the power goes through the windings. Since 1/3 of the total power is associated with each phase, the windings in each autotransformer must handle

(c) The voltages across each phase of the autotransformer are 13.8 / 3 = 7967 V and 13.2 / 3 = 7621V. The voltage across the common winding (N C ) is 7621 kV, and the voltage across the series winding (N SE) is 7967 kV – 7621 kV = 346 V. Therefore, a single phase of the autotransformer connected as an ordinary transformer would be rated at 7621/346 V and 30.3 kVA.

Transformer Problem: A 5000-VA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 120-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V. (a) Sketch the transformer connection that will do the required job. (b) Find the kilovolt ampere rating of the transformer in the configuration. (c) Find the maximum primary and secondary currents under these conditions.

SOLUTION (a) For this configuration, the common winding must be the smaller of the two windings, and NSE = 4 Nc = . The transformer connection is shown below:

Transformer (b) The kVA rating of the autotransformer can be found from the equation

(c) The maximum primary current for this configuration will be

and the maximum secondary current is

Transformer Practice Problems A 450-kVA, 460-V:7.97-kV transformer has an efficiency of 97.8 percent when supplying a rated load of unity power factor. If it is connected as a 7.97:8.43-kV autotransformer, calculate its rated terminal currents, rated kVA, and efficiency when supplying a unity-power-factor load.

Transformer (Three-Phase)

Transformer (Three-Phase) Single Phase vs. Three Phase Power Systems Up to this now, we have focused primarily upon single-phase transformers. Singlephase meaning. (2) power lines as an input source; therefore, only (1) primary and (1)secondary winding is required to accomplish the voltage transformation Basically, the power company generators produce electricity by rotating (3) coils or windings through a magnetic field within the generator . These coils or windings are spaced 120 degrees apart. As they rotate through the magnetic field they generate power which is then sent out on three (3) lines as in three-phase power. Three-Phase transformers must have (3) coils or windings connected in the proper sequence in order to match the incoming power and therefore transform the power company voltage to the level of voltage we need and maintain the proper phasing or polarity.

Transformer (Three-Phase) Three phase electricity powers large industrial loads more efficiently than singlephase electricity. When single-phase electricity is needed, It is available between any two phases of a three-phase system, or in some systems , between one of the phases and ground. By the use of three conductors a three-phase system can provide 173% more power than the two conductors of a single-phase system. Three-phase power allows heavy duty industrial equipment to operate more smoothly and efficiently. Three-phase power can be transmitted over long distances with smaller conductor size. Three phase transformers is a common and popular method for electric power transmission. There are many benefits to three phase power like it occupies less floor space for same ratings, weighs less, costs about 15% less. First, all three wires can carry the same current. Secondly, power transfer is constant into a linear and balanced load.

Transformer (Three-Phase) Three phase transformers can be constructed in two different ways i.e. :1. A three phase bank consists of three single phase transformers. 2.

Three windings wrapped around a common core.

Transformer (Three-Phase) Construction A three phase transformer is constructed by winding three single phase transformers on a single core. (Whether the winding sets share a common core assembly or each winding pair is a separate transformer, the winding connection options are the same) These transformers are put into an enclosure which is then filled with dielectric oil. Since it is a dielectric, a nonconductor of electricity, it provides electrical insulation between the windings and the case. It also used to help provide cooling and to prevent the formation of moisture, which can deteriorate the winding insulation. Three Phase Transformer Connections: There are only 4 possible transformer combinations: Delta to Delta -: industrial application Delta to Wye -: most common, commercial and industrial Wye to Delta -: high voltage transmissions Wye to Wye -: rare, causes harmonics and balancing problems.

Transformer (Three-Phase) Three phase transformers are connected in delta or wye configurations. A wye-delta transformer has its primary winding connected in a wye and its secondary winding connected in a delta. A delta-wye transformer has its primary winding connected in delta and its secondary winding connected in a wye. The reasons for choosing a Y or ∆ configuration for transformer winding connections are the same as for any other three-phase application: Y connections provide the opportunity for multiple voltages, while ∆ connections enjoy a higher level of reliability (if one winding fails open, the other two can still maintain full line voltages to the load). Probably the most important aspect of connecting three sets of primary and secondary windings together to form a three-phase transformer bank is paying attention to proper winding phasing (the dots used to denote “polarity” of windings).

Transformer (Three-Phase) 1. Delta to Delta When there is no need for a neutral conductor in the secondary power system, ∆-∆ connection schemes are preferred because of the inherent reliability of the ∆ configuration.

This connection is economical for large low voltage transformers in which insulation problem isn’t so urgent because it increases number/phase.

VLP = VΦP

VLS = VΦS

a

+

N S2

V LP

-

a’

V φP

N P1

+

N S3

V φS

V LS b’

N S1 b c

c’

Transformer (Three-Phase) 2.Wye to Delta (Y) The center point of the “Y” must tie either all the “-” or all the “+” winding points together. (∆) The winding polarities must stack together in a complementary manner ( + to -). Inputs A1, A2, A3 may be wired either “∆” or “Y”, as may outputs B1, B2, B3. Three individual transformers are to be connected together to transform power from one three-phase system to another.

Transformer (Three-Phase) The Y- ∆ connection is commonly used in stepping down from a high voltage to a medium or low voltage. One reason is that a neutral is thereby provided for grounding on the high-voltage side, a procedure which can be shown to be desirable in many cases. Ya ∆ a’

b

NP

NS

1

NP

Vφ VL P

2

P

2

3

c

VLP = 3VΦP

b’ V φP

N P1

N S1 V φS

NS NS

3

a’ V LP

b’

1

NP

b

Vφ

VL

S

S

VLS = VΦS

a

N P2

N S2 c’

c’

c

N P3

N S3

Transformer (Three-Phase) The voltage ratio of each phase V V

Φ P

=

a

Φ S

The line voltage ratio V LP 3V Φ P = = V Ls V ΦS

a =

V LP 3V

3a

V

= LS

a =

LP

V

3 LS

V = V

Φ P LS

V LP 3 V LS

Transformer (Three-Phase) Y-∆. The secondary voltage is shifted 30° relative to the primary voltage. Because of the phase shift. transformers can not be connected in parallel unless they have proper phase sequence. (i.e. They are in phase with each other). This connection has no problem with 3rd harmonic components in its voltages, since they consumed in circulating current on the ∆ side.

Transformer (Three-Phase) 3. Delta to Wye

VLP = VΦP

V φS

VLS = 3VΦS

VLP V a 3VLP = ΦP = ⇒a= VLS VLS 3VΦs 3 a =

V LP 3 V LP = V LS V LS

= 3

VΦP VΦS

This connection has the same advantages, and the same phase shift as the Y-∆

+ V LP

-

V φP

a’ V LS

b’ c’

Transformer (Three-Phase) V LP VΦP = = a V LS VΦS This transformer has no phase shift associated with it because ratio of transformation is exactly same in primary as in secondary and hence no problems with unbalanced loads or harmonics.

No difficulty is experienced from unbalanced loading as was in caes of wye - wye connection.

Transformer (Three-Phase) 4. Wye to Wye This connection is most economical for small and medium scale voltage transformation. Because the number/phase and amount of insulation required is minimum. Ratio of line voltages on the primary and secondary side is the same as the transformation ratio of each transformer. However there is a phase shift of 30 degrees between the phase voltages and line voltages on the primary and secondary sides.

Transformer (Three-Phase) 3 V φ P V LP = V LS 3 V φ S The Y-Y connection has two very serious

=

a

problems. 1. If loads on the transformer circuit are unbalanced, then the voltages on the phases of the transformer can become severely unbalanced. 2. Third harmonic voltages can be large. Third harmonic components of the each of the three phases will be in phase with each other. However, the third harmonic components of each cycle of the fundamental frequency. These are always third harmonic components in a transformer because of nonlinearity of the core, and these components add up. The result is a very large third harmonic component of the voltage on the top of fundamental voltage

Transformer Harmonics Harmonics are voltage and current frequencies riding on top of the normal sinusoidal voltage and current waveforms. Usually these harmonic frequencies are in multiples of the fundamental frequency, which is 50 hertz (Hz) in the Pakistan, 60 Hz in U.S.and Canada and 40Hz in Japan.. The most common source of harmonic distortion is electronic equipment using switch-mode power supplies, such as computers, adjustable-speed drives, and highefficiency electronic light ballasts. Harmonics are created by these “switching loads” (also called “nonlinear loads,” because current does not vary smoothly with voltage as it does with simple resistive and reactive loads): Each time the current is switched on and off, a current pulse is created.

Transformer The resulting pulsed waveform is made up of a spectrum of harmonic frequencies, including the 50 Hz fundamental and multiples of it. This voltage distortion typically results from distortion in the current reacting with system impedance. (Impedance is a measure of the total opposition—resistance, capacitance, and inductance— to the flow of an alternating current.) The higher-frequency waveforms, collectively referred to as total harmonic distortion (THD), perform no useful work and can be a significant nuisance.

Harmonic waveforms are characterized by their amplitude and harmonic number. In the U.S. and Canada, the third harmonic is 180 Hz—or 3 x 60 Hz—and the fifth harmonic is 300 Hz (5 x 60 Hz). The third harmonic (and multiples of it) is the largest problem in circuits with single-phase loads such as computers and fax machines. Figure shows how the 60-Hz alternating current (AC) voltage waveform changes when harmonics are added.

Transformer

Transformer The Problem with Harmonics Any distribution circuit serving modern electronic devices will contain some degree of harmonic frequencies. The harmonics do not always cause problems, but the greater the power drawn by these modern devices or other nonlinear loads, the greater the level of voltage distortion. Potential problems (or symptoms of problems) attributed to harmonics include: ‰Flickering lights ‰Very high neutral currents ‰Overheated phase conductors, s, and transformers ‰Premature failure of transformers and uninterruptible power supplies (UPSs) ‰Reduced power factor ‰Reduced system capacity (because harmonics create additional heat, transformers and other distribution equipment cannot carry full rated load)

Transformer Identifying the Problem Without obvious symptoms such as nuisance breaker trips or overheated transformers, how do you determine whether harmonic current or voltages are a cause for concern? Here are several suggestions for simple, inexpensive measurements that a facility manager or staff electrician could take, starting at the outlet and moving upstream: ‰Measure the peak and root mean square (RMS) voltage ‰Inspect distribution s. ‰Measure phase and neutral currents at the transformer secondary with clamp-on current probes. ‰Compare transformer temperature and loading with nameplate temperature rise and capacity ratings.

Transformer Solutions to Harmonics Problems The best way to deal with harmonics problems is through prevention: choosing equipment and installation practices that minimize the level of harmonics in any one circuit or portion of a facility. Many power quality problems, including those resulting from harmonics, occur when new equipment is haphazardly added to older systems. However, even within existing facilities, the problems can often be solved with simple solutions such as fixing poor or nonexistent grounding on individual equipment or the facility as a whole, moving a few loads between branch circuits, or adding additional circuits to help isolate the sensitive equipment from what is causing the harmonic distortion. If the problems cannot be solved by these simple measures, there are two basic choices: to reinforce the distribution system to withstand the harmonics or to install devices to attenuate or remove the harmonics.

Transformer

Transformer Parallel Operation of Transformer

Transformer The essential requirement for the parallel operation of two or more transformers are the following. 1. The polarity should be the same. 2. The voltage ratios should be same. 3. The percentage impedance should be equal. 4. The phase rotation/phase sequence should be same. 5. The vector diagrams and the phase displacement should be the same.

Transformer Polarity should be the same The term polarity in parallel operation of transformers refers to a phase relationship b/w two or more units. It can be applied to indicate the directional relationship of primary and secondary terminal voltages of a single unit. Any to single phase transformers have the same polarity when their instantaneous terminal voltages are in phase.

Correct Connection

Wrong Connection

Transformer Voltage ratios should be same. If the voltage ratios aren’t identical, for the same primary , the secondary voltage would be different. Thus the different secondary voltages would result circulating current within the secondary circuits and heating of the transformers even on no load.

Percentage impedance should be equal If the percentage impedance are different, the load sharing will not be proportional to VA ratings. As percentage impedance cannot be exactly equal, it is preferable to operate transformers of the same ratings in parallel.

Vector group A Vector group is the International Electrotechnical Commission (IEC) method of categorizing the primary and secondary winding configurations of threephase transformers. Within a poly phase system power transformer it indicates the windings configurations and the difference in phase angle between them. The phase windings of a poly phase transformer can be connected together internally in different configurations, depending on what characteristics are needed from the transformer. For example, in a three-phase power system, it may be necessary to connect a three-wire system to a four-wire system, or vice versa. Because of this, transformers are manufactured with a variety of winding configurations to meet these requirements. Different combinations of winding connections will result in different phase angles between the voltages on the windings. This limits the types of transformers that can be connected between two systems, because mismatching phase angles can result in circulating current and other system disturbances.

Vector group Symbol designation The vector group provides a simple way of indicating how the internal connections of a particular transformer are arranged. In the system adopted by the IEC, the vector group is indicated by a code consisting of two or three letters, followed by one or two digits. The letters indicate the winding configuration as follows: D: Delta winding, also called a mesh winding. Each phase terminal connects to two windings, so the windings form a triangular configuration with the terminals on the points of the triangle. Y: Wye winding, also called a star winding. Each phase terminal connects to one end of a winding, and the other end of each winding connects to the other two at a central point, so that the configuration resembles a capital letter Y. The central point may or may not be connected outside of the transformer.

Vector group Z: Zigzag winding, or interconnected star winding. Basically similar to a star winding, but the windings are arranged so that the three legs are "bent" when the phase diagram is drawn. Zigzag-wound transformers have special characteristics and are not commonly used where these characteristics are not needed. III: Independent windings. The three windings are not interconnected inside the transformer at all, and must be connected externally. In the IEC vector group code, each letter stands for one set of windings. The HV winding is designated with a capital letter, followed by medium or low voltage windings designated with a lowercase letter. Phase displacement zero = 0 Letter Y represents star connected H.V Phase displacement 180

= 0

Letter y represents star connected L.V

Phase displacement 30 lag = 0

Letter D represents star connected H.V

Phase displacement 30 lead= 0

Letter d represents star connected L.V

Vector group The digits following the letter codes indicate the difference in phase angle between the windings, with HV winding is taken as a reference. The number is in units of 30 degrees. For example, a transformer with a vector group of Dy1 has a deltaconnected HV winding and a wye-connected LV winding. The phase angle of the LV winding lags the HV by 30 degrees. The point of confusion is in how to use this notation in a step-up transformer. As the IEC60076-1 standard has stated, the notation is HV-LV in sequence. For example, a step-up transformer with a delta-connected primary, and star-connected secondary, is not written as 'dY11', but 'Yd11'. The 11 indicates the LV winding leads the HV by 30 degrees. Transformers built to ANSI standards usually do not have the vector group shown on their nameplate and instead a vector diagram is given to show the relationship between the primary and other windings.

Transformer Voltage Regulation and Efficiency Output Voltage of Transformer Varies with Load Due to Voltage Drop on Series Impedance of Transformer Equivalent Model Full Load Regulation Parameter, compares output no-load Voltage with its Full Load Voltage: V.R. =

P , pu

in per unit: V.R. = For Ideal Transformer V.R.=0

V S ,F .L .

× 100 %

(V P / a ) − V F . L . × 100 % V F .L .

At no load VS= VP / a thus :V.R.=

V

V S , N .L . − V S ,F .L .

− V S , FL

V S , FL

, pu

, pu

× 100 %

Transformer Voltage Regulation and Efficiency THE TRANSFORMER PHASOR DIAGRAM To determine the voltage regulation of a transformer: The voltage drops should be determined In below a Transformer equivalent circuit referred to the secondary side shown:

Transformer Voltage Regulation and Efficiency Since current which flow in magnetizing branch is small can be ignored Assuming secondary phasor voltage as reference VS with an angle of 0◦ Writing the KVL equation:

VP = V S + R eq I S + jX a

From this equation the phasor diagram can be shown: At lagging power factor:

eq

IS

Transformer Voltage Regulation and Efficiency If power factor is unity, VS is lower than VP so V.R. > 0 V.R. is smaller for lagging P.F. With a leading P.F., VS is larger VP Î V.R.<0 P.F. =1 Î

P.F. leading Î

Transformer Voltage Regulation and Efficiency Table Summarize possible Value for V.R. vs Load P.F.:

Lagging P.F.

VP/ a > VS

V.R. > 0

Unity P.F.

VP / a < VS

V.R. <0 (smaller)

Leading P.F.

VS > VP/ a

V.R. < 0

Transformer Voltage Regulation and Efficiency Transformer Efficiency (as applied to motors, generators and motors)

Pout η= x100 % Pin

η =

Pout x 100 % Pout + Ploss

Losses in Transformer: 1- Copper I²R losses 2- Core Hysteresis losses 3- Core Eddy current losses Transformer efficiency may be determined as follows:

η=

V S I S cos θ x100 % PCu + Pcore + V S I S cos θ

Per-Unit System † In power systems there are so many different elements such as Motors, Generators and Transformers with very different sizes and nominal values. † To be able to compare the performances of a big and a small element, per unit system is used.

Per Unit System † Power system quantities such as voltage, current and impedance are often expressed in per unit or percent of specified values. † Per unit quantities are calculated as:

Actual Value Per Unit Value = Base Value

Per Unit System Per Unit Values S pu =

S S base

I pu =

I I base

Vpu =

V Vbase

Z pu =

Z Z base

Conversion of Per Unit Values Z pu =

Z Z base

Sbase = 2 Z Vbase

Z = Z base Z pu

2 Vbase = Z pu S base

Per Unit System † Usually, the nominal apparent power (S) and nominal voltage (V) are taken as the base values for power (Sbase) and voltage (Vbase). † The base values for the current (Ibase) and impedance (Zbase) can be calculated based on the first two base values.

Per Unit System Vbase I base

Sbase Sbase = Vbase 2

Z base

Vbase V base = = I base Sbase

Per Unit System

Z actual Z% = × 100% Z base † The percent impedance † e.g. in a synchronous generator with 13.8 kV as its nominal voltage, instead of saying the voltage is 12.42 kV, we say the voltage is 0.9 p.u.

Transformer Voltage Base

Vb 2

⎛ V2 = ⎜⎜ ⎝ V1

⎞ ⎟⎟ •V b1 ⎠

Per Unit in 3φ Circuit † Simplified: „ Concerns ing phase or line voltages are removed in the per-unit system „ Actual values of R, XC and XL for lines, cables, and other electrical equipment typically phase values. „ It is convenient to work in of base VA (base volt-amperes)

Per Unit in 3φ Circuit • Usually, the 3-phase SB or MVAB and line-to-line VB or kVB are selected • IB and ZB dependent on SB and VB

SB = IB =

3V B I B SB , 3V B

VB =

3I B Z B

V B / 3 (V B ) ZB = = IB SB

2

Change of Base † The impedance of individual generators & transformer, are generally in of percent/per unit based on their own ratings. † Impedance of transmission line in ohmic value † When pieces of equipment with various different ratings are connected to a system, it is necessary to convert their impedances to a per unit value expressed on the same base.

Change of Base old be the per unit impedance on the power base Sold Z pu B old B

& voltage base V Z

old pu

⎛ S old ZΩ B ⎜ = = ⎜ V old Z Bold ⎝ B

(

)

2

⎞ ⎟• ZΩ ⎟ ⎠

1

new be the new per unit impedance on the new power base S new Z pu B

& new voltage base VBnew

Z

new pu

⎛ S new ZΩ B = new = ⎜ ⎜ V new ZB ⎝ B

(

)

2

⎞ ⎟ • ZΩ ⎟ ⎠

2

Change of Base From (1) and (2), the relationship between the old and the new per unit value

Z

new pu

=Z

old pu

⎛S ⎜⎜ ⎝S

new B old B

⎞⎛ V ⎟⎟⎜⎜ ⎠⎝ V

old B new B

If the voltage base are the same,

Z

new pu

new ⎛ S old ⎜⎜ Bold = Z pu ⎝ SB

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

2

Advantages † Transformer equivalent circuit can be simplified by properly specifying base quantities. „ Give a clear idea of relative magnitudes of various quantities such as voltage, current, power and impedance. „ Avoid possibility of making serious calculation error when referring quantities from one side of transformer to the other.

Advantages † Per-unit impedances of electrical equipment of similar type usually lie within a narrow numerical range when the equipment ratings are used as base values. „ Manufacturers usually specify the impedances of machines and transformers in per-unit or percent in nameplate rating.

Advantages † The circuit laws are valid in per unit systems, and the power and voltage equation are simplified since the factor √3 and 3 are eliminates in the per-unit systems. † Ideal for the computerized analysis and simulation of complex power system problems.

Procedure for Per Unit Analysis † Pick S Base for the system. † Pick VBase according to line-to-line voltage. † Calculate Z Base for different zones. † Express all quantities in p.u. † Draw impedance diagram and solve for p.u. quantities. † Convert back to actual quantities if needed.

How to Choose Base Values ? † Divide circuit into zones by transformers. † Specify two base values out of I B ,VB , Z B , S B ; for example, S Base and VBase † Specify voltage base in the ratio of zone line to line voltage.

How to Choose Base Values ? V1 :V2

V2 : V3

V3 : V4

Source

I Base1 =

SBase VBase1

VBase1 ZBase1 =

VBase1 I Base1

Zone 2

Zone 3

Zone 4

VBase2

VBase3

VBase4

Example Given a one line diagram,

†

Vg = 13.2kV

~

Z line = 10 + j100Ω

Ig

Z load = 300Ω

5 MVA

10 MVA

13.2 ∆ – 132 Y kV

138 Y - 69 ∆ kV

X l1 = 0.1p.u.

X l 2 = 0.08p.u.

Find I g I t -line I load Vload and Pload

Step 1, 2, and 3: Base Values V g = 13 . 2 kV

~

Z line = 10 + j100 Ω

Ig

Z load = 300 Ω

5 MVA

10 MVA

13.2 ∆ – 132 Y kV

138 Y - 69 ∆ kV

X l 1 = 0 . 1p.u.

X l 2 = 0 . 08 p.u.

S B = 10MVA Zone 1 VB1 = 13.8kV Z B1 = I B1 =

2

VBl1−l

SB S B31Φ l-l B1

3V

(13.8k ) =

Zone 2 VB2 = 138kV

2

10M

= 19.04Ω

10M = = 418.4 3 ⋅13.8k

Z B2 = I B2 =

2

VBl−2 l

SB S B3Φ2 l-l B2

3V

(138k ) =

Zone 3 VB3 = 69kV

2

10M

= 1904Ω

10M = = 41.84 3 ⋅138k

Z B3 = I B3 =

VBl−3 l

2

SB S B3Φ3 3 VBl-3l

2 ( 69k ) =

10M

=

= 476Ω

10 M = 83.67 3 ⋅ 69k

Step 4: All in Per Unit Quantities X l 2 = 0.08p.u.

+ -

new Z p.u. =

old old Z p.u. ZB

Z

new B

(

Zload,p.u. =

)

0.1× (13.2k ) 5M X l1,p.u. = = 0.183 19.04Ω

Vg,p.u. =

Vg VB1

=

2

Zload 300Ω = = 0.63 Z B3 476Ω

13.2kV = 0.913∠0° 13.8kV Zline,p.u. =

Zline 10 + j100Ω = 5.25×10−3 (1 + j10) = Z B2 1904

Step 5: One Phase Diagram & Solve X l1,p.u. = 0.183

+ -

Z line,p.u. = 5.25 ×10 −3 (1 + j10)

X l 2 = 0.08

Vg, p.u. = 0.913∠0°

Z load,p.u. = 0.63

0.96∠0° = = 1.35∠ − 26.4° I load,p.u. = Z total,p.u. 0.709∠26.4° Vload,p.u. = I load,p.u. Z load,p.u. = 0.8505∠ − 26.4° Vg,p.u.

* S load,p.u. = Vload,p.u. I load, p.u. = 1.148

I g,p.u. = I t -line,p.u. = I load,p.u. = 1.35∠ − 26.4°

Step 6: Convert back to actual quantities Vg = 13.2kV

~

Z line = 10 + j100Ω

Ig

Z load = 300Ω

5 MVA

10 MVA

13.2 ∆ – 132 Y kV

138 Y - 69 ∆ kV

X l1 = 0.1p.u.

X l 2 = 0.08p.u.

I g, p.u. = I t -line,p.u. = I load,p.u. = 1.35∠ − 26.4° Zone 1

I g = I g, p.u. I B1

Zone 2

I t -line = I t -line,p.u. I B2

Zone 3

I load = I load,p.u. I B3 Vload = Vload,p.u.VB3 S load = S load,p.u. S B

Advantage of per unit calculation † Simplify calculation by eliminating transformers. † Helps to spot data errors „ p.u. is more uniform compare to actual impedance value of different sizes of equipment. † Helps to detect abnormality in the system „ Operator at control center can spot over/under voltage/current rating easily.

Transformer Voltage Regulation and Efficiency Table Summarize possible Value for V.R. vs Load P.F.: Lagging P.F.

VP/ a > VS

V.R. > 0

Unity P.F.

VP / a < VS

V.R. <0 (smaller)

Leading P.F.

VS > VP/ a

V.R. < 0

Since transformer usually operate at lagging P.F., a simplified method is introduced Simplified Voltage Regulation Calculation For lagging loads: the vertical components related to voltage drop on Req & Xeq partially cancel each other Îangle of VP/a very small