Full representation of the real transformer


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1 TRASFORMERS EQVALET CRCT OF TWOWDG TRASFORMER TR Dots show the points of higher potential. There are applied following conventions of arrow directions: for primary circuit the passive sign convention (receiver, the power is absorbed), for secondary circuit  the active sign convention (source, the power is released). Full representation of the real transformer Features: Real transformer with two windings (primary & secondary) of different numbers of turns (usually). Sinusoidal voltage supply. The windings are characterised by their resistances R and R. The magnetic core with power losses and with the main (mutual) flux Φ and leaage fluxes Φ l and Φ l. Primary current and secondary current flow. Electromotive forces due to mutual flux are induced accordingly: E 4.44 fφ E 4.44 fφ Emfs due to leaage fluxes: E x 4.44 fφ l E x 4.44 fφ l Leaage fluxes are directly proportional to currents and and emfs corresponding to leaage fluxes can be represented by the voltage drops at leaage reactances X and X Representation of the transformer by semiideal transformer (with no leaage fluxes and zeroresistance winding)
2 REFERRG SECODARY WDG TO PRMARY Secondary circuit referred (recalculated) to primary so as to have the same potential differences Assume that we recalculate secondary emf into new value that will be called secondary emf referred to primary E E 444. fφ 444. fφ E and in the same manner secondary output voltage referred to primary ow the beginnings of primary and secondary windings have the same potentials (E E ) TR. Connection of two bottom points of the circuit can be done without any consequence as they belong to two separated circuits. The only result of such connection is the equalisation of their potentials.. After recalculation of emfs (E E ) the potentials of the upper points become equal each other. Their connection doesn t lead to the flow of current between them! All referred values are indicated with primes for example R ; X Referred secondary circuit has to represent the real quantities of energy conversion. t means that the values of power and losses cannot be changed in the new model of transformer. Therefore: secondary current referred to primary: secondary resistance referred to primary: R R R R ( ) ( ) and in the same manner secondary reactance referred to primary: X X ( ) The points having the same potentials (E E ) can be connected without any current flow. We have now only one electrical circuit without electrical separation between primary and secondary circuits. How we can represent (by means of electrical circuit) the semiideal magnetic core with two identical emfs induced in the windings and with power losses in magnetic circuit?
3 EQVALET (ELECTRCAL) CRCT OF TWOWDG TRASFORMER TR 3 o  noload current f  magnetizing current (reactive component of noload current) ow  active component of noload current Parallel branch of this Ttype circuit represents the magnetic core of the transformer: ironcore resistance R Fe  the resistance having the value corresponding to power loss in the magnetic circuit, according to relation: P P P + P R o Fe h e Fe ow magnetizing reactance X f  the reactance of primary circuit corresponding to mutual flux and representing primary emf due to relation: E f X f or E jx f f E where: + or + o f ow o f ow Phasor diagram for transformer equivalent circuit: and voltage balance equations: R + jx + E ( R + jx ) + ( R + jx ) +
4 OLOAD OPERATO AD OLOAD TEST At noload state the secondary circuit of the transformer is opened: 0; o Active power supplied to the transformer at noload state: P o P Cuo + P Fe R o + P Fe oload loss (power loss not depending on the value of current): P o P Fe P o  R o oload test of the transformer is easy to be made in laboratory or even in a substation: TR 4 R induction regulator (voltage source of variable output), A ammeter, V voltmeter, W  wattmeter Measuring circuit diagram During test the supplying voltage is varied from 0 up to (sometimes more). The following quantities are measured: ; ; o ; P o Appropriate characteristics are drawn in the function of primary voltage: Observe the course of o characteristic! Doesn t it loo as magnetizing curve? Try to explain why? Observe also very interesting course of noload power factor curve. How it can be explain from the physical meaning point of view? otice that noload power factor value for rated voltage is very low! Basic noload characteristics From the noload test results we can determine some parameters of the transformer being tested or its equivalent circuit: P o cosϕ ow o cosϕo f o sinϕo o P RFe X f P P o ow o Fe Rated noload current (i.e. the value of noload current at rated voltage) expressed in p.u. or in percent o o or or o % 00% is rather of small value: from a few % in large power transformers to 030% in small transformers. f The value of P Fe (iron core loss at nominal voltage) is always given in ratings of transformers or at the nameplates.
5 MAGETZG CRRET HARMOCS Assumption: transformer (at noload electromagnetic coil) is supplied with the voltage of sinusoidal waveform. This voltage must be balanced by sinusoidal emf induced in the primary winding, hence the flux (flux density) has to be of sine waveform. Determination of noload current waveform TR 5 OCDEF magnetic flux density waveform B(t) sinusoidal Φf(i) and Bf(i) must be determined from the characteristic of electromagnetic circuit withsaturation of the core andhysteresis phenomena taen into consideration. C C O from risingup part of hysteresis loop; E E OL from fallingdown part of hysteresis loop From the figure: i 0 is a sum of active (i 0w ) and reactive (i f magnetising current) components. i f is of nonsinusoidal waveform: i f i f +i f3 +i f5 +i f (odd harmonics, the 3 rd one is the most important from higher order harmonics). Conclusion: n the transformer supplied from sinusoidal voltage source the waveform of magnetizing current absorbed from the same source tends to be disturbed from the sinusoidal pattern at least 3 rd harmonic is required! Question: What can happen when the source cannot provide higher harmonics of current to flow? n such situation the flux (flux density) waveform contains 3 rd and other odd harmonics, therefore emf e waveform is affected by higher harmonics (first of all by 3 rd harmonic). See next figure:
6 SHORTCRCT STATE, SHORTCRCT TEST AD PARAMETERS At short circuit the secondary terminals of transformer are shortcircuited. is equal to zero and and can be very high (depending on voltage). They are much higher than o, therefore, after assuming o 0 the equivalent circuit can be simplified: TR 6 Due to o 0 the parallel branch of the circuit can be neglected. Only series resistances and reactances appear in equivalent circuit. We define the following parameters: shortcircuit resistance and reactance R R +R and X X +X shortcircuit impedance (sometimes called equivalent impedance of transformer) Z R + jx Z R + X ( Z in short  circuited transf.) Phasor diagram for shortcircuited transformer: As X corresponds to leaage fluxes, the saturation effect doesn t affect its value and we can assume that X is constant, independently on current or voltage. t means also that Z const. can be supplemented by R, X, Z triangle (shortcircuit impedance) triangle. n laboratory or during field tests the shortcircuit tests are very often performed. For the voltage varied from zero up to the value providing the flow of current not too much higher than they are measured: P,, cosϕ and their characteristics drawn:
7 For we determine and specify: shortcircuit voltage (having two components: R and X ); P P shortcircuit power; cosϕ cosϕ shortcircuit power factor. Shortcircuit voltage is the value of supplying voltage in shortcircuited transformer that provides a flow of rated currents ( and ). ts value is always shown in ratings of transformers or at the nameplates (usually in p.u. or in %). TR 7 During test the value of cosϕ can be measured directly by power factor meter or calculated from other values readings: cos ϕ P P S Application of per unit (or relative) values: P Z r ; r ; Pr ; Zr where Z S Z yields: Z r Zr Z Typical percent value of shortcircuit voltage is of the following level:  several % for small and medium power transformers,  0 to 0% for large power transformers (hundreds of MVA). Shortcircuit power, i.e. the power absorbed by shortcircuited transformer is totally lost in the transformer (the output power P 0). The iron losses in such transformer are much smaller when compared to copper loss and can be neglected. Hence, shortcircuit power (absorbed by the transformer with rated current) can be calculated as follows: P R + ( ) R PCu The value of P Cu (rated copper loss, or in other words copper loss at rated load) is always shown in ratings of transformers or at the nameplates. t is easy to prove that: P Cu P 00% Cu % R % PCu % S and other interesting relation concerning the value of shortcircuit current flowing in shortcircuited transformer supplied with rated voltage (rated shortcircuit current): r Z Z r Try to prove it! Practical meaning of this relation is as follows: imagine the transformer having shortcircuit voltage0%. n such transformer the shortcircuit current at nominal supplying voltage is 0.
8 TRASFORMER DER LOAD TR 8 For simplified consideration of the transformer operation under load we can apply its simplified equivalent circuit. Simplification of equivalent circuit (EC) leads to some error much lower when o is smaller in comparison to or currents. Hence, we can use such EC for discussing power transformers fully loaded. sing such model we can consider the problem of output voltage variation for different value and character of the load current. Let us consider the following problem: transformer is supplied with constant primary voltage and the loading current varies. The character of load current is given by its phase angle ϕ varying at closed interval [90 o, +90 o ]. For 0 the output voltage o. For >0 and some ϕ At this figure the voltage drops triangle is drawn not to the scale. The hypotenuse DC in case of corresponds to the value of and can be few percent of. we observe the voltage drop (difference in voltage values): o CA CB R cosϕ + X sinϕ Assuming and expressing in per units + r o o r r sin o o ( R cosϕ X ϕ ) and for r Rr cosϕ + X r sinϕ Rr cosϕ + Xr sinϕ and final expression for the value of output voltage o r n similar manner, for any value of supply voltage and corresponding value of o we can calculate the output voltage o r or S o r S o Rr R r Xr X r o is noload output voltage when o is noload output voltage for
9 The function f(cosϕ ) for is shown in following figure TR 9 lead leading character of the current, i.e. the current of capacitive character, or current phasor leads the voltage phasor; lag lagging character of the load, inductive load, current phasor lags the voltage phasor. t is seen that max( ) occurs at cosϕ cosϕ i.e. for very special value being a nown parameter of the transformer (shortcircuit power factor). Some interesting practical observations can be concluded:. in the transformer fully loaded ( it is very much required not to overload the transformer above its rated current) the voltage drop at internal impedances of the transformer can be maximum (or in p.u. r or % ). For example in the transformer having % 8% the maximum voltage drop (assuming ) can be 8%, i.e. max o 00% 8% o from where it follows that minimum output voltage could be of the value of 8% of o and it would appear for cosϕ cosϕ. For some capacitive loads (leading character of the current) the voltage drop can be of negative value, what means the output voltage at such load can be higher than at no load. magine your transformer having rated voltages 5000/0 V. These voltages are voltage values at noload. Let your transformer operates as stepdown and let it to be supplied with 5 V and fully loaded. Would you lie to have your home appliances to be supplied with the voltage V? Summarising above we can conclude that in the transformer having high value of shortcircuit voltage one can expect correspondingly large variation of output voltage with varying load.
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