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Chapter 10: Practical Transformers

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Introduction

• In the real world, transformers are not ideal– windings have resistance

– the cores are not infinitely permeable

– the flux produced by the primary is not completely captured by the secondary

• leakage flux must be accounted

– iron core produces eddy-current and hysteresis losses

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Imperfect Cores

• What happens when an infinitely permeable core is replaced by an iron core having hysteresis and eddy-current losses?– core imperfections are represented

by Rm and Xm in parallel with the primary winding

• Rm models the iron losses

• Xm models the permeability

– the current Im flowing in Xm is the magnetizing current that creates the flux Φm

– the total current I0 needed to produce the flux is called the exciting current

– Rm andXm can be measured experimentally by

• the power values are measured under no-load conditions

mm

mm Q

EX

P

ER

21

21 ==

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Loose Couplings

• Consider now a perfect core but the windings are loosely coupled– the primary and secondary coils

have negligible resistance

– the primary is connected to a source Eg

• the coil draws no current to drive a mutual flux Φm1a

• the flux produces a counter voltage Ep that equals Eg

– the flux produces a voltage E2

on the secondary coil

• under no-load conditions, I2

is zero and no mmf exist to drive any leakage flux

Nf

Eg

44.4max =Φ

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Loose Couplings

• Now a load Z is connected across the secondary– currents I1 and I2 immediately

begin to flow, and are related by

– I2 produces an mmf N2I2 and I1

produces an mmf N1I1 in the opposite direction

– mmf N2I2 forms a flux Φ2, consisting of a mutually coupled flux Φm2 and a leakage flux Φf2

– mmf N1I1 forms a flux Φ1 , consisting of a mutually coupled flux Φm1 and a leakage flux Φf1

• The new mmf’s upset Φm1abalance– resolving the modeling conflicts

• combine Φm1 and Φm2 into a single mutual flux Φm

• Es consists of two parts: E2(N2Φm) and Ef2(N2Φf2)

• Ep consists of two parts: E1(N1Φm) and Ef1(N1Φf1)

2211 ININ =

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Leakage Reactance

• The four induced voltages can be rearranged– the rearrangement does not

change the induced voltages

– Ef2 is a voltage drop across a reactance

Xf2 = Ef2/I2

– Ef2 is a voltage drop across a reactance

Xf1 = Ef1/I1

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Imperfect Transformers

• Example– a large 120 V, 60 Hz transformer with no loads draws an exciting current

I0 of 5 A at rated voltage

– a wattmeter test shows the iron losses to be 180 W

– find

• the reactive power absorbed by the core

• Rm, Xm, If, and Im

• Example– the secondary winding consist of 180 turns, and under load the winding

draws 18 A, producing 20 mWb of mutual flux and 3 mWb of leakage flux

– calculate

• the voltages induced in the secondary winding

• the secondary leakage reactance

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Equivalent Circuit

• The primary and secondary windings are composed of copper or aluminum conductors– conductors exhibit resistance to

the current flow

– the leakage reactance can also be modeled as a series inductance

• the core excitation and losses are modeled as a shunt circuit– combining all elements with the

ideal transformer forms an equivalent circuit for practical transformers

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Losses

• As in all machines, a transformer has losses– I2R losses in the primary and secondary windings

– hysteresis losses and eddy-current losses in the core

– stray losses due to currents induced in the tank and metal supports by the primary and secondary leakage fluxes

• Losses appear in the form of heat– produces an increase in termperature

– drop in efficiency• iron losses depend on the mutual flux and hence the applied

voltage

• the winding losses depend on the current drawn by the load

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Losses

• Example– the nameplate of a distribution transformer indicates

250 kVA, 60 Hz, 4160 V primary, 480 V secondary

– calculate• the nominal primary and secondary currents

• if the core losses are 1200 W and the full-load copper losses are 1800 W what is the transformer efficiency?

• when is the transformer most efficient?

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Voltage Regulation

• Important attribute of a transformer– constantly held primary voltage

– impact of secondary voltage due to changing loads

• Example– a single-phase transformer rated at 3 MVA, 69 kV/4160 V

with an internal impedance of 127 ohms as seen on the primary

– calculate the rated primary and secondary currents

– the voltage regulation from no-load to full load at unity power factor

FL

FLNL

E

EEVR

−=

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Measuring Impedances

• We can measure using two tests the actual values of Xm, Rm, Xp and Rp for a given transformer– voltages, currents, and real powers are

measured

– open-circuit test, secondary opened

• rated voltage applied to the primary

–

–

–

– short-circuit test, secondary shorted

• rated current applied to primary

–

–

220 mmmpm PSQIES −=⋅=

mpmmpm QEXPER 22 ==

2scscpscscp IPRIEZ ==

22ppp RZX −=

sp EEa=

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Measuring Impedances

• Example– 500 kVA, 69 kV/4160V, 60 Hz transformer with terminals X1

and X2 shorted

– measurements: Esc = 2600 V, Isc = 4 A, Psc = 2400 W

– find the HV leakage reactance and resistance values

• Example– terminals H1 and H2 are opened, voltage applied to terminals

X1 and X2

– measurements: Es = 4160 V, I0 = 2 A, Pm = 5000 W

– find HV magnetization impedance

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Practical Transformers

• Homework– Problems: 10-18, 10-23, 10-25, 10-30, 10-31