Chapter 10: Practical Transformers

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

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 m

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

Rm and Xm can be measured experimentally by

the power values are measured under no-load conditions

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

o the coil draws no current to drive a mutual flux m1a

o 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

Loose Couplings

• Now a load Z is connected across the secondary

currents I1 and I2 immediately begin to flow, and are related by

N1 I1 = N2 I2

I2 produces an mmf N2 I2 and I1 produces an mmf N1I1 in the opposite direction

mmf N2I2 forms a flux F2, consisting of a mutually coupled flux m2 and a leakage flux f2

mmf N1I1 forms a flux F1, consisting of a mutually coupled flux m1 and a leakage flux f1

• The new mmf’s upset m1a balance

resolving the modeling conflictso combine m1 and m2 into a single mutual

flux m

o Es consists of two parts: E2(N2m) and Ef2(N2f2)

o Ep consists of two parts: E1(N1m) and Ef1(N1f1)

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

Ef1 is a voltage drop across a reactance

Xf1 = Ef1/I1

Imperfect TransformersExample• 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

a. the reactive power absorbed by the coreb. 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

a. the voltages induced in the secondary windingb. the secondary leakage reactance

Equivalent Circuit• The primary and secondary

windings are composed of copper or alluminum conductors conductors exhibit resistance to the

current flow the leakage reactance can also be

modelled 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

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

o iron losses depend on the mutual flux and hence the applied voltageo the winding losses depend on the current drawn by the load

Losses

Example• the nameplate of a distribution transformer indicates 250 kVA,

60 Hz, 4160 V primary, 480 V secondary

• calculatea. the nominal primary and secondary currents

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

c. when is the transformer most efficient?

Voltage Regulation

• Important attribute of a transformer constantly held primary voltage impact of secondary voltage due to changing loads

Voltage Regulation

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

with an internal impedance of 127 ohms as seen on the primarya. calculate the rated primary and secondary currents

b. the voltage regulation from no-load to full load for a 2000 kW resistive load (at unity power factor) knowing that the primary supply voltage is fixed at 69 kV

c. The primary and secondary currents if the secondary is accidentally short-circuited

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

o rated voltage applied to the primary

short-circuit test, secondary shortedo rated current applied to primary

Measuring Impedances

Example

• 500 kVA, 69 kV/4160V, 60 Hz transformer with terminals X1 and X2 shorted

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

b. find the HV leakage reactance and resistance values

Example

• terminals H1 and H2 are opened, voltage applied to terminals X1 and X2

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

b. find HV magnetization impedance

Homework

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