Download - Chapter 10 - Practical Transformer
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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 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
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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
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
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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
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)
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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
Ef1 is a voltage drop across a reactance
Xf1 = Ef1/I1
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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
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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
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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
o iron losses depend on the mutual flux and hence the applied voltageo 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
• 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?
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Voltage Regulation
• Important attribute of a transformer constantly held primary voltage impact of secondary voltage due to changing loads
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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
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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
o rated voltage applied to the primary
short-circuit test, secondary shortedo rated current applied to primary
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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
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Homework
• Problems: 10-18, 10-23, 10-25, 10-30, 10-31