chapter 10: practical transformersfrick/ee4220-em_dynamics/lecture1… · flux Φm2 and a leakage...
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10/9/2003 Electromechanical Dynamics 1
Chapter 10: Practical Transformers
10/9/2003 Electromechanical Dynamics 2
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
10/9/2003 Electromechanical Dynamics 3
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 ==
10/9/2003 Electromechanical Dynamics 4
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
10/9/2003 Electromechanical Dynamics 5
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 =
10/9/2003 Electromechanical Dynamics 6
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
10/9/2003 Electromechanical Dynamics 7
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
10/9/2003 Electromechanical Dynamics 8
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
10/9/2003 Electromechanical Dynamics 9
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
10/9/2003 Electromechanical Dynamics 10
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?
10/9/2003 Electromechanical Dynamics 11
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
−=
10/9/2003 Electromechanical Dynamics 12
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=
10/9/2003 Electromechanical Dynamics 13
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
10/9/2003 Electromechanical Dynamics 14
Practical Transformers
• Homework– Problems: 10-18, 10-23, 10-25, 10-30, 10-31