ee212 passive ac circuits
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EE212 Passive AC Circuits. Lecture Notes 3 Transformers. Magnetic Circuit. What is the relationship between magnetic flux and magnetomotive force, F m ? What is the relationship between electric current, i , and magnetomotive force? Magnetic field strength, H = F m / l. - PowerPoint PPT PresentationTRANSCRIPT
EE212 Passive AC Circuits
Lecture Notes 3Transformers
1EE 212 2010-2011
Magnetic Circuit
• What is the relationship between magnetic flux and magnetomotive force, Fm?
• What is the relationship between electric
current, i, and magnetomotive force?
Magnetic field strength, H = Fm / l 2EE 212 2010-2011
Magnetic Circuit
• What is Faraday’s Law? – The voltage induced in an electric circuit is
proportional to the rate of change of the magnetic flux linking the circuit
3EE 212 2010-2011
Magnetic Circuit
Consider a coil around a magnetic core. If a current i flows through the coil, a magnetic flux is generated in the core.
N = number of turns in the coilRm = constant known as reluctance(depends on the magnetic path of the flux)
a
b
i
Lvab = L didt
= N i in webers (Wb)mR
1
Direction of flux by
Right-Hand Rule
Fingers curled around coil – direction of current
Thumb – direction of flux
Rm = in ampere-turns/WbA.l
l = length of magnetic path A = cross-section area = permeability
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B-H Curve
Flux Density, B = in teslas, TA
Magnetic field strength, H = in AT/mliN.
B = H
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Coupled Circuits
Circuits that affect each other by mutual magnetic fields
a
b
i1
1
L1vab
c
d
vcd
2
L2 i2
The flux 2 generated by current i2 in Coil 2
induces a voltage in Coil 1, and vice-versa.
vab = L1 dtdi1 ± M
dtdi2
vcd = L2 dtdi2 ± M
dtdi1
± depending on whether the fluxes add or oppose each other
L1, L2: self inductance
M: mutual inductance ratio of induced voltage in one circuitto the rate of change of current in another circuit
6EE 212 2010-2011
Coupled Circuits in Phasors
If input signals are sinusoidal waveforms, coupled circuits can be in phasor representation
vab = L1 dtdi1 ± M
dtdi2
vcd = L2 dtdi2 ± M
dtdi1
± depending on flux directions
a
b
Vab
I1
L1
c
d
Vcd
I2
L2
M
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Equivalent Circuit with Dependent Sources
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Dot Convention
Dots are placed at one end of each coil, so that currents entering the dots produce fluxes that add each other.
a
b
i1
1
L1
c
d
2
L2 i2
a
b
I1
L1
c
dI2
L2
(currents entering the dots produce upward fluxes)
The dots provide information on how the coils are wound with respect to each other.
A current i entering a dotted terminal in one coil
induces a voltage M with a positive polarity
at the dotted terminal of the other coil.dtdi
Vab = (jwL1) I1 + (jwM) I2 for the currents as shown
(+) if both currents enter the dotted terminals (or the undotted terminals).
(-) if one current enters a dotted terminal and the other current enters an undotted terminal.
9EE 212 2010-2011
Coefficient of Coupling, k
k = 0 ≤ k ≤ 1
k depends on the magnetic properties of the flux path.
When k = 0, no couplingk = 0.01 to 0.1, loosely coupledk > 0.5, close coupled, e.g. air corek ≈ 1.0, e.g. power transformer
all the flux generated by one coil is linked to the other coil (i.e. no leakage flux)
k = 1.0 ideal transformer
21LLM
10EE 212 2010-2011
E1
N2
E2
N1
I+
-
V1
+
-
V1 = -E1
I
E1
E2
Transformer
Faraday’s Law
i
v1 N1
+
-e2N2
+
-
e1
primarywinding
secondarywinding
dtde2 = N2
When a voltage V1 is applied to the primary winding, an emf e2 is induced in the secondary winding. The induced emf lags the inducing flux by 900.
11EE 212 2010-2011
Transformer Application in Power System
Instrument Transformers:CT (current transformer)PT or VT (voltage transformer)
- V/I step up/down- Y/D conversion- circuit (dc) isolation- Z matching (for max power transfer, min. reflection from load)
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No leakage flux
dtd and e2 = N2
2
1
ee
= 2
1
NN
= a (turns ratio)
1
2
ii
= 2
1
ee
= 2
1
NN
= a
Ideal Transformer
No Losses No voltage drops in the windings: V1 = - e1
Instantaneous powers in primary and secondary are equal (i.e. all the energy from the primary is transferred to the secondary winding).
e1 i1 = e2 i2 Therefore,
Coupling Coefficient, k = 1 , i.e. the same flux
goes through both windings
e1 = N1 dtd
13
=>
Turns ratio (a or n) is also known as the transformation ratio.
EE 212 2010-2011
Transformer Loading• a secondary current I2 is drawn by the load• I2 generates a flux that opposes the mutual flux
(Lenz’s Law: effect opposes the cause)• reduction in would reduce induced emf e1
• since source voltage V1 is constant, and V1 = - e1, must remain constant• the primary winding must draw an additional current I’1 from the source
to neutralize the demagnetizing effect from the secondary
e1
N2
e2
N1
I1
V1
+
-
ZLV2
I2
Primary current I1 = I’1 + I14EE 212 2010-2011
Ideal Transformer
• No leakage flux, i.e., k = 1• Self-inductance, L1 = L2 = ∞ i.e., magnetizing current = 0• Coil losses are negligible• With polarities, dots and currents as shown:
V2 = V1 /aI2 = a I1
Z1 = a2 Z2 where Z2 is the load impedance
15
a
b
Vab
I1
L1
c
d
Vcd
I2
L2
M
● ●+
-
+
-EE 212 2010-2011
Actual Transformer i1
v1
N1
+
-e2
N2
+
-e1 ZL
R1 R2
v2
- resistance in primary and secondary windings- leakage reactance in pr. and sec. windings- voltage drops in both windings (leakage impedance)
- losses- copper loss primary: I1
2·R1 secondary: I22·R2
- iron loss (core loss) Core loss depends on voltage and frequency
eddy current loss hysteresis loss 16EE 212 2010-2011
Eddy Current Loss:
emf induced in core generates eddy currents which circulate in the core material, generating heat.
laminations (silica sheets between core layers) – to reduce eddy current, and minimize loss
Iron (Core) Losses
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Hysteresis Loss:
The direction of the magnetic flux in the core changes every cycle. Power is consumed to move around the magnetic dipoles in the core material, and energy is dissipated as heat.
Hyst. loss (vol. of core) x (area of hyst. loop)
Iron (Core) Losses
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Transformer Construction
• Coil Winding • Core Assembly • Core-Coil Assembly • Tank-up • Accessories Mounting and Finishing
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Core Assembly
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Core-Coil Assembly• Core vertical sides – limbs, top horizontal side – yoke • Yoke is removed to insert the coils into the limbs • LV coil is first placed on the insulated core limbs • Insulating blocks are placed at the top and bottom of the
LV coil • Cylinder made out of corrugated paper is placed over
the LV Coil • HV coil is placed over the cylinder • The top yoke is fixed in position • LV and HV windings are connected as required
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Tank-up
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Accessories Mounting• Connections of LV and HV coil ends to the terminal
bushings are made• Transformer tap changer and protection accessories
(e.g. Buchholz relay, Conservator, Breather, temperature indicator, etc.) are installed
• Tank is closed
Functions of Transformer Oil • Cooling• muffle noise• displace moisture (avoid insulation degradation)
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Transformer Cooling
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Typical Power Transformers
Pole-mounted Single-phase Transformer
Three-phase Transformer
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Transformer Rating• Rated kVA• Rated Voltage primary and secondary: (transformers are normally operated
close to their rated voltages)
• Rated Current (FL Current) is the maximum continuous current the transformer can withstandFor single phase transformer:Rated primary current = Rated VA / Rated Pr Voltage
When rated current flows through a transformer, it said to be fully loaded.
The actual current through a transformer varies depending on the load connected at different times of the day.
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Equivalent Circuit• Represent inductively coupled circuits by a conductively connected circuit• Equivalence in terms of loop equations• Assume a load impedance is connected to the secondary.
V1
+
-L22
+
-
R1 R2
V2
M
N2N1
L11I1 I2
inductively coupled circuits
KVL at Loop 1:
KVL at Loop 2:
-V1 + R1I1 + jwL11I1 - jwM I2 = 0
R2I2 + V2+ jwL22I2 - jwM I1 = 0
The loop equations are:turns ratio, 2
1NN = a
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Equivalent Circuit (continued)
conductively connected circuit
Consider following substitutions:
The loop equations remain the same:
V1
- V2
R1 + jwL11 - jwM
- jwM R2 + jwL22
=I1
I2
V1
+
-
+
-aV2
aMI1
R1 L1 a2R2a2L2
M a·M L2 a2·L2 R2 a2·R2
V2 a·V2 I2 a
I2
Let L11 = L1 + aM and L22 = L2 + M/a
KVL at Loop 1:
KVL at Loop 2:
-V1 + R1I1 + jwL1I1 + jwaM (I1-I2/a) = 0
jwaM (I2/a - I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0
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Transformer Equivalent Circuit
equivalent circuit referred to the primary side
a2ZLV1
+
-
I1
R1 X1 a2R2a2X2
Ie
IIc
Rc X
R1, R2= primary, secondary winding resistanceX1, X2= primary, secondary leakage reactanceI1, I2= primary, secondary current
ZL = load impedance
a2I
Ie = excitation currentI = magnetizing currentX = magnetizing reactance
Rc = core loss resistance (equivalent resistance contributing to core loss)Ic = core loss equivalent current 29EE 212 2010-2011
Example: Transformer Equivalent CircuitA 50-kVA, 2400/240-V, 60-Hz distribution transformer has a leakage impedance of (0.72 + j0.92) W in the high voltage (HV) side and (0.0070 + j0.0090) W in the low voltage (LV) winding. At rated voltage and frequency, the admittance of the shunt branch of the equivalent circuit is (0.324 – j2.24)x10-2 S [siemens] when viewed from the LV side. Draw the circuit:
a) viewed from the LV sideb) viewed from the HV side
Turns ratio, a = 2400/240 = 10
0.72/a2 W j0.92/a2 W 0.0070 W j0.0090 W
j44.6429 W308.642 W
a)
0.72 W j0.92 W 0.0070a2 W j0.0090a2 W
308.642a2 W j44.6429 a2 W
b)
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Transformer: Approximate Equivalent CircuitRe Xe
Rc X
Re Xe
Xe
Re = R1 + a2R2
Xe = X1 + a2X2
where, R2 and X2 are referred to the primary side
Re and Xe obtained from Short Circuit TestRc and X obtained from Open Circuit Test
For transformers operating close to full load
For transformers in a large power network analysis31EE 212 2010-2011
Short Circuit Test
LVHV
Vsupply
+
-
W A
V
Connect meters on HV side as shownShort circuit LV sideEnergize HV side with a variable voltage source and increase voltage gradually to get rated current reading on the AmmeterTake the V, I and P readings from meters
Re Xe
Rc X
I
V
+
-
Equivalent circuit referred to HV side
P = copper loss = I2 Re Re = 2IP
Ze = Xe = IV 2
e2
e RZ -
R1 ≈ a2 R2 ≈ 2
Re X1 ≈ a2 X2 ≈ 2Xe
32EE 212 2010-2011
Open Circuit TestConnect meters on LV side as shownOpen circuit HV sideEnergize LV side with rated voltageTake the V, I and P readings from meters
Equivalent circuit referred to LV side
Vsupply
W A
V+
-LVH
V
Rc X
I
V
+
-
IIe
Ic
P = core loss = Rc = c
2
RV
PV2
Ie = I , Ic = cR
V
I = X = 2c
2e II -
φIV
33EE 212 2010-2011
Transformer Efficiency
Efficiency, h = Power Output / Power Input
Power Output = |VL| |IL| cos = |IL|2 RL
Power Input = Power Output + Cu losses + Core lossesCu Loss – varies with load current
Core Loss – depends on voltage (usually a constant for practical purposes)
Transformer efficiency is maximum when Cu loss = core loss
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Transformer Efficiency (continued)
Power transformers: - usually operate at rated capacity, - designed to have max. h at full load
Distribution transformers: carry a widely varying load - designed to have max. h at less than full load - always energized despite load levels – designed to have low core loss
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Example: TransformerFind the equivalent circuit of a 50-kVA, 2400/240-V transformer. The following readings were obtained from short circuit and open circuit tests:
If rated voltage is available at the load terminals, calculate the transformer efficiency at (a) full load with 0.8 p.f. lagging
(b) 60% load with 0.8 p.f. lagging. What is the voltage at the HV terminals of the transformer?
Test V-meter (V) A-meter (A) W-meter (W)
Short Circuit 48 20.8 617 Open Circuit 240 5.41 186
If rated voltage is applied at the primary terminals, calculate the transformer efficiency at (c) full load with 0.8 p.f. lagging
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The transformer is used to step down the voltage at the load end of a feeder whose impedance is 0.3 + j1.6 ohm. The voltage at the sending end of the feeder is 2400 V. Find the voltage at the load terminals when the connected load is
(a) Zero
(b)60% load at 0.8 p.f. lagging
(c) Full load at 0.8 p.f. laging
(d)Full load at 0.5 p.f. leading
What will be the current on the low voltage side if a short circuit occurs at the load point?
Transformer Example 2
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When a constant rated voltage is applied to the primary,
At no load- no current, and therefore no voltage drop in transformer- secondary voltage, Vsec = rated voltage
As load (resistive or inductive) increases- voltage drop in transformer (Vdrop) increases
- secondary voltage decreases (reverse is the case with capacitive load increase)
Voltage Regulation
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To maintain rated voltage at the secondary,
At no load, primary voltage required, Vpr = rated voltage
At a certain load, required Vpr = rated voltage + Vdrop in transformer
Voltage regulation is the change in primary voltage required to keep the secondary voltage constant from no load to full load, expressed as a percentage of rated primary voltage.
Load p.f. has a big effect on voltage regulation.
Voltage Regulation (continued)
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A 50-kVA, 2400/240-V, 60-Hz transformer has a leakage impedance of (1.42+j1.82) ohms on the HV side. The transformer is operating at full load in all 3 cases below.
1. Calculate the voltage regulation ata) 0.8 p.f. lagging b) unity p.f. c) 0.5 p.f. leading
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Example
- Use of transformer tap changers: on-load/off-load- Injecting reactive power (Vars): series/shunt compensation
Voltage ControlThe allowable voltage variation at the customer load point is usually ± 5% of the rated (nominal) value. Control measures are taken to maintain the voltage within the limits.
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Autotransformers• Used where electrical isolation between primary
side and secondary side is not required.• Usually relatively low power transformers• Can be readily made to be variable.• Can think of it as two separate windings
connected in series.• Usually it is a single winding with a tap point.• Can be used as either step-up or step-down
transformer
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Autotransformers (continued)
• Schematic diagrams:
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From J.D. Irwin, “Basic Engineering Circuit Analysis”, 3rd ed. Macmillan
Autotransformers (continued)
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From Jackson et al. “Introduction to Electric Circuits” 8th ed., Oxford
Autotransformers (continued)
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Assume N1 = 200and N2 = 100
Let Vsource = 120 V
Then what is V2?
Assume I2 is 15 A. Then Sload = ?, and Ssource = ?
Autotransformers (continued)
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What is the node equation at the tap?
Node equation: I1 + IZY = I2 therefore, IZY =
From transformer action:N1I1 = N2IZY
therefore,
That is, only part of the load current is due to the magnetomotive force that the current in the primary coil exerts on the secondary coil. The remaining portion is direct conduction of the source current to the load.