EE212 Passive AC Circuits

1 / 46

# EE212 Passive AC Circuits - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## EE212 Passive AC Circuits

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### EE212 Passive AC Circuits

Lecture Notes 3

Transformers

EE 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

EE 212 2010-2011

Magnetic Circuit
• The voltage induced in an electric circuit is proportional to the rate of change of the magnetic flux linking the circuit

EE 212 2010-2011

Rm = in ampere-turns/Wb

i

a

di

L

vab = L

dt

b

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 iin webers (Wb)

N = number of turns in the coil

Rm = constant known as reluctance

(depends on the magnetic path of the flux)

Direction of flux by

Right-Hand Rule

l = length of magnetic path

A = cross-section area

m = permeability

Fingers curled around coil

– direction of current

Thumb – direction of flux

EE 212 2010-2011

Flux Density, B = in teslas, T

Magnetic field strength, H = in AT/m

B-H Curve

B = m H

EE 212 2010-2011

f1

i1

a

vab

L1

b

f2

c

vcd

L2

i2

d

Coupled Circuits

Circuits that affect each other

by mutual magnetic fields

± depending on whether the fluxes add or oppose each other

L1, L2: self inductance

• M: mutual inductance
• ratio of induced voltage in one circuit
• to the rate of change of current
• in another circuit

The flux f2 generated by

current i2 in Coil 2

induces a voltage in Coil 1,

and vice-versa.

EE 212 2010-2011

I1

I2

M

a

c

L1

Vab

Vcd

L2

b

d

Coupled Circuits in Phasors

If input signals are sinusoidal waveforms,

coupled circuits can be in phasor representation

• ± depending on flux directions

EE 212 2010-2011

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.

f1

i1

a

L1

b

f2

c

L2

I1

i2

a

c

d

L1

L2

d

b

I2

Dot Convention

Dots are placed at one end of each coil, so that currents entering the dots produce fluxes that add each other.

The dots provide information on how the coils are wound with respect to each other.

(currents entering the dots

produce upward fluxes)

(+) 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.
• Vab = (jwL1) I1 + (jwM) I2 for the currents as shown

EE 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 coupling
• k = 0.01 to 0.1, loosely coupled
• k > 0.5, close coupled, e.g. air core
• k ≈ 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

EE 212 2010-2011

f

if

+

+

e1

N1

N2

e2

v1

-

-

secondary

winding

primary

winding

e2 = N2

If

+

+

E2

E1

V1

-

-

N1

N2

V1 = -E1

If

f

E1

E2

Transformer

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.

EE 212 2010-2011

Transformer Application in Power System
• V/I step up/down
• Y/D conversion
• circuit (dc) isolation
• Z matching
• (for max power transfer,

Instrument Transformers:

CT (current transformer)

PT or VT (voltage transformer)

EE 212 2010-2011

=

= a (turns ratio)

=

= a

=

Ideal Transformer

• No leakage flux

Coupling Coefficient, k = 1, i.e. thesame flux f goes through both windings

e1 = N1

and e2 = N2

=>

Turns ratio (a or n) is also known as the transformation ratio.

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,

EE 212 2010-2011

I1

I2

+

V2

ZL

e2

e1

V1

-

N1

N2

• a secondary current I2 is drawn by the load
• I2 generates a flux that opposes the mutual flux f

(Lenz’s Law: effect opposes the cause)

• reduction in f would reduce induced emf e1
• since source voltage V1 is constant, and V1 = - e1, f must remain constant
• the primary winding must draw an additional current I’1 from the source to neutralize the demagnetizing effect from the secondary

Primary current I1 = I’1 + If

EE 212 2010-2011

I1

I2

M

a

c

L1

Vab

Vcd

L2

b

d

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/a

I2 = a I1

Z1= a2Z2 where Z2 is the load impedance

+

+

-

-

EE 212 2010-2011

f

i1

R1

R2

+

+

e1

v1

ZL

e2

v2

-

-

N2

N1

Actual Transformer

• - resistance in primary and secondary windings
• - leakage reactance in pr. and sec. windings
• - voltage drops in both windings (leakage impedance)
• - losses
• - copper loss primary: I12·R1 secondary: I22·R2
• - iron loss (core loss)
• Core loss depends on voltage and frequency
• eddy current loss hysteresis loss

EE 212 2010-2011

Iron (Core) Losses

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

EE 212 2010-2011

Iron (Core) Losses

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)

EE 212 2010-2011

Transformer Construction
• Coil Winding
• Core Assembly
• Core-Coil Assembly
• Tank-up
• Accessories Mounting and Finishing

EE 212 2010-2011

Core Assembly

EE 212 2010-2011

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

EE 212 2010-2011

Tank-up

EE 212 2010-2011

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)

EE 212 2010-2011

Transformer Cooling

EE 212 2010-2011

Typical Power Transformers

Pole-mounted

Single-phase Transformer

Three-phase Transformer

EE 212 2010-2011

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 withstand

For 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.

EE 212 2010-2011

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.

turns ratio,

= a

R2

R1

M

+

+

V2

V1

L11

L22

I2

I1

-

-

N1

N2

KVL at Loop 1:

-V1 + R1I1 + jwL11I1 - jwM I2 = 0

KVL at Loop 2:

R2I2 + V2+ jwL22I2 - jwM I1 = 0

inductively coupled circuits

The loop equations are:

EE 212 2010-2011

Equivalent Circuit (continued)

L1

R1

V1

- V2

I1

I2

a2L2

a2R2

+

+

aM

V1

I1

aV2

-

-

R1 + jwL11- jwM

- jwMR2 + jwL22

=

Consider following substitutions:

M a·ML2a2·L2R2a2·R2

V2a·V2I2

conductively connected circuit

KVL at Loop 1:

-V1 + R1I1 + jwL1I1 + jwaM (I1-I2/a) = 0

KVL at Loop 2:

jwaM (I2/a - I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0

Let L11= L1 + aM

and L22= L2 + M/a

The loop equations remain the same:

EE 212 2010-2011

Transformer Equivalent Circuit

a2X2

X1

a2R2

R1

+

I1

Ie

If

a2ZL

Ic

V1

Xf

Rc

-

Ie = excitation current

If = magnetizing current

Xf = magnetizing

reactance

equivalent circuit referred to the primary side

R1, R2= primary, secondary winding resistance

X1, X2= primary, secondary leakage reactance

I1, I2= primary, secondary current

Rc = core loss resistance (equivalent resistance contributing to core loss)

Ic = core loss equivalent current

EE 212 2010-2011

Example: Transformer Equivalent Circuit

a)

b)

j0.92/a2 W

j0.0090 W

0.72/a2 W

0.0070W

j44.6429 W

308.642 W

j0.92W

j0.0090a2W

0.72W

0.0070a2 W

j44.6429 a2W

308.642a2 W

A 50-kVA, 2400/240-V, 60-Hzdistribution transformer has a leakage impedance of (0.72 + j0.92) Win 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)x10-2S [siemens]when viewed from the LV side. Draw the circuit:

a) viewed from the LV side

b) viewed from the HV side

Turns ratio, a = 2400/240 = 10

EE 212 2010-2011

Transformer: Approximate Equivalent Circuit

Xe

Re

Xf

Rc

Xe

Re

Xe

Re = R1 + a2R2

Xe = X1 + a2X2

where, R2 and X2 are referred

to the primary side

Re and Xe obtained from Short Circuit Test

Rc and Xf obtained from Open Circuit Test

For transformers operating close to full load

For transformers in a large power network analysis

EE 212 2010-2011

Short Circuit Test

P = copper loss = I2 Re Re =

Ze =  Xe =

+

R1 ≈ a2 R2 ≈

Vsupply

X1 ≈ a2 X2 ≈

-

HV

LV

W

V

A

Xe

Re

I

+

V

Xf

Rc

-

• Connect meters on HV side as shown
• Short circuit LV side
• Energize HV side with a variable

voltage source and increase voltage

on the Ammeter

• Take the V, I and P readings from meters

Equivalent circuit referred to HV side

EE 212 2010-2011

Open Circuit Test

P = core loss =  Rc =

+

Ie = I , Ic =

Vsupply

If =  Xf =

-

HV

LV

W

V

A

I

+

Ie

If

V

Ic

Rc

Xf

-

• Connect meters on LV side as shown
• Open circuit HV side
• Energize LV side with rated voltage
• Take the V, I and P readings from meters

Equivalent circuit referred to LV side

EE 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 losses

Cu 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

EE 212 2010-2011

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

EE 212 2010-2011

Example: Transformer

Find 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
• 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?
• If rated voltage is applied at the primary terminals, calculate the transformer efficiency at
• full load with 0.8 p.f. lagging

EE 212 2010-2011

Transformer Example 2

• 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
• Zero
• 60% load at 0.8 p.f. lagging
• Full load at 0.8 p.f. laging
• What will be the current on the low voltage side if a short circuit occurs at the load point?

EE 212 2010-2011

Voltage Regulation

• When a constant rated voltage is applied to the primary,

- 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)

EE 212 2010-2011

Voltage Regulation (continued)

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.

EE 212 2010-2011

Example

• 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 at
• 0.8 p.f. lagging
• unity p.f.

EE 212 2010-2011

Voltage Control

The 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.

• Injecting reactive power (Vars): series/shunt compensation

EE 212 2010-2011

Autotransformers
• Used where electrical isolation between primary side and secondary side is not required.
• Usually relatively low power transformers
• 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

EE 212 2010-2011

Autotransformers (continued)
• Schematic diagrams:

From J.D. Irwin, “Basic Engineering Circuit Analysis”, 3rd ed. Macmillan

EE 212 2010-2011

Autotransformers (continued)

From Jackson et al. “Introduction to Electric Circuits” 8th ed., Oxford

EE 212 2010-2011

Autotransformers (continued)

Assume N1 = 200

and N2 = 100

Let Vsource = 120 V

Then what is V2?

Assume I2 is 15 A. Then Sload = ?, and Ssource = ?

EE 212 2010-2011

Autotransformers (continued)

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.

EE 212 2010-2011