Power System Power System FundamentalsFundamentalsPower System Power System FundamentalsFundamentals

EE 317 Power System FundamentalsEE 317 Power System Fundamentals Lecture 6Lecture 6

06 October 2010

AimsAims

Chapter 3 – Transformers Chapter 4 – AC Machinery Fundamentals

TransformersTransformersA transformer is a device that transfers

electrical energy from one circuit to another through inductively coupled electrical conductors. A changing current in the first circuit (the primary) creates a changing magnetic field; in turn, this magnetic field induces a changing voltage in the second circuit (the secondary).

TransformersTransformersBy adding a load to the secondary circuit, one can

make current flow in the transformer, thus transferring energy from one circuit to the other.

The secondary induced voltage VS is scaled from the primary VP by a factor ideally equal to the ratio of the number of turns of wire in their respective windings:

s s

p p

V N

V N

Step-up or Step-downStep-up or Step-down

By appropriate selection of the numbers of turns, a transformer thus allows an alternating voltage to be stepped up — by making NS more than NP — or stepped down, by making it less.

s s

p p

V N

V N

Step-up or Step-downStep-up or Step-down

Three-phase pole-mounted step-down transformer.

TransformersTransformersTransformers are some of the most efficient

electrical 'machines',[1] with some large units able to transfer 99.75% of their input power to their output.[2] Transformers come in a range of sizes from a thumbnail-sized coupling transformer hidden inside a stage microphone to huge units weighing hundreds of tons used to interconnect portions of national power grids. All operate with the same basic principles, though a variety of designs exist to perform specialized roles throughout home and industry.

Basic principles Basic principles The transformer is based on two principles: firstly

that an electric current can produce a magnetic field (electromagnetism) and secondly that a changing magnetic field within a coil of wire induces a voltage across the ends of the coil (electromagnetic induction). By changing the current in the primary coil, it changes the strength of its magnetic field; since the changing magnetic field extends into the secondary coil, a voltage is induced across the secondary.

Transformer CoreTransformer CoreA simplified transformer design is shown to the left. A current passing through the primary coil creates a magnetic field. The primary and secondary coils are wrapped around a core of very high magnetic permeability, such as iron; this ensures that most of the magnetic field lines produced by the primary current are within the iron and pass through the secondary coil as well as the primary coil.

Transfer of EnergyTransfer of Energy

Transformer WEBTransformer WEB

http://en.wikipedia.org/wiki/Transformer#Basic_principles

Induction law Induction law The voltage induced across the secondary coil may be

calculated from Faraday's law of induction, which states that :

S S

dV N

dt

Induction lawInduction law

where VS is the instantaneous voltage, NS is the number of turns in the secondary coil and Φ equals the magnetic flux through one turn of the coil. If the turns of the coil are oriented perpendicular to the magnetic field lines, the flux is the product of the magnetic field strength B and the area A through which it cuts.

S S

dV N

dt

Induction lawInduction law The area is constant, being equal to the cross-

sectional area of the transformer core, whereas the magnetic field varies with time according to the excitation of the primary. Since the same magnetic flux passes through both the primary and secondary coils in an ideal transformer,[1] the instantaneous voltage across the primary winding equals:

P P

dV N

dt

Induction lawInduction lawTaking the ratio of the two equations for VS

and VP gives the basic equation[5] for stepping up or stepping down the voltage:

s s

p p

V N

V N

Ideal power equation Ideal power equation If the secondary coil is attached to a load that allows

current to flow, electrical power is transmitted from the primary circuit to the secondary circuit. Ideally, the transformer is perfectly efficient; all the incoming energy is transformed from the primary circuit to the magnetic field and thence to the secondary circuit. If this condition is met, the incoming electric power must equal the outgoing power. Some energy is lost as heat so therefore it is not 100% efficient:- VPIP=VSIS

Ideal power equationIdeal power equation

s s P

p p S

V N I

V N I

Thus, if the voltage is stepped up (VS > VP), then the current is stepped down (IS < IP) by the same factor. In practice, most transformers are very efficient (see below), so that this formula

is a good approximation

Ideal power equationIdeal power equation The impedance in one circuit is transformed by the

square of the turns ratio.[1] For example, if an impedance ZS is attached across the terminals of the secondary coil, it appears to the primary circuit to have an impedance of . This relationship is reciprocal, so that the impedance ZP of the primary circuit appears to the secondary to be .

2( )PSS

NZ

N OR 2( )SP

P

NZ

N

Chapter 3Chapter 3

TransformersBenefits of Transformers (Completed)

Types and Construction, The Ideal TransformerTransformer Efficiency and Voltage RegulationTransformer TapsAutotransformers3- Transformer connections– Y-Y, Y-, -Y, -

Types and ConstructionTypes and Construction Types:

POWER transformerVoltage samplingCurrent SamplingImpedance transformation

Construction:Core formShell form

Take it apartTake it apartCores are constructed of laminations electrically

isolated from each other to minimize eddy currentsPrimary and secondary windings are wrapped on

top of each other (in some smaller size units they can be wound bifilar with center taps)

In larger units: it simplifies insulation of Hi-V from core and reduces leakage flux

Names of TypesNames of Types POWER: Unit Transformer: output of generation steps-up to

transmission voltages (110kV or higher) Substation Transformer: lowering transmission to

distribution voltage (2.3-34.5 kV) Distribution Transformer: lowers distribution voltage to

household or business levels (110, 208, 240/480V) SPECIAL-PURPOSE Instrument Txs: Potential Transformer: samples hi-voltage for instruments Current Transformer: samples hi-currents for instruments

The Ideal TransformerThe Ideal Transformer Lossless device w/ Input winding and output winding

ati

ti

tiNtiN

N

Na

aN

N

tv

tv

S

P

SSPP

S

P

S

P

S

P

1

)(

)(

)()(

)(

)(

Ideal Transformer and PhasorsIdeal Transformer and Phasors

Magnitudes change, V & I phase angles do not

Power in the Ideal TransformerPower in the Ideal Transformer

Real Transformer EfficiencyReal Transformer Efficiency

In a real transformer current flows in the primary even when there is no load (or an open circuit) on the secondary…

Why?

Magnetization and Core LossesMagnetization and Core Losses

Magnetization current: the current required to produce a flux in the core of the transformer

Core-Loss current: the current required to make up for the hysteresis and eddy current losses

Transformer lossesTransformer losses

Copper (I2R) losses in both coils Eddy current losses in the transformer core Hysteresis losses in magnetic domains of

laminations Leakage flux that escapes the core and pass

through only one winding which produce self inductance

Transformer efficiencyTransformer efficiency

Transformer TapsTransformer Taps Typical Installation: (load usually disconnected)

+5.0% tap+2.5% tapNominal Rating-2.5% tap-5.0% tap

Which has more or less turns in the secondary?

Voltage RegulationVoltage Regulation TCUL – tap changing under load transformer

A.K.A. - the voltage regulator Uses built-in voltage sensing circuitry which

automatically changes taps to keep system voltage supported

Very common in modern power systems

AutotransformersAutotransformers Sharing a common winding with either: additional turns in series for a step-up

autotransformer or a tap located before the terminal ends of

the common coil for a step-down autotransformer

3- Transformation options3- Transformation options Almost all major power systems are three phase

AC systems… Transformers that serve three phase circuits are

constructed in one of two ways:3 single-phase transformers (3 cores)1 three-phase transformer wound on a single three-legged core

Preferred Option todayPreferred Option today Lighter, cheaper and slightly more efficient the

three-phase approach is preferred. Older method used three units which allowed for

individual replacement if there was a fault Many installations today still have three single

phase transformers in operation

3- Transformer connections3- Transformer connections Y-Y connection

VLP / VLS = aVery rarely used due to stability problems

Y- connectionVLP / VLS =3 a (phase shifts exist)

-Y connectionVLP / VLS = a/3 (phase shifts exist)

- connectionVLP / VLS = aNo problems with unbalanced loads or phase shift

Example No 1Example No 1A ferromagnetic core is shown in Figure . The depth of the core is 5 cm. The other dimensions of thecore are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. Withthis current, what is the flux density at the top of the core? What is the flux density at the right side of thecore? Assume that the relative permeability of the core is 1000.

Solution Example 1Solution Example 1There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms a third region. If we assume that the mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the

corners of the core, then the path lengths are 1L = 2(27.5 cm) =

55 cm, 2 L= 30 cm, and 3 L = 30 cm. The reluctances of

these regions are:

Solution Example 1Solution Example 11 1

1 70

0.5558.36 . /

(1000)(4 10 / )(0.05 )(0.10 )r

mkA t Wb

A A x H m m mm mm p -Â = = = =l l

2 22 7

0

0.3047.75 . /

(1000)(4 10 / )(0.05 )(0.10 )r

mkA t Wb

A A x H m m mm mm p -Â = = = =l l

3 33 7

0

0.395.49 . /

(1000)(4 10 / )(0.05 )(0.05 )r

mkA t Wb

A A x H m m mm mm p -Â = = = =l l

Solution Example 1Solution Example 1

(0.005 )(201.6 . / ) 1008 .Wb kA t Wb A tfÁ = Â = =

1008 .2.016

500

A ti A

N t

Á= = =

1 2 3 58.36 47.75 95.49 201.6 . /tot kA t Wb =  + + = + + =

The total reluctance is thus

And the magnetomotive force required to produce a flux of 0.005 Wb is

And the required current is

Thus flux density on the top of the core is:

0.0050.67

(0.15 )(0.05 )

WbB T

A m m

f= = =

Assignment No 1Assignment No 1A ferromagnetic core with a relative permeability of 1500 is shown in Figure P1-3. The dimensions are as shown in the diagram, and the depth of the core is 7 cm. The air gaps on the left and right sides of the core are 0.070 and 0.020 cm, respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. If there are 400 turns in the coil wrapped around the center leg of the core and if the current in the coil is 1.0 A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap?

P1-3 daigramP1-3 daigram