Power System Fundamentals EE 317 Power System Fundamentals Lecture 6 Lecture 6 06 October 2010

Aims  Chapter 3 – Transformers  Chapter 4 – AC Machinery Fundamentals

Transformers  A 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 ). electrical energycircuitinductively coupledelectrical conductorscurrent induces

Transformers  By adding a load to the secondary circuit, one can make current flow in the transformer, thus transferring energy from one circuit to the other.load  The secondary induced voltage V S is scaled from the primary V P by a factor ideally equal to the ratio of the number of turns of wire in their respective windings:

Step-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.alternating voltage

Step-up or Step-down Three-phase pole-mounted step-down transformer.

Transformers  Transformers 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.efficient[1][2]microphonepower grids

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.electric currentmagnetic fieldelectromagnetismelectromagnetic induction

Transformer Core A 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.magnetic fieldmagnetic permeabilityiron

Transfer of Energy

Transformer WEB

Induction law  The voltage induced across the secondary coil may be calculated from Faraday's law of induction, which states that :Faraday's law of induction

Induction law  where V S is the instantaneous voltage, N S 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.voltagemagnetic fluxmagnetic field

Induction 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:[1]

Induction law  Taking the ratio of the two equations for V S and V P gives the basic equation[5] for stepping up or stepping down the voltage:[5]

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:- V P I P =V S I S magnetic fieldelectric power

Ideal power equation Thus, if the voltage is stepped up (V S > V P ), then the current is stepped down (I S < I P ) by the same factor. In practice, most transformers are very efficient (see below), so that this formula is a good approximation

Ideal power equation  The impedance in one circuit is transformed by the square of the turns ratio.[1] For example, if an impedance Z S 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 Z P of the primary circuit appears to the secondary to be.[1] OR

Chapter 3  Transformers Benefits of Transformers (Completed) Types and Construction, The Ideal Transformer Transformer Efficiency and Voltage Regulation Transformer Taps Autotransformers 3-  Transformer connections –Y-Y, Y- ,  -Y,  - 

Types and Construction  Types: POWER transformer Voltage sampling Current Sampling Impedance transformation  Construction: Core form Shell form

Take it apart  Cores are constructed of laminations electrically isolated from each other to minimize eddy currents  Primary 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 Types  POWER:  Unit Transformer: output of generation steps-up to transmission voltages (110kV or higher)  Substation Transformer: lowering transmission to distribution voltage ( 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 Transformer  Lossless device w/ Input winding and output winding

Ideal Transformer and Phasors  Magnitudes change, V & I phase angles do not

Power in the Ideal Transformer

Real 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 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 losses  Copper (I 2 R) 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 efficiency

Transformer Taps  Typical Installation: (load usually disconnected) +5.0% tap +2.5% tap Nominal Rating -2.5% tap -5.0% tap  Which has more or less turns in the secondary?

Voltage 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

Autotransformers  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 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 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 connections  Y-Y connection V LP / V LS = a Very rarely used due to stability problems  Y-  connection V LP / V LS =  3 a (phase shifts exist)   -Y connection V LP / V LS = a/  3 (phase shifts exist)   -  connection V LP / V LS = a No problems with unbalanced loads or phase shift

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

Solution Example 1 There 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 1 L = 2(27.5 cm) = 55 cm, 2 L = 30 cm, and 3 L = 30 cm. The reluctances of these regions are:

Solution Example 1

The total reluctance is thus And the magnetomotive force required to produce a flux of Wb is And the required current is Thus flux density on the top of the core is:

Assignment No 1 A 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 and 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 daigram