Real-World Transformers: Imperfect Cores & Losses

Objectives

• Review the ideal transformer with an imperfect core and loose coupling effects
• Understand power transformer construction
• Review standard terminal marking and polarity tests
• Discuss losses, transformer ratings, and cooling methods
• Analyze voltage regulation
• Study energy-efficient single-phase transformers

Transformers in The Real World

• Real-world transformers deviate from the ideal model and require modified analysis
• Windings possess resistance
• Cores aren't infinitely permeable
• Leakage flux needs consideration
• Iron cores generate eddy currents
• Hysteresis losses are present
• Practical transformer properties are describable by an equivalent circuit having an ideal transformer with resistances and reactance

Ideal Transformer with Imperfect Core

• Replace the perfect core of an ideal transformer with an iron core that has hysteresis and eddy-current losses and low permeability
• Imperfections can be represented by two circuit elements in parallel with the primary terminals of the ideal transformer
• Resistance Rm represents iron losses and resultant heat
• Magnetizing reactance measures the permeability of the transformer core
• Current flowing through represents the magnetizing current required to create flux in the core
• Connecting the transformer to an AC source under no-load conditions and measuring active and reactive power allows calculation of Rm and Xm values through the following formulas: Rm = E₁²/Pm and Xm = E₁²/Qm
  • Rm = resistance representing iron losses [Ω]
  • Xm = magnetizing reactance of the primary winding [Ω]
  • E₁ = primary voltage [V]
  • Pm = iron losses [W]
  • Qm = reactive power needed to set up the mutual flux Фm [var]
• Total exciting current needed to produce flux in an imperfect core is the phasor sum
• The peak value of mutual flux: Φm = E₁/(4.44fN₁)

Example 1

• A large transformer operating at no-load draws an exciting current I₀ of 5 A when the primary is connected to a 120 V, 60 Hz source, with iron losses of 180 W
• Reactive power absorbed by the core is 572 var
• The value of the impedance corresponding to the iron losses (Rm) is 80 Ω
• The value of the magnetizing reactance Xm is 25.2 Ω
• The value of the current If needed to supply the iron losses is 1.5 A
• The value of the magnetizing current Im is 4.8 A
• The exciting current I₀ is 5 A

Ideal Transformer with Loose Coupling

• Assume a transformer with a perfect core but slight loose coupling between its primary and secondary windings
  • Primary and secondary windings have negligible resistance with turns N1 and N2
• Across the primary is Ep, setting up a mutual flux Φm1a in the core
  • Flux lags 90° behind Ep, and its peak value is Φm1a = Ep / 4.44 fN₁
  • Since magnetic core is infinitely permeable, lossless, no-load current I₁ = 0
• No mmf is available to drive flux through the air, so no leakage flux links with the primary
• Voltage E₂ is: E2 = (N2/N₁) Ep

Connecting Load Z Across Secondary

• Currents I₁ and I₂ begin to flow in primary and secondary windings, with the following relationship: I₁/I₂ = N₂/N₁ or N₁I₁ = N₂I₂
• I₂ produces a mmf N₂I₂, , I₁ produces a mmf N₁I₁
• The mmf N₂I₂ produces a total AC flux Φ₂
  • A portion of Φ1 (Φm1) links with the secondary winding, another portion (Φf1) does not
  • Flux Φf2 is called the secondary leakage flux
• The mmf N₁I₁ produces a total AC flux Φ₁
  • A portion of Φ₂ (Φm2) links with the primary winding, another portion (Φf2) does not.
  • Flux Φf1 is called the primary leakage flux
• Magnetomotive forces due to I₁ and I₂ upset the magnetic field Φm1a that existed in the core
• The total flux produced by I1 is composed of a new mutual flux Φm1 and a leakage flux Φf1
• The total flux produced by I₂ is composed of a mutual flux Φm2 and a leakage flux Φf2
• Φm1 and Φm2 combine into single mutual flux Φm
• Primary leakage flux Φf1 is created by N₁I₁, while the secondary leakage flux is created by N₂I₂
  • Leakage flux Φf1 is in phase with I₁ and leakage flux Φf2 is in phase with I₂

Primary and Secondary Induced Voltages

• Voltage Es induced in the secondary has two parts:
• A voltage Ef2 induced by leakage flux Φf2 , E_f2 = 4.44 fN_2Φ_f2
• A voltage E₂ induced by mutual flux Φm , E_2 = 4.44 fN_2Φ_m
• E_f2 and E_2 are not in phase
• Voltage Ep induced in the primary has two parts:
  • A voltage Ef1 induced by leakage flus: Φf1
  • A voltage E₁ induced by mutual flux Φm : , E_1 = 4.44 fN_1Φ_m
  • Ef1 = 4.44 fN1Φf1
• Induced voltage Ep = applied voltage Eg

Primary and Secondary Leakage Reactance

• Four induced voltages E₁, E₂, Ef1, and Ef2 are better identified by transformer circuit rearrangement
• Ef2 is a voltage drop across reactance Xf2 , X_f2 = E_("f2")/I_2
• Primary winding is shown twice to show E1 from Ef1
• Efficiency of Efl is a voltage drop across reactance Xfl, X_f1 = E_("fl")/I_1

Resistance and Leakage Reactance of Primary and Secondary

• A circuit is composed of resistive and inductive elements (R₁, R₂, Xf1, Xf2, Z) coupled together by a mutual flux Φm
• The leakage-free magnetic coupling enclosed in a dotted square is an ideal transformer

Complete Equivalent Circuit of Practical Transformer

• Add circuit elements to represent a practical core, the complete equivalent circuit of a practical transformer
  • The value of all the circuit elements that make up a practical transformer can be found through appropriate tests

Example 2

• The secondary winding of a transformer has 180 turns
  • Under load, the secondary current is 18 A at 60 Hz
  • Mutual flux has a peak value of 20 mWb
  • Secondary leakage flux has a peak value of 3 mWb
• The voltage induced in the secondary winding by its leakage flux is 143.9 V
• The value of the secondary leakage reactance is 8 Ω
• The value of E₂ induced by the mutual flux Om is 959 V

Construction of a Power Transformer

• Power transformers are designed so that their characteristics approach those of an ideal transformer
• Leakage reactances Xf1 and Xf2 are minimized by winding the primary and secondary coils on top of each other
• The coils are carefully insulated from each other and from the core
• Tight coupling between the coils means that the secondary voltage at a no-load is almost exactly equal to N2/N₁ times the primary voltage
• Good voltage regulation guarantees when a load is connected to the secondary terminals
• Winding resistances R₁ and R₂ are kept low, both to reduce the I²R loss and resulting heat and to ensure high efficiency
• Primary and secondary coils are distributed over both core legs inorder to reduce the amount of copper
• The number of turns on the primary and secondary windings depends upon their respective voltages
• A transformer is reversible, meaning either winding can be used as the primary (winding that is connected to the source)

Standard Terminal Markings

• The polarity of an instrument transformer is shown by dots on the primary and secondary terminals
• Power transformer terminals are designated H₁ and H₂ for the high-voltage (HV) winding and X₁ and X₂ for the low-voltage (LV) winding
• By convention, H₁ and X₁ have the same polarity
• Power transformers mount the four terminals on the transformer tank in a standard way so the transformer has either additive or subtractive polarity
• Additive and subtractive polarity will depend upon the location of the H₁-X₁ terminals
• Knowing that a power transformer has additive (or subtractive) polarity, there is no need to identify the terminals by symbols
• Subtractive polarity is standard for single-phase transformers above 200 kVA, provided that the high-voltage winding is rated above 8660 V
• All other transformers have additive polarity

Polarity Tests

• These tests determine whether a transformer possesses additive or subtractive polarity
• Method 1: Involves connecting a high-voltage winding to a low-voltage (ex. 120 V) AC source and requires:
  • connecting a jumper J between any two adjacent HV and LV terminals,
  • connect a voltmeter Ex between the other two adjacent HV and LV terminals,
  • connecting another voltmeter Ep across the HV winding.
  • If Ex returns a higher reading than Ep, the polarity is additive (This says that H₁ and X₁ are diagonally opposite)
  • If Ex returns a lower reading than Ep, the polarity is subtractive, and terminals H₁ and X₁ are adjacent
• Method 2: Requires connecting a DC source, in series with an open switch to the LV winding of the transformer, while noting the following:
  • The transformer terminal is connected to the positive side of the source and is marked X₁
  • A DC voltmeter is connected across the HV terminals
  • When the switch is closed, a voltage is momentarily induced in the HV winding, and where direction is:
    • If pointer of the voltmeter moves upscale, the transformer terminal connected to the (+) terminal of the voltmeter is marked H₁ and the other is marked H₂

Example 3

• A 500 kVA, 69 kV/600 V transformer has polarity test readings of Ep = 118 V, Ex = 119 V
• The polarity is additive because Ex is greater than Ep
• The HV and LV terminals connected by the jumper must be labeled H₁ and X₂ (or H₂ and X₁)

Transformer Taps

• Due to voltage drops in transmission lines, the voltage in a particular region of a distribution system may be consistently lower than normal
• To correct this problem taps are provided on the primary windings of distribution transformers i, with the following:
• Taps enable changing turns ratio and raise secondary voltage by 4.5, 9, or 13.5 percent
• This maintains a satisfactory secondary voltage, even if the primary voltage is below normal values.
• Certain transformers change the taps automatically whenever the secondary voltage is above or below a pre-set level.

Losses and Transformer Rating

• Transformer losses come from the following:
• I²R losses in the windings
• Hysteresis and eddy-current losses in the core
• Stray losses due to currents induced in metal supports and tank by primary and secondary leakage fluxes
• Heat produced by iron losses depends upon the peak value of the mutual flux Φm, depends upon the applied voltage
• Heat dissipated in the windings depends upon the current they carry
• Transformer temperature must be at an acceptable level, by setting limits to both the applied voltage and the current drawn by the load
• There limits determine the nominal voltage Enp and nominal current Inp of the transformer winding (primary or secondary)
• Power rating of a transformer equals the product of the nominal voltage times the nominal current of the primary or secondary winding
• The power-handling capacity of a transformer can be expressed in:
  • Volt-amperes (VA)
  • Kilovolt-amperes (kVA)
  • Megavolt-amperes (MVA)

Example 4

• Nameplate indicates a 250 kVA, 60 Hz distribution transformer (primary 4160 V, secondary 480 V)
• The nominal current of the 4160 V winding is 60 A If 2000 V is applied to the 4160 V primary, the transformer flux and iron losses will be lower and core cooler
• The load current should not exceed its nominal value, otherwise the windings will overheat
• The maximum power output using the far lower voltage is 120 kVA
• The nominal current of the 480 V winding has 521A;

Cooling Methods

• Adequate cooling of the windings and core is needed to prevent rapid deterioration of the insulating materials within a transformer
• Indoor transformers below 200 kVA are directly cooled by the natural flow of the surrounding air
• Larger transformers can be built with forced circulation of clean air
• Distribution transformers below 200 kVA are usually immersed in mineral oil and enclosed in a steel tank
• Oil is a much better insulator than air and is invariably used on high-voltage transformers
• As the power rating increases, external radiators are added to increase cooling of the oil-filled tank
• For transformers in the megawatt range, cooling can be effected by an oil-water heat exchanger
• Type of transformer cooling symbols:
  • AA-dry-type, self-cooled
  • AFA-dry-type, forced-air cooled
  • OA-oil-immersed, self-cooled
  • OA/FA-oil-immersed, self-cooled/forced-air cooled

Simplifying Equivalent Circuit

• The complete equivalent circuit of the transformer gives far more detail than is needed
• The circuit simplifies when the transformer operates 1) at no-load and 2) at full-load
  • At no-load, I₂ is zero and so is I₁ because T is an ideal transformer
  • Only the exciting current lo flows in R₁ and Xf1, so the impedance is negligible
  • The current in R₂ and Xf2 is zero
  • Meaning that we can neglect these four impedances
  • At full-load, Ip is at least 20 times larger than lo, so the magnetizing branch is neglible
• We can further simplify the circuit by shifting everything to the primary side, thus eliminating transformer T

Voltage Regulation

• An important attribute of a transformer is its voltage regulation
• With the primary impressed voltage constant at its rated value, the voltage regulation, in percent, is the secondary voltage at no-load at full-load and is defined as:
  • (ENL - EFL) / EFL × 100
• Voltage regulation depends upon the power factor of the load

Example 5

• A 3000 kVA, 69kV/4.16 kV, 60 Hz single-phase transformer has a total internal impedance of 127 Ω, referred to the primary side
• The rated primary current is 43.5 A
• The rated secondary current is 721 A
• Because the transformer exceeds 500 kVA, the windings have negligible resistance compared to their leakage reactance, so Zp = Xp = 127 Ω
• For a 2000 kW resistive load with a fixed primary supply voltage fixed at 69kV then voltage regulation is:
  • Approximate impedance of the 2000 kW load on the secondary side is Z = 4160²/2000000 = 8.65Ω
• Since theprimary voltage is held constant at 69 kV, the secondary voltage at no-load is 4160 V

Measuring Transformer Impedances

• For a given transformer, we can determine the actual value of the:
  • Open-circuit test, which rates voltage is applied to the primary winding, and current, voltage, and active power are measured
  • Short-circuit test , where secondary winding is short-circuited and a voltage much lower than normal (less than 5 percent of rated voltage) is applied to the primary
• Open circuit test will give us information for; Active power absorbed by the core, Apparent power absorbed by the core and Reactive power absorbed by the core with can be determined by:
• Qm = √(Sm² - Pm²)
• the Resistance can be then determined by the following formula:
• Rm= Ep²/Pm
• Magnetizing Reactance use the following formula,
  • Xm²= Ep²/Qm, also Turns ratio
  • A= N1/N2=Ep/Es is used
• To assess total transformers in pedance use the following formula,
  • Zp=Esc/Iso Total transfer resistance is
  • Rp=PsC/lso²

Example 6 -Calculating impedance of a trns Former.

• In case a short circuit test is preformed the following formulas should be used to calculate the valve of Reactants, Resistance.
• Transformer impedance referred to primary
• Zp = Esc/lse
• Resistance referred to primary
• Rp is the resulting formula
• Rp=PsC/lso² leakage reactants referred to primary is
  • Xp = √(Z2-R2)
• to indicate the value the internal impendence ZP is always indicated on Name plate* it is shown as percentage of Normal Load.
• Z= V/A

Example 7

• This Formula calculate the total interoperated of the former is used:

• Zmp= 5% of 69°(Total value you can derive for calculations)= this is 3.52

• Short Circuit Current

• internal impediment Zp can be used to calculate the amount of current in short current

Example 8

to calculate short circut current you should calculate this parameters that is short

• Isc =VA/E%(Z) * where current rated = VA * E

Energy Efficiency from Single Base Transformers

• Energy efficient is the energy efficiency of distribution transformers which is very high between, 96 % to 99 %

In order to prevent large distribution in the system of the large distribution we can implement the Following

• no load iron the core can be reduced through improvement of the design through the process and the make properties are the core

• Load loss copper loss can be reduce increasing the Cross Section of the winding

• Cooling loss and be reduce the decrease of other types of former losses To prevent This the following needs to be done!

• Increase in Transformers and also the replacing Silicon steel cores with the 1970,

• the material is produced so there is no low energy loss

• The reduction of no load los is estimated about 70% 80%.High efficiency and be reduces by 1,5 Million's

• The increasing amount of transfers reduces both energy energy and greenhouse gas transmission

• The energy savin annual product and three nuclear points