Fundamentals of Electrical Engineering Lecture 7: Transformer PDF

FUNDAMENTALS OF ELECTRICAL ENGINEERING LECTURE 7: TRANSFORMER...

FUNDAMENTALS OF ELECTRICAL ENGINEERING LECTURE 7: TRANSFORMER MAGNETISM, MATERIALS Prof. Dr.-Ing. Saša Bukvić-Schäfer (Hochschule Hamm-Lippstadt) Prof. Dipl.-Ing. Volker Wachenfeld (Hochschule Biberach) 04.12.2024 Fundamentals of Electrical Engineering 11 AGENDA 1. Fundamentals Fundamentals––Electromagnetism Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 22 BASIC CONCEPT: MAGNETS Magnets: have a long history (thousands of years) and they've been mainly used as compasses. S N S N N S N S ATTRACTION REPULSION S N N S N S S N Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 33 BASIC CONCEPT: MAGNETS Magnets: have a long history (thousands of years) and they've been mainly used as compasses. To describe magnets, we have to accept that: You can not „buy“ a magnet with north or N S south pole only! N No magnetic monopoles exist - the two S N N S magnetic poles S always come in N a pair S N S N S Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 44 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 55 BASIC CONCEPT: MAGNETIC FIELD Magnetic Field: no sources or sinks – Maxwell: ර 𝐵 ∙ 𝑑 𝑠Ԧ = 0 ! 𝑆 The magnetic field lines do not originate and terminate on poles – they form closed loops. Although magnetic field lines appear to originate at north pole and terminate on south pole, they run within the magnet between the poles. Magnetic field lines can never cross – if the fields from two or more magnets overlap at N S 𝐵 the same location, they add (as vectors) to produce a single, total field at that point. The net magnetic field at any point is the vector sum of all magnetic field lines present at that point Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 66 BASIC CONCEPT: MAGNETIC FIELD Electromagnetism describes the relation between 𝐵 magnetism and electricity Magnetism and electricity cannot be separated without “robbing” each of a large part of its usefulness. They are interdependent forces. The experimental work of Ørsted (in 1820) 𝐼 showed that a magnetic needle is deflected by an adjacent electric current and aligns itself Quelle: Famous Scientists. Auf: https://www.famousscientists.org/hans-christian- oersted/ - zuletzt abgerufen am 07.10.2020 perpendicularly to a current-carrying wire. Ampère extended the experimental work of Ørsted (working with Quelle: Google. Auf: https://sites.google.com/site/historyofelectricity98765/andre-marie- ampere-1775-1836 – zuletzt abgerufen am 07.10.2020 parallel wired carrying current etc.) and start developing a mathematical and physical theory to understand the relationship between electricity and magnetism. Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 77 BASIC CONCEPT: MAGNETIC FIELD – AMPÈRE For any closed loop path, the sum of the length elements times the magnetic field strength (𝐻) in the direction of the length element is equal to the all electric current enclosed in the loop. ර 𝐻 ∙ 𝑑 𝑠Ԧ = ෍ 𝐼𝑒𝑛𝑐 𝑑𝑠Ԧ 2𝜋 𝐼 𝑟 ර 𝐻 ∙ 𝑟 ∙ 𝑑𝜑 = 𝐼 ⇒ 𝐻 ∙ 𝑟 ∙ 2𝜋 = 𝐼 ⇒𝐻= 0 2𝜋 ∙ 𝑟 Ampère showed that current through a conductor will produce a circular magnetic field around it. 𝐼 𝐻~𝐼 !! Line integral of the magnetic field around the conductor equals the current through it Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 88 FUNDAMENTALS OF ELECTROMAGNETISM – MAGNETIC PARAMETERS 𝐼 𝐼 Cause? 𝑉 Effect? 𝐼 𝑉 𝑉 ~ 𝑅 Direction? 𝑉 𝑅𝑚 𝐼= 𝑅 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 99 FUNDAMENTALS OF ELECTROMAGNETISM – MAGNETIC PARAMETERS Cause = „driving“ parameter, depending on 𝐼 Current 𝑉 Idea: The current is conducted as an effect of the voltage (Ohm's law). The voltage itself has no direct impact on the magnetic field. Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 10 10 FUNDAMENTALS OF ELECTROMAGNETISM – MAGNETIC PARAMETERS Cause = „driving“ parameter, depending on 𝐼 Current and Number of windings of the solenoid (= coil) 𝑉 𝑤 𝛩 Magnetic parameter: Magnetomotive Force! 𝛩 =𝐼∙𝑤 𝐼 𝐼 𝐼 Idea: It does not seem to matter whether I use three individual coils with one winding each, through which the current 𝐼 is conducted, or one coil with three windings, through which the current 𝐼 is conducted, to generate a magnetic field! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 11 11 FUNDAMENTALS OF ELECTROMAGNETISM – MAGNETIC PARAMETERS Magnetic flux 𝛷 = Sum of all field lines penetrating the cross section 𝑆 𝛷 𝛩 =𝑤∙𝐼 𝐼 𝑆 𝑆 𝑤 𝛷 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 12 12 FUNDAMENTALS OF ELECTROMAGNETISM – MAGNETIC PARAMETERS Magnetic parameter: Magnetomotive Force! 𝐼 𝛷 𝛩 =𝐼∙𝑤 𝑉 𝑤 𝛩 Effect = Field inside the solenoid depending on cause (magnetomotive force) and material properties Magnetic parameter: Magnetic flux! 𝑅𝑚 𝛩 = 𝛷 ∙ 𝑅𝑚 𝛩 resp. 𝛷= 𝑅𝑚 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 13 13  FUNDAMENTALS OF ELECTROMAGNETISM – ELECTRICAL EQUIVALENT CIRCUIT Magnitude of the magnetic flux 𝐼 𝛷 𝑅𝑚,𝐹𝑒 depending on what? the less conductive the material, the...… the longer the path of the field lines, the… 𝑉 𝛩 the smaller the cross-section, the… A … greater the reluctance (magnetic resistance)! 𝑅𝑚,gap 𝑙𝑔𝑎𝑝 = 𝛿 Reluctance: 1 1 𝑙 𝑅𝑚 ~ ∙𝑙 ∙ resp. 𝑅𝑚 = 𝑙𝑖𝑟𝑜𝑛 𝜇0 ∙ 𝜇𝑟 𝑆 𝜇 0 ∙ 𝜇𝑟 ∙ 𝑆 𝛩 𝛷= 𝑅𝑚 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 14 14 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 15 15 THE MAGNETIC FIELD “STRENGTH” – MAGNETIZING FORCE 𝐻 … Magnetizing force: Magnetomotive Force 𝛩 related to the length of the field lines Magnetomotive force: 𝛩 = 𝑤 ∙ 𝐼 What is impacting the magnetizing force? with 𝐻 … Magnetizing force 𝛩 𝑤∙𝐼 𝛩 … Magnetomotive force (1.31) 𝐻 = = ~𝐼 𝑙 … Length of the solenoid 𝑙 𝑙 𝑤 … Number of windings 𝐼 … Current Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 16 16 THE MAGNETIC FLUX DENSITY – MAGNETIC INDUCTION Equal number of field lines = magnetic flux is constant! 𝑙 𝑅𝑚 = 𝜇0 ∙ 𝜇𝑟 ∙ 𝑆 ∙𝑆 𝑙 𝑅𝑚 ∙ 𝑆 = 𝜇0 ∙ 𝜇𝑟 𝛩 small cross section 𝛷= 𝑅𝑚 great cross section 𝛷 𝛩 Magnetic induction: 𝐵 = = great magn. induction 𝑆 𝑅𝑚 ∙ 𝑆 small magn. induction 𝛩 𝐻= 𝑙 𝛩 𝐵 = 𝜇0 ∙ 𝜇𝑟 ∙ 𝐻 𝐵 = 𝜇0 ∙ 𝜇𝑟 ∙ 𝑙 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 17 17  MAGNETIC FIELD OF A SOLENOID – ENFORCEMENT IN THE IRON CORE Reference: www.abi-physik.de/buch/das-magnetfeld/dauer--und-elektromagnete/ - download October 12th, 2021 Coil/Solenoid with iron core The propagation of the magnetic field depends on the material of the core and its shape through which the magnetic field is conducted! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 18 18 MAGNETIC FIELD OF A SOLENOID – EFFECT OF THE CORE MATERIAL Material Permeability 𝜇𝑟 Classification Silver 0,99998 Copper 0,999991 diamagnetic Lead 0,999983 Water 0,999991 𝑅𝑚,𝐹𝑒 Vacuum 1 𝛷 Air 1,0000004 paramagnetic Aluminium 1,00003 Cobalt 250 𝛩 Nickel 600 ferromagnetic Iron MU-Metal 5.000 100.000 𝛷 𝑅𝑚,𝑔𝑎𝑝 𝜇 = 𝜇0 ∙ 𝜇𝑟 ~ 𝛩 𝑉𝑠 𝜇0 = 4 ∙ 𝜋 ∙ 10−7 𝐴𝑚 𝑅𝑚,𝐹𝑒 ≪ 𝑅𝑚,𝑔𝑎𝑝 Reluctance: 𝑙 𝑅𝑚 = 𝜇 0 ∙ 𝜇𝑟 ∙ 𝑆 𝑅𝑚 = 𝑅𝑚,𝐹𝑒 + 𝑅𝑚,𝑔𝑎𝑝 ≈ 𝑅𝑚,𝑔𝑎𝑝 In a magnetic circuit with air gap, the air gap dominates the reluctance! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 19 19 FERROMAGNETIC MATERIAL PROPERTIES Ferromagnetic materials have like paramagnetic only partially filled (inner) orbitals. Main difference to the paramagnetics is that ferromagnetic materials show so called domain-structures. That means that several single magnetic moments are aligned parallel (so called Weiss mean field). The direction of alignment varies from domain to domain in a more or less random manner, so that some of the materials do not have to show any magnetic field on its own – commonly the word „unmagnetized“ is used to describe it. If the material is exposed to a magnetic field the domain will reorient so that the dipoles in domains are aligned with the external field. The domains will even remain somehow aligned when the external field is removed, creating a magnetic field of their own! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 20 20 FERROMAGNETIC MATERIAL PROPERTIES As soon as a magnetic field is applied, the energetically „cheapest“ process in the ferromagnetic material begins: Re-orientation and the domains that are oriented in the direction of the field start to grow at the expense of the other domains, which is called domain wall motion. 𝐵 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 21 21 MAGNETIZING FORCE AND MAGNETIC INDUCTION The remanence induction 𝐵𝑟 is Magnetic the value of the magnetic induction 𝐵 Saturation of material induction that is permanently maintained in the iron after magnetization has occurred. Principle: The greater 𝐵𝑟 , the Remanence- 𝐵 for 𝝁𝒓 = 𝒄𝒐𝒏𝒔𝒕. Induction 𝑟 "stronger" the magnet. Magnetizing 𝐻~𝐼 force 𝐻 𝐻𝐶 Coercivity Coercivity 𝐻𝐶 is the value of the magnetizing force strength required to demagnetize the iron magnetized with remanence. Saturation of material 𝐵 = 𝜇0 ∙ 𝜇𝑟 ∙ 𝐻 = 𝜇0 ∙ 𝜇𝑟 (𝐻, 𝐵) ∙ 𝐻 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 22 22 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 23 23  INDUCTION AC power generation is based on Faraday’s Law of induction: “The induced electromotive force in any closed circuit is equal to the negative of the time rate Faraday‘s circuit of change of the magnetic flux enclosed by the circuit” w, e.g. w=4 S N + – N  S Reference: Wikipedia – doenload October 16th, 2023 S N 𝛷(𝑡) Battery N S 𝑑𝛷(𝑡) 𝑉 𝑡 𝑣 𝑡 𝜀 = 𝑣 𝑡 = −𝑤 ∙ 𝑑𝑡 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 24 24 INDUCTION – FARADAY‘S LAW Factors influencing the induced electromotive force (EMF, 𝜺) or voltage: The number of windings in the coil: 𝑤 of individual conductors cutting through the magnetic field, the amount of induced voltage will be the sum of all the individual loops of the coil. The speed of the relative motion between the coil and the magnet: The wire will cut the lines of flux at a faster rate so more induced voltage. The strength of the magnetic field: If the same coil of wire is moved at the same speed through a stronger magnetic field, there will be more emf produced because there are „more lines“ to cut. 𝑑𝛷(𝑡) 𝑑 𝑑 𝜀 = 𝑣 𝑡 = −𝑤 ∙ = −𝑤 ∙ 𝛷(𝑡) = −𝑤 ∙ ෡ ∙ 𝑠𝑖𝑛 (𝜔 ∙ 𝑡) 𝛷 𝑑𝑡 𝑑𝑡 𝑑𝑡 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 25 25 INDUCTION – FARADAY‘S LAW Magnetic flux 𝛷 = Sum of all field lines penetrating the cross section 𝑆 𝛷 𝛩 =𝑤∙𝐼 𝐼 𝑆 𝑆 𝑤 𝛷 𝛷 = 𝐵 ∙ 𝑆Ԧ resp. 𝑑𝛷 = 𝐵 ∙ 𝑑𝑆Ԧ Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 26 26 INDUCTION – FARADAY‘S LAW Actually, the magnetic flux density may be constant in time and we can still induce voltage It’s the change in flux that matters! 𝑆 𝑑𝛷 𝑑 𝐵 ∙ 𝑆Ԧ 𝜀 = 𝑣𝑖𝑛𝑑 𝑡 = −𝑤 ∙ = −𝑤 ∙ with 𝐵 ∙ 𝑆Ԧ = 𝐵 ∙ 𝑆 ∙ 𝑐𝑜𝑠𝛼 𝑑𝑡 𝑑𝑡 𝐵 𝑆Ԧ Partially differentiated: 𝛼 𝑑 𝐵 ∙ 𝑆 ∙ 𝑐𝑜𝑠𝛼 𝑑𝐵 𝑑 𝑆 ∙ 𝑐𝑜𝑠𝛼 𝜀 = −𝑤 ∙ = −𝑤 ∙ 𝑆 ∙ 𝑐𝑜𝑠𝛼 ∙ −𝑤∙𝐵∙ 𝑑𝑡 𝑑𝑡 𝑑𝑡 If the coil is stationary and the magnetic If the coil is moving and the magnetic flux changes: transformer principle flux is constant: generator principle Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 27 27 INDUCTION – LENZ‘S LAW Assume there is a magnetic field which induces voltage in a conductor. Reference: Wikipedia – doenload October 16th, 2023 What is the direction of the induced current? 𝐼>0 𝑑𝐵 >0 𝑑𝑡 The direction will be to minimize the change in magnetic flux (Think of inertia, opposing the change in velocity) Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 28 28 SELF INDUCTANCE Magnetomotive Force: 𝛩 = 𝑤 ∙ 𝐼 Reluctance: 𝑅𝑚 = 𝑙 Τ 𝜇0 ∙ 𝜇𝑟 ∙ 𝑆 Assume that the change of the magnetic flux is a result of the change in (inducing) current Due to change in magnetic flux, voltage induced: 𝑑𝛷 𝑡 𝑑 𝛩 𝑡 Τ𝑅𝑚 𝜀 = 𝑣𝑖𝑛𝑑 = −𝑤 ∙ = −𝑤 ∙ 𝑑𝑡 𝑑𝑡 Flux change due to current change 𝑤 ∙ 𝑖 𝑡 ൗ 𝑙ൗ 𝜇 𝑅∙ 𝑚 𝑑 𝑤 0 𝜇𝑟 ∙ 𝑆 𝜀 = −𝑤 ∙ 𝑖(𝑡) 𝑑𝑡 2 𝑆 𝑑𝑖 𝑡 𝑑𝑖 𝑡 𝜀 = −𝑤 ∙ 𝜇𝑟 ∙ 𝜇0 ∙ ∙ = −𝐿 ∙ 𝑙 𝑑𝑡 𝑑𝑡 Inductance 𝐿 is the name given to the property of a component that opposes the change of current flowing through it Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 29 29 INDUCTANCE 𝐷 𝑙 Inductance: Solenoid Solenoid Solenoid (international) with iron core („German“) 𝑤2 𝐿 … Inductance 𝐿= 𝑅𝑚 𝑤 … Number of windings 𝑆 𝜇𝑟 … Permeability of the core material 𝐿= 𝑤2 ∙ 𝜇 𝑟 ∙ 𝜇0 ∙ 𝜇0 … Permeability of vacuum 𝑙 (magnetic constant) Reluctance: 𝑙 Geometry: 𝑆 … Cross section of solenoid core 𝑅𝑚 = 𝐷2 𝑙 … Length of the solenoid 𝜇 0 ∙ 𝜇𝑟 ∙ 𝑆 𝑆= ∙𝜋 𝐷 … Diameter of the solenoid 4 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 30 30 INDUCTANCE Inductance: 𝐷 𝑤2 2 𝑆 𝐿= = 𝑤 ∙ 𝜇 𝑟 ∙ 𝜇0 ∙ 𝑅𝑚 𝑙 𝑙 The basic unit of measurement for inductance is called the Henry 𝐻 : A circuit will have an inductance value of one Henry when an emf of one volt is induced in the circuit were the current flowing through the circuit changes at a rate of one ampere/second. Inductance 𝐿: measured value of an inductor’s “reaction” to the change of the current being conducted through the circuit – the larger its value in Henries, the higher will be the induced voltage at the given rate of current change! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 31 31 MUTUAL INDUCTANCE Current 𝑖1 𝑡 „creates“ a magnetic flux 𝛷1 The second coil is exposed to the major part of this flux (𝛷12 , 𝛷12 penetrating both, coil 1 and 2) The induced voltage in the second coil is only depending on the change of flux seen by the second coil: 𝜀2 (𝑡) 𝑤2 𝑑𝛷12 𝑡 𝜀2 𝑡 = −𝑤2 ∙ 𝑑𝑡 As the changing flux is result of changing current in the first coil, the formula can be re-written: 𝑤1 𝑑𝑖1 𝑡 𝑖1 (𝑡) 𝜀2 𝑡 = −𝑀12 ∙ 𝛷1𝜎 𝑑𝑡 With the coupling inductance 𝑀12 = 𝐿1 ∙ 𝐿2 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 32 32 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 33 33 TRANSFORMER Reference: dimitrisvetsikas1969/licence free from Pixabay: https://pixabay.com/de/photos/transformator-elektrizit%C3%A4t-1446784/ - download October 16th, 2023 A transformer is a static device that by electromagnetic induction allows for changing the alternating voltage form one to another level without changing a frequency. Transformers are important electrical-electrical energy conversion components One important reason we use AC is because we can easily change the voltage levels Transformers enable this conversion of voltage level with high efficiency (up to 99%) and no moving parts (low maintenance) Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 34 34 TRANSFORMER Magn. Energy Magn. Flux Induction Electric Electric Energy Energy Magn. Energy Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 35 35 TRANSFORMER PARTS Basically, a transformer consists of: two or more windings wrapped around a common core The windings are called: primary winding – connected to the AC- electric power source and secondary winding – having the desired voltage level, connected to the loads. Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 36 36 TRANSFORMER PARTS – MAGNETIC CORE The core magnetically couples the windings which are electrically separated Therefore, the core material should have high permeability („conductivity“ for the flux) Ferromagnetic and ferrite materials suitable However, hysteresis curve to be taken into account: describes the behavior as well as the losses Hysteresis losses occur each time the magnetic field is reversed In order to align magnetic domains with the magnetic field energy is needed. Alignment itself produces molecular friction (and in the end: heat = thermal losses). Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 37 37 TRANSFORMER PARTS – MAGNETIC CORE: HYSTERESIS LOSSES Classification of soft or hard magnetic materials based on their hysteresis characteristics 𝐵 Soft magnetic materials: used in devices that are subjected to alter- nating magnetic fields and in which energy losses must be low (transformer core) Hard magnetic materials: HARD 𝐻 utilized in permanent magnets, which must have a high resistance to demagnetization. SOFT Hysteresis losses: 1 𝑃ℎ𝑦𝑠𝑡 = 𝑉𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 ∙ ∙ න 𝐻 ∙ 𝑑𝐵 𝑇 Frequency!! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 38 38 TRANSFORMER PARTS - WINDINGS The winding receiving electrical energy from the source is called the “primary” winding. The winding, which receives energy from the primary winding, via the magnetic field, is called the “secondary” winding. Either the high- or low-voltage winding can be the primary or the secondary. Main property of the winding is a good conductivity: Resistivity of the material will result in so called winding losses, copper losses or 𝐼 2 ∙ 𝑅 losses Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 40 40 TRANSFORMER PARTS - WINDINGS There is some loss of energy in a transformer due to resistance of the primary winding to the magnetizing current, even when no load is connected to the transformer. When a load is connected to a transformer and electrical currents are conducted in both primary and secondary windings, further losses of electrical energy occur. The only possibility to reduce these losses is either to reduce the current (which is actually not possible, as the current is determined by the load requirements) or by using a material that has a low resistance per cross-sectional area (and of course without adding significantly to the costs of the transformer). From the technical and economical point of view mainly two materials seems to be suitable: copper and aluminum Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 41 41 TRANSFORMER ARRANGEMENTS Core-type: the windings surround the core. Shell-type: the steel magnetic circuit (core) forms a shell surrounding the windings. In both cases the low voltage winding is wound directly on the core and the high voltage winding is wound over the low voltage winding. Core-type transformer Shell-type transformer high voltage winding low voltage winding Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 42 42 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 43 43 IDEAL SINGLE-PHASE TRANSFORMER Ideal means no losses. Therefore, following assumptions made: 𝑙 No core losses: 𝑅𝑚𝐹𝑒 = 𝜇0 ∙ 𝜇𝑟 ∙ 𝐴 Near infinite core permeability: 𝜇𝑟 → ∞ No eddy currents 𝑣1 𝑡 𝑣2 𝑡 Ideal winding: No winding resistance: 𝑅𝐶𝑢 → 0 Ideal design: 𝛷12 No flux leakage – only (main) magnetic flux, which magnetically links both windings: 𝛷1 = 𝛷2 = 𝛷12 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 44 44 IDEAL SINGLE-PHASE TRANSFORMER 𝑖1 𝑡 Primary coil with 𝑤1 windings Secondary coil with 𝑤2 windings 𝑣1 𝑡 𝑤1 𝑤2 𝑣2 𝑡 𝛷12 𝑡 is generated by the varying current 𝑖1 𝑡 in the primary winding 𝛷12 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 45 45 IDEAL SINGLE-PHASE TRANSFORMER Same flux passes through both coils 𝛷1 𝑡 = 𝛷2 𝑡 = 𝛷12 𝑡 𝑖1 𝑡 Therefore: 𝑑𝛷12 𝑡 𝑣1 𝑡 𝑑𝛷12 𝑡 𝑣1 𝑡 = −𝑤1 ∙ ⇒ =− 𝑑𝑡 𝑤1 𝑑𝑡 𝑣1 𝑡 𝑤1 𝑤2 𝑣2 𝑡 and 𝑑𝛷12 𝑡 𝑣2 𝑡 𝑑𝛷12 𝑡 𝑣2 𝑡 = −𝑤2 ∙ ⇒ =− 𝑑𝑡 𝑤 2 𝑑𝑡 𝑣1 𝑣2 𝑣1 𝑤1 ⇒ = 𝑜𝑟 = 𝛷12 𝑤1 𝑤2 𝑣2 𝑤2 Coil with more windings will provide the greater voltage Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 46 46 IDEAL SINGLE-PHASE TRANSFORMER Now a load 𝑅𝐿 is connected to the secondary coil 𝑖1 𝑡 𝑖2 𝑡 A current 𝑖2 𝑡 is supplying the load with a power 𝑝2 = 𝑣2 ∙ 𝑖2 Still, the same magnetic flux 𝛷12 𝑣1 𝑡 𝑤1 𝑤2 𝑣2 𝑡 𝑅𝐿 passes through both coils In case of an ideal transformer no losses can occur, therefore: 𝑝1 = 𝑣1 ∙ 𝑖1 = 𝑣2 ∙ 𝑖2 = 𝑝2 𝛷12 We can expect 𝑖2 𝑣1 𝑤1 𝑣1 ∙ 𝑖1 = 𝑣2 ∙ 𝑖2 ⇒ = = 𝑖1 𝑣2 𝑤2 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 47 47 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 48 48 NON-IDEAL SINGLE-PHASE TRANSFORMER Non-ideal means, we have to take into account: Core losses 𝑤1 Winding losses (primary and secondary) 𝑣1 𝑡 𝑣2 𝑡 𝑅𝐿 Flux leakage 𝑤2 𝑖2 𝑡 𝑖1 𝑡 𝛷12 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 49 49 NON-IDEAL SINGLE-PHASE TRANSFORMER – CORE LOSSES Open-circuit (no load) condition (so we can neglect the losses in the primary winding) The current 𝒊𝟏 in the primary winding 𝑤1 is only covering the losses (pure resistive core losses) and magnetizing current 𝑣1 𝑡 𝑅𝐿 (reactive part). 𝑣2 𝑡 𝑖1 𝑡 = 𝑖𝑚 𝑡 + 𝑖𝐶 𝑡 = 𝑖𝜇 𝑡 + 𝑖𝐶 𝑡 𝑤2 The magnetizing current 𝑖𝜇 𝑡 is actually the “cause” for the main flux 𝛷12 (both are 𝑖1 𝑡 in-phase), which on the other side induces 𝛷12 both voltages in the primary and secondary winding. Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 50 50 NON-IDEAL SINGLE-PHASE TRANSFORMER – CORE LOSSES The current 𝐼 1 is a phasor sum of reactive part 𝐼 𝜇 and pure resistive part 𝐼 𝐶 : 𝐼𝜇 + 𝐼𝐶 = 𝐼1 Voltage 𝑉 1 is in phase with the resistive part 𝑤1 and leads the magnetization current (90°) Induced voltage in the 𝑉1 𝑉2 secondary winding 𝑉 2 will be under no load conditions in phase 𝑉1 𝑤2 with 𝑉 1 𝑉2 𝐼1 𝛷12 𝐼1 𝐼𝐶 𝐼𝜇 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 51 51 NON-IDEAL SINGLE-PHASE TRANSFORMER – WINDING LOSSES The primary winding is supplied by AC voltage and the current in the 𝛷1 𝛷2 primary winding creates the 𝑤1 magnetic flux. This magnetic flux 𝛷1 will consist of 𝑣1 𝑡 𝑅𝐿 the one part, which is called main 𝑣2 𝑡 flux 𝛷12 and links magnetically the 𝛷2𝜎 both windings (and induces 𝛷1𝜎 𝑤2 voltages), and the other part called 𝑖2 𝑡 leakage flux 𝛷1𝜎. 𝑖1 𝑡 𝛷1 = 𝛷12 +𝛷1𝜎 𝛷12 The same on secondary side 𝛷2 = 𝛷12 + 𝛷2𝜎 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 52 52 NON-IDEAL SINGLE-PHASE TRANSFORMER – WINDING LOSSES The leakage flux represents the flux’ paths that do not include both 𝛷1 𝛷2 windings and leak out of the core. 𝑤1 𝑑𝛷1𝜎 𝑣1 𝑡 = −𝑤1 ∙ 𝑑𝑡 𝑣1 𝑡 𝑉 1𝜎 = 𝑗𝑋1𝜎 ∙ 𝐼 1 𝑣2 𝑡 𝑅𝐿 𝛷2𝜎 What about the secondary winding? 𝛷1𝜎 𝑉 2𝜎 = 𝑗𝑋2𝜎 ∙ 𝐼 2 𝑤2 𝑖2 𝑡 Additionally, we also have to take 𝑖1 𝑡 𝛷12 into account the resistance of the windings! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 53 53 EQUIVALENT CIRCUIT Primary side: Copper losses of the winding - 𝛷1 𝛷2 represented by 𝑅1 𝑤1 𝑅1 The leakage losses related to the primary winding only- represented 𝑣1 𝑡 𝑣2 𝑡 𝑅𝐿 with 𝑋1𝜎 𝛷2𝜎 𝛷1𝜎 𝐼1 𝑤2 𝑖2 𝑡 𝑅1 𝑋1𝜎 𝑋1𝜎 𝑉1 𝛷12 𝑖1 𝑡 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 54 54 EQUIVALENT CIRCUIT Core: Resistive losses of core – 𝛷1 𝛷2 represented by 𝑅𝐶 𝑤1 Magnetizing reactive part - represented with 𝑋1𝜇 𝑣1 𝑡 𝑣2 𝑡 𝑅𝐿 𝛷2𝜎 𝛷1𝜎 𝐼1 𝑤2 𝑖2 𝑡 𝑅1 𝑋1𝜎 𝑖1 𝑡 𝑉1 𝑅𝐶 𝑋1𝜇 𝑉 1𝜇 𝛷12 𝑑𝛷12 𝑉 1𝜇 = 𝑗𝑋1𝜇 ∙ 𝐼 𝜇 resp. 𝑣1𝜇 𝑡 = −𝑤1 ∙ 𝑑𝑡 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 55 55 EQUIVALENT CIRCUIT Secondary side – as the core losses are „attached“ to the primary side, on the 𝛷1 𝛷2 secondary side we additionally have: 𝑤1 Copper losses of the winding - 𝑅2 represented by 𝑅2 𝑣1 𝑡 𝑣2 𝑡 𝑅𝐿 The leakage losses only related to the 𝛷2𝜎 secondary winding - represented by 𝑋1𝜎 𝛷1𝜎 𝐼2 𝑤2 𝑋2𝜎 𝑅2 𝑋2𝜎 𝑖1 𝑡 𝑖2 𝑡 𝑉 2𝜇 𝑉2 𝑅𝐿 𝛷12 𝑑𝛷12 Induced voltage: 𝑣2𝜇 𝑡 = −𝑤2 ∙ 𝑑𝑡 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 56 56 EQUIVALENT CIRCUIT Adding all parts together: 𝐼1 𝐼2 𝑅1 𝑋1𝜎 𝑅2 𝑋2𝜎 𝑉1 𝑅𝐶 𝑋1𝜇 𝑉 1𝜇 𝑤2 ′ 𝑉 2 𝑍𝐿 𝑉 2𝜇 = ∙ 𝑉 2𝜇 𝑤1 𝑑𝛷12 𝑑𝛷12 𝑉 1𝜇 = 𝑗𝑋1𝜇 ∙ 𝐼 𝜇 resp. 𝑣1𝜇 𝑡 = −𝑤1 ∙ Induced voltage: 𝑣2𝜇 𝑡 = −𝑤2 ∙ 𝑑𝑡 𝑑𝑡 … connected via an ideal transformer 𝑑𝛷12 𝑣1𝜇 𝑡 𝑤1 −𝑤1 ∙ 𝑤1 𝑤2 ′ = 𝑑𝑡 = ⟹ 𝑣1𝜇 𝑡 = ′ ∙ 𝑣2𝜇 𝑡 = 𝑣2𝜇 𝑡 ⟹ 𝑣2𝜇 𝑡 = ∙𝑣 𝑡 𝑣2𝜇 𝑡 −𝑤2 ∙ 𝑑𝛷12 𝑤2 𝑤2 𝑤1 2𝜇 𝑑𝑡 … and defining 𝑤1 𝑤2 ′ ′ 𝑣2 𝑡 = ∙ 𝑣2 𝑡 ⇒ 𝑣2 𝑡 = ∙ 𝑣2 𝑡 𝑤2 𝑤1 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 57 57 EQUIVALENT CIRCUIT Non-ideal transformer – complete equivalent circuit: 𝑤1 ′ 𝐼2 = ∙𝐼 𝐼1 𝑤2 2 𝑅1 𝑋1𝜎 𝑅2 𝑋2𝜎 𝑤2 ′ 𝑉1 𝑅𝐶 𝑋1𝜇 𝑉 1𝜇 𝑤2 ′ 𝑤2 𝑍𝐿 ∙𝑉 ∙ 𝑉 2𝜇 = ∙ 𝑉 1𝜇 𝑤1 2 𝑤1 𝑤1 𝑤1 Ideal transformer: 𝑣2′ 𝑡 = ∙ 𝑣2 𝑡 𝑤2 Looking at the currents 𝑖1 𝑡 𝑤2 𝑤2 𝑤1 ′ = ⟹ 𝑖1 𝑡 = ∙ 𝑖2 𝑡 = 𝑖2′ 𝑡 ⟹ 𝑖2 𝑡 = ∙ 𝑖2 𝑡 𝑖2 𝑡 𝑤1 𝑤1 𝑤2 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 58 58 EQUIVALENT CIRCUIT For omitting the ideal transformer, we have to match the voltages and currents: 𝐼1 𝐼 ′2 𝑅1 𝑋1𝜎 𝑅2′ ′ 𝑋2𝜎 𝑉1 𝑅𝐶 𝑋1𝜇 𝑉 1𝜇 𝑉 1𝜇 𝑉 2′ 𝑍𝐿 To transform the secondary impedances, we have to consider: The power dissipation on the resistance 𝑅2 and the transformed resistance 𝑅2′ must be the same: ! = 𝑅2 ! 2 2 2 2 𝑃𝑅2 = 𝑅2 ∙ 𝐼22 = 𝑅2′ ∙ 𝐼2′ 𝑤2 ′ 𝑤2 ′ 𝑤1 𝑤2 ′ 𝑅2 ∙ ∙ 𝐼2 = 𝑅2 ∙ ∙ 𝐼22 ⇒ 𝑅2 = ∙ 𝑅2 ′ with: 𝐼2 = ∙ 𝐼2 𝑤1 𝑤1 𝑤2 𝑤1 Same approach for the inductive leakage reactance 𝑋2 and the transformed 2 ′ ′ 𝑤1 value 𝑋2𝜎 : 𝑋2𝜎 = ∙ 𝑋2 𝑤2 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 59 59 OPEN CIRCUIT – NO LOAD CONDITIONS In an open circuit situation, the secondary current is zero: 𝐼 110 𝐼 ′2 𝑅1 𝑋1𝜎 𝑅2′ ′ 𝑋2𝜎 𝑉1 𝑅𝐶 𝑋1𝜇 ′ 𝑉 2′ 𝑉 1𝜇 = 𝑉 2𝜇 The current in the primary winding is only magnetizing the core and covering the losses in core and primary winding. 𝐼 𝐴 Typically, this current 𝐼 10 is very small 𝑡 𝑠 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 60 60 SHORT CIRCUIT CONDITIONS In case of a short circuit, the secondary voltage is zero: ′ 𝐼 1𝑆𝐶 1 𝐼 2𝑆𝐶 2 𝑅1 𝑋1𝜎 𝑅2′ ′ 𝑋2𝜎 𝑉1 𝑅𝐶 𝑋1𝜇 ′ 𝑉 2𝜇 0 = 𝑉 2′ 𝑡 The short-circuit behavior (losses and the short-circuit voltage) are significant for the overall performance of the transformer. Typically, the core resistance 𝑅𝐶 and main inductive reactance 𝑋1𝜇 are much greater than the other resistances and inductive leakage reactances.  can be neglected The short-circuit voltage is an important criterion especially during parallel operations of the transformers! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 61 61 SHORT CIRCUIT CONDITIONS In case of a short circuit, the core resistance 𝑅𝐶 and main reactance 𝑋1𝜇 can be neglected! 𝐼 𝑆𝐶 𝐼 𝑆𝐶 𝑅1 𝑋1𝜎 𝑅2′ ′ 𝑋2𝜎 𝑅𝑇 𝑋𝑇 𝑉1 0 = 𝑉 2′ 𝑡 𝑉1 Under typical load conditions (e.g., > 30 % rated power), the short circuit equivalent circuit is sufficient to describe the behavior of the transformer! The resulting impedances can be calculated as 2 𝑤1 𝑅𝑇 = 𝑅1 + 𝑅2′ = 𝑅1 + 𝑤2 ∙ 𝑅2 2 ′ 𝑤1 𝑋𝑇 = 𝑋1𝜎 + 𝑋2𝜎 = 𝑋1𝜎 + 𝑤2 ∙ 𝑋2𝜎 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 62 62 TRANSFORMER UNDER LOAD CONDITIONS Under typical load conditions, we can use this equivalent circuit diagram! 𝐼 1 𝑉𝑅 𝑉𝑋 𝑅𝑇 𝑋𝑇 𝑉1 𝑉 2′ 𝑍𝐿 𝑉𝑋 𝑉𝑅 To draw the phasor diagram, we have to calculate 𝑉 𝑅 = 𝑅𝑇 ∙ 𝐼 1 𝑉1 𝑉 2′ 𝑉 𝑋 = 𝑗𝑋𝑇 ∙ 𝐼 1 And with Kirchhoff’s law: 𝜑 𝐼1 𝑉 1 = 𝑉 𝑅 + 𝑉 𝑋 + 𝑉 2′ Phasor diagram, e.g., for an inductive load! Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 63 63 AGENDA 1. Fundamentals – Electromagnetism 1.1 Magnetic field 1.2 Magnetizing force and magnetic induction 1.3 Induction 2. Transformer 2.1 Ideal single-phase transformer 2.2 Non-Ideal single-phase transformer 2.3 Three-phase transformer Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 64 64 THREE-PHASE TRANSFORMER Three sets of primary windings (one per phase) and three sets of secondary windings wound on the same core. Both the primary windings and the secondary windings can be connected in one of several ways. 𝐿1 𝐿3 𝐿3 𝐿2 𝐿2 𝐿2 𝑁 𝐿1 𝐿1 𝐿3 𝑁 Y-connection (star connection): one end of each D-connection (delta connection): Z-connection (interconnected star of the three phases is connected together in one the three phase windings are connected in or zigzag connection) point called the neutral or the star point. series and form a ring Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 65 65 CONNECTION SYMBOLS > The windings can be configured in a y/wye or delta connection. > If a winding is distributed over 2 “legs”, this is referred to as a zig-zag circuit. > In addition, the primary and secondary side circuits can be different. Symbol Autoformer High voltage (HV) Y D Z Y Low voltage (LV) y d z a High High Single Beneficial for Special rated rated phase use with applications voltages currents loads Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 66 66 WIRING OF THE WINDINGS 𝐿1 𝐿2 𝐿3 𝐿1 𝐿2 𝐿3 𝐿1 𝐿2 𝐿3 HV HV HV LV LV LV Yy6 Dy5 Yd5 𝐿1 𝐿2 𝐿3 𝐿1 𝐿2 𝐿3 𝐿1 𝐿2 𝐿3 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 67 67 WIRING OF THE WINDINGS 𝑉12(𝑈𝑆) 𝑉 23 𝑉2 In phase:𝑉12 (HV) and 𝑉1 (LV) 𝑉3 𝐿1 𝐿2 𝐿3 𝑵 5 ∙ 30° = 𝑈 1(𝑈𝑆) 150° 𝑉 12 (HV) HV 𝑉 12 (HV) 𝑉12(𝑈𝑆) 𝑉 31 𝑉1(𝐿𝑉) LV 𝑉12(𝑂𝑆) 𝑉1 (LV) Dy5 𝐿1 𝐿2 𝐿3 𝑉 12 (HV) Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 68 68 THREE-PHASE Y-CONNECTION Three-phase system in the low-voltage grid: > The local network transformer supplies 3 symmetrical AC voltage sources > The AC voltage sources form a common star/middle point (y or wye-connection) 𝑉𝐿3𝑁 = 𝑉3 𝐼 𝐿3 = 𝐼 3  𝑉 𝐿3𝐿1 𝑉 𝐿1𝐿2 𝐿3 𝐿3 𝑉𝐿2𝑁 = 𝑉2 𝑉 𝐿1𝑁 Y/Star 𝑉 𝐿2𝐿3 𝑉𝐿3𝐿1 = 𝑉31 𝑉 𝐿3𝑁 N  𝐼 𝐿2 = 𝐼 2 𝐿2 𝐿2 𝑉 𝐿2𝑁 𝑉𝐿1𝑁 = 𝑉1 𝐼 𝐿1 = 𝐼 1 𝑉𝐿1𝐿2  𝑉 𝐿2𝐿3 𝐿1 𝐿1 Y (Wye) 𝑁 𝐼𝑁 𝑁 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 69 69 TRANSFORMER CONFIGURATION ◼ Choice of connection of a 3-phase transformer depends on several factors: ◼ 3 or 4 wire network used (neutral needed?) – Y-connection preferred for grounding ◼ Load asymmetry: think about the distribution grid – single phase and three-phase loads ◼ Economic reasons (cost of construction and connection method) 𝐿1 𝐿3 𝐿3 𝐿2 𝐿2 𝐿2 𝑁 𝐿1 𝐿1 𝐿3 𝑁 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 70 70 TRANSFORMER CONFIGURATION ◼ Consider the difference between Y and △: ◼ In a △-connected winding, the current trough each phase winding is the line current divided by √3. ◼ On the other hand, the △-connected winding requires √3 times as many windings as a Y- connected winding for the same voltage. 𝐿1 𝐿3 𝐿3 𝐿2 𝐿2 𝐿2 𝑁 𝐿1 𝐿1 𝐿3 𝑁 Lecture 7: Transformer 04.12.2024 Fundamentals of Electrical Engineering 71 71  SINGLE-PHASE EQUIVALENT CIRCUIT DIAGRAM OF A THREE-PHASE TRANSFORMER Typical local grid transformer: 𝐷𝑦𝑛5 Balanced transformer: 𝐼3 𝑅𝑇 𝑗𝑋𝑇 𝑉1 = 𝑉𝐿 = 𝑉3 = 𝑉 W w Super- 𝐼2 𝑅𝑇 𝑗𝑋𝑇 V v ordinate 𝐼1 𝑅𝑇 𝑗𝑋𝑇 grid U u HV LV Conclusion: 𝑉1 Σ𝐼 = 0! 𝑉2 1. The same happens 𝑉 𝑍1 𝑍2 𝑍3 𝐼= 𝑉3 in all three circuits 𝑍 2. The total current in the neutral 𝑖3 𝑡 conductors is zero! 𝑖2 𝑡 𝑖1 𝑡 Σ𝐼 = 0! Σ𝐼 = 0! 𝐼1 = 𝐼2 = 𝐼3 = 𝐼 For symmetric loads: 𝑍1 = 𝑍2 = 𝑍3 = 𝑍