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1 EE0 Electrical Energy Systems Lecture : Transformer and Per Unit Analysis Panida Jirutitijaroen Department of Electrical and Computer Engineering /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
2 Detailed Syllabus 0/0/0 Three-phase circuit analysis: Introduction to three-phase circuit. Balanced three-phase systems. 0/0/0 Three-phase circuit analysis: Delta-Wye connection, Relationship between phase and line quantities 7/0/0 Three-phase circuit analysis: Per-phase analysis, Three-phase power calculation 7/0/0 Guest Lecture Energy & Environment, Smart Grid & Challenges Ahead by Prof. J Nanda (IIT Delhi, IEEE Fellow) 30/0/0 Generator modeling: Simple generator concept 03/0/0 Generator modeling: Equivalent circuit of synchronous generators 03/0/0 Generator modeling: Operating consideration of synchronous generators, i.e. Excitation voltage control, real power control, and loading capability 06/0/0 Generator modeling: Principle of asynchronous generators 0/0/0 Transmission line modeling: Overhead VS Underground cable 0/0/0 Transmission line modeling: Four basic parameters of transmission line 3/0/0 Transmission line modeling: Long transmission line model, Medium-length transmission line model, Short transmission line model 7/0/0 Transmission line modeling: Operating consideration of transmission lines i.e. voltage regulation, line loadability, efficiency 7/0/0 Transformer and per unit analysis: Principle of transformer, Single-phase transformer 7/0/0 Transformer and per unit analysis: Single-phase per unit analysis 0/03/0 Transformer and per unit analysis: Three-phase transformer, Three-phase per unit analysis 0/03/0 Review : if time permits. /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
3 Learning outcomes Outline Reference IN THIS LECTURE /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 3
4 Learning Outcomes Formulate equivalent circuits of various components in electrical energy systems. Equivalent circuit of transformer Explain basic operations of different components in electrical energy systems. Short circuit/ open circuit test /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 4
5 Outline Fundamental concept of transformer Single Phase Transformer Ideal Transformer Reflected load Maximum power transfer Practical Transformer Transformer operation Short circuit test Open circuit test /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 5
6 References Glover, Sarma, and Overbye, Power System Analysis and Design. Chapter 3 /9/0 EE0: Three-phase circuit by P. Jirutitijaroen 6
7 Magnetic flux Electromagnetic induction Dot notation Ampere s Law Faraday s Law FUNDAMENTAL CONCEPT OF TRANSFORMER /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 7
8 Magnetic Flux DC /AC DC source Constant magnetic flux AC source Varying magnetic flux What will happen if we have another coil to link the varying magnetic flux? Source: ndex.php/fuseaction/663. article/d/0085/id/376 /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 8
9 Electromagnetic Induction Recall Faraday s law: When we link Coil to the magnetic flux generated by coil, if the flux is varying, there will be induced electromotive force (EMF) at Coil. The voltage, V, will be generated by the magnetic force across wire. Source logspot.com/0/07/intr oduction-to-coupledcircuit.html /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 9
10 Dot Notation The direction of induced EMF depends on the direction of magnetic flux i.e. location that the coil links magnetic flux. Dot notation is used to indicate the direction of current out of Coil in the equivalent circuit /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 0
11 Magnetic Core We can better link the magnetic flux by using magnetic core. Magnetic flux Ф is now confined in the core and links both windings. Note that N refers to number of turns. We are now interest to relate V and V, and relate i and i. /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
12 Faraday s Law Recall that: Let Then We can write the above equation in a Phasor form. BA cos Since magnetic flux, B = flux density (Weber/m²), A = cross-sectional area (m²). We can write: t t N cos e N sin t 90 E E N N j jba /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
13 Faraday s Law For ideal transformer, we assume that the flux linkage at coil and coil is the same i.e. there is no flux linkage loss. We can now find relationship between the voltage at two sides of the transformer as follows. j N jba V N j N jba V N V V N N /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 3
14 Ampere s Law Current passing through a conductor creates magnetic field around it Hdl I enclosed B = μh B = Magnetic flux density (Weber/m² or Tesla) H = Magnetic field intensity (A/m) μ = Magnetic core permeability (H/m) /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 4
15 Flux Flux Ampere s Law Applied to Transformer Magnetic flux along the path equals the net current enclosed by the path Hl I path enclosed Hl path i N i N i₁n₁ -i₂n₂ Bl path Magnetic permeability i N i N /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 5
16 Magnetic Core Permeability Magnetic core permeability represent the resistance that the magnetic core will allow the magnetomotive force to pass through. For ideal transformer, the ideal value of the permeability is infinity. We can now see the relationship of the current from both sides of the transformers. Bl path i N in i N i N /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 6
17 Ideal transformer Reflected load Impedance matching Practical transformer SINGLE PHASE TRANSFORMER /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 7
18 Ideal Transformer Assumptions:. No resistance in both windings.. No leakage flux around the core. 3. No core resistive loss. 4. Core permeability is infinite. Primary side V V N N Secondary side i N i N /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 8
19 An Ideal Transformer Model We represent an equivalent circuit of an ideal transformer as shown below. Define turn ratio as: From Faraday s and Ampere s Law: /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 9 av V N N V N N a ai i N N i a:
20 Complex Power Complex power at primary side, S * * I * V I av VI S is the same as the complex power at secondary side. This means that ideal transformer has no real/reactive power losses. a /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 0
21 Example From the circuit below, what is the current at the secondary and primary side? 00 V (rms) 0: V V Z = 00 Ω V 0 V 0 i V 00 i a i 0.A 0.0 A 0 V /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
22 Reflected Load We can reflect the load from one side of a transformer to the other side of a transformer. This trick allows us to combine the two separate primary/secondary circuits for easy(?) calculation. Interest to find reflected load Z₁ Z V av a a Z i i i a a: V Z₂ Z₁ /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen
23 Impedance Matching When a voltage source V with internal resistance Rs is connected to a load R, the amount of power at the load depends on the value of load resistance R. Maximum power transfer occurs when R = Rs. In the case that we need to connect the voltage source to a load that does not satisfy the above condition, we can use transformer to match impedance for maximum power transfer. To find an appropriate transformer, we let Rs = a²r and find a transformer turn ratio. Rs Rs a²r a: ~ R When Rs R ~ R /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 3
24 Ideal VS Practical Transformer Ideal transformer. Zero resistance in the both windings.. No leakage flux around the core. 3. No core resistive loss. 4. Core permeability is infinite Practical transformer. Winding losses (copper losses) represented as resistance in both windings.. Leakage flux around the core. 3. Core resistive losses (hysteresis loss + eddy current loss) 4. Magnetic core permeability is finite. How can we represent this effect in the circuit? /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 4
25 Finite Magnetic Core Permeability For practical transformer, Bl path i N in E N Recall that, i jba E j i path i N N We call this Magnetizing current. The ratio between the voltage across the coil (E₁) and magnetizing current can be written as jω(..). Thus we use an inductor to represent the effect of finite magnetic core permeability in the equivalent circuit. i N N Bl N /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 5
26 A Practical Transformer Model Copper losses Leakage reactance Reflected copper losses and leakage reactance of secondary winding. Iron losses (core) Magnetizing susceptance Note that in Chapter 3 [Glover, Sarma, and Overbye, Power System Analysis and Design ], the core losses are represented as shunt admittance, Y = G jb where G and B is positive. The imaginary part is negative to represent inductive property. /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 6
27 A Simplified Model Z₁ and Z₂ are series impedances representing the resistive loss and flux linkage loss in the two windings. Y is a shut admittance representing iron core loss and magnetizing susceptance. Typically Y is very small i.e. resistance is very large. This means that the currents flowing through Z₁ and a²z₂ are almost the same. We can simply combine Z₁ and a²z₂ to Zeq, the equivalent series impedance. Z₁ a²z₂ a: Zeq = Z₁+a²Z₂ a: Y simplified Y /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 7
28 Transformer Parameter Tests Short circuit test To find equivalent series impedance. Short circuit the secondary side. Apply rated current at the primary side. Measure real power and voltage at the primary side. Zeq a: Open circuit test To find equivalent shunt admittance. Open circuit the primary side. Apply rated voltage at the secondary side. Measure real power and current at the secondary side. Zeq a: P₁, V₁ ~ Y Y ~ P₂,I₂ /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 8
29 Example : Short Circuit Test Consider a single-phase 0kVA, 480/0 V 60 Hz transformer. During short circuit test, rated current is applied to the primary side. The voltage of 35 V and real power of 300 W are measured. Find equivalent series impedance of this transformer. 3 P₁ = 300 W V₁ = 35 V Zeq = Req + jxeq ~ Y 480:0 I S rated, rated V, rated R Z eq eq X eq I P, rated I V, rated A /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 9
30 Example 3: Open Circuit Test Consider the same transformer as Example. During open circuit test: rated voltage applied to secondary side, then I₂ = A and P₂ = 00 W. Find equivalent shunt admittance Y of this transformer. Zeq Y = G-jB 480:0 ~ P₂ = 00 W I₂ = A Y V, rated 0 V, V, rated 480 V P 00 Geq S V 480 V I, rated, rated I V a, rated S B S /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 30
31 Saturation In practical transformer model, we assume constant core permeability and linear relationship between B and H follows. In fact, the B-H curve for ferromagnetic materials used for transformer core is nonlinear and has multiple values. As H increases, the core become saturated i.e. the magnetic flux density B increase at a much lower rate. This effect is NOT included in the equivalent circuit. B-H curve is approximated by a dashed line. /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 3
32 Summary A turn ratio (a:) is the ratio between the number of turns on the primary side of transformer and that on the secondary side. A load on the secondary side of transformer can be reflected to the primary side of the transformer. We can use transformer for impedance matching by choosing the turn ratio that makes reflected load equal to internal resistance of voltage source. A practical transformer contains series impedance and shunt admittance. Series impedance represents winding losses and flux leakage losses Shunt admittance represents iron core losses and magnetizing susceptance. Short circuit test is used to find seires impedance by short circuit the secondary side. Open circuit test is used to find shunt admittance by open circuit the primary side. Zeq = Z₁+a²Z₂ Y a: /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 3
33 Next Lecture Single-phase per unit analysis We use this analysis to eliminate transformer model in the circuit. /9/0 EE0: Transformer and Per Unit Analysis by P. Jirutitijaroen 33
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