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Single Phase Transformer© By Nafees Ahmed Asstt. Prof. Department of Electrical Engineering DIT, University, Dehradun, Uttarakhand 1. 2. 3. 4. 1 References: V. Del Toro “Principles of electrical engineering” Prentice Hall International B. L Theraja “ Electrical Technology” S. Chand www.google.com www.wikipedia.org Prepared By: Nafees Ahmed www.eedofdit.weebly.com It is a static device which transfers energy form one electrical circuit to another electrical circuit without change in frequency. Very high efficiency (up to 96% or up to 98%, Because there is no rotating part) Types of transformer: Depending on types of phase 1. 1-ϕ transformer 2. 3- ϕ transformer Depending on use a. Step up transformer: Output voltage is higher then applied voltage b. Step down transformer: Output voltage is less then applied voltage Depending of construction 1. Shell type transformer 2. Corel type transformer HL VV L H V V HL VV L H V V LV LV HV HV LV LV HV HV LV LV Core type Shell type Figure 1 Construction and working principle: Construction: Laminated Core Flux Secondary Secondary (HV/LV) (HV/LV) Primary (LV/HV) 2 Figure 2 Laminated core of Si-Steel (Soft Steel) Soft Steel is used to reduce hysteresis losses Prepared By: Nafees Ahmed www.eedofdit.weebly.com Laminated core is to reduced eddy current losses Principle: Based on low of electromagnetic induction. Ideal Transformer: Conditions for ideal transformer are 1. Copper losses are zero: => Winding resistance = 0 2. Iron (Core) losses are zero: => Hysteresis & eddy current losses =0 3. Leakage flux is zero: => Permeability of iron is infinity => 4. Magnetizing current is zero S 1 l 0 A EMF Equation: (Prove that emf per turn for a transformer is constant) ϕ v1 e1 N1 N2 e2 v2 Figure 3: Core type 1-Ph transformer Let an alternating voltage (v1) is applied to the primary side So an alternating flux will produce in the core Let m Sint (1) Because of this an emf will induce in primary as well as in secondary windings Primary induced emf e1 N1 d dt d m Sint dt e1 N1 m Cost e1 N 1 e1 N1 m Sint / 2 Or Where 3 e1 E m1 Sint / 2 (2) Prepared By: Nafees Ahmed www.eedofdit.weebly.com E m1 N1 m Max Value of induced emf So its RMS value E1 Em1 N1m N1m 2f 4.44 fm N1 2 2 2 E1 4.44 f m N1 (3) Similarly RMS values of secondary induced emf E 2 4.44 f m N 2 (4) For ideal transformer Induced emf in primary (E1) = Applied Voltage (V1) Induced emf in secondary (E2) = Terminal Voltage (V2) From equations (3) & (4) E1 E 2 4.44 f m Constant N1 N 2 Hence emf/turn = constant Voltage and current transformation ratio: Again from equations (3) & (4) V2 E 2 N 2 Constant say K V1 E1 N 1 K= Voltage (Current) transformation ratio For ideal transformer Input VA = Output VA V1I1 = V2I2 V2 I 1 V1 I 2 Hence V2 E 2 N 2 I 1 K V1 E1 N 1 I 2 Transformer on NO load: Consider an ideal transformer with magnetizing current: Ideal transformer without iron loss: Im & ϕ will be in same phase V1 & E1 will be equal and opposite 4 Prepared By: Nafees Ahmed www.eedofdit.weebly.com V1= -E1 ϕ Im E1 V1 N1 N2 E2 ϕ Im V2 E1 E2 Figure 4b: Phasor diagram of ideal transformer Without Iron loss Figure 4a Ideal transformer with iron loss: + a little copper loss No load current (I0) will be very less (2% - 5% of full load current) No load power factor angle (ϕ0) will be very high ( 780 to 870) Im= Magnetizing or wattles component of no load current (I0) Ie= Active or wattfull component of no load current (I0) V1= -E1 ϕ I0 Ie=I0Cosϕ0 I0 Φ0 ϕ Im=I0Sinϕ0 V1 E1 N1 N2 E2 V2 E1 Figure 4c No load power E2 Figure 4d: Phasor diagram of ideal transformer with Iron loss P0 V1I oCos0 = Total iron loss I0 Neglecting copper loss I 02 R I e I 0Cos0 I m I 0 Sin0 V R0 1 Ie V X0 1 Im I0 Im Ie V1 X0 R0 E1 Figure 4e: Equivalent Circuit Note: No load power factor is very less because ϕ0 is very high. (ϕ0 is high because Ie<<Im) 5 Prepared By: Nafees Ahmed www.eedofdit.weebly.com Transformer on load: ϕ I1 =I0 + I1 ’ I2 L V1 E1 N1 E2 N2 V2 O A D Figure 5a MMF induced in primary winding =MMF induced in secondary winding N1I1' N 2 I 2 N I1' 2 I 2 KI 2 N1 Note: I1’ and I2 will be in phase opposition Load may be pure resistive, inductive or capacitive resulting in unity, lagging and leading power factor respectively. So the phasor diagram may be on unity, lagging and leading power factor. I1 V1= -E1 I ’ =KI 1 V1= -E1 I1 ’ V1= -E1 I1’ =KI2 I1 I1 =KI2 Φ1 Φ1 Φ0 Φ0 I0 ϕ 2 Φ1 Φ0 ϕ I0 I0 ϕ Φ2 I2 Φ2=0 E1 E2=V2 Φ2 I2 E1 E2=V2 E1 E2=V2 I2 Figure 5b: Phasor diagram at Unity PF Figure 5c: Figure 5d: Phasor diagram at lagging PF Phasor diagram at leading PF Resistance and leakage reactance: In transformer windings do have some resistance Let R1 = Resistance of primary winding R2 = Resistance of secondary winding The flux which is linked with primary as well as secondary windings is known as common flux. 6 Prepared By: Nafees Ahmed www.eedofdit.weebly.com There is some flux which does not link with primary or secondary winding know as leakage flux as shown in figure 6a. Let ϕL1 = leakage flux of primary winding (Flux which does not link with secondary winding) Common Flux I1 R1 R2 I2 Leakage Flux L O V1 A V2 D Figure 6a ϕL2 = leakage flux of secondary winding (Flux which does not link with primary winding) L1I1 and L 2I 2 So at no load I1 & I2 both are negligible hence leakage flux is negligible. Leakage flux produces a self induced back emf in their respective windings. They are therefore equivalent to small reactance in series with the respective windings. This reactance is known as leakage reactance. Let X1 = Leakage reactance of primary winding X2 = Leakage reactance of secondary winding Common Flux I1 X1 R1 R2 X2 I2 L V1 N1 E1 N2 O E2 A V2 D Figure 6b I1 X1 I1’ R1 R2 X2 I2 I0 Im Ie V1 R0 X0 L E1 E2 O A V2 D Figure 6c: Equivalent Circuit Note: 7 V1≠ E1 & V2 ≠ E2 as we are taking previously. Prepared By: Nafees Ahmed www.eedofdit.weebly.com Phasor diagram of actual transformer ON Load: Load may be pure resistive load( Unity power factor), inductive load (Lagging power factor) and capacitive load (Leading power factor) so as the phasor diagram. Common Flux I1 X1 R1 R2 X2 I2 L V1 N2 N1 E1 E2 O A V2 D Figure 7a 1. Phasor diagram for pure resistive load (at unity power factor): V1 I 1X 1 I1R1 I1’ =KI2 I1 -E1 Φ1 Φ0 I0 ϕ Φ2=0 I2 I2R2 V2 E1 I2X2 E2=V2 Figure 7b: Phasor diagram at Unity PF Note: a. Resistive drop I2R2 is parallel to I2 and inductive drop I2X2 is perpendicular (leading) to I2. b. Similarly resistive drop I1R1 is parallel to I1 and inductive drop I1X1 is perpendicular (leading) to I1. 8 Prepared By: Nafees Ahmed www.eedofdit.weebly.com c. This is also true for rest of following two diagrams (lagging and leading pf). If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then Φ1 = Φ2 = Φ1’= 0 or say Φ E2 V2 I 2 R2 2 I 2 X 2 2 E2 V2 I 2 R2 E1 V1 E2 K E1 I1 R1Cos I1 R1 Sin 2 I1 X 1Cos I1 R1 Sin 2 Neglecting I1 X 1Cos I1R1Sin 2 V1 E1 I1 R1Cos I1 R1 Sin 0 V1 E1 I1 R1 2. Phasor diagram for inductive load (at lagging power factor): V1 I 1X 1 I1R1 -E1 I1’ =KI2 I1 Φ1 Φ0 I0 ϕ Φ2 I2 I2R2 V2 E1 I2X2 E2=V2 Figure 7c: Phasor diagram at lagging PF 9 Prepared By: Nafees Ahmed www.eedofdit.weebly.com If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then Φ1 = Φ2 = Φ1’= Φ (say) E2 V2 I 2 R2Cos2 I 2 X 2 Sin2 2 I 2 X 2Cos2 I 2 R2 Sin2 2 E2 V2 I 2 R2Cos2 I 2 X 2 Sin2 E1 V1 neglecting I 2 X 2Cos2 I 2 R2 Sin2 E2 K E1 I1 R1Cos I1 R1 Sin 2 I1 X 1Cos I1 R1 Sin 2 Neglecting I1 X 1Cos I1R1Sin 2 V1 E1 I1 R1Cos I1 R1 Sin 3. Phasor diagram for capacitive load (at leading power factor): I1 I1X1 I1’ =KI2 V1 I1R1 -E1 Φ1 Φ0 ϕ I0 Φ2 I2 E1 V2 I2X2 E2=V2 I2R2 Figure 7d: Phasor diagram at leading PF If we neglect I1R1, I1X1 and I2R2, I2X2 drops, then Φ1 = Φ2 = Φ1’= Φ (say) E2 V2 I 2 R2Cos2 I 2 X 2 Sin2 2 I 2 X 2Cos2 I 2 R2 Sin2 2 E2 V2 I 2 R2Cos2 I 2 X 2 Sin2 10 Prepared By: Nafees Ahmed neglecting I 2 X 2Cos2 I 2 R2 Sin2 www.eedofdit.weebly.com E1 V1 E2 K E1 I1R1Cos I1R1Sin 2 I1 X1Cos I1R1Sin 2 Neglecting I1 X 1Cos I1R1Sin 2 V1 E1 I1R1Cos I1R1Sin Note: The above results can be obtained directly if we replace Φ with – Φ in the case of lagging power factor. Equivalent circuit of transformer: Let R1 & X1 are resistance and leakage reactance of primary windings and R2 & X2 are resistance and leakage reactance of secondary windings respectively. I1 X1 I1’ R1 X2 I2 R2 I0 Im Ie R0 V1 X0 L O E2 E1 A V2 D Figure 8a: Transformer Equivalent Circuit We know V2 E2 N 2 I1 K V1 E1 N1 I 2 V1 N 1 V2 1 V2 K N2 I1 X1 2 N1 N R2 2 I1’ R1 I0 V1 R0 2 Im Ie X0 N2 N I 2 1 N E1 1 E2 N2 N1 N X2 L 2 O A N1 N V2 2 D Figure 8b: Transformer Equivalent Circuit Referred to primary side 11 Prepared By: Nafees Ahmed www.eedofdit.weebly.com 2 X1 I1 R1 N1 N R2 2 I1’ I0 Ie R0 V1 N2 N I 2 1 2 Im N E1 1 E2 N2 X0 N1 N X2 L 2 O A N1 N V2 2 D Figure 8c: Approximately Equivalent Circuit Referred to primary side after shifting parallel branch 2 I1 X1 N1 N R2 2 R1 N2 N I 2 1 2 N1 N X2 2 L O V1 A N1 N V2 2 D Figure 8d: Approximately Equivalent Circuit Referred to primary side after neglecting parallel branch Equivalent resistance and reactance of transformer: Let R1 = Resistance of primary winding R2 = Resistance of secondary winding X1 = Leakage reactance of primary winding X2 = Leakage reactance of secondary winding I1R1 = Resistive drop in primary winding I2R2 = Resistive drop in secondary winding I1X1 = Reactive drop in primary winding I2X2 = Reactive drop in secondary winding 1. Equivalent resistance and reactance referred to secondary side: K2X1 K2R1 R2 X2 I2 I1/K L KV1 V2 O A D Figure 9a: Approximately Equivalent Circuit Referred to secondary side 12 Prepared By: Nafees Ahmed www.eedofdit.weebly.com The above circuit is obtained after considering the following relations V2 N 2 I1 K V1 N1 I 2 V2 KV1 I2 & Total resistive drop I 22 K 2 R1 I 22 R2 I 22 K 2 R1 R2 I 22 R02 Where I1 K R02 = Equivalent resistance referred to secondary side 2 2 2 Total reactive drop I 2 K X1 I 2 X 2 I 22 K 2 X1 X 2 I 22 X 02 Where X02 = Equivalent reactance referred to secondary side X02 R02 I2 L KV1 V2 O A D Figure 9b: Approximately & simplified Equivalent Circuit Referred to secondary side KV1 V2 ϕ I2 I2X02 I2R02 Figure 9c: Phasor Diagram (Assuming lagging load) KV1 V2 I 2 R02Cos I 2 X 02Sin 2 I 2 X 02Cos I 2 R02Sin 2 KV1 V2 I 2 R02Cos I 2 X 02Sin If ϕ = 0 13 Neglecting I 2 X 02Cos I 2 R02Sin V2 KV1 I 2 R02Cos I 2 X 02Sin V2 KV1 I 2 R02 Prepared By: Nafees Ahmed www.eedofdit.weebly.com If ϕ = 900 V2 KV1 I 2 X 02 If ϕ is leading Replace ϕ with – ϕ in above equation, so V2 KV1 I 2 R02Cos I 2 X 02Sin V2 KV1 I 2 R02Cos I 2 X 02Sin 2. Equivalent resistance and reactance referred to primary side: X1 I1 R1 R2 / K 2 KI 2 X2 / K2 L O V1 A V2 / K D Figure 9d: Approximately Equivalent Circuit Referred to primary side The above circuit is obtained after considering the following relations V2 N 2 I1 K V1 N1 I 2 V1 Total resistive drop 1 V2 & I1 KI 2 K I12 R1 I12 R2 / K 2 I12 R1 R2 / K 2 I12 R01 Where R01 = Equivalent resistance referred to primary side 2 2 2 Total reactive drop I1 X 1 I 2 X 2 / K Where 14 I12 X 1 X 2 / K 2 I12 X 01 X01 = Equivalent reactance referred to primary side Prepared By: Nafees Ahmed www.eedofdit.weebly.com X01 R01 KI2 L V1 V2/K O A D Figure 9e: Approximately & simplified Equivalent Circuit Referred to primary side V1 V2/K ϕ I1 I1X01 I1R01 Figure 9f: Phasor Diagram (Assuming lagging load) V1 V2 / K I1R01Cos I1 X 01Sin I1 X 01Cos I1R01Sin V1 V2 / K I1R01Cos I1 X 01Sin Neglecting I1 X 01Cos I1R01Sin 2 2 If ϕ = 0 V1 V2 I1 R01 K If ϕ = 900 V1 V1 I1 X 01 K If ϕ is leading Replace ϕ with – ϕ in above equation, so V1 V2 / K I1R01Cos I1 X 01Sin V1 V2 / K I1R01Cos I1 X 01Sin Voltage regulation: It is change in the voltage of secondary side from no load to full load. Let V1 = Full load secondary voltage E2 = No load secondary voltage So % voltage regulation 15 Prepared By: Nafees Ahmed www.eedofdit.weebly.com No load voltage Full load voltage 100 No load voltage E V 2 2 100 E2 Voltage drop in sec winding 100 E2 I R Cos I 2 R02Sin 2 02 100 (1) E2 Where Cos ϕ = Power factor R02 = Equivalent resistance referred to secondary side X02 = Equivalent reactance referred to secondary side +Ve sign for lagging power factor and –Ve sign for leading power factor. Condition for zero voltage regulation: From equation (1) it is clear that voltage regulation will be zero for leading power factor only. So condition of zero voltage regulation I 2 R02Cos I 2 R02Sin 0 E2 R tan 02 X 02 Losses in transformer: The losses are of two type core loss and copper loss (1) Core or iron losses: It is also known as fixed losses, again of two type (a) Eddy current losses: Given by Pe Ke Bm2 f 2t 2v Watt (1) Where Ke = Constant Bm = Maxi value of flux density f = frequency t = thickness of laminations v = Volume of core We know E 4.44 fm N m E 4.44 f N Area ( A) Area ( A) E 4.44 fBm N Area ( A) Bm E/ A (2) 4.44 fN Put the value of Bm in equation (1) 16 Prepared By: Nafees Ahmed www.eedofdit.weebly.com 2 E/ A 2 2 f t v Pe K e 4.44 fN 2 1 22 Pe K e Et v 4.44 AN For a given transformer N, A, t and V are constants so Pe E 2 Pe V 2 E V (Voltage ) So eddy current losses are independent of frequency. (b) Hysteresis losses: Given by Ph Bmx fv Where Watt (3) η = Constant Again put the value of Bm from equation (2) to equation (3) x E/ A fv Ph 4.44 fN x 1 x 1 x Ph E f v 4.44 AN For a given transformer N, A and V are constants so Ph E x f 1 x Ph V 2 f 1 x E V (Voltage ) So eddy current losses dependent of voltage and frequency both. (2) Copper Losses: Also know as variable losses because they depend on load current. R1 = Resistance of primary winding R2 = Resistance of secondary winding I1 = Full load current in primary winding I2 = Full load current in secondary winding Full load copper losses PC I12 R1 I 22 R2 Copper losses at x time of full load PCx x 2 PC 17 Where Prepared By: Nafees Ahmed x Any load current 1 ; for half load x etc Full load current 2 www.eedofdit.weebly.com Tests on transformer 1. Open Circuit (OC) test or No load test: By OC test we can find out Iron losses (Pi) No load current (I0) Cosϕ0, Ie, Im, R0 & X0 Wattmeter 1-Phase Auto Auto Transformer M V CC COM PC L A V Rated Voltage LV HV Figure: 10 Iron losses No load current Let Note: (i) (ii) Pi = Reading of wattmeter (P0) I0 = Reading of Ammeter V = Reading of voltmeter P0 = Pi = VI0Cosϕ0 P Cos0 i VI 0 Ie = I0Cosϕ0 Im = I0Sinϕ0 V V R0 & X0 Im Ie Rated voltage is applied at LV side. This test is generally done on LV side (Why?) 2. Short Circuit (SC) test: By OC test we can find out Copper losses (PC) Equivalent resistance or leakage reactance (R01 & X01 OR R02 & X02) referred to metering side. Full load Cu losses PC = Reading of wattmeter (Wsc) Let Short circuit current Isc = Reading of Ammeter Wsc I sc2 Req Wsc I sc Z eq 18 Prepared By: Nafees Ahmed R Z R01 or R02 eq Z 01 or Z 02 eq www.eedofdit.weebly.com X eq Z eq2 Req2 X eq X 01 or X 02 Wattmeter CC V V Shorted COM PC L d 1-Phase Auto Auto Transformer M A Rated Current HV LV Figure: 11 Note: (i) (ii) (iii) Rated Current is applied at HV side. This test is generally done on HV side (Why?) Why the position of ammeter and voltmeter is changed as compared to OC test? Efficiency of transformer: O / P power 100 I / P power O / P power 100 O / P power losses O / P power 100 O / P power Iron loss Cu loss Efficiency at full load V2 I 2Cos2 100 V2 I 2Cos2 Pi PC P2 100 P2 Pi PC Where P2 V2 I 2 Cos2 Rated VA Cos2 Efficiency at x time of full load 19 xP2 100 (1) xP2 Pi x 2 PC Prepared By: Nafees Ahmed Here Cos2 Load PF www.eedofdit.weebly.com Condition for maximum efficiency Differentiating equation (1) w. r. t “x” and putting dη/dx=0 xP2 Pi x 2 PC P2 xP P 2 xPC d 100 0 2 dx xP2 Pi x 2 PC xP2 Pi x 2 PC P2 xPP 2 xPC 0 x 2 PC Pi Cu loss Iron loss x or Variable loss Constant loss Pi PC Example 1: A 50 KVA, 4400/220 V transformer has R1 = 3.45 Ω , R2 = 0.009 Ω. The values of reactance are X1 = 5.2 Ω, X2 = 0.015 Ω. Calculate a. Equivalent resistance as referred to primary b. Equivalent resistance as referred to secondary c. Equivalent reactance as referred to primary d. Equivalent reactance as referred to secondary e. Equivalent impedance referred to both side f. Total copper loss first using individual resistance of two windings and secondly as referred to each sides. Solution: K N 2 V2 220 1 0.05 N1 V1 4400 20 a. Equivalent resistance as referred to primary R01 R1 R2 / K 2 R01 3.45 0.009 / 1 / 20 2 R0 1 7.05 Ω Ans b. Equivalent resistance as referred to secondary R02 K 2 R1 R2 R02 1 / 20 3.45 0.009 2 R0 2 0.0176Ω 20 Ans Prepared By: Nafees Ahmed www.eedofdit.weebly.com c. Equivalent reactance as referred to primary X 01 X 1 X 2 / K 2 X 01 5.2 0.015 / 1 / 20 2 X 0 1 11.2 Ω Ans d. Equivalent reactance as referred to secondary X 02 K 2 X 1 X 2 X 02 1 / 20 5.2 0.015 2 X 0 2 0.028Ω Ans e. Equivalent impedance referred to both side Z 0 1 7.05 j11.2Ω Ans Z 0 2 0.0176 j0.028Ω Ans f. Total copper loss first using individual resistance of two windings and secondly as referred to each sides. I1 VA Rating 50 103 11.3636 A V1 4400 I2 VA Rating 50 103 227.2727 A V2 220 Copper loss I12 R1 I 22 R2 11.3636 3.45 227.2727 0.009 910.38 W Ans Taking referred values 2 2 W Copper loss I12 R01 11.3636 7.05 W 910.38 W Ans Copper loss I 22 R02 2 227.2727 0.0176 W 909.09 W Ans Error due to approxim at ion 2 21 Prepared By: Nafees Ahmed www.eedofdit.weebly.com Example 1: Following results were obtained on a 100 KVA, 11000/220 V, single phase transformer (i) OC Test (LV Side) 220 V, 45 A, 2 KW (ii) SC Test (HV Side) 500 V, 9.09 A, 3 KW Determine equivalent circuit parameter of transformer referred to low voltage side and efficiency at full load unity power factor. Solution: Equivalent circuit parameters: From OC test: Pi = P0 = 2 KW, I0 = 45 A, V2 = 220 V P0 = Pi = V2I0Cosϕ0 P Cos0 i 0.202 V2 I 0 Ie = I0Cosϕ0=9.09 A Im = I0Sinϕ0=44.07 A R0 From SC test: PC = Wsc = 3 KW, V2 24.20 Ω Ie Isc = 9.09 A, Ans & X 0 V2 5 Ω Im Ans Vsc = 500 V Wsc I sc2 R01 R 0 1 36.31 Ω Ans Vsc I sc Z 01 Z 0 1 55 Ω Ans 2 2 X 01 Z 01 R01 X 0 1 41.31 Ω Efficiency at full load and unity power factor: Ans x RatedVA Cos2 100 x RatedVA Cos2 Pi x 2 PC 1100 1 100 1100 1 2 12 3 η 95.24% Ans Note: 100 1000 9.09 A 11000 Here SC Test is done on full load, so 3 KW is the full load cu loss. Full load HV side current 22 Prepared By: Nafees Ahmed I1 www.eedofdit.weebly.com