Transformer Agus Purwadi, Qamaruzzaman & Nana Heryana Laboratorium Penelitian Konversi Energi Elektrik Institut Teknologi Bandung Transformer A transformer is a device that convert ac electric energy at one voltage level at another voltage level. Transformer are important to modern life 1. • Thomas A. Edison (1882) Power Distribution System : New York-USA ; 120 Vdc Problems : Losses & Voltage Drop = inefficient Example : 1000 A Rs=0,05 ohm/10km 50 V 120 V 120 kVA 70V 50 kVA 70 kVA 70 kVA 58% High Voltage 1A Rs=0,05 ohm/10km 0,05 V 70 kV ~ 70 kVA 70kV 0,05 VA 70 kVA 70 kVA Transformer are important to modern power system Low voltage High Voltage Real Application on Power System PM G 380 V/20 kV TR TR M 20 kV/380 V Transformer 1. Power Frequency , 50/60 Hz 2. Radio frequency, > 30 kHz 1. Power Transformer 2. Instrument Transformer 2. Type and Construction Of Transformer 3. The Ideal Transformer IP(t) IS(t) + VP(t) + NP NS - vP (t ) N P a vS (t ) N S VS(t) - Turn ratio a a NP NS The Ideal Transformer IP(t) IS(t) + VP(t) + NP NS - N P .iP (t ) N S .I S (t ) VS(t) - Turn ratio a iP (t ) 1 iS (t ) a The Ideal Transformer IP(t) IS(t) + + VP(t) NP NS VS(t) - - In term of phasor quantities VP a VS IP 1 IS a Power in ideal Transformer Power input ideal PIN VP I P cos P Power input POUT VS I S cos P S Turn ratio equation POUT VS I S cos S VS VP / a and I S aIP POUT VP aI P cos PIN a Power in ideal Transformer Active power POUT VP I P cos PIN QOUT VS I S sin VP I P sin QIN reactive power Apparent power SOUT VS I S VP I P S IN Evaluasi • Untuk memahami konsep trafo ideal berikan ke mahasiswa, contoh sederhana misalnya : – Cara mendapatkan tegangan sekunder y volt jika tegangan primernya x volt. – Arus sekunder b ampere pada tegangan y volt, cari arus primer a ampere pada tegangan X volt. – Ambil nilai x dan y yang bedanya sangat besar – Soal pada sistem 1 fasa dulu saja. Impedance Transformer Impedance = ratio of the phasor voltage across it ti the phasor current IL VL ZL VL ZL IL Impedance Transformer IP VP Z 'L IP + + VP VS - if VP aVS IS VS ZL IS ZL - IP IS / a aVS VP 2 VS Z 'L a IP IS / a IS Z 'L a2 Z L Theory of operation of real singlephase transformers IP(t) IS(t) + + VP(t) NP NS VS(t) - - The basis transformer operation can be derived from Faraday’s law Flux linkage N i i 1 N eind d dt eind d N dt EMF equation of a transformer Faraday law of electromagnetic induction max cost e Nmax sin t RMS Value E Nm 2 4,44 fNm d e N dt Transformer Losses • Copper (I2R) losses • Eddy current losses • Hysteresis losses • Leakage flux The equivalent circuit of a transformer IP RP XP XS + VP RS IS + Rc jXM NP NS - VS Ideal transformer the model of a real transformer The equivalent circuit of a transformer IP IS /a + + RP j j XP VP Rc a2 XS a2 RS aVS jXM - Referred to primary a IP j XP /a2 IS j XS + + RP VP / a /a2 RS RC /a2 VS j XM /a2 - Referred to secondary Phasor diagram corresponding – referred to primary V1 j I1X1 E E 1=a 2 I1R1 Im I0 I2/a IX aI2R2 I1 2 ja 2 aV2 2 IC The equivalent circuit of a transformer IP Req + j Xeq IS /a + Ih+e Im aV VP Rc jXM - Approximate transformer models S Test on transformers • Open-Circuit Test wattmeter Ip (t) A + v(t) ~ V Vp (t) - transformer • Result : VOC IOC POC Open-Circuit Test Result Conductance of the core-loss resistor 1 GC RC 1 XM Susceptance of the magnetizing inductor BM Admittance 1 1 YE j RC XC YE GC jBM Magnitude admittance I OC YE VOC Open-Circuit Test Result Power Factor Power Factor angle Admittance POC PF cos VOC .I OC POC cos VOC .I OC 1 I OC YE V OC I OC YE cos1 PF V OC Test on transformers Short-Circuit Test wattmeter Ip (t) IS (t) A + v(t) ~ V Vp (t) - transformer Result : VSC ,ISC ,POC Short-Circuit Test Result Z SE Series impedance VSC I SC Power Factor of the current PSC PF cos VSC .I SC Lagging - Current angle (-),Impedance angle (+) Therefore, Z SE VSC 0 0 VSC I SC I SC PSC cos VSC .I SC 1 Short-Circuit Test Result Series impedance Z SE Req jX eq Admittance Z SE (RP a 2 RS ) j( X P a 2 X S ) 2.2 The equivalent circuit impedances of a 20 kVA, 8000 / 240 V, 60 Hz transformer are to be determined. The open circuit test were performed on the primary side of the transformer, and the following data were taken : Open-circuit test (on primary) VOC = 8000 V Shots-circuit test (on primary) VOC = 489 V IOC = 0,214 A IOC = 2,5 A POC = 400 W POC = 240 W Ip + Vp - Req 38.4 I h e Im Rc 159 k j Xm j 38.4 k j X eq Is a j 192 + a Vs - The per-unit system of measurement actual quantity Quantity per unit base valueof quantity Pbase , Qbase or Sbase Vbase I base In single-phase system : Z base Vbase I base Ybase I base Vbase Z base (Vbase ) 2 S base Voltage Regulation & Efficiency Voltage Regulation VR VS ,no load VS , full load VS , full load x100% Efficiency Pout x100% Pin Pout x100% Pout Ploss Autotransformer IH Step-down autotransformer ISE VC N C VSE N SE NSE IL VH IC NC VL NC VH N SE N C VL N C I C N SE I SE VL VC VH VC VSE I H I SE I L I SE I C I L N SE N C IH NC Power rating -autotransformer S IO N SE NC SW N SE SIO =Input and Output apparent powers SW = apparent power in the transformer windings For example, a 5000 kVA autotransformer connecting a 110 kV system to a 138 kV system would have an NC/NSE turn of ratio of 110 : 28. Such an autotransformer would actually have windings rated at : SW S IO N SE 28 5000kVA 1015kVA NC N SE 28 110 Three-phase transformer NP1 NS1 NS2 NP2 NP3 NS3 A three-phase transformer bank composed of independent transformer NP1 NP2 NP3 NS1 NS2 NS3 A three-phase transformer wound on a single three-legged core Three-phase transformer connections • • • • Wye-wye (Yy) Wye-delta (Yd) Delta-wye (Dy) Delta-delta (Dd) Wye-wye (Yy) Connection V P V S a VLP VLS 3 V P 3 V S a Wye-delta (Yd) Connection VLP VLS 3 V P V S VLP 3a VLS Delta-wye (Dy) Connection V P VLP VLS 3 V S VLP a VLS 3 Delta-delta (Dd) Connection VLP V P a VLS V S A 50 kVA 13.800 / 208 V, Dy distribution transformer has a resistance of 1 percent and reactance of 7 percent per unit. (a) What is the transformer’s phase impedance referred to the high-voltage side? (b) Calculate this transformer’s voltage regulation at full load and 0,8 PF lagging, using the calculated high-side impedance. (c) Calculate this transformer’s voltage regulation under the same conditions, using the per-unit system. Instrument transformers • Potential transformer (PT) • Current transformers (CT) CT PT S V W A