Lecture 4 Three Phase Transformer (Part 2) PDF

Three Phase Transformer Lecture 4  Review on Real Transformer Primary Copper loss Secondary Copper loss and Flux Leakage and Flux Leakage Core loss and Magnetizing loss  Review on Parameters To find Rc and...

Three Phase Transformer Lecture 4  Review on Real Transformer Primary Copper loss Secondary Copper loss and Flux Leakage and Flux Leakage Core loss and Magnetizing loss  Review on Parameters To find Rc and XM: Open circuit (OC) test or No-load test POC VOC VOC  OC = cos −1 Rc = XM = VOC I OC I OC cos OC I OC sin  OC To find Req = R1 + R2 and Xeq = X1 +X2 Short circuit (SC) test PSC VSC VSC VSC  SC = cos−1 Z eq =  SC Req = cos SC X eq = sin  SC VSC I SC I SC I SC I SC  Review on other theories Voltage Regulation VNL − VFL VS , NL − VS , FL VR = = 100% VFL VS , FL Efficiency Pout = 100% Pout + Pcoreloss + Pcopperloss Maximum Efficiency d  max  =0 dI S Pcu = Pc  Example 3-12 A 1000kVA, 6.6kV/250V, 50Hz, single phase transformer gave the following Open Circuit and Short Circuit Test results: – Open circuit test – Voc = 250V, Ioc = 160A, Poc = 4000W ≈ Pc_rated – (OC at HV, Analysis at LV side) – Short circuit test Vsc = 400V, Isc = 150A, Psc = 6000W ≈ Pcu_rated (SC at LV, Analysis at HV side)  Example 3-12 Determine: – The approximate equivalent circuit referred to the primary – The voltage regulation when operating at full load at a power factor of 0.8 lagging – The efficiency when operating at full load at 0.8 power factor lagging  Solution (1) Refer all parameters to Primary Side a = 6600 = 26.4 250  Solution (1) (i) Open Circuit Test (Analysis at LV) POC 4000  OC = cos−1 = cos −1 = 84.26 VOC I OC 250(160) I OC 160 YOC = =  − 84.26 = 0.64 − 84.26 VOC 250 0.064 − j 0.634 POC 4000 Gc = 2 = 2 = 0.064 S V OC 250 BM = Y 2 OC − Gc2 = 0.64 2 − 0.064 2 = 0.638 S  Induced to Primary Side 1 Rc = = 15.62 Rc = a 2 Rc = 10.89 k Gc 1  Xm = = j1.57  = 1.094 k X m = a2 X m Bm  Solution (1) SC at LV, analysis at HV PSC 6000 Xeq  SC = cos−1 = cos −1 = 84.26 Req VSC I SC 400(150) VSC 400 Z SC = = 84.26 = 2.6784.26 = 0.267 + j 2.66 I SC 150  Overall Equivalent Circuit Rc = 10.89 k X m = 1.094 k R1 = 0.1335 R2 = 0.1335 Z eq = Req + jX eq = 0.267 + j 2.66 X l1 = 1.33 X l2 = 1.33  Solution (2) Ignore the Excitation branch Full Load Current Full Load Voltage No-load Voltage S FL −1 1000k I FL =  − cos pf =  − cos −1 0.8 = 152 − 36.87  VFL 6600 VP = V2, FL + I 2, FL Z eq = 6600 + 152  − 36.87  (2.67 84.26  ) = 6880 2.48  V VNL − VFL 6880 − 6600 %VR = = 100% VFL 6600 = 4.24%  Solution (3) (iii) V p2 66002 Pcoreloss = = = 4000W Rc 10.89k Pcopperloss = I FL 2 Req = 152 2  0.267 = 6169W Pout =  100% Pout + Pcoreloss + Pcopperloss 1000k  0.8 = 100% 1000k  0.8 + 4000 + 6169 = 98.7% To be continued… Why 3φ? Why not 4, 5 or 6 phase? Almost all power generation and distribution systems are 3 phase; therefore, we need three phase transformers 3–phase is chosen because it is less complex, and hence less costly than the higher phase order systems Advantages – Constant power delivery and it leads to three phase motors in machinery running more smoothly – Easier motor wiring ( three-phase induction motors don’t required brushes and startup mechanism.) – Less wires required for connecting three phase equipment since the current is divided among the three phases  Three Phase Transformers Two ways to construct 3φ transformer – A three phase bank transformer by connecting each phase individually – A three phase transformer by wrapping three sets of windings on a common core Three Single Phase Transformers Three Phase Transformers 3φ Transformers type Core type Shell type  Evolution of Core Type Three Phase Transformers If the transformer is balanced and induced Emf’s are sinusoidal, the magnetic field in each phase is sinusoidal. The three legs are merged and the total flux in the combined leg is zero. Then the center leg will be eliminated. However, this is not convenient in terms of stacking and laminations  Evolution of Shell Type Three Phase Transformers It was started with stacking three single phase, shell type units. The winding direction of the centre unit b is made opposite to that of units a and c. If the system is balanced with the sequence abc, the fluxes will also be balanced with angle 120 degrees.  3φ Connections Delta Connections: – A delta system is a good short-distance distribution system. – It is used for neighbourhood and small commercial loads close to the supplying substation. – The voltage would be the same between any two wires   V = V L Wye Connections: – Good protection circuitry as it is connected in 4 wires with neutral line. – The voltage between any two power conductors will always be 1.732 (√3) times the voltage between the neutral and any one of the power phase conductors. VLY = 3VY To be continued…  4 possible transformer combinations Wye to Wye (Yy) – Usage: rare, seldom use because it causes harmonics and balancing problems; Used in solving equivalent circuit Wye to Delta (Yd) – Usage: high voltage transmissions; use to step down from HV to LV Delta to Wye (Dy) – Usage: most common, commercial and industrial; Use to step up from LV to HV Delta to Delta (Dd) – Usage : industrial applications  The key of analyzing 3φ transformer Draw the per-phase equivalent circuit Apply the same techniques for single phase transformer ( equivalent circuit, VR, η, impedances) Watch out for the factor of √3 and the ±30° phase shift!!  V = V L  I L = 3I V = 3V L Y Y I L = I Y Y 3 φ transformers  Y-Y connection (1) VL , P = 3V , P VL , P 3V , P n= = =a VL , S 3V , S  Y-Y connection (2) Two serious problems – Phase voltages become severely unbalanced if the loads are unbalanced – Large third harmonics components which will damage transformer Solutions: – Solidly ground the neutrals of transformer especially in primary side – Use third winding connection in Δ to trap the 3rd harmonics  Harmonics Harmonics are voltages or currents that operate at a frequency that is an integer (whole-number) multiple of the fundamental frequency.  Y-Y connection (3) Primary Primary: VAN, VBN, VCN Secondary : Van, Vbn, Vcn  Y-Y connection (4) Balanced load Four-wire configuration  Y-Y Equations VAN VBN VCN = = =a Van Vbn Vcn I A I B IC 1 = = = I a Ib Ic a VAB = VAN − VBN = 3VAN  A + 30  VBC = VBN − VCN = 3VBN  B + 30  VCA = VCN − VAN = 3VCN  C + 30   Y-Δ connection(1) VL , P = 3V , P VL , S = V , S VL , P 3V , P n= = = 3a VL , S V , S  Y-Δ connection(2) Advantages – No problem with third harmonic voltage – More stable with unbalanced loads Problems: – The secondary line voltage is shifted by 30° relative to primary line voltage – Problems in paralleling secondary of two transformer banks Used in stepping down from HV to medium or LV – Neutral wire at HV side – Naturally decrease the secondary voltage by 1/√3 Y-Δ connection(3) Y-Δ connection(4)  Y-Δ Phasor diagram Vab lag VAB by 30 degree (Yd1)  Δ-Y connection (1) VL , P = V , P VL , S = 3V , S VL , P V , P a n= = = VL , S 3V , S 3  Δ-Y connection (1) Problem: – The secondary line voltage lags the primary by 30° Advantages are similar to Y –Δ connection Used to step up the voltage since it is naturally provides a √3 increment for the secondary voltage level Δ-Y connection (2)  Δ-Y connection (3) Vab lag VAB by 30 degree (Dy1)  Δ- Δ connection (1) VL , P = V , P VL , S = V , S VL , P V , P = =a VL , S V , S  Δ- Δ connection (1) If three-phase banks are used, one transformer can be removed for repair or maintenance while the remaining two transformers continue to function. This is known as the “open-delta” or “V“ connection with the rating reduced to 58 percent of the original bank. Used in moderate voltage systems because the windings operate at full line-to-line voltage. Δ- Δ connection (2) Δ- Δ connection (3)  Δ- Δ connection (4) VAB VBC VCA = = =a Vab Vbc Vca I AB I BC I CA I A I B I C 1 = = = = = = I ab I bc I ca I a I b I c a I A = I AB − I CA = 3I AB  A − 30 I B = I BC − I AB = 3I BC  B − 30 I C = I CA − I BC = 3I CA C − 30   Δ- Δ connection (5) Open Delta or V-V connection Two transformer operate in three phase operation Economy To be continued…  Circuit Computation Analyze circuit with “per-phase equivalent circuit” To draw the “per-phase equivalent” circuit – Draw the equivalent Y-Y connection of transformer – Draw all Δ–connected load in Y configuration (using the Y-Δ transformation) – Draw all Δ–connected sources (if any!) in Y configuration – Therefore, the equivalent Y-Y three phase circuit can be obtained. – Draw the equivalent circuit for one phase to neutral.  per-phase equivalent circuit Δ-Y transformer n= a = N 3 ,1 Δ-connected load 3 N 3 , 2 aI 3 aI 3 V a 3V a Δ-1 Δ-2  Δ-Y Transformation Convert Δ-circuit into Y-circuit Power Source  VL = Y VLN 3 Load  Z Z = Y 3  Simplified Circuit aI 3 V a 3V a Δ-Y formulae a N1 n= = 3 3N 2 OR V a V V1 3= a = V 2 V 3 a Single phase number of turns ratio; Simplified circuit and formulas  Example 3-9 Three 1φ transformers, 50kVA, 2400:240V are connected in Y-Δ in a 3φ 150kVA bank to step down the voltage at the load end. The voltage at the sending end of the feeder is 4160V line to line and its feeder’s impedance is 0.15+j1.00 Ω/phase. On their secondary sides, the transformers supply a balanced 3φ load through a feeder whose impedance is 0.0005+j0.002 Ω/phase. Find the line to line voltage at the load when the load draws the rated current from the transformers at a power factor of 0.80 lagging. ( Z eqHV,Trans = 1.42 + j1.82  )   Solution (1) Induced parameters to HV Transformer bank HV ( Z eq ,Trans = 1.42 + j1.82  )  Overall transformer diagram Feeder is the cable’s transmission Assuming load is Y-connection if the question does not mention the connection  Solution (2) Y-Δ formula 3  2400 n = 3a = = 17.32 VL , P = 3V , P 240 VL , P 3V , P VL , P = = 3a V , P = = 4160 3 = 2400V VL , S V , S 3 50k I , P = = 20.8 A at 0.8 lagging 2400  Solution (3) Per-phase equivalent circuit (referred to HV) Z f 2 = (n) 2  (0.0005 + j 0.002) = 0.15 + j 0.6   V2 = 24000 − (20.8 − 36.7  )(1.72 + j 3.42) = 24000 − (20.8 − 36.7  )(3.8363.3 ) = 2329 − 0.88  To be continued… Solution (4) a: Y-> Y OR n:Y -> Δ -> Y 3V2Y VLY2 = = 2329 = 232.9V 3a 10  Example 3-10 The three transformers of example 3-9 are connected Δ-Δ load and supplied with power through a 2400V (line to line) three phase feeder whose reactance is 0.8 Ω/phase. The equivalent reactance of each transformer referred to its HV side is 1.82 Ω. All circuit resistance are neglected. At its sending end, the feeder is connected to the secondary terminals of a three phase Y-Δ–connected transformer whose 3φ rating is 500kVA, 24000:2400V (line to line voltages). The equivalent reactance of the sending end transformers is 2.76 Ω/phase Δ referred to 2400V side. The voltage applied to the primary terminals is 24000V line to line. A three phase SC occurs at the 240V terminals of the receiving-end transformers. Compute the steady-state SC current in the 2400V feeder wires, in the primary and secondary windings of the receiving-end transformers, and at the 240-V terminals.  Solution (1) LV ,   ) ( Z eq ,Trans = j 2.76  Transformer bank with similar power rating in Example 3-9 HV ,   ) ( Z eq ,Trans = j1.82  Y-Δ transformer at input side Δ-Δ transformer at output side  Solution (2) Per phase equivalent circuit ( referred to 2400 V) side Zeq,T2 = 0.61 Ω/φ VL , P V , P = = 2400 3 = 1385V 3 j 2.76 Transform Impedance Z eq ,T 1 = jX eq ,T 1 = = j 0.92   from ZΔ to ZY- connection 3 j1.82 Z eq ,T 2 = jX eq ,T 2 = = j 0.61  3 Z f = jX f = j 0.8    Solution (3) V 1385 I HV SC = = = 594 − 90 A /  Z total j 2.33 LV I 2 I SC N a= = HV = 1 Use Δ-Δ single phase number of turns ratio I1 I SC N2 LV I SC = aI SC HV , 240V 2400 =  594 240 = 5940 A   Inrush Current An inrush of exciting current frequently occurs upon energizing a power transformer. Inrush current occurs in a power transformer due to saturation and nonlinear iron characteristic. It is eight times the rated current (8Iφ) of the excited winding even with all other windings open. Due to winding losses and magnetic losses, inrush current eventually drops to the normal value of the excitation current ( about 2-5% of rated current). Serious inrush current happens when the transformer is energized at the instant the voltage goes through zero immediately following which the polarity of the voltage is such that the flux increases in the direction of the residual flux (φ residual). Inrush current, input surge current or switch-on surge refers to the maximum, instantaneous input current drawn by an electrical device when first turned on. For example, light bulbs have high inrush currents until their filaments warm up and their resistance increases.  Inrush current d d v(t ) = Vm sin wt = =N dt dt t Vm At time t = 0,  = Φ(0) = Φr N 0 sin wtdt +  (0) V = m (1 − cos wt ) + r wN Vm Vm =− cos wt + + r wN wN 2Vm At time wt = π, max,inrush = + r wN Doubling φmax will result in an enormous Iφ = I mag  Solution to Inrush current Use Negative Temperature Coefficient (NTC) thermistor in series with the primary. NTC thermistor offers high resistance at the beginning of switching and limits the inrush current. After a short time, the NTC thermistor resistance decreases to a low value due to self heating and does not affect normal operation To be continued…  Zigzag Transformer (Interconnected Star) The transformer coils are connected as follows: – The outer coil of phase a1-a is connected to the inner coil of phase c2-N – The outer coil of phase b1-b is connected to the inner coil of phase a2-N – The outer coil of phase c1-c is connected to the inner coil of phase b2-N – The inner coils are connected together to form the neutral and our tied to ground – The outer coils are connected to phases a1,b1,c1 of the existing delta system Used to suppress triplen harmonic (3rd, 9th, 15th, etc.) or provide a neutral connection as an earthing transformer, and to obtain a phase angle shift. Rarely used for typical industrial or commercial use, because they are more expensive to construct than conventional wye-connected transformers. But zigzag connections are useful in special applications where conventional transformer connections aren't effective. Wire Configuration  Vector Groups IS: 2026 (Part 1V)-1977 gives 26 sets of connections star-star, star-delta, and star zigzag, delta-delta, delta star, delta-zigzag, zigzag star, zigzag-delta. Displacement of the low voltage winding vector varies from zero to -330° in steps of -30°, depending on the method of connections. –Digit 0 =0° that the LV phasor is in phase with the HV phasor –Digit 1 =30° lagging (LV lags HV with 30°) because rotation is anti-clockwise. –Digit 11 = 330° lagging or 30° leading (LV leads HV with 30°) –Digit 5 = 150° lagging (LV lags HV with 150°) –Digit 6 = 180° lagging (LV lags HV with 180°) The point of confusion is occurring in notation in a step-up transformer. As the IEC60076-1 standard has stated, the notation is HV-LV in sequence. Grp Primary Secondary For example, a step-up transformer with a delta-connected primary, and star-connected Yd1 Vp = VLL∟θp Vs = VLL∟θp-30˚ secondary, is not written as ‘dY11’, but ‘Yd11’. The 11 indicates the LV winding leads the HV Yd11 Vp = VLL∟θp Vs = VLL∟θp+30˚ by 30 degrees.  3φ Transformer Configuration Configuration Primary Secondary Ratio 𝑉𝑃𝑌 𝑉𝐿𝐿 𝑌 Yd11 𝑉𝑃𝑌 = 𝑉𝐿𝐿 𝑌 ∠𝜃𝑃 𝑉𝑆Δ = 𝑉𝐿𝐿 Δ ∠𝜃𝑃 + 30∘ = ∠ − 30 ∘ 𝑉𝑆Δ 𝑉𝐿𝐿 Δ 𝑉𝑃Δ 𝑉𝐿𝐿 Δ Dy11 𝑉𝑃Δ = 𝑉𝐿𝐿 Δ ∠𝜃𝑃 𝑉𝑆𝑌 = 𝑉𝐿𝐿 𝑌 ∠𝜃𝑃 + 30∘ = ∠ − 30 ∘ 𝑉𝑆𝑌 𝑉𝐿𝐿 𝑌 𝑉𝑃𝑌 𝑉𝐿𝐿 𝑌 Yd1 𝑉𝑃𝑌 = 𝑌 𝑉𝐿𝐿 ∠𝜃𝑃 𝑉𝑆Δ = Δ 𝑉𝐿𝐿 ∠𝜃𝑃 − 30∘ Δ = Δ ∠ + 30∘ 𝑉𝑆 𝑉𝐿𝐿 𝑉𝑃Δ 𝑉𝐿𝐿 Δ Dy1 𝑉𝑃Δ = 𝑉𝐿𝐿 Δ ∠𝜃𝑃 𝑉𝑆𝑌 = 𝑉𝐿𝐿 𝑌 ∠𝜃𝑃 − 30∘ = ∠ + 30 ∘ 𝑉𝑆𝑌 𝑉𝐿𝐿 𝑌 𝑌 𝑌 𝑌 ⋇ Δ Δ ⋇ 𝑽𝑌𝑃 3𝑽1𝜙 𝑛1 𝑺3𝜙 = 3𝑽1𝜙 𝑰1𝜙 = 3𝑽1𝜙 𝑰1𝜙 𝑁𝑌−∆ = Δ = Δ = 3∠ ± 30∘ 𝑽𝑆 𝑽1𝜙 𝑛2 𝑌 Δ ⋇ 𝑽1𝜙 𝑰1𝜙 𝑽Δ𝑃 Δ 𝑽1𝜙 𝑛1 1 Δ = 𝑌 𝑁∆−𝑌 = 𝑌= = ∠ ± 30 ∘ 𝑽1𝜙 𝑰1𝜙 𝑽𝑆 𝑌 3𝑽1𝜙 𝑛2 3 Transformer Vector Groups Thank you