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Chapter Two TRANSFORMERS GWCET, NAGPUR

Chapter Two TRANSFORMERS GWCET, NAGPUR. Three Phase Transformers. Almost all major generation & Distribution Systems in the world are three phase ac systems Three phase transformers play an important role in these systems Transformer for 3 phase cct.s is either:

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Chapter Two TRANSFORMERS GWCET, NAGPUR

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  1. Chapter Two TRANSFORMERS GWCET, NAGPUR

  2. Three Phase Transformers • Almost all major generation & Distribution Systems in the world are three phase ac systems • Three phase transformers play an important role in these systems • Transformer for 3 phase cct.s is either: (a) constructed from 3 single phase transformers, or (b) another approach is to employ a common core for the three sets of windings of the three phases • The construction of a single three phase transformer is the preferred today, it is lighter, smaller, cheaper and slightly more efficient • There is an advantage that each unit in the bank could be replaced individually in the event of a fault, however this does not outweigh the other advantages of combined 3 ph. unit

  3. Three Phase Transformers • How the core of compact three phase is built • φa+φb+φc=0 and central leg can be removed

  4. Three Phase Transformers • The two constructions

  5. Three Phase Transformers • 3 phase transformer connections • The windings of primary and secondary (in any construction) can be connected in either a wye (Y) or delta (Δ) • This provides a total of 4 possible connections for 3 phase transformer (if Neutral is not grounded): (a) Wye-wye Y-Y (b) Wye-delta Y-Δ (c) Delta-wye Δ-Y (d) Delta-Delta Δ-Δ

  6. Three Phase Transformers • To analyze a 3-phase transformer, each single transformer in the bank should be analyzed • Any single phase in bank behaves exactly like 1 phase transformer just studied • impedance, V.R., efficiency, & similar calculations for 3 ph. are done on per phase basis, using the same technique already used in single phase Transformer • The applications, advantages and disadvantages of each type of three phase connections will be discussed next

  7. Three Phase Transformers • WYE-WYE connection • In Y-Y connection, primary voltage on each phase is VφP=VLP/√3 • Primary phase voltage is related to secondary phase voltage by turns ratio of transformer • Phase voltage of secondary is related to Line voltage of secondary by VLS=√3 VφS • Overall the voltage ratio of transformer is: • Y-Y

  8. Three Phase Transformers • Two serious concerns on Y-Y connection 1- if loads on transformer cct. are unbalanced, voltages on phases of transformer severely unbalanced, also source is loaded in an unbalanced form 2- Third harmonic voltages can be large(there is no path for passage of third harmonic current) • Both concerns on unbalance load condition & large 3rd Harmonic voltages can be rectified as follows:

  9. Three Phase Transformers • Solidly grounding the neutrals of windings specially primary winding, this connection provide a path for 3rd harmonic current flow, produced and do not let build up of large 3rd voltages . Also provides a return path for any current imbalances in load • Adding a third winding (tertiary) connected in Δ (a) 3rd harmonic components of voltage in Δ will add up, causing a circulating current flow within winding (b) tertiary winding should be large enough to handle circulating currents(normally 1/3 of power rating of two main windings) One of these corrective techniques should be employed with Y-Y, however normally very few transformer with this type of connection is employed (others can do the same job)

  10. Three Phase Transformers • WYE-DELTA CONNECTION • VLP=√3 VφP, while : VLS= VφS • Voltage ratio of each phase : VφP/ VφS=a • VLP/ VLS= √3 VφP/ VφS= √3 a  Y-Δ • Y-Δ doesn’t have shortcomings of Y-Y regarding generation of third harmonic voltage since the Δ provide a circulating path for 3rd Harmonic • Y-Δ is more stable w.r.t. unbalanced loads, since Δ partially redistributes any imbalance that occurs • This configuration causes secondary voltage to be shifted 30◦ relative to primary voltage • If secondary of this transformer should be paralleled with secondary of another transformer without phase shift, there would be a problem

  11. Three Phase Transformers • WYE-DELTA CONNECTION

  12. Three Phase TransformersY-Δ Connection • The phase angles of secondaries must be equal if they are to be paralleled, it means that direction of phase shifts also should be the same • In figure shown here, secondary lags primary if abc phase sequence applied, • However secondary leads primary when acb phase sequence applied

  13. Three Phase TransformersΔ-Y Connection • DELTA-WYE CONNECTION • In Δ-Y primary line voltage is equal to primary phase voltage VLP=VφP, in secondary VLS=√3VφS • Line to line voltage ratio ; • VLP/ VLS = VφP/ [√3 VφS ]=a/√3  Δ-Y • This connection has the same advantages & phase shifts as Y- Δ • And Secondary voltage lags primary voltage by 30◦ with abc phase sequence

  14. Three Phase TransformersΔ- Δ Connection • DELTA-DELTA CONNECTION • In Δ-Δ connection VLP= VφP and VLS= VφS • Voltage ratio : VLP/VLS= VφP / VφS =a Δ-Δ • This configuration has no phase shift and there is no concern about unbalanced loads or harmonics

  15. Three Phase TransformersΔ- Δ Connection

  16. THREE PHASE TRANSFORMERSPER UNIT • In 3 phase, similarly a base is selected • If Sbase is for a three phase system, the per phase basis is : S1φ,base= Sbase/3 • base phase current, and impedance are: • Iφ,base= S1φ,base/ Vφ,base= Sbase /(3Vφ,base) • Zbase=(Vφ,base)²/ S1φ,base • Zbase=3(Vφ,base)²/ Sbase • Relation between line base voltage, and phase base voltage depends on connection of windings, if connected in Δ ; VL,base=Vφ,base and • if connected in Wye: VL,base= √3Vφ,base • Base line current in 3 phase transformer: • IL,base= Sbase/ (√3 VL,base)

  17. THREE PHASE TRANSFORMERSPER UNIT • A 50 kVA 13800/208 V Δ-Y distribution transformer has a resistance of 1 percent & a reactance of 7 percent per unit (a) what is transformer’s phase impedance referred to H.V. 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

  18. THREE PHASE TRANSFORMERSPER UNIT • SOLUTION • (a) Base of High voltage=13800 V, Sbase=50 kVA • Zbase=3(Vφ,base)²/Sbase=3(13800)²/50000=11426Ω • The per unit impedance of transformer is: • Zeq=0.01+j 0.07 pu • Zeq=Zeq,pu Zbase =(0.01+j0.07 pu)(11426)= 114.2 + j 800 Ω • (b)to determine V.R. of 3 phase Transformer bank, V.R. of any single transformer can be determined • V.R. =(VφP-a VφS)/ (aVφS) x 100% • Rated phase voltage on primary 13800 V, rated phase current on primary: Iφ=S/(3 Vφ) =50000/(3x13800)=1.208 A

  19. THREE PHASE TRANSFORMERSPER UNIT • Example … • Rated secondary phase voltage: 208 V/√3=120V • Referred to H.V.  V’φS=a VφS13800 V • At rated voltage & current of secondary: VφP=a VφS+Req Iφ + j Xeq Iφ = 13800/_0◦ +(114.2)(1.208/_-36.87)+(j800)(1.208)/_-36.87)= 13800+138/_-36.87+966.4/_53.13= 13800+110.4-j82.8+579.8+j773.1= 14490+j690.3= 14506/_2.73◦ V • V.R. = (VφP-a VφS )/ (a VφS ) x 100%= (14506-13800)/13800 x 100% = 5.1%

  20. THREE PHASE TRANSFORMERSPER UNIT • Example … • (c) V.R. using per unit system • output voltage 1/_0◦ & current 1/_-36.87◦ pu • VP=1/_0◦ +(0.01) (1/_-36.87◦)+(j0.07)(1/_-36.87◦)=1+0.008-j0.006+0.042+ j0.056 =1.05+j0.05=1.051/_2.73◦ • V.R.= (1.051-1.0)/1.0 x100% = 5.1%

  21. Apparent Power Rating of a Transformer • Apparent power rating & Voltage rating set current flow of windings • Current flow important as it controls I²R losses in turn control heating of coils Heating is critical, since overheating the coils reduce insulation life • Actual VA rating of a transformer may be more than a single value: In real Transformer: (a) may be a VA rating for transformer by itself, (b) another (higher) rating for transformer with forced cooling • If a transformer’s voltage reduced for any reason (i.e. operating with lower frequency than normal)then transformer VA rating must reduced by an equal amount, otherwise current exceed permissible level & cause overheating

  22. Inrush Current • This is caused by applied voltage level at energization of transformer, or due to residual flux in the transformer core • Suppose that voltage is : v(t)=VM sin(ωt+θ) V • The maximum flux reached during first half-cycle of applied voltage depends on θ • If θ=90◦ or : v(t)=VM cos(ωt) & no residual flux in core  max. flux would be : φmax=Vmax/(ωNP) • However if θ=0 the max. flux would be φmax=2Vmax/(ωNP) and is twice the steady-state flux

  23. Inrush Current • With this high maximum flux if the magnetization curve examined it shows passage of enormous magnetizing current, (looks like short circuit in part of cycle) • In these cases that θ is not 90◦ this excess current exist, therefore power system & transformer must be able to withstand these currents

  24. Transformer Nameplate • Example:

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