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L18 / Three-phase transformers

Three-Phase Transformers II

Apply phase conventions and delta–wye terminal relationships to transformer phasors.

Available37 slides
Three-phase transformer network and taps

01 / UNDERSTAND & PREDICT

Understand the model, then predict the result

Finalized lecture slides

Open / download original PDF ↗

Follow the original explanations, diagrams, derivations, and examples in slide order, then use the companion experiment below.

L18 original slide 1 of 37
L18 · 1 / 37
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Figures and page order follow the student PDF for this lecture.

The L18 bank is rated 60 MVA, 138/13.8 kV, Y–Δ. Its terminal ratio is 10, but its winding ratio is 10/√3. On a 100 MVA system base, does changing base change its physical impedance?
  • Convert winding impedance into an equivalent Y phase model.
  • Check per-unit base changes by recovering physical ohms.
  • Distinguish a physical tap, fixed voltage bases and directional complex ratios.
Three-phase transformer network and taps: baseline circuit or quantitative model illustration
Model illustration at baseline inputs. Change the parameters in the next section to explore the live diagram.

Balanced branch with phase shift

Z_{Y,L}=Z_{\Delta,L}/3,\quad Z_{Y,H}/Z_{Y,L}=k_{LL}^2

Use a positive-sequence model, retaining series resistance and leakage and neglecting excitation. The L-side physical delta becomes an equivalent Y impedance ZΔ/3. That equivalent neutral is a reference, not a grounding claim.

Bank bases and physical impedance

Z_B=V_{B,LL}^2/S_B,\quad z_{new}=z_{old}\frac{S_{B,new}}{S_{B,old}}\left(\frac{V_{B,old}}{V_{B,new}}\right)^2

Use total three-phase Sbase and line-to-line Vbase. The L18 rated voltage bases stay fixed when Sbase changes. The two compatible side bases give the same pu impedance; a delta winding has three times the equivalent Y impedance.

Directional complex tap

v_H=z_{eq}i_H+t_{HL}v_L,\quad i_L=t_{HL}^*i_H,\quad t_{HL}=\tau e^{j\theta_{HL}}

For the L18 Y–Δ abc connection δLH=−30°, so θHL=+30°. H current enters and L current leaves. Fix the voltage bases and define tHL=τexp(jθHL); then iL=tHL* iH. Increasing active H turns increases τ and reduces ideal LV voltage at fixed HV voltage.

Autotransformer power paths

S_{magnetic}/S_{terminal}=(V_H-V_L)/V_H

The source slides show a series-aiding step-down autotransformer: VH=(Ns+Nc)e per turn and VL=Nc e per turn. Power is transferred through both a direct electrical path and magnetic coupling. The common connection removes galvanic isolation; a conventional two-winding bank model cannot represent that isolation difference.

Baseline example: check each step

  1. a=(138/√3)/13.8=5.77350; rated terminal ratio is 10. On compatible rated voltage bases, ideal magnitude ratio in pu is one.
  2. z100=(0.006+j0.090)×100/60=0.010+j0.150 pu. ZB,H=138²/100=190.44 Ω.
  3. Recover ZY,H=1.9044+j28.566 Ω. On L, ZY,L=0.019044+j0.28566 Ω and ZΔ,L=0.057132+j0.85698 Ω.
  4. At no load and τ=1, vL=1∠−30° pu, corresponding to 13.8∠−30° kV line voltage with VAB,H at 0°. At τ=1.05, |VLL,L|=13.8/1.05=13.14286 kV.
  5. With loading, solve vL=(vH−z iH)/tHL and iL=tHL* iH. Input minus output complex power equals z|iH|²; inspect both balance checks.
Original slide headings for this lecture37
  1. 1Three-Phase Transformers II
  2. 2Lectureoutline
  3. 3PART 1 From single-phase to three-phase model
  4. 4Quickrecap: single-phasetransformer
  5. 5Quickrecap: reducedseries model
  6. 6Three-phaseY–Δconnection
  7. 7Seriesimpedance on Hor L
  8. 8L-sideΔ→Yequivalent
  9. 9Phasevoltages and displacement
  10. 10Voltagenormalization
  11. 11Phaseimpedance normalization
  12. 12Currentnormalization
  13. 13Powernormalization
  14. 14Assemblethe balancedY–Δbranch
  15. 15Fourconnections: networksummary
  16. 16PART 2 Bank example and checks
  17. 17Bankexample: data
  18. 18Bankexample: solutionroute
  19. 19Bankexample: voltageratios
  20. 20Bankexample: no-loadphasor
  21. 21Bankexample: changeof base
  22. 22Bankexample: recoverohms
  23. 23Bankexample: ratedcurrents
  24. 24PART 3 Autotransformer concept
  25. 25Two-windingand autotransformer
  26. 26Seriesand common sections
  27. 27Autotransformerpower paths
  28. 28Autotransformerapplications
  29. 29Three-phaseautotransformer
  30. 30PART 4 Tap ratio concept
  31. 31Tappedwinding construction
  32. 32Threedifferentratios
  33. 33Off-nominalmodel
  34. 34Tapmagnitudeand voltage
  35. 35Tap-changingequipment
  36. 36Magnitudeand phasein one model
  37. 37Summary
Cross-check the original slides

02 / EXPLORE

Change one input and explain the response

Set input-current loading to 1 and vary τ. Explain voltage drop and check power balance. Then change only Sbase: physical ohms, line voltages and currents should stay unchanged.

Advanced parameters / test readings

Preparing the model.

Active winding ratio—
System-base resistance—
System-base reactance—
LV terminal line voltage—
LV angle relative to HV—
Physical HV phase resistance—
Physical LV delta winding resistance—
Series real-power loss—

LV voltage versus H-side tap

LV voltage versus input-current loading

Current intermediate values and numerical checks
Balanced positive sequence, Y–Δ abc with δLH=−30°, fixed rated voltage bases and H-side series impedance; excitation neglected. HV voltage is fixed at 1 pu. Loading prescribes HV input-current magnitude and lagging power factor, not a constant-power LV load. Tap changes active H turns. Autotransformer construction is covered in slides and concepts, outside this two-winding numerical branch.


03 / EDIT & COMPUTE
Edit code to reproduce the model independently
Reproduce the baseline, then modify the parameter scan. The source contains reusable independent model functions; edit the current function and inspect numerical checks.







  
  
case is a snapshot of the controls when you press Run. Call solve(case) and assign the final solution to result to plot it.

  
  
  
  
Download teaching models

  
The first run needs internet access to download Python. Computation stays in your browser; the solver uses only the standard library.

  
Ready to run.

  
  
Output appears here.

  
  
  
Inspect and edit the model source (advanced)
Edit this module's function and run again. case.module selects the module; solve(case) returns values, plots, and checks. The parameter experiment keeps the original JavaScript reference for comparison.

  





04 / CHECK & EXPLAIN
Companion experiment practice and feedback
Fixed practice inputs
L18 example: 60 MVA, 138/13.8 kV, Y–Δ abc; equipment z=0.006+j0.090 pu. System base is 100 MVA with rated line-voltage bases. τ=1, no load.
Practice uses fixed baseline inputs independently of the controls. Each field displays its tolerance.


±0.001 
±0.001 pu
±0.0001 Ω



At no load with fixed voltage bases and HV voltage, what does τ=1.05 do?


Finally, explain in your own words
  1. What are the inputs, references, and main assumptions?
  2. Set input-current loading to 1 and vary τ. Explain voltage drop and check power balance. Then change only Sbase: physical ohms, line voltages and currents should stay unchanged.
  3. Did your code edit change physical parameters, the method, or representation bases? Which check helps identify that?
Passing numerical and understanding checks records this lecture’s companion practice as “practice checks passed.”









    

  




  
    
    




  
  
    

      
ECE 685 · L18
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