[PSA] Chapter 5 - Transformer Modeling and Per Unit System

5.1 Single-phase transformer model

• Two-winding transformer

• $\Phi_m$ : Mutual flux, contained within magnetic core that links all the turns of primary, secondary windings
• $\Phi_{l1}, \Phi_{l2}$ : Primary / Secondary leakage flux (Only link primary/secondary circuit itself)
• $X_1, X_2$ : Usually lowvoltage marking notation
• $H_1, H_2$ : High voltage
• $\cdot$ : indicates mutual flux components’ direction. due to current $i_1, i_2’$, tend to add the mutual flux when currents enter the dotted terminals

Ideal transformer

1. No losses
2. No leakage fluxes
3. Magnatic core has infinite permeability
   • Practical transformer is close to ideal transformer
   • 0.5% order of loss (of the transformer power rating)
   • 5% order of leakage fluxes compared to mutual flux
   • high permeability of special alloy steels

Calculations on ideal transformer

• Flux linkages : $\lambda_i = N_i \Phi_m$
• Terminal voltages : $v_i = {d\lambda_i \over dt} = N_i {d\Phi_m \over dt}$
• Voltage gain

\[{v_2 \over v_1} = {N_2 \over N_1} = n = {1 \over a}\]

• $a$ : Transformer turns ratio
• Magnetomotive Force(MMF ; 기자력) : # of turns times current passing thru that line
• $R$ : reluctance of core, works as constant factor btw MMF and flux
  • Hopkinson’s Law : $MMF = R \Phi$
• $F = N_1i_1 + N_2 i_2 = R\Phi_m = 0 \times \Phi_m$ (Can neglect Reluctance for infinite permeability)

\[{i_2' \over i_1} = -{N_1 \over N_2} = -{1\over n} = -a\]

• To simplify the equation, define $i_2 = -i_2’$

\[{i_2 \over i_1} = {N_1 \over N_2} = {1\over n} = a\]

• ABCD parameter representation of ideal transformer

\[\mathbf T_{ideal} = \begin{bmatrix}a & 0 \\ 0 & {1\over a}\end{bmatrix}\]

• cf) ABCD for series impedance / shunt admittance

\[\mathbf T_{Z} = \begin{bmatrix}1 & Z \\ 0 & 1\end{bmatrix} \ \ \mathbf T_{Y} = \begin{bmatrix}1 & 0 \\ Y & 1\end{bmatrix}\]

Implementation of physical transformer

• Considering leakage fluxes

\[\begin{aligned} \lambda_1 = \lambda_{l1} + N_1 \Phi_m \\ \lambda_2 = \lambda_{l2} + N_2 \Phi_m \end{aligned}\]

• Leakage inductance can be defined as : $\lambda_{li} = L_i {di_i \over dt}$
• Can get voltage equation

\[\begin{aligned} v_1 &= r_1i_1 + {d\lambda_1 \over dt} = r_1i_1 + L_{l1}{di_1\over dt} + N_1 {d\Phi_m \over dt} \\ v_2 &= r_2i_2+ {d\lambda_2 \over dt} = r_2i_2 + L_{l2}{di_2\over dt} + N_2 {d\Phi_m \over dt} \end{aligned}\]

• $r_1, r_2$ : Resistance of primary/secondary winding

Finite permeability

• Suppose $i_2’ = 0$ (case of open circuit) → still $i_m$ (magnetization current) flows

\[i_m = {R \Phi_m \over N_1}\]

• Can get relationship with $i_1, i_m$ and $i_2’$

\[i_1 = i_m - {N_2 \over N_1}i_2' = i_m + {N_2 \over N_1}i_2\]

• Voltage relation

\[e_1 \triangleq N_1 {d\Phi_m \over dt} = L_m {di_m\over dt}\]

where magnetizing inductance

\[L_m \triangleq {N_1^2 \over R}\]

• Neglecting phenomena of hysteresis and eddy current, (to take care of it, add parallel resistor with $L_m$

More simplified transformer model

• small value of series impedance, high value of parallel (shunt) impedance can be neglected

• Phasor diagram for transformer
• For instance, transformer supplies lagging load($I$ lags $V$), the phasor diagram would be like the figure above.
• Using the relationship, $I_m = V_2 / jX_m, I_1 = I_m + I_2, V_1 = V_2 + jX_lI_1$

5.2 Three-phase transformer connections

• Four possible connections : Wye-Wye, Wye-Delta, Delta-Wye, Delta-Delta
• Favored connection : Delta-Wye on with Wye on the high-voltage side.
• High-voltage $Y$ advantages :
  • Each winding sees $1/\sqrt 3$ portion of compared to line-line voltage.
  • Fewer turns required to provide line voltage
  • Neutral point available.
  • Ground-fault protection and easy voltage stabilization
• Low-voltage $\Delta$ advantages :
  • Provides closed path for triplen-harmonic currents (3x times natural frequency matters)
  • Blocks Zero-sequence current from passing
• $\Delta-\Delta$
  • No $30^\circ$ phase shift
  • Open-delta mode (If one transformer is lost, it still can generate voltage with same phase)

Voltage gain among $\Delta-Y$ connection

• For Wye-Wye, Delta-Delta : Only voltage gain applied (phase unchanged)
• For Wye-Delta, Delta-Wye, we get

\[V_{a'n'} = nV_{ab} = n(V_{an}-V_{bn})= \sqrt 3 ne^{j\pi/6}V_{an}\]

• We can seperate complex voltage gain $K_1 \triangleq \sqrt 3 n e^{j\pi/6}$
• For current formula, we get

\[I_a = I_{ab}-I_{ca} = n(I'_a-I'_c) = \sqrt 3 n e^{-j\pi/6}I_a' = K_1^*I'_a\]

• Thus, $I’_a = I_a / K_1^*$
• Total complex power conserves

\[S' = V_{a'n'}(I_a')^* = K_1 V_{an}({I_a \over K_1^*})^* = V_{an}I_a^* = S\]

5.3 Per-phase analysis

• Assuming balanced condition and symmetric network, there’s no zero-sequence voltage (or common mode) → Can assume neutral points are at the same potential

Per-phase diagram (Wye-Wye)

Per-phase diagram (Delta-delta)

• Two figures above are equivalent circuits
• Can prove two circuits are equivalent by open/shortening secondary circuit
• Secondaries open circuited
  1. No current flows in secondary circuit (opened)
  2. No current in primary circuit($I$), Works as voltage division ($L_l +L_m$)
  3. Using Y-Delta transformation, equivalent branch impedance is $j\omega (L_l+L_m)/3$
• Secondaries short circuited
  1. No secondary voltage → no primary voltage.
  2. Primary transformer part is short-circuited → need only to compare $L_l$
  3. Using Y-Delta transformation, equivalent leakage inductance for Wye is $L_l /3$

Per-phase Diagram (Delta-Wye)

5.4 Normal systems

• Large circuling current may occur if transformers are parallely interconnected without matching turn ratios of the transformers

Normal systems

• A system is normal if i the per-phase equivalent circuit, the product of the complex ideal transformer gains around very loop is $1$
• Can be broken down into two sub-conditions (Magnitude / Phase condition)

5.5, 5.6 Per-unit normalization (plus Three-phase quantities)

• Normalize certain quantity by dividing it into base value of quantity
• For three-phase quantities, may also be normalized by picking appropriate three-phase bases

\[S_B^{3\phi} \triangleq 3S_B\] \[V_{iB}^{ll}\triangleq \sqrt 3 V_{iB}\]

• By defining bases like this, complex power and Voltage per unit is same (Constants cancels out)

5.8 Per unit analysis of normal system

1. Pick a voltamphere base for the whole system. (e.g. 10MVA)
2. Pick one voltage base (e.g. 138kV)
3. Find all impedance bases for the different sections express all impedances in consistent pu terms
4. Draw impedance diagram for the entire system.

5.9 Regulating Transformers for Voltage and Phase Angle Control

• Add a small component of voltage, typically less than 0.1 p.u. ( to adjust line or phase voltages)

Voltage-Magnitude regulator

• By adjusting ‘tap’ values in exciting transformer, Can add additional amout of voltage
• Each node gains $\Delta V_{an}$ by passing through transformer

\[V_{a'n} = V_{an} + \Delta V_{an}\]

Phase-angle regulating transformer

\[\begin{aligned}V_{cc'} &= \rho V_{ab} \\ V_{aa'} &= \rho V_{bc} \\V_{bb'} &= \rho V_{ca} \end{aligned}\]

• Windings shown in parallel are the priamry and secondary of single-phase transformer.
• $\rho$ is a small positive number, changed by adjusting tap

\[V_{a'b'} = V_{a'a}+V_{ab}+V_{b'b}\]

• by considering the sign convention

\[\begin{aligned}V_{a'b'}&= V_{ab}+\rho (V_{ca}-V_{bc}) \\ &= V_{ab} +\rho(e^{j2\pi/3}-e^{j 4\pi/3})V_{ab} \\ &= (1+\rho j\sqrt 3) V_{ab}\end{aligned}\]

• Resultant phase diagram

Open-circuit voltage ratio mismatch

• Can make equivalent circuit with ratio $\bar n = n’ / n$, leaving pu reactance $X_1$

5.10 Autotransformers

• Primary, secondary windings are electrically connected and mutually coupled.

5.11 Transmission Line and transfomers

• Transmission line can be modeled as impedance diagram by following :