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Analysis of electric power and energy systems

Lecture 4: Transformers, power flow analysis part 2



Bertrand Cornélusse
[email protected]

What will we learn today?

  • The power transformer
  • The next part of power flow analysis: how to include transformers and transformers with tap changers

You will be able to do exercises 6.2, 6.3, 6.4 from Ned Mohan's book.

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The transformer

A (single phase) transformer is made of two magnetically coupled coils or windings. An ideal transformer is a two-port represented as

.grid[ .kol-1-2[.width-80[]] .kol-1-2[ with $$u_2 = n u_1$$ $$i_2 = -\frac{1}{n}i_1$$ ]]

In power systems, transformers are mainly used to transmit power over long distances by changing the voltage level, thus decreasing the current for a given power level. The voltage level of a synchronous generator is around 20kV.

Voltage is changed around five times between generation and load.

It is also used to measure currents and voltages, electrically isolate parts of a circuit (not the auto-transformer we will see), and match impedances.

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<iframe width="600" height="450" src="https://www.youtube.com/embed/vh_aCAHThTQ" frameborder="0" allowfullscreen></iframe>

Non-ideal model

.center.width-90[]

The ideal model is complemented by elements

  • $X_m$ that models the magnetizing inductance
  • $X_{leakage, i}$ that models the flux not captured by the core on side $i$
  • $R_{core}$ that models eddy current and hysteresis losses, i.e., losses in the iron core
  • $R_{1}$ and $R_{2}$ that model (coil) copper losses

Parameters are either given in the datasheet or obtained by open-circuit and short-circuit tests.

Laminated core to decrease losses.

The excitation current, the sum of the currents in $R_{core}$ and $X_m$, is often neglected, leading to a simpler non-ideal model, and the series impedances can be transferred from one side to the other:

.grid[ .kol-1-2[.center.width-90[]] .kol-1-2[ with $$Z_p = R_1 + jX_{leakage, 1}$$ and $$Z_s = R_2 + jX_{leakage, 2}$$]]

Per unit representation

Let's consider the rated voltages and currents on both sides of the (ideal) transformer as base values. As $$V_{s, base} = n V_{p, base} $$ and $$ I_{p, base} = n I_{s, base},$$ the MVA base is the same on both sides, and thus $$Z_{s, base} = n^2 Z_{p, base}$$

.grid[ .kol-1-2[ Hence, in per unit, the transformer can be replaced by a single impedance
$$Z_{tr} = \frac{Z_{p}}{Z_{p, base}}+\frac{Z_{s}}{Z_{s, base}}.$$ ] .kol-1-2[ .center.width-100[] ]]

Thus we have also that $$\begin{aligned}Z_{tr} &= \frac{Z_{p}+ Z_s/n^2}{Z_{p, base}} \\ &= \frac{n^2 Z_{p} + Z_s}{Z_{s, base}}\end{aligned}$$ i.e. the impedance is the same whether we see it from the primary or the secondary side, although the voltage bases differ.

Also, if the three-phase transformer is wye-delta connected, a 30° phase shift must be applied (more on this later).

Example 6.1

Consider the one-line diagram

.width-90[]

with

  • a 200 km line with $R = 0.029 \Omega/km$, $X=0.326 \Omega/km$, neglected shunt impedances
  • two transformers with a leakage reactance of $0.2 pu$ in the (500 kV, 1000 MVA) base, and losses neglected.

What is the equivalent model in a (345 kV, 100 MVA) base?

In the (500 kV, 1000 MVA) base:

  • $Z_{line, pu} = 200 \times (0.029 + j 0.326) / (500^2/1000) = 0.0232 + j 0.2608 pu$

  • hence, the total impedance between buses 1 and 2 is $$Z_{12} = 0.0232 + j 0.2608 + 2 * j 0.2pu = 0.0232 + j 0.6608 pu $$

In the (345 kV, 100 MVA) base:

  • the pu value of the impedance is the same in the (500 kV, 1000 MVA) and (345 kV, 1000 MVA) bases,

  • since we can transfer the impedance from one side of each transformer to the other, cf. a previous remark

  • if we now change the MVA base to 100 MVA, $$Z_{12} = (0.0232 + j 0.6608) \times (100/1000) pu = 0.00232 + j 0.06608 pu$$ since the base impedance is proportional to the inverse of the MVA base.

Efficiency

The efficiency expressed in % is $$100 \times \frac{P_{output}}{P_{input}} = 100 \times \left(1 - \frac{P_{losses}}{P_{input}}\right) $$

  • maximal when loaded such that copper losses = iron losses (cancel derivative of efficiency w.r.t current)
  • Around 99.5 % in large power transformers at full load.

Tap changers

  • Some transformers are equipped with a system allowing to change the $1:n$ ratio
  • The ability to change the tap under load is called load tap changer (LTC) or on-load tap changer (OLTC)
  • This is mainly used for voltage control
  • It is usually implemented using auto-transformers
  • We will see later on how to include this in the power flow analysis

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<iframe width="600" height="450" src="https://www.youtube.com/embed/R_NxRDXOEFk" frameborder="0" allowfullscreen></iframe>

Auto-transformers

The two windings (of the same phase) are connected in series, without galvanic insulation. They are commonly used when the ratio is limited.

Advantages:

  • Physically smaller
  • less costly (less copper)
  • higher efficiency
  • easy to implement tap changes
  • "solid" earth grounding

Disadvantages:

  • no electrical insulation
  • higher short circuit current
  • full voltage at secondary if it breaks (in case of a step down)

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<iframe width="600" height="450" src="https://www.youtube.com/embed/lltVwhoPvh0" frameborder="0" allowfullscreen></iframe>

Phase shift in delta-star transformers

The star part has $n$ times the number of turns of the delta part (primary side).

Let's reason on phase $a$,

  • Voltage $\bar{V}_{a,s}$ is on the same core as $\bar{V}_{AC,p} = \sqrt{3}\bar{V}_{a,p} \angle{-30^\circ}$ where $\bar{V}_{a,p}$ is the (virtual) phase-neutral voltage on the primary side.
  • Since $\bar{V}_{a,s} = n \bar{V}_{AC,p}$, $\bar{V}_{a,s} = n\sqrt{3} \bar{V}_{a,p} \angle{-30^\circ}$

.center.width-50[]

We gain a $\sqrt{3}$ factor in the amplification, but a lagging phase shift of 30°.

The same reasoning holds for phases $b$ and $c$.

Power flow regulation by phase shifting

We have seen in lecture 2 that active power flows are dictated by the voltage magnitudes but also the sine of the angle difference between buses: .grid[ .kol-1-2[ .center.width-100[]] .kol-1-2[ $$\begin{aligned} S_r &= \bar{V}_r\bar{I}^* = V_r \left(\frac{V_s \angle -\delta - V_r}{-jX}\right) \\ &= \frac{V_s V_r \sin \delta }{X} +j \frac{V_s V_r \cos \delta - V^2_r}{X} \end{aligned}$$ $\delta$ is the angle between $\bar{V}_r$ and $\bar{V}_s$]]

If we have a device that can generate an adjustable phase shift, we can control the power flows. This is the purpose of phase-shifting transformers.

In practice phase shifting is achieved by "combining the signal with a fraction of itself shifted by 90°". For the details of how it is implemented or modeled, see

Example: phase shifting transformers on the borders of Belgium

.center.width-50[]

380/380 kV: in series with:

  1. line Zandvliet (B) - Borssele (NL) and Zandvliet (B) - Geertruidenberg (NL)
  2. line Meerhout (B) - Maasbracht (NL)
  3. line Gramme (B) - Maasbracht (NL)
  • nominal power 3VN Imax = 1400 MVA
  • phase shift adjustment: 35 positions, +17/-17 × 1.5° (at no load)

.footnote[From ELEC0014.]

.center.width-50[]

220/150 kV :

  • in series with the Chooz (F) - Monceau (B) line nominal power: 400 MVA
  • in-phase adjustment: 21 positions, +10/-10 × 1.5 %
  • quadrature adjustment: 21 positions, +10/-10 × 1.2°

.footnote[From ELEC0014.]

Remarks

In three-phase operation,

  • either there are three separate single-phase transformers (easier to fix when there is a problem on a phase, more modular)
  • or a three-phase transformer, that is a single core with three auto-transformers on it, cf. the video at the beginning of this presentation (cheaper, lighter core and less copper).

Some transformers called three-winding transformers are equipped with a third winding (not to be confused with a three-phase transformer) that is used for auxiliary purposes (feeding auxiliary devices e.g., fans, providing reactive power support, ...).

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Transformers in the power flow analysis

Transformer without regulation

A transformer, in the per-unit representation, can thus be represented

  • as a two-port if the shunt admittance is considered
  • as a simple series leakage impedance if the shunt admittance is neglected

Representing taps and phase shifts

Let $Y_l$ be the leakage admittance and $t$ be the off-nominal turns ratio: .grid[ .kol-1-2[

  • if $ 0 < t \leq 1$, this corresponds to a simple tap-changer
  • if $ 0 < |t| \leq 1$ but is complex, then this is a phase-shifter ($\angle{t} &lt; \pi/2$)] .kol-1-2[.center.width-90[]]]

We have $$\bar{I}_1 = \left(\bar{V}_1 - \frac{\bar{V}_2}{t}\right) Y_l$$ and since $\frac{\bar{V}_2}{t} \bar{I}_1^\star = -\bar{V}_2 \bar{I}_2^\star$ by energy conservation $$\bar{I}_2 = - \frac{\bar{I}_1}{t^\star} = - \bar{V}_1 \frac{Y_l}{t^\star} + \bar{V}_2 \frac{Y_l}{|t|^2}$$

Thus tap and phase shift can be represented by the admittance matrix

$$\left[\begin{array}{c} \bar{I}_1 \\ \bar{I}_2\end{array}\right] = \left[\begin{array}{c c} Y_l & -\frac{Y_l}{t} \\ -\frac{Y_l}{t^\star} & \frac{Y_l}{|t|^2} \end{array}\right] \left[\begin{array}{c} \bar{V}_1 \\ \bar{V}_2\end{array}\right]$$

  • if $ 0 < t \leq 1$, this can be represented as a $\pi$ two-port
  • if $ 0 < |t| \leq 1$ but is complex, this is not the case

In the power flow analysis, you must pay attention to this when constructing the system-wide admittance matrix.

Pandapower example

See python notebook / video recording.

References

  • Mohan, Ned. Electric power systems: a first course. John Wiley & Sons, 2012. Chapter 6.
  • Weedy, Birron Mathew, et al. Electric power systems. John Wiley & Sons, 2012. Section 3.8, Section 5.7.
  • Course notes of ELEC0014 by Pr. Thierry Van Cutsem.

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The end.