Computational Technique

The standard definition of complex power is $S = VI^* = P + jQ$. With this convention, a lagging (inductive) load absorbs positive $Q$. Using $S = V^*I$ gives the same $P$ but reverses the sign of $Q$.

Derivation

Let $V = |V|\angle\theta_v$ and $I = |I|\angle\theta_i$, with $\phi = \theta_v - \theta_i$ (positive for a lagging current).

$$\begin{aligned} VI^* &= |V||I|\angle(\theta_v-\theta_i) = |V||I|\cos\phi + j|V||I|\sin\phi\\ V^*I &= |V||I|\angle(\theta_i-\theta_v) = |V||I|\cos\phi - j|V||I|\sin\phi \end{aligned}$$

So $V^*I = (VI^*)^* = P - jQ$.

Phasor diagrams (inductive load, current lags)

 S = V I*                 S = V* I
      Q>0  S                 V
     ^    /                 /
     |   / phi             / phi  (V ref)
     |  /                 /------> P
     | /                   \
     |/-------> P           \  S
                             v  Q<0
 Inductive load:           Inductive load:
 Q positive (absorbs)      Q appears negative

Consequences of using $S = V^*I$

• The real power $P$ is unchanged, since $\cos\phi$ is even.
• The sign of $Q$ is reversed: an inductor (lagging current) would appear to supply reactive power, and a capacitor to absorb it.
• This contradicts the standard convention (IEEE/IEC) that inductive loads absorb VAr. Generator, motor and load data would be read wrongly.
• The power factor angle and the power triangle are flipped, so lagging and leading would be confused.
• In load flow and compensation studies, capacitor banks would be sized in the wrong direction.

Example: Take $V = 100\angle0^\circ$ V and $I = 10\angle-30^\circ$ A (inductive load).

• $VI^* = 1000\angle30^\circ = 866 + j500$: the load absorbs 500 var, which is correct.
• $V^*I = 1000\angle-30^\circ = 866 - j500$: this wrongly shows the inductive load as a source of 500 var.

$S = VI^*$ is used so that inductive Q is positive.

The statement is true. The per-unit impedance of a transformer is the same on both sides, provided the base voltages on the two sides are in the ratio of the transformer turns (with the same VA base).

Proof

Consider a transformer of rating $S$ (VA), with voltages $V_1$ (LV) and $V_2$ (HV) and turns ratio $a = V_1/V_2 = N_1/N_2$. Choose:

• the same base VA, $S_b$, on both sides
• base voltages $V_{b1} = V_1$ and $V_{b2} = V_2$, so that $V_{b1}/V_{b2} = a$

Base impedances:

$$Z_{b1} = \frac{V_{b1}^2}{S_b},\qquad Z_{b2} = \frac{V_{b2}^2}{S_b} \;\Rightarrow\; \frac{Z_{b1}}{Z_{b2}} = a^2$$

Let $Z_1$ be the equivalent impedance referred to the LV side. Referred to the HV side, it becomes:

$$Z_2 = Z_1\left(\frac{N_2}{N_1}\right)^2 = \frac{Z_1}{a^2}$$

Per-unit values:

$$Z_{1,pu} = \frac{Z_1}{Z_{b1}},\qquad Z_{2,pu} = \frac{Z_2}{Z_{b2}} = \frac{Z_1/a^2}{Z_{b1}/a^2} = \frac{Z_1}{Z_{b1}} = Z_{1,pu}$$

So $Z_{pu}$ is the same whichever side it is referred to.

Numerical check

Take a 10 kVA, 200/400 V transformer with $Z = 0.4\ \Omega$ referred to the LV side.

• LV side: $Z_b = 200^2/10000 = 4\ \Omega$, so $Z_{pu} = 0.4/4 = 0.1$.
• HV side: $Z = 0.4\times(400/200)^2 = 1.6\ \Omega$ and $Z_b = 400^2/10000 = 16\ \Omega$, so $Z_{pu} = 1.6/16 = 0.1$.

Significance

The ideal transformer disappears from the per-unit equivalent circuit, leaving only a series impedance. Manufacturers can therefore quote a single % or p.u. impedance without naming a side. The result holds only when the bases follow the turns ratio. If they do not, an off-nominal tap must be included.

In the per-unit system, every quantity (voltage, current, power, impedance) is expressed as a fraction of a chosen base value of the same unit:

$$\text{per-unit value} = \frac{\text{actual value}}{\text{base value}}$$

Usually the base MVA ($S_b$) and the base kV ($V_b$) are chosen. The other bases follow from them:

$$I_b = \frac{S_b}{\sqrt3 V_b},\qquad Z_b = \frac{V_b^2}{S_b}$$

For example, at 100 MVA and 132 kV, $Z_b = 174.24\ \Omega$, so a 50 Ω line has $Z = 0.287$ p.u.

Advantages

1. Transformers disappear: The p.u. impedance of a transformer is the same on both sides, so the ideal transformers drop out of the equivalent circuit.
2. Compact ranges: The p.u. impedances of machines and transformers of similar type lie in a narrow range whatever their size. This makes it easy to check data and to estimate missing values.
3. Manufacturers' data: Impedances are given in % or p.u. on the equipment's own rating, so they can be used directly after a simple change of base.
4. Three-phase and single-phase are the same: In p.u. the line and phase values are equal and the factor $\sqrt3$ disappears from the calculations.
5. Easier hand and computer calculation: The numbers are of order 1, so errors are easy to spot. Load flow and fault programs use p.u. throughout.
6. Meaningful comparison: A voltage of 0.95 p.u. immediately shows a 5 % drop at any voltage level.

Advantages of the per-unit system

1. The p.u. impedance of a transformer is the same on the HV and LV sides, so ideal transformers vanish from the equivalent circuit.
2. The p.u. impedances of similar equipment lie in a narrow band, whatever the rating. This makes it easy to check data and to assume typical values.
3. Manufacturers give impedances in % or p.u. on the rating of the equipment, and these can be used directly.
4. Line and phase quantities have the same p.u. value, and $\sqrt3$ factors disappear. Single-phase and three-phase calculations look alike.
5. The numbers are near 1, which reduces arithmetic errors. It is easy to see abnormal conditions, for example $V = 0.9$ p.u.
6. It is ideal for computer studies (load flow, fault and stability analysis).

Choosing base power and base voltages

1. Base power ($S_b$, MVA): Choose one value for the whole system. It is usually the rating of the largest machine or a round figure such as 100 MVA.
2. Base voltage ($V_b$, kV): Choose it in one section, usually the generator or line section, often equal to that section's rated voltage.
3. Other sections: Carry the base voltage through each transformer by its line-to-line voltage ratio:

$$V_{b,new} = V_{b,old}\times\frac{V_{rated,\ new\ side}}{V_{rated,\ old\ side}}$$

For single-phase units in a Y–Δ bank, use the line voltage ratio ($\sqrt3 V$ on the Y side).

4. Derived bases:

$$I_b = \frac{S_b}{\sqrt3V_b},\qquad Z_b = \frac{V_b^2}{S_b}$$

5. Change of base: Convert each equipment value to the common base:

$$Z_{pu,new} = Z_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{b,old}}{V_{b,new}}\right)^2$$

Example: Take $S_b = 100$ MVA and 11 kV in the generator circuit, with an 11/132 kV transformer. The line base is 132 kV and $Z_b = 132^2/100 = 174.24\ \Omega$.

Complex power $S$ is the product of the voltage phasor and the conjugate of the current phasor. Its real part is the active power and its imaginary part is the reactive power:

$$S = VI^* = |V||I|\angle\phi = P + jQ$$

where:

• $P = |V||I|\cos\phi$ (W)
• $Q = |V||I|\sin\phi$ (var)
• $|S| = |V||I|$ (VA) is the apparent power
• $\phi = \theta_v - \theta_i$

For a three-phase system, $S_{3\phi} = 3V_{ph}I_{ph}^* = \sqrt3 V_LI_L\angle\phi$.

Power triangle

        S (VA)
       /|
      / |  Q (var)
     /phi|
    /____|
      P (W)

Significance in circuit and power system analysis

• One complex number gives both $P$ and $Q$, together with their signs, so power flow is described completely.
• Sign convention: With $S = VI^*$, an inductive load absorbs $+Q$ and a capacitor absorbs $-Q$ (that is, it supplies Q). A source delivers power when $S$, computed with current leaving its + terminal, is positive.
• Conservation: The sum of complex power over all elements is zero (Tellegen's theorem), so $\sum P$ and $\sum Q$ each balance. This is the basis of load flow equations.
• It gives the power factor directly, $\cos\phi = P/|S|$, and so the size of the compensation needed.
• Equipment such as generators, transformers and cables is rated in VA (that is, $|S|$), because heating depends on $|I|$.
• With $S = I^2Z = |V|^2/Z^*$ it gives the losses and the reactive drop in lines directly.

Example: If $V = 230\angle0^\circ$ V and $I = 10\angle-36.87^\circ$ A, then $S = 2300\angle36.87^\circ = 1840 + j1380$ VA. The load absorbs 1840 W and 1380 var at a power factor of 0.8 lagging.

Use $S = EI^*$. The current $I_{na}$ flows from n to a inside the source, so it leaves the positive terminal a. A positive $P$ or $Q$ therefore means the source is delivering that power.

Simplify the voltage

$$E_{an} = -120\angle210^\circ = 120\angle(210^\circ - 180^\circ) = 120\angle30^\circ\ \text{V}$$

Complex power

$$\begin{aligned} S &= E_{an}I_{na}^* = (120\angle30^\circ)(10\angle-60^\circ)\\ &= 1200\angle-30^\circ\\ &= 1200(\cos30^\circ - j\sin30^\circ)\\ &= 1039.2 - j600\ \text{VA} \end{aligned}$$

Interpretation

The current leads the voltage by 30°, so the source sees a leading (capacitive) load. That load returns reactive power to the source.

Answer: $P = 1039.2$ W, delivered by the source. $Q = 600$ var, absorbed by the source (delivered $Q = -600$ var).

(a) Impedance diagram (no values)

Each generator is shown as an EMF behind its internal impedance. Each transformer is shown as series leakage impedance with a shunt magnetising branch. The line is a series $R + jX$ (short line), and the load is an impedance.

  G1        T1                  T2
 (~)-Zg1-+-Zt1-+--R+jX line--+-Zt2-+-- Load
         |     |             |     |   Z_L
 (~)-Zg2-+    Ym             Ym     |
  G2     |     |             |     |
 ref ----+-----+-------------+-----+---

(b) Selecting the bases

• $S_b = 5000$ VA for the whole system.
• Generator section: $V_b = 250$ V.
• Line section: through T1 (250/800 V), $V_b = 250\times800/250 = 800$ V.
• Load section: through T2 (rated 1000 V on the line side), $V_b = 800\times500/1000 = 400$ V.

Per-unit values

Use $Z_{new} = Z_{old}\,\dfrac{S_{b,new}}{S_{old}}\left(\dfrac{V_{old}}{V_{b,new}}\right)^2$.

$$\begin{aligned} G_1&: j0.2\times\frac{5000}{2000}\times1 = j0.5\\ G_2&: j0.3\times\frac{5000}{2000} = j0.75\\ T_1&: j0.2\times\frac{5000}{4000}\times\left(\frac{250}{250}\right)^2 = j0.25 \end{aligned}$$

Line: the base impedance is

$$Z_b = \frac{800^2}{5000} = 128\ \Omega$$ $$Z_{line} = \frac{40 + j150}{128} = 0.3125 + j1.1719\ \text{p.u.}$$

T2 (rated 1000 V, while the base on its side is 800 V):

$$j0.06\times\frac{5000}{8000}\times\left(\frac{1000}{800}\right)^2 = j0.0586$$

Reactance (per-unit) diagram

  j0.5         j0.25   0.3125+j1.172   j0.0586
 (G1)--+------/\/\---------/\/\--------/\/\----Load
       |                                       |
 (G2)--+ j0.75                                 |
       |                                       |
 ------+------------ reference ----------------+

The magnetising branches are neglected in the reactance diagram. The load impedance is not given, so it is shown as a block. Its p.u. value would be $Z_L/Z_b$ with $Z_b = 400^2/5000 = 32\ \Omega$.

Choice of bases

• $S_b = 90$ MVA and $V_b = 11$ kV in the generator circuit.
• Line: through T1 (10/132 kV), $V_b = 11\times\dfrac{132}{10} = 145.2$ kV.
• T2 is a Y–Δ bank of three single-phase 66/10 kV units. Its line-voltage ratio is $\sqrt3\times66 = 114.32$ kV / 10 kV, and its three-phase rating is $3\times30 = 90$ MVA.
• Motor circuit: $V_b = 145.2\times\dfrac{10}{114.32} = 12.70$ kV.

Per-unit reactances

Use $X_{new} = X_{old}\,\dfrac{S_{b}}{S_{rated}}\left(\dfrac{V_{rated}}{V_b}\right)^2$.

$$\begin{aligned} G&: 0.15\ \text{(same base)}\\ T_1&: 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\\ \text{Line}&: Z_b = \frac{145.2^2}{90} = 234.26\ \Omega,\quad X = \frac{100}{234.26} = 0.4269\\ T_2&: 0.05\times\frac{90}{90}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\\ M_1&: 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\\ M_2&: 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789 \end{aligned}$$

(T2 gives the same value on its HV side: $0.05\times(114.32/145.2)^2 = 0.0310$.)

Reactance diagram

  j0.15    j0.0446   j0.4269   j0.031
 +-/\/\-----/\/\------/\/\-----/\/\---+-----+
 |                                     |     |
(Eg)                              j0.2231 j0.2789
 |                                     |     |
 |                                   (Em1) (Em2)
 +------------- reference --------------+-----+

The transformer winding connections (Y, Δ, earthing) do not change the positive-sequence p.u. reactances. They only introduce a phase shift, which is ignored in the reactance diagram.

Significance of the reactance diagram

A reactance diagram is the per-phase equivalent circuit of the system, drawn from the single line diagram, in which:

• resistances, magnetising branches and line charging are neglected
• all values are in p.u. on a common base

It is used for short-circuit (fault) calculations and stability studies, where the reactances dominate. It gives a simple network that can be reduced quickly.

Base values

• $S_b = 100$ MVA, and $V_b = 13.8$ kV at the generator (bus 1).

Both paths give 11 kV at bus 4, so the bases are consistent.

Per-unit values

$$\begin{aligned} X_G &= 0.18\times\frac{100}{90} = 0.2\\ X_{T1} = X_{T2} = X_{T3} = X_{T4} &= 0.1\times\frac{100}{50} = 0.2\\ X_{L1} &= \frac{50}{484} = 0.1033\\ X_{L2} &= \frac{70}{174.24} = 0.4017\\ X_M &= 0.2\times\frac{100}{80}\times\left(\frac{10.45}{11}\right)^2 = 0.2256 \end{aligned}$$

Load: $S = 57/100 = 0.57$ p.u. at a power factor of 0.8 lagging, and $V = 10.45/11 = 0.95$ p.u. As a series impedance:

$$Z_L = \frac{|V|^2}{S^*} = \frac{0.95^2}{0.57}\angle36.87^\circ = 1.5833\angle36.87^\circ = 1.2667 + j0.95\ \text{p.u.}$$

Reactance diagram

               T1     L1      T2
          +--j0.2--j0.1033--j0.2--+
  j0.2    |                       |
 +-/\/\---+ bus1            bus4  +---+------+
 |        |                       |   |      |
(Eg)      +--j0.2--j0.4017--j0.2--+ j0.2256 1.2667
 |             T3     L2      T4     |   +j0.95
 |                                  (Em)  |
 +------------- reference -----------+----+

Base values

• $S_b = 100$ MVA, and $V_b = 132$ kV in the transmission circuit, so $Z_b = 132^2/100 = 174.24\ \Omega$.
• G1 side, through T1 (12.2/161 kV): $V_b = 132\times\dfrac{12.2}{161} = 10.0025$ kV.
• G2 side, through T2 (161/13.8 kV): $V_b = 132\times\dfrac{13.8}{161} = 11.3143$ kV.

Per-unit impedances

Use $Z_{new} = Z_{old}\dfrac{S_b}{S_{rated}}\left(\dfrac{V_{rated}}{V_b}\right)^2$.

$$\begin{aligned} G_1&: 0.15\times\frac{100}{50}\left(\frac{13.8}{10.0025}\right)^2 = j0.5710\\ T_1&: 0.1\times\frac{100}{80}\left(\frac{161}{132}\right)^2 = j0.1860\\ T_2&: 0.1\times\frac{100}{40}\left(\frac{161}{132}\right)^2 = j0.3719\\ G_2&: 0.15\times\frac{100}{20}\left(\frac{13.8}{11.3143}\right)^2 = j1.1157 \end{aligned}$$

Lines:

$$\begin{aligned} Z_{40+j160} &= \frac{40 + j160}{174.24} = 0.2296 + j0.9183\\ Z_{20+j80} &= \frac{20 + j80}{174.24} = 0.1148 + j0.4591\ \text{(each section)} \end{aligned}$$

Load (50 MVA, 0.8 lagging, 154 kV): $V = 154/132 = 1.1667$ p.u. and $S = 0.5$ p.u.

$$Z_L = \frac{|V|^2}{S^*} = \frac{1.1667^2}{0.5}\angle36.87^\circ = 2.7222\angle36.87^\circ = 2.1778 + j1.6333\ \text{p.u.}$$

Impedance diagram

 j0.571  j0.186      0.2296+j0.9183      j0.3719  j1.1157
+-/\/\----/\/\--+-------/\/\/\/\-------+--/\/\----/\/\--+
|               |                      |               |
(E1)            +-0.1148+j0.4591-+-0.1148+j0.4591-+   (E2)
|                                |                     |
|                        2.1778+j1.6333 (load)         |
|                                |                     |
+---------------- reference -----+---------------------+

Assumption: 11 kV is the line-to-line EMF. The per-phase EMFs are therefore $E_1 = 6.351\angle0^\circ$ kV and $E_2 = 6.351\angle40^\circ$ kV. The current $I$ is taken as flowing from machine 1 to machine 2 through $Z = j15\ \Omega$.

Current

$$\begin{aligned} I &= \frac{E_1 - E_2}{jX} = \frac{6351\angle0^\circ - 6351\angle40^\circ}{j15}\\ &= 289.6\angle-160^\circ\ \text{A} \end{aligned}$$

Complex power (three-phase)

Machine 1, with power taken out of its terminal:

$$S_1 = 3E_1I^* = -5.185 + j1.887\ \text{MVA}$$

Machine 2, with power taken into it (since $I$ enters it):

$$S_2 = 3E_2I^* = -5.185 - j1.887\ \text{MVA}$$

Check with the standard formulas:

$$\begin{aligned} P &= \frac{E_1E_2\sin\delta}{X} = \frac{11\times11\times\sin40^\circ}{15} = 5.185\ \text{MW}\\ Q &= \frac{E^2(1-\cos\delta)}{X} = \frac{121(1-\cos40^\circ)}{15} = 1.887\ \text{Mvar} \end{aligned}$$

(i) Real power

• Machine 1: $P_1 = -5.185$ MW out of it, so it consumes 5.185 MW and acts as a motor.
• Machine 2: $P_{2,in} = -5.185$ MW, so it generates 5.185 MW and acts as a generator.

Machine 2's EMF leads by 40°, so real power flows from 2 to 1. The line is lossless ($R = 0$), so the two powers are equal.

(ii) Reactive power

• Machine 1: $Q_1 = +1.887$ Mvar out of it, so it delivers 1.887 Mvar.
• Machine 2: $Q_{2,in} = -1.887$ Mvar, so it also delivers 1.887 Mvar.

Both machines supply reactive power to the line, which absorbs:

$$3|I|^2X = 3\times289.6^2\times15 = 3.774\ \text{Mvar}$$

Answer: Machine 1 is a motor consuming 5.185 MW, and machine 2 is a generator producing 5.185 MW. Each machine supplies 1.887 Mvar, a total of 3.774 Mvar absorbed by the line reactance.

Single line diagram

  G         T1 (3 x 1-ph)     line       T2 (3 x 1-ph)
 (~)--||--[ D | Y ]----------j180----[ Y | Y ]----> Load
 20 MVA     5 MVA each                  5 MVA each   15 MVA
 11 kV      11/127.02 kV                127.02/      20 kV
 X"=10%     X=15%                       12.702 kV    0.9 lag
                                        X=20%

Transformer bank ratings

• T1 (Δ–Y): $3\times5 = 15$ MVA. Line voltages: 11 kV / $\sqrt3\times127.02 = 220$ kV.
• T2 (Y–Y): 15 MVA, 220 kV / $\sqrt3\times12.702 = 22$ kV.

Base values

Assumption: the 15 kV base is taken in the generator circuit, with $S_b = 25$ MVA.

Per-unit reactances

$$\begin{aligned} X_G &= 0.10\times\frac{25}{20}\left(\frac{11}{15}\right)^2 = 0.0672\\ X_{T1} &= 0.15\times\frac{25}{15}\left(\frac{11}{15}\right)^2 = 0.1344\\ X_{line} &= \frac{180}{3600} = 0.05\\ X_{T2} &= 0.20\times\frac{25}{15}\left(\frac{220}{300}\right)^2 = 0.1793 \end{aligned}$$

Load

15 MVA at 20 kV and 0.9 lagging: $S = 15/25 = 0.6$ p.u., $V = 20/30 = 0.6667$ p.u.

$$S = 0.6\angle25.84^\circ = 0.54 + j0.2615\ \text{p.u.}$$

As a series impedance:

$$Z_L = \frac{|V|^2}{S^*} = \frac{0.6667^2}{0.6}\angle25.84^\circ = 0.7407\angle25.84^\circ = 0.6667 + j0.3229\ \text{p.u.}$$

Reactance diagram

  j0.0672   j0.1344    j0.05    j0.1793
 +-/\/\------/\/\------/\/\------/\/\----+
 |                                       |
(Eg)                              0.6667 + j0.3229
 |                                     (load)
 +----------------- reference -----------+

A single line diagram (SLD), or one-line diagram, shows a balanced three-phase power system using one line to represent all three phases. Standard symbols show the generators, transformers, lines, breakers, buses and loads, together with their ratings and connections.

Typical SLD

  G1                                  M
 (~)--[x]--8|8--+===== line =====+--8|8--[x]--(M)
       CB   T1  |                |   T2
               Bus 1           Bus 2
                |                |
               Load            Load
  (~) generator   8|8 transformer   [x] breaker

Information shown

• Ratings: MVA, kV and % reactance of machines and transformers
• Transformer winding connections (Y, Δ) and neutral earthing
• Line impedances, buses, and the location of breakers, CTs and PTs
• The points where loads and generators connect

Significance in power system analysis

1. Simplicity: A balanced three-phase system is solved on a per-phase basis, so one line gives all the needed information without clutter.
2. Starting point for analysis: The impedance and reactance diagrams, the bus admittance matrix, load flow, fault studies and stability studies are all built from the SLD.
3. Planning and operation: It shows the system layout clearly to planners, operators and protection engineers.
4. Protection coordination: It shows the locations of breakers and relays and the fault paths.
5. Data in one place: All the ratings needed for p.u. calculation are on one drawing.

A simple example is a generator, a step-up transformer, a transmission line, a step-down transformer and a load in series, each shown by its symbol and rating.

Base values

• $S_b = 30$ MVA, and $V_b = 6.6$ kV in the G1 circuit.
• Line section (through T1, 6.6/115 kV): $V_b = 115$ kV, so $Z_b = 115^2/30 = 440.83\ \Omega$.
• G2 circuit (through T2, 115/6.6 kV): $V_b = 6.6$ kV.
• T3 is a Y–Y bank of single-phase 69/6.9 kV units. Its line ratio is $\sqrt3\times69 = 119.5$ kV / $\sqrt3\times6.9 = 11.95$ kV, and its rating is $3\times10 = 30$ MVA.
• G3 circuit: $V_b = 115\times\dfrac{11.95}{119.5} = 11.5$ kV.

Per-unit reactances

$$\begin{aligned} G_1&: 0.2\times\frac{30}{25} = 0.24\\ G_2&: 0.15\times\frac{30}{15} = 0.30\\ G_3&: 0.15\times\frac{30}{30}\left(\frac{13.2}{11.5}\right)^2 = 0.1976\\ T_1&: 0.1\times\frac{30}{30} = 0.10\\ T_2&: 0.1\times\frac{30}{15} = 0.20\\ T_3&: 0.1\times\frac{30}{30}\left(\frac{119.5}{115}\right)^2 = 0.1080\\ \text{Line } j120&: \frac{120}{440.83} = 0.2722\\ \text{Line } j90&: \frac{90}{440.83} = 0.2042 \end{aligned}$$

Reactance diagram

 j0.24  j0.10   j0.2722        j0.2042  j0.108  j0.1976
+-/\/\--/\/\----/\/\----+(J)+---/\/\----/\/\----/\/\--+
|                        |                            |
(E1)                   j0.20 (T2)                   (E3)
|                        |                            |
|                      j0.30 (G2)                     |
|                        |                            |
|                      (E2)                           |
+------------------- reference -----------------------+

Here J is the junction bus. Winding connections and earthing do not affect these positive-sequence reactances.

Base values

• $S_b = 100$ MVA, and $V_b = 11$ kV in the generator circuits.
• Line section (through 11/220 kV transformers): $V_b = 220$ kV, so $Z_b = 220^2/100 = 484\ \Omega$.
• G3 circuit (through T3, 220/11 kV): $V_b = 11$ kV.

Conversion to the common base

$$X_{pu,new} = X_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{b,old}}{V_{b,new}}\right)^2$$

Every component is rated at 100 MVA, and its rated voltage equals the base voltage of its section. Both correction factors are therefore 1, and each p.u. value equals its percentage reactance divided by 100.

In ohms, each line is $0.10\times484 = 48.4\ \Omega$.

Reactance diagram

 j0.25  j0.06          j0.10 (L1)
+-/\/\--/\/\--+------+--/\/\--+------+ j0.07  j0.20
|             |      |        |      +--/\/\--/\/\--+
(E1)          |      +--/\/\--+                     |
|             |        j0.10 (L2)                  (E3)
|  j0.20 j0.07|                                     |
+-/\/\--/\/\--+                                     |
|                                                   |
(E2)                                                |
+----------------------- reference -----------------+

Simplified values (useful for fault studies)

• The two lines in parallel: $j0.10\,\|\,j0.10 = j0.05$.
• G1–T1 branch: $j0.31$. G2–T2 branch: $j0.27$.
• These two branches in parallel: $\dfrac{0.31\times0.27}{0.58} = j0.1443$.
• G3–T3 branch: $j0.27$.

In a per-unit reactance diagram every element is expressed on one common MVA base, and the base voltage changes from zone to zone in the ratio of the transformer line-to-line voltages. Resistances, magnetising branches and the grounding impedances are neglected (they carry no current under balanced conditions).

Step 1: Base values in each zone

• Base MVA (all zones): $S_b = 300$ MVA
• Generator zone: $V_{b1} = 20$ kV (given)
• Line zone (T1 is 20/230 kV): $V_{b2} = 20 \times \frac{230}{20} = 230$ kV
• T2 is a Y–Δ bank of single-phase units 127/13.2 kV, so its line-to-line ratio is $\sqrt{3}\times127 / 13.2 = 220/13.2$ kV. Motor zone:

$$V_{b3} = 230 \times \frac{13.2}{220} = 13.8\ \text{kV}$$ $$Z_{b,\text{line}} = \frac{V_{b2}^2}{S_b} = \frac{230^2}{300} = 176.33\ \Omega$$

Step 2: Change of base

$$X_{pu,new} = X_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{old}}{V_{b,new}}\right)^2$$

Step 3: Per-unit reactances

$$\begin{aligned} X_G &= 0.20\ \text{pu (already on its own rating)}\\ X_{T1} &= 0.10\times\frac{300}{350} = 0.0857\ \text{pu}\\ X_{line} &= \frac{0.8\times 65}{176.33} = \frac{52}{176.33} = 0.2949\ \text{pu}\\ X_{T2} &= 0.10\times\frac{300}{300}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0915\ \text{pu}\\ X_{M1} &= 0.20\times\frac{300}{200}\times\left(\frac{13.2}{13.8}\right)^2 = 0.2744\ \text{pu}\\ X_{M2} &= 0.20\times\frac{300}{100}\times\left(\frac{13.2}{13.8}\right)^2 = 0.5488\ \text{pu} \end{aligned}$$

(The three-phase rating of T2 is $3\times100 = 300$ MVA.)

Step 4: Reactance diagram

     j0.0857   j0.2949   j0.0915
 +---/\/\/-----/\/\/-----/\/\/---+--------+
 |    T1        line       T2    |        |
j0.2                          j0.2744  j0.5488
 |                               |        |
(Eg)                           (Em1)    (Em2)
 |                               |        |
 +-------------------------------+--------+
            reference (neutral) bus

Answer: On 300 MVA base, $X_G = 0.2$, $X_{T1} = 0.0857$, $X_{line} = 0.2949$, $X_{T2} = 0.0915$, $X_{M1} = 0.2744$ and $X_{M2} = 0.5488$ pu, with base voltages 20 kV, 230 kV and 13.8 kV in the three zones.

All reactances are converted to a common base of 50 MVA; base voltages follow the transformer line-to-line ratios starting from 13.8 kV at G1.

Base values

• G1 zone: $V_b = 13.8$ kV
• Line zone (T1 13.8/220 kV): $V_b = 220$ kV, $Z_b = \frac{220^2}{50} = 968\ \Omega$
• G2 zone: T2 is a Y–Δ bank of 127/18 kV units, line-to-line ratio $\sqrt{3}\times127/18 = 220/18$ kV, so $V_b = 220\times\frac{18}{220} = 18$ kV
• G3 zone (T3 220/22 kV): $V_b = 22$ kV

Formula: $X_{new} = X_{old}\,\frac{S_{b,new}}{S_{old}}\left(\frac{V_{old}}{V_{b,new}}\right)^2$

Per-unit values

$$\begin{aligned} X_{G1} &= 0.2\times\frac{50}{20} = 0.5\ \text{pu}\\ X_{G2} &= 0.2\times\frac{50}{30}\times\left(\frac{18}{18}\right)^2 = 0.3333\ \text{pu}\\ X_{G3} &= 0.2\times\frac{50}{30}\times\left(\frac{20}{22}\right)^2 = 0.2755\ \text{pu}\\ X_{T1} &= 0.1\times\frac{50}{25} = 0.2\ \text{pu}\\ X_{T2} &= 0.1\times\frac{50}{3\times10} = 0.1667\ \text{pu}\\ X_{T3} &= 0.1\times\frac{50}{35} = 0.1429\ \text{pu}\\ X_{L1} &= \frac{80}{968} = 0.0826\ \text{pu}\\ X_{L2} &= \frac{100}{968} = 0.1033\ \text{pu} \end{aligned}$$

Reactance diagram

   j0.2   j0.0826        j0.1033   j0.1429
 +-/\/\/--/\/\/---+------/\/\/------/\/\/--+
 |  T1     L1     |        L2         T3   |
j0.5           j0.1667 (T2)              j0.2755
 |                |                        |
(E1)           j0.3333                   (E3)
 |                |                        |
 |              (E2)                       |
 |                |                        |
 +----------------+------------------------+
            reference (neutral) bus

Answer: $X_{G1}=0.5$, $X_{G2}=0.333$, $X_{G3}=0.276$, $X_{T1}=0.2$, $X_{T2}=0.167$, $X_{T3}=0.143$, $X_{L1}=0.0826$ and $X_{L2}=0.1033$ pu on 50 MVA, with base voltages 13.8, 220, 18 and 22 kV.

Complex power delivered by a source whose current leaves its positive terminal is $S = VI^*$ (generator convention). A positive P or Q then means the source delivers it; a negative value means it absorbs it.

$$\begin{aligned} S &= VI^* = (100\angle0^\circ)(10\angle-10^\circ)\\ &= 1000\angle-10^\circ\ \text{VA}\\ &= 1000\cos10^\circ - j\,1000\sin10^\circ\\ &= 984.81 - j\,173.65\ \text{VA} \end{aligned}$$

• $P = +984.81$ W: positive, so the source delivers real power.
• $Q = -173.65$ var: negative, so the source absorbs 173.65 var (equivalently it delivers $-173.65$ var).

The current leads the voltage by 10°, so the load connected to the source is capacitive; a capacitive load supplies reactive power, which flows back into the source.

Answer: P = 984.8 W delivered; Q = 173.6 var absorbed by the source.

Base: 90 MVA and 11 kV in the generator circuit (generator rating). Base voltages in other zones follow the line-to-line turns ratios.

Base voltages

• Generator zone: $V_{b1} = 11$ kV
• Line zone (T1 is 10/132 kV): $V_{b2} = 11\times\frac{132}{10} = 145.2$ kV
• T2: three single-phase 66/10 kV units connected Y (line side) – Δ (motor side), so line-to-line ratio is $66\sqrt{3}/10 = 114.32/10$ kV. Motor zone:

$$V_{b3} = 145.2\times\frac{10}{114.32} = 12.70\ \text{kV}$$ $$Z_{b2} = \frac{145.2^2}{90} = 234.26\ \Omega$$

Per-unit reactances (on 90 MVA)

$$\begin{aligned} X_G &= 0.25\ \text{pu}\\ X_{T1} &= 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\ \text{pu}\\ X_{line} &= \frac{100}{234.26} = 0.4269\ \text{pu}\\ X_{T2} &= 0.05\times\frac{90}{3\times30}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\ \text{pu}\\ X_{M1} &= 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\ \text{pu}\\ X_{M2} &= 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789\ \text{pu} \end{aligned}$$

Reactance diagram

     j0.0446   j0.4269   j0.0310
 +---/\/\/-----/\/\/-----/\/\/---+--------+
 |    T1        line       T2    |        |
j0.25                         j0.2231  j0.2789
 |                               |        |
(Eg)                           (Em1)    (Em2)
 |                               |        |
 +-------------------------------+--------+
            reference bus

Answer: On 90 MVA base: $X_G = 0.25$, $X_{T1} = 0.0446$, $X_{line} = 0.4269$, $X_{T2} = 0.0310$, $X_{M1} = 0.2231$, $X_{M2} = 0.2789$ pu; base voltages 11 kV, 145.2 kV and 12.70 kV.

The magnitude of $E_2$ is not printed; it is taken as 100 V, i.e. $E_2 = 100\angle30^\circ$ V (the usual textbook data).

Current I flows from machine 1 to machine 2. With this direction:

• $S_1 = E_1 I^*$ is the power supplied by machine 1 (current leaves its + terminal).
• $S_2 = E_2 I^*$ is the power absorbed by machine 2 (current enters its + terminal).

Current

$$\begin{aligned} E_1 - E_2 &= 100 - (86.60 + j50) = 13.40 - j50\\ I &= \frac{13.40 - j50}{0.5 + j5} = 10.30\angle-159.29^\circ\ \text{A}\\ &= -9.635 - j3.643\ \text{A} \end{aligned}$$

(i) and (ii) Machine powers

$$\begin{aligned} S_1 &= E_1I^* = 100(-9.635 + j3.643) = -963.57 + j364.31\ \text{VA}\\ S_2 &= E_2I^* = -1016.63 - j166.29\ \text{VA} \end{aligned}$$

(iii) Power absorbed by the impedance

$$\begin{aligned} |I|^2 &= 10.30^2 = 106.12\\ S_Z &= |I|^2 Z = 106.12(0.5 + j5) = 53.06 + j530.59\ \text{VA} \end{aligned}$$

Check: P: $1016.63 - 963.57 = 53.06$ W. Q: $364.31 + 166.29 = 530.6$ var. Both balance.

Real power flows from machine 2 to machine 1 because $E_2$ leads $E_1$. Both machines supply reactive power, and all of it is used up in the line reactance.

Answer: Machine 1 consumes 963.6 W and supplies 364.3 var. Machine 2 generates 1016.6 W and supplies 166.3 var. The impedance absorbs 53.1 W and 530.6 var.

Base: 25 MVA and 13.8 kV in the generator circuit. T2 is read as three single-phase units of 69/13.2 kV connected Y (line side) – Δ (motor side).

Base voltages

• Generator zone: $V_{b1} = 13.8$ kV
• Line zone (T1 13.2/115 kV): $V_{b2} = 13.8\times\frac{115}{13.2} = 120.23$ kV
• T2 line-to-line ratio: $69\sqrt{3}/13.2 = 119.51/13.2$ kV. Motor zone: $V_{b3} = 120.23\times\frac{13.2}{119.51} = 13.28$ kV

$$Z_{b2} = \frac{120.23^2}{25} = 578.18\ \Omega$$

Per-unit values (25 MVA base)

$$\begin{aligned} X_{G1} &= 0.15\ \text{pu}\\ X_{T1} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0839\ \text{pu}\\ Z_{line} &= \frac{20 + j65}{578.18} = 0.0346 + j0.1124\ \text{pu}\\ X_{T2} &= 0.11\times\frac{25}{30}\times\left(\frac{13.2}{13.28}\right)^2 = 0.0906\ \text{pu}\\ X_{M1} &= 0.15\times\frac{25}{15}\times\left(\frac{13}{13.28}\right)^2 = 0.2396\ \text{pu}\\ X_{M2} &= 0.15\times\frac{25}{10}\times\left(\frac{13}{13.28}\right)^2 = 0.3594\ \text{pu} \end{aligned}$$

Impedance diagram

    j0.0839  0.0346+j0.1124  j0.0906
 +--/\/\/-------/\/\/--------/\/\/--+-------+
 |   T1          line          T2   |       |
j0.15                          j0.2396  j0.3594
 |                                  |       |
(Eg)                             (Em1)   (Em2)
 |                                  |       |
 +----------------------------------+-------+
              reference bus

Answer: $X_{G1}=0.15$, $X_{T1}=0.0839$, $Z_{line}=0.0346+j0.1124$, $X_{T2}=0.0906$, $X_{M1}=0.2396$, $X_{M2}=0.3594$ pu (25 MVA; 13.8 / 120.23 / 13.28 kV).

Complex power is the phasor product of voltage and the conjugate of current. Its real part is the active power and its imaginary part is the reactive power:

$$S = VI^* = |V||I|\angle(\theta_v - \theta_i) = P + jQ$$

• $P = |V||I|\cos\phi$ (W), the average power converted to work or heat.
• $Q = |V||I|\sin\phi$ (var), the power that oscillates between source and the fields of L and C.
• $|S| = \sqrt{P^2+Q^2}$ (VA), the apparent power. Equipment is rated in VA.
• $\phi = \theta_v - \theta_i$ is the power-factor angle. Q is positive for a lagging (inductive) current.

Sign conventions

Load (motor) convention: the current enters the positive terminal. $S = VI^*$ is then the power absorbed.

• P > 0: the element absorbs real power (resistor, motor).
• Q > 0: it absorbs reactive power (inductor, lagging load).
• Q < 0: it supplies reactive power (capacitor).

Source (generator) convention: the current leaves the positive terminal. $S = VI^*$ is then the power delivered.

• P > 0: the source delivers real power (generator action). P < 0 means it absorbs P (motor action).
• Q > 0: it delivers reactive power (over-excited machine). Q < 0 means it absorbs Q.

 Load convention          Source convention
   I -->  +                 +  --> I
   +------o                 o------+
   | Load |  S absorbed     | Src  |  S delivered
   +------o                 o------+
          -                 -

Example: $V = 100\angle0^\circ$ V and the current leaving the + terminal is $I = 5\angle-30^\circ$ A. Then $S = 500\angle30^\circ = 433 + j250$ VA, so the source delivers 433 W and 250 var.

No figure is printed, so the usual arrangement is assumed: G1–Tr1 and G2–Tr2 feed a common 66 kV bus; from it the line TL and Tr3 (66/400 kV) supply the load. The two generator emfs are taken as equal, so the load current divides between the two generator branches in inverse ratio of their impedances. The 1200 A load current is taken as referred to the 11 kV level, since its voltage level is not stated.

 G1--Tr1(11/66)--+
                 |66 kV bus
 G2--Tr2(12.5/75)+--TL(10 ohm)--Tr3(66/400)--Load

Base values ($S_b$ = 1000 kVA)

• G1 side: 11 kV (given). 66 kV bus: $11\times\frac{66}{11} = 66$ kV.
• G2 side through Tr2: $66\times\frac{12.5}{75} = 11$ kV.
• Load side: $66\times\frac{400}{66} = 400$ kV.
• $Z_b$ at 66 kV $= \frac{66^2}{1} = 4356\ \Omega$. $I_b$ at 11 kV $= \frac{1000}{\sqrt3\times11} = 52.49$ A.

Per-unit reactances

$$\begin{aligned} X_{G1} &= 0.025\ \text{pu}\\ X_{G2} &= 0.001\times\frac{1000}{500}\times\left(\frac{11}{11}\right)^2 = 0.002\ \text{pu}\\ X_{Tr1} &= 0.02\ \text{pu}\\ X_{Tr2} &= 0.02\times\frac{1000}{500}\times\left(\frac{12.5}{11}\right)^2 = 0.0517\ \text{pu}\\ X_{TL} &= \frac{10}{4356} = 0.0023\ \text{pu}\\ X_{Tr3} &= 0.02\times\frac{1000}{1500} = 0.0133\ \text{pu} \end{aligned}$$

Reactance diagram

 j0.025  j0.02
+-/\/\/--/\/\/--+
|  G1     Tr1   |   j0.0023  j0.0133
(E1)            +---/\/\/----/\/\/---> Load
|  G2     Tr2   |     TL      Tr3
+-/\/\/--/\/\/--+
 j0.002  j0.0517
(E1, E2 return to the common reference bus)

Current shared by G1 and G2

Branch impedances: $Z_1 = 0.025 + 0.02 = 0.045$ pu and $Z_2 = 0.002 + 0.0517 = 0.0537$ pu.

$$\begin{aligned} I_{G1} &= I_L\frac{Z_2}{Z_1+Z_2} = 1200\times\frac{0.0537}{0.0987} = 1200\times0.5439 = 652.6\ \text{A}\\ I_{G2} &= I_L\frac{Z_1}{Z_1+Z_2} = 1200\times0.4561 = 547.4\ \text{A} \end{aligned}$$

(In per unit the load current is $1200/52.49 = 22.86$ pu. G1 carries 12.43 pu and G2 carries 10.43 pu.)

Answer: G1 supplies about 652.6 A and G2 about 547.4 A (54.4 % and 45.6 % of the load current).

Note: if the G2 reactance of 0.1 % is a misprint for 10 %, then $X_{G2} = 0.2$ pu, $Z_2 = 0.2517$ pu, and the shares become 1018.0 A (G1) and 182.0 A (G2).