GTU 3150911 Power System- II Summer 2026 Solved Paper & Solutions

The per unit (pu) system expresses each quantity (voltage, current, power, impedance) as a fraction of a base value of the same kind.

Advantages:

1. The transformer ratio disappears, so a network with many voltage levels is solved as one circuit.
2. The values for equipment of the same type lie in a narrow range (a transformer reactance is 0.05 to 0.15 pu), so a wrong value is seen at once.
3. The factor $\sqrt{3}$ and the line or phase difference do not appear.
4. The pu value is the same for single phase and three phase forms, and on either side of a transformer.

Let the voltage be $v = V_m\sin\omega t$ and the current be $i = I_m\sin(\omega t - \phi)$, where $\phi$ is the phase angle between them.

Instantaneous power:

It has a constant part and a part that varies at twice the supply frequency.

Real (active) power is the average of $p$. It is the power that is changed to work or heat:

Reactive power is the power that moves back and forth between the source and the inductors or capacitors. It does no net work:

Apparent power is the product of the rms voltage and current:

In complex form, $S = P + jQ = VI^*$. The three powers form the power triangle. The power factor is $\cos\phi = \dfrac{P}{S}$. For an inductive load, $Q$ is positive and the current lags. For a capacitive load, $Q$ is negative.

A long line (more than about 250 km) cannot be treated with lumped values, because the voltage and the current change along the line, and the shunt current flows all along it. The rigorous method is the exact one. It uses the line parameters as distributed along the whole length, and not as lumped values. The assumptions are a balanced three phase line, which is analysed for one phase (phase to neutral), with uniform parameters and a steady sinusoidal state. Let $z$ be the series impedance and $y$ the shunt admittance per unit length, and $l$ be the length of the line. Take $x$ as the distance measured from the receiving end.

Differential equations

For a small element $dx$ at the distance $x$, with the voltage $V$ and the current $I$:

Differentiate the first, and put in the second:

where $\gamma = \sqrt{zy}$ is the propagation constant. The solution is

where $Z_c = \sqrt{\dfrac{z}{y}}$ is the characteristic (surge) impedance.

Constants

At the receiving end, $x = 0$, so $V = V_r$ and $I = I_r$:

Sending end

At $x = l$, the voltage and the current are $V_s$ and $I_s$:

Using $\cosh\gamma l = \dfrac{e^{\gamma l} + e^{-\gamma l}}{2}$ and $\sinh\gamma l = \dfrac{e^{\gamma l} - e^{-\gamma l}}{2}$:

Meaning of the terms

• $\gamma = \alpha + j\beta$ has the attenuation constant $\alpha$ (the decay of the wave along the line, in nepers per km) and the phase constant $\beta$ (the phase shift, in radians per km).
• $Z_c$ is the impedance that the line shows to a wave. A line that ends in a load of $Z_c$ has no reflection, and it is "matched".
• The terms with $\cosh$ and $\sinh$ are the sum of a forward wave and a backward (reflected) wave.

Line constants

Comparing with $V_s = AV_r + BI_r$ and $I_s = CV_r + DI_r$:

and $AD - BC = \cosh^2\gamma l - \sinh^2\gamma l = 1$.

Ferranti effect is the rise of the voltage at the receiving end of a long, lightly loaded or open-circuited transmission line, above the sending end voltage.

A long line has a large shunt capacitance. When the load is small, the charging current $I_c$ of this capacitance flows through the series inductance of the line. The voltage drop $I_c X_L$ across the inductance is in phase with the receiving end voltage, so it adds to it. So $V_r > V_s$.

For an open-ended lossless line, $V_r = \dfrac{V_s}{\cos\beta l}$. For a 300 km line at 50 Hz, $\beta l \approx 0.314$ rad, so $V_r \approx 1.05\,V_s$, a rise of about 5 percent. It is controlled by shunt reactors.

Given: $Z_1 = 50 + j40\ \Omega$, $Z_2 = 80 - j30\ \Omega$, in parallel across $V = 200$ V.

Check: $S_1 = \dfrac{V^2}{Z_1^<em>} = 487.8 + j390.2$ and $S_2 = \dfrac{V^2}{Z_2^</em>} = 438.4 - j164.4$. Their sum is $926.2 + j225.9$.

A medium line (about 80 to 250 km) is solved with the lumped parameters. The shunt capacitance is no longer negligible, as in a short line, but it is also not necessary to use the distributed values. The total series impedance is $Z = R + jX_L$ and the total shunt admittance is $Y = j\omega C$. In the nominal T method, the total series impedance $Z$ of the line is split into two equal halves, $\dfrac{Z}{2}$ on each side, and the total shunt admittance $Y$ is placed at the middle.

Let $V_c$ be the voltage at the middle point (across $Y$) and $I_c$ be the current in the shunt branch.

Check: $AD - BC = \left(1 + \dfrac{YZ}{2}\right)^2 - YZ\left(1 + \dfrac{YZ}{4}\right) = 1$. This holds for any passive, bilateral network. Also $A = D$, since the network is symmetrical.

Meaning of the constants: $A$ is a pure number, the ratio $V_s/V_r$ at no load. $B$ is the transfer impedance, in ohm. $C$ is the transfer admittance, in siemens. $D$ is a number, the ratio $I_s/I_r$ with the receiving end short circuited.

Example: $Z = 20 + j60\ \Omega$ and $Y = j4\times10^{-4}$ S. Then $YZ = -0.024 + j0.008$, and

Why the T circuit: it places the whole capacitance at the middle of the line, so the charging current is carried by half the impedance on the sending side. It is a better model than a single lumped capacitance at one end. It has more elements to calculate than the $\pi$ model for the load flow, as it adds an extra middle node. The same constants can be used to find the voltage regulation and the efficiency of the line, with the receiving end values given. So the constants completely describe the line as a two port network.

A set of unbalanced three phase voltages or currents can be replaced by three balanced sets, called the symmetrical components (Fortescue's theorem).

1. Positive sequence: three phasors of equal magnitude, 120 degrees apart, with the same phase order (abc) as the original.
2. Negative sequence: three phasors of equal magnitude, 120 degrees apart, with the opposite phase order (acb).
3. Zero sequence: three phasors of equal magnitude, in phase with each other.

With the operator $a = 1\angle120^\circ$:

They are used to analyse unsymmetrical faults.

A circuit breaker is selected after the fault level and the system data are known. The main points are:

1. Rated voltage: equal to or more than the highest system voltage.
2. Rated normal current: the continuous current that it carries without too much heating, more than the maximum load current.
3. Breaking capacity: the highest short circuit current (rms, at the instant of the contact separation) that it can break. It is found from the fault study, and is in MVA or kA. It must be more than the fault level.
4. Making capacity: the peak current it can close on a fault, taken as about 2.55 times the symmetrical breaking current.
5. Short time rating: the current that it can carry for a short time (1 or 3 seconds) without damage.
6. Operating duty and speed: the opening time and the auto-reclosing duty.
7. Type: it is chosen by the voltage level and use, such as air circuit breaker or miniature type for low voltage, vacuum breaker for medium voltage, and SF$_6$ breaker for high voltage.

In the zero sequence network, the zero sequence currents $I_0$ are in phase in all three phases. They flow only if there is a path through the neutral and the earth. So the zero sequence impedance diagram of a transformer depends on its winding connection and on the grounding of the neutral.

Let $Z_0$ be the zero sequence impedance of the transformer, which is its leakage impedance. The buses $P$ and $Q$ are the two sides, and the reference bus is the earth.

Reason for $3Z_n$: the neutral current is $3I_0$, so the voltage drop is $3I_0Z_n$. It is seen by a single phase as $I_0 \times 3Z_n$.

Use: these diagrams are joined to make the zero sequence network of a whole system, for the analysis of the earth faults (line to ground and double line to ground).

A surge absorber is a protective device that absorbs the energy of a travelling surge and reduces its steepness and magnitude. A steep wave front is dangerous for the winding of a transformer or a machine, since the voltage is not shared uniformly among the turns.

• It is connected between the line and the earth, near the equipment to be protected.
• Types are an RC or surge capacitor (the resistor dissipates the energy), the Ferranti surge absorber (an air cored inductor coupled with an earthed metal sheet, like a short circuited transformer secondary), and a ZnO surge arrester, which also limits the voltage.

Insulation coordination is the selection of the insulation levels of the different parts of a power system, and of the protective devices, so that a failure or flashover, if it comes, occurs at the chosen place, which is cheap and easy to repair, such as the surge arrester or the line insulator, and not in the costly equipment, such as the transformer. The protective level of the arrester is kept below the insulation strength of the equipment, with a margin.

Basic Impulse Insulation Level (BIL): the crest value of a standard lightning impulse voltage that the insulation of the equipment can withstand without failure. The standard impulse wave has a rise time (front) of $1.2\ \mu$s and falls to half its value in $50\ \mu$s. It is written as 1.2/50 $\mu$s.

Example: a 132 kV system has a standard BIL of 550 kV.

The BIL of the equipment is chosen with the margin above the protective level of the arrester. It is used to coordinate the insulation of the equipment, so that a surge cannot damage it.

Line to line fault between the phases $b$ and $c$, through a fault impedance $Z_f$. Let the phase $a$ be the reference.

The phase $a$ is the healthy phase. The phases $b$ and $c$ touch each other (through $Z_f$), so no current goes to the earth. The operator $a = 1\angle120^\circ = -\dfrac{1}{2} + j\dfrac{\sqrt{3}}{2}$ has the properties $1 + a + a^2 = 0$ and $a - a^2 = j\sqrt{3}$, which are used below.

Conditions at the fault:

The current leaves through $b$ and returns through $c$, so there is no zero sequence current.

So $I_{a0} = 0$ and $I_{a2} = -I_{a1}$. Also $I_b = -j\sqrt{3}\,I_{a1}$.

The condition $V_b - V_c = Z_fI_b = -j\sqrt{3}\,Z_fI_{a1}$ gives

Connection of the sequence networks

The positive and the negative sequence networks are joined in parallel (as a loop), with $Z_f$ in series. The zero sequence network is not used, since $I_{a0} = 0$.

For a solid fault, $Z_f = 0$, so the magnitude of the fault current is $\dfrac{\sqrt{3}\,E_a}{Z_1 + Z_2}$. If $Z_1 = Z_2$, it is $\dfrac{\sqrt{3}}{2}$ times the three phase fault current $\dfrac{E_a}{Z_1}$. So this fault is less severe than the three phase fault, but it is more common than it.

Use: the line to line fault current is used to set the phase fault relays, and the voltage of the healthy phase stays near normal.

Special cases: if the two networks have equal impedance and the fault is solid, the positive and the negative sequence voltages at the fault point are equal, $V_{a1} = V_{a2} = \dfrac{E_a}{2}$. The voltage of the faulted phases is then $V_b = V_c = -\dfrac{E_a}{2}$.

The arcing ground effect is a phenomenon in an ungrounded neutral system. When a line to ground fault occurs, the fault current is the capacitive charging current. If it is small, the arc at the fault point may be extinguished and re-struck again and again.

The line capacitance $C$ and the system inductance $L$ form an oscillating circuit. The repeated striking and extinction cause high frequency oscillations, which give severe overvoltages on the healthy phases (up to 5 to 6 times the normal phase voltage).

Remedy: ground the neutral, preferably through an arc suppression coil (Peterson coil), which gives an inductive current that cancels the capacitive current, so the arc dies out.

Corona is the ionisation of the air around a conductor, when the voltage gradient at its surface exceeds the breakdown strength of the air. It has a faint violet glow and a hissing sound. The factors affecting it are:

1. Atmospheric conditions: a lower air density (low pressure and high temperature, so a lower $\delta$) and stormy or wet weather reduce the breakdown voltage, so corona is more.
2. Conductor size: a larger radius gives a lower field at the surface, and a higher critical voltage, so corona is less.
3. Spacing between conductors: a larger spacing raises the critical voltage, but if the spacing is made very large, the effect is small.
4. Surface of the conductor: a rough, dirty or stranded surface has points where the field is high, so corona starts earlier than on a smooth surface.
5. Line voltage: corona starts when the voltage is above the critical disruptive voltage, and it rises fast with the voltage.
6. Frequency of the supply: the corona loss increases with the frequency.

A travelling wave (surge) on a line of the surge impedance $Z_1$ meets a T-junction, where two lines of the surge impedance $Z_2$ and $Z_3$ are connected in parallel.

This is the case where a line branches into two. A cable and an overhead line, or a transformer, also change the impedance at a joint. At the point where the impedance changes, the wave cannot go on unchanged, so a part is reflected back, and a part is refracted (transmitted) forward. For the incident wave, let $e_1$ and $i_1 = \dfrac{e_1}{Z_1}$ be the voltage and the current. The reflected wave is $e_1'$, $i_1' = \dfrac{e_1'}{Z_1}$, and the wave that goes ahead (refracted or transmitted) into the lines 2 and 3 is $e_2$.

Equivalent impedance

The two outgoing lines are in parallel. So the junction sees the load

Boundary conditions

At the junction, the voltage and the current must be continuous:

Here the reflected current has the opposite direction, so it carries a minus sign.

Solution

From the voltage equation $e_1' = e_2 - e_1$. Put it in the current equation:

Refraction (transmission) coefficient:

Reflection coefficient:

The refracted voltage $e_2$ goes into both the lines 2 and 3 equally. The currents are $i_2 = \dfrac{e_2}{Z_2}$ and $i_3 = \dfrac{e_2}{Z_3}$.

Special cases

So at a T-junction with two lines of the same impedance as line 1, $Z_t = \dfrac{Z_1}{2}$, so $\tau = \dfrac{2}{3}$ and $\rho = -\dfrac{1}{3}$. The voltage of the surge is reduced.