(SEM VI) THEORY EXAMINATION 2024-25 POWER SYSTEM-II

BEE601 – POWER SYSTEM-II

Section-Wise Solved Answers (2024–25)

SECTION A

Attempt all questions – 2 × 7 = 14 marks

(a) During three-phase fault which sequence component should be calculated? Explain with reason.

During a three-phase fault, only the positive sequence component is calculated. This is because a three-phase fault is a balanced fault, and in a balanced system, the negative and zero sequence components are zero. All three phase currents are equal in magnitude and displaced by 120°, which corresponds only to the positive sequence network. Hence, for three-phase fault analysis, only the positive sequence network is considered.

(b) Explain the need of load flow analysis.

Load flow analysis is required to determine the steady-state operating condition of a power system. It provides information about bus voltages, power flows, reactive power distribution, and system losses. Load flow studies are essential for planning, operation, system expansion, voltage control, and economic operation of power systems.

(c) Enumerate different types of buses and their significance in power system.

In a power system, buses are classified as slack bus, PV bus, and PQ bus. The slack bus maintains system voltage and supplies losses. The PV bus controls voltage magnitude and real power, usually representing generators. The PQ bus represents load buses where real and reactive power are specified. Each bus type helps simplify load flow calculations.

(d) A 100 kV surge travels on an overhead line with surge impedance 500 Ω and is terminated by a cable of surge impedance 50 Ω. Find reflected and transmitted voltage.

Incident voltage = 100 kV
Surge impedance of line (Z₁) = 500 Ω
Surge impedance of cable (Z₂) = 50 Ω

Reflection coefficient:

Γ=Z2−Z1Z2+Z1=50−50050+500=−0.818\Gamma = \frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{50 - 500}{50 + 500} = -0.818Γ=Z2​+Z1​Z2​−Z1​​=50+50050−500​=−0.818

Reflected voltage:

Vr=Γ×Vi=−0.818×100=−81.8 kVV_r = \Gamma \times V_i = -0.818 \times 100 = -81.8 \text{ kV}Vr​=Γ×Vi​=−0.818×100=−81.8 kV

Transmitted voltage:

Vt=Vi+Vr=100−81.8=18.2 kVV_t = V_i + V_r = 100 - 81.8 = 18.2 \text{ kV}Vt​=Vi​+Vr​=100−81.8=18.2 kV 

(e) Enumerate five factors affecting transient stability of power system.

Transient stability is affected by fault type, fault clearing time, system inertia, initial operating condition, and network configuration. Faster fault clearing improves stability, while higher inertia resists sudden rotor angle changes.

(f) Classify relays based on their application.

Relays are classified as overcurrent relays, distance relays, differential relays, thermal relays, and directional relays based on their application. Each relay type is designed to detect specific fault conditions in power systems.

(g) Differentiate between restriking voltage and recovery voltage.

Restriking voltage appears across circuit breaker contacts immediately after current interruption and may cause arc re-ignition. Recovery voltage appears after the arc is fully extinguished and the system regains normal voltage. Restriking voltage is transient, whereas recovery voltage is steady.

SECTION B

Attempt any three – 7 × 3 = 21 marks

(a) Determine equivalent reactance of power network during three-phase fault.

Given:
Generator reactance = 0.25 pu
Transformer reactance = 0.12 pu
Transmission line reactance = 0.28 pu
Fault current = 5 pu

Total reactance seen during fault:

Xeq=1If=15=0.2 puX_{eq} = \frac{1}{I_f} = \frac{1}{5} = 0.2 \text{ pu}Xeq​=If​1​=51​=0.2 pu

Reactance excluding network:

X=0.25+0.12+0.28=0.65 puX = 0.25 + 0.12 + 0.28 = 0.65 \text{ pu}X=0.25+0.12+0.28=0.65 pu

Let network reactance = Xn

1X+Xn=5⇒X+Xn=0.2⇒Xn=−0.45 pu\frac{1}{X + X_n} = 5 \Rightarrow X + X_n = 0.2 \Rightarrow X_n = -0.45 \text{ pu}X+Xn​1​=5⇒X+Xn​=0.2⇒Xn​=−0.45 pu

Thus, network reactance is –0.45 pu.

(b) Explain ZBUS building algorithm with all types of modifications.

ZBUS building algorithm is used to construct the bus impedance matrix by step-by-step addition of elements. Modifications include adding a branch to reference bus, adding a branch to existing bus, adding a link between buses, and adding mutual coupling. This method avoids repeated matrix inversion.

(c) Explain Bewley’s Lattice diagram and prove open-circuited line behaves as leading PF network.

Bewley’s lattice diagram graphically represents multiple reflections and transmissions of travelling waves. In an open-circuited line, reflected voltage is positive and current reflection is negative, causing current to lead voltage, hence behaving as a leading power factor network.

(d) Derive power flow equations and prove reactive power proportionality.

Active and reactive power flow equations are derived using bus admittance matrix. Reactive power depends on voltage magnitude difference between buses. Mathematically:

Q∝∣Vs∣−∣Vr∣Q \propto |V_s| - |V_r|Q∝∣Vs​∣−∣Vr​∣

Hence, reactive power flow is proportional to the voltage magnitude difference.

(e) Explain working of overcurrent relay with neat diagram.

An overcurrent relay operates when current exceeds a preset value. It consists of a current coil and operating mechanism. When fault current flows, the relay coil energizes and closes the trip circuit of the circuit breaker.

SECTION C

Q3. Attempt any one

(a) Derive fault current for double line-to-ground fault and draw sequence networks.

In double line-to-ground fault, positive, negative, and zero sequence networks are connected in parallel. Fault current is derived using symmetrical components and is higher than single line-to-ground fault current.

(b) Design sequence networks and calculate fault current for SLG fault.

For single line-to-ground fault, positive, negative, and zero sequence networks are connected in series. Fault current is:

If=3VZ1+Z2+Z0I_f = \frac{3V}{Z_1 + Z_2 + Z_0}If​=Z1​+Z2​+Z0​3V​

Sequence networks are drawn accordingly.

Q4. Attempt any one

(a) Formation of YBUS matrix and its modification.

YBUS matrix is formed using self and mutual admittances. When a new line is added, corresponding elements are modified without recomputing entire matrix.

(b) Application of Newton-Raphson method in load flow analysis.

Newton-Raphson method uses Jacobian matrix and iterative corrections. It converges faster and is more accurate than Gauss-Seidel but requires higher computation.

Q5. Attempt any one

(a) Travelling wave problem – surge transmission and reflection.

Surge impedance of line and cable are calculated from L and C. Reflection and transmission coefficients are used to find transmitted and reflected voltages and currents.

(b) Explain Surge Impedance Loading (SIL).

SIL is the loading at which reactive power generated equals reactive power absorbed. Maximum power at SIL:

P=V2Z0P = \frac{V^2}{Z_0}P=Z0​V2​

SIL improves voltage profile and stability.

Q6. Attempt any one

(a) Explain equal area criterion for transient stability.

Equal area criterion states that a system is stable if accelerating area equals decelerating area. It is used to analyze stability during sudden load or fault changes.

(b) Effect of clearing time on stability.

Longer clearing time increases rotor angle deviation and may cause instability. Faster fault clearing improves transient stability margin.

Q7. Attempt any one

(a) Arc formation, extinction and quenching methods in circuit breakers.

Arc is formed during contact separation. Arc extinction methods include high resistance method and current zero method. Proper quenching ensures safe interruption.

(b) Explain fault clearing time, relay time, breaker time, overreach and underreach.

Fault clearing time is sum of relay and breaker time. Relay time is sensing duration, breaker time is mechanical opening time. Overreach and underreach refer to relay operating beyond or within set limits.