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POWER SYSTEM ANALYSIS (EEE601)
Section-wise Solved Answers & Notes
SECTION – A (10 × 2 = 20 Marks)
Short, direct answers
(a) Line and phase voltages in per-unit are equal
In per-unit system, base voltage is chosen such that
VL(base)=3VP(base)V_{L(base)}=\sqrt{3}V_{P(base)}VL(base)=3VP(base)
Hence, when actual line and phase voltages are divided by their respective base values, per-unit values become equal.
(b) Relation between zero sequence current and neutral current
In=3I0I_n = 3I_0In=3I0
Neutral current is three times the zero-sequence current.
(c) Per-unit reactance conversion
Given:
Rated: 30 MVA, 11 kV, X = 20% = 0.2 pu New base: 50 MVA, 10 kV
Xnew=Xold×SnewSold×(VoldVnew)2X_{new} = X_{old}\times \frac{S_{new}}{S_{old}} \times \left(\frac{V_{old}}{V_{new}}\right)^2Xnew=Xold×SoldSnew×(VnewVold)2 Xnew=0.2×5030×(1110)2=0.403 puX_{new}=0.2\times\frac{50}{30}\times\left(\frac{11}{10}\right)^2 = \boxed{0.403\ pu}Xnew=0.2×3050×(1011)2=0.403 pu
(d) Short circuit capacity
It is the maximum fault MVA a system can deliver during short circuit.
Short circuit MVA=3VIfault\text{Short circuit MVA} = \sqrt{3}V I_{fault}Short circuit MVA=3VIfault
(e) Zero sequence network
Zero-sequence network consists of zero-sequence reactances of generators, transformers and lines, connected according to grounding and winding connections.
(f) Load flow analysis
It determines bus voltages, power flows, losses, and reactive power in a power system under steady-state condition.
(g) Transient stability
Ability of power system to maintain synchronism after a large disturbance like fault or sudden load change.
(h) Critical clearing time
Maximum time allowed to clear a fault without losing system stability.
(i) Causes of voltage surge
• Lightning strokes • Switching operations
(j) Bewley’s lattice diagram
Graphical method used to analyze travelling wave reflections and transmissions in transmission lines.
SECTION – B (Any 5 × 10 = 50 Marks)
(a) Impedance diagram on per-unit basis
Steps: Select base MVA and base voltage
Convert all reactances to common base Neglect resistance
Draw reactance-only diagram Purpose: Simplifies fault and stability analysis.
(b) Three-phase fault calculation
Steps: Convert all impedances to per-unit
Draw reactance diagram Find equivalent reactance
Calculate fault current If=1XeqI_f = \frac{1}{X_{eq}}If=Xeq1
Fault MVA:
=SbaseXeq= \frac{S_{base}}{X_{eq}}=XeqSbase
(c) Zero sequence network for L-L fault
Zero-sequence network does not participate in line-to-line fault.
Sequence network connection: • Positive and negative sequence in parallel
• Zero sequence open If=VZ1+Z2I_f = \frac{V}{Z_1+Z_2}If=Z1+Z2V
(d) Load flow bus classification • Slack Bus: Voltage & angle known
• PV Bus: Voltage & power known • PQ Bus: Power known
Gauss-Seidel Method: • Assume voltages
• Calculate bus currents • Update voltages
• Repeat till convergence Acceleration factor improves speed.
(e) Critical clearing angle (Equal area criterion)
Given:
Pmax1=2.0, Pfault=0.5, Ppost=1.5P_{max1}=2.0,\ P_{fault}=0.5,\ P_{post}=1.5Pmax1=2.0, Pfault=0.5, Ppost=1.5
Equal area criterion: Areaacc=Areadec\text{Area}_{acc} = \text{Area}_{dec}Areaacc=Areadec
Graphically determine critical clearing angle δc.
(f) Step-by-step solution of swing equation
Swing equation: d2δdt2=PaM\frac{d^2\delta}{dt^2}=\frac{P_a}{M}dt2d2δ=MPa
At discontinuity: Pa=Pm−PeP_a=P_m-P_ePa=Pm−Pe
Numerical integration used for transient stability studies.
(g) Travelling waves
Characterized by: • Surge impedance
• Velocity of propagation • Reflection & transmission
Wave equation:
∂2V∂x2=LC∂2V∂t2\frac{\partial^2 V}{\partial x^2}=LC\frac{\partial^2 V}{\partial t^2}∂x2∂2V=LC∂t2∂2V
(h) Effect of cable on surge
Given: Z1=400Ω, Z2=50Ω, Vi=100kVZ_1=400\Omega,\ Z_2=50\Omega,\ V_i=100kVZ1=400Ω, Z2=50Ω, Vi=100kV
Transmitted voltage:
Vt=Vi2Z2Z1+Z2V_t = V_i \frac{2Z_2}{Z_1+Z_2}Vt=ViZ1+Z22Z2 Vt=100×100450=22.2 kVV_t = 100\times\frac{100}{450} = \boxed{22.2\ kV}Vt=100×450100=22.2 kV
SECTION – C (Any 2 × 15 = 30 Marks)
Q3 (a) Single line diagram & per-unit system
Single line diagram: Simplified representation of 3-phase system.
Per-unit system advantages: • Eliminates transformer turns ratio
• Simplifies calculations • Reduces numerical errors
(b) Symmetrical components Balanced star load:
Before fuse removal → only positive sequence
After two fuses removed → • Positive sequence
• Negative sequence • Zero sequence present
Q4 (a) Transient current in R-L circuit
At switching: i=Imax(1−e−t/τ)i = I_{max}(1-e^{-t/\tau})i=Imax(1−e−t/τ)
Maximum transient current: imax=2Isteadyi_{max}=2I_{steady}imax=2Isteady
(b) Formation of Y-bus matrix
Steps: • Self admittance = sum of connected admittances
• Mutual admittance = negative of line admittance Y-bus is symmetrical matrix.
Q5
(a) Assumptions in stability studies
• Constant mechanical input • Neglect damping
• Constant voltage • Classical generator model
(b) Equal area criterion
Used to determine transient stability of SMIB system by equating accelerating and decelerating areas.
(c) Protection against surges
• Lightning arresters • Ground wires
• Surge absorbers • Insulation coordination