📘 Power Systems – Mega Revision Formula Sheet (Page 20)

Final rapid revision formulas for GATE + PSU.

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🔹 Per Unit System

Base Current: I_base = S_base / (√3 V_base) Base Impedance: Z_base = V_base² / S_base Per Unit Value: PU = Actual / Base

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🔹 Fault Analysis

3-Phase Fault Current: I_f = V / Z_total SLG Fault Current: I_f = 3V / (Z1 + Z2 + Z0) Fault MVA: MVA_fault = √3 V I If impedance reduces by x%, fault current increases by 1/(1-x)

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🔹 Transmission Lines

Surge Impedance: Z_s = √(L/C) SIL = V² / Z_s Short line → Ignore capacitance Medium line → Nominal π Long line → Hyperbolic equations

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🔹 Stability

Power Equation (SMIB): P = (EV/X) sinδ Maximum Power: P_max = EV / X Swing Equation: M d²δ/dt² = P_m - P_e Maximum power at δ = 90°

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🔹 Economic Dispatch

Incremental Cost: dC/dP = λ With Losses: (dC/dP) × Penalty Factor = λ Penalty Factor: 1 / (1 - ∂P_L/∂P_i)

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🔹 Load Flow

Gauss-Seidel → Linear convergence Newton-Raphson → Quadratic convergence Slack Bus → V & angle fixed PV Bus → P & V fixed PQ Bus → P & Q specified

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🔹 HVDC

Power = V × I Advantages:

• No reactive power flow
• No stability limit
• Lower losses for long distance

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🔹 Protection

Distance Relay → Measures impedance Zone 1 → 80% line Differential Relay → Based on KCL PSM = Fault Current / Relay Setting

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🔹 Load Factors

Load Factor = Average Load / Peak Load Diversity Factor = Sum of Individual Max / System Max

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Page 20 – Mega Revision Formula Sheet Completed

🔥 Power Systems Complete Master Module (20 Pages) Achieved 🔥

 

📘 Distance Protection – Zone Setting & Numerical Problems

Distance relay measures impedance (Z = V/I) and operates when measured impedance is less than preset zone impedance.

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🔹 1️⃣ Basic Principle

Z = V / I

• Short circuit → Current increases
• Measured impedance decreases
• If Z_measured < Z_set → Relay trips

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🔹 2️⃣ Zone Settings

• Zone 1 → 80% of protected line (Instantaneous)
• Zone 2 → 100% line + 50% next line (Time delay)
• Zone 3 → Backup (Further reach)

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🔹 Problem 1: Zone 1 Setting

Line impedance = 0.5 + j4 Ω Length = 100 km Find Zone 1 setting.

Solution

Total impedance magnitude = √(0.5² + 4²) = √(0.25 + 16) = √16.25 = 4.03 Ω

Zone 1 = 80% × 4.03 = 3.22 Ω

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🔹 Problem 2: Fault at 60 km

Fault occurs at 60 km on same line. Will Zone 1 operate?

Step 1: Fault Impedance

Impedance per km = 4.03 / 100 = 0.0403 Ω/km Fault impedance = 60 × 0.0403 = 2.418 Ω

Since 2.418 < 3.22 → Zone 1 trips instantly.

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🔹 Problem 3: Zone 2 Setting

Adjacent line impedance = 3 Ω Calculate Zone 2 setting.

Solution

Zone 2 = Line impedance + 50% next line = 4.03 + (0.5 × 3) = 4.03 + 1.5 = 5.53 Ω

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🔹 Problem 4: Measured Voltage & Current

Measured V = 11 kV Measured I = 1000 A Find measured impedance.

Solution

Z = V / I = 11000 / 1000 = 11 Ω

If Zone setting = 5 Ω:

Since 11 > 5 → Relay does NOT operate.

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📊 Types of Distance Relays

• Impedance Relay
• Reactance Relay
• Mho Relay (most common)

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🎯 GATE & Interview Points

• Distance relay independent of fault current magnitude
• Zone 1 is instantaneous
• Zone 2 and 3 are time delayed
• Mho relay has circular characteristic

Distance Protection = Impedance Based Intelligent Protection

 

📘 Derivations of Unsymmetrical Fault Currents

                                         

We derive LG, LL and LLG fault currents using symmetrical components. Assume prefault voltage V = 1 pu.

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🔹 1️⃣ Single Line-to-Ground (LG) Fault Derivation

Assume phase 'a' is grounded: Va = 0 Ib = Ic = 0 Using symmetrical components: Ia = I1 + I2 + I0 For LG fault: I1 = I2 = I0 Why? Because boundary condition forces equal sequence currents. Now: Ia = 3I1 Sequence network connection: Z1, Z2, Z0 connected in series. Voltage equation: V = I1 Z1 + I2 Z2 + I0 Z0 Since I1 = I2 = I0: V = I1 (Z1 + Z2 + Z0) Therefore: I1 = V / (Z1 + Z2 + Z0) Total fault current: Ia = 3I1

I_fault = 3V / (Z1 + Z2 + Z0)

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🔹 2️⃣ Line-to-Line (LL) Fault Derivation

Assume phase b and c shorted: Ib = -Ic Ia = 0 Symmetrical component condition: I0 = 0 I1 = -I2 Zero sequence current absent. Sequence network: Z1 and Z2 in series. Voltage equation: V = I1 Z1 + I2 Z2 Since I2 = -I1: V = I1 (Z1 + Z2) Thus: I1 = V / (Z1 + Z2) Phase fault current: I_fault = √3 I1

I_fault = √3 V / (Z1 + Z2)

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🔹 3️⃣ Double Line-to-Ground (LLG) Fault Derivation

Assume b and c shorted to ground. Boundary conditions: Va ≠ 0 Vb = 0 Vc = 0 Sequence currents satisfy: I1 + I2 + I0 = 0 Sequence network: Z2 and Z0 in parallel Then in series with Z1. Equivalent impedance: Z_eq = Z1 + (Z2 Z0)/(Z2 + Z0) Thus: I1 = V / Z_eq Total fault current: Ia = 3I1

I_fault = 3V / (Z1 + (Z2 Z0)/(Z2 + Z0))

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📊 Key Comparison from Derivation

Fault Type	Sequence Network Connection	Formula
LG	Z1 + Z2 + Z0 (Series)	3V/(Z1+Z2+Z0)
LL	Z1 + Z2	√3V/(Z1+Z2)
LLG	Z1 + (Z2 || Z0)	3V/(Z1+(Z2Z0)/(Z2+Z0))

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🎯 Important Conceptual Insights

• LG fault → all three sequence currents equal
• LL fault → zero sequence absent
• LLG fault → mixed parallel connection
• Grounding strongly affects Z0

Fault Derivation = Boundary Conditions + Sequence Network Logic