Power System-I (EX-404) - Unit 3 Short Notes

1.0 Power System Economics & Load Characteristics

1.1 Load Curve & Load Duration Curve (LDC)

• Load Curve: Graph of load (kW/MW) versus time (hours/days). Shows chronological variation.

  • Daily Load Curve: Variation within 24 hours.

  • Annual Load Curve: Average daily load for each month/season.

• Load Duration Curve (LDC): Load values arranged in descending order vs. cumulative time. Derived from Load Curve.

  • Utility: Determines capacity factor, plant factor, and scale of operation. Essential for economic dispatch and unit commitment.

  • Key Insight: Area under LDC = Total energy consumed (kWh).

[!TIP] Exam Focus: Be able to plot both curves from given tabular data and calculate load factor from them.

1.2 Key Performance Factors & Definitions

Inter-relationship Proof:

• Max. Demand = $$\displaystyle \frac{\text{Connected Load}}{\text{Demand Factor}} $$

• Avg. Load = Max. Demand × Load Factor

• Annual Energy = Avg. Load × 8760 h

• Improvement in Load Factor with higher Diversity Factor:

  • Simultaneous Max. Demand = $$\displaystyle \frac{\sum \text{Individual Max. Demands}}{\text{Diversity Factor}} $$

  • For same individual loads, higher Diversity Factor → Lower Simultaneous Max. Demand → Higher Load Factor (for same total energy).

1.3 Economic Aspects of Generation

• Effect of Load Factor on Cost:

  • High Load Factor → More uniform generation → Lower capital cost per unit (fixed cost spread over more kWh).

  • Low Load Factor → Idle capacity during off-peak → Higher cost/kWh.

  • Cost per kWh ∝ 1 / Load Factor (approx., if fixed costs dominate).

• Effect of Diversity Factor on Overall Cost:

  • High Diversity Factor → Lower simultaneous peak → Smaller required plant capacity → Lower capital cost.

  • Reduces reserve requirements.

• Kelvin's Law for Economic Conductor Size:

  • Statement: The most economical conductor size is that for which the annual cost of energy wasted equals the annual interest & depreciation on the capital cost of the conductor.

  • Derivation:

    Let:

    • $P$ = Power transmitted (W)

    • $l$ = Length of line (km)

    • $r$ = Resistance per km per conductor ($\Omega$/km)

    • $I$ = Load current (A)

    • $a$ = Cross-sectional area (cm²)

    • $$\displaystyle K_1 $$ = Annual interest & depreciation cost per unit volume (Rs/cm³-year)

    • $$\displaystyle K_2 $$ = Cost of energy wasted per unit (Rs/kWh)

    • $n$ = Number of conductors

    • $ρ$ = Resistivity ($\Omega$-cm)

    • $J$ = Current density (A/cm²)

    • $$\displaystyle LF_{loss} $$ = Loss load factor (often ≈ Load Factor² for constant impedance load)

    • $T$ = Annual operating hours (8760 h)

    Capital Cost/year = $$\displaystyle K_1 \times \text{Volume} = K_1 \times (a \times l \times n) $$

    Energy Wasted/year = $$\displaystyle I^2 r l \times T \times LF_{loss} = \left(\frac{I}{a}\right)^2 \times (ρ l) \times l \times T \times LF_{loss} = J^2 ρ l^2 T LF_{loss} $$

    Cost of Wasted Energy/year = $$\displaystyle K_2 \times J^2 ρ l^2 T LF_{loss} $$

    For economy: Capital Cost = Cost of Wasted Energy

$$ K_1 a l n = K_2 J^2 ρ l^2 T LF_{loss} $$

    Substitute $$\displaystyle a = I/J $$:

$$ K_1 \frac{I}{J} l n = K_2 J^2 ρ l^2 T LF_{loss} $$

$$ \boxed{J^3 = \frac{K_1 I n}{K_2 ρ l T LF_{loss}}} \quad \Rightarrow \quad \boxed{J = \sqrt[3]{\frac{K_1 n}{K_2 ρ l T LF_{loss}}} \times \sqrt[3]{I}} $$

    **Simplified for given system:** $J \propto \sqrt[3]{I}$. Economical current density is **constant** for a given system.

*   **Assumptions**: Constant load, same interest/depreciation for conductor & energy cost, energy cost based on wasted energy.

*   **Limitations**:

    1.  Ignores technical factors (voltage drop, stability).

    2.  Assumes constant load; actual load varies.

    3.  Interest/depreciation rates for conductor vs. energy may differ.

    4.  Does not account for future load growth.

    5.  Does not consider cost of supports, insulation, etc.

1.4 System Configuration & Interconnection

• Radial System: Simple, low cost, but poor reliability (single point of failure).

• Ring Main System: Improved reliability, easier maintenance, but higher cost & complex protection.

• Interconnected System: Multiple stations/grids connected.

  • Advantages:

    • Economy: Shared reserve, optimal dispatch (α, β controls), load diversity.

    • Reliability: Alternate power paths, reduced outage impact.

    • Stability: Mutual support during disturbances.

  • Disadvantages: High initial cost, complex protection & control, synchronization issues.

• Methods of Power Transfer (α, β Controls):

  • α-control (Angle Control): Change power angle δ between sending/receiving end voltages. $P \propto \sin δ$.

  • β-control (Voltage Magnitude Control): Change magnitude of sending/receiving end voltages. $$\displaystyle P \propto |V_S||V_R| \sin δ $$.

  • In practice, both are used via excitation control (affects $|V|$) and turbine governor (affects δ via frequency).

1.5 Generation Types

• Conventional: Thermal (coal, gas, nuclear), Hydro. Base/Mid-load.

• Non-Conventional (Renewable): Solar PV, Wind, Geothermal, Biomass, Tidal. Often intermittent, require storage/grid support.

• Distributed Generation (DG): Small-scale generation (kW-MW) located near load centers.

  • Advantages: Reduced T&D losses, improved reliability, deferred upgrades, voltage support.

  • Challenges: Grid integration, protection coordination, power quality, reverse power flow.

2.0 Transmission Line Parameters: Overhead Lines

2.1 Fundamental Concepts

• Symmetrical Spacing: Conductors equidistant (e.g., equilateral triangle). Inductance & capacitance per phase equal.

• Unsymmetrical Spacing: Conductors at unequal distances. Inductance & capacitance per phase unequal → unbalanced voltages.

• Transposition:

  • Need: To balance inductance & capacitance of unsymmetrical lines over full length.

  • Procedure: Conductors are periodically swapped at transposition towers so each occupies each position for equal length. Makes line electrically symmetrical on average.

2.2 Inductance Calculations

• Single-Phase Two-Wire Line:

  • Flux linkage per conductor: $$\displaystyle \psi = \frac{\mu_0}{2\pi} I \left( \frac{1}{4} + \ln\frac{D}{r'} \right) $$ Wb-turns/km

  • Where $D$ = distance between conductors (m), $$\displaystyle r' = r e^{-1/4} = 0.7788r $$ (GMR).

  • Inductance per conductor: $$\displaystyle L = \frac{\psi}{I} = \frac{\mu_0}{2\pi} \left( \ln\frac{D}{r'} \right) + \frac{\mu_0}{8\pi} $$ H/km.

  • Loop Inductance ($$\displaystyle L_{loop} $$): $2 \times L$ (for two conductors in series).

• Three-Phase Lines:

  • Symmetrical Spacing: $$\displaystyle L = \frac{\mu_0}{2\pi} \ln\frac{D}{r'} $$ H/km/phase (same as single-phase with $D$ = GMD).

  • Unsymmetrical Spacing: Use Geometrical Mean Distance (GMD).

$$ \text{GMD} = \sqrt[3]{D_{12} D_{23} D_{31}} \quad \text{(for equilateral, GMD = D)} $$

$$ L = \frac{\mu_0}{2\pi} \ln\frac{\text{GMD}}{r'} \text{ H/km/phase} $$

• GMD & Self-GMD (GMR):

  • GMD: For mutual inductance between bundles/phases. $$\displaystyle \text{GMD} = (D_1 \cdot D_2 \cdot ... \cdot D_n)^{1/n} $$.

  • GMR (Self-GMD): For self inductance of a composite conductor. Accounts for internal flux linkage. $$\displaystyle r' = r e^{-1/4} $$ for solid round conductor.

  • For bundled conductors (n sub-conductors per phase, spacing $d$):

$$ \text{GMR}_{\text{bundle}} = (n \cdot r' \cdot d^{n-1})^{1/n} $$

• Effect of Earth: Negligible for lines > 10m height. Can be accounted for by method of images (imaginary conductors below ground), but usually ignored.

2.3 Capacitance Calculations

• Single-Phase Line:

  • Capacitance to neutral: $$\displaystyle C_n = \frac{2\pi \varepsilon_0}{\ln\frac{D}{r}} $$ F/km.

  • Line capacitance (between conductors): $$\displaystyle C = \frac{\pi \varepsilon_0}{\ln\frac{D}{r}} $$ F/km.

• Three-Phase Lines:

  • Symmetrical: $$\displaystyle C_n = \frac{2\pi \varepsilon_0}{\ln\frac{\text{GMD}}{r}} $$ F/km/phase to neutral.

  • Unsymmetrical: Use GMD.

• Effect of Earth: Increases capacitance slightly (image method).

• Bundled Conductors:

$$ C_n = \frac{2\pi \varepsilon_0 n}{\ln\frac{\text{GMD}}{\text{GMR}_{\text{bundle}}}} \text{ F/km/phase} $$

*   Bundling **increases GMR** → **decreases inductance** → **increases capacitance** → reduces ** surge impedance loading (SIL)**.

2.4 Conductor Characteristics

• Skin Effect: Non-uniform current distribution at AC (higher density near surface). Increases effective resistance ($$\displaystyle R_{ac} > R_{dc} $$). More pronounced at higher frequency & larger conductor size.

• Proximity Effect: Magnetic field from adjacent conductors causes further current crowding. Increases $$\displaystyle R_{ac} $$. Significant in tightly packed conductors (buses, cables).

2.5 Comparison & Special Cases

• Double Circuit 3-Phase Line: Two 3-phase circuits on same tower.

  • Inductance: Use GMD considering all conductors. Mutual inductance between circuits reduces net inductance.

  • Capacitance: Higher due to reduced equivalent GMD.

• Bundled Conductor (2 per phase) vs. Double Circuit Line:

  | Feature | Bundled Conductor (2/ph) | Double Circuit Line | | :--- | :--- | :--- | | Purpose | Reduce inductance, increase capacitance, reduce corona | Increase power transfer capacity, reliability | | Configuration | 2 conductors close together per phase | 2 separate 3-phase circuits | | Inductance | Lower than single circuit | Lower than single circuit (due to mutual) | | Capacitance | Higher than single circuit | Higher than single circuit | | 应用 | EHV/UHV transmission (≥ 220 kV) | High power transfer, corridor constraints |

• Tuned Power Lines: Line with added series capacitors & shunt reactors to make surge impedance loading (SIL) match transmitted power. Minimizes voltage variation with load change (flat voltage profile). $$\displaystyle SIL = \frac{V^2}{Z_0} $$.

3.0 Transmission Line Modeling & Performance

3.1 Classification of Lines

3.2 Equivalent Circuits & Constants

• Short Line:

$$ V_S = V_R + I_R (R + jX) $$

$$ I_S = I_R $$

**A, B, C, D Constants**: $$\displaystyle A=D=1 $$, $$\displaystyle B=Z $$, $$\displaystyle C=0 $$.

• Medium Line - Nominal-π Model:

  • Half shunt admittance at each end: $$\displaystyle Y/2 = j\omega C l / 2 $$.

  • Series impedance: $$\displaystyle Z = (R + j\omega L)l $$.

  • Phasor Diagram (π-configuration):

    
    V_S ---[Z]--- V_R
    
      |          |
    
     Y/2        Y/2
    
      |          |
    
     GND        GND
    
    

  • Generalized Constants (ABCD):

$$ A = 1 + \frac{YZ}{2}, \quad B = Z, \quad C = Y\left(1 + \frac{YZ}{4}\right), \quad D = A $$

    Where $$\displaystyle Y = j\omega C l $$, $$\displaystyle Z = (R + j\omega L)l $$.

• Medium Line - Nominal-T Model:

  • Half series impedance in middle: $Z/2$ at each end.

  • Full shunt admittance in middle: $Y$.

  • Constants: $$\displaystyle A = 1 + \frac{YZ}{2} $$, $$\displaystyle B = Z\left(1 + \frac{YZ}{4}\right) $$, $$\displaystyle C = Y $$, $$\displaystyle D = A $$.

  • Note: Both models give similar results; π is more common for calculation.

• Long Line - Rigorous Solution:

  • Differential Equations:

$$ \frac{dV}{dx} = (R + j\omega L)I = Z'I $$

$$ \frac{dI}{dx} = (G + j\omega C)V = Y'V $$

*   **Propagation Constant**: $$\displaystyle \gamma = \sqrt{Z'Y'} = \alpha + j\beta $$

    *   $\alpha$: Attenuation constant (Np/km)

    *   $\beta$: Phase shift constant (rad/km)

*   **Characteristic Impedance**: $$\displaystyle Z_c = \sqrt{Z'/Y'} $$ (Surge impedance, purely real for lossless line).

*   **Solution**:

$$ V(x) = V_R \cosh(\gamma x) + I_R Z_c \sinh(\gamma x) $$

$$ I(x) = \frac{V_R}{Z_c} \sinh(\gamma x) + I_R \cosh(\gamma x) $$

    At $$\displaystyle x = l $$ (sending end):

$$ V_S = V_R \cosh(\gamma l) + I_R Z_c \sinh(\gamma l) $$

$$ I_S = \frac{V_R}{Z_c} \sinh(\gamma l) + I_R \cosh(\gamma l) $$

*   **Equivalent π/T**: Can be derived with $$\displaystyle Z = Z_c \sinh(\gamma l) $$, $$\displaystyle Y = \frac{2}{Z_c} \tanh(\gamma l / 2) $$ (for π).

3.3 Performance Analysis

• Voltage Regulation (VR):

$$ \text{VR (\%)} = \frac{|V_S|_{nl} - |V_S|_{fl}}{|V_S|_{fl}} \times 100\% $$

*   $$\displaystyle |V_S|_{nl} $$: Sending voltage at no-load (receiving end open).

*   $$\displaystyle |V_S|_{fl} $$: Sending voltage at full-load.

*   For short line: $$\displaystyle \text{VR} \approx \frac{I_R R \cos\phi_R + I_R X \sin\phi_R}{V_R} \times 100\% $$.

• Transmission Efficiency ($\eta$):

$$ \eta = \frac{P_R}{P_S} \times 100\% = \frac{V_R I_R \cos\phi_R}{V_S I_S \cos\phi_S} \times 100\% $$

For short line: $$\displaystyle \eta \approx \frac{\cos\phi_R}{\cos\phi_S} \times \frac{V_R}{V_S} $$.

• Ferranti Effect:

  • Phenomenon: Voltage at sending end ($$\displaystyle V_S $$) exceeds receiving end ($$\displaystyle V_R $$) for a lightly loaded or open-circuited long line.

  • Cause: Capacitive charging current flows through line inductance, causing voltage drop that adds to $$\displaystyle V_R $$.

  • Derivation for Open-Circuited Line ($$\displaystyle I_R = 0 $$):

$$ V_S = V_R \cosh(\gamma l) \approx V_R \left(1 + \frac{\gamma^2 l^2}{2}\right) \text{ (for small } \gamma l) $$

    Since $$\displaystyle \gamma^2 = Z'Y' = (R+j\omega L)(G+j\omega C) \approx -\omega^2 LC $$ (if $R,G$ small), $$\displaystyle \cosh(\gamma l) > 1 $$ → $$\displaystyle V_S > V_R $$.

*   **Mitigation**: Shunt reactors at receiving end or intermediate points.

• Power Circle Diagram:

  • Concept: Graphical representation of receiving-end power ($$\displaystyle P_R + jQ_R $$) for constant $$\displaystyle V_S $$, $$\displaystyle V_R $$, and line parameters.

  • Equation (from ABCD): $$\displaystyle P_R + jQ_R = \frac{V_S V_R}{B} - \frac{V_R^2}{B} A^* $$ (for lossless line, $$\displaystyle A=D $$, $$\displaystyle B=jX $$, $$\displaystyle C=0 $$).

  • Construction:

    1. Plot circle with diameter $$\displaystyle 2V_S V_R / |B| $$ centered at $$\displaystyle (V_S V_R / |B|, 0) $$ in $P$-$Q$ plane.

    2. For given power factor, draw line from origin at angle $$\displaystyle \phi_R $$; intersection gives $$\displaystyle P_R, Q_R $$.

  • Interpretation: Shows limits of power transfer ($$\displaystyle P_{max} = V_S V_R / |B| $$), reactive power requirements, stability margin.

3.4 Medium Line Numerical Problems

• Standard Approach (Nominal-π):

  1. Calculate $$\displaystyle Z = (R + jX)l $$, $$\displaystyle Y = j\omega C l $$.

  2. Find $A, B, C, D$.

  3. Given $$\displaystyle V_R $$, $$\displaystyle P_R $$, $$\displaystyle \cos\phi_R $$ → find $$\displaystyle I_R = \frac{P_R}{\sqrt{3} V_R \cos\phi_R} $$ (3-phase) or single-phase equivalent.

  4. Calculate $$\displaystyle V_S = A V_R + B I_R $$, $$\displaystyle I_S = C V_R + D I_R $$.

  5. Sending-end power factor: $$\displaystyle \cos\phi_S = \cos(\angle V_S - \angle I_S) $$.

  6. Efficiency: $$\displaystyle \eta = \frac{P_R}{P_S} = \frac{P_R}{\sqrt{3} V_S I_S \cos\phi_S} $$.

  7. Voltage Regulation: Need $$\displaystyle V_S $$ at no-load. For π-model, set $$\displaystyle I_R=0 $$ → $$\displaystyle V_S^{(nl)} = A V_R $$ (if $$\displaystyle C=0 $$ approx) or solve with $$\displaystyle I_R=0 $$ exactly.

4.0 Underground Cables

4.1 Construction & Types

• Single-Core Cable: One conductor + insulation + sheath + armoring + outer sheath.

• Multi-Core Cable: Multiple insulated conductors in common sheath.

• Insulating Materials: PVC, XLPE, Paper-impregnated (oil-filled, gas-filled), Rubber.

4.2 Parameter Calculations

• Capacitance of Single-Core Cable:

$$ C = \frac{2\pi \varepsilon_0 \varepsilon_r}{\ln\frac{r_2}{r_1}} \text{ F/km} $$

Where $$\displaystyle r_1 $$ = conductor radius, $$\displaystyle r_2 $$ = internal sheath radius, $$\displaystyle \varepsilon_r $$ = relative permittivity of insulation.

• Dielectric Stress (Electric Field):

  • At radius $x$ ($$\displaystyle r_1 < x < r_2 $$):

$$ E(x) = \frac{V}{x \ln\frac{r_2}{r_1}} \text{ V/m} $$

*   **Maximum Stress**: At $$\displaystyle x = r_1 $$ (conductor surface).

$$ E_{max} = \frac{V}{r_1 \ln\frac{r_2}{r_1}} $$

*   **Minimum Stress**: At $$\displaystyle x = r_2 $$ (sheath inner surface).

$$ E_{min} = \frac{V}{r_2 \ln\frac{r_2}{r_1}} $$

*   **Ratio**: $$\displaystyle \frac{E_{max}}{E_{min}} = \frac{r_2}{r_1} $$.

• Grading of Cables:

  • Purpose: To make $E(x)$ more uniform, reduce $$\displaystyle E_{max} $$, utilize insulation better.

  • Methods:

    1. Intersheath Grading: Use multiple thin conductive intersheaths at specific potentials. Each layer has different $$\displaystyle \varepsilon_r $$.

    2. Capacitance Grading: Use multiple layers of different dielectrics ($$\displaystyle \varepsilon_{r1}, \varepsilon_{r2}, ... $$) such that $$\displaystyle \frac{\varepsilon_{r1}}{r_1} = \frac{\varepsilon_{r2}}{r_2} = ... $$ for uniform stress.

    3. Longitudinal Grading: Vary insulation thickness along cable length (not common).

4.3 Comparison with Overhead Lines

5.0 Mechanical Design of Overhead Lines

5.1 Line Supports & Sag

• Supports: Wooden poles (low voltage), Steel tubular poles (medium), RCC poles (medium), Steel towers (high/EHV).

• Sag & Tension:

  • Assumptions: Parabolic shape (for sag << span), uniform load $w$ (kg/m).

  • Derivation (Level supports, span $L$, tension $$\displaystyle T_0 $$ at lowest point):

$$ y = \frac{w}{2T_0} x^2 \quad \text{(parabola)} $$

    **Sag** ($S$) at support ($$\displaystyle x = L/2 $$):

$$ \boxed{S = \frac{w L^2}{8 T_0}} $$

*   **Maximum Tension** ($$\displaystyle T_{max} $$) at support:

$$ T_{max} = \sqrt{T_0^2 + (wS)^2} \approx T_0 + \frac{w^2 L^2}{8T_0} \text{ (if } S \ll T_0/w) $$

*   **Effect of Ice & Wind**:

    *   **Total effective load**: $$\displaystyle \vec{w}_{eff} = \vec{w}_g + \vec{w}_i + \vec{w}_w $$

    *   $$\displaystyle w_g $$: weight of conductor (kg/m)

    *   $$\displaystyle w_i $$: weight of ice coating (kg/m) = $$\displaystyle 9.81 \times \rho_{ice} \times \pi t_i (2r + t_i) $$

    *   $$\displaystyle w_w $$: wind pressure (kg/m²) × projected area = $$\displaystyle P_w \times (2r + 2t_i) $$ (approx. diameter)

    *   **Resultant sag**: $$\displaystyle S_{eff} = \frac{w_{eff} L^2}{8 T_0} $$ (along direction of $$\displaystyle w_{eff} $$).

    *   **Horizontal tension** $$\displaystyle T_0 $$ is reduced due to increased load → sag increases.

• Safety Factor: $$\displaystyle SF = \frac{\text{Ultimate Strength}}{T_{max}} $$. Typical: 2 to 2.5.

5.2 Sag Template & String Chart

• Sag Template: Full-scale drawing of parabolic curves for different tensions & temperatures. Used for tower spotting on profile map. Ensures minimum ground clearance.

• String Chart: Graph of sag vs. temperature for fixed span & tension. Used to determine stringing tension during installation for given conditions.

• Difference: Template is for design/planning (ground clearance). String chart is for construction/erection (tension setting).

5.3 Conductor Material & Selection

• Materials: Copper (high conductivity, heavy, costly), Aluminium (light, cheap, lower conductivity), ACSR (Aluminium Conductor Steel Reinforced - combines Al conductivity & steel strength).

• Selection Criteria: Conductivity, strength, weight, cost, thermal expansion, corona performance.

• Sag Calculation with Ice/Wind: Use $$\displaystyle w_{eff} $$ as above. Calculate $$\displaystyle S_{eff} $$ and $$\displaystyle T_{max} $$. Ensure $$\displaystyle T_{max} < \text{Ultimate Strength} / SF $$ and $$\displaystyle S_{eff} + \text{Support height} > \text{Ground clearance} $$.

6.0 Insulators & Insulation Coordination

6.1 Types of Insulators

6.2 Voltage Distribution in String

• Capacitance Model:

  • $C$: Self-capacitance of each insulator (to ground via metal fittings).

  • $mC$: Capacitance from each insulator pin to ground (fittings to tower).

  • $$\displaystyle m = 0.1 $$ to $0.2$ typical.

• Derivation (for n identical insulators):

  Let $$\displaystyle V_1, V_2, ..., V_n $$ = voltage across each unit (top to bottom).

  $$\displaystyle V_1 + V_2 + ... + V_n = V_{string} $$ (line voltage to ground).

  Using KCL at each node (except top/bottom), solve recurrence:

$$ V_k = V_1 \left[ \cosh((n-k)\theta) + m \sinh((n-k)\theta) \right] $$

where $$\displaystyle \cosh\theta = \frac{2+m}{2\sqrt{m}} $$.

**Bottom unit voltage** is maximum.

• String Efficiency:

$$ \boxed{\text{String Efficiency (\%)} = \frac{V_{string}}{n \times V_1} \times 100\% = \frac{V_{string}}{n \times V_{max\ unit}} \times 100\%} $$

*   Measures effectiveness of string. 100% = uniform voltage.

*   Decreases with more units (non-uniformity increases).

6.3 Methods to Improve String Efficiency

1. Grading of Insulator Units: Use insulators with different capacitances (e.g., larger diameter discs at bottom, smaller at top) to make voltage distribution more uniform.

2. Guard Ring: Metal ring electrically connected to bottom of string and grounded via a grading ring. Provides additional capacitance to ground for top units, equalizing voltage.

   • Working: Guard ring capacitance $$\displaystyle C_g $$ adds shunt capacitance to top units, increasing their voltage drop.

6.4 Testing & Other Concepts

• Flash-over Voltage: Minimum voltage causing insulator surface flashover (arc). Tested by applying increasing voltage until flashover.

• Insulator Pollution: Deposition of salt, dust, industrial fumes → conductive layer → leakage current → flashover at lower voltage. Mitigation: periodic washing, greasing, longer creepage distance.

7.0 System Configuration & Substation Equipment

7.1 Distribution System Configuration

• Why 3-φ, 3-Wire for Transmission?

  • Saves one conductor (25% material saving vs. 3-φ, 4-wire).

  • No neutral needed as loads are balanced.

  • Lower right-of-way, tower cost.

• Why 3-φ, 4-Wire for Distribution?

  • Provides neutral for single-phase loads (residential, commercial).

  • Allows two voltage levels: Line-to-Line ($$\displaystyle V_{LL} $$) and Line-to-Neutral ($$\displaystyle V_{LN} = V_{LL}/\sqrt{3} $$).

• Copper Efficiency Comparison:

  • For same power $P$, voltage $$\displaystyle V_{LL} $$, power factor $\cos\phi$, and same losses:

  • 3-φ, 3-wire: $$\displaystyle I = P/(\sqrt{3} V_{LL} \cos\phi) $$. Total conductor cross-section $$\displaystyle A_{3w} \propto I $$.

  • 3-φ, 4-wire: Neutral carries unbalanced current. For balanced load, neutral current = 0. But conductor size often same as phase. Material ≈ 33% more than 3-wire.

  • Conclusion: 3-wire more economical for transmission; 4-wire necessary for distribution diversity.

7.2 Bus Bar Arrangements

• Single Bus Bar with Sectionalization: One bus, divided by circuit breakers. Allows isolation of faulty section without total shutdown.

• Sectionalized Double Bus Bar System: Two parallel buses with coupler breaker. Maintenance possible on one bus without interrupting supply. More flexible, reliable.

• Ring Mains: Bus bars arranged in ring. High reliability, but complex protection.

7.3 Substation Equipment (List & Brief)

1. Bus Bars: Conductors for power collection/distribution.

2. Circuit Breakers (CB): Make/break normal & fault currents.

3. Isolators/Disconnect Switches: Isolate equipment for maintenance (no load breaking).

4. Current Transformers (CT): Step down current for metering/protection.

5. Potential Transformers (PT): Step down voltage for metering/protection.

6. Lightning Arresters (LA): Protect against surges (lightning, switching).

7. Wave Traps: Block high-frequency carrier signals (for power line carrier communication).

8. Capacitor Voltage Transformers (CVT): For high voltage metering (cheaper than PT).

9. Relays: Protection logic (overcurrent, differential, distance).

10. Control & Monitoring Panels.

7.4 Substation Types: AIS vs. GIS

8.0 Special Topics & Problem Solving

8.1 Conductor Cross-Section & System Comparison

• DC 2-Wire vs. Single-Phase AC (Equal Power $P$, Equal Losses, Same Length $l$):

  • DC 2-Wire: $$\displaystyle P = 2 I_{dc} V $$, Loss $$\displaystyle = 2 I_{dc}^2 R_{dc} = 2 I_{dc}^2 \frac{\rho l}{a_{dc}} $$.

  • Single-Phase AC: $$\displaystyle P = 2 I_{ac} V \cos\phi $$ (two conductors). Loss $$\displaystyle = 2 I_{ac}^2 R_{ac} = 2 I_{ac}^2 \frac{\rho l}{a_{ac}} \frac{R_{ac}}{R_{dc}} $$.

  • Equal Power: $$\displaystyle I_{dc} = \frac{P}{2V} $$, $$\displaystyle I_{ac} = \frac{P}{2V \cos\phi} $$.

  • Equal Losses: $$\displaystyle I_{dc}^2 \frac{1}{a_{dc}} = I_{ac}^2 \frac{1}{a_{ac}} \frac{R_{ac}}{R_{dc}} $$.

  • Skin Effect Ratio: $$\displaystyle k = \frac{R_{ac}}{R_{dc}} > 1 $$.

  • Result: $$\displaystyle \boxed{\frac{a_{ac}}{a_{dc}} = \frac{1}{k \cos^2\phi}} $$. AC requires larger area due to skin effect & power factor.

• Single-Phase to 3-Phase Conversion (Same $$\displaystyle V_{LL} $$, Same Losses):

  • Single-Phase: $$\displaystyle P_1 = V I_1 \cos\phi_1 $$, Loss $$\displaystyle = 2 I_1^2 R_1 $$.

  • 3-Phase: $$\displaystyle P_3 = \sqrt{3} V I_3 \cos\phi_3 $$, Loss $$\displaystyle = 3 I_3^2 R_3 $$.

  • Same voltage $V$ (line-to-line for 3-φ, line-to-neutral for 1-φ? Clarify assumption). Typically: Same line-to-line voltage $$\displaystyle V_{LL} $$.

  • For 1-φ: $$\displaystyle V_{phase} = V_{LL} $$? No, 1-φ has two wires: voltage $V$. Assume same potential difference $V$ between conductors.

  • Equal losses: $$\displaystyle 2 I_1^2 R_1 = 3 I_3^2 R_3 $$. If same conductor material/length: $$\displaystyle R_1 \propto 1/a_1 $$, $$\displaystyle R_3 \propto 1/a_3 $$.

  • Equal power: $$\displaystyle V I_1 \cos\phi_1 = \sqrt{3} V I_3 \cos\phi_3 \Rightarrow I_1 = \sqrt{3} I_3 \frac{\cos\phi_3}{\cos\phi_1} $$.

  • Assume same $\cos\phi$ for simplicity (common in problems).

  • Then: $$\displaystyle 2 (\sqrt{3} I_3)^2 \frac{1}{a_1} = 3 I_3^2 \frac{1}{a_3} \Rightarrow \frac{a_3}{a_1} = \frac{3}{2 \times 3} = \frac{1}{2} $$? Wait:

    $$\displaystyle 2 \times 3 I_3^2 / a_1 = 3 I_3^2 / a_3 \Rightarrow 6/a_1 = 3/a_3 \Rightarrow a_3/a_1 = 1/2 $$.

  • But conductor count differs: 1-φ uses 2 conductors, 3-φ uses 3.

  • Total conductor volume (or cross-section * number):

    • 1-φ total area $$\displaystyle A_1 = 2 a_1 $$.

    • 3-φ total area $$\displaystyle A_3 = 3 a_3 = 3 \times (a_1/2) = 1.5 a_1 $$.

  • So 3-φ uses less total conductor material for same power & losses? Contradicts common knowledge.

  • Correct Approach (from standard texts):

    For same power $P$, voltage $V$ (line-to-line for 3-φ, line-to-neutral for 1-φ? Let's use same voltage between conductors $V$):

    • 1-φ: $$\displaystyle P = V I_1 \cos\phi $$, Loss $$\displaystyle = 2 I_1^2 R $$.

    • 3-φ: $$\displaystyle P = \sqrt{3} V I_3 \cos\phi $$, Loss $$\displaystyle = 3 I_3^2 R $$.

    • Equal losses: $$\displaystyle 2 I_1^2 = 3 I_3^2 \Rightarrow I_1/I_3 = \sqrt{3/2} \approx 1.225 $$.

    • Equal power: $$\displaystyle V I_1 = \sqrt{3} V I_3 \Rightarrow I_1/I_3 = \sqrt{3} \approx 1.732 $$.

    • Conflict: Cannot satisfy both with same $R$. So adjust conductor area.

    • Let $$\displaystyle a_1 $$, $$\displaystyle a_3 $$ be phase conductor areas. $R \propto 1/a$.

    • Loss equality: $$\displaystyle 2 (P/(V\cos\phi))^2 \frac{1}{a_1} = 3 (P/(\sqrt{3} V \cos\phi))^2 \frac{1}{a_3} $$

    • Simplify: $$\displaystyle 2 \frac{P^2}{V^2 \cos^2\phi} \frac{1}{a_1} = 3 \frac{P^2}{3 V^2 \cos^2\phi} \frac{1}{a_3} = \frac{P^2}{V^2 \cos^2\phi} \frac{1}{a_3} $$

    • $$\displaystyle \Rightarrow \frac{2}{a_1} = \frac{1}{a_3} \Rightarrow a_3 = \frac{a_1}{2} $$.

    • Total conductor cross-section:

      • 1-φ: $$\displaystyle A_1 = 2 a_1 $$

      • 3-φ: $$\displaystyle A_3 = 3 a_3 = 3 \times (a_1/2) = 1.5 a_1 $$

    • Percentage additional load when converting 1-φ to 3-φ with same conductor material (same $$\displaystyle a_1 = a_3 $$? No, problem says "additional conductor of same size").

    • Problem statement (JUN 2024): "An existing single-phase AC system comprising of two overhead conductors is to be converted into a 3-phase, 3-wire system by providing an additional conductor of same size. Calculate the percentage of additional load that can be transmitted by the three-phase system if the operating line voltage and percentage line losses remain the same in both the systems."

    • Interpretation: Same conductor size $a$ for all conductors. Same voltage $V$ (between any two conductors? For 1-φ, voltage $V$ across two conductors; for 3-φ, line-to-line voltage $$\displaystyle V_{LL} = V $$). Same loss percentage → same loss in absolute terms? Or same loss factor? Usually means same losses as fraction of transmitted power.

    • Let $$\displaystyle P_1 $$, $$\displaystyle P_3 $$ be powers transmitted.

    • Losses: $$\displaystyle Loss_1 = I_1^2 R_{total,1} $$, $$\displaystyle Loss_3 = I_3^2 R_{total,3} $$.

    • $$\displaystyle R_{total,1} = 2 R $$ (two conductors, each resistance $R \propto 1/a$).

    • $$\displaystyle R_{total,3} = 3 R $$ (three conductors, same $R$ per conductor).

    • Currents: $$\displaystyle I_1 = P_1/(V \cos\phi) $$, $$\displaystyle I_3 = P_3/(\sqrt{3} V \cos\phi) $$ (assuming same $\cos\phi$).

    • Loss equality (same absolute loss? Or same % loss? "percentage line losses remain the same" means $$\displaystyle \frac{Loss}{P} $$ is same for both systems.

    • So: $$\displaystyle \frac{Loss_1}{P_1} = \frac{Loss_3}{P_3} = k $$ (same loss fraction).

    • $$\displaystyle Loss_1 = k P_1 = I_1^2 \cdot 2R = \left(\frac{P_1}{V\cos\phi}\right)^2 2R $$

    • $$\displaystyle Loss_3 = k P_3 = I_3^2 \cdot 3R = \left(\frac{P_3}{\sqrt{3} V \cos\phi}\right)^2 3R = \frac{P_3^2}{3 V^2 \cos^2\phi} 3R = \frac{P_3^2}{V^2 \cos^2\phi} R $$

    • From 1-φ: $$\displaystyle k = \frac{2R}{V^2 \cos^2\phi} P_1 $$

    • From 3-φ: $$\displaystyle k = \frac{R}{V^2 \cos^2\phi} P_3 $$

    • Equate: $$\displaystyle \frac{2R}{V^2 \cos^2\phi} P_1 = \frac{R}{V^2 \cos^2\phi} P_3 \Rightarrow P_3 = 2 P_1 $$.

    • Additional load = $$\displaystyle P_3 - P_1 = P_1 $$.

    • Percentage additional load = $$\displaystyle \frac{P_3 - P_1}{P_1} \times 100\% = 100\% $$.

    • Answer: 100% more load can be transmitted.

    • Check: With same conductor size, 3-φ carries twice the power of 1-φ for same voltage and same loss fraction.

8.2 Numerical Problem Integration

• Load Data & Installed Capacity:

  • Max. Demand = $$\displaystyle \frac{\sum \text{Individual Max.}}{\text{Diversity Factor}} $$

  • Avg. Load = Max. Demand × Load Factor

  • Annual Energy = Avg. Load × 8760

  • Installed Capacity = Max. Demand × (1 + Reserve %) → Reserve based on reliability, maintenance.

• Inductance/Capacitance:

  • Always compute GMD/GMR first.

  • For bundled: $$\displaystyle GMR_{bundle} = (n \cdot r' \cdot d^{n-1})^{1/n} $$, $$\displaystyle GMD_{phase} = \sqrt[3]{D_{12} D_{23} D_{31}} $$ (for horizontal, use actual distances).

• Sag with Ice/Wind:

  • Compute $$\displaystyle w_g $$, $$\displaystyle w_i $$, $$\displaystyle w_w $$ separately.

  • $$\displaystyle w_{eff} = \sqrt{(w_g + w_i)^2 + w_w^2} $$ (vector sum).

  • Use $$\displaystyle S = \frac{w_{eff} L^2}{8 T_0} $$.

  • Given $$\displaystyle T_{max} $$ and $SF$: $$\displaystyle T_0 = \frac{T_{max}}{SF} - \text{approx correction} $$? Actually $$\displaystyle T_{max} = \sqrt{T_0^2 + (w_{eff} S)^2} $$. Iterate or use approximation $$\displaystyle T_{max} \approx T_0 + \frac{w_{eff}^2 L^2}{8 T_0} $$.

• Insulator String:

  • Given $$\displaystyle m = C_{pin-ground}/C_{self} $$.

  • For 3 units: $$\displaystyle V_1 : V_2 : V_3 = 1 : (1+m) : (1+3m+m^2) $$? Standard result for 3 units:

    $$\displaystyle V_1 = V_{string} \cdot \frac{1}{1 + (1+m) + (1+3m+m^2)} $$? No.

    Actually: $$\displaystyle V_1 = V_{string} \cdot \frac{1}{1 + (1+m) + (1+3m+m^2)} $$? That denominator is sum of ratios.

    Let $$\displaystyle V_1 = V $$, then $$\displaystyle V_2 = V(1+m) $$, $$\displaystyle V_3 = V(1+3m+m^2) $$.

    $$\displaystyle V_{string} = V[1 + (1+m) + (1+3m+m^2)] = V(3 + 4m + m^2) $$.

    So $$\displaystyle V_1 = \frac{V_{string}}{3+4m+m^2} $$, $$\displaystyle V_2 = \frac{V_{string}(1+m)}{3+4m+m^2} $$, $$\displaystyle V_3 = \frac{V_{string}(1+3m+m^2)}{3+4m+m^2} $$.

    String Efficiency = $$\displaystyle \frac{V_{string}}{3 V_3} \times 100\% $$ (since $$\displaystyle V_3 $$ is max).

• Cable Stress & Dimensions:

  • Given $$\displaystyle E_{max} $$, $$\displaystyle E_{min} $$, $$\displaystyle r_1 $$ → find $$\displaystyle r_2 $$ and $V$.

  • $$\displaystyle E_{max} = \frac{V}{r_1 \ln(r_2/r_1)} $$, $$\displaystyle E_{min} = \frac{V}{r_2 \ln(r_2/r_1)} $$.

  • $$\displaystyle \frac{E_{max}}{E_{min}} = \frac{r_2}{r_1} \Rightarrow r_2 = r_1 \times \frac{E_{max}}{E_{min}} $$.

  • Then $$\displaystyle V = E_{max} \cdot r_1 \ln(r_2/r_1) $$.

• Transmission Efficiency & Volume of Conductor:

  • Given: $P$, $V$, $\cos\phi$, $\eta$, $l$, $\rho$.

  • For 3-φ: $$\displaystyle P = \sqrt{3} V I \cos\phi \Rightarrow I = P/(\sqrt{3} V \cos\phi) $$.

  • Loss $$\displaystyle = P \left(\frac{1}{\eta} - 1\right) = 3 I^2 R $$.

  • $$\displaystyle R = \frac{\rho l}{a} $$ (per phase).

  • So $$\displaystyle 3 \left(\frac{P}{\sqrt{3} V \cos\phi}\right)^2 \frac{\rho l}{a} = P \left(\frac{1}{\eta} - 1\right) $$.

  • Solve for $a$: $$\displaystyle a = \frac{3 \rho l P}{3 V^2 \cos^2\phi \cdot P \left(\frac{1}{\eta} - 1\right)} = \frac{\rho l}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$? Check:

    $$\displaystyle 3 I^2 R = 3 \times \frac{P^2}{3 V^2 \cos^2\phi} \times \frac{\rho l}{a} = \frac{P^2 \rho l}{V^2 \cos^2\phi a} $$.

    Set equal to $$\displaystyle P \left(\frac{1}{\eta} - 1\right) $$:

    $$\displaystyle \frac{P^2 \rho l}{V^2 \cos^2\phi a} = P \left(\frac{1}{\eta} - 1\right) \Rightarrow a = \frac{P \rho l}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

  • Volume (3-φ): $$\displaystyle Vol_3 = 3 \times a \times l = \frac{3 P \rho l^2}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

  • For 1-φ 2-wire: $$\displaystyle P = 2 V I \cos\phi \Rightarrow I = P/(2V\cos\phi) $$.

    Loss $$\displaystyle = 2 I^2 R = 2 \times \frac{P^2}{4 V^2 \cos^2\phi} \times \frac{\rho l}{a} = \frac{P^2 \rho l}{2 V^2 \cos^2\phi a} $$.

    Set equal to $$\displaystyle P \left(\frac{1}{\eta} - 1\right) $$:

    $$\displaystyle \frac{P^2 \rho l}{2 V^2 \cos^2\phi a} = P \left(\frac{1}{\eta} - 1\right) \Rightarrow a = \frac{P \rho l}{2 V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

    Volume: $$\displaystyle Vol_1 = 2 \times a \times l = \frac{P \rho l^2}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

  • Ratio: $$\displaystyle \frac{Vol_3}{Vol_1} = \frac{3 P \rho l^2 / (...)}{P \rho l^2 / (...)} = 3 $$.

    So 3-φ requires 3 times the conductor volume of 1-φ for same power, voltage, losses, length? That seems off. Let's recalc carefully.

    For 1-φ: $$\displaystyle a_1 = \frac{P \rho l}{2 V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$, $$\displaystyle Vol_1 = 2 a_1 l = \frac{P \rho l^2}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

    For 3-φ: $$\displaystyle a_3 = \frac{P \rho l}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$, $$\displaystyle Vol_3 = 3 a_3 l = \frac{3 P \rho l^2}{V^2 \cos^2\phi \left(\frac{1}{\eta} - 1\right)} $$.

    Yes, $$\displaystyle Vol_3 = 3 Vol_1 $$. But standard result: 3-φ uses less conductor than 1-φ for same power & distance? Wait, for same power and voltage (line-to-line for 3-φ, line-to-neutral for 1-φ?).

    Actually, typical comparison: For same power transmitted and same voltage between conductors (i.e., $$\displaystyle V_{LL} $$ for 3-φ, $V$ for 1-φ where $$\displaystyle V = V_{LL} $$), 3-φ uses less conductor.

    Let's assume same voltage $V$ means same potential difference between any two conductors.

    For 1-φ: $$\displaystyle P_1 = V I_1 \cos\phi $$.

    For 3-φ: $$\displaystyle P_3 = \sqrt{3} V I_3 \cos\phi $$.

    For same $P$, $$\displaystyle I_3 = I_1 / \sqrt{3} $$.

    Losses: 1-φ: $$\displaystyle Loss_1 = 2 I_1^2 R_1 $$; 3-φ: $$\displaystyle Loss_3 = 3 I_3^2 R_3 = 3 (I_1^2/3) R_3 = I_1^2 R_3 $$.

    Set $$\displaystyle Loss_1 = Loss_3 $$: $$\displaystyle 2 I_1^2 R_1 = I_1^2 R_3 \Rightarrow R_3 = 2 R_1 $$.

    Since $R \propto 1/a$, $$\displaystyle a_3 = 2 a_1 $$.

    Volume: $$\displaystyle Vol_1 = 2 a_1 l $$, $$\displaystyle Vol_3 = 3 a_3 l = 3 \times 2 a_1 l = 6 a_1 l $$.

    So 3-φ uses 3 times the volume? That can't be right.

    Ah! The mistake: In 1-φ, the voltage $V$ is across the two conductors. In 3-φ, the phase voltage is $V/\sqrt{3}$ if $V$ is line-to-line. But the problem says "operating line voltage" – for 3-φ, line voltage is $$\displaystyle V_{LL} $$; for 1-φ, line voltage is just $V$. So if we say "same operating line voltage", for 1-φ it's $V$, for 3-φ it's $$\displaystyle V_{LL} = V $$. Then phase voltage for 3-φ is $V/\sqrt{3}$.

    But in power formula for 3-φ: $$\displaystyle P_3 = \sqrt{3} V_{LL} I_3 \cos\phi = \sqrt{3} V I_3 \cos\phi $$.

    For 1-φ: $$\displaystyle P_1 = V I_1 \cos\phi $$.

    So for same $P$, $$\displaystyle I_3 = I_1 / \sqrt{3} $$.

    Now resistance per conductor: $$\displaystyle R = \rho l / a $$.

    Losses:

    1-φ: $$\displaystyle Loss_1 = 2 I_1^2 R_1 = 2 I_1^2 \frac{\rho l}{a_1} $$.

    3-φ: $$\displaystyle Loss_3 = 3 I_3^2 R_3 = 3 (I_1^2/3) \frac{\rho l}{a_3} = I_1^2 \frac{\rho l}{a_3} $$.

    Equal losses: $$\displaystyle 2 I_1^2 \frac{\rho l}{a_1} = I_1^2 \frac{\rho l}{a_3} \Rightarrow a_3 = \frac{a_1}{2} $$.

    Volume:

    1-φ: $$\displaystyle Vol_1 = 2 a_1 l $$.

    3-φ: $$\displaystyle Vol_3 = 3 a_3 l = 3 \times (a_1/2) l = 1.5 a_1 l $$.

    So 3-φ uses less total conductor volume (1.5 vs 2). Percentage additional load when converting 1-φ to 3-φ with same conductor size ($$\displaystyle a_1 = a_3 $$):

    With same $a$, losses will differ. But problem says "percentage line losses remain the same" – meaning same loss fraction of transmitted power.

    Let's solve that version (JUN 2024 question):

    Given: Same conductor size $a$, same voltage $V$ (line-to-line for 3-φ, line-to-neutral? Actually "operating line voltage" for 1-φ is just $V$, for 3-φ is $$\displaystyle V_{LL}=V $$). Same loss fraction $$\displaystyle k = Loss/P $$.

    For 1-φ: $$\displaystyle P_1 = V I_1 \cos\phi $$, $$\displaystyle Loss_1 = 2 I_1^2 R $$, $$\displaystyle R = \rho l / a $$.

    So $$\displaystyle k = \frac{2 I_1^2 R}{P_1} = \frac{2 I_1^2 \rho l / a}{V I_1 \cos\phi} = \frac{2 I_1 \rho l}{a V \cos\phi} $$.

    Thus $$\displaystyle I_1 = \frac{k a V \cos\phi}{2 \rho l} $$.

    Then $$\displaystyle P_1 = V \cdot \frac{k a V \cos\phi}{2 \rho l} \cdot \cos\phi = \frac{k a V^2 \cos^2\phi}{2 \rho l} $$.

    For 3-φ: $$\displaystyle P_3 = \sqrt{3} V I_3 \cos\phi $$, $$\displaystyle Loss_3 = 3 I_3^2 R $$, same $k$:

    $$\displaystyle k = \frac{3 I_3^2 R}{P_3} = \frac{3 I_3^2 \rho l / a}{\sqrt{3} V I_3 \cos\phi} = \frac{\sqrt{3} I_3 \rho l}{a V \cos\phi} $$.

    So $$\displaystyle I_3 = \frac{k a V \cos\phi}{\sqrt{3} \rho l} $$.

    Then $$\displaystyle P_3 = \sqrt{3} V \cdot \frac{k a V \cos\phi}{\sqrt{3} \rho l} \cdot \cos\phi = \frac{k a V^2 \cos^2\phi}{\rho l} $$.

    Ratio: $$\displaystyle \frac{P_3}{P_1} = \frac{k a V^2 \cos^2\phi / (\rho l)}{k a V^2 \cos^2\phi / (2 \rho l)} = 2 $$.

    So 3-φ can transmit twice the power of 1-φ with same conductor size, same voltage, same loss fraction.

    Additional load = $$\displaystyle P_3 - P_1 = P_1 $$.

    Percentage additional = $$\displaystyle \frac{P_3 - P_1}{P_1} \times 100\% = 100\% $$.

    Final Answer: 100%.

    This matches the standard result I recall.

[!TIP] Common Pitfall: Confusing line voltage vs. phase voltage in 3-φ systems. Always clarify: "operating line voltage" means $$\displaystyle V_{LL} $$ for 3-φ, and the voltage between the two conductors for 1-φ. In the conversion problem, we add one conductor to make 3-φ 3-wire, so line voltage remains the same potential difference between any two outer conductors.