18.3 Distribution Systems, Substation Layouts & Power Flow Control

18.3 Distribution Systems, Substation Layouts & Power Flow Control

Power distribution engineering, substation busbar schemes, load flow formulation, and reactive power compensation are primary operational subjects evaluated on the PRC Registered Electrical Engineer (REE) Licensure Examination. Distribution networks transport electric energy from primary bulk substations to commercial, industrial, and residential end consumers. System engineers must maintain voltage regulation within statutory boundaries (typically $\pm 5\%$ mandated by the Philippine Distribution Code) while optimizing feeder efficiency and operational continuity.

---

1. Distribution System Topologies & Feeder Voltage Drop Calculations

Electric distribution networks are divided into Primary Distribution ($13.8\ \text{kV}, 34.5\ \text{kV}$ in Meralco/electric cooperative grids) and Secondary Distribution ($230\ \text{V}$ single-phase 2-wire/3-wire, $230/400\ \text{V}$ or $480/277\ \text{V}$ 3-phase 4-wire).

Distribution Network Topologies

1. Radial System: Feeders branch outward from a single substation source without downstream loop ties. Low initial capital cost, simple protective relaying. Disadvantage: Any fault on the main feeder de-energizes all downstream customers (zero redundancy).
2. Ring Main (Loop) System: Feeders originate from a substation, loop through load centers, and return to the same or adjacent substation. Closed ring operation with isolator switches permits isolating fault sections while maintaining continuous dual-feed supply to all healthy loads.
3. Interconnected (Mesh) System: Multiple substations feed a interconnected grid network. Offers maximum service reliability and minimal voltage drop; utilized in high-density commercial business districts.

Feeder Voltage Drop Calculations

Uniformly Loaded Distributor Fed at One End:

Consider a distributor of length $L$ carrying uniform current loading $i$ (A/m) with line impedance $z$ ($\Omega$/m). Total current $I_{\text{total}} = i \cdot L$ and total impedance $Z_{\text{total}} = z \cdot L$.

The voltage drop at distance $x$ from the feeding end is:

$$V_{\text{drop}}(x) = \int_0^x i (L - y) z\ dy = i z \left( L x - \frac{x^2}{2} \right)$$

Maximum voltage drop at the far end ($x = L$):

$$V_{\text{drop, max}} = \frac{1}{2} i z L^2 = \frac{1}{2} (i L) (z L) = \frac{1}{2} I_{\text{total}} Z_{\text{total}}$$

REE Board Exam Principle: A uniformly loaded distributor fed at one end has a total voltage drop equal to an equivalent single concentrated load $I_{\text{total}}$ connected at its midpoint ($L/2$).

Uniformly Loaded Distributor Fed at Both Ends (Equal Voltage $V_0$):

Minimum voltage occurs at the exact midpoint ($x = L/2$). Maximum voltage drop is:

$$V_{\text{drop, max}} = \frac{1}{8} i z L^2 = \frac{1}{8} I_{\text{total}} Z_{\text{total}}$$

---

2. Power System Load Characteristics & Demand Metrics

Distribution transformers and feeder conductors are sized using statistical load indices.

---

3. Substation Layouts, Switchgear & Grounding Design

Substations transform voltage levels, route power, and protect grid apparatus.

Primary Substation Busbar Schemes

1. Single Busbar Scheme: Lowest cost, single bus. Breaker maintenance requires completely de-energizing all connected feeders.
2. Double Busbar with Single Breaker: Main and Transfer buses connected via bus-coupler breaker. Permits transferring feeders between buses without power interruption, but breaker maintenance requires feeder outage.
3. Double Busbar with Double Breaker: Each feeder has two circuit breakers connected to independent buses. Highest reliability, allows servicing any breaker without feeder interruption. High capital cost.
4. Breaker-and-a-Half Scheme: Three circuit breakers shared between two feeders (1.5 breakers per feeder). Offers supreme security and flexibility; any breaker can be isolated for maintenance without disconnecting any feeder line.
5. Ring Bus Scheme: Breakers arranged in a closed ring. Each feeder connected between two breakers. Compact footprint, no main busbar required; opening any single breaker maintains all feeders in service.

Substation Grounding Grid (IEEE Std 80 Safety Limits)

A metallic grounding grid (copper mesh buried $0.5\ \text{m}$ deep) maintains low earth resistance ($R_g < 1.0\ \Omega$). Ground grid design prevents fatal electric shocks during single-line-to-ground fault conditions by enforcing safety thresholds:

• Touch Voltage ($V_{\text{touch}}$): Potential difference between grounded metal equipment structure and human feet standing $1\ \text{m}$ away.
• Step Voltage ($V_{\text{step}}$): Potential difference between human feet separated by $1\ \text{m}$ distance on the ground surface without touching any structure.

---

4. Power Flow Analysis & Voltage / VAR Control Methods

Power flow (load flow) calculations determine steady-state voltage magnitudes ($|V_i|$), phase angles ($\theta_i$), active power ($P_i$), and reactive power ($Q_i$) across all system buses.

Bus Classification in Load Flow Analysis

Nodal Admittance Matrix ($Y_{\text{bus}}$) & Power Flow Equations

The $N$-bus system nodal admittance matrix equation is $\mathbf{I}_{\text{bus}} = \mathbf{Y}_{\text{bus}} \mathbf{V}_{\text{bus}}$. The non-linear complex power equation at bus $i$ is:

$$S_i^* = P_i - j Q_i = V_i^* \sum_{j=1}^N Y_{ij} V_j$$

• Gauss-Seidel Method: Iterative nodal formulation. Low memory requirements, but slow linear convergence rate (increases with large bus counts).
• Newton-Raphson Method: Iterative method based on Taylor series expansion utilizing the Jacobian Matrix ($[J]$) of partial derivatives. Exhibits fast quadratic convergence, making it the industry standard for large multi-bus grids.

Reactive Power (VAR) & Voltage Control

1. Under-Load Tap-Changing (ULTC) Transformers: Mechanically adjusts primary-to-secondary turns ratio $a$ under energized load to raise distribution secondary voltage.
2. Shunt Capacitor Compensation: Connects parallel capacitor banks at load substations to supply leading reactive power ($Q_c$), offsetting lagging inductive load current. Required capacitor bank rating for power factor correction from $\text{pf}_1 = \cos \theta_1$ to $\text{pf}_2 = \cos \theta_2$: $$Q_c = P \left( \tan \theta_1 - \tan \theta_2 \right) \quad (\text{kVAR})$$
   • Benefits: Elevates feeder voltage magnitude ($V_R$), reduces total feeder current ($I = \frac{P}{\sqrt{3} V \cos \phi}$), and dramatically cuts line copper losses ($P_{\text{loss}} \propto I^2 R$).
3. Series Capacitors: Connected in series with transmission line conductors to cancel line inductive reactance ($X_{\text{net}} = X_L - X_C$). This increases maximum steady-state power transfer capability: $$P_{\max} = \frac{|V_1| |V_2|}{X_L - X_C} \sin \delta$$
4. Static VAR Compensators (SVC / STATCOM): Power electronics-based dynamic reactive power compensators providing rapid sub-cycle VAR absorption or injection.

---

Solved Board Exam Examples

Example 1: Feeder Voltage Drop Calculation

Problem: A $3$-phase, $13.8\ \text{kV}$ primary distribution feeder carries three concentrated loads along a $6\ \text{km}$ route from the main substation:

• Load 1 at $2\ \text{km}$: $100\ \text{A}$ at $0.80$ pf lagging.
• Load 2 at $4\ \text{km}$: $150\ \text{A}$ at $0.85$ pf lagging.
• Load 3 at $6\ \text{km}$: $80\ \text{A}$ at $0.90$ pf lagging.

Feeder line impedance is $z = 0.25 + j 0.35\ \Omega/\text{km/phase}$. Calculate total line-to-line voltage drop across the entire $6\ \text{km}$ feeder.

Solution:

1. Calculate branch currents (working backwards from far end $6\ \text{km}$ to source):
   • Branch 3 ($4 \text{ to } 6\ \text{km}$, length $2\ \text{km}$): Carries Load 3 current: $$I_3 = 80 (0.90 - j 0.4359) = 72.0 - j 34.87\ \text{A}$$
   • Branch 2 ($2 \text{ to } 4\ \text{km}$, length $2\ \text{km}$): Carries Load 3 + Load 2 current: $$I_2 = (72.0 - j 34.87) + 150(0.85 - j 0.5268) = (72.0 - j 34.87) + (127.5 - j 79.02) = 199.5 - j 113.89\ \text{A}$$
   • Branch 1 ($0 \text{ to } 2\ \text{km}$, length $2\ \text{km}$): Carries total current (Load 3 + Load 2 + Load 1): $$I_1 = (199.5 - j 113.89) + 100(0.80 - j 0.60) = (199.5 - j 113.89) + (80.0 - j 60.0) = 279.5 - j 173.89\ \text{A}$$
2. Branch impedance for each $2\ \text{km}$ section ($Z_{\text{sec}}$): $$Z_{\text{sec}} = 2 \times (0.25 + j 0.35) = 0.50 + j 0.70\ \Omega$$
3. Calculate line-to-neutral voltage drop for each section: $$\Delta V_1 = (279.5 - j 173.89)(0.50 + j 0.70) = (139.75 + 121.72) + j(195.65 - 86.95) = 261.47 + j 108.70\ \text{V}$$ $$\Delta V_2 = (199.5 - j 113.89)(0.50 + j 0.70) = (99.75 + 79.72) + j(139.65 - 56.95) = 179.47 + j 82.70\ \text{V}$$ $$\Delta V_3 = (72.0 - j 34.87)(0.50 + j 0.70) = (36.00 + 24.41) + j(50.40 - 17.44) = 60.41 + j 32.96\ \text{V}$$
4. Calculate total line-to-neutral voltage drop ($\Delta V_{\text{LN}}$): $$\Delta V_{\text{LN}} = \Delta V_1 + \Delta V_2 + \Delta V_3 = (261.47 + 179.47 + 60.41) + j(108.70 + 82.70 + 32.96) = 501.35 + j 224.36\ \text{V}$$ $$\left| \Delta V_{\text{LN}} \right| = \sqrt{501.35^2 + 224.36^2} = \sqrt{251,351.8 + 50,337.4} = \sqrt{301,689.2} = 549.26\ \text{V}$$
5. Calculate total line-to-line voltage drop: $$\Delta V_{L-L} = 549.26 \times \sqrt{3} = 951.34\ \text{V}$$

---

Example 2: Shunt Capacitor Compensation & Loss Reduction

Problem: An industrial facility connected to a $3$-phase, $13.8\ \text{kV}$ distribution system draws an active load $P = 1,200\ \text{kW}$ at a power factor of $0.72$ lagging. Calculate: (a) Required capacitor bank kVAR rating to elevate power factor to $0.95$ lagging, (b) Percentage reduction in line current, and (c) Percentage reduction in feeder $I^2 R$ copper losses.

Solution:

1. Calculate power factor angles and initial/target reactive power: $$\theta_1 = \arccos(0.72) = 43.945^\circ \implies \tan \theta_1 = 0.9640$$ $$\theta_2 = \arccos(0.95) = 18.195^\circ \implies \tan \theta_2 = 0.3287$$ $$Q_1 = P \tan \theta_1 = 1200 \times 0.9640 = 1,156.8\ \text{kVAR}$$ $$Q_2 = P \tan \theta_2 = 1200 \times 0.3287 = 394.4\ \text{kVAR}$$
2. Calculate required shunt capacitor rating ($Q_c$): $$Q_c = Q_1 - Q_2 = 1,156.8 - 394.4 = 762.4\ \text{kVAR}$$
3. Calculate initial current ($I_1$) and improved current ($I_2$): $$I_1 = \frac{1,200,000}{\sqrt{3} \times 13,800 \times 0.72} = 69.74\ \text{A}$$ $$I_2 = \frac{1,200,000}{\sqrt{3} \times 13,800 \times 0.95} = 52.86\ \text{A}$$
4. Calculate percentage current reduction: $$\% \text{Current Reduction} = \frac{69.74 - 52.86}{69.74} \times 100\% = \frac{16.88}{69.74} \times 100\% = 24.21\%$$
5. Calculate percentage reduction in $I^2 R$ line losses: $$\% \text{Loss Reduction} = \left[ 1 - \left( \frac{I_2}{I_1} \right)^2 \right] \times 100\% = \left[ 1 - \left( \frac{52.86}{69.74} \right)^2 \right] \times 100\% = \left[ 1 - 0.5744 \right] \times 100\% = 42.56\%$$