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11.2 Utility Systems, Transformers & Switchgear
PRC Registered Electrical Engineer REE
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1 Chapter 1: Overview, RA 7920 & Electrical Engineering Ethics
1.1 PRC Board of Electrical Engineering & REE Examination Overview 1.2 Republic Act No. 7920 (New Electrical Engineering Law) & License Classes 1.3 Code of Ethics & Professional Practice for Electrical Engineers
2 Chapter 2: Math — Algebra, Trig & Geometry
2.1 Algebraic Equations, Polynomials & Complex Numbers 2.2 Trigonometry — Identities, Triangles & Spherical Trig 2.3 Analytic & Solid Geometry
3 Chapter 3: Engineering Mathematics — Calculus & Differential Equations
3.1 Differential Calculus, Limits, Derivatives & Applications 3.2 Integral Calculus, Integration Techniques & Area/Volume Applications 3.3 Ordinary Differential Equations & Initial-Value Problems
4 Chapter 4: Engineering Mathematics — Advanced Math & Transform Methods
4.1 Laplace Transforms, Inverse Transforms & Circuit Applications 4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis 4.3 Vector Analysis, Linear Algebra, Matrices & Complex Analysis
5 Chapter 5: Engineering Mathematics — Probability, Statistics & Engineering Data Analysis
5.1 Probability Theory, Counting Techniques & Bayes' Theorem 5.2 Engineering Data Analysis I — Descriptive Statistics & Probability Distributions 5.3 Engineering Data Analysis II — Confidence Intervals, Hypothesis Testing & Regression
6 Chapter 6: Engineering Mathematics — Numerical Methods & Analysis
6.1 Numerical Root-Finding & Solution of Linear Systems 6.2 Interpolation, Numerical Integration & Numerical Solution of ODEs
7 Chapter 7: ESAS — Physics, Chemistry & Computer Fundamentals
7.1 Physics for Engineers 7.2 General Chemistry & Materials Science 7.3 Computer Fundamentals, Programming & Data Logic 7.4 Logic Circuits, Switching Theory & Microprocessor Systems
8 Chapter 8: ESAS — Mechanics & Thermal-Fluid Engineering
8.1 Engineering Mechanics — Statics & Dynamics 8.2 Strength of Materials & Mechanics of Deformable Bodies 8.3 Thermal-Fluid Engineering — Thermodynamics & Fluid Mechanics
9 Chapter 9: ESAS — Engineering Economics, Management & Contracts
9.1 Engineering Economics & Depreciation 9.2 Engineering Management & Ethics 9.3 Philippine Electrical Engineering Law, Code & Contracts 9.4 Technopreneurship 101 & Management of Engineering Projects
10 Chapter 10: PEC 1 — Safety, Sizing & Protection
10.1 PEC Fundamentals, Safety Standards & Grounding/Bonding 10.2 Conductor Sizing, Ampacity Adjustment & Voltage Drop 10.3 Overcurrent Protection, Branch Circuit & Feeder Sizing 10.4 Basic Occupational Safety & Health (BOSH) and Electrical Safety Standards
11 Chapter 11: PEC 1 & 2 — Equipment & Utility Systems
11.1 PEC Equipment Installation & Wiring Methods 11.2 Utility Systems, Transformers & Switchgear 11.3 Special Occupancies & Hazardous Locations
12 Chapter 12: Electrical Professional — DC & AC Circuit Fundamentals
12.1 DC Circuit Analysis & Network Theorems 12.2 Single-Phase AC Circuits & Phasor Analysis 12.3 Balanced & Unbalanced Three-Phase AC Circuits
13 Chapter 13: Electrical Professional — EM Fields & Transients
13.1 Electrostatics, Magnetostatics & Maxwell's Equations 13.2 Electromagnetic Waves & Transmission Line Fields 13.3 First-Order & Second-Order Circuit Transients
14 Chapter 14: Electrical Professional — Electronics, Industrial Electronics & Communications
14.1 Electronics 1 & 2 — Semiconductor Devices & Analog Circuits 14.2 Electrical Apparatus & Devices — Industrial Electronics & Power Converters 14.3 Fundamentals of Electronic Communications
15 Chapter 15: Electrical Professional — Transformers & DC Machines
15.1 Single-Phase & Three-Phase Transformers 15.2 Transformer Connections & Special Transformers 15.3 DC Generators & DC Motors
16 Chapter 16: Electrical Professional — AC Synchronous & Induction Machinery
16.1 Synchronous Generators (Alternators) 16.2 Synchronous Motors & Special AC Machines 16.3 Three-Phase & Single-Phase Induction Motors
17 Chapter 17: Electrical Professional — Power Plants & Energy Conversion
17.1 Hydroelectric, Thermal & Combined Cycle Power Plants 17.2 Renewable Energy Systems & Geothermal Power Plants 17.3 Power Plant Economics, Load Curves & Energy Conversion
18 Chapter 18: Electrical Professional — Transmission & Distribution
18.1 Transmission Line Parameters & Performance Models 18.2 Mechanical Overhead Line Design, Underground Cables & Insulators 18.3 Distribution Systems, Substation Layouts & Power Flow Control
19 Chapter 19: Electrical Professional — Power System Analysis
19.1 The Per-Unit System, Base Conversion & Impedance Diagrams 19.2 Symmetrical Components & Unbalanced Fault Analysis 19.3 Load Flow, Economic Dispatch & Power System Stability
20 Chapter 20: Electrical Professional — Protection, Control, Instrumentation & Illumination
20.1 Power System Protection & Protective Relaying 20.2 Industrial Motor Control, Logic Circuits & PLC Automation 20.3 Electrical Instrumentation, Transducers & Measurement Systems 20.4 Illumination Engineering, Lighting Design & Photometrics 20.5 Feedback Control Systems & Research Methods
11.2 Utility Systems, Transformers & Switchgear
Key Takeaways Explain This Topic AI
PEC Part 2 specifies electric utility safety clearances for overhead lines over roads (minimum 5.5 m 5.5\ \text{m} 5.5 m for primary voltages up to 22 kV 22\ \text{kV} 22 kV ) and joint-use pole vertical separations ( 1.0 m 1.0\ \text{m} 1.0 m for low voltage, 1.5 m 1.5\ \text{m} 1.5 m for high voltage).
Transformer primary overcurrent protection under PEC Article 4.50 is restricted to a maximum setting of 125% of rated primary full-load current ( I 1 I_1 I 1 ) when secondary protection is omitted for systems ≤ 1000 V \le 1000\ \text{V} ≤ 1000 V , or up to 300% for primary fuses on medium-voltage utility systems ( > 1000 V >1000\ \text{V} > 1000 V ).
Transformer vaults require a minimum 2-hour fire-rated construction (3-hour if no automatic sprinkler system exists), liquid-tight door sills of at least 100 mm 100\ \text{mm} 100 mm height to contain oil leaks, and net ventilation area of 0.006 m 2 0.006\ \text{m}^2 0.006 m 2 per kVA of transformer capacity.
Substation busbar configurations balance reliability and cost, ranging from single bus to breaker-and-a-half and double bus double breaker arrangements where circuit breaker maintenance can occur without interrupting load feeder continuity.
Substation grounding grids employ buried bare copper mesh with a crushed rock surface layer (high resistivity ρ ≈ 3000 Ω ⋅ m \rho \approx 3000\ \Omega\cdot\text{m} ρ ≈ 3000 Ω ⋅ m ) to reduce touch ( V touch V_{\text{touch}} V touch ) and step ( V step V_{\text{step}} V step ) potentials below lethal thresholds during single line-to-ground faults.
Last updated: August 2026
11.2 Utility Systems, Transformers & Switchgear
While PEC Part 1 governs interior building electrical systems, PEC Part 2 (Electrical Safety Code for Utilities) regulates overhead power distribution, transmission line clearances, pole construction, joint-use structures, and utility substation installations. This section explores utility line engineering, high-voltage switchgear bus configurations, transformer protection rules under PEC Article 4.50, transformer vault design, surge arresters, and substation grounding grids.
1. PEC Part 2 Overview & Electric Utility Line Clearances
Electric utility supply lines must maintain strict vertical and horizontal clearances above ground, roads, railways, and buildings to prevent accidental contact and electrical flashover.
Minimum Overhead Conductor Ground Clearances (PEC Part 2)
Clearances are specified at maximum conductor operating temperature ( 60 ∘ C 60^\circ\text{C} 6 0 ∘ C or higher) and worst-case conductor sag:
Crossing / Location Surface Low-Voltage Lines ( ≤ 600 V \le 600\ \text{V} ≤ 600 V ) Primary Lines ( > 600 V >600\ \text{V} > 600 V to 22 kV 22\ \text{kV} 22 kV ) High-Voltage Distribution ( > 22 kV >22\ \text{kV} > 22 kV to 50 kV 50\ \text{kV} 50 kV )
Thoroughfares, Highways, and Main Streets 4.6 m 4.6\ \text{m} 4.6 m (15 ft) 5.5 m 5.5\ \text{m} 5.5 m (18 ft) 6.0 m 6.0\ \text{m} 6.0 m (20 ft)
Residential Driveways & Commercial Parking Lots 4.6 m 4.6\ \text{m} 4.6 m (15 ft) 5.5 m 5.5\ \text{m} 5.5 m (18 ft) 5.8 m 5.8\ \text{m} 5.8 m (19 ft)
Spaces Accessible to Pedestrians Only 3.7 m 3.7\ \text{m} 3.7 m (12 ft) 4.6 m 4.6\ \text{m} 4.6 m (15 ft) 5.0 m 5.0\ \text{m} 5.0 m (16.5 ft)
Railroads (Track Rails) 8.2 m 8.2\ \text{m} 8.2 m (27 ft) 8.5 m 8.5\ \text{m} 8.5 m (28 ft) 9.0 m 9.0\ \text{m} 9.0 m (30 ft)
Joint-Use Pole Structure Clearances
Joint-use poles accommodate both electric supply conductors (top position) and communication lines (bottom position):
Vertical Clearance on Poles: Minimum 1.0 m 1.0\ \text{m} 1.0 m ( 40 in 40\ \text{in} 40 in ) vertical clearance between low-voltage supply conductors ( ≤ 8.7 kV \le 8.7\ \text{kV} ≤ 8.7 kV ) and communication cables. For primary distribution ( > 8.7 kV >8.7\ \text{kV} > 8.7 kV to 22 kV 22\ \text{kV} 22 kV ), vertical separation must be at least 1.5 m 1.5\ \text{m} 1.5 m ( 60 in 60\ \text{in} 60 in ) .
Overhead Line Sag and Tension Mechanics
For a flat horizontal span of length L L L , uniform conductor weight per unit length w w w (N/m), and horizontal conductor tension T T T (N), conductor sag S S S is calculated using the parabolic approximation: S = w L 2 8 T S = \frac{w L^2}{8 T} S = 8 T w L 2
Total conductor length S L S_L S L including elastic elongation and temperature sag: S L = L + 8 S 2 3 L S_L = L + \frac{8 S^2}{3 L} S L = L + 3 L 8 S 2
2. Substation Layout and High-Voltage Switchgear
Electric utility and industrial substations convert transmission/distribution voltage levels, route power across grid feeders, and isolate fault conditions using circuit breakers, disconnect switches, and busbar configurations.
Substation Busbar Topologies
Bus Topology Description Reliability / Flexibility Cost
Single Bus All feeders and transformers connect to a single common busbar Low (Bus fault shuts down entire substation) Lowest
Main and Transfer Bus Includes a main bus plus a transfer bus for servicing individual breakers Moderate (Breaker maintenance without outage) Moderate
Double Bus, Single Breaker Two main buses with selector switches connecting each feeder breaker High (Feeders can switch between buses) High
Ring Bus Bus sections configured in a continuous ring with breakers between segments High (Breaker maintenance requires no feeder outage) High
Breaker-and-a-Half Three breakers for every two feeder circuits connected between two main buses Highest (Complete redundancy; single bus fault loses no feeders) Highest
High-Voltage Circuit Breakers & Disconnect Switches
Circuit Breakers ( SF 6 \text{SF}_6 SF 6 , Vacuum, Air-Blast): Capable of breaking full rated continuous current as well as severe short-circuit fault currents ( I sc I_{\text{sc}} I sc up to 40 – 63 kA 40\text{--}63\ \text{kA} 40 – 63 kA ). Rated by interrupting capacity, operating time (typically 2–3 cycles), and Basic Impulse Insulation Level (BIL).
Air-Break Disconnect Switches: Provide visible isolation point during maintenance. Must never break load current unless specifically equipped with arc-chutes or load-break interrupter attachments.
- 3. Utility & Industrial Transformer Protection & Vault Sizing
- Overcurrent Protection Rules (PEC Article 4.50 / NEC 450.3)
Transformer primary and secondary overcurrent protection (fuses or circuit breakers) prevents thermal damage from overload and internal winding short circuits.
Nominal Voltages ≤ 1000 V \le 1000\ \text{V} ≤ 1000 V :
Primary Protection Only: Overcurrent device setting ≤ 125 % \le 125\% ≤ 125% of primary full-load current ( I 1 I_1 I 1 ). If 125 % × I 1 125\% \times I_1 125% × I 1 does not match a standard fuse/breaker size, the next higher standard rating is permitted up to 168 % 168\% 168% (for I 1 ≥ 9 A I_1 \ge 9\ \text{A} I 1 ≥ 9 A ).
Primary & Secondary Protection: Primary overcurrent device setting ≤ 250 % \le 250\% ≤ 250% of I 1 I_1 I 1 , provided secondary overcurrent protection is installed and set at ≤ 125 % \le 125\% ≤ 125% of secondary full-load current ( I 2 I_2 I 2 ).
Nominal Voltages > 1000 V > 1000\ \text{V} > 1000 V (Medium Voltage Distribution):
Overcurrent Device Type Primary Protection Only Primary Setting (With Secondary Protection) Secondary Setting (With Secondary Protection)
Circuit Breaker 300 % × I 1 300\% \times I_1 300% × I 1 600 % × I 1 600\% \times I_1 600% × I 1 300 % × I 2 300\% \times I_2 300% × I 2
Fuse 250 % × I 1 250\% \times I_1 250% × I 1 300 % × I 1 300\% \times I_1 300% × I 1 250 % × I 2 250\% \times I_2 250% × I 2
Transformer Vault Construction (PEC Article 4.50)
Liquid-filled transformers installed inside buildings must be housed in fireproof transformer vaults:
Fire Resistance: Walls and roofs must have a minimum 2-hour fire-rated construction (3-hour if no automatic water spray fire suppression system exists).
Door Sills (Kerb): Door openings must feature a liquid-tight sill or curb at least 100 mm 100\ \text{mm} 100 mm ( 4 in 4\ \text{in} 4 in ) high to retain spilled transformer dielectric oil.
Ventilation: Net area of ventilation openings must be not less than 0.006 m 2 0.006\ \text{m}^2 0.006 m 2 ( 1 sq ft 1\ \text{sq ft} 1 sq ft ) per kVA of transformer capacity, with a minimum opening area of 1.0 m 2 1.0\ \text{m}^2 1.0 m 2 .
- 4. Surge Arresters and Substation Grounding Grids
- Metal-Oxide Varistor (MOV) Surge Arresters
Surge arresters protect transformer insulation and switchgear from high-voltage transient surges caused by direct or indirect lightning strikes and switching operations. Connected between phase conductors and ground close to transformer bushings.
Substation Grounding Grid Design (IEEE 80 / PEC Part 2)
During a single line-to-ground fault in a high-voltage substation, fault current flowing into the earth elevates the Substation Ground Potential Rise (GPR): GPR = I g × R g \text{GPR} = I_g \times R_g GPR = I g × R g
where I g I_g I g is the grid fault current and R g R_g R g is the substation ground grid resistance.
Touch Voltage ( V touch V_{\text{touch}} V touch ): Maximum potential difference between a grounded metallic structure touched by a person's hand and the earth surface under their feet during a fault.
Step Voltage ( V step V_{\text{step}} V step ): Maximum potential difference between a person's feet spaced 1.0 m 1.0\ \text{m} 1.0 m apart on the ground surface without touching any grounded structure.
Crushed Rock Layer: A surface layer of crushed rock ( 80 – 150 mm 80\text{--}150\ \text{mm} 80 – 150 mm deep) with high electrical resistivity ( ρ s ≈ 3000 Ω ⋅ m \rho_s \approx 3000\ \Omega\cdot\text{m} ρ s ≈ 3000 Ω ⋅ m ) is spread over outdoor substation yards to increase body contact resistance and limit shock currents below dangerous levels.
- Solved Board Exam Examples
- Example 1: Overhead Distribution Line Sag Calculation
Problem: A 13.8 kV 13.8\ \text{kV} 13.8 kV 3-phase overhead utility distribution line uses ACSR conductors with a weight of w = 8.5 N/m w = 8.5\ \text{N/m} w = 8.5 N/m per conductor across a horizontal span of L = 120 m L = 120\ \text{m} L = 120 m between support poles. If maximum allowable horizontal conductor tension is T = 3600 N T = 3600\ \text{N} T = 3600 N , calculate: (a) the conductor sag S S S at mid-span, and (b) total conductor length S L S_L S L along the span.
Solution:
Calculate mid-span sag S S S using parabolic equation: S = w L 2 8 T = 8.5 × ( 120 ) 2 8 × 3600 S = \frac{w L^2}{8 T} = \frac{8.5 \times (120)^2}{8 \times 3600} S = 8 T w L 2 = 8 × 3600 8.5 × ( 120 ) 2 S = 8.5 × 14 , 400 28 , 800 = 122 , 400 28 , 800 = 4.25 m S = \frac{8.5 \times 14,400}{28,800} = \frac{122,400}{28,800} = 4.25\ \text{m} S = 28 , 800 8.5 × 14 , 400 = 28 , 800 122 , 400 = 4.25 m
Calculate total conductor length along the parabolic curve ( S L S_L S L ): S L = L + 8 S 2 3 L = 120 + 8 × ( 4.25 ) 2 3 × 120 S_L = L + \frac{8 S^2}{3 L} = 120 + \frac{8 \times (4.25)^2}{3 \times 120} S L = L + 3 L 8 S 2 = 120 + 3 × 120 8 × ( 4.25 ) 2 S L = 120 + 8 × 18.0625 360 = 120 + 144.5 360 = 120 + 0.4014 = 120.4014 m S_L = 120 + \frac{8 \times 18.0625}{360} = 120 + \frac{144.5}{360} = 120 + 0.4014 = 120.4014\ \text{m} S L = 120 + 360 8 × 18.0625 = 120 + 360 144.5 = 120 + 0.4014 = 120.4014 m
Example 2: Medium-Voltage Transformer Primary Fuse Sizing & Fault Calculation
Problem: A 1000 kVA 1000\ \text{kVA} 1000 kVA , 13.8 kV / 480 V 13.8\ \text{kV} / 480\ \text{V} 13.8 kV /480 V , 3-phase delta-wye utility transformer has an impedance of Z = 5.75 % Z = 5.75\% Z = 5.75% . The primary side is protected by fuses (primary protection only). Calculate: (a) rated primary full-load current ( I 1 I_1 I 1 ), (b) maximum allowable primary fuse rating under PEC Article 4.50 ( > 1000 V >1000\ \text{V} > 1000 V primary fuse limit = 250 % = 250\% = 250% ), and (c) maximum 3-phase symmetrical short-circuit current ( I sc I_{\text{sc}} I sc ) on the secondary 480 V 480\ \text{V} 480 V busbar assuming an infinite primary utility source.
Solution:
Calculate rated primary full-load current ( I 1 I_1 I 1 ): I 1 = S 3 × V 1 = 1 , 000 , 000 3 × 13 , 800 = 1 , 000 , 000 23 , 899.79 = 41.84 A I_1 = \frac{S}{\sqrt{3} \times V_1} = \frac{1,000,000}{\sqrt{3} \times 13,800} = \frac{1,000,000}{23,899.79} = 41.84\ \text{A} I 1 = 3 × V 1 S = 3 × 13 , 800 1 , 000 , 000 = 23 , 899.79 1 , 000 , 000 = 41.84 A
Calculate maximum primary fuse rating ( 250 % 250\% 250% of I 1 I_1 I 1 for > 1000 V >1000\ \text{V} > 1000 V primary fuse only): I fuse, max = 2.50 × 41.84 A = 104.6 A I_{\text{fuse, max}} = 2.50 \times 41.84\ \text{A} = 104.6\ \text{A} I fuse, max = 2.50 × 41.84 A = 104.6 A
Standard compliant fuse size selection: 100 A 100\ \text{A} 100 A primary fuse.
Calculate rated secondary full-load current ( I 2 I_2 I 2 ): I 2 = S 3 × V 2 = 1 , 000 , 000 3 × 480 = 1 , 000 , 000 831.384 = 1202.8 A I_2 = \frac{S}{\sqrt{3} \times V_2} = \frac{1,000,000}{\sqrt{3} \times 480} = \frac{1,000,000}{831.384} = 1202.8\ \text{A} I 2 = 3 × V 2 S = 3 × 480 1 , 000 , 000 = 831.384 1 , 000 , 000 = 1202.8 A
Calculate 3-phase short-circuit current on 480 V 480\ \text{V} 480 V secondary busbar ( I sc I_{\text{sc}} I sc ): I sc = I 2 % Z / 100 = 1202.8 0.0575 = 20 , 918.3 A ≈ 20.92 kA I_{\text{sc}} = \frac{I_2}{\%Z / 100} = \frac{1202.8}{0.0575} = 20,918.3\ \text{A} \approx 20.92\ \text{kA} I sc = % Z /100 I 2 = 0.0575 1202.8 = 20 , 918.3 A ≈ 20.92 kA