⚡AC Power System Fundamentals
Understanding Power Components
Real vs Apparent Power
Real Power (P): Actual work done, measured in watts
Apparent Power (S): Total power, measured in VA
Reactive Power (Q): Energy stored/released, measured in VAR
Example: Motor draws 10 kVA, does 8 kW work
Reactive component: √(10² - 8²) = 6 kVAR
Power factor = 8/10 = 0.8
Why Apparent Power Matters
Conductors sized for current, not just real power
Transformers rated in kVA, not kW
Utility billing often includes demand charges
Poor power factor = higher operating costs
Circuit breakers must handle full current
Design all equipment for apparent power
Common Power Factor Scenarios
Resistive Loads
Incandescent lighting: PF = 1.0
Electric heaters: PF = 1.0
Apparent power = Real power
No reactive component
Ideal but rare in practice
Inductive Loads
Motors: PF = 0.7-0.9 lagging
Fluorescent lights: PF = 0.5-0.95
Transformers: PF = 0.95-0.98
Current lags voltage
Most common in buildings
Capacitive Loads
Power factor correction caps
Long cables (charging current)
Some electronic ballasts
Current leads voltage
Used to improve overall PF
🏭Industrial Power Applications
Motor Drive Systems
Variable Frequency Drive
100 HP motor (74.6 kW)
Drive efficiency: 96%
Drive power factor: 0.95
Input power: 74.6 ÷ 0.96 = 77.7 kW
Apparent power: 77.7 ÷ 0.95 = 81.8 kVA
Size supply for 82 kVA, not 75 kW
Direct-on-Line Motor
50 HP motor, full load PF = 0.85
Full load current: 65A at 480V
Starting current: 6 × 65 = 390A
Starting kVA: √3 × 480 × 390 ÷ 1000 = 324 kVA
Running kVA: √3 × 480 × 65 ÷ 1000 = 54 kVA
Transformer must handle starting surge
Power Distribution Systems
Plant Electrical Load
Connected load: 2,500 kVA
Demand factor: 0.75
Power factor: 0.82
Maximum demand: 2,500 × 0.75 = 1,875 kVA
Real power: 1,875 × 0.82 = 1,538 kW
Size transformer for 2,000 kVA
Feeder Cable Sizing
Load: 500 kVA at 480V, 3-phase
Current: 500,000 ÷ (√3 × 480) = 601A
Derating factors: 0.8 (temperature, conduit)
Required capacity: 601 ÷ 0.8 = 751A
Cable selection: 500 kcmil (755A)
Based on apparent power, not real power
🔧Power Factor Correction Applications
Economic Benefits
Utility Cost Savings
Plant load: 1,000 kW at 0.70 PF
Original demand: 1,000 ÷ 0.70 = 1,429 kVA
After correction to 0.95 PF: 1,000 ÷ 0.95 = 1,053 kVA
Demand reduction: 1,429 - 1,053 = 376 kVA
Monthly savings: 376 kVA × $8/kVA = $3,008
Annual savings: $36,096
System Capacity Release
1,500 kVA transformer at 0.75 PF
Real power capacity: 1,500 × 0.75 = 1,125 kW
After PF correction to 0.95:
New capacity: 1,500 × 0.95 = 1,425 kW
Additional capacity: 1,425 - 1,125 = 300 kW
27% increase without new transformer
Capacitor Bank Design
Fixed Capacitor Bank
Load: 800 kW, improve from 0.75 to 0.90 PF
Required kVAR: 800 × (tan(41.4°) - tan(25.8°))
kVAR needed: 800 × (0.882 - 0.484) = 318 kVAR
Capacitor rating: 350 kVAR (standard size)
Voltage: 480V, 3-phase
Install automatic switching for varying loads
Automatic Power Factor Controller
6-step controller with 50 kVAR steps
Total capacity: 300 kVAR
Target PF: 0.95 ± 0.02
Response time: 30-60 seconds
Prevents hunting and over-correction
Maintains optimal PF automatically
🔋Generator and UPS Applications
Emergency Generator Sizing
Critical Load Analysis
Emergency lighting: 25 kW
Fire pumps: 200 kW at 0.8 PF
HVAC (minimum): 150 kW at 0.85 PF
Data center: 300 kW at 0.95 PF
Total real power: 675 kW
Weighted PF: 0.87
Required capacity: 675 ÷ 0.87 = 776 kVA
Size for 900 kVA generator
Motor Starting Consideration
75 HP fire pump motor starting
Running load: 776 kVA
Motor LRA: 75 × 746 ÷ (0.8 × 480 × 1.73) = 84A per phase
Starting kVA: 6 × 84 × 480 × 1.73 ÷ 1000 = 418 kVA
Total during start: 776 + 418 = 1,194 kVA
Need 1,250 kVA generator for starting
UPS System Design
IT Load Sizing
Server load: 180 kW IT power
UPS efficiency: 94%
UPS power factor: 0.9 output
UPS kW rating needed: 180 ÷ 0.94 = 191 kW
UPS kVA rating: 191 ÷ 0.9 = 213 kVA
Select 250 kVA UPS (includes growth)
Battery Runtime Calculation
UPS load: 150 kW at 0.9 PF
DC power required: 150 ÷ 0.94 = 160 kW
Battery voltage: 480V DC nominal
DC current: 160,000 ÷ 480 = 333A
For 15-min runtime: 333A × 0.25hr = 83 Ah
Size for 100 Ah battery bank
📊Power Measurement and Monitoring
Power Quality Meters
Key Measurements
Voltage THD: Should be < 5%
Current THD: Varies by load type
Power Factor: Both displacement and true
Demand: 15-minute rolling average
Energy: kWh and kVARh
Monitor trends, not just instantaneous values
Data Analysis Example
Peak demand: 1,250 kVA at 2:30 PM
Average PF during peak: 0.78
Peak real power: 1,250 × 0.78 = 975 kW
Improvement opportunity: Raise PF to 0.95
New demand: 975 ÷ 0.95 = 1,026 kVA
Potential savings: 224 kVA demand reduction
Energy Management Systems
Demand Response
Peak demand warning at 1,400 kVA
Current load: 1,450 kVA
Shed non-critical loads: 100 kVA
Reduced demand: 1,350 kVA
Avoided peak charge: 50 kVA × $15 = $750/month
Automated load shedding saves money
Load Forecasting
Historical peak: 1,200 kVA (last year)
Planned additions: 200 kVA
Energy efficiency projects: -50 kVA
Growth factor: 5% annually
Projected peak: (1,200 + 200 - 50) × 1.05 = 1,418 kVA
Plan infrastructure for 1,500 kVA
💡Apparent Power Conversion Tips
Quick Calculations
Power Factor Rules
Unity PF (1.0): kW = kVA
0.9 PF: kVA = kW ÷ 0.9
0.8 PF: kVA = kW × 1.25
Motor rule: kVA ≈ HP × 1.0 to 1.2
Current Calculations
Single-phase: I = kVA × 1000 ÷ V
Three-phase: I = kVA × 1000 ÷ (√3 × V)
480V, 3-phase: I = kVA × 1.2
Design Guidelines
Sizing Factors
Transformers: Size for kVA, not kW
Generators: Include 25% margin
UPS systems: Consider efficiency losses
Cables: Size for full load current
Typical Power Factors
LED lighting: 0.9-0.95
Induction motors: 0.75-0.85
Electronics: 0.6-0.9
Mixed commercial: 0.85-0.90