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Three Phase Electrical

Reference data and engineering information about three phase electrical for electrical applications.

threephaseelectrical

Overview

Engineering reference data for Three Phase Electrical in electrical engineering.

Key Formulas

Ohm's Law

V=IRV = IR

Voltage = Current × Resistance.

Power

P=VI=I2R=V2/RP = VI = I^2R = V^2/R

Electrical power.

Energy

E=PtE = Pt

Energy = Power × Time.

Variables

SymbolDescriptionUnit
VVVoltageV
IICurrentA
RRResistanceΩ
PPPowerW

Power Factor Reference Data

9 rows
Typical Power Factors for Common Three-Phase Devices
Device
Power Factor
Lamp, fluorescent uncompensated0.5
Lamp, fluorescent compensated0.93
Lamp, incandescent1
Motor, induction 100% load0.85
Motor, induction 50% load0.73
Motor, induction 0% load0.17
Motor, synchronous0.9
Oven, resistive heating element1
Oven, induction compensated0.85

Source: engineeringtoolbox.com

Additional Formulas

Real power calculations for three-phase systems:

Using line-to-line voltage: Wapplied=3UllIcos(ϕ)=3UllIPFW_{\text{applied}} = \sqrt{3} \, U_{ll} \, I \, \cos(\phi) = \sqrt{3} \, U_{ll} \, I \, \text{PF}

Using line-to-neutral voltage: Wapplied=3UlnIcos(ϕ)W_{\text{applied}} = 3 \, U_{ln} \, I \, \cos(\phi)

Total power in a balanced three-phase system: W=3UIW = \sqrt{3} \, U \, I

Brake horsepower (actual shaft power): WBHP=3UIPFμ746W_{BHP} = \frac{\sqrt{3} \, U \, I \, \text{PF} \, \mu}{746} where μ\mu is device efficiency (decimal).

Important Definitions and Properties

Power Factor (PF) is defined as cos(ϕ)\cos(\phi), where ϕ\phi is the phase angle between voltage and current. It ranges from 0 to 1:

  • Pure resistive loads (e.g., incandescent lamps, resistive heaters) have PF=1\text{PF} = 1 (current and voltage are in phase).
  • Inductive loads (e.g., motors, solenoids, relays) have PF<1\text{PF} < 1 (current lags voltage).

Load Type Characteristics:

  • Resistive loads convert current directly into other forms of energy, primarily heat.
  • Inductive loads use magnetic fields for operation (e.g., motors, solenoids, relays).

Example Calculation: For a pure resistive load (PF=1\text{PF} = 1) in a 400 V (line-to-line) / 230 V (line-to-neutral) system with 20 A current: Wapplied=3×400V×20A×1=13,856W13.9kWW_{\text{applied}} = \sqrt{3} \times 400 \, \text{V} \times 20 \, \text{A} \times 1 = 13,856 \, \text{W} \approx 13.9 \, \text{kW}

References