3-Phase Motor Power Calculator

3-Phase Power Formula:

\[ P = \sqrt{3} \times V \times I \times PF \times \eta / 1000 \]

Voltage (V):

Current (I):

Power Factor (PF):

(0-1)

Efficiency (η):

(0-1)

Power (P):

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1. What is 3-Phase Power Calculation?

The 3-phase power calculation determines the real power consumed by a 3-phase motor or electrical system. It accounts for voltage, current, power factor, and efficiency to provide accurate power measurement in kilowatts (kW).

2. How Does the Calculator Work?

The calculator uses the 3-phase power formula:

\[ P = \sqrt{3} \times V \times I \times PF \times \eta / 1000 \]

Where:

\( P \) — Power in kilowatts (kW)
\( V \) — Line-to-line voltage (V)
\( I \) — Current (A)
\( PF \) — Power factor (0-1)
\( \eta \) — Motor efficiency (0-1)
\( \sqrt{3} \) — Constant for 3-phase systems (≈1.732)

Explanation: The formula calculates real power by accounting for both the apparent power (V×I) and the power factor (cosine of phase angle between voltage and current).

3. Importance of Power Calculation

Details: Accurate power calculation is essential for sizing electrical components, determining energy consumption, calculating operating costs, and ensuring proper motor selection for applications.

4. Using the Calculator

Tips: Enter line-to-line voltage in volts, current in amps, power factor (typically 0.8-0.95), and motor efficiency (typically 0.85-0.95 for most industrial motors).

5. Frequently Asked Questions (FAQ)

Q1: What's the difference between real and apparent power?
A: Real power (kW) is the actual power consumed, while apparent power (kVA) is the product of voltage and current. The ratio is the power factor.

Q2: Why divide by 1000 in the formula?
A: This converts the result from watts to kilowatts (1 kW = 1000 W) for more practical engineering units.

Q3: What's a typical power factor for motors?
A: Induction motors typically have power factors between 0.80-0.95 at full load, lower when lightly loaded.

Q4: How does efficiency affect power calculation?
A: Efficiency accounts for losses between electrical input and mechanical output power. Higher efficiency means more input power is converted to useful work.

Q5: Can this be used for single-phase systems?
A: No, single-phase systems use P = V × I × PF / 1000 (without the √3 factor).