Three power calculations from single phase and three phase loads

Three-phase power — single-phase + three-phase loads explained (clean, compact & worked examples)

Key rules 

• Power adds. Whether a load is single-phase or three-phase, its real power (kW) adds algebraically.

• A statement like “three-phase load = 90 kW” means total three-phase real power = 90 kW (i.e., 30 kW per phase in a balanced system), not 90 kW per phase.

• For balanced three-phase loads, you can convert total ↔ per-phase by dividing or multiplying by 3:

• $P_{\text{phase}} = P_{\text{total}}/3$

• $P_{\text{total}} = 3 \times P_{\text{phase}}$

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Important formulas

Total three-phase real power (line-to-line voltage)

$$P_{\text{3φ}}\;(\text{kW})=\dfrac{\sqrt{3}\;V_{L-L}\;I\;PF}{1000}$$

where $V_{L-L}$ is line-to-line RMS voltage, $I$ is line current (A), and $PF$ is power factor.

Total three-phase real power (line-to-neutral / per-phase voltage)

$$P_{\text{3φ}}\;(\text{kW})=\dfrac{3\;V_{L\!-\!N}\;I_{\text{ph}}\;PF}{1000}$$

where $V_{L-N}=V_{L-L}/\sqrt{3}$ and $I_{\text{ph}}$ is phase current.

Single-phase apparent power and current (per phase)

• Apparent power per phase (kVA): $S_{\text{ph}}=\dfrac{P_{\text{total}}}{3\times PF}$ (kVA per phase)

• Phase current (A), using phase voltage $V_{L-N}$:

$$I_{\text{ph}}=\dfrac{S_{\text{ph}}\times 1000}{V_{L-N}}=\dfrac{P_{\text{total}}}{3\times PF \times V_{L-N}}$$

Apparent power and PF reminders

• Apparent power $S$ (kVA) = $P$ (kW) / PF.

• For resistive loads PF ≈ 1. For induction motors PF ≈ 0.8 (full load typical) — values vary by machine and loading.

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Worked numeric examples (step-by-step)

Example A — Using line-to-line voltage formula

Given: total three-phase real power $P_{\text{3φ}}=90\ \text{kW}$, $PF=0.8$, $V_{L-L}=415\ \text{V}$.
Solve for line current $I$.

Use $I=\dfrac{P_{\text{3φ}}\times 1000}{\sqrt{3}\;V_{L-L}\;PF}$.

Compute digit-by-digit:

• $\sqrt{3}\approx 1.732$

• Denominator = $1.732 \times 415 \times 0.8 = 1.732 \times 332 = 574.624$ (approx)

• $I = \dfrac{90{,}000}{574.624} \approx 156.5\ \text{A}$.

So line current ≈ 156.5 A.

Example B — Per-phase view (same result)

Total $P=90\ \text{kW}$ → per-phase real power $P_{\text{ph}}=90/3=30\ \text{kW}$.
Assume phase voltage $V_{L-N}=240\ \text{V}$ (so $V_{L-L}\approx 415\ \text{V}$).
Phase current $I_{\text{ph}} = \dfrac{P_{\text{ph}}\times 1000}{V_{L-N}\times PF}=\dfrac{30{,}000}{240\times0.8}=\dfrac{30{,}000}{192}=156.25\ \text{A}$.

Matches Example A (rounding differences aside).

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How to combine single-phase and three-phase loads

1. Compute real power (kW) for each load (single-phase loads: $P = V \times I \times PF$; three-phase loads: use three-phase formulas above).

2. Sum real powers to get total system kW: $P_{\text{total}} = P_{3\phi\_loads} + P_{1\phi\_loads}$.

   • Example: three-phase equipment = 90 kW, single-phase equipment spread across phases = 15 kW total → system total = 105 kW.

3. If you need currents per phase, convert the total real power assigned to each phase to per-phase kW:

   • If single-phase loads are connected unevenly across phases, the system becomes unbalanced; compute each phase’s power/current separately and include neutral current by vector (or by arithmetic if you use magnitudes and phase angles).

4. Neutral current: for unbalanced single-phase loads, neutral current is the vector sum of phase currents (can be significant if loads are unbalanced).

5. Billing / demand meters: utilities typically measure total demand (kW/kVA) differently (e.g., three-phase demand meter), so you are billed on total demand — you won’t be billed 3× the total just because each phase carries some load. (But check local tariff rules for peak demand measurement method.)

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Balanced vs Unbalanced — practical notes

• Balanced assumption (equal load per phase): simplifies calculations by dividing by 3. Valid for large motors, balanced three-phase loads, and many industrial setups.

• Unbalanced conditions (typical in domestic/commercial where many single-phase loads exist): you must compute per-phase currents from actual single-phase loads. Neutral currents and phase voltage drops become important; protection settings and conductor sizing must be verified for unbalance.

• For accurate protection, conductor sizing, and harmonic analysis, treat the single-phase loads individually (don’t just divide totals by 3).

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Short checklist for calculation

1. Convert each load to kW (and note PF).

2. Sum kW to get $P_{\text{total}}$.

3. If you need line currents: use $I=\dfrac{P_{\text{total}}\times1000}{\sqrt{3}\;V_{L-L}\;PF_{\text{equiv}}}$ where $PF_{\text{equiv}}$ is either a known system PF or compute apparent power precisely.

4. If loads are unbalanced, find per-phase kW and compute per-phase currents separately using $I_{\text{ph}} = \dfrac{P_{\text{ph}}\times1000}{V_{L-N}\times PF_{\text{ph}}}$.

5. Check neutral current (vector sum of phase currents) and thermal limits.

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Advantages of three-phase over single-phase (clean list)

1. For a given power, three-phase machines are smaller and lighter.

2. Three-phase motors are self-starting; single-phase often needs auxiliary start.

3. Higher efficiency and typically better power factor for three-phase motors.

4. Lower torque pulsation — smoother torque in polyphase systems.

5. For transmission of the same power, three-phase requires less conductor material (more economical).

6. Easier parallel operation of generators in three-phase systems.

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