4.2 Three-Phase Relationships and Real, Apparent, and Reactive Power

4.2 Three-Phase Relationships and Real, Apparent, and Reactive Power

Quick Summary: Three-phase questions are guaranteed marks in Sections C and D. Fix the √3 relationships (star: $V_L = \sqrt{3}\, V_{ph}$; delta: $I_L = \sqrt{3}\, I_{ph}$), the √3 power formulas, and the power triangle ($S^2 = P^2 + Q^2$, $\cos\phi = P/S$) in memory, and every motor-current, maximum-demand, and cable-sizing calculation in this guide becomes mechanical.

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1. Star (Y) and Delta (Δ) Relationships

Singapore's 400 V / 230 V system is exactly this star relationship: each phase-to-neutral voltage is 230 V, and $\sqrt{3} \times 230 = 400\text{ V}$ between any two lines.

2. Single-Phase and Three-Phase Power Formulas

• Single-phase: $P = V \times I \times \cos\phi$; $S = V \times I$.
• Three-phase (star or delta, using line values):

$$P = \sqrt{3} \times V_L \times I_L \times \cos\phi \qquad S = \sqrt{3} \times V_L \times I_L \qquad I_L = \frac{P}{\sqrt{3}\, V_L \cos\phi}$$

Sanity check: 45 kVA at 400 V gives $I_L = 45{,}000 / (\sqrt{3} \times 400) = 64.95\text{ A}$ — the Electrician licence ceiling expressed in current terms (Chapter 1).

3. Real, Apparent, and Reactive Power — The Power Triangle

• Real (active) power $P$ (kW): does useful work (heat, torque, light).
• Reactive power $Q$ (kVAr): sustains magnetic fields in motors and transformers; oscillates between source and load without doing useful work.
• Apparent power $S$ (kVA): the vector sum that the supply and cables must actually carry.

$$S^2 = P^2 + Q^2 \qquad \cos\phi = \frac{P}{S} \qquad \tan\phi = \frac{Q}{P}$$

Cables, breakers, and transformers are thermally limited by current, hence by $S$ — two loads with equal kW but different power factors draw different currents. This is why the licence scope is defined in kVA, not kW.

4. Worked Motor Calculations (EMA Sample-Question Style)

Question (from EMA's published sample list): for a 3-phase 400 V, 2 kW motor with power factor 0.8, find (a) the running current, (b) the direct-on-line starting current, and (c) whether the cable is sized for starting or running current.

(a) Running current:

$$I_{run} = \frac{P}{\sqrt{3} \times V_L \times \cos\phi} = \frac{2{,}000}{1.732 \times 400 \times 0.8} = \frac{2{,}000}{554.3} = 3.61\text{ A}$$

(b) DOL starting current: induction motors draw 6–8 × full-load current at standstill (Chapter 3.2):

$$I_{start} \approx 6 \times 3.61 \approx 21.7\text{ A} \;\text{to}\; 8 \times 3.61 \approx 28.9\text{ A}$$

(c) Cable sizing basis: the cable is sized for the running current (with correction factors), not the starting current — the inrush lasts only seconds, and the protective device is chosen (Type C/D curve or a suitable fuse) to withstand it without tripping. The overload relay, set at $1.0 \times I_{FLC}$ (or $0.58 \times I_{FLC}$ inside a star-delta loop), protects the motor thermally during run.

Efficiency variant: if the 2 kW is shaft (output) power with efficiency $\eta = 0.85$, first convert to input: $P_{in} = 2{,}000 / 0.85 = 2{,}353\text{ W}$, then $I_{run} = 2{,}353 / 554.3 = 4.25\text{ A}$. Read whether the question gives output or input power — EMA long questions mark the distinction.

5. Per-Phase Analysis and Metering

For unbalanced loads, compute each phase separately: per-phase $P_{ph} = V_{ph} \times I_{ph} \times \cos\phi = 230 \times I_{ph} \times \cos\phi$, then sum the three phases. Metering follows the same structure: whole-current kWh meters connect directly for small loads, while CT-operated meters (e.g., 100/5 A) step down larger currents — the SLD must show the arrangement of protective devices before and after the meter (Chapter 6.1).

6. Reverse Calculations and Common Three-Phase Traps

Written-test long questions frequently work backwards from current to power. Example: a balanced 3-phase heater bank draws 20 A per line at unity power factor from 400 V. Its real power is $P = \sqrt{3} \times 400 \times 20 \times 1.0 = 13{,}856\text{ W} \approx 13.9\text{ kW}$, and its apparent power is $S = \sqrt{3} \times 400 \times 20 = 13.9\text{ kVA}$ (identical because $\cos\phi = 1$).

Traps EMA Examiners Exploit

1. Forgetting √3 in 3-phase work: using $P = V_L \times I_L$ instead of $\sqrt{3} \times V_L \times I_L$ underestimates power by 42%. If an answer looks 1.73× too small, the √3 was dropped.
2. Mixing line and phase values in delta loads: a delta motor winding sees the full 400 V, but its winding current is only $I_L/\sqrt{3}$ — the basis of the 0.58 × FLC overload-relay rule (Chapter 3.2).
3. Using 230 V with √3: the √3 formulas require the line-to-line voltage (400 V); using 230 V with √3 produces line/phase-confused answers (e.g., 26 kVA instead of 45 kVA).
4. kW versus kVA confusion: licence scope, breaker thermal duty, and supply capacity are kVA (current-based); appliance nameplates are usually kW. Convert through $\cos\phi$ before comparing them.