11.3 Power Factor, Efficiency & Transformer Theory

11.3 Power Factor, Efficiency & Transformer Theory

Exam Focus: True/reactive/apparent power and the power triangle, power factor arithmetic, transformer turns and current ratios, primary and secondary full-load current, and efficiency.

Load calculation questions give you volt-amperes. Nameplate questions often give you watts and a power factor. Knowing how to move between them is what separates a correct feeder size from a wrong one.

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Three Kinds of Power

                        THE POWER TRIANGLE
                                  /|
                                /  |
              S (VA)          /    |  Q (VAR)
            Apparent        /      |  Reactive
                          /  angle |
                        /__________|
                           P (W)
                          True Power

              P = S x cos(angle)          PF = P / S
              S = sqrt(P^2 + Q^2)

Power Factor

$PF = \frac{P\ (\text{watts})}{S\ (\text{volt-amperes})}$

Worked Example

A single-phase 240 V motor draws 20 A at a power factor of 0.85.

• Apparent power: $S = 240 \times 20 = 4{,}800\ \text{VA}$
• True power: $P = 4{,}800 \times 0.85 = 4{,}080\ \text{W}$

Notice the direction of travel: the conductor and overcurrent device are sized from the 4,800 VA / 20 A, not from the 4,080 W. Sizing from watts would undersize the circuit.

[!CAUTION] Do not divide watts by volts for a motor. For any load with a power factor below unity, $I = \dfrac{P}{E \times PF}$ for single-phase and $I = \dfrac{P}{1.732 \times E \times PF}$ for three-phase. On the exam, Article 430 sidesteps this entirely by giving you full-load current directly in Tables 430.248 and 430.250 — use those tables for motor circuit sizing, and reserve the power-factor formula for theory items.

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Efficiency

$\%\ \text{Efficiency} = \frac{P_{output}}{P_{input}} \times 100$

A motor delivering 10 hp of mechanical output while drawing 8,500 W:

• Output in watts: $10\ \text{hp} \times 746 = 7{,}460\ \text{W}$
• Efficiency: $7{,}460 / 8{,}500 = 0.878 = \mathbf{87.8%}$

1 horsepower = 746 watts. Memorize it; it appears in motor questions constantly.

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Transformer Theory

A transformer transfers energy between two coils by mutual induction. It changes voltage and current, but not power (apart from losses) and not frequency.

The Three Ratios

$\frac{E_P}{E_S} = \frac{N_P}{N_S} = \frac{I_S}{I_P}$

Read carefully: voltage and turns move together, while current moves opposite. A step-down transformer has more primary turns, less secondary voltage, and more secondary current.

$E_P \times I_P = E_S \times I_S \quad (\text{ideal, no losses})$

Worked Example — Turns and Current

A single-phase transformer has 800 primary turns and 100 secondary turns, with 480 V applied to the primary.

1. Turns ratio: $800 : 100 = 8 : 1$ (step-down)
2. Secondary voltage: $480 / 8 = \mathbf{60\ \text{V}}$
3. If the secondary delivers 40 A, primary current is $40 / 8 = \mathbf{5\ \text{A}}$
4. Check: $480 \times 5 = 2{,}400\ \text{VA}$ and $60 \times 40 = 2{,}400\ \text{VA}$ ✔

Full-Load Current from kVA

The kVA rating is the same on both sides. Use the voltage of the side you want.

Example: a 75 kVA, three-phase transformer with a 208 V secondary. $I_S = \frac{75{,}000}{1.732 \times 208} = \frac{75{,}000}{360.3} = \mathbf{208.2\ \text{A}}$

And its 480 V primary: $I_P = \frac{75{,}000}{1.732 \times 480} = \frac{75{,}000}{831.4} = \mathbf{90.2\ \text{A}}$

These two numbers feed directly into the Article 450 overcurrent protection rules covered in Chapter 8.

Transformer Losses

• Core (iron) loss — hysteresis and eddy currents in the steel. Essentially constant whenever the transformer is energized, regardless of load.
• Copper loss — $I^2R$ in the windings. Varies with the square of the load current.

Because core loss is constant, a lightly loaded transformer is an inefficient transformer, and an energized but unloaded transformer still consumes power.

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Quick Reference