§02 / 39

Conversions & Equations

Foundation · The math toolkit

Every section that follows references one of these formulas. Memorize them through reps. Once they're reflex, the rest of electrical design becomes choosing which one to apply.

The Power Triangle

The single most important visual in electrical engineering. Real power (P) does work. Reactive power (Q) supports magnetic fields in motors and transformers. Apparent power (S) is what the conductors actually carry. Conductors and transformers are sized for S (apparent), not P (real).

Conductors carry S. Loads consume P. Utilities bill on both.

Pythagorean relationship

S² = P² + Q²

Apparent power is the vector sum. Always.

Power factor

PF = cos θ = P / S

PF = 1.0 means all power is real (resistive load). PF = 0.8 means 80% of apparent power is real, 60% is reactive.

Solving for any side

P = S · PF

Q = S · sin θ

The Six Core Formulas — 1φ vs 3φ

Memorize these six. They cover ~90% of the conversion math you'll do for the rest of your career.

Quantity	Single-phase (1φ)	Three-phase (3φ)	Why the difference
Apparent S (kVA)	`V × I / 1000`	`√3 × V_LL × I / 1000`	3-phase has three conductors carrying current — the √3 captures the geometry of three sinusoids 120° apart. V_LL = line-to-line voltage.
Real P (kW)	`V × I × PF / 1000`	`√3 × V_LL × I × PF / 1000`
Reactive Q (kVAR)	`V × I × sin θ / 1000`	`√3 × V_LL × I × sin θ / 1000`
Solve for I from kVA	`I = kVA × 1000 / V`	`I = kVA × 1000 / (√3 × V_LL)`	Most common reverse — sizing conductors from a known load
Solve for I from kW	`I = kW × 1000 / (V × PF)`	`I = kW × 1000 / (√3 × V_LL × PF)`	If you only know P, you must include PF to get I
HP → kW	`kW = HP × 0.746` · `HP = kW × 1.341`		Output (mechanical) — does NOT account for motor efficiency or PF

√3

Why √3 in three-phase formulas?

Three sinusoidal voltages, equal magnitude, 120° apart. When you measure line-to-line voltage (between two phase conductors), the geometry of two phasors 120° apart yields √3 ≈ 1.732 × the line-to-neutral voltage. So V_LL = √3 × V_LN.

Examples that should be reflex: 277 × √3 = 480 · 120 × √3 = 208 · 2400 × √3 = 4160 · 7200 × √3 = 12,470.

The HP → kW → FLA Chain

The most-used calculation in motor work. Three steps. Each step needs one new piece of nameplate data.

For NEC purposes, FLA comes from NEC Table 430.250 not the nameplate (the table is conservative)

Conversions Worth Memorizing

HP → kW

×0.746

1 HP = 0.746 kW exactly. Reverse: × 1.341.

3φ Line-to-line

×√3

V_LL = √3 × V_LN. 277 → 480, 120 → 208, 2400 → 4160.

kVA at 480V 3φ

×1.203

kVA × 1.203 ≈ FLA. Useful sanity check on 480V loads.

kW → BTU/hr

×3,412

Server heat load → cooling requirement.

tons → kW (cooling)

×3.517

1 ton refrigeration ≈ 3.517 kW heat removed.

Δ% drop, 1φ

2VIL/CM

VD = (2 × L × I × R) / 1000. Use NEC Ch9 Table 9 ohms.

Worked Example 1 — Atlas DC1 Chiller Motor

Atlas DC1 has four 750-ton centrifugal chillers. Each is driven by a single induction motor. The mechanical engineer's MEL gives you the chiller, the manufacturer's cutsheet gives you the rest. You need FLA to start branch-circuit design.

Example 01 · Atlas DC1 spine Chiller CH-1 · 750-ton centrifugal · Trane CenTraVac equivalent

Given (from cutsheet)

Motor HP

450 HP nominal

Voltage

480V, 3φ, 60Hz

Efficiency η

96.0% (NEMA Premium)

Power factor

0.91 at full load

Step-by-step

HP → kW (mechanical output)

kW_out = 450 HP × 0.746 = 335.7 kW
kW (output) → kW (input) — divide by efficiency to get the electrical power the wire must deliver

kW_in = 335.7 / 0.96 = 349.7 kW
kW → kVA — divide by PF to get apparent power (what conductors carry)

kVA = 349.7 / 0.91 = 384.3 kVA
kVA → FLA — divide by √3 × V_LL

FLA = (384.3 × 1000) / (√3 × 480) = 384,300 / 831.4 = 462 A
Cross-check against NEC Table 430.250 — for 450 HP at 460V the table lists FLC = 480 A (close to our 462; NEC table is slightly conservative)

For NEC sizing use FLC = 480 A from NEC 430.250, not the calculated 462 A. NEC requires the table value (430.6).

NEC 430.6(A)(1) — the key trap

For motor sizing under NEC, the table values in 430.247–430.250 are used, NOT the nameplate FLA. The nameplate is for actual current measurement; the table is for design. They will differ by 5–15%, and the test/exam will check that you used the table value.

Worked Example 2 — Apartment Service Calculation

The math doesn't change for residential. Smaller voltages, single-phase split, but the same conversions. This example shows you the same tools applied at the other end of the spectrum.

Example 02 · Alternate scale A single 200A 120/240V apartment unit · NEC 220 Standard Method

Given

Service

200A, 120/240V, 1φ-3-wire

Connected load

28.4 kVA (after NEC 220 demand factors)

Find: actual service current

Single-phase formula: I = kVA × 1000 / V

I = (28.4 × 1000) / 240 = 118.3 A

Service breaker = 200 A is correctly sized — the demand load fits below the breaker, with margin for code-required 125% on continuous portions.
Why 240, not 120? Most large appliances (range, dryer, A/C, heat) use 240V. The service voltage for current calculation is the line-to-line value, even on split-phase.
Sanity check with HP-equivalent: 28.4 kVA at PF ≈ 1.0 = 28.4 kW = ~38 HP equivalent. A reasonable apartment runs ~5–8 kW peak; the 28.4 here represents NEC 220 demand load including diversity for a fully-equipped 2-bedroom unit.

Same formula, different scale

Atlas DC1's chiller calculation and this apartment calculation use the identical I = kVA / (k × V) formula. The only difference is k = 1 (single-phase) vs k = √3 (three-phase). Once the formula is muscle memory, scale is just substitution.

Per-Unit Math — The Foundation of Fault Analysis

Per-unit (pu) normalizes every quantity to a common base. Once normalized, you can add transformer impedances directly across voltage levels without conversion. Used in fault analysis (§12), load flow (§16), and protection coordination (§17).

Choose a base — typically

S_base = pick (e.g., 100 MVA or transformer kVA)

V_base = system voltage at the bus

Once chosen, every other base derives from these two.

Derived bases

I_base = S_base / (√3 × V_base)

Z_base = V_base² / S_base

Convert any quantity to pu

Z_pu = Z_actual / Z_base

Transformer %Z = 100 × Z_pu on the transformer's own base.

Convert pu impedance to a different base

Z_pu,new = Z_pu,old × (S_new / S_old) × (V_old / V_new)²

Critical when combining transformers of different sizes.

Example · Atlas DC1 spine Per-unit conversion of TX-A into a system-wide base

Pick a system base

S_base

100 MVA (arbitrary, chosen for clean numbers)

V_base at MV

12.47 kV

V_base at LV

480V

Step-by-step

TX-A nameplate impedance: 5.75% on TX-A's own base of 2,500 kVA.

Z_pu,TX on own base = 0.0575 pu
Convert to 100 MVA system base:

Z_pu,TX,new = 0.0575 × (100,000 / 2,500) × (12.47/12.47)² = 0.0575 × 40 × 1 = 2.30 pu
Compute base current at 480V bus:

I_base,LV = 100,000 / (√3 × 0.480) = 120,300 A
Fault current at 480V (assuming infinite source feeding TX-A):

I_fault,pu = 1.0 / 2.30 = 0.435 pu
I_fault,actual = 0.435 × 120,300 = 52,300 A

Same answer as the simple FLA / %Z approximation in §09 — because in this case, the utility was assumed infinite. With real utility impedance (per §12), the fault drops to ~50.3 kA.

Why per-unit matters

In multi-voltage systems (utility 12.47 kV → 480V → 415V → 240V), per-unit lets you add impedances without converting voltages back and forth. Every transformer's %Z just becomes a pu number on the system base, and you sum them. The MVA method (§12) is a shortcut version of per-unit math.

If You See THIS, Think THAT

If you see…	Think / use…
"480V" or "480Y/277V" on a cutsheet	3-phase. Use √3 in formulas. V_LL = 480, V_LN = 277.
"208Y/120V" or "120/208V"	3-phase wye. Most common commercial. V_LL = 208, V_LN = 120.
"120/240V"	Single-phase 3-wire (residential). NO √3. V_LL = 240.
HP given on motor nameplate	Mechanical output. Multiply by 0.746 to get kW; then divide by η × PF for kVA.
"Calculate motor branch circuit"	Use FLC from NEC Table 430.250, NOT the nameplate FLA. (Per NEC 430.6.)
kVA given but you need amps	3φ: `I = kVA × 1000 / (√3 × V_LL)` · 1φ: `I = kVA × 1000 / V`
kW given (no PF mentioned)	Stop. You need PF before you can find kVA or amps. If load is purely resistive (heaters, incandescent), PF = 1.0 and kW = kVA.
"Per-unit" in a problem	Quantities are normalized to a base. Always find S_base and V_base first; then I_base = S_base / (√3 × V_base).
Server load in kW for cooling sizing	Multiply kW × 3,412 for BTU/hr, or kW / 3.517 for cooling tons. (Atlas DC1: 2.5 MW IT load = 711 tons of cooling.)
"Tons" on chiller cutsheet	Refrigeration tons. 1 ton = 3.517 kW heat removed. Motor input is different — see HP→FLA chain.

Drill — Quick Self-Check

Five problems. Hide answers; work them mentally; reveal to check. The goal is reflex, not deliberation.

Drill 01 · 1φ → amps

A 5 kVA, 240V single-phase load. What is the current?

Drill 02 · 3φ → amps

A 75 kVA load at 480V 3φ. What is the current?

Drill 03 · HP → kW

A 100 HP motor. What is mechanical output in kW?

Drill 04 · kW → kVA

A 100 kW load at PF = 0.85. What is kVA?

Drill 05 · Atlas DC1 quick math

Atlas DC1 IT load = 2.5 MW. PF for the IT load = 0.95 (modern PSU). What kVA does the UPS see?

Drill 06 · cooling math

Atlas DC1 IT load = 2.5 MW. How many tons of cooling does it need?