Rappel pour compagnons Refresher 10 of 10 · about 5 min

Journeyman formula card

What you’ll learn

Every formula from the third-year, fourth-year and refresher material on one page.
Which variable is which — line or phase, per winding or per line.
A tool and a lesson for each group, when you need the working or the why.

FLA = kVA × 1000 ÷ (√3 × V)	three-phase, each side
FLA = kVA × 1000 ÷ V	single-phase
Turns ratio = V winding pri ÷ V winding sec	per winding, not line to line
Open delta ≈ 57.7% of the three-can bank	1 ÷ √3

Isc = FLA × 100 ÷ %Z	infinite bus, at the terminals
Multiplier = 100 ÷ %Z	e.g. 5% → 20 × FLA

Ns = 120 × f ÷ P	synchronous speed, rpm
Slip % = (Ns − N) ÷ Ns × 100	N = actual speed
f rotor = slip × f	rotor current frequency
I = HP × 746 ÷ (√3 × V × eff × PF)	three-phase current from HP (use the nameplate for real work)

kVAR = kW × (tan φ1 − tan φ2)	φ = arccos PF
C = Q ÷ (2π f V²)	per capacitor; Q in VAR, V across it

% = max deviation from average ÷ average × 100

three line-to-line readings

10 HP, 460 V, eff 0.90, PF 0.85: I = 10 × 746 ÷ (1.732 × 460 × 0.90 × 0.85) = 7,460 ÷ 609.5 = 12.2 A (a physics estimate — the nameplate FLA is what you use).
Nameplate speed 1760 rpm on 60 Hz: 4-pole, Ns 1800, slip = 40 ÷ 1800 = 2.2%.
On a drive at 50 Hz: shaft ≈ 120 × 50 ÷ 4 − 40 = 1460 rpm; output ≈ 460 ÷ 60 × 50 = 383 V.

Print it double-sided and keep it with your code book tabs while you study.
Before using a formula, say out loud what V and I it wants — line or phase.
Work each answer back the other way; a formula that does not close was used wrong.

Using 3 instead of √3 with line values.
Using the line ratio as the turns ratio on a delta–wye bank.
Dividing by %Z as a whole number without the 100 (5 instead of 0.05).
Using a calculated motor current where your code book tells you to use its own values.

0 right of 0 answered 5 questions · pick one answer each

1.What is the synchronous speed of a 6-pole motor on 60 Hz?
2.A 300 kVA transformer with 5% impedance and a 208 V secondary. Infinite-bus fault current at its terminals?
3.A 4–20 mA pressure transmitter is ranged 0–100 psi. What does 8 mA represent?
4.kVAR to correct a 50 kW load from PF 0.80 to 0.95?
5.Three equal 12 Ω impedances in delta. The equivalent wye has arms of…

Answers

B. 1200 rpm — Ns = 120 × f ÷ P = 120 × 60 ÷ 6 = 1200 rpm.
C. About 16,650 A — FLA = 300,000 ÷ (1.732 × 208) = 832.7 A. Isc = FLA × 100 ÷ %Z = 832.7 × 20 = 16,654 A.
B. 25 psi — PV = LRV + (mA − 4) ÷ 16 × span = 0 + (8 − 4) ÷ 16 × 100 = 25 psi.
A. About 21 kVAR — kVAR = kW × (tan φ1 − tan φ2) = 50 × (0.750 − 0.329) = 50 × 0.421 = 21.1 kVAR.
D. 4 Ω — Balanced: Z wye = Z delta ÷ 3 = 12 ÷ 3 = 4 Ω.

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