2.4 Electrical Units, Formulas & Conversions

Electrical Units, Formulas & Conversions

Precise unit conversions, conductor area calculations, and voltage drop evaluations are essential competencies for journeyman electricians. System performance, equipment longevity, and fire safety depend on selecting conductors capable of handling circuit continuous loads while restricting voltage drop within acceptable operating limits. This section provides a comprehensive technical reference for SI electrical units, thermal/mechanical conversions, conductor geometry (circular mils), exact voltage drop calculations, and equipment efficiency formulas.

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1. Master Electrical Units & Derived Standards

Every electrical calculation utilizes fundamental units defined by the International System of Units (SI) and derived electromagnetic relationships.

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2. Power, Mechanical & Thermal Conversions

In commercial and residential trade work, electricians frequently interface with mechanical equipment rated in horsepower (HP) and HVAC/heating equipment rated in British Thermal Units (BTU).

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|                                ESSENTIAL CONVERSION CONSTANTS                           |
|                                                                                         |
|   1. Mechanical Power to Electrical Power:                                              |
|      • 1 Horsepower (HP) = 746 Watts = 0.746 Kilowatts                                  |
|      • Watts = HP × 746                                                                 |
|      • HP = Watts / 746                                                                 |
|                                                                                         |
|   2. Thermal Energy to Electrical Power:                                                |
|      • 1 Kilowatt (kW) = 3,412 BTU/hr                                                   |
|      • 1 Kilowatt-hour (kWh) = 3,412 BTU of thermal heat                                |
|      • Heating Watts = (BTU/hr) / 3.412                                                 |
|                                                                                         |
|   3. Air Conditioning & Refrigeration Capacity:                                         |
|      • 1 Ton of Refrigeration = 12,000 BTU/hr                                           |
|      • 1 Ton of Refrigeration = 12,000 / 3,412 = 3.517 kW = 3,517 Watts                 |
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Conversion Example: Commercial Duct Heater

A commercial building blueprint specifies a supplemental electric duct heater rated at $68,240\text{ BTU/hr}$.

• Electrical power required in Kilowatts: $P = \frac{68,240\text{ BTU/hr}}{3,412\text{ BTU/kWh}} = 20.0\text{ kW} = 20,000\text{ Watts}$.
• If operating at $240\text{V}$ single-phase: Current $I = \frac{20,000\text{ W}}{240\text{V}} = 83.33\text{ A}$.

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3. Conductor Geometry, Mils & Circular Mils (cmil)

In North American electrical engineering and NEC tables, wire sizes are expressed in American Wire Gauge (AWG) from 18 AWG up to 4/0 AWG, and in kcmil (thousands of circular mils) for conductors $250\text{ kcmil}$ and larger.

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|                              CIRCULAR MIL DEFINITIONS & AREA                            |
|                                                                                         |
|   • 1 Mil = 0.001 inch (one-thousandth of an inch)                                      |
|   • Diameter (d) in Mils = Diameter in inches × 1,000                                   |
|   • Circular Mil Area (cmil) = d²  (where d is diameter in mils)                        |
|                                                                                         |
|   Comparison between Square Mils and Circular Mils:                                     |
|   • Area in Square Mils = (π / 4) × d² ≈ 0.7854 × cmil                                  |
|   • Area in Circular Mils = Area in Square Mils / 0.7854 ≈ 1.2732 × Square Mils         |
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NEC Chapter 9, Table 8 Conductor Properties (Key Copper Standards):

• 14 AWG: $4,110\text{ cmil}$ (Solid diameter $\approx 64.1\text{ mils}$)
• 12 AWG: $6,530\text{ cmil}$ (Solid diameter $\approx 80.8\text{ mils}$)
• 10 AWG: $10,380\text{ cmil}$ (Solid diameter $\approx 101.9\text{ mils}$)
• 8 AWG: $16,510\text{ cmil}$
• 6 AWG: $26,240\text{ cmil}$
• 4 AWG: $41,740\text{ cmil}$
• 3 AWG: $52,620\text{ cmil}$
• 2 AWG: $66,360\text{ cmil}$
• 1 AWG: $83,690\text{ cmil}$
• 1/0 AWG: $105,600\text{ cmil}$
• 2/0 AWG: $133,100\text{ cmil}$
• 3/0 AWG: $167,800\text{ cmil}$
• 4/0 AWG: $211,600\text{ cmil}$
• $250\text{ kcmil}$: $250,000\text{ cmil}$
• $500\text{ kcmil}$: $500,000\text{ cmil}$

Rectangular Copper Busbar Calculation:

A copper switchboard busbar measures $1/4\text{ inch} \times 2\text{ inches}$ in cross-section.

• Convert dimensions to mils: $0.250\text{ in} = 250\text{ mils}$; $2.000\text{ in} = 2,000\text{ mils}$.
• Area in square mils: $A_{\text{sq mils}} = 250 \times 2,000 = 500,000\text{ sq mils}$.
• Convert square mils to circular mils: $\text{cmil} = \frac{500,000}{0.7854} = 636,618\text{ cmil} \approx 636.6\text{ kcmil}$.

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4. Comprehensive Voltage Drop Calculations & NEC Guidelines

When electrical current travels through a circuit conductor, the internal resistance of the wire creates an unavoidable voltage drop ($I \times R$). If conductors are undersized or circuit runs are excessively long, the reduced operating voltage causes motor overheating, contactor chattering, dim lighting, and electronic device failure.

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|                                 VOLTAGE DROP FORMULAS                                   |
|                                                                                         |
|   1. Single-Phase Voltage Drop:      VD = ( 2 × K × I × D ) / cmil                      |
|                                                                                         |
|   2. Three-Phase Voltage Drop:       VD = ( √3 × K × I × D ) / cmil                     |
|                                         = ( 1.732 × K × I × D ) / cmil                  |
|                                                                                         |
|   3. Sizing Conductor Area (cmil):                                                      |
|      • Single-Phase:                 cmil = ( 2 × K × I × D ) / VD_allowable            |
|      • Three-Phase:                  cmil = ( 1.732 × K × I × D ) / VD_allowable        |
|                                                                                         |
|   Where:                                                                                |
|   • K = Specific Conductor Resistance at 75°C:                                          |
|         - Copper (Cu):   K = 12.9 Ω·cmil/ft                                             |
|         - Aluminum (Al): K = 21.2 Ω·cmil/ft                                             |
|   • I = Load Current in Amperes                                                         |
|   • D = One-Way Circuit Distance in Feet                                                |
|   • cmil = Conductor Cross-Sectional Area in Circular Mils (NEC Ch 9, Table 8)          |
|   • Factor '2' in single-phase accounts for out-and-back conductor loop                  |
|   • Factor '√3' (1.732) in three-phase accounts for 120° vector phase relationship      |
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National Electrical Code Voltage Drop Standards (Informational Notes):

• NEC 210.19(A) Informational Note No. 4 (Branch Circuits): Recommends that branch circuit conductors be sized to limit voltage drop to no more than $3%$ of nominal circuit voltage at the farthest outlet supplying power, heating, or lighting loads.
• NEC 215.2(A)(1)(b) Informational Note No. 2 (Feeders): Recommends that feeder conductors be sized to limit voltage drop to no more than $3%$.
• Total Combined System Limit: The maximum total combined voltage drop on both the feeder and the branch circuit combined should not exceed $5%$ from the service disconnecting means to the final load outlet.

Maximum Allowable Voltage Drops by System Voltage:

• $120\text{V}$ Circuit: $3% = 3.60\text{V}$; $5% = 6.00\text{V}$
• $208\text{V}$ Circuit: $3% = 6.24\text{V}$; $5% = 10.40\text{V}$
• $240\text{V}$ Circuit: $3% = 7.20\text{V}$; $5% = 12.00\text{V}$
• $277\text{V}$ Circuit: $3% = 8.31\text{V}$; $5% = 13.85\text{V}$
• $480\text{V}$ Circuit: $3% = 14.40\text{V}$; $5% = 24.00\text{V}$

Step-by-Step Worked Problems:

Problem 1: Evaluating Single-Phase Voltage Drop & Upsizing

A $120\text{V}$, single-phase parking lot lighting branch circuit carries a continuous load of $16\text{ A}$ copper conductors over a one-way distance of $175\text{ feet}$ using $12\text{ AWG}$ THHN copper ($6,530\text{ cmil}$).

1. Calculate Actual Voltage Drop: $\text{VD} = \frac{2 \times K \times I \times D}{\text{cmil}} = \frac{2 \times 12.9 \times 16\text{ A} \times 175\text{ ft}}{6,530\text{ cmil}} = \frac{72,240}{6,530} = 11.06\text{V}$
2. Calculate Percentage Drop: $\%\text{VD} = \left(\frac{11.06\text{V}}{120\text{V}}\right) \times 100\% = 9.22\% \quad (\text{Fails 3\% limit of 3.60V})$
3. Calculate Minimum Circular Mils to Meet 3% Limit ($3.60\text{V}$ max drop): $\text{cmil}_{\text{required}} = \frac{2 \times 12.9 \times 16 \times 175}{3.60\text{V}} = \frac{72,240}{3.60} = 20,066.67\text{ cmil}$
4. Select Conductor from NEC Chapter 9, Table 8:
   • $8\text{ AWG} = 16,510\text{ cmil}$ (Too small)
   • $6\text{ AWG} = 26,240\text{ cmil}$ (Complies; actual drop with $6\text{ AWG} = 2.75\text{V} = 2.29%$).

Problem 2: Sizing Three-Phase Feeder for Voltage Drop

A $480\text{V}$, 3-phase commercial building feeder supplies an industrial rooftop chiller drawing $75\text{ A}$ over a one-way distance of $350\text{ feet}$ using copper conductors. Sizing for a maximum $3%$ voltage drop ($14.40\text{V}$ maximum allowable drop):

1. Calculate Required Circular Mil Area: $\text{cmil} = \frac{1.732 \times K \times I \times D}{\text{VD}_{\text{allowable}}} = \frac{1.732 \times 12.9 \times 75\text{ A} \times 350\text{ ft}}{14.40\text{V}} = \frac{586,482}{14.40} = 40,727.9\text{ cmil}$
2. Select Conductor from NEC Chapter 9, Table 8:
   • $6\text{ AWG} = 26,240\text{ cmil}$ (Too small)
   • $4\text{ AWG} = 41,740\text{ cmil}$ (Properly sized to comply with $3%$ threshold).

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5. Electrical Equipment Efficiency Formulas

Efficiency ($\eta$) measures how effectively electrical equipment converts electrical input power into useful mechanical or thermal output power without wasting energy as heat.

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|                                 EQUIPMENT EFFICIENCY FORMULAS                           |
|                                                                                         |
|   1. Efficiency (η) = ( Power Output / Power Input ) × 100%                             |
|                     = ( Watts_Out / Watts_In ) × 100%                                   |
|                                                                                         |
|   2. Input Power (Watts_In) = Watts_Out / η                                             |
|                                                                                         |
|   3. Single-Phase Motor Input Current:                                                  |
|      I = ( HP × 746 ) / ( Voltage × PF × η )                                            |
|                                                                                         |
|   4. Three-Phase Motor Input Current:                                                   |
|      I = ( HP × 746 ) / ( √3 × Voltage × PF × η )                                       |
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Motor Efficiency Worked Example:

A $15\text{ HP}$, $230\text{V}$, single-phase motor has an efficiency rating of $86%$ ($0.86$) and operates at a power factor of $0.88$.

1. Mechanical Output Power: $P_{\text{out}} = 15\text{ HP} \times 746\text{ W/HP} = 11,190\text{ Watts}$.
2. Electrical Input Power: $P_{\text{in}} = \frac{11,190\text{ W}}{0.86} = 13,011.63\text{ Watts}$.
3. Full-Load Current Drawn: $I = \frac{13,011.63\text{ W}}{230\text{V} \times 0.88} = \frac{13,011.63}{202.4} = 64.29\text{ Amperes}$