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PE Electrical and Computer

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1.2 Units, Conversions, Engineering Constants & Fundamental Power Equations

Key Takeaways

  • Power engineering calculations require seamless conversion between SI units (W, J, N·m, Pa) and US Customary / English units (hp, BTU, ft·lbf, psi), with foundational identities such as 1 hp = 746 W and 1 kWh = 3,412.14 BTU.
  • Fundamental electromagnetic constants include vacuum permittivity ε0 ≈ 8.854 × 10^-12 F/m, permeability μ0 = 4π × 10^-7 H/m, speed of light c ≈ 3.0 × 10^8 m/s, and intrinsic impedance of free space η0 ≈ 377 Ω.
  • Standard North American power grids operate at f = 60 Hz with an angular frequency of ω = 2πf ≈ 376.99 rad/s (often rounded to 377 rad/s), while 50 Hz systems operate at ω ≈ 314.16 rad/s.
  • The complex power triangle relates real power (P in Watts), reactive power (Q in VAR), and apparent power (S in VA) via S = P + jQ = V_rms * (I_rms)*, where power factor is PF = cos(θ) = P / |S|.
  • The Passive Sign Convention (PSC) dictates that power entering the positive terminal of an element represents absorbed power (P > 0, load), whereas power exiting the positive terminal represents delivered power (P < 0, source).
Last updated: August 2026

1.2 Units, Conversions, Engineering Constants & Fundamental Power Equations

Power systems engineering demands absolute precision across multiple unit systems, physical constants, electromagnetic relationships, and circuit theorems. An error in converting mechanical horsepower to electrical kilowatts, confusing line-to-line with line-to-neutral voltages, or misapplying sign conventions can invalidate complex calculations. This section consolidates the mathematical, electromagnetic, and thermodynamic foundations tested across all domains of the NCEES PE Power examination.

1. Unit Systems & Engineering Conversion Factors

The PE Power examination incorporates both the International System of Units (SI) and US Customary (English) Units. Candidates must execute conversions rapidly and without ambiguity.

+-----------------------------------------------------------------------------+
|                     CORE POWER ENGINEERING CONVERSIONS                      |
|                                                                             |
|   POWER:    1 hp = 746 W = 0.746 kW                                         |
|             1 kW = 1.34048 hp = 3,412.14 BTU/hr = 737.56 ft-lbf/s          |
|             1 Ton of Refrigeration = 12,000 BTU/hr = 3.51685 kW             |
|                                                                             |
|   ENERGY:   1 kWh = 3.6 × 10^6 J = 3,412.14 BTU = 2.655 × 10^6 ft-lbf       |
|             1 BTU = 1,055.06 J = 778.17 ft-lbf = 0.29307 W-hr               |
|             1 Calorie (thermochemical) = 4.184 J                            |
|                                                                             |
|   TORQUE:   1 N-m = 0.73756 lb-ft = 8.8507 lb-in                           |
|             1 lb-ft = 1.35582 N-m                                           |
|                                                                             |
|   LENGTH:   1 mil = 0.001 in = 2.54 × 10^-5 m                               |
|             1 circular mil (cmil) = (π/4) × (1 mil)^2 = (π/4) × 10^-6 sq in |
|             1 kcmil (MCM) = 1,000 cmil = 0.5067 mm^2                        |
+-----------------------------------------------------------------------------+

Comprehensive Conversion Matrix

DimensionPrimary SI UnitEnglish / Customary UnitExact / Standard Conversion Factor
Electrical Power ($P$)Watt ($\text{W} = \text{J/s}$)Horsepower ($\text{hp}$)$1\text{ hp} = 746.0\text{ W} = 0.746\text{ kW}$
Thermal Power ($\dot{Q}$)Kilowatt ($\text{kW}$)$\text{BTU/hr}$$1\text{ kW} = 3,412.142\text{ BTU/hr}$
Cooling CapacityKilowatt ($\text{kW}$)Ton of Refrigeration ($\text{TR}$)$1\text{ TR} = 12,000\text{ BTU/hr} = 3.51685\text{ kW}$
Energy ($W, E$)Joule ($\text{J} = \text{N}\cdot\text{m}$)Kilowatt-hour ($\text{kWh}$)$1\text{ kWh} = 3.60 \times 10^6\text{ J} = 3.6\text{ MJ}$
Thermal EnergyJoule ($\text{J}$)British Thermal Unit ($\text{BTU}$)$1\text{ BTU} = 1,055.056\text{ J}$
Mechanical Torque ($T$)Newton-meter ($\text{N}\cdot\text{m}$)Pound-foot ($\text{lb}\cdot\text{ft}$)$1\text{ lb}\cdot\text{ft} = 1.355818\text{ N}\cdot\text{m}$
Rotational Speed ($n, \omega$)Radians/sec ($\text{rad/s}$)Revolutions/min ($\text{RPM}$)$\omega = \frac{2\pi \cdot n}{60} = \frac{\pi \cdot n}{30} \approx 0.10472 \cdot n$
Magnetic Flux ($\Phi$)Weber ($\text{Wb} = \text{V}\cdot\text{s}$)Maxwell ($\text{Mx}$) / Line$1\text{ Wb} = 10^8\text{ Maxwells}$
Flux Density ($B$)Tesla ($\text{T} = \text{Wb/m}^2$)Gauss ($\text{G}$) / $\text{Lines/in}^2$$1\text{ T} = 10^4\text{ Gauss} = 64.516\text{ kLines/in}^2$
Magnetic Field ($H$)Ampere-turns/m ($\text{A-t/m}$)Oersted ($\text{Oe}$)$1\text{ Oe} = \frac{1000}{4\pi} \approx 79.577\text{ A-t/m}$

Mechanical Power & Torque Relationships

In rotating machine analysis (motors and generators), mechanical shaft power is converted to and from electrical power:

SI Units: Pmech=T⋅ωm=T⋅(2πn60)[Watts]\text{SI Units: } P_{\text{mech}} = T \cdot \omega_m = T \cdot \left(\frac{2\pi n}{60}\right) \quad [\text{Watts}]

English Units: Pmech[hp]=T[lb⋅ft]×n[RPM]5252.113≈T×n5252\text{English Units: } P_{\text{mech}} [\text{hp}] = \frac{T [\text{lb}\cdot\text{ft}] \times n [\text{RPM}]}{5252.113} \approx \frac{T \times n}{5252}

Derivation of Constant 5252: 33,000 ft⋅lbf/min per hp2π rad/rev=33,0006.283185=5252.113\text{Derivation of Constant 5252: } \frac{33,000\text{ ft}\cdot\text{lbf/min per hp}}{2\pi\text{ rad/rev}} = \frac{33,000}{6.283185} = 5252.113

2. Fundamental Electromagnetic & Grid Constants

Power systems depend on physical constants that govern transmission line capacitance, inductance, wave propagation, and transformer core magnetization.

+-----------------------------------------------------------------------------+
|                      PHYSICAL & ELECTROMAGNETIC CONSTANTS                   |
|                                                                             |
|   Permittivity of Free Space (ε0):    8.8541878 × 10^-12 F/m                |
|   Permeability of Free Space (μ0):    4π × 10^-7 H/m ≈ 1.256637 × 10^-6 H/m |
|   Speed of Light in Vacuum (c):       2.99792458 × 10^8 m/s ≈ 3.0 × 10^8 m/s|
|   Speed of Light in Miles/sec:        186,282 miles/s                       |
|   Intrinsic Wave Impedance (η0):      sqrt(μ0 / ε0) ≈ 120π ≈ 376.73 Ω       |
|   Electron Charge (q_e):              1.60217663 × 10^-19 C                 |
|                                                                             |
|   Standard 60 Hz Grid:                f = 60 Hz ==> ω = 2π(60) ≈ 376.99 rad/s|
|   Standard 50 Hz Grid:                f = 50 Hz ==> ω = 2π(50) ≈ 314.16 rad/s|
+-----------------------------------------------------------------------------+

Electromagnetic Wave Relationships:

c=1μ0ε0=2.998×108 m/sc = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 2.998 \times 10^8\text{ m/s}

Wavelength on a 60 Hz line: λ=cf=3.0×108 m/s60 Hz=5.0×106 m=5,000 km≈3,107 miles\text{Wavelength on a 60 Hz line: } \lambda = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m/s}}{60\text{ Hz}} = 5.0 \times 10^6\text{ m} = 5,000\text{ km} \approx 3,107\text{ miles}

Quarter-Wavelength Line: λ4=1,250 km≈777 miles\text{Quarter-Wavelength Line: } \frac{\lambda}{4} = 1,250\text{ km} \approx 777\text{ miles}

3. The AC Power Triangle & Complex Power Formulations

In alternating current (AC) single-phase and three-phase circuits, power is decomposed into real, reactive, and apparent components.

                  COMPLEX POWER TRIANGLE (LAGGING / INDUCTIVE LOAD)
                  
                        +......................................+
                        |                           .          |
                        |                  .                   |
                        |          .   APPARENT POWER (S)      |
                        |  .           [Volt-Amperes, VA]      |
                        +------------------------+             |
                        |   REAL POWER (P)       | REACTIVE    |
                        |   [Watts, W]           | POWER (Q)   |
                        |   P = |S|*cos(θ)       | [VAR]       |
                        |   P = V*I*cos(θ)       | Q = |S|*sin(θ)|
                        +------------------------+-------------+
                                  Load Angle θ = θ_v - θ_i

Complex Power Equations

For a single-phase system with RMS voltage phasor $\mathbf{V} = |V|\angle\theta_v$ and current phasor $\mathbf{I} = |I|\angle\theta_i$:

S=VI∗=(∣V∣∠θv)(∣I∣∠−θi)=∣V∣∣I∣∠(θv−θi)=P+jQ\mathbf{S} = \mathbf{V} \mathbf{I}^* = (|V|\angle\theta_v)(|I|\angle -\theta_i) = |V||I|\angle(\theta_v - \theta_i) = P + jQ

Real (Active) Power: P=Re{S}=∣V∣∣I∣cos⁡(θv−θi)=∣S∣cos⁡θ[Watts, W]\text{Real (Active) Power: } P = \text{Re}\{\mathbf{S}\} = |V||I|\cos(\theta_v - \theta_i) = |S|\cos\theta \quad [\text{Watts, W}]

Reactive Power: Q=Im{S}=∣V∣∣I∣sin⁡(θv−θi)=∣S∣sin⁡θ[Volt-Amperes Reactive, VAR]\text{Reactive Power: } Q = \text{Im}\{\mathbf{S}\} = |V||I|\sin(\theta_v - \theta_i) = |S|\sin\theta \quad [\text{Volt-Amperes Reactive, VAR}]

Apparent Power: ∣S∣=∣S∣=P2+Q2=∣Vrms∣∣Irms∣[Volt-Amperes, VA]\text{Apparent Power: } |S| = |\mathbf{S}| = \sqrt{P^2 + Q^2} = |V_{\text{rms}}||I_{\text{rms}}| \quad [\text{Volt-Amperes, VA}]

Power Factor: PF=P∣S∣=cos⁡θ=cos⁡(θv−θi)\text{Power Factor: } \text{PF} = \frac{P}{|S|} = \cos\theta = \cos(\theta_v - \theta_i)

Reactive Factor: RF=Q∣S∣=sin⁡θ\text{Reactive Factor: } \text{RF} = \frac{Q}{|S|} = \sin\theta

tan⁡θ=QP  ⟹  Q=Ptan⁡(arccos⁡(PF))\tan\theta = \frac{Q}{P} \implies Q = P \tan(\arccos(\text{PF}))

Inductive vs. Capacitive Load Conventions

ParameterInductive Load (Lagging PF)Capacitive Load (Leading PF)
Current PhaseCurrent lags voltage ($\theta_i < \theta_v$)Current leads voltage ($\theta_i > \theta_v$)
Phase Angle $\theta$$\theta = \theta_v - \theta_i > 0$ (Positive)$\theta = \theta_v - \theta_i < 0$ (Negative)
Reactive Power ($Q$)$Q > 0$ (Absorbs inductive VARs)$Q < 0$ (Supplies VARs / Absorbs leading VARs)
Complex Impedance$\mathbf{Z} = R + jX_L = R + j\omega L$$\mathbf{Z} = R - jX_C = R - j\frac{1}{\omega C}$
Physical EquipmentInduction motors, transformers, ballastsCapacitor banks, synchronous condensers (overexcited)

[!TIP] Conjugate Current Reminder: When computing complex power $\mathbf{S} = \mathbf{V}\mathbf{I}^*$, always take the complex conjugate of the current. Forgetting the conjugate flips the sign of reactive power $Q$, incorrectly converting an inductive load into a capacitive load.

4. Fundamental Circuit Laws & Sign Conventions

+-----------------------------------------------------------------------------+
|                        PASSIVE SIGN CONVENTION (PSC)                        |
|                                                                             |
|        Load / Absorbing Element              Source / Delivering Element    |
|               i --->                                    i --->              |
|         +-----------------+                       +-----------------+       |
|     +---| (Current Enters |---+               +---| (Current Leaves |---+   |
|   v     | Positive Terminal)  |             v     | Positive Terminal)  |   |
|     ----|   P = v * i > 0     |---+           ----|   P = - v * i < 0   |---+
|         +-----------------+                       +-----------------+       |
+-----------------------------------------------------------------------------+

Kirchhoff's Laws

  • Kirchhoff's Current Law (KCL): The algebraic sum of all currents entering any node (or closed boundary) equals zero: ∑k=1NIk=0  ⟹  ∑Iin=∑Iout\sum_{k=1}^N \mathbf{I}_k = 0 \implies \sum \mathbf{I}_{\text{in}} = \sum \mathbf{I}_{\text{out}}
  • Kirchhoff's Voltage Law (KVL): The algebraic sum of all branch voltages around any closed loop equals zero: ∑k=1MVk=0\sum_{k=1}^M \mathbf{V}_k = 0

Joule Heating & Electrical Loss Formulation

When current $I$ flows through an impedance $\mathbf{Z} = R + jX$, active power is dissipated strictly in the resistive element as heat:

Ploss=Irms2R=VR,rms2R[Watts]P_{\text{loss}} = I_{\text{rms}}^2 R = \frac{V_{R,\text{rms}}^2}{R} \quad [\text{Watts}]

Qstored=Irms2X[VAR]Q_{\text{stored}} = I_{\text{rms}}^2 X \quad [\text{VAR}]

5. Heat Dissipation & HVAC Equipment Sizing Calculations

Electrical losses in switchgear, uninterruptible power supply (UPS) systems, variable frequency drives (VFDs), and power transformers are converted entirely into thermal heat. Electrical engineers must calculate this heat dissipation to specify heating, ventilation, and air conditioning (HVAC) cooling equipment.

+-----------------------------------------------------------------------------+
|                     HEAT DISSIPATION CALCULATION WORKFLOW                   |
|                                                                             |
|   [Equipment Losses P_loss (kW)]                                            |
|                 |                                                           |
|                 v  Multiply by 3,412.142 BTU/hr per kW                      |
|   [Thermal Heat Rate Q_dot (BTU/hr)]                                        |
|                 |                                                           |
|                 v  Divide by 12,000 BTU/hr per Ton                          |
|   [Required Cooling Capacity (Tons of Refrigeration, TR)]                   |
+-----------------------------------------------------------------------------+

Step-by-Step Worked Example: Substation Transformer Room HVAC

Problem: A continuous-duty industrial substation transformer room contains:

  1. One $1,500\text{ kVA}$ transformer operating at $80%$ load with an efficiency of $98.6%$.
  2. Two $150\text{ hp}$ variable frequency drive (VFD) cabinets operating at full load with an efficiency of $97.0%$ each.
  3. Auxiliary battery charger and lighting load dissipating a constant $4.2\text{ kW}$.

Calculate the total heat dissipation rate in $\text{BTU/hr}$ and the minimum required HVAC cooling capacity in Tons of Refrigeration (TR).

Solution:

Step 1: Transformer Loss Operating Load S=0.80×1,500 kVA=1,200 kVA\text{Operating Load } S = 0.80 \times 1,500\text{ kVA} = 1,200\text{ kVA} Assuming PF=0.90  ⟹  Pout=1,200×0.90=1,080 kW\text{Assuming } \text{PF} = 0.90 \implies P_{\text{out}} = 1,200 \times 0.90 = 1,080\text{ kW} Pin=Poutη=1,080 kW0.986=1,095.335 kWP_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1,080\text{ kW}}{0.986} = 1,095.335\text{ kW} Ploss, xfmr=Pin−Pout=1,095.335−1,080=15.335 kWP_{\text{loss, xfmr}} = P_{\text{in}} - P_{\text{out}} = 1,095.335 - 1,080 = 15.335\text{ kW}

Step 2: VFD Inverter Cabinet Losses (2 units) Motor Mechanical Output per drive: Pmech=150 hp×0.746 kW/hp=111.90 kW\text{Motor Mechanical Output per drive: } P_{\text{mech}} = 150\text{ hp} \times 0.746\text{ kW/hp} = 111.90\text{ kW} Drive Electrical Input: Pin, VFD=111.90 kW0.970=115.361 kW\text{Drive Electrical Input: } P_{\text{in, VFD}} = \frac{111.90\text{ kW}}{0.970} = 115.361\text{ kW} Ploss, per VFD=115.361−111.90=3.461 kWP_{\text{loss, per VFD}} = 115.361 - 111.90 = 3.461\text{ kW} Ploss, VFDs (total)=2×3.461 kW=6.922 kWP_{\text{loss, VFDs (total)}} = 2 \times 3.461\text{ kW} = 6.922\text{ kW}

Step 3: Total Electrical Heat Generation Ptotal loss=Ploss, xfmr+Ploss, VFDs+Paux=15.335+6.922+4.200=26.457 kWP_{\text{total loss}} = P_{\text{loss, xfmr}} + P_{\text{loss, VFDs}} + P_{\text{aux}} = 15.335 + 6.922 + 4.200 = 26.457\text{ kW}

Step 4: Convert to Thermal Heat Rate ($\text{BTU/hr}$) Q˙=26.457 kW×3,412.142 BTU/(hr⋅kW)=90,275.04 BTU/hr\dot{Q} = 26.457\text{ kW} \times 3,412.142\text{ BTU/(hr}\cdot\text{kW)} = 90,275.04\text{ BTU/hr}

Step 5: Convert to Cooling Tons Cooling Capacity=90,275.04 BTU/hr12,000 BTU/(hr⋅Ton)=7.523 Tons\text{Cooling Capacity} = \frac{90,275.04\text{ BTU/hr}}{12,000\text{ BTU/(hr}\cdot\text{Ton)}} = 7.523\text{ Tons}

A standard design would specify an $8.0\text{-Ton}$ or $10.0\text{-Ton}$ commercial cooling unit to provide safety margin and handle ambient solar building gains.

6. Common Exam Traps & Pitfalls

+-----------------------------------------------------------------------------+
|                   FUNDAMENTAL EQUATION PITFALL CHECKLIST                    |
|                                                                             |
|   [!] Motor Horsepower is SHAFT Output: When sizing branch circuits or      |
|       heat loads, never multiply hp by 746 and stop. You MUST divide by     |
|       efficiency (η) and power factor (PF) to get electrical kVA/kW input.  |
|   [!] Line-to-Line vs. Per-Phase in 3-Phase Equations:                      |
|       - 3-Phase Power: S = sqrt(3) * V_LL * I_L                             |
|       - Per-Phase Power: S_1phi = V_LN * I_L = (1/3) * S_3phase             |
|       Using S = 3 * V_LL * I_L produces an error by a factor of 1.732 (sqrt3)|
|   [!] Frequency Sensitivity in Reactance:                                   |
|       - X_L = 2πfL = 377 * L at 60 Hz, but 314.16 * L at 50 Hz.             |
|       - X_C = 1 / (2πfC) = 1 / (377 * C) at 60 Hz.                          |
|   [!] Sign of Power Factor Angle: Leading PF has θ < 0 (negative Q);        |
|       Lagging PF has θ > 0 (positive Q).                                    |
+-----------------------------------------------------------------------------+
Test Your Knowledge

A continuous-duty industrial electrical equipment room contains a 750 kVA dry-type transformer operating at 75% load with an efficiency of 98.2%, and a 480V motor control center (MCC) with total continuous internal resistive I^2*R and control circuit losses of 8.5 kW. What is the total rate of heat dissipation into the room, and what minimum cooling capacity in tons of refrigeration (TR) is required solely to remove this electrical heat load (assume 1 TR = 12,000 BTU/hr and power factor = 0.88)?

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Test Your Knowledge

A 460V, 3-phase, 60 Hz, 4-pole induction motor delivers a rated mechanical shaft output of 50 hp while operating at a speed of 1,760 RPM. The motor operates with an electrical efficiency of 92.5% and a lagging power factor of 0.86. What is the full-load mechanical output torque in lb-ft, and what is the total electrical input active power in kW?

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Test Your Knowledge

A single-phase AC branch circuit delivers power to a load whose terminal voltage is v(t) = 170 cos(377t + 30°) V and whose drawn current is i(t) = 14.14 cos(377t - 15°) A. Using the passive sign convention, which set of values represents the RMS voltage, real power (P), reactive power (Q), and load power factor?

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1.3 Measurement & Metering Principles (Power, Energy, Power Factor)

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