Low-Voltage Feeder Study

The system this study analyses

Drawn from the study’s own definition and annotated with the example values below. Every quantity on it is one the study entered or computed.

Every shared value is entered once

The system voltage is not typed into four calculators. It is defined once for the study and carried into every step that needs it — so a change reaches all of them, and no two steps can describe different systems.

The 5 steps

Worked through below with an example low-voltage feeder study. Every number shown was produced by the same calculators the library already ships — nothing on this page is illustrative.

1. Step 01

   Source transformer

   Transformer Sizing & Full-Load Current Calculator

   Computes the transformer's primary and secondary full-load currents from its rating and the two voltages, and reports the nearest standard rating.

   Inputs and their sources

   • SystemThree-phaseFixed by the study
   • Transformer rating500 kVAYou enter
   • Primary voltage13.8 kVYou enter
   • Secondary voltage480 VYou enter

   Example results

   • Secondary full-load current601.4A
   • Primary full-load current20.92A
   • Nearest standard size500kVA
   ›Why this step is here

   The study starts at the source because the secondary voltage established here is the same voltage every downstream step works from. The transformer's own full-load current is reported for reference; it is a property of the transformer, not of the load, so it is not carried into the feeder calculations.

   ›Worked solution

   1. Primary full-load current

      I_FL,P = S / √3 V_P$$I_{FL,P} = \dfrac{S}{\sqrt{3}\,V_P}$$
      I_FL,P = (5.000e+5) / (√3 × 1.380e+4)$$I_{FL,P} = \dfrac{5.000e+5}{\sqrt{3} \times 1.380e+4}$$
      I_FL,P = 20.92 A$$I_{FL,P} = 20.92\ \text{A}$$

   2. Secondary full-load current

      I_FL,S = S / √3 V_S$$I_{FL,S} = \dfrac{S}{\sqrt{3}\,V_S}$$
      I_FL,S = (5.000e+5) / (√3 × 480)$$I_{FL,S} = \dfrac{5.000e+5}{\sqrt{3} \times 480}$$
      I_FL,S = 601.4 A$$I_{FL,S} = 601.4\ \text{A}$$

   Calculator assumptions

   • Self-cooled (ONAN / dry-type AA) nameplate rating; forced-air or other cooling stages would raise the FLA basis.
   • Balanced loading on both sides.
   • Standard-size list reflects common IEEE/ANSI distribution transformer ratings — verify against the manufacturer's actual offered sizes.

   References

   • IEEE Std C57.12.00 — General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
   • NEC Table 450.3 — Overcurrent Protection for Transformers

2. Step 02

   Load power and line current

   Three-Phase Power Calculator

   Computes apparent power, reactive power and the line current the load draws, from its real power, power factor and the system voltage.

   Inputs and their sources

   • SystemThree-phaseFixed by the study
   • Real power200 kWYou enter
   • Power factor0.9You enter
   • System voltage480 VYou enter
   • Load typeLagging (inductive)You enter

   Example results

   • Apparent power222.222kVA
   • Reactive power (lagging)96.864kVAR
   • Line current267.29A
   ›Why this step is here

   This is the step the rest of the study depends on. The line current computed here is the current the feeder actually carries, so it is the value the voltage-drop and ampacity steps both take — entered nowhere, derived once.

   ›Worked solution

   1. Apparent power from real power and power factor

      S = P / pf$$S = \dfrac{P}{\text{pf}}$$
      S = 200 / 0.9 kW$$S = \dfrac{200}{0.9}\ \text{kW}$$
      S = 222.2 kVA$$S = 222.2\ \text{kVA}$$

   2. Reactive power from the power triangle

      Q = √(S² - P²)$$Q = \sqrt{S^2 - P^2}$$
      Q = √(222.2² - 200²) kVA$$Q = \sqrt{222.2^2 - 200^2}\ \text{kVA}$$
      Q = 96.86 kVAR$$Q = 96.86\ \text{kVAR}$$

   3. Line current (three-phase)

      I_L = S / √3 V_LL$$I_L = \dfrac{S}{\sqrt{3}\,V_{LL}}$$
      I_L = (2.222e+5) / (√3 × 480)$$I_L = \dfrac{2.222e+5}{\sqrt{3} \times 480}$$
      I_L = 267.3 A$$I_L = 267.3\ \text{A}$$

   Calculator assumptions

   • Balanced three-phase system assumed in three-phase mode.
   • Voltage entered is line-to-line for three-phase, or the single-phase RMS voltage for single-phase.
   • Power factor is treated as a magnitude; leading vs. lagging only changes the sign convention, not the values shown.

   References

   • IEEE Std 100, The Authoritative Dictionary of IEEE Standards Terms — power triangle definitions
   • Glover, Sarma, Overbye, Power System Analysis and Design, 6th ed., Ch. 2

3. Step 03

   Feeder voltage drop

   Voltage Drop Calculator

   Computes the voltage drop over the feeder run for the load current derived in step 02, and the resulting receiving-end voltage.

   Inputs and their sources

   • SystemThree-phaseFixed by the study
   • Conductor materialCopperYou enter
   • Conductor size500 kcmilYou enter
   • Load current267.3 ADerivedStep 02 · Load power and line current
   • One-way run length250 ftYou enter
   • Source voltage480 VYou enter
   • Load power factor0.9You enter
   • Maximum acceptable drop3 %You enter

   Example results

   • Voltage drop (percent)0.97%
   • Voltage drop (volts)4.66V
   • Receiving-end voltage475.3V
   ›Why this step is here

   The current is not re-entered here. It arrives from step 02, so changing the load real power or the system voltage moves this result without anyone retyping a number — and the provenance line records where the value came from.

   ›Worked solution

   1. Conductor properties (per foot)

      R, X from conductor table$$R,\ X \text{ from conductor table}$$
      R = 0.0000258 Ω/ft, X = 0.000039 Ω/ft$$R = 0.0000258\ \Omega/\text{ft},\quad X = 0.000039\ \Omega/\text{ft}$$

      Representative values for THWN/THHN-class conductors in non-metallic conduit.

   2. Effective impedance per foot

      Z_eff = R cos θ + X sin θ$$Z_{eff} = R\cos\theta + X\sin\theta$$
      Z_eff = 0.0000258 × 0.9 + 0.000039 × sin (cos⁻¹(0.9))$$Z_{eff} = 0.0000258 \times 0.9 + 0.000039 \times \sin(\cos^{-1}(0.9))$$
      Z_eff = 0.00004022 Ω/ft$$Z_{eff} = 0.00004022\ \Omega/\text{ft}$$

   3. Voltage drop (three-phase, k = √3)

      VD = k × I × L × Z_eff$$VD = k \times I \times L \times Z_{eff}$$
      VD = 1.732 × 267.3 × 250 × 0.00004022$$VD = 1.732 \times 267.3 \times 250 \times 0.00004022$$
      VD = 4.655 V$$VD = 4.655\ \text{V}$$

   4. Voltage drop as a percentage

      VD% = VD / V × 100$$VD\% = \dfrac{VD}{V} \times 100$$
      VD% = 4.655 / 480 × 100$$VD\% = \dfrac{4.655}{480} \times 100$$
      VD% = 0.9698%$$VD\% = 0.9698\%$$

   5. Receiving-end voltage

      V_R = V - VD$$V_R = V - VD$$
      V_R = 480 - 4.655$$V_R = 480 - 4.655$$
      V_R = 475.3 V$$V_R = 475.3\ \text{V}$$

   Calculator assumptions

   • Lagging (inductive) power factor assumed; a leading power factor would produce slightly less drop than shown.
   • Conductor resistance and reactance are representative THWN/THHN values in non-metallic conduit, not a substitute for the governing code table.
   • Balanced loading assumed for the three-phase case; steady-state conditions (not motor starting).

   References

   • NEC 210.19(A) Informational Note No. 4 and 215.2(A) Informational Note No. 2 — recommended voltage drop limits
   • IEEE Std 141 (Red Book), Ch. 3 — Voltage Considerations

4. Step 04

   Conductor ampacity derating

   Cable Ampacity & Sizing Assistant

   Safety-sensitive

   Applies the ambient-temperature and grouping adjustment factors to the tabulated ampacity, and reports the derated ampacity beside the load current derived in step 02.

   Inputs and their sources

   • Load current267.3 ADerivedStep 02 · Load power and line current
   • Conductor materialCopperYou enter
   • Ambient temperature30 °CYou enter
   • Current-carrying conductors in raceway3You enter

   Example results

   • Derated ampacity of suggested conductor285A
   • Table ampacity (before derating)285A
   • Required table ampacity267.3A
   • Ambient temperature factor1
   • Grouping adjustment factor1
   • Suggested size (reference)300kcmil
   ›Why this step is here

   This step reports calculated ampacity values for the stated conditions: the tabulated ampacity, the two adjustment factors, and the derated result. It draws no conclusion. The derated ampacity and the load current are two numbers presented side by side, and reading one against the other is the engineer's work rather than the study's. This step's own model-boundary notes are printed with it and are worth reading first.

   ›Worked solution

   1. Correction factors

      k_a (ambient), k_g (grouping)$$k_a \text{ (ambient)}, \ k_g \text{ (grouping)}$$
      looked up from ambient temperature and conductor count$$\text{looked up from ambient temperature and conductor count}$$
      k_a = 1, k_g = 1$$k_a = 1, \quad k_g = 1$$

   2. Required table ampacity

      I_req = I / (k_a × k_g)$$I_{req} = \dfrac{I}{k_a \times k_g}$$
      I_req = 267.3 / (1 × 1)$$I_{req} = \dfrac{267.3}{1 \times 1}$$
      I_req = 267.3 A$$I_{req} = 267.3\ \text{A}$$

   3. Smallest standard size clearing the requirement

      smallest size where I_table ≥ I_req$$\text{smallest size where } I_{table} \geq I_{req}$$
      size = 300 kcmil$$\text{size} = \text{300 kcmil}$$
      I_table = 285 A$$I_{table} = 285\ \text{A}$$

   4. Derated (allowable) ampacity of the suggested size

      I_allow = I_table × k_a × k_g$$I_{allow} = I_{table} \times k_a \times k_g$$
      I_allow = 285 × 1$$I_{allow} = 285 \times 1$$
      I_allow = 285 A$$I_{allow} = 285\ \text{A}$$

   Calculator assumptions

   • Representative 75°C-column ampacity, ambient-temperature correction, and grouping-adjustment values, consistent in shape with NEC Table 310.16 and 310.15(B)(1)/(C)(1) — verify against the governing code edition and jurisdiction.
   • Does not apply 125% continuous-load sizing — apply that separately if the load is continuous.
   • Does not check voltage drop, conduit fill, or termination temperature rating — verify each separately.

   References

   • NEC Table 310.16 and Table 310.15(B)(1) / 310.15(C)(1) (or the equivalent CEC Table 2/5A/5C for Canadian installations)
   • IEEE Std 835 — Power Cable Ampacity Tables

5. Step 05

   Power-factor correction

   Power Factor Correction Capacitor Sizing

   Computes the reactive power a capacitor bank would have to supply to move the load from its existing power factor to the target, and the line current before and after.

   Inputs and their sources

   • Real power200 kWYou enter
   • Existing power factor0.9You enter
   • Target power factor0.95You enter
   • System voltage480 VYou enter
   • Frequency60 HzYou enter
   • Bank connectionWyeYou enter

   Example results

   • Required capacitor bank31.13kVAR
   • Capacitance per phase358.37µF
   • Current before correction267.3A
   • Current after correction253.2A
   • Line current reduction5.3%
   ›Why this step is here

   Real power and power factor are the SAME shared study inputs step 02 used, not a second copy — so this step and the load-current calculation can never describe two different loads.

   ›Worked solution

   1. Reactive power at existing power factor

      Q_1 = P tan (cos⁻¹(pf_1))$$Q_1 = P \tan(\cos^{-1}(\text{pf}_1))$$
      Q_1 = 200 × tan (cos⁻¹(0.9))$$Q_1 = 200 \times \tan(\cos^{-1}(0.9))$$
      Q_1 = 96.86 kVAR$$Q_1 = 96.86\ \text{kVAR}$$

   2. Reactive power at target power factor

      Q_2 = P tan (cos⁻¹(pf_2))$$Q_2 = P \tan(\cos^{-1}(\text{pf}_2))$$
      Q_2 = 200 × tan (cos⁻¹(0.95))$$Q_2 = 200 \times \tan(\cos^{-1}(0.95))$$
      Q_2 = 65.74 kVAR$$Q_2 = 65.74\ \text{kVAR}$$

   3. Required capacitor bank

      Q_c = Q_1 - Q_2$$Q_c = Q_1 - Q_2$$
      Q_c = 96.86 - 65.74$$Q_c = 96.86 - 65.74$$
      Q_c = 31.13 kVAR$$Q_c = 31.13\ \text{kVAR}$$

   4. Equivalent per-phase capacitance (wye)

      C = Q_c / (ω V_LL²)$$C = \dfrac{Q_c}{\omega V_{LL}^2}$$
      using ω = 2π f$$using \ \omega = 2\pi f$$
      C = 358.4 µF$$C = 358.4\ \mu\text{F}$$

   Calculator assumptions

   • Balanced three-phase load and capacitor bank.
   • Bank is sized for the stated real power and existing power factor at that operating point — verify against the load's minimum demand, not only its peak.
   • Capacitor bank connection (wye or delta) only affects the per-phase capacitance value, not the required kVAR.

   References

   • IEEE Std 1036, IEEE Guide for the Application of Shunt Power Capacitors
   • IEEE Std 141 (Red Book), Ch. 8 — Power Factor Capacitors

Why these steps, in this order

What this study reports, and what it leaves to you