Economic Conductor Sizing (I2R Payback)
Whether upsizing a conductor pays for itself in reduced I^2R heat loss.
Example
You enter
- Per-phase load current (A) 100
- Smaller conductor resistance (ohm, run) 0.2
- Larger conductor resistance (ohm, run) 0.125
- Annual run hours 4000
- Electricity rate ($/kWh) 0.12
- Added upsize cost ($) 800
You get
- Annual saving 1080
- Payback (yr) 0.74
Details, formula, and sources
Three-phase loss = 3 x I^2 x R, so the annual saving = (loss at the small size - loss at the large) x run hours x rate, and payback = the added copper cost / that saving. A 100 A feeder from 0.20 to 0.125 ohm over 4,000 hr at $0.12 saves $1,080/yr, paying back an $800 upsize in 0.7 years; at 40 A the same upsize saves only $173/yr, a 4.6-year payback -- upsizing only pays on heavily loaded, long-hour feeders. A screening estimate; the code minimum still governs the conductor.
loss = 3 x I^2 x R / 1000 (kW, per size); annual_saving = (loss_small - loss_big) x hours x rate; payback = upsize_cost / annual_saving.
Economic (loss-based) conductor sizing, standard energy-engineering practice (NEC Informative Annex D / IEEE economic-conductor methods) by name; first-principles I^2R loss.
The three-phase I^2R loss and the payback arithmetic are first-principles; conductor resistances are in NEC Chapter 9 Table 8/9.
Estimate. AHJ and licensed professional govern.
Field names used by the API: current_a, r_small_ohm, r_big_ohm, hours, rate_kwh, upsize_cost, annual_saving, payback_yr
- I^2R loss three-phase loss = 3 x I^2 x R per conductor sizefirst-principles resistive loss
- Payback saving = (loss_small - loss_big) x hours x rate; payback = cost / savingeconomic arithmetic
- Code still governs NEC minimum ampacity, voltage drop, and fill govern the actual conductorscope of this tile