Electrical Power Systems & Energy Engineering
Transformer All-Day Efficiency: Core vs Copper Loss Formula
Calculate distribution transformer peak-efficiency load fraction x_opt = SQRT(P_core / P_cu,fl), peak instantaneous efficiency, and 24-hour duty-cycle All-Day Energy Efficiency.
Why distribution transformers peak at 50% load rather than 100% rated kVA!
Utility distribution transformers stay energized 24 hours a day, incurring Constant Core (Iron Hysteresis + Eddy-Current) Loss P_fe every hour regardless of load, plus Winding I^2*R Copper Loss x^2 * P_cu,fl proportional to the SQUARE of per-unit kVA load fraction x = S / S_rated! Differentiating instantaneous efficiency eta(x) = (x * S_rated * PF) / (x * S_rated * PF + P_fe + x^2 * P_cu,fl) proves maximum efficiency occurs precisely where variable copper loss equals constant iron loss: x_opt = SQRT(P_fe / P_cu,fl)!
Verify on Rated_Capacity S_r = 100 kVA (B2), Constant_Core_Loss P_fe = 1.00 kW (B3), Full_Load_Copper_Loss P_cu,fl = 4.00 kW (B4), Load_Power_Factor PF = 0.80 (B5), and a 24-hour duty cycle of 8 hr at 100% load (B6), 8 hr at 50% load (B7), and 8 hr energized at 0% idle no-load (B8): Optimal Load Fraction is x_opt = SQRT(1.00 / 4.00) = 0.5000 (50% kVA load!), where Peak Efficiency reaches 40 / (40 + 1.00 + 0.25*4.00) = 40 / 42 = 0.9524 (95.24%, beating 100%-load efficiency 80 / 85 = 94.12%)! Across the 24-hour cycle, Daily Output Energy is 8*80 + 8*40 + 8*0 = 960.00 kWh, Daily Loss Energy is 24*1.00 (core) + 8*4.00 + 8*1.00 (copper) = 64.00 kWh, and 24-Hour All-Day Efficiency is 960 / (960 + 64) = 960 / 1,024 = 0.9375 (93.75%)!
| Transformer Energy Metric (100 kVA, P_fe=1 kW, P_cu=4 kW, PF=0.80) | Formula | Verified Output |
|---|---|---|
| Peak-Efficiency Load Fraction (x_opt) | SQRT(P_fe / P_cu,fl) | 0.5000 (50.0% Rated kVA) |
| Maximum Instantaneous Efficiency | (x_opt*S_r*PF) / (x_opt*S_r*PF + 2*P_fe) | 0.9524 (95.24%) |
| 24-Hour Delivered Energy (W_out) | (h_100*1.0 + h_50*0.5) * S_r * PF | 960.00 kWh / day |
| 24-Hour All-Day Energy Efficiency | W_out / (W_out + 24*P_fe + W_cu) | 0.9375 (93.75%; 64 kWh Loss) |
Peak-Efficiency Load Fraction & Instantaneous Peak Efficiency Formula
Notice why utilities specify low-core-loss amorphous-metal transformers (small P_fe / P_cu,fl) for residential feeders that spend most of the night lightly loaded!
=LET(sr_kv, B2, pf_fe, B3, pc_fl, B4, pf_ld, B5, IF(OR(sr_kv <= 0, pf_fe <= 0, pc_fl <= 0, pf_ld <= 0, pf_ld > 1, B6 < 0, B7 < 0, B8 < 0, B6 + B7 + B8 <> 24), "Invalid Transformer Cycle", ROUND(SQRT(pf_fe / pc_fl), 4)))24-Hour All-Day Energy Efficiency Formula
Enforce h_full + h_half + h_idle = 24 hours so core loss is integrated across the complete 24-hour energized day!
=LET(sr_kv, B2, pf_fe, B3, pc_fl, B4, pf_ld, B5, h_fl, B6, h_hl, B7, IF(ISTEXT(E2), E2, LET(w_ot, (h_fl * 1 + h_hl * 0.5) * sr_kv * pf_ld, w_ls, 24 * pf_fe + (h_fl * 1 + h_hl * 0.25) * pc_fl, ROUND(w_ot / (w_ot + w_ls), 4))))Try the example
Use the original values or change them to test your understanding.