--- title: "Efficiencies of 120 kVA Y/Y transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:120kva-yy-1600w-max-efficiency-3" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/120kva-yy-1600w-max-efficiency-3 site: BCT Engineering --- # Efficiencies of 120 kVA Y/Y transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2069 Poush, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/120kva-yy-1600w-max-efficiency-3 - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 120 kVA, 6000/400 V, Y/Y 3-phase, 50 Hz transformer has an iron loss of 1600 W. The maximum efficiency occurs at ¾ full load. Find the efficiency that occurs at 3/4 full load. Find the efficiencies of the transformer at: (i) Full-load and 0.8 power factor (ii) Half-load and unity power factor. ## Short answer Full-load copper loss = 1600/(0.75)² = 2844.44 W. Maximum efficiency (¾ FL, upf) = 96.57%. (i) Full load, 0.8 pf: efficiency = 95.58%. (ii) Half load, unity pf: efficiency = 96.29%. ## Answer Given: $S = 120$ kVA, $P_i = 1600$ W, maximum efficiency at $x = \frac34$ full load. #### Full-load copper loss At maximum efficiency, copper loss = iron loss: $$ \begin{aligned} x^2P_{cu,FL} &= P_i \\ P_{cu,FL} &= \frac{1600}{(0.75)^2} = 2844.44\ \text{W} \end{aligned} $$ #### Efficiency at ¾ full load (maximum efficiency) Power factor not stated for this part, so **unity pf** is assumed: $$ \begin{aligned} \text{Output} &= 0.75 \times 120 \times 1 = 90\ \text{kW} \\ \text{Losses} &= 1600 + 1600 = 3200\ \text{W} \\ \eta_{max} &= \frac{90000}{90000 + 3200}\times 100 = 96.57\% \end{aligned} $$ #### (i) Full load, 0.8 pf $$ \begin{aligned} \text{Output} &= 120 \times 0.8 = 96\ \text{kW} \\ \text{Losses} &= 1600 + 2844.44 = 4444.44\ \text{W} \\ \eta &= \frac{96000}{96000 + 4444.44}\times 100 = 95.58\% \end{aligned} $$ #### (ii) Half load, unity pf $$ \begin{aligned} \text{Output} &= 0.5 \times 120 \times 1 = 60\ \text{kW} \\ P_{cu} &= 0.5^2 \times 2844.44 = 711.11\ \text{W} \\ \eta &= \frac{60000}{60000 + 1600 + 711.11}\times 100 = 96.29\% \end{aligned} $$ | Load | pf | Output | Cu loss | Efficiency | |---|---|---|---|---| | ¾ FL | 1.0 | 90 kW | 1600 W | 96.57% (max) | | Full | 0.8 | 96 kW | 2844.44 W | 95.58% | | Half | 1.0 | 60 kW | 711.11 W | 96.29% | The Y/Y connection and voltage rating do not affect the answer because efficiency depends only on total power and losses. **Answer:** $\eta_{max}$ (¾ FL, upf) = **96.57%**; (i) **95.58%**; (ii) **96.29%**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Maximum efficiency occurs at 3/4 full load with iron loss 1600 W. The full-load copper loss is: A. 1600 W B. 2133 W C. 2844.44 W D. 3200 W Answer: C. x^2 Pcu,FL = Pi gives Pcu,FL = 1600/0.5625 = 2844.44 W. (Key point: Finding full-load copper loss) 2. At maximum efficiency (3/4 load) the total losses are: A. 1600 W B. 3200 W C. 4444 W D. 2844 W Answer: B. At maximum efficiency copper loss equals iron loss, so total loss = 1600 + 1600 = 3200 W. (Key point: Losses at maximum efficiency) 3. At full load and 0.8 pf the output and losses used are: A. 120 kW output, 3200 W loss B. 96 kW output, 4444.44 W loss C. 96 kW output, 3200 W loss D. 120 kW output, 4444.44 W loss Answer: B. Output = 120 x 0.8 = 96 kW; losses = 1600 + 2844.44 = 4444.44 W, giving 95.58%. (Key point: Efficiency at full load 0.8 pf) 4. Why do the Y/Y connection and 6000/400 V ratings not affect the efficiency calculation? A. Efficiency depends only on total output power and losses B. Star connection has no copper loss C. Voltage ratings change the iron loss to zero D. The ratio cancels the power factor Answer: A. Efficiency = output/(output + losses) in total watts, independent of connection or voltage. 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