How is all-day efficiency defined?

Energy output in 24 h divided by energy output plus energy losses in 24 h

Output kW over input kW at full load

Average of efficiencies at each load

Maximum efficiency during the day

How is iron loss energy counted?

0.4 kW for all 24 hours = 9.6 kWh, as the core is always excited

How is copper loss at each load found?

1.7 × (kVA/150)² kW, using kVA = kW/pf

What is the all-day efficiency?

About 98.45%, from 990/(990 + 9.6 + 5.957)

2081 Chaitra (new course) · 6 marks

All-day efficiency of 150 kVA transformer

Short answer

Total energy output is 990 kWh. Iron loss over 24 h is 9.6 kWh, total copper loss 5.957 kWh, sum 15.557 kWh. All-day efficiency = 990 / (990 + 15.557) = 98.45%.

Daily loading of a 150 kVA transformer having iron loss of 400 W and full load copper loss of 1700 W is given below.

Time	Load (kW)	Power factor
6:00 – 10:00	50	0.9
10:00 – 17:00	60	0.85
17:00 – 21:00	70	0.8
21:00 – 6:00	10	0.95

Determine the all-day efficiency of the transformer.

All-day efficiency = (energy output in 24 h) / (energy output + energy losses in 24 h). Iron loss lasts all 24 hours; copper loss varies as the square of kVA load.

Data: 150 kVA, $P_i = 0.4$ kW, $P_{cu,fl} = 1.7$ kW.

Energy table

Copper loss at each load $= 1.7\times(\text{kVA}/150)^2$ .

Period	Hours	kW	pf	kVA	Cu loss (kW)	Cu energy (kWh)	Output (kWh)
6–10	4	50	0.90	55.56	0.2332	0.933	200
10–17	7	60	0.85	70.59	0.3765	2.635	420
17–21	4	70	0.80	87.50	0.5785	2.314	280
21–6	9	10	0.95	10.53	0.0084	0.075	90
Total	24					5.957	990

Losses and efficiency

\begin{aligned} \text{Iron energy} &= 0.4\times24 = 9.6\ \text{kWh} \\ \text{Total losses} &= 9.6 + 5.957 = 15.557\ \text{kWh} \\ \eta_{all\text{-}day} &= \frac{990}{990 + 15.557}\times100 = 98.45\% \end{aligned}

Answer: all-day efficiency ≈ 98.45%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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