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T-06c: Transformer Efficiency

Section: A | Priority: 🟠 HIGH | Exam Frequency: 4/7 years Sources: Theraja Ch-32 (Art. 32.29–32.34), VK Mehta Ch-7 (Art. 7.22–7.24), Slides L-10 S19

Why This Topic Matters

Efficiency calculations appeared in 4/7 papers (2017, 2018, 2019, 2023). The question types include: (1) calculate efficiency at full-load and half-load for different power factors, (2) prove the condition for maximum efficiency, and (3) calculate all-day efficiency with a load schedule. All-day efficiency appeared in 3/7 papers (2018, 2019, 2023). These are pure formula-plugging questions. Free marks.

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📝 Key Definitions

Transformer efficiency ($\eta$): "The efficiency of a transformer is defined as the ratio of output power to input power. Since input = output + losses: $\eta =\text{Output}/(\text{Output}+\text{Losses})$." — VK Mehta, Art. 7.22

All-day efficiency (${\eta }_{\text{all-day}}$): "The all-day efficiency (or energy efficiency) is defined as the ratio of total energy output (kWh) to total energy input (kWh) over a 24-hour period. It is of particular importance for distribution transformers which remain connected to the supply round the clock but deliver varying loads throughout the day." — VK Mehta, Art. 7.24

Maximum efficiency condition: "Efficiency of a transformer is maximum when copper loss equals iron loss, i.e., variable loss = constant loss. This condition yields the load fraction $x=\sqrt{P_{Fe}/P_{Cu,FL}}$." — VK Mehta, Art. 7.23

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The Efficiency Formula

A transformer has exactly two types of losses:

1. Iron loss ($P_{Fe}$): Constant at all loads. Measured from OC test.
2. Copper loss ($P_{Cu}$): Proportional to load current squared. At fraction $x$ of full load: $P_{Cu}=x^2{P}_{Cu,FL}$.

$$\boxed{\eta =\frac{x\cdot S\cdot \cos \varphi }{x\cdot S\cdot \cos \varphi +P_{Fe}+x^2{P}_{Cu,FL}}\times 100\%}$$

where:

• $x$ = fraction of full load (1 = full load, 0.5 = half load)
• $S$ = rated VA (in watts, not kVA!)
• $\cos \varphi$ = load power factor
• $P_{Fe}$ = iron loss (from OC test)
• $P_{Cu,FL}$ = full-load copper loss (from SC test)

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Condition for Maximum Efficiency

Differentiate $\eta$ with respect to $x$ and set to zero:

$$\frac{d\eta }{dx}=0⟹x^2{P}_{Cu,FL}=P_{Fe}$$

$$\boxed{\text{At maximum efficiency: Copper loss = Iron loss}}$$

$$x_{\text{max}\eta }=\sqrt{\frac{P_{Fe}}{P_{Cu,FL}}}$$

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All-Day (Energy) Efficiency

$$\boxed{{\eta }_{\text{all-day}}=\frac{\text{Total kWh output}}{\text{Total kWh output}+\text{Total kWh losses}}\times 100\%}$$

Procedure:

1. Energy output: For each load period, $\text{kWh}=x\cdot S\cdot \cos \varphi \times \text{hours}$
2. Iron loss energy: $P_{Fe}\times 24$ kWh (runs 24 hours, always!)
3. Copper loss energy: For each period, $x_i^2\times P_{Cu,FL}\times t_i$ hours
4. Sum everything and divide.

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🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Prove that maximum efficiency occurs when copper loss = iron loss.

Appeared: 2019 Q3(a) — 4 marks

Full Answer:

Derivation:

Transformer efficiency: $\eta =\frac{P_{out}}{P_{out}+P_{Fe}+P_{Cu}}$

At fraction $x$ of full load: $P_{out}=xS\cos \varphi$, $P_{Cu}=x^2{P}_{Cu,FL}$, $P_{Fe}$ is constant.

$$\eta =\frac{xS\cos \varphi }{xS\cos \varphi +P_{Fe}+x^2{P}_{Cu,FL}}$$

Let $A=S\cos \varphi$ (constant for fixed pf). Then:

$$\eta =\frac{xA}{xA+P_{Fe}+x^2{P}_{Cu,FL}}$$

For maximum $\eta$, set $d\eta /dx=0$. Using the quotient rule:

$$\frac{d\eta }{dx}=\frac{A(xA+P_{Fe}+x^2{P}_{Cu,FL})-xA(A+2x{P}_{Cu,FL})}{(xA+P_{Fe}+x^2{P}_{Cu,FL}{)}^2}=0$$

Numerator must be zero:

$$A(xA+P_{Fe}+x^2{P}_{Cu,FL})-xA(A+2x{P}_{Cu,FL})=0$$

$$x{A}^2+A{P}_{Fe}+A{x}^2{P}_{Cu,FL}-x{A}^2-2A{x}^2{P}_{Cu,FL}=0$$

$$A{P}_{Fe}-A{x}^2{P}_{Cu,FL}=0$$

$$\boxed{P_{Fe}=x^2{P}_{Cu,FL}}$$

That is: iron loss = copper loss at the operating load.

The load fraction for maximum efficiency:

$$x=\sqrt{\frac{P_{Fe}}{P_{Cu,FL}}}$$

For example, if $P_{Fe}=350$ W and $P_{Cu,FL}=400$ W:

$$x=\sqrt{350/400}=\sqrt{0.875}=0.935$$

Maximum efficiency occurs at 93.5% of full load.

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🎯 Q2: 25 kVA transformer, iron loss = 350 W, full-load Cu loss = 400 W. Find efficiency at FL and HL, at upf and 0.8 pf lag.

Appeared: 2017 Q5(c) — 4 marks

Full Answer:

Given: $S=25$ kVA = 25000 W, $P_{Fe}=350$ W, $P_{Cu,FL}=400$ W.

Condition	$x$	Output (W)	$P_{Fe}$ (W)	$P_{Cu}=x^2\times 400$ (W)	Input (W)	$\eta$
FL, upf	1.0	25000	350	400	25750	$25000/25750=\boxed{97.1\%}$
FL, 0.8 lag	1.0	20000	350	400	20750	$20000/20750=\boxed{96.4\%}$
HL, upf	0.5	12500	350	100	12950	$12500/12950=\boxed{96.5\%}$
HL, 0.8 lag	0.5	10000	350	100	10450	$10000/10450=\boxed{95.7\%}$

Why is efficiency lower at 0.8 pf? The losses (350 + 400 W) stay the same. But the output drops (multiplied by 0.8). So losses take a bigger fraction of the input.

Why is HL efficiency less than FL at same pf? At half load, the iron loss (350 W) is proportionally larger compared to the output (12.5 kW) than at full load (25 kW). Maximum efficiency occurs at $x=\sqrt{350/400}=0.935$ (93.5% of full load), which is close to FL.

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🎯 Q3: 100 kVA transformer, $P_{Fe}$ = 1 kW, $P_{Cu,FL}$ = 1 kW. Load profile: 4h no-load, 12h half-load, 8h full-load. Find all-day efficiency.

Appeared: 2023 Q2(a) — 6 marks

Full Answer:

Given: $S=100$ kVA, $P_{Fe}=1$ kW, $P_{Cu,FL}=1$ kW, unity pf assumed.

Energy output:

Period	Load fraction $x$	kW Output	Hours	kWh
No-load	0	0	4	0
Half-load	0.5	50	12	600
Full-load	1.0	100	8	800
Total			24	1400

Iron loss energy (runs 24 hours, always, even at no-load): $1\times 24=24$ kWh

Copper loss energy:

Period	$x^2\times P_{Cu,FL}$ (kW)	Hours	kWh
No-load	$0^2\times 1=0$	4	0
Half-load	$0.25\times 1=0.25$	12	3
Full-load	$1\times 1=1$	8	8
Total Cu			11

Total losses = $24+11=35$ kWh

$${\eta }_{\text{all-day}}=\frac{1400}{1400+35}=\frac{1400}{1435}=\boxed{97.6\%}$$

Why is all-day efficiency less than full-load efficiency? During the 4 hours of no-load, the transformer still consumes 4 kWh of iron losses for zero output. These are "wasted" hours from the efficiency perspective. Distribution transformers are designed with low iron loss for this reason.

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🎯 Q4: Define all-day efficiency.

Appeared: 2018 Q2(a) — 1 mark

Full Answer:

"All-day efficiency (or energy efficiency) is defined as the ratio of total energy output (kWh) to total energy input (kWh) over a 24-hour period." — VK Mehta, Art. 7.24

$${\eta }_{\text{all-day}}=\frac{\text{Total kWh output in 24h}}{\text{Total kWh input in 24h}}\times 100=\frac{\text{Total kWh output}}{\text{Total kWh output}+\text{Total kWh losses}}\times 100$$

It is important for distribution transformers that remain connected to the supply 24 hours a day but supply varying loads. The iron loss accumulates over all 24 hours regardless of load, while copper loss varies with the load profile.

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⚡ Exam Tips & Common Mistakes

1. Iron loss runs 24 hours. Even during no-load periods. This is the single most common error in all-day efficiency calculations.
2. Copper loss scales with $x^2$, not $x$. At half load: $P_{Cu}=0.25P_{Cu,FL}$, not $0.5P_{Cu,FL}$.
3. Use watts, not kVA. Output power = $x\times S\times \cos \varphi$ (watts). Don't forget the power factor.
4. At maximum efficiency, $x=\sqrt{P_{Fe}/P_{Cu,FL}}$. If $P_{Fe}=350$ W and $P_{Cu}=400$ W, then $x=0.935$. Max η is at 93.5% of full load.
5. All-day efficiency is always less than full-load efficiency because iron losses accumulate during idle periods.

🔗 Related Topics

• T-06a: OC Test — Measures $P_{Fe}$
• T-06b: SC Test — Measures $P_{Cu,FL}$
• T-05: Voltage Regulation — Often calculated alongside efficiency
• T-11: Miscellaneous — Hysteresis and eddy current loss formulas

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