Transformer Efficiency and Autotransformers

Both losses are known from the tests of Chapter 46 — \(P_i\) from the open-circuit wattmeter and \(P_c\) from the short-circuit wattmeter — or from Chapter 47's Sumpner test, halved.

If \(x\) is the fraction of full load, the output scales with \(x\) and the copper loss with \(x^{2}\), while the iron loss does not change at all.

Take the full-load expression and divide numerator and denominator by \(I_2\):

This is Chapter 33's condition for the DC machine, unchanged. It carries over because it depends only on the shape of the loss expression — one term fixed and one proportional to the square of the current — and not on whether the machine rotates.

Writing the condition in terms of the load fraction, with \(P_c\) the full-load copper loss:

Since \(P_i\) is usually less than \(P_c\), the ratio is below one and the peak falls below full load — typically between 70 % and 90 %.

• Power transformers, at a generating station, are switched in or out of the circuit depending on the load to be handled.

• A distribution transformer at a substation is never switched off, and remains energised irrespective of the load.

• In that case the constant loss continues to be dissipated throughout the day, whether or not any output is being delivered.

The ordinary power efficiency is then a misleading figure of merit, and a concept of energy-based efficiency arises — the all-day efficiency.

The design consequence, already met in Chapter 44, is that a distribution transformer is built with a deliberately low iron loss, even at the cost of higher copper loss — which moves its maximum-efficiency point down to perhaps half load.

• A transformer with one winding only.

• Part of the winding is common to both primary and secondary.

• There is no electrical isolation, as there is in a two-winding transformer.

• The theory and operation remain similar — the EMF equation, the turns ratio and the ampere-turn balance all apply unchanged.

• Because there is only one winding, it uses less copper and is therefore cheaper.

• It is used where the transformation ratio varies little from unity.

Applying this to the two sections of the autotransformer winding:

If an ordinary two-winding transformer were to perform the same task, its primary would need \(N_1I_1\) and its secondary \(N_2I_2\), so its total weight would be proportional to \(N_1I_1 + N_2I_2\). Taking the ratio:

In a two-winding transformer every watt reaching the load must first cross the air gap magnetically. In an autotransformer only part of it does; the rest flows straight through the metal.

• Efficiency at a load fraction \(x\) is \(xS\cos\phi/(xS\cos\phi + P_i + x^{2}P_c)\): output rises as \(x\), copper loss as \(x^{2}\), iron loss not at all.

• Dividing through by \(I_2\) and minimising the denominator gives the condition \(P_i = P_c\) — constant loss equals variable loss.

• The load fraction is \(x = \sqrt{P_i/P_c}\) and the kVA is \(S\sqrt{P_i/P_c}\). The power factor does not affect where the peak lies, only its value.

• All-day efficiency is a ratio of energies over 24 hours. The iron-loss energy is \(24P_i\) regardless of duty, so the figure is always lower than the commercial efficiency.

• An autotransformer has one winding, part common to both sides, and gives no isolation.

• Copper weight \(\propto\) turns × current, giving \(W_a = (1-K)W_0\) and a saving of \(KW_0\), greatest as \(K \to 1\).

• Power divides into \(P_{ind} = \text{Input}(1-K)\) and \(P_{cond} = \text{Input}\,K\).

• A two-winding transformer reconnected as an autotransformer is uprated by \(1/(1-K)\).

• Advantages: cheaper, better regulation, lower losses. Disadvantages: no isolation, the broken-winding hazard, and a high short-circuit current.

• Letting the iron loss vary with load. It is fixed by the supply voltage.

• Scaling the copper loss linearly. It varies as \(x^{2}\).

• Thinking the power factor moves the maximum-efficiency load. It cancels; only the peak value changes.

• Using kVA in the all-day numerator. Energy is kWh, so the power factor must be applied.

• Omitting the no-load hours from the iron-loss energy. It runs the full 24 hours.

• Defining \(K\) upside down in the saving formula. Here \(K = V_2/V_1\) and the saving is \(KW_0\).

• Expecting a large saving at a large ratio. The saving is greatest when \(K\) is near unity.

• Forgetting that an autotransformer gives no isolation. This is its defining limitation.

• Overloading the common winding. It carries \(I_2 - I_1\), not \(I_2\).

• Assuming the uprating is free of consequence. The percentage impedance falls by the same factor, so the fault current rises.

For efficiency questions find \(x = \sqrt{P_i/P_c}\) first. For autotransformer questions identify which physical winding becomes the series and which the common.

1. P48.1 A 100 kVA transformer has \(P_i = 900\) W and \(P_c = 1600\) W. Find the kVA at maximum efficiency.

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   \[x = \sqrt{\frac{900}{1600}} = 0.750 \quad\Longrightarrow\quad 75.0~\mathrm{kVA}\]

2. P48.2 For P48.1, find the maximum efficiency at 0.9 power factor and the full-load efficiency.

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   \[\eta_{\max} = \frac{(75\,000)(0.9)}{67\,500 + 1800} = \frac{67\,500}{69\,300} = 97.40\,\%\]

   \[\eta_{FL} = \frac{90\,000}{90\,000 + 900 + 1600} = \frac{90\,000}{92\,500} = 97.30\,\%\]

3. P48.3 A 50 kVA transformer with \(P_i = 400\) W and \(P_c = 700\) W supplies full load at 0.8 pf for 8 h, half load at 0.8 pf for 8 h, and nothing for 8 h. Find the all-day efficiency.

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   \[\text{output} = (50)(1.0)(0.8)(8) + (50)(0.5)(0.8)(8) = 320 + 160 = 480~\mathrm{kWh}\]

   \[\text{copper} = \frac{(700)(1)(8) + (700)(0.25)(8)}{1000} = 5.60 + 1.40 = 7.00~\mathrm{kWh}\]

   \[\text{iron} = \frac{(400)(24)}{1000} = 9.60~\mathrm{kWh}\]

   \[\eta_{\text{all-day}} = \frac{480}{480 + 7.00 + 9.60} = \frac{480}{496.6} = 96.66\,\%\]

4. P48.4 An autotransformer has \(K = 0.8\). Find the copper saving and the ratio of its weight to that of an equivalent two-winding transformer.

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   \[\frac{W_a}{W_0} = 1 - K = 0.200, \qquad \text{saving} = KW_0 = 80\,\%\]

   It uses one fifth of the copper.

5. P48.5 A 10 kVA, 1000/100 V two-winding transformer is reconnected as a 1100/1000 V autotransformer. Find its new rating.

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   \[\text{100 V winding: } \frac{10\,000}{100} = 100~\mathrm{A} \quad\text{(series)}\]

   \[\text{Rating} = \frac{(1100)(100)}{1000} = 110~\mathrm{kVA}\]

   \[K = \frac{1000}{1100} = 0.9091, \qquad \frac{1}{1-K} = 11 \quad\Longrightarrow\quad (10)(11) = 110 \quad\checkmark\]

6. P48.6 For P48.5, find the conductively and inductively transferred power.

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   \[P_{ind} = (110)(1 - 0.9091) = 10.0~\mathrm{kVA}, \qquad P_{cond} = (110)(0.9091) = 100.0~\mathrm{kVA}\]

   The inductive part equals the original two-winding rating, as it must.

7. P48.7 Derive the condition for maximum efficiency, and explain why the power factor does not appear in it.

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   Start from \(\eta = V_2I_2\cos\phi_2/(V_2I_2\cos\phi_2 + P_i + I_2^{2}R_{02})\) and divide numerator and denominator by \(I_2\):

   \[\eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + P_i/I_2 + I_2R_{02}}\]

   For a given power factor the numerator is constant, so \(\eta\) is greatest when the denominator is least. Differentiating with respect to \(I_2\):

   \[-\frac{P_i}{I_2^{2}} + R_{02} = 0 \quad\Longrightarrow\quad P_i = I_2^{2}R_{02} = P_c\]

   Why the power factor is absent: the term \(V_2\cos\phi_2\) appears in both numerator and denominator and does not depend on \(I_2\), so it plays no part in the differentiation. Changing the power factor scales the efficiency but does not move the current at which it peaks.

8. P48.8 Derive the copper saving of an autotransformer.

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   The weight of copper is proportional to length × cross-section. Length is proportional to the number of turns, and cross-section to the current the conductor carries, so weight \(\propto NI\).

   \[\text{section AC} \propto \left(N_1 - N_2\right)I_1, \qquad \text{section BC} \propto N_2\left(I_2 - I_1\right)\]

   An ordinary two-winding transformer doing the same duty needs \(N_1I_1 + N_2I_2\). Taking the ratio and using \(N_1I_1 = N_2I_2\):

   \[\frac{W_a}{W_0} = \frac{N_1I_1 - N_2I_1 + N_2I_2 - N_2I_1}{N_1I_1 + N_2I_2} = \frac{2N_1I_1 - 2N_2I_1}{2N_1I_1} = 1 - \frac{N_2}{N_1} = 1 - K\]

   \[\text{Saving} = W_0 - W_a = KW_0\]

   The saving grows as \(K\) approaches unity, which is why autotransformers suit ratios close to 1 : 1.

9. P48.9 Explain the broken-winding hazard and why it rules out some applications.

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   Because the two sides share a winding, there is no electrical isolation. Consider a 2200/220 V autotransformer supplying a 220 V load from the common section.

   If the common winding breaks, the load is left connected across the series section, and the full 2200 V appears at its terminals — a tenfold overvoltage on equipment insulated for 220 V.

   Worse, no protective device need operate: the current may not rise until the equipment itself fails. In a two-winding transformer the same break simply disconnects the load, harmlessly.

   Consequence: autotransformers are not used to supply low-voltage consumer circuits from high-voltage systems, however attractive the copper saving. They are confined to duties where the two voltages are close, so that the worst-case overvoltage is modest.

10. P48.10 Why is a distribution transformer designed with lower iron loss than a power transformer of the same rating?

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    Because of the duty cycle. A power transformer at a generating station is switched in and out with the load and runs near full rating whenever it is energised; a distribution transformer is never switched off and is lightly loaded for much of the day.

    The iron loss therefore runs for all 24 hours in a distribution transformer, while output accumulates only during the loaded hours. In Example 48.2 the iron loss contributed 7.20 kWh against the copper loss's 1.70 kWh — more than four times as much.

    The design response is better core steel, thinner laminations and a lower working flux density, accepting a larger core and higher copper loss in exchange.

    The side effect is that the maximum-efficiency point moves down, since \(x = \sqrt{P_i/P_c}\) falls as \(P_i\) falls — typically to about half load, which is where a distribution transformer actually spends its time. The peak is placed where the machine lives, not where its nameplate says.