Losses and Efficiency in DC Machines

These occur generally due to current in the various windings.

The armature copper loss is typically 3–6 % of rated output and the field losses 1–5 %.

These are the losses within the armature due to its rotation within the magnetic field of the poles.

Chapter 27 supplied the frequency: the armature iron is magnetised alternately north and south at

so both components rise with speed — hysteresis in proportion, eddy current as the square. Iron loss is therefore a function of speed and flux, and not of load current at all, which is what puts it among the constant losses of Section 33-5.

The qualification is that "constant" means constant with respect to load. A machine run at a different speed, or with its field weakened, has a different iron loss — which is why the speed-control methods of Chapter 37 change the efficiency as well as the speed.

For the efficiency analysis that follows, the losses are regrouped by how they behave as the load changes.

The commercial efficiency is the product of the other two, and it is the only one a purchaser cares about.

For a generator delivering current \(I\) at terminal voltage \(V\):

If \(I_{sh}\) is negligible compared with the load current, then \(I_a \approx I\) and

The result is general. It holds for transformers, for AC machines, and indeed for any device whose losses split into a term independent of load and a term proportional to its square — a fact worth remembering, because it will reappear in Part 3.

• Copper losses are \(I_a^{2}R_a\) (3–6 %), \(I_{sh}^{2}R_{sh}\) and \(I_{se}^{2}R_{se}\) (1–5 %). Brush contact loss \(2v_bI_a\) is conventionally grouped with the armature loss.

• Iron losses are \(K_hB_{max}^{1.6}fV\) and \(K_eB_{max}^{2}f^{2}V\), with \(f = PN/120\). They depend on speed and flux, not on load.

• Mechanical losses — friction and windage — depend on speed. With iron losses they total 3–15 %.

• Stray losses arise from armature reaction and commutation, and are assigned a value of 1 % of full-load output.

• Constant losses: iron, mechanical and shunt field. Variable losses: armature and series-field copper.

• Power flow, generator: mechanical input \(\to\) \(E_gI_a\) \(\to\) \(VI_a\) \(\to\) \(VI_L\). For a motor the same chain runs backwards with \(E_b\).

• Efficiency \(\eta = VI/(VI + I^{2}R_a + W_c)\), and \(\eta = \eta_m\eta_e\).

• Maximum efficiency occurs when variable loss equals constant loss, at \(I = \sqrt{W_c/R_a}\), where the total loss is \(2W_c\).

• The efficiency curve is steep at low load and flat near the maximum, so part-load operation is costly but the exact peak location is not critical.

• Using \(I_L\) in the armature copper loss. It is \(I_a^{2}R_a\), and \(I_a = I_L + I_{sh}\) for a generator, \(I_L - I_{sh}\) for a motor.

• Treating the brush loss as \(I_a^{2}R_b\). The contact drop is roughly constant, so the loss is \(2v_bI_a\).

• Putting the shunt field loss among the variable losses. It is \(V^{2}/R_{sh}\) and does not involve the load.

• Putting the series field loss among the constant losses. It carries the load current and varies as \(I^{2}\).

• Calling iron loss constant in an absolute sense. It is constant with respect to load, but changes with speed and flux.

• Taking the rotational losses off the wrong side. They belong on the mechanical side of \(E_gI_a\), not the electrical side.

• Using \(\eta = \text{output}/(\text{output} + \text{losses})\) for a motor with the electrical output. A motor's output is mechanical; its input is electrical.

• Assuming maximum efficiency occurs at full load. It occurs where \(I^{2}R_a = W_c\), which may be well above or below rated load.

• Thinking the condition tells you the value of the maximum. It locates the peak only; reducing a loss is what raises it.

• Forgetting that efficiency collapses at light load. The constant loss is unchanged while the output shrinks.

Find \(I_a\) first — with the correct sign for a generator or a motor — then list the losses by class.

1. P33.1 A 10 kW, 200 V shunt generator has \(R_a = 0.15~\Omega\), \(R_{sh} = 100~\Omega\) and iron plus mechanical losses of 400 W. Find the full-load efficiency, neglecting brush and stray losses.

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   \[I_L = 50~\mathrm{A}, \quad I_{sh} = 2~\mathrm{A}, \quad I_a = 52~\mathrm{A}\]

   \[I_a^{2}R_a = (2704)(0.15) = 405.6~\mathrm{W}, \quad VI_{sh} = 400~\mathrm{W}\]

   \[\text{total loss} = 405.6 + 400 + 400 = 1205.6~\mathrm{W}\]

   \[\eta = \frac{10\,000}{11\,205.6} = 89.24\,\%\]

2. P33.2 For P33.1, find the load current at which efficiency is maximum and the value there.

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   \[W_c = 400 + 400 = 800~\mathrm{W}, \qquad I = \sqrt{\frac{800}{0.15}} = 73.03~\mathrm{A}\]

   \[P_{\text{out}} = (200)(73.03) = 14\,606~\mathrm{W}, \qquad \text{loss} = 1600~\mathrm{W}\]

   \[\eta_{\max} = \frac{14\,606}{16\,206} = 90.13\,\%\]

   This is 146 % of full load, so again the machine never reaches its own peak.

3. P33.3 A machine has constant losses of 900 W and an armature resistance of 0.05 \(\Omega\). At what current is efficiency maximum, and what is the total loss there?

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   \[I = \sqrt{\frac{900}{0.05}} = \sqrt{18\,000} = 134.2~\mathrm{A}\]

   \[\text{total loss} = 2W_c = 1800~\mathrm{W}\]

4. P33.4 A 240 V shunt motor takes 40 A. With \(R_a = 0.25~\Omega\), \(R_{sh} = 120~\Omega\) and rotational losses of 350 W, find the output and efficiency. Neglect brush drop.

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   \[I_{sh} = 2~\mathrm{A}, \qquad I_a = 40 - 2 = 38~\mathrm{A}\]

   \[P_{\text{in}} = (240)(40) = 9600~\mathrm{W}\]

   \[E_b = 240 - (38)(0.25) = 230.5~\mathrm{V}, \qquad P_{\text{conv}} = (230.5)(38) = 8759~\mathrm{W}\]

   \[P_{\text{out}} = 8759 - 350 = 8409~\mathrm{W}, \qquad \eta = \frac{8409}{9600} = 87.59\,\%\]

5. P33.5 A machine with \(W_c = 1000\) W and \(R_a = 0.08~\Omega\) runs at 60 A and at 200 V. Find its efficiency, and say whether it is above or below the maximum-efficiency current.

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   \[\eta = \frac{(200)(60)}{(200)(60) + (3600)(0.08) + 1000} = \frac{12\,000}{13\,288} = 90.31\,\%\]

   \[I_{\max} = \sqrt{\frac{1000}{0.08}} = 111.8~\mathrm{A}\]

   The machine is running well below the maximum-efficiency current, so its efficiency would improve if it were loaded further.

6. P33.6 The machine of P33.5 is run at quarter of that load, 15 A. Find the efficiency and comment.

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   \[\eta = \frac{3000}{3000 + 18 + 1000} = \frac{3000}{4018} = 74.66\,\%\]

   A fall of nearly 16 percentage points. The variable loss has dropped from 288 W to 18 W, but the constant 1000 W is now a third of the output. Light-load running is where efficiency is destroyed.

7. P33.7 Why is the shunt field loss classed as constant but the series field loss as variable?

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   The shunt field is connected across the terminals, so \(I_{sh} = V/R_{sh}\) and its loss is \(V^{2}/R_{sh}\) — determined by the terminal voltage alone, which is held roughly constant.

   The series field carries the load current itself, so its loss \(I^{2}R_{se}\) rises as the square of the load.

   The classification is by behaviour with load, not by which winding the copper happens to be in.

8. P33.8 A designer proposes raising the armature resistance so that maximum efficiency falls at rated load. Comment.

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   The condition \(I^{2}R_a = W_c\) would indeed be satisfied at rated load, so the peak would move there. But the efficiency at that peak would be lower than the efficiency the machine already achieves at rated load, because a loss has been added rather than removed.

   The condition locates the maximum; it says nothing about its height. The only useful change is to reduce the constant losses until the condition is met at the wanted load — thinner laminations, better bearings, a more efficient field winding.

9. P33.9 Explain why the iron loss appears on the mechanical side of the power-flow diagram.

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   Because it is supplied by the shaft, not by the terminals. The armature iron is dragged through the alternating magnetisation by whatever is turning it, and the hysteresis and eddy-current losses appear as an extra mechanical torque opposing rotation.

   In a generator the prime mover supplies it before any electrical power exists; in a motor it is subtracted from the converted power \(E_bI_a\) before the shaft output is reached.

   The product \(E_gI_a\) or \(E_bI_a\) is the dividing line: everything above it is electrical loss, everything below it is mechanical.

10. P33.10 A station has four identical 500 kW generators and a load of 900 kW. Should it run two machines or four? Justify with the shape of the efficiency curve.

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    Two machines, each at 450 kW, which is 90 % of rating.

    Four machines would each carry 225 kW, or 45 % of rating. From the shape of the curve, efficiency at 45 % load is several points below that at 90 %, because each machine still incurs its full constant losses while producing half the output.

    Running four machines means paying four sets of constant losses instead of two, for the same total output. This is exactly the argument Chapter 32 gave for paralleling several units and shutting some down as the load falls.