Types of DC Machines and Applications

DC generators are classified according to the methods of their field excitation.

Chapter 20 set out the family tree; this chapter works through each branch with its circuit equations. The organising question, as before, is where the field current comes from — an external supply, or the machine itself.

Besides the excited types, there are also permanent magnet DC generators. In these, no field winding is placed around the poles.

A DC generator in which current is supplied to the field winding from an external DC source is called a separately excited DC generator.

The flux produced by the poles depends upon the field current within the unsaturated region of the magnetic material of the poles — that is, \(\phi \propto I_f\) — but in the saturated region the flux remains constant.

The factor of two on the brush drop is not optional: the armature current passes through one brush on the way in and another on the way out, so both drops appear in series. Chapter 22 gave \(v_b \approx 1\) V for carbon brushes, making the total about 2 V — negligible in a 500 V machine, serious in a 24 V one.

A DC generator whose field winding is excited by the current supplied by the generator itself is called a self-excited DC generator.

The field coils may be connected in parallel with the armature, in series with the armature, or partly in series and partly in parallel with the armature winding. Accordingly, self-excited generators are classified as:

• Shunt wound generators

• Series wound generators

• Compound wound generators

In a shunt wound generator the field winding is connected across the armature winding, forming a parallel or shunt circuit, so that the full terminal voltage is applied across the field winding.

• A very small current \(I_{sh}\) flows through it, because this winding has many turns of fine wire and therefore a very high resistance \(R_{sh}\) — of the order of 100 \(\Omega\).

• \(I_{sh}\) is practically constant at all loads; therefore the DC shunt machine is considered to be a constant-flux machine.

The field winding is connected in series with the armature winding, forming a series circuit, so that the full line current \(I_L\) — equal to the armature current \(I_a\) — flows through it.

Since the series field winding carries the full-load current, it has a few turns of thick wire with low resistance — usually less than one ohm.

There are two sets of field windings on each pole. One is connected in series, having a few turns of thick wire; the other is connected in parallel, having many turns of fine wire.

A compound wound generator may be long-shunt or short-shunt.

Short Shunt

The shunt field winding is connected in parallel with the armature winding only.

\[I_{se} = I_L\]

\[I_{sh} = \frac{V + I_LR_{se}}{R_{sh}} = \frac{E_g - I_aR_a}{R_{sh}}, \qquad I_a = I_L + I_{sh}\]

\[V = E_g - I_aR_a - I_LR_{se}\]

Including brush contact drop,

\[V = E_g - I_aR_a - I_LR_{se} - 2v_b\]

\[\text{Power developed} = E_gI_a, \qquad \text{Power output} = VI_L\]

In compound wound DC generators, the field is produced by the shunt as well as the series winding. Generally the shunt field is stronger than the series field.

Differential compounding sounds perverse until one asks what a drooping characteristic is good for. The answer is welding: a steeply falling voltage limits the current when the electrode touches the work, and stabilises the arc. Chapter 29 develops all these characteristics properly.

Every generator type has a motor counterpart, using the same machine with the same connections. Only two things change: the sign of the armature drop, and the fact that in a shunt motor the line current feeds both armature and field rather than the armature feeding the field.

In every case the converted power is \(E_bI_a\) and the brush drop is subtracted as \(2v_b\), exactly as for the generator. Chapters 34 and 35 develop the torque and speed characteristics that follow.

Cumulative compound motors combine the two: the series field gives good starting torque while the shunt field guarantees a definite no-load speed, so the machine cannot run away. This is the standard choice wherever a heavy intermittent load must be started but the drive may also run light — presses, shears, rolling mills and lifts.

• DC generators are classified by their method of field excitation: permanent magnet, separately excited, or self-excited as shunt, series or compound.

• Permanent-magnet machines are compact with a constant field, but the magnet flux deteriorates with time and cannot be adjusted.

• Separately excited: \(I_a = I_L\), \(V = E_g - I_aR_a - 2v_b\). Flux under independent control, with \(\phi \propto I_f\) below saturation.

• Self-excited machines rely on residual magnetism to start building up.

• Shunt: \(I_{sh} = V/R_{sh}\) (small, so a constant-flux machine), \(I_a = I_L + I_{sh}\), and the drop uses \(I_a\) — not \(I_L\).

• Series: \(I_a = I_{se} = I_L\), \(V = E_g - I_a(R_a + R_{se}) - 2v_b\). Flux \(\propto I_{se}\) only below saturation.

• Compound, long shunt: \(I_{sh} = V/R_{sh}\), series field carries \(I_a\). Short shunt: series field carries \(I_L\) and \(I_{sh} = (V + I_LR_{se})/R_{sh} = (E_g - I_aR_a)/R_{sh}\).

• Cumulative compounding has the series field assisting the shunt field; differential has it opposing. Cumulative machines may be flat, over- or under-compounded.

• Motors use the same connections with \(E_b = V - I_aR_a\) and, for a shunt motor, \(I_a = I_L - I_{sh}\).

• Series motors for traction and hoists; shunt for constant speed; cumulative compound where both are needed.

• Writing the armature drop as \(I_LR_a\) in a shunt machine. It is \(I_aR_a\), and \(I_a = I_L + I_{sh}\) in a generator.

• Forgetting that the shunt equations are coupled. \(I_{sh}\) depends on \(V\), which depends on \(I_a\), which depends on \(I_{sh}\) — solve simultaneously.

• Using one brush drop instead of two. The current passes through a brush at each end, so the total is \(2v_b\).

• Applying the terminal voltage to a short-shunt field. There the field sees \(V + I_LR_{se}\).

• Putting \(I_a\) through the series field of a short-shunt machine. It carries \(I_L\); only in long shunt does it carry \(I_a\).

• Assuming \(\phi \propto I\) always. True only below saturation; beyond it the flux is essentially constant.

• Counting the field loss twice in a series machine. The series-field \(I^{2}R\) is already inside \(I_a^{2}(R_a + R_{se})\).

• Treating a negative voltage regulation as an error. Over-compounded machines are designed for it.

• Keeping the generator current relation for a motor. A shunt motor has \(I_a = I_L - I_{sh}\).

• Expecting a permanent-magnet machine to hold its rating for ever. The magnets weaken with time, and there is no field rheostat to compensate.

Identify the type and the connection first, then write the field current, then the armature current, then the voltage — in that order, every time.

1. P26.1 A separately excited generator has \(E_g = 300\) V and delivers 75 A with \(R_a = 0.20~\Omega\) and 1 V per brush. Find \(V\) and the output power.

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   \[V = 300 - (75)(0.20) - 2 = 300 - 15 - 2 = 283~\mathrm{V}\]

   \[P_{\text{out}} = (283)(75) = 21\,225~\mathrm{W}\]

2. P26.2 A shunt generator delivers 60 A at a terminal voltage of 200 V, with \(R_a = 0.12~\Omega\), \(R_{sh} = 100~\Omega\) and 1 V per brush. Find \(E_g\).

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   \[I_{sh} = \frac{200}{100} = 2.00~\mathrm{A}, \qquad I_a = 60 + 2.00 = 62.0~\mathrm{A}\]

   \[E_g = V + I_aR_a + 2v_b = 200 + (62.0)(0.12) + 2 = 200 + 7.44 + 2 = 209.4~\mathrm{V}\]

   Here \(V\) is given, so no simultaneous solution is needed — a useful simplification worth spotting.

3. P26.3 A shunt generator has \(E_g = 260\) V and delivers 120 A, with \(R_a = 0.05~\Omega\), \(R_{sh} = 130~\Omega\) and 1 V per brush. Find \(V\).

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   Now \(V\) is unknown and the equations are coupled:

   \[V\left(1 + \frac{0.05}{130}\right) = 260 - (120)(0.05) - 2 = 252\]

   \[V = \frac{252}{1.000385} = 251.90~\mathrm{V}\]

   \[I_{sh} = \frac{251.90}{130} = 1.938~\mathrm{A}, \qquad I_a = 121.94~\mathrm{A}\]

4. P26.4 A series generator delivers 40 A at 230 V with \(R_a = 0.14~\Omega\), \(R_{se} = 0.06~\Omega\) and 1 V per brush. Find \(E_g\) and the developed power.

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   \[E_g = 230 + (40)(0.20) + 2 = 230 + 8 + 2 = 240~\mathrm{V}\]

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

5. P26.5 A long-shunt compound generator delivers 100 A at 250 V, with \(R_a = 0.08~\Omega\), \(R_{se} = 0.03~\Omega\), \(R_{sh} = 125~\Omega\) and 1 V per brush. Find \(E_g\).

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   \[I_{sh} = \frac{250}{125} = 2.00~\mathrm{A}, \qquad I_a = 102.0~\mathrm{A}\]

   \[E_g = 250 + (102.0)(0.11) + 2 = 250 + 11.22 + 2 = 263.2~\mathrm{V}\]

6. P26.6 Repeat P26.5 for a short-shunt connection.

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   \[I_LR_{se} = (100)(0.03) = 3.00~\mathrm{V}, \qquad I_{sh} = \frac{250 + 3.00}{125} = 2.024~\mathrm{A}\]

   \[I_a = 102.02~\mathrm{A}\]

   \[E_g = 250 + (102.02)(0.08) + 3.00 + 2 = 250 + 8.162 + 3.00 + 2 = 263.2~\mathrm{V}\]

   Practically identical to P26.5, but the series field carries 100 A rather than 102 A.

7. P26.7 A generator gives 240 V on no load and 228 V on full load. Find the regulation and classify it.

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   \[\frac{240 - 228}{228} \times 100 = \frac{12}{228} \times 100 = 5.26\,\%\]

   Positive regulation, so the machine is a shunt or under-compounded generator.

8. P26.8 A 220 V shunt motor takes 45 A from the line, with \(R_{sh} = 110~\Omega\), \(R_a = 0.15~\Omega\) and 1 V per brush. Find \(I_a\), the back EMF and the converted power.

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   For a motor the line current splits:

   \[I_{sh} = \frac{220}{110} = 2.00~\mathrm{A}, \qquad I_a = I_L - I_{sh} = 45 - 2.00 = 43.0~\mathrm{A}\]

   \[E_b = 220 - (43.0)(0.15) - 2 = 220 - 6.45 - 2 = 211.55~\mathrm{V}\]

   \[P_{\text{conv}} = E_bI_a = (211.55)(43.0) = 9097~\mathrm{W}\]

9. P26.9 Which generator type would you choose to feed a load at the end of a long cable, and why?

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   An over-compounded cumulative generator. Its terminal voltage rises as the load increases, and that rise is designed to match the \(IR\) drop in the cable.

   The result is that the voltage at the far end — where the load actually is — stays constant, even though the generator terminals do not. A flat-compounded machine would hold its own terminals constant and let the far end sag, which is not what the customer cares about.

10. P26.10 Why is a series motor unsuitable for a lathe, and a shunt motor unsuitable for a crane hoist?

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    Series motor on a lathe: a lathe needs a definite, nearly constant speed for a given cut, and it is often run with little load while setting up. A series motor's speed varies steeply with load and it races dangerously when unloaded, since \(N \propto 1/\phi\) and \(\phi \propto I_a\).

    Shunt motor on a hoist: a hoist must break a heavy load away from rest. A shunt motor gives only \(T \propto I_a\), so reaching the required starting torque means drawing a very large current — whereas a series motor's \(T \propto I_a^{2}\) delivers the same torque for far less.

    A cumulative compound motor serves both duties, which is why it is so common on cranes: series torque for starting, shunt field for a safe no-load speed.