Parallel Operation of DC Generators

A power station rarely runs one large machine. It runs several smaller ones on a common pair of bus bars, and switches them in and out as the load demands.

1. Run the incoming machine up to speed with its main switch open.
2. Excite it and let it build up, following Chapter 28 — field rheostat at minimum first, then adjusted.
3. Check the polarity across the open switch with a voltmeter. A near-zero reading confirms correct connection; a reading of about twice the bus voltage means the machine is reversed.
4. Adjust the field until the machine's voltage is equal to, or a fraction of a volt above, the bus voltage.
5. Close the switch. The machine is now on the bus but carrying almost no load.
6. Raise its excitation to make it take up load, lowering the excitation of the machine being relieved to keep the bus voltage constant.

Step 6 is the one students find counter-intuitive. A generator takes up load when its own EMF is raised, not when the load is somehow directed to it — and since the bus voltage would rise too, the other machine's field must be weakened by a corresponding amount.

Once paralleled, both machines are connected to the same bus, so they must have the same terminal voltage. That single constraint determines everything.

Suppose machine A momentarily takes a little more than its share — because of a speed fluctuation, say.

This is why shunt generators parallel readily: their characteristics droop naturally. Series generators and over-compounded machines do not, and need the special measures of Sections 32-8 and 32-9.

Since sharing depends on the EMFs, and each EMF depends on its field current, the operator controls the division with the field rheostats.

The third row is the operator's working method. Load transfer and voltage control are separate objectives, and adjusting one field alone changes both at once.

What happens when the external load is removed but both machines remain connected? Their EMFs are unequal, so current still flows — around the loop formed by the two armatures.

This is not merely inefficient. If B's prime mover is a diesel engine, it is now being motored; if a steam turbine, it may be damaged by windage heating with no steam flow. Every paralleled machine is therefore protected by a reverse-current relay, which trips its breaker if the current reverses.

Compound generators are widely used because of the flat or rising characteristic of Chapter 29. That same characteristic makes them unstable in parallel.

The instability is a direct consequence of Section 32-5. A cumulative compound machine's characteristic is flat or rising, so \(\mathrm{d}V/\mathrm{d}I \ge 0\) and the stabilising negative feedback is absent.

The whole difficulty lies in one fact: each series field carries its own machine's current, so any imbalance is amplified. The remedy removes exactly that.

Series generators, whose characteristics rise steeply, are unstable in parallel for the same reason and to a far greater degree. They too require an equaliser bar, and in practice are seldom paralleled at all.

• Generators are paralleled for continuity, part-load efficiency, maintenance, growth, handling and reserve.

• The conditions are equal terminal voltage, correct polarity and drooping characteristics. There is no frequency or phase to match.

• Reversed polarity puts the two EMFs in series around the loop, giving a fault current of \((E_A + E_B)/(R_A + R_B)\).

• Once paralleled, \(V = E_A - I_AR_A = E_B - I_BR_B\) and \(I_A + I_B = I_L\) determine the sharing completely.

• Sharing is governed chiefly by the no-load EMFs, and is adjusted by the field rheostats.

• To shift load without changing the bus voltage, strengthen one field and weaken the other together.

• A drooping characteristic gives negative feedback and stable sharing; a rising one gives a runaway.

• On light load a machine with the lower EMF motors, absorbing a circulating current \((E_A - E_B)/(R_A + R_B)\). Reverse-current relays protect against this.

• Compound generators are unstable in parallel, because extra current strengthens a machine's own series field. Stability requires \(k_{se} \lt R_a + R_{se}\).

• The equaliser bar puts the series fields directly in parallel, so each sees a share of the total current and the feedback loop is broken. It must be short and of very low resistance.

• Assuming the machines share equally because they are identical in rating. They share according to their EMFs and resistances, not their nameplates.

• Thinking the machine with the lower resistance takes more load. The EMF difference usually dominates, as Example 32.1 shows.

• Adjusting one field alone to shift load. That changes the bus voltage too; both rheostats must move together.

• Forgetting the sign when a machine motors. For an absorbing machine \(V = E + IR\), not \(E - IR\).

• Omitting the polarity check before closing the switch. A reversed connection is a dead short across both armatures.

• Believing a rising characteristic merely shares badly. It does not share at all — one machine takes everything.

• Connecting the equaliser bar to the wrong side of the series field. It must join the armature-side terminals, so the two windings end up in parallel.

• Using a light conductor for the equaliser. Its resistance must be negligible compared with the series fields, or it does nothing useful.

• Expecting an equaliser to fix unequal load sharing. It secures stability; the division is still set by the fields.

• Paralleling a series generator without special measures. Its steeply rising characteristic makes it the least stable of all.

Write the common bus voltage two ways, then use \(I_A + I_B = I_L\). Watch the sign for any machine that is absorbing.

1. P32.1 Two shunt generators supply 250 A in parallel. A has \(E = 240\) V, \(R_a = 0.06~\Omega\); B has \(E = 235\) V, \(R_a = 0.05~\Omega\). Find the bus voltage and each current.

   Show answer

   \[I_A = \frac{E_A - E_B + I_LR_B}{R_A + R_B} = \frac{240 - 235 + (250)(0.05)}{0.11} = \frac{17.5}{0.11} = 159.1~\mathrm{A}\]

   \[I_B = 250 - 159.1 = 90.9~\mathrm{A}\]

   \[V = 240 - (0.06)(159.1) = 240 - 9.55 = 230.5~\mathrm{V}\]

   Check: \(235 - (0.05)(90.9) = 230.5\) V \(\checkmark\)

2. P32.2 For P32.1, what must \(E_B\) become for equal sharing at 125 A each?

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   \[V = 240 - (0.06)(125) = 232.5~\mathrm{V}\]

   \[E_B = 232.5 + (0.05)(125) = 232.5 + 6.25 = 238.75~\mathrm{V}\]

3. P32.3 For P32.1, find the circulating current if the load is removed, and identify which machine motors.

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   \[I_c = \frac{240 - 235}{0.11} = 45.5~\mathrm{A}\]

   \[V = 240 - (0.06)(45.5) = 237.3~\mathrm{V}\]

   Check: \(235 + (0.05)(45.5) = 237.3\) V \(\checkmark\)

   Machine B motors, since it has the lower EMF. It absorbs about \((237.3)(45.5) = 10.8\) kW.

4. P32.4 Two generators have equal no-load EMFs of 220 V, with \(R_A = 0.04~\Omega\) and \(R_B = 0.06~\Omega\). They supply 400 A. Find each current.

   Show answer

   With equal EMFs the sharing is inversely proportional to the resistances:

   \[\frac{I_A}{I_B} = \frac{R_B}{R_A} = \frac{0.06}{0.04} = 1.5\]

   \[I_A = 400\left(\frac{1.5}{2.5}\right) = 240~\mathrm{A}, \qquad I_B = 160~\mathrm{A}\]

   \[V = 220 - (0.04)(240) = 210.4~\mathrm{V}\]

5. P32.5 A generator is connected to a 250 V bus with reversed polarity. Both armature resistances are 0.05 \(\Omega\) and both EMFs 250 V. Find the fault current.

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   \[I = \frac{E_A + E_B}{R_A + R_B} = \frac{500}{0.10} = 5000~\mathrm{A}\]

   Twenty times a typical rated current — hence the voltmeter check across the open switch, which would have read 500 V instead of zero.

6. P32.6 A compound generator has \(R_a + R_{se} = 0.025~\Omega\) and a series field contributing 0.035 V per ampere. Is it stable in parallel? What would make it so?

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   \[\frac{\mathrm{d}V}{\mathrm{d}I} = 0.035 - 0.025 = +0.010~\Omega\]

   The characteristic rises, so it is unstable in parallel.

   Remedies: fit an equaliser bar, or reduce the series excitation — by fitting a diverter resistor across the series field — until \(k_{se} \lt 0.025\). The first is preferred, because the second sacrifices the compounding the machine was built for.

7. P32.7 Explain why a drooping characteristic gives stable load sharing.

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   If a machine momentarily takes more than its share, a drooping characteristic makes its terminal voltage fall. Since all machines are tied to one bus voltage, it cannot sustain the excess and falls back.

   The response opposes the disturbance — negative feedback — so the division of load is a stable equilibrium.

   With a rising characteristic the response reinforces the disturbance: more current gives more voltage, which lets the machine take more still. The same distinction decided the stability of the operating point in Chapter 28, where saturation supplied the falling return.

8. P32.8 Why must a paralleled machine have a reverse-current relay?

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   If the machine's EMF falls below the bus voltage — through loss of excitation, loss of speed, or simply because the load has dropped — current flows into it and it runs as a motor.

   It then drives its own prime mover, which may be damaged: a diesel engine is motored against compression, and a steam turbine overheats from windage with no steam flow to cool it. The station also wastes the power absorbed.

   The relay detects the reversal of current direction and trips the machine's breaker, disconnecting it from the bus.

9. P32.9 Where exactly is the equaliser bar connected, and why there?

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   It joins the terminals on the armature side of each series field — the junction between each armature and its own series winding.

   Connected there, the two series windings have one end common (through the bar) and the other end common (through the positive bus), so they are directly in parallel. They then divide the total current in inverse proportion to their resistances.

   Neither machine's series excitation depends on its own armature current any more, which is precisely the feedback path that caused the runaway.

10. P32.10 Two identical machines share a 500 A load unequally at 300 A and 200 A. Their series fields are each 0.008 \(\Omega\). Find the series-field currents with an equaliser fitted, and the current in the bar.

    Show answer

    Equal resistances, so with the bar the total divides equally:

    \[I_{se,A} = I_{se,B} = \frac{500}{2} = 250~\mathrm{A}\]

    \[I_{\text{bar}} = 300 - 250 = 50~\mathrm{A}\]

    flowing from machine A's junction to machine B's.

    Without the bar the fields would carry 300 A and 200 A, a ratio of 1.5, and A's stronger excitation would drive the imbalance further. With the bar both fields see 250 A whatever the armatures do.