Losses, Efficiency and Testing

Start with the currents, because every loss in the machine is a function of one of them. The shunt field sits directly across the terminals, and the armature must supply both the load and the field:

The generated e.m.f. is the terminal voltage plus what the armature resistance takes on the way out:

A 4 V drop on 250 V — deliberately small, because it is dissipated as heat at 200 A.

The copper losses, one term per winding:

Note that the field loss is the larger of the two here, and it does not change with load.

Close the ledger. The useful output is the terminal power; adding every loss gives the mechanical power the prime mover must supply:

The output of the prime mover and the input to the generator are the same 51.75 kW — they are the same shaft.

The power developed in the armature is the hinge between the mechanical and the electrical halves of the machine, and it can be found from either side:

Mechanical input less the stray loss equals the electrical power generated. That agreement is the check that the ledger balances.

Now the three efficiencies, each a ratio of two numbers already on the table:

Check that the three are consistent. The commercial efficiency spans both boundaries, so it must be the product of the other two:

Work backwards from the efficiency to get the total loss, which is the only route into this problem:

Split that loss into constant and variable parts. The constant part is everything that does not follow the load current:

What remains must be the armature copper loss, and the armature current is known:

Maximum efficiency is reached when the variable loss has risen to meet the constant loss — not at full load, and not at any particular fraction of it:

That is 60% of full load. Machines are routinely designed this way, because they spend far more hours at part load than at rated load.

Confirm that it really is a maximum. Writing the efficiency as a function of load current, with \(P_c\) the constant loss,

Neglecting the small difference between \(I_a\) and \(I_L\), the bracket is \(I R_a/V_t + P_c/V_tI\), whose derivative vanishes when \(I^2R_a = P_c\) — the condition used above.

Identify the currents from the connection. In a long-shunt machine the shunt field is connected across the terminals, outside the series field, so the series winding and the armature both carry \(I_a\):

The drop in the series winding is \(102 \times 0.06 = 6.12\ \text{V}\), which with the armature drop sets the generated e.m.f.

One copper-loss term per winding:

The shunt-field loss can equally be written \(V_tI_f = 220 \times 2 = 440\ \text{W}\) — the same number, since the field is across the terminals.

The stray loss is what the efficiency leaves over. The overall efficiency fixes the input:

The torque comes from the input, not the output. The prime mover has to supply the whole 25 kW at 1000 rpm:

Using the 22 kW output instead would give 210.1 N·m and understate the driving engine by 12%.

The load fixes the terminal voltage, which is the quantity the field winding sees:

Field and armature currents:

Add every drop between the conductors and the terminals. There are two brushes in the circuit, one per polarity, so the contact drop is counted twice:

Only the electrical efficiency can be found, because no iron or friction loss is given — and without it the mechanical input is unknown:

Low, because this is a small machine: the 12 V armature drop is 6% of 200 V, where Problem 1's was 1.6%.

Where the missing 1136 W goes, as a check on the figure:

The field takes 800 W to deliver 4000 W — a 20% excitation burden that only a small machine would tolerate.

Collect the resistances that carry the armature current into a single armature-circuit resistance. The armature, series field and interpole field are all in the same series path:

The shunt field is not in this path — it is across the terminals and is treated separately.

The currents, with the long-shunt connection again putting \(I_a\) through the series and interpole windings:

The loss ledger:

The efficiency, with the input formed by adding the losses to the known output:

Roughly 42% of the loss is armature-circuit copper and varies with the square of the load; the other 58% — field copper, iron, windage and friction — is fixed. That split is what makes the maximum-efficiency point of Problem 2 fall well below full load.

The point of the no-load run is to measure everything that will not change when the machine is loaded. At no load the machine develops almost no torque, so the whole input goes into losses:

Subtract the one loss that is load-dependent — the no-load armature copper loss — and what remains is the constant loss:

This 1246.4 W contains the shunt field loss \(250 \times 2 = 500\ \text{W}\) together with 746.4 W of iron, friction and windage loss. It stays the same at every load and in either mode of operation.

aAs a motor taking 50 A. The field still draws 2 A, so the armature takes the rest:

bAs a generator delivering 50 A. Now the armature must supply the field as well as the load, so the armature current is larger and the copper loss higher:

Why the generator figure is the higher of the two, despite its larger copper loss. In the motor the 50 A is the input and the losses are subtracted from it; in the generator the 50 A is the output and the losses are added to it. The same loss therefore appears against a larger denominator:

What the test cannot tell you. The machine was never loaded, so the iron loss is measured at no-load flux rather than at the flux the load produces, stray load loss is missed entirely, and nothing at all is learnt about commutation or temperature rise. Swinburne's test is a prediction, not a demonstration — which is exactly the gap Problem 7 closes.

Find the two armature currents first. The generator armature feeds the motor armature directly; the supply makes up only the shortfall, so at the junction

The motor must carry more current than the generator delivers, because it has to make good the losses of both machines.

The supply pays for the losses and nothing else. The power drawn into the armature circuits is therefore the total armature-circuit loss of the pair:

The stray loss is the remainder, shared equally between two machines assumed identical and running at the same speed and flux:

This covers iron, friction and windage. The field copper losses are not in it, because the field currents were excluded from the 25 A reading.

Motor efficiency. Its input is everything the supply and the generator push into it — armature and field alike:

Generator efficiency. Its output is the armature power it delivers; its input is the shaft power plus its own excitation:

Check the shaft. The two machines are bolted together, so the motor's mechanical output must be the generator's mechanical input — and the generator's input above includes 750 W of electrical excitation that never crosses the coupling:

A second check: the total supplied power \(250(25 + 3 + 2.5) = 7625\ \text{W}\) equals the sum of all six loss terms, \(1012.5 + 800 + 625 + 750 + 4437.5\).

Why the test is worth the trouble. Two 50 kW machines have been run at full load and full temperature on a supply of 7.6 kW, under 8% of the 100 kW the pair is handling. Nothing is dissipated in a load bank, the machines can be left running for a heat run, and the losses are measured at true load flux rather than extrapolated from no load.