Swinburne's test on a DC shunt machine
Swinburne's test on a DC shunt machine: no-load readings, constant-loss calculation, efficiency as motor and as generator, advantages and limitations, plus viva questions with answers.
Aim
To conduct Swinburne's test on a DC shunt machine and predetermine its efficiency when running as a motor and as a generator, without loading the machine.
Apparatus required
| Apparatus | Specification | Qty |
|---|---|---|
| DC shunt machine | 220 V, 5 hp, 1500 rpm | 1 |
| Three-point starter | Matched to the machine | 1 |
| Voltmeter (MC) | 0–300 V | 1 |
| Ammeter (MC) | 0–5 A, line current | 1 |
| Ammeter (MC) | 0–2 A, field current | 1 |
| Field rheostat | As specified for the machine | 1 |
| Tachometer | Digital or contact type | 1 |
Theory
A DC machine's losses split into two groups. The constant losses — iron loss, friction and windage, and shunt field copper loss — depend on flux and speed, both of which barely change from no load to full load on a shunt machine. The variable loss is the armature copper loss I_a²R_a, which changes with everything.
Swinburne's test measures the constant losses once, by running the machine as a motor on no load. All the input then goes into losses, and subtracting the small no-load armature copper loss leaves the constant group.
With the constant loss known and R_a measured, the efficiency at any load can be computed. As a motor, I_a = I_L − I_sh and the output is the input less both losses. As a generator, I_a = I_L + I_sh and the input is the output plus both losses — the losses sit outside the ratio rather than inside it, which is why the same machine at the same line current shows a slightly higher efficiency as a generator.
It is the cheapest efficiency test there is: only no-load power is drawn, so a 50 kW machine can be tested from a socket. What it buys in convenience it gives up in honesty — see the limitations below.
Circuit connections
Check every point below against your board before switching on. There is no diagram here on purpose — a wrong diagram is worse than none, and this is the list a demonstrator actually walks through with you.
- Supply through the three-point starter to the armature, with the line ammeter in series.
- Shunt field across the supply through the field rheostat, with its own ammeter in series to read I_sh.
- Voltmeter across the supply terminals.
- Field rheostat at minimum resistance, i.e. maximum field current, before starting.
- Armature resistance is measured separately by the volt–ampere method with the machine at rest and the field disconnected.
Procedure
- 1Check the connections and set the field rheostat to minimum resistance.
- 2Start the motor with the starter handle moved slowly and steadily through all studs. Let it run on no load until the speed settles.
- 3Adjust the field rheostat until the machine runs at its rated speed.
- 4Record the supply voltage V, the no-load line current I₀, the shunt field current I_sh, and the speed.
- 5Switch off. With the machine at rest and the field circuit disconnected, apply a low DC voltage across the armature and read the current, to get R_a by the volt–ampere method. Take the reading at a current close to rated for a representative hot value.
- 6Compute the constant losses, then the efficiency as a motor and as a generator at the rated line current.
Work out your readings
Type in the numbers off the meters. This fills the tabular column, works the calculation through step by step, plots the characteristic — and tells you when a reading cannot physically be right, which is the part a manual can't do. Everything stays on this device, and it works with the network off.
Nameplate and machine data
Fill in the machine data to see the results, the worked calculation and the curve.
Precautions
- The field rheostat must be at minimum resistance when starting. Starting with a weak field lets a shunt motor run away on no load.
- Never open the field circuit while the machine is running — the flux collapses and the speed rises dangerously.
- Move the starter handle steadily. Snatching it to the last stud puts the full starting current through the armature.
- Measure R_a with a DC source and meters, not with an ordinary multimeter — the resistance is a fraction of an ohm and contact resistance swamps it.
- The no-load reading must be taken with nothing coupled to the shaft.
Sources of error
Every record asks for these, and every record gets the same three lines copied from the one before. These are the errors this particular experiment actually has.
- Stray load loss is never measured. At full load it is real, so the efficiency this test predicts is slightly optimistic.
- R_a is measured cold. Running at rated current heats the winding, and the copper loss at operating temperature is higher than calculated.
- The constant losses are assumed to stay constant with load. Armature reaction at heavy load changes the flux slightly, so they do not.
- The test cannot reveal sparking at the commutator or the temperature rise under load, because the machine is never loaded.
Viva questions with answers
What are the advantages of Swinburne's test?
It is economical — only the no-load losses are drawn from the supply, so even a large machine can be tested from a small source. It gives efficiency at any load from a single set of readings, and it takes minutes rather than hours.
What are its limitations?
Stray load loss is not accounted for, so the answer is optimistic. It cannot check the temperature rise or commutation under load, because the machine is never loaded. And it does not apply to series motors.
Why can Swinburne's test not be performed on a DC series motor?
A series motor's flux comes from the load current itself. On no load the current is small, the flux is small, and the speed becomes dangerously high — the machine would run away before any reading could be taken. There is also no meaningful constant flux to hold the 'constant losses' assumption together.
What are the constant losses, and why are they constant?
Iron loss, friction and windage, and shunt field copper loss. On a shunt machine the applied voltage and the flux hardly change from no load to full load, and the speed changes only a few percent, so all three stay essentially fixed.
Why is efficiency as a generator higher than as a motor at the same line current?
For a motor, η = (input − losses)/input; for a generator, η = output/(output + losses). With the same V·I_L and similar losses, the second ratio is the larger of the two. The machine is also carrying a slightly different armature current in each case, since I_a = I_L ∓ I_sh.
Which test would you use instead, to check the machine under real load?
Hopkinson's test — two identical machines coupled back to back, one motoring and one generating, with the supply making up only the losses. It loads both machines fully while drawing very little power, and it does reveal temperature rise and commutation.
Why is the field current subtracted from the line current for a motor and added for a generator?
In both cases the field is across the supply. As a motor the line feeds both armature and field, so I_a = I_L − I_sh. As a generator the armature feeds both the line and its own field, so I_a = I_L + I_sh.