Testing of DC Machines

i Learning Objectives

By the end of this chapter you should be able to:

Explain why efficiency is usually predicted from losses rather than measured directly.
Distinguish direct from indirect tests.
Perform Swinburne's test and compute the constant losses from one reading.
Predict the efficiency at any load as both generator and motor from that one test.
State the three limitations of Swinburne's test.
Explain Hopkinson's test and why the supply provides only the losses.
Compute both machines' efficiencies from the Hopkinson readings.
Use the retardation test to find the rotational losses and separate iron from friction.

Section 39-1

Why Not Simply Load the Machine

The obvious way to find efficiency is to measure input and output and divide. For a small machine that is exactly what is done. For anything large it becomes impractical.

Energy cost

A full-load test on a 500 kW machine draws 500 kW for the duration of the heat run — usually several hours.

Where does the output go?

A loading device — a brake, a water rheostat, a load bank — must absorb the whole output and dissipate it as heat.

Accuracy

Efficiency is 90 % or better, so it is the small difference between two large measured quantities. A 1 % error in each gives a 10 % error in the loss.

The last point is the decisive one. If input and output are each measured to 1 %, their difference — the loss — carries an error of roughly ten times that. Measuring the losses directly and computing the efficiency from them is far more accurate, as well as far cheaper, because the small quantity is then measured directly rather than inferred from two large ones.

Section 39-2

Direct and Indirect Tests

Table 39.1 — The tests of this chapter.
Test	Type	Measures	Supply must provide
Brake or dynamometer	Direct	Output directly	Full rated power
Swinburne	Indirect	Constant losses on no load	Only the no-load losses
Hopkinson	Indirect, regenerative	Total losses of two machines	Only the losses — but at full load
Retardation	Indirect	Rotational losses from the deceleration	Nothing during the test

All three indirect tests rest on Chapter 33's division of the losses. Once the constant losses \(W_c\) and the armature resistance are known, the efficiency at any load follows without ever applying that load.

Section 39-3

Swinburne's Test

The machine is run as a motor on no load at rated voltage and rated speed. That single condition yields everything needed.

🔍

The Reasoning

On no load, the input is almost all loss

With no mechanical output, the entire input is dissipated. Subtracting the one loss that will change with load — the no-load armature copper loss — leaves the constant losses.

\[W_c = VI_{L0} - I_{a0}^{2}R_a \qquad\text{where}\qquad I_{a0} = I_{L0} - I_{sh}\]

\(W_c\) comprises the iron, friction, windage and shunt field losses, and none of these changes appreciably with load.

Run the machine as a motor on no load at rated voltage, with the field rheostat set for rated speed.
Read the line current \(I_{L0}\) and the shunt field current \(I_{sh}\).
Measure the armature resistance \(R_a\) separately, and correct it to the working temperature.
Compute \(W_c\), then the efficiency at any desired load.

As a generator delivering \(I_L\)

\[I_a = I_L + I_{sh}\]

\[\eta = \frac{VI_L}{VI_L + I_a^{2}R_a + W_c}\]

As a motor taking \(I_L\)

\[I_a = I_L - I_{sh}\]

\[\eta = \frac{VI_L - I_a^{2}R_a - W_c}{VI_L}\]

Note that \(W_c\) as defined above already includes the shunt field loss, so it must not be added again.

1 Worked Example 39.1 — A Complete Swinburne Calculation

Problem. A 250 V shunt machine takes 5 A on no load, of which 2 A is the shunt field current. The armature resistance is 0.30 \(\Omega\). Find the efficiency when it runs (a) as a generator delivering 60 A and (b) as a motor taking 60 A from the line.

Constant losses.

\[I_{a0} = 5 - 2 = 3~\mathrm{A}, \qquad VI_{L0} = (250)(5) = 1250~\mathrm{W}\]

\[W_c = 1250 - (3)^{2}(0.30) = 1250 - 2.7 = 1247.3~\mathrm{W}\]

Of which the shunt field accounts for \((250)(2) = 500\) W, leaving about 747 W of iron, friction and windage.

(a) As a generator delivering 60 A.

\[I_a = 60 + 2 = 62~\mathrm{A}, \qquad I_a^{2}R_a = (3844)(0.30) = 1153.2~\mathrm{W}\]

\[\text{total loss} = 1247.3 + 1153.2 = 2400.5~\mathrm{W}\]

\[\eta = \frac{15\,000}{15\,000 + 2400.5} = \frac{15\,000}{17\,400.5} = 86.20\,\%\]

(b) As a motor taking 60 A.

\[I_a = 60 - 2 = 58~\mathrm{A}, \qquad I_a^{2}R_a = (3364)(0.30) = 1009.2~\mathrm{W}\]

\[\text{total loss} = 1247.3 + 1009.2 = 2256.5~\mathrm{W}\]

\[\eta = \frac{15\,000 - 2256.5}{15\,000} = \frac{12\,743.5}{15\,000} = 84.96\,\%\]

Comment. One no-load reading has produced the efficiency of the same machine in both roles, at any load. That economy is the whole appeal of the method.

Note why the generator figure is the higher of the two. At the same 60 A of line current the generator's armature carries 62 A and the motor's only 58 A — yet the generator's efficiency is output over a larger input, while the motor's is a smaller output over the same input. The two are not comparable at equal line current; they would be at equal armature current.

2 Worked Example 39.2 — Maximum Efficiency From the Same Data

Problem. For the machine of Example 39.1 running as a generator, find the load current at which the efficiency is greatest, and its value.

Condition. From Chapter 33, variable loss equals constant loss:

\[I^{2}R_a = W_c \quad\Longrightarrow\quad I = \sqrt{\frac{1247.3}{0.30}} = \sqrt{4157.7} = 64.48~\mathrm{A}\]

Efficiency there. The total loss is \(2W_c\):

\[P_{\text{out}} = (250)(64.48) = 16\,120~\mathrm{W}\]

\[\eta_{\max} = \frac{16\,120}{16\,120 + 2494.6} = 86.60\,\%\]

Comment. The maximum lies at 64.5 A, only a little above the 60 A of Example 39.1 — and the efficiency there is 86.60 % against 86.20 %. Four-tenths of a percentage point for a 7 % change in load, which is the flat top of Chapter 33's efficiency curve seen again.

This is a genuinely useful by-product of Swinburne's test: the same single reading locates the machine's best operating point.

Section 39-4

Merits and Limitations

Merits

Very economical: the supply provides only the no-load losses, a few percent of rating.
Efficiency can be predicted at any load from a single test.
The same data serves the machine as generator or motor.
Quick, and needs no loading equipment.

Limitations

No check on commutation at full load, nor on temperature rise. The machine is never actually loaded.
Stray load losses are not accounted for, so the predicted efficiency is optimistic.
The iron loss on load differs from the no-load value, because armature reaction distorts the flux.
Not applicable to series motors, which cannot be run on no load at all.

! The Limitation That Matters Most

Swinburne's test predicts an efficiency but proves nothing about whether the machine can survive the load. Chapter 31 showed that commutation deteriorates in proportion to armature current, and a machine that sparks unacceptably at full load will pass a Swinburne test without complaint.

Nor does the test reveal the temperature rise, which is what actually determines the rating. For an acceptance test on a new design, a real heat run is required — and that is precisely what Hopkinson's test provides at almost no energy cost.

The series-motor exclusion follows from Chapter 35: with no load the field current vanishes, the flux with it, and \(N \propto 1/\Phi\) sends the speed towards destruction. A series motor cannot be run on no load, so it cannot be tested by Swinburne's method.

Section 39-5

Hopkinson's Test

Also called the back-to-back or regenerative test. Two identical machines are coupled mechanically and connected electrically, one running as a motor and driving the other as a generator, whose output is fed back to the supply.

Hopkinson's back-to-back test. Both machines run at full load while the supply provides only \(V(I_1 + I_3 + I_4)\).

♻

Why the Supply Provides Only the Losses

The output of one machine is the input of the other

The generator's electrical output feeds the motor's armature; the motor's mechanical output drives the generator. The power circulates round the loop, and the supply need only make up what is lost on each pass.

\[\text{Total losses of both machines} = V\left(I_1 + I_3 + I_4\right)\]

where \(I_1\) is the armature current drawn from the supply and \(I_3\), \(I_4\) are the two field currents.

Working Through the Losses

\[\begin{aligned} \text{Motor armature current} &= I_1 + I_2 \\[4pt] \text{Generator armature current} &= I_2 \\[4pt] \text{Motor armature Cu loss} &= \left(I_1 + I_2\right)^{2}R_a \\[4pt] \text{Generator armature Cu loss} &= I_2^{2}R_a \\[4pt] \text{Field losses} &= VI_3 + VI_4 \end{aligned}\]

Subtracting all of these from the supply power leaves the stray losses — iron, friction and windage — for the two machines together. Since they are identical and equally loaded, each is credited with half.

3 Worked Example 39.3 — A Hopkinson Test Worked Through

Problem. Two identical 250 V machines are tested back to back. The supply delivers 12 A to the armature circuit and the generator delivers 50 A to the motor. The motor field takes 2 A, the generator field 2.5 A, and each armature resistance is 0.25 \(\Omega\). Find the efficiency of each machine.

Total losses.

\[V\left(I_1 + I_3 + I_4\right) = 250\left(12 + 2 + 2.5\right) = (250)(16.5) = 4125~\mathrm{W}\]

Identified losses.

\[\text{motor armature Cu} = (12 + 50)^{2}(0.25) = (3844)(0.25) = 961~\mathrm{W}\]

\[\text{generator armature Cu} = (50)^{2}(0.25) = 625~\mathrm{W}\]

\[\text{motor field} = 500~\mathrm{W}, \qquad \text{generator field} = 625~\mathrm{W}\]

Stray losses.

\[4125 - (961 + 625 + 500 + 625) = 4125 - 2711 = 1414~\mathrm{W}\]

\[\text{per machine} = 707~\mathrm{W}\]

Motor efficiency.

\[\text{input} = V\left(I_1 + I_2\right) + VI_3 = 15\,500 + 500 = 16\,000~\mathrm{W}\]

\[\text{losses} = 961 + 500 + 707 = 2168~\mathrm{W}\]

\[\eta_m = \frac{16\,000 - 2168}{16\,000} = \frac{13\,832}{16\,000} = 86.45\,\%\]

Generator efficiency.

\[\text{output} = VI_2 = 12\,500~\mathrm{W}\]

\[\text{losses} = 625 + 625 + 707 = 1957~\mathrm{W}\]

\[\eta_g = \frac{12\,500}{12\,500 + 1957} = \frac{12\,500}{14\,457} = 86.46\,\%\]

Comment. The two come out within 0.01 of each other, which is a useful check on the arithmetic — identical machines at comparable loading should agree closely. They are not required to be exactly equal, because the motor's armature carries 62 A against the generator's 50 A, and their field currents differ.

Note that the generator field takes more current than the motor's. That is expected: the generator must be over-excited relative to the motor so that its EMF exceeds the bus voltage and it can deliver current — the regenerative condition of Chapter 38.

4 Worked Example 39.4 — What the Method Saves

Problem. Each machine in Example 39.3 is rated at 250 V, 50 A. Compare the supply capacity needed for the Hopkinson test with that needed to test both machines by direct loading.

Machine ratings.

\[(250)(50) = 12\,500~\mathrm{W} = 12.5~\mathrm{kW\ each}, \qquad 25~\mathrm{kW\ for\ the\ pair}\]

Supply actually required.

\[4125~\mathrm{W} = 4.125~\mathrm{kW}\]

Ratio.

\[\frac{25}{4.125} = 6.06 \quad\Longrightarrow\quad \text{the supply provides } 16.5\,\% \text{ of the combined rating}\]

Comment. A 5 kW supply tests two 12.5 kW machines at full load, for as long as you like. That is what makes a proper heat run affordable, and it is the decisive advantage over Swinburne's test — the machines really are loaded, so commutation and temperature rise can both be observed.

The saving grows with size. On a pair of 500 kW machines the same arithmetic would call for a supply of perhaps 80 kW instead of 1 MW, which is the difference between a routine works test and an impossibility.

Section 39-6

Merits and Limitations

Merits

Both machines are tested at full load while the supply provides only the losses.
A genuine heat run is possible, so temperature rise can be measured.
Commutation can be observed under real load conditions.
The load is easily varied by adjusting the two field rheostats.
Machines can be tested at overload just as cheaply.

Limitations

Two identical machines are required, which is often impossible.
The stray losses are assumed equally divided, which is only approximately true.
The two machines are not equally loaded — the motor armature carries more current than the generator's.
More equipment and a more elaborate setup than Swinburne's test.

The two tests are complementary, not rival. Swinburne's is a quick, cheap check on a single machine and the natural choice for routine work. Hopkinson's needs a matched pair but delivers what Swinburne's cannot — a full-load heat run and a sight of the commutator doing its real job. A manufacturer uses Swinburne's on every machine and Hopkinson's on the type test.

Section 39-7

The Retardation Test

Also called the running-down test. It measures the rotational losses — iron, friction and windage — from the rate at which the machine slows when its supply is removed.

⏱

The Principle

The stored kinetic energy pays for the losses

Once the armature is disconnected, the only thing keeping it turning is its own kinetic energy, and the only thing slowing it is the rotational loss. Differentiating \(\tfrac{1}{2}J\omega^{2}\):

\[W = -\frac{\mathrm{d}}{\mathrm{d}t}\left(\tfrac{1}{2}J\omega^{2}\right) = J\omega\left|\frac{\mathrm{d}\omega}{\mathrm{d}t}\right|\]

So the loss at any speed follows from the slope of the speed-time curve at that speed, provided \(J\) is known.

Run the machine as a motor a little above the speed of interest.
Disconnect the armature from the supply, leaving the field excited, and record speed against time as it runs down.
Take the slope of that curve at the speed of interest.
Compute \(W = J\omega\left|\mathrm{d}\omega/\mathrm{d}t\right|\).

In practice the slope is found from the time taken to fall through a small band of speed straddling the point of interest — say 1030 to 970 rev/min for the loss at 1000 rev/min. This is the same exponential run-down that Chapter 38 derived for rheostatic braking, with the braking resistor removed so that only the rotational losses remain.

! Finding the Moment of Inertia

The method needs \(J\), which is rarely known accurately. It is found by repeating the run-down with a known extra inertia — a flywheel of calculated \(J'\) — bolted to the shaft.

If the same speed band now takes time \(t'\) instead of \(t\), then since the loss is unchanged,

\[\frac{J}{t} = \frac{J + J'}{t'} \quad\Longrightarrow\quad J = J'\frac{t}{t' - t}\]

Section 39-8

Separating Iron and Mechanical Losses

The retardation test as described gives the total rotational loss. The two components are separated by a second run-down.

Run 1 — field excited

The iron is magnetised as in normal running, so the machine loses energy to hysteresis and eddy currents as well as to friction and windage.

\[W_1 = P_{\text{iron}} + P_{\text{fric}}\]

Run 2 — field unexcited

With no flux there is no alternating magnetisation, so the iron loss vanishes and only the mechanical loss remains.

\[W_2 = P_{\text{fric}}\]

\[P_{\text{iron}} = W_1 - W_2\]

Because the same \(J\) and the same \(\omega\) apply to both runs, the losses are simply in the inverse ratio of the run-down times:

\[\frac{W_1}{W_2} = \frac{t_2}{t_1}\]

so the separation can be made from the two times alone, once either loss is known absolutely.

5 Worked Example 39.5 — Retardation Test With Separation

Problem. A machine of moment of inertia 25 kg·m² is run up and allowed to slow through the band 1030 to 970 rev/min. With the field excited this takes 40 s; with the field unexcited it takes 60 s. Find the total rotational loss at 1000 rev/min, the mechanical loss and the iron loss.

Angular velocity and its rate of change.

\[\omega = \frac{2\pi(1000)}{60} = 104.72~\mathrm{rad/s}\]

\[\Delta\omega = \frac{2\pi(1030 - 970)}{60} = \frac{2\pi(60)}{60} = 6.283~\mathrm{rad/s}\]

Run 1 — field excited.

\[\left|\frac{\mathrm{d}\omega}{\mathrm{d}t}\right| = \frac{6.283}{40} = 0.15708~\mathrm{rad/s^{2}}\]

\[W_1 = J\omega\left|\frac{\mathrm{d}\omega}{\mathrm{d}t}\right| = (25)(104.72)(0.15708) = 411.2~\mathrm{W}\]

Run 2 — field unexcited.

\[\left|\frac{\mathrm{d}\omega}{\mathrm{d}t}\right| = \frac{6.283}{60} = 0.10472~\mathrm{rad/s^{2}}\]

\[W_2 = (25)(104.72)(0.10472) = 274.2~\mathrm{W}\]

Iron loss.

\[P_{\text{iron}} = 411.2 - 274.2 = 137.0~\mathrm{W}\]

Check by the ratio of times: \(W_1/W_2 = 60/40 = 1.500\), and \(411.2/274.2 = 1.500\) \(\checkmark\)

Comment. Friction and windage account for two-thirds of the rotational loss and the iron for one third. That split is typical, and it is one no other test in this chapter can provide — Swinburne's lumps the two together in \(W_c\), and Hopkinson's lumps them together as the stray loss.

The separation matters in design. If the iron loss dominates, thinner laminations or better steel will help; if friction dominates, the bearings and the cooling fan are where to look. The retardation test is the only one of the three that tells you which.

Section 39-9

Comparison and Selection

Table 39.2 — The three indirect tests compared.
	Swinburne	Hopkinson	Retardation
Machines needed	One	Two identical	One
Supply required	No-load losses only	Losses of both, at full load	None during the run
Gives	\(W_c\), hence \(\eta\) at any load	\(\eta\) of both machines	Rotational losses
Machine actually loaded	No	Yes	No
Heat run possible	No	Yes	No
Commutation checked	No	Yes	No
Separates iron from friction	No	No	Yes
Suits series motors	No	Yes	Yes
Typical use	Routine test on every machine	Type test, heat run	Loss analysis in design

Each answers a different question. Swinburne's asks how efficient will this machine be? — cheaply, on any single machine, but on trust. Hopkinson's asks will it survive full load? — which needs the machine genuinely loaded, and a matched pair to make that affordable. The retardation test asks where is the loss going?, which neither of the others can answer because both lump the rotational losses into a single figure.

Section 39-10

Summary and Key Formulas

Direct loading of a large machine is costly and inaccurate, because efficiency is the small difference between two large measured quantities.
Swinburne's test runs the machine as a motor on no load. \(W_c = VI_{L0} - I_{a0}^{2}R_a\), which includes the iron, friction, windage and shunt field losses.
From that one reading the efficiency follows at any load, as generator (\(I_a = I_L + I_{sh}\)) or motor (\(I_a = I_L - I_{sh}\)).
Its limitations: no full-load check on commutation or temperature rise; stray load losses ignored; iron loss differs on load because of armature reaction; and it cannot be used on series motors.
Hopkinson's test couples two identical machines back to back. The supply provides only \(V(I_1 + I_3 + I_4)\), the total losses of both, so a full-load heat run costs a fraction of the rating.
Motor armature carries \(I_1 + I_2\); generator armature carries \(I_2\). Stray losses are found by subtraction and divided equally.
Retardation test: \(W = J\omega\left|\mathrm{d}\omega/\mathrm{d}t\right|\) from the run-down curve. \(J\) is found by repeating with a known added inertia.
Running down with and without field excitation separates iron loss from friction and windage.

Table 39.3 — Formulas of this chapter.
Quantity	Formula	Notes
Swinburne constant losses	\(W_c = VI_{L0} - I_{a0}^{2}R_a\)	includes field loss
No-load armature current	\(I_{a0} = I_{L0} - I_{sh}\)	—
Efficiency as generator	\(\eta = \dfrac{VI_L}{VI_L + I_a^{2}R_a + W_c}\)	\(I_a = I_L + I_{sh}\)
Efficiency as motor	\(\eta = \dfrac{VI_L - I_a^{2}R_a - W_c}{VI_L}\)	\(I_a = I_L - I_{sh}\)
Maximum efficiency	\(I = \sqrt{W_c/R_a}\)	from Chapter 33
Hopkinson total losses	\(V\left(I_1 + I_3 + I_4\right)\)	both machines
Motor armature current	\(I_1 + I_2\)	generator carries \(I_2\)
Stray loss per machine	\(\tfrac{1}{2}\left[V(I_1+I_3+I_4) - \sum\text{Cu}\right]\)	assumed equal
Retardation loss	\(W = J\omega\left\|\dfrac{\mathrm{d}\omega}{\mathrm{d}t}\right\|\)	at the speed of interest
Moment of inertia	\(J = J'\dfrac{t}{t' - t}\)	added flywheel \(J'\)
Loss ratio between runs	\(\dfrac{W_1}{W_2} = \dfrac{t_2}{t_1}\)	same \(J\) and \(\omega\)
Iron loss	\(P_{\text{iron}} = W_1 - W_2\)	excited minus unexcited

Section 39-11

Common Mistakes

Adding the field loss to \(W_c\) again. The Swinburne \(W_c\) already contains it.
Using \(I_{L0}\) for the no-load armature copper loss. It is \(I_{a0} = I_{L0} - I_{sh}\).
Getting the sign wrong on \(I_a\). Generator \(I_L + I_{sh}\); motor \(I_L - I_{sh}\).
Applying Swinburne's test to a series motor. It cannot be run on no load.
Believing Swinburne's test validates the machine. It predicts efficiency; it proves nothing about commutation or temperature rise.
Taking the Hopkinson motor and generator armature currents as equal. The motor carries \(I_1 + I_2\) and the generator only \(I_2\).
Forgetting the field currents in the Hopkinson supply power. It is \(V(I_1 + I_3 + I_4)\), not \(VI_1\).
Expecting the two Hopkinson efficiencies to be identical. The machines are identical but not identically loaded.
Using \(\mathrm{d}N/\mathrm{d}t\) in rev/min directly. The formula needs \(\omega\) in rad/s.
Leaving the field excited in the second retardation run. It must be unexcited to remove the iron loss.

Section 39-12

Chapter Review

Practice Problems

For Swinburne problems find \(W_c\) first. For Hopkinson, find the total losses from the supply, then subtract everything identifiable.

P39.1 A 220 V shunt machine takes 6 A on no load, of which 1.5 A is field current, with \(R_a = 0.25~\Omega\). Find \(W_c\).

Show answer

\[I_{a0} = 6 - 1.5 = 4.5~\mathrm{A}, \qquad VI_{L0} = (220)(6) = 1320~\mathrm{W}\]

\[W_c = 1320 - (4.5)^{2}(0.25) = 1320 - 5.06 = 1314.9~\mathrm{W}\]
P39.2 For P39.1, find the efficiency as a motor taking 50 A.

Show answer

\[I_a = 50 - 1.5 = 48.5~\mathrm{A}, \qquad I_a^{2}R_a = (2352.25)(0.25) = 588.1~\mathrm{W}\]

\[\text{input} = (220)(50) = 11\,000~\mathrm{W}, \qquad \text{loss} = 1314.9 + 588.1 = 1903.0~\mathrm{W}\]

\[\eta = \frac{11\,000 - 1903.0}{11\,000} = 82.70\,\%\]
P39.3 For P39.1, find the efficiency as a generator delivering 50 A.

Show answer

\[I_a = 51.5~\mathrm{A}, \qquad I_a^{2}R_a = (2652.25)(0.25) = 663.1~\mathrm{W}\]

\[\eta = \frac{11\,000}{11\,000 + 1314.9 + 663.1} = \frac{11\,000}{12\,978.0} = 84.76\,\%\]
P39.4 For P39.1, find the load current for maximum efficiency as a generator.

Show answer

\[I = \sqrt{\frac{1314.9}{0.25}} = \sqrt{5259.6} = 72.52~\mathrm{A}\]

\[\eta_{\max} = \frac{(220)(72.52)}{(220)(72.52) + 2(1314.9)} = \frac{15\,955}{18\,585} = 85.85\,\%\]
P39.5 A Hopkinson test on two 220 V machines gives: supply armature current 15 A, generator output 60 A, motor field 2 A, generator field 3 A, \(R_a = 0.20~\Omega\) each. Find the total losses and the stray loss per machine.

Show answer

\[\text{total} = 220(15 + 2 + 3) = (220)(20) = 4400~\mathrm{W}\]

\[\text{motor arm Cu} = (75)^{2}(0.20) = 1125~\mathrm{W}, \quad \text{gen arm Cu} = (60)^{2}(0.20) = 720~\mathrm{W}\]

\[\text{fields} = 440 + 660 = 1100~\mathrm{W}\]

\[\text{stray} = 4400 - (1125 + 720 + 1100) = 1455~\mathrm{W}, \quad \text{per machine } 727.5~\mathrm{W}\]
P39.6 For P39.5, find the generator efficiency.

Show answer

\[\text{output} = (220)(60) = 13\,200~\mathrm{W}\]

\[\text{losses} = 720 + 660 + 727.5 = 2107.5~\mathrm{W}\]

\[\eta = \frac{13\,200}{13\,200 + 2107.5} = \frac{13\,200}{15\,307.5} = 86.23\,\%\]
P39.7 A machine with \(J = 18\) kg·m² slows from 1020 to 980 rev/min in 25 s with the field excited. Find the rotational loss at 1000 rev/min.

Show answer

\[\omega = 104.72~\mathrm{rad/s}, \qquad \Delta\omega = \frac{2\pi(40)}{60} = 4.189~\mathrm{rad/s}\]

\[\left|\frac{\mathrm{d}\omega}{\mathrm{d}t}\right| = \frac{4.189}{25} = 0.16755~\mathrm{rad/s^{2}}\]

\[W = (18)(104.72)(0.16755) = 315.8~\mathrm{W}\]
P39.8 The same machine takes 45 s through the same band with the field unexcited. Find the iron loss.

Show answer
Since \(W \propto 1/t\) for the same \(J\) and \(\omega\):
\[W_2 = (315.8)\left(\frac{25}{45}\right) = 175.4~\mathrm{W}\]

\[P_{\text{iron}} = 315.8 - 175.4 = 140.4~\mathrm{W}\]
P39.9 Why is Swinburne's test not applicable to a series motor, and what would you use instead?

Show answer
Swinburne's test requires the machine to be run as a motor on no load. A series motor's field carries the armature current, so on no load the current is small, the flux is small, and since \(N \propto 1/\Phi\) the speed rises without limit — Chapter 35's runaway. The machine would destroy itself before any reading could be taken.
Instead: a Hopkinson test if a matched pair is available, since both machines stay loaded throughout; or direct loading with a brake for a small machine; or a retardation test for the rotational losses alone, which needs no sustained no-load running.
P39.10 A manufacturer must decide which test to apply. Which for a routine check on every machine, which for a type test, and which to investigate a machine running hot?

Show answer
Routine check on every machine: Swinburne's. It needs one machine, a few minutes and a supply capable only of the no-load losses. Efficiency is predicted at every load from a single reading.
Type test on a new design: Hopkinson's. Only this one loads the machines genuinely, so temperature rise can be measured over a proper heat run and commutation observed under full-load current — and it does so at about a sixth of the combined rating.

Machine running hot: retardation. The other two lump iron and mechanical losses together, so neither can say which is excessive. Running down with the field excited and again unexcited separates them, pointing either at the laminations or at the bearings and fan.

Multiple-Choice Questions

MCQ 1. Efficiency is usually found from the losses because direct measurement is:
(a) illegal (b) costly and inaccurate (c) impossible (d) too fast

Show answer
(b) costly and inaccurate — the loss is a small difference between two large measured quantities.
MCQ 2. In Swinburne's test the machine is run:
(a) as a generator on full load (b) as a motor on no load (c) at standstill (d) back to back

Show answer
(b) as a motor on no load, at rated voltage and speed.
MCQ 3. The constant losses from Swinburne's test include:
(a) armature copper only (b) iron, friction, windage and shunt field (c) stray load loss (d) nothing

Show answer
(b) iron, friction, windage and shunt field. The field loss must not be added again.
MCQ 4. Swinburne's test cannot be used on:
(a) shunt motors (b) series motors (c) compound motors (d) generators

Show answer
(b) series motors, which cannot be run on no load.
MCQ 5. The chief defect of Swinburne's test is that it gives no check on:
(a) efficiency (b) commutation and temperature rise (c) armature resistance (d) speed

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(b) commutation and temperature rise — the machine is never actually loaded.
MCQ 6. In Hopkinson's test the supply provides:
(a) full rated power (b) only the losses of both machines (c) nothing (d) half the rating

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(b) only the losses of both machines, since the power circulates round the loop.
MCQ 7. In Hopkinson's test the motor armature carries:
(a) \(I_1\) (b) \(I_2\) (c) \(I_1 + I_2\) (d) \(I_2 - I_1\)

Show answer
(c) \(I_1 + I_2\) — supply current plus generator output. The generator carries only \(I_2\).
MCQ 8. The main practical drawback of Hopkinson's test is that it needs:
(a) a large supply (b) two identical machines (c) a brake (d) a flywheel

Show answer
(b) two identical machines, which is often not possible.
MCQ 9. In the retardation test the loss is:
(a) \(J\omega^{2}\) (b) \(J\omega\,\mathrm{d}\omega/\mathrm{d}t\) (c) \(\tfrac{1}{2}J\omega\) (d) \(J\,\mathrm{d}\omega/\mathrm{d}t\)

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(b) \(J\omega\,\mathrm{d}\omega/\mathrm{d}t\), from differentiating the stored kinetic energy.
MCQ 10. Iron loss is separated from friction by running down:
(a) at two speeds (b) with and without field excitation (c) with two flywheels (d) in reverse

Show answer
(b) with and without field excitation. Unexcited, there is no flux and so no iron loss.

Conceptual Questions

Explain why efficiency is predicted from losses rather than measured directly on a large machine.
Describe Swinburne's test and derive the expression for the constant losses.
Show how the efficiency as generator and as motor follows from a single Swinburne reading.
Give the merits and the four limitations of Swinburne's test.
Describe Hopkinson's test and explain why the supply provides only the losses.
Derive the loss breakdown in Hopkinson's test and explain how the stray losses are apportioned.
Describe the retardation test and derive \(W = J\omega\left|\mathrm{d}\omega/\mathrm{d}t\right|\).
Explain how the moment of inertia is determined, and how iron and mechanical losses are separated.

Looking Ahead

Part 2 is complete. Twenty chapters have taken the DC machine from its construction and winding through the generator in all its forms, then through the motor, its starting, its speed control, its braking and finally its testing. Every one of them rested on two equations established in Chapter 27 — \(E = K_a\Phi\omega\) and \(T = K_a\Phi I_a\), sharing one constant.

Part 3 turns to the transformer, and the ground prepared in Part 1 comes back into use. The magnetic circuit of Chapters 2 to 5, the hysteresis and eddy-current losses of Chapters 6 and 7, the AC excitation of Chapter 8 and Faraday's law of Chapter 10 are the whole of a transformer's theory — with the crucial simplification that nothing rotates, so there is no commutator, no brushes, no armature reaction and no mechanical loss.

Much will nonetheless look familiar. The EMF equation will reappear in the form \(E = 4.44 f N\Phi_m\) first met in Chapter 8; voltage regulation will be defined exactly as in Chapter 29; and the condition for maximum efficiency — that variable loss equals constant loss — carries over from Chapter 33 unchanged, because it depends only on the shape of the loss expression and not on the machine.