22-Elec-B8 Power Electronics and Drives · Undated paper

Question 6 of 6: PROBLEM 4 — Three-Phase Bridge Drive for a Separately Excited d.c. Motor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2019 — 16-Elec-B8 Power Electronics and Drives. Three hours; open book; any non-communicating calculator, whose make and model must be written on the first inside sheet of the work book. The paper is in two parts and the candidate must attempt all parts: Part 1-A is ten five-point short-answer items, Part 1-B is ten five-point multiple-choice items each requiring a written explanation, and Part 2 is four thirty-point problems. Page-1 Note 4 states that the maximum total score is 220 points and that 150 points is a full mark (100 per cent), so there are 70 points of built-in bonus. Note 5 warns that an answer without its working scores nothing. All a.c. voltages and currents below are rms unless stated otherwise, and three-phase voltages are line-to-line. Every part of every question is solved below, in the exam's own order, with the six sections mapping to Part 1-A, Part 1-B and PROBLEMs 1 to 4.

Reference texts.

M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. — the primary reference for this exam code (devices ch. 4, controlled rectifiers ch. 3 and 10, choppers ch. 5, a.c. voltage controllers ch. 11, d.c. drives ch. 15).
N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed. — PWM, switching-converter waveforms, snubbers and utility applications.
C. W. Lander, Power Electronics, 3rd ed. — compact treatments of phase control and of the chopper current equations used in PROBLEM 3.
B. K. Bose, Modern Power Electronics and AC Drives, and R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control — drive context for PROBLEMs 2 and 4.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — transformer core loss (Part 1-A item b) and the separately excited d.c. machine (PROBLEM 4).
J. J. Grainger and W. D. Stevenson, Power System Analysis, and IEEE Std 519 — harmonics, HVDC economics and static VAR compensation.

Two printing anomalies in the paper itself, both worth a line in the answer book. First, Part 1-B Question 9 repeats Question 5 word for word (“AC voltage controllers convert…”) — ten of the fifty Part 1-B points are the same item asked twice. Both are answered below, the second with the quantitative detail the first does not need. Second, the arithmetic of the cover page does not close in the candidate's favour by accident: 50 + 50 + 120 = 220 points are on offer against a full mark of 150, so a candidate should attempt every part and let the surplus absorb the inevitable slips rather than budgeting time as if 100 points were the target.

Question 6: PROBLEM 4 — Three-Phase Bridge Drive for a Separately Excited d.c. Motor (30 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A three-phase fully controlled (six-pulse) bridge supplies the armature of a separately excited d.c. motor from a 220 V line-to-line source; the field is constant and the armature current is held at 150 A at both operating points.

Given data
Quantity	Symbol	Value
Line-to-line source voltage	$V_{LL}$	220 V
Armature current (both points)	$I_{a}$	150 A
Operating point 1	$\alpha_{1}$, $N_{1}$	45°, 1750 rpm
Operating point 2	$\alpha_{2}$, $N_{2}$	55°, 1200 rpm
Field	—	separately excited, held constant

Find. (a) the armature terminal voltage at 45°; (b) the armature circuit resistance, the developed output power and the torque at 1200 rpm with $\alpha=55^\circ$.

PROBLEM 4: the three-phase fully controlled bridge output Va = 1.35047 x 220 x cos(a). The two operating points sit on the same cosine, and the 39.67 V that separates them is exactly the back-e.m.f. change between 1750 rpm and 1200 rpm because the armature current is held at 150 A.

Approach. The bridge output is a pure cosine function of firing angle, so each operating point gives one terminal voltage; the pair of armature equations $V_{a}=E+I_{a}R_{a}$ written at two speeds with the same current then separates the machine constant from the resistance, after which power and torque follow from the back e.m.f.

Part (a) — write the mean output of the six-pulse bridge. A three-phase fully controlled bridge conducts two devices at a time from the line-to-line voltages, giving six pulses per cycle and a mean $$V_{a}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha=1.35047\,V_{LL}\cos\alpha .$$ The ideal no-load d.c. voltage is therefore $V_{do}=1.35047\times220=297.10$ V, and commutation overlap and device drops are neglected as usual.
Evaluate the armature voltage at the first operating point. With $\alpha_{1}=45^\circ$, $\cos 45^\circ=0.707107$: $$V_{a1}=297.07\times0.707107\quad\Longrightarrow\quad\boxed{V_{a1}=210.08\ \mathrm{V}} .$$ Note that the stated speed of 1750 rpm plays no part in this answer — the terminal voltage of the converter depends only on $V_{LL}$ and $\alpha$. The speed is supplied because part (b) needs the pair of operating points.
Part (b) — evaluate the second armature voltage. With $\alpha_{2}=55^\circ$, $\cos 55^\circ=0.573576$: $$V_{a2}=297.07\times0.573576=170.41\ \mathrm{V}.$$ The converter has therefore lowered the armature voltage by 39.67 V between the two points.
Write the armature equation at each point and subtract. For a separately excited machine with constant field, $V_{a}=E+I_{a}R_{a}$ with $E=k\phi\,\omega$. The angular speeds are $\omega_{1}=2\pi\times1750/60=183.260$ rad/s and $\omega_{2}=2\pi\times1200/60=125.664$ rad/s. Because $I_{a}$ is the same 150 A at both points, the $I_{a}R_{a}$ terms cancel on subtraction: $$V_{a1}-V_{a2}=k\phi\left(\omega_{1}-\omega_{2}\right) \;\Longrightarrow\; k\phi=\frac{210.08-170.41}{183.260-125.664}=\frac{39.672}{57.596}=0.68881\ \frac{\mathrm{V\,s}}{\mathrm{rad}}.$$ That is the whole trick of the question: holding the current constant turns two unknowns into one subtraction.
Recover the armature circuit resistance. Substituting $k\phi$ back into the first operating point, the back e.m.f. there is $E_{1}=k\phi\,\omega_{1}=0.68881\times183.260=126.23$ V, so $$R_{a}=\frac{V_{a1}-E_{1}}{I_{a}}=\frac{210.08-126.23}{150} \quad\Longrightarrow\quad\boxed{R_{a}=0.559\ \Omega}.$$ Checking at the second point, $E_{2}=0.68881\times125.664=86.56$ V and $E_{2}+I_{a}R_{a}=86.56+83.85=170.41$ V, which reproduces $V_{a2}$ exactly.
Compute the developed output power at 1200 rpm. The power converted from electrical to mechanical form is the product of back e.m.f. and armature current: $$P_{out}=E_{2}I_{a}=86.558\times150\quad\Longrightarrow\quad \boxed{P_{out}=12.98\ \mathrm{kW}}.$$ For comparison the armature input is $V_{a2}I_{a}=25.56$ kW, of which $I_{a}^{2}R_{a}=12.58$ kW is lost in the armature resistance — an efficiency of only 51 per cent, which is what deep phase-back on a resistive armature costs.
Compute the developed torque. With constant field the torque constant equals the e.m.f. constant in SI units, so $$T=k\phi\,I_{a}=0.68881\times150\quad\Longrightarrow\quad\boxed{T=103.3\ \mathrm{N}\cdot\mathrm{m}},$$ and the cross-check $T=P_{out}/\omega_{2}=12983.7/125.664=103.3\ \mathrm{N}\cdot\mathrm{m}$ agrees to four figures. Because both the field and the armature current are held constant, this same torque is developed at 1750 rpm as well; only the power differs.
Sanity-check the drive against its own data. At 1750 rpm the developed power is $E_{1}I_{a}=18.93$ kW, so the drive is a constant-torque, variable-power system between the two points — the signature of armature-voltage control below base speed. The rectifier is also operating well inside its range: at $\alpha=55^\circ$ the bridge could still deliver 297.10 V at $\alpha=0$, so there is ample headroom for the transient voltage needed to accelerate the load.

PROBLEM 4 — results
Quantity	Symbol	Value
Ideal no-load bridge output	$V_{do}=1.35047V_{LL}$	297.10 V
(a) Armature voltage at $\alpha=45^\circ$	$V_{a1}$	210.08 V
Armature voltage at $\alpha=55^\circ$	$V_{a2}$	170.41 V
Machine constant	$k\phi$	0.6888 V·s/rad
Back e.m.f. at 1750 / 1200 rpm	$E_{1}$, $E_{2}$	126.23 V, 86.56 V
(b) Armature circuit resistance	$R_{a}$	0.559 $\Omega$
(b) Output (developed) power at 1200 rpm	$P_{out}$	12.98 kW
(b) Developed torque	$T$	103.3 N·m

Check: “output power” is taken as the developed (electromagnetic) power $E I_{a}$, because the question supplies no data on friction, windage or core loss; the shaft power is smaller by those rotational losses. Commutation overlap, device forward drops and armature-reaction weakening of the field are all neglected, which is standard for this level of question and is why $V_{a}=1.35047V_{LL}\cos\alpha$ is used unmodified.

Topic. Armature-voltage speed control of a separately excited d.c. motor from a three-phase fully controlled bridge.

Key relations. Six-pulse bridge mean output $V_{a}=(3\sqrt{2}/\pi)V_{LL}\cos\alpha$; armature equation $V_{a}=E+I_{a}R_{a}$; $E=k\phi\,\omega$ and $T=k\phi\,I_{a}$; $\omega=2\pi N/60$; developed power $P=EI_{a}=T\omega$.

Why this works. Two operating points at the same armature current give two linear equations in the two unknowns $k\phi$ and $R_{a}$, and because the current is identical the resistive drop is identical too, so subtracting the equations removes $R_{a}$ outright and delivers $k\phi$ in one line. Substituting back gives $R_{a}$. This constant-current pairing is the standard examination device for a machine whose nameplate constants are not given, and recognising it converts an apparently under-determined problem into two subtractions. Once $k\phi$ is known, torque and power need no further circuit analysis at all.

Common pitfall. Using the half-controlled or three-pulse constant ($1.17V_{ph}$ or $0.675V_{LL}$) instead of $1.35047V_{LL}$, or applying $V_{a}=1.35V_{LL}\cos\alpha$ to a phase voltage. The second is quoting the armature input $V_{a}I_{a}$ as the output power: at $\alpha=55^\circ$ almost half the armature input is dissipated in $R_{a}$, so the two differ by a factor of two here and the error is glaring.

Source. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed., §3.9 and ch. 15 (d.c. drives); S. J. Chapman, Electric Machinery Fundamentals, 5th ed., ch. 8 (separately excited d.c. motors); R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control, ch. 3.

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