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22-Elec-A6 Power Systems and Machines · May 2016

Question 3 of 7: DC Shunt Motor

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

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.

Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.

Question 3: DC Shunt Motor (20 marks)

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.

Given data — Question 3
Terminal voltageV = 220 V
Armature / field resistanceRa = 0.1 Ω, Rf = 100 Ω
Rotational lossesnegligible
Condition 1n1 = 1600 rpm, It = 100 A
Condition 2developed torque T = 100 N·m (same V, field)

Find. For condition 1: (a) output power, (b) output torque, (c) efficiency. For condition 2: (d) total current, (e) speed, (f) efficiency.

Approach. Split the line current into constant field current and armature current, find the back-emf from the armature loop, and use $P=E_aI_a$ and $T=P/\omega$. Since the field is unchanged, the machine constant $K\Phi=E_a/\omega=T/I_a$ is fixed and links the two conditions.

  1. Field, armature current and back-emf. $I_f=\dfrac{V}{R_f}=\dfrac{220}{100}=2.2$ A, so $I_a=I_t-I_f=100-2.2=97.8$ A and $$E_a=V-I_aR_a=220-97.8(0.1)=210.22\text{ V}.$$
  2. Output power (a). With negligible rotational loss the developed power equals the shaft output: $$\boxed{P_{\text{out}}=E_aI_a=210.22(97.8)=20\,560\text{ W}=20.56\text{ kW}.}$$
  3. Output torque (b). With $\omega_1=1600\cdot\dfrac{2\pi}{60}=167.55$ rad/s, $$\boxed{T=\frac{P_{\text{out}}}{\omega_1}=\frac{20\,560}{167.55}=122.7\text{ N}\cdot\text{m}.}$$
  4. Efficiency (c). The input is $P_{\text{in}}=VI_t=220(100)=22\,000$ W, hence $$\boxed{\eta=\frac{20\,560}{22\,000}=93.45\%.}$$
  5. Machine constant. $K\Phi=\dfrac{E_a}{\omega_1}=\dfrac{210.22}{167.55}=1.2547\ \text{V}\cdot\text{s}$; check: $T=K\Phi\,I_a=1.2547(97.8)=122.7$ N·m ✓.
  6. New armature and total current (d). The field is unchanged, so $I_{a2}=\dfrac{T}{K\Phi}=\dfrac{100}{1.2547}=79.70$ A and $$\boxed{I_{t2}=I_{a2}+I_f=79.70+2.2=81.90\text{ A}.}$$
  7. New speed (e). $E_{a2}=V-I_{a2}R_a=220-79.70(0.1)=212.03$ V, so $\omega_2=\dfrac{E_{a2}}{K\Phi}=\dfrac{212.03}{1.2547}=168.99$ rad/s, i.e. $$\boxed{n_2=168.99\cdot\frac{60}{2\pi}=1614\text{ rpm}.}$$ The lighter torque load lets the motor speed up slightly.
  8. New efficiency (f). $P_{\text{out}2}=T\omega_2=100(168.99)=16\,899$ W and $P_{\text{in}2}=220(81.90)=18\,018$ W, so $\eta_2=\dfrac{16\,899}{18\,018}=93.79\%.$
Final results — Question 3
QuantityCondition 1 (1600 rpm)Condition 2 (T = 100 N·m)
Total current100 A81.90 A
Speed1600 rpm1614 rpm
Output power / torque20.56 kW / 122.7 N·m16.90 kW / 100 N·m
Efficiency93.45%93.79%