Question 4 of 6: Separately-Excited / Shunt DC Motor — Field Weakening
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Closed-book written exam, 3 hours. Six questions of equal value (20 marks each); five constitute a complete paper. All six are solved here as a study resource. AC quantities are rms; three-phase voltages are line-to-line and power is total power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics; T. Wildi, Electrical Machines, Drives, and Power Systems.
Question 4: Separately-Excited / Shunt DC Motor — Field Weakening (20 marks)
Find. (a)–(c) output power, developed torque, efficiency at the original operating point; (d)–(g) armature current, line current, speed and efficiency after field weakening.
Approach. Split the line current into field and armature parts, find the back-emf, then power and torque. For field weakening, hold the developed torque constant (constant-torque load) with flux proportional to field current to get the new armature current and speed.
Field, armature current, back-emf.$I_f=V_t/R_f=240/150=1.6$ A, so $I_a=I_L-I_f=41.6-1.6=40$ A and
$$E_a=V_t-I_aR_a=240-40(0.4)=224\ \text{V}.$$
(a) Output power.$P_{dev}=E_aI_a=224(40)=8960$ W, so
$$P_{out}=P_{dev}-P_{rot}=8960-600=\boxed{8360\ \text{W}.}$$
(b) Developed torque. With $\omega_1=1600\cdot 2\pi/60=167.55$ rad/s,
$$T_{dev}=\frac{P_{dev}}{\omega_1}=\frac{8960}{167.55}=\boxed{53.5\ \text{N}\!\cdot\!\text{m}.}$$
(c) Efficiency.$P_{in}=V_tI_L=240(41.6)=9984$ W, so
$$\eta=\frac{8360}{9984}=\boxed{83.7\%.}$$
(d) New armature current. New field $I_f'=240/200=1.2$ A, so flux $\Phi\propto I_f$ falls to 0.75 of its value. A constant developed torque $T=K_a\Phi I_a$ requires
$$I_a'=I_a\frac{I_f}{I_f'}=40\cdot\frac{1.6}{1.2}=\boxed{53.3\ \text{A}.}$$
(e) New line current. $$I_L'=I_a'+I_f'=53.3+1.2=\boxed{54.5\ \text{A}.}$$
(f) New speed.$E_a'=240-53.3(0.4)=218.7$ V. Since $E_a=K_a\Phi\,\omega$,
$$n_2=n_1\frac{E_a'}{E_a}\frac{\Phi}{\Phi'}=1600\cdot\frac{218.7}{224}\cdot\frac{1.6}{1.2}=\boxed{2083\ \text{rpm}.}$$
(g) New efficiency.$P_{dev}'=E_a'I_a'=218.7(53.3)=11\,662$ W, $P_{out}'=11\,662-600=11\,062$ W, $P_{in}'=240(54.5)=13\,087$ W:
$$\eta'=\frac{11\,062}{13\,087}=\boxed{84.5\%.}$$
