Question 4 of 5: DC Shunt Motor with Field Weakening
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2015. Closed-book; formula sheets supplied. FIVE questions constitute a complete paper and all are of equal value (20 marks each). All ac quantities are rms; three-phase voltages are line-to-line and power is total unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — Ch. 2 transformers, Ch. 6–7 induction machines, Ch. 5 synchronous machines, Ch. 8–9 DC machines; A. E. Fitzgerald, C. Kingsley & S. Umans, Electric Machinery, 6th ed.; P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed.
Question 4: DC Shunt Motor with Field Weakening (20 marks)
Given. A 250 V, 1600 rpm DC shunt motor on a constant-torque load, drawing line current 20 A at rated voltage, with $R_a=0.4\ \Omega$, $R_f=250\ \Omega$, and negligible rotational losses. A linear magnetic circuit means flux is proportional to field current, $\phi\propto I_f$.
Given data
$V$
$n_1$
$I_{L1}$
$R_a$
$R_f$
new $I_{f2}$
250 V
1600 rpm
20 A
0.4 Ω
250 Ω
0.8 A
Find. Armature current, output power, torque and efficiency at rated; then, after weakening the field to 0.8 A at constant load torque, the inserted resistance (value and rating), new armature current, new speed and new efficiency.
Figure Q4. Shunt connection: field winding R_f and armature (R_a in series with the back-emf E_a) are in parallel across the 250 V supply.
Approach. The shunt field current is set by the mains and $R_f$; the armature current is the remainder of the line current. Back-emf gives developed power and torque; efficiency is output over input. For field weakening use two invariants of a linear machine on a constant-torque load: $\tau\propto\phi I_a$ (so $I_fI_a$ is constant) and $E_a\propto\phi\,n$ (so $E_a\propto I_f n$).
Field and armature current (part a). The shunt field draws $I_f=V/R_f=250/250=1.0$ A, so$$I_a=I_L-I_f=20-1=\boxed{19\ \text{A}}.$$
Back-emf and output power (part b). $E_a=V-I_aR_a=250-19(0.4)=242.4$ V. With rotational losses neglected the mechanical output equals the developed power:$$P_{out}=E_aI_a=242.4(19)=\boxed{4606\ \text{W}}.$$
Developed torque (part c). With $\omega_1=2\pi(1600)/60=167.55$ rad/s,$$\tau=\frac{P_{out}}{\omega_1}=\frac{4606}{167.55}=\boxed{27.5\ \text{N}\cdot ext{m}}.$$
Efficiency (part d). The input from the mains is $P_{in}=VI_L=250(20)=5000$ W, so$$\eta=\frac{P_{out}}{P_{in}}=\frac{4606}{5000}=\boxed{92.1\%}.$$
Inserted field resistance (part e). To reduce the field current to 0.8 A the total field-circuit resistance must be $V/I_{f2}=250/0.8=312.5\ \Omega$, so the added resistance and its dissipation are$$R_{ext}=312.5-250=\boxed{62.5\ \Omega},\qquad P_{ext}=I_{f2}^2R_{ext}=0.8^2(62.5)=\boxed{40\ \text{W}}.$$
New armature current (part f). Constant load torque with $\phi\propto I_f$ gives $I_{f1}I_{a1}=I_{f2}I_{a2}$:$$I_{a2}=\frac{I_{f1}I_{a1}}{I_{f2}}=\frac{1.0(19)}{0.8}=\boxed{23.75\ \text{A}}.$$Weakening the field forces more armature current to hold the same torque.
New speed (part g). The new back-emf is $E_{a2}=250-23.75(0.4)=240.5$ V. Since $E_a\propto I_f n$,$$n_2=n_1\frac{E_{a2}}{E_{a1}}\frac{I_{f1}}{I_{f2}}=1600\left(\frac{240.5}{242.4}\right)\left(\frac{1.0}{0.8}\right)=\boxed{1984\ \text{rpm}}.$$ Field weakening raises the speed, as expected.
New efficiency (part h). The line current is now $I_{L2}=I_{a2}+I_{f2}=24.55$ A, so $P_{in}=250(24.55)=6138$ W and $P_{out}=E_{a2}I_{a2}=240.5(23.75)=5712$ W:$$\eta=\frac{5712}{6138}=\boxed{93.1\%}.$$
