22-Elec-A6 Power Systems and Machines · December 2016
Question 4 of 6: 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, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).
Given. A 30 hp, 240 V, 1150 rpm dc shunt motor, $\eta=88.5\%$.
Given data
Rated output
Terminal V
Rated speed
Ra
Rf
η
30 hp (22.38 kW)
240 V
1150 rpm
0.096 Ω
93.6 Ω
88.5 %
Find. Armature current, output and developed torque, developed power, series resistances for a starting-torque limit and for speed control, and the current/speed at reduced load.
DC shunt motor: field ($R_f$) and armature ($R_a$, emf $E_a$) branches across the 240 V supply.
Approach. Get line current from rated power and efficiency, subtract field current for armature current, then use $E_a=V-I_aR_a$ and $T\propto\Phi I_a$ (constant field flux). Starting and speed-control resistances follow from the armature loop equation with the appropriate $E_a$.
Armature current (part a). Rated output $P_{out}=30(746)=22\,380$ W, so input $P_{in}=P_{out}/\eta=25\,288$ W and line current $I_L=P_{in}/V=105.4$ A. Field current $I_f=V/R_f=240/93.6=2.56$ A, hence
$$\boxed{I_a=I_L-I_f=105.4-2.56=102.8\ \text{A}.}$$
Output torque (part b). $\omega_m=1150(2\pi/60)=120.4$ rad/s, so $T_{out}=P_{out}/\omega_m=22\,380/120.4=185.8$ N·m.
Developed power and torque (parts c, d). $E_a=V-I_aR_a=240-102.8(0.096)=230.1$ V, so
$$P_{dev}=E_a I_a=230.1(102.8)=23\,658\ \text{W},\qquad T_{dev}=\frac{P_{dev}}{\omega_m}=\frac{23\,658}{120.4}=196.5\ \text{N}\cdot\text{m}.$$
Starting resistance (part e). With constant field flux, torque is proportional to armature current, so a starting torque of 200% rated needs $I_{a,\text{start}}=2 I_a=205.6$ A. At standstill $E_a=0$, so
$$\boxed{R_{ext}=\frac{V}{I_{a,\text{start}}}-R_a=\frac{240}{205.6}-0.096=1.071\ \Omega.}$$
Reduced load — new armature current (part f). At 40% rated torque, $I_a'=0.40 I_a=41.1$ A (torque $\propto I_a$).
New speed (part g). $E_a'=V-I_a'R_a=240-41.1(0.096)=236.0$ V. With flux constant, speed $\propto E_a$:
$$\boxed{n'=n\,\frac{E_a'}{E_a}=1150\,\frac{236.0}{230.1}=1180\ \text{rpm}.}$$
Series resistance to hold rated speed (part h). To run at 1150 rpm the emf must return to $E_a=230.1$ V while carrying $I_a'=41.1$ A:
$$\boxed{R_{ext}=\frac{V-E_a}{I_a'}-R_a=\frac{240-230.1}{41.1}-0.096=0.144\ \Omega.}$$
