Question 5 of 6: DC Shunt Motor Performance

Question 5: DC Shunt Motor Performance (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. DC shunt motor, 240 V, 20 hp, 2000 rpm rated, $\eta_{FL}=76\%$, $R_A=0.2\ \Omega$, field current $I_f=1.8$ A.

Find. (a) full-load line current; (b) full-load shaft torque; (c) series starting resistance for $I_{start}=1.5I_{a,FL}$; (d) no-load speed at $I_{L,NL}=23$ A; (e) speed regulation.

Approach. Line current follows from input power ($=P_{out}/\eta$). Torque from $P_{out}/\omega_m$. Starting resistance limits the armature current at zero back-emf. No-load speed scales with back-emf $E_a$ (constant flux), and speed regulation compares no-load to full-load speed.

1. Full-load line current (part a). $P_{out}=20(746)=14\,920$ W; $P_{in}=P_{out}/\eta=14\,920/0.76=19\,632$ W. $$\boxed{I_L=\frac{P_{in}}{V}=\frac{19\,632}{240}=81.8\ \text{A}}$$ Armature current $I_{a,FL}=I_L-I_f=81.8-1.8=80.0$ A.
2. Full-load shaft torque (part b). $\omega_m=2000\cdot\dfrac{2\pi}{60}=209.4\ \text{rad/s}$, $$\boxed{T=\frac{P_{out}}{\omega_m}=\frac{14\,920}{209.4}=71.24\ \text{N}\!\cdot\!\text{m}}$$
3. Starting resistance (part c). At standstill $E_a=0$, so $I_{a,start}=\dfrac{V}{R_A+R_{ext}}$. Setting $I_{a,start}=1.5(80.0)=120$ A, $$R_A+R_{ext}=\frac{240}{120}=2.0\ \Omega\ \Rightarrow\ \boxed{R_{ext}=2.0-0.2=1.8\ \Omega}$$
4. No-load speed (part d). No-load armature current $I_{a,NL}=23-1.8=21.2$ A. Back-emfs (flux constant, since $I_f$ fixed): $$E_{a,NL}=V-I_{a,NL}R_A=240-21.2(0.2)=235.76\ \text{V},\quad E_{a,FL}=240-80(0.2)=224.0\ \text{V}$$ Since $n\propto E_a$: $\ \boxed{n_{NL}=2000\cdot\dfrac{235.76}{224.0}=2105\ \text{rpm}}$.
5. Speed regulation (part e). $$\boxed{\text{SR}=\frac{n_{NL}-n_{FL}}{n_{FL}}=\frac{2105-2000}{2000}=5.25\%}$$