Question 1 of 5: DC Shunt Motor — Rating and Field Weakening

Question 1: DC Shunt Motor — Rating and Field Weakening (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. A separately-shunt-excited DC motor operating from rated terminal voltage, first at its nameplate operating point and then with a weakened field at constant load torque.

Find. Armature current, output power, developed torque and efficiency at the initial point; then the armature current, speed, efficiency and the size and rating of the field rheostat after the field is weakened to 0.8 A at constant load torque.

Approach. Split the line current into field and armature parts, find the back-emf from the armature loop, and use $P_{dev}=E_aI_a$ and $T=P_{dev}/\omega_m$; then exploit the linear magnetic circuit ($K_a\Phi \propto I_f$) with the torque held constant to track the new operating point.

1. Split the line current. The field is directly across the supply, so $I_{f1}=V_t/R_f = 250/250 = 1\ \text{A}$, and by KCL $$I_{a1}=I_{L1}-I_{f1}=21-1=\boxed{20\ \text{A}}.$$
2. Back-emf and developed (output) power. From the armature loop $E_{a1}=V_t-I_{a1}R_a = 250-(20)(0.4)=242\ \text{V}$. With rotational losses negligible the developed power equals the shaft output: $$P_{out}=P_{dev}=E_{a1}I_{a1}=(242)(20)=\boxed{4840\ \text{W}}\ (\approx 6.5\ \text{hp}).$$ The motor is drawing well above its 3-hp nameplate at this point.
3. Developed torque. With $\omega_{m1}=2\pi n_1/60 = 2\pi(1600)/60 = 167.55\ \text{rad/s}$, $$T_{dev}=\frac{P_{dev}}{\omega_{m1}}=\frac{4840}{167.55}=\boxed{28.89\ \text{N}\cdot\text{m}}.$$
4. Efficiency. Input power $P_{in}=V_tI_{L1}=(250)(21)=5250\ \text{W}$, so $$\eta_1=\frac{P_{out}}{P_{in}}=\frac{4840}{5250}=\boxed{92.19\%}.$$
5. Field-weakening constant $K_a\Phi$. Since $E_a=K_a\Phi\,\omega_m$, the initial flux constant is $K_a\Phi_1 = E_{a1}/\omega_{m1}=242/167.55 = 1.4443\ \text{V}\cdot\text{s}$. The magnetic circuit is linear, so $\Phi\propto I_f$ and $K_a\Phi_2 = K_a\Phi_1\,(I_{f2}/I_{f1}) = 1.4443(0.8/1.0)=1.1555\ \text{V}\cdot\text{s}$.
6. New armature current (torque constant). With $T=K_a\Phi\,I_a$ held constant, $$I_{a2}=\frac{T_{dev}}{K_a\Phi_2}=\frac{28.89}{1.1555}=\boxed{25.0\ \text{A}}\qquad\left(=I_{a1}\tfrac{I_{f1}}{I_{f2}}=\tfrac{20}{0.8}\right).$$ Weakening the field forces more armature current to hold the same torque.
7. New speed. New back-emf $E_{a2}=V_t-I_{a2}R_a = 250-(25)(0.4)=240\ \text{V}$, hence $$\omega_{m2}=\frac{E_{a2}}{K_a\Phi_2}=\frac{240}{1.1555}=207.7\ \text{rad/s}\;\Rightarrow\; n_2=\boxed{1983\ \text{rpm}}.$$ The machine speeds up, as expected when the field is weakened.
8. New efficiency. The new line current is $I_{L2}=I_{a2}+I_{f2}=25+0.8=25.8\ \text{A}$, so $P_{in2}=V_tI_{L2}=(250)(25.8)=6450\ \text{W}$ while the shaft output is $P_{out2}=E_{a2}I_{a2}=(240)(25)=6000\ \text{W}$. Thus $$\eta_2=\frac{6000}{6450}=\boxed{93.02\%}.$$
9. Field rheostat value and rating. To pass 0.8 A the total field resistance must be $V_t/I_{f2}=250/0.8=312.5\ \Omega$, so the inserted resistance is $$R_{ext}=312.5-250=\boxed{62.5\ \Omega}.$$ Its power dissipation is $P_{ext}=I_{f2}^{2}R_{ext}=(0.8)^2(62.5)=\boxed{40\ \text{W}}$, which sets the minimum wattage rating of the rheostat.