Shunt DC motor with field weakening: 22-Elec-A6 Power Systems and Machines, May 2017, Q4

Question 4: Shunt DC motor with 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 250 V, 1700 rpm shunt motor on a constant-torque load, with the data below and negligible rotational loss; the field circuit is then re-set from 250 $\Omega$ to 200 $\Omega$.

Given data
Quantity	Value
Ratings	250 V, 1700 rpm, shunt
Line current (initial)	41.6 A
$R_a$ / $R_f$	0.4 Ω / 250 Ω
Load	constant torque
New field resistance	200 Ω

Find. Armature current, output power, developed torque and efficiency at rated field; then armature current, line current, speed and output power after the field resistance is lowered.

Shunt DC motor: field circuit R_f across the 250 V line, armature (R_a in series with back-EMF E_a) in parallel.

Approach. Split line current into field and armature parts, get $E_a=V_t-I_aR_a$ and hence power and torque; for the weakened field use flux $\propto I_f$ (linear), torque constant, and $E_a=K_a\Phi\omega$.

(a) Armature current. Field current $I_f=V_t/R_f=250/250=1.0$ A, so $I_a=I_L-I_f=41.6-1.0=40.6$ A.
(b) Output power. Back-EMF $E_a=V_t-I_aR_a=250-40.6(0.4)=233.76$ V. With rotational losses negligible, $P_{out}=P_{dev}=E_aI_a=233.76(40.6)=9491$ W.
(c) Developed torque. $\omega_m=2\pi(1700)/60=178.0$ rad/s, so $$\boxed{\,T_{dev}=P_{dev}/\omega_m=53.31\ \text{N}\cdot\text{m}\,}$$
(d) Efficiency. $P_{in}=V_tI_L=250(41.6)=10{,}400$ W, so $\eta=9491/10{,}400=91.26\%$.
New field. $I_{f2}=250/200=1.25$ A. A linear magnetic circuit gives $\Phi_2/\Phi_1=I_{f2}/I_{f1}=1.25$ (the field is strengthened, not weakened, because $R_f$ was reduced).
(e) New armature current. Constant torque means $K_a\Phi I_a$ is fixed, so $I_{a2}=I_{a1}\,\Phi_1/\Phi_2=40.6/1.25=32.48$ A.
(f) New line current. $I_{L2}=I_{a2}+I_{f2}=32.48+1.25=33.73$ A.
(g) New speed. $E_{a2}=250-32.48(0.4)=237.01$ V. From $E_a=K_a\Phi\omega$, $n_2=n_1\dfrac{E_{a2}}{E_{a1}}\dfrac{\Phi_1}{\Phi_2}=1700\dfrac{237.01}{233.76}\dfrac{1}{1.25}=1378.9$ rpm (the stronger field slows the motor).
(h) New output power. $P_{out2}=E_{a2}I_{a2}=237.01(32.48)=7698$ W (equivalently $T\omega_2$ with the same 53.31 N·m torque).

Question 4 results
Quantity	Rated field	Field $R_f$=200 Ω
Armature current	40.6 A	32.48 A
Line current	41.6 A	33.73 A
Back-EMF $E_a$	233.76 V	237.01 V
Speed	1700 rpm	1378.9 rpm
Output power	9491 W	7698 W
Developed torque	53.31 N·m	53.31 N·m
Efficiency	91.26 %	—

Topic: Shunt DC motor and field control

Key relations: $I_L=I_f+I_a$; $E_a=V_t-I_aR_a=K_a\Phi\omega$; $T=K_a\Phi I_a$; linear circuit $\Phi\propto I_f$.

Why this works: On a shunt motor the field is fixed by the line voltage and $R_f$. Reducing $R_f$ raises $I_f$ and flux; a constant load torque then needs less armature current ($T=K_a\Phi I_a$), and since $E_a$ hardly changes the speed drops in proportion to flux ($\omega\propto E_a/\Phi$). Field weakening (larger $R_f$) does the opposite — it speeds the motor up.

Common pitfall: Assuming a smaller $R_f$ weakens the field; it strengthens it. Also treating speed as $\propto E_a$ alone, forgetting the flux term, or forgetting to remove $I_f$ from the line current.

Source: Chapman, Electric Machinery Fundamentals, 5th ed., ch. 8 (shunt DC motor, speed control by field resistance).

Question 4 of 5: Shunt DC motor with field weakening

Question 4: Shunt DC motor with field weakening (20 marks)