DC Shunt Motor — Field Weakening: 22-Elec-A6 Power Systems and Machines, May 2013, Q1

Question 1: DC Shunt Motor — 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 DC shunt motor operating from a fixed 440 V supply, whose flux is then weakened and whose developed torque is forced up.

Given data
Supply voltage	$V_t = 440\ \text{V}$
Initial armature current	$I_{a1} = 20\ \text{A}$
Initial speed	$N_1 = 500\ \text{rpm}$
Armature resistance	$R_a = 0.6\ \Omega$
New flux	$\phi_2 = 0.70\,\phi_1$
New torque	$T_2 = 1.40\,T_1$

Find. The new armature current $I_{a2}$ and the new speed $N_2$.

Approach. Developed torque obeys $T = k\,\phi\,I_a$, which fixes the new current; the back-emf $E_b = V_t - I_a R_a$ together with $E_b = k\,\phi\,N$ then fixes the new speed.

New armature current from the torque ratio. Since $T = k\,\phi\,I_a$, the ratio of the two operating points is $$\frac{T_2}{T_1} = \frac{\phi_2 I_{a2}}{\phi_1 I_{a1}} \;\Rightarrow\; 1.40 = \frac{(0.70\,\phi_1)\,I_{a2}}{\phi_1\,(20)}.$$ Solving, $I_{a2} = \dfrac{1.40 \times 20}{0.70}$, so $$\boxed{I_{a2} = 40\ \text{A}}$$
Back-emf before the change. With the armature-circuit KVL $E_b = V_t - I_a R_a$, $$E_{b1} = 440 - (20)(0.6) = 428\ \text{V}.$$
Back-emf after the change. Using the new current, $$E_{b2} = 440 - (40)(0.6) = 416\ \text{V}.$$
New speed from the emf relation. Because $E_b = k\,\phi\,N$, the speed ratio is $\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}\cdot\dfrac{\phi_1}{\phi_2}$. Substituting, $$N_2 = 500 \times \frac{416}{428} \times \frac{1}{0.70} = 694\ \text{rpm}.$$ $$\boxed{N_2 \approx 694\ \text{rpm}}$$

The speed rises by roughly 39 % even though the terminal voltage is unchanged: weakening the field is the classic way a shunt motor runs above base speed. The current has to double so that the weaker flux can still deliver the demanded 40 % more torque.

Question 1 — results
Quantity	Value
New armature current $I_{a2}$	40 A
New speed $N_2$	694 rpm

Topic. Speed–torque control of the DC shunt motor by field weakening.

Key relations. $T = k\,\phi\,I_a$; $E_b = V_t - I_a R_a$; $E_b = k\,\phi\,N$.

Why this works. In a separately-excited or shunt machine the two machine constants ($k$) are identical, so ratios of torque and of back-emf eliminate the constant and the geometry entirely. Torque is set by the product $\phi I_a$, so a 30 % flux cut demands a proportionally larger current to raise torque. Speed follows $N \propto E_b/\phi$; the numerator drops only slightly (the extra $I_a R_a$ drop is small against 440 V) while the denominator falls 30 %, so the net effect is a substantial speed increase.

Common pitfall. Forgetting that $E_b$ itself changes when $I_a$ changes — students often hold $E_b$ fixed and get the wrong speed. Always recompute $E_b = V_t - I_a R_a$ with the new current before applying the speed ratio.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 9 (DC motor speed control).

Question 1 of 6: DC Shunt Motor — Field Weakening

Question 1: DC Shunt Motor — Field Weakening (20 marks)