Question

A 50 KW, 230 V DC shunt motor has an armature resistance of $0.1\Omega$ and a field resistance of $200\Omega$ . It runs on no load at a speed of 1400 rpm, drawing a current of 10 A from the mains. When delivering a constant load, the motor draws a current of 200 A from the mains. Assume that the armature reaction cause a reduction in flux/pole of 4% of its no-load value. Then the torque developed at this load in N-m is

• A. 298.4
• B. 810.28
• C. 292.8
• D. 316.42

Answer 1

The correct option is A 298.4

$E_a\propto {\varphi }_1{N}_1$

$\Rightarrow$ $I_f=\frac{230}{200}=1.15A$

$(V-IR)\propto {\varphi }_1{N}_1$

$[230-(10-1.15\left)\right(0.1\left)\right]\propto 1400{\varphi }_1....\left(1\right)$

$[230-(200-1.15\left)\right(0.1\left)\right]\propto N_2{\varphi }_2....\left(2\right)$

$\text{Equation (2) divided by (1),}$

$\frac{210.1}{229.1}=\frac{N_2}{1400}\times 0.96$

$\Rightarrow$ $N_2=1337rpm$

$\therefore \text{Torque developed}(T_d\text{)}=\frac{210.115\times (200-1.15)}{\left(\frac{2\pi \times 1337}{60}\right)}$$=298.4Nm$