Shunt motor: field resistance to raise speed to 1100 rpm

Assume the flux is proportional to the field current (unsaturated). The load torque is constant, so $T\propto\Phi I_a$ is constant. Let $R_x$ be the added field resistance.

Initial condition

$$I_{f1} = \frac{250}{125} = 2\ \text{A}, \qquad I_{a1} = 25 - 2 = 23\ \text{A}, \qquad E_{b1} = 250 - 23(0.5) = 238.5\ \text{V}$$

New condition (1100 rpm)

Let $I_{f2} = s$ (A).

Constant torque: $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $2\times 23 = s\,I_{a2}$:

$$I_{a2} = \frac{46}{s}$$

Back emf:

$$E_{b2} = 250 - 0.5I_{a2} = 250 - \frac{23}{s}$$

Speed ratio: $E_b\propto\Phi N$:

$$E_{b2} = E_{b1}\times\frac{\Phi_2}{\Phi_1}\times\frac{N_2}{N_1} = 238.5\times\frac{s}{2}\times\frac{1100}{900} = 145.750\,s$$

Equating:

$$250 - \frac{23}{s} = 145.750\,s \;\Rightarrow\; 145.750\,s^2 - 250s + 23 = 0$$ $$s = \frac{250 + \sqrt{250^2 - 4(145.750)(23)}}{2(145.750)} = 1.6177\ \text{A}$$

(The larger root is taken; the smaller one gives an unrealistically weak field.)

Added resistance

$$R_f + R_x = \frac{250}{s} = \frac{250}{1.6177} = 154.54\ \Omega$$ $$R_x = 154.54 - 125 = 29.54\ \Omega$$

Check: $I_{a2} = 46/1.6177 = 28.44$ A, $E_{b2} = 250 - 0.5(28.44) = 235.78$ V, and $145.750(1.6177) = 235.78$ V.

Answer: additional field resistance $\approx 29.54\ \Omega$.