Field diverter to raise series motor speed

A diverter is a resistance $R_d$ in parallel with the series field. It takes part of the armature current, so the field current and flux fall and the speed rises.

Given: $V = 500$ V, $N_1 = 500$ rpm, $I_1 = 60$ A, $R_{se} = 0.3\ \Omega$, $R_a = 0.2\ \Omega$, $N_2 = 600$ rpm, torque constant, no saturation ($\phi \propto I_{se}$).

          +--Rse--+
  + o-----|       |----Ra---(M)----o -
          +--Rd---+
   Ia2 splits: Ise2 in Rse, Ia2-Ise2 in Rd

Step 1: Initial back emf

$$E_{b1} = 500 - 60 \times (0.3 + 0.2) = 470\ \text{V}$$

Step 2: Torque condition

$T \propto \phi I_a \propto I_{se} I_a$:

$$I_{se2} I_{a2} = 60 \times 60 = 3600$$

Step 3: Speed condition

$$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{I_{se1}}{I_{se2}} = 1.2$$

with $E_{b2} = 500 - 0.2 I_{a2} - 0.3 I_{se2}$ (drop across field and diverter $= 0.3 I_{se2}$).

Substituting $I_{se2} = 3600/I_{a2}$:

$$\begin{aligned} E_{b2} \times \frac{60\, I_{a2}}{3600} &= 1.2 \times 470 = 564 \\ E_{b2} I_{a2} &= 33\,840 \\ 500 I_{a2} - 0.2 I_{a2}^2 - 1080 &= 33\,840 \\ 0.2 I_{a2}^2 - 500 I_{a2} + 34\,920 &= 0 \\ I_{a2} &= 71.91\ \text{A} \end{aligned}$$

(The other root, about 2428 A, is not practical.)

$$I_{se2} = \frac{3600}{71.91} = 50.06\ \text{A}, \quad E_{b2} = 470.60\ \text{V}$$

Check: $N_2 = 500 \times \dfrac{470.60}{470} \times \dfrac{60}{50.06} = 600$ rpm.

Step 4: Diverter resistance

$$\begin{aligned} I_d &= I_{a2} - I_{se2} = 71.91 - 50.06 = 21.84\ \text{A} \\ R_d &= \frac{I_{se2} R_{se}}{I_d} = \frac{50.06 \times 0.3}{21.84} = 0.688\ \Omega \end{aligned}$$

Answer: Diverter resistance $R_d \approx 0.69\ \Omega$ (armature current becomes 71.9 A, field current 50.1 A).