Series motor voltage to drive fan at 300 rpm

For a series motor the field current equals the armature current, so (below saturation) $\phi \propto I_a$. Then $T \propto \phi I_a \propto I_a^2$ and $N \propto E_b/\phi \propto E_b/I_a$.

Given: $V_1 = 400$ V, $I_1 = 20$ A, $N_1 = 200$ rpm, $R_a = 0.5\ \Omega$, $R_{se} = 0.5\ \Omega$, so the total series resistance is $R = R_a + R_{se} = 1\ \Omega$. $N_2 = 300$ rpm. Load torque $T \propto N^2$.

Step 1: New current

The motor torque must equal the fan torque:

$$T \propto I^2 \quad \text{and} \quad T \propto N^2 \;\Rightarrow\; I \propto N$$ $$\begin{aligned} I_2 &= I_1 \times \frac{N_2}{N_1} = 20 \times \frac{300}{200} = 30\ \text{A} \end{aligned}$$

Step 2: Back emf at 200 rpm

$$\begin{aligned} E_{b1} &= V_1 - I_1 (R_a + R_{se}) \\ &= 400 - 20 \times 1 = 380\ \text{V} \end{aligned}$$

Step 3: Back emf at 300 rpm

Since $N \propto E_b / \phi$ and $\phi \propto I$:

$$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2}$$ $$\begin{aligned} E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \times \frac{I_2}{I_1} \\ &= 380 \times \frac{300}{200} \times \frac{30}{20} \\ &= 380 \times 1.5 \times 1.5 = 855\ \text{V} \end{aligned}$$

Step 4: Required applied voltage

$$\begin{aligned} V_2 &= E_{b2} + I_2 (R_a + R_{se}) \\ &= 855 + 30 \times 1 = 885\ \text{V} \end{aligned}$$

Assumption: the magnetic circuit is unsaturated ($\phi \propto I$) and brush drop is neglected.

Answer: To drive the fan at 300 rpm the motor needs an applied voltage of 885 V and takes a current of 30 A.