This is homework. Given a 500W, 200V compound wound motor with constant shunt field and a series field that varies with armature current, how do we calculate the change in motor speed.
I know the equation for motor speed \$n\$
$$n \propto \frac{E}{\Phi} $$
therefore
$$\frac{n_{2}}{n_{1}} = \frac{E_{2}}{E_{1}}\frac{\Phi_{1}}{\Phi_{2}} $$
where the indices 1, 2 refer to the values in the initial and final states respectively. Here \$E\$ is induced emf and \$\Phi\$ is the net flux. Given the original speed \$n_{1}\$, the two armature currents \$I_{a, 1}\$ and \$I_{a, 2}\$ and the resistance values of the armature windings \$R_{a}\$ and series coils \$R_{series}\$, I can work out the induced emf with the motor equation $$E = V_{t} - I_{a}(R_{a} + R_{series})$$ where \$V_{t}\$ is the terminal voltage of 200V.
But I cannot lose the unknown flux \$\Phi\$. Assuming cumulative compound \$\Phi = \Phi_{shunt} + \Phi_{series}\$ and
$$\Phi_{2} = \Phi_{1} + \Delta\Phi = \Phi_{shunt, 1} + \Phi_{series, 1} + \Delta\Phi_{series} $$ If we are told that \$I_{a, 2} = 2I_{a, 1}\$, then \$\Delta\Phi_{series} = \Phi_{series, 1}\$ and $$\frac{n_{2}}{n_{1}} = \frac{E_{2}}{E_{1}}\frac{\Phi_{shunt, 1} + \Phi_{series, 1}}{\Phi_{shunt, 1} + 2\Phi_{series, 1}} $$
Without knowing the fluxes, how do we can solve this?
EDIT: possible solution
$$\frac{n_{2}}{n_{1}} = \frac{E_{2}}{E_{1}}\frac{\Phi_{1}}{\Phi_{2}} = \frac{E_{2}}{E_{1}}\frac{I_{1}}{I_{2}} = \frac{E_{2}}{E_{1}}\frac{I_{1}}{I_{1} + \Delta I} $$
My assumption is that the net flux will remain proportional to the line current, even though the shunt flux is constant.
