The power output of an ideal Magnetohydrodynamic (MHD) power generator is related to the velocity of the conducting fuel through fundamental electromagnetic principles.

MHD Generator Power Derivation

In an MHD generator, a conducting fluid (like plasma) moves with velocity $u$ through a magnetic field $B$. This induces an electromotive force (EMF) across the fluid, perpendicular to both $u$ and $B$. The induced EMF ($\mathcal{E}$) can be expressed as:

$ \mathcal{E} = B L u $

where $L$ is the characteristic length of the generator duct perpendicular to $B$ and $u$.

Assuming the conducting fluid has an internal resistance $r$, and it is connected to an external load resistance $R_L$, the current $I$ flowing in the circuit is given by Ohm's Law:

$ I = \frac{\mathcal{E}}{r + R_L} $

The power $P$ delivered to the load resistance $R_L$ is:

$ P = I^2 R_L = \left(\frac{\mathcal{E}}{r + R_L}\right)^2 R_L $

To find the maximum power output ($P_{\text{max}}$), we need to find the condition where $P$ is maximized with respect to $R_L$. This occurs when the load resistance is equal to the internal resistance ($R_L = r$).

Substituting $R_L = r$ into the power equation gives the maximum power:

$ P_{\text{max}} = \left(\frac{\mathcal{E}}{r + r}\right)^2 r = \left(\frac{\mathcal{E}}{2r}\right)^2 r = \frac{\mathcal{E}^2}{4r} $

Now, substitute the expression for $\mathcal{E}$ back into the $P_{\text{max}}$ equation:

$ P_{\text{max}} = \frac{(B L u)^2}{4r} = \frac{B^2 L^2 u^2}{4r} $

For a given ideal MHD generator, $B$, $L$, and $r$ are constants. Therefore, the maximum power output $P_{\text{max}}$ is directly proportional to the square of the velocity ($u$) of the conducting fuel.

$ P_{\text{max}} \propto u^2 $