EMI BYU 3

Solution 3: MHD Generator Power

3. Power in MHD Generator

Step 1: Induced EMF and Internal Resistance

The conducting liquid moves with velocity $v$ in field $B$. The motional EMF across the plates (distance $d$) is:

$$ \mathcal{E} = vBd $$

The liquid between the plates acts as an internal resistor. The resistance is determined by length $d$, area $A$, and conductivity $\sigma$:

$$ r = \frac{1}{\sigma} \frac{\text{length}}{\text{area}} = \frac{d}{\sigma A} $$

Step 2: Circuit Current

The equivalent circuit is a voltage source $\mathcal{E}$ with internal resistance $r$ connected to load $R$.

$$ I = \frac{\mathcal{E}}{R + r} = \frac{vBd}{R + \frac{d}{\sigma A}} = \frac{vBd \sigma A}{\sigma A R + d} $$

Step 3: Power Dissipated

The power dissipated in the load $R$ is $P = I^2 R$.

$$ P = \left( \frac{vBd \sigma A}{\sigma A R + d} \right)^2 R $$ $$ P = \frac{(vBd)^2 \sigma^2 A^2 R}{(\sigma A R + d)^2} $$

Step 4: Source of Power

The current $I$ flowing through the liquid in the presence of magnetic field $B$ experiences a magnetic braking force $F_m = I d B$ opposing the flow. The external agent maintaining the liquid velocity must do mechanical work against this force. This mechanical work is converted into electrical energy.

$$ P = \frac{(vBd)^2 \sigma^2 A^2 R}{(\sigma A R + d)^2} $$

Power comes from the agency maintaining the liquid flow.