How do I predict volume loss due to evaporation when boiling water?

Question

Suppose I have a pot with diameter $D$ containing a volume of water $V$, being heated by a flame under it. If the ambient air temperature is $T$ and relative humidity is $R$, how can I calculate the expected rate of loss due to evaporation over a period of time, $t$? Assume the period of time begins once the water begins boiling.

Edit: For simplicity's sake, also assume the flame is just hot enough to get the water boiling.

Answer 1

You need to know the rate of heat given by the flame to the water.

Suppose the flame transfers $h$ kJ/s to the water. The latent heat of evaporation of water is $2260\ \mathrm{kJ/kg}$.

For energy balance, the heat given to the water must be equal to the amount of heat required to convert water into steam.

$$ h = \dot{m} \times 2260 \\ \therefore \dot{m} = \frac{h}{2260}\ \mathrm{kg/s} $$

If you say the rate of heat transfer doesn't matter (ignore the flame and assume the water is at a certain temperature and stays at that temperature throughout the experiment),

$$ \dot{m} = \frac{\Theta A (x_\mathrm s - x)}{3600}\ \mathrm{kg/s} $$

Where
$\Theta = (25 + 19 v)$ is the evaporation coefficient ($\mathrm{kg/(m^2\cdot h)}$). This is an empirical equation, so you can't derive it from first principles.

$v$ is the velocity of air just above the surface of the water ($\mathrm{m/s}$)

$A$ is the surface area of the water ($\mathrm{m^2}$)

$x_\mathrm s$ is the humidity ratio in saturated air at the same temperature as the water surface ($\mathrm{kg}$ $\mathrm{H_2O}$ in $\mathrm{kg}$ dry air)

$x$ is the humidity ratio in the air ($\mathrm{kg}$ $\mathrm{H_2O}$ in $\mathrm{kg}$ dry air)

It's fairly straightforward to find $x$ and $x_\mathrm s$ from the relative humidity and the Mollier chart.