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

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.

• I don't know any formula for this, but wouldn't the heat of the flame and the material of the pot also play a role? A conductive material would heat the water faster and so will a stronger flame. – Jerry Dec 30 '13 at 15:15
• Ah, yes, the strength of the flame certainly would, but I'm not sure if the material would matter. If the water is just hot enough to boil and maintain that boil, then whether or not the pot's material conducts heat well shouldn't matter - it's already conducting enough heat. – Michael Moussa Dec 30 '13 at 15:44
• Right, but I also said to assume $t$ begins once the water starts boiling. So whether it takes 10 minutes or an hour to get up to the boil, my scenario is concerned only with the volume loss once the boil begins. – Michael Moussa Dec 30 '13 at 15:48
• The most important thing is the rate of heat input to the boiling water. The air temperature, humidity, and diameter are mostly irrelevant. – DumpsterDoofus Dec 30 '13 at 16:42
• The vaporization enthalpy of liquid water is approximately 2200 kJ/liter, which you may find helpful. – DumpsterDoofus Dec 30 '13 at 16:47

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 $2260$ 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.

$$h = \dot{m} \times 2260 \\ \therefore \dot{m} = \frac{h}{2260} \text{ kg/sec}$$

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),

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

Where
$\Theta = (25 + 19 v)$ is the evaporation coefficient ($kg/m^2\cdot hr$). 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 ($m/s$)

$A$ is the surface area of the water ($m^2$)

$x_s$ is the humidity ratio in saturated air at the same temperature as the water surface (kg H2O in kg Dry Air)

$x$ is the humidity ratio in the air (kg H2O in kg Dry Air)

It's fairly straightforward to find $x$ and $x_s$ from the relative humidity and the Mollier chart

• Since I knew my burner was rated for 22,000 BTU/hr, I went with your first equation. 22,000 BTU/hr = ~6.45kJ/s, and 6.45 kJ/s / 2,260 = ~0.00285kg/s. That came out to roughly 10.274 kg/h (or 2.7gal/hr). That seemed a little high, but I remember reading somewhere that I could realistically expect ~50% heat loss. That would bring the gal/hr down to 1.35, and looks a lot more reasonable. Thank you for your detailed answer! – Michael Moussa Jan 2 '14 at 16:34
• Nice! What is BTU and how do you find it? Thank you! – Ciprian Tomoiagă Dec 16 '19 at 11:58