How do I calculate rate of water evaporation at various water temps?

Question

I am trying to figure out how many gallons of fresh water will evaporate per hour in a large tank at various water temperatures?

Specifically,

• surface area is 250,000 sq ft surface area (500' x 500')
• depth is 10 feet
• wind speed is 10 mph
• air temp is 70 deg F
• humidity is 30%

The water temp scenarios I'd like to compare are 80 deg F, 160 deg F, and 212 deg F.

My question is, "what is the formula for calculating the rate of water evaporation (e.g., gallons per day) given the above scenario"?

Answer 1

According to this engineering website, the rate of water evaporation is given by $$g_s=\frac{(25+19v) A(x_s - x)}{3600}$$ where $v$ is the air velocity above the water, $A$ is the surface area of the water, $x_s$ is the maximum humidity ratio of saturated air, $x$ is the humidity ratio for air.

Note that this equation gives the rate of evaporation in kg s$^{-1}$ and if you need to get this rate in terms of kg day$^{-1}$ you would need to multiply the RHS by $(3600\times 24)$ or for kg hr$^{-1}$ multiply by $3600$.

Note that you have included the water temperature, but that and air temperature is taken care of in the humidity ratios above. So you will need to obtain the values for $x_s$ and $x$ (see link above and links therein to find out how to obtain these values).