Saturday, July 20, 2013

Water Evaporation Rate Per Surface Area

The rate of evaporation of water surface area per meter per second or evaporation flux can be calculated by using the Irving Langmuir formula.

The units is kg/m2/second. If the evaporation rate is only in units of kg/m2, not per second, then it is called as the evaporation rate.

It is also known as Hertz–Knudsen equation, and also known as Knudsen-Langmuir equation.

The actual formula was used by Irving Langmuir to measure vapor pressure by measuring the rate of evaporation, in this article the formula is used in reverse to determine the evaporation rate with a known vapor pressure.

According to Irving Langmuir, the rate at which water molecules are lost due to evaporation into atmosphere is equal to the rate at which water vapor molecules from air hit the surface of water if it is in equilibrium. In equilibrium, evaporation rate and condensation rate will be the same.

dm / dt = mass flow rate (kg) in a certain area (m-2) in one second (second, s-1), so that the unit is kg/m2/s.

Pv = vapor pressure at a given temperature, or boiling point pressure at certain temperature, in units of pascal (pa).

Pp = vapor partial pressure of the substance in the gas mixture, eg water vapor pressure in the air at a given temperature, in units of pascal (pa).

m = molecular weight (0,01801528 kg/mol).

R = ideal gas constant or constants Mendeleev = 8.314 Joules / (mol Kelvin).

As seen in the formula, if the vapor pressure (Pv) is greater than the partial pressure (Pp), there will be evaporation. Conversely, if the partial pressure (Pp) is greater than the vapor pressure (Pv), there will be condensation.

Langmuir formula does not calculate wind speed factor above water surface. Langmuir formula is using pressure and temperature parameters only, to calculate the rate of evaporation and condensation through water surface.

Example of calculation
For example, your house has a swimming pool with a water surface area of ​​100 square meters, temperature 30 degrees Celsius, the molecular weight of the swimming pool water is 0.018 kg / mol. Known air humidity is around 55%. Need to be calculated how much water evaporates every second.

Some ways to determine the vapor pressure at certain temperature
Vapor pressure (Pv) is also referred to as the boiling point, in which a liquid or solid substance at certain temperature will boil when given a certain pressure. Water will boil at a temperature of 30 degrees Celsius if the pressure is reduced or vacuumed, see the video here.

Pressure at which water boils can be calculated by Antoine formula:

where P is the pressure in units of Torr, and Tb is the temperature in Celsius.

For the temperature 30 degrees celsius we will get log10 P is equal to 31.74 Torrs. Because the Langmuir formula vapor pressure (Pv) in units of pascal, so the value is equal to 4,231.68 pascals.

Vapor pressure can also be estimated by Psychrometric diagram below. You do this by finding the humidity ratio (kg/kg') on the psychrometric diagram with a temperature of 30 degrees celsius and 100% humidity, and humidity ratio values ​​obtained at 0.027 kg/kg'. How to read the psychrometric diagram described below. Then the vapor pressure (Pv) is:

Pv = x / (0.62198 + x) * (atmospheric pressure, 101325 pa)

= 0.027 / (0.62198 + 0.027) * 101325

Pv = vapor pressure = 4,215.5 pascals

To crosscheck the vapor pressure can be compared to the Boiling Point Calculator, 4128 pascals obtained results are slightly different, possibly due to rounding.

Calculate the partial pressure (Pp) of water vapor in the air
Obtained from Psychrometric chart below, for air with a temperature of 30 degrees Celsius and 55% humidity, the humidity ratio obtained = x = 0.015 kg/kg' (kg water / kg dry air, or g water / g dry air).

How to read the psychrometric diagram is by finding the temperature 30 degrees Celsius on the bottom, drag an arrow to the top to reach the line of 55% humidity, and drag an arrow to the right which will indicate moisture ratio (x).

Partial pressure of water vapor in the air can be calculated with humidity ratio data:

Partial pressure = Pp = x / (0.62198 + x) . (atmospheric pressure; 101,325 pa)
= 0.015 / (0.62198 + 0.015) . 101325
Partial pressure = Pp = 2,386.064 pascals

Using the Irving Langmuir formula
With a temperature of 30 degrees Celsius is equal to 30 + 273.15 = 303.15 degrees Kelvin. Relative humidity is 55%, then the calculations by Irving Langmuir formula will be as below:

With a partial pressure of 2386.064 pascals, the evaporation flux will be 1.97 kg/m2/s, this is the maximum evaporation that can occur in that condition. So as the swimming pool has water surface area of ​​100 square meters, the water will evaporate rate is 197 kg per second. With the density of water 1kg/liter then water evaporate volume is 197 liters per second.

But the actual evaporation is lower, which is around 1 / 100,000 up to 1 / 1,000,000 smaller than the calculation. This is due to the presence of a layer of water vapor on the surface of pond water, which is almost saturated vapor. So that evaporation is much smaller. As quoted from Frank E. Jones's book titled Evaporation of Water with Emphasis on Applications and Measurements. Jones quoted this opinion from De Boer in The Dynamical Character of Adsorption book.

In Jones's book it is also stated that evaporation of sea water in the tropical region around 2 meters per year, or around 5.5 mm per day. As quoted from Patrick Paroline, evaporation of pond water is about 1/4 - 1/2 inch per day, or around 6.4 - 12.7 mm per day.

What if the air humidity increases?
Evaporation rate will be slower if the air contains moisture, or there is a partial pressure of water vapor in the air. Once water of swimming pool evaporates, the air at the surface of the pool soon becomes saturated aka humidity reaches 100%, thereby reducing the rate of evaporation to zero. If the surface of the pool in the breeze that swept humid air at the surface of the pool water, the evaporation rate will increase.

When the air is saturated by water vapor, the vapor pressure of water will be equal to its partial pressure in the air, so that evaporation rate will be equal to condensation rate as revealed by Irving Langmuir.

If the partial pressure greater than the vapor pressure of water, then the calculation will be negative indicating that condensation is occurring or condensation of moisture from the air into the pool, and the pool water will increase. Obvious example of this condition is in the morning, when air temperature is cool and become saturated by water vapor therefore causing condensation.

What happens if the air pressure is 0 or vacuum?
To broaden our perspective let's try to calculate the rate of evaporation if no atmospheric pressure alias vacuum . If the atmospheric pressure is reduced to vacuum, then the partial pressure of water vapor in the air at that point is zero aka no moisture, maximum evaporation will occur.

Thus obtained the rate of evaporation per square meter per second is 4.51 kg/m2/s, this is the maximum evaporation that can occur in that vacuum condition. Evaporation will be reduced if area above water surface is filled by vapor, evaporation will be zero if area above water surface is already saturated by vapor or maximum partial pressure above water surface.

Video of water boils at a temperature of about 28-30 degrees Celsius by reducing the air pressure or vacuumed, please see here.

Units of measurement will be very confusing for calculations using derived formulas as above. The following unit conversions can help:

1 atmosphere = 101325 pascals = 14.69595 psi = 29.92126 inches hg = 760 mm hg = 760 torrs

Temperature Fahrenheit to Celsius: (°F - 32) x 5/9 = °C