Mist Desalination

Back to Contents

I live in central California and, like my neighbors, I have a farmer’s appreciation of the value of water, especially during a drought. In those years when the rain falls too lightly and too little snow bestows its blessing upon the Sierra Nevada, we worry about water. True, we have a vast reservoir of water one hundred miles to the west, but if we were to take water directly from the Pacific Ocean, pump it over the coastal mountains, and distribute it across the San Joaquin Valley, we would quickly convert one of the most productive agricultural regions on Earth into a sterile, even toxic, desert. Before we can use water from the ocean, we must remove the salt and other minerals that make it unusable: we need a fast and cheap method of desalination.

At its most basic the obvious method of desalting seawater consists of evaporating a portion of the seawater and then condensing the vapor in a separate container. We want to imagine modifications to that concept until we have in mind the design of the most efficient evaporative desalinator that we can conceive. Put most simply, we want to imagine a device that gives us the maximum rates of evaporation and condensation of the water for minimum applied effort. The idea behind evaporative desalinization leads to the use of a fine mist for both evaporation and condensation.

In this version of desalination the fundamental part of the desalinator consists of two chambers, one above the other. In the lower chamber evaporation draws fresh water from seawater as vapor and in the upper chamber that vapor condenses into liquid fresh water. We need only elaborate that fundamental design to make it work.

The rate at which water evaporates and condenses conforms to Langmuir’s evaporation equation (see Appendix I). That equation tells us that the factors affecting the evaporation rate consist of the surface area of the liquid, the difference between the vapor pressure of the liquid and the partial pressure of the vapor pressure on the liquid, and the inverse square root of the absolute temperature of the liquid. We can ignore the latter factor because it gets swamped by changes in the vapor pressure due to changes in the temperature in conformity with the Clausius-Clapeyron equation (see Appendix II).

The first factor comes easily under our control. We need only pump seawater into the evaporation chamber through a fog nozzle, of the kind used on fire hoses, to break the stream of water into a mist of small droplets. A spherical drop containing one cubic meter of water has a diameter of 1.24 meters and has a surface area of 4.83 square meters, but if we break it up into droplets that each have a diameter of 100 microns (0.1 millimeter, the thickness of a sheet of 20-lb copier paper), that cubic meter of water has a surface area of 60,000 square meters. The simple act of spraying the water into the evaporation chamber increases its surface area by four orders of magnitude, which increase, according to Langmuir’s equation, increases the evaporation rate by four orders of magnitude.

We can also use that ploy in the condenser. There we spray cold fresh water into the chamber so that the vapor from the condensation will condense on it. As with evaporation, the increased surface area on the particles increases the rate of condensation.

Evaporation and condensation rates also depend on the driving force between the liquid and vapor phases. In this case we want to look at what we can do with the difference between the vapor pressure in the seawater and the partial pressure of the water vapor pressing on that liquid.

In the evaporator we want dry air, air containing as little water vapor as possible, to rise through the falling mist coming from nozzles set in the ceiling of the chamber. The dry air coming from the bottom of the chamber, with minimum partial pressure of water vapor, first encounters the droplets that have cooled the most through evaporation and, thus, have the lowest vapor pressure. As the air rises through the chamber it encounters progressively warmer droplets, in which the vapor pressure pushes progressively harder, so more vapor outgasses into the air. If the air rises sufficiently fast, the increasing partial pressure of the water vapor remains below the vapor pressure of the liquid in the droplets and vapor continues to come out of the droplets.

In the condenser we want wet air to rise through a falling fresh-water mist, on whose droplets the water vapor will condense. In this case the vapor-laden air encounters the warmest water first and the coldest water last. Here we want to ensure that, throughout the column, the partial pressure of the vapor exceeds the vapor pressure in the liquid. Eventually, of course, the system will satisfy that criterion, but we want that satisfaction to come at the earliest moment in the moist air’s traverse of the condenser.

The dried air passes out of the top of the condenser and into a duct where fans drive it back to the bottom of the evaporator. For maximum efficiency we design the desalinator with the condenser directly on top of the evaporator, so the duct simply runs along one wall of that stack. Because the air enters each chamber through the bottom, where liquid accumulates to get pumped out, we have the air enter each chamber through a set of flues mounted in the floor of the chamber. On each flue we put a conical cap to prevent liquid from falling into the flue.

At the top of each chamber we have two sets of vanes by each of the exit holes. The vanes in the first set tilt so that the air coming through them spins and flings any droplets of liquid outward to strike baffles mounted around the vanes and drip back into the chamber. The vanes in the second set then stop the spin so that the air passes straight into the next stage of its traverse through the device.

Now we have the problem of misplaced heat. The device has cold seawater coming into it, but we want to have seawater as hot as the boiling point of water. On the other hand we have the fresh water coming out of the condenser hot and we want it cold. Here the engineers have a straightforward solution: they pass the hot fresh water and the cold seawater in opposite directions through a heat exchanger. A standard design begins with each pipe segueing from a circular cross section to a square cross section and then splitting the square into thin rectangular channels separated from one another by a space as wide as each channel. The channels of the two pipes are then interleaved, thereby giving us alternating channels carrying seawater and fresh water, providing maximum thermal contact between the two streams. Inside each channel, helping to support it, the engineers put spirals like those in Archimedean screws: the spirals induce turbulence in the flow, thereby ensuring that all parts of the water come into contact with the walls and further enhance heat flow from the fresh stream into the salt stream.

If we make the heat exchanger long enough, we can make the temperature of the fresh water come arbitrarily close to that of the incoming seawater. But the seawater won’t reach the boiling temperature, so the seawater-carrying tubes must pass through a heat exchanger through which the system circulates hot oil. The oil, in turn, gains its heat from an external source, such as solar heating.

After leaving the heater, the hot seawater gets sprayed into the lower chamber of the desalinator. We want to makes the droplets of mist as small as possible in order to maximize the rate of evaporation. On the other hand, we want to make the droplets big enough to descend through the updraft of air that evaporates the water. Thus, we need the weight of each droplet to exceed the drag force exerted by the upward moving air. Although water droplets tend to flatten out as they fall, we can approximate them with a sphere for our preliminary estimates, so we have

(Eq’n 1)

in which v represents the speed at which the droplet moves relative to the air. For a sphere the drag coefficient, CD, is about 90%. We thus calculate the minimum radius of droplets as

(Eq’n 2)

At average salinity seawater has a density of 1025.5 kilograms per cubic meter and at one atmosphere and 15 Celsius air has a density of 1.225 kilograms per cubic meter. At higher temperatures the density of air decreases, so we can write Equation 2 as

(Eq’n 3)

with v in meters per second and r in meters. That formula lets us calculate the minimum size that the droplets must have in cold air, though they could be smaller in hotter air due to the lower density of air. So for the air moving through the desalinator at one meter per second we need a minimum radius of about four hundredths of a millimeter for our droplet. It would take a drop with a radius of one millimeter to fall through air rising at five meters per second (about eleven miles per hour).

Of course we don’t have to rely only on gravity to move the water. We can blow air into a cylindrical container in such a way as to create a vortex and then give the water its initial centrifugal impetus by spraying it through nozzles mounted on a spinning ring that has a diameter much less than that of the container. Using centrifugal force rather than gravity to move the water enables the device to operate with increased rates of evaporation and condensation.

So how big do we want to make these things? The California Aqueduct carries about nine and one-half million acre-feet of water from Northern California to Southern California every year. Taking that flow as an order of magnitude estimate of what any given desalination project requires, we choose one hundred thousand acre-feet per year (about 11.4 acre-feet or 14,069 tonnes per hour) as the standard output of a unit that can be mass-produced. By using many small mass-produced units rather than one or a few larger custom-built units, we can lower the cost of building a desalination system.

To gain an impression of what this problem requires for its solution, let’s consider a system that produces 100,000 acre-feet of fresh water per year on a continuous basis. That system must yield 3.9 tonnes of fresh water per second. If that water flows through a pipe with a cross section of one square meter, it moves at 14.4 kilometers per hour (9.6 miles per hour). So we need first to figure out what we need to make water vapor come off of hot seawater at that rate.

At 150 degrees Fahrenheit (65 Celsius) water has a vapor pressure equal to one quarter of an atmosphere (25 kiloPascals). Under a pressure of one atmosphere (101 kiloPascals) at that temperature air can hold 160 grams of water vapor in each cubic meter. At 68 degrees Fahrenheit (20 Celsius) that same cubic meter holds 17 grams of water vapor, so for every kilogram of fresh water we want our system to yield we must run seven cubic meters of saturated air through our condenser. To obtain an output of 3.9 tonnes per second, we must push 27,300 cubic meters of saturated air through our desalinator every second. We must also run something else through our system and do so quickly and efficiently.

For every kilogram of water that we evaporate at one atmosphere of pressure we must provide 2.257x106 watt-seconds (joules) of heat. For our 3.9 tonnes of water per second that we want to evaporate every second that’s power moving at the rate of 8.8 million kilowatts. That’s the water’s heat of vaporization and when we condense the water we must remove that much heat as heat of condensation. Ideally, that heat would not leave the evaporation chamber, because the loss of that heat cools the evaporating water. Practically, we design a system in which the heat follows the shortest possible path to return to the evaporation chamber. A compressor squeezes the vapor-laden air, heating it adiabatically, and then the compressed air flows through narrow pipes in the evaporation chamber, where the condensing water releases its heat.

That latter process would not work well in the mist-based system. But there is an alternative. Instead of dropping a mist through hot air, drive small bubbles of hot air through a body of seawater at the bottom of the evaporator. If the bubbles have the same size as do the droplets described above, we get the same amount of evaporation. So how big does our desalination plant have to be to process 27,300 cubic meters of air per second?

If we blow air through the evaporator at a speed of one meter per second, the floor of the evaporator will need to have the area of a square a little over 165 meters on a side, a little smaller than the footprint of the Los Angeles Coliseum. In this case, the floor of the evaporation chamber will be a solid plate with many small holes drilled into it. Air rising through the holes will get pinched off as bubbles as they emerge from the plate into the water. If the holes are small enough, surface tension will prevent water from leaking through the plate. A screen mounted above the water will prevent drops of water and bits of foam from being blown into the condenser.

Of course, no system is entirely leakproof, especially with regard to heat. The builders of the desalinators will certainly insulate them to minimize heat loss, but they will have to add something else as well. Solar collectors on the roof will heat oil at a rate of 27,300 kilowatts to make up for thermal losses in the system. The oil will go through pipes into the evaporator and thence back to the roof.

Now, what can we do with these things once we have them? Santa Monica Bay is quite shallow, so a large number of them could be built on an array of artificial islands. Five such units would produce enough fresh water for the City of Los Angeles that the city could stop taking water from the Owens Valley, thereby allowing that valley’s agricultural community to recover from the loss of its water over the last century.

Currently the Central Valley Project provides seven million acre-feet of water to the San Joaquin Valley every year to augment the natural rainfall and snowmelt from the Sierra Nevada Mountains. If Southern California could get its fresh water from desalination plants and we could distribute the water flowing in the California Aqueduct throughout the San Joaquin Valley, then one of the most productive agricultural areas on Earth could become even more productive. Of course, that statement depends upon the halting, even reversal, of the paving of the valley floor with subdivisions.

Many years ago my uncle Frank Rose told me a strange story he learned during the Second World War (in which he participated as a Marine). The Japanese, enjoying elaborate fantasies of conquering the western United States, had already made detailed plans for rebuilding California. Their basic plan for the Central Valley consisted of uprooting the cities and towns and rebuilding them in the foothills of the mountains, thereby making virtually every acre of fertile soil on the valley floor available for agricultural use.

Raised in a culture that evolved in a country with severely limited agricultural land, where the hills have been reshaped into terraces that support rice paddies, the Japanese, as a society, appreciate the full value of agricultural land as we, as a society, no longer do. In what looks like a sick Malthusian joke, we increase our population and then build living space for the extra people on top of the land that we need to feed them.

Appendix I: Deriving Langmuir’s Evaporation Equation

Irving Langmuir (1881 Jan 13 – 1957 Aug 16) once derived a neat equation that describes the evaporation of a liquid. And he used an astonishingly elegant argument to get it.

We begin to follow Langmuir’s logical path by assuming that we have a body of some mono-molecular liquid, such as water or ethyl alcohol, with a suitable atmosphere of gas, whose pressure prevents the liquid from boiling, lying atop it. We also assume that the gas contains a certain amount of the liquid’s vapor, which exerts a partial pressure upon the gas-liquid interface. We want to calculate the rate at which the molecules of the vapor strike a small area of the interface.

Establish a minuscule area dA on the interface and establish two angular coordinates. We measure the polar angle θ with respect to a line drawn perpendicular to dA and the azimuthal angle ϕ with respect to an arbitrary circle lying parallel to dA. Consider only those vapor molecules that float arbitrarily close to the interface and that have velocities v with speeds in the range between v and v+dv oriented in directions whose angular coordinates lie between θ and θ+dθ and between ϕ and ϕ+dϕ.

Define a minuscule interval of time dt such that vdt is very much shorter than the mean free path of the vapor molecules, thereby ensuring that only a negligible number of molecules get deflected by collisions before they can reach the interface. We thus have a set of vapor molecules contained within a skewed cylinder of cross section dA and length vdt (whose direction makes an angle θ with the normal to the interface) which strike the interface in the interval dt. That cylinder contains a volume dA(vdt)cosθ and contains vapor molecules of the designated speed in the density f(v)d3v. The product of multiplying those two numbers together tells us how many vapor molecules strike the given area of the interface in the given elapse of time.

If we divide that product by dA and dt, we get

(Eq’n A-1)

the rate at which vapor molecules with velocities between v and v+dv strike a unit area of the interface. To get the total rate at which vapor molecules strike a unit area of the interface we must integrate that function over all relevant velocities, so we get

(Eq’n A-2)

in which v+ represents vcosθ>0; those vapor molecules for which v+<0 won’t hit the interface, so we dismiss them from our calculation as irrelevant.

We want to carry out that integration over the appropriate region of velocity space using the same angular coordinates that we established for location space. The element of volume in that space comes out as

(Eq’n A-3)

in which dΩ represents the minuscule solid angle on a spherical shell of radius v. We thus get Equation A-2 as

(Eq’n A-4)

In the last line of that equation n represents the density of the vapor molecules and the barred vee represents the average speed of the molecules (which we calculate explicitly in the appendix below). Using the ideal gas law in the form p=nkT to substitute for n in Equation A-4 and using the explicit calculation of the average speed in terms of the temperature gives us

(Eq’n A-5)

If we multiply that equation by the mass of each vapor molecule, we get

(Eq’n A-6)

the flux of vapor molecules in kilograms per square meter per second onto the interface. Having thus tacitly assumed the use of the MKS system of units, we must now require that we express pressure in newtons per square meter, molecular mass in kilograms, and thermal energy in joules.

Now we come to Langmuir’s genius-level flash of inspiration. He conceived the liquid as a gas in which the vapor was dispersed in accordance with the liquid’s vapor pressure. Thus, for example, he would conceive gasoline as a gas of substance X with gasoline vapor dispersed within it. He could then use Equation A-6 to calculate the rate at which the vapor would strike the interface from the liquid side. Because any molecule that strikes the interface passes through it, that calculation represents a flux of vapor from the liquid. If we then subtract from that calculation the flux of vapor coming into the liquid from the gas, we get Langmuir’s evaporation equation,

(Eq’n A-7)

In that equation pv represents the vapor pressure of the liquid and pp represents the partial pressure of the vapor in the gas. We can see that if pv exceeds pp, we will have net evaporation and that if pp exceeds pv, the net mass flux comes out as a negative number and we thus have net condensation.

Appendix II: The Clausius-Clapeyron Equation

In 1834 Benoit Paul Émile Clapeyron (1799 Feb 26 – 1864 Jan 28) published his "Memoir on the Motive Power of Heat" in the Journal de l’Ecole Polytechnique and in 1843 he had it republished in Poggendorf’s Annalen der Physik. In Section IV of that paper he derived a simple equation that describes a liquid’s latent heat (he called it caloric) of vaporization as a function of differential changes in the liquid’s vapor pressure in relation to differential changes in the liquid’s temperature.

In 1850 Rudolf Clausius published in Poggendorf’s Annalen der Physik a long paper "On the Motive Power of Heat, and on the Laws which can be Deduced from it for the Theory of Heat". He derived something close to the modern expression of the Clausius-Clapeyron equation in Section II in devising his Equation 21.

The modern derivation of the equation follows a different route, one based more on the work of Josiah Willard Gibbs and the use of thermodynamic potentials, which neither Clapeyron nor Clausius knew.

(Eq’n B-1)

In that equation the Greek letter mu represents the chemical potential of the fluid.

In those situations in which we want to use temperature and pressure as our independent variables, the factors that we can control, we use the Gibbs free energy as our master equation,

(Eq’n B-2)

in accordance with the Legendre transformation of the system’s energy content. That gives us the differential as

(Eq’n B-3)

But we also have the differential Gibbs free energy as

(Eq’n B-4)

Comparing those two equations gives us

(Eq’n B-5)

which expresses the Gibbs-Duhem relation and tells us how to change the chemical potential by changing the pressure and temperature of the system.

We want to apply that relation to a study of a system whose active part consists of a certain amount of a single chemical species. We have the system in contact with reservoirs that keep it at a constant temperature and pressure. Under those circumstances our target substance exists in two phases and we want to discover the conditions of those phases existing in equilibrium with each other.

In that situation the system reaches equilibrium by so evolving that the Gibbs free energy of the system achieves the minimum possible value. If we plot the substance’s phases on a temperature versus pressure grid, then the points representing the temperatures and pressures at which the system reaches equilibrium between the two phases form a curve across the plot: the conditions under which all of the system consists of phase A lie on one side of that curve and the conditions under which all of the substance consists of phase B lie on the other side. In order to have a minimum value at equilibrium the Gibbs free energy must have the same value on both sides of that curve; if N particles change from one phase to the other, crossing one given point on the equilibrium curve, then, in accordance with Equation B-2, we must have

(Eq’n B-6)

If we impose minuscule changes, dT and dp, in the temperature and pressure upon our system in such a way that the system remains in equilibrium, then Equation B-6 becomes

(Eq’n B-7)

which we transform via the Gibbs-Duhem relation to

(Eq’n B-8)

Rearranging that equation, we get

(Eq’n B-9)

Now we assume that the phase change of our chosen substance occurs between the liquid (phase B) and the gas (phase A) phases. In that case

(Eq’n B-10)

in which HV represents the heat of vaporization of the substance (measured in joules per mole of material). We also assume that we have made the pressure of the system light enough that we can make the molar volume of the gas very much larger than the molar volume of the liquid, VA>>VB, so that we can write Equation B-9 as

(Eq’n B-11)

in which I have used the ideal gas law, pV=RT, in its form for one mole of gas, to replace the molar volume (with the universal gas constant, R=8.314 joule per mole per degree Kelvin). That equation rearranges and then integrates readily to

(Eq’n B-12)

The first of those equations displays one form of the Clausius-Clapeyron equation.

Because we based that equation on the equilibrium between the liquid and gas phases of a substance, the pressures in that equation correspond to the vapor pressure of the liquid at two different temperatures. Thus, if we know the vapor pressure of a liquid at one temperature, Equation B-12 enables us to calculate the vapor pressure at any other temperature. Because a liquid boils when its vapor pressure equals or exceeds the pressure of the atmosphere pressing down on it, we can take the boiling temperature of a liquid at normal atmospheric pressure (14.7 pounds per square inch or 101,325 newtons per square meter) as our known point. As an example let’s calculate the vapor pressure of water at ten-degree intervals between 0 C and 100 C.

For water we have the heat of vaporization at 100 Celsius (373 Kelvin, which we must use in our calculation) as 539.55 calories per gram, which corresponds to HV=2,257,477 joules per kilogram. The gas constant, 8.314 joule per mole per degree Kelvin, applied to water (18 grams per mole or 55.55... moles per kilogram) takes the value 461.89 joules per kilogram per degree Kelvin. To calculate the vapor pressure p at the temperature T we use a two-step process, in the first step of which we calculate an intermediary number X. First we have Equation B-12 in the form

(Eq’n B-13)

and then we have

(Eq’n B-14)

Applying those formulae gives us Column I in the following table (with vapor pressure expressed in pounds per square inch):

 T I II III 0 C/273 K 0.1211 0.0886 0.0912 10 C/283 K 0.2280 0.1781 0.1781 20 C/293 K 0.4110 0.3392 0.3325 30 C/303 K 0.7127 0.6155 0.5957 40 C/313 K 1.1931 1.0701 1.0282 50 C/323 K 1.9346 1.7893 1.7156 60 C/333 K 3.0472 2.8893 2.7760 70 C/343 K 4.6741 4.5203 4.3674 80 C/353 K 6.9979 6.8684 6.6970 90 C/363 K 10.2467 10.1693 10.0301 100 C/373 K 14.7000 14.7000 14.7000

(Table B-1)

Column II lays out the measured values of the vapor pressure and we see a serious discrepancy between those values and the calculated values in Column I. However, our calculations come close enough to the measured values that we cannot legitimately dismiss the Clausius-Clapeyron equation as invalid: we can only state that it stands before us flawed and/or incomplete. We know that we left the equation flawed: we dismissed the volume of the liquid phase as negligible and used the ideal gas law rather than the more accurate van der Waals version. And we know that we may have missed some subtle aspect of phase change in our original analysis and therefore failed to take it into account. Correcting those errors provides the subject of a subsequent essay in this series.

Column III shows vapor pressures calculated through a revised version of Equation B-13. In this case I modified Equation B-13 by dividing 14.7 psi by the measured vapor pressure at 283 Kelvin and working backward through Equations B-14 and B-13. Thus I got Equation B-13 in the form

(Eq’n B-15)

That semi-empirical result gives us figures good enough for some preliminary engineering calculations, but we still see discrepancies which indicate the need for a deeper understanding of vapor pressure and phase change and a suitable revision of the Clausius-Clapeyron equation.

efefef

Back to Contents