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 Subject: Ocean evaporation rate calculation Category: Science > Earth Sciences Asked by: centure7-ga List Price: \$20.05 Posted: 22 Apr 2005 05:28 PDT Expires: 22 May 2005 05:28 PDT Question ID: 512638
 ```How would I calculate evaporation rate of ocean water, in terms of volume per time per area (gallons per minute per acre for example) given the following: - Relative humidity - Wind speed - Water temperature - Air temperature Would air pressure make a significant difference? Would there be a significant difference from small waves to large waves? Are there other factors I am not mentioning here that make a big difference?```
 Subject: Re: Ocean evaporation rate calculation Answered By: hedgie-ga on 20 May 2005 00:40 PDT Rated:
 ```centure7-ga You have interesting but complex questions. It would be helpful (and it may increase the rate at which your questions are answered) if you would rate the answers. Rating helps researcher to gage how useful was his/her the contribution to the answer in those situations when full, exhaustive answer is not possible. (This is one of those situations ;-) I will focus on the physical processes which determine the rate of evaporation. That will provide a handle on, what the units are and will provide a complement to the empirical study referenced in the comment. But first, some actual numbers for the ocean: ------------------------------- " The average amount of evaporation F.e from the sea surface is about 120 cm/yr, i.e. the equivalent of the sea surface sinking by that amount. Local values range from an annual minimum of as little as 30 to 40 cm/yr in high latitudes to maxima of 200 cm/yr in the tropics associated with the trade winds. This decreases to about 130 cm/yr at the equator where the mean wind speeds are lower. How is the evaporation rate Fe determined? A direct method is to determine the rate of water loss from a pan of water, but this has serious practical difficulties. For large area estimates and for prediction, a formula using easily measured parameters is desirable. Evaporation is basically a diffusive process that depends on how water vapor concentration changes with height above the sea surface and on the processes that cause diffusion. In section 5.1.3, we discussed "eddy diffusion" which is analogous to molecular diffusion, except that the turbulence in the air or water is considered to be the process that diffuses properties, rather than movement of individual molecules. The air turbulence is the process that controls the diffusion that creates evaporation, and this air turbulence depends on wind speed. Therefore we expect the evaporation rate to depend on wind speed. (This explains why we feel cooler when the wind is blowing which creates the effect known as 'wind chill' " This is quote from section 5.4.6 of excellent text: CHAPTER 5 Water, Salt and Heat Budgets of the Oceans http://www-pord.ucsd.edu/~ltalley/sio210/pickard_emery/chapter_5_917.htm whole book http://www-pord.ucsd.edu/~ltalley/sio210/pickard_emery/ and books http://oceanworld.tamu.edu/ocean410/ocng410_text_book.html and projects described on the site , such as http://www-pord.ucsd.edu/po/research/TS.html may be of interest for this and the for the related question http://answers.google.com/answers/threadview?id=514897 I now want to focus on the theoretical aspect of the evaporation: ---------------------------------------------- What does it mean " Evaporation is basically a diffusive process?" The search term for the fundamental mechanism is: Stefan Problem named after Slovenian physicist ( J. Stephan) http://www.answers.com/topic/joseph-stefan The quote from his book, given in the above link: "There always something will remain, that we shall not know, why?" is appropriate comment (sigh) to your question. One reason 'why' is that even if we would know all the physical parameters of the phase transition (evaporation, melting, .. ) that is latent heat of the transition, diffusion rates,heat conductivity, ... ,we still would need to solve the transport equations, to determine the rate of transition, speed of the boundary. OK - let me try to put it more simply: The (textbook) example of the Stefan problem is an ice cube, melting in the glass of water. Melting is phase transition (just like evaporation) and the boundary between the phases (surface of the cube) is moving (in either direction, depending on temperatures). The speed of the moving boundary depends on how quickly the fluid phase can supply the heat (and in same cases material) needed for the transition. This is how winds into play. Heat comes by conduction, radiation (sun rays) and convection and water vapour is carried away by wind (convection) and diffusion (condiction). When you look at the 'big' picture: http://www.physicalgeography.net/fundamentals/8b.html you will notice that while water evaporates and vapor is carried away by the wind, some of that water returns to the ocean right away. Only small part comes back through rains over land and runoff. How much? That depends on the local conditions -- it is different near coast and in the open sea. For large scale (global) model (cumulative) numbers are known: 361 Tm^3 evaporates and 324 returns directly (by rain over the ocean), according to http://ww.mines.edu/Academic/ courses/dc/dcgn101/StreamsRivers.doc Here Tm^3 means Tera-cubic-meters; T= E12 is SI prefix for million of millions as defined here http://physics.nist.gov/cuu/Units/prefixes.html So, at one spot, water may evaporate and be carried away by winds; at another spot it may rain all the time and that same water is returned to the ocean. This distances (from one sot to the other spot) affects the 'net evaporation': The scale of the model and the grid (cell size) needs to be specified before 'net evaporation rate' is well defined. So, while local studies, such as the one cited in the comment are possible, and constants determined empirically may reflect local miroclima, and have some temporary and limited validity, true answers are tied to global circulation models (both water and atmosphere), which are a topic of an ongoing research, are far from completion and very complex. That is one reason why Stephan's sigh is still valid. Hope this is useful. Questions welcome, rating appreciated. Hedgie``` Request for Answer Clarification by centure7-ga on 28 May 2005 02:51 PDT ```I found that information to be very useful, but am hoping you can actually clarify someone else's comment. Since hfshaw hasn't responded to my comment below, I am hoping that you can provide the answer to the questions I asked hfshaw, which are (in regards to the equation provided with the comment): 1) What are the units of measurement units expected to be used for each variable given? 2) What should I plug in for k(v)? 3) Please provide an example equation using specific variables so I can be sure that I understand. Thanks!``` Clarification of Answer by hedgie-ga on 01 Jun 2005 06:57 PDT ```Hello again centure7-ga, and thanks for the rating and tip on the other question. That equation hfshaw mentioned is explained on page 28 of the reference he gave http://wrri.nmsu.edu/research/rfp/studentgrants03/reports/herting.pdf E is in units mm/day K.e in 1/kPa is a complex function of geometry conversion constant is given as 86.4 * E6 ...etc Please do look at the referenced paper, particularly pages 20-30. and if anything is not clear after that, post another RFC. Hanon's eq. on page 26 is much more simple - and more credible then 'Bulk-Aerodynamic' on page 28. Please, bear in mind this is students research, not a peer reviewed article. Hedgie```
 centure7-ga rated this answer: and gave an additional tip of: \$4.95 `Thanks for going out of your way yet again to thoroughly answer my question.`

 ```There are a number of ways to parameterize the evaporation rate from a water surface as a function of "observable" variables. See the appendix of the document at for a good summary of the various approaches. The set of variables you have chosen correspond to the ones used in what is called the "bulk aerodynamic" method, the equation for which can be written as: E = a * k(v), * v *(e_sat(T) - e) where E is the evaporation rate a is a constant that accounts for the units used to express the other variables v is the wind speed measured at a specified height above the air-water surface k(v) is an "efficiency factor" that parameterizes the the extent to which wind eddies near the surface are effect vertical transport of water vapor away from the water-air interface. k depends on v, the densities of the air and water, the atmospheric pressure, and it also takes into account the surface roughness of the interface due to waves. e_sat is the saturation water vapor at the water surface, which is a function of temperature (T) e is the actual water vapor pressure of the air above the surface. Note that 100*e/e_sat is the relative humidity, expressed as a percentage. The URL given above will provide you more detail on the numerical values for the constants, as well as an expression for k.```
 ```hfshaw, I found your link to be very helpful. Could you give an example of what the math would be for humidity of 80%, wind speed of 15mph, a k(v) of a typical value for the ocean along the coastline (air pressure of perhaps 1ATM, and water density of maybe 1.03g/cc), and a water temperature of 30C? The part I don't understand is mostly just what unit measurements are used. In other words, how you would calculate with any specified units, and what units must be specified. I would imagine that the equation requires metric units and a Celcius temperature. I'm also not sure what to plug in for k(v). Again, thanks for the information you have provided!```