Tank Water Evaporation Estimates

A. Terry,
I think you're asking the wrong question.

For starters, evaporation depends on more than just temperature. The humidity, air pressure, air flow, etc. all play a role.

Second, the amount of evaporation at 5°C with an ambient air temperature of -20 to 0°C is bound to be very small. There will certainly be heat loss, but evaporation will be far from the primary source of it.

I would suggest, rather than trying to work through the complex physics of it all, a practical experiment would more easily demonstrate the required information, such as total energy required to maintain the water at 5°C, and if the loss of water to evaporation over time is even measurable.

A. Hi Douglas,
I was intrigued with your mention of 'fog' sprays ... and, obviously, as you pointed out, water consumption.

People tend to euphemistically promote products. FOG spray. ATOMIZATION of droplets. But if you research manufacturers' tech. data, you'll notice that with 'jets', the smaller the jet size (for a given gpm), the smaller the droplets will be.

And from a viewpoint of mass transfer and lower water consumption you need small droplets. The best are the so-called air atomizing nozzles but, ah, the best they can do is around 60 microns (going by memory) but that's a big, big step compared to the 300 micron sized droplets of, say, most smaller jets.

My own preference (considering potential jet blockage, spray angle and droplet size) was the solid cone wide angle injection moulded (plastic) jets, 1/8" sizing.

Food for thought, eh?

A. For still tanks, the following equation works fairly well.

E = exp ^(7.2-.03236T) where E = the evaporation rate in gallons/ft2/hour, and T = temperature in degrees Fahrenheit.

For air agitated tanks, the evaporation is much higher and the equation is

E = exp ^(5.95-.0265T)

A. 'exp' stands for the exponential function, so everything in the parentheses is the exponent and the base is "e", which is roughly 2.718.

Also, when dealing with evaporation the volume of the tank is a non-issue. To get the evaporation rate for a specific tank you would multiply your result by the area of water-air contact (the water exposed to air).

A. For still tanks with still air above them.

E=(exp)-(7.2-0.03236T)

 

For air agitated tanks or tanks with a high air velocity across them.

E=(exp)-(5.95-0.0266T)

 

E = Evaporation in US gallons/square foot/hour

(exp) is the exponential function,

T = Temperature in in degrees F

The equations work very well for 120-200 °F.

A. Hi David. I could be wrong, but I believe that this particular formula is just an empirical solution with no grounding in the physical world -- just a mathematical "best-fit" approach that tracks observed data pretty well within a certain range.

There have been attempts to do calculations based on real physical derivations, such as the Langmuir Equation and Stelling's Formula, if you wish to search for those terms, but they are too complicated for my personal interest level :-)

Regards,

A. The formula will only give you a starting point. Air currents over the top of the tank increase evaporation as does a hood. Relative humidity and air temperature have an extreme effect of the evaporation. If the tank has agitation, evaporation will be massively higher. Parts moving in/out will effect it also as well as the tank temperature.

A. If you use a program such as Microsoft Excel, simply type in as written:

EXP((exp)- (5.95 - 0.266 0.0266T))

This should get you where you want to be.

Best of Luck

Ira

A. The correct factor is 0.0266, not 0.266. For your tank at 140 °F the estimated evaporation rate is 0.107 gallons per square foot per hour. This works fairly well for agitated and exhausted tanks, but not if there is a blanket of foam or other cover.

A. The loss of water by evaporation is dependent on lots of factors, such as the rate of air extraction over the tank (you do have extraction don't you....?), the relative humidity, the ambient temperature, the degree of agitation, the presence (or otherwise) of foams, suppressants, choffles, etc. However, assuming there is no agitation, no surface coverings and the ambient temperature is 20 °C with a RH of 60%, the amount of water loss by evaporation from your 50 ft2 tank will be: No extraction - 42.5 lb; extraction at 1.5ft/sec - 66 lbs; extraction at 3ft/sec - 75 lbs; extraction at 6ft/sec - 110 lbs. For a bath temperature of 60 °C (140 °F), the equation is Y=0.9586e^^(0.1441x), where Y is the water loss in lbs/ft2 and x is the flow rate of the extraction in ft/sec.

A. Hi Michelle. It sounds to me like after one day you would have .99 of the amount of water you started with; and after two days you'd have .99 of the amount you had after the first day, i.e., .99*.99 of what you started with. After 3 days you'd have .99*.99*.99 of what you started with, and after n days you'd have .99ⁿ.

Regards,

A. The equation is a measure of the rate of water loss due to evaporation. However, I am not convinced of its accuracy because it doesn't take into account any air movement. This is discussed in letter 19871 [ed. note, that discussion is now on this page, above].

In 1944 Metal Finishingmagazine published an article on water loss due to extraction and they gave figures of loss based on air flow rates and water loss in terms of lbs/ft^^2 of tank surface area. For no exhaust, the relationship is Y=0.0291e^^0.0537x, where Y is the water loss and x is the temperature in °C. For 1.5 ft/sec air flow rate the relationship is Y=0.065e^^0.0495x; for 3 ft/sec it is Y=0.0445e^^0.0593x; for 6 ft/sec it is Y=0.065e^^0.0629x

In 1965 the National Physical Labs in the UK produced similar data where their relationship was Y=0.1557e^^0.0288x.

These equations are calculated from the raw data given by the experimentalists and are not given in the original documents. Although the data is old, the evaporation process has not altered, so I think it is sound information. No doubt a plating plant manufacturer will be able to give similar data.

A. The units are based on a per hour loss rate. Sorry for the omission.

A. The evaporation rate will depend primarily on the air/liquid surface area and the temperature, and the agitation rate. This assume the absence of an insulating foam layer. The volume has no influence.

For a still tank, I have used an empirically derived equation to estimate this.

Evaporation in gallons/hour/square foot =
EXP -(7.2-0.3236 T)

Where T is the temperature in degrees F.

Air agitation or air bubbles have the effect of increasing the air/liquid interface area and will result in much higher evaporation rates. Likewise, high air velocity at the surface will also result in a much higher rate of evaporation.

A. Yes, you will need to convert from US gallons to liters and from square feet to square meters. The exponential function is the one in EXCEL and on scientific calculators.

This is only a starting point, and I use a different equation for air agitated tanks or tanks with high velocity air movement over them. Either of these could result in 50-100% more evaporation.

LK

Thank you Mr. Kirman, Lyle.

I was thinking ' X to the power y' function - cleared up, and gives me a starting point.

Again, thank you for sharing your knowledge, and thank you to Finishing.com .

Regards,

Not too surprising, but Dr. Cook's estimate and my estimate are virtually identical. 11.4 inches is 7.1 gallons per ft2 per 24 hrs. or 0.296 gallons per ft2 per hour.

A. Hi, H.R.

Keeping the surface as still as possible with a wind-break of trees or walls probably helps, and may also provide some shade which will help. Unfortunately, though, aeration of some sort may be necessary anyway for the health of the body of water.

Regards,

A. Hi Marc
Unless there is a significant air current through the vent to carry away the vapour, I would expect the loss to be close to zero.
Why is this important?

 Hi
Thanks for your answer.
I have to evaluate the evaporation rate of this tank because I need to calculate the energy consumption of this phenomenon.
BR.
Marc

A. Hi Glenn. I'm not certain that I'm clear on the question, but we've now merged a few threads which may help answer it for you ...
If you are talking about Lyle Kirman's equations, and using his factors, he has told us that the units of E are "Evaporation in US gallons/square foot/hour".
If you are looking at some other equations elsewhere, I don't think there is any way anyone can tell you what units the author was using though.

Regards,

A. Hi Gopinatha, the 'formula' is not e^^{(7.2-.03236T)};
it is e^^{-(7.2-.03236T)} ... note the minus sign.
So, as the temperature rises, the exponent becomes less negative, so the estimated evaporation is greater.

I get 0.124 gallons/ft2/hr at 158 °F and 0.0693 gallons/ft2/hr at 140 °F.

But as far as I know this 'equation' has no grounding at all in physics -- it's just mathematical 'curve-fitting' that approximates measurements of evaporation losses for certain conditions. In other words, evaporation was measured at different temperatures and then constants were picked which best fit those observations. I've also seen 'formulas' for evaporation of the form E = wT + xT^2 + yT^3 +w zT^4, which are neither more nor less accurate than this exponential formula, neither more nor les grounded in physical laws ... just a different mathematical approach to the curve-fitting.

Regards,

 Dear sir,

Thanks for the clarification. We were calculating the water consumption in plating process. During the calculations, we looked for the evaporation losses and arrived at the figures. Now, we modified the same. Now, the losses seems to be reasonable.

It was mentioned that 40 lit/m2 to 50 lit/m2 is the consumption of water in the plating process from finishing.com site. From the calculations, it is observed that the consumption is approximately 50 lit/m2.

Thanks with Regards

Hi Santosh,

I guess it will also depend on humidity, wind speed, altitude, etc.

Regards,
David

A. The only way of getting a truly accurate time for evaporation is to sit by the lake with a stop watch, or perhaps a calendar. Oh and take a picnic.
Best regards
Mark

A. Hi Kee. I don't think there are any such 'formulas' because the problem is too complex for a solution or calculated result based on first principles.

Rather, I feel that a lot of observations have been made and recorded, and what can be done is to find tables which include those relevant observations, and then devise a least squares or other curve fitting method to generate the desired curves.

But if you have one particular air velocity, water temperature, air temperature, and pressure which you cannot find in a table and cannot interpolate from data you can find in a table, I think measurement rather than calculation will be your only alternative. Happy to be further educated on the subject though.

Regards,

A. You might get a maximum estimate by assuming that the air is saturated when it leaves and is only 50-60 % RH when it enters. You'll need to know the air temperature leaving for this calculation.