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# "Tank Water Evaporation Estimates"

Current postings:

February 1, 2022Q. I'm trying to provide fresh water during cold periods to animals. I have a rectangle container that is open at the top to the environment. The ambient temperature can range from -20 to 0 °C. Is it possible to estimate the heat loss due to evaporation. The water temperature is to be maintained at 5 °C. Most tables do not provide information at the low range of temperature for estimating the energy loss due to evaporation. Thank you.

Terry StreetEngineer, Retired - Fairport, New York

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February 2, 2022

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.

Ray Kremer

Stellar Solutions, Inc.

McHenry, Illinois

Closely related Q&A's, oldest first:

**"Non-equilibrium Evaporation and Condensation Processes"**

by *Yuri B. Zudin, *

from **Abe Books**

or

Affiliate Link

(commissions from your purchases make finishing.com possible)

Q. I am working on a study to minimize wastewater in an aircraft plant metal finishing line. I am considering fog rinses above hot process tanks to minimize dragout losses. Are there standard calculations for the evaporation losses from hot process tanks (150 to 190 °.) that can reliably predict the evaporation that could be offset by the fog spray. Are there problems with this approach other than the obvious problems of too much water used and the physical arrangement of the nozzles?

Douglas Mullengineers - Salina, Kansas

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2002

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, l/8" sizing.

Food for thought, eh?

Freeman Newton [dec]

*(It is our sad duty to advise that Freeman passed away*

April 21, 2012. R.I.P. old friend).

April 21, 2012. R.I.P. old friend).

^

2002

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)

Lyle Kirmanconsultant - Cleveland Heights, Ohio

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**"Evaporation of Water"**

by *Frank E. Jones*

from **Abe Books**

or

Affiliate Link

(commissions from your purchases make finishing.com possible)

Questions for Lyle Kirkman's equation: E = exp ^(7.2-.03236T)

1. What does 'exp' stands for? Want to figure out how to calculate after substituting temperature...

2. Don't you consider the volume of the tank in the equation?

- San Diego

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July 30, 2014

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).

- Minneapolis, Minnesota

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July 11, 2017

Q. Hi,

If the hot water tank in question is a covered tank (12 ft diameter) with a 12 inch exhaust duct - natural draft

Would you presume that the only exit for the water vapor is the area of the 12" exhaust duct (0.785ft2) instead of the 113 ft2 top surface area?

- Chicago Heights, Illinois USA

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To minimize search efforts and to offer multiple viewpoints, we combined previously separate threads onto this page. Please forgive any resultant repetition, failures of chronological order, or what may look like readers disrespecting previous responses -- those other responses may not have been on the page at the time :-)

2002

Q. I'm just wandering if the equations you use to estimate tank evaporation rate are only applicable for closed tank. Is there any formulas for open tank? Normally we use the formula for the swimming pool for open tank.

Regards,

Stephen Chou- Vancouver, B.C. Canada

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2002

- 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.

Lyle Kirmanconsultant - Cleveland Heights, Ohio

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2003

Q. Thanks for the great rule of thumb on calculating tank evaporation. Can you provide the full formula however, as I cannot follow the units in this estimated calc. and am unsure how you got to the gal/sf/hr unit. What is the swimming pool evaporation formula anyway?

David Gibbons- San Diego, California

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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 interest level :-)

Regards,

Ted Mooney, P.E.

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

^

To minimize search efforts and to offer multiple viewpoints, we combined previously separate threads onto this page. Please forgive any resultant repetition, failures of chronological order, or what may look like readers disrespecting previous responses -- those other responses may not have been on the page at the time :-)

## How to calculate the exponential function for water evaporation problems

2003Q. I want to calculate evaporation rate of water from my nickel tank. I have the formula but I do not know how to calculate the exponential function.

Formula is

E = (exp)- (5.95-0.266T)

E= EVAPORATION RATE, T= TEMPERATURE IN °F

Out tank 5049 gals

Size of the tank is 120" x 60" x 162"

TANK TEMP. IS 140 °F

Popatbhai B. Patel

electroplating consultant - Roseville, Michigan

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2003

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.

James Watts- Navarre, Florida

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2003

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

Ira Donovan, M.S.F.Kansas City, Missouri

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2003

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.

Lyle Kirmanconsultant - Cleveland Heights, Ohio

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2003

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.

Trevor Crichton

R&D practical scientist

Chesham, Bucks, UK

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March 28, 2013

Q. Exponential Decay (Evaporation of water in a swimming pool).

A swimming pool contains 100KL of water. If one percent of the water
evaporates everyday:

A) Determine the equation of the amount of water in the pool on a particular day.

B) find the amount of water after 2 weeks.

C) how long will it take to have 10KL of water?

- Calgary, Alberta, Canada

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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 .99^{n}.

Regards,

Ted Mooney, P.E.

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

^

To minimize search efforts and to offer multiple viewpoints, we combined previously separate threads onto this page. Please forgive any resultant repetition, failures of chronological order, or what may look like readers disrespecting previous responses -- those other responses may not have been on the page at the time :-)

## Rate of evaporation

2005Q. I would like to know the name of this equation and a reference for it and how accurate is it.

The equation:

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

Shant Mardirossian- Noord, Holland

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2005

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.

Trevor Crichton

R&D practical scientist

Chesham, Bucks, UK

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2005

Q. This is a question to TREVOR CRICHTON from U.K.

Many thanks for your answer to my question about rate of evaporation of a tank.

My question is about the units of your equation -- is it pound/square foot/hour?

I do not see hour in your units.

Please give an answer to this question.

Shant Mardirossian [returning]- Holland

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2005

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

Trevor Crichton

R&D practical scientist

Chesham, Bucks, UK

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January 7, 2014

Q. Hello Finishing.com

I came across this thread surfing the net and it seemed close to what I'm trying to calculate - although maybe of topic for this particular portal - if so, pardon me.

I have a cylindrical container - height 2.9 m - radius 1.1 m - surface area/top of liquid 3.8 m sq.(11 cu. meters/11000 litre capacity - although will fill to 10,000 liter). The container is baffled with a down pumping axial impeller and will be fitted with a heating/cooling pump or coil to bring a liquid water based sugar/sucrose mix from a starting point of 2- 5% sugar concentration to a concentration of approximately 11% (give or take depending on other sugar based additives at a later stage).

The tank will be fitted with a hood and Lyne-Arm & Coil configuration (to be chosen to spec) to retrieve stilled/evaporated water to another holding tank as well as to bring liquid mix to spec - approx. 11%. All this will take place indoors at room temp.

I would like to be able to calculate, with some precision if possible, the rate per hour in liters of evaporated water from the liquid sugar tank to the water holding tank with variability of temperature and sugar concentration (although sugar concentration can be measured at critical spec using gravity/refractometer equip).

Considerations:

Temperature - water boils at 100 °CC - I may want to boil, however lower temps are more likely (preservation of any nutrients is important).

Pressure - may/will be slightly higher in evaporation tank- but never reach equilibrium - Lyne-Arm and coil will be a factor- will affect boiling point.

Surface Area - although not static, turbulent flows at the surface will probably aid the evaporation process - however, it may over complicate the calculations as compared to a flat/static surface - also, at boil, surface would be turbulent, but as stated, boiling is not likely.

* also, any horizon point dip to the center, caused by the rotation of the impeller will increase the liquid surface area - this should also aid evaporation rate - but again, this may over complicate the calculation (rotational speed around 100 rpm - its a guess - not sure just how much of a "vortex trough" this would make if any).

This will aid me in my plant efficiency considerations from increasing capacity to parameters of bottling equipment. I would appreciate any nomenclature your could provide.

Thank you.

David

juice/pop manufacture - Cornwall, Ontario, Canada

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January 15, 2014

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.

consultant - Cleveland Heights, Ohio

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January 19, 2014

Q. Thank you Mr. Kirman, Lyle

I am basically looking for a best "guestimation" and your post and others have helped considerably as I am looking for something to work off of and will perform measurements during the stills operation.

Referring to your answer:

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.

I would prefer to use metric and celsius - liters/hour/meter sq. - I assume you are using US gallons - so I will need to do some conversions.

A link on the same thread regards a problem with this equation that I am having as well - ref. 19871 - the semantics of the nomenclature - where the exponential function comes from and how to calculate it - which is not clear for me in that thread either.

Would appreciate any further assistance you could provide.

Regards,

David

juice/pop manufacture - Cornwall, Ontario, Canada

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January 22, 2014

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

consultant - Cleveland Heights, Ohio

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January 26, 2014

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,

beverage manufacturer - Cornwall Ontario Canada

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## Air agitation effects on evaporation

2006Hi.does the air flow rate into an air agitated tank not affect the evaporation equation below

E=exp-(5.95-0.265T).

- Nelspruit, South Africa

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2006

Check out letters ~~17485 & 34192~~ *[ed. note, those discussions are now in this page, above]*. They may be able to help you out some. The first has some formulas that might help some (one, in particular, for air agitated tanks) and the second has a reference for some original data and more formulas.

Good luck!

Accurate Anodizing Inc.

Compton, California, USA

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2006

Since, a long time ago, I was the source of this equation, I feel obligated to comment.

The equation is just an approximation that works fairly well for typical air agitated tanks at 1-2 CFM of air per square foot of surface area. In reality, the air flow rate does affect the evaporation rate since the air bubbles should be near saturation at the tank temperature when they leave the tank. But twice as much air doesn't necessarily give twice as much evaporation, since some of the effect of agitation is to increase the surface area at the top of the tank. Air velocity over the surface of the tank is also an important factor that affects the evaporation rate.

So, use the equation as an estimate, but if you really need to know, measure the evaporation rate for your tank by how fast the level drops.

consultant - Cleveland Heights, Ohio

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## Evaporation Rate in Water Tank with A Roof and Open Vent

August 23, 2010Q. I have a large water tank with 85 °C temperature with a closed roof but it has a 14 inch vent. Please introduce me a formula to calculate the evaporation rate.

Fred Koomaleki- Canada

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August 24, 2010
A. Sir: Galvanizing Consultant - Hot Springs, South Dakota, USA ^ August 25, 2010 A. Since this is a covered tank, the water vapor escaping will depend a lot upon the ventilation rate, which at 85 °C will be mostly water vapor leaving the vent. consultant - Cleveland Heights, Ohio ^ |

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.

Lyle Kirmanconsultant - Cleveland Heights, Ohio

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August 27, 2010

Q. In this connection I wonder if water loss due to evaporation from large water bodies have been estimated. Temperatures would be lower and fluctuating and wind conditions variable. But water loss, especially in tropical countries, could be significant. It would perhaps be possible to take measures to reduce evaporation from reservoirs and conserve water.

H.R. Prabhakara - Consultantbangaloreplasmatek.com - Bangalore Karnataka India

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August 30, 2010

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,

Ted Mooney, P.E.

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

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June 9, 2016

Q. Hi,

I have a very similar problem.

I have to evaluate the evaporation on a tank.

The tank is filled with water at 95 °C

There is permanent incoming and outgoing flows of about 60 cubic meters per hour each

The water is warmed in the tank by a water bath heat exchanger.

The incoming flow arrives above the water surface.

The tank is covered and there is a vent (a tube of about 4 inches diameter).

Would you have some idea to evaluate the evaporation rate?

Mathematical is better even if not so precise -- the high end of the vent pipe is not so easy to reach.

Regards

- Lorient, Britanny, France

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June 12, 2016

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?

Geoff Smith

Hampshire, England

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June 13, 2016

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

- Lorient, Britanny, France

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September 23, 2017

Q. Open stock tank evaporation --

what are units of E?

- Fredericksburg Texas

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September 2017

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,

Ted Mooney, P.E. RET

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

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## Tank water loss estimates in a plating plant

November 14, 2017Q. Good Morning.

This is with regard to the tank water loss estimates in a plating plant.

When we use the formulas for calculating water loss by using the following formula,

E = exp^(7.2-0.03226T), as the temperature is coming down, the evaporation of water is increasing. But the evaporation of water will increase as the temperature of the bath goes up by maintaining the other variables constant.

For example, at 158 degrees Fahrenheit, the loss is approximately, 8.06 gallons/ft2/hr and at 140 degrees Fahrenheit, the loss is approximately,14.43 gallons/ft2/hr

Please clarify on this.

Environmental Engineer - Tirupati, Andhra Pradesh and India

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November 2017

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,

Ted Mooney, P.E. RET

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

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November 16, 2017

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

AMARA RAJA BATTERIES LIMITED - Tirupati, Andhra Pradesh and India

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January 26, 2018

Q. I want to know how much water will evaporate in a day when T will be 40 °C and the surface area will be 16000 sq ft? Can I have the formula?

Santosh Hingane- Pune,India

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April 8, 2018

Hi Santosh,

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

Regards,

David

David Shiu

- Singapore

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April 10, 2018

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 now cloudy and rainy rock in the irish sea

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September 25, 2018

Q. Trying to estimate evaporation rate in a closed tank. The tank will have a fan blowing in the air at 100 CFM, water surface area is 50 sq.ft, the atmospheric pressure will be 74 kpa at 7900 ft AMSL.The empirical equations predict evap at gal/hr, but doesn't take into account any external air flow.

Is there an equation that takes into account the forced air flow, surface area of water, temperature of the air and temperature of the water?

Thanks in advance for the response

- Denver, Colorado, USA

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September 2018

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,

Ted Mooney, P.E. RET

Striving to live Aloha

finishing.com - Pine Beach, New Jersey

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October 20, 2018

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.

Lyle Kirmanconsultant - Cleveland Heights, Ohio

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