A question for the engineers at UPS.....

[dannyboy]

There is no real place for this question on this site, but I am interested in the answer, so any help would be appriciated. If it needs moved, please feel free to do so!

I do know there are formulas out there on the rate of water evaporation that is not in motion, but in my case there are several wild cards to add to the mix.

Case in point.

Pond has aprox 560 SF of surface area. It has a depth of 6 feet. There is a major water fall that is 6 feet wide, three feet long, with 22,000 gallons of water flowing over it per hour. That water flows into the upper pond, then into the lower pond via two more falls, aprox 3 feet wide and two feet long each.

Then there is a skimmer return that roils the water surface of just the lower pond and keeps it in a circular motion, the flow of which is 3800 gallons per hour.

I am aware of many of the situations that affect evaporation rates, temp, wind etc. Could there be a formula developed that I could use that would give me at least a rough idea of what to expect?

While not uneducated, I am somewhat math challenged (wish like heck I had paid more attention in high school and college!)

Thank you in advance!

d

[freeloader]

Originally Posted by dannyboy

There is no real place for this question on this site, but I am interested in the answer, so any help would be appriciated. If it needs moved, please feel free to do so!

I do know there are formulas out there on the rate of water evaporation that is not in motion, but in my case there are several wild cards to add to the mix.

Case in point.

Pond has aprox 560 SF of surface area. It has a depth of 6 feet. There is a major water fall that is 6 feet wide, three feet long, with 22,000 gallons of water flowing over it per hour. That water flows into the upper pond, then into the lower pond via two more falls, aprox 3 feet wide and two feet long each.

Then there is a skimmer return that roils the water surface of just the lower pond and keeps it in a circular motion, the flow of which is 3800 gallons per hour.

I am aware of many of the situations that affect evaporation rates, temp, wind etc. Could there be a formula developed that I could use that would give me at least a rough idea of what to expect?

While not uneducated, I am somewhat math challenged (wish like heck I had paid more attention in high school and college!)

Thank you in advance!

d

Hmmm I think that's a question on the PT supervisor test. Remember - google is your friend. I am sure there is a formula to calculate the rate of evaporation out there.

[dannyboy]

There is, until you factor in the water movement and water falls. And that is where the greatest amount of water will be lost. Then when you figure out the days where you have a stiff dry wind, you can multiply that by a factor of at least 10 or more.

Did a lot of googling, but so far, no answers.

d

[rod]

I think that would be impossible to answer. Your only hope would be for you keep track of it on a daily basis noting how much water you had to add to keep it at a specified level every day. The temperature, humidity and wind speed would have to be included on a daily basis. After a few years of gathering figures you would have a chart that says if its 90 degrees with 85% humidity and 0 to 5 mile an hour winds you will have to add X amount of gallons on that day. (But then again wouldn't it be easier just to turn the hose on , pop a beer and kick back in a lawn chair until the pool is full again?)

[dannyboy]

It is when its you and I. But for the customer that buys the pond, not so much so.
Then add to that the pond owner that claims the pond is leaking, but only when the water falls are on, and you have a three ring circus.

Actually the numbers can be pretty close, if you figure in the extra variables.

Im just not good enough for the task.

Actually, the water refil is automated. It is set to where it always remains within 1/4 inch of full.

Shots of the pond in question.

d

[preloader]

OK, ima say it. Im glad my dumb butt just has to hump packages. Fortunately, i can't even remember all the parameters so it wont make my head hurt thinking about it.

[feeder53]

(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )

Here is the formula, and there are a good number of variables.

[feeder53]

A real situation involves the fact that the humidity near the interface is much higher than even a short distance away, and that the water vapor must diffuse away. This effect will slow the evaporation down quite a a lot because the evaporation rate is proportional to the difference between the vapor pressure and the partial pressure of the substance, and diffusion can only take water away so fast. As the water evaporates, the partial pressure of water in the gas right over the water will be nearly equal to the vapor pressure, and then it will drop as you go away from the surface, and how steeply this drops (which depends on the airflow rate and how long the water has been there evaporating) determines the rate at which water will diffuse away. Even with a fan blowing air past the surface, the process is limited by diffusion very close to the surface because a thin layer of air (called the "boundary layer") right next to the surface does not move relative to the surface. I won’t do the work on the diffusion because it depends too much on the details of the setup. The diffusion constant for Nitrogen is 0.185 cm**2/sec at room temperature and 1 atm.

Lowering the partial pressure of water will raise the evaporation rate as mentioned above. Lowering the air pressure will increase the diffusion rate. The partial pressure of water at 100 degrees F is so high in fact, that bubbles will form spontaneously in the fluid and cause it to boil rapidly if the water is placed in a vacuum (upsetting the surface area because of all of the bubbles). Increasing the temperature will increase the evaporation rate. It appears in the denominator (in the square root) above, but a much more important dependence comes in with the vapor pressure.

vapor pressure is proportional to exp(-latent heat/RT)

[dannyboy]

Thanks feeder. That is as far as I got. The problem is those variables, as you mentioned. A flow of 20,000 gallons an hour churning over a rock surface will make the answer to the above meaningless.

As far as the surface, when you have a stiff breeze, or have the water roiling around, that too affects the formula stated above. To what extent, I have no idea, but I do know that it can raise the answer 10 fold plus.

d

[dilligaf]

Originally Posted by dannyboy

It is when its you and I. But for the customer that buys the pond, not so much so.
Then add to that the pond owner that claims the pond is leaking, but only when the water falls are on, and you have a three ring circus.

Actually the numbers can be pretty close, if you figure in the extra variables.

Im just not good enough for the task.

Actually, the water refil is automated. It is set to where it always remains within 1/4 inch of full.

Shots of the pond in question.

d

Way cool D!!!!!!!!!!!!!!Wanna build a pond for me? LOL

[tieguy]

No engineer here.

I would imagine your formula would also have to account not only for evaporation but also splash. Water in motion would tend to generate water droplets that would either splash or mist outside of the confines of the pond. This misting would also increase or decrease depending on temperature and wind.

This formula should work:


Everything I see tells me there are no easy formulas to calculate and quantify.

You would probably have to hire an engineer to perform the calculation. There are some web sites out there where people bid on jobs of this type, I'll try to get you a link if you're interested.

if the design of your ponds is fairly consistent and you can measure water loss on a daily basis you could perform your own measurements to use as a rough estimate.

P.S. Gorgeous pond. You are soooooooo talented.

[BLACKBOX]

Why don't you pose that question to Marilyn Vos Savant. She's the lady with the incredible I.Q. that answers questions from readers in the "Parade" section of your Sunday newspaper.

[trplnkl]

Why not just print out Feeders worded explanation, give it to your client and tell him to read it while drinkin a cold beer in the shade?

[JustTired]

Originally Posted by trplnkl

Why not just print out Feeders worded explanation, give it to your client and tell him to read it while drinkin a cold beer in the shade?

LOL. That was my exact thought.

After reading those explanations, the old saying "If you can't dazzle them with brilliance...baffle them with bull****." came to mind.

Some one who can afford a nice pond like that is worried about a "potential" small leak? As long as the price of water doesn't reach the price of gas, I don't think I'd worry about it.

[rod]

E=mc2 ?

It's a relatives rates problem that means you need to know Differential Calculus

[mathematics]

you won't be able to get an answer simply because you don't have exact volumes or surface areas. no amount of calculus will help you.

[UpstateNYUPSer]

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A Great Night's Sleep

(JK)

[drewed]

Originally Posted by mathematics

you won't be able to get an answer simply because you don't have exact volumes or surface areas. no amount of calculus will help you.

Yea, theyre too many variables in this to make absolutely no fun out of it lol

[helenofcalifornia]

Just Tired, you don't live in California where the price of water almost equals the price of gas!!! Ain't nothing cheap here.

[Big Babooba]

Simple engineering problem. If you have 10 men standing around the pond and they each consume 6 - 12 oz cans of beer per hour, then the water evaporation is a moot point. 8 men will do the trick, but I added 2 more for a margin of safety. But if they drink more than 6 beers per hour, the Law of Diminishing Returns comes into effect. Those poor boys may pee their pants thus decreasing the actual amount entering the pond.