How fast does water evaporate?

Factual Questions
1

I have a recently installed backyard pond. I suspect it might have a small leak but the pond installers claim the water loss is due to evaporation.

I’m skeptical.

The pond has approximately 137 square feet of surface area, plus another 20 square feet of water running down a small “stream” which includes a small waterfall at the source.

Here is my specific question. On a sunny 80 degree (F) day with light (5-10 mph) breezes and 25 percent humidity, how much would the water level drop after 12 hours if the only water loss was due to evaporation?

How much would evaporate if the air temperature was only 65 degrees (F) but with 60 percent humidity?

Other items that might come into consideration… the water temperature is about 65-70 degrees (F) and there is (as yet) minimal plant life.

2

I’m not going to attempt to re-write the equation I found but see if the link below might be helpful.

http://www.eren.doe.gov/rspec/equations.html

3

Thanks Whack-a-mole for this link. I think I could calculate this, but I need a little more guidance. For example, I don’t have any idea what the value for Y would be. (Y = latent heat required to change water vapor at surface water temperature, BTU/lb)

Also, how do I convert relative humidity to these P values?
(P[sub]DP[/sub] = saturation pressure at room air dewpoint, in.Hg.)
(P[sub]W[/sub] = saturation vapor pressure taken at the surface water temperature, in.Hg.)

Then of course there is the issue of how to calculate “active” surface area for my little stream… pretend there are lots of miniature swimmers? I’m guessing the additional evaporation from this area is relatively small anyway, so I’ll just treat it like “quiet” water.

4

Glad to help.

I would love to be the cool guy that could calculate this for you but you’ll note I wasn’t prepared to write out the equation (I’ve never bothered figuring out how to use the Code function of this board) for you much less solve it. It’s not that I’m lazy but rather that I am severely mathematically challenged. I’ll admit given all the proper numbers to plug in this equation I suppose I’d be up to the task as well but I likely have even less of a clue about the necessary values than you do (using the proper units and understanding how they relate in the answer is almost always my major stumbling block).

Fortunately I think there are people on this board who would have these answers (or know where to find them). Hopefully one of them will be along shortly.

5

~grin~ I think you’re already cool for getting me this far.

I really didn’t mean to imply that I was expected you to provide the answers to my follow-up questions. I was really just explicitly presenting them to the Teeming Millions[sup]TM[/sup].

6

saturation vapor pressure table:
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/watvap.html
You need to pck some temperatures and dew points.

Latent Heat of Vaporization of Water at 0°C lv = 2.50 x 106 J kg-1
Latent Heat of Vaporization of Water at 100°C lv = 2.25 x 106 J kg-1. (That’s 540 cal/g)
(http://www.met.psu.edu/dept/computers/third_order/bwpage4.html)

The value you want will be between these two numbers. Maybe someone else feels like converting all the numbers to the same units ?

7

Thanks Squink. Now what I have to do is convert joules-per-kilogram to BTUs-per-pound, and convert mmHg to inHg, and then find the dew point to get the value for P[sub]DP[/sub].

Then if I plug in all the values into the original formula supplied by Whack-a-mole it looks like all the units cancel out nicely to leave me with an evaporation rate in pounds-per-hour.

Of course, then I’ll have to convert pounds-per-hour into a cubic-inches-per-hour times number of hours, divided by the surface area in square-inches to get the number of inches the water level will fall.

Whew. I didn’t think it’d be this much work.

8

I have two ponds; one in a greenhouse at 25C, the other outside at about 12C. The greenhouse pond has a good sized waterfall with 4 splashy drops over about 3 feet of height. I’m in Alberta, at 3400 feet too, so evaporation doesn’t get a whole lot higher than here. My ponds are no bigger than 30 sq feet, and around 2.5 feet deep.

I don’t see more than a 2-inch drop in water level per week in the warm one, which gets no rain like the cold one; this is with some water actually splashing out from the waterfall too. If you feel the need to top up the pond more than once a week, chances are you may have a small leak. You could let the water go down until it stops to find it (if it does), or check out some of the pond book or website on how to find leaks.

Over 12 hours for your pond, just by experience and not looking at equations, I’d expect the water level to drop by mere millimeters if it’s only due to evaporation.

9

Ahhhh… the voice of experience. Thanks mmmiiikkkeee.

I’m at nearly sea level and I’m losing about 3/4 to an inch a day. At 150 square feet of surface area that’s equivalent to… well… damn, more formulas… let’s just say that’s a lot of water.

I’m still going to do the calculations to satisfy my own curiosity, but my gut and mmmiiikkkeee’s experience tells me I have a leak.

10

According to Whack-a-mole’s formula, you should be losing about one-tenth of an inch per day. Being less conservative, you might get ten times that much, but over twelve hours I’d expect less than half an inch, for sure.

11

According to Whack-a-mole’s formula, you should be losing thousandths of an inch per day. Being less conservative, you might get ten times that much, but over twelve hours I’d expect millimeters, for sure.

12

That first post had a calculation error in it; I didn’t think it had posted; sorry.

13

[Off Topic]
Sorry for the off topic post but I was looking at the link Algernon provided ( Humidity Formulas ) and couldn’t believe what I saw:

What deranged mind dreamed up that beauty? It’s not that it looks all that complex to actually work through but it is an ugly beast! Whoever he was that invented that thing he must have been suffering from heat stroke after smoking a fatty and downing a pint of Jack. I can only assume those numbers are some sort of constants derived from something else because if they truly were pulled out of some guys ass then I don’t think my stoned/drunken/heat stroke theory is too far off.

I sometimes marvel at the wonder of numbers and what you can do with them but occasionally I see things like this and then remember why I stayed away from mathematics.
[/Off Topic]

14

(150 square feet * 144 square inches/square foot * 0.75")/231 = about 70 gallons you’re losing per day… yep, that’s one hell of a lot of evaporation!!!

15

Nametag, if you come back to this thread and see this, did you actually do the full calculation? Or did you do some back of the envelope estimations?

I have to laugh because I had the identical reaction when I saw that calculation. Somebody had waaaaaay too much time on their hands.

mmmiiikkkeee, I didn’t believe your calculation and had to find the conversion factors and do the calculation myself. And sure 'nuff… slightly over 70 gallons a day. I’m sorry I doubted you. Based on this I think I need to rename my “small leak” to at least a “medium leak”.

16

Algernon, how about posting your final formula, with all the units normalized and such ?

17

I have to confess Squink, that I’m hoping Nametag did all the heavy lifting.

The formula I have so far looks a lot like the heat index formula we’ve been disparaging.

For example, here is the calculation for the dew point at 70 degrees (F), at 25 percent relative humidity… this is required to get the variable P[sub]DP[/sub].

DP = (-430.22+237.7ln(.25(6.1110**(7.5(5/9*(70-32))/(237.7+(5/9*(70-32)))))/(-ln(.25*(6.1110**(7.5(5/9*(70-32))/(237.7+(5/9*(70-32))))+19.08)

(admittedly, I’ve intentionally made this look more complex by combining a bunch of steps into a single formula, but hey, you get the idea…)

I’ll work on the series of equations and post them here. Perhaps later tonight, or tomorrow. I’m really curious about the result.

18

[Sorry for the delay. I’ve been busy with an important client IRL.]

I was hoping that either Nametag (due to his earlier posts in this thread) or kabbes (due to his mathematical skills) would come to my rescue, but alas, I had to do this myself.

At the request of Squink, I promised to post the final answer. So here it is.

Over the course of 24 hours, under the given conditions, my pond will drop 0.7 inches per every 24 hours (a loss of 68.8 gallons). Coincidently, just yesterday, I received a brochure from my city water commission in which it stated that 190 gallons of water is lost in a single day by a faucet dripping one drop per second.
One not-so-slight disclaimer: the solution I arrived at is assuming no errors in my calculations, which is something I’m not sure I’d bet on, especially with the unit conversions.

Here are the given conditions…
TA = air temperature = 80 F = 26.7 C
TW = water temperature at surface = 68 F = 20 C
A = surface area of pond = 155 square feet
V = wind velocity = 10 mph
H = relative humidity = 25 percent = 0.25

Here is the calculation for the dew point…
SVP = saturation vapor pressure = 6.11 * 10**(7.5*TA/(237.7+TA)) [formula]
If TA = 26.7, SVP = 34.89
VP = vapor pressure = H * SVP [formula]
If H = .25 then VP = 8.72
DP = dew point = (-430.22 + (237.7 * ln(VP))) / (-ln(VP)+19.08) [formula]
At conditions given, DP = 5.06 degrees C

Here is the formula for the evaporation rate…
R = (A(C1+(C2*V))/Y)(PW-PDP) [formula]
where R = evaporation rate in lbs per hour
C1 = 69.4 btu / hr ft[sup]2[/sup] in.Hg [constant]
C2 = 30.8 btu / hr ft[sup]2[/sup] in.Hg [constant]
Y = latent heat required to change water vapor at surface water temperature (in btu/lb)
PW = saturation vapor pressure @ surface water temperature (in in.Hg)
PDP = saturation vapor pressure @ dew point of air (in in.Hg)

Here is the calculation for Y…
Y = 2.50 * 106 joules per kg @ 0 C [from table]
Y = 2.25 * 106 joules per kg @ 100 C [from table]
Y = 2.45 * 106 joules per kg @ 20 C [extrapolation assuming linear “curve”]
Y = 2.32 * 103 btu per kg @ 20 C [converting 1054.35 joules per btu]
Y = 1.05 * 10**3 btu per lb @ 20 C [converting 2.2 lbs per kg]

Here are the calculations for PW and PDP…
PW = 17.54 mm.Hg @ 20 C [from table]
PW = 0.69 in.Hg @ 20 C [converting 0.03937 in. per mm]
PDP = 6.54 mm.Hg @ DP (5 C) [from table]
PDP = 0.26 in.Hg @ DP (5 C) [converting 0.03937 in. per mm]

With substitutions…
R = (A(C1+(C2V))/Y)(PW-PDP) [formula]
R = (155(69.4+(30.810))/1.0510**3)(0.69-0.26)
R = (155(377.4))/1050)(.43)
R = 55.7.43
R = 23.9 lb / hr

Converting to water level drop over the course of 24 hours…
E = evaporated water volume in lbs = 23.9 lbs * 24 hrs = 573.6 lbs per day
EG = evaporated water volume in US gallons = 68.8 gallons [ converting 0.12 USgallons per lb]
EI = evaporated water volume in cubic inches = 15,892 [converting 231 cubic inches per USgallon]

AI = surface area of pond in cubic inches = 155 square feet * 144 square inches per square foot = 22,230
L = water level drop = EI / AI = 15,892 / 22,230
L = 0.7 inches

19

Dammit, dammit, dammit! This is the second time I had to write this &#@**!

Dammit, dammit, dammit; I must have gotten the BTU/joule conversion factor from the ONLY SITE IN THE UNIVERSE where it’s wrong. My calculation was therefore off by a factor of 1000. Sorry Other than that (and a little rounding), your calculation checks mine.

As for the heat index equation, it’s probably NOT derived; such things are usually done experimentally. You determine from theory (and a little data) that you have a second-degree equation in x and y, and that you need terms of x, y, xy, and their squares. Then you juggle data until the numbers work out.

20

Thanks Nametag for checking my calculations. I was sweating it a bit there, seeing how my number came out so much different than your earlier post.

And thanks for coming back and posting. I was afraid I’d do all that work to write out my calculations for naught.

Regarding my OP… nearly three quarters of an inch of evaporation from my pond on a hot and dry day is more than I expected, but it’s in line with my actual experience. Maybe I don’t have a leak after all. I may owe the pond installers an apology.