I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures between
75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a day.

One cubic inch = 0.00432900431 US gallons

So, measure the drop in water level (in inches) after 24 hours and multiply the
result by the area of your pond (in square inches) then by 0.00432900431 US
gallons, and you'll have your answer.

For instance, my pond is 4 feet by 12 feet. On a typical hot summer day, I lose
anywhere from 0.25 inch, to 0.5 inch. Using the lower number, which is more
typical for my pond, I get

48 in x 144 in x 0.25 in x 0.00432900431 US gallons = 7.4805 gallons per day.

Our pond can lose an inch in a day, but we have two falls and two streams to
promote evaporation. It loses more like 1/4 to 1/2" per day if the pump is
off. Both surface and circulation count.

Jim

--
____________________________________________
See our pond at: home.bellsouth.net\p\pwp-jameshurley
Ask me about Jog-A-Thon fundraiser (clears $120+ per child) at: jogathon.net

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures between
75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a day.

One cubic inch = 0.00432900431 US gallons

So, measure the drop in water level (in inches) after 24 hours and multiply the
result by the area of your pond (in square inches) then by 0.00432900431 US
gallons, and you'll have your answer.

For instance, my pond is 4 feet by 12 feet. On a typical hot summer day, I lose
anywhere from 0.25 inch, to 0.5 inch. Using the lower number, which is more
typical for my pond, I get

48 in x 144 in x 0.25 in x 0.00432900431 US gallons = 7.4805 gallons per day.

Our pond can lose an inch in a day, but we have two falls and two streams to
promote evaporation. It loses more like 1/4 to 1/2" per day if the pump is
off. Both surface and circulation count.

Jim

--
____________________________________________
See our pond at: home.bellsouth.net\p\pwp-jameshurley
Ask me about Jog-A-Thon fundraiser (clears $120+ per child) at: jogathon.net

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

Thanks for the information.

I am new to ponds and I was not sure if I had a leak in the pond or it
was evaporation. My 120 sqft pond is in three pools. There are two
sluices from the top to the middle and one from the middle to lower.
Sluice is about 9 inches wide. The upper ones has about 1 inch of water
flow through the sluices from the top pond and the middle has about 2
inches of water. I am loosing about 20 gallons per day so from what you
say I may not have a serious leak.

Jim and Phyllis Hurley wrote:
Our pond can lose an inch in a day, but we have two falls and two streams to
promote evaporation. It loses more like 1/4 to 1/2" per day if the pump is
off. Both surface and circulation count.

Jim

Keith Nuttle wrote in message ...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

Google will reveal this to be a complex thermodynamic question which
has been given much attention. One answer I have found is:

"References:
I work with an equipment for measuring
Evaporation called Evaporimeter. We have found
evaporation rates of water in 20 degrees Celsius
to be around 40 g/m2/hr.

That completely ignores humidity [or dewpoint], unfortunatly. That
quoted rate works out to about very roughly 1/8 inch/hour at 68F.

If you do not get a satisfactory answer here, google around for a good
scientific answer, besides, "Couple of inches," or inchaso."
--
Crashj

Keith Nuttle wrote in message ...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

Google will reveal this to be a complex thermodynamic question which
has been given much attention. One answer I have found is:

"References:
I work with an equipment for measuring
Evaporation called Evaporimeter. We have found
evaporation rates of water in 20 degrees Celsius
to be around 40 g/m2/hr.

That completely ignores humidity [or dewpoint], unfortunatly. That
quoted rate works out to about very roughly 1/8 inch/hour at 68F.

If you do not get a satisfactory answer here, google around for a good
scientific answer, besides, "Couple of inches," or inchaso."
--
Crashj

There is a nomograph program located at
http://www.kcmmb.org/FAQ/faq-default.htm which is used for determining the
evaporation rate from concrete to determine when protective measures need to
be taken to prevent drying shrinkage cracks. The nomograph is a zip program
which is downloaded and installed on your computer. The inputs are
atmospheric temperature, relative humidity, concrete temperature (pond
temperature) and wind velocity. The output is in pounds per square foot per
hour. Using average high and low temperature, average humidity and wind
velocity you could compute the 24 hour evaporation loss. A loss of 1 pound
per square foot would translate to about 1 1/2 inches.
--
RichToyBox
http://www.geocities.com/richtoybox/pondintro.html
"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

Kieth,

There are some empirical ways to find out about your rate of evaporation.
There is a sheet on it (that we wrote a few years ago). I forgot who has it
now.

Briefly, put a cut off milk jug in the pond on a solid base. Fill it to the
same level as the pond. Let both sit. They face essentially the same
evaporation rates. If the pond goes down at a markedly faster rate than the
jug, you have a leak. In this experiment, both the jug and the pond need to
be still. You can't have the pond running a lot of water over the falls,
etc, increasing the evaporation of the pond.

Let us know if you try it.

Jim

--
____________________________________________
See our pond at: home.bellsouth.net\p\pwp-jameshurley
Ask me about Jog-A-Thon fundraiser (clears $120+ per child) at: jogathon.net

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.

Kieth,

There are some empirical ways to find out about your rate of evaporation.
There is a sheet on it (that we wrote a few years ago). I forgot who has it
now.

Briefly, put a cut off milk jug in the pond on a solid base. Fill it to the
same level as the pond. Let both sit. They face essentially the same
evaporation rates. If the pond goes down at a markedly faster rate than the
jug, you have a leak. In this experiment, both the jug and the pond need to
be still. You can't have the pond running a lot of water over the falls,
etc, increasing the evaporation of the pond.

Let us know if you try it.

Jim

--
____________________________________________
See our pond at: home.bellsouth.net\p\pwp-jameshurley
Ask me about Jog-A-Thon fundraiser (clears $120+ per child) at: jogathon.net

"Keith Nuttle" wrote in message
...
I live in central Indiana, and we have experienced daytime temperatures
between 75 and 90 degrees with dew points of 55 to 65 degrees.

My pond has about 120 square feet of surface area.

Can someone give me an estimate, in gallons, of the evaporation during a
day.