I am in the planning stages of digging out an exsisting 1 acre pond from 4' to a 2 acre pond with a maximum depth of 8'. I am considering having a 4" well dug to supply the pond with well water during the heated months in central Louisiana. My concerns are the cool water being pumped into the pond and affecting the water temperture causing problems for the fish. Can anyone shine some light on this concern?

Welcome to the Pond Boss forum! It's always good to see another Louisianian on board.

Well water is not a problem! It's the best thing there is for a fish pond. The water is usually hard, and it often clears up the pond without adding chemicals.

Most well water has little dissolved oxygen. If you have already stocked fish, be careful not to run too much at a time with that big 4" well.

Welcome to the forum!

If you run the water over some bio-balls or something like that, and get it bouncing pretty well, it will aerate itself. Cecil has a pretty good system using three 5 gallon buckets and running the water from one bucket to another. I think if you ask him nicely he will post a pic and a how-to.

Talk with the well driller and see if he can put two lengths of screen on the end of the 4" casing. It will allow more water to flow, and the screen will last longer. Any idea of how many gallons of water a minute the proposed well will flow?

I pump 28 gpm of well water into my ¾ acre pond 24 hours a day during the summer months, my CC, LMB and HBG seem to love it. During the heat of the day they hang around the cooler water.

What a great dock setup! Looks great!

I fill my pond as needed with a 2 inch , 3 hp pump rated at 3600 gallons per hour. The water goes into a 5 gallon bucket that has holes drilled into the bottom and lower side. This bucket never over flows and I have enough oxygen with this manner of fill that I never have dead fish. I watch to make sure at all times that the fish are taking this water and it is not causing them any problems. My pond was filled with this well.

Have any of you guys calculated the electricity costs associated with running these well pumps? Do they run continuously or are they on a timer? My pond is in a remote area and I would probably have to pay to have utility poles run to it unless anyone knows if solar power is an option? Any feedback will be appreciated.

Mine is the well for the house, and I have a manual valve that will divert 100% to 0% of the water to the pond. I calculated for a 1 hp motor, $.07/hr cost. I don't know if I did it correctly or not tho. My electric rate is $.095/kwh

I just don't know if a well can keep up with my seepage and evaporation. It's a 1/2 acre pond. I guess I need to get with a well guy in Ellis county.

George, here are a couple of useful conversions for calculating energy cost.

1 KWH = 1.33 HP
1 HP = 0.75 KWH

You must first determine how much you pay per KWH.

The basic calcuation for cost is:

Horsepower X 0.75 X number of hours of operation X price per KWH

Example: Pump = 1 HP, 24 hour operation, Price = $0.10/KWH.
Daily operation cost = 1 X 0.75 X 24 X 0.10 = $1.80
Hourly operation cost = $1.80 / 24 = $.075

It looks like you were right on esshup.

Keep in mind that the price you end up paying is based on actual power usage as a result of amp draw. This is referred to as brake horsepower which is typically lower than the nameplate horsepower on your pump unless your running at the upper end of the pumps abilities. So this calculation provides a general idea of the cost.

I hope that helps.

using 1/4" loss per day you'll want a well/pump that can produce at least 28 GPM 24/7 for 1/2 acre.

Richard/Essup does about $54/mo cost sound about right?

Essup, could you link me to previous posts about your setup?

Correction: works out to 2.36 GPM (Thanks Chris)

21780 Surface sq ft
0.0208333333333333 depth ft
453.75 Cu. Ft.
0.133680556 Cu. ft/gallon
3394.28570300082 gallons/day needed
1440 min. day
2.35714284930612 GPM Needed

Ryan I came up with 2.36 GPM 24/7.

325828 Gallons per acre foot/ 12"= 27152 gallons per acre inch

27152 Gallons * .5 acre= 13576 gallons

13576 Gallons * .25" evaporation= 3394 gallons

3394 gallons / 1440 minutes per day= 2.36 GPM

I don't have any pics of my set-up, although I'll take a few tomorrow and post them. Either my pond is a lot smaller than I thought (could be, it's shrunk a lot in the past few months), or my well pumps out a lot more than I thought. I'm electrically challenged, and math isn't my forte, so I contacted a friend who is better than I am, who has a friend that is better than him, and what he sent me is posted below.

Is that clear as mud?

My pond is very irregular shaped, and I'm at a loss how to figure out the area until Google updates their satellite map. I was guessing that the pond when at a level that is 2' higher than it is now, was around an acre and a half.

Well pump is 1 hp, rated at 28 gpm. I didn't pump for 24 hr, and the level dropped 1". I pumped for 12 hr. and the level rose 3/8". The pond slope varies around the pond from 2:1 - 4:1. Ambient temp that day was 91* during the day, down to 70* at night. Daytime Rh was 60% I have a yardstick on a stake, so I can measure the water level pretty accurately.

Answer part 1:
Pretty sure there's enough info to solve this, which I'll attempt later this evening. We know the inflow rate (Qi) is 28 gpm and we know that after 24 hours we can raise the pond level 3/4 inch if we pump continuously.

Since the pond drains 1" in 24 hours absent our pumping, we therefore know that the inflow rate is 1.75 times the outflow rate. Now we have all flows in terms of a known flowrate; so far so good.

If we neglect evaporation (probably pretty reasonable over a single day) and we assume that the slope of the pond bank is also negligible over an inch of level change (seems pretty reasonable), we can approximate the shape of the pond by an easily computed volume, say a slice of a cylinder. Now we've got a formula for Volume in terms of area and level and it's off to the races.

From here, it's a matter of writing the equation for water balance, dV/dt=Qin-Qout, substituting pond height for level, rearranging to solve for area and Bob's your uncle!

I think.

Part 2

Data:

Pump discharge = 28 gpm
Pond level subsidence rate in absence of pumping = 1"/day Pond level increase rate with pumping = 3/8" per 12 hrs = 3/4" per day

Assumptions:
1. Near the surface, the sides of the pond have insignificant slope (makes it easier to solve, although still amenable to solution if we assume some arbitrary slope).

2. The shape of the pond's surface can be approximated as a circle.

3. Water loss due to evaporation is negligible (once we estimate the surface area, we can backtest this assumption by choosing a range of evaporation rates per unit area based on temperature and wind speeds).

3. Water temperature during the time period in question is constant, so volume per unit mass of water is constant.

4. Inflow due to rain or other natural causes is negligible.

Solution:

1) Water balance for pond

Rate of change of pond volume equals the inflow (pump discharge) minus the outflow (leakage through the bottom and sides of the pond), in symbols:

dV/dt = Qi - Qo

where
V = pond volume, gallons
Qi = rate of water flow in = pump discharge rate = 28 gpm Qo = rate of water leakage

Qo is not given explicitly, but we know that in the absence of pumping, the pond level subsides 1" per day. When the pump is running, the pond level increases 3/8" in 12 hours, or 3/4" per day. Therefore, the pump provides not only the volume lost to leakage, but an additional 3/4" per day. From this information we can say that the pump discharge is 1-3/4 (or 7/4) times as great as the leakage rate, or:

Qi = Qo / (7/4)
Qo = 4/7*28 gpm = 16 gpm

so:

dV/dt = 28 gpm - 16 gpm = 12 gpm

Our model of the pond near its surface is a cylinder (circular surface with straight sides), so the volume of a section is:

V = Area * Height

so:

dV/dt = d(Area * Height)/dt

Since the Area is constant, remove it from the differential and solve for it:

(dV/dt)/(dH/dt) = Area

we know dH/dt = 3/4"/day, so, solving and using unit conversions as necessary, we have:

12 gallons | 1440 min | day | 1 acre-ft | 12 in | = 0.85 acre
min | day | 0.75 in | 325,851 gallons | ft |

Great info, but I've got to go lay down - I've got a headache...

Here's the promised pics. The ball valve with the red handle is 2". The pipe coming out of the building is 1.5". I bumped up the 2" to 3" because I had the pipe here. There is a 20' section of 3" PVC, that transitions to 3" lay-flay hose, (so I can drive over it with the lawn tractor) back to 3" PVC, then at the pond it transitions to a 50' section of 2" lay flat hose, which is stuck inside of a 3" section of PVC so it won't erode the pond bank. I can also move/drive over the lay-flat near the pond to mow. The other ball valves are for future sprinkler usage. I have a shut-off with a drain inside the building so I can clear the lines for winter. While it's not the most aestetic, I just have the pipe/hose laying on the ground, nothing is permanent yet. From the ball valve to the end of the pipe at the pond is a run of 260'.




When I'm not in the house I just crank the valve all the way open, and right before I go to bed. The pump is a 1 hp 7 stage, and the head is only 20' max.
performance chart

Well, (no pun intended), it sounds like the daily operational costs are minimal. I'd gladly pay $60/month or even twice that if I can make a real difference in my ponds water level during the hot/dry months here in North Central Texas.

Thanks to everyone for their time and effort in answering this question. Mythbusters has nothing on you guys!
George Spann

Thanks for the welcome into the forum all. I greatly appreciate the advise as well as conversation on the post. This will keep the mind thinking.

I am still in the planning stages until the fall.

The pond was dug up about 40 years ago with a bulldozer for a watering hole for the cows and horses we raised at the time. It was about 5' on the deepest end and about 2' on the shallow end. The area of the pond is a natural "slew" that was dammed off to hold the water. The pond runs North & South with drainage canals North & South of the pond for drainage back into the Waxia bayou. There was a few fish at one time from frinds stocking from a fishing trip. Since then, the pond is about 3' on the deep side and about 1' on the shallow side. Fish wise, may have a "Shoepick" or two but nothing else.

I had a contractor that built my home come by to give me an estimate on digging it out and building up the banks around the pond. His estimate is about 4 days worth of digging and leveling the remaining dirt into the surrounding area. I have crops in the field now but should have it harvested by October. Once the milo has been harvested, then I will bring him in to begin the work.

I will most probably utilize Dunn's Fish Farm to stock the pond as well as my other supplies. I am planning a feeding program once stocked and hopefully will assist in producing some really nice fish. I will stock for a 2 acre pond in accordance with their recommendations.

I am isolated from most of civilization back in this area and have a well for water supply to our home. With a large family, I do not think my 3" well can produce for the family as well as the pond so I am considering having a well dug just for the pond. I did state a 4" well but still uncertain of what will be necessary to keep up with the pond. The soil is a "Silty Clay Loam" so I am not too concerned about the pond retaining the water. My water level is at 24' and will have a continues slope from the well down to the pond so I am assuming the most hydrastic head will be the 24'. Please correct me if I am wrong, anyone.

One question I do have is what is the amount of well water I can supply to the pond without it affecting the fish? I don't guess attempting to keep the pond at a constant level would be the correct approach???

Unless you plan to keep a fish that is temperature sensitive, I see no problem just using the well to keep your pond at a particular level. If you do something to aerate the water (preferably prior to introducing it to the pond), it sounds like all should be well. Even ignoring this would probably be ok if the volume of well water entering the pond wasn't particularly large relative to the overall size of the pond. Another potential problem from heavy use of well water is that well water can be oversaturated in gases (e.g., nitrogen), which can be bad for the fish. However, the limited volume of water you'll be adding should not be a problem, particularly if you aerate similar to what Cecil does with his buckets.

Thanks so much TheMoMule for the advise. I will procede as planned.

Ok all, I have been thinking quit a bit since I started this post and looking into another project but uncertain if it is obtainable.

Back ground info:
What I have is a water well that was hand dug many, many, a moon ago by my grandfather. The well casing is galvanised steel and has an ID of 1.50" and the tubing is 3/4" galvanized steel. The well was utilized, back then, with an above ground pump with a galvanized line running into a water trough for the cattle & horses. When he retired and sold off the cattle and horses, the well was abandoned.

Since then, I removed the steel tubing and utilizing what was available in my shed, I ran a steel weight on my open face and dropped it in the well. Water level is at 24' and the well total depth is 48'. I replaced the steel tubing with 3/4" PVC and a new foot valve. The foot valve is at 40' from surface and have about 2.5' stick up above ground. I worked the tubing up & down by hand and she produces water with each stroke. So far so good......I think...

As of today, I had all trees cleared out except the cypress, a u shaped levee facing the watershed, had a 65k # sheep’s foot to compact the clay loam levee as it was built, purchased a diffuser aeration system from Ted Lea, (Great guy, great product, good people there)which is 3 - 2 diffuser stations connected to a 5 station manifold system. I will have a 1.25" underground trunk line running from the pump to the manifold.

With that said, my little simple mind is thinking I can tee off of the trunk line near the pump and run a 1.25" trunk line 200' to my refurbished water well. Once the air line is at the well, I can seal off the casing from the tubing at surface, tie in the air line to the casing side of the well, (with a regulator to control air flow) apply air to the casing side and force water through the foot valve, up the tubing and have the flow gravity feed to the pond.

Questions that are floating in my head is, 1.) Will this work? 2.) How many gpm can I expect to get out of a 3/4" tubing? 3.) How do you calculate how much pressure it would require to move 8.33 ppg water up 24'from the present water level to surface? If this will work efficiently, then I will have to identify how much the water aquifer will be able to supply in a given time. I am assuming that the small volume I will be removing will not affect the recovery of the water aquifer.

Any help, thoughts or comments are welcomed. My beautiful and precious wife says I have to much time on my hands but can never find the time to complete all the projects I have running.

The math used on here is amazing. You folks really got ponds down to a science. The issue I have with my pond is that the drainage basin into the pond is pretty much sand. It takes a lot of rain to totally saturate the sand and leak into the pond. I get 3 inches of rain, the pond level goes up 3 inches.

A 5hp pump using 900 KWH will cost you approximately $67/mo.

Use this site to help you determine your specific costs:

http://www.consumerspower.org/home_energy/billestimator.php

Okie, I have some of the same problems. When I get a decent rain, all kinds of sand and debris wash in.

My average Kwh rate is $0.11.

That is some pretty cheap electricity.

900Kwh comes out to about $128.00 here.

Energy efficient three phase pumps running on single phase thru an Inverter/VFD will have a quick payback.

I remember in 04' when electric cost made a huge jump around here. I had a 15,500 sq ft shop full of equipment running all the time, and my sisters household electric bills were about the same as mine. Difference is, we were more energy efficient utilizing 3 phase.