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Water loss in a storage pond 4

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EngineerofSteel

Structural
May 18, 2005
156
A dairy pumps water into a large lagoon. Typically, the pressure head averages 10'.

Currently, they lose a lot of water due to permeable soil at the base.

They want to empty the lagoon and emplace a clay liner. Soil samples have been sent, and a K value is requested from the lab.

My task is to calculate the before and after water loss as part of an economic feasibility assessment.

However, in reviewing Darcy's law, it looks like I need to take two hydraulic head measurements in the field. Am I correct?

I have this equation: Q= -KAi
i= (h1 - h2)/L

h1 and h2 are the two hydraulic head measurements, and L the distance between them.

Is there a way to calc this with just the K value?
 
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CarlB,

Thanks for the link. Yes, this is a steady state situation.

How do I account for the 10 feet of water? This will accelerate infiltration.

Is there another equation?
 
The 10' pond depth affects the rate initially (say when first filled) but over time the gradient approaches 1.
The gradient may be less than 1 if the groundwater table is shallow & "mounding" occurs. This is discussed in that document I referenced.
 
There is always a surprise, it seems.

Today I learned there is a very high water table on this dairy. They install drain pipe at 8' below grade to drain water (which otherwise stymies crop growth).

So, there will not be a static state. water is constantly moved off-site by the drains.

So, how now to calculate i for my Darcy's equation, Q=-KAi ?

Alternate methods to calculate water loss will be appreciated.

Thanks.
 
Evaporation will consist of approximately 30 inches of water loss per year. When you say pressure head, I assume you mean depth. A measurement of the pressure of the water in the soil is needed to find the change of head between the bottom of the pond and the soil voids. I would solve a flow net for a unit width slice.
 
civilperson,

We have exact evaporation rates calculated per local weather averages and extremes.

Yes, the pressure head in this case is depth.

Can you please elaborate on " a flow net for a unit width slice" ?

If you can give the reference for this procedure also, it is appreciated. Thanks.

 
Let me take a stab at this: Let's say you have a pond and the depth to the mudline is 10 ft. Let's say the underlying natural soil has a hydraulic conductivity of 4x10E-3 ft/min (2x10E-3 cm/sec). Let's say that there is some subdrain that's 8 ft below the mudline that maintains an elevation head only (i.e., there is no pressure head). Let's say that the pond is 1 acre in plan dimension.

The volume of water that would move vertically from the mudline to the drain would be calculated using Darcy's law as such:

Q=kiA

Using the drain pipe elevation as the datum, the gradient would be:

DeltaH/DeltaL where DeltaH is the difference in head between the mudline and the pipe. The head at the mudline is 18 ft (8 ft of elevation head and 10 ft of pressure head) and the head at the pipe would be zero (there is no pressure head and it's the datum). Delta L would be 8 ft. So the gradieint is 18/8, which is 2.25. The flow would then be:

Q=4x10E-3(2.25)45,560 = 392 cfm (about 3,000 gpm!)

Let's now say that you place a 6-in thick clay liner with a hydraulic conductivity of 4xE-6 ft/min. Under this scenario, you would then use Darcy's law for the liner thickness, which would have a completly different gradient - you'd have all of the head loss in the liner. Using the bottom of the liner as the datum, you'd have 10.5 ft of pressure head attenuated in 0.5 ft, which is a gradient of 21. For this case:

Q= 4xE-6(21)43,560 = 3.66 cfm (27 gpm)

Hope this helps.

f-d



¡papá gordo ain’t no madre flaca!
 
Darcy's law

v = k*i
where i = (H+d)/d

v = specific discharge
k = hydraulic conductivity (given by the lab)
H = total head (water depth)
d = liner thickness
 
fattdad & MEM1,

Great posts. This is what I needed. Thanks very much.

-DD
 
Draining the pond is unnecessary. The bentonite clay can be broadcast from a boat or from shore and automatically travels/flows to the leaks in the bottom. 6" depth is way more than is commonly used in gold course ponds or decorative water displays.
 
civilperson,

You make a good point. The local inspector and my boss are likely thinking of the industry standard for effluent lagoons.

However, this lagoon is fresh water. I don't know if 6" thickness will be required.

However, there is not a specific "leak". The entire area drains well. The liner is intended to prevent water loss, and thereby reduce the cost of pumping water into it.

Thanks,
-DD
 
You can actually use bentomat if you want to. It's a sheet product that may be less than 1/2-thick. I wouldn't use the broadcast method as you are trying to fix a leak over the entire pond area. I'd drain the pond and constructt a liner.

You can follow my example and use whatever thickness you want to.

f-d

¡papá gordo ain’t no madre flaca!
 
The bottom of this pond is not lined. The sides have a synthetic liner. The bottom is compacted native soil. The entire pond bottom drains water.

This is a freshwater storage pond. The original designer did not account for the loss.

The pond will be drained and a clay bottom will be placed and compacted.

-DD
 
I'm curious, why is the dairy pumping "fresh" water into this pond?
 
A Dairy uses a lot of water to clean the cows' hooves and undersides before they enter the milking parlor. More water is used to spray utters, and the largest volume is used to "flush" the spray pen, drip area and alleys. That effluent water enters other settling lagoons in series.

All dairies need water storage. This is usually done with tanks. This dairy has a fresh water pond.
 
Thank you for posting this forum. I am currently working on a very similar problem where I need to find the required clay pond liner thickness. All of the responses have helped clarify. I just have one question.

Darcy's equation can be re-written for this situation as:

d=H*k/(v-k), where
d = pond liner thickness
H = water depth in pond
k = hydraulic conductivity (coefficient of permeability)
v = specific discharge

If k>v, then the answer is a negative thickness. What do you (or I) do with a negative answer? Is it valid? Should (v-k) be an absolute value?

-RR
 
Just to have a look a the big picture: Have you looked at a cost analysis of a PVC liner vs the clay liner you're planning to install? If you don't have a source of clay nearby, shipping costs can make it prohibitive.

Tim Grote - The Irrigation Engineers.
 
Rookierick

The ponds I deal with are runoff control ponds and are required to have a minimum of 12" of compacted soil liner. If your pond is a runoff control pond, check with you local environmental protection for required standards. If your pond is a fresh water pond as in the above situation, then there are probably no required construction standards. If your coefficient of permeability is low enough, then theoretically a liner thickness is not required and you would have very little seepage. That being said, the coefficient of permeability was determined by compacting the liner material to a specified compaction and moisture content. Your liner no matter the thickness should be constructed to the same compaction and moisture. The thicker the liner, the lower the seepage rate. If I were you, I would still recommend at least a 6" lift, and make sure you have proper compaction and moisture to match what the lab tests indicated.
 
If you are in a area where the pond will freeze over in winter, a common way to help control leakage after you have it built, then you spread bentonite on the ice and let it do the job in the spring.

Draining a pond and doing the liner can run into problems with mud causing significant construction problems. I'd look very carefully at this draining option before I jump in and do it. Asking you local contractors about this idea may help you decide.
 
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