HONR 297Environmental ModelsChapter 2: Ground Water2.5: The Interstitial Velocity Equation

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What we Know so FarRecall that we are trying to answer

two general questions related to ground water flow.◦How much water flows through an

aquifer?◦The velocity of water in the aquifer.

So far, we’ve seen how to answer the first question via Darcy’s Law.

For the second question, let’s consider some “experiments”!

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Experimental Set-up

In the experiments that follow, we will use the set-up above as our base case.

The tube through which ground water flows will have a square cross section, with dimensions 1 m by 1 m, with cross-sectional area one square meter.

For the end view, we are looking at water flowing out at us!

Ground Water Flow Direction

Side View End View

1 m

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Experiment 1

Suppose water is flowing through the tube at a rate of ten cubic meters per minute.

Then the total amount of water in a ten-meter long piece of tube would have to flow out of the right end in one minute.

Question: What would the water’s speed (or velocity) need to be for this to be accomplished?

Answer: The water would have to be moving at a speed of 10 m/min.

Ground Water Flow Direction

Side View End View

1 m

10 m

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Experiment 2

Suppose water is flowing through the tube at a rate of ten cubic meters per minute, but half of the original flow pathway has been blocked.

Now, the total amount of water in a 20-meter long piece of tube would have to flow out of the right end in one minute.

Question: What would the water’s speed (or velocity) need to be for this to be accomplished?

Answer: The water would have to be moving at a speed of 20 m/min.

Side View End View

1 m

20 m

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Experiment 3

Suppose instead of the half of the original flow pathway being blocked, as in the second experiment, now we fill the tube with geologic material (such as gravel or sand).

If the open spaces or voids in the tube comprise 50 percent of the available open space, then again half of the pathway will be blocked.

Thus, if water is flowing through the tube at a rate of ten cubic meters per minute, we are in essentially the same situation as in the second experiment!

Again, the total amount of water in a 20-meter long piece of tube would have to flow out of the right end in one minute, so the water’s velocity would have to be 20 m/min.

Here, we are talking about net flow of all of the water in the flow direction – due to the gravel or sand, some of water droplets may not move to the right in straight lines, so they’d have to actually be moving faster, as all the water in the 20-foot piece of tube must make it out in 1 minute!

Side View End View

1 m

20 m

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PorosityDefinition

◦The porosity of a geologic material is the fraction of the bulk volume that is actually open spaces or voids. We denote porosity with the Greek letter eta (η).

Note: Examples of open spaces or voids include spaces between particles in soil or open fractures in solid rock.

The geologic material in Experiment 3 would have porosity 50% or 0.5.

Typical porosity values are shown in Table 2.1 in our text, p. 26.

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Experiment 4

Suppose we put sand in the tube with a porosity of 0.25.

Question: What would the water’s velocity need to be if the flow rate is measured to be 8 m^3/min?

Answer: ◦ In order to get 8 cubic meters of water in the tube’s open

spaces, since only 25% of the tube has open space, we’d need the tube’s length to be 4*8 = 32 m.

◦ It follows that the water’s velocity would need to be 32 m/min.

Side View End View

1 m

32 m

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Fluid VelocityLooking at the examples above,

we see that in order to determine fluid velocity in an aquifer, all we need to do is look at the volumetric flow rate per unit cross-sectional area (for example, one square meter) and divide by the porosity.

Check that this works in each case above!

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FluxDefinition

◦ The volumetric fluid flow rate per unit cross-sectional area is known as flux and denoted q.

From Darcy’s Law, flux is given byq = Q/A = (K i A)/A = K i. (1)

Note that units of flux q are [q] = (length^3 per time) per length^2

which reduce to length/time … (check!). It follows from (1) and our experimental

examples that the velocity of ground water in an aquifer is given by

ν = q/η = K i /η. (2)

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Interstitial Velocity EquationDefinition:

◦The interstitial velocity equation is

ν = (K i)/ηwhere ν = the effective fluid velocity along the axis of the flow pathway;K = the hydraulic conductivity of the aquifer material;

η = the porosity of the medium.

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Example 1Note that to compute interstitial velocity, all

we need to know about an aquifer are its hydraulic gradient (which depends on measurements taken from the aquifer via monitoring wells) and its hydraulic conductivity and porosity, which depend on the aquifer’s material make-up.

Recall Example 1 from the section on Darcy’s Law where we found the volumetric flow rate for an aquifer – let’s try to find the velocity of the ground water in this aquifer, assuming that the aquifer’s porosity is 40%.

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Example 1Example 1 (from Darcy’s

Law Notes): Imagine an underground sand aquifer that is 40 feet thick, 200 feet wide, and has a hydraulic conductivity of 25 ft/day. Two test wells 750 feet apart are drilled into the aquifer, along the axis of flow, and the measured head values at these wells were found to be 120 and 114 feet, respectively.

Solution:From this information,

we see thatK = 25 ft/day i = Δh/L =

(120 ft -114 ft)/750 ft = (6 ft)/(750 ft)

η = 0.40Thus,

ν = K i/ η =

= ((25 ft/day)(6 ft)/(750 ft))/(0.40)

= 0.5 ft/day

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Example 2Since velocity = distance/time, it follows

that time = distance/velocity.Thus if we know the interstitial velocity

of an aquifer’s ground water, we can figure out how long it takes for the ground water to travel a given distance!

For example, the ground water in the Example 1 aquifer will travel 2 miles in

((2 miles)/(0.5 ft/day))(5280 ft/mile)) = 21,120 days = (21,120 days)/(365 days/yr) ≈ 58 years

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HomeworkRead Sections 2.5 and 2.6 in

Hadlock.

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ReferencesCharles Hadlock, Mathematical

Modeling in the Environment, Chapter 2, Section 5.