Lecture 5. Darcy’s Law. Dmitriy Leykekhman Spring 2010Darcy’s Law. Dmitriy Leykekhman Spring...

Transcript of Lecture 5. Darcy’s Law. Dmitriy Leykekhman Spring 2010Darcy’s Law. Dmitriy Leykekhman Spring...

MATH 1050QMathematical Modeling in the Environment

Lecture 5. Darcy’s Law.

Dmitriy Leykekhman

Spring 2010

D. Leykekhman - MATH 1050Q Mathematical Modeling in the Environment Course info – 1
Experimental setup for pushing water through a geologicalsample

figure 2.7 from C. Hadlock’s book

Two-dimensional simplification from the previous figure


Doubling the pressure


Question: what do you think would happen to the rate at which waterwould be forced through the sample?

Doubling the pressure


Question: what do you think would happen to the rate at which waterwould be forced through the sample?

Doubling the length of the sample


Question: what do you think would happen to the flow rate in suchexperimental setup?

Doubling the length of the sample


Question: what do you think would happen to the flow rate in suchexperimental setup?

Doubling the pressure and the length of the sample


Question: what do you expect the net effect of the flow rate to be?

Doubling the pressure and the length of the sample


Question: what do you expect the net effect of the flow rate to be?

Doubling the cross-section area of the sample


Question: what do you expect to happen to the total amount of waterforced through the apparatus?

Doubling the cross-section area of the sample


Question: what do you expect to happen to the total amount of waterforced through the apparatus?

Summary of principles

1. More pressure will increase movement (flow rate) of water.

2. Longer length of aquitard substances will decrease the flow rate.

3. Larger cross section will increase flow rate, but not the speed!

4. Flow rate will depend on the properties of the geologic mediumunder consideration.

Revision of the experimental apparatus

figure 2.13 from C. Hadlock’s bookD. Leykekhman - MATH 1050Q Mathematical Modeling in the Environment Course info – 9
Summary of previous principles in mathematical language

1. The flow rate is proportional to the net driving pressure, representedby ∆h

2. The flow rate is inversely proportional to the length L of the sample

3. The flow rate is proportional to the cross-sectional area A of flowpathway

4. The constant of proportionality for the above relationships willnaturally depend on the specific geological medium

Physical parameters and their relationship

I h1, h2 are the heights of the water above some arbitrary plane.

I ∆h = h1 − h2I L is the length of the medium through which the water is moving.

I A is the cross sectional area.

I Q is the total flow rate, measure in volume per unit time.

I K is the corresponding constant of proportionality (hydraulicconductivity)

Q = K ×∆h× 1L×A

Typical ranges of values of porosity and hydraulicconductivity

Geological Medium Porosity, η Hydraulic Conductivity K (ft/day)

Gravel 0.25-0.40 100-100,000

Sand 0.25-0.50 0.01-1000

Silt 0.35-0.50 0.001-0.1

Clay 0.40-0.70 0.0000001-0.001

Sandstone 0.05-0.30 0.00001-0.1

Limestone 0.001-0.20 0.0001-0.1

Granite (fractured) 0.0001-0.10 0.0001-10

Darcy’s Law

Let

i =∆hL

i is called the hydraulic gradient.

Theorem (Darcys Law)

Q = KiA.

Two monitoring wells

figure 2.14 from C. Hadlock’s bookD. Leykekhman - MATH 1050Q Mathematical Modeling in the Environment Course info – 14

Question: What information do you need to apply the Darcy’s Law?

i =h1 − h2L

L is given. From the picture we can compute

h1 = 100 ft − 16 ft = 84 ft

andh2 = 96 ft − 18 ft = 78 ft

Hencei = (84− 78)/350 ≈ 0.17

We also need to know K. This can be obtained from the knowledge ofthe types of soil or rock aquifer consists of.

We also need to know A (not given in the picture.)



i =h1 − h2L


h1 = 100 ft − 16 ft = 84 ft

andh2 = 96 ft − 18 ft = 78 ft

Hencei = (84− 78)/350 ≈ 0.17





i =h1 − h2L


h1 = 100 ft − 16 ft = 84 ft

andh2 = 96 ft − 18 ft = 78 ft

Hencei = (84− 78)/350 ≈ 0.17





i =h1 − h2L


h1 = 100 ft − 16 ft = 84 ft

andh2 = 96 ft − 18 ft = 78 ft

Hencei = (84− 78)/350 ≈ 0.17



Block diagram of idealized aquifer


Computing the total flow rate

If we assume the value for the hydraulic conductivity K = 100 ft/day,picture gives all the info we need to compute the total flow rate.

i =h1 − h2L

=65 ft − 54 ft

1200 ft

A = 40 ft × 200 ft

Hence the total flow rate

Q = 100 ft/day× 65 ft − 54 ft1200 ft

× (40 ft × 200 ft) ≈ 7, 333 ft3/day

Question: How much in terms of gallons per day?



i =h1 − h2L

=65 ft − 54 ft

1200 ft

A = 40 ft × 200 ft


Q = 100 ft/day× 65 ft − 54 ft1200 ft

× (40 ft × 200 ft) ≈ 7, 333 ft3/day




i =h1 − h2L

=65 ft − 54 ft

1200 ft

A = 40 ft × 200 ft


Q = 100 ft/day× 65 ft − 54 ft1200 ft

× (40 ft × 200 ft) ≈ 7, 333 ft3/day




i =h1 − h2L

=65 ft − 54 ft

1200 ft

A = 40 ft × 200 ft


Q = 100 ft/day× 65 ft − 54 ft1200 ft

× (40 ft × 200 ft) ≈ 7, 333 ft3/day


Concentration of the contaminant



Let C denote the concentration of the contaminant ( measured inweight/volume ). Then,

C =contaminant inflow rate

Q, the water flow rate


ExampleSuppose the flow rate in the aquifer Q = 1200 ft3/day and thecontaminant is leaking at the rate of 2 lb3/day.Thus the concentration of contaminant is

C =2 lb/day

1200 ft3/day≈ 0.00167lb/ft3.

Problem: Convert the result to kilogram per cubic meter kg/m3, to gramper cubic centimeter g/cm3, and to parts per million ppm.


ExampleSuppose the flow rate in the aquifer Q = 1200 ft3/day and thecontaminant is leaking at the rate of 2 lb3/day.Thus the concentration of contaminant is

C =2 lb/day

1200 ft3/day≈ 0.00167lb/ft3.

Problem: Convert the result to kilogram per cubic meter kg/m3, to gramper cubic centimeter g/cm3, and to parts per million ppm.

Summary of physical parameters

I h1, h2 are the heights of the water above some arbitrary plane.

I ∆h = h1 − h2I L is the length of the medium through which the water is moving.

I A is the cross sectional area.

I Q is the total flow rate, measure in volume per unit time.

I K is the hydraulic conductivity.

I i = ∆h/L is the hydraulic gradient.I C is the concentration of the contaminant.

Lecture 5. Darcy’s Law. Dmitriy Leykekhman Spring 2010Darcy’s Law. Dmitriy Leykekhman Spring...

Documents

Transcript of Lecture 5. Darcy’s Law. Dmitriy Leykekhman Spring 2010Darcy’s Law. Dmitriy Leykekhman Spring...