# Groundwater GLY 265 - Weebly Exercise 2 Sieve Size Grain size(mm) Mass retained in each sieve (g)...

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1

Groundwater GLY 265

Practical Session.2

Mr. Y Van Wyk

Department of Geology

University of Pretoria

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Exercise 1

The following parameters were measured in a constant head

permeameter test:

h = - 5 cm

l = 10 cm

Cross sectional area A = 100 cm2

Duration of experiment t = 1 h

Volume of percolated water after 1 h: V = 3,6 liter

a. What is the hydraulic conductivity of the sample?

b. What is the darcy velocity and the linear velocity, if the sample

(porosity 0.2) is representative for an aquifer (hydraulic gradient

dh/dL = -0.0025)?

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Solution

Q = kf I A kf = Q/(I*A)

Q = 3.6 l in 1 h = 3.6 x 10-3 m/60 x 60s = 1 * 10-6 m3/s

A = 100 cm2 = 0,01 m2

kf = 1 x10-6 m3/s/0.5 x 0.01m2 = 2 * 10-4 m/s sand

Darcy velocity

vf = Q/f vf = kf * I vf = 2 * 10-4 m/s * 0.0025 = 5.0 * 10-7

m/s

Linear velocity:

vn = va = vf /ne= 2.5 * 10-6 m/s

I = Dh/Dl = 5cm/10cm = 0,5

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Exercise 2

Using the tabulated results of a grain size distribution test on a field

sample, perform the following tasks:

a. Prepare a grain size distribution curve for this sample.

b. Calculate the hydraulic conductivity and the uniformity coefficient. Use the

Hazen and Kozeny – Carmen correlations to estimate K. (c = 0.0116m.sec-1)

c. Is this a well-graded or poorly graded sample?

NB: for the Kozeny-Carmen equation assume ρwg =9810 N/m3 and μ = 1.4 x 10-3

N.s/m2 , η = 0.38

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Exercise 2

Sieve Size Grain size(mm) Mass retained in

each sieve (g)

Cumulative Mass

Passed (g)

Total Percent Passed

¾ in. 19 191.0 830.5 81.3%

5/8 in. 15.9 - - -

½ in. 12.7 131.7 698.8 68.4%

3/8 in. 9.5 - - -

¼ in. 6.4 184.5 514.3 50.3%

No.4 4.75 51.5 462.8 45.3%

No.10 2.00 152.0 310.8 30.4%

No. 20 0.841 - - -

No. 40 0.420 - - -

No.60 0.25 191.8 119.0 11.6%

No. 100 0.150 66.2 52.8 5.16%

No. 200 0.075 49.1 3.7 0.36%

Pan <0.075 3.7 0.0 0.0%

Sum 1021.5

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Solution

• Hazen Method

K= C(d10)2

= 0.0116 m.sec-1x (0.25mm)2

= 7.25 x 10-6 m/s

Kozeny – Carmen Equation

1801

250

2

3 d

n

ngK w

ρwg/μ = unit weight/viscosity of water

n = porosity

d50 = median grain diameter (50%

passing)

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Solution

K = (7.0 x 106m.sec-1)*(0.1425)*(2.37 x 10-7)m2

= 0.234m/s

Orders of magnitude overestimation!!!

Cu = d60/d10 = 10.5mm/.025mm = ranges btw (30-40) Very poorly

sorted

1801

2

50

2

3 d

n

ngK w

8

0.25mm

6.5mm

10.5mm

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Exercise 3

If the elevation of h1 is 35m and the elevation of h2 is 0m, what is

the hydraulic gradient of the distance from h1 to h2 is 5.6km?

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Solution

• Equation:

• I=(h2-h1)/L

• What we know:

• h2-h1= 35m L=5.6 km

• Answer: I=6.25m/km

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Exercise 3

Find the velocity of the water in the picture if the hydraulic conductivity is

114m/day.

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Solution

• Equation: V=KI or V=K(h2-h1/L)

• What we know:

• K=114m/day (h2-h1)=60m L=1000m

• Hydraulic gradient (I) = .06

• Solution:

• V=114m/day * .06

V=6.84 m/day

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Exercise 4

• An aquifer is 2045m wide and 28m tall. Its Hydraulic

gradient is 0.05 and its K is 145m/day.

a. Calculate the velocity of the groundwater as well as

the amount of water that passes through the end of the

aquifer in a day if the porosity of the aquifer is 32%.

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Solution

Equation:

V=KI or V=K(h2-h1/L) and Q=V*W*D

What we know:

• K=145m/day I=.05

• W=2045 D=28 Porosity =32%

• Solution:

First we must solve for V

• V=145m/day * .05

V=7.25m/day• Now that we know V we can determine the discharge (Q) of water through

the end of the aquifer

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Solution

• Q= 7.25m/day * 2045m * 28m

• Q=415,135 m3/day.

This means that each day, if the aquifer had a porosity of

100%, like a river, that 415,135 m3/day.

However, most of the aquifer is not made up of empty

space, only 32% is (porosity of 32%).

So we must multiply 415,135 m3/day by .32 to get the

amount of water that flows through the aquifer with a

porosity of 32%.

• 415,135 m3/day * .32 = 132843.2 m3/day