An aquifer has a porosity of 0.37 and a specific yield of 0.21. A cone of depres
ID: 113460 • Letter: A
Question
An aquifer has a porosity of 0.37 and a specific yield of 0.21. A cone of depression formed around a pumping well has a volume of 950 m3.
a. What volume of water will drain from the cone of depression? [4]
b. What is the degree of saturation in the pores after gravity drainage? [4]
3. Information for a column packed with sand
length of sand column: 50 cm
grain size distribution shown as Curve 1 in problem 1
inside diameter of (circular) column: 6 cm
head change from inlet to outlet: 30 cm
hydraulic conductivity: 0.01 cm/s
porosity: 0.29
tortuosity: 0.75
a) what is the specific discharge, linear velocity, and interstitial velocity? [6]
b) what is the Reynold’s number, and is the flow turbulent or laminar? [4]
c) what is the intrinsic permeability of the sand? [3]
Explanation / Answer
a. Given, porosity (n)= 0.37
Specific yield (Sy) = 0.21
Volume of pumping well (V) = 950m3
Sy = Vw / V .....(i)
where Vw= volume of drained water
From Equation (i), Vw = Sy * V
= 0.21 * 950
= 199.5 m3
b. Degree of saturation ( Sd) = Vw / Vv ...(ii)
Vv= volume of void
Now, n = Vv / Vt
Vt = Total volume = 950+ 199.5 = 1149.5 m3
=> Vv= 0.37 * 1149.5
= 425.31 m3
From eq. (ii)
Sd = 199.5/ 425.31
= 0.46
= 46%
3. a) Specific Discharge (q) = K * dh/ dl
where, K = hydraulic conductivity
dh/ dl= hydraulic gradient
q = 0.01 * (30/6)
= 0.05
Linear velocity (V) = q/ n
= 0.05/ 0.29
=0.17 cm/s
Interstitial velocity = Linear velocity = 0.17 cm/s
b) Reynold number ( Re ) = ( V * l ) / v
where l= length of the sample column = 50cm
v= kinematic viscosity of fluid = tortuosity= 0.75
Re = (0.17 * 50 ) / 0.75
= 11.49
So the flow is laminar.
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