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# H1 (cm) H2 (cm) HV Q (cm3/min) 14.1 1000 D D= 15cM Trial 1 L = ISCM Small Column - Trial 2 13.5 13 0.5 950 sandy sediment Trial 3 17.6 917 D= 23...

Hi there,

I need help calculating my hydraulic conductivity and the base flow from these values. The values needed to calculate the hydraulic conductivity for both columns is in the picture I have attached to this message. We use Darcys law (rearranged) to calculate for hydraulic conductivity (k). The rearranged Darcys law I used was k= Q * L / A (deltaH)

Would A (cross sectional area) be pi * r^2? so for the small column would A = pi * 7.5^2? I don't know what they mean by cross sectional area but I worked out A by halving the diameter for the columns and then using the equation to get the area of the circle to determine A. I may be completely wrong. Please help me calculate the hydraulic conductivity of the two columns.

After hydraulic conductivity is calculated, we are required to calculate base flow using Darcys equation Q = -K*A(deltaH/deltaA).

Our supervisor has given us the values of (deltaH/deltaA) =0.055m/m and cross sectional area measured on the field site = 3600m^2 ( this is different to the cross sectional area of the two columns mentioned above). I keep getting the answers wrong for base flow. Also, the base flow MUST be in m^3/s units, so the calculated hydraulic conductivity above is cm/min and therefore must be converted to m/s. Please help me with this  H1 (cm)
H2 (cm)
HV
Q (cm3/min)
14.4
8.3
6.1
1000
D
D= 15cM
Trial 1
L = ISCM
Small Column -
Trial 2
13.5
13
0.5
950
sandy sediment Trial 3
17.3
16.7
0.6
917
D= 23
Large Column -
Trial 1
30.6
29.4
1.2
1887
L
L=28%
Gravel rich
Trial 2
46.4
44.3
2.1
2188
sample
Trial 3
37.9
36.6
1.3
2055
2250
2200
RZ = 0.7793
2150
2100
k =
QXL
Ax AH
Flow Rate (cm^3/min)
2050
2000
Where:
k = hydraulic conductivity (cm/min)
1950
L = length between two manometers (cm)
Q = the outflow rate (cm3/min)
1900
A = cross-sectional area (cm?)
1850
AH = the total head difference (cm)
0
0.5
1
1.5
2
2.5
Head Change (cm)

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