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A sequence of horizontally lying sandstones and shales is known to exist beneat...

May 6, 2024

Solution by Steps

step 1

Calculate the hydraulic head at each borehole using the formula: hydraulic head = elevation of the top of the bore - depth to water inside the bore

step 2

For borehole A: hydraulic head = 213 m - 33 m = 180 m

step 3

For borehole B: hydraulic head = 275 m - 85 m = 190 m

step 4

For borehole C: hydraulic head = 337 m - 63 m = 274 m

step 5

For borehole D: hydraulic head = 240 m - 40 m = 200 m

step 6

Calculate the differences in hydraulic head between boreholes

step 7

The hydraulic gradient between two points is the difference in hydraulic head divided by the distance between those points

step 8

Use the coordinates of the boreholes to calculate the distances between them

step 9

For boreholes A and B, the distance is sqrt((1500 - 1500)^2 + (4000 - 2000)^2) = 2000 m

step 10

The hydraulic gradient between A and B is (180 m - 190 m) / 2000 m = -0.005

step 11

For boreholes A and C, the distance is sqrt((1500 - 2000)^2 + (4000 - 3000)^2) = sqrt(500^2 + 1000^2) = sqrt(1250000) = 1118.03 m

step 12

The hydraulic gradient between A and C is (180 m - 274 m) / 1118.03 m = -0.084

step 13

For boreholes A and D, the distance is sqrt((1500 - 3232)^2 + (4000 - 3000)^2) = sqrt(1732^2 + 1000^2) = sqrt(3990624) = 1997.66 m

step 14

The hydraulic gradient between A and D is (180 m - 200 m) / 1997.66 m = -0.01

step 15

Determine the direction of flow by comparing the hydraulic heads; water flows from higher to lower hydraulic head

step 16

Calculate the average hydraulic gradient using the results from steps 10, 12, and 14

step 17

The average hydraulic gradient is (0.005 + 0.084 + 0.01) / 3 = 0.033

step 18

Calculate the flow through a 1000 m wide section of the sandstone unit using Darcy's Law: Q = T * i * w, where Q is the flow rate, T is the transmissivity, i is the hydraulic gradient, and w is the width of the section

step 19

Using the given transmissivity of 100 m^2/day and the average hydraulic gradient of 0.033, the flow rate is Q = 100 m^2/day * 0.033 * 1000 m = 3300 m^3/day

[question c] Answer

The hydraulic gradient is approximately 0.033, and the direction of flow is from higher to lower hydraulic head.

[question d] Answer

The flow through a 1000 m wide section of this sandstone unit is 3300 m^3/day.

Key Concept

Hydraulic gradient and Darcy's Law

Explanation

The hydraulic gradient is a measure of the change in hydraulic head per unit distance in the direction of the highest rate of decrease of hydraulic head. Darcy's Law is used to calculate the flow rate through a porous medium, given the hydraulic gradient, transmissivity, and width of the section.

Solution by Steps

step 1

To determine the direction of groundwater flow, we need to look at the equipotential lines and the hydraulic head values

step 2

Groundwater flows from high hydraulic head to low hydraulic head

step 3

The hydraulic head values decrease from left to right, indicating that groundwater flows from left to right

[question a] Answer

Groundwater in this aquifer flows from left to right.

Key Concept

Groundwater Flow Direction

Explanation

Groundwater flows from areas of high hydraulic head to areas of low hydraulic head, which can be determined by the orientation of the equipotential lines and their values.

Solution by Steps

step 1

To assess the specific discharge in the central section, we consider the spacing of the equipotential lines

step 2

Specific discharge is related to the hydraulic gradient, which is the change in hydraulic head over distance

step 3

In the central section, the equipotential lines are spaced further apart, indicating a lower hydraulic gradient

step 4

A lower hydraulic gradient implies a decrease in specific discharge, assuming hydraulic conductivity remains constant

[question b] Answer

The specific discharge is decreasing in the central section of the plan view.

Key Concept

Specific Discharge and Hydraulic Gradient

Explanation

The specific discharge in an aquifer is directly related to the hydraulic gradient. A decrease in the gradient, as indicated by wider spacing of equipotential lines, suggests a decrease in specific discharge.

Solution by Steps

step 1

To find the hydraulic conductivity in the central section, we use the fact that the cross-sectional area and lengths are similar for all sections

step 2

Hydraulic conductivity is proportional to the specific discharge and inversely proportional to the hydraulic gradient

step 3

Given that the specific discharge is lower in the central section and the hydraulic gradient is also lower, the hydraulic conductivity can be assumed to remain constant

[question c] Answer

The hydraulic conductivity in the central section is also

10 \mathrm{~m} / \text{day}

Key Concept

Hydraulic Conductivity Constancy

Explanation

If the cross-sectional area and lengths of the aquifer sections are similar, and the specific discharge changes proportionally with the hydraulic gradient, the hydraulic conductivity remains constant across the sections.