7. Consider a confined aquifer which is tilted so that it forms a 41° angle with

Question

7. Consider a confined aquifer which is tilted so that it forms a 41° angle with the horizontal overlying aquiclude. The aquifer pinches out stratigraphically with depth. Two wells penetrate the aquifer and are separated by a horizontal distance of 1700 meters. The surface elevation at both wells is 300 meters. Well A is 200 meters deep and well B is 75 meters deep. Pieziometric head in well A is 5 meters below the surface. Pieziometric head in well B is 25 meters below the surface. The thickness of the aquifer is 49 meters. The hydraulic conductivity of the aquifer is 103 m/sec. Find: (a) elevation of hydraulic head in each well, (b) direction of flow, (e) hydraulic gradient, (d) Darcy Flux in days, (e) discharge

Explanation / Answer

As per norms I should answer for first 4 subdivision only

a) elevation of hydraulic head = surface elevation - depth to water

For well A = 300 - 200 = 100 m

For well B = 300 - 75 = 225 m

b) Direction of flow = h dh/dL = Steepest composite gradient along plane formed by input data (using h2) for an unconfined aquifer [L]. Value will be negative since groundwater flows from high head to low head.

200-75 * (5-25) / 1700 = -1.47

Then ground water flow is from high head to low head ie - A to B

c) Verticle Hydraulic gradient between Well A and Well B = (head B - head A) / (depth well A - depth well B)

= (25 - 5)/ (200-75) = 0.16

Horizontal Hydraulic gradient between Well A and Well B = head B - head A / horizontal distance between wells

= 25-5 / 1700 = 0.0118

d) Darcy Flux = - K dh/ dl

where K - Hydraulic conductivity (m/s) (10^-5 m/s= 0.864 m/d

dh - Difference between hydraulic head ( head B- Head A) (25-5 = 20 m)

dl - Distance between two well (m) 1700 m

Flux = (0.864 * 20) / 1700 = 0.0102 day _1

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