Question

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m apart are 52.35 m and 50.90 m. If the horizontal hydraulic conductivity of the aquifer is 25 m/day. Determine the flow rate per unit width of the aquifer, specific discharge, and average linear velocity of the flow assuming steady unidirectional flow. How long would it take for a tracer to travel the distance between the observation wells?

Answer 1

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Step-by-step explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

`q = -Kb(dh)/(dl)`

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

`V = (Q)/(A) = (q)/(b) = -K(dh)/(dl)\\\\V = -(25 m/d).((-1.45 m)/(1000 m)) = 0.0363 m/day`

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

`= 4132.23 days *(24 .hrs)/(1.day) = 99173.52, hours`