A) A fully penetrating unconfined well of 12 in. The diameter is pumped at a rate of 1 ft^3/sec. The coefficient of permeability is 750 gal/day per square foot. The drawdown in an observation well located 200 ft away from the pumping well is 10 ft below its original depth of 150 ft. Find the water level in the well.
A) A fully penetrating unconfined well of 12 in. The diameter is pumped at a rate of 1 ft^3/sec. The coefficient of permeability is 750 gal/day per square foot. The drawdown in an observation well located 200 ft away from the pumping well is 10 ft below its original depth of 150 ft. Find the water level in the well.
B) In addition to the requested information, calculate the radius of influence for the well (in ft). You will get an extremely large and unreasonable value (in the millions...pinky to corner of mouth required).
C) Then develop a plot of the water surface profile for the well during pumping. Orient the plot so that the x-axis represents the distance from the production well (and looks like the bottom of the aquifer (i.e., the datum)) and the pumping well is represented by the y-axis and shows the water depth (in feet) in the aquifer. I recommend selecting small intervals (10-20 feet) of x-axis values between the y-axis and 500 feet. Then use increments of 1,000-1,000,000 as you approach the radius of influence. Your final data point should be at x = influence. You can use y-axis values between ~130-150 feet. Show two plots for this problem – one that is a normal arithmetic plot and another that has a log scale on the x-axis. To
To obtain full credit for this problem, the plot must be computer generated using a program such as MS Excel.
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 3 images
