Confined Aquifer Pumping Test Analysis
This problem requires calculating the hydraulic conductivity ($K$) of a confined aquifer using steady-state radial flow principles derived from a pumping test. We are provided with the constant pumping rate ($Q$), aquifer thickness ($b$), and drawdowns ($s$) observed at two different distances ($r$) from the pumping well.
Steady-State Radial Flow Equation
Under steady-state conditions in a confined aquifer, the flow rate ($Q$) is related to hydraulic conductivity ($K$), aquifer thickness ($b$), and the difference in potentiometric head ($\Delta h$) between two radial distances ($r_1$ and $r_2$) by:
$ Q = \frac{2 \pi K b (h_2 - h_1)}{\ln(r_2/r_1)} $
In this context, drawdowns ($s$) represent the decrease in potentiometric head from the static (undisturbed) head ($h_0$), so $h = h_0 - s$. For points $r_1$ and $r_2$ where $r_2 > r_1$, and given that drawdown increases with proximity to the well ($s_1 > s_2$), the head difference $(h_2 - h_1)$ can be expressed as $(s_1 - s_2)$.
The governing equation thus becomes:
$ Q = \frac{2 \pi K b (s_1 - s_2)}{\ln(r_2/r_1)} $
Given Data
- Pumping rate, $Q = 0.06 \text{ m}^3/\text{s}$
- Aquifer thickness, $b = 30 \text{ m}$
- Drawdown at $r_1 = 60 \text{ m}$, $s_1 = 3.6 \text{ m}$
- Drawdown at $r_2 = 120 \text{ m}$, $s_2 = 2.0 \text{ m}$
Calculating Hydraulic Conductivity ($K$)
First, compute the differences and ratios from the given data:
- Drawdown difference: $s_1 - s_2 = 3.6 \text{ m} - 2.0 \text{ m} = 1.6 \text{ m}$
- Ratio of distances: $r_2/r_1 = 120 \text{ m} / 60 \text{ m} = 2$
- Natural logarithm of the distance ratio: $\ln(2) \approx 0.6931$
Rearrange the flow equation to solve for $K$:
$ K = \frac{Q \ln(r_2/r_1)}{2 \pi b (s_1 - s_2)} $
Substitute the known values into the rearranged equation:
$ K = \frac{(0.06 \text{ m}^3/\text{s}) \times \ln(2)}{2 \pi \times (30 \text{ m}) \times (1.6 \text{ m})} $
$ K = \frac{0.06 \times 0.6931}{2 \pi \times 30 \times 1.6} $
$ K = \frac{0.041586}{301.5929} $
$ K \approx 0.00013789 \text{ m/s} $
Determining the Value of 'n'
The problem states that the hydraulic conductivity is $K = n \times 10^{-4} \text{ m/s}$. To find $n$, equate the calculated $K$ value:
$ 0.00013789 \text{ m/s} = n \times 10^{-4} \text{ m/s} $
Solve for $n$:
$ n = \frac{0.00013789}{10^{-4}} $
$ n = 1.3789 $
Rounding the value of $n$ to two decimal places gives $1.38$.
Important Questions from Steady Flow Through Wells
- A $40$ cm diameter tubewell is constructed in a $10$ m thick confined aquifer having hydraulic conductivity of $25$ m day$^{-1}$. The peizometric surface is observed to be $40$ m high from the impervious stratum at the radius of influence of $500$ m. The drawdown in the tubewell is $30$ m. If the thickness of aquifer is doubled and diameter of tubewell is reduced to half, keeping all other parameters and conditions same, the change in discharge from the well is [Take $ \pi = 3.14$]
- In an area, one tubewell of 15 cm diameter was operating such that groundwater table is 40 m above the bottom of screen. The groundwater formation has permeability (K) as 20 m.d$^{-1}$ and well has radius of influence of 980 m with a maximum drawdown of 12 m. After some time, a second tubewell of same size was installed at a distance of 500 m from the first tubewell. If both the tubewells are running simultaneously and have equal discharge, the loss of discharge in the first tubewell due to installation of the second tubewell, in %, is _________. (Rounded off to 2 decimal places)
- A $0.30$ m diameter well penetrates an unconfined aquifer with a saturated depth of $40$ m. After $8$ hours of pumping at a steady rate of $0.03$ $m^3$ $s^{-1}$, the drawdown in two observation wells located at $20$ m and $50$ m away from the pumping well are found to be $3$ m and $2$ m, respectively. The drawdown in the pumping well in m is ________ (rounded off to $1$ decimal place, Consider $\pi = 3.14$).
Water from a confined aquifer having transmissivity of 1000 $m^2 day^{-1}$ is pumped through a fully penetrating well of 300 mm diameter at a rate of 1200 $m^3 day^{-1}$. If the radius of influence is 400 m, the drawdown in the well under steady-state flow condition in meter is __________ [round off to two decimal places]
(Take $\pi = 3.14$)- The drawdown at the well face of a 30 cm diameter fully penetrating well in a confined aquifer is 3 m. The radius of influence of this well is 3 km. For the same discharge and aquifer condition, the drawdown at 300 m away from the centre of the well is ________ m.
