The discharge ($Q$) from a tubewell in a confined aquifer under steady-state conditions can be calculated using the formula for radial flow:
$ Q = \frac{2 \pi K b \Delta h}{\ln(R/r_w)} $
Where:
- $K$ = Hydraulic conductivity
- $b$ = Aquifer thickness
- $\Delta h$ = Head difference across the aquifer section (head at radius of influence minus head at well face)
- $R$ = Radius of influence
- $r_w$ = Radius of the tubewell
Initial Discharge Calculation
Given parameters:
- Initial tubewell diameter $d_1 = 40$ cm, so initial radius $r_{w1} = 20$ cm $= 0.2$ m
- Initial aquifer thickness $b_1 = 10$ m
- Hydraulic conductivity $K = 25$ m/day
- Radius of influence $R = 500$ m
- Piezometric head at $R$, $h_R = 40$ m
- Drawdown in the tubewell $s_{w1} = 30$ m
- Head at the well face $h_{w1} = h_R - s_{w1} = 40 - 30 = 10$ m
- Head difference $\Delta h = h_R - h_{w1} = 40 - 10 = 30$ m
- $ \pi = 3.14 $
Using the formula:
$ Q_1 = \frac{2 \times 3.14 \times 25 \text{ m/day} \times 10 \text{ m} \times 30 \text{ m}}{\ln(500 \text{ m} / 0.2 \text{ m})} $
$ Q_1 = \frac{47100 \text{ m}^3/\text{day}}{\ln(2500)} $
Since $ \ln(2500) \approx 7.824 $:
$ Q_1 \approx \frac{47100}{7.824} \approx 6020.00 \text{ m}^3/\text{day} $
New Discharge Calculation
The parameters are changed as follows:
- New tubewell diameter $d_2 = 40 \text{ cm} / 2 = 20$ cm, so new radius $r_{w2} = 10$ cm $= 0.1$ m
- New aquifer thickness $b_2 = 10 \text{ m} \times 2 = 20$ m
- Other parameters ($K$, $R$, $\Delta h$) remain the same for comparison. So, $\Delta h = 30$ m.
Calculating the new discharge ($Q_2$):
$ Q_2 = \frac{2 \times 3.14 \times 25 \text{ m/day} \times 20 \text{ m} \times 30 \text{ m}}{\ln(500 \text{ m} / 0.1 \text{ m})} $
$ Q_2 = \frac{94200 \text{ m}^3/\text{day}}{\ln(5000)} $
Since $ \ln(5000) \approx 8.517 $:
$ Q_2 \approx \frac{94200}{8.517} \approx 11059.94 \text{ m}^3/\text{day} $
Percentage Change in Discharge
The change in discharge is $ Q_2 - Q_1 $. The percentage change is calculated relative to the initial discharge $Q_1$.
$ \text{Percentage Change} = \frac{Q_2 - Q_1}{Q_1} \times 100\% $
$ \text{Percentage Change} = \frac{11059.94 \text{ m}^3/\text{day} - 6020.00 \text{ m}^3/\text{day}}{6020.00 \text{ m}^3/\text{day}} \times 100\% $
$ \text{Percentage Change} = \frac{5039.94}{6020.00} \times 100\% \approx 0.8372 \times 100\% $
$ \text{Percentage Change} \approx 83.72\% $
Since $Q_2 > Q_1$, the discharge has increased.
Important Questions from Steady Flow Through Wells
- A well fully penetrates a $30\text{ m}$ thick confined aquifer. After a long period of pumping at a constant rate of $0.06\text{ m}^3\text{.s}^{-1}$, drawdowns of $3.6\text{ m}$ and $2.0\text{ m}$ are found at distances $60\text{ m}$ and $120\text{ m}$ away from the well, respectively. The hydraulic conductivity of the aquifer is $n \times 10^{-4}\text{ m.s}^{-1}$. The value of $n$ is ________ (Rounded off to two decimal places)
- 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.
