Problem Solution: Tube Well Diameter Calculation
To find the diameter of the tube well in a confined aquifer, we use Thiem's Equation for steady-state flow towards a fully penetrating well.
1. Given Data
- Thickness of the confined aquifer (\( B \)) = \( 30 \, \text{m} \)
- Hydraulic conductivity (\( k \)) = \( 4 \times 10^{-4} \, \text{m/s} \)
- Discharge (\( Q \)) = \( 50 \, \text{L/s} = 50 \times 10^{-3} \, \text{m}^3/\text{s} = 0.05 \, \text{m}^3/\text{s} \)
- Drawdown at the well (\( s_w \)) = \( 5 \, \text{m} \)
- Radius of influence (\( R \)) = \( 250 \, \text{m} \)
2. Formula
For a confined aquifer, the discharge is given by:
$$ Q = \frac{2\pi k B s_w}{\ln\left(\frac{R}{r_w}\right)} $$
Where \( r_w \) is the radius of the tube well.
3. Step-by-Step Calculation
Step 1: Rearrange the formula to solve for \( \ln\left(\frac{R}{r_w}\right) \):
$$ \ln\left(\frac{R}{r_w}\right) = \frac{2\pi k B s_w}{Q} $$
Step 2: Substitute the known values:
$$ \ln\left(\frac{250}{r_w}\right) = \frac{2 \times \pi \times (4 \times 10^{-4}) \times 30 \times 5}{0.05} $$ $$ \ln\left(\frac{250}{r_w}\right) = \frac{2 \times 3.14159 \times 0.06}{0.05} $$ $$ \ln\left(\frac{250}{r_w}\right) = \frac{0.37699}{0.05} $$ $$ \ln\left(\frac{250}{r_w}\right) \approx 7.5398 $$
Step 3: Solve for \( r_w \):
$$ \frac{250}{r_w} = e^{7.5398} $$ $$ \frac{250}{r_w} \approx 1881.45 $$ $$ r_w = \frac{250}{1881.45} \approx 0.13287 \, \text{m} $$
Step 4: Calculate the diameter (\( d \)) in mm:
$$ d = 2 \times r_w = 2 \times 0.13287 = 0.26574 \, \text{m} $$ $$ d = 0.26574 \times 1000 \, \text{mm} $$ $$ d \approx 265.74 \, \text{mm} $$
Summary
The diameter of the tube well is approximately \( 265.74 \, \text{mm} \).
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)
- 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 $\pi = 3.14$)
