This problem involves calculating the loss of discharge in one tubewell due to the interference effect of a second, identical tubewell operating simultaneously nearby.
Calculate Logarithmic Terms for Interference Analysis
First, determine the values of the logarithmic terms required for interference calculations. The well radius is half its diameter.
- Well radius ($r_w$): $15 \text{ cm} / 2 = 7.5 \text{ cm} = 0.075 \text{ m}$
- Radius of influence ($R$): $980 \text{ m}$
- Distance between wells ($d$): $500 \text{ m}$
- Calculate $\ln(R/r_w)$:
$ \ln(R/r_w) = \ln(980 \text{ m} / 0.075 \text{ m}) = \ln(13066.67) \approx 9.4775 $
- Calculate $\ln(R/d)$:
$ \ln(R/d) = \ln(980 \text{ m} / 500 \text{ m}) = \ln(1.96) \approx 0.6729 $
Determine the Ratio of Discharges
When two identical wells interfere and pump at the same rate ($Q_{new}$), the ratio of this discharge to the discharge of a single well operating alone ($Q_{alone}$) is influenced by the radii and the distance between wells. The discharge of a well operating alone is proportional to the drawdown, and the interference increases the total drawdown for a given discharge. The ratio of the new discharge to the original discharge is given by:
$ \frac{Q_{new}}{Q_{alone}} = \frac{1}{1 + \frac{\ln(R/d)}{\ln(R/r_w)}} $
Substitute the calculated logarithmic values:
$ \frac{Q_{new}}{Q_{alone}} = \frac{1}{1 + \frac{0.6729}{9.4775}} = \frac{1}{1 + 0.0710} = \frac{1}{1.0710} \approx 0.9337 $
Calculate the Percentage Loss of Discharge
The loss of discharge in the first tubewell is the difference between its discharge when operating alone and when operating with the second well. This is calculated as a percentage of the original discharge ($Q_{alone}$).
Percentage Loss $= \left(1 - \frac{Q_{new}}{Q_{alone}}\right) \times 100\%$
$ \text{Percentage Loss} = (1 - 0.9337) \times 100\% = 0.0663 \times 100\% = 6.63\% $
The calculated loss of discharge is approximately 6.63%, which falls within the given range of 6.6 to 6.7.
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$]
- 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.
