Question

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 _____.

Hint:

In steady-state flow, drawdown is inversely proportional to transmissivity and directly proportional to the logarithm of the ratio of the radius of influence to the well radius.

Answer 1

Correct Answer: 1.48

The formula for drawdown \( s \) at a fully penetrating well under steady-state conditions is given by the Theis equation:
\[ s = \frac{Q}{4 \pi T} \ln\left(\frac{r}{r_0}\right), \] where:
- \( Q \) is the discharge (1200 m$^3$ day$^{-1}$),
- \( T \) is the transmissivity (1000 m$^2$ day$^{-1}$),
- \( r \) is the radius of influence (400 m),
- \( r_0 \) is the well radius (300 mm = 0.3 m).
Substitute the values:
\[ s = \frac{1200}{4 \pi \times 1000} \ln\left(\frac{400}{0.3}\right). \] First, calculate the logarithmic term:
\[ \ln\left(\frac{400}{0.3}\right) = \ln(1333.33) \approx 7.2. \] Now substitute:
\[ s = \frac{1200}{4 \pi \times 1000} \times 7.2 \approx \frac{1200 \times 7.2}{4 \pi \times 1000} \approx \frac{8640}{12566.37} \approx 0.688 \, \text{m}. \] Thus, the drawdown is approximately \(\boxed{0.69} \, \text{m}\).