A fully penetrating well is installed in a homogenous and isotropic confined aquifer. The aquifer has uniform thickness of 16 m and hydraulic conductivity of 25 m/d. Water is being pumped out from the well at a constant rate of $0.1 \text{ } m^3/s$ till steady state condition is reached. If a drawdown of 3.5 m is observed at a distance of 75 m from the well then the drawdown at a distance of 150 m from the well will be________ (in m, rounded off to two decimal places).
Aquifer Drawdown Calculation
This problem requires calculating the drawdown at a specific distance from a pumping well in a confined aquifer under steady-state conditions, given initial drawdown information and aquifer properties.
Aquifer Properties and Conditions:
- Aquifer thickness, b = 16 m
- Hydraulic conductivity, K = 25 m/d
- Pumping rate, Q = 0.1 m3/s
- Observed point: Distance r1 = 75 m, Drawdown s1 = 3.5 m
- Target point: Distance r2 = 150 m
Drawdown Formula and Relation
In a confined aquifer at steady state, the drawdown s at a distance r from a fully penetrating well is described by:
$s = \frac{Q}{2 \pi K b} \ln\left(\frac{R}{r}\right)$
Here, R is the radius of influence. This formula indicates that drawdown decreases as the distance r increases from the well (assuming R is constant and larger than r).
A relationship between drawdowns at two different radial distances (r1, s1) and (r2, s2) can be derived by eliminating the unknown radius of influence R and the constant factor $\frac{Q}{2 \pi K b} \ln(R)$:
$s_2 = s_1 + \frac{Q}{2 \pi K b} \ln\left(\frac{r_1}{r_2}\right)$
Steady State Calculation Steps
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Pumping Rate Conversion:
Convert the pumping rate Q from cubic meters per second (m3/s) to cubic meters per day (m3/d).
$Q = 0.1 \frac{m^3}{s} \times 86400 \frac{s}{d} = 8640 \frac{m^3}{d}$
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Constant Factor Calculation:
Calculate the constant factor $C = \frac{Q}{2 \pi K b}$.
$C = \frac{8640 \frac{m^3}{d}}{2 \pi (25 \frac{m}{d}) (16 \text{ } m)} = \frac{8640}{800 \pi} \frac{m}{d}$
$C = \frac{10.8}{\pi} \frac{m}{d} \approx 3.4377 \text{ } m$
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Drawdown Determination:
Substitute the calculated constant factor and given values into the derived relationship for s2.
$s_2 = 3.5 \text{ } m + \left(\frac{10.8}{\pi} \text{ } m\right) \times \ln\left(\frac{75 \text{ } m}{150 \text{ } m}\right)$
$s_2 = 3.5 + \frac{10.8}{\pi} \times \ln(0.5)$
$s_2 \approx 3.5 + 3.4377 \times (-0.693147)$
$s_2 \approx 3.5 - 2.3849$
$s_2 \approx 1.1151 \text{ } m$
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Final Drawdown Result:
Rounding the calculated drawdown to two decimal places yields the final answer.
s2 = 1.12 m.
Important Questions from Groundwater Occurrence
An unconfined aquifer of areal extent $20 \text{ km} \times 20 \text{ km}$ has hydraulic conductivity of $4 \text{ m/day}$, porosity of $0.32$, and storage coefficient (specific yield) of $0.18$. If the initial saturated thickness of the aquifer is $30 \text{ m}$, and $4\times10^8 \text{ m}^3$ of water is extracted from the aquifer, then the decline in the saturated thickness is _____________ (rounded off to two decimal places) m.
An unconfined aquifer extends over an area of $1 \text{ km}^2$ and has hydraulic conductivity, total porosity and specific retention of $20 \text{ m per day}, 30\%$ and $10\%$, respectively. After pumping some groundwater from this aquifer, the water table dropped to a depth of $20 \text{ m}$ from the ground level. If the water table was initially at $14.5 \text{ m}$ below the ground level, the change in groundwater storage in million cubic meters would be_______
Match the following items between Column I and Column II with the most appropriate combinations:
Column I Column II P Properties of confined and semi- confined aquifers 1 Perched aquifer Q Property of an unconfined aquifer 2 Coefficient of storage R Property of semi-confined aquifer 3 Specific yield S A special case of an unconfined aquifer 4 Leakage factor - One of the key assumptions made by Dupuit-Forchheimer about the flow behavior in the porous medium towards subsurface drains is
An underground hazardous waste storage tank is leaking. The contaminant concentration directly beneath the site is 0.5 mg/L. The contaminant is travelling at an effective rate of 0.4 m per day towards a water well which is 2 km away.
Assume that the degradation of the contaminant follows a first order reaction, and the initial concentration of the contaminant becomes half in 10 years.
In this case, the contaminant concentration expected at the well under steady state conditions is ________ mg/L (rounded off to two decimal places).
