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QB_IDE_ADV_108
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NAT • 2 Marks
Section 5: Irrigation and Drainage Engineering
Wells and Pumps
Subtopic: Steady flow through wells
Question Statement:
The total drawdown in a pumping well is given by Jacob's well loss equation: $s_w = B Q + C Q^2$, where $B = 0.0025\text{ day/m}^2$ is the aquifer loss coefficient and $C = 5.0 \times 10^{-7}\text{ day}^2/\text{m}^5$ is the well loss coefficient. The well is pumped at a steady discharge of $Q = 2000.0\text{ m}^3/\text{day}$. The well efficiency ($\eta_w = \frac{B Q}{s_w} \times 100\%$) in percent ($\text{\%}$) is ________ (round off to one decimal place).
Step-by-Step Solution & Explanation
**Method 1: Jacob's Well Efficiency Equation**
1. Aquifer formation loss ($s_{\text{aq}} = B Q$):
$s_{\text{aq}} = 0.0025\text{ day/m}^2 \times 2000.0\text{ m}^3/\text{day} = 5.00\text{ m}$
2. Turbulent well loss ($s_{\text{well}} = C Q^2$):
$s_{\text{well}} = (5.0 \times 10^{-7}\text{ day}^2/\text{m}^5) \times (2000.0\text{ m}^3/\text{day})^2$
$s_{\text{well}} = (5.0 \times 10^{-7}) \times (4.0 \times 10^6) = 2.00\text{ m}$
3. Total well face drawdown ($s_w$):
$s_w = s_{\text{aq}} + s_{\text{well}} = 5.00\text{ m} + 2.00\text{ m} = 7.00\text{ m}$
4. Well efficiency ($\eta_w$):
$\eta_w = \frac{s_{\text{aq}}}{s_w} \times 100\% = \frac{5.00}{7.00} \times 100\% = 71.428\% \approx 71.4\%$
**Method 2: Factored Discharge Form**
$\eta_w = \frac{1}{1 + \frac{C}{B} Q} \times 100\%$
$\frac{C}{B} = \frac{5.0 \times 10^{-7}}{2.5 \times 10^{-3}} = 2.0 \times 10^{-4}$
$\frac{C}{B} Q = (2.0 \times 10^{-4}) \times 2000 = 0.40$
$\eta_w = \frac{1}{1 + 0.40} \times 100\% = \frac{1}{1.40} \times 100\% = 71.43\% \approx 71.4\%$
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