Question Bank • QB_IDE_ADV_108 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).

Answer Key:

71.4

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** $\eta_w = \frac{1}{1 + \frac{C}{B} Q} \times 100\%$ $\frac{C}{B} = \frac{5.0 \times 10^{-7}}{2.5 \times 10^{-3}} = 2.0 \times 10^{-4}$ $\frac{C}{B} Q = (2.0 \times 10^{-4}) \times 2000 = 0.40$ $\eta_w = \frac{1}{1 + 0.40} \times 100\% = \frac{1}{1.40} \times 100\% = 71.43\% \approx 71.4\%$
← Previous (QB_IDE_PUMP_003) Next (QB_IDE_ADV_114) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test