Question

Two reservoirs differ by 10 m in water surface elevation and are connected by a 50 m long, 10 cm diameter pipeline (friction factor $f = 0.02$). If the last 25 m is replaced with a 20 cm diameter pipe of the same material, the increase in discharge (in %, rounded to one decimal place) is:

Hint:

Replacing a high–friction small pipe segment with a larger diameter pipe dramatically reduces headloss and increases discharge.

Answer 1

Correct Answer: 39

Using Darcy–Weisbach equation: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Let $Q$ be discharge, $A$ area, $V = Q/A$. Original pipe (50 m, $D = 0.10$ m): \[ h_f = 0.02 \cdot \frac{50}{0.10} \cdot \frac{Q^2}{2gA_1^2} \] \[ A_1 = \frac{\pi}{4}(0.1)^2 = 0.007854\ \text{m}^2 \] \[ h_f = 10 \cdot \frac{Q^2}{2gA_1^2} \] Total head = 10 m: \[ 10 = 10 \cdot \frac{Q^2}{2gA_1^2} \] \[ Q_1 = A_1\sqrt{2g} = 0.007854 \times 4.427 = 0.0348\ \text{m}^3/s \] Modified pipe: First 25 m has $D = 0.10$ m, last 25 m has $D = 0.20$ m. Areas: \[ A_1 = 0.007854,\quad A_2 = \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2 \] Headloss: \[ h_f = 0.02\left[\frac{25}{0.10}\frac{Q^2}{2gA_1^2} + \frac{25}{0.20}\frac{Q^2}{2gA_2^2}\right] \] Simplifying: \[ h_f = \frac{Q^2}{2g}\left( \frac{50}{A_1^2} + \frac{12.5}{A_2^2} \right)0.02 \] Compute: \[ \frac{50}{A_1^2} = 811,000,\quad \frac{12.5}{A_2^2} = 12,650 \] \[ h_f = \frac{Q^2}{2g}(823,650)(0.02) \] \[ 10 = Q^2 \cdot \frac{823,650 \cdot 0.02}{19.62} \] \[ Q_2 = 0.0487\ \text{m}^3/s \] Increase in discharge: \[ %\,\text{increase} = \frac{Q_2 - Q_1}{Q_1} \times 100 = \frac{0.0487 - 0.0348}{0.0348} \times 100 \approx 39.9% \] Thus: \[ \boxed{39.9%} \]