Pipe over hill with cavitation limit

Data. $L = 1000$ m, $D = 0.3$ m, $f = 0.032$, level difference $H = 13$ m, summit 5 m above the upper reservoir level, minimum absolute pressure 1.2 m of water. Atmospheric pressure $= 760$ mm Hg $= 0.76(13.6) = 10.336$ m of water. Head losses are taken as pipe friction only (bend losses neglected; entry and exit losses would change the result by less than 1%).

Discharge

$$H = f\frac{L}{D}\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.032\left(\frac{1000}{0.3}\right) = 106.67$$ $$\frac{V^2}{2g} = \frac{13}{106.67} = 0.12187\ \text{m} \;\Rightarrow\; V = 1.546\ \text{m/s}$$ $$Q = \frac{\pi}{4}(0.3)^2(1.546) = 0.1093\ \text{m}^3/\text{s} = 109.3\ \text{l/s}$$

Length of pipe from the upper reservoir to the summit

The critical point is the summit S, where absolute pressure head is 1.2 m. Energy equation from the upper reservoir surface (datum, atmospheric) to the summit:

$$\frac{p_a}{\gamma} = 5 + 1.2 + \frac{V^2}{2g} + f\frac{l_1}{D}\frac{V^2}{2g}$$ $$10.336 = 6.2 + 0.12187 + f\frac{l_1}{D}(0.12187) \;\Rightarrow\; f\frac{l_1}{D}\,(0.12187) = 4.0141$$ $$l_1 = \frac{4.0141\times 0.3}{0.032\times 0.12187} = 308.8\ \text{m}$$

Answer: The summit may be at most about 309 m of pipe from the upper reservoir; the discharge is $0.1093$ m³/s ($109.3$ l/s).