Problem Statement
Known
Water at \(60^\circ\mathrm{F}\) flows through a constant-diameter siphon from a large tank, over a hill crest (point 2), and out to atmosphere at point 3.
Cavitation begins when the local absolute pressure drops to the saturation pressure of the liquid. The saturation pressure is the pressure at which liquid water at a given temperature starts to vaporize and form vapor bubbles.
For water at \(60^\circ\mathrm{F}\), \[ p_{\mathrm{sat}} = 0.256~\mathrm{psia}. \] In this example, point 2 is the most likely location for the minimum pressure, so the cavitation condition is taken as \[ p_{2,\mathrm{abs}} = p_{\mathrm{sat}} = 0.256~\mathrm{psia}. \]
Point 1 is taken inside the hose rather than at the tank free surface, so for steady, inviscid flow in a uniform-diameter hose the Energy Grade Line (EGL) and Hydraulic Grade Line (HGL) can be shown consistently along the hose.
Find
Determine the maximum hill elevation \(H\) such that cavitation does not occur at the crest.
Analysis
Apply Bernoulli along a streamline from point (2) to point (3):
\[ p_2 + \rho g z_2 + \frac{1}{2}\rho V_2^2 \;=\; p_3 + \rho g z_3 + \frac{1}{2}\rho V_3^2 \]Because the hose diameter is constant, \(V_2 = V_3\). Also, the outlet is a free jet so \(p_3 = 0\) (gage). Therefore,
\[ z_2 = z_3 - \frac{p_2}{\rho g} \]With the tank bottom as the datum, \(z_2 = H\) and \(z_3 = -h_2\). Thus,
\[ H = -h_2 - \frac{p_2}{\rho g} \]Convert the given cavitation pressure to gage pressure:
\[ p_2 = p_{2,\mathrm{abs}} - p_{\mathrm{atm}} = 0.256 - 14.7 = -14.44~\mathrm{psig} \]Substituting this into the hill-height relation gives the maximum allowable crest elevation.
\[ H_{\max} = -h_2 - \frac{p_2}{\rho g} \]