Question 4.17: Water at 20^◦ C is siphoned from an open tank through a 2-cm...

(a) In order to determine the ideal velocity of the flowas it leaves the siphon at point 3, the Bernoulli equation is applied between points 1 and 3. Because the cross-sectional area of the tank at point 1 is much larger than the cross-sectional area of the tubing, the velocity at point 1 is assumed to be zero. Assuming the datum is at the bottom of the second tank as follows:

Z_{1}: = 7 m                 Z_{3}: = 1.5 m                 P_{1}: = 0 \frac{N}{m^{2}}                 P_{3}: = 0 \frac{N}{m^{2}}                 V_{1}: = 0 \frac{m}{sec}
\rho : = 998 \frac{kg}{m^{3}}                 g: = 9.81 \frac{m}{sec^{2}}                 \gamma : = \rho .g = 9.79 \times 10^{3} \frac{kg}{m^{2}s^{2}}

Guess value:                  V_{3}: = 10 \frac{m}{sec}

Given

\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2.g} = \frac{P_{3}}{\gamma } + Z_{3} + \frac{V^{2}_{3} }{2.g}
V_{3} : = Find ( V_{3} ) = 10.388 \frac{m}{sec}

Application of the Bernoulli equation between points 1 and 3 illustrates the conversion of potential energy stored in the difference in elevation between points 1 and 3 to kinetic energy stored in the ideal velocity of the flow as it leaves the siphon at point 3.
(b) In order to determine the ideal maximum allowable height of point 2 without causing cavitation, the Bernoulli equation is applied between points 2 and 3, assuming that the pressure at point 2 is the vapor pressure of the water at 20^{\circ} C . However, in order to determine the relationship between the ideal velocities at points 2 and 3, the continuity equation is applied between points 2 and 3. Thus, from Table A.2 in Appendix A, for water at 20^{\circ} C , the vapor pressure, p_{\upsilon } is 2.34 \times 10^{3} N/m^{2} abs. However, because the vapor pressure is given in absolute pressure, the corresponding gage pressure is computed by subtracting the atmospheric pressure as follows: p_{gage} = p_{abs} – p_{atm}, where the standard atmospheric pressure is 101.325 10^{3} N/m^{2} abs.

P_{V}: = 2.34 \times 10^{3} \frac{N}{m^{2}} – 101.325 \times 10^{3} \frac{N}{m^{2}} = -9.899 \times 10^{4} \frac{N}{m^{2}}
P_{2}: = P_{V} = – 9.899 \times 10^{4} \frac{N}{m^{2}}                     D_{2}: = 2 cm                    D_{3}: = 2 cm

Guess value:                  V_{2}: = 10 \frac{m}{sec}                   Z_{2}: = 15 m

Given

\frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2} }{2.g} = \frac{P_{3}}{\gamma } + Z_{3} + \frac{V^{2}_{3} }{2.g}
V_{2}. \frac{\pi . D^{2}_{2} }{4} = V_{3}. \frac{\pi . D^{2}_{3} }{4}
\left ( \begin{matrix} Z_{2} \\ V_{2} \end{matrix} \right ) : = Find (Z_{2}, V_{2})
Z_{2} = 11.61 m                    V_{2} = 10.388 \frac{m}{s }

Thus, the maximum allowable height of point 2 should be less than 11.61 m in order to avoid cavitation.