Question 3.14: Consider the water siphon shown in Fig. E3.14. Assuming that...

• Assumptions: Frictionless, steady, incompressible flow. Write Bernoulli’s equation starting from where information is known (the surface, z_1) and proceeding to where information is desired (the tube exit, z_2). See Figure 3.14a.

Note that the velocity is approximately zero at z_1, and a streamline goes from z_1 to z_2. Note further that p_1 and p_2 are both atmospheric, p = p_{atm}, and therefore cancel. (a) Solve for the exit velocity from the tube:

V_2=\sqrt{2g(z_1 – z_2)}                  Ans. (a)

The velocity exiting the siphon increases as the tube exit is lowered below the tank surface. There is no siphon effect if the exit is at or above the tank surface. Note that z_3 and z_4 do not directly enter the analysis. However, z_3 should not be too high because the pressure there will be lower than atmospheric, and the liquid might vaporize. (b) For the given numerical information, we need only z_1 and z_2 and calculate, in SI units,

V_2 = \sqrt{2(9.81  m/s^2) [0.6  m – (-0.25)  m]} = 4.08 m/s

Q = V_2A_2 = (4.08  m/s)(\pi/4)(0.01  m)^2 = 321  E – 6  m^3/s = 321  cm^3/s               Ans. (b)

• Comments: Note that this result is independent of the density of the fluid. As an exercise, you may check that, for water (998 kg/m^3), p_3 is 11,300 Pa below atmospheric pressure. In Chap. 6 we will modify this example to include friction effects.