We would like to use the Bernoulli equation to relate the flow quantities between the labeled points. The conditions necessary for the Bernoulli equation to apply have been reasonably assumed. (We feel least confident in our assumption of negligible viscous losses in the siphon, and in Chapter 20 we will discuss a way to characterize the importance of viscosity in a given flow.)

We must have a streamline on which to apply the Bernoulli equation, and so we assume that the one sketched below exists. This streamline connects points 1 and A, and A and 2.

We will make one more assumption in order to solve this problem. By inspection, the reservoir is much larger than the siphon diameter. That is,

A_1 \gg A_2 .

So, if we conserve mass from point 1 to point 2, we will have

\rho V_1 A_1=\rho V_2 A_2 .

And, with A_1 \gg A_2 , we must have V_1 \ll V_2 . We approximate this very small velocity at 1 by saying V_1 ≈ 0.

We can now apply Bernoulli’s equation between points 1 and 2, to find the unknown V_2:

\frac{p_1}{\rho}+\frac{V_1^2}{2}+g z_1=\frac{p_2}{\rho}+\frac{V_2^2}{2}+g z_2

where, as we have just said, V_1 ≈ 0, and where p_1 = p_2 = p_{atm}. (If we are using gage pressures, this means p_1 = p_2 = 0.) We note from the figure that z_1 = 0, and z_2 = −7 m, so

\begin{aligned} g z_1 & =\frac{V_2^2}{2}+g z_2 ,\\ V_2^2 & =2 g\left(z_1-z_2\right) ,\\ V_2 & =\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(7 \mathrm{~m})}=11.7 \mathrm{~m} / \mathrm{s} . \end{aligned}

Our assumed streamline also goes through point A, so that the Bernoulli equation is

\frac{p_1}{\rho}+\frac{V_1^2}{2}+g z_1=\frac{p_A}{\rho}+\frac{V_A^2}{2}+g z_A .

If we conserve mass within the constant-area siphon, we must have

\rho V_A A_A=\rho V_2 A_2 .

Or, since A_A = A_2, \ V_A = V_2 = 11.7 m/s. We are now equipped to solve the Bernoulli equation for the unknown p_A:

We have used standard atmospheric pressure, p_{atm} = 101.325 kPa.