## Q. 18.5

For the given water siphon, find(a) the speed of water leaving as a free jet at point 2 and (b) the water pressure at point A in the flow. State all assumptions. Heights $h_1$ = 1 m, and $h_2$ = 8 m. Drawing is not to scale.

Given: Length of siphon used to remove water from large tank.
Find: Speed at 2 and pressure at A
Assume: Steady flow (all transient effects associated with flow initiation have died down), incompressible (water has constant, uniform density, equal to 1000 kg/m³), and negligible viscous effects.

## Verified Solution

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 $p_A$:

\begin{aligned} \frac{p_1}{\rho}+g z_1 & =\frac{p_A}{\rho}+\frac{V_2^2}{2}+g z_A \\ p_A & =p_1+\rho\left(g z_1-\frac{V_2^2}{2}-g z_A\right) \\ & =p_{\mathrm{atm}}+\rho g\left(z_1-z_A\right)-\rho \frac{V_2^2}{2} \\ & =p_{\mathrm{atm}}+\left(1000 \mathrm{~kg} / \mathrm{m}^3\right)\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(0-1 \mathrm{~m})-\left(1000 \mathrm{~kg} / \mathrm{m}^3\right) \frac{(11.7 \mathrm{~m} / \mathrm{s})^2}{2}, \end{aligned}
\begin{aligned} p_A & =22.8 \mathrm{~kPa}(\mathrm{abs}) \\ & =-78.5 \mathrm{~kPa} \text { (gage). } \end{aligned}

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