Question 10.5: For the water siphon in Figure 10.15, find (a) the speed of ...
For the water siphon in Figure 10.15, 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.
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 the next chapter we 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, so we assume that the one sketched in Figure 10.16 exists. This streamline connects points 1 and A, and A and 2.
We make one more assumption 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 have
ρV_1A_1 = ρV_2A_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 \approx 0 .
We can now apply Bernoulli’s equation between points 1 and 2, to find the unknown V_2 :
\frac{P_1}{ρ} +\frac{V^{2}_{1}}{2}+gz_1=\frac{P_2}{ρ} +\frac{V^{2}_{2}}{2}+gz_2 ,
where, as we have just said, V_1 \approx 0 , and where p_1 = p_2 = p_{atm} . (If we are using gauge pressures, this means p_1 = p_2 = 0 .) We note from the figure that z_1 = 0 , and z_2 = –7 m, so
gz_1=\frac{V^{2}_{2}}{2}+gz_2V^{2}_{2}=2g(z_1-z_2)
V_2 = \sqrt{2(9.81 \frac{\textrm{m}}{s^2})(7 \textrm{ m})}=11.7 \frac{\textrm{m}}{s} .
Our assumed streamline also goes through point A so that the Bernoulli equation is
\frac{P_1}{ρ} +\frac{V^{2}_{1}}{2}+gz_1=\frac{P_A}{ρ} +\frac{V^{2}_{A}}{2}+gz_A .
If we conserve mass within the constant area siphon, we must have
\rho V_A A_A = ρV_2 A_2or, 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 :
\frac{P_1}{ρ} +gz_1=\frac{P_A}{ρ} +\frac{V^{2}_{2}}{2}+gz_AP_A=P_1+ρ\left\lgroup gz_1-\frac{V^{2}_{2}}{2}-gz_A\right\rgroup
= P_{atm} + ρ g(z_1-z_A) -ρ \frac{V^{2}_{2}}{2}
= P_{atm} + (1000 \frac{\textrm{kg}}{{m^3}})(9.81 \frac{\textrm{m}}{s^2})(0-1 \textrm{ m})-(1000 \frac{\textrm{kg}}{{m^3}}) \frac{(11.7 \frac{\textrm{m}}{s})^2}{2}
P_A= 22.8 kPa (abs)
= –78.5 kPa (gage).
We have used standard atmospheric pressure, p_{atm} =101.325 kPa.
