2
$\begingroup$

I've been trying to derive the Torricelli's law by using Newton's first and second law but I'm not coming up with an exact answer. Here's the problem.

Consider a large cylindrical vessel with a very small orifice of surface area $S$ at the bottommost circular surface. We want to find the velocity of the water coming out of the orifice ,if water is filled to a height $h$ above the bottommost surface.

I chose the system to be the whole liquid minus a small fraction of liquid just above the orifice. Since the container has a vey large area as compared to the orifice , liquid must have negligible velocity and near to zero accelaration . Therefore by using Newton's first law of motion , the net forces on the system must be zero. The system is acted upon by atmospheric pressure, gravity, and the forces due to the container. If the liquid is approximately at rest then the normal forces must exactly balance the weight of liquid on all regions except the one just above the orifice ; the unbalanced force being equal to ${\rho}Shg$.

The small cup just above the orifice is ejected with a velocity $V_{rel}$ and exerts a thrust force ${\rho}SV_{rel}^2$. Hence applying the condition for equilibrium we get :

$${\rho}Shg={\rho}SV_{rel}^2$$ Or $$V_{rel}={\sqrt{hg}}$$ Am I wrong with elementary concepts or concepts of fluid mechanics? Thanks in advance.

Improve this question
$\endgroup$
10
  • $\begingroup$ It is a balance between potential and kinetic energy, your 'thrust force' in this case is missing a factor $0.5$ $\endgroup$
    – nluigi
    Commented Dec 15, 2017 at 10:34
  • $\begingroup$ Could you please elaborate a bit more? $\endgroup$
    – user150098
    Commented Dec 15, 2017 at 10:41
  • $\begingroup$ kinetic energy per unit volume is defined as $\frac{1}{2}\rho v^2$ not $\rho v^2$ $\endgroup$
    – nluigi
    Commented Dec 15, 2017 at 10:54
  • $\begingroup$ But that's not the thrust it imparts. Thrust is defined to be $dm/dt V_rel^2$. It turns out to be ${\rho}v^2$. $\endgroup$
    – user150098
    Commented Dec 15, 2017 at 11:03
  • 1
    $\begingroup$ The forces by the other parts of the vessel wall are not quite hydrostatic when the fluid is flowing out, and the direction of the velocity vector near the orifice is not vertical. The flow streamlines in the tank near the orifice are converging radially (in a hemispherical sense) toward the orifice. So the basic premise of the analysis is not quite correct. $\endgroup$
    – Chet Miller
    Commented Dec 15, 2017 at 12:45

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.