# Hydrodynamic force (conservation of momentum)

I have an open container which is tilted with 45 degress. The container has an orifice on one of the sides. The water that flows from the container through the orifice hits a plate and the water then seperates into two flows (upwards and downwards). I want to know how much Force the fluid excerts on the plate when the water flows through the orifice.

Can someone help me set up the equation for the linear conservation of momentum for the control volume marked with red in the figure?

My hypothesis is that the hydrodynamic force must be the equivalent to the external force in the opposite direction

$$F_{hydro}=-F_{ext}$$

I Think the Force F1 can be calculated using hydrostatic Force but I am not sure how to determine the remaining forces (not even sure if F2 and F3 is relevant )

Does somebody know how to do this?

OBS: This is not a homework question, I simply want to improve my understanding within the field of fluids mechanics.

• Is this a homework question? Commented Dec 5, 2022 at 0:26
• Also, can you provide a bit more information/some basic assumptions about the system. Are the $F_i$, $i=1,2,3$ givens? Is there a flow rate given through the openings? Or is the situation static? Is the container pressurized? Commented Dec 5, 2022 at 0:28
• Thanks for the Comment! This is not for homework(I am not a student) and I did an Update on the question.
– Nil
Commented Dec 5, 2022 at 1:55
• Are you only interested in the force $F^{ext}$? Commented Dec 5, 2022 at 11:03
• Well I am actually interested in the dynamic force caused by moving fluid but if this is just the reaction force to the external right?
– Nil
Commented Dec 5, 2022 at 12:07

Let's look into this, and first add some assumptions. Let's presume that at the openings you marked with $$F_2$$ and $$F_3$$ there are pipes attached whose outflow is controlled by valves that we can close, and that the influx into the container is equal to the outflux at those valves such that the water level in the container remains constant. Let's further assume that the pressure at the free surface in the container is constant (atmospheric pressure, for example), that the fluid is incompressible and that we can neglect viscous effects in the volume element you are interested in.
Now, you can follow the following with Landau & Lifshitz, Volume 6, I'll refer to the equations therein. We use Bernoulli's equation for an incompressible fluid (10,4 in L&L) $$\frac{v^2}{2} + \frac{p}{\rho} + g z = \text{const.}$$ wherein $$v$$ is fluid velocity, $$p$$ pressure, $$\rho$$ density, $$g$$ gravitational acceleration and $$z$$ height in some cartesian coordinate system. Compare the pressures $$p_1$$ and $$p_2$$ at one specific point of the plate for two different velocities $$0 = v_1 < v_2$$ and you'll find that the pressure $$p_1$$ when there is no fluid flow is higher. I.e. switching on the fluid flow, say by opening the valves I wished for, we actually lower the pressure at the plate and thus the external force. Which we can calculate directly by integrating the force density $$p \vec{n} + \rho \vec{v}\left(\vec{v}\cdot\vec{n}\right)$$ (7,4 in L&L) over the plate. $$\vec{n}$$ denotes the normal vector to the plate. And by a no-flux boundary condition $$\vec{v}\cdot\vec{n}=0$$ and we only have to integrate the pressure.