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Liquid exiting a nozzle and exerting a pressure force on an inclined plate

I'm attempting a fluid mechanics questions that requires me to sketch a pressure distribution along the surface of the plate and explain why it shaped the way it is. I'm struggling to comprehend how to sketch this distribution.

Does it vary linearly/non-linearly? Is it continuous or constant along the plate?

My general thoughts are that it would be it triangular shaped and would decrease linearly as you went further away from where the water jet strikes the plate.

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  • $\begingroup$ HInt: stop trying to guess the answer, and apply Newton's laws of motion. $\endgroup$
    – alephzero
    Apr 28 '17 at 14:37
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This task is not trivial at all. In order to get a crude estimate of the pressure distribution, we will assume inviscid flow. The flow is separated by a streamline in the middle (red line in the picture, sp = stagnation point). We will try to get solutions for the right half of the plate and for the left half of the plate separately by applying potential theory. To be more precise we will use the power law for the potential flow. The formula that we will apply is given by:

$$\phi = |A|\mathrm{e}^{\varphi_A}r^n \cos n\theta.$$

The radius $r$ is given as the distance to the stagnation point (sp) and the angle $\theta$ is the angle in radians with respect to the horizontal $x$-axis. The parameter $|A|$ needs to be obtained by additional data (e.g. by assuming uniform flow for $Q=Au$). The angle $\varphi_A$ is the angle in radians with which we rotate the base flow to get the same orientation as in the problem.

enter image description here

The right half of the flow is a $60°$-corner flow rotated ($n=3$) with an additional rotation of $30°$ ($\varphi_A=\frac{\pi}{6}$). This will result in a potential $\phi$:

$$\phi = |A|\mathrm{e}^{i\frac{\pi}{6}}r^3 \cos3\theta.$$

The velocities can be obtained by using the relationships:

$$u_r = \frac{\partial \phi}{\partial r} \qquad u_{\theta}=\frac{1}{r}\frac{\partial \phi}{\partial \theta}.$$

Analogous, we will use $n=3/2$ for the left-hand side and $\varphi_A = \frac{\pi}{2}$ got get:

$$\phi = |A|\mathrm{e}^{i\frac{\pi}{2}}r^3 \cos(3\theta/2).$$

The velocities can be obtained by the previous relationships.

Finally, to get our pressure distribution, we need to apply Bernoulli's equation to the right side of the flow and for the left side of the flow separately.

$$p_1+\frac 12 \rho u^2_1+\rho g h_1 = p(r,\theta) + \frac{1}{2}\rho \left[u^2_r+u^2_{\varphi} \right]+\rho g h$$

$$p(r,\theta)-p_1=\frac 12 \rho \left[u^2_1-u^2_r-u^2_{\varphi} \right]+\rho g (h_1-h)$$

The height $h_1=4.85 \text{ m}$ and $u_1= 12.2 \frac {\text{m}}{\text{s}} $ and $p_1$ is the pressure at the exit of the nozzle. If they are not given take any positions for which you know $h_1, p_1$ and $u_1$.

On the inclined plate the height $h$ is given by $h=r\sin(30°)$.

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