Hydrostatic Pressure Curved Surface

[member 428835]

Hi PF!

I have been reading and reading for a clean explanation for the hydrostatic force of a curved submerged 2-D (or 3-D if you're up to it) surface. I really don't care what the curve looks like: quadratic, circular, partway sinusoidal, etc. All articles I read involve the centroid, but could someone help me through a derivation where we do all the math ourself? If so you choose the geometry (or I can) and let's start into it!

Thanks either way!

Josh

 

[Chestermiller]

The pressure increases with depth according to the hydrostatic equation, correct? The pressure acts on every surface in the direction normal to that surface, correct? So the pressure force on a differential area dA is $p\vec{n}dA$, where $\vec{n}$ is a unit normal to the surface. You need to add up (integrate) all these pressure forces vectorially.

Chet

 

[member 428835]

The pressure increases with depth according to the hydrostatic equation, correct?
Chet

Correct. So the pressure at any submerged point is $p = \rho g d$ where $p$ is pressure, $\rho$ is density, $g$ is acceleration of gravity, and $d$ is the vertical distance submerged?

The pressure acts on every surface in the direction normal to that surface, correct? So the pressure force on a differential area dA is $p\vec{n}dA$, where $\vec{n}$ is a unit normal to the surface. You need to add up (integrate) all these pressure forces vectorially.

So pressure is then $$\vec{p} = \iint_A \rho g d \cdot \vec{n} dA$$ (I used the $\cdot$ so my depth variable doesn't look like a differential, not to mistake it with a dot product).

 

[Chestermiller]

Correct. So the pressure at any submerged point is $p = \rho g d$ where $p$ is pressure, $\rho$ is density, $g$ is acceleration of gravity, and $d$ is the vertical distance submerged?
So pressure is then $$\vec{p} = \iint_A \rho g d \cdot \vec{n} dA$$ (I used the $\cdot$ so my depth variable doesn't look like a differential, not to mistake it with a dot product).

Correction to your equation:
$$\vec{F} = \int_A \rho g z \vec{n} dA$$
OK. Let's take a simple example. Let the area of interest be parallel to the y axis, and have a width w. Let the equation for the surface be x = x(z). Then $dA=w\sqrt{(dx)^2+(dz)^2}$. What is the equation for the unit normal to the contour?

 

[member 428835]

Correction to your equation:
$$\vec{F} = \int_A \rho g z \vec{n} dA$$

Of course, not sure why I didn't put a force.

OK. Let's take a simple example. Let the area of interest be parallel to the y axis, and have a width w. Let the equation for the surface be x = x(z). Then $dA=w\sqrt{(dx)^2+(dz)^2}$. What is the equation for the unit normal to the contour?

I think $\vec{n} = - \hat{j}$. I can't really picture the surface you describe though. Is it something like a cylinder in the $y$ direction with an arbitrary cross section that is a function of $x$ and $z$?

 

[Chestermiller]

Of course, not sure why I didn't put a force.I think $\vec{n} = - \hat{j}$. I can't really picture the surface you describe though. Is it something like a cylinder in the $y$ direction with an arbitrary cross section that is a function of $x$ and $z$?

At every value of y, x is the same function of z. On the surface of interest, y runs from 0 to W.

The unit normal to the surface is not j. The unit tangent to the surface is $\vec{t}=\frac{(dx)\vec{i}+(dz)\vec{k}}{\sqrt{(dx)^2+(dz)^2}}$. The unit normal to the surface is then $$\vec{n}=\frac{(dz)\vec{i}-(dx)\vec{k}}{\sqrt{(dx)^2+(dz)^2}}$$

So, if we call p(z) is the pressure at depth z, what is the resultant force vector acting on the surface, according to our integral?

 

[member 428835]

Ohhhhhh I think I see the shape you're describing now, and you're talking about the sides? Is the force then $$\vec{F} = \int_0^W \int_{z_0}^{z_1} \rho g z \left( \hat{i} - x'(z) \hat{k} \right) \, dz dw$$

 

[Chestermiller]

Ohhhhhh I think I see the shape you're describing now, and you're talking about the sides? Is the force then $$\vec{F} = \int_0^W \int_{z_0}^{z_1} \rho g z \left( \hat{i} - x'(z) \hat{k} \right) \, dz dw$$

Very nice. Now the w can be integrated immediately, and the x component of the force can be obtained as an exact differential. The z component should be partially handled using integration by parts. Check out the x component carefully. You should be able to see some kind of center of mass thing from this. BTW, if you are integrating from z0 to z1, you need to include the pressure a z0 in the pressure expression at depth z.

Chet

 

[member 428835]

$$\vec{F} = \int_0^W \int_{z_0}^{z_1} \rho g z \left( \hat{i} - x'(z) \hat{k} \right) \, dz dw =\\ W g \rho \hat{i}\int_{z_0}^{z_1} z \, dz - W g \rho \hat{k}\int_{z_0}^{z_1}\rho z x'(z) \, dz =\\ W g \rho \hat{i}\int_{z_0}^{z_1} z \, dz - W g \rho \hat{k} \left( z_1 x(z_1)-z_0 x(z_0) - \int_{z_0}^{z_1}x \, dz\right)$$
where I believe the $z \, dz$ integral can be rewritten as a center of mass. You agree with this?

 

[Chestermiller]

$$\vec{F} = \int_0^W \int_{z_0}^{z_1} \rho g z \left( \hat{i} - x'(z) \hat{k} \right) \, dz dw =\\ W g \rho \hat{i}\int_{z_0}^{z_1} z \, dz - W g \rho \hat{k}\int_{z_0}^{z_1}\rho z x'(z) \, dz =\\ W g \rho \hat{i}\int_{z_0}^{z_1} z \, dz - W g \rho \hat{k} \left( z_1 x(z_1)-z_0 x(z_0) - \int_{z_0}^{z_1}x \, dz\right)$$
where I believe the $z \, dz$ integral can be rewritten as a center of mass. You agree with this?

Yes. It's going to be $\Delta z$ times the pressure at the center of mass.