Hydrostatic Pressure

Hydrostatic Pressure Definition

So what is hydrostatic pressure?

Hydrostatic pressure varies depending on the fluid and the depth.

Hydrostatic Pressure Formula

You will need some variables going forward:

• \(P\) is the hydrostatic pressure in \(\text{N}/\text{m}^2\);

• \(\rho\) is the density of the fluid in \(\text{kg}/\text{m}^3\);

• \(g\) is the acceleration due to gravity in \(\text{m}/\text{s}^2\); and

• \(h\) is the height of the water column in \(\text{m}\).

Then the formula for hydrostatic pressure is

\[P = \rho gh.\]

Notice that pressure increases as the height of the water column (also called the depth) increases. That fact is illustrated in the image below.

Fig. 2: The hydrostatic pressure, ranked from lowest to highest, would go \(A, B, C\).

The density of water is \(1000\, \text{kg}/\text{m}^3\), and for gravity on Earth you can use \(9.81 \, \text{m}/\text{s}^2 \).

Hydrostatic Force

You are probably wondering what the difference between hydrostatic force and pressure is.

If you think about the example of the diver at the start of the article, the hydrostatic force on the diver is the force of the water on their submerged body. As you have seen, pressure increases as the height of the water column increases. In other words, the deeper the diver goes, the higher the pressure is, which in turn means the hydrostatic force acting on their body increases as well.

Hydrostatic Force Formula

The hydrostatic force formula is

\[F = PA,\]

where:

• \(F\) is the hydrostatic force in \(\text{N}\);

• \(P\) is they hydrostatic pressure in \(\text{N}/\text{m}^2\); and

• \(A\) is the area of the submerged surface in \(\text{m}^2\).

Notice that the formula works for very complicated surfaces, like that of a diver's body. However figuring out the exact hydrostatic force on a diver is a bit of a big problem to start with, so instead let's look at submerging a flat plate in a fluid instead.

Fig. 3: A flat plate submerged \(d\) units under the water.

In the picture above, there is a flat hexagonal plate submerged \(d\) units under the water. The rectangle shown has height \(\Delta y\) and width \(x_i\), so the area of the rectangle is \(x_i\Delta y\). That means the hydrostatic force on the rectangle is

\[ \begin{align} F_{\text{rectangle}} &= PA \\ &= (\rho gh)(x_i\Delta y ) \\ &= (1000\cdot 9.81 d_i)(x_i\Delta y) \\ &= 9810d_ix_i\Delta y \, \text{N}. \end{align}\]

An approximation of the hydrostatic force on the entire plate would be the sum of the hydrostatic forces on all of the rectangles, so if there are \(n\) rectangles,

\[ F_{\text{plate}} \approx \sum\limits_{i=i}^n 9810d_ix_i\Delta y .\]

This should look familiar to you since finding areas this way is one of the applications of integrals!

Hydrostatic Forces on Submerged Surfaces

Let's continue on with the plate example started in the previous section. The first step is to make a properly labeled diagram with the \(x\) and \(y\)-axis included. This plate is vertically symmetric, so it makes sense to center it on the \(y\)-axis with the bottom of the plate on the \(x\) axis.

The top of the plate is \(6\) meters below the surface of the water. The plate can be thought of in two sections: the upper section where \(y \ge 2\); and the lower section where \(y \le 2\).

Fig. 4: The plate placed on the coordinate axis with the depth and equation of the upper line of the plate labeled.

Then the total area of the plate is the sum of the upper section area and the lower section area. Therefore when integrating you can split the integral at \(y=2\) to make calculations easier. In fact, since the area is symmetric across the \(y\) axis, you can find the area of the right half of the plate and then multiply by \(2\) to get the total area of the plate.

Upper Plate Calculations

For the upper part of the plate, the depth below the surface is \(d_i\). The equation of the line describing the edge of the plate is \(y = -0.8x + 8\).

The summation for the hydrostatic force on the upper part of the plate is

\[ F_{\text{upper}} \approx \sum\limits_{i=i}^n 9810d_ix_i\Delta y ,\]

which has a \(\Delta y\) in it. This means the integration will be done with respect to \(y\) and not with respect to \(x\). Therefore solving the equation of the line for \(x\) in terms of \(y\) gives

\[x = -1.25y+10.\]

You will need to be careful since writing it this way only gives you the right half of the plate! Since the plate is symmetric, to find the area from the left side of the plate to the right side of the plate, you can multiply the summation by \(2\). Substituting the line equation into the summation and multiplying by \(2\) gives

\[ F_{\text{upper}} \approx 2\sum\limits_{i=i}^n 9810d_i(-1.25y_i+10)\Delta y .\]

That just leaves writing \(d_i\) in terms of \(y_i\). As you can see from the picture above,

\[y_i = 12-d_i,\]

and now the summation becomes

\[ F_{\text{upper}} \approx 2\sum\limits_{i=i}^n 9810(12-y_i)(-1.25y_i+10)\Delta y .\]

Then the total force on the upper half of the plate is

\[\begin{align} F_{\text{upper}} &= 2\lim\limits_{n\to \infty} \sum\limits_{i=1}^n 9810(12-y_i)(-1.25y_i+10) \Delta y \\ &= 2\int_2^6 9810(12-y)(-1.25y+10) \, \mathrm{d}y \\ &= \left. 2(9810)\left(\frac{5y^3-150y^2+1440y}{12}\right)\right|_{y=2}^{y=6} \\ &= 2(9810)\left(\frac{500}{3}\right) \\ &= 3270000 \, \text{N}.\end{align} \]

Lower Plate Calculations

Let's look at the lower plate. The diagram below shows the equation of the line for that part of the plate.

Fig. 5: Lower half of the submerged plate.

The equation of the line is \(y=0.4x-1\). As in the calculations for the upper half of the plate, you need to write this equation in terms of \(x\) so it can be substituted in. Rewriting the equation of the line in terms of \(x\) you get \(x = 2.5y+2.5\). Since \(d_i = 12-y_i\), the hydrostatic force on a horizontal rectangular strip running from the \(y\)-axis to the line is

\[ \begin{align} F_i &= P_iA_i \\ &= 9810d_i(2.5y_i+2.5)\Delta y \\ &= 9810(12-y_i)(2.5y_i+2.5 ) \Delta y. \end{align}\]

The total force on the upper half of the plate is then

\[\begin{align} F_{\text{lower}} &= 2\lim\limits_{n\to \infty} \sum\limits_{i=1}^n 9810(12-y_i)(2.5y_i+2.5 ) \Delta y \\ &= 2(2.5)(9810)\int_0^2 (12-y)(y+1 ) \, \mathrm{d}y \\ &= \left. (49050)\frac{-2y^3+33y^2+72y}{6}\right|_{y=0}^{y=2} \\ &=(49050)\left(\frac{130}{3}\right) \\ &= 2125500\, \text{N}.\end{align} \]

As you would expect, the hydrostatic force on the lower part of the plate is smaller than that on the upper part of the plate because the area on the lower plate is smaller than the area on the upper part of the plate.

Total Hydrostatic Force on the Plate

Then to get the total hydrostatic force on the plate, you just need to add the hydrostatic force on the upper part of the plate to the hydrostatic force on the lower part of the plate:

\[ \begin{align} F_{\text{total}} &= F_{\text{upper}} + F_{\text{lower}} \\ &= 3270000 + 2125500 \\ &= 5395500 \, \text{N}.\end{align} \]

As you would expect, the hydrostatic force on the lower part of the plate is smaller than that on the upper part of the plate because the area on the lower plate is smaller than the area on the upper part of the plate.