What is the basic relationship between depth and pressure?
How are absolute and relative pressure different?
How can use our knowledge of centroids to compute the equivalent point forces of fluids?
Pressure is the term used for a force distributed over an area
\begin{equation}
P =\frac{F}{A}\tag{7.9.1}
\end{equation}
. We will consider the effect of fluid pressure on underwater surfaces, including slanted or curved objects. In all cases we will simply ask the question: what is the pressure at each point and how does it change along the surface?
Pressure can be measured in two different ways
Absolute pressure is the pressure measured above an absolute or perfect vacuum. The absolute pressure of the surrounding atmosphere is approximately \(\kPa{101.3}\) or \(\psinch{14.7}\text{,}\) and a perfect vacuum is 0 psi or 0 kPa.
Gage pressure is the pressure indicated by a standard pressure gage. The gage reads zero when exposed directly to the atmosphere, positive when the pressure is higher than atmospheric pressure, and negative pressure indicates a vacuum. In effect, pressure gages ignores the pressure of the atmosphere which surrounds us.
We will use gage pressure for the remainder of the chapter.
Commonly used pressure units include:
1 pascals (Pa) = 1 \(\text{N}/\text{m}^2\)
1 kilopascal (kPa) = 1000 \(\text{N}/\text{m}^2\)
1 pound per square inch (psi) = \(\text{lb}/\text{in}^2\)
1 kip per square inch (ksi) = 1000 \(\text{lb}/\text{in}^2\)
1 pound per square foot (psf) = 1 \(\text{lb}/\text{ft}^2\)
Subsection7.9.1Principles of Fluid Statics
A fluid, like water or air exerts a pressure on its surroundings. This pressure applies a distributed load on surfaces surrounding the fluid, like the face of a dam, an irrigation control gate, a teakettle, or the drum of a steam boiler.
When you dive underwater, the pressure you feel in your ears increases with depth. At the surface, the gage pressure is zero no matter which unit system you use. As you descend, the fluid pressure \(P\) increases with depth according to the equation
\begin{equation}
P = \rho g h\tag{7.9.2}
\end{equation}
, where:
\(\rho\) is the density of a fluid,
\(g\) is gravitational acceleration, and
\(h\) is the height of fluid above the point of interest.
Since fluid pressure increases linearly with depth, it behaves as a distributed load which increases linearly from 0 at the surface to \(\rho g h\) at depth \(h\text{,}\) acting normal to the surface. The pressure can be replaced with an equivalent force acting through the centroid of the triangular loading, with a magnitude equal to the triangular area. The pressure on horizontal surfaces is constant, and it is normal to all surfaces.
(a)Distributed pressure.
(b)Equivalent force.
(c)Pressure is perpendicular to the surface.
Figure7.9.1.Pressure on submerged surfaces.
Some points to remember when solving fluid pressure problems.
The pressure due to the fluid always acts perpendicular the surface.
A particle underwater will feel the same pressure from all directions.
Pressure increases linearly with depth. \(P = \rho g h\)
\(P = \rho g h\) assumes a constant density and thus is valid only for incompressible fluids like water or oil, but not for compressible fluids like air.
In English units, specific weight \(\gamma\) is often used instead of density \(\rho\) to describe fluids. Specific weight is the weight per unit volume of a substance, while density is its mass per unit volume. The two properties are related by \(\gamma = \rho g\text{.}\) The specific weight of freshwater at room temperature is about \(\pqf{62.4}\text{.}\)
Gage pressure is the pressure above the surrounding atmospheric pressure. Atmospheric pressure is approximately \(\psinch{14.7}\) or \(101.3~\text{kPa}\text{,}\) but since this pressure acts on everything equally and from all directions, the pressure scale can be offset to make the pressure of the surroundings \(\psinch{0}\text{,}\) gage.
Question7.9.2.
Does fluid pressure depend on the surface area of the container? For instance, is the pressure below the Atlantic Ocean less than the pressure below the Pacific Ocean since the Pacific is larger?
, approximately 9% greater than atmospheric pressure.
At \(\ft{30}\) below the surface, the pressure is 10 times higher, \(\psinch{13.0}\) which is nearly twice atmospheric pressure.
Subsection7.9.2Fluid Statics Applications
Example7.9.4.Force on a submerged window.
An aquarium tank has a \(\m{3}\times \m{1.5}\) window AB for viewing the inhabitants. The tank contains water with density \(\rho = \kgqm{1000}\text{.}\)
Find the force of the water on the window, and the location of the equivalent point load.
This force may also be visualized as the volume of a trapezoidal prism with a \(\m{1.5}\) depth into the page.
The line of action of the equivalent force passes through the centroid of the trapezoid, which may be calculated using composite areas, see Section 7.5.
Dividing the trapezoid into a triangle and a rectangle and measuring down from the surface of the tank, the distance to the equivalent force is
\begin{align*}
d \amp= \frac{\sum A_i \bar{y}_i}{\sum A_i}\\
d \amp = \frac{\big[P_A(\m{3})\big](\m{3.5}) + \left[\dfrac{1}{2}(P_B-P_A)(\m{3})\right](\m{4})} { \big[P_A(\m{3})\big] +\left[\dfrac{1}{2}(P_B-P_A)(\m{3})\right]}\\
d \amp = \m{3.71}
\end{align*}
If you prefer, you may use the formula from the Centroid table to locate the centroid of the trapezoid instead.
For the gate to tip, the force of the water must act at or above A. That happens when the centroid of the load intensity diagram from the water has its equivalent point force at or above A, so
A gate at the end of a freshwater channel is fabricated from three \(\kg{125}\text{,}\)\(\m{0.6}\times \m{1}\) rectangular steel plates. The gate is hinged at \(A\) and rests against a frictionless support at \(D\text{.}\) The depth of the water \(d = \m{0.75}\text{.}\)
Draw the free-body diagram and determine the reactions at \(A\) and \(D\text{.}\)
A free-body diagram of a cross section of the gate is shown. For simplicity the thickness of the steel plates has been ignored. You should ensure that sufficient distances are provided to locate the loads.
The easiest way to solve this is to apply the principle of transmissibility: slide the lower trapezoid left until it aligns with the upper triangle and makes a triangular loading.