Fluid Pressure at a Point

Pressure is used to indicate the normal force per unit area at a given point acting on a given plane within the fluid mass of interest. How does the pressure at a point vary with the orientation of the plane passing through the point?

Consider the free-body diagram below, obtained by removing a small triangular wedge of fluid from some arbitrary location within the fluid mass:

Since there are no shearing stresses, the only external forces on the wedge are due to pressure and weight.

• For simplicity the forces in the x direction are not shown, and the z axis is taken as the vertical axis so the weight acts in the negative z direction.
• Although we are primarily interested in fluids at rest, to make the analysis as general as possible, we will allow the fluid element to have accelerated motion.
• The assumption of zero shearing stresses will still be valid so long as the fluid element moves as a rigid body; that is, there is no relative motion between adjacent elements.

Newton’s 2nd law in the $y$ and $z$ axes give:

$$\begin{array}{rl}\sum F_y& =p_y\delta x\delta z-p_s\delta x\delta s\sin \theta =\rho \frac{\delta x\delta y\delta z}{2}{a}_y\\ \sum F_z& =p_z\delta z\delta y-p_s\delta x\delta s\cos \theta -\gamma \frac{\delta x\delta y\delta z}{2}=\rho \frac{\delta x\delta y\delta z}{2}{a}_z\end{array}$$

where:

• $p_s$, $p_y$, and $p_z$ are the average pressures on the faces
• $\gamma$ and $p$ are the fluid specific weight and density
• $a_y$ and $a_z$ are the accelerations

Clarifying above equations

• Essentially we can just think of the terms as forces in the given direction. Since the $\delta x,\delta y,\delta z$ terms are lengths, something like $p_y\delta x\delta z$ is just force in the $y$ direction acting on the $xz$ plane, since pressure = force / area, such that $p_y=F_{xz}/A_{xz}$, where $A_{xz}=\delta x\delta z$.
• Similarly, $\rho \frac{\delta x\delta y\delta z}{2}$ gives mass, since mass = volume * density.

The equations of motion can be rewritten as:

$$\begin{array}{rl}{p}_y-p_s& =\rho a_y\frac{\delta y}{2}\\ p_z-p_s& =(p{a}_z+y)\frac{\delta z}{2}\end{array}$$

Since we are really interested in what is happening at a point, we take the limit as $\delta x,\delta y,\delta z$ approach zero while maintaining the angle, and it follows that

$$\begin{array}{r}{p}_y=p_s\\ p_z=p_s\end{array}$$

or just $p_y=p_z=p_s$. The angle was arbitrarily chosen so we can conclude that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present. This is known as Pascal’s Law.