Basic Equation for Pressure Field

How does the pressure in a fluid in which there are no shearing stresses vary from point to point? There are two types of forces acting on this element: surface forces due to to the pressure and a body force equal to the weight of the element.

Let’s consider the pressure change in the $y$ direction for the element below.

The force on the left face is $p_y\delta x\delta z$ and the pressure on the right face is $(p_y+\delta p_y)\cdot {\delta }_x{\delta }_z$. Thus, the net force in the $y$-direction is

$$\begin{array}{rl}\delta F_y& =p_y\delta x\delta z-(p_y+\delta p_y)\delta x\delta z\\ & =-\delta p_y\delta x\delta z\end{array}$$

Note that the pressure at any point is a function of spatial coordinates:

$$p=p(x,y,z)$$

By considering the Taylor series expansion, the pressure change in any direction is given by:

$$dp=\frac{\partial p}{\partial x}dx+\frac{\partial p}{\partial y}dy+\frac{\partial p}{\partial z}dz$$

For an isolated direction, we can write

$$\delta p_y=\frac{\partial p}{\partial y}\delta y$$

where $\delta p_y$ is the small pressure difference between two points separated by $\delta y$, $\frac{\partial p}{\partial y}$ represents how pressure changes per unit distance in the $y$-direction, and $\delta y$ is a small displacement in the $y$-direction.

We can substitute into our previous $\delta F_y=-\delta p_y\delta x\delta z$ to get:

$$\delta F_y=-\frac{\partial p}{\partial y}\delta y\delta x\delta z$$

Similarly, we can derive the other directions:

$$\begin{array}{r}\delta F_x=-\frac{\partial p}{\partial x}\delta y\delta x\delta z\\ \delta F_z=-\frac{\partial p}{\partial z}\delta y\delta x\delta z\end{array}$$

The total surface force on the element can be expressed in vector form:

$$\begin{array}{rl}\delta {\mathbf{F}}_s& =\delta F_x\mathbf{i}+\delta F_y\mathbf{j}+\delta F_z\mathbf{k}\\ & =-\left(\frac{\partial p}{\partial x}\mathbf{i}+\frac{\partial p}{\partial y}\mathbf{j}+\frac{\partial p}{\partial z}\mathbf{k}\right)\delta y\delta x\delta z\\ & =-\nabla p\cdot \delta y\delta x\delta z\end{array}$$

The body force or weight of the element is given by:

$$\delta {\mathbf{F}}_b=-\delta w\mathbf{k}=-\delta m\cdot g\mathbf{k}=-\gamma \delta x\delta y\delta z\mathbf{k}$$

Then, the total force on the element is given by:

$$\begin{array}{rl}\delta \mathbf{F}& =\delta {\mathbf{F}}_s+\delta {\mathbf{F}}_b\\ & =(-\nabla p-\gamma \mathbf{k})\delta x\delta y\delta z\\ & =\rho \delta x\delta y\delta z\cdot \mathbf{a}\end{array}$$

where $\mathbf{a}=a_x\mathbf{i}+a_y\mathbf{j}+a_z\mathbf{k}$ is the acceleration of the element.

Therefore, we can also write

$$-\nabla p-\gamma \mathbf{k}=\rho \mathbf{a}$$

and

$$\begin{array}{r}-\frac{\partial p}{\partial x}=\rho a_x\\ -\frac{\partial p}{\partial y}=\rho a_y\\ -\frac{\partial p}{\partial z}=\rho a_z\end{array}$$