# Is gravity included in "static pressure"?

In the context of Bernoulli, does "static pressure" typically refer to $p$, or to $p + \rho gz$?

• Could explain a bit more what the context actually is? Depending on the context it could "seem" that the answer is YES or NO, and actually the answer is more likely NO. Anyhow, please provide the context and we will figure it out. Oct 3 '15 at 14:46

$p_0$ is the static pressure

$\rho g z$ is the hydrostatic pressure

$\frac{1}{2} \rho v^2$ is the dynamic pressure

Hence

$p(z) = p_0 + \rho g z + \frac{1}{2} \rho v^2$

Without any further information on where you want to apply "Bernoulli" it is hard to answer the question. But in general the static pressure $p$ does not include $\rho g z$.

Lets recall again Bernoulli's principle which states:

Along a streamline $\frac{v^2}{2} + gz + \frac{p}{\rho}$ is constant. Here $g$ is the gravity which is the force field acting on the fluid. This equation/model is true for small fluid speeds (with respect to its speed of sound, see Mach Number for further information on compressibility effects). If one multiplies the equation with $\rho$ all parts of the equation have the dimension of a pressure and the contstant is referred to as total pressure, $p_\mathrm{t}$.

$\rho\frac{v^2}{2} + \rho gz + p =$ Total Pressure $= p_\mathrm{t}$

And if one assumes an isentropic process, it is possible to convert one form of pressure into another form of pressure (always along the streamline).

Most of the times $\Delta \rho$ and $\Delta z$ are small along the streamline which is investigated that it can be neglected. Then Bernoulli's principle can be simplified to:

$\rho\frac{v^2}{2} + p = p_\mathrm{t}$

The static pressure, $p$, in this equation therefore does not include the gravity. A drastic change of the altitude along the streamline needs to be addressed by including an additional term in the equation.

I think there is no unique technical term to define pressures in different situations. As a reference states here: $$p_\text{static}=\rho g h$$

If we consider the data computed from any incompressible CFD solver, the data of pressure term indeed includes the hydrostatic part when gravity $$g$$ is available.

I do prefer to think the fluid pressure (incompressible) in this way for $$p+\frac{\rho v^2}{2}+\rho g z=p_0$$

1. $$\rho g z$$ is the potential at $$z_A$$ not the hydrostatic pressure. If zero is the reference position, the hydrostatic pressure or pressure due to acceleration should be $$p_{hs}=\rho g (-z)=-\rho g z$$, since gravity has a negative direction.
2. $$p$$ raised in the previous equation has two components: hydrodynamic pressure $$p_{hd}$$ and hydrostatic pressure $$p_{hs}$$, which is to say $$p=p_{hd}+p_{hs}$$.
3. Hence, it is quite obvious that static pressure $$p_{s}=p_{hd}=p_0-\frac{\rho v^2}{2}$$
4. Dynamic pressure $$p_{d}=\frac{\rho v^2}{2}$$
5. $$p_s+p_d=p_0$$(total pressure)

As for the common situation, pressure $$p$$ is the static pressure $$p_s$$ (hydrodynamic pressure $$p_{hd}$$) plus hydrostatic pressure $$p_{hs}$$ (or pressures due to inertial force). Excluding the rest balance, $$p_s$$ or $$p_{hd}$$ is the real reason to make flow move (to balance dynamic pressure $$p_d$$).