Can I apply $\rho g h$ in fluid dynamics

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Solution 1

Take a small cube of water. Since, there are no shear forces in a static fluid, pressure at any place is same in all directions.

The net force in the cube exerted in the faces perpendicular to $x$ axis parallel to the base of the cube (neglecting higher order derivatives of $p$) is given by $(p - \left(p + \partial_x p~\Delta x\right))\Delta y\Delta z$ where '$\partial_x$' denotes the partial derivative with respect to $x\,.$

Similarly, taking into consideration all the other faces, the resultant force is $$-~(\partial_x p + \partial_y p + \partial_z p)~\Delta x\Delta y\Delta z = -~\nabla p ~\Delta x\Delta y\Delta z\,.~~~~~~~~~\textrm{definition of gradient operator}$$

Then for the cube to be in equilibrium, $$-\nabla p + \mathrm F_\textrm{others} = 0\,.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\textrm{Newton's Second Law of Motion}$$

If the force can be expressed as the gradient of a scalar potential $\varphi,$ per unit mass, then

$$\underbrace{-\nabla p - \rho\nabla \varphi = 0}_\textrm{equation of hydrostatics}\,.\tag{I}$$

If $\rho$ is constant, then the second term is a pure gradient term. Hence,

$$\begin{align}\nabla(p +\rho\cdot \varphi) &= 0\\ \implies~~~~~~~~ p + \rho \cdot \varphi &= \textrm{constant} \end{align}$$

When the other force is gravity, then the equation of motion has the solution $$ p + \rho g h = \textrm{constant}\,.$$


In the general case, the shear stress is non-zero for there exists internal force in the flowing fluid. So, there exists viscous force $\mathbf F_\textrm{viscous force}$. Hence, the equation of motion for an element of fluid per unit volume is given by $$\rho \times (\textrm{acceleration}) = -\nabla p - \rho \nabla \varphi + \mathrm F_\textrm{viscous force} ~~~~\textrm{Newton' s Second Law of Motion}\tag{II}$$

where

$\begin{align}\textrm{acceleration} \equiv \dot{\mathbf v}(x,y,z,t) & = v_x ~\partial_x\mathbf v + v_y ~\partial_y \mathbf v + v_z~\partial_ z \mathbf v + \partial_t \mathbf v \\ & = (\mathbf v\cdot\nabla) ~\mathbf v + \partial_t \mathbf v\tag{III}\,.\end{align}$

Putting $\rm (III)$ in $\rm (II),$ we get

$$\underbrace{\frac{\partial \mathbf v}{\partial t} +(\mathbf v\cdot\nabla )~\mathbf v = -\frac{\nabla p}{\rho} -\nabla \varphi + \frac{\mathbf F_\textrm{viscous force}}{\rho}}_\textrm{General equation of motion of a fluid element per unit volume}\,.\tag{IV}$$


Can I apply $\rho gh$ in fluid dynamics

When the other force is gravitational force, then $\varphi = gh\,.$

However, the crux of the post is that in case, the fluid is moving, the equation of motion is not $\rm (I)$ anymore; then you have to use the general $\rm(IV)$ for further analysis of the problem.

References:

$\bullet$ Lectures on Physics Vol. 2 by Feynman, Leighton, Sands.

Solution 2

Yes. See Bernoulli's Equation :
$$p+\frac12\rho v^2+\rho gh = constant$$
along a streamline, assuming the fluid is incompressible (constant density) and there is no viscosity, and flow is laminar (not turbulent) and steady (there is no acceleration along a streamline).

This equation contains the term $\rho g h$ and applies to fluids in motion. Note that the kinetic energy of the fluid must also be taken into account, in the term $\frac12 \rho v^2$. It is a statement of the conservation of energy (PE+KE).

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Updated on August 01, 2022

Comments

  • Qmechanic
    Qmechanic over 1 year

    The pressure in a fluid at a depth $h$ below the free surface (due to liquid itself) is $\rho g h$ where $\rho$=density of the fluid. Does this apply to case when the liquid is in motion also by which I mean that I know that it is applicable in fluid statics but is it equally applicable in fluid dynamics.

  • Emilio Pisanty
    Emilio Pisanty about 7 years
    ...assuming [...] and that the flow is laminar, i.e. not turbulent.
  • sammy gerbil
    sammy gerbil about 7 years
    @EmilioPisanty : I think that assumption is included in "along a streamline" because turbulent flow has no streamlines.
  • Emilio Pisanty
    Emilio Pisanty about 7 years
    @sammy It bears making that explicit, given the tone of the OP - it may not be quite clear just how important that assumption is.
  • sammy gerbil
    sammy gerbil about 7 years
    @EmilioPisanty : Fair point. Thank you for your advice.
  • NIRANJAN S PATTANSHETTI
    NIRANJAN S PATTANSHETTI about 7 years
    So in fluid dynamics also, the pressure at a depth $h$ from free surface is $\rho g h$ when fluid is not accelerating.
  • sammy gerbil
    sammy gerbil about 7 years
    ... When the fluid is not moving at all. Note the $\frac12\rho v^2$ term. Motion of fluid $(v>0)$reduces pressure $p$ if $h$ is constant.
  • Emilio Pisanty
    Emilio Pisanty about 7 years
    Given the tone of the OP (and therefore its likely future audience), this answer will probably benefit from a layman's summary of what the math means.
  • Admin
    Admin about 7 years
    Thanks @Emilio, for the response; could you tell me explicitly what point I need to expand? The purpose of the post was to show that when the other force is gravity, $\rho g h$ can be used; but the equation of motion won't be same. Both the cases have been briefly discussed here prior to the jotting down of the equation of motion which is basically just the second law. I'll be happy to add more if you tell me what I should add specifically.
  • Emilio Pisanty
    Emilio Pisanty about 7 years
    That's for you to evaluate and think about critically. The OP (and future visitors along with him) is asking about $\rho gh$, which is a highschool-physics concept. Do you think the equation-of-motion formalism as you have developed it here is accessible and understandable to that demographic? We're not here to hold your hand; you're here to hone your communication skills, and that includes thinking carefully about who you're writing for.