1
$\begingroup$

I recently came across the chapter of hydrostatics and read about Pascal's Law,which states that: "a pressure exerted in a confined liquid is transmitted equally and undiminished in all directions."

There was an experiment given along with that to show it.

But that experiment did not satisfy me and so I wanted to know if there is a theoretical explanation (maybe by using formulae) to this about how the pressure is samely transmitted no matter what the area of cross-section of different segments of the confined container is.

I would be highly obliged if somebody could kindly help me out regarding this.

Thank you!

$\endgroup$
1
  • $\begingroup$ Take a look at Bernoulli's principle en.wikipedia.org/wiki/…. $\endgroup$ – user220805 Aug 6 '20 at 12:31
1
$\begingroup$

You can satisfactorily explain Pascal's law, using the Euler equation for incompressible fluid flow. We can rewrite the equations in the following form

$$\rho \frac{\mathrm d\mathbf v}{\mathrm dt}=\mathbf f - \nabla p\tag{1}$$

where $\rho$ is the density of the fluid, $p$ is the pressure (scalar) function and $\mathbf f$ is the volume density of mass forces i.e. $\mathrm d\mathbf F/\mathrm dV$, where $\mathrm d\mathbf F$ is the net external force acting on the infinitesimal element inder consideration, and $\mathrm dV$ is its volume. In most of the cases, gravity is only the external force acting on the fluid, thus $\mathbf f$ becomes $\rho\mathbf g$.

Now, initially, let's assume the pressure varies as the function of the location of the infitesimal element. Thus we can represent it as $p(\mathbf r)$, where $\mathbf r$ is the position vector of the element. Now since the state is steady, thus $\mathrm d\mathbf v/\mathrm dt=0$. Thus, using equation $(1)$ we get

$$\nabla p(\mathbf r)=\rho \mathbf g\tag{2}$$

Now, let's say that the pressure function changes to another function, $p'(\mathbf r)$. Rewriting equation $(1)$, we get

$$\nabla p'(\mathbf r)=\rho \mathbf g\tag{3}$$

Comparing equation $(2)$ with equation $(3)$, we get

\begin{align} \nabla p(\mathbf r)&=\nabla p'(\mathbf r)\\ \nabla \big(p(\mathbf r)-p'(\mathbf r)\big)&=0\tag{for all \(\mathbf r\))(4} \end{align}

The equation $(4)$ can be true only when

$$p(\mathbf r)-p'(\mathbf r)=\text{constant}=\Delta p\tag{for all \(\mathbf r\)}$$

Thus the new pressure function must have increased by the same value everywhere. This is equivalent to saying that the pressure was transmitted everywhere equally.

$\endgroup$
0
$\begingroup$

The pressure is given by the force that acts on the surface by the water. The pressure gradient would then mean, that molecules of the liquid would be acted on by bigger force from one side than from the other. This means, they would start to move until the equilibrium, in which pressure is everywhere the same, is reached.

For ordinary liquids, this equilibrium is reached fairly quickly, thus the law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.