Hydrostatic Pressure | Sahithyan's Notes

At a point,

P = \lim_{\delta A \to 0} \frac{\delta F}{\delta A}

Pascal's Law

The hydrostatic pressure at a point is the same from all directions.

Proof

Consider the fluid element shown, containing the point $A$ .

From the image: $\sin{\theta}=\frac{\Delta{z}}{\Delta{s}}\;\land\;\cos{\theta}=\frac{\Delta{x}}{\Delta{s}}$

For equilibrium:

P_x(\Delta{y}\Delta{z})-P_s(\Delta{y}\Delta{s})\sin{\theta}=0 \implies P_x=P_s

P_z(\Delta{x}\Delta{y})-P_s(\Delta{y}\Delta{s})\cos{\theta}-\frac{1}{2}\Delta{x}\Delta{y}\Delta{z}\rho g=0 \implies P_z=P_s+\frac{1}{2}\Delta{z}\rho g

As all $\Delta{x},\Delta{y},\Delta{z}$ approaches $0$ : $P_z=P_s$ . Therefore $P_x=P_z=P_s$

Let $p$ be the pressure at the point $A\equiv (x,y,z)$ .

p = f(x,y,z)

dp = \frac{\partial p}{\partial x} \text{d}x + \frac{\partial p}{\partial y} \text{d}y + \frac{\partial p}{\partial z} \text{d}z

By considering equilibrium of this fluid element containing $A$ .

Fluid element containing point A

In the $x$ direction,

\bigg( p - \frac{\partial p}{\partial x} \frac{\Delta{x}}{2} \bigg) \Delta{y} \Delta{z} - \bigg( p + \frac{\partial p}{\partial x} \frac{\Delta{x}}{2} \bigg) \Delta{y} \Delta{z} = 0

\frac{\partial p}{\partial x} = 0

Similarly $\frac{\partial p}{\partial y} = 0$ can be proven.

In the $z$ direction,

\bigg( p - \frac{\partial p}{\partial z} \frac{\Delta{z}}{2} \bigg) \Delta{x} \Delta{y} - \bigg( p + \frac{\partial p}{\partial z} \frac{\Delta{z}}{2} \bigg) \Delta{x} \Delta{y} - \Delta{x} \Delta{y} \Delta{z} \rho g = 0

\frac{\partial p}{\partial z} = -\rho g

dp = -\rho g\,\text{d}z

p = -\int{\rho g\,\text{d}z}

Surface of constant pressure.

Pressure exerted by atmospheric air.

Measured in respect to atmospheric pressure.

Measured in respect to perfect vaccum.

\text{Absolute Pressure} = \text{Atmospheric Pressure} + \text{Guage Pressure}

P = -\rho gz + c

P + \rho gz = c = P^{*}

A diagram showing the variation of pressure along a submerged surface.

Height of a particular fluid column that will produce the pressure at a point.

\text{Pressure head} = h = \frac{p}{\gamma}

Suppose there are 2 points $\text{A},\text{B}$ with a height difference of $h$ .

$P_1 = -\rho gz_1 + c$

$P_2 = -\rho gz_2 + c$

$P_2 - P_1 = -\rho g(z_2 - z_1) = -\rho g (-h) = h \rho g$

$P_2 = P_1 + h \rho g$