Pressure and Hydrostatics

Pressure on a Fluid Parcel

• Acts in compression and uniformly
• Pressure acts normal to a surface

Relation to Acceleration

The pressure gradient is $\frac{Force}{Volume}=\frac{ma}{V}=\rho a$

Also note that $d{F}_{press}\Rightarrow \int =P$

Net Force Due to Pressure

On a square parcel…

• On the left… $p_{1}dy$
• On the bottom… $p_{2}dx$
• On the right… $\left(p_1+\frac{dp}{dx}\right)dy$
• On top… $\left(p_2+\frac{dp}{dy}\right)dx$

$d{F}_{press}=d{F}_{x}i+d{F}_{y}j+d{F}_{z}k$ = $\left(-\frac{\partial P}{\partial x}i-\frac{\partial P}{\partial y}j-\frac{\partial P}{\partial z}k\right)dxdydz$ = $-\nabla P$

Hydrostatics

If friction is negligible and the fluid is at rest… $\nabla p=pg$
Pressure due to the weight of a liquid of constant density is given by $p=\rho gh$

As an aside, we can use this relationship to find the gage pressure at a point due to hydrostatic effects of gravity which is moving straight down from a free surface

Since $P=\frac{F}{A}$ then $F_h=\gamma h_{}A$

Gauge Pressure

An add-on to atmospheric pressure…
If gauge pressure is 1.5 Pa, then we need to add 1.5 Pa to our $p_{atm}$

Using Gauge Pressure

Only the pressure difference from atmospheric pressure matters in momentum balances, where $p_{gauge}=p_{absolute}-p_{atm}$

Specific Weight and Gravity

Specific weight:

• $\gamma =\rho g$ [$\frac{\mathrm{kg}}{m^3}$] if we examine units we can get $p=\gamma h$

Specific Gravity:

• $S{G}_{gas}=\frac{{\rho }_{gas}}{{\rho }_{air}}$
• $S{G}_{liquid}=\frac{{\rho }_{liquid}}{{\rho }_{water}}$

Manometer

Using the concept of gauge pressure and specific weights and gravity…
The manometer uses a no-slip surface (velocity at the surface is 0 $p_1$ and $p_2$ are “static” pressures)

We expect the pressure at the left tube to be greater than the pressure at the right tube. When we move up volume wise, pressure decreases.

$\rho g-\nabla P+f_{friction}=\rho a$ Or… $p_{left}=p_{right}+{\rho }_{w}gh$

In the manometer, the fluid is not moving

• The velocity profile is a parabola with a minimum to the right (kind of like how you would cap a rocket)

Manometer Solving Tips

• We basically solve these questions by determining pressure difference relationships
• If we decrease in height to a new fluid we increase in pressure, otherwise we decrease in pressure
  • Out relationship here is $p=\pm \gamma h$

If we are given the pressure at a station and it is = to atmospheric pressure, then we can use gauge pressure from Pressure and Hydrostatics.