CHAPTER. 2. Fluid Statics
- Physics of a Fluid at rest or moving without relative motion
- No relative motion (deformation) = No shearing stress!
Basic questions:
1. Is the pressure inside fluid a constant or a function of region of interest?
2. How is the fluid weight balanced?
Pressure at a Point in Fluid
Q: What is the pressure at a point in fluid?
Basic information from the figure:
No shearing stress on the planes of the wedge (Why?)
External forces acting on the planes
1. Forces by the pressure = Pressure Area
Direction: Perpendicular to the wedge plane (Why?)
2. Weight of the fluid in the wedge (Direction = )
= Specific weight Volume
Let’s use the equations of motion (along y and z axes)
y comp.: (1)
zcomp.: (2)
where py, pz, and ps : Average pressures on the faces of wedge
: Specific weight of fluid
: Density of fluid
From the geometry of wedge,
and(3)
By inserting Eqs. (3) into Eq. (1) and Eq. (2),
and
To obtain the pressure at a point (, , and 0)
py = ps pz = ps
Pressure at a point inside fluid at rest (No shearing stress, )
Isotropic (= Independent to direction) (Pascal’s law)
However, what is the pressure at different point?
The pressuredepends on the location of point
Pressure Variation
- How does the pressure in a fluid at rest vary from point to point?
Consider a small rectangular element of fluid at arbitrary position shown
External forces:
1. Forces by the pressures
: Perpendicular to the faces
2. Weight of the fluid inside the element
: (Downward, )
1. All forces due to the pressure (Surface forces) along yaxis
where p: the pressure at the center of element
By using a similar manner, and
Total surface force:
(where : Volume of the element)
Vector operator “Del” or “Gradient”
e.g. + + = : Pressure gradient
2. Weight of the fluid element
Finally, Equation of motion of a fluid element
By eliminating the volume,
: General equation of fluid without shearing stress
The fluid is at rest ()
→(Pressure variation)
1. Horizontal direction (x and y) = 0 = 0: No change 2. Vertical direction (z)
Go deeper (z)pressure (p)
Case. 1 Incompressible Fluid (Liquid) (i.e. Specific weight Constant)
Q. Determine the pressure difference betweenp1 and p2
From the situation shown
: Constant and
Using the equation above,
If : Known reference pressure *
(Defined from bottom to surface) = (depth)
: Hydrostatic pressureat depth h(Linear variation)
- Pressure in homogeneous and incompressible fluid at rest
- Independent of the size or shape of container
c.f. : Pressure head
*Popular choice of reference: = Atmospheric pressure at sea level
101.3 kPa (kN/m2)= 1013 hPa = 760 Torr = 2116.2 lb/ft2
of air with a density, 1.225 kg/m3 at 288.15 K (15 oC)
Example: Hydraulic jacks, brakes, and lifts
How can we stop (or lift) a car of several tons by applying a small force?
Consider a liquid in closed container
At the same height (depth),
or[: Applied (Resultant) force]
e.g. If (Magnification of force)
Case. 2 Compressible Fluid (Gases)
= : No longer constant, i.e. depend on the pressure and temperature
What is the relation between , p, and T?
Ideal gas law:
Then, Eq. of Hydrostatic pressure for Gas,
: We have to know the temperature as function of depth.
Example. Pressure of atmosphere above the Sea level
(1) Troposphere (Altitude, z= 0 ~ 10 km): Temperature decreases linearly
i.e. where : Temperature at the Sea level
: Lapse rate (6.5 K/km)
By calculating the equation,
where : Standard atmospheric pressure
(2) Stratosphere (Altitude, z= 10 ~ 20 km): Constant T