Post on 02-Jan-2016
Main Topics
The Basic Equations of Fluid Statics
Pressure Variation in a Static Fluid
Hydrostatic Force on Submerged Surfaces
Buoyancy
2.3.1 Pressure and head
In a liquid with a free surface the pressure at any depth h measured from the free surface can be found by applying equation (2.3) to the figure.
From equation (2.3): P1 – P2= g (ya-y)
But ya-y = h , and P2 = Patm (atmospheric pressure since it is at free surface).
Thus,P1 – Patm= gh
or P1 = Patm + gh (abs) (2.4) or in terms of gauge pressure (Patm= 0),:
P1 = gh = h (2.5)
h
P1
P2 = Patm
y
ya
Free surface
Pressure Variation in aStatic Fluid
Compressible Fluid: Ideal Gas
Need additional information, e.g., T(z) for atmosphere
Differential Manometer The liquids in manometer
will rise or fall as the pressure at either end changes.
P1 = PA + 1ga
P2 = PB + 1g(b-h) + mangh
P1 = P2 (same level)
PA + 1ga = PB + 1g(b-h) + mangh
or PA - PB = 1g(b-h) + mangh - 1ga
PA- PB = 1g(b-a) + gh(man - 1)
Figure 2.13: Differential manometer
Center of Pressure
Line of action of resultant force FR=PCA lies underneath where the pressure is higher.
Location of Center of Pressure is determined by the moment.
Ixx,C is tabulated for simple geometries.
,xx Cp C
c
Iy y
y A
Hydrostatic Force on Submerged Surfaces
Plane Submerged Surface
We can find FR, and y´ and x´,by integrating, or …
Hydrostatic Force on Submerged Surfaces
Plane Submerged Surface• Algebraic Equations – Total Pressure Force
Hydrostatic Force on Submerged Surfaces
Plane Submerged Surface• Algebraic Equations – Net Pressure Force
Table 2.1 Second Moments of Area
G G
h
b
G
G
hh/3
G
d
G
Rectangle
Triangle
d4/64d2/4Circle
bh3/36bh/2
bh3/12bh
IgAreaShape
G G
h
h/3G h
G
b
GG
d
b
A 6-m deep tank contains 4 m of water and 2-m of oil as shown in the diagram below. Determine the pressure at point A and at the tank bottom. Draw the pressure diag.
Pressure at oil water interface (PA)
PA = Patm + Poil (due to 2 m of oil)
= 0 + oilghoil = 0 + 0.98 x 1000 x 9.81 x 2
= 15696 PaPA = 15.7 kPa (gauge)
Pressure at the bottom of the tank;PB = PA + waterghwater
PB = 15.7x1000 + 1000 x 9.81 x 4
= 54940 PaPB = 54.9 kPa (gauge)
water = 1000 kg/m3
SG of oil = 0.98
Patm = 0
4 m
2 m
PA
PA=15.7 kPa
B
Aoil
water
PB = 54.9 kPA
Pressure Diagram
Hydrostatic Force on Submerged Surfaces
Curved Submerged Surface• Horizontal Force = Equivalent Vertical Plane Force• Vertical Force = Weight of Fluid Directly Above
(+ Free Surface Pressure Force)
Hydrostatic Forces on Curved Surfaces
FR on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface.
Easiest approach: determine horizontal and vertical components FH and FV separately.
Submerged Curved SurfaceResultant force:Horizontal and vertical components Horizontal component:
• FH = s*w*(h + s/2),Where, h = Depth to the top of rectangle (beginning of curve
surface)s = projected rectangle heightw = projected rectangle length or width
• Center of pressurehp = hC + IC/(hCA)hC = h + s/2
Vertical Component FV = VolumeA*w Where,
A = entire area of fluid
w = projected rectangle length or width
F F F
F
F
R H V
V
H
2 2
1 tan
Hydrostatic Buoyant Force Archimedes’ principle
When a body is submerged or floating, the resultant force by the fluid is called
the buoyancy force. This buoyancy force is acting vertically upward
The buoyancy force is equal to the weight of the fluid displaced by body.
The buoyancy force acts at the centroid of the displaced volume of fluid.
A floating body displaces a volume of fluid whose weight - body weight
For equilibrium: + ΣFy = 0 Fb – W = 0 or Fb = W
Therefore we can write ;
Fb = weight of fluid displaced by the body
Or Fb = W = mg = g
Where Fb = buoyant force
= displaced volume of fluid
W = weight of fluid
W = mg
Fb= W Fb = W
GB
GB
W = mg
Volume of displaced fluid
Buoyancy and Stability
Buoyancy is due to the fluid displaced by a body. FB=fgV.
Archimedes principal : The buoyant force = Weight of the fluid displaced by the body, and it acts through the centroid of the displaced volume.
Buoyancy and Stability
Buoyancy force FB is equal only to the displaced volume fgVdisplaced.
Three scenarios possible1. body<fluid: Floating body
2. body=fluid: Neutrally buoyant
3. body>fluid: Sinking body
Stability of Immersed Bodies
Rotational stability of immersed bodies depends upon relative location of center of gravity G and center of buoyancy B.• G below B: stable• G above B: unstable • G coincides with B: neutrally stable.