pharos univ. me 259 fluid mechanics static forces on inclined and curved surfaces

Pharos Univ.ME 259 Fluid Mechanics

Static Forces on Inclined and Curved Surfaces

Main Topics

The Basic Equations of Fluid Statics

Pressure Variation in a Static Fluid

Hydrostatic Force on Submerged Surfaces

Buoyancy

The Basic Equationsof Fluid Statics

Body Force


Surface Force


Total Force


Newton’s Second Law


Pressure-Height Relation

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

Incompressible Fluid: Manometers

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


Plane Submerged Surface

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


Plane Submerged Surface

We can find FR, and y´ and x´,by integrating, or …


Plane Submerged Surface• Algebraic Equations – Total Pressure Force


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


Curved Submerged Surface


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.

Forces on Curved Surfaces

h1

h2

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

Buoyancy

For example, for a hot air balloon

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.

Stability of Floating Bodies

If body is bottom heavy (G lower than B), it is always stable.

Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment.

Measure of stability is the metacentric height GM. If GM>1, ship is stable.

pharos univ. me 259 fluid mechanics static forces on inclined and curved surfaces

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