me2342 sec2 statics

7/29/2019 ME2342 Sec2 Statics

1/43

P. S. Krueger ME/CEE 2342 2 - 1

ME/CEE 2342:

Paul S. Krueger

Associate Professor

Department of Mechanical Engineering

Southern Methodist UniversityDallas, TX 75275

[email protected]

(214) 768-1296Office: 301G Embrey

Fluid Mechanics

Section 2 Fluid Statics

[Chapter 3 in the text book]
mailto:[email protected]:[email protected]


2/43


Definition: In fluid statics, there is no relative motion

between fluid particles.

This means there is no shear stress in the fluid and we only

need to deal with pressure.

Pressure at a Point:

The pressure at a point in a fluid experiencing no shear

stress is independent of orientation.

This means that fluid pressure depends only on location in

fluid statics!


3/43


Proof:


4/43P. S. Krueger ME/CEE 2342 2 - 4

0F =

x

Similarly,pz=ps.

Thus,pz=px =ps =p. Since is arbitrary,p is independent of

orientation.



Pressure Variation Between Points: Basic

Equation of Fluid StaticsConsider a stationary (static) fluid:

Note: we are no longer looking at just one point; we areconsidering pressure variation from point to point.



0F =

x

No pressure variation in the horizontal direction means that

p =p(z) (depends onzonly). Mathematically we write

0=

=

y

p

x

p

z

But, m = xyz



Substituting form we have

Now, define

And divide by xyz

12 ppp

Thus,

sgdz

dp ==

This (along with the conditionp =p(z)) is the basic equation

of fluid statics.


8/43


Basic Result for Liquids:

Pressure increases linearly with depth!


9/43


Examples:

1) Manometer

From our previous results


10/43


But,

because s for a gas is small.

We will assumep is constantfor a static gas.Thus,

So, a manometer is a pressure gage (measures gage

pressure directly)


11/43


2) Liquid Pressure

Find the (gage) pressure in the water at (3).Note: ( ) wwgfgf SGSGgg ===


12/43


Begin by writing the relationship between pressures at

different points:

Combine:


13/43


Alternative method:

Start at a location wherep is known. Then follow a path

along the tube to unknownp. Add sh when going in thedirection of gravity (down) and subtract when going against

gravity (up).

If we apply this method to the previous problem we get


14/43


Alternative method applied to an example from the text

book:


15/43


Note: If the tube is inclined, use the verticaldistance, not

the distance along the tube to find the pressure difference.


16/43


3) Hydraulic Jack

At equilibrium

Thus, if there is a large difference in area, a large load canbe lifted/supported with a small force.


17/43


Hydrostatic Force on Submerged Surfaces

Conceptually, think of a fish tank:


18/43

G l R l h f h i f i


19/43


Goal: Replace the pressure forces on the region of interest

(plate/window) by a net resultant force,FR, located at

(xp,yp) and depth hp on the plate.

Note: We knowFR is perpendicular to the plate because the

pressure acts perpendicular to the plate.Begin by with the net force on dA:

By definition, F=FR

O


20/43


Or,

where

A

c ydA

A

y1

Therefore

AhF csR =

Interpretation: s

hc

= gage pressure at the depth of thecentroid (pc). So,FR is the net force form the average

pressure on the area of interest.

L ti f F F M


21/43


Location ofFR: ypFR = Mz

But

where

Th f


22/43


Therefore

Using our previous result forFR

we have

Ay

Iyy

c

cxx

cp

,

+=

Similarly,

Ay

Ixx

c

cxy

cp

,+=

Notes:


23/43


(xp,yp) is called the center of pressure. It is notat the

centroid in general. Rather,FR is located belowthecentroid because pressure increases with depth.

If the pressure above the liquid (p1) and outside the tank

(p0) are not the same, there is an additional forceFR =(p1p0)A on the plate located at the plate centroid.

Notes:

Special Cases:1) = 0: hc is the depth of the whole surface. ThenFR acts

through the centroid of the surface

2) = /2: hc =yc

E l


24/43


Examples:

1) Window in a Tank

Find the magnitude and location of the force on the window.

From the textbook (p 91)


25/43


From the textbook (p. 91)

The window is vertical, so

Then,

Also


26/43


Also,

So,

Graphically,

2) Gate


27/43


2) Gate

The gate is hinged atB and rests on a smooth wall atA. The width of the gate into the page is b = 4 ft.

The weight of the gate is negligible

The pressure below the gate is atmospheric

Find the forcePexerted on the gate at pointA by the wall.

FBD of the gate Dont forget to do this!!


28/43


FBD of the gate Don t forget to do this!!

Since the pressure outside the gate is atmospheric


29/43


Since the pressure outside the gate is atmospheric,

Also,

Alternatively,

Finally for equilibrium we require


30/43


Finally, for equilibrium we require

Cylindrical Curved Surfaces


31/43


Cylindrical Curved Surfaces

Integrating pressure over the surface is hard.Instead, deal with a volume of fluid defined by the curved

surface and the horizontal and vertical projections of the

surface.

FBDs of the fluid volume and surface:


32/43


FH

,FV

= horizontal and vertical components of the fluid force

on the curved surface.

Fy = fluid static force on the horizontal projection of the

curved surfaceFx = fluid static force on the vertical projection of the curved

surface

W= weight of the fluid in the specified volume.

For these FBDs, statics requires


33/43


, q

1) FH= Fx

2) FV=Fy + W

3) The moment ofFHandFVabout an arbitrary point O is the

same as the moment ofFx,Fy, and Wabout point O.

Special Case: Circular Surface.

In this case the resultant must act through the center of thecircle since pressure is always perpendicular to the surface.

Example: Forces on a Curved Gate


34/43


Example: Forces on a Curved Gate

Find the forcePrequired to hold the gate shut.

FBDs:


35/43


Method 1: Method 2:

Method 1:


36/43


Method 2:


37/43


Buoyancy


38/43


Buoyancy

Physics: Pressure on the top is less than pressure on the

bottom of the object, so there is a net upward force exerted

by the fluid on the object. This force is called buoyancy.

Magnitude and location of buoyant force:


39/43


FB = weight of the displaced fluid = WDFB acts through the CG of the displaced fluid.

If the fluid is incompressible

DsD VW =

where VD is the volume of the displaced fluid.

Example:


40/43


p

Determine the specific gravity (SG) of the object ifh = 1 ft.

FBD:

Now,


41/43


For equilibrium,

Some Notes on Stability


42/43


y

Note: For a floatingobject, the CG can be above the center

f b d th hi till b t bl Thi i t t


43/43


of buoyancy and the ship can still be stable. This is nottrue

for immersed objects.

[Source: www.flyingchair.net]

me2342 sec2 statics

Documents