1

Lec 4: Fluid statics, buoyancy and stability, pressure

2

• For next time:– Read: § 3-1 to 3-4– HW 2

• Outline:– Zeroth law of thermodynamics– Pressure and resulting forces– Buoyancy and stability

• Important points:– How to calculate pressure force– How to calculate application point of pressure force– How to analyze stability

3

Fluid statics

• Fluid statics deals with non-flow situations--fluids at rest.

• It is particularly applicable with pressure measurements in terms of fluid column heights.

4

TEAMPLAYTEAMPLAY

• You accidentally drive your car into a lake and it submerges but does not admit a significant amount of water into the passenger compartment.

• A. Can you open a door?• B. How will you get out?

5

Fluid statics

• The car door may be regarded as a plane surface of area about 10 square feet.

• In order to study the force on the submerged car door resisting attempts to open it, we must delve into– Force magnitude– Force application point, known as center

of pressure.

6

Fluid statics

• Consider the effect of a constant pressure at the top of the liquid. This could be Patm or some other pressure P0.

– We can neglect P0 as long as it acts on both sides.

7

Fluid statics• Consider an arbitrary flat shape and

orientation:

0

0

P P gh

P gysin

The pressure at any point on the shape

8

Fluid statics

• The resultant force FR is given by

PdAFR

ydAgsinAP

dA)ysinP(

0

0

g

9

Fluid statics

• The integral is related to the y coordinate of the centroid (center)

ydAAy orydAA

1y CC

AP)AsingyP(F AVGC0R

10

TEAMPLAYTEAMPLAY

• Your pickup, named Bigfoot, has a door which is 4 ft high by 3.5 ft wide and all windows are stuck in the closed position. The bottom of the door is 4 ft off the ground. You accidentally drive into a stock tank where it comes to rest on its wheels in water 10 ft deep. Assume the bottom of the stock tank is flat. What is the force on the door? Can you open it?

11

Fluid statics

• Now that we know the resultant force on a submerged plane body is

• where yc is the y-coordinate of the centroid.

• it is necessary to know where the center of pressure is, that is, the point through which it acts.

AP)AsingyP(F AVGC0R

12

Fluid statics

• In general the location yP of the center of pressure isbelow the location of the centroid yC because the pressure increases with depth.

13

Fluid statics• Equate the moment of the resultant force FR to

the moment of the distributed pressure force about the x-axis.

• Where is the second moment of area (area moment of inertia).

P R 0

20

0 C XX,0

y F yPdA y(P ysin )dA

P ydA gsin y dA

P y A gsin I

g

dAyI 2XX,0

14

Fluid statics

• Most area moments of inertia are given about the centroid of the shape (IXX,C).

• They are relate to the moment IXX,0 about the x-axis by

• Area moments of inertia about the centroid are in Fig. 10-5 for some common shapes. Centroids are also given there.

AyII 2CXX,CXX,0

15

TEAMPLAYTEAMPLAY

• Solve Problem 10-13

16

Buoyancy

• A buoyant force FB is caused by increasing pressure with depth, so

depthgAF f

h)gA(sF

and

gAsF

fBOTTOM

fTOP

17

Buoyancy

• The upward force from the bottom is obviously greater, and so the net buoyancy force is

• where is the density of the fluid, not the body, and V is the volume of the body.

gVgAhF ffNET

f

18

TEAMPLAYTEAMPLAY

• The previous equation does not depend on the density of the submerged body.

• What changes about a submarine as it goes up and down (with zero propulsive thrust)?

• What is the upward force on a submarine as it holds a constant depth?

• Does this force change as it changes depth (with zero thrust)?

19

Buoyancy and stability

• The buoyant force for a constant volume system is equal to the weight W of the displaced fluid.

WgVF fB

20

Buoyancy

• The gravity force downward on a submerged body acts through the centroid.

• Similarly, the buoyant force upward must act through the centroid or there would be a rolling moment.

• Thus, we have Archimedes’ Principle:The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.

21

Buoyancy

• For floating bodies, the buoyant force is given by the weight of the displaced fluid, or

DISPLACEDDISPLACEDfB WgVF

22

Stability

• Immersed bodies must be bottom-heavy to be stable. Thus the center of gravity G must below the center of buoyancy B so that any disturbance will provide a restoring moment about G.

23

Stability

• Model a submarine as a horizontal tube with the top half empty and the bottom half filled with engines, crew quarters, and weaponry. Neglect the mass of the shell (tube). Where is the center of gravity G? Where is the center of buoyancy B? Do the two forces act to restore the sub to an upright condition if it starts to roll, or increase its rolling tendency?

24

Stability

• Rotational stability criteria are similar for floating bodies.

• However, if the center of buoyancy shifts during rolling motion, it may be possible to have the center of gravity G above the center of buoyancy and still achieve stability.

25

Stability• The metacenter M is required to be

above G. The metacenter height is the vertical distance between G and M.

• For many hull shapes the metacenter is almost a fixed point for rolling angles up to about 20°.

26

Pressure

• The normal force exerted on a (small) area. [Small enough that changes over the area are unimportant, and large enough that molecular effects also are unimportant.]

27

Pressures

• For pressures above atmospheric

• For pressures below atmospheric

P1

Patm

P=0

Pabsolute

Pgage

P1

Patm

P=0Pabsolute

PvacP2

28

In the SI system we use

• 1 Pa = 1 N/m2

• 1 kPa = 1,000 N/m2

• 1 bar = 100,000 N/m2

• 1 MPa = 1,000,000 N/m2

29

In the USCS system we use

• lbf/in2 or psi

• psi is usually written with an “a”suffix (psia) or a “g” suffix, for absolute or gage (psig)

30

Atmospheric pressure is

1 atm = 14.696 psia = 101.325 kPa= 1.01 bar

0 psig = 14.696 psia

Absolute pressure (Pabs) = gage pressure (psig) +

atmospheric pressure (Patm)

31

For example

A gage pressure of 20.0 psig is an absolute pressure at standard sea level conditions of:

Pabs = Patm + Pg = 14.7 psia + 20.0 psig

= 34.7 psia

32

TEAMPLAYTEAMPLAY

• Consider the bicycle tire, which produces a pressure reading of 30 psig. What is the absolute pressure at sea level and at 10,000 ft altitude where Patm ~ 10 psia?

33

Pressure Measurement

G as at

ppressure L

atmp

M anom eterliquid

LgρΔP

atm P pp

34

TEAMPLAYTEAMPLAY

What pressure (above atmospheric) is exerted on your ears at the bottom of a 12 foot deep swimming pool?

12 ft

Assume the density of water is:

62.4 lbm/ft3