Week2 Chapter 2 Fluid Statics

Chemical Engineering Fluid Mechanics (CDB 1033)

May 2015

Chapter 2

Fluid Statics CLO2: Describe the concept of fluid statics in pressure measurement and buoyancy force determination

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Outline • Definition of fluid statics • Hydrostatic Equilibrium • Barometric equation • Pressure measurement: Principle and devices • Buoyancy


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Learning Outcome • At the end of this chapter, you should be able to: – Define the term “fluid statics” statics” – Calculate – Calculate pressure using manometers – Determine – Determine buoyancy force – Apply – Apply the fluid statics principle to the chemical engineering unit operations.


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What Is Fluid Statics? Fluid statics (also called hydrostatics) hydrostatics) is the science of fluids dealing with forces applied by fluids at rest or in rigidbody motion*  

No shear stresses involved The only forces develop on the surfaces of the particles will be due to pressure.

Application of fluids statics 1. Manometers 2. Cont Contin inuo uous us grav gravit ity y deca decant nter er 3. Centrifu ifugal dec decanter ter 4. De Dete term rmin inat atio ion n bu buoy oyan ancy cy fo forc rcee 5. others (*Rigid (*Rigid body motion - fluid that that is moving moving in such a manner manner that there there is no


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Fluid Statics

• Example: – Water – Water in a tank. – Water – Water in a lake (Water actually move very slowly in the lake. However the movement of water relative to each other is nearly zero that water is seen as “static”) “st atic”)


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Pressure It can be defined as a normal force exerted by a fluid per unit area. It has the unit of Newtons Newtons per square meter (N/m2) which which is called called a pasca pascall (Pa) Types of pressure defined in fluid mechanics: • Absolute pressure (P (Pabs): actual pressure at given position • Gage pressure (P (Pgage): Pabs - Patm ( Pvac): Patm-Pabs • Vacuum pressure (P


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Absolute, Gage and Vacuum pressures

Pgage Pabs

Pvac Patm

Patm Pabs

Absolute vacuum, P abs = 0

Pabs = Patm + Pgage

Absolute vacuum, Pabs = 0

Pabs = Patm – P – Pvac


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Atmospheric Pressure

• Highest recorded Patm 108.6 kPa (Tosontsengel, Mongolia, 19 December 2001)

• Lowest Lowest reco recorde rded d non-tor non-tornad nadic ic Patm  87.0 kPa (Western Pacific during Typhoon Tip, 12 October 1979)

In Fluid Mechanics: • Patm = 1 atm atm = 10 101. 1.32 325 5 kPa kPa = 14 14.7 .7 psi psi • Some devices devices uses uses 100 100 kPa (or 1 bar) as atmospheric pressure.


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Class Example 1

a) A vacu vacuum um gag gage e conn connect ected ed to to a cham chamber ber reads reads 24 24 kPa at location location where where the local local atmospher atmospheric ic pressure is 92 kPa. Determine Pabs. b) If the the abso absolut lute e press pressure ure in in a tank tank is 20 20 psi psi at at norma normall atmospheric pressure, determine Pgage. c) Gage Gage pres pressu sure re rea readin dings gs show showss a val value ue of – of – 60 60 kPa. What does it means? And, determine Pabs.


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Solution

Class Example 1: Solution

a) Pgauge = 24 kPa, Patm = 92 kPa Pabs = 24 + 92 = 116 kPa b) Pgauge

= Pabs - Patm = 20-14.7 = 5.3 psi c) Pgage = – 60 – 60 kPa It means vacuum since the value is negative. (The same if mention as Pvac = 60 kPa) Pabs = Pgauge+Patm = 101.3 – 101.3 – 60 60 = 41.3 kPa


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• In a stationary fluid the pressure is exerted equally in all directions and is referred to as the static pressure.

P1

P3

P4

P2

An element of fluid

Fluid


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Pressure Variation in a Fluid at Rest • •

Fluid pressure at a point is the same in all directions (Pascal’s Law) In a fluid at rest the pressure intensities in a horizontal plane are equal

Since fluid is at rest, there is no accelerations, and hence, Newton’s Laws of motion sum of the forces f orces on any part of the fluid in any direction is zero Let us consider the z direction, direction, opposite the direction of gravity. gravity. The sum of forces (positive upward):

Pz= z z = z

F bottom – F top – F weight = 0 We can write,

z=0  x

(P z=0)  x  y - (P z=z )  x y – ρg x  y  z = 0 ………(1)

W

y

Pz=0 forces acting acting Fig. Surface and body forces


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Dividing equation 1 by x y z and rearranging, we find

P z=z ─ P z=0 z

Pz= z

ρg = ─ ρg

z = z

If we now let z approach zero, then

limit z

dP P = ─ ρg ρg = ─ = ─ γ = dz 0 z

z=0  x W

This is the basic equation of fluid statics, called the Barometric equation

y

Pz=0


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Pressure Pres sure-- depth relations relationships hips for incompres incompressible sible fluids fluids What is incompressible fluids??

In fluid mechanics, mechanics, an incompressible fluid is a fluid that does not change the volume (e.g., density) of the fluid due to external pressure.. Incompress pressure Incompressible ible fluids fluids are a hypothetical type of fluids, fluids , which are introduced for the convenience of calculations. The compressibility of an incompressible fluid is always zero

Example: water

What is the difference between Compressible Fluids and Incompressible Fluids ?


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PressurePressur e- depth relationships relationships for incompressible incompressible fluids Assumption: Assumption: g is constant We can write barometric equation can be directly integrated



2

1

2

  dz     dz

dP  

1

P1 = Patm 1

z1

z

2

to yield

h y

1

P2  P1   z 2  z1  or

P2  P1   ( z 1  z 2 )

or

P 2

 P 1   h

z2

2

P2

x

v ariation in a Fig. Notation for pressure variation fluid at rest with a free surface

Pressure in a liquid at rest increases linearly with distance from the free surface


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Pressure Pres sure-- depth relations relationships hips for incompres incompressible sible fluids fluids Ps z=h



Ps

P

Ps

h

dP

    dz 0

 P   h P  P  h s

z=0

• For free surface  Ps = Patm • Normally take Ps = 0 (Pgage definition) • P = h = Pgage • P increase as one go downward. • P decrease as one go upward.

P


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Pressure at the same elevation

At same elevation, pressure is the same for the same fluid at rest.


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Class Example 2

Find the pressure at the bottom of a tank containing glycerin under pressure shown (glycerin = 12.34 N/m3)

Solu Solutti on: Assumption: Assumption: glycerin is an incompressible fluid

Pbottom = Ps + h = 50 + (12.34)(2) = 74.68 kPa

50 kPa

2m


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Class Example 3 Determine the pressure at the bottom of an open tank containing water at atmospheric pressure. Ps = Patm

Solution: Ass A ssum ump pti on: water is an incompressible fluid

P bottom = Ps + gh = Ps + (1000)(9.81)(10) If we measure the pressure relative to atmospheric pressure (gage pressure),It follows that Ps = 0, therefore, P bottom = 98. 98.1 1 kPa kPa (gag (gage) e)

10 m


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Compressible Fluid  Density  change with pressure

Example; Ideal/perfect gases

The equation of state for ideal gas is 



P RT



PM R u T

P M Ru R T

= absulate absulate pressure pressure (kPa) (kPa) = molar mass (kg/kmol) = Universal Universal gas gas constant constant (J/mol (J/mol K) = Gas constant (kJ/kg K) = absolute temperature (K)

Substituting above equation for density in Barometric equation, we find

dP dz



PM R u T

g …………….(1)


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If the temperature is constant, equation 1 can be separated and integrated as follows:



2

1

ln

dP P P 2 P 1

 

gM R u T gM w R u T



2

1

dz

( z 2  z 1 )

 gM w  z 2  z1  P2  P1 exp   R u T 


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Pressure Measurement • Objective: • Understand the principles of manometer u sing • Learn how to calculate pressure using manometer


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In measuring pressure: • Pressure increases as one go downward • In calculation, has to “plus” the pressure

A

B

• Pressure decreases as one go upward • In calculation, has to “minus” the pressure


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Measurement of Pressure Barometer : (Mercury Barometer) A barometer is used to measure atmospheric pressure

• A simple barometer consists of: – a tube more than 30 inch (760 mm) long – inserted in an open container of mercury – a closed and evacuated end pointing upward – open end in the mercury pool – mercury extending from the container up into the tube. – It contains mercury vapor at its saturated vapor pressure

P2

P1


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P vapor

Measurement of atmospheric pressure

A

In reference to barometric equation, we can write,

P atm= γHgh + P vapor P atm Since pressure exerted by mercury vapor is very small, therefore,

B Mercury

P atm= γHgh


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Manometer: They use vertical or inclined liquid

columns to measure pressure A standard technique for measuring pressure involves the use of liquid columns in vertical or inclined tubes.

 Piezometer

tube  U-tube manometer  Inverted U-tube manometer  Inclined tube manometer


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Piezometer Piezomet er Tube Tube Manometer Ma nometer The simplest manometer is a tube, open at the top, which is attached to the top of a vessel containing liquid at a pressure (higher than atmospheric) to be measured. Patm

As the tube is open open to the atmosphere atmosphere the pressure measured is relative to atmospheric so is gauge pressure h1

Pressure at A = pressure due to column of liquid above A

A

0

h2

P A = Patm + ρgh1 = ρgh1 Pressure at B = pressure due to column of liquid above B P B = Patm + ρgh2 = 0 + ρgh2

B Liquid

This method can only be used for liquids (i.e., not for gases) and only when the liquid height is convenient to measure. It must not be too small or too large and pressure changes must be detectable


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The “U”-Tube “U”-Tube Manometer Using a “U”-Tube “U”-Tube enables the pressure of both of both liquids and gases to be measured with the same instrument. instrument. The “U” is connected as in the figure below and filled with a fluid called called the manometric fluid. The The fluid whose pressure is being measured should have a mass mass density density less than that of the manometric fluid and the two fluids should not be able to mix readily - that is, they must must be immiscible. immiscible. Fluid density, ρ D h2

A h1

B

Manometric Manometric fluid density, density, ρman

C


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Pressure in a continuous static fluid is the same at any horizontal level so, PB= PC ……….(1) For the left hand arm, Pressure at B = pressure at A + pressure due to height hei ght h 1 of measured fluid PB= P A + ρgh1

For the right hand arm Pressure Pressure at C = pressure pressure at D + pressure due to height h of manometric manometric fluid

PC= Patm + ρmangh2 = ρmangh2 ( Since we we are measuring measuring gage gage pressure, pressure, Patm = 0 )

Putting the values of PB and PC in equation 1, we find, PA =ρmangh2 - ρgh1

If the fluid being measured is a gas, the density will probably be very low in comparison to the density of the manometric manometric fluid i.e., i.e., ρ man>> ρ. In this case the term ρgh1can be neglected, and the gauge pressure give by PA =ρmangh2


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Measurement of Pressure Difference Using a “U”-Tube “U” -Tube Manometer If the “U”-tube “U”-tube manometer is connected to a pressurized vessel at two points the pressure difference between these two points can be measured. B

If the manometer is arranged as in the figure, then we say Pressure at C = Pressure at D Fluid density, ρ

PC = PD ………(1)

hb

E

PC= PA +ρgha ……. (2)

h

PD= PB +ρg(h b – h) h) + ρmangh ……(3)

A ha

Combining equation 1, 2 & 3, PA +ρgha = PB +ρg(h b – h) h) + ρmangh or, PA-PB = ρg(hb – ha) + (ρman – ρ) gh

C

Manometric Manometric fluid density, density, ρman

if the fluid whose pressure difference is being measured is a gas and ρ man>> ρ, then the terms involving ρ can be neglected, so

P -P

=ρ

gh

D


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At the lower part of the pipe, require another manometer manometer liquid to measure P

Flow

2

1

h

Manometer liquid, ( ρman )

ρ

Prove that pressure difference P1-P2= P = g h ( ρman - ρ)


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h

Flow

At the upper part part of the pipe,it does not require another manometer liquid to measure P


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Inclined-tube manometer

Fluid C

Liquid B

1

h3 b

a

h1 h2

c

d L

Fluid A

Prove that P1-P2= γBLsin Lsinθ θ

2 e


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Selection of Manometer Care must be taken when attaching the manometer to vessel, no burrs must be present around this joint. Burrs would alter the flow causing local pressure variations to affect the measurement.

Advantages Advan tages and and disadvanges disadvanges of Manometer Manometer Advantages •

small pressure differences can be measured

•

They are very simple

•

No calibration is required; the pressure difference can be calculated from first principles.

Disadvantages • Not for measuring larger pressure differences • Some liquids are unsuitable for use. Surface tension can also cause errors due to capillary rise; this can be avoided if the diameters of the tubes are sufficiently large - preferably not not less than 15 mm diameter • Slow response; unsuitable for measuring fluctuating pressures.


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Class example 1 A simple U-tube manometer is installed across an orifice meter. The manometer is filled with mercury (specific gravity 13.6), and the liquid above the mercury is carbon tetrachloride (specific gravity 1.6). The manometer reads 200 mm. What is the pressure difference over the manometer?

Solution: Pressure at X = pressure at X' Pressure at X, P X= P1 + ρCCl4g (a+h)

Pressure at X‘, PX'= P2 + ρmer g h +ρCCl4g a We can write, P1 + ρCCl4g (a+h) = P2 + ρmer g h +ρ +ρCCl4g a or, P1-P2 = ρmer g h - ρCCl4g h =ghρ =ghρwater (SG mercury- SG CCl4) = 9.81x 0.2 x1000 (13.6-1.6) = 23544 Pa The pressure difference is 23544 Pa Pa


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Class example 2 The fluid shown shaded in the manometer is ethyl iodide with a specific gravity of 1.93. The heights are h 1 = 100 cm and h 2 = 20 cm. a)

What What is the the gag gage e pre press ssur ure e in in the the tan tank? k?

b)

What What is the the abs absol olut ute e pre press ssur ure e in in the the tank tank? ?

Tank full of air

h1

h2


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Ptank + air gh2 – EIg(h1 + h2)= Patm Since we’re taking gage pressure, Patm = 0 Ptank = air gh2 + EIg(h1 + h2) = – (1.2)(9.81)(0.2) – (1.2)(9.81)(0.2) + (1.93)(1000)(9.81)(1.00 + 0.200) = 22717.6 Pa = 22.72 kPag Assuming the density of air is i s too small Ptank = air gh2 + EIg(h1 + h2) = (1.93)(1000)(9.81)(1.00 + 0.200) = 22719.96 Pa = 22.72 kPag


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Example

Self assessment Assignment Assignment

• The mercury manometer below indicates a differential reading of 0.30 m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure in pipe B if the specific gravity of the oil and mercury is 0.91 and 13.6, respectively.

Ans: 33.47 kPa


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Mechanical and Electronic Pressure Measuring Device 

Measure of high pressures



Pressure gage Used where only a visual indication is needed at the site where the pressure is being measured.



Pressure transducer Pressure is measured at a point, and the value is displayed at another point.


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Bourdon Pressure Gage •

When pressure acts on an elastic structure, the structure will deform.

•

This deformation relates to the magnitude of pressure

• As the pressure within tube increases  tube straighten  translated into pointer motion on dial. •

The pressure indicated 

the difference between that communicated by the system to the external (ambient) pressure 

Pgage


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Pressure transducer •

The sensed pressure is converted to electrical signal – generated, and transmitted at another location such as central control station – Continuously monitor pressure changes with time (for rapidly changing pressure)

•

Types of pressure transducers: – Strain gage pressure transducer – Linear-variable differential transformer (LVDT) pressure transducer – Piezoelectric pressure transducers – Quartz resonator pressure transducers


Objectives:

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Buoyancy



To derive the equation of buoyant force



To analyze the case of bodies floating on a fluid



Use the principle of static equilibrium to solve forces involved in buoyancy problems.


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What is Buoyancy When an object is submerged or floating in a static fluid the resultant force exerted on it by the fluid is called buoyancy for ce. ce. In a column of fluid, fluid , pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The magnitude of buoyancy force is proportional to the difference in the pressure between the top and the bottom of the column, c olumn, and (as explained by Archimedes' principle) is also equivalent equivalent to the weight of the fluid that would otherwise occupy the column, column, i.e. the displaced fluid.

Archimedes’ Principle: Any Principle: Any Object, wholly or partially immersed immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object


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Buoyant force – force – Principle Principle

P1

a b

P2

• The buoyant force is caused by the increase of pressure in a fluid with depth. • Consider a body with a thickness, b is submerged in a liquid of density, f. • The hydrostatic forces: F1 = P1A acting downward F2 = P2A acting upward


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Buoyant force – force – Principle (Cont’d) (Cont’d)

P1

a

• Buoyant force = the difference between the two forces i.e the net upward force

b

FB P2

= P2A – P – P1A = f g(a+b)A – g(a+b)A – f gaA = f gbA = f gVb


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The buoyant force on the solid object is equal to the weight of the fluid displaced (Archimedes' principle) principle)


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For full fully y subme submerge rged d or imme immersed rsed body body in a fluid fluid , Archimedes’ principle is restated as: The buoyant force on a completely submerged submerged body is equal to the weight of fluid displaced

FB = W = f gVb = f Vb


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Buoyant force = Weight of fluid displaced With respect to the fluid dispersed:

FB = W = bgVf = bVf FB

 ρf gVbody  mbody    ρf g  ρbody     ρf    mbody g  ρbody   


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Example • A piece of irregularly shaped metal weighs 300.0 N in air. When the metal is completely submerged in water, it weighs 232.5 N. find the volume of the metal. Solution: FB = W = 300 – 300 – 232.5 232.5 = 67.5 N FB = fgV = (1000 (1000)( )(9. 9.81) 81) V = 67. 67.5 5 V = 0.00688 m3


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For floating bodies, Archimedes’ principle is restated as: A floating body displaces a volume volume of fluid whose weight is exactly equal to its own The weight of the entire body must be equal to the buoyant force (Weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body).


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• For floating bodies,

W  ρ f gVbody,sub  ρavg, bodygVtotal FB

Vbody,sub Vtotal



ρ avg, body ρ f


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Example • Determine the submerged depth of a cube of steel 0.30 m on each side floating in mercury. Given SGsteel = 7.8 SGHg = 13.6

steel D mercury


Solution • FB = W Vbody,total = 0.3  0.3  0.3 = 0.027 m3 f gVbody,sub = body gVbody,total

(13.6)(1000)(9.81)(0.3  0.3  D) = (7.8)(1000)(9.81)(0.027) D = 0.172 m

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Example • A solid block (SG = 0.9) floats such that 75% of its volume is in water and 25% of its volume is in liquid X, which is layered above the water. Determine the density of liquid X.


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Solution • FB = W bgVf = fgVb bg(Vw + VX) = wgVb,sub in w + XgVb,sub in X bVb = wVb,sub in w + XVb,sub in X

0.9Vb = (1.0)(0.75Vb) + (SGX)(0.25Vb) SGX = 0.60  X = 600 kg/m3


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Sink or Float

An object with an average specific weight less than that of the fluid tend to float because W < FB


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Sink or Float

An object with an average specific weight greater than that of the fluid tend to sink because W > FB.


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Sink or Float

An object whose average specific weight is equal to that of the fluid is neutrally buoyant.


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What we leant in this chapter (Outcomes)



What is fluid statics and hydrostatic equilibrium?



Pascal Law



How to derive Barometric equation



Definition of incompressible and compressible fluids?



Pressure depth relationships for fluids

Application  Application

of fluids statics statics

(a) pressure measurement using manometer (b) Buoyancy force determination determination End of Chapter 2

Week2 Chapter 2 Fluid Statics

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