Chapter 4

fluidsReferences: 1- Physics in Biology and Medicine, 3rd E, Paul Davidovits.2- College Physics, 6th E, Serway3- Web Sites

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Objectives• 1- Understand the static behavior of fluids

• 2-Illustrate the properties of fluid pressure, buoyant force in liquids, and surface tension

• 3-Understand the behaviors of fluids in motion

• 4-Study some examples from biology and zoology.

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Introduction• The differences in the physical properties of solids, liquids, and gases are

explained in terms of the forces that bind the molecules.

• 1 – Solids: the molecules are rigidly bound; a solid therefore has a definite shape and volume.

• 2-Liquids: The molecules constituting a liquid are not bound together with sufficient force to maintain a definite shape, but the binding is sufficiently strong to maintain a definite volume.

• 3- Gases: the molecules are not bound to each other. Therefore a gas has neither a definite shape nor a definite volume

• Fluids are liquids and gases• Fluids and solids are governed by the same laws of mechanics.

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1- Static FluidsForce and Pressure in a Fluid

• When a force is applied to• 1-A solid: this force is transmitted to the other

parts of the solid with its direction unchanged.

• 2- A fluid: Because of a fluid’s ability to flow, it transmits a force uniformly in all directions. Therefore, the pressure at any point in a fluid at rest is the same in all directions.

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Pressure• A fluid in a container exerts a force on all parts of the container in contact with the

fluid.

• Fluid also exerts a force on any object immersed in it. P= F/A• F is always perpendicular to A.

• The pressure in a fluid increases with depth because of the weight of the fluid above.

P2 −P1 = ρgh

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Pressure units

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1 torr =1mm Hg =13.5mm water =1.33×103 dyn/cm2

=1.32×10−3 atm =1.93×10−2 psi =1.33×102 Pa (N/m2)

4-2-b PRESSURE MEASUREMENTS

• The open-tube manometer• P = P0 - rgh.

• P is called the absolute pressure,• P - P0 is called the gauge pressure.

• If P > P0 h is +ve

• If P > P0 h is -ve

•

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Barometer

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Pascal’s Principle

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In an incompressible liquid, the increase in the pressure at any point is transmitted undiminished to all other points in the liquid. This is known as Pascal’s principle.

Archimedes’ Principle

• Archimedes’ principle states that • a body partially or wholly submerged in a fluid is

buoyed upward by a force that is equal in magnitude to the weight of the displaced fluid.

B=(Pa-Pb)A=(fluid.g.h)A=fluid g.V=Mfluid.g

4-2-d Archimedes’ Principle

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B=(Pa-Pb)A=(fluid.g.h)A =fluid g.V=Mfluid.g

Archimedes’ principle states that a body partially or wholly submerged in a fluid is buoyed upward by a force that is equal in magnitude to the weight of the displaced fluid.

Case 1: Totally Submerged ObjectB=fluid g.Vfluid

Fg=Mobject.g= Vobject object.g

B-Fg= fluid g.Vfluid- Vobject object.g, Vfluid= Vobject=V

B-Fg =g.Vfluid – object)

Case 2 Floating Object

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• rfluid g.Vfluid=Vobject robject.g

the fraction of the volume of a floating object that is below the fluid surface is equal to the ratio of the density of the object to that of the fluid.

Problem 1• In a huge oil tanker, salt water has flooded an oil tank to a depth of 5.00 m. On

top of the water is a layer of oil 8.00 m deep, as in the cross-sectional view of the tank in Figure. The oil has a density of 0.700 g/cm3. Find the pressure at the bottom of the tank. (Take 1 025 kg/m3 as the density of salt water.)

Solution• P1=P0 +oil gh1= 1.01x105 Pa+(7.00x102 kg/m3)(9.80m/s2)

(8.00m)• = 1.56x105 Pa• so P bott= P1+water gh2= 1.56x105 Pa+ (1.025x103 kg/m3) )

(9.80m/s2)(5.00m)• = 2.06x105 Pa

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Problem 2• Estimate the net force exerted on your eardrum- A~1 cm2 - due to the

water above when you are swimming at the bottom of a pool that is 5.0 m deep.

• Solution• • DP=P-P0=gh

• =(1.00x103 kg/m3)(9.80m/s2)(5.00m)= 4.9x104 Pa• Fnet = A P=(1x10-4 m2)( 4.9x104 Pa)~ 5N

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4-3 The human brain

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The human brain is immersed in a fluid (the cerebrospinal fluid) of density 1 007 kg/m3, which is slightly less than the average density of the brain, 1 040 kg/m3.

Most of the weight of the brain is supported by the buoyant force of the surrounding fluid.

4-4 Buoyancy of Fish

• The bodies of some fish contain: • either porous bones • or air-filled swim bladders • that decrease their average density and

allow them to float in water without an expenditure of energy.

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4-4 Buoyancy of Fish• EX. Cuttlefish• contains a porous bone that has a density of 0.62 g/cm3

• its body has a density of 1.067 g/cm3.• the percentage of the body volume occupied by the porous bone that

makes the average density of the fish be the same as the density of sea water (1.026 g/cm3) by using the following equation

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-The cuttlefish lives in the sea at a depth of about 150 m. At this depth, the pressure is 15 atm .

-The spaces in the porous bone are filled with gas at a pressure of about 1 atm.

-Therefore, the porous bone must be able to withstand a pressure of 14 atm.

The cuttlefish alters its density by injecting or withdrawing fluid from its porous bone.

In fish that possess swim bladders

• The decrease in density is provided by the gas in the bladder.

• To achieve the density reduction calculated in the preceding example, the volume of the bladder is only about 4% of the total volume of the fish

• Fish with swim bladders alter their density by changing the amount of gas in the bladder.

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4-5 Surface Tension

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The surface of a liquid contract and behave somewhat like a stretched membrane.

This contracting tendency results in a surface tension that resists an increase in the free surface

That surface tension is force acting tangential to the surface, normal to a line of unit length on the surface

FT = TL

capillary action.

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The surface molecules near the wall are attracted to the wall. This attractive force is called adhesion.

These molecules are also subject to the attractive cohesive force exerted by the liquid

So If the adhesive force is greater than the cohesive force, the liquid wets the container wall, and the liquid surface near the wall is curved upward.

If the opposite is the case, the liquid surface is curved downwardFm = 2πRT

capillary action.

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W= πR2hg

Fm = 2πRT

Fm = 2πRT cos in Y-direction

2πRT cos θ = πR2 hρg

h =2T cos θ / Rρggh= 2T/R if =0

Pout

Pin

Air

Problem 3

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Find the height to which water would rise in a capillary tube with a radius equal to 5.0 x 10-5 m. Assume that the contact angle between the water and the material of the tube is small enough to be considered zero.

Solution

h=2Tcos / gR =2(0.073 N/m) / (1.00x103 kg/m3)(9.80 m/s2)(5.0x10-5 m) =0.30 m

Assignment

• Solve the following problems• 1,3,5,6

• Surfactants are molecules that lower surface tension of liquids. (The word is an abbreviation of surface active agent.)

• Write a short account on the effect of surfactants that lowers the surface tension

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