fluids Chapter 4 References: 1- Physics in Biology and Medicine, 3rd E, Paul Davidovits. 2- College Physics, 6th E, Serway 3- Web Sites 16 April 2017

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. 16 April 2017

Introduction Fluids are liquids and gases 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. 16 April 2017

1- Static Fluids Force 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. 16 April 2017

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 16 April 2017

Pressure units 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) 16 April 2017

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 16 April 2017

Barometer 16 April 2017

Pascal’s Principle 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. 16 April 2017

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=(rfluid.g.h)A=rfluid g.V=Mfluid.g

4-2-d 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=(rfluid.g.h)A =rfluid g.V=Mfluid.g Case 1: Totally Submerged Object B= rfluid g.Vfluid Fg=Mobject.g= Vobject robject.g B-Fg= rfluid g.Vfluid- Vobject robject.g , Vfluid= Vobject=V B-Fg =g.V (rfluid – robject) 16 April 2017

Case 2 Floating Object 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. 16 April 2017

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 +roil gh1= 1.01x105 Pa+(7.00x102 kg/m3)(9.80m/s2)(8.00m) = 1.56x105 Pa so P bott= P1+rwater gh2= 1.56x105 Pa+ (1.025x103 kg/m3) )(9.80m/s2)(5.00m) = 2.06x105 Pa 16 April 2017

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=rgh =(1.00x103 kg/m3)(9.80m/s2)(5.00m)= 4.9x104 Pa Fnet = A DP=(1x10-4 m2)( 4.9x104 Pa)~ 5N 16 April 2017

4-3 The human brain 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. 16 April 2017

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. 16 April 2017

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 -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. 16 April 2017

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. 16 April 2017

4-5 Surface Tension FT = TL 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 16 April 2017

capillary action. 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 downward Fm = 2πRT 16 April 2017

capillary action. h =2T cos θ / Rρg rgh= 2T/R if q=0 Air W= πR2hrg Fm = 2πRT Fm = 2πRT cos q in Y-direction Air 2πRT cos θ = πR2 hρg h =2T cos θ / Rρg rgh= 2T/R if q=0 Pin Pout 16 April 2017

Problem 3 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 00 / rgR =2(0.073 N/m) / (1.00x103 kg/m3)(9.80 m/s2)(5.0x10-5 m) =0.30 m 16 April 2017

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 16 April 2017