Fluids PDF

Fluids In this course, we will discuss the behavior of liquids and gases, both of which play an important role in the life sciences. The differences in the physical properties of solids, liquids, and gases are explained in terms of the forces that bind the molecules. In a solid, the molecules a...

Fluids In this course, we will discuss the behavior of liquids and gases, both of which play an important role in the life sciences. The differences in the physical properties of solids, liquids, and gases are explained in terms of the forces that bind the molecules. In a solid, the molecules are rigidly bound; a solid therefore has a definite shape and volume. 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. A liquid adapts its shape to the vessel in which it is contained. In a gas, the molecules are not bound to each other.  Dynamic Fluids Fluids Incompressible (Volume and Density are constant) Ex: Liquids Compressible (Volume and Density are variable) Ex: Gases Fluids Non-Viscous (Low viscosity) Ex: Water, Gases Viscous (High viscosity) Ex: Honey, Oil  Fluids Flow Laminar Flow (Steady, Irrotational and the speed is constant at each point in a uniform pipe) Turbulent Flow (Unsteady flow, rotational and the speed is variable at each point in a uniform pipe) Ideal Fluild (It is the fluid which is incompressible, non-viscous and has a steady flow) Ex: Water  Force and Pressure in a Fluid Solids and fluids transmit forces differently. When a force is applied to one section of a solid, this force is transmitted to the other parts of the solid with its direction unchanged. 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. The force exerted by a fluid at rest on any area is perpendicular to the area. A fluid in a container exerts a force on all parts of the container in contact with the fluid. A fluid also exerts a force on any object immersed in it.  Force and Pressure in a Fluid P2 − P1 = ρgh The pressure in a fluid increases with depth because of the weight of the fluid above. In a fluid of constant density ρ, the difference in pressure, P2 − P1, between two points separated by a vertical distance h is P2 − P1 = ρgh Fluid pressure is often measured in millimeters of mercury, or torr [after Evangelista Torricelli (1608–1674), the first person to understand the nature of atmospheric pressure]. One torr is the pressure exerted by a column of mercury that is 1 mm high. Pascal, abbreviated as Pa is another commonly used unit of pressure. The relationship between the torr and several of the other units used to measure pressure follows: Example 1: A certain pressure can support a column of pure water 0.7 m high. The same pressure will support a column of saline solution 0.6 m high. What is the density of the saline solution? Pw = Ps ρwghw= ρsghs ρs=(hw*ρw)/hs ρs=(0.7*1000)/0.6 = 1.17*103 kg/m3 Example 2: Estimate the force exerted on your eardrum due to the water above, when you are swimming at the bottom of a pool that is 5 m deep and the surface area of the eardrum is 1 cm2 (neglect the atmospheric pressure)? Example 3:  Pressure Measurements During the weather report on a television news program, the barometric pressure is often provided. This is the current pressure of the atmosphere, which varies over a small range from the standard value provided earlier. How is this pressure measured? One instrument used to measure atmospheric pressure is the common barometer, invented by Evangelista Torricelli (1608–1647). A long tube closed at one end is filled with mercury and then inverted into a dish of mercury. The closed end of the tube is nearly a vacuum, so the pressure at the top of the mercury column can be taken as zero. The pressure at point A, due to the column of mercury, must equal the pressure at point B, due to the atmosphere. If this were not the case, there would be a net force that would move mercury from one point to the other until equilibrium is established. Therefore, it follows that P0=ρHggh, where ρHg is the density of the mercury and h is the height of the mercury column. As atmospheric pressure varies, the height of the mercury column varies, so the height can be calibrated to measure atmospheric pressure. Let us determine the height of a mercury column for one atmosphere of pressure, P0 = 1 atm = 1.013 X 105 Pa: one atmosphere of pressure is defined to be the pressure equivalent of a column of mercury that is exactly 0.760 0 m in height at 0°C. A device for measuring the pressure of a gas contained in a vessel is the open- tube manometer illustrated in next figure. One end of a U-shaped tube containing a liquid is open to the atmosphere, and the other end is connected to a system of unknown pressure P. The pressures at points A and B must be the same (otherwise, the curved portion of the liquid would experience a net force and would accelerate), and the pressure at A is the unknown pressure of the gas. Therefore, equating the unknown pressure P to the pressure at point B, we see that P = P0 + ρHggh. The difference in pressure P & P0 is equal to ρHggh. The pressure P is called the absolute pressure, while the difference P & P0 is called the gauge pressure.  BLOOD PRESSURE MEASUREMENTS BY CANNULATION In many experiments with anesthetized animals, the blood pressure in an artery or vein is measured by the direct insertion into the blood vessel a cannula, which is a small plastic tube containing saline solution plus an anti clotting agent. The saline solution, in turn, is in contact with the fluid in the manometer and does not mix. BLOOD PRESSURE MEASUREMENTS BY SPHYGMOMANOMETER The pressure in the bag is first increased until the brachial artery is closed entirely. The pressure in the sack is then slowly reduced, while a stethoscope is used to listen for noises in the brachial artery below the sack. When the pressure is slightly below the systolic (peak) pressure produced by the heart, the artery will open briefly. Because it is only partially opened, the flow velocity is high and turbulent and therefore noisy. This resulting noise heard as a tapping sound. When the pressure in the sack is lowered further, the artery remains open during longer portions of the heart cycle but is still closed during the diastolic (minimum) pressure portion of the cycle. Hence, sounds are heard, but they are interrupted by periods of silence. When the pressure in the sack reaches the diastolic pressure, the artery remains open during the entire heart cycle. Blood pressures are usually presented as systolic diastolic ratios. Typical readings for a resting healthy adult are about 120/80 in torr and 16/11 in kpa.  The borderline for high blood pressure (hypertension) is usually defined to by 140/90 in torr and 19/12 in kpa. Pressures appreciably above that level require medical attention, because prolonged high blood pressure can lead to serious damage of the heart or other organs before a person is aware of any problem.  The Motion of Fluids The study of fluids in motion is closely related to biology and medicine. In fact, one of the foremost workers in this field, L. M. Poiseuille (1799– 1869), was a French physician whose study of moving fluids was motivated by his interest in the flow of blood through the body. In this part, we will review briefly the principles governing the flow of fluids and then examine the flow of blood in the circulatory system. BERNOULLI’S EQUATION If frictional losses are neglected, the flow of an incompressible fluid is governed by Bernoulli’s equation, which gives the relationship between velocity, pressure, and elevation in a line of flow. Bernoulli’s equation states that at any point in the channel of a flowing fluid the following relationship holds: Here P is the pressure in the fluid, h is the height, ρ is the density, and v is the velocity at any point in the flow channel. The first term in the equation is the potential energy per unit volume of the fluid due to the pressure in the fluid. (Note that the unit for pressure, which is dyn/cm2, is identical to erg/cm3, which is energy per unit volume.) The second term is the gravitational potential energy per unit volume, and the third is the kinetic energy per unit volume. Bernoulli’s equation follows from the law of energy conservation. Because the three terms in the equation represent the total energy in the fluid, in the absence of friction their sum must remain constant no matter how the flow is altered. Blood flowing at 4 m/s passes through a Venturi tube as shown. If h=12 cm , what is the velocity of the blood in the constriction? Example 4: A dam springs a leak at a point 20 m below the surface. What is the emergent velocity? We will illustrate the use of Bernoulli’s equation with a simple example. Consider a fluid flowing through a pipe consisting of two segments with cross sectional areas A1 and A2, respectively. The volume of fluid flowing per second past any point in the pipe is given by the product of the fluid velocity and the area of the pipe, A × v. If the fluid is incompressible, in a unit time as much fluid must flow out of the pipe as flows into it. Therefore, the rates of flow in segments 1 and 2 are equal; that is, Consider an incompressible fluid fills a cylindrical pipe, like an artery. Then if more fluid enters one end of the pipe, an equal amount leaves the other end. This means that the volume of fluid enters the pipe per unit time, must equal the volume of fluid leaves the pipe per unit time. The volume of fluid per unit time flows through a pipe or an artery is called flow rate Q. Flowrate  Q  V m s 3 1 t If Q1 is the rate at which the fluid enters the pipe and Q2 is the rate at which the fluid leaves the pipe, then; Q1  Q2 This equation is known as equation of continuity. Now assume a fluid flows through the pipe with a constant velocity v and the cross-sectional area of this pipe is A In a time ∆t, the fluid moves a distance ∆x = v ∆t, and the volume of fluid leaving the tube is ∆V = A ∆x = A v ∆t. Alternatively, ∆V = Q ∆t, substitute for ∆V Q  Av If the cross –sectional area of the pipe is changed from A1 to A2 ,this will give rise to another form of continuity equation where In our case A1 is larger than A2 so we conclude that the velocity of the fluid in segment 2 is greater than in segment 1. Bernoulli’s equation states that the sum of the terms at any point in the flow is equal to the same constant. Therefore the relationship between the parameters P, , h, and v at points 1 and 2 is where the subscripts designate the parameters at the two points in the flow. Because in our case the two segments are at the same height (h1 = h2), last equation can be written as This relationship shows that while the flow velocity in segment 2 increases, the pressure in that segment decreases.  The Mass Flow Rate (Rate of Mass) Rate of Mass = m/t (kg/s)  “The mass of the fluid pass in a unit time“ Rate of Mass = ρV/t Rate of Mass = ρQ Rule for multiple flow paths for incompressible fluids  Special Case: If The are N tubes with equal area A and fluid speed v then: A1v1= NAv or r21v1=Nr2v Example 5: The velocity of blood in the aorta of an adult person is 0.33 m/s. Calculate the number of branches veins knowing that the velocity of the blood through it is 0.0044 m/s and that the radius of each is 0.35 cm while that of aorta is 0.7 cm? r12v1=Nr2v (0.7)2*0.33 = N*(0.35)2*0.044 N= 30 Example 6: Example 7: A blood vessel of radius r splits into 100 vessels, each with radius r/10. If the average velocity in the larger vessel is v, find the average velocity in each of the smaller vessels? Q1=Q2 , r1=r , r2=r/10 r12v1=Nr2v2 r2*v1 = 100*(r/10)2*v2 v2 = v1  Example 8: Blood flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m higher than point A. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B. Next find the absolute pressure at Point B. Example 9: Example 10:  VISCOSITY AND POISEUILLE’S LAW Frictionless flow is an idealization. In a real fluid, the molecules attract each other; consequently, relative motion between the fluid molecules is opposed by a frictional force, which is called viscous friction. Viscous friction is proportional to the velocity of flow and to the coefficient of viscosity for the given fluid. As a result of viscous friction, the velocity of a fluid flowing through a pipe varies across the pipe. Laminar flow. The length of the arrows indicates the magnitude of the velocity of the fluid. The velocity is highest at the center and decreases toward the walls; at the walls of the pipe, the fluid is stationary. Such fluid flow is called laminar. The figure shows the velocity profile for laminar flow in a pipe. The lengths of the arrows are proportional to the velocity across the pipe diameter. If viscosity is taken into account, it can be shown that the rate of laminar flow Q through a cylindrical tube of radius R and length L is given by Poiseuille’s law, which is where P1 − P2 is the difference between the fluid pressures at the two ends of the cylinder and η is the coefficient of viscosity measured in units of dyn (sec/cm2), which is called a poise. The viscosities of some fluids are listed below. In general, viscosity is a function of temperature and increases as the fluid becomes colder. There is a basic difference between frictionless and viscous fluid flow. A frictionless fluid will flow steadily without an external force applied to it. This fact is evident from Bernoulli’s equation, which shows that if the height and velocity of the fluid remain constant, there is no pressure drop along the flow path. But Poiseuille’s equation for viscous flow states that a pressure drop always accompanies viscous fluid flow. We can express the pressure drop as The expression P1 − P2 is the pressure drop that accompanies the flow rate Q along a length L of the pipe. The product of the pressure drop and the area of the pipe is the force required to overcome the frictional forces that tend to slow down the flow in the pipe segment. Note that for a given flow rate, the pressure drop required to overcome frictional losses decreases as the fourth power of the pipe radius. Thus, even though all fluids are subject to friction, if the area of the flow is large, frictional losses and the accompanying pressure drop are small and can be neglected. In these cases, Bernoulli’s equation may be used with little error.  Application on POISEUILLE’S LAW Example: Consider the flow of blood in a uniform blood vessel, so it flows from a high pressure to low pressure. High Pressure Low Pressure (Systolic Pressure) (Diastolic Pressure) (Turbulent Flow) (Laminar Flow) Normal Person High Pressure: 120 torr = 120 mmHg Low Pressure: 80 torr = 80 mmHg Example 11: In giving a transfusion, blood drips from a sealed storage bag with a 1 m pressure head through capillary tubing of 2 mm inside diameter, passing through a hypodermic needle that is 4 cm long and has an inside diameter of 0.5 mm. If the blood pressure within the vein into which the blood is being transfused is at a gauge pressure of 18 torr, find how long it will take to give the patient 1 L of blood. How long will it take if the inside diameter of the needle is only 0.4 mm? P1-P2= PA-PB PA=ρgh=1.06*103*9.8*1=10388 N/m2 A B PB=18*1.013*105/760=2400 N/m2 For 1 L= 10-3 m3 the time t= 3.6 hours Example 12: Example 13: A patient receives a blood transfusion through a needle of radius 0.20 mm and length 2.0 cm. The density of blood is 1,050 kg/m3. The bottle supplying the blood is 0.500 m above the patient's arm. What is the rate of flow through the needle? Example 14:  TURBULENT FLOW If the velocity of a fluid is increased past a critical point, the smooth laminar flow shown in next figure is interrupted. Laminar flow The flow becomes turbulent with eddies and whirls disrupting the laminar flow as shown in next figure. In a cylindrical pipe the critical flow velocity vc above which the flow is turbulent, is given by Turbulent fluid flow. Here D is the diameter of the cylinder, is ρ the density of the fluid, and η is the viscosity. The symbol ℜ is the Reynold’s number, which for most fluids has a value between 2000 and 3000. The frictional forces in turbulent flow are greater than in laminar flow. Therefore, as the flow turns turbulent, it becomes more difficult to force a fluid through a pipe. Laminar Flow: ℜ PB , ρBlood= ρ , h=hB-hH PF = PH = PB , ρBlood= ρ PF=PH+hHρg PF= 13.1 kPa PF=PB+hBρg PB= 13.2 kPa PH=PB+hρg PH= 13.3 kPa The heart pressure (PH) doesn’t change either the man is standing or in a reclining position. Hence the gauge pressures at the heart pH, at the foot pF and at the brain PB are pF  p H  ghH  pB  ghB related by: ρ is the density of blood. In discussions of the circulatory system, it is convenient to measure pressures in kilopascals, where 1 kpa =103 pa. Many discussions of the circulatory system use an older unit, the torr; 1 torr = 0.1333 kpa. Typical values for adults are hH= 1.3 m, and hB = 1.7 m with ρ = 1.595 x 103 pF  p H  ghH kg m-3, one finds: = (1.0595 x 103 kg m-3 ) (9.8m s-2) (1.3m) = 1.35 x 104 pa = 13.5 kpa. PH is typically 13.3 kpa, PF = 26.8 kpa. In a similar way we find PB = 9.3KPA. This explains why the pressures in the lower and upper parts of the body are very different when the person is standing, although they are about equal when reclining. This situation poses several problems. The most important are the tendency for blood to drain out of the venous side of the upper back to the heart and the difficulty of lifting blood from the lower extremities up to the heart. To retard drainage from the venous side of the upper body, particularly from the brain, where constant volume and flow rate are extremely important, the muscles surrounding the veins contract and cause constriction. When a person stands motionless, the blood pressure is barely adequate to force the blood from the feet back to the heart. Thus when a person sits or stands without muscular movement, blood gathers in the veins of the legs. This increases the pressure in the capillaries and may cause temporary swelling of the legs. With some exceptions, the blood pressure of most animals is in the same range as that of humans. For example, the systolic blood pressure of a pig, cat, and dog is for all three about 120 torr. The giraffe with its head positioned high above the heart is an exception. Its blood pressure is significantly higher, typically 240/160. Example 19: Find the blood pressure at the brain and the feet for an adult person, where the brain and the heart are above the feet by 1.7 m and 1.3 respectively. Given that: The blood pressure at the heart is 13.3 kpa and the blood density is 1050 kg/m3. B h PF = ? H hB PH =13.3 kpa PB = ? ρBlood= ρ= 1050 kg/m3 , h=hB-hH=1.7-1.3=0.4 m hH PH=PB+hρg F PB=PH-hρg=(13.3*10^3)-(0.4*1050*9.8)=9.2*10^3 pa=9.2 kpa PF=PH+hHρg =(13.3*10^3)+(1.3*1050*9.8)=26.7*10^3 pa=26.7 kpa Example 20: When a man stands, his brain is 0.5 m above his heart. If he bends so that his brain is 0.4 m below his heart, show how the blood pressure change in his brain? B H h=0.5 h= 0.4 H hB B’ hB hH hH F F Man Standing Up Man Bending PH=PB+hρg PH=PB’+hρg PB=PH-0.5ρg PB’=PH+0.4ρg Pressure Change PB’-PB= PH+0.4ρg – (PH-0.5ρg) PB’-PB= 0.9ρg Example 21: If the brain of a giraffe is 2m above the heart. Find the pressure difference between the heart and the brain. ( Assume the velocity of the blood is the same in the both locations)? B h=2 H hB hH F PH=PB+hρg PH-PB=hρg = 2ρg PH  PB  ghB  hH   1059.59.82  2.08 x10 4 Pa Example 22: The height difference between the brain and the foot for an adult is 1.8 m. If the pressure in the foot is 26.8 kPa. What is the pressure in the brain? (Given: ρ for blood= 1051 kg/m3 and g=10 m/s2) B h H hB = 1.8 hH F PF=PB+hBρg PB=PF - hBρg PB=26.8*1000 – (1.8*1050*10) = 7900 Pa = 7.9 kPa Example 23: At what acceleration would you expect the blood pressure in the brain to drop to zero for an erect person? a The blood pressure in the heart doesn’t change while the human body accelerates upward by an B acceleration (a). h=0.5 H hB PH=13.2 kPa = 13200 Pa and ρ= 1050 kg/m3 hH PH=PB+hρa At PB = 0 What a = ????? F PH=hρa a = PH /hρ = 13200/(0.5*1050) = 25 m/s2 a/g= 25/10= 2.5 Then a = 2.5 g Power Produced by the Heart The energy in the flowing blood is provided by the pumping action of the heart. We will now compute the power generated by the heart to keep the blood flowing in the circulatory system. The power PH produced by the heart is the product of the flow rate Q and the energy E per unit volume of the blood; that is, Example 24: At rest, when the blood flow rate is 5 liter/min, or 83.4 cm3/sec, the kinetic energy of the blood flowing through the aorta is 3.33×103 erg/cm3. The energy corresponding to the systolic pressure of 120 torr is 160×103 erg/cm3. The total energy is 1.63×105 erg/cm3—the sum of the kinetic energy and the energy due to the fluid pressure. Calculate the power produced by the heart?