Capter 6. Fluid Mecanics Notes: Most of te material in tis capter is taken from Young and Freedman, Cap. 12. 6.1 Fluid Statics Fluids, i.e., substances tat can flow, are te subjects of tis capter. But before we can delve into tis topic, we must first define a few fundamental quantities. 6.1.1 Mass Density and Specific Gravity We ave already encountered te mass density (often abbreviated to density) in previous capters. Namely, te mass density is simply te ratio of te mass m of an object to its volume V ρ = m V (6.1) wit units of kg/m 3. Evidently, te density of objects can vary greatly depending of te materials composing tem. For example, te density of water is 1,000 kg/m 3 at 4 C, tat of iron is 7,800 kg/m 3, wile a neutron star as a mean density of approximately 10 18 kg/m 3! Te specific gravity of a substance is defined as te ratio of te mass density to tat of water at 4 ο C (i.e., 1,000 kg/m 3 ). It would probably be more precise to use te term relative density instead of specific gravity, but suc is not te custom 6.1.2 Pressure and Buoyance A fluid is composed at te microscopic level by molecules and/or atoms tat are constantly wiggling around. Wen te fluid is contained in a vessel tese particles will collide wit te walls of te container, a process tat will ten cange teir individual momenta. Te cange of momentum tat a particle experiences will impart an impulse over te time interval during wic te collision takes place, as a result te walls of te vessel will feel a force. Te pressure p at a given point on a wall is defined as te force component perpendicular to te wall at tat point per unit area. Tat is, if df is tis elemental perpendicular force applied to an infinitesimal area da on a wall, ten te pressure on tat area is p df da. (6.2) Wen te pressure is te same at all points of a macroscopic, plane surface of area A, ten te perpendicular force F must also be te same everywere on tat surface and - 123 -
Te pascal (Pa) is te unit of pressure wit p = F A. (6.3) 1 Pa = 1 N/m 2. (6.4) Related to te pascal is te bar, wic equals 10 5 Pa, and, accordingly, te millibar, wic equals 100 Pa. Te atmosperic pressure p a, i.e., te average atmosperic pressure at sea level, is 1 atmospere (atm) wit 1 atm = 101,325 Pa = 1,103.25 millibar. (6.5) It is important to note tat motion of te particles tat cause te pressure is random in orientation and pressure is terefore isotropic. Tat is, pressure at one point is te same in all directions. Also, since te pressure at a point is directly proportional to te force effected at tat point, it sould be clear tat weigt can be a source of pressure. For example, te pressure in te eart s atmospere decreases as one goes to iger altitude as te weigt of te, or te amount of, fluid above is reduced. Similarly, an increase in pressure is felt by a diver wo descends to greater depts in a body of water. We can quantify tis effect by studying ow pressure varies witin a fluid contained in a vessel. Accordingly, referring to Figure 1, we consider a fluid of uniform density ρ under te effect of gravity g and consider a fluid element of tickness dy and area A. We assume tat te bottom of te vessel is located at y = 0 and te position of te fluid element is at y ( > 0 ; y tus increases upwards). If te pressure at te bottom of te element is p, ten te pressure immediately on top of it will be p + dp. If we furter assume tat te fluid is in equilibrium, ten tis fluid element must be static and te different forces, say, at te bottom of te element must cancel eac oter out. Tat is, pa were dw is te weigt of te fluid element ( p + dp)a + dw = 0, (6.6) dw = ( ρady)g. (6.7) Te quantity between parenteses in equation (6.7) is simply te mass of te fluid element. Equation (6.6) ten becomes or dp A = ρg Ady, (6.8) - 124 -
Figure 1 Te pressure as function of eigt in a fluid. dp = ρg. (6.9) dy Equation (6.9) is often called te equation of ydrostatic equilibrium. Tis result sows tat pressure decreases as one moves upward in te vessel, as expected. We can integrate tis equation to find te difference in pressure between two points y 1 and y 2 ( y 2 > y 1 ) wit p 2 = 1 2 dp = ρg dy 1 = ρg( y 2 y 1 ), 2 (6.10) wic we rewrite as (wit Δy = y 2 y 1 > 0 ) = p 2 + ρgδy. (6.11) If we set y 2 at te top surface of te fluid (i.e., near te opening of te vessel), ten p 2 p 0, were 0 means zero dept, equals te pressure at te exterior of te fluid. For example, if te vessel is located at sea level, ten and p 0 = 1 atm, (6.12) = p 0 + ρgδy. (6.13) - 125 -
It is ten convenient to tink of Δy > 0 as te dept in te fluid were te pressure is encountered. Equation (6.13) also implies tat increasing p 0 by some amount will increase te pressure at any point witin te fluid by te same amount. Tis is te socalled Pascal s Law Pressure applied to an enclosed fluid is transmitted undiminised to every portion of te fluid and te walls of te containing vessel. We can use equation (6.11) to explain te beavior of objects submerged (sometimes not completely) in a fluid, suc as water. Let us consider Figure 2 were an objet of mass m, orizontal area A, and eigt is immersed in a fluid of density ρ ; te wole apparatus is subject to gravity. We denote by and p 2 te pressures at te bottom and top surfaces of te object, respectively, likewise te force components perpendicular to tose surfaces are F 1 and F 2. But we know from equation (6.11) tat A p 2 A = F 1 F 2 = ρga, (6.14) or, wile defining te volume of te object wit V = A, we ave F 1 F 2 = ρvg. (6.15) Since te F 1 F 2 is net buoyancy force acting on te body and ρv is te mass of fluid displaced by te presence of te body, we are ten led to Arcimedes Principle Te net upward, buoyancy force acting on a partially or completely immersed body equals te weigt of fluid displaced by te body. It is important to note tat te buoyancy force is independent of te weigt of te object. Also, altoug we derived tis result for an object of rectangular volume, it sould be clear tat it applies to any possible sape since only te net perpendicular forces on te areas spanned by te top and bottom surfaces of te object come into play. p = F /A 2 2 1 = F 1 /A Figure 2 - An object immersed in a fluid. - 126 -
6.1.3 Exercises 1. Hydraulic lift. Consider te representation of a ydraulic lift sown in Figure 3. Use Pascal s Law to explain te lift s functioning. Solution. According to equation (6.11), two points in fluid located at te same eigt are subjected to te same pressure. If we consider points 1 and 2 in Figure 3 bot located at te surface of te fluid, ten we can write and terefore = p 2 F 1 = F 2 (6.16) A 2 F 2 = A 2 F 1. (6.17) It follows tat we can multiply at point 2 te effect of te force F 1 applied at point 1 over an area (wit a piston, for example) if we use a surface A 2 > at point 2. Te resulting force is multiplied by te ratio of te areas, as sown in equation (6.17). 2. Arcimedes Principle and Buoyancy. We drop a rectangular piece of cork of mass m, volume A, and density ρ m = 240 kg/m 3 in a vessel containing water (of density ρ = 1000 kg/m 3 ). Determine te dept at wic te object s bottom surface settles once equilibrium is reaced. Figure 3 A ydraulic lift. - 127 -
Assume tat an atmosperic pressure p 0 = 1 atm is present at te surface of te water and tat ρ air = 1.2 kg/m 3. Solution. If denser tan water, an object would sink to te bottom of te vessel. We can understand tis by considering Newton s Second Law for te forces acting on te object A p 2 A mg = ma. (6.18) But by replacing A p 2 A wit te rigt-and side of equation (6.14) and setting m = ρ m A we get or and te object sinks to te bottom. ( ρ ρ m )ga = ρ m Aa, (6.19) ρ a = g 1 ρ m < 0 (6.20) If on te oter and te object is less dense tan water, as is te case ere, ten te object will float, as sown in Figure 4. We denote by 1 > 0 te dept at wic te object s bottom surface is located relative to te surface of te water and by 2 > 0 te portion above te water (i.e., = 1 + 2 ). Using Newton s Second Law we write A p 0 + dp 0 dy 2 A mg = 0 (6.21) since we assume equilibrium (i.e., a = 0 ), and were is te pressure at te object s bottom surface, m = ρ m A (as before), and 2 dp 0 dy is te cange in atmosperic pressure from te water surface to te top of te object. p = 1 atm 0 2 ρ m 1 ρ 1 Figure 4 An object partially immersed in water. - 128 -
From Arcimedes buoyancy relation we ave p 0 A p 0 + dp 0 dy 2 A = ρ A air 2g, (6.22) or, as expected for ydrostatic equilibrium, dp 0 dy = ρ airg. (6.23) Also from te equation of ydrostatic equilibrium (i.e., equation (6.13)), we can write = p 0 + ρg 1. (6.24) Inserting equations (6.23) and (6.24) into equation (6.21), wile setting 2 = 1, we find tat and ( p 0 + ρg 1 ) p 0 ρ air g( 1 ) ρ m g = 0, (6.25) 1 = ρ m ρ air ρ ρ air ρ m ρ 0.24, (6.26) since te density of air is completely negligible compared to tose of water and cork. It sould be clear from tis discussion tat an object tat is completely immersed in a fluid, and at equilibrium (as in Figure 2), must ave te same density of te fluid. 3. (Prob. 12.33 in Young and Freedman.) A rock is suspended by a ligt string. Wen te rock is in te air, te tension in te string is 39.2 N. Wen te rock is totally immersed in water, te tension in te string is 28.4 N. Wen te rock is totally immersed in an unknown liquid, te tension is 18.6 N. Wat is te density of te unknown liquid? Solution. According to Arcimedes Principle te buoyancy force acting on te rock equals te weigt of te displaced volume of liquid. Tat is, F b = ρvg, (6.27) - 129 -
were V is te volume of te rock. According to Newton s Second Law T + F b mg = 0, (6.28) wit T te tension in te string and m te mass of te rock. Applying equation (6.28) to te cases of air and water and ten equating tem, we ave and T air + ρ air Vg = T water + ρ water Vg (6.29) V = T air T water ( ρ water ρ air )g ( 39.2 28.4)N = ( 1000 1.2)kg/m 3 9.81 m/s 2 = 1.10 10 3 m 3. (6.30) One way to proceed is to insert tis result in equation (6.28) to find te mass of te rock (let us coose te case were it is immersed in water) m = T water g + ρ waterv = 28.4 N 9.81 m/s 2 +1000 kg/m3 1.10 10 3 m 3 = 4.00 kg. (6.31) Equation (6.28) for te unknown liquid ten yields ρ u = mg T u Vg = 4.00 kg 9.81 m/s2 18.6 N 1.10 10 3 m 3 9.81 m/s 2 = 1907 kg/m 3. (6.32) Alternatively, we could ave written and, wit a similar outcome, T air + ρ air Vg = T u + ρ u Vg, (6.33) ρ u = ρ air + T air T u Vg. (6.34) - 130 -
6.2 Fluid Flows Te kinematics and motions of fluids can be a very complicated affair. In wat follows, we will limit our studies to ideal fluids, i.e., tose tat are incompressible and nonviscous (tat ave no internal friction). We will also concentrate on steady flows, wen motions ave taken place for a long enoug period tat no transient beavior remains. Finally, we will not consider turbulence (random and caotic flow patterns) and limit ourselves to laminar flows were layers of fluid can smootly flow next to one anoter witout te creation of turbulence. 6.2.1 Te Continuity Equation As a fluid is in steady motion and its flow progresses wit time, it must be tat te amount of mass flowing per unit time across to areas perpendicular to te flow is conserved. Tat is, if we consider te flow configuration sown in Figure 5, ten we can write te following dm dt = ρ dx 1 dt = ρ v 1 (6.35) for te amount of mass dm passing troug area in te time interval dt ( ρ is te density of te fluid). If te flow is steady, ten te amount of mass flowing troug some oter area A 2 during te same interval dt. More specifically, we can write or simply ρ v 1 = ρa 2 v 2, (6.36) v 1 = A 2 v 2. (6.37) Figure 5 Continuity and mass conservation in a flow. - 131 -
Equation (6.37) is te so-called continuity equation for steady flows (of incompressible fluids). Generally, we can state tat te volume flow rate is conserved dv dt = Av = constant. (6.38) 6.2.2 Bernoulli s Equation We now seek to apply te mass continuity equation wile taking into account any canges of pressure tat can accompany te flow of fluids. Suc canges in pressure are to be expected wenever te cross-sectional area A canges along a flow. Tis is because as te area varies, te velocity must also cange according to te continuity equation (6.37); if te flow speed canges, ten tere must be forces acting on te flow to cause tis acceleration. Finally, pressure variations must also occur since pressure is defined as te force per cross-sectional area. To derive te equation tat relates tese quantities, we will use te work-energy teorem defined in Capter 2, wic we write er for convenience W oter = ΔK + ΔU grav, (6.39) were W oter is te work done by all forces oter tan gravity, ΔK is te cange in kinetic energy, and ΔU grav is te cange in gravitational potential energy. In our case we will substitute W oter W pressure W p. Let us consider te tube of canging cross-section sown in Figure 6. We first concentrate on te section on te left of widt v 1 dt and crosssection, troug wic te flow speed is v 1 and te pressure. We can ask wat amount of work dw 1 was done by te pressure on a fluid tat as traveled from te entrance to te exit of tat section? Te important fact to remember is tat pressure is isotropic, meaning tat te force at te entrance as te same magnitude as te force at te exit but of opposite direction. It terefore follows tat dw p,1 = F net,1 dr 1 ( ) = v 1 dt + p A v dt 1 1 1 = 0. entrance ( ) exit (6.40) Te pressure does no work wen cross-section of te flow is constant. Te same result dw p,2 = 0 would be found for te section, of widt v 2 dt and cross-section A 2, troug wic te flow speed is v 2 and te pressure p 2, on te rigt of te tube. Te same cannot be said for te middle section of te tube, were te cross-section canges from to A 2. In tis case we find - 132 -
v dt 1 A2 v dt 2 p 1 p 2 p 2 p 2 Figure 6 A tube of canging cross-section, troug wic an incompressible and nonviscous fluid is flowing. dw p,12 = F net,12 dr 12 ( ) = v 1 dt + p 2A 2 v 2 dt, (6.41) entrance ( ) were dt is an infinitesimal time interval suc tat v 1 dt and v 2 dt are muc smaller tan te widt of te section. We now use equation (6.37) for mass continuity and transform equation (6.41) to exit dw p,12 = ( p 2 )dv, (6.42) wit dv te volume element spanned in te interval dt (see equation (6.38)). Tis equation can be integrated over between any two points along a tube, and generalized to (for an incompressible and non-viscous fluid) W p = ( p 2 )dv. (6.43) We can now write down te corresponding canges in kinetic and gravitational potential energies (if tere is a cange in vertical position y between point 1 and 2) ΔK = 1 2 ρ v 2 2 ( 2 v 1 )dv ΔU grav = ρg( y 2 y 1 )dv. (6.44) Combining equations (6.39), (6.43), and (6.44) we get te so-called Bernoulli s Equation + 1 2 ρv 2 1 + ρgy 1 = p 2 + 1 2 ρv 2 2 + ρgy 2. (6.45) - 133 -
Since equation (6.45) applies for any two points along te flow, we can write p + 1 2 ρv2 + ρgy = constant. (6.46) It is interesting to note tat, altoug it is defined as an applied force per unit area, pressure can also be tougt of as energy per unit volume. Finally, we note tat wen te fluid is not moving v 1 = v 2 = 0 in equation (6.45) we recover = p 2 + ρg( y 2 y 1 ), (6.47) wic is te same as equation (6.11), previously derived for cases of ydrostatic equilibrium. 6.3 Exercises 4. (Prob. 12.40 in Young and Freedman.) Artery blockage. A medical tecnician is trying to determine wat percentage of a patient s artery is blocked by plaque. To do tis, se measures te blood pressure just before te region of blockage and finds tat it is 1.20 10 4 Pa, wile in te region of blockage it is 1.15 10 4 Pa. Furtermore, se knows tat te blood flowing troug te normal artery just before te point of blockage is traveling at 30.0 cm/s, and tat te specific gravity of tis patient s blood is 1.06. Wat percentage of cross-sectional area of te patient s artery is blocked by te plaque? Solution. We use Bernoulli s equation (i.e., equation (6.45)) wile assuming tat y 1 = y 2, were 1 and 2 correspond to regions before and at te blockage, respectively. We first determine te speed of te blood in te blockage wit v 2 = 2 ( p 2 ) 2 + v 1 ρ = 2 ( 1.20 104 1.15 10 4 )Pa + ( 0.300 m/s) 2 1,060 kg/m 3 = 1 m/s. (6.48) We ten use te continuity equation to find A 2 = v 1 v 2 = 0.30. (6.49) Te artery is terefore 70% blocked by te plaque. - 134 -
5. (Prob. 12.55 in Young and Freedman.) A dam as te sape of a rectangular solid. Te side facing te lake as an area A and a eigt. Te surface of te fres water lake beind te dam is at te top of te dam. (a) Sow tat te net orizontal force exerted by te water on te dam is ρga 2 tat is, te average gauge pressure across te face of te dam times te area. (b) Sow tat te torque exerted by te water about an axis along te bottom of te dam is ρga 2 6. Solution. (a) Te pressure at a given dept y ( y = 0 is at te bottom of te dam) is given by p = p 0 + ρg( y) (6.50) wit p 0 te atmosperic pressure at te top of te lake (and dam). Te force exerted by te water on a orizontal strip of te dam of widt dy at tat dept is df = p A dy ( ) = A p + ρg y 0 dy, (6.51) wit A te widt of te dam since it is rectangular. Te total force exerted by te water on te dam will ten be F = 0 df ( ) dy = A p 0 + ρg ρg ydy 0 0 = A ( p 0 + ρg) 1 2 ρg2 (6.52) = p 0 A + 1 2 ρga. However, te dam feels a force equal to p 0 A on its side opposing te lake from te atmospere. Te total force on te dam is terefore F total = 1 ρga. (6.53) 2 (b) Te torque about te axis at y = 0 acting on a orizontal strip of te wall at dept y is - 135 -
A dτ = df p 0 dy y = A (6.54) ρgy( y)dy. Te total torque on te dam is tus τ = 0 dτ ( ) = A ρg ydy y 2 dy 0 0 = A 2 3 3 ρg 3 (6.55) = 1 6 ρga2. 6. (Prob. 12.98 in Young and Freedman.) A sipon, as sown in Figure 7, is a convenient device for removing liquids from containers. To establis te flow, te tube must be initially filled wit fluid. Let te fluid ave a density ρ, and let te atmosperic pressure be p atm. Assume tat te cross-sectional area of te tube is te same at all points along it. (a) If te lower end of te sipon is at a distance below te surface of te liquid in te container, wat is te speed of te fluid as it flows out of te lower end of te sipon? (Assume tat te container as a very large diameter, and ignore any effect of viscosity.) (b) A curious feature of a sipon is tat te fluid initially flows upill. Wat is te greatest eigt H tat te ig point of te tube can ave if flow is still to occur? Solution. (a) Te pressure at te top of te liquid in te container is p 0 = p atm, te same as it is at te lower end of te tube. Applying Bernoulli s equation wit points 1 and 2 at te top of te liquid in te container and te lower end of te tube, respectively, we ave p 0 + 1 2 ρv 2 1 + ρg = p 0 + 1 2 ρv 2 2, (6.56) or But wit a very large container we can assume tat v 1 0 and v 2 2 = v 1 2 + 2g. (6.57) v 2 = 2g. (6.58) - 136 -
(b) Te pressure at te ig point of te tube can be related to tat at its low end wit p = p 0 ρg( H + ). (6.59) Te liquid will not flow if te absolute pressure is negative anywere in te tube, since a zero absolute pressure would imply tat a perfect vacuum be present at tat point. We ten write or Figure 7 A sipon, for removing liquids from a container. p 0 ρg( H + ) > 0, (6.60) H < p 0. (6.61) ρg We note tat equation (6.61) also implies a limitation on H +, wic for water and normal atmosperic pressure yields H + < p atm ρg < 10.3 m. (6.62) - 137 -