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# Fluid Mechanics AME 30331

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This 0 page Class Notes was uploaded by Yesenia Hansen on Sunday November 1, 2015. The Class Notes belongs to AME 30331 at University of Notre Dame taught by Philippe Sucosky in Fall. Since its upload, it has received 33 views. For similar materials see /class/232721/ame-30331-university-of-notre-dame in Aerospace Engineering at University of Notre Dame.

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Date Created: 11/01/15

AME 30331 Fall 09 NAVIERSTOKES AND EULER EQUATIONS DISCUSSION ON THE EFFECTS OF COMPRESSIBILITY AND VISCOSITY We have derived 2 complete sets of equations for 1 Newtonian viscous incompressible model 2 lnviscid incompressible model In both cases we ignore compressibility while in 2 we also ignore viscosity Under what conditions is this reasonable Consider compressibility In general p pPT lncompressible means i a u 0 p Dt 6x 6y 62 This requires small changes in P and T relative to reference conditions Slowly varying flow Small temperature differences between elements of the flow Small length scale relative to length scale required for a significant pressure change due to gravity Low Mach number see proof below Proof The Mach number is defined as MUc where U is a representative speed and c is the speed of sound The requirement of a small change in pressure can be written mathematically as APP ltlt1 When this condition is satisfied AP pUZ Additionally if we consider the case of an ideal gas P pRT pUZ U2 yUZ pRT RT zyRT Therefore where 7 is the ratio of specific heats 2 AP This expression can be rewritten y Z yMZ 6 Therefore APP ltlt1 can also be expressed in terms of the Mach number as gM2 ltlt1 Generally Mlt 03 is used as a working criterion to neglect compressibility AME 30331 Fall 09 Consider viscosity The classic example is highspeed flow past a solid body outerflow inner or boundary layer flow edge of the boundary layer We want to compare the order of magnitude of the shear stress p Le a typical viscous y term in the outer and inner flow regions The inner flow region is de ned by a thickness 6 In the boundary layer the characteristic velocity is u and the characteristic dimension in the y direction is 6 6 ltlt1 Therefore in terms of scales g 1 1 dy 6 In the outer region the characteristic velocity is u and the characteristic dimension in the y direction is the chord of the solid body maximum thickness Therefore in terms of scales y H y u 2 dy chord Using 1 and 2 du d7 3 y boundarylayer N 6 Chord gtgt1 u 6 u outer ow Chord Therefore a typical viscous term in the boundary layer is much larger than one in the outer ow Thus viscous effects are neglected in the outer flow AME 30331 Fall 09 MEASURES OF FLUID MASS AND WEIGHT DENSITY Density is defined as mass per unit volume p6 units pkgm3 For a differential volume element of mass 5m and volume 5 1 density can be expressed as p 6m5 l For an ideal gas P pRT where P is the absolute pressure R is the gas constant and T is the thermodynamic temperature SPECIFIC VOLUME The specific volume is the reciprocal of density and is defined as volume per unit mass units u m3kg u i p SPECIFIC WEIGHT The weight of a unit volume of substance is called specific weight and is expressed as 7pg units y Nm3 where g is the gravitational acceleration SPECIFIC GRAVITY The specific gravity is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature usually water at 4 C for which pHZO 1000 kgm3 SGL units dimensionless szo4 c AME 30331 Fall 09 SUPERPOSITION OF BASIC PLANE POTENTIAL FLOWS Source in uniform stream o nmhined velocitv potential and Uniform flow Source Streamfunction 11 Uy 11 16 27239 Velocity potential Q Ux Q 1lnr 27239 Therefore the combination of a uniform flow and a doublet is expressed in cylindrical coordinates as Streamfunction Velocity potential 0 Velocity field AME 30331 Fall 09 The velocity magnitude is V2 o Stagnation goint At b along the xaxis the velocity of the flow due to the source cancels that due to the uniform flow atrb and 6717 11 2 vrrb67r0ltgt o Equationofthe passinqthrouqhthe quot point At the stagnation point I Therefore the streamline equation through the stagnation point is AME 30331 Fall 09 7rbU gt stagnation point The combination of a uniform flow and a source can be used to describe the flow around a streamlined body placed in a uniform stream Width ofthe streamlined body The width ofthe body can be obtained as t9 gt 0 Le at infinity along the xaxis Pressure field The pressure field can be determined using the Bernoulli equation between any two points in the flow AME 30331 Fall 09 Doublet A doublet is the combination of a source and sink 02 source a 0 Combined velociy potential and streamfunction Sink Source Streamfunction w w 6 27239 27239 Velocity potential Q 11n r Q 11n r 27239 2 coordinates as Therefore the combination of a uniform flow and a doublet is expressed in cylindrical Streamfunction Velocity potential From the combined streamfunction Using the identity tan61 62 tan 61 tan 62 1 tan 61 tan 62 the expression can be rewritten AME 30331 Fall 09 where tan 191 tant92 2 Therefore 1ant91 192 The streamfunction is obtained by taking the inverse tangent The expression above is the streamfunction for the combination of a source and a sink ma A doublet Is obtained as m gt00 a gt0 and gtK 7r Therefore the streamfunction and velocity potential for a doublet can be expressed as where K is the strength of the doublet AME 30331 Fall 09 Streamlines for a sourcesink pair Streamlines for a doublet AME 30331 Fall 09 Flow around a circular cylinder The flow around a circular cylinder can be represented by combining a doublet with a uniform flow 0 nmhined velocity potential and Uniform flow Doublet K 39 6 Streamfunction 11 Uy i r K 6 Velocity potential Q Ux Q amp r Therefore the combination of a uniform flow and a coordinates as doublet is expressed in cylindrical Streamfunction Velocity potential 0 Boundary condition In order to represent a cylinder of radius a the perimeter of the cylinder must be a streamline in the flow Therefore the streamfunction must be constant along the perimeter of the cylinder AME 30331 Fall 09 n1 1 o and velocity potential of flow around a circular cylinder of radius a Streamfunction Velocity potential o Velocityfield The velocity field can be derived using the definition of the streamfunction in cylindrical coordinates 1 6 6 V v and V6 v r 66 6r Points where the maximum velocity is attained on the surface of the cylinder atra 11 V 9 3 v6 max 2 AME 30331 Fall 09 AME 30331 Fall 09 Circulatorv flow around a cylinder in a uniform stream Circulatory flow around a cylinder in a uniform stream can be obtained by combining a vortex a uniform flow and a doublet Le a vortex and a flow around a cylinder 0 nmhined velocity potential and Vortex Flow around cylinder f 2 Streamfunction 112 1nr zUr1 a 2sin6 r F 2 Velocity potential Q E6 Q Ur 1 2 cos 6 r where F is the circulation Therefore the streamfunction and velocity potential for the circulatory flow around a cylinder in a uniform stream can be written Streamfunction Velocity potential 0 Velocity field o Stagnation points on the cylinder Stagnation points are points where the velocity is zero AME 30331 Fall 09 9 H e 5 2 v6ra0cgt Therefore the number of stagnation points on the cylinder depends on the value of F U r0 gt A F lt 47tUa ltl F gt I 47rUa V V AME 30331 Fall 09 VISCOSITY DEPENDENCE ON PRESSURE AND TEMPERATURE In general u is a very weak function of pressure but a reasonable function of temperature d In liquids ult0 dT d lngases ugt0 dT Temperature C 10 4 2 1xluquot 6 v4 72 lxinl a N a 4 239 a 1 5 g ll 5 quot 392 g Mercury 3 95 Elxllll V lt H 4 g 72 1x10 6 74 72 1x10 in Temperature F Explanations o In liquids molecules are close together and are bound by strong cohesive forces Heat produces more energetic motion which decreases cohesive forces and in turn viscosity 0 ln gases molecules are further apart Heat produces more motion and collisions which make molecules interact with one another and increase cohesive forces AME 30331 Fall 09 VISCOSITY AND FLUID DEFORMATION Consider a fluid layer between two large parallel plates The top plate is moving at a velocity V under the action of a force E while the bottom plate is xed V F The force E used to translate the top plate is transmitted to the fluid layer through a shear stress 1 defined as a force per unit area Shear stress acting on the fluid layer At the same time the translation of the top plate produces a velocity pro le in the fluid Question are these two quantities shear stress and velocity profile related Determination of the velociy pro le Since the distance between the two plates 1 is small it can be assumed that the variations of the velocity in the ydirection are linear Boundary conditions AME 30331 Fall 09 Velocity pro le An uar dis lacement and rate ofdeformation During a small time increment 6t the line AB rotates by an angle d to become the line AB The angle of rotation d is the angular displacement and is also referred to as shear strain Shear strain The rate of deformation also referred to as shearing rate and noted 7 can be obtained by taking the derivative of the angular displacement with respect to time Shearing rate Experiments reveal that for most common fluids the rate of deformation or shearing rate is directly proportional to the shear stress Relation between shearing rate and shear stress AME 30331 Fall 09 AME 30331 Fall 09 EXAMPLE PROBLEM PIPE FLOW AND ENERGY LOSS A pump delivers water y 980gtlt103 Nm3 v 112gtlt10 6 m2s at a gage pressure P1 550 kPa to a pair of inclined pipes made of commercial steel roughness 8 0045 mm The pipes make an angle 6 30 with respect to the horizontal The characteristics ofthe pipes are Ll L2 3m D1 25 cm D2 50cm Determine the velocity of water being ejected to the atmosphere at the upper end ignoring minor losses Hint since the velocity is to be determined this problem requires you to use iterations Use a first guess of V1 9ms for the speed in the smallerdiameter pipe and perform two additional iterations Does the process appear to be converging Step 1 Derivinq an expression for the velocity in the pipe Start with the energy equation between the entrance of the first pipe and the exit of the second pipe 2 2 i L pg 2g pg 2g To write this equation we have assumed turbulent flow in each pipe ie oz1 a2 1 and nearly uniform velocity profile in each pipe ie T71 V1 and 17212 The head loss term must be decomposed into a head loss in pipe 1 and a head loss in pipe 2 Therefore the energy equation can be rewritten 2 2 gmuy z 151ng pg 2g D1 2g D2 2g Since P2 P at39m then P1 P2 ie gage pressure delivered by the pump PW Also given the geometry of the pipe Z1 Z2 L1 L2 sinH Therefore the previous equation can be simplified as Pg 1 L V2 L V2 amp Vf V L L2s1n6f1 f2 2 2 pg 2g D1 2g D2 2g Instead of expressing this relation in terms of two different velocities V1 and V2 we can use conservation of mass to keep only one unknown velocity Continuity Q1 Q2 mm A12 2 312214 Q D2 Substituting back into the energy equation K1 I LlL2Sin6 f QJ4 pg 2g which can be rewritten as Pg W 2g L2sin6 Pg V1 2 D 4 L D 4 1 71 f1 f2 72 71 D2 D1 D2 D2 Using the numerical values provided 12 1042 VI Equation 1 09375 120f 375f2 where f1 and f2 the friction factors in pipe 1 and 2 respectively are unknown Step2fquot quot39 quot quot quess Using the initial guess provided in the problem we would like to find the values of the unknown friction factors and update the velocity value We start with VI 9 ms 2 D1 UsIng contInuIty V2 V1 3 225 ms 2 With the knowledge of the velocity in each pipe the Reynolds number can be calculated leDl VIDI Re1 200893 u v Re2 100446 u We obtain a first estimate for the friction factor by using the approximate Colebrook formula 1325 g 574 2 1n T 37D Re 7 fl 239x10 2 Therefore f2 220 gtlt10392 Using the Colebrook formula we obtain a refined value for the friction factor 20sD J J7 37 Re f1 237x10 2 f2 218gtlt10 2 Therefore Now we can substitute these frictions factors into the expression that was derived for V1 Equation 1 12 VI i 2288m 09375 120f1 375f2 2 Using continuity V2 Vl 572 m 2 This is the end of the guess step Step 3 First iteration We start with the new values of V1 and V2 and we calculate the corresponding Reynolds numbers Re1 VIDI 510714 v V2D Re2 2 255357 V Using the approximate Colebrook formula f1 233 gtlt10 2 fl 205 gtlt10 2 Using the Colebrook formula f1 232x10392 f2 203 x10392 Once again substituting these values into the expression for V1 equation 1 yields 12 VI i 2328m 09375 120f1 375f2 Using continuity V2 Vl 2 This is the end of the 1St iteration Step 4 Second iteration Based on the new velocity values Re 519643 v 1 VD Re2 2 2 v 259821 Re1 519643 gtgt 2300 Note that Therefore the Initial assumption that the flow is turbulent in Re2 259821 gtgt 2300 each pipe is satisfied Using those updated Reynolds numbers in the approximate Colebrook formula f1 233 gtlt10 2 fl 205 gtlt10 2 Using the Colebrook formula f1 232x10392 f2 203 x10392 Substitute back into the expression for V1 equation 1 12 VI i 2328m 09375 120f1 375f2 2 Using continuity V2 Vl 582 m 2 This is the end of the 2quotd iteration Step 5 Study of the convergence Let s look at the variations of the estimated values V1 and V2 along the iterative process V V In terms of percent change V1 change V2 change Therefore by the end of the second iteration the iterative process has already converged

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