Fluid Dynamics MAE 643
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Review of undergraduate material August 277 2007 2 The Basics of Fluid Dynamics Graduate students come with varied backgrounds It is assumed here that all have had one undergraduate course in uid mechanics but the perspective in such courses and what students have retained from them is also quite diverse This document supplemented by Sections 41 and 42 of the Lec ture Notes is intended as a summary of methods and ideas assumed to be known with a few ne points thrown in for good measure Very little class time is devoted to this chapter to be covered in independent study mode Undegraduate textbooks are on reserve at the library Bernoulli7s equation if applicable dimensional analysis Ch4 and control volume analysis are mathematically simple and students in this eld must be able to rely on them no excuses 01 Bernoulli s equation Obtained from streamwise momentum assume no viscous forces incom pressible steady state Project the momentum equations on the streamline and integrate between two points A and B The result is 2 2 pQZApAWZB 193 1 This equation and its modi ed form including losses below is one of the most useful relations for undergraduate uid mechanics One of the reasons is that it is algebraic rather than di erential It also contains the hydrostatic equation as a particular case But its limitations should be noticed and a few details about its proper use will be emphasized Let us consider Fig 01 the example of a water tank with an ori ce B located a distance H below the free surface A Assume the bulk of the water in the tank to be motionless how good an approximation is this and take the free surface as reference level 2A 0 at reference pressure pA 0 The jet issues into ambient air and is therefore at p3 0 neglecting change in pressure between A and B is this accurate Then Bernoulli7s equation gives VB V 29H 2 known as Torricelli7s formula 01 BERNO ULLI S EQUATION 3 A Figure 1 Torrice11igts ori ce hat is ohvious with the ori ce is less ohvious with a pipe of length L Fig 2 for which the entrance is distinct from its exit Pressure at the pipe entrance is not known a priori First neg1ect friction so1ve easier cases rst then add complexity and note the differences rnost common undergraduate rnistake starts by app1ying Bernou11igts equation along the pipe ie from O to B the error is to ignore t at PC 74 99H 3 since conditions at point c are not hydrostatic In the absence of friction the correct so1ution starts with ArborB resu1ting in Torricelli s formula again Then ArborC or OrtorB gives 1 PC 99H EPVBZ 4 which requires the knowledge of V3 to give pg Note that the rnean ve1ocity is the same at all points in the pipe in order to consewe mass at steady state Another point of confusion can he re1ated to the restriction of Bernou11igts equations to strearnhnes When strearnhnes u1tirnate1y reach a reservoir Figure 2 A simple pipe ow setiup where static conditions prevail one can switch streamlines But if a manome ter tube is added to the discharge pipe Fig 3 Bernoulli s equation cannot apply from A B or C to point D because of the shear layer at the base of the tube More on this in Ch 5 as for so many things overriding assumptions without knowing eXactl what you are doing can lead to mist es Although the absence of viscous effects is also an important restriction in the derivation of Bernoulli s equation its interpretation as an energy equation allows for adihoc corrections for viscosity which would result in pressure losses The equation is V2 V2 pTApngpA p 23p31osses 5 1 2 L losses EpV B ReD ZKW 6 where the Darcy friction factor fR5 can be found from Moody s diagram Fig 4 or equivalent formulae once Re VD 1 is known The correspond ing losses are distributed along the pipe wall the pressure drop is balanced by the wall stress Poiseuille ow ln contrast the minor losses are lo calized phenomena for which a suitable coef cient Km is found in standard 9 WWW am BERN O ULLI S EQ UATI ON A D NOT a airrw iua B gCA H u q 7 q aYVCAuACw i L Figure 3 A pipe ow with a manometer m m mm mm Figure 4 Sketch of the Moody diagram tables entrance exit bends ttings etc Note that exit loss Km 1 is equivalent to the exit kinetic energy use one or the other not both depend ing on whether point B is the pipe exit or a point at rest in some discharge reservoir Thus Bernoulli7s equation with losses gives a modi ed Torricelli relation as 29H 1 1 E K where the hidden dependence of f on V3 requires iterative solution For turbulent ow the dependence is weak but if we assume laminar pipe ow and ignore entrance e ects then VB 7 i 64 764 V 8 Re VBD f See below for control volume analysis of Poiseuille ow A summary of relevant ideas could start with the mind map of Fig 5 build your own add ideas express them in your own way 02 Reynolds Transport Theorem Consider a generic property 73 scalar of vector with speci c value per unit mass p Pppdv 9 In very few instances the evolution of the property is governed by an equation of the type it RHS 10 where the right hand side rhs RHS f Th8 p dV represents the various mechanisms by which the value of 73 can be modi ed For example in a bank account there can be deposits and withdrawals and assorted fees and the rhs would re ect these operations and account for changes in daily balance Any perplexity at apparent leakage is due to our incomplete accounting of the various mechanisms at play This is the spirit of conservation laws in physics Within classical physics ie excluding the implications of E mcz mass is conserved through motion and chemical reactions and we have dtmdtpdV0 11 02 REYNOLDS TRANSPORT THEOREM 7 Eamole Wmerssme N Nvusclb ALou STREAM LINE 7 MoMENTuM epgtbba bo g 23 dirtd39Ccu M00643 dx grw mmcumk CASES V a drunks IFS 39 Y W pr 414 wogum 85 MQZZERId CEM Figure 5 Some Bernoullirelated ideas add your own with a trivial rhs Such conservation laws seem to be easier to formulate for systems including a xed collection of particles closed7 systems in ther modynamics terminology Such conservation laws are very special they require the matching of a property with a relevant rhs mechanisms in such a way that the evolution equation holds with hardly any restriction Very few properties are conserved in this sense Newton7s second law mgmdtg 12 is a remarkable vector relation Conservation of angular momentum is de rived from it and should not be considered an independent conservation law Then there is the rst law of thermodynamics th Q i w 13 for changes in total energy under the effect of heat and work exchanges engineering sign convention is adopted here and the entropy balance dtS 397 14 where the mechanisms of entropy production can be made explicit That is all There are no other known conservation laws that apply to all classical physical systems In the scope of their applicability they are very different from process or material relations or empirical formulae But because of their general applicability all common engineering systems obey them and you can only ignore them at your own risk These fundamental laws in their original form dealing with closed sys tems need to be generalized to apply to control volumes open systems This was done by O Reynolds Fig 6 The proof can be found in most undergraduate uids textbooks and involves the accounting of uxes associ ated with the mass exchange through various parts of the systems boundary The result is worth remembering involving a correction to the closed system formulation BtppdVjfppQMWh8dl 15 The surface integral must be over the entire surface of the system with the unit normal pointing outward Fig 7 We will use the notation dA a lt16 02 REYNOLDS TRANSPORT THEOREM 9 LAGRWGIAN mm 0 ELF n s quota CLOSED 9mm 4 momtum E x from 10 CONSERVATION uws F02 I 3 tdu quot EU 16mm TRANS Parr Uh or CDMTKoLVoLUHE 1P K g he39s d1 M INCH diver u N Whn e uAtCm at we lro wt Figure 6 Fundamental laws and Reynolds transport theorem 10 Figure 7 Control volume and flux through the boundary combining the element area and the direction of its normal Then the dot product with velocity captures the relation between the directions and mag nitudes of the ow and the surface in the vicinity of a given point it measures how much volume leaves the system That volume carries the properties of its contents with it A differential form of the conservation laws for flow systems is easily obtained We will assume that the control volume itself is xed ie the control surface is not moving a restriction that can be removed at the cost of additional terms Then the time derivative can move in and out of the volume integral Making use of the divergence theorem Gauss we obtain axm p ms v ppm dv 0 W Since this holds for any xed volume the integrand vanishes 6tppVprprh8 18 021 Application a simple 2D ow Consider the right angled corner dividing a uniform steady ow as shown Fig 8 Assume uniform flow at the exit points The ow is exposed to ambient pressure except at the points of contact with the wedge The control surface should include the entrance and exit sections con nected by arbitrary surfaces across Which there is no ow Applying Bernoulli s 02 REYNOLDS TRANSPORT THEOREM 11 V2 A2 V3 Figure 8 Control Volume and ow around a wedge e uations between 1 and 2 and 1 and 3 one nds that V V2 V3 so that single V is adequate notation in this case The rnass ux integral gets velocity and the total rnass is constant for the control Volume atmfp1w0 pg dA0 19 u u u Because the ux term is a dot product it is independent of any choice of es as yet unspeci edl furthermore the dot product is always negative at an entrance p int and positive at an exit point nally only the velocity component normal to the surface contributes to the dot product Fig 9 Thus We have PA1V1 A2V2 Mm o 20 and therefore A1 A2 As 21 e turn balance requires a little more care ignore frictlox What role does it play7 A free body diagrarn re ects the forces applied by the Figure 9 FED for wedge aw the force is applied to the uid its reaction to the wedge 39 39 L e and a 39 39 axes is also required Fig 9 For steady ow the RTT reduces to Af MMpll siAE 22 which can be broken into 17 and yrcumpunents AfJszaawaz 23gt P y 7 i y MJrMJrMwa dA F 24 with the signs as for mass balance ie independent of axes whereas the vector for momentum is spht into components for which the value and sign reminder Therefore we have th7A1V1pV2 cos etA2V2 st sin MAJa Fe 25 02 REYNOLDS TRANSPORT THEOREM 13 Figure 10 Flat plate boundary layer con guration and p0iA1V1pV25inc A2V2 9V5 wsaAaVa Fy 26 which can be reduced easily to give F and Fy 022 Application at plate boundary layer an also The at plate boundary layer wrurpressure gradient steady ow c be studied at this level the resulting information is important and should be familiar to all stu Th goration is shown on Fig 10 Axes xy e e co at the leading edge of the plate are omitte 39 st mass ance manipulations fullcrw the same idea as above and are presented more concisely aimfptm A MampA p O 7U A Viz Ud Z7 p newA3 A e Indeed there is no mass ux through the impermeable wall 1 Note the negative sign at 2 positive at 3 and 4 consistently with the diagram Since 14 U at 4 is less than Ultgt0 and the areas are equal the ux at 3 must be positive Vd UwiUd gt0 28 AS z A gt y lt gt This result should gradually be seen as common sense as the boundary layer develops less and less mass progresses in the immediate vicinity of the wall The mass ux de cit in the wall region is balanced by a net mass ux away from the wall ie by a de ection of the freestream away from the at plate Fig 10 This is a good opportunity to review or learn the concept of displacement thickness Similarly for X momentum balance the only applied force is the wall shear stress Tm opposing the motion We have prQMAlmdfv A1A2Aa44pUQ39M 0 pkaoAg Udez Uzdy 29 A3 A4 Note the di erent treatment of ux term dot product and component of momentum As a result 7 ax iUfoAg UOO U00 7 Udy Uzdy Um 7 Udy A1 A4 A4 A4 30 The friction force is proportional to the de cit in momentum ux at the exit section See the end of chapter problem about de ning diplacement and momentum thickness and their interpretation The same considerations apply to wakes jets and similar con gurations see Ch 8 023 Poiseuille ow in Circular pipes Fully developed ow in a circular pipe is a classic problem approachable by control volume analysis It also lays the groundwork for evaluating losses in Bernoulli7s equation an obvious link to the previous section Fully developed means that the velocity distribution across the pipe has become independent of the location along the pipe in standard notations u depends on the radial coordinate distance from centerline 7 not on axial 02 REYNOLDS TRANSPORT THEOREM 15 Figuie 11 Contiol volume in fully developed pipe ow distance 2 fully developed oi on ang ulem position 9 by symmetiy Note that cylindxical cooidinates do simplify the ploblam Steady ow is assumed Then Considel a cylindxical element of ladius 7 and length dz centeied in the pipe Fig One Consequence of tully developed ow is that theie can he no iadial component of velocity lndeed assume an outwaxd ow at any point by symmetiy it must apply along the Cilcumfelence which would inciease mass ow iate ie velocity outside and deciease it inside if this weie tiue u would depend on 2 which means the ow is evolving see entiance ow next chaptei Then since the mass ow iates Ammw d7 cancel out thiough the hases of the cylindxical cv and is zeio thiough the side sulface mass is Conserved Sim elly momentum halance gi dz 7 27 7 Anomalypr deTWWMMQW d7 7 7w dadz 27w deny 31 The viscous stiess is given by Newton s folmula c my 32gt The momentum uxes cancel out fully developed again and we ale lett with it d 2mm 33 w take a deiivative with iespect to 7 which amounts to taking two cvgts of slightly di elant 75 taking the di elence and keeping the leading teim EN DE u owp RZ Figure 12 Poiseuille pro le in a circular pipe an annular cv of thickness dr We get 1 idzp ip udTTdTu 34 Which expresses the balance of pressure drop and viscous resistance For fully developped ow dzp 1 must be a constant and negative if the motion is in the direction of increasing Then integration is straight forward and gives 2 p 71urAlnrB 35 The boundary conditions are that u should be nite at the centerline and vanishes no slip at r R Then the solution takes the form of the Poiseuille solution Fig 12 pRZ 72 um 4 lt17 E lt36 It is easy to verify that the average velocity is half the centerline velocity this is di erent in the case of the plane channel make sure you understand Why We can also calculate the wall stress as ipR Tm 2 37 02 REYNOLDS TRANSPORT THEOREM 17 39FTuLLv DEVELoPeh I no wwscrwe 7 Teams 5 3 L LOOKS LIKE ours FLOW llc Sec 0 s chnon 1 c LRCULA39S 1in T2M61HT Poisson39s Mo 39FIPvF memz Re 39I otseucuf u39 mama c F 5 NJ I PROFOLE 5 UV Figure 13 About fully developed pipe ows 18 Over a length of pipe L7 the balance of pressure and friction gives 6p m 72 38 L R so the Darcy friction factor takes the familiar form 2Dp 64 7 39 f Uavp RigD where the Reynolds number is based on the average velocity Fig 13 03 And there is more In addition to the topics covered in this packet7 it is expected that all student will be pro cient with the basics of dimensional analysis see Sections 41 and 42 of the Lecture Notes7 and related problems Undergraduate level control volume analysis is also included in Ch 8 Boundary Layers 04 Food for thought A great way to learn is to study variants making appropriate changes to familiar material and observing these changes rippling all the way to the answer One useful type of variant is the spiral approach of revisiting a given problem while adding complexity and relaxing assumptions However7 this provides a learning opportunity only if the context of the familiar solution is present Create your own variants to the examples presented in the notes Problems 1 As shown on Fig 147 a plane inclined at an angle or to the horizontal supports a owing layer thickness d of uid of density p and viscosity u Find the fully developed velocity pro le What is the average ve locity What is the mass ux per unit width of the plane Adapted from Tritton7 p469 N In a Couette ow Adapted from Tritton7 p4697 two parallel walls form a channel of width 10 One of the walls is at rest7 and the other 03 Hgt 53gt 1 00 9 FOOD FOR THOUGHT 19 moves in its own plane with speed U A uid of density p and viscosity 1 lls the channel A pressure gradient G is applied parallel to the walls Poiseuille Under which circumstances will the uid velocity exceed U How does the wall stresses depend on U and on the mean uid velocity u A water jet pump works as a coaxial jet of diameter 1 speed V inside a pipe of diameter D and upstream speed 1 Using control volume analysis determine the pressure rise The exit pipe from the water tank shown on Fig 14 is equipped with a manometer tube Select one of the two designs explain the reason lgnoring friction determine the manometer reading h for this design In the spraying nozzle shown on Fig 14 the draw through the vertical tube does not signi cantly affect the mass ow rate of the same liquid through the nozzle The nozzle exit and the reservoir are at ambi ent pressure Neglect all losses Determine how the upward velocity through the vertical tube depends on the system parameters Repeat the previous problem by adding friction and minor losses in the vertical tube The mixing box Fig 14 is characterized by the intake and discharge areas and velocities as shown Under steady ow conditions determine the force required to hold the box in place note V4 is not given assume it to be normal to the box boundary The exit pipe from the water tank shown on Fig 14 is equipped with a manometer tube lgnoring friction determine the height of tube required to avoid spills A plane liquid jet of thickness h speed V and density p impinges on a at plate inclined at an angle or as shown on Fig 14 lgnore friction on the plate The jet splits into two of thicknesses hl and hg Determine hlh and the force on the plate as a function of the parameters A liquid of density p is at rest in an open tank Fig 14 Some of the water ows through a siphon of diameter D with its highest point a height h above the free surface and its discharge point a distance H 20 Figure 14 Figures relative to the problems as labeled H H FOOD FOR THOUGHT 21 below the surface A manometer tube of diameter d is connected to the top of the siphon7 as shown Determine the manometer level E relative to the free surface Neglect friction A jet of liquid density p and area A strikes a block and splits into two jets7 as on Fig 14 Neglect friction The upper jet of area a exits at an angle 5 while the lower jet is turned 90 7 downward Neglecting uid weight7 derive the formula for the components of force applied ot the block A curved nozzle assembly that discharges to the atmosphere is shown on Fig 14 The nozzle weight is W Determine the relation between the force applied by the nozzle on the coupling to the inlet pipe and the problem parameters Consider the siphon pictured on Fig 14 Determine the relation be tween the discharge velocity and the problem parameters