Viscous Flow in Pipes.pdf

Chapter 8Viscous Flow in Pipes

8.1 General Characteristics ofPipe Flow• We will consider round pipe flow, which is completely filledwith the fluid.

8.1.1 Laminar or Turbulent Flow• The flow of a fluid in a pipe may be laminar flow or itmay be turbulent flow.• Osborne Reynolds experiment --typical dye streaksV8.2 Laminar/turbulent pipe flow

Instantaneous velocity fluctuation• Time dependence of fluid velocity at a point.V8.3 Intermittent turbulent burst in pipe flowTurbulentRe>4000Transitional2100

8.1.2 Entrance Region and FullyDeveloped Flow• Consider a pipe flow with uniform inlet velocityA boundary layer in which viscous effect areimportant is produced along the pipe wall such thatthe initial velocity profile changes with distancealong the pipe.

Entrance length• Typical entrance lengths are given byle= 0.06 ReDfor laminar flowle= 4.4( Re) 16Dfor turbulent flowe.g. Re = 1.0 le= 0.6DRe = 2000 l = 120DFor practical engineering problems4 510 < Re < 10 so that 20D< l < 30De• Calculation of the velocity profile and pressure distributionwithin the entrance region is quite complex.e

8.1.3 Pressure and Shear Stress• Fully developed steady flow in a constant diameter pipemay be driven by gravity and/or pressure force.• For horizontal pipe flow, gravity has no effect except for ahydrostatic pressure variation across the pipe, that isusually negligible.• Viscous effects provide the restraining force that exactlybalances the pressure force, allowing the fluid to flowthrough the pipe with no acceleration.• In non-fully developed flow, the fluid accelerates ordecelerates as it flows, therefore there is a balancebetween pressure, viscous and inertia (acceleration) flow.

Integral Analysis for Increased Pressure Drop in theEntrance Region(From S.R. Turns, Thermal-Fluid Sciences, Cambridge Univ. Press, 2006)The increased pressure drop in the entrance region result from tworeasons: 1. flow acceleration, 2. increased wall shear stress.1. Flow accelerationIntegral momentum balance for the control volume below leads to4P0A x-sec −P2A x-sec − ∫ τAwdAw = mv &avg−mv&avgw 3Increased momentum flowor flow acceleration

2. Increased Wall Shear Stress:⎡ τ dA ⎤ > ⎡ τ dA⎢∫⎥ ⎢∫⎤⎥⎣ Aw w w ww ⎦developing⎣ Aw⎦regionfullydevelopedIncreased wall shear stress

Pressure distribution• Pressure distribution drag a horizontal pipe∂p∂xentranceregion∂p> =∂xfully developed regionconstant

Nature of flow• The nature of the pipe flow is strongly dependent onwhether the flow is laminar or turbulent, which is a directconsequence of the differences in the nature of the shearstress.• Flow rate as a function of pressure drop in a givenhorizontal pipe.VQΔp⋅D=32⋅μ⋅l4π D Δp= −128μlHagen-Poiseuille flowLaminar flow Re

8.2 Fully Developed Laminar Flow• Consider a fully developed flow in long straight, constantdiameter sections of a pipe.• There are numerous ways to derive important resultspertaining to fully developed laminar flow.

8.2.1 From F=ma Applied Directly to aFluid Element• Consider a element within a pipe• Free body diagram of a cylinder of fluid• The force balance becomes, (a=0, the fluid is not accelerating)2 2Δp2τp πr − p −Δp πr − τ 2πrl= 0 → =( ) ( ) ( )1 1lr

• Since neither Δpnor are functions of the radialcoordinate, r , it follows that 2τr must also beindependent of r .i.e.,τ = Crlwhere C is a constant2τwC = whereτwis the wall shear stressD2τwr→ τ =D• Thus the pressure drop and wall shear stress arerelated by,4 τ wΔ p = l Dl >> 1 D• A small shear stress can produce a large pressuredifference if the pipe is relatively long

• The above analysis is valid for both laminar andturbulent flow.• To proceed further we must determine therelationship between shear stress and velocity.• For laminar Newtonian flow,uτ = μ ∂∂ydudrΔp 2 τ du ⎛ Δp ⎞= , =− ⎜ ⎟rl r dr ⎝2μl⎠Δp∫du=− rdr2μl∫For pipe flow τ = − μ ( τ > 0)Then⎛ Δp⎞⎜ ⎟⎝2μl⎠2u =− r + C1 C1, where is a constant2DΔpDB.C.: u = 0., r = , so that C1=2 16μl

• Hence the velocity profile can be written as,( )u r( )u r22 2⎡ ⎤ ⎡ ⎤ΔpD ⎛2r ⎞ ⎛2r⎞= ⎢1− ⎜ ⎟ ⎥ = Vc⎢1−⎜ ⎟ ⎥16μl⎢⎣ ⎝ D ⎠ ⎥⎦ ⎢⎣ ⎝ D ⎠ ⎥⎦2ΔpDwhere Vc= is the centerline velocity16μlIn terms of wall shear stress,2τ D ⎡ ⎛1 r ⎞ ⎤= w ⎢ − ⎜ ⎟ ⎥ where R=D is the pipe radius.4μ⎢⎣⎝ R ⎠ ⎥⎦2• Volume flow rate2r=RR⎡⎛ r ⎞ ⎤Q = ∫udA = ∫ u ( r)2πrdr = 2πV 1r 0c ⎢ − ⎥rdr=∫0⎜ ⎟⎢⎣⎝ R ⎠ ⎥⎦π=2RVc2

• Average velocityVQ2 2π RVcVcΔp⋅D= = =22πR2 32⋅μ⋅l4π D Δp= − Hagen-Poiseuille flow128μlLaminar flow Re

• For nonhorizontal pipeThus instead ofΔp−γ lsinθ 2τ=l randVΔ p =l2πr( Δ −γlsinθ) π( Δ −γlsinθ)p D p D= , Q=32μl128μlIf V = 0, Δ p= γlsinθ= γΔz2 4

8.2.2 From the Navier-Stokes Equation• Basic equationsur∇⋅ V = 0,ur∂Vur ur ∇pur ur2+ V ⋅∇ V =− + g+ v∇V∂tρ• For steady fully developed flow in a pipeur∇⋅ V = 0,r ur2∇ p+ ρgk = μ∇V• In terms of polar coordinate∂p1 ∂ ⎛ ∂u⎞+ ρgsinθ = μ ⎜r⎟∂x r ∂r⎝∂r⎠( ) = ( )f x g r= constant,

8.2.3 From Dimensional Analysis• Consider pressure drop in the horizontal pipe( , l, , μ )Δ p=F V Dk− r = 5− 3=2D⋅Δp⎛ l ⎞= φ ⎜ ⎟μ ⋅V⎝D⎠dimensionless groups• If we assume that the pressure drop is directly proportionalto the pipe length.2D ⋅Δp ClΔp CμV Δp D= , = ⇒ V =2μ ⋅V D l D l CμorQ( π 4 )4 4C Δp⋅D π D p= AV = =μl⋅Δ128μlThe value of C must be determined by theory or experiment.C = 32 for round pipe.

• It is advantageous to describe in terms ifdimensionless quantities.32μlVΔ p =2DΔpμlV D μ l ⎛ l ⎞= = = ⎜ ⎟V V VD D ⎝ D⎠232 64641 2 1 22ρ2ρ ρ Re2i.e.,l ρVΔ p= f ⋅ ⋅D 2( ) ( )where f =Δp D l ρV 2 2 friction factor or Darcy friction factorFor laminar fully developed pipe flow• Alternatively, sincef =64Re4⋅lτwΔ p =DΔp⋅( D l)4⋅lτD 1 8τwf = = =ρV D ρV ρV 2w1 2 1 22l2(8.19)(8.20)

8.2.4 Energy Consideration• Consider the energy equation for incompressible, steady flowbetween flow locations.p V p Vγ 2gγ 2g2 21 1 2 2+ α1 + z1 = + α2 + z2 + h2• For fully developed flowα = α , V = V1 2 1 2p1 ⎛ p2⎞+ z1− ⎜ + z2⎟=hLγ ⎝ γ ⎠Δ p 2τ2τl4lτ= ∴whL= , hL=l r γ r γ D(Recall p = p +Δp, z − z =l sinθ)1 2 2 1EX. 8.3 Laminar pipe flow properties⎛Δp−γ l sin θ 2τ⎞⎜=l r⎟⎜⎟Δpl sin θ 2τ− =⎜ γrγ⎟⎝ l ⎠Which indicates that it is the shear stress at the wall ( which isdirectly related to the viscosity and the shear stress throughoutthe fluid ) that is responsible for the head loss.

8.3 Fully developed turbulent flow• Turbulent pipe flow is actually more likely to occur thanlaminar flow in practical situation.• Consider flow in a pipe

• Velocity variation at a point.• Turbulent processes and heat and mass transfer processes areconsiderably enhanced in turbulent flow compared to laminarflow.Mix a cup of coffee (turbulent flow)Mix two colors of a viscous paint (laminar flow)V8.4 Stirring color into paintV8.5 Laminar and turbulent mixingV8.7 Turbulence in a bowl

8.3.2 Turbulent shear stress• Turbulent flow is random and chaotic and can only becharacterized using stately terms.u = u x, y, zt ,• Mean part( )1=T∫t0+ Ttime average u ( , , , )t0u x y z t dtwhere T is longer than the period of the longestfluctuations, but is shorter than any unsteadiness of theaverage velocity.

Turbulence intensity• Fluctuating partu = u + u′ or u′= u−u1 1′ = ( − ) = ⎡TT ⎣−1= ( Tu − Tu ) = 0T′1 t + T=T∫′ > 0t + T t + T t + T∫ ∫ ∫0 0 0u u u dt udt u dtt0 ⎢ t0 t02 0 2u u dtt0• Turbulence intensity⎤⎥⎦Iu= =u⎡1 t0+ T22u′⎤dt′ ⎢T∫t⎣⎥0⎦u12Well designed wind tunnel: I ≈ 0.01 (may be down to 0.0002)

Time-averaged Navier-Stokes EquationsLet u = u + u′ , υ = υ + υ′ , w= w+w′and then take time average2 2 2⎛∂u ∂u ∂u′ ∂u ∂u′ ∂u ∂u′ ⎞ ∂p ⎛∂ u ∂ u ∂ u ⎞x-dir:ρ⎜ + u + u′ + υ + υ′ + w + w′⎟=− + ρgx+ μ⎜ + +2 2 2 ⎟⎝ ∂t ∂x ∂x ∂y ∂y ∂z ∂z ⎠ ∂x ⎝ ∂x ∂y ∂z⎠2∂u′ ∂u′ ∂u′ ∂u′ ∂u′ υ′ ∂υ′ ∂u′ ∂uw ′ ′ ∂w′But u′ = − u′ ; υ′ = − u′ ; w′ = − u′∂x ∂x ∂x ∂y ∂y ∂y ∂z ∂z ∂z∂u′ ∂υ′ ∂w′and + + = 0 (from continuity equation)∂x ∂y ∂z⎛2 2 2 2∂u ∂u ∂u′ ∂u ∂u′ υ′ ∂u ∂uw ′ ′ ⎞ ∂p ⎛∂ u ∂ u ∂ u ⎞→ ρ + u + + υ + + w + =− + ρgx+ μ + +⎜2 2 2t x x y y z z ⎟⎜ ⎟⎝∂ ∂ ∂ ∂ ∂ ∂ ∂⎠∂x ⎝ ∂x ∂y ∂z⎠Terms with fluctuations moved and merged to the r.h.s.Finally,⎛∂u ∂u ∂u ∂u ⎞ ∂p ∂ ⎛ ∂u 2 ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u⎞ρ⎜ + u + υ + w ⎟=− + ρgx+ ⎜μ − ρu′ ⎟+ ⎜μ − ρu′ υ′ ⎟+ ⎜μ −ρuw′ ′ ⎟⎝ ∂t ∂x ∂y ∂z ⎠ ∂x ∂x⎝ ∂x ⎠ ∂y⎝ ∂y ⎠ ∂z⎝ ∂z⎠⎛∂υ ∂υ ∂υ ∂υ ⎞ ∂p∂ ⎛ ∂υ ⎞ ∂ ⎛ ∂υ 2 ⎞ ∂ ⎛ ∂υ⎞ρ⎜ + u + υ + w ⎟=− + ρgy+ ⎜μ − ρu′ υ′ ⎟+ ⎜μ − ρυ′ ⎟+ ⎜μ −ρυ′ w′⎟⎝ ∂t ∂x ∂y ∂z ⎠ ∂y ∂x⎝ ∂x ⎠ ∂y⎝ ∂y ⎠ ∂z⎝∂z⎠⎛∂w ∂w ∂w ∂w⎞ ∂p ∂ ⎛ ∂w ⎞ ∂ ⎛ ∂w ⎞ ∂ ⎛ ∂w2 ⎞ρ⎜ + u + υ + w ⎟=− + ρgz+ ⎜μ − ρuw ′ ′ ⎟+ ⎜μ − ρυ′ w′ ⎟+ ⎜μ −ρw′⎟⎝ ∂t ∂x ∂y ∂z ⎠ ∂z ∂x⎝ ∂x ⎠ ∂y⎝ ∂y ⎠ ∂z⎝ ∂z⎠

Origin of shear stress• The shear stress for turbulent flow can not be evaluatedas for laminar flow, i.e.,τt≠uμ ∂∂y• Laminar flow : random Turbulent flow: randommotion of molecules. motion of 3D eddies.The eddystructuregreatlyenhancesand promotesmixing withinthe fluid.

Total stress• Total stressduτ = μ − ρu′ υ′= τlam+ τdyif the flow is laminar, u′ = υ′ = 0, u′ υ′= 0−ρu′ υ′ ( −ρυ′ w′, K)--Reynolds stresses• Structure of turbulent flow in a pipeturb

• Typically the value of τ turb is 100 to 1000 timesgreater than τ lam in the outer region, while theconverse is true in the in the viscous sublayer.• The viscous sublayer is usually a very thin layer.(cf. Ex. 8.4)• An alternate way of expressing turbulent shear stressτturb• Prandtl mixing length2 duη = ρlmdyτdu= ηdywhere η : eddy viscosity which is a function of both the fluidand flow conditions.22 ⎛du⎞turb= ρlmlm= ?⎜⎝dy⎟⎠

8.3.3 Turbulent Velocity Profile• Typical Structure of the turbulent velocity profile in a pipe* law of the walluu *uu *=yu*ν⎛ yu*⎞= 2.5 ln ⎜ ⎟+5.0⎝ ν ⎠where the coefficients aredetermined experimentallyviscous sublayer 0 ≤ yu * ν ≤ 5τwu*oru τ=ρfriction velocityV − u ⎛ R ⎞ = ⎜ ⎟u*⎝ y ⎠cIn the central region, 2.5ln is often used

Appendix: Derivation of the law of the wallFor the sublayer, Prandtl suggested in 1930 that u beindependent of sublayer thickness, andu = f( μ, τ , ρ, y)wUsing the pi method:a b c1 1Π1= uμτwρ→ a= 0, b=− , c=2 2d e f1 1Π2= yμτwρ→ d =− 1, e= , c=2 2⇒Π = u/ τ / ρ = u/ u*, where τ / ρ ≡u*1ywyu*νΠ2= τw/ ρ =νu ⎛ yu*⎞→ = f ⎜ ⎟u * ⎝ ν ⎠u yu*Experimentally, it is found that = .u * νw

Turbulent Velocity Profile• Empirical power–law velocityuVc⎛= ⎜1−⎝r ⎞⎟R⎠1 nthe value of n is a function of theReynolds number.EX. 8.4 Turbulent pipe flowpropertiesV8.8 Laminar to turbulent flow from a pipeV8.9 Laminar/turbulent velocity profiles

8.3.4 Turbulence ModelingTime-averaging approach: need to solve the closureproblem for the Reynolds stresses, such as −ρuυ′ ′, etc.Different methods for Reynolds stress closure:• Algebraic (zero-equation) models—by modeling the eddyviscosity or mixing length• One-equation models—solve one additional transport equation• Two-equation models—solve two more additional transportequations, such as k-ε model, k-ω model• Reynolds stress transport models—solve the transport equationfor Reynolds stressesLarge eddy simulations (LES)—explicitly solve for thelarge eddies in a calculation and implicitly account for the smalleddies by using a subgrid-scale model (SGS model).Direct numerical simulations (DNS)—directly solve thetransient 3-D Navier-Stokes equations with very fine grids andtime steps

8.3.5 Chaos and TurbulenceChaos theory involves the behavior of nonlineardynamical systems and their response to initial andboundary conditions. The flow of viscous fluid,governed by the complex nonlinear Navier-Stokesequations, is such a system.The chaotic behaviorassociated with thelogistic equation withincrease of r:Xnext = rX(1 − X)r=3(from Chaos, by James Gleick)

The Butterfly EffectThe phrase refers to the idea that a butterfly’s wings mightcreate tiny changes in the atmosphere that may ultimatelyalter the path of a tornado or delay, accelerate or even preventthe occurrence of a tornado in a certain location. The flappingwing represents a small change in the initial condition of thesystem, which causes a chain of events leading to large-scalealterations of events. Had the butterfly not flapped its wings,the trajectory of the system might have been vastly different.Of course the butterfly cannot literally cause a tornado. Thekinetic energy in a tornado is enormously larger than theenergy in the turbulence of a butterfly. The kinetic energy of atornado is ultimately provided by the sun and the butterfly canonly influence certain details of weather events in a chaoticmanner. (from Wikipedia)

About the difficulty to understand turbulenceOn his death bed, Heisenberg is reported to havesaid, "When I meet God, I am going to ask him twoquestions: Why relativity? And why turbulence? Ireally believe he will have an answer for the first."A similar witticism has been attributed to Horace Lamb(who had published a noted text book onHydrodynamics)—his choice being quantum mechanics(instead of relativity) and turbulence. Lamb was quoted assaying in a speech to the British Association for theAdvancement of Science in 1932, "I am an old man now,and when I die and go to heaven there are two matters onwhich I hope for enlightenment. One is quantumelectrodynamics, and the other is the turbulent motionof fluids. And about the former I am rather optimistic."

8.4 Dimensional Analysis of PipeFlow• Most turbulent pipe flow analyses are based on experimentaldata and semi-empirical formulas.h L =h Lmajor +h Lminor• Major losses : loss associated with the straight portion of thepipe Δ p=F( V, D, l,ε, μ,ρ)• pressure drop and head loss in a pipe• For laminar flow the pressure drop isfound to be independent of theroughness of the pipe.• For turbulent flow, the pressuredrop is expected to be a functionof the roughness.

Pressure drop• For turbulent flow, the pressure drop is expected to bea function of the roughness.

Pressure drop in pipe• We will considerk− r = 7− 3=412Δpρ⎛ ρVDε ⎞= % lφ ⎜ , , ⎟⎝ μ ⎠2V D D0≤ε D ≤ 0.05• Experiments indicate that pressure drop is proportional tothe pipe length.12fΔpl ⎛ ε ⎞= φ ⎜Re , ⎟ρ ⎝ ⎠, Δ p = for2V D D DΔpD⎛ ε ⎞φ Re,= =1 2 ⎜ ⎟2ρVl ⎝ D⎠lρV22- valid for horizontal pipefriction factor

Friction coefficient• For laminar flowf =64Reεindependent of D• For turbulent flowεf = φ ⎛ Re,⎞⎜ ⎟⎝ D ⎠

Moody chart64f =Re

• Energy Equation2 2p1 V1 p2 V2+ α1 + z1 = + α2 + z2+ hγ 2gγ 2gif D 1= D 2, z z 1=2 (horizontal pipe) with fully developedflow α1 = α2thusl ρVΔ p= p1− p2= γ hL= f D 22l VhL= f D 2 g2L(cf. Eq. 5.89 in Section 5.3.4)- Darcy – Weisbach equationvalid for fully developed steadyincompressible pipe flow horizontal or on a hill.

Determination of friction coefficient• In general with• Moody chart1 210% accuracy is best expected• Colebrook formulaV= Vl ρVp1− p2 = γ ( z2 − z1) + γhL= γ ( z2 − z1)+ f D 2Part of the pressure change is due to the elevationchange and part is due to the heat loss associated withfriction effects.1 ⎛εD 2.51=− 2.0 log+f ⎜⎝ 3.7 Re f⎞⎟⎠2

8.4.2 Minor Losses• Major losses : loss associated with the straight portion ofthe pipe• Minor losses : loss associated with valves, bends,tees,………..Minor losses

Loss coefficients for minor loss• The head loss information for essentially all components isgiven in dimensionless form and based on experimentaldata.• loss coefficientKLso thathorLKLh= =L2 12( V 2g)2V= KL⋅2g= φLΔpρV12( geometry,Re)2Δ p = K ⋅ ρV2

Loss coefficients for minor lossFor many practical applications, Re is large, theflow is dominated by inertia effect.thusΔpKL12= φρV2( geometry)• Equivalent length2 2V leq VhL= KL= f2g D 2gKDLor leq=f• loss coefficientsAn obvious way to reduce loss is to round the entranceregion.

Entrance and Exit FlowsFigure 8.22 (p. 438)Entrance flow conditions and losscoefficient (Refs. 28, 29). (a) Reentrant, K L= 0.8, (b) sharp-edged, K L= 0.5, (c) slightlyrounded, K L= 0.2 (see Fig. 8.24), (d) wellrounded,K L= 0.04 (see Fig. 8.24).Figure 8.25 (p. 440)Exit flow conditions and loss coefficient.(a) Reentrant, K L= 1.0, (b) sharp-edged, K L= 1.0, (c) slightly rounded, K L= 1.0, (d)well-rounded, K L= 1.0.V8.10 Entrance/exit flows

Pressure loss• Flow patter and pressure distribution for a sharp edgeentrance.The majority of the loss is due to inertia effects that are eventuallydissipated by the shear stresses within the fluid.Only a small portion of the loss is due to the wall shear stresswithin the entrance region.

Loss coefficient• Loss coefficients

Loss coefficient• Loss coefficients

Loss coefficient• Loss coefficients

Loss coefficient• Loss coefficients for a sudden expansionp = p = p = pAa b c=A2 31AV1 1 3 3( )1 3 3 3 3 3 3 122p1 V1p3 V3+ = + + hγ 2gγ 2gLAVp A − p A = ρ AV V −VK=h ⎛ A= = 1−V g ⎝ AL12 ⎜122L⎞⎟⎠2V8.11 Separated flow in a diffuser

Loss coefficient

Loss coefficient

Loss coefficient

8.4.3 Noncircular Conduits• For noncircular ductfCρVDh= where Reh= , Dh=Reμh4AP2l VεhL= f , relative roughness,D 2 g Dhh

Noncircular Conduits• For noncircular duct• C value2C l Vεf = , hL= f , relative roughness,RehDh 2 g Dh