Introduction Advanced Transport Phenomena Introduction • (Video) lectures (3 hrs) + tutorial sessions (3 hrs) • Book: ‘Transport Phenomena’, second edition by Bird, Stewart and Lightfoot BSL: SLIDES + ADDITIONAL MATERIAL AVALABLE ON CANVAS • Assessment • Exam/assignments (75%) + assignment on generalized Fourier methods (25%) performed in groups of 2 persons • follow lectures (development of mathematical toolkit, CANVAS) • serious preparation of tutorials essential (CANVAS) J.A.M. Kuipers • Prior knowledge necessary: • Introduction to Physical Transport Phenomena • Mathematics: knowledge about and ability to solve ordinary differential equations Introduction • Further study • ‘Boundary Layer Theory’ (Schlichting) • ‘Conduction of Heat in Solids’ (Carslaw/Jager) • ‘Mathematics of Diffusion’ (Crank) Introduction • BSL: structure of content in table 0.2-1 (p. 4) • Columns: quantity that is transported • Momentum • Energy • Mass • Rows: type of transport • • • • • • • • Transport by molecular motion Transport in one dimension Transport in arbitrary continua Transport with two independent variables Transport in turbulent flow Transport across phase boundaries Transport in large systems Transport by other mechanisms BSL 1.1: Newton’s law of viscosity • Viscosity characterizes the resistance of a fluid against flow • Fluid contained between two flat plates. • At t=0, the lower plate is suddenly brought into motion • Definition for μ: Advanced Transport Phenomena BSL 1: Viscosity and the mechanisms of momentum transport J.A.M. Kuipers dvx dy • Only valid for Newtonian fluids τ yx = − µ BSL 1: Non-Newtonian fluids BSL 1.1: Newton’s law of viscosity Rheological models: two types of empirical models • two-parameter models • three-parameter models Newton’s law from a different point of view Shear stress Momentum flux Rheological behavior of most Newtonian fluids can be described as: τ yx = −η dvx dy where η is the effective viscosity which depends on dvx/dy or τyx 6 BSL 1: Non-Newtonian fluids BSL 1: Non-Newtonian fluids • Pseudoplastic materials: η decreases with increasing shear rate (-dvx/dy) • Dilatant materials: η increases with increasing shear rate (-dvx/dy) BSL 1: Non-Newtonian fluids Two-parameter models: • Bingham model (e.g. suspensions): dv 0 dy x τ yx = −µ ± τ if τ yx > τ 0 0 dvx = 0 if τ yx < τ 0 dy • Ostwald-De Waele model (power-law model): τ yx = −m dvx dy n −1 dvx dy BSL 1: Non-Newtonian fluids • Eyring-model: 1 dvx B dy τ yx = A ⋅ arcsinh − this model follows from Eyring’s kinetic theory for fluids Three-parameter models: • Ellis-model: α −1 dv − x = ϕ0 + ϕ1 τ yx τ yx dy ( ) BSL 1: Non-Newtonian fluids • Reiner-Philippoff model: dv 1 − x= τ dy µ∞ + ( µ0 − µ∞ ) / (1 + (τ yx / τ s ) 2 ) yx • Rheological behavior in unsteady state: Advanced Transport Phenomena BSL 2: Shell momentum balances and velocity distributions in laminar flow J.A.M. Kuipers especially polymers often exhibit time-dependent rheological behaviour BSL 2: Velocity distributions in laminar flow • In general: theory has in fact already been treated in Introduction to Physical Transport Phenomena assumed to be known BSL 2.1: Shell momentum balance and BC’s • Momentum balance in words (Newton’s 2nd law): rate of accumulation of momentum in system sum of rate of rate of the forces = momentum − momentum + in out acting on the system • Boundary conditions (S: solid wall, L: liquid and G: gas) • SL-interface: fluid velocity equals the velocity of the solid wall (noslip condition) • LG-interface: stress-tensor components are assumed to be zero • LL-interface: continuity of velocity (no-slip condition) and the stress components BSL 2.2-2.4 BSL 2.5: Flow of two adjacent non-miscible fluids • BSL 2.2 Flow of a falling film: Introduction to Physical Transport Phenomena • BSL 2.3 Flow through a circular tube: Introduction to Physical Transport Phenomena • BSL 2.4 Flow through an annulus: Introduction to Physical Transport Phenomena Differential z-momentum balance (valid for phase I and II): dτ xz dp p0 − pL = − = dx dz L BSL 2.5: Flow of two adjacent non-miscible fluids Integration of differential momentum balance: p − pL I τ xzI 0 x + C1 L p − pL II τ xzII 0 x + C1 L Continuity of stress at the L-L interface: = x 0= : τ xzI τ xzII In other words: I II C= C= C1 1 1 BSL 2.5: Flow of two adjacent non-miscible fluids Combination with (2.5-2) and (2.5-3) and applying Newton’s viscosity law gives: dvzI p0 − pL −µ = x + C1 dx L dvzII p0 − pL − µ II = x + C1 dx L Integration of (2.5-5) and (2.5-6) gives: I C p −p vzI = − 0 I L x 2 − 1I x + C2I µ 2µ L C p −p vzII = − 0 II L x 2 − II1 x + C2II µ 2µ L BSL 2.5: Flow of two adjacent non-miscible fluids Three integration constants three additional BC’s: : vzI vzII = x 0= : vzI 0 = x -b= 0 x= +b : vzII = Application of the boundary conditions gives: C2I = C2II C p −p − 0 I L b 2 + 1I b + C2I 0= µ 2µ L C p −p 0= − 0 II L b 2 − II1 b + C2II µ 2µ L BSL 2.5: Flow of two adjacent non-miscible fluids Velocity profiles: 2 p0 − pL ) b 2 2 µ I µ I − µ II x x ( v = + − 2 µ I L µ I + µ II µ I + µ II b b 2 p0 − pL ) b 2 2 µ II µ I − µ II x x ( v = + I − I II II II 2µ L This results in: I II p − pL µ − µ C1 = − 0 b I II 2L µ + µ p − p 2µ I C2I = C2II + 0 I L b2 I = II 2µ L µ + µ Momentum flux or shear stress profile: p − pL x 1 µ I − µ II = τ xz 0 b − I II L b 2 µ + µ BSL 2.6: Flow around a sphere • From (2.6-13) follows for the kinetic force: I z II z BSL 2.5: Flow of two adjacent non-miscible fluids µ + µ µ + µ b b Limiting case of idential phase viscosities: Velocity profiles in I and II are the same (parabolic profile) Fk = 6πµ Rv∞ which is Stokes’s law, introduced during Introduction to Physical Transport Phenomena • Condition: Re < 0.1 • Importance of Stokes’ law: • Colloidal systems • Fluidization • Sedimentation BSL 3: Introduction Advanced Transport Phenomena BSL 3: The equations of change for isothermal systems • Derivation of general microbalances for isothermal systems consisting of a single component very important chapter! • Non-isothermal systems? BSL 11 J.A.M. Kuipers • Reactive (multi-component) systems? BSL 19 • In BSL 3 derivation of • Micro balances for mass • Micro balances for momentum • Micro balances for mechanical energy BSL 3.1: Equation of continuity • Mass balance over a volume element ΔV=ΔxΔyΔz, fixed in space: • Mass balance in words: rate of rate of rate of increase = mass − mass of mass in out BSL 3.1: Equation of continuity BSL 3.1: Equation of continuity • Mass balance: ∂ρ =∆y∆z [( ρ vx ) |x − ( ρ v ) |x+∆x ] + ∂t ∆x∆z ( ρ v y ) | y − ( ρ v y ) | y +∆y + ∆x∆y [( ρ vz ) |z − ( ρ vz ) |z +∆z ] ∆x∆y∆z • Dividing by ΔV=ΔxΔyΔz and taking the limit as Δx, Δy and Δz approach zero gives: ∂ρ ∂ ∂ ∂ = − ( ρ vx ) + ( ρ v y ) + ( ρ vz ) ∂t ∂y ∂z ∂x BSL 3.1: Equation of continuity • Alternative way of writing the equation of continuity using the substantial derivative towards time: ∂v ∂v ∂ ∂ρ ∂ρ ( ρ ) + vx + ρ x + v y + ρ y ∂t ∂x ∂x ∂y ∂y ∂vz ∂ ∂ρ + vz + ρ= ⋅v ) 0 ( ρ ) + ( v ⋅∇ρ ) + ρ ( ∇= ∂z ∂z ∂t or: Dρ 1 Dρ = − ρ ( ∇ ⋅ v ) or: = − (∇ ⋅ v ) Dt ρ Dt • Continuity equation for incompressible fluid: (∇ ⋅ v ) = 0 BSL 3.1: Equation of continuity • Written in compact vector notation: ∂ρ = − (∇ ⋅ ρv ) ∂t with (∇ ⋅ ρ v ) the divergence of ρ v • Physical meaning: −(∇ ⋅ ρ v ) is the net rate of mass addition per unit volume by convection BSL 3.2: Equation of motion Momentum balance over a volume element ΔV=ΔxΔyΔz, fixed in space: • Momentum balance in words: accumulation of momentum in ∆V per unit of time inflow of momentum = per unit of time outflow of sum of momentum the forces − + per unit acting of time on ∆V BSL 3.2: Equation of motion BSL 3.2: Equation of motion • Note: momentum is a vectorial quantity thus momentum balance is a vector equation with x-,y- and z-components • We consider the x-component only (analogy for y- and zdirection) • Net convective transport of x-momentum: ∆y∆z ( ρ vx vx |x − ρ vx vx |x +∆x ) + ∆x∆z ( ρ v y vx | y − ρ v y vx | y +∆y ) + ∆x∆y ( ρ vz vx |z − ρ vz vx |z +∆z ) BSL 3.2: Equation of motion • Net molecular transport of x-momentum: ∆y∆z (τ xx |x −τ xx |x +∆x ) + ∆x∆z (τ yx | y −τ yx | y +∆y ) + ∆x∆y (τ zx |z −τ zx |z +∆z ) • Net force in the x-direction acting on ΔV: • Pressure force: ∆y∆z ( p |x − p |x+∆x ) • Force due to gravity ρ g x ∆x∆y∆z BSL 3.2: Equation of motion • Accumulation of momentum per unit of time: ∂ ∆x∆y∆z ρ vx ∂t • Filling in all the terms, dividing by ΔV and taking the limit as Δx, Δy and Δz approach zero gives: ∂ ∂ ∂ ∂ − ( ρ vx vx ) + ( ρ v y vx ) + ( ρ vz vx ) ( ρ vx ) = ∂t ∂y ∂z ∂x ∂ ∂ ∂ ∂p − (τ xx ) + (τ yx ) + (τ zx ) − + ρ g x ∂y ∂z ∂x ∂x BSL 3.2: Equation of motion • Analogously for the y- and z-direction gives: ∂ ∂ ∂ ∂ ρ vy ) = − ( ρ vx v y ) + ( ρ v y v y ) + ( ρ vz v y ) ( ∂t ∂y ∂z ∂x BSL 3.2: Equation of motion • Notation in compact vector notation + physical interpretation of the terms (BSL (3.2-9)): ∂ ∂ ∂ ∂p − (τ xy ) + (τ yy ) + (τ zy ) − + ρ g y ∂y ∂z ∂x ∂y and: ∂ ∂ ∂ ∂ − ( ρ vx vz ) + ( ρ v y vz ) + ( ρ vz vz ) ( ρ vz ) = ∂t ∂y ∂z ∂x ∂ ∂ ∂ ∂p − (τ xz ) + (τ yz ) + (τ zz ) − + ρ g z ∂y ∂z ∂x ∂z BSL 3.2: Equation of motion • Alternative way of writing the equation of x-momentum using the substantial derivative towards time: ρ ∂τ Dvx ∂τ ∂p ∂τ = − − xx + yx + zx + ρ g x Dt ∂x ∂x ∂y ∂z • Analogous equations are valid for the y- and z-direction • Compact vector notation + physical interpretation: BSL 3.2: Equation of motion • From this form it is clear that the Navier-Stokes equations are a representation of Newton’s 2nd law: mass per unit of volume x accelaration = sum of the forces per unit of volume • For closure of the micro momentum balance a specification of the components of the tensor τ is required BSL 3.2: Equation of motion BSL 3.2: Equation of motion Components of τ (for Newtonian fluids): • Normal components: • Tangential components: ∂v 2 τ xx = −2 µ x + µ ( ∇ ⋅ v ) ∂x 3 ∂v 2 τ yy = −2 µ y + µ ( ∇ ⋅ v ) ∂y 3 ∂v 2 τ zz = −2 µ z + µ ( ∇ ⋅ v ) ∂z 3 τ yz = τ zy = −µ y + z ∂z ∂y ∂v ∂v Combination of the equations for τ with the micro balances for x-, y- and z-momentum gives: • x-component ∂v ∂ ∂ ∂ ∂ ρ vx + ( ρ vx vx ) + ( ρ v y vx ) + ( ρ vz vx ) = ∂t ∂x ∂y ∂z ∂v τ xy = τ yx = −µ x + y ∂y ∂x ∂v ∂x τ zx = τ xz = −µ z + ∂vx ∂z − ∂p ∂ ∂vx 2 ∂ ∂v ∂v + 2µ − µ ( ∇ ⋅ v ) + µ x + y ∂x ∂x ∂x 3 ∂y ∂y ∂x + ∂ ∂vx ∂vz µ + + ρ gx ∂z ∂z ∂x 16 BSL 3.2: Equation of motion • y-component ∂ ∂ ∂ ∂ ρ v y + ( ρ vx v y ) + ( ρ v y v y ) + ( ρ vz v y ) = ∂t ∂x ∂y ∂z − ∂p ∂ ∂v y ∂vx ∂ ∂v y 2 + µ + + 2µ − µ ( ∇ ⋅ v ) ∂y ∂x ∂x ∂y ∂y ∂y 3 + ∂ ∂vz ∂vx µ + + ρgy ∂z ∂x ∂z BSL 3.2: Equation of motion • z-component ∂ ∂ ∂ ∂ ρ vz + ( ρ vx vz ) + ( ρ v y vz ) + ( ρ vz vz ) = ∂t ∂x ∂y ∂z ∂p ∂ ∂vz ∂vx ∂ ∂vz ∂v y µ µ + + + + ∂z ∂x ∂x ∂z ∂y ∂y ∂z ∂ ∂v 2 + 2µ z − µ ( ∇ ⋅ v ) + ρ g z ∂z ∂z 3 − • In practice, drastic simplification of the general micro balances is often allowable BSL 3.2: Equation of motion Limiting situations • Constant density (ρ) and viscosity (μ): ∂ ( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p + µ∇ 2v + ρ g ∂t (Navier-Stokes equations) with ∇ 2 the Laplace-operator: 2 2 2 ∂ ∂ ∂ + 2+ 2 2 ∂x ∂y ∂z Notation of convective transport term: 2 ∇= ρ vx vx ρ vv = ρ v y vx ρ vz vx ρ vx v y ρ vy vy ρ vz v y ρ vx vz ρ v y vz ρ vz vz BSL 3.3: Equation of mechanical energy Micro balance for momentum in vector form: D v = −∇p − ( ∇ ⋅τ ) + ρ g Dt Take the dot product with velocity v : ρ D 1 2 v = − ( v ⋅∇p ) − ( v ⋅ [∇ ⋅τ ]) + ρ ( v ⋅ g ) Dt 2 Physical meaning: ρ distance force per coverered ) unit of × =( v= volume per unit of time work done per unit of time and volume BSL 3.2: Equation of motion • Ideal fluid [ ∇ ⋅τ ] = 0 ∂ ( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p + ρ g (Euler equation) ∂t • Description of the compressible flow of a isothermal flowing fluid: • • • • Micro balance for mass Micro balance for momentum Equation of state p = p ( ρ ) Equation for pressure dependency of viscosity µ = µ ( ρ ) BSL 3.3: Equation of mechanical energy Using the definition of the substantial derivative towards time and manipulating ( v ⋅ [∇ ⋅τ ]) and ( v ⋅∇p ) : BSL 3.3: Equation of mechanical energy For Newtonian fluids, the term ( −τ : ∇v ) can be written as a sum of squares and is therefore always positive: 1 3 3 ∂v ∂v 2 ( −τ : ∇v ) = µΦ v = µ=i∑1 =j∑1 i + j − ( ∇ ⋅ v ) δ ij 2 ∂x j ∂xi 3 2 Describing the degradation of mechanical energy into thermal energy (viscous dissipation heating) • Importance at flow with very high velocities (“space shuttle”) and other situations in which very large velocity gradients occur • If the term ( −τ : ∇v ) would not be there, full reversible interconversion of the various forms of energy would be possible BSL 3: Micro balances for momentum and mass in curvilinear coordinates Important coordinates for practical applications: • Cartesian coordinates • Cylindrical coordinates • Spherical coordinates Mathematical details of the transformation of coordinates BSL Appendix A BSL 3.3: Equation of mechanical energy The term( p ( −∇ ⋅ v ) ) describes the reversible conversion of mechanical energy into thermal energy per unit of volume and time • Important for compressible fluids that undergo sudden and/or large pressure changes (f.i. compressors and turbines) BSL 3: Micro balances for momentum and mass in curvilinear coordinates What is needed in order to transform the equations? • Equations for the gradient-operator • Equations for the derivatives of the unit vector towards the new coordinates Micro balances in different coordinate systems • Micro balance for mass BSL §B.4 • Micro balance for momentum in terms of τ BSL §B.5 • Micro balance for momentum in terms of velocity gradients BSL §B.6 • Components of the viscous tension tensor τ BSL §B.1 • Viscous dissipation term ( −τ : ∇v ) =µΦ v for Newtonian fluids BSL §B.7 BSL 3: Analysis of flow phenomena using general micro balances BSL 3: Applications BSL 2: analysis of flow phenomena based on differential micro balances • Flow of a falling film Disadvantages • Flow of a Newtonian fluid through a circular tube • Momentum balance has to be set up for every single problem • Limited application (rectilinear flow patterns!) Alternative: analysis of flow phenomena based in (reduced) micro balances Advantages: • Generally applicable • Reduction of general micro balances ‘automatically’ generates a list of assumptions made • Flow of a Newtonian between two concentric cylinders (Couette viscosity meter) • Flow of an Newtonian fluid between two concentric spheres BSL 3: Flow of a falling film Use z-momentum balance based on geometry: ∂v ∂v ∂v ∂v ∂p z +v z +v z= ρ z +v − + ∂t x ∂x y ∂y z ∂z ∂z ∂ 2v ∂ 2v ∂ 2v z+ z+ z + ρg µ z 2 2 ∂x 2 ∂y ∂z BSL 3: Flow of a falling film Reduced momentum balance: 0= µ ∂ 2 vz ∂ 2 vz g + = + ρ g cos ( β ) ρ µ z ∂x 2 ∂x 2 Since vz is merely a function of x, the reduced momentum balance can be rewritten into: 0 µ = d 2 vz + ρ g cos ( β ) dx 2 Boundary conditions: dvz = at x 0:= 0 dx = at x δ= : vz 0 Solution: ρ gδ 2 cos ( β ) x 2 = vz 1 − 2µ δ 31 BSL 3: Flow of a Newtonian fluid through a circular tube Use z -momentum balance based on geometry: ∂v ∂v ∂v ∂v z ∂p z +v z +v z= +v − + ∂t θ ∂θ r ∂r z ∂z ∂z ∂ 2v ∂ 2v ∂ v ∂ 1 1 z z + ρg z + 2 + µ r z r ∂r ∂r r ∂θ 2 ∂z 2 BSL 3: Flow of a Newtonian fluid through a circular tube Reduced momentum balance: 1 ∂ ∂vz ∂p 0= − +µ r + ρ gz ∂z r ∂r ∂r ρ Since vz is merely a function of r, the reduced momentum balance can be rewritten into: 1 d dvz dP − +µ 0= r dz r dr dr Boundary conditions: dvz = at r 0:= 0 dr = at r R= : vz 0 Solution: 2 P0 − PL ) R 2 r ( = vz 1 − 4µ L R 33 BSL 3: Flow between two concentric cylinders • Applicable for e.g. bearings, Couette viscosity meter BSL 3: Flow between two concentric cylinders r -momentum balance: ∂vr ρ + vr 2 ∂vr vθ ∂vr vθ ∂v ∂p 1 ∂ 2 v 2 ∂v ∂ 2 v ∂ 1 ∂ + − + vz z = − +µ ( rvr ) + 2 2r − 2 θ + 2r + ρ g r ∂r r ∂θ r ∂z ∂r r ∂θ ∂z r ∂θ ∂r r ∂r ∂t reduced r -momentum balance: vθ2 ∂p = − r ∂r θ -momentum balance: −ρ ∂v ∂vθ ∂v v ∂v v v 1 ∂p 1 ∂ 2v 2 δ v ∂ 2v ∂ 1∂ − + µ + vr θ + θ θ + r θ + vz θ = ( rvθ ) + 2 θ2 + 2 r + 2θ + ρ gθ ∂z r r ∂θ r δθ ∂z ∂r r ∂θ r ∂θ ∂t ∂r r ∂r reduced θ -momentum balance: ρ d 1 d ( rvθ ) dr r dr z -momentum balance: 0= ∂vz ∂v v ∂v ∂v ∂p 1 ∂ ∂vz 1 ∂ 2 vz ∂ 2 vz + vr z + θ z + vz z = − +µ + 2 + ρ gz r + 2 2 ∂z ∂r r ∂θ ∂z ∂z ∂t r ∂r ∂r r ∂θ ρ • Flow in the θ-direction where vθ is a function or radius r reduced z -momentum balance: ∂p 0= − + ρ gz ∂z BSL 3: Flow between two concentric cylinders Boundary conditions for the θ -momentum balance: = R: vθ 0 at r κ= at r = R: vθ = Ω0 R With Ω0 the rotation speed of the outer cylinder Integration of the reduced θ -momentum balance gives: κR − r r κR vθ = Ω0 R κ − 1 κ BSL 3: Flow between two concentric cylinders Application: Couette viscosity meter • From measurements of M and at known geometry (κ, R and L) and known rotation speed Ω0, the viscosity of the fluid can be determined Pressure as a function of r and z? • Integration of the equation for the total differential of p in which first the reduced momentum balances in the rand z-direction are substituted BSL 3: Flow between two concentric cylinders BSL gives the momentum density in the radial direction: ∂ v τ rθ = − µ r θ ∂r r Substitution in the equation for vθ gives: 1 κ 2 τ rθ = −2 µΩ0 R 2 2 r 1− κ 2 Moment M that must be applied on the outer cylinder: = M 2π RL ( −τ rθ |r = R ) R BSL 3: Flow through two concentric spheres assumption: creeping flow conditions BSL 3: Flow between two concentric spheres θ -momentum balance: 2 vφ ∂vθ ∂v v ∂v ∂vθ vr vθ vφ cot (θ ) + vr θ + θ θ + + − = sin θ θ φ ∂ ∂ ∂ ∂ t r r r r r ( ) ρ BSL 3: Flow between two concentric spheres Equation of continuity (with vr = 0): = 0 ∂ 1 ∂ 2 ∂vr 1 ( vθ sin (θ ) ) r + 2 r ∂r ∂r r sin (θ ) ∂θ ∂ 1 1 ∂ 2 ∂vθ 1 ∂ 1 = ∂ = ( vθ sin (θ ) ) 0 sin θ v ( ) ( ) + 2 2 r θ ∂ r θ θ sin ( ) ∂ ∂ ∂ ∂ sin θ θ θ r r r r ( ) 1 ∂p − +µ 2 Note: the quantity vθ sin (θ ) is independent of θ and is therefore r ∂θ ∂ vθ 2 ∂vr 2 cot (θ ) ∂vφ 1 + + − + g ρ 2 2 θ 2 merely a function of r! r 2 ∂θ r 2 sin 2 (θ ) ∂φ r sin (θ ) ∂φ = u ( r ) vθ ( r ,θ ) ⋅ sin (θ ) and substitute this result in the Define reduced θ -momentum balance: 1 ∂ ∂v 1 ∂p − + µ 2 r2 θ 0= r ∂θ r ∂r ∂r 1 ∂ 1 ∂ ( vθ sin (θ ) ) + 2 r ∂θ sin(θ ) ∂θ Note: vθ is function or r as well as θ! θ -momentum balance: 1 ∂p 1 d 2 du − +µ 2 0= r r ∂θ r sin (θ ) dr dr BSL 3: Flow between two concentric spheres In this situation, p is merely a function of θ and u is merely a function of r which means that the reduced momentum balance can be separated! Separated equations: dp sin (θ ) =B dθ and: µ d 2 du r =B r dr dr where B is a separation constant that has yet to be determined BSL 3: Flow between two concentric spheres Integration of the latter two equations using cos (θ ) = x gives: 1 − cos ( ε ) 1- x ∆p = B ⋅ ln = ⋅ ln B 1 + cos ε = − B ⋅ E ( ε ) ( ) 1 + x Boundary conditions: at θ = ε : p = p0 and at θ = π − ε : p = p1 ( ∆p = p0 − p1 ) and: = u R ⋅ ∆p r R 1 − + κ 1 − 2 µ E ( ε ) R r Boundary conditions: at r κ R= : u 0 and at r R= :u 0 = = BSL 3: Flow between two concentric spheres Equation for the volume flow rate ( ∑i ( 2π ri ∆r ) ui ): π π R 3 ∆p 3 2π rdr 1− κ ) Q ∫= vθ r ,θ = = ( κR 2 6µ E (ε ) R Advanced Transport Phenomena BSL 4: Velocity distributions with more than one independent variable J.A.M. Kuipers BSL 4: Velocity distributions with more than one independent variable In practice, velocity profiles depend on more than one independent variable! •Unsteady flow •Two-dimensional flow Contents of this chapter: • Unsteady flow: unsteady flow near a wall suddenly set in motion and unsteady laminar flow between two parallel plates • Stationary viscous flow in two dimensions (stream function) • Stationary ideal flow in two dimensions (potential flow) • Boundary-layer theory BSL 4: Velocity distributions with more than one independent variable By looking at certain examples, some of the widely used techniques for solving (complex non-linear) partial differential equations (PDE) are presented: •Method of combination of variables (similarity solutions) •Method of separation of variables •Stream function and velocity potential •Boundary layer approaches (approximate solutions) BSL 4.1: Time-dependent flow of Newtonian fluids Flow near a wall suddenly set in motion Initially the fluid and the wall are at rest. At t=0, the wall is suddenly set in motion. BSL 4.1: Time-dependent flow of Newtonian fluids Reduced x-momentum balance: ∂vx ∂ 2 vx =ν 2 ∂t ∂y with ν = kinematic viscosity Initial and boundary conditions: at t ≤ 0, vx = 0 for all y = vx V for all t > 0 at y 0,= ∞, v x = at y = 0 for all t > 0 BSL 4.1: Time-dependent flow of Newtonian fluids Try a solution of the form: vx ( y , t ) y φ= = (η ) with η V 4ν t Transformation of the derivative towards time and derivative towards spatial co-ordinate: ∂ vx ∂ 1η = (φ ) = − φ ' ∂t V ∂t 2t and: ∂ 2 vx ∂ 2 1 = = φ φ" ( ) ∂y 2 V ∂y 2 4ν t BSL 4.1: Time-dependent flow of Newtonian fluids Substitution of this result in in the PDE: 0 φ "+ 2ηφ ' = Integration using φ ' = F : F= φ=' C1e −η 2 and again integration: η 2 φ C1 ∫ e −η dη + C2 = o Application of the boundary conditions: η ∫e −η 2 1 − ∞0 2 φ= ∫e 0 −η dη dη 2 η −η 2 1− erfc (η ) = ∫ e dη = π 0 BSL 4.1: Time-dependent flow of Newtonian fluids BSL 4.1: Time-dependent flow of Newtonian fluids or: vx y y = 1 − erf erfc = V 4ν t 4ν t In correspondence with solution presented in Introduction to Physical Transport Phenomena Graphical solution in which the power-law model is also displayed with m =µ ρ BSL 4.1: Time-dependent flow of Newtonian fluids Unsteady laminar flow between two parallel plates Initially the fluid is at rest. At t=0, a constant pressure dp p0 − pL gradient − = dx L is imposed over the plate. The fluid will accelerate and eventually, a stationary velocity profile will be established BSL 4.1: Time-dependent flow of Newtonian fluids Use dimensionless quantities: Dimensionless velocity φ: ∂v ∂v ∂p ρ x= − + µ 2x ∂t ∂x ∂y vx vx = φ = 2 ∞ ( p0 − pL ) d ( vx )max 2µ L Initial condition: Dimensionless time τ : at= t 0, - d ≤ y ≤ d : = vx 0 v ⋅t τ= 2 d Boundary conditions Dimensionless y -coordinate ξ : at t > 0, y = -d : vx = 0 y at t > 0, y = + d : vx = 0 ξ= d Reduced x-momentum balance: 2 BSL 4.1: Time-dependent flow of Newtonian fluids Combine with the dimensionless quantities gives: steady state ∂φ ∂φ = 2+ 2 ∂τ ∂ξ 2 φ → φ∞ d 2φ∞ 0 +2= dξ 2 with dimensionless initial and boundary conditions: at τ = 0, − 1 ≤ ξ ≤ 1: φ = 0 BSL 4.1: Time-dependent flow of Newtonian fluids Try a solution of the following form: φ (= ξ ,τ ) φ∞ (ξ ) − φt (ξ ,τ ) solution = "steady solution" - "unsteady solution" Substitute this “trial” solution in the dimensionless PDE • ODE for steady solution d 2φ∞ 0 +2= dξ 2 • PDE for unsteady solution ∂φt ∂ 2φt = ∂τ ∂ξ 2 −1: φ = at τ > 0, ξ = 0 +1: φ = at τ > 0, ξ = 0 BSL 4.1: Time-dependent flow of Newtonian fluids BSL 4.1: Time-dependent flow of Newtonian fluids The boundary conditions for the stationary solution φ∞ and the unsteady solution φt remain unchanged, however……. Solution of PDE for φt: try a solution of the form: Initial condition for φt: Combination with dimensionless PDE for φt and division by gives: at τ = 0, − 1 ≤ ξ ≤ 1: φt = φ∞ F ' (τ ) G " (ξ ) = F (τ ) G (ξ ) Solution of ODE for φ∞: φ∞ = 1 − ξ 2 check: φ (ξ= ,τ ) F (τ ) ⋅ G (ξ ) d 2φ∞ + 2 =−2 + 2 =0 dξ 2 Note: left side of the equation is independent of ξ and identically equal to the right side of the equation which is independent of τ BSL 4.1: Time-dependent flow of Newtonian fluids BSL 4.1: Time-dependent flow of Newtonian fluids Left and right side of the equation equal a constant (the so-called separation constant): Because of symmetry: C1=0 F ' (τ ) G " (ξ ) = = −µ 2 F (τ ) G (ξ ) Separated equations: Based on the boundary conditions for ξ=±1: 1 C2 cos ( µ ) =0 → µn = n + π 2 with: n = 0,1, 2 and µn the so-called eigenvalues and F '+ µ 2 F = 0 G "+ µ 2G = 0 the corresponding eigenfunctions cos(µnξ ) General solution for φt: e − µ τ [C1 sin ( µξ ) + C2 cos ( µξ )] where C1 and C2 are integration constants φt 2 BSL 4.1: Time-dependent flow of Newtonian fluids Resulting solution for φt : (φt )n K n e (( ) ) 2 − n+ 1 π τ 2 1 cos n + πξ 2 1 with cos n + πξ the eigenfunctions and K n constants 2 that follow "in theory" from the initial condition: 1 1 ξ 2 =K n cos n + πξ φ∞ =− 2 Problem: the initial condition φ∞ = 1 − ξ 2 can not be described 1 by a single eigenfunction cos n + πξ ! 2 BSL 4.1: Time-dependent flow of Newtonian fluids Solution: try to realize a description of the initial condition based on the sum of all available eigenfunctions: n =∞ 1 φ∞ =− 1 ξ 2 =∑ K n cos n + πξ n =0 2 1 Multiply the right and left side by cos m + πξ d ξ ( m = 0,1, 2,..) 2 and integrate from ξ = −1 to ξ = +1: 1 ∫ -1 = (1 − ξ ) cos m + 12 πξ dξ 2 1 1 1 K n ∫ cos n + πξ ⋅ cos m + πξ d ξ n =0 2 2 −1 n =∞ ∑ BSL 4.1: Time-dependent flow of Newtonian fluids BSL 4.1: Time-dependent flow of Newtonian fluids Evaluation: use orthogonality of eigenfunctions: Solution to the original problem: 1 1 ∫ cos n + πξ ⋅ cos m + πξ d ξ =0 if m ≠ n −1 2 2 and: 2 n 1 4 ( −1) − n + π τ n= ∞ 1 2 2 φ (ξ ,τ ) =(1 − ξ ) − ∑ e cos n + πξ 3 n =0 2 1 n+ π 2 1 1 1 cos n + πξ ⋅ cos m + πξ d ξ =1 if m =n −1 2 2 Combination of results gives the following expression for K n : 1 ∫ Note: for t→∞, the stationary solution (parabolic velocity profile) is obtained! 4 ( −1) 1 Kn = ∫ (1 − ξ ) cos n + πξ d ξ = 3 −1 2 1 n + π 2 1 n 2 BSL 4.1: Time-dependent flow of Newtonian fluids BSL 4.4: Boundary-layer theory Absence of molecular momentum transport (μ=0) in the bulk of a real fluid is often a reasonable approximation but is not applicable for the analysis of transport phenomena in the immediate neighborhood of a wall boundary-layer theory is used Boundary-layer theory: molecular transport dominates in the immediate neighborhood of the wall and the complete velocity gradient is situated in a very thin boundary layer near the wall (thickness=δ) BSL 4.4: Boundary-layer theory BSL 4.4: Boundary-layer theory Unsteady boundary-layer flow Viscous effects only play a role inside the boundary layer A fluid is set into motion by a suddenly moving solid wall Assumption:the dimensionless velocity profiles are “self similar” during development in time t: Qualitative development of the velocity profile and approximation with the boundary-layer approach: vx y = φ= (η ) with: η V δ (t ) Reduced x-momentum balance: ∂vx ∂ 2v = ν 2x ∂t ∂y 23 BSL 4.4: Boundary-layer theory Evaluation of the derivatives in terms of the yet to be determined functions φ and η: ∂vx y dδ η dδ = V φ ' − 2 = V φ ' − ∂t δ dt δ dt and: ∂ vx 1 = V φ " 2 2 ∂y δ Combination with the reduced x-momentum balance gives: 2 {δ δ }{ ηφ } d dt − ' = vφ " BSL 4.4: Boundary-layer theory Integrate this equation from η=0 to η=1: { }{ δ dδ dt 1 ∫ −ηφ 'dη 0 } 1 = v ∫ φ "dη 0 Define the constants M and N as follows: 1 1 1 1 0 0 0 M= −η φ + ∫ φ dη = ∫ −ηφ ' dη = ∫ φ dη 0 N = 1 η ∫ φ " d= 0 1 ' φ ' (1) − φ ' ( 0 ) φ= 0 Combine this with the found ODE for δ gives after integration with initial condition δ=0 at t=0: δ = 2 ( N M )ν t BSL 4.4: Boundary-layer theory BSL 4.4: Boundary-layer theory Before proceeding, an assumption must be made regarding the dimensionless velocity distribution ! vx V = φ (η ) Try a polynomial expression of the following form: vx =φ (η ) =a0 + a1η + a2η 2 + a3η 3 V Combination of the equation for φ(η) with the equations for M and N gives M=3/8 and N=3/2 and finally: δ = 2 2ν t Compare this result to the results found at Introduction to Physical Transport Phenomena: Boundary conditions: δ = 2 πν t = φ ( 0 ) 1,= φ (1) 0, = φ ' (1) 0 and = φ "( 0) 0 tutorials: extension of boundary layer approach to a power-law fluid This gives: φ (η ) =1 − 32 η + 12 η 3 ( 0 ≤ η ≤ 1) dv τ yx = −m x dy BSL 4.4: Boundary-layer theory n −1 dvx dy BSL 4.4: Boundary-layer theory Steady two-dimensional boundary layer flow along a flat plate Mathematical description • Equation of continuity ∂vx ∂v y + = 0 ∂x ∂y • x-momentum balance vx ∂vx ∂v ∂ 2v ∂ 2v + v y x = ν 2x + 2x ∂x ∂y ∂y ∂x This leads to 2 equations for vx and vy which can in principle be solved after specification of suitable boundary conditions: In this steady system, the thickness of the boundary layer increases with increasing x 29 = at x 0:= vx v∞ = y 0:= vx 0 at = at y δ= ( x ): vx v∞ BSL 4.4: Boundary-layer theory BSL 4.4: Boundary-layer theory Using the equation of continuity, vy can be expressed in terms of vx using vy=0 at y=0: Note: PDE is non-linear and 1st order in x and 2nd order in y find an approximate solution! y ∂v v y = − ∫ x dy 0 ∂x Substitution of this in in the x-momentum balance gives: Assumption: the dimensionless velocity profiles are “self similar” with increasing distance x: vx ∂vx y ∂vx ∂vx ∂ 2 vx − ∫ = dy ν ∂x 0 ∂x ∂y ∂y 2 vx y φ= = (η ) with: η v∞ δ ( x) in which convective transport of momentum in the xdirection is assumed to dominate the molecular transport of x-momentum in the x-direction BSL 4.4: Boundary-layer theory Evaluation of the terms from the x-momentum balance in terms of φ(η): ∂vx y dδ η dδ v∞φ ' − 2 v∞φ ' − = = ∂x δ dx δ dx ∂vx ∂ 2 vx 1 1 ' v= v∞φ " 2 φ ∞ 2 ∂y ∂y δ δ Substitution of these equations in the x-momentum balance: η dδ η η dδ 1 v∞φ v∞φ ' − − ∫ v∞φ ' − δ dη v∞φ ' δ dx 0 δ dx δ 1 = ν v∞φ " 2 δ BSL 4.4: Boundary-layer theory or: ν φ ' η∫ ηφ ' dη − ηφφ ' δ d δ = φ" 0 dx v ∞ Integrate this equation with respect to η: ν 1 1∫ φ ' η∫ uφ 'du dη − 1∫ ηφφ ' dη δ d δ = φ " dη 0 0 dx v∞ 0∫ 0 or in compact notation: dδ ν C ( B − A) δ = dx v∞ 1 1 with: = A ∫ ηφφ ' dη C ∫ φ= "dη φ ' (1) − φ ' ( 0 ) = 0 0 η η η =1 1 1 1 B =∫ φ ' ∫ uφ ' du dη =φ ∫ uφ ' du − ∫ ηφφ ' dη =∫ ηφ ' dη − A 0 0 0 0 η =0 0 BSL 4.4: Boundary-layer theory Integration of the equation for the boundary layer thickness with boundary condition δ=0 at x=0: C ν x B − A v∞ δ ( x ) = 2 Before proceeding, an assumption regarding the dimensionless velocity vx v∞ = φ (η ) must be made! Try a polynomial expression of the following form: vx =φ (η ) =a0 + a1η + a2η 2 + a3η 3 v∞ Boundary conditions: = φ ( 0 ) 0,= φ (1) 1,= φ ' (1) 0 and = φ "( 0) 0 BSL 4.4: Boundary-layer theory Using the approximate solution for vx, the drag force Fx can be obtained from: Fx =− 2 ∫ ∫ ( τ yx ) W L 0 0 y =0 dxdz = 1.292 ρµ LW 2 v∞3 Exact numerical solution (Blasius): replace the constant 1.292 by 1.328!!! BSL 4.4: Boundary-layer theory This gives: φ (η= ) 3 2 η − 12 η 3 Substitution of this equation for φ(η) in the equations for A, B and C gives A=9/35, B=33/280 and C=-3/2 Boundary layer thickness: = δ 280 ν x νx = 4.64 13 v∞ v∞ Velocity profile within the boundary layer (0≤y≤δ(x)): vx = v∞ 1 y y 3 − 2 2 4.64 ν x v∞ 4.64 ν x v∞ 3 BSL 4.4: Boundary-layer theory Generalization of the boundary-layer theory to flows with a more complex geometry (e.g. airplane wing) BSL Fig. 4.4-1 Orthogonal co-ordinate system ! x in tangential direction along the surface y in normal direction to the surface 38 BSL 4.4: Boundary-layer theory Qualitative: in contrast (due to curvature) to the flat plate (no curvature), pressure changes occur outside the boundary layer that are imposed on the boundary layer dp∞ − ∂p = − ∂x boundary layer dx y =δ ( x ) Mathematical description on basis of micro-balances •equation of continuity ∂vx ∂v y + = 0 ∂x ∂y •x-momentum balance ∂vx ∂vx 1 ∂p ∂ 2 vx ∂ 2 vx vx + vy = − +ν 2 + 2 ∂x ∂y ρ ∂x ∂y ∂x BSL 4.4: Boundary-layer theory Eliminate vy from the x-momentum balance by integration of the equation of continuity and neglect molecular transport in comparison with convective transport in the x-direction: ∂vx y ∂vx ∂vx 1 ∂p ∂ 2 vx vx − ∫ = − +ν 2 dy ρ ∂x ∂x 0 ∂x ∂y ∂y Outside the boundary layer: ideal flow for which the Bernoulli: equation holds: 1 2 ρ v∞2 ( x ) + p∞ ( x ) = constant Differentiation with respect to x gives: dv dp ρ v∞ ∞ + ∞ = 0 dx dx BSL 4.4: Boundary-layer theory Combination of these equations with the x-momentum balance in the boundary layer gives the Prandtl boundary layer equation: ∂vx y ∂vx ∂vx dv∞ 1 ∂p ∂ 2 vx ∂ 2 vx vx v∞ − ∫ = − +ν 2 = +ν 2 dy ρ ∂x dx ∂x 0 ∂x ∂y ∂y ∂y Limitation: curvature of surface should not be too large because in this case separation occurs and therefore the boundary layer concept will break down This is due to the fact that the velocity profile in the boundary layer ceases to satisfy the “similarity” condition (flow reversal occurs) BSL 4.4: Boundary-layer theory Integration of Prandtl boundary layer equation with respect to y gives the famous von Kármán momentum balance: 1 − τ yx ρ y =0 = dv d δ 2 v∞2 ) + δ1v∞ ∞ ( dx dx with: δ1 = ∞ 0 ∫ 1 − vx dy ("displacement thickness") v∞ and: = δ2 vx vx 1 − v dy ("momentum thickness") 0 v ∞ ∞ ∞ ∫ BSL 4.4: Boundary-layer theory • Solution of Prandtl boundary layer equation “exact” solution • Solution of von Kármán momentum balance based on uniformity of velocity profiles approximate solution • Further study: “Boundary Layer Theory” (Schlichting) BSL 5: velocity distributions in turbulent flow • Theory treated so far only valid for laminar flow • Laminar flow: streamlines do not intersect (“orderly movement”) • Description of laminar flow: Pressure and velocity profiles can be derived by solving the micro balances for momentum and mass by analytical or numerical techniques (Computational Fluid Dynamics) • In practice in process equipment, turbulent flow (“eddies”) often occurs! • Turbulent flow: the streamlines intersect continuously, movement of fluid particles takes place in all directions (“chaotic movement”) Advanced Transport Phenomena BSL 5: Velocity distributions in turbulent flow J.A.M. Kuipers BSL 5: velocity distributions in turbulent flow coherent structures in a turbulent channel flow BSL 5: velocity distributions in turbulent flow presence of vortical flow structures does not necessarily imply a turbulent flow condition: vortices can also prevail in laminar flows BSL 5: velocity distributions in turbulent flow Consequence of this way of describing is that the micro balances of (time-averaged or time-smoothed) momentum contain terms for the “turbulent momentum flux”: (τ ) (τ ) ρ v v (τ ) (τ ) = ρ v v (τ ) (τ ) ρ v v • Local pressure and velocity components contain a strongly fluctuating component • Description of turbulent flow: equation of continuity and the Navier-Stokes equations are valid! • Solving conservation laws gives the pressure and velocity components as a function of (x,y,z) and t • Practical problem: the scale (in space and time) on which changes occur in turbulent flow are very small! Directly solution of the conservation laws is not possible Use approximation methods von Karman vortex street behind cylinder in cross-flow operation (Das et al., 2018) (τ xxt ) t (τ yx ) τt ( zx ) BSL 5: velocity distributions in turbulent flow t xy t xz ' ' x x ρ vx' v 'y ρ vx' vz' t yy t yz ' ' y x ρ v 'y v 'y ρ v 'y vz' t zy t zz ' ' z x ρ vz' v 'y ρ vz' vz' which have to be determined empirically in terms of the time-smoothed velocity components (“closure problem”) ! BSL 5.1: Comparisons of laminar and turbulent flows Time-smoothed velocity profiles • Radial velocity profile for laminar flow in a circular tube: r 2 v 1 dp and: Φ v ≈ − = 1 − and: z = vz ,max R vz ,max 2 dz Condition: Re<2100 • Time-smoothed radial velocity profile for turbulent flow in a circular tube: vz vz vz ,max r = 1 − R 17 and: vz vz ,max Condition: 104 < Re < 105 4 dp = and: Φ v ≈ − 5 dz 47 BSL 5.1: Comparisons of laminar and turbulent flows Qualitative comparison of radial velocity profiles for flow in a circular tube Turbulent flow compared with laminar flow: velocity profile is flatter Eddy transport: turbulent momentum transport •With increasing distance to the tube wall, we can distinguish: Laminar sublayer Buffer zone Fully developed turbulent flow zone Velocity fluctuations in turbulent flows where t0 is large in comparison with time scale for turbulent fluctuations but small in comparison with time scale at which the macroscopic flow patterns change Velocity distribution in a circular tube near the solid wall: •Center of the tube: random velocity fluctuations •Near the tube wall: velocity fluctuations in axial direction exceed velocity fluctuations in radial direction •Turbulent flow has a structure that changes gradually BSL 5.1: Comparisons of laminar and turbulent flows Definition of time-smoothed velocity: 1 t + t0 vz = ∫ vz dt t0 t BSL 5.1: Comparisons of laminar and turbulent flows = vz 1 t + t0 vz dt O= (v z ) v z ∫= t0 t BSL 5.1: Comparisons of laminar and turbulent flows The velocity in terms of the mean value and the fluctuation: v= vz + vz' z where = vz' 0 and ( vz' ) ≠ 0 2 Intensity of turbulence I: I= (v ) ' z 2 vz Flow in a circular tube: I=1% ↔ I=10% BSL 5.1: Comparisons of laminar and turbulent flows Turbulent flow in ducts BSL 5.2: Time-smoothed equations of change for incompressible fluids Every quantity is composed of the sum of its mean value and its fluctuating component (Reynolds decomposition) •Equation of continuity: ∂ ∂ ∂ vx + vx' ) + ( v y + v 'y ) + ( vz + vz' ) = 0 ( ∂x ∂y ∂z •Equation of motion for x-component (y- and z-component in similar way): ∂ ∂ ∂ ρ ( vx + vx' ) ) = − ( p + p ' ) − ( ρ ( vx + vx' )( vx + vx' ) ) ( ∂t ∂x ∂x ∂ ∂ − ( ρ ( v y + v 'y ) ( vx + vx' ) ) − ( ρ ( vz + vz' )( vx + vx' ) ) ∂y ∂z + µ∇ 2 ( vx + vx' ) + ρ g x BSL 5.2: Time-smoothed equations of change for incompressible fluids O( ∂ ∂ ∂ vx + vx' ) + ( v y + v 'y ) + ( vz + vz' )) = O(0) ( ∂x ∂y ∂z ∂ ∂ ∂ O(vx + vx' ) ) + ( O(v y + v 'y ) ) + ( O (vz + vz' ) ) = 0 ( ∂x ∂y ∂z ∂ ∂ ∂ O(vx ) + O(vx' ) ) + ( O(v y ) + O(v 'y ) ) + ( O (vz ) + O(vz' ) ) = 0 ( ∂x ∂y ∂z ∂ ∂ ∂ 0 ( vx ) + ( v y ) + ( vz ) = ∂x ∂y ∂z BSL 5.2: Time-smoothed equations of change for incompressible fluids ∂ ∂ ∂ ρ ( vx + vx' ) )) =O(− ( p + p ' ) − ( ρ ( vx + vx' )( vx + vx' ) ) ( ∂t ∂x ∂x ∂ ∂ − ( ρ ( v y + v 'y ) ( vx + vx' ) ) − ( ρ ( vz + vz' )( vx + vx' ) ) ∂y ∂z O( + µ∇ 2 ( vx + vx' ) + ρ g x ) focus on this term: ∂ ∂ ' ' ρ ( vx + vx' )( vx + v= ρ O(( vx + vx' )( vx + v= ( ( x ) )) x )) ) ∂x ∂x ∂ ∂ vx' vx' ) ) ρ O(vx vx + 2vx vx' += ρ (O(vx vx ) + O(2vx vx' ) + O(vx' vx' )) ) ( ( ∂x ∂x ∂ ∂ similar for two other ρ vx vx + ρ vx' vx' = ρ vx vx + τ xxt = convection terms ∂x ∂x O( ( ) ( ) BSL 5.2: Time-smoothed equations of change for incompressible fluids O( ∂ ∂ ' ' ρ ( v y + v 'y ) ( vx + v= ρ O(( v y + v 'y ) ( vx + v= ( ( x ) )) x )) ) ∂y ∂y ∂ ρ O(v y vx + v y vx' + v 'y vx + v 'y vx' ) ) = ( ∂y ∂ ∂ ρ v y vx + ρ v 'y vx' = ρ v y vx + τ yxt ) ( ∂y ∂y ( ∂ ∂ ∂ 0 ( vx ) + ( v y ) + ( vz ) = ∂x ∂y ∂z ∂ ∂ ' ' O( ( ρ ( vz + vz' )( vx + v= ρ O(( vz + vz' )( vx + v= ( x ) )) x )) ) ∂z ∂z ∂ ρ O(vz vx + vz vx' + vz' vx + vz' vx' ) ) = ( ∂z ∂ ∂ ρ vz vx + ρ vz' vx' = ρ vz vx + τ zxt ) ( ∂z ∂z ) BSL 5.2: Time-smoothed equations of change for incompressible fluids Vector notation • Equation of continuity ∂ ∂p ∂ ∂ ∂ − − ( ρ vx vx ) − ( ρ v y vx ) − ( ρ vz vx ) ( ρ vx ) = ∂t ∂x ∂x ∂y ∂z ∂ ∂ ∂ − ρ vx' vx' + ρ v 'y vx' + ρ vz' vx' + µ∇ 2 vx + ρ g x ∂y ∂z ∂x ( ) ( ) ( ) BSL 5.2: Time-smoothed equations of change for incompressible fluids 1, 2 and 3 vi ↔ vi i = p ↔ p 1, 2 and 3 τ ij ↔ τ ij(l ) + τ ij(t ) i, j = • Equation of motion ∂ ( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p − ∇ ⋅τ (l ) − ∇ ⋅τ (t ) + ρ g ∂t with τ(t) the turbulent momentum flux tensor: τ (t ) • Time-smoothed equation of motion for x-component (yand z- component in a similar way): Correspondence rules: ((∇ ⋅ v )) = 0 (τ ) (τ ) (τ ) t (τ yx ) (τ yyt ) (τ yzt ) = τt t t ( zx ) (τ zy ) (τ zz ) t xx Time-smoothed equations • Time-smoothed equation of continuity: ) ( BSL 5.2: Time-smoothed equations of change for incompressible fluids t xy t xz ρv v ' ' ρ v y vx ' ' ρ vz vx ' ' x x Equations in BSL Table B.5 can be adapted for timesmoothed turbulent flow systems by changing all ' ' x y ρv v ρv v ρ v 'y v 'y ρ v 'y vz' ρv v ' z ' y ' ' x z ρ v v ' ' z z • these quantities are usually referred to as the Reynolds stresses vi to vi and p to p as well as τ ij to τ= τ ij(l ) + τ ij(t ) ij in any of the coordinate systems given Components of τ(t) need to be specified in terms of timesmoothed velocity components empirical approach behavior of turbulent flow cannot be predicted a priori BSL 5.4: Empirical expressions for the turbulent momentum flux BSL 5.4: Empirical expressions for the turbulent momentum flux Solving turbulent flow problems specification of turbulent stress model is needed empirical information necessary Boussinesq-model: 4 empirical models in which μ(t) represent the so-called turbulent viscosity (or eddy viscosity) (in contrast with μ, this is not at property of the fluid) τ yx(t ) = − µ (t ) •Boussinesq-model •Prandtl-model •Von Karman-model •Deissler-model Qualitative: μ(t) is large in the center of the turbulent flow and small near the solid walls BSL 5.4: Empirical expressions for the turbulent momentum flux τ yx dvx dvx = − ρl dy dy 2 BSL 5.4: Empirical expressions for the turbulent momentum flux Von Karman-model: s Prandtl-model: (t ) dvx dy wall (t ) τ sz dvz dvz = − ρl = −τ rz(t ) ds ds (t ) τ yx = − ρκ r 2 R where l is the so-called mixing length for which Prandtl proposed the following expressions: Wall turbulence: l = κ1 y (y =distance from wall) Free turbulence: l = κ 2b (b =width of mixing zone) The model assumes that the behavior of eddies is similar to that of molecules in a low-density gas (l ↔ λ) r=0 2 2 3 ( dvx dy ) dvx (d 2 2 vx dy 2 ) dy where κ2 is a “universal” constant (κ2≈0.36-0.40) Deissler-model: n 2v z s n 2v x y − − dv ν dvz ν x τ sz(t ) = − ρ n 2 vz s 1 − e − ρ n 2 vx y 1 − e τ yx(t ) = ds dy where n is an empirical constant (Deissler: n=0.124) In contrast to Prandt-model and Von Karman-model, the Deissler model is applicable in the area near the wall BSL 5.4: Velocity distributions in turbulent pipe flow Derivation of the radial velocity profile for turbulent in a circular tube BSL 5.4: Velocity distributions in turbulent pipe flow ∂vz ∂v v ∂v ∂v + vr z + θ z + vz z = ∂r r ∂θ ∂z ∂t ρ •far from wall region (turbulent transport dominates) z-momentum balance 1 ∂ ∂p 1 ∂ ∂ (rτ rz ) − (τ θ z ) − (τ zz ) + ρ g z − r ∂θ ∂z r ∂r ∂z − Distinguish two different areas: BSL §B.5 apply correspondence rules stress components are now •near wall region (laminar transport dominates) sum of laminar and turbulent ∂v ∂v v ∂v ∂v ρ z + vr z + θ z + vz z = terms due to time-smoothing ∂r r ∂θ ∂z ∂t = τ ij τ ij( l ) + τ ij( t ) − BSL 5.4: Velocity distributions in turbulent pipe flow • • • • • 1 ∂ ∂p 1 ∂ ∂ (τ rz ) − (τ θ z ) − (τ zz ) + ρ g z − r ∂θ ∂z r ∂r ∂z “steady state” time-smoothed flow rotation symmetry in time-smoothed flow fully developed flow in z-direction pressure gradient driven pipe flow no gravity acting in the z-direction simplification p − pL 1 d ∂p 1 ∂ 0= (τ rz ) ⇒ 0 =0 (rτ rz ) − − − L r dr ∂z r ∂r reduced z-momentum equation BSL 5.4: Velocity distributions in turbulent pipe flow velocity profile “far from wall” Time-smoothed equation of motion for the steady flow in a circular tube in separated form: ∂v ∂v v ∂v ∂v ρ z + vr z + θ z + vz z = ∂r r ∂θ ∂z ∂t − 1 ∂ ∂p 1 ∂ ∂ (τ rz ) − (τ θ z ) − (τ zz ) + ρ g z − ∂z r ∂r r ∂θ ∂z d (rτ rz ) = ( p0 − pL ) rdr L = rτ rz ( p0 − pL ) 1 r 2 + K L 2 1 τ rz(l ) + τ rz(t ) with: τ= rz τ rz ≈ τ rz(t ) Far from wall: turbulent transport dominates for which we take the Prandtl mixing length model: (t ) τ rz dv = ρκ s z ds 2 2 1 2 s= R − r (distance from the wall) BSL 5.4: Velocity distributions in turbulent pipe flow Integration of the equation of motion with boundary condition τrz=0 at r=0 gives: r s ( p0 − pL ) R= = τ rz τ 0 1 − 2L R R where τ0 is the shear wall stress at the tube wall (s=0): ( p0 − pL )π R 2 = τ 0 2π RL integral force balance z-direction Center of the turbulent flow turbulent momentum flux dominates so: 2 s 2 2 dv z ρκ1 s = τ 0 1 − R ds Prandtl’s assumption: s τ 0 1 − ≈ τ 0 R BSL 5.4: Velocity distributions in turbulent pipe flow Calibration of turbulence-model: Experimentally determined radial velocity profile fits the model best if: κ1 0.36 = and: s1+ 26 for= which: v1+ 12.85 This gives the so-called logarithmic velocity distribution which for Re>2·104 describes the measured velocity profiles for turbulent flow reasonably: + v= 1 ln ( s + ) + 3.8 s + ≥ 26 0.36 BSL 5.4: Velocity distributions in turbulent pipe flow Using this approximation gives: τ 1 dvz dvz 1 v* = ± 0 = >0 because : ds ds ρ κ1 s κ1 s Integration from s=s1 (end of the buffer layer seen from the wall) to an arbitrary position s: s = vz − vz ,1 ln s ≥ s1 κ1 s1 v* turbulent core s or in dimensionless variables: = v + − v1+ buffer zone wall viscous s=s1 sub-layer s+ ρ v* s vz = ln + s + ≥ s1+ = with: s + and: v + κ1 s1 µ v* 1 BSL 5.4: Velocity distributions in turbulent pipe flow velocity profile “near wall” Superposition of the laminar and turbulent momentum flux: dv dv τ rz = τ rz( l ) + τ rz( t ) = τ rz( l ) − τ sz( t ) = − µ z − τ sz( t ) = µ z − τ sz( t ) dr ds dv dv = µ z + ρ n 2 vz s (1 − exp {− n 2 vz s ν } ) z ds ds combined with: ( p0 − pL ) R r = τ 1 − s ≈ τ τ rz = 0 0 2L R R wall gives with s=R-r: (1) τ0 = µ buffer zone dvz dv + ρ n 2 vz s (1 − exp {− n 2 vz s ν } ) z ds ds turbulent core s viscous s=s1 sub-layer BSL 5.4: Velocity distributions in turbulent pipe flow Integration from the tube wall (s=0) to an arbitrary value of s gives in terms of the already mentioned dimensionless distance s+ and dimensionless velocity v+: s+ ds + + 0 ≤ s + ≤ 26 v (2) ∫ 2 + + 2 + + 0 1 + n v s (1 − exp {− n v s } ) with n=0.124 for long smooth tubes. Note that we have found an implicit equation for v+ For small values of s+, the implicit equation reduces into: + v= s + 0 ≤ s+ ≤ 5 BSL 5.4: Empirical expressions for the turbulent momentum flux Graphical representation of the velocity profiles of the two examples (1) (2) (3) (3) which follows from integration of Newton’s law of viscosity over the viscous sublayer BSL 7: Macroscopic balances for isothermal flow systems Advanced Transport Phenomena BSL 7: Macroscopic balances for isothermal flow systems J.A.M. Kuipers BSL 3: micro balances for conservation of mass, momentum and mechanical energy BSL 7: macro balances for conservation of mass, momentum and mechanical energy Derivation of macro balances from micro balances: • integration over an arbitrary macroscopic system volume • macroscopic system with one inflow and one outflow integral theorems required as tools BSL 7: Macroscopic balances for isothermal flow systems BSL 7: Macroscopic balances for isothermal flow systems Integral Theorems Integral Theorems f Volume V enclosed by surface S n S unit outward normal on S f ⋅n V vs Volume V(t) enclosed by surface S(t) ρ = ρ ( x, y , z , t ) n S (t ) V (t ) f = f ( x, y , z , t ) dV ∫∫ ( f ⋅ n )dS ∫∫∫ (∇⋅ f )= V d ρdV= dt V∫∫∫ (t ) S Gauss divergence theorem Gottfried Wilhelm Leibnitz Carl Friedrich Gauss BSL 7: Macroscopic balances for isothermal flow systems Averaging A2 r θ a x b A= ∫ f ( x)dx =< f > (b − a) a extension to 2D: A: area below the curve defined by f(x) < f ∫∫ f ( x, y)dxdy = ∫∫ f (r ,θ )rdrdθ = ∫∫ fdS = S1 S1 Leibnitz Theorem (1) ∂ρ + (∇ ⋅ ρv ) = 0 ∂t (2) ∂ ( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g = 0 ∂t b x ∂ρ dV + ∫∫ ρ (vs ⋅ n )dS ∫∫∫ ∂ t V (t ) S (t ) BSL 7: Macroscopic balances for isothermal flow systems average value of f ( x) on interval [a,b] ≡< f > A1 y surface expands with local velocity v s micro balances for mass (1), momentum (2) and mechanical energy (3) A1 = A2 f ( x) vs ⋅ n unit outward normal on S(t) S1 > S1 (3) kg / (m3 s ) N / m3 ∂ 1 2 2 1 ( 2 ρ v ) = − ( ∇ ⋅ 2 ρ v v ) − ( ∇ ⋅ pv ) − p ( − ( ∇ ⋅ v ) ) ∂t − ( ∇ ⋅ (τ ⋅ v ) ) − ( −τ : ∇v ) + ρ ( v ⋅ g ) W / m3 BSL 7.1: Macroscopic mass balance Conservation law for mass on micro scale: ∂ρ + (∇ ⋅ ρv ) = 0 ∂t Integration of the micro balance for mass: ∫∫∫ V { • Convection term: (using Gauss’s divergence theorem): ∫∫∫ V } ∂ρ + ( ∇ ⋅ ρ v ) dV = 0 ∂t d ∂ρ dV = V ( t ) ∂t dt { ∫∫∫ V (t ) } ρ dV − assume constant volume ∫∫ S (t ) n )dS ρ (vs ⋅ = d ( mtot ) dt BSL 7: Macroscopic balances for isothermal flow systems surface integral evaluation for mass convection S ∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS = Sw S1 S2 S2 S1 ∫∫ (ρ v ⋅ n )dS =ρ ∫∫ −vdS =−ρ 1 S1 1 +ρ ρ ∫∫ +vdS = ∫∫ (ρ v ⋅ n )dS = S2 < v1 > S1 S1 2 S2 ∆ = "2"− "1" BSL 7.2: Macroscopic momentum balance Conservation law of momentum on micro scale: ∂ ( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g = 0 ∂t Integration of the microscopic momentum balance: ∂ ∫∫∫ ( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g dV = 0 V ∂t only valid Working out the terms: for constant volume • Accumulation term: } { ∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS S1 d = ( mtot ) ρ1 v1 S1 − ρ 2 v2 S2 dt d ( mtot ) = w1 − w2 = −( w2 − w1 ) = −∆w dt with mtot the total mass in the system at time t ∫∫ (ρ v ⋅ n )dS = ( ∇ ⋅ ρ v )dV = ∫∫S ( ρ v ⋅ n )dS = − ρ1 v1 S1 + ρ 2 v2 S2 Combination of the results gives the final expression for the macroscopic mass balance: Working out the terms: • Accumulation term (Leibnitz theorem): ∫∫∫ BSL 7.1: Macroscopic mass balance 2 < v2 > S 2 n S2 v ∫∫∫ V v n { } ∂ d ( ρ v= ) dV ∂t dt { ∫∫∫ ρ vdV V }− ∫∫ S (t ) ρ v (vs ⋅= n )dS d ( Ptot ) dt where Ptot is the total amount of momentum in the system at time t BSL 7.2: Macroscopic momentum balance • Convection term (using Gauss’s divergence theorem): ∫∫∫ V ( ∇ ⋅ ρ vv= )dV ∫∫ S ( ρ vv ⋅ n )dS BSL 7: Macroscopic balances for isothermal flow systems surface integral evaluation for momentum convection ∫∫ (ρ vv ⋅ n )dS= ∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS= S = − ρ1 v12 S1 + ρ 2 v22 S 2 ∫∫ ( ρ v ) vn dS = S where S1 and S 2 are vectorial quantities that are defined as follows: Sw S1 S2 ∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS S1 S2 S1 ∫∫ (ρ vv ⋅ n )dS =ρ e ∫∫ −v dS =−ρ 2 1 1 • Magnitude: = S1 S= S2 1 and S 2 S1 • Direction: corresponds with the direction of the flow at S1 and S2 respectively 1 ρ 2 e2 ∫∫ +v 2 dS = + ρ 2 < v22 > S 2 ∫∫ (ρ vv ⋅ n )dS = n S2 v S2 e1 and e2 : unit vectors in direction of the flow BSL 7.2: Macroscopic momentum balance • Pressure force (using Gauss’s divergence theorem): ∫∫∫ V − p1S1 + p2 S 2 + Fp ∫∫ p ndS = {∇p}dV =⋅ S • Viscous friction force: ∫∫∫ V ( ∇ ⋅τ )dV = ∫∫S (τ ⋅ n )dS = + Fv • Gravitational force: ∫∫∫ V ρ g )dV g= mtot g ∫∫∫ ρ dV (= V pressure force exerted by fluid on walls between S1 and S2 viscous force exerted by fluid on walls between S1 and S2 e1 < v > S1 S1 S2 v 2 1 e2 n BSL 7.2: Macroscopic momentum balance • Substitution of the results gives the macroscopic momentum balance in vector notation: d = ( Ptot ) ρ1 v12 S1 − ρ 2 v22 S2 + p1S1 − p2 S2 − F + mtot g dt F = Fp + Fv or: d ( P= tot ) dt v12 v1 w1 − v22 v2 w2 + p1S1 − p2 S 2 − F + mtot g where: w1 = ρ1 v1 S1 w2 = ρ 2 v2 S 2 vectorial mass flow rates (kg/s) ! BSL 7.2: Macroscopic momentum balance Notation of the macroscopic momentum balance in terms of Δ: v2 d ∆ = "2"− "1" ( Ptot ) = −∆ w + pS − F + mtot g dt v For steady systems: v2 F = −∆ w + pS + mtot g v This equation is useful for the calculation of forces exerted on walls (structures) by fluids Approximation for turbulent flowing fluids in comparison with “flat” velocity profile: v2 v 2 ∆ ≈ ∆ =∆ v v v BSL 7.4: Macroscopic mechanical energy balance (Bernoulli’s law) W is amount of work exerted by the fluid on the surroundings per unit of time EV is the amount of mechanical energy that is converted (dissipated) into thermal energy per unit of time: EV = ∫∫∫ V ( −τ : ∇v )dV Reduction of the general macroscopic energy balance: Steady system d ( Ktot + Φ tot + Atot ) = 0 dt from the mass balance): w= w= w 1 2 BSL 7.4: Macroscopic mechanical energy balance (Bernoulli’s law) Conservation law for mechanical energy on micro scale: ∂ 1 2 2 1 ( 2 ρ v ) = − ( ∇ ⋅ 2 ρ v v ) − ( ∇ ⋅ pv ) − p ( − ( ∇ ⋅ v ) ) ∂t − ( ∇ ⋅ (τ ⋅ v ) ) − ( −τ : ∇v ) + ρ ( v ⋅ g ) Integration over macroscopic volume of flow system gives: 1 v 3 d ˆ + Gˆ w − W − EV +Φ ( Ktot + Φ tot + Atot ) = −∆ dt 2 v With Ktot, Φ tot and Atot the total kinetic, potential and (Helmholtz) free energy respectively of the system: ˆ ˆ dV Atot = ∫∫∫ ρ AdV K tot = ∫∫∫ 12 ρ v 2 dV Φ tot = ∫∫∫ ρΦ V V V ˆ , Aˆ and Gˆ represent respectively potential energy, free energy and Φ free enthalpy per unit of mass BSL 7.4: Macroscopic mechanical energy balance (Bernoulli’s law) • Isothermal system p2 dp p1 ρ ∆G = ∫ • Define: EV W and: EˆV = Wˆ = w w Using these assumptions/definitions, the final form of the steady macroscopic mechanical energy balance is: p2 1 v3 ˆ + ∫ dp + Wˆ + EˆV = 0 Bernoulli’s equation ∆ + ∆Φ p1 ρ 2 v BSL 7: Macroscopic balances for isothermal flow systems Summary of macroscopic balances in ∆ formulation • Macroscopic mass balance: d ( mtot ) = −∆w dt • Macroscopic momentum balance: BSL 7: Applications Applications of macroscopic balances • Pressure rise and friction loss in a sudden enlargement • Performance of a liquid-liquid ejector v2 d ( Ptot ) = −∆ w + pS − F + mtot g dt v • Thrust on a pipe-bend • Macroscopic mechanical energy balance: 1 v 3 d ˆ + Gˆ w − W − EV +Φ ( Ktot + Φ tot + Atot ) = −∆ dt 2 v BSL 7: Applications BSL 7: Applications Pressure rise and friction loss in a sudden enlargement Macroscopic mass balance: = w1 w= ρ 2 v2 S2 2 or: ρ1 v1 S1 for an incompressible fluid (β=S1/S2): v1 1 = v2 β Macroscopic momentum balance for the main flow direction: Medium: fluid in turbulent flow Question: find an expression for the pressure rise between “1” and “2” and the friction loss 0 = v1w1 − v2 w2 + p1S1 − p2 S 2 − F Components of the force F: • Viscous force on the cylindrical surface negligible • Pressure force acting on (S2-S1): (S2-S1)p1 21 BSL 7: Applications Substitution of result in the momentum balance gives: p2 − = p1 ρ v2 ( v1 − v2 ) Combining the momentum balance with the mass balance: 1 p2 −= p1 ρ v22 − 1 β since β<1 p2>p1 BSL 7: Applications Eliminating the pressure difference using the momentum balance: 2 1 1 2 1 v2 ˆ EV = − − 1 ρ v2 + 2 − v22 β ρβ 2 1 2 1 2 1 1 1 2 = v2 1 − + 2 − 1 = 2 v2 − 1 β β β Mechanical energy balance: 1 2 (v 2 2 − v12 ) + 1 ρ 0 ( p2 − p1 ) + EˆV = 23 BSL 7: Applications Performance of a liquid-liquid ejector Medium: fluid in turbulent flow Total area in plane “1” and “2” equals S (constant) v= 0 and: S1 1 3 ρ 2 v2 S2 Macroscopic mass= balance: w1 w= 2 or: ρ1 v1 S1 1 2 1 2 for an incompressible fluid: v1 = v2 = 3 v0 + 3 ( 2 v0 ) = 3 v0 Macroscopic momentum balance: v2 General form: F = −∆ w + pS + mtot g v Working out the general form: v22 v12 0= w2 + p2 S 2 − w1 + p1S1 v2 v1 Assumption: force F acting on the system wall is negligible Central stream at “1”: v1 BSL 7: Applications S Annular stream at “1”: v0 and: S1 23 S v1 = 2 24 25 BSL 7: Applications Evaluation of average quantities: v22 = v2 and: v12 = v1 v= 2 2 3 v0 v + 23 ( ) = 1 2 v0 v + 3 0 3( 2 ) 1 3 + 16 v0 = 2 1 3 3 4 v0 3 Using this result in the momentum balance gives: 2 1 p2 − p1 = 18 ρ v0 Pressure rise occurs due to the mixing of the two fluids: kinetic energy decreases from “1” to “2” 26 Substitution in macroscopic mechanical energy balance: 1 0 ( 92 − 165 ) v02 + ( 181 ρ v02 ) + EˆV = ρ EˆV = ( 165 − 92 − 181 ) v22 = 5 144 v22 Results are valid for liquid-liquid ejectors Gas-gas ejectors: ρ varies considerably and the overallenergy balance and an equation of state must be used 28 v3 v3 1 Application of general form: 12 2 − 12 1 + [ p2 − p1 ] + EˆV = 0 v1 ρ v2 v23 Evaluation of average quantities:= v2 1 2 v13 1 and:= 2 v1 v + 32 ( v20 ) = v + 32 ( v20 ) 3 1 3 1 3 0 2 1 3 0 + 2 = v0 1 1 12 1 3 2 2 3 5 16 1 2 2 v2 ) 12= ( 23 v0 ) (= 2 2 9 v02 v02 27 BSL 7: Applications or: Macroscopic mechanical energy balance: p2 dp 1 v3 ˆ + ∆Φ + + Wˆ + EˆV = 0 General form: ∆ 2 ∫ p1 ρ v v0 2 2 2 0 BSL 7: Applications BSL 7: Applications Thrust on a pipe bend Water at 95 °C is flowing at a rate of 2.0 ft3/s through a 60° bend, in which there is contraction from 4 inch to 3 inch internal diameter Force exerted on bend if downstream pressure is 1.1 atm ? ρwater=0.962 g/cm3 μwater=0.299 cp Conversion factors: 1ft3 = 28.3·10-3 m3 1 inch = 0.0254 m ft = 0.305 m 1 cp = 10-3 kg/(m·s) 29 BSL 7: Applications Calculation of Re: ρ v D 4Q ρ 4 ⋅ ( 2.0 ⋅ 28.3 ⋅10 ) ⋅ 962 = = = = Re 3.04 ⋅106 −3 µ π Dµ π ⋅ 0.0762 ⋅ 0.299 ⋅10 −3 assumption of flat velocity profiles OK (turbulent flow) Macroscopic mass balance: v S w1 =w2 → 1 = 2 =β ( β < 1) v2 S1 Macroscopic momentum balance: v2 F = −∆ w + pS + mtot g v BSL 7: Applications Working out the momentum balance for the x- and ycomponent respectively: v12 v22 Fx= w1x + p1S1x − w2 x + p2 S 2 x v1 v2 = v1 ( ρ v1S1 ) + p1S1 − v2 ( ρ v2 S 2 cos (θ ) ) − p2 S 2 cos (θ ) = ρ v22 S 2 ( β − cos (θ ) ) + ( p1 − p2 ) S1 + p2 ( S1 − S 2 cos (θ ) ) and: v12 v22 Fy= w1 y + p1S1 y − w2 y + p2 S 2 y − mtot g v1 v2 = −v2 ( ρ v2 S 2 sin (θ ) ) − p2 S 2 sin (θ ) − mtot g = − ρ v22 S 2 ( sin (θ ) ) − p2 S 2 sin (θ ) − mtot g 30 BSL 7: Applications Mechanical energy balance: 1 v23 1 v13 1 0 −2 2 + [ p2 − p1 ] + EˆV = v v ρ 1 2 1 2 v − 12 v + 2 1 1 ρ Inserting this into the mechanical energy balance: 1 p1 − = p2 ρ v22 ( 12 − 12 β 2 + = ρ v22 ( 107 − 12 β 2 ) 5) Inserting this into the x-momentum balance gives: Fx = ρ v22 S 2 ( β − cos (θ ) ) + ρ v22 ( S 2 β ) ( 107 − 12 β 2 ) + p2 S 2 (1 β − cos (θ ) ) Simplification: 2 2 BSL 7: Applications [ p2 − p1 ] + EˆV = 0 Estimation of ÊV using friction charts or tables and the definition for ÊV : 2 1 2 2 1 2 Eˆ v 12 ( v ) ev gives: Eˆ= 5 v2 v ≈ 2 v2 ( 5 ) Summary: x-momentum balance: = Fx ρ ( Q 2 S 2 ) {107 β −1 − cos (θ ) + 12 β } + p2 S 2 ( β −1 − cos (θ ) ) y-momentum balance: Fy = − ρ ( Q 2 S 2 ) sin (θ ) − p2 S 2 sin (θ ) − π R 2 L ρ g where R and L are the radius and length of a cylinder with the same volume as the chosen control volume 33 BSL 7: Applications Now, the magnitude and direction of the force F can be derived from the given equations: •Magnitude: = F Fx2 + Fy2 BSL 9 & 10: Thermal conductivity and the mechanism of energy transport & shell energy balances and temperature distributions in solids and laminar flow J.A.M. Kuipers •Direction: α Advanced Transport Phenomena arctan ( Fx − Fy ) where alpha is the angle that the force makes with the vertical BSL 10: Shell heat balance + BC’s BSL 9.1: Fourier’s law of heat conduction Development of steady-state temperature profile for a solid slab exposed to temperature difference d ( ρ C pT ) dT k d ( ρ C pT ) qy = −k = − = −α dy ρC p dy dy Fourier’s law of heat conduction Heat balance in words: rate of heat accumulation in system net heat heat flow heat flow = − + production rate in rate out in system boundary conditions required + Prescribed temperature (Dirichlet boundary condition) T = Tw Jean-Baptiste Joseph Fourier + Prescribed heat flux (Neumann boundary condition) isotropic material ∂T ∂T ∂T ∂T ∂T ∂T , ky , kz q = ( qx , q y , qz ) = − k x = − K ⋅∇T = −k , , = −k ∇T ∂y ∂z ∂x ∂x ∂y ∂z Anisotropic materials: thermal conductivity is different in x-, y- and z-direction. Note: anisotropic thermal conductivity also occurs in disperse systems −k ∂T ∂n w = qw + Mixed boundary condition (Robin boundary condition) −k ∂T ∂n w =h(T∞ − Tw ) BSL 10.2: Heat conduction with an electrical heat source System: copper wire with radius R (cross sectional area S), length L and electrical conductivity ke=(1/ρe) [Ω-1cm-1] through which an electrical current with current density J [A/cm2] is fed Qualitative: electrical energy is converted into heat a radial temperature profile is present inside the wire: T=T(r) Amount of heat produced per unit of volume and time: I 2 Re ( JS ) 2 S L J2 2 ρ Se = Re J 2 ρ= J = = = (W / m3 ) e e SL SL L S ke Re: electrical resistance of wire in Ω, ρe is the specific electrical resistance (material property) Assumptions: • thermal conductivity k is independent of T and the surface temperature (at r=R) of the wire is constant and equals T0 • there is no heat transport (conduction) in axial direction BSL 10.2: Heat conduction with an electrical heat source Inserting this into the heat balance and dividing by dV gives: 0 ( rqr ) r − ( rqr ) r +∆r + Se r ∆r Take the limit as Δr → 0: 1d 0= − ( rqr ) + Se r dr Boundary conditions: rqr Integration:= 1 2 = at r 0= : qr 0 = at r R= : T T0 Se r + C1 2 BSL 10.2: Heat conduction with an electrical heat source Description of the temperature profile ? set up a differential heat balance over a differential volume element dV=2πrLΔr Net conductive radial heat transport: ( 2π rL ⋅ qr ) r − ( 2π rL ⋅ qr ) r +∆r Heat production in dV: ( 2π rL∆r ) Se BSL 10.2: Heat conduction with an electrical heat source Based on the boundary conditions at r=0, it follows that C1=0. Inserting Fourier’s law gives: −k dT Se r = 2 dr Integration using the boundary condition at r=R gives: 2 Se R 2 r T −= T0 1 − 4k R Note that the temperature profile is parabolic (analogous to laminar tube flow): correspondence rules are given in BSL BSL 10.4: Heat conduction with a viscous heat source System: steady incompressible flow between two concentric cylinders Qualitative: due to friction between the rotating fluid layers, mechanical energy will be converted into internal energy (ΔT) (viscous dissipation) BSL 10.4: Heat conduction with a viscous heat source Sv: amount of heat produced per unit of volume and time due to viscous dissipation: Sv=Sv(μ,(dvz/dx)): dv Sv =− ( τ : ∇v ) =µ z dx 2 check with BSL §B.7 ! Velocity profile in idealized system: x vz = V b Idealization: neglect the curvature (flat plate approach) V Sv = µ b Differential heat balance over a volume element dV=A·dx gives: 2 V 0= qx x A − qx x + dx A + µ Adx b Dividing by A·dx en taking the limit as dx → 0 gives after combination with Fourier’s law: 2 d d dT V − k µ ( qx ) = = dx dx dx b = − kT 2 1 2 V µ x 2 + C1 x + C2 b 0=µ d 2 vz dx 2 Combining these two relations: BSL 10.4: Heat conduction with a viscous heat source Integration gives: follows from reduced z-momentum equation 2 = vz K1 x + K 2 = x 0= : vz 0 = x b= : vz V BSL 10.4: Heat conduction with a viscous heat source Integration constants follow from the boundary conditions: at x 0= : T T0 and= at x b= : T Tb = Solution: T − T0 x 1 x x = + 2 Br 1 − Tb − T0 b b b where Br is the dimensionless Brinkman number: = Br µV 2 internal rate of heat production = k (Tb − T0 ) heat flow by conduction In general: viscous heating is not important, exceptions are encountered in system with extremely large velocity gradients: •Bearings •Extruders (rheology) •High velocity flow around objects BSL 10.7: Heat conduction in a cooling fin BSL 10.7: Heat conduction in a cooling fin Aim of applying of cooling fins: enlarging the heat transfer area Differential heat balance over control volume (2BW)dz: Problem: efficiency is not 100% due to heat transfer limitation in the cooling fin Analogy: effectiveness of a porous catalyst Dividing by 2BWdz en taking the limit as dz→0 gives: dq h − z = (T − Ta ) dz B Combination with Fourier’s law gives: d 2T h = (T − Ta ) dz 2 kB Boundary conditions: = at z 0= : T Tw qz z 2 BW − qz Actual situation Model representation T is a function of x,y and z but the dependence on z is dominant T is function of z alone Some heat is lost from the fin at the end (area 2BW) and at the edges (2BL+2BL) No heat is lost from the end or from the edges The heat transfer coefficient is a function of position External heat flux is given by q=h(T-Ta), where h is constant and T depends on z BSL 10.7: Heat conduction in a cooling fin Make the equations dimensionless using: T − Ta • Dimensionless temperature: Θ = Tw − Ta z • Dimensionless distance: ξ = L hL2 • Dimensionless heat transfer coefficient: N = kB The differential equation then takes the form: d 2Θ N 2Θ = 2 dξ see: matfys.doc (canvas) Θ =e mξ try m = ±N = Θ K1e Nξ + K 2 e − Nξ z + dz 2 BW − h (T − Ta ) 2Wdz = 0 dT = at z L= : 0 dz BSL 10.7: Heat conduction in a cooling fin Dimensionless boundary conditions: e x + e− x at ξ= 0 : Θ = 1 cosh( x) = 2 dΘ x at ξ 1:= 0 = e − e− x sinh( x) = dξ 2 Solution: sinh( x) cosh ( N (1 − ξ ) ) tanh( x) = Θ= cosh( x) cosh ( N ) The relation for the “effectiveness” of the cooling fin follows from the (dimensionless) temperature Definition of η: η= actual rate of heat loss from the fin rate of heat loss from an isothermal fin at Tw BSL 10.7: Heat conduction in a cooling fin BSL 10.8: Forced convection or: W L η ∫ ∫h (T − Ta ) dzdy = (Tw − Ta ) dzdy 0 o W L ∫ ∫h 0 o 1 ∫ Θd ξ 0 1 ∫ dξ 0 Inserting the expression for Θ gives after integration: = η { } tanh ( N ) 1 1 1 − sinh ( N (1 − ξ ) )= 0 cosh ( N ) N N Qualitative comparison between forced and free convection Application: error in thermocouple measurement due to parasitic heat conduction BSL 10.8: Forced convection Example: steady laminar flow of a viscous fluid in a circular tube with radius R Assumption: constant physical properties (ρ, μ, k and Cp) • z<0: uniform fluid temperature T0 • z>0: constant radial heat flux q1 imposed at the wall Radial velocity profile: r 2 = vz vz ,max 1 − R with: p0 − pL ) R 2 ( vz ,max = 4µ L BSL 10.8: Forced convection Analysis of heat transport in developed laminar flow Qualitative: T=T(r,z) set up a differential energy balance over an annular ring with volume dV=2πrΔrΔz Thermal energy balance in words: net convective net conductive accumulation = inflow of heat + inflow of heat of heat in dV in z -direction in r- and z-direction BSL 10.8: Forced convection BSL 10.8: Forced convection Net heat conduction in axial direction: Net convective inflow of energy: {ρ h z − ρ h z +∆z } vz 2π r ∆r where h is the enthalpy per unit of mass: T h= h0 + ∫ C p dT T0 Net heat conduction in radial direction: ( 2π r ∆zqr ) r − ( 2π r ∆zqr ) r +∆r BSL 10.8: Forced convection Inserting Fourier’s law and the expression for the radial velocity profile gives: r 2 ∂T 1 ∂ ∂T ∂ 2T ρ C p vz ,max 1 − = k r + 2 r ∂r ∂r ∂z R ∂z where k is the (constant) thermal conductivity Reduction of micro balance: axial heat transport is dominated by convection: r 2 ∂T 1 ∂ ∂T ρ C p vz ,max 1 − = k r R ∂ z r ∂ r ∂r ( 2π r ∆rqz ) z − ( 2π r ∆rqz ) z +∆z Combination with the energy balance in words gives after dividing by dV=2πrΔrΔz : ρ vz 0= h z − h z +∆z ( rqr ) r − ( rqr ) r +∆r qz z − qz + + ∆z ∆z r ∆r z +∆z Taking the limit as Δr→0 and Δz→ 0 gives: ρ vz ∂q ∂h ∂T 1∂ = ρ C p vz = − ( rqr ) − z ∂z ∂z r ∂r ∂z BSL 10.8: Forced convection Boundary conditions: r = 0: for all z: T =finite ∂T = −q1 r= R: for all z: − k ∂r = z 0:= for all r: T T0 Introducing dimensionless quantities: • Dimensionless temperature: T − T0 Θ= q1 R k r • Dimensionless radial coordinate: ξ = R • Dimensionless axial coordinate: λ = zk ρ C p vz ,max R 2 BSL 10.8: Forced convection • Dimensionless energy balance: 1 ∂ ∂Θ = ξ (1 − ξ 2 ) ∂Θ ∂λ ξ ∂ξ ∂ξ • Dimensionless boundary conditions: ξ 0 : for all λ: Θ=finite ∂Θ ξ 1:= for all λ: 1 ∂ξ λ 0 : for all ξ : Θ=0 Solution: R. Siegel et al., Applied Scientific Research, A7, 386-392 (1958) using separation of independent variables: ˆ (ξ , λ ) G ' + µ 2G = Θ(ξ , λ ) = Θ∞ (ξ , λ ) − Θ 0 ˆ (ξ , λ ) = (ξ F ' )' + µ 2ξ (1 − ξ 2 ) F = 0 Θ F (ξ )G (λ ) BSL 10.8: Forced convection Asymptotic solution for large λ (thermally developed flow): Θ = Θ (ξ , λ ) = C0 λ + Ψ (ξ ) where C0 is a yet to be determined constant Physical motivation: for large λ, Θ increases linearly with λ and the shape of the dimensionless (radial) temperature profiles do not change with increasing λ (self-similarity) Problem: boundary condition at tube inlet can not be met replace it therefore with an integral energy balance: = 2π Rzq1 R ∫ 0 ρ C p (T − T0 ) vz ( r ) 2π rdr Sturm-Liouville problem for F: see matfys.doc (canvas) BSL 10.8: Forced convection or in dimensionless form: λ= ∫ Θ (ξ , λ ) (1 − ξ ) ξ d ξ 1 2 0 Inserting the asymptotic solution into the dimensionless energy balance: 1 d dΨ C0 (1 − ξ 2 ) ξ = ξ dξ dξ Integration gives: dΨ ξ = C0 ( 12 ξ 2 − 14 ξ 4 ) + C1 dξ Based on the “natural boundary condition” (Θ and therefore Ψ must be finite) C1=0 BSL 10.8: Forced convection Further integration: = Ψ C0 ( 14 ξ 2 − 161 ξ 4 ) + C2 Total asymptotic solution: = Θ C0 λ + C0 ( 14 ξ 2 − 161 ξ 4 ) + C2 From the boundary condition at the tube wall, it follows that C0=4 and from the integral energy balance, it follows that C2=-7/24 Expression for the dimensionless temperature profile: Θ = 4λ + ξ 2 − 14 ξ 4 − 247 BSL 10.8: Forced convection BSL 10.9: Free convection Using the expression for the temperature profile, all important secondary quantities can be derived: • Radial arithmetic average temperature: 2π R T = ∫ ∫T 0 0 ( r , z ) rdrdθ 2π R ∫ ∫ rdrdθ collect fluid in stirred vessel and measure temperature 0 0 • Flow-averaged or “cup-mixing” temperature: vzT = vz 2π R ∫ ∫ vz 0 0 ( r ) T ( r , z ) rdrdθ 2π R ∫ ∫ vz 0 0 ( r ) rdrdθ vz ( r ) T (r , z ) ∞ Fluid with (variable) density ρ and dynamic viscosity μ is located between two parallel plates a distance 2B apart Left plate: T=T2 Right plate: T=T1 Qualitative: fluid rises near left (hot) wall en descends near the (cool) right wall: however no net convection in de z-direction BSL 10.9: Free convection Thermal energy balance and reduced form: ∂T ∂ 2T ∂ 2T ρ C p vz ( = y) k 2 + 2 ∂z ∂z temperature profile ∂y is fully developed Boundary conditions: = y -b= : T T2 T2 = − K1b + K 2 + K1b + K 2 y= +b : T = T1 T1 = Solution of the thermal energy balance: 1 y T = Tm − ∆T 2 b with: ∆T = (T2 − T1 ) and Tm = T2 + T1 2 BSL 10.9: Free convection d 2T k 2 =0 dy T K1 y + K 2 = density variations are rather small ∂vx ∂v y ∂vz ∂ρ ∂ ∂ ∂ + ( ρ vx ) + ( ρ v y ) + ( ρ vz ) = 0 ⇒ + + =0 ∂t ∂x ∂y ∂z ∂x ∂y ∂z z-momentum balance with “constant” physical properties ∂v ∂v ∂v ∂p ∂vz ∂ 2v ∂ 2v ∂ 2v + vx z + v y z + vz z = − + µ 2z + 2z + 2z + ρ g z ∂x ∂y ∂z ∂z ∂y ∂z ∂t ∂x ρ • • • • • steady state flow only flow in z-direction no x-dependence (infinite size in x-direction) flow in z-direction is fully developed gravity acting in negative z-direction reduced z-momentum equation BSL 10.9: Free convection BSL 10.9: Free convection Reduced momentum balance: d 2 vz ∂p 0= − + µ 2 − ρg ∂z dy where ρ is the temperature dependent ρ(T) for which the following Taylor-expansion is used: ∂ρ ρ = ρ m + (T − Tm ) + = ρ m − ρ m β m (T − Tm ) + ∂T T =Tm where ρm= ρ(Tm) and β the volumetric expansion coefficient 1 ∂ρ 1 defined as: β = − ⇒ β = for ideal gas ρ ∂T p T Regarding the pressure in the z-momentum balance, we assume that the pressure is only a function of z Inserting these results into the z-momentum balance gives: µ d 2 vz dp = + ρ m g − ρ m β m (T − Tm ) g 2 dy dz Substitute the expression for the temperature profile: d 2 vz dp y 1 µ 2= + ρ m g − ρ m β m g − ∆T dy dz b 2 Boundary conditions: y= −b : vz = 0 y= +b : vz = 0 ∂p dp → ∂z dz BSL 10.9: Free convection Solution: vz A B ρ m β m gb 2 ∆T 3 b 2 dp η + + ρ m g η 2 + C1η + C2 η = y b 12 µ 2 µ dz vz = Aη 3 + Bη 2 + C1η + C2 ) vz (1=) 0 Boundary conditions: vz ( −1= 0 =− A + B − C1 + C2 0 =+ A + B + C1 + C2 vz BSL 10.9: Free convection Solution: ρ m β m gb 2 ∆T 3 b 2 dp vz η − η ) + + ρ m g (η 2 − 1) = ( 12 µ 2 µ dz Demand: net volume flow rate of the fluid equals zero: 1 ∫ C2 = − B C1 = −A ρ m β m gb 2 ∆T 3 b 2 dp 2 η − η + ( ) + ρ m g (η − 1) 12 µ 2 µ dz −1 vz (η ) dη = 0 Inserting the expression for the velocity profile gives: b 2 dp 4 dp = − ρm g + ρm g × − = 0 → 2 µ dz 3 dz BSL 10.9: Free convection Final result: vz 3 ρ m β m gb 2 ∆T y y − 12 µ b b or in terms of the dimensionless velocity φ = bvz ρ m µ and the dimensionless length η = y b Gr 3 φ = (η − η ) 12 where Gr is the dimensionless Grashof number gb3 gb3 ∆ρ Gr βm∆ T = = vm2 vm2 ρ m and: ∆ρ = ρ1 − ρ 2 BSL 11: The equations of change for nonisothermal systems BSL 10: analysis of heat transport based on simple balances temperature profiles + related quantities BSL 11: derivation of energy equations (PDE that describes transport and interconversion of different forms of energy) When are equations of energy needed? for transport in nonisothermal systems (reactive systems with heat effects) Equations of energy • Mechanical energy equation (BSL 3) • Total energy equation (BSL 11) • Thermal energy equation (BSL 11) Advanced Transport Phenomena BSL 11: The equations of change for nonisothermal systems J.A.M. Kuipers BSL 11.1: The energy equation Law of conservation of energy in words: accumulation of internal and kinetic = energy per unit of time conductive inflow + of internal energy per unit of time convective inflow of convective outflow of internal and kinetic − internal and kinetic energy per unit of time energy per unit of time conductive outflow − of internal energy per unit of time net work done by the − system on the surroundings per unit of time Dimension of all terms: (J/s=W) Note: the potential energy term is part of the work term BSL 11.1: The energy equation BSL 11.1: The energy equation ρ evx x+∆x ρ evx x Balance for a differential volume element with size ∆x, ∆y and ∆z in the x-, y- and z-direction respectively and fixed in space Accumulation of internal and kinetic energy: e ∂ ∂ of energy ∆x∆y∆z ( ρUˆ + 12 ρ v 2 ) =∆x∆y∆z ( ρ e ) accumulation per unit of volume ∂t ∂t Net convective inflow of internal and kinetic energy per unit of time: ∆y∆z vx ( ρUˆ + 12 ρ v 2 ) − vx ( ρUˆ + 12 ρ v 2 ) + { { ∆x∆z v y ( ρUˆ + 12 ρ v 2 ) − v y ( ρUˆ + 12 ρ v 2 ) y { ∆x∆y vz ( ρUˆ + 12 ρ v 2 ) − vz ( ρUˆ + 12 ρ v 2 ) z {−∇ ⋅ ( ρUˆ + BSL 11.1: The energy equation x + dx {−(∇ ⋅ q )} ∆x∆y∆z } + ∆x∆z {q y y − qy y + dy } + ∆x∆y {q z z − qz z + dz }= Net work done by the system on surroundings consists of: }= net convective transport of e per unit of volume and time { net work done against gravity per unit of volume and time } ∆y∆z {vx p x+dx − vx p x } + ∆x∆z v y p y +dy − v y p y + ∆x∆y {vz p z +dz − vz p z } = {(∇ ⋅ pv )} ∆x∆y∆z net work done against pressure forces per unit of volume and time Work done by viscous forces per unit of time: { ∆y∆z (τ xx vx + τ xy v y + τ xz vz ) x+dx − (τ xx vx + τ xy v y + τ xz vz ) x { Work done against forces acting on the control volume (body forces: gravity) Work done against forces acting on the walls of the control volume (surface forces: pressure forces + viscous forces) − ρ∆x∆y∆z {vx g x + v y g y + vz g z } =− ρ (v ⋅ g )∆x∆y∆z z + dz }+ Work done by pressure forces per unit of time: net conductive heat transport per unit of volume and time Gravitational work per unit of time: ρ v 2 )v } ∆x∆y∆z y + dy BSL 11.1: The energy equation Net contribution due to conduction per unit of time: ∆y∆z {qx x − qx 1 2 } x + dx x } ∆x∆z (τ yx vx + τ yy v y + τ yz vz ) y +dy − (τ yx vx + τ yy v y + τ yz vz ) y { ∆x∆y (τ zx vx + τ zy v y + τ zz vz ) z +dz − (τ zx vx + τ zy v y + τ zz vz ) z = {( ∇ ⋅ (τ ⋅ v ) )} ∆x∆y∆z net work done against viscous forces per unit of volume and time } } BSL 11.1: The energy equation Inserting this into the balance and taking the limit as Δx→0, Δy→0 and Δz→0 gives after dividing by ΔxΔyΔz: ∂ ∂ ∂ ∂ ∂q ∂q ∂q − ( ρ evx ) + ( ρ ev y ) + ( ρ evz ) − x + y + z ( ρe) = ∂t ∂y ∂z ∂x ∂x ∂y ∂z ∂ ∂ ∂ + ρ ( g x vx + g y v y + g z vz ) − ( pvx ) + ( pv y ) + ( pvz ) ∂y ∂z ∂x ∂ ∂ ∂ − (τ xx vx + τ xy v y + τ xz vz ) + (τ yx vx + τ yy v y + τ yz vz ) + (τ zx vx + τ zy v y + τ zz vz ) ∂y ∂z ∂x where e is the sum of internal and kinetic energy per unit of mass: e= Uˆ + 12 v 2 Alternative compact vector notation by removing ΔxΔyΔz from all terms in vector representation: ∂ ( ρ e ) = − ( ∇ ⋅ ρ ev ) − ( ∇ ⋅ q ) + ρ ( v ⋅ g ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) ) ∂t BSL 11.1: The energy equation This equation describes the change in total energy per unit of volume and time that a stationary observer would notice Using definition of substantial derivative towards time gives: ρ This equation describes the change in total energy per unit of volume and time from the perspective of an observer moving with the local fluid velocity Corresponding equation of mechanical energy (BSL 3): D ρ ( 12 v 2= ) p ( ∇ ⋅ v ) − ( ∇ ⋅ pv ) + ρ ( v ⋅ g ) − ( ∇ ⋅ (τ ⋅ v ) ) + (τ : ∇v ) Dt BSL 11.1: The energy equation Subtracting this balance of mechanical energy from the balance of total energy gives the thermal energy balance: D ρ (Uˆ ) = − ( ∇ ⋅ q ) − p ( ∇ ⋅ v ) − (τ : ∇v ) Dt The terms p ( ∇ ⋅ v ) and (τ : ∇v ) are present in both equations but with opposite signs these terms describes therefore the interconversion of mechanical and thermal energy! Term p ( ∇ ⋅ v ) can be either >0 or <0 for an incompressible fluid and therefore describes reversible interconversion of energy Term ( −τ : ∇v ) is always >0 and therefore describes an irreversible conversion of mechanical into thermal energy D ˆ 1 2 (U + 2 v ) = − ( ∇ ⋅ q ) + ρ ( v ⋅ g ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) ) Dt BSL 11.1: The energy equation Potential energy in the equation of change for total energy ˆ where Φ̂ is the External force per unit of mass g = −∇Φ potential energy per unit of mass. Using the definition of the substantial derivative towards time gives: D ˆ ∂ ˆ ˆ )= ∂ ( Φ ˆ ) − (v ⋅ g ) Φ )= Φ ) + ( v ⋅∇Φ ( ( Dt ∂t ∂t If Φ is independent of time, the equation of total energy can be rewritten as: ρ D ˆ ˆ 1 2 (U + Φ + 2 v ) = − ( ∇ ⋅ q ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) ) Dt BSL 11.1: The energy equation Modification of equation of change for thermal energy Formulation in terms of temperature and heat capacity instead of internal energy: = Uˆ Uˆ (Vˆ , T ) → ∂Uˆ ˆ ∂Uˆ ∂p dUˆ = dV + dT = − p + T dVˆ + CˆV dT ∂T Vˆ ∂Vˆ T ∂T Vˆ This gives: D DT ∂p DVˆ ρ (Uˆ ) = − p + T ρ + ρ CˆV Dt Dt ∂T Vˆ Dt Using the continuity equation gives (BSL 3): DVˆ D 1 1 Dρ ρ = ρ = − = (∇ ⋅ v ) Dt Dt ρ ρ Dt BSL 11.1: The energy equation Inserting the results gives the equation of change for thermal energy in terms of T: ∂p DT ρ CˆV = − ( ∇ ⋅ q ) − T ( ∇ ⋅ v ) − (τ : ∇v ) Dt ∂T Vˆ Simplified forms of the equation of thermal energy Newtonian fluid with constant heat conductivity: ∂p DT ρ CˆV = k ∇ 2T − T ( ∇ ⋅ v ) + µΦV Dt ∂T ρ where ΦV is the viscous dissipation function (usually neglected): ∂vx 2 ∂v y 2 ∂v y 2 ∂v y ∂vx 2 = ΦV 2 + + + + ∂x ∂y ∂y ∂x ∂y 2 ∂vz ∂v y ∂vx ∂vz 2 ∂vx ∂v y ∂vz + + + + + − + ∂y ∂z ∂z ∂x 3 ∂x ∂y ∂z 2 BSL 11.1: The energy equation Other source terms for the equation of change of thermal energy: • Chemical reactions • Nuclear reactions • Electrical phenomena 2 BSL 11.2: Special forms of the energy equation i. Ideal gas DT ρ CˆV = k ∇ 2T − p ( ∇ ⋅ v ) Dt ii. Fluid flowing in constant pressure system DT ρ Cˆ p = k ∇ 2T Dt iii. Fluid with constant density DT ρ Cˆ p = k ∇ 2T Dt iv. Stationary solid ∂T ρ Cˆ p = k ∇ 2T ∂t Energy equation Energy equation for curvilinear coordinate systems Important coordinate systems: see matfys.doc and appendix A of BSL for details of transformation to general orthogonal co-ordinates BSL 11.3: The equation of motion for forced and free convection Forced convection Flow occurs due to action of some external force such as a pressure force or a gravitational force ρ • Cartesian coordinates Dv = −∇p − ( ∇ ⋅τ ) + ρ g Dt with • Cylindrical coordinates • Spherical coordinates See BSL §B.8 and §B.9 for detailed tables BSL 11.3: The equation of motion for forced and free convection Free convection Flow occurs due to internally generated forces due to density differences (“buoyant forces”) Free convection requires a temperature dependent density Consider a medium of which the temperature varies around a certain average Tm: Pressure gradient in stagnant medium: ∇p =ρ m g {( ) 2 3 τ = − µ ( ∇v ) + ( ∇v ) − µ ( ∇ ⋅ v ) I T } Non-isothermal flow: ρ and μ must be given as function of pressure p and temperature T linking equation of motion and equation of change for thermal energy BSL 11.3: The equation of motion for forced and free convection Assumption: velocity gradients due to ΔT are small and pressure gradient is virtually constant in comparison with stagnant medium: Dv ρ = − ρ m g − ( ∇ ⋅τ ) + ρ g Dt Using ρ=ρm in the left side and ρ- ρm=-ρmβm(T-Tm) in the right side gives: Dv ρm = − ( ∇ ⋅τ ) − ρ m β m (T − Tm ) g Dt Where βm is the cubic expansion coefficient at Tm: 1 ∂ρ β m = − ρ ∂T p mean BSL 11.3: The equation of motion for forced and free convection Forced convection ↔ Free convection Forced convection: direct effect of pressure and gravitational forces BSL 11.4: Use of the equations of change to solve steady-state problems Formulation in terms of different reference frames extensive table on BSL p. 340-341 BSL Example 11.4-2: Tangential flow in an annulus with viscous heat generation Free convection: indirect effect of pressure and gravitational forces via the density Situations are limiting cases: in reality there is a smooth transition between the two transport mechanisms Assumption: physical properties ρ, μ and κ are constant BSL 11.4: Use of the equations of change to solve steady-state problems ρ CˆV DT ∂p = −(∇ ⋅ q ) − T ( ∇ ⋅ v ) − (−τ : ∇v ) Dt ∂T ρ • incompressible flow • Newtonian fluid with constant viscosity • isotropic heat conduction with constant k DT ρ CˆV = k ∇ 2T + µΦ v Dt ∂T ∂T vθ ∂T ∂T 1 ∂ ∂T 1 ∂ 2T ∂ 2T k ρ CˆV + vr + = + vz r ∂r r ∂r + r 2 ∂θ 2 + ∂z 2 + µΦ v ∂r r ∂θ ∂z ∂t • steady state conditions, only flow in θ-direction • rotational symmetry, no θ-direction of state variables • infinite size in z-direction: no z-dependence BSL 11.4: Use of the equations of change to solve steady-state problems = 0 k 1 ∂ ∂T r + µΦ v r ∂r ∂r reduced thermal energy equation • steady state conditions, only flow in θ-direction • rotational symmetry, no θ-dependence of state variables • infinite size in z-direction: no z-dependence of state variables ∂vr 2 1 ∂vθ vr 2 ∂vz 2 ∂ vθ 1 ∂vr = Φ v 2 + + + + r + ∂r r ∂θ r ∂z ∂r r r ∂θ 2 1 ∂vz ∂vθ ∂vr ∂vz 2 + + + + − [ (∇ ⋅ v ) ] r ∂θ ∂z ∂z ∂r 3 2 2 ∂ v Φ v = r θ ∂r r 2 reduced viscous dissipation function 2 BSL 11.4: Use of the equations of change to solve steady-state problems Velocity profile (BSL 3): r −κR κ R r vθ = Ω0 R 1 −κ κ Equation of change for thermal energy: 2 1 d dT d vθ 0 k r + µ r r dr dr dr r Substitution of the equation for vθ(r) in the thermal energy equation gives: 0 1 d dT 4 µΩ02 R 4κ 4 1 k r + r dr dr (1 − κ 2 )2 r 4 BSL 11.4: Use of the equations of change to solve steady-state problems with: N Dimensionless quantities: • Dimensionless radius ξ: r ξ= R • Dimensionless temperature Θ: T − Tκ Θ= T1 − Tκ • Dimension thermal energy equation: = 0 1 d dΘ 1 ξ + 4N 4 ξ dξ dξ ξ BSL 11.4: Use of the equations of change to solve steady-state problems Boundary conditions: µΩ02 R 2 κ4 κ4 Br = 2 k (T1 − Tκ ) (1 − κ 2 )2 (1 − κ 2 ) and Br the dimensionless Brinkman number: Br = BSL 11.4: Use of the equations of change to solve steady-state problems µΩ02 R 2 k (T1 − Tκ ) Integration (twice) of the dimensionless thermal energy equation: Θ=− N ξ2 + C1 ln (ξ ) + C2 ξ κ Θ = = 0 ξ= 1 Θ = 1 Solution (radial temperature profile): N N ln (ξ ) Θ= ( N + 1) − 2 − ( N + 1) − 2 ξ κ ln (κ ) For sufficiently high values of N (for sufficiently strong viscous dissipation) a maximum in the temperature distribution occurs: 2 ln(1/ κ ) ξm = (1/ κ 2 ) − (1 + 1/ N ) BSL 12.1: Unsteady heat conduction in solids Energy equation for solids: ∂T = − ( ∇ ⋅ q ) = ( ∇ ⋅ k ∇T ) ρC p ∂t For constant thermal conductivity k the heat diffusion equation is obtained with α=k/(ρCp): Advanced Transport Phenomena BSL 12: Temperature distributions with more than one independent variable k ∂T = ∇ 2T =α∇ 2T ∂t ρ C p J.A.M. Kuipers Reference for solving heat diffusion equation for different geometries and boundary conditions: Conduction of Heat in Solids by H.S. Carslaw and J.C. Jaeger (CJ) Laplace transformation is used in CJ to obtain analytical solutions, we will use separation of independent variables requiring concept of orthogonality of functions Orthogonality of functions Orthogonality • ORTHOGONALITY OF VECTORS ⋅ g) (f = n g ∑ f= i =1 j j b 0 • FUNCTION VALUES OF f AND g AT n EQUALLY SPACED POINTS [ f ( h) f (nh) ] [ g ( h) g (nh) ] h = • ORTHOGONALITY RELATION FOR f AND g n ∑ f ( jh) g ( jh) = 0 multiply by h j =1 n h∑ f ( jh) g ( jh) = 0 j =1 h→0 n→∞ a f and g are orthogonal on the interval [a,b] ∫ f ( x) g ( x)dx = 0 a b dx ∫ φ ( x)φ ( x)= i j 0 i≠ j a b−a n +1 b ∫ f ( x) g ( x)dx = 0 orthogonality of functions f(x) and g(x) and functions of the same set (basis functions) b orthogonality of functions with respect to a weight function w(x) ∫ f ( x) g ( x)w( x)dx = 0 a b )dx ∫ φ ( x)φ ( x)w( x= i j 0 i≠ j a concept of orthogonality is very important for the solution of PDE’s where a prescribed function needs to be represented in basis functions satisfying the spatial part of the PDE leading to Sturm-Liouville problems Sturm-Liouville Theory d dy 0 p ( x) + (q ( x) + λ w( x)) y = dx dx Orthogonality + Sturm-Liouville Theory i =∞ differential equation for y(x) on interval [a,b] β1 y (b) + β 2 y (b) = 0 b ∫ boundary conditions a solution exists only for distinct values of parameter λi= ( i 1,.., ∞ ) termed the eigenvalues, the corresponding solutions are the = eigenfunctions yi ( x) which comprise an orthogonalset with respect to the weight function w( x) b )dx 0 ∫ yi ( x) y j ( x)w( x= a BSL 12.1: Unsteady heat conduction in solids BSL Example 12.1-2: Heating of a finite solid slab Heat diffusion equation ∂T ∂T =α 2 ∂t ∂y 2 Initial condition = t 0 -b ≤ y ≤ b: = T T0 Boundary conditions t > 0 y= -b : T= T1 t >0 y= +b : T = T1 b i =∞ f ( x)φ j ( x) w( x)dx = ∫ ∑ K iφi ( x) φ j ( x)( w( x)dx a i =1 b b 2 ( ) ( ) ( ) K φ x φ x w x dx K = ∑ i ∫ i j i ∫ φi ( x ) w( x ) dx i =1 a a i =∞ b i≠ j note: parameter λ=µ2 is the separation constant introduced through the method of separation of independent variables in the original PDE the Sturm-Liouville problem description i =1 α1 y ( a ) + α 2 y ' ( a ) = 0 homogeneous two-point ' expand f ( x) in eigenfunctions φi ( x) satisfying f ( x) = ∑ K iφi ( x) ∫ f ( x)φ ( x)w( x)dx i a ⇒ Ki = b ∫φ 2 i ( x) w( x)dx multiply with φ j ( x) w( x)dx and integrate over[a,b] use orthogonality of eigenfunctions with respect to w(x) integrals can be evaluated using standard expressions for the specific eigenfunctions: see book on Fourier Analysis by M.R. Spiegel (canvas) a BSL 12.1: Unsteady heat conduction in solids Dimensionless temperature: T −T Θ= 1 T1 − T0 Dimensionless distance: η= y b Dimensionless time: τ= αt b2 BSL 12.1: Unsteady heat conduction in solids BSL 12.1: Unsteady heat conduction in solids Solving the problem using separation of variables Dimensionless problem description ∂Θ ∂ 2 Θ = ∂τ ∂η 2 = Θ f (η ) ⋅ g (τ ) Try a solution of the following form: Inserting this in the dimensionless PDE and dividing by = Θ f (η ) ⋅ g (τ ) gives after separation: Dimensionless initial & boundary conditions: = Θ 1 τ 0 -1 ≤ η ≤ 1:= dg = −µ 2 g dτ and: and: τ >0= η -1: = Θ 0 τ > 0 η = +1: Θ = 0 d2 f = −µ 2 f 2 dη where µ2 is the separation constant BSL 12.1: Unsteady heat conduction in solids Solving the ordinary DE’s gives the general solution: = Θ exp ( − µ 2τ ) [ B ⋅ sin ( µη ) + C ⋅ cos ( µη )] Based on the symmetry of the problem, B must equal 0! Based on the boundary condition at η=1: C ⋅ cos ( µ ) = 0 From these equations, the eigenvalues µn emerge: µn= ( n + ) π n= 0,1, 2,…, ∞ 1 2 and the corresponding eigenfunctions: cos = ( µnη ) cos ( ( n + 12 ) πη ) BSL 12.1: Unsteady heat conduction in solids Applying the superposition principle: = Θ ∞ ∑ n =0 ( ) K n exp − ( n + 12 ) π 2τ ⋅ cos ( ( n + 12 ) πη ) 2 Applying the initial condition: = 1 ∞ ∑ n =0 K n cos ( ( n + 12 ) πη ) Using the Sturm-Liouville’s theorem gives: 1 ∫ cos ( ( n + 12 ) πη )dη 2 ( −1) = Kn = 2 1 ( n + 12 ) π ∫ cos ( ( n + 2 ) πη )dη 0 1 0 n BSL 12.1: Unsteady heat conduction in solids BSL 12.1: Unsteady heat conduction in solids Inserting this gives the following solution in terms of the original variables: T1 − T T1 − T0 ( −1) exp − n + 1 2 π 2 α t ⋅ cos n + 1 π y 2∑ ( ( 2) 2) n =0 ( n + 1 ) π b2 b 2 ∞ n Derived quantities like average temperature and the heat flux at the wall can be obtained respectively by integration and differentiation term by term of this series BSL 12.1: Unsteady heat conduction in solids BSL 12.1: Unsteady heat conduction in solids Unsteady heat conduction in an infinitely long cylinder Dimensionless temperature: Heat diffusion equation: T −T Θ= 1 T1 − T0 ∂T 1 ∂ ∂T =α r ∂t r ∂r ∂r Dimensionless distance: Initial condition: η= t= T0 0 r ≤ R: T = Dimensionless time: Boundary conditions: t > 0= r 0: τ= T (r ) ≤ M t > 0 r= R : T= T1 r R this boundary condition is equivalent to: ∂T ∂r =0 r =0 αt R2 BSL 12.1: Unsteady heat conduction in solids BSL 12.1: Unsteady heat conduction in solids Solving the problem using separation of variables Dimensionless problem description: ∂Θ 1 ∂ ∂Θ η = ∂τ η ∂η ∂η = Θ f (η ) ⋅ g (τ ) Try a solution of the following form: Inserting this in the dimensionless PDE and dividing by = Θ f (η ) ⋅ g (τ ) gives after separation: Dimensionless boundary conditions: = Θ 1 τ 0 η ≤ 1: = dg = −µ 2 g dτ and: and: 1 d df η = −µ 2 f η dη dη τ ≥ 0 for all η ≤ 1: Θ ≤ N τ ≥0 = η 1: = Θ 0 where µ2 is the separation constant BSL 12.1: Unsteady heat conduction in solids Solving the ordinary DE’s gives the general solution: = Θ exp ( − µ 2τ ) [ B ⋅ J 0 ( µη ) + C ⋅ Y0 ( µη )] where J0 is the 0th order Bessel function of the 1st kind and Y0 the 0th order Bessel function of the 2nd kind (Neumann function) Bessel’s differential equation 0 x 2 y '' + xy ' + ( µ 2 x 2 − n 2 ) y = non-integer n: y(x)=K1 J n ( µ x) + K 2 J − n ( µ x) for integer n: y(x)=K1 J n ( µ x) + K 2Yn ( µ x) see book of M.R. Spiegel “Fourier Analysis” (canvas) BSL 12.1: Unsteady heat conduction in solids Based on the natural boundary condition, C equals 0 Based on the boundary condition at η=1 we obtain: B ⋅ J0 ( µ ) = 0 These equations give the eigenvalues µn: = µn β n = n 1, 2,3,…, ∞ and the corresponding eigenfunctions: J 0 ( β nη ) = 0 where βn are the roots of J0 J0 ( x) = 1− x2 x4 x6 + − + ⋅⋅⋅ 22 22 42 22 42 62 first six roots of J0 ∆β n ≈ π for large n BSL 12.1: Unsteady heat conduction in solids This result can easily be obtained using standard relations for Bessel functions (handout + book of Spiegel) Applying the superposition principle: Θ ∞ ∑ n =1 K n exp ( − β n2τ ) J 0 ( β nη ) Inserting gives the following solution in terms of the original variables: Applying the initial condition: 1= ∞ ∑ n =1 K n J 0 ( β nη ) Use orthogonality of Bessel functions: 1 Kn ∫η J0 ( β nη )dη 2 = 2 β n J1 ( β n ) ∫ η J 0 ( β nη )dη 0 1 0 BSL 12.1: Unsteady heat conduction in solids 1 ( µ x)dx ∫ xJ = 0 2 n d n ( x J n ( x) ) = x n J n−1 ( x) dx d −n ( x J n ( x) ) = − x − n J n+1 ( x) dx 1 ' n2 2 2 ( J n ( µ )) + (1 − 2 )( J n ( µ )) 2 µ ∞ T1 − T αt β r 1 = 2∑ exp − β n2 2 ⋅ J 0 n n =1 β J ( β ) T1 − T0 R R n 1 n Derived quantities like average temperature and the heat flux at the wall can be obtained respectively by integration and differentiation term by term of this series see “Fourier Analysis” by M.R. Spiegel BSL 12.1: Unsteady heat conduction in solids BSL 12.1: Unsteady heat conduction in solids Unsteady heat conduction in a sphere Heat diffusion equation: ∂T 1 ∂ ∂T = α 2 r2 ∂t r ∂r ∂r Initial condition: t= T0 0 r ≤ R: T = Boundary conditions: t > 0 for all r ≤ R : t > 0 r= R : T= T1 T (r ) ≤ M BSL 12.1: Unsteady heat conduction in solids Dimensionless temperature: T −T Θ= 1 T1 − T0 Dimensionless distance: η= r R Dimensionless time: τ= αt R2 BSL 12.1: Unsteady heat conduction in solids Solving the problem using separation of variables Θ f (η ) ⋅ g (τ ) Try a solution of the following form:= Inserting this in the dimensionless PDE and dividing by = Θ f (η ) ⋅ g (τ ) gives after separation: dg = −µ 2 g dτ and: 1 d 2 df η = −µ 2 f 2 η dη dη where µ2 is the separation constant BSL 12.1: Unsteady heat conduction in solids Dimensionless problem description: ∂Θ 1 ∂ 2 ∂Θ = η ∂τ η 2 ∂η ∂η Dimensionless boundary conditions: = τ 0 η ≤ 1: = Θ 1 and: τ ≥ 0 for all η ≤ 1: Θ ≤ N τ ≥0 = η 1: = Θ 0 BSL 12.1: Unsteady heat conduction in solids Solving the ordinary DE’s gives the general solution: [ B ⋅ sin ( µη ) + C ⋅ cos ( µη )] = Θ exp ( − µ 2τ ) η Based on the natural boundary condition, C equals 0 Based on the boundary condition at η=1 we obtain: B ⋅ sin ( µ ) = 0 These equations give the eigenvalues µn: µn = n ⋅ π n = 1, 2,3,…, ∞ and the corresponding eigenfunctions: sin ( nπη ) η BSL 12.1: Unsteady heat conduction in solids Applying the superposition principle: = Θ ∞ ∑ n =1 ( K n exp − ( nπ ) τ 2 ) sin (ηnπη ) Applying the initial condition: 1= ∞ ∑ n =1 Kn BSL 12.1: Unsteady heat conduction in solids Inserting gives the following solution in terms of the original variables: ∞ ( −1) T1 − T 2∑ = n =1 T1 − T0 nπ n +1 R nπ r 2 αt exp − ( nπ ) 2 sin R r R Derived quantities like average temperature and the heat flux at the wall can be obtained respectively by integration and differentiation term by term of this series sin ( nπη ) η Using Sturm-Liouiville’s theorem gives: 1 Kn ∫ η ⋅ sin 0 ( nπη )dη ( −1) 2 = 1 2 nπ ∫ sin ( nπη )dη 0 n +1 BSL 12.1: Unsteady heat conduction in solids BSL 12.2: Steady heat conduction in laminar, incompressible flow General procedure for analysis of heat conduction in flowing fluids •Solve microscopic momentum balance velocity profile •Combine velocity profile with thermal energy equation and solve the resulting equation temperature profile Condition: temperature dependency of viscosity can be neglected. Due to the T-dependence of viscosity, this is a critical condition for liquids BSL 12.2: Steady heat conduction in laminar, incompressible flow Heat conduction in laminar tube flow BSL 12.2: Steady heat conduction in laminar, incompressible flow Solution of this PDE: three solutions can be distinguished: 1. Heat conduction in laminar tube flow due to a stepwise change in the wall temperature (Graetz-Nusselt problem) a. Complete solution of the PDE by separation of variables (BSL Example 12.2-1) 2. Heat conduction in laminar tube flow due to a constant heat flux imposed at the wall b. Asymptotic solution of the PDE for short distances down the tube by combination of variables (BSL Example 12.22) We will focus on problem type 2. BSL 12.2: Steady heat conduction in laminar, incompressible flow Asymptotic solution of the PDE for short distances down the tube by combination of variables (BSL Example 12.2-2) Microscopic balance for thermal energy in original form: r 2 ∂T 1 ∂ ∂T vz ,max 1 − α = r R z r r ∂ ∂ ∂r where vz,max is the velocity in the centre of the tube: vz ,max p0 − pL ) R 2 ( = 4µ L c. Asymptotic solution for large distances down the tube (BSL §10.8) BSL 12.2: Steady heat conduction in laminar, incompressible flow Simplification for working out the problem: a. Curvature is neglected; instead of radial coordinate r, here s=R - r (distance to the tube wall) is used as a new variable b. Fluid is unbounded in the s-direction (s=0 to s=∞) c. The velocity profile near the tube wall is linearized, due to the small heat penetration depth in the s-direction: 2s s dv vz ( s ) = vz ( 0 ) + z s = vz ,max = v0 R R ds s =0 BSL 12.2: Steady heat conduction in laminar, incompressible flow Inserting the approximations in the microscopic balance for heat: ∂T ∂ 2T s =α 2 v0 R ∂z ∂s with boundary conditions: ( ) s= T0 ∞: T = ∂T 0= : -k q1 ∂s : T T0 z 0= = s Note that two out of three boundary conditions are identical which is essential for the method of combination of independent variables to work! BSL 12.2: Steady heat conduction in laminar, incompressible flow Dimensionless problem description Dimensionless thermal energy equation: ∂Θ ∂ 2 Θ η = ∂λ ∂η 2 Dimensionless boundary conditions: η =∞: Θ =0 ∂Θ η 0= : 1 = ∂η λ 0: = = Θ 0 BSL 12.2: Steady heat conduction in laminar, incompressible flow Dimensionless variables Dimensionless temperature: Θ= T − T0 q1 R k Dimensionless radial coordinate: η= s R Dimensionless axial coordinate: zα λ= v0 R 2 BSL 12.2: Steady heat conduction in laminar, incompressible flow For this problem it is easier to generate the solution in terms of the dimensionless heat flux Ψ: ∞ Θ(∞, λ ) − Θ(η , λ ) = −Θ(η , λ ) = − ∫ Ψ (η , λ ) dη η Inserting this into the dimensionless PDE: η ∂ ∞ − ∂Ψ ∫ Ψ (η , λ ) dη = ∂λ η ∂η BSL 12.2: Steady heat conduction in laminar, incompressible flow Dividing by η and differentiation to η gives: ∂Ψ ∂ 1 ∂Ψ = ∂λ ∂η η ∂η with the boundary conditions: assuming = ω η a λ b ⇒ a + 3= b 0 to obtain ODE for F(ω ) Since F=1 at ω=0 it follows that K2=1 Using the boundary condition F=0 at ω=∞ gives: −1 K1 = ∞ 3 ∫ u ⋅ exp ( −u ) du 0 ∫ u ⋅ exp ( −u ) du − ∫ u ⋅ exp ( −u ) du ∫ u ⋅ exp ( −u ) du ω 3 ( −u ) du ∫ u ⋅ exp ( −u ) du ∫ u ⋅ exp F = Ψ = ω∞ 0 3 3 0 ∞ F = K1 ∫ u ⋅ exp ( −u 3 ) du + K 2 ω 0 BSL 12.2: Steady heat conduction in laminar, incompressible flow Rewriting in standard form using gamma-functions (x=u3): ∞ Ψ= ∫ ω 3 x 2 3−1e − x dx ∞ 2 3−1 − x e dx ∫x = 1− 3 3 Γin ( 23 , ω 3 ) Γ ( 23 ) where Γ(z) is the complete gamma-function and Γin(z,y) is the incomplete gamma-function: 0 ∞ or: Separation and integration yields: 0 Inserting the constants gives: 0 dF =K1ω ⋅ exp ( −ω 3 ) dω where K1 is an integration constant BSL 12.2: Steady heat conduction in laminar, incompressible flow F= ω F " + ( 3ω 3 − 1) F ' = 0 G= Combination of independent variables: Try a solution of the form: Ψ =F (ω ) where ω is given by: ∞ Transformation of PDE gives the following ODE: Lowering the order F ' = G and integration of the ODE gives: = λ 0: = Ψ 0 = η 0: = Ψ 1 η =∞: Ψ =0 η ω= 13 ( 9λ ) BSL 12.2: Steady heat conduction in laminar, incompressible flow ∞ z −1 − x Γ( z) = ∫ x e dx 0 y z −1 − x Γin ( z , y ) = ∫ x e dx 0 BSL 12.2: Steady heat conduction in laminar, incompressible flow Resulting expression for the dimensionless heat flux: BSL 12.2: Steady heat conduction in laminar, incompressible flow Working out the integral gives: 2 η3 Γ in 3 , qs 9λ Ψ= = 1− 2 q1 Γ( 3) −η 3 2 η 3 exp 9λ η Γ in 3 , 9λ − 3 9λ = Θ 1 − 3 2 2 Γ ( 3 ) 9λ Γ( 3) Derivation of the expression for the temperature profile can easily be obtained from the expression for Ψ by integration (Fourier’s law): This is the final expression for the dimensionless temperature profile in the entrance zone of the tube 2 u3 use integration by Γin ( , ) ∞ ∞ parts and Leibnitz 3 due to “thermal saturation” the wall 3 9λ du formulae 9λ Θ(η , λ ) = ∫ Ψ ( u , λ ) du =− ∫ 1 to differentiate Θ (0, λ ) = temperature (at η=0) must increase in order 2 η η the incomplete Γ ( 23 ) Γ( ) to maintain the imposed constant heat flux gamma function 3 BSL 12.2: Steady heat conduction in laminar, incompressible flow The solution for the case that the tube wall is at a constant temperature T1 (at z>0) can be derived analogously: 1 η3 Γin 3 , T1 − T 9λ =Θ= T1 − T0 Γ ( 13 ) The heat flux at the tube wall: −k ∂T ∂r r=R k k 3 ∂Θ −1 3 = − (T1 − T0 ) = − (T1 − T0 ) 1 ( 9λ ) R ∂η η =0 R Γ(3) BSL 12.2: Steady heat conduction in laminar, incompressible flow Using the definition of the Nusselt number gives (with Γ(1/3)=2.679): ∂T −1 3 2R −k 6 241 3 α z ∂r r = R −1 3 Nu= = = ( 9λ ) k (T1 − T0 ) Γ ( 13 ) Γ ( 13 ) v d 2 z = 1.08 Gz ⋅ L −1 3 which is the well-known solution for the Nusselt number in the thermally non-developed regime BSL 12.4: Boundary layer theory for nonisothermal flow Extension of the boundary layer theory to simultaneous momentum and heat transport Model system: laminar flow along a flat heated plate BSL 12.4: Boundary layer theory for nonisothermal flow Quantitative description Continuity equation: ∂vx ∂v y + = 0 ∂x ∂y Momentum equation x-direction: ∂vx ∂vx ∂ 2 vx vx + vy = ν 2 ∂x ∂y ∂y Thermal energy equation: ∂T ∂T ∂ 2T vx + vy = α 2 ∂x ∂y ∂y BSL 12.4: Boundary layer theory for nonisothermal flow Solving the micro balances using a boundary layer approach Dimensionless velocity φ: v0 − vx vx y η = = φ (η ) with: = v0 − v∞ v∞ δ ( x) Dimensionless temperature Θ: T0 − T y = Θ (ηT ) with: ηT = T0 − T∞ δT ( x ) where δ is the thickness of the momentum boundary layer and δT is the thickness of the thermal boundary layer BSL 12.4: Boundary layer theory for nonisothermal flow Assume δ T = δ∆ and distinguish Δ≤1 and Δ ≥1 Solve the heat transfer problem for Δ≤ 1: combination of the continuity equation and the thermal energy equation (elimination of vy): vx φ { } { } ∂T y ∂vx ∂T ∂ 2T − ∫ = α 2 ( x, yˆ )dyˆ ∂x 0 ∂x ∂y ∂y ∂Θ y ∂φ ∂Θ α ∂ 2 Θ − ∫ ( x, yˆ )dyˆ = ∂x 0 ∂x ∂y v∞ ∂y 2 BSL 12.4: Boundary layer theory for nonisothermal flow Evaluation of terms (chain rule) η dδ 1 dδT ∂Θ = −Θ' (ηT ) T T = −ηT Θ' (ηT ) ∂x δ T dx δ T dx η dδ 1 dδ ∂φ = −φ ' (η ) = −ηφ ' (η ) δ dx δ dx ∂x ∂Θ 1 ' = Θ (ηT ) ∂y δ T BSL 12.4: Boundary layer theory for nonisothermal flow Insertion into the thermal energy equation: −φ (η ) {ηT Θ' (ηT )} { } 1 dδT η ' dδ 1 ' α 1 " + ∫ uφ du Θ (ηT ) = Θ (ηT ) 0 dx δ T v∞ δ T2 δ T dx Multiply by δTδT and use: δT = δ∆ and: η = ηT ∆ Insertion gives: −φ (ηT ∆ )ηT Θ' (ηT )∆ 2δ ∂Θ 1 " = Θ (ηT ) ∂y 2 δ T2 2 dδ + dx { ηT ∆ ∫ 0 } uφ ' du Θ' (ηT )∆δ dδ α " = Θ (ηT ) dx v∞ BSL 12.4: Boundary layer theory for nonisothermal flow BSL 12.4: Boundary layer theory for nonisothermal flow Multiplication of equation with dηT and integration of the equation over the thermal boundary layer gives: Earlier obtained result for hydrodynamics (BSL 3) ( E − D)δ dδ α F = dx v∞ in which the numbers D, E and F follow from: 1 E =∆ ∫ Θ 0 ' { ηT ∆ ∫ 0 } uφ du dηT ' and: dδ ν C = dx v∞ in which the numbers A, B and C follow from: 1 B = ∫φ' and: 0 { uφ du}dη η ∫ 1 0 1 A = ∫ φφ 'η dη 0 and: 1 " ∫Θ 0 dηT ' 0 1 D =∆ 2 ∫ φ (ηT ∆ ) Θ'ηT dηT and F= ( B − A) δ C = ∫ φ "dη 0 BSL 12.4: Boundary layer theory for nonisothermal flow BSL 12.4: Boundary layer theory for nonisothermal flow Evaluation of the numbers A to F Division of the left hand and right hand side of the two ODE’s: dδ α F dx = v∞ dδ ν C ( B − A) δ dx v∞ ( E − D )δ ( E − D) F C α 1 = f (∆) = = ( B − A) ν Pr rate of momentum diffusion Pr = rate of heat diffusion The ratio between the thicknesses of the boundary layers is a function of the Prandtl-number only !!! BSL 12.4: Boundary layer theory for nonisothermal flow Inserting into the equation for A to F gives: Hydrodynamic boundary layer thickness δ: 1260 ν x νx δ = 5.84 37 v∞ v∞ Ratio parameter Δ: 1 3 3 5 1 6 37 1 ∆ − ∆ + ∆ = 15 280 360 630 Pr Approximating equation for ratio parameter Δ: ∆ =Pr −1 3 Temperature profile in thermal boundary layer: 3 T0 − T y y y =2 − 2 + T0 − T∞ δ∆ δ∆ δ∆ 4 polynomial representation of the dimensionless velocity and temperature profiles within their respective boundary layers: coefficients follow from the boundary conditions and PDE’s! Dimensionless velocity distribution: φ =2η − 2η 3 + η 4 Dimensionless temperature distribution: = Θ 2ηT − 2ηT3 + ηT4 BSL 12.4: Boundary layer theory for nonisothermal flow Local heat transfer coefficient αx: −k α= x ∂T ∂y 1 1 2k 2k v∞ x 2 ν 3 k 12 13 = = = 0.342 Re x Pr T0 − T∞ x δ∆ 5.84 x ν α y =0 Local Nusselt-number Nux: Nu = x αxx 1 1 = 0.342 Re 2x Pr 3 k Extension to more complicated situations: • “Boundary layer theory”, Schlichting • “Convective Heat and Mass Transfer”, Kays & Crawford