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CHAPTER 1 BASIC CONCEPTS 1.1 Convection Heat Transfer Examine thermal interaction between a surface and an adjacent moving fluid. 1.2 Important Factors in Convection Heat Transfer Three factors play major roles in convection heat transfer: (1) Fluid motion (2) Fluid nature (3) Surface geometry. 1.3 Focal Point in Convection Heat Transfer The determination of temperature distribution in a moving fluid can be expressed as T T ( x, y, z, t ) (1.1) 1.4 The Continuum and Thermodynamic Equilibrium Concepts Continuum model: Material is composed of continuous matter. Behavior of individual molecules is ignored. Valid when molecular man free path is small relative to the characteristic dimension of the system (small Knudsen number, Kn 10 1 ). Knudson number: defined as Kn De (1.2) Thermodynamic equilibrium: Depends molecular interaction with an adjacent surface. At thermodynamic equilibrium: (1) Fluid and adjacent surface have the same velocity (no velocity slip). (2) Fluid and adjacent surface have the same temperature (no temperature jump). Valid for Kn 10 3 1.5 Fourier’s Law of Conduction Gives a relationship between heat transfer rate or heat flux and temperature: (1.3b) 2 qx k T , x qy k T , y qz k T z (1.8) Used in the formulation of the governing equation for temperature distribution. 1.6 Newton's Law of Cooling This law defines the heat transfer coefficient h in terms of heat flux and surface temperature qs h (Ts T ) (1.9) Used in the formulation of boundary conditions. Used to determine the rate of transfer or heat flux when temperature distribution is difficult to obtain analytically. 1.7 The Heat Transfer Coefficient h Defined in (1.9). h is not a property. It depends on: (1.10) h f (geometry, fluid motion, fluid properties, T ) It is determined analytically using (1.8) and (1.9). This requires the determination of temperature distribution. h k T ( x,0, z ) y Ts T (1.12) It is determined experimentally using Newton’s law of cooling (1.9). Table 1.1 gives typical values for h. Use this table as a guide only. 1.8 Radiation: Stefan-Boltzmann Law Emissivity : radiation surface property Absorptivity : surface property defined as the fraction of radiation energy incident on a surface which is absorbed by the surface q12 is the net heat exchanged by radiation between two bodies. Special case: a small surface which is completely enclosed by a much larger surface, q12 is given by Stefan-Boltzmann radiation law q12 1 A1 (T14 T24 ) (1.14) 3 Subscript 1 in (1.14) refers to the small surface. Temperature is in absolute degrees. 1.9 Differential Formulation of Basic Laws Various formulation procedures: Differential Integral Variational Finite difference Differential formulation: Based on the assumption of continuum. The basic laws are applied to an infinitesimal element. The result is a partial differential equation which is valid at every point. 1.10 Mathematical Background (a) Velocity Vector V V (b) Velocity Derivative ui v j V x u i x (1.15a) wk v j x w k x (1.15b) k (1.16) (c) The Operator Cartesian: x i y j z Cylindrical: r ir 1 r 1 r i i z (1.17) iz Spherical: r (d) Divergence of a Vector ir 1 r sin i (1.18) 4 div.V V u x v y w z (1.19) u x v y w z (1.20) (e) Derivative of the Divergence V x x or V x V x (1.21) (f) Gradient of Scalar. The gradient of a scalar, such as temperature T, is a vector given by Grad T T i x T T j y T k z (1.22) (g) Total Differential and Total Derivative Consider the variable f: f Total differential of f: and t: f ( x, y, z, t ) This is the total change in f resulting from changes in x, y, z f dx x df f dy y f dz z f dt t Total derivative df dt Df Dt u f x v f y f z f y w f z convective derivative w f t (1.23) where u f x v f t Example: f local derivative u in (1.23) gives du dt Du Dt u u x v u y w u z u t (1.24) Cylindrical coordinates: dv r dt Dv r Dt vr vr r v r vr v2 r vz vr z vr t (1.25a) 5 dv dt Dv Dt dv z dt vr Dv z Dt v r vr v r v vz r v r T x v vrv r vz vz vz vz z v z v t vz t (1.25b) (1.25c) Example: f = T dT dt DT Dt u T y w T z T t (1.26) 1.11 Units Basic SI units: Length (L): meter (m) Time (t): second (s) Mass (m): kilogram (kg) Temperature (T): kelvin (K) Derived units: Force (newton): N = kg-m /s2 Energy (joules): J = N m = kg-m2 /s2 Power (watts): W = J/s = N m/s = kg-m2 /s3 1.12 Problem Solving Format Convection problems lend themselves to a systematic solution procedure. Use the following format when solving problems. Read details of each step and take advantage of illustrative examples and posted solutions to homework problems to develop skills in using this problem solving methodology. The following is an outline of this method: (1) Observations (2) Problem Definition (3) Solution Plan (4) Plan Execution (i) Assumptions (ii) Analysis (iii) Computations (iv) Checking (5) Comments CHAPTER 2 DIFFERENTIAL FORMULATION OF THE BASIC LAWS 2.1 Introduction Differential formulation of basic laws: Conservation of mass Conservation of momentum Conservation of energy 2.2 Flow Generation (i) Forced convection. Motion is driven by mechanical means. (ii) Free (natural) convection. Motion is driven by natural forces. 2.3 Laminar vs. Turbulent Flow Laminar flow: no fluctuations in velocity, pressure, temperature, … Turbulent flow: random fluctuations in velocity, pressure, temperature, … Transition from laminar to turbulent flow: Determined by the Reynolds number: Flow over a flat plate: Ret V xt /ν Flow through tubes: Ret = u D v 500,000 2300 u u turbulent laminar t t Fig. 2.1 2.4 Conservation of Mass: The Continuity Equation 2.4.1 Cartesian Coordinates 2 The principle of conservation of mass is applied to an element dxdydz Rate of mass added to element - Rate of mass removed from element = Rate of mass change within element my y dy dx mx ( my ) y mx x dy ( mx ) dy y my (a ) (b ) Fig. 2.2 (2.1) Expressing each term in terms of velocity components gives continuity equation t x u v y w z (2.2a) 0 This equation can be expressed in different forms: t u x v y w u x z v y w z 0 (2.2b) or D Dt or t V 0 (2.2c) V 0 (2.2d) For constant density (incompressible fluid): V 2.4.2 Cylindrical Coordinates 0 (2.3) 3 t 1 r r 1 r r vr v vz z (2.4) 0 2.4.3 Spherical Coordinates 1 t r 2 r 1 r sin r 2 vr 1 r sin v sin v 0 (2 .5) 2.5 Conservation of Momentum: The Navier-Stokes Equations of Motion 2.6 2.5.1 Cartesian Coordinates Application of Newton’s law of motion to the element shown in Fig. 2.5, gives F ( m)a dz y dy (a) dx Application of (a) in the x-direction, gives x Fx (b) ( m)a x z Fig. 2.5 Each term in (b) is expressed in terms of flow field variables: density, pressure, and velocity components: Mass of the element: m (c) dxdydz Acceleration of the element a x : ax du dt Du Dt Substituting (c) and (d) into (b) Du dxdydz Dt Fx (e) Forces: (i) Body force Fx body (ii) Surface force g x dxdydz (g) u u x v u y w u z u t (d) 4 Summing up all the x-component forces shown in Fig. 2.6 gives xx δFx surface x yx zx y z dxdydz (h) Combining the above equations Du Dt gx y-direction: Dv Dt gy z-direction: Dw Dt gz x-direction: yx zx x y z xy yy zy xx (2.6a) By analogy: x y xz yz x y (2.6b) z zz (2.6c) z IMPORTANT THE NORMAL AND SHEARING STRESSES ARE EXPRESSED IN TERMS OF VELOSICTY AND PRESSURE. THIS IS VALID FOR NEWTOINAN FLUIDS. (See equations 2.7a-2.7f). THE RESULTING EQUATIONS ARE KNOWN AS THE NAVIER-STOKES EQUAITONS OF MOTION SPECIAL SIMPLIFIED CASES: (i) Constant viscosity DV Dt g p V 1 3 V 2 (2.9) (2.9) is valid for: (1) continuum, (2) Newtonian fluid, and (3) constant viscosity (ii) Constant viscosity and density DV Dt g V p 2 (2.10) (2.10) is valid for: (1) continuum, (2) Newtonian fluid, (3) constant viscosity and (4) constant density. The three component of (2.10) are x: u t u u x v u y w u z gx p x 2 u x 2 2 u y 2 2 u z 2 (2.10x) 5 v t y- z- w t u u v x v v y w v z gy p y w x v w y w w z gz p z 2 2 v 2 v v x2 y2 z2 2 2 2 w x2 w y2 (2.10y) w z2 (2.10z) 2.5.2 Cylindrical Coordinates The three equations corresponding to (2.10) in cylindrical coordinates are (2.11r), (2.11 ), and (2.11z). 2.5.3 Spherical Coordinates The three equations corresponding to (2.10) in spherical coordinates are (2.11r), (2.11 ), and (2.11 ). 2.6 Conservation of Energy: The Energy Equation dz y 2.6.1 Formulation dy dx The principle of conservation of energy is applied to an element dxdydz x A B Rate of change of internal and kinetic energy of element z Net rate of internal and kinetic Fig. 2.7 energy transport by convection C Net rate of heat addition by conduction The variables u, v, w, p, T, and (2.14) _ D Net rate of work done by element on surroundin gs are used to express each term in (2.14). Assumptions: (1) continuum, (2) Newtonian fluid, and (3) negligible nuclear, electromagnetic and radiation energy transfer. Detailed formulation of the terms A, B, C and D is given in Appendix A The following is the resulting equation DT Dp k T T Dt Dt (2.15) is referred to as the energy equation cp is the coefficient of thermal expansion, defined as (2.15) 6 1 (2.16) T p The dissipation function is associated with energy dissipation due to friction. It is important in high speed flow and for very viscous fluids. In Cartesian coordinates is given by 2 2 3 u 2 x u x v y v y w z 2 w 2 z u y 2 v x v z w y 2 w x u 2 z (2.17) 2 2.6.2 Simplified Form of the Energy Equation Cartesian Coordinates (i) Incompressible fluid. Equation (2.15) becomes cp DT Dt (2.18) k T (ii) Incompressible constant conductivity fluid. Equation (2.18) is simplified further if the conductivity k is assumed constant cp DT Dt T y w k 2 (2.19a) T or cp T t u T x v T z 2 T x 2 k 2 T 2 y 2 z2 T (2.19b) (iii) Ideal gas. (2.15) becomes cp DT Dt k T Dp Dt or cv DT Dt k T p (2.22) V (2.23) Cylindrical Coordinates. The corresponding energy equation in cylindrical coordinate is given in (2.24) Spherical Coordinates. The corresponding energy equation in cylindrical coordinate is given in (2.26) 7 2.7 Solutions to the Temperature Distribution The flow field (velocity distribution) is needed for the determination of the temperature distribution. IMPORTANT: Table 2.1 shows that for constant density and viscosity, continuity and momentum (four equations) give the solution to u, v, w, and p. Thus for this condition the flow field and temperature fields are uncoupled (smallest rectangle). For compressible fluid the density is an added variable. Energy equation and the equation of state provide the fifth and sixth required equations. For this case the velocity and temperature fields are coupled and thus must be solved simultaneously (largest rectangle in Table 2.1). TABLE 2.1 Basic law No. of Equations Unknowns p u v w 1 u v w Momentum 3 u v w Equation of State 1 T p Viscosity relation ( p, T ) 1 T p 1 T p Energy 1 Continuity Conductivity relation k k ( p, T ) TT k p k 2.8 The Boussinesq Approximation Fluid motion in free convection is driven by buoyancy forces. Gravity and density change due to temperature change give rise to buoyancy. According to Table 2.1, continuity, momentum, energy and equation of state must be solved simultaneously for the 6 unknowns: u. v, w, p, T and Starting with the definition of coefficient of thermal expansion 1 T p , defined as (2.16) 8 or 1 T (f) T This result gives (2.28) (T T ) Based on the above approximation, the momentum equation becomes DV 1 gT T p p v 2V Dt (2.29) 2.9 Boundary Conditions (1) No-slip condition. At the wall, y 0 V ( x,0, z, t ) (2.30a) 0 or u( x,0, z, t ) v( x,0, z, t ) w( x,0, z, t ) (2) Free stream condition. Far away from an object ( y 0 (2.30b) ) (2.31) u( x, , z, t ) V Similarly, uniform temperature far away from an object is expressed as (2.32) T ( x, , z, t ) T (3) Surface thermal conditions. Two common surface thermal conditions are used in the analysis of convection problems. They are: (i) Specified temperature. At the wall: T ( x,0, z , t ) Ts (2.33) (ii) Specified heat flux. Heated or cooled surface: k T ( x,0, z, t ) y qo (2.34) 2.10 Non-dimensional Form of the Governing Equations: Dynamic and Thermal Similarity Parameters Express the governing equations in dimensionless form to: (1) identify the governing parameters (2) plan experiments (3) guide in the presentation of experimental results and theoretical solutions Dimensional form: Independent variables: x, y, z and t 9 Unknown variables are: u, v , w, p and T . These variables depend on the four independent variables. In addition various quantities affect the solutions. They are p , T , V , Ts , L, g , p , and Fluid properties c p , k, , , and Geometry 2.10.1 Dimensionless Variables Dependent and independent variables are made dimensionless as follows: V* V , V p * (p p ) V 2 , T * (T T ) , (Ts T ) g* g , g (2.35) V x y z , y* , , t* t x* z* L L L L Using (2.35) the governing equations are rewritten in dimensionless form. 2.10.2 Dimensionless Form of Continuity D Dt * V* 0 (2.37) No parameters appear in (2.37) 2.10.3 Dimensionless Form of the Navier-Stokes Equations of Motion DV * Gr * * 1 *2 * * T g P V* Re Dt * Re 2 (2.38) Constant (characteristic) quantities combine into two governing parameters: V L Re Gr V L Reynolds number (viscous effect) (2.39) , Grashof number (free convection effect) (2.40) v g Ts T L3 v2 , 2.10.4 Dimensionless Form of the Energy Equation Consider two cases: (i) Incompressible, constant conductivity 10 DT * Dt 1 RePr * *2 T* Εc Re * (2.41a) Constant (characteristic) quantities combine into two additional governing parameters: Pr Εc cp / k V c p (Ts k / cp v, Prandtl number (heat transfer effect) (2.42) 2 T ) , Eckert number (dissipation effect – high speed, large viscosity) (2.43) (ii) Ideal gas, constant conductivity and viscosity DT * 1 RePr Dt *2 T* Εc Dp * Dt * Εc Re * (2.41b) No new parameters appear. 2.10.5 Significance of the Governing Parameters Dimensionless temperature solution: T* f ( x * , y * , z * , t * ; Re, Pr , Gr , Ec ) (2.45) NOTE: Six quantities: p , T , Ts , V , L, g and five properties c p , k, , , and replaced by four dimensionless parameters: Re, Pr, Gr and Ec. , are Special case: negligible free convection and dissipation: Two governing parameters: T* f ( x * , y * , z * , t * ; Re, Pr ) (2.46) Geometrically similar bodies have the same solution when the parameters are the same. Experiments and correlation of data are expressed in terms of parameters rather than dimensional quantities. Numerical solutions are expressed in terms of parameters rather than dimensional quantities. 2.10.6 Heat Transfer Coefficient: The Nusselt Number Local Nusselt number: Nu x = f ( x* ; Re, Pr , Gr , Ec ) Special case: negligible buoyancy and dissipation: (2.51) 11 Nu x = f ( x* ; Re, Pr ) (2.52) Free convection, negligible dissipation Nu x = f ( x* ; Gr , Pr ) (2. 53) For the average Nusselt number, x * is eliminated in the above. 2.9 Scale Analysis A procedure used to obtain order of magnitude estimates without solving governing equations. CHAPTER 3 EXACT ONE-DIMENSIONAL SOLUTIONS 3.1 Introduction Exact solutions for simple cases are presented. Objective is to: Understand the physical significance of each term in the equations of continuity, Navier-Stokes, and energy. Identify the conditions under which certain terms can be neglected. General procedure: first determine the flow field and then the temperature field. 3.2 Simplification of the Governing Equations Simplifying Assumptions (1) Laminar flow (2) Constant properties (3) Parallel streamlines (fully developed flow): v (3.1) 0 Continuity (two-dimensional, constant density): u x 0 (3.2) 0 (3.3) It follows that 2 u x2 (3.1)-(3.3) result in significant simplifications. (4) Negligible axial variation of temperature. For axial flow, this condition leads to T x 0 (3.4) (3.4) is exact for certain channel flows and a reasonable approximation for others. The following are conditions that may lead to the validity of (3.4): (i) Parallel streamlines. (ii) Far away from the entrance region of a channel (infinitely long channels). 2 (iii) Uniform surface conditions. If (3.4) is valid everywhere: 2 T x2 0 Rotating flows, Fig. 3.2: The streamlines are concentric circles T 0 (3.6) 0 (3.7) and 2 T 2 3.3 Exact Solutions Applications of simplifications of Section 3.2. 3.3.1 Couette Flow This is shear driven flow Fluid is set in motion by moving channel surface Streamlines are parallel No axial variation Many terms in the Navier-Stokes equations and energy equation drop out Example 3.1: Couette Flow with Dissipation Upper plate moves with velocity U o Moving plate at temperature To Taking into consideration dissipation Determine temperature distribution and the rate of heat transfer at the moving plate Assume laminar flow Solution Review all assumptions Analysis Starting with the energy equation (3.5) 3 T t cp u T x T y v 2 T z w k 2 T x2 2 T y2 T (2.19b) z2 Dissipation: 2 2 3 u 2 x u x 2 v y v y w z w 2 z u y v x 2 v z w y 2 u 2 z w x (2.17) 2 Simplification of the above equations: (1) Incompressible fluid: t x y (a) 0 z (2) Infinite plates: x z (b) w 0 (3) Continuity and no-slip condition: v (f) 0 Navier-Stokes equation (2.10x): u t u u x v u y w u z p x gx 2 u x 2 2 u 2 y 2 z2 u (2.10x) Introduce the above simplifications into (2.10x): d 2u dy 2 (j) 0 Solution to (j) is u C1 y C 2 (k) Use boundary conditions on u to determine the two constants. Solution becomes u Uo y H (3.8) Dissipation function simplifies to: u y 2 (n) 4 Use (3.8) U o2 (o) H2 Energy equation (2.19b) simplifies to k d 2T U o2 dy 2 H2 0 (p) Solution is T To U o2 k 1 1 2 y2 H2 (3.9) Heat flux at the moving plate: q (H ) k dT ( H ) dy Use (3.9) q (H ) U o2 H Example 3.2: Flow in a Tube at Uniform Surface Temperature Study this example with attention to physical conditions and how they lead to simplifications of the governing equations. Follow the procedure of Example 3.1 above 3.3.3 Rotating Flow Note that all angular variation of velocity, pressure and temperature vanish in Concentric rotating flows. Example 3.3: Lubrication Oil Temperature in Rotating Shaft Study this example with attention to physical conditions and how they lead to simplifications of the governing equations. Follow the procedure of Example 3.1 (3.10) CHAPTER 4 BOUNDARY LAYER FLOW: APPLICATION TO EXTERNAL FLOW 4.1 Introduction Navier-Stokes equations and the energy equation are simplified using the boundary layer concept. Under special conditions certain terms in the equations can be neglected. Two key questions: (1) What are the conditions under which terms in the governing equations can be dropped? (2) What terms can be dropped? 4.2 The Boundary Layer Concept: Simplification of the Governing Equations 4.2.1 Qualitative Description Consider convection over a semiinfinite plate (Fig. 4.1). Under certain conditions the effect of viscosity is confined to a thin region near the surface called the velocity or viscous boundary layer, . Under certain conditions the effect of thermal interaction is confined to a thin region near the surface called the thermal boundary layer t . Conditions for the formation of the two boundary layers: Velocity boundary layer conditions: (1) Slender body (2) High Reynolds number (Re > 100) Thermal boundary layer conditions: (1) Slender body (2) High product of Reynolds and Prandtl numbers (Re Pr > 100) Peclet Number Pe RePr V L cp k c pV L k (4.1) 2 NOTE: Fluid velocity at the surface vanishes. Large changes in velocity across . Large changes in temperature across t. Viscosity plays no role outside . 4.2.2 The Governing Equations Assumptions: (1) Steady, (2) two-dimensional, (3) laminar, (4) uniform properties, (5) no dissipation, and (6) no gravity. Governing equations: u x u u v y u y 1 p x ν v v y 1 p y ν cp u T x T y k u x v v x v (2.3) 0 2 u 2 x 2 y2 2 2 v x2 2 T x2 u v y2 2 T y2 (2.10x) (2.10y) (2.19) 4.2.3 Mathematical Simplification Simplify the above equations based on boundary layer approximations. 4.2.4 Simplification of the Momentum Equations (i) Intuitive Argument Follow the intuitive argument leading to: 2 u 2 x 2 y2 u (4.2) and p y 0 (4.3) Therefore p p(x) It follows that p x dp dx dp dx (2.10x) simplifies to the following boundary layer x-momentum equation (4.4) 3 u u x 1 dp x u y v 2 ν u y2 (4.5) (ii) Scale Analysis We start by assuming that (4.6) 1 L Follow scale analysis leading to equations (4.2)-(4.4) Follow scale analysis leading to: 1 L (4.14b) Re L (4.14b) shows that (4.6) is valid when Re L 1. (4.14b) is generalized as 1 x (4.16) Re x 4.2.5 Simplification of the Energy Equation The energy equation for two-dimensional constant properties flow is c u T x v 2 T y k 2 T x2 T y2 (2.19) (2.19) is simplified for boundary layer flow using two arguments: (i) Intuitive Argument Follow the intuitive argument leading to: 2 2 T x2 T y2 (4.17) (2.19) simplifies to the following boundary layer energy T u x 2 T v y T y2 (4.18) (ii) Scale Analysis We start by assuming that t L 1 Follow scale analysis leading to equations (4.17) and (4.18) Follow scale analysis for the validity of (4.19). Two cases are considered: (4.19) 4 Case (1): t . Follow the argument leading to: 1 t L (4.24) PrRe L Thus t The criterion for t PrRe L 1 when L 1 (4.25) : Taking the ratio of (4.24) to (4.14b) gives 1 t (4.27) Pr Thus Case (2): t Pr when t 1 (4.28) . Follow the argument leading to: 1 t L Pr 1/3 (4.31) Re L Thus t 1 when Pr 1/3 Re L L The criterion for t 1 (4.32) : Taking the ratio of (4.31) to (4.14b) 1 Pr 1/3 t (4.33) Thus when Pr 1/3 t (4.34) 1 4.3 Summary of Boundary Layer Equations for Steady Laminar Flow Review all assumptions leading to the following boundary layer equations: Continuity: u x v y (2.3) 0 x-Momentum: u u x 1 dp dx u y v ν 2 u y2 (4.13) Energy: u T x Note the following: The continuity is not simplified. v T y 2 T y2 (4.18) 5 Solution to inviscid flow outside the boundary layer gives pressure gradient needed in (4.13). To include buoyancy effect, add [ g (T T ) ] to (4.13). 4.4 Solutions: External Flow For constant properties, velocity distribution is independent of temperature. First obtain the flow field solution and then use it to determine temperature distribution. 4.4.1 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Uniform Surface Temperature The basic problem is shown in Fig. 4.5. The plate is at uniform temperature T s . Upstream temperature is T . Apply all the assumptions summarized in Section 4.3. Governing equations: (continuity, momentum, and energy) are given in (2.3), (4.13), and (4.18). (i) Velocity Distribution. Determine: Velocity and pressure distribution. Boundary layer thickness (x) . Wall shear o (x ). (a) Governing equations and boundary conditions u x u u x v u y v y (2.3) 0 1 dp dx ν 2 u y2 (4.13) The velocity boundary conditions are: u(x,0) 0 (4.35a) v(x,0) 0 (4.35b) u( x, ) V (4.35c) u (0, y ) V (4.35d) (b) Scale analysis: boundary layer thickness, wall shear and friction coefficient. We showed that 6 1 x (4.16) Re x Define Darcy friction coefficient C f o Cf (4.37a) (1 / 2) V 2 Follow scale analysis leading to 1 Cf (4.37b) Re x (c) Blasius solution: similarity method Equations (2.3) and (4.13) are solved analytically by Blasius. For inviscid flow over flat plate u = V , v = 0, p = p = constant (4.38) Thus the pressure gradient is dp dx (4.39) 0 (4.39) into (4.13) u u x u y v ν 2 u (4.40) y2 (2.3) and (4.40) are solved by the method of similarity transformation. The basic approach is combining the two independent variables x and y into a single variable (x, y) and postulate that u/V depends on only. For this problem the correct form of the transformation variable is ( x, y) V y (4.41) νx Assume u V df d (4.42) Using (4.41) and (4.42), integration of the continuity gives 1 ν 2 V x v V df d Transform all derivatives in terms of f and 2 d3 f d 3 f( ) f (4.43) , substitute (4.42), (4.43) d2 f d 2 Boundary conditions (4.35a-4.35d) transform to 0 (4.44) 7 df (0) 0 d f (0) 0 (4.45a) (4.45b) df ( ) 1 d df ( ) 1 d (4.45c) (4.45d) NOTE: The momentum is transformed into an ordinary differential equation. Table 4.1 Blasius solution [1] Boundary conditions (4.35c) and (4.35d) coalesce into a single condition. (4.44) is solved by power series. The solution is presented in Table 4.1. From Table 4.1 we obtain x 5.2 Re x (4.46) Scaling gives 1 x (4.16) Re x From Table 4.1 we obtain Cf 0.664 Re x (4.48) Scaling gives Cf 1 Re x y 0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.0 5.2 5.4 5.6 6.0 7.0 8.0 (4.37b) (ii) Temperature Distribution. Determine: Temperature distribution. Thermal boundary layer thickness t. Heat transfer coefficient h(x). Nusselt number Nu(x). (a) Governing equation and boundary conditions V f x 0.0 0.02656 0.10611 0.23795 0.42032 0.65003 0.92230 1.23099 1.56911 1.92954 2.30576 2.69238 3.08534 3.28329 3.48189 3.68094 3.88031 4.27964 5.27926 6.27923 df d u V 0.0 0.13277 0.26471 0.39378 0.51676 0.62977 0.72899 0.81152 0.87609 0.92333 0.95552 0.97587 0.98779 0.99155 0.99425 0.99616 0.99748 0.99898 0.99992 1.00000 d2 f d 2 0.33206 0.33147 0.32739 0.31659 0.29667 0.26675 0.22809 0.18401 0.13913 0.09809 0.06424 0.03897 0.02187 0.01591 0.01134 0.00793 0.00543 0.00240 0.00022 0.00001 8 u T x v 2 T y T y2 (4.18) The boundary conditions are: T ( x,0) Ts (4.49a) T ( x, ) T (4.49b) T (0, y) T (4.49c) (b) Scale analysis: Thermal boundary layer thickness, heat transfer coefficient and Nusselt number Return to the results of Section 4.2.5: Case (1): t (Pr <<1) 1 t x Case (2): t PrRe x (4.50) (Pr >>1) 1 t x Pr 1/3 (4.51) Re x Scale analysis for h. Begin with h T ( x,0) y k Ts T (1.10) Using the scales, the above gives h k (4.52) t Case (1): t (Pr <<1). Substituting (4.50) into (4.52) h k PrRe x , x for Pr <<1 (4.53) Defining the local Nusselt number Nu x as Nu x hx k (4.54) Substituting (4.53) into (4.54) Nu x Case (2): t Pr 1/2 Rex , for Pr <<1 (4.55) (Pr >>1). Substituting (4.51) into (4.52) h k 1/3 Pr Re x , for Pr >>1 x The corresponding Nusselt number is (4.56) 9 Nu x Pr 1/3 Re x , for Pr >>1 (4.57) (c) Pohlhausen’s solution: Temperature distribution, thermal boundary layer thickness, heat transfer coefficient, and Nusselt number Equation (4.18) is solved analytically by Pohlhausen using similarity transformation. Define T T Ts Ts (4.58) (4.58) into (4.18) 2 u x v (4.59) y2 y Boundary conditions (4.49) become 0 (4.60a) (x, ) 1 (4.60b) (0, y) 1 (4.60c) (x,0) Combine x and y into a single variable (x, y) given by ( x, y) V y (4.41) νx Assume ( x, y) ( ) Velocity components u and v in (4.59) are given by Blasius solution u V v V df d 1 ν 2 V x (4.42) df d f (4.43) 0 (4.61) (4.41)-(4.43) into (4.59) d2 d 2 Pr d f( ) 2 d Using (4.41), the three boundary conditions (4.60a-4.60c) transform to 0 (4.62a) ( ) 1 (4.62b) ( ) 1 (4.62c) (0) Integration details of (4.61) are found in Appendix B. The temperature solution is 10 d2 f ) d 2 ( ) 1 d f d 2 0 Surface temperature gradient is 0.332 d2 f d 2 d d 1 .0 Pr Pr (4.63) Pr 2 d (0) d Pr 100 10 1 0.7(air) 0.8 (4.64) 0.1 0.6 d T Ts T Ts 0.4 The integrals in (4.63) and (4.64) are evaluated numerically. Boundary layer thickness t is determined from Fig. 4.6. The edge of the thermal layer is defined as the distance y where T T . This corresponds to T T Pr 0.2 0 2 4 0.01 6 8 y V 10 12 Fig. 4.6 Pohlhausen ' s solution for temper ature distribution for laminar flow over a semi - infinte isothermal flat plate Ts Ts 1 , at y t (4.65) The heat transfer coefficient h is determined using equation (1.10) T ( x,0) y k Ts T h (1.10) Using (4.41) and (4.58) into the above h( x ) k V d (0) νx d (4.66) Average heat transfer coefficient L 1 L h h( x )dx (2.50) 0 Substituting (4.66) into (2.50) and integrating h 14 x 2 k d (0) Re L L d The local Nusselt number is obtained by substituting (4.66) into (4.54) (4.67) 11 d (0) Re x d Nu x (4.68) The corresponding average Nusselt number is d (0) Re L d Table 4.2 gives d (0) / d for various values of Pr. Nu L (4.69) 2 d (0) d d (0) d 0.564 Pr 1/ 2 , 0.332 Pr 1/ 3 , d (0) d Pr < 0.05 (4.71a) 0.6 < Pr < 10 (4.71b) Pr >10 (4.71c) 0.339 Pr 1/ 3 , Table 4.2 d (0) Pr d 0.001 0.01 0.1 0.5 0.7 1.0 7.0 10.0 15.0 50 100 1000 0.0173 0.0516 0.140 0.259 0.292 0.332 0.645 0.730 0.835 1.247 1.572 3.387 4.4.2 Applications: Blasius Solution, Pohlhausen’s Solution, and Scaling Review Examples 4.1, 4.2, and 4.3. They illustrate the application of Blasius solution, Pohlhausen’s solution, and scaling to the solution of convection problems. 4.4.3 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Variable Surface Temperature Surface temperature varies with axial distance x according to Ts ( x) T Cx n (4.74) Assumptions: see Section 4.3. (i) Velocity Distribution. Blasius flow field solution is applicable to this case: u V v V where the similarity variable 1 ν 2 V x is defined as df d (4.42) df d f (4.43) 12 ( x, y) V y (4.41) νx (ii) Governing Equations for Temperature Distribution. Based on the assumptions listed in Section 4.3, temperature is governed by energy equation (4.18) u T x 2 T y v T (4.18) y2 The boundary conditions for this problem are: T ( x,0) (iii) Solution. Define Ts Cx n T (a) T ( x, ) T (b) T (0, y) T (c) as T T Ts Ts (4.58) Assume ( x, y) (4.75) ( ) Using (4.41)-(4.43), (4.58), (4.74) and (4.75), energy equation (4.18) transforms to (see Appendix C for details) d2 d 2 nPr df (1 d Pr d f( ) 2 d ) 0 (4.76) Boundary conditions (a)-(c) become 0 (4.77a) ( ) 1 (4.77b) ( ) 1 (4.77c) (0) Local heat transfer coefficient and Nusselt number are determined using (1.10) h T ( x,0) y k Ts T (1.10) Using (4.41), (4.58) and (4.72) into the above T ( x,0) y Cx n V d (0) νx d Substituting into (1.10) h( x ) k V ν d (0) x d (4.78) The average heat transfer coefficient for a plate of length L is defined in equation (2.50) 13 L 1 L h h( x)dx (2.50) 0 (4.78) into (2.50) h k d (0) Re L L d 2 (4.79) (4.78) into (4.54) gives the local Nusselt number d (0) Re x d Nu x (4.80) The corresponding average Nusselt number is d (0) Re L d Key factor: surface temperature gradient d (0) / d . Nu L (4. 81) 2 2.0 (iv) Results. (4.76) is solved numerically subject to boundary conditions (4.77). Fig. 4.8 gives d (0) / d for three Prandtl numbers. 30 d (0) d 10 1.0 Pr 0 Fig.4.8 0.5 n 1.0 0.7 1.5 d (0) for plate with varying surface temperature, d Ts T Cx n [4] 4.4.4 Laminar Boundary Layer Flow over a Wedge: Uniform Surface Temperature Assumptions: listed in Section 4.3. x-momentum equation: u u x v u y 1 dp dx ν 2 u y2 (4.13) Inviscid flow solution for V (x) is V ( x) Cx m C is a constant and m is defined as (4.82) 14 m (4.83) 2 Application of (4.13) to the inviscid flow outside the viscous boundary layer, gives 1 dp dx V x V Substituting into (4.13) u u x v 2 u V y V x u(x,0) 0 (4.85a) v (x,0) 0 (4.85b) ν u (4.84) y2 The boundary conditions are u( x, ) V ( x) Cx m (4.85c) Solution to the velocity distribution is obtained by the method of similarity. Define ( x, y) V ( x) νx y y C ν x (m as 1) / 2 (4.86) Assume u V ( x) dF d (4.87) Continuity equation (2.3), (4.86), and (4.87) give the vertical velocity component v v V ( x) ν m 1 1 m F xV ( x) 2 1 m dF d (4.88) 0 (4.89) (4.82) and (4.86)-(4.88) into (4.84) d 3F d 3 m 1 d 2F F 2 d 2 dF m d 2 m This is the transformed momentum equation. Boundary conditions (4.85) transform to dF (0) d F (0) dF ( ) d 0 0 1 (4.90a) (4.90b) (4.90c) Solution. (4.89) is integrated numerically. The solution gives F ( ) and dF / d . These in turn give the velocity components u and v. Temperature distribution. Start with the energy 15 2 u x v (4.59) y2 y Boundary conditions 0 (4.60a) (x, ) 1 (4.60b) (0, y) 1 (4.60c) (x,0) Where is defined as T T Ts Ts (4.58) ( ) (4.75) Assume is defined in (4.86). (4.86)-(4.88) and (4.75) into (4.59) and (4.60) d2 d 2 Pr d (m 1) F ( ) 2 d (0) 0 0 (4.91) (4.92a) ( ) 1 (4.92b) ( ) 1 (4.92c) Solution. Separating variables in (4.91), integrating twice and applying boundary conditions (4.92), gives the temperature solution as exp ( ) 1 exp 0 (m 1) Pr 2 (m 1) Pr 2 F ( )d d 0 (4.93) F ( )d d 0 (4.93) gives the temperature gradient at the surface d (0) d exp 0 (m 1)Pr 2 1 F ( )d d 0 The integrals in (4.93) and (4.94) are evaluated numerically. Results for d (0) / d are given in Table 4.3 (4.94) 16 Table 4.3 Surface temperature gradient d (0) and surface velocity d gradient F (0) for flow over an isothermal wedge m 0 0.111 0.333 1.0 wedge angle d (0) / d F (0) 0 0.3206 0.5120 0.7575 1.2326 o / 5 (36 ) / 2 (90o) (180o) at five values of Pr 0.7 0.8 1.0 5.0 10.0 0.292 0.331 0.384 0.496 0.307 0.348 0.403 0.523 0.332 0.378 0.440 0.570 0.585 0.669 0.792 1.043 0.730 0.851 1.013 1.344 Heat transfer coefficient h and Nusselt number Nu. Equation (1.10) gives h T ( x,0) y k Ts T h (1.10) Using (4.58), (4.75) and (4.86) into (1.10) gives h( x ) k V ( x) d (0) νx d (4.95) (4.95) into (4.54) gives the Nusselt number Nu x d (0) Re x d (4.96) where Re x is the local Reynolds number defined as Re x xV (x) ν (4.97) CHAPTER 5 APPROXIMATE SOLUTIONS: THE INTEGRAL METHOD 5.1 Introduction Seek approximate solution when: Exact solution is unavailable. Form of exact solution is not suitable or convenient. Solution requires numerical integration. The integral method gives approximate solutions. 5.2 Differential vs. Integral Formulation Example: boundary layer flow, Fig. 5.1. Differential formulation, Fig. 5.1a: the basic laws are formulated for a differential element dx dy. Solutions satisfy the basic laws exactly (at every point). Integral formulation, Fig. 5.1b: the basic laws are formulated for the element . dx Solutions satisfy the basic laws in an average sense (for section ). 5.3 Integral Method Approximation: Mathematical Simplification Reduction in the number of independent variables. Reduction of the order of the governing differential equation may result 5.4 Procedure Integral solutions are obtained for the velocity and temperature fields. The following procedure is used in obtaining integral solutions: 2 (1) Integral formulation of the basic laws: Conservation of mass, momentum, and energy. (2) Assumed velocity and temperature profiles: Several options. Polynomials are used in Cartesian coordinates. Assumed velocity and temperature profiles should satisfy known boundary conditions Assumed profile contains an unknown parameter or variable. (3) Determination of the unknown parameter or variable: Integral form of the basic law gives the unknown parameter or variable. 5.5 Accuracy of the Integral Method Different assumed profiles give different solutions and accuracy. Errors are acceptable in many engineering applications. Accuracy is not very sensitive to the form of an assumed profile. No procedure is available for identifying assumed profiles that will result in the most accurate solutions. 5.6 Integral Formulation of the Basic Laws 5.6.1 Conservation of Mass Boundary layer flow over porous plate of porosity P with mass injection. Conservation of mass for element dx , shown in Fig. 5.2 and enlarged in Fig. 5.3, gives dm x (a) dme dx dmo dx dme ( x) dme d dx u dy dx (5.1) v o Pdx mx dmx dx dx mx 0 dme is the external mass flow rate into element. dmo Fig. 5.3 5.6.2 Conservation of Momentum Application of the momentum theorem in the x-direction to the element Fx M x (out ) M x (in) Axial velocity u varies with x and y. dx (a) 3 Pressure p varies with x only (boundary layer approximation). dp )d 2 (p V ( x) me p p d ( p )dx dx Mx Mx dx o (1 dM x dx dx dx P)dx (a) forces (b) x momentum Fig. 5.4 M x = x-momentum, given by ( x) u 2 dy Mx (c) 0 o is wall shear, given by u x,0 y o (d) Equation (a) gives ( x) dp dx u x,0 1 P y d dx ( x) u 2 dy V d x dx 0 u dy V x Pv o (5.2) 0 NOTE: (1) (2) (3) (4) Equation (5.2) is the integral formulation of conservation of momentum. Equation (5.2) applies to laminar as well as turbulent flow. Although u is a function of x and y. Evaluating the integrals in (5.2) results in a first order ordinary differential equation with x as the independent variable. Special Cases: (i) Case 1: Incompressible fluid Boundary layer approximation gives dp dx dp dx The x-momentum equation for boundary layer flow is (4.12) 4 u x u v 1 dp x u y ν 2 u (4.5) y2 Applying equation (4.5) at y dp dx dp dx V ( x) Substituting (5.3) into (5.2) and noting that dV dx (5.3) is constant ( x) dV V ( x) dx ν u x,0 y 1 P d dx ( x) u 2 dy V x d dx u dy V 0 x Pv o (5.4) 0 (ii) Case 2: Incompressible fluid and impermeable flat plate For boundary layer flow over a flat plate dV dx dp dx dp dx 0, P (e) 0 For an impermeable plate vo (f) 0 (e) and (f) into (5.4) x v u x,0 y d V dx x udy 0 d dx u 2 dy (5.5) 0 5.6.3 Conservation of Energy Application of conservation of energy to the element t dx , neglecting changes in kinetic and potential energy, axial conduction, and dissipation: dE x dx dEe dx t Ex Ex Based on these assumptions, conservation of energy for the element gives dEc dEe dx (a) dEo dEx dx dx dEc dEe = energy added by external mass dEo = energy added by injected mass dEo Fig.5.6 E x = energy convected with boundary layer flow Formulating each term in (a) dEc k (1 P) T x,0 dx y (b) 5 dEe t ( x) d c pT dx u dy dx c p T (c) vo Pdx 0 dEo (d) c pTo vo Pdx t ( x) Ex (e) c p uT dy 0 Substituting (b)-(e) into (a) T x,0 y k1 P t d dx ( x) c p uTdy t d c pT dx 0 ( x) udy c p v o P To T (5.6) 0 NOTE: Equation (5.6) is integral formulation of conservation of mass and energy. Although u and T are functions of x and y, evaluation of the integrals gives a first order ordinary differential equation with x as the independent variable. Special Case: Constant properties and impermeable flat plate Setting P 1 in (5.6) T x,0 y t ( x) d dx u (T T )dy (5.7) 0 5.7 Integral Solutions Flow field solution. Temperature field solution. 5.7.1 Flow Field Solution: Uniform Flow over a Semi-Infinite Plate Integral form of governing equation: x v u x,0 y V d dx udy x d dx 0 u 2 dy (5.5) 0 Assumed velocity profile u( x, y) Boundary conditions a0 ( x) a1 ( x) y a2 ( x) y 2 a3 ( x) y 3 (a) 6 (1) u(x,0) 0 (2) u( x, ) V (3) u ( x, ) y 2 (4) u ( x,0) y2 0 0 NOTE: The fourth boundary condition is obtained by setting y = 0 in (2.10x) Boundary conditions give the four coefficients. Thus u V 3 y 2 3 1 y 2 (5.9) Note that the assumed velocity is in terms of the unknown variable (x). Boundary layer thickness. Use (5.5) to determine (x). Substituting (5.9) into (5.5) 3 vV 1 2 39 2 d V 280 dx Separating variables, integrating and noting that (0) 140 v 13 V d 0 (b) 0 x dx 0 Evaluating the integrals and rearranging 280 / 13 4.64 Re x Re x x (5.10) Friction coefficient. (5.10) into (5.9) gives u as a function of x and y. With the velocity distribution determined, friction coefficient C f is obtained using (4.36) and (4.37a) u x,0 y 3v o Cf 2 2 V x V /2 V /2 Using (5.10) to eliminate (x) Cf Compare with Blasius solution: 0.646 Re x (5.11) 7 5.2 x Re x , Blasius solution (4.46) , Blasius solution (4.48) and Cf 0.664 Re x Note the small error in prediction C f . 5.7.2 Temperature Solution and Nusselt Number: Flow over a Semi-Infinite Plate (i) Temperature Distribution A leading section of the plate of length x o is insulated and the remaining part is at uniform temperature Ts . Assume laminar, steady, two-dimensional, constant properties boundary layer flow and neglect axial conduction and dissipation. Determine t , h(x), and Nu(x). Must determine flow field u( x, y) and temperature T(x,y). Flow field solution of Section 5.7.1 applies to this case, equation (5.9). Equation (5.7) gives the integral formulation of conservation of energy for this problem t ( x) T x ,0 y d dx u (T T )dy (5.7 0 Assumed temperature profile T ( x, y ) Boundary conditions (1) T ( x,0) (2) T ( x, (3) T ( x, y 2 (4) Ts t) t) T ( x,0) y2 T 0 0 b0 ( x) b1 ( x) y b2 ( x) y 2 b3 ( x) y 3 (a) 8 NOTE: The fourth condition is obtained by setting y 0 in the energy equation (2.19). Boundary conditions give the four coefficients. Thus T ( x, y ) Ts (T Ts ) 1 y3 2 t3 3 y 2 t (5.13) (5.9) and (5.13) into (5.7), evaluating the integral, gives 3 2 T Ts t d (T dx 2 3 20 Ts ) V t 3 280 4 t (5.14) Simplification of (5.14). Note that t 1 , for Pr (5.15) 1 It follows that 3 280 2 3 20 t t (5.14) simplifies to 2 d V dx 10 t t (b) where 280 x 13 V (c) Boundary condition t ( xo ) 0 (h) Solution to (b) t 4.528 x Pr 1/3 Re x 1/2 1 xo x 3/ 4 1/ 3 (5.17b) (ii) Nusselt Number. Local Nusselt number is defined as Nu x hx k (j) h is the local heat transfer coefficient given by T ( x,0) y Ts T k h (k) Using (5.13) into (k) h( x ) 3 k 2 t (5.19) 9 Eliminating t by using (5.17b) k h( x) 0.331 1 x xo x 1/ 3 3/ 4 Pr 1/3 Re x1/2 (5.20) Substituting into (j) Nu x 0.331 1 xo x 3/ 4 1/ 3 Pr 1/3 Re x1/2 (5.21) (iii) Special Case: Plate with no Insulated Section Set x o 0 in (5.17b), (5.20) and (5.21) 4.528 t x Pr 1/3 (5.23) Re x 1/2 k h( x) 0.331 Pr 1/3 Re x1/2 x (5.24) Nu x 0.331Pr 1/3 Re x 1/2 (5.25) Examine the accuracy of the local Nusselt number. For Pr 10 equation (4.72c) gives Pohlhausen’s solution Nu x 0.339 Pr 1/ 3 Re x , for Pr 10 (4.72c) Comparing this result with integral solution (5.25) shows that the error is 2.4%. Example 5.1: Laminar Boundary Layer Flow over a Flat Plate: Uniform Surface Temperature This is a repeat of the Section 5.7.1 and 5.72, using assume linear velocity and temperature profiles. A linear profile gives less accurate flow and heat transfer results. The procedure of the previous sections is repeated in this example. The following is a summary of the results. Assumed velocity u Boundary conditions (1) u(x,0) 0 (2) u( x, ) V Velocity solution a0 a1 y (b) 10 u V y (c) Integral solution to x 12 Re x (5.26) b1 y (f) Assumed temperature T b0 Boundary conditions (1) T ( x,0) Ts (2) T ( x, t ) T Temperature solution T Ts (T Ts ) y (g) t Integral solution to t 12ν 1 1/3 V Pr t x 1 ( xo / x) 3 / 4 1/ 3 (o) Solution to local Nusselt number Nu x 0.289 Pr 1/3 Re x 1 ( xo / x) 3 / 4 Special Case: no insulated section, set xo Nu x 1/ 3 (5.27) 0 in (5.27) gives 0.289 Pr 1/3 Re x (5.28) / x and Nu x / Pr 1 / 3 Rex1/2 with integral Comments. Table 5.1 compares exact solutions for results for the case of a plate with no insulated section based on assumed linear and polynomial profiles Note that the integral method gives more accurate prediction of Nusselt number than of the boundary layer thickness . Table 5.1 Solution Exact (Blasius/ Pohlhausen) 3rd degree polynomial Linear x Re x 5.2 4.64 3.46 Nu x Pr 1 / 3 Re1 / 2 0.332 0.339 0.289 11 5.7.3 Uniform Surface Flux Plate with an insulted leading section of length x o . Plate is heated with uniform flux q s along its surface x xo . Steady, two-dimensional, laminar flow. Determine surface temperature and the local Nusselt number. Surface temperature is unknown. Solution Newton’s law of cooling gives h( x ) qs Ts ( x ) T Introducing the definition of the Nusselt number Nu x qs x k Ts ( x ) T (b) Need surface temperature Ts (x). Use the integral form of the energy equation to determine Ts (x) T x,0 y t ( x) d dx u (T T )dy (5.7) 0 u( x, y) for a third degree polynomial is given by (5.9) u V 3 y 2 1 y 2 3 (5.9) Assume temperature T ( x, y) T Boundary conditions (1) k (2) T x, T x,0 y t T qs b0 b1 y b2 y 2 b3 y 3 (c) 12 (3) T x, y 2 (4) t 0 T x,0 y2 0 Application of boundary conditions give the coefficients in (c) 2 3 T ( x, y ) T Surface temperature. Set y 1 y 3 qs 3 t2 k y t (5.29) 0 in the above Ts ( x ) T ( x,0) 2 qs 3 k T (5.30) t (5.30) into (b) 3 2 Nu x Must determine x t (5.31) x t. (5.9) and (5.29) into (5.7), evaluate the integral d dx V 2 t 1 10 Simplify for Prandtl numbers larger than unity, 10 t t / (e) 1 3 d dx V 3 1 140 t t Thermal boundary layer thickness. Integrating and use boundary condition t ( x o ) 0 , gives 1/ 3 t 10 V (x xo ) (j) Use (5.10) to eliminate in (j), rearrange t 3.594 x Pr 1/3 Re 1/2 x xo x 1 1/ 3 (5.32) Surface temperature. (5.32) into (5.30) gives Ts ( x ) T q 2.396 s 1 k xo x Local Nusselt number. (5.32) into (5.31) gives 1/ 3 x Pr 1/3 Re 1/2 x (5.33) 13 Nu x 0.417 1 xo x 1/ 3 Pr 1/3 Re 1/2 x Special Case: Plate with no insulated section, set xo 0 in (5.33) and (5.34) q x Ts ( x ) T 2.396 s 1/3 k Pr Re1/2 x Nu x 0.417 Pr 1/3 Re1/2 x (5.34) (5.35) (5.36) Compare with differential formulation solution: Nu x 0.453 Pr 1/3 Re1/2 x (5.37) CHAPTER 6 HEAT TRANSFER IN CHANNEL FLOW 6.1 Introduction (1) Laminar vs. turbulent flow Flow through tubes, transition Reynolds number Re D t is Re Dt uD 2300 (6.1) (2) Entrance vs. fully developed region Classification based on velocity and temperature profiles: (i) Entrance region (ii) Fully developed region (3) Surface boundary conditions Two common boundary conditions:: (i) Uniform surface temperature (ii) Uniform surface heat flux (4) Objective. Objective depends on surface thermal boundary condition: (i) Uniform surface temperature. Determine axial variation of: (1) Mean fluid temperature (2) Heat transfer coefficient (3) Surface heat flux (ii) Uniform surface flux. Determine axial variation of: (1) Mean fluid temperature (2) Heat transfer coefficient (3) Surface temperature 6.2 Hydrodynamic and Thermal Regions: General Features Fluid enters with uniform velocity and temperature. (1) Entrance region. Extends from the inlet to the section where the boundary layer thickness reaches the center of channel. (2) Fully developed region. This zone follows the entrance region. 2 6.2.1 Flow Field (1) Entrance Region (Developing Flow, 0 x Lh ). Name: hydrodynamic entrance region. Length: L h (hydrodynamic entrance length). Streamlines are not parallel. Core velocity u c increases with distance Pressure decreases with distance ( dp / dx 0 ). D/ 2. r uc Vi x u u Lh (2) Fully Developed Flow Region. x fully developed Lh Fig. 6.1 Streamlines are parallel ( vr 0). u / x 0 for two-dimensional incompressible fluid. 6.2.2 Temperature Field (1) Entrance Region (Developing Temperature, 0 x Lt ) Name: Thermal entrance region. Length: Lt (thermal entrance length). Core temperature Tc is uniform, Tc Ti . D/2 t (2) Fully Developed Temperature Region. x r Tc Ts Ts T Vi Ti Lt x Ts t Temperature varies radially and axially, T / x 0. fully developed Lt Fig. 6.2 6.3 Hydrodynamic and Thermal Entrance Lengths 6.3.1 Scale Analysis (1) Hydrodynamic Entrance Length Lh . Starting with external flow result (4.16) 1 x Applying (4.16) to a tube at x (4.16) Re x Lh : D and Re Lh Substituting (b) into (4.16) and rearranging Re D Lh D (b) 3 1/ 2 Lh / D Re D (6.2) ~1 (2) Thermal Entrance Length Lt . Starting with external flow result (4.24) t Applying (4.24) at x ~ L Re L1 / 2 Pr 1/ 2 (4.24) Lt : t D and Re Lt Re D Lt D (b) Substituting (b) into (4.24) and rearranging Lt / D Re D Pr 1/ 2 (6.3) ~1 (6.2) and (6.3) give Lt ~ Pr Lh (6.4) 6.3.2 Analytic and Numerical Solutions: Laminar Flow (1) Hydrodynamic Entrance Length L h . Results for L h : Lh De Table 6.1 Entrance length coefficients C h and C t [1] (6.5) C h Re D e Ct Table 6.1 gives C h Compare with scaling: Lh / D Re D geometry uniform surface flux uniform surface temperature 0.056 0.043 0.033 0.09 0.066 0.041 0.085 0.057 0.049 0.075 0.042 0.054 0.011 0.012 0.008 1/ 2 (6.2) ~1 a a b Rewrite (6.5) a/b =1 a 1/ 2 Lh / De Re De Ch Ch 1/ 2 (a) Example: Rectangular channel, aspect ratio 2, Table 6.1 gives C h 0.085. Substituting this value into (a), gives a/b = 2 b a b a/b = 4 4 1/ 2 Lh / De Re D e 0.085 1 / 2 (b) 0.29 Scaling replaces 0.29 by unity. (2) Thermal Entrance Length Lt . Lt depends on surface boundary conditions: Two cases: (i) Uniform surface temperature. (ii) Uniform surface flux. Solution Lt De (6.6) C t PrRe D Table 6.1 gives C t for both cases. Compare with scaling. Rewrite (6.6) Lt / De PrRe D 1/ 2 Ct 1/ 2 (c) Scaling gives 1/ 2 Lt / D Re D Pr (6.3) ~1 Example: Rectangular channel, aspect ratio 2, Table 6.1 gives Ct this value into (c), gives 0.049. Substituting 1/ 2 Lt / De PrRe De 0.049 1 / 2 (d) 0.22 Scaling replaces 0.22 be unity. Turbulent flow: L Lh Lt L D (6.7) 10 6.4 Channels with Uniform Surface Heat Flux q s Inlet mean temperature: Tmi Tm (0) . L Determine: (1) Total heat transfer rate q s . (2) Mean temperature variation Tm (x). (3) Surface temperature variation Ts (x ). Total heat transfer rate Tmi 0 x qs Fig. 6.3 Tm (x) 5 qs q s As (6.8) qs P x As = surface area P = perimeter Mean temperature Tm (x) . Conservation of energy between inlet and section x: qs qs P x mcp [Tm ( x) Tmi ] or Tm ( x) qs P x mc p Tmi (6.9) Surface temperature Ts (x) . Newton’s law of cooling gives qs h( x) Ts ( x) Tm ( x) or Ts ( x) qs h( x ) Tm ( x) Using (6.9) Ts ( x) Tm i qs Px mc p 1 h( x ) (6.10) NOTE: Determining Ts (x) requires knowing h(x). To determine h(x): Must know if: Flow is Laminar or turbulent. Entrance or fully developed region 6.5 Channels with Uniform Surface Temperature Inlet mean temperature: Tmi Tm (0) . Ts Determine: (1) Mean temperature variation Tm (x). (2) Total heat transfer rate q s between x location x. (3) Surface heat flux variation q s (x). Tmi 0 dx dqs Tm Tm Conservation of energy to element m c p dTm m 0 and Mean temperature variation Tm (x). dqs Tm (x) x dx (a) Fig. 6.4 dTm dx dx 6 Newton's law: dqs h( x) Ts Tm ( x) Pdx (b) P h( x)dx m cp (c) Combine (a) and (b) dTm Ts Tm ( x) Integrating (c) x T ( x) Ts ln m Tmi Ts P m cp Definite h (6.11) h( x)dx 0 x h 1 x h( x)dx (6.12) 0 (6.12) into (6.11), solve Tm (x) Tm ( x) (Tmi Ts ) exp[ Ts Ph x] mcp (6.13) NOTE: Determining Tm (x) requires knowing h(x). To determine h(x): Must know if: Flow is laminar or turbulent. Entrance or fully developed region Heat transfer rate. Conservation of energy: qs (6.14) m c p [Tm ( x) Tmi ] Surface heat flux.: Newton’s law: q s ( x) h( x)[Ts Tm ( x)] (6.15) 6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number Nu D 6.6.1 Scale Analysis Estimate h(x) and Nu D . Tube: radius ro , surface temperature T s , mean temperature Tm . Fourier’s law and Newton’s law: h Scaling (6.16) T (ro , x) k r Tm Ts r qs ro 0 Tm (6.16) Fig. 6.5 Ts 7 Tm k Ts t h~ Tm Ts or k h~ (6.17) t The Nusselt number hD k Nu D Use (6.17) D Nu D ~ (6.18) t Fully developed region: t (x ) ~ D, equation (6.18) gives Nu D ~ 1 (fully developed) Entrance region: Need to scale t ( t (x ) < (6.19) D). For external flow t ~ x Pr 1/ 2 Re x 1/ 2 (4.24) (4.24) into (6.18) Nu D ~ D 1 / 2 1/2 Pr Re x x (c) Expressing Re x in terms of Re D Re x ux ν uD x ν D Re D x D (d) Substitute (d) into (c) D Nu D ~ x 1/2 Pr 1 / 2Re1/2 D (6.20a) Rewrite Nu D PrRe D x/D 1/ 2 ~1 Scaling estimates (6.19) and (6.20) will be compared with exact solutions. (6.20b) 8 6.6.2 Basic Considerations for the Analytical Determination of Heat Flux, Heat Transfer Coefficient and Nusselt Number r qs Need to determine velocity and temperature distribution. Assume: fully developed velocity Neglect axial conduction Section outline: ro 0 Tm Definitions Governing equations for determining: Ts Fig. 6.5 (i) Surface heat flux (ii) Heat transfer coefficient (iii) Nusselt number (1) Fourier’s law and Newton’s law. Surface heat flux. Fourier’s law gives surface heat flux q s qs k T x, ro r (a) Define dimensionless variables T Ts , Ti Ts x/D , Re D Pr r , ro R vx v k Ts ro Ti vx , u vr vr , u Re D uD ν (6.21) Substitute into (a) qs ( ) 0( ,1) R (6.22) Heat transfer coefficient. Define h h q" s Tm Ts (6.23) Combine (6.22) and (6.23) h( ) where m k (Ts Ti ) ro (Tm Ts ) ( ,1) R k ro 1 m( ) ( ,1) R (6.24) is defined as m Tm Ts Ti Ts (6.25) Nusselt number. Define: Nu ( ) h( ) D k h( )2ro k (6.26) ( ,1) R (6.27) (6.24) into (6.26) Nu ( ) 2 m( ) 9 (2) The Energy Equation. Review assumptions on energy equation (2.24). T r c p vr T z vz k 2 1 T r r r r T (2.24) z2 Replace z by x, use dimensionless variables: vx 2 Re D Pr vr 4 R R R R R 1 ( Re D Pr) 2 2 2 (6.28) where Pe Re D Pr , Peclet number (6.29) Pe (6.30) Neglect conduction for PrRe D 100 Thus, under such conditions, (6.28) becomes vx 2 Re D Pr vr 4 R R R R R (6.31) (3) Mean (Bulk) Temperature Tm . Define: ro mc p Tm c p v x T 2 rdr (a) v x 2 rdr (b) 0 Mass flow rate m is given by ro m 0 (b) into (a), assume constant properties ro v x Trdr Tm 0 (6.32a) ro v x rdr 0 Dimensionless form: 1 v x R dR m Tm Ts Ti Ts 0 1 v x R dR 0 6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region 6.7.1 Definition of Fully Developed Temperature Profile (6.32b) 10 Far away from the entrance ( x / d fully developed. 0.05Re D Pr ), temperature profile becomes To define fully developed temperature, introduce the dimensionless temperature Ts ( x ) T ( r , x ) Ts ( x) Tm ( x) (6.33) Fully developed temperature is defined as a profile in which is independent of x: (r ) (6.34) 0 (6.35) (6.34) gives x (6.33) and (6.35) give Ts ( x) T (r , x) x Ts ( x) Tm ( x) x Expand and use the definition of (6.36a) 0 in (6.33) dTs dx T x (r ) dTs dx dTm dx 0 (6.36b) 6.7.2 Heat Transfer Coefficient and Nusselt Number Examine h and Nu in the fully developed region. Fourier’s and Newton’s law: T (ro , x) r Tm Ts k h (6.16) Use (6.33) to eliminate T (ro , x) / r . (6.16) gives h k d (ro ) = constant dr (6.37) IMPORTANT CONCLUSOIN: THE HEAT TRANSFER COEFFICIENT IN THE FULLY DEVELOPED REGION IS CONSTANT INDEPENDET OF LOCATION. Nusselt number Nu D hD k D d (ro ) dr (6.38) Scaling estimate based on limiting case of entrance region: Nu D ~ 1 (fully developed) Scale estimate based on fully developed region: (6.19) 11 Scale T (ro , x) / r as T (ro , x) Ts Tm ~ r D Substitute into (6.16) h~ k D (6.39) Substitute (6.39) into (6.38) (fully developed) Nu D ~ 1 (6.40) 6.7.3 Fully Developed Region for Tubes at Uniform Surface flux r Determine: qs T (i) Surface temperature Ts (x). (ii) Heat transfer coefficient. u 0 D x qs Fig. 6.6 Newton’s law qs h Ts ( x) Tm ( x) (a) Since q s and h are constant it follows that Ts ( x) Tm ( x) constant (b) Differentiate dTs dx dTm . dx (c) dTs dx (d) (c) into (6.36b) T x (c) and (d) T x dTs dx dTm (for constant q s ) dx (6.41) Unknowns: T (r, x), Tm (x) and Ts (x) Conservation of energy: q s Pdx mc pTm or mc p Tm m dTm dx dx Tm Tm dx qs Fig. 6.7 dTm dx dx 12 dTm dx qs P = constant mc p (6.42) dTs dx dTm q s P = = constant dx mc p (6.43) Substitute (6.42) into (6.41) T x Integrate(6.43) qs P x C1 mc p Tm ( x) (e) Use inlet condition Tm (0) Tmi (f) Solution (e) becomes Tm ( x) qs P x mc p Tmi (6.44) Need to determine T (r, x) and Ts (x). This requires solving the differential form of the energy equation. Set vr 0 in energy equation (2.24) T x c p vx k T r r r r (6.45) Fully developed flow axial velocity vx r2 2u 1 ro2 (6.46) (6.43) and (6.46) into (6.45) c p 2u 1 However, m ro2 u and P r2 2 ro qs P m c p k T r r r r (g) 2 ro , equation (g) becomes 4q s r2 1 2 ro ro k T r r r r (6.47) Boundary conditions are: T (0, x) 0 r T (ro , x) k qs r Integrate (6.47) (6.48a) (6.48b) 13 4 r2 qs ro 2 r4 T r kr 2 4ro Boundary condition (6.48a) gives f (x) (h) f x 0. Equation (h) becomes 4q s kro T r r3 r 2 4ro2 Integrate again r2 4 4q s kro T (r , x) r4 (6.49) g ( x) 16ro2 The integration “constant” is g (x) . Use Tm (x) to determine g (x). Substitute (6.46) and (6.49) into (6.32a) 7 ro q s (6.50) Tm ( x) g ( x) 24 k Equate (6.44) and (6.50) gives g (x) g ( x) Tmi 7 ro q s 24 k Pq s x mc p r4 16ro 7 ro q s 24 k 11 ro q s 24 k Pq s x mc p (6.51) (6.51) into (6.49) T (r , x) Tmi Set r 4q s kro r2 4 2 Pq s x mc p (6.52) ro in (6.52) to obtain Ts (x) Ts ( x) Tmi (6.53) (6.44), (6.52) and (6.53) into (6.33) gives (r ) 24 1 2 (r ) 1 r 11 ro2 r4 24 Pq s 7 x x 11 mc p 11 2 4ro (6.54) Differentiate (6.54) and substitute into (6.38) gives Nu D 48 11 4.364 (6.55) NOTE: (6.55) applies to laminar fully developed velocity and temperature in tubes with uniform surface heat flux. The Nusselt number is independent of Reynolds and Prandtl numbers. Scaling gives Nusselt as Nu D ~ 1 (6.40) 14 This compares favorable with (6.55). 6.7.4 Fully Developed Region for Tubes at Uniform Surface Temperature Determine: Nusselt number Solve the energy equation for the fully developed region Neglect axial conduction and dissipation. Energy equation: set vr 0 in (2.24) c p vx T x k T r r r r (6.45) Boundary conditions T (0, x) 0 r T (ro , x) Ts (6.56a) (6.56b) Axial velocity for fully developed flow is vx 2u 1 r2 (6.46) ro2 Use (6.36a) to Eliminate T / x in (6.45) Ts ( x) T (r , x) x Ts ( x) Tm ( x) x For uniform Ts ( x) 0 (6.36a) Ts , above gives T x Ts T (r , x) dTm Ts Tm ( x) dx (6.57) (6.46) and (6.57) into (6.45) 2 cpu 1 r 2 Ts T (r , x) dTm ro2 Ts Tm ( x) dx k T r r r r (6.58) Solution: (6.58) was solved using an infinite power series. Solution gives the Nusselt number as Nu D 3.657 (6.59) 6.7.5 Nusselt Number for Laminar Fully Developed Velocity and Temperature in Channels of Various Cross-Sections Table 6.2 lists Nusselt numbers for channels of various cross-sections. Two cases: (1) uniform surface heat flux and (2) uniform surface temperature. Nusselt number of Non-circular channels is based on the equivalent diameter. Scaling estimate: Table 6.2 Nu D ~ 1 (fully developed) (6.40) Nusselt number for laminar fully developed conditions in channels of various cross-sections [3] Nusselt number Nu D Table 6.2: Nusselt number ranges from 2.46 to 8.235. a b Channel geometry Uniform surface flux Uniform surface temperature 4.364 3.657 1 3.608 2.976 2 4.123 3.391 4 5.331 4.439 8 6.49 5.597 8.235 7.541 3.102 2.46 6.8 Thermal Entrance Region: Laminar Flow through Tubes a 6.8.1 Uniform Surface Temperature: Graetz Solution b a Laminar flow. b Fully developed inlet velocity. b a Neglect axial conduction (Pe > 100). a Uniform surface temperature Ts . b Fully developed flow: vr 0 15 (3.1) Axial velocity vz 1 dp 2 2 (r ro ) 4 dz (3.12) r Ts (3.12) expressed in dimensionless form vx vx u 2 2(1 R ) T Ti (6.61) u 0 x t (3.1) and (6.61) into energy equation (6.31) 1 1 R2 2 1 R R R R Fig. 6.8 (6.62) Boundary conditions ( ,0) R 0 (6.63a) ( ,1) 0 (6.63b) (0, R) 1 (6.63c) Analytic and numerical solutions to this problem have been obtained. Review analytic solution leading to: 16 (i) Mean temperature, m( ) Gn 8 m 2 2 exp( 2 (6.66) ) n n n 0 (ii) Local Nusselt number, Nu ( ) G n exp( 2 2 n ) n 0 Nu (66.7) Gn 2 2 n 0 exp( 2 2 n ) n (iii) Average Nusselt number, Nu ( ) h ( )D k (f) and G n for 0 n 10. Table 6.4 gives Nu ( ) and Nu ( ) RESULTS Table 6.3 lists values of n Nu ( ) at selected values of the axial distance . Fig. 6.9 gives the variation of Nu ( ) and Nu ( ) along a tube. Table 6.4 Table 6.3 Uniform surface temperature [4] n 0 1 2 3 4 5 6 7 8 9 10 n 2.70436 6.67903 10.67338 14.67108 18.66987 22.66914 26.66866 30.66832 34.66807 38.66788 42.66773 Gn 0.74877 0.54383 0.46286 0.41542 0.38292 0.35869 0.33962 0.32406 0.31101 0.29984 0.29012 Local and average Nusselt number for tube at uniform surface temperature [5] = x/D Re D Pr 0 0.0005 0.002 0.005 0.02 0.04 0.05 0.1 Nu ( ) Nu ( ) 12.8 8.03 6.00 4.17 3.77 3.71 3.66 3.66 19.29 12.09 8.92 5.81 4.86 4.64 4.15 3.66 17 Average Nu Nusselt number Local Nu Fig. 6.9 x/D ReD Pr Local and average Nusselt number for tube at uniform surface temepratu re [4] NOTE: (1) The average Nusselt number is greater than the local Nusselt number. (2) Asymptotic value of Nusselt number of 3.657 is reached at 0.05 . Thus (6.69) Nu( ) 3.657 (3) Evaluate fluid properties at the mean temperatures Tm , defined as Tm Tmi Tmo 6.8.2 Uniform Surface Heat Flux Repeat Graetz entrance problem replacing the uniform surface temperature with uniform heat flux. (6.70) 2 r qs T Ti u 0 D x t qs Inlet velocity is fully developed. Fig. 6.10 Energy equation is 1 1 R2 2 1 R R R R (6.62) 18 Boundary conditions ( ,0) ) R (6.71a) 0 q s ro k (Ti Ts ) ( ,1) R (6.71b) (6.71c) (0, R) 1 Solution. Local Nusselt number: 1 Nu ( ) hx k 11 48 1 2 An exp( 2 2 Table 6.5 (6.72] ) n n 1 n The average Nusselt number is given by Nu ( ) hx k 11 48 1 2 The eigenvalues in Table 6.5 Limiting case: An n 1 2 n 1 exp( 2 2 2 n 1 ) 2 n (6.73] and the constant An are listed (fully developed) Nu ( ) Uniform surface flux [4] 48 4.364 11 1 2 3 4 5 6 7 8 9 10 2 n 25.6796 83.8618 174.1667 296.5363 450.9472 637.3874 855.8495 1106.3290 1388.8226 1703.3279 (6.74) Average Nu Nusselt number Local Nu x/D ReD Pr Fig. 6.11 Local and average Nusselt number for tube at uniform surface heat flux [4] An 0.198722 0.069257 0.036521 0.023014 0.016030 0.011906 0.009249 0.007427 0.006117 0.005141 CHAPTER 7 FREE CONVECTION 7.1 Introduction 7.2 Features and Parameters of Free Convection (1) Driving Force. Requirements (i) Gravitational field (ii) Density change with temperature (2) Governing Parameters. Two parameters: (i) Grashof number T ) L3 g (Ts Grashof number = GrL 2 (7.1) (ii) Prandtl number is the Coefficient of thermal expansion, also known as compressibility factor. For ideal gases it is given by 1 , for ideal gas T (2.21) Rayleigh number Ra L GrL Pr g (Ts T ) L3 ν2 Pr = g (Ts T ) L3 ν (7.2) (3) Boundary Layer. Flow: Laminar, turbulent, or mixed. Boundary layer approximations are valid for Ra x 10 4. (4) Transition from Laminar to Turbulent Flow. For vertical plates: transition Rayleigh number, Rax t , is Rax t 109 (7.3) (5) External vs. Enclosure Free Convection. (i) External free convection: surface is immersed in infinite medium. (ii) Enclosure free convection. Free convection takes place inside closed volumetric regions. (6) Analytic Solutions. Velocity and temperature fields are coupled. Momentum and energy equation must be solved simultaneously. 2 7.3 Governing Equations Approximations: (1) Constant density, except in evaluating gravity forces. (2) The Boussinesq approximation (relates density change to temperature change). (3) No dissipation. Continuity, momentum, and energy equations are obtained from equations (2.2), (2.29) and (2.19), respectively u x u x u u y v u v x v y u x 1 y T x v (p (7.4) 0 1 g (T T ) v v y (p p ) v( 2 T y v( p ) T x2 2 v x2 2 u 2 x 2 y2 2 v y2 u ) ) T y2 (7.7) Continuity equation (7.4) is unchanged x-component of the Navier-Stokes equations simplifies to u x v u y v βg T T 2 u y2 (7.8) Energy equation (7.7) u T x v T y 2 α T y2 (7.9) (7.4), (7.8), and (7.9) contain three unknowns: u, v, and T. Momentum and energy are coupled. 7.4 Laminar Free Convection over a Vertical Plate: Uniform Surface Temperature Uniform temperature T s (Fig. 7.1). Infinite fluid at temperature T . Determine: velocity and temperature distribution. 7.4.1 Assumptions. Note all assumptions listed in this section. 7.4.2 Governing Equations (7.6) 2 7.3.1 Boundary Layer Equations u (7.5) 3 u x u u x v v y u y (7.4) 0 2 v βg T T u y2 (7.8) 2 u where v x α y (7.10) y2 is defined as T T Ts T (7.11) 7.4.3 Boundary Conditions. Velocity: (1) (2) (3) (4) u(x,0) 0 v(x,0) 0 u(x, ) 0 u(0, y) 0 Temperature: (5) (x,0) 1 (6) (x, ) 0 (7) (0, y) 0 7.4.4 Similarity Transformation. Introduce the similarity variable Grx 4 1/ 4 y x (7.14) where g (Ts T ) x 3 Grx (7.15) ν2 Let ( ) (7.16) Grx d x d (7.20) ( x, y) u 2v v (Grx )1/ 4 x Continuity gives v 1/ 4 (4) d d (7.21) 3 (7.20) and (7.21) into (7.8) and (7.10) and using (7.11) and (7.16), gives d3 d 3 3 d2 d 2 d 2 d 2 0 (7.22) 4 d2 d 2 3Pr d d 0 (7.23) Transformation of boundary conditions: Velocity: d (0) 0 d (2) (0) 0 d ( ) (3) 0 d d ( ) (4) 0 d (1) Temperature: (1) (0) 1 (2) ( ) 0 (3) ( ) 0 The problem is characterized by a single parameter which is the Prandtl number. 7.4.5 Solution. (7.22) and (7.23) and their five boundary conditions are solved numerically. The solution is presented graphically in Figs. 7.2 and 7.3. Fig. 7.2 gives the 5 7.4.6 Heat Transfer Coefficient and Nusselt Number. Fourier’s law and Newton’s law: T ( x,0) y Ts T k h Express in terms of (7.24) and h k dT d (0) T d d y Ts Use(7.11) and (7.14) h 1/ 4 k Grx x 4 d (0) d (7.25) Local Nusselt number Nu x Grx 4 hx k 1/ 4 d (0) d (7.26) Average h 1 L h Table 7.1 [1,2] L h( x)dx (2.50) 0 (7.25) into (2.50), and performing the integration h 4 k GrL 3L 4 1/ 4 d (0) d (7.27) Average Nusselt number is Nu L hL k 4 GrL 3 4 1/ 4 d (0) d (7.28) Solution depends on a single parameter which is the Prandtl number. d (0) Numerical solution gives , listed in Table 7.1. d Special Cases Very small and very large Prandtl numbers: Nu x Nu x 0.600 ( PrRa x )1/4 , 0.503( PrGrx )1/4 , Pr 0 Pr (7.29a) (7.29b) 7.5 Laminar Free Convection over a Vertical Plate: Uniform Surface Heat Flux Assumptions: Same as constant temperature plate. Pr _ d (0) 0.01 0.03 0.09 0.5 0.72 0.733 1.0 1.5 2.0 3.5 5.0 7.0 10 100 1000 0.0806 0.136 0.219 0.442 0.5045 0.508 0.5671 0.6515 0.7165 0.8558 0.954 1.0542 1.1649 2.191 3.9660 d d 2 ( 0) d 2 0.9862 0.676 0.6741 0.6421 0.5713 0.4192 0.2517 0.1450 6 Surface boundary conditions k T ( x,0) y (7.30) qs Surface flux is specified. Determine: Surface temperature Ts (x) and local Nusselt number Nu x . Solution by similarity transformation. Solution: Surface temperature Ts ( x) T 5 ν 2 (q s ) 4 gk 4 1/ 5 x ( 0) (7.31) Local Nusselt number g qs Nu x 5ν 2 k 1/ 5 x 4 1 (0) (7.32) Table 7.2 [4] (0) is a dimensionless parameter which depends on the Prandtl number and is given in Table 7.2 [4]. Correlation equation for (0) (0) 4 9 Pr 1/ 2 5Pr 10Pr 10 1/ 5 2 , 0.001 Pr 1000 (7.33) Properties at the film temperature T f Tf T Pr 0.1 1.0 7.34) Ts (L / 2) / 2 7.6 Inclined Plates Vertical plate solutions of Sections 7.4 and 7.5 apply to inclined plates, with g replaced by g cos . This approach is recommended for 60o . 7.7 Integral Method 7.7.1 Integral Formulation of Conservation of Momentum Assume: t (a) 100 (0) - 2.7507 - 1.3574 - 0.76746 - 0.46566 7 Application of the momentum theorem in the x-direction to the element 7.6 Fx dx , Fig. (b) M x (out ) M x (in) dx is enlarged in Fig. 7.7 dM x dx dx Mx p d ( p ) dx dx dy o dx dx gdxdy M x dW ( p dp / 2)d p Fig. 7.7 p p dp d 2 d p dx dx p o dx Mx dM x dx dx Mx (c) Simplify dp o dx dM x dx dx dW (d) Wall shearing stress u x,0 y o (e) Weight of element dW dx g dy (f) 0 The x-momentum of the fluid entering element ( x) u 2 dy Mx (g) 0 (e), (f) and (g) into (d) u x,0 y dp dx d dx gdy 0 u 2 dy (h) 0 Combine pressure and gravity terms dp dx dp dx g (i) 8 Multiply by and rewrite as integral dp dx g (j) gdy 0 (j) into (h) u x ,0 y g ( d dx )dy 0 u 2 dy (k) 0 Express density difference in terms of temperature change (T (2.28) T ) (2.28) into (k) ν u ( x,0) y g (T T ) dy 0 d dx u 2 dy (7.35) 0 (7.35) applies to laminar as well as turbulent flow. 7.7.2 Integral Formulation of Conservation of Energy Assume: (1) (2) (3) (4) No changes in kinetic and potential energy Negligible axial conduction Negligible dissipation Properties are constant Forced convection formulation of conservation of energy, (5.7), is applicable to free convection T x,0 y ( x) d dx u (T T )dy (7.36) 0 7.7.3 Integral Solution Vertical plate, Fig. 7.6. Uniform surface temperature T s . We assumed t . Thus we have two equations, (7.35) and (7.36) for the determination of a single unknown . Since both (7.35) and (7.36) must be satisfied, we introduce another unknown as follows: Assumed Velocity Profile: u x, y a0 ( x) a1 ( x) y a2 ( x) y 2 Boundary conditions on the velocity (1) u(x,0) 0 (2) u(x, ) 0 a3 ( x) y 3 (a) 9 (3) u( x, ) y 2 (4) u ( x,0) y 2 0 g (Ts T ) Applying the four boundary conditions gives a n . Equation (a) becomes g (Ts T ) 4ν u 2 y 2 y 1 (b) Let g (Ts T ) 4ν u o ( x) 2 (c) (b) becomes u y uo ( x) 1 y 2 (7.37) Treat uo (x) as the second unknown function, independent of . Assumed Temperature Profile: T ( x, y) b0 ( x) b1 ( x) y b2 ( x) y 2 (d) The boundary conditions are (1) T ( x,0) Ts (2) T ( x, ) T T ( x, ) (3) 0 y Application of the above boundary conditions gives T ( x, y) T (Ts T ) 1 y 2 (7.38) Heat Transfer Coefficient and Nusselt Number T ( x,0) y Ts T k h (7.24) (7.38) into (7.24) h 2k ( x) (7.39) Thus the local Nusselt number is Nu x Must find and uo (x) and (x). hx k 2 x ( x) (7.40) 10 Solution (7.37) and (7.38) into (7.35) ν uo g (Ts T ) 2 y 1 d u o2 dx 2 dy 0 y y2 1 4 dy (e) 0 Evaluate the integrals 1 d 2 uo δ 105 dx 1 βg Ts T 3 δ ν uo δ (7.41) (7.37) and (7.38) into (7.36) ( x) 2 (Ts T ) 1 d uo T ) dx (Ts y y 1 4 dy (f) 0 Evaluate the integrals 1 d uo 60 dx 1 (7.42) (7.41) and (7.42) are two equation for (x) and uo (x). Assume a solution of the form u o ( x) Ax m (7.43) ( x) Bx n (7.44) A, B, m and n are constants. substitute (7.43) and (7.44) into (7.41) and (7.42) 2m n 2 2m A Bx 105 n 1 m n ABx m 210 1 g To T Bx n 3 1 x B n 1 A m vx B n n (7.45) (7.46) Exponents in each equation must be identical. Thus 2m n 1 n m n 1 m n (g) n (h) 1 4 (i) Solve (g) and (h) for m and n gives m 1 , n 2 (i) into (7.45) and (7.46) gives A and B A 5.17v Pr and 20 21 1/ 2 g (Ts T ) v2 1/ 2 (l) 11 20 21 B 3.93 Pr -1/2 Pr 1/ 4 g (Ts T ) 1/ 4 (m) v2 (i) and (m) into (7.44) 20 1 3.93 1 21 Pr x 1/ 4 ( Ra x ) 1/ 4 (7.47) (7.47) into (7.40) 1/ 4 20 1 0.508 1 21 Pr Nu x ( Ra x )1/ 4 (7.48) 7.7.4 Comparison with Exact Solution for Nusselt Number (7.26) is the exact solution to the local Nusselt number 1/ 4 Grx 4 Nu x d (0) d (7.26) Rewrite (7.26) as Grx 4 1/ 4 d (0) d Nu x (7.49) Rewrite (7.48) Grx 4 1/ 4 Nu x 20 1 0.508 1 21 Pr The right hand side of (7.49) and (7.50) are compared in Table 7.3. The exact solution for Pr 0 Nu x exact 0.600 ( PrRa x )1/4 , Pr integral 0.514( PrRa x )1 / 4 , Pr 0 Exact and integral solutions for Pr Nu x exact Nu x integral 0.503( Ra x )1/4 , 0 (7.51a) are Pr (7.29b) 0.508( Ra x )1 / 4 , Pr (7.51b) NOTE: The error ranges from 1% for Pr 14% for Pr 0 . (4 Pr )1/ 4 (7.50) Table 7.3 Pr _ d (0) 0.01 0.03 0.09 0.5 0.72 0.733 1.0 1.5 2.0 3.5 5.0 7.0 10 100 1000 0.0806 0.136 0.219 0.442 0.5045 0.508 0.5671 0.6515 0.7165 0.8558 0.954 1.0542 1.1649 2.191 3.9660 d 0.508 20 1 1 21 Pr (7.29a) 0 Applying integral solution (7.47) to Pr Nu x 1/ 4 to 0.0725 0.1250 0.2133 0.213 0.4627 0.5361 0.5399 0.6078 0.7031 0.7751 0.9253 1.0285 1.1319 1.2488 2.2665 4.0390 1/ 4 (4 Pr )1/4 Chapter 8: Convection in External Turbulent Flow 8.1. Introduction Turbulent flow is disordered, with random and unsteady velocity fluctuations; hence, exact predictions cannot be determined. Turbulence affects local velocity distribution, drag force, and heat transfer in both natural and industrial processes. Understanding of turbulent flow leads to the ability to make design improvements to either reduce or enhance turbulent effects Our understanding still relies on empirical data and rudimentary conceptual drawings, and, more recently, computer simulations. Exact solutions are not possible. Chapter Focus: Wall-bounded shear flows 8.1.1. Examples of Turbulent Flows (a) Mixing Processes Combustion processes – proper mix of fuel and air is one of the requirements for combustion efficiency Chemical processing, such as the production of polymers Laminar mixing occurs when liquids are viscous and/or slowly mixed, and can be problematic (b) Free Shear Flow Jet Flows (refer to fig. 8.1a): jet energy is dissipated to the surrounding fluid Turbulent wake (refer to fig. 8.1b): transfers energy between an object and the ambient flow, contributes to an object’s drag Smoke stack exhaust is dispersed by turbulence (refer to figure 8.c) (c) Wall-Bounded Flows Flow of air over a flat plate or airfoil Flow of fluid in a pipe Irregular or random motions cause the shape of the velocity profile and the boundary layer edge location to change with time Instantaneous velocity profiles are time-averaged for simplicity: u Turbulent velocity can be decomposed into steady (mean, u ) and unsteady (fluctuating, u ) components Note: Refer to fig. 8.2 for details on the velocity profiles and unsteady components The mixing of velocity fluctuations in turbulent flow creates a steeper profile than that of a laminar flow, with a larger boundary layer and higher wall shear stress. Turbulent fluctuations enhance momentum transfer between the surface and the flowing fluid, resulting in higher skin friction; this suggests that surface fluctuations similarly enhance heat transfer. Note: See figure 8.3 for a comparison of laminar and turbulent velocity profiles. By understanding turbulence, we can alter designs to take advantage of turbulent effects, such as increased heat transfer and reduced drag. Note: See the golf ball example in fig. 8.4 for an example of drag reduction via turbulence. 8.1.2. The Reynolds Number and the Onset of Turbulence The Reynolds number is the ratio of inertial to viscous forces, and indicates the onset of turbulence. Reynolds equation: uD Re D (8.1) The onset of turbulence for flow through tubes is approximately Ret uD / 2300 . For uniform flow over a semi-infinite flat plate, the onset of turbulence is approximately Ret V xt / 500,000 . Viscous forces dominate at low flow velocity. At high velocity, inertial forces acting on individual particles dominate; the flow amplifies these disturbances, creating a more chaotic flow. Turbulence initiates near the wall in wall-bounded flows. Note: See fig. 8.5 for the development of turbulent flow over a semi-infinite flat plate. 8.1.3. Eddies and Vorticity Eddies are regions of intermittent, swirling patches of fluid An eddy is a particle of vorticity, : V (8.2) Eddies form in regions of velocity gradient. Example: for 2D flow over a flat plate: z v x High shear stress results in high vorticity. u y The formation and behavior of eddies within the boundary layer is not fully understood. Vortex stretching increases the kinetic energy of the vortex, and is thought to be a major mechanism for the main flow to transfer energy to the turbulence. Note: Refer to fig. 8.6 for details of eddy formation, vortex formation and vortex stretching in a wall-bounded flow. Eddies provide bulk motion (advection) and mixing within the boundary layer. Advection differentiates turbulent flow from laminar flow; laminar flow has no bulk motion and relies solely on viscous diffusion to transfer momentum. 8.1.4. Scales of Turbulence Viscous effects play a vital role in turbulence. Most turbulence originates from shear flow. Rotation of a fluid element occurs under the action of viscous shear. Lewis Richardson proposed the concept of energy cascade in 1922: Turbulence is comprised of a range of different eddy sizes. Treating the eddy as a distinct fluid structure with: Characteristic size: Characteristic velocity: u Eddy turnover time: t /u Reynolds number for the eddy: Re u / Inertial forces in a turbulent flow cause large, unstable eddies with high kinetic energy and high Reynolds numbers (on the scale of the main flow) to break up into smaller and smaller eddies. Energy is transferred as each eddy breaks into smaller ones. The breaking up process continues until the Reynolds numbers of the eddies approach unity; at this time, viscous forces dissipate the energy of the small eddies into heat. Note: Fig. 8.7 illustrates the cascade process. Eddies decrease in size much faster than in velocity; therefore, the smaller eddies experience very high velocity gradients and have high vorticity. The largest eddies contain the bulk of the kinetic energy in a turbulent flow; the smallest eddies contain the bulk of vorticity and therefore the mechanism of dissipation. Andrey Kolmogorov proposed a model in 1942 based on the ideas above: The largest eddies contain the bulk of the kinetic energy The smallest scales reach a Reynolds number of unity prior to dissipating into heat Kolmogorov’s relations: / ~ Re 3/ 4 (8.3a) v / u ~ Re 1/ 4 (8.3b) 1/ 2 (8.3c) / t ~ Re / is the ratio of the length scales of the largest and smallest eddies v / u is the ratio of their velocities / t is the ratio of their time scales The Reynolds number is that of the largest eddy: Re The variables , v and u / are called the Kolmogorov microscales The variables , u and t are called the integral scales Important points from Kolmogorov’s model: There is a vast range of eddy sizes, velocities and time scales in turbulent flow, making modeling difficult The smallest eddies are not infinitely small, since they are dissipated into heat by viscous forces. The scale of the smallest eddies is determined by the scale of the largest eddies through the Reynolds number. Turbulent flow responds to an increase in velocity by producing smaller eddies, hence, viscous dissipation is increased. Faster turbulent flows have finer turbulent structure; this is observed in the real world. 8.1.5. Characteristics of Turbulence Turbulence is comprised of irregular, chaotic, three-dimensional fluid motion, but containing coherent structures. Turbulence occurs at high Reynolds numbers, where instabilities give way to chaotic motion. Turbulence is comprised of many scales of eddies, which dissipate energy and momentum through a series of scale ranges. The largest eddies contain the bulk of the kinetic energy and break up by inertial forces. The smallest eddies contain the bulk of the vorticity and dissipate into heat. Turbulent flows are not only dissipative, but also dispersive through the advection mechanism. 8.1.6. Analytical Approaches The continuum hypothesis still applies to the governing equations of fluid mechanics for turbulent flow. This is based on the example in section 8.1.4, where the microscale for air flowing over a 10 cm diameter cylinder at 10 m/s was approximately 20 microns. The mean free path of air at standard conditions is 10-8 m; this is three orders of magnitude smaller. The computing power required to process a direct numerical simulation (DNS) using Computational Fluid Dynamics (CFD) is enormous and exceeds practical limitations, due to the vast range of scales of turbulence between the largest and smallest eddies. An analytical approach is used to predict macroscale properties, such as velocity, drag force and heat transfer. Microscopic flow structure is ignored; turbulent fluctuations are analyzed statistically. Important idealizations are used to simplify statistical analysis: Homogeneous turbulence: turbulence whose microscale motion does not, on average, change from location to location or from time to time Isotropic turbulence: turbulence whose microscale motion does not, on average, change as the coordinate axes are rotated These idealizations are not necessarily realistic, but they can be approximated in the laboratory, so experimental data can be compared to the simplified statistical analytic flow models. 8.2. Conservation Equations for Turbulent Flow 8.2.1. Reynolds Decomposition Reynolds proposed that turbulent flow can be considered as the superposition of a timeaveraged and a fluctuating component. Reynolds Decomposition Approach: Each fluctuating property in the governing equations is decomposed into a time averaged and a fluctuating component. The entire equation is time averaged. For an arbitrary property g: g g g (8.4) Where g , the time-averaged component, is determined by: g 1 g t dt 0 By definition, the time average of the fluctuating property, g , is 0: (8.5) 1 g g t dt (8.6) 0 0 Useful averaging identities: a a (a ) 2 ab (a ) 2 ab a b a t ab a b 0 (8.7a) ab ab (8.7b) (8.7c) aa 0 (8.7d) (8.7e) a2 (a ) 2 (8.7g) a x (8.7i) a t (a ) 2 a x 0 (8.7f) (8.7h) (8.7j) An example proving identities (8.7a) and (8.7g) is given. 8.2.2. Conservation of Mass Reynolds decomposition is applied to the Cartesian conservation of mass equation: t ( v) y ( u) x ( w) z 0 (2.2a) Analysis is limited to incompressible, two-dimensional flow. Substituting the Reynolds decomposed velocities, u u u , and v v v : (u u ) x (v v ) y 0 (a) u x u x v y v y 0 (b) u x v y v y 0 (c) This is expanded: The equation is time averaged: u x Invoking identity (8.7h), (c) becomes: u x u x v y v y 0 (d) Using identity (8.7a), u u and v v , and by (8.6), u v 0 , so (d) reduces to the time averaged turbulent flow continuity equation, identical in form to (2.2): v y u x (8.8) 0 Subtracting (8.8) from (b) demonstrates that the divergence of the fluctuating term is 0: v y u x (8.9) 0 8.2.3. Momentum Equations Reynolds decomposition is applied to the Cartesian momentum equations: u t 2 2 2 u x v u y w u z gx p x u x2 u y2 u z2 v v u t x v y w v z gy p y v x2 v y2 2 v u 2 2 v z2 (2.10x) (2.10y) Assumptions (1) Two-dimensional flow (2) Incompressible flow (3) Constant properties (4) Body forces are neglected (5) Flow is, on average, steady-state, so u and v are constant Formulation Equations (2.10x) and (2.10y) become: u t u u x v u y p x v t u v x v v y p y 2 u 2 x 2 y2 2 v 2 x 2 y2 u v (8.10x) (8.10y) From the product rule of differentiation, u( u / x) and v( u / y) in the xmomentum equation become: u u x u2 x v u y (u v) y u x u u (a) v y Substituting (a) into the x-momentum equation (8.10x): (b) u2 x u t (u v) y u x u ( A) u v y 2 p x 2 u x2 u (c) y2 ( B) The terms (A) and (B) in (c) above are combined, and by conservation of mass: v y u x u (d) 0 The x-momentum equation reduces to: (u v) y u2 x u t 2 p x 2 u x2 u (8.11) y2 The y-momentum equation reduces in a similar fashion Reynolds decomposition is now performed on the simplified x- and y-momentum equations, resulting in the turbulent x- and y-momentum equations for turbulent flow: 2 u u x v u y p x u x2 v x v v y p y 2 u v x2 2 u y2 (u ) 2 x v uv x 2 y2 uv y (8.12x) (v ) 2 y (8.12y) Equations (8.12) are identical to equations (8.10) with two notable exceptions: The transient terms u / t and v / t disappear. More importantly, new terms indicating fluctuating velocity are introduced. 8.2.4. Energy Equation Reynolds decomposition is applied to the Energy equation for incompressible flow: cp T T u t x v T y w T z 2 k T x2 2 T y2 2 T z2 (2.19b) Assumptions (1) Heat generation is neglected (2) Thermal properties are considered constant (3) Steady-on-average flow (4) Two dimensional flow (5) The dissipation function Φ is neglected; this is appropriate as long as the flow is not highly viscous or compressible Formulation The Energy equation reduces to: T t cp u T x v 2 T y k 2 T x2 T y2 (8.13) After Reynolds decomposition and time averaging is applied (left as an exercise, Problem 8.7), equation (8.13) becomes: cp u T x v 2 T y 2 T x2 k T y2 cp uT x cp vT y (8.14) Equation (8.14) is almost identical to the steady-state Energy equation (8.10) with two notable exceptions: The transient term c p T / t disappears Two terms indicating fluctuating velocities and temperature are introduced, and come out of the convective terms on the left side of the equation. 8.2.5. Summary of Governing Equations for Turbulent Flow Continuity: v y u x (8.8) 0 x-momentum: 2 u u x v u y p x u x2 v x v v y p y 2 u 2 u y2 (u ) 2 x v uv x uv y (8.12x) (v ) 2 y (8.12y) vT y (8.14) y-momentum: v x2 2 y2 Energy: cp u 8.3. T x v T y 2 k T x2 2 T y2 cp uT x cp Analysis of External Turbulent Flow Goal: Solve the governing equations for turbulent flow for forces of interest, such as drag force and heat transfer. Focus: Flow along a surface Boundary layer concept is invoked Similar development of boundary layer equations to those of laminar flow (Chapter 4), with two important differences: Turbulent flow governing equations contain time-averaged quantities (i.e. u , T ), which are handled exactly like u and T previously Turbulent equations contain fluctuating terms (i.e. u and T ) that require additional consideration 8.3.1. Turbulent Boundary Layer Equations (i) Turbulent Momentum Boundary Layer Equation Turbulent flow over a flat plate is considered (refer to Fig. 8.8) Assumptions (1) Steady-state (2) Incompressible flow (3) Constant properties (4) The boundary layer is thin, L , so: (8.15) 1 L Formulation Following the arguments used for the laminar boundary layer, the following scalar arguments are made: u ~V (8.16a) x~L (8.16b) y~ (8.16c) Following an analysis similar to that in Section 4.2, the viscous dissipation terms in (8.12x) compare as follows: 2 u 2 x2 y2 u (8.17) The pressure gradient in the y-direction is negligible: p y (8.18) 0 The pressure gradient in the x-direction is expressed as: p x The fluctuation terms dp dx (u ) 2 and x dp dx uv require additional scaling arguments. y (8.19) There is no preferred direction to the fluctuations; this is equivalent to assuming the turbulence is isotropic, therefore u ~v 2 u (8.20) ~uv (8.21) Comparing the fluctuating terms using scale analysis: Since (u ) 2 (u ) 2 ~ x L (a) u v u v (u ) 2 ~ ~ y (b) L , we can conclude that (u ) 2 x uv y (8.22) Using the simplifications (8.17) and (8.22), the x-momentum equation for the turbulent boundary layer reduces to: u x u v u y dp dx 2 u y2 uv y (8.23) (ii) Turbulent Energy Equation The derivation for laminar boundary layer equations is again followed. Formulation Scaling arguments for the thermal boundary layer: (8.16b) x~L y~ (8.24) t T ~ Ts T (8.25) The scale for the velocity sizes depend on t and , as in Chapter 4. The second derivative terms compare as follows 2 T 2 2 y2 x The fluctuation terms cp x uT T and (8.26) cp y vT require additional scaling arguments: There is no preferred direction for the fluctuations, so: u ~v (8.20) uT ~vT (8.27) The fluctuating terms compare as follows (it is left to the reader to derive this): vT y uT x (8.28) Applying simplifications (8.26) and (8.28), the energy equation for the turbulent boundary layer reduces to: cp u T x v T y 2 T y2 k cp vT y (8.29) 8.3.2. Reynolds Stress and Heat Flux Equations (8.23) and (8.29), the x-momentum and energy equations for the turbulent boundary layer, are written as follows to provide physical insight and a way to model the time-averaged fluctuation terms, u v and v T : u u x cp u u y v T x v dp dx T y y y k T y u y uv cp v T (8.30) (8.31) The terms in parenthesis on the right side of equation (8.30) represent shear stress The first term represents the molecular shear stress from the time-averaged velocity Boussinesq suggested (1877) that the second term can be viewed as shear imposed by the time-averaging velocity fluctuations. Note: Refer to Fig. 8.9 and to the related discussion in the text for a visualization of Boussinesq’s idea. The time average fluctuation term u v is proportional to the velocity gradient, just like for the viscous shear stress, suggesting u v behaves like shear in the flow. uv u y The apparent shear stress experienced by the flow is made up of two parts Molecular shear imposed by the time-averaged velocity component Turbulent shear stress velocity fluctuations u v , or Reynolds stress, imposed by the time-averaged The apparent energy flux in the energy equation contains a turbulence-induced heat flux, c p v T , called the turbulence heat flux or Reynolds heat flux. 8.3.3. The Closure Problem of Turbulence Summary of turbulence boundary equations: Continuity v y u x (8.8) 0 x-momentum u u x v u y dp dx y u y uv (8.30) Energy cp u T x v T y y k T y cp v T (8.31) The boundary conditions for these equations are: u (x,0) 0 (8.31a) v(x,0) 0 (8.31b) u ( x, ) V (8.31c) u (0, y) V (8.31d) T ( x,0) Ts (8.31e) T ( x, ) T (8.31f) T (0, y ) (8.31g) T If we know the velocity field outside of the boundary layer, the pressure can be determined: The pressure gradient is expressed as dp dx dp dx (8.32) From inviscid flow theory, outside the boundary layer: V dV dx 1 dp dx (8.33) We are left with three equations, (8.8), (8.30) and (8.31), and five unknowns, u , v , T , u v , and v T ; this is the closure problem of turbulence. The two terms u v and v T are nonlinear terms There is no exact solution to the turbulence boundary layer equations Modeling will facilitate numerical and approximate solutions 8.3.4. Eddy Diffusivity Reynolds Stress, based on Boussinesq’s hypothesis, is modeled as uv u y M (8.34) is the momentum eddy diffusivity M M is referred to as the eddy viscosity The Reynolds heat flux is modeled as cp v T H cp H T y (8.35) is the thermal eddy diffusivity cp H is referred to as the eddy conductivity The boundary layer momentum and energy equations are written as: u u x cp u u y v T x v dp dx T y y These are simplified by dividing (8.36) through by u u x u T x v v u y y T y y k cp H and (8.37) by u y T y (8.36) (8.37) cp : u y M H y M T y (8.38) (8.39) The terms in brackets in (8.38) represent the apparent shear stress: u y app M (8.40) The terms in brackets in (8.39) represent the apparent heat flux: qapp cp H T y (8.41) The negative sign in (8.41) assigns the correct direction to heat transfer. and H are properties of the flow and are dependent on the velocity and temperature fields, respectively M The number of unknowns has not been reduced; the velocity fluctuation terms have been replaced with expressions containing different unknowns Finding ways to evaluate 8.4. M and H is one of the main goals of turbulence research Momentum Transfer in External Turbulent Flow Section Focus: Finding ways to evaluate M and H Energy and momentum equations are decoupled Energy equation solution requires knowledge of both velocity and temperature fields Momentum equation requires knowledge of velocity field, so solving for momentum transfer requires solutions of continuity and momentum equations only This still leaves two equations and three unknowns, but the situation is simplified; now only M requires modeling 8.4.1. Modeling Eddy Diffusivity: Prandtl’s Mixing Length Theory Boussinesq postulated that M is constant This does not allow u v to approach zero at the wall, which is unrealistic, since turbulent fluctuations are expected to dampen out near the wall Prandtl’s model reasons that fluid particle behavior is analogous to that of molecules in the kinetic theory of gases. Note: Refer to Fig. 8.10 for visual details, (i.e. directions) of variables discussed below. In a two-dimensional flow, Prandtl defines: v as a velocity fluctuation forcing a particle towards the wall , the mixing length, as the distance the particle travels as a result of the velocity fluctuation The resulting velocity fluctuation u is approximated using Taylor series expansion: u final So, with u uinitial u dy y (a) u final uinitial, u ~ u y (b) Again assuming that the velocity fluctuations have no preferred direction, u ~ v , we have v ~ u y (c) By the above scales, one argues that the scale of the turbulent stress term u v ~ (u )( v ) ~ 2 u y u v is: 2 (d) Therefore, solving equation (8.34) for eddy viscosity gives: -u v u ~ 2 u/ y y M (8.42) The absolute value on the derivative in equation (8.42) ensures the eddy diffusion remains positive. Note that the mixing length itself must still be modeled, and is dependent on the type of flow. Prandtl’s model for mixing length for flow over a flat plat is: where (8.43) y is some constant This implies that the mixing length approaches zero as y approaches zero. Thus equation (8.42) becomes Prandtl’s mixing-length model: 2 2 M y u y However: is unknown No single value for is effective throughout the boundary layer. Before developing a suitable model for understood. This is presented next. , boundary layer behavior must be further 8.4.2. Universal Turbulent Velocity Flow Profile In Chapter 5, an approximate solution for the integral form of the boundary layer equations for laminar flow was found using a universal velocity profile u y ; this approach is used here. This approach also provides physical insight that can be applied to other solution techniques. Section Goal: Find a universal velocity profile function (i) Large-Scale Velocity Distribution: “Velocity Defect Law” Variables making up the velocity distribution are normalized: y becomes one axis (8.44) u / V becomes the other axis, where V is the velocity outside of the boundary layer Figure 8.11 shows velocity curves at different values of wall friction; if the plot were truly universal, all curves would collapse into one. Data is normalized by the wall friction factor, C f o (1 / 2) V 2 , where 0 is the shear stress at the wall. Further manipulations to the data are performed: For convenience, we cast the data relative to the velocity outside the boundary layer: (8.45) (u V ) A friction velocity is defined: u* o / (8.46) Friction velocity can also be written as: u* V Cf /2 (8.47) Note: u * has the same dimension as velocity. The velocity difference is normalized by u * to produce the velocity defect: (u V ) u* (8.48) The velocity defect is plotted against y / in Figure 8.12, which also shows the range of experimental data. Note that the curves from Fig. 8.11 collapse into a single curve Note: The defect plot does not show enough detail near the wall. An improved plot will probably be logarithmic, and will require an entirely new set of coordinates. (ii) Wall Coordinates The following coordinates will collapse the boundary layer velocity data into a single curve reasonably well: u y Similarly, v v / u* , and x u u* yu* xu * / . (8.49) (8.50) All of the above variables are dimensionless, and are called wall coordinates. Note: Refer to Figure 8.13 to see a plot of data produced by Clauser using these coordinates. The boundary layer extends out to approximately 2000 to 5000 The plot includes data obtained from flow in a pipe as well as flow along a flat plate, so the profile appears to be universal. (iii) Near-Wall Profile: Couette Flow Assumption The turbulent boundary layer momentum equation, (8.38), is invoked to develop a model of the velocity profile near the wall. Assumption: Flow is over a flat plate, so dp / dx 0. The momentum equation is simplified by recognizing that the flow is nearly parallel to the wall, so v ~ 0 . Conservation of mass implies that u / x ~ 0 ; therefore the u component of velocity does not change significantly along the wall. This scaling argument suggests that the convective terms in the momentum equation, u u / x and v u / y , are each approximately zero, therefore, in (8.38): u y M y ~ 0 near the wall. The bracketed term is the apparent shear stress, app / (Eqn. 8.40) The above relationship implies that the apparent stress is approximately constant (with respect to y), resulting in the Couette Flow Assumption: app M u ~ constant y (8.51) Recall from Chapter 3: this result is similar to Couette Flow. Since the local shear is constant in (8.51), we can replace app with its value at the wall, o. The presence of the eddy diffusivity makes (8.51) different from Couette flow, implying that the shear stress is constant, but not necessarily linear. The Couette Flow Assumption is used to develop a velocity profile at the wall: Substituting the definitions of u and y into (8.51), it can be shown that: 1 M u y After rearranging and integrating: 1 (8.52) y dy u 1 M (8.53) / 0 This is a general expression for the universal velocity profile in wall coordinates This integral can be evaluated more simply by dividing the boundary layer into two near-wall regions: (1) a region very close to the wall where viscous forces dominate (2) a region where turbulent fluctuations dominate (iv) Viscous Sublayer The wall tends to damp out or prevent turbulent fluctuations, so viscous forces dominate very close to the wall: M . The Couette Flow Assumption reduces to u 1. y Integrating, with boundary condition u u 0 at y y , 0 y 0 yields: 7 (8.54) This relation compares well to experimental data from y+ 0 to 7, which is called the viscous sublayer. Equation (8.54) is illustrated in Figure 8.13. Note that the curvature in the plot results from the semi-logarithmic coordinates. (v) Fully Turbulent Region: “Law of the Wall” Further away from the wall, turbulent fluctuations (i.e., Reynolds stresses) dominate, so . M The Couette Flow Assumption (8.52) therefore becomes: M u (8.55) 1 y As before, is constant; this is the same value as in the viscous sublayer (which is the value at the wall), so discontinuity between regions is avoided. Substituting wall coordinates into Prandtl’s mixing length theory (8.44) yields: 2 M ( y )2 Equation (8.56) is substituted into (8.55): u y (8.56) 2 (y ) 2 2 u 1 y Solving the above for the velocity gradient: u y 1 y (8.57) Integrating the above results in the Law of the Wall: u 1 ln y (8.58) B is called von Kármán’s constant, and experiments show that 0.41 . B is a constant of integration, and is found as follows: The viscous sublayer and the Law of the Wall region appear to intersect at roughly y u ~ 10.8 . Using the above as a boundary condition, B 5.0 . An approximation for the Law of the Wall region is: u 2.44 ln y 5.0, 50 y 1500 (8.59) (vi) Other Models The velocity profile presented in the previous sections is a two-layer model, and is called the Prandtl-Taylor model. Three-layer models, like that of von Kármán, also exist. van Driest’s continuous law of the wall is a single equation model that illustrates how single equation models work: The eddy diffusion must diminish as y approaches zero, so van Driest proposed a mixing length model of this form: y1 e y/ A (8.60) The term in parentheses is damping factor that makes (8.60) approach zero at the wall. Using (8.60) with (8.42) in (8.51) yields: app 2 2 y 1 e y/ A 2 u y (8.61) As y approaches zero, the eddy diffusivity approaches zero, leaving pure viscous shear. Transforming (8.61) into wall coordinates, and solving for u / y , yields: u y 2 1 2 1 4 y 2 1 e y /A (8.62) 2 For flow over a smooth, flat plate, van Driest used κ = 0.4 and A+ = 26. When integrated numerically, Equation (8.62) produces the curve shown in Fig. 8.13. D.B. Spalding’s model is commonly used for both flat plates and pipe flow: y where Spalding used u e B e u 1 ( u )2 2 u ( u )3 6 (8.63) = 0.40 and B = 5.5 Note that that u+ is implicit in (8.63), so there is no closed-form solution and one must numerically solve for u+ Reichardt developed a profile frequently applied to pipe flow: u where 1 ln 1 y C1 e y /X y e X 0.33 y (8.64) = 0.40, C = 7.8, and X = 11. (vii) Effect of Pressure Gradient The velocity profiles modeled up to this point assume the pressure gradient is zero. Figure 8.14 depicts how the velocity profile is affected by a pressure gradient. The plot, for flow over a flat plate, shows that in the presence of an adverse pressure gradient, the velocity profile beyond y 350 deviates from the Law of the Wall model. The deviation is referred to as a ―wake.‖ The region y 350 is commonly referred to as the wake region. The region where the data continue to adhere to the Wall Law is called the overlap region. The wake increases with adverse pressure gradient, until separation, where the velocity profile deviates even from the overlap region. A slight wake exists for zero or even a strong favorable pressure gradient, although the difference between the two sets of data is negligible. Wake models are not addressed in this text. The Law of the Wall-type models developed earlier model flat plate flow reasonably well in the presence of zero pressure gradient. A favorable pressure gradient is approximately what we encounter in pipe flow; this is one reason why the models developed here apply as well to pipe flow. 8.4.3. Approximate Solution for Momentum Transfer: Momentum Integral Method To obtain drag force on the surface of a body, the momentum integral equation is invoked, as in Chapter 5. In independent works, both Prandtl and von Kármán used this approach to estimate the friction factor on a flat plate. (i) Prandtl-von Kármán Model Considering a flat, impermeable plate exposed to incompressible, zero-pressure-gradient flow, the integral momentum equation reduces to equation (5.5): ( x) u ( x,0) d V y dx ( x) u dy 0 d dx u 2 dy (5.5) 0 This equation applies to turbulent flows if the behavior of the flow on average is considered, and the flow properties are interpreted as time-averaged values Recall that the integral method requires an estimate for the velocity profile in the boundary layer. Prandtl and von Kármán used Blasius’ model for the shear at the wall of a circular pipe, based on dimensional analysis and experimental data: Cf where C f 2 o 0.07910 Re D1 / 4 , 4000 Re D 105 (a) / um2 , and um is the mean velocity over the pipe cross-section. Based on the above, the velocity profile in the pipe could be modeled as u uCL y ro 1/ 7 (b) where y is the distance from the wall, and u CL is the centerline velocity This is the well-known 1/7th Law velocity profile, further discussed in Chapter 9. In an external flow a mean velocity is not defined, nor is ro,, so the following adjustments are made: ro is approximated by the edge of the boundary layer u CL is approximated as representing the free-stream velocity V The velocity profile in the boundary layer is then modeled as: u V y 1/ 7 (8.65) The fundamental problem with using this model in the integral equation is that the velocity profile gradient goes to infinity as y goes to zero. Equation (8.65) cannot be used directly to estimate the wall shear in equation (5.5). Prandtl and von Kármán modeled the wall shear differently: Since the characteristics of the flow near the surface of the plate are similar to that of pipe flow, the Blasius correlation was adapted to find an expression for the wall shear on a flat plate: Recasting (a) in terms of the wall shear and the tube radius: 1/ 4 0.03326 o 2 um roum It can be shown that for the 1/7th velocity profile, u m 0.8167uCL . Recall, u CL is modeled as V The expression for 0 is written as: u ( x ,0 ) y o 1/ 4 0.02333V 2 (8.66) V In terms the friction factor, this is: Cf o 2 V2 0.02333 V 1/ 4 (8.67) This can now be used in the momentum equation. An example (Example 8.2) is presented in which the expressions for the boundary layer thickness and the friction factor for flow over a flat plate are developed using the 1/7th power law for the velocity profile (8.65) and the expression for friction factor (8.67). The following equations are found within the example: 7 d 72 dx o V2 4 5 5/ 4 72 V 0.02333 7 (8.68) 1/ 4 x C (8.69) The expression for the boundary layer thickness is: 0.3816 x Re1x / 5 The expression for the friction factor is: (8.70) Cf 0.02968 2 Re1x / 5 (8.71) Note: Refer to Fig. 8.15 for additional considerations for the boundary layer over a flat plate. Note that, according to this model, the turbulent boundary layer Re x 1/ 5 /x varies as , as does the friction factor C f ; this is contrast to laminar flow, in which /x and C f vary as Re x 1 / 2 . (ii) Newer Models A limitation of the Prandtl-von Kármán model is that the approximation for the wall shear, Eqn. (8.66), is based on limited experimental data, and is of limited applicability even for pipe flow White presents a method that makes use of the Law of the Wall velocity profile (8.59): Because the wall coordinates u and y can be expressed as functions of Cf /2 , an expression for the friction factor can be developed from the Law of the Wall (a technique which is also seen in analysis of pipe flow – see Section 9.5). Substituting the definitions of u and y , as well as u * , into the Law of the Wall expression (8.59): u V 2 Cf 2.44 ln Cf yV 2 5.0 Any y value within the wall law layer would satisfy this expression, but a useful value to choose is the edge of the boundary layer, where u ( y ) V , so: 1 Cf /2 where Re V 2.44 ln Re Cf 2 5.0 (8.72) / Equation (8.72) relates the skin friction to the boundary layer thickness, and can be used in the integral momentum equation, though it’s cumbersome. By curve-fitting values obtained from (8.72) over a range of values from Re 104 to 107 , yielding the approximate relation: Cf 0.02Re 1/ 6 (8.73) The above is used to estimate wall shear using the integral method, along with the 1/7th power law for the velocity profile, resulting in the following solutions to the integral momentum equation: 0.16 x (8.74) Re1x / 7 Cf 0.0135 2 Re1x / 7 (8.75) The above equations replace the less accurate Prandtl-von Kármán correlations; White recommends these for general use. Kestin and Persen used Spalding’s law of the wall for the velocity profile to develop a more-accurate, but more cumbersome correlation; White simplified their model to: Cf 0.455 (8.76) 2 ln (0.06Re x ) According to White, the above relation is accurate to Kestin and Persen’s model to within 1%. (iii) Total Drag Total drag is found by integrating the wall shear along the entire plate. Assumptions Laminar flow exists along the initial portion of the plate. The plate has width w. Formulation The total drag over the entire plate is: xcrit FD L o lam wdx 0 Dividing by 1 2 V 2A 1 2 o turb wdx (8.77) xcrit V 2 wL , the drag coefficient C D is: xcrit CD 1 L L C f ,lamdx 0 C f ,turbdx (8.78) xcrit Substituting Equation (4.48) for laminar flow and using White’s model (8.75) for turbulent flow, we obtain with some manipulation: CD The above assumes xcrit 0.0315 Re1L/ 7 5 105 . 1477 Re L (8.79) 8.4.4. Effect of Surface Roughness on Friction Factor The previous models have all assumed flow over smooth walls. The interaction between the turbulent flow and the complex, random geometric features of a rough wall is the subject of advanced study and numerical modeling. Crude modeling and experimental study give some physical insight. k is defined as the average height of roughness elements on the wall k is transformed into wall coordinates: k ku* / For small values of k , ( k 5 ), experiments show that the velocity profile and friction factor are unaffected by roughness. Roughness is contained within the viscous sublayer; disturbances are likely dampened out by the viscosity-dominated flow. For k 10 , the roughness extends beyond the viscous sublayer and the viscous sublayer begins to disappear, likely due to enhanced mixing the roughness provides. For k 70 , viscous effects are virtually eliminated and the flow is fully rough. The shape of the velocity profile changes little after this roughness value, so it is expected that increasing the roughness will not change the friction factor. Refer to Figure 8.16 for an illustration of how the near-wall velocity profile is affected by roughness. Roughness tends to shift the Law of the Wall down and to the right The shift in velocity profile means that the velocity gradient at the wall is greater, and therefore the friction factor increases The slope of the wall law curve is not affected by roughness Correlations for friction factor on rough plates are highly dependent on roughness geometry; there are few models. For fully rough flow over sand-rough plates, White [14] suggests the following correlation, based on a wall law velocity profile developed from experimental data Cf 1.4 3.7 log10 x k 2 , x k Re x 1000 (8.80) Note that this correlation does not include the Reynolds number 8.5. Energy Transfer in External Turbulent Flow From Chapter 2, the heat transfer for flow over a geometrically similar body like a flat plate can be correlated through dimensionless analysis by Nu x f ( x * , Re, Pr ) where x * is the dimensionless location along the body (2.52) This neglects buoyancy and viscous dissipation The momentum and thermal eddy diffusivities introduced in equations (8.38) and (8.39), M and H , are used to create the turbulent Prandtl number, a new dimensionless parameter: M Prt (8.81) H There are several options to develop suitable models for turbulent heat transfer: Find an analogy between heat and mass transfer Develop a universal temperature profile and then attempt to obtain an approximate solution for heat transfer using the integral method The universal temperature profile may also lend itself to a simple algebraic method for evaluating the heat transfer More advanced methods, including numerical solutions to the boundary layer flow, are not covered in this text 8.5.1. Momentum and Heat Transfer Analogies Reynolds theorized that heat transfer and the frictional resistance in a pipe are proportional, based on his study of turbulent flow in steam boilers This implies that if the friction along a pipe wall is measured or predicted, heat transfer can be determined by using a multiplying factor, allowing one to solve directly for heat transfer. (i) Reynolds Analogy The Reynolds analogy for external flow is developed: Assumptions Flow is parallel and is over a flat plate The pressure gradient dp/ dx is zero Formulation The boundary layer momentum and energy equations reduce to: u u x u T x v v u y y T y y 0, T (y M u y (8.82a) T y (8.82b) 0) Ts , (8.83a) H The boundary conditions are u (y 0) u(y T (y ) V , (8.83b) ) T The variables are normalized as follows: v , V u ,V V U T T Ts , X Ts x , and Y L y L Equations (8.82) and boundary conditions (8.83) become: U U U X x U (Y U Y V V 0) U (Y 1 M V L Y U Y (8.84a) 1 y 0, ) 1, H V L Y (Y (Y (8.84b) Y (8.85a) 0) 0 (8.85b) ) 1 Note that normalizing the boundary conditions has made them identical. Equations (8.84) can then be made identical if ( possible under two conditions: M) ( H), which is The kinematic viscosity and thermal diffusivity are equal: (8.86) This condition limits the analogy to fluids with Pr = 1 This also suggests that the velocity and thermal boundary layers are approximately the same thickness, t The eddy diffusivities are equal M H (8.87) This is justified by arguing that the same turbulent mechanism—the motion and interaction of fluid particles—is responsible for both momentum and heat transfer; Reynolds essentially made this argument, so equation (8.87) is sometimes referred to as the Reynolds Analogy. This also means that the turbulent Prandtl number Prt is equal to 1 The analogy is now complete; the normalized velocity and temperature profiles, U ( X , Y ) and ( X , Y ) are equal. Refer to Figure 8.17 for an illustration. Derivation of a relationship between the shear stress and heat flux at the wall: The ratio of the apparent heat flux and shear stress (equations. 8.40 and 8.41) is qapp / c p ( ( app / H) T/ y M) u/ y (8.88) By imposing conditions (8.86) and (8.87), the terms in parenthesis cancel and, after substituting the dimensionless variables into (8.88): qapp c p (Ts T ) V app / Y U/ Y (8.89) Since the dimensionless velocity and temperature profiles are identical, their derivates cancel The ratio qapp / app is constant throughout the boundary layer, so this ratio can be represented by the same ratio at the wall , and equation (8.89) becomes: qo c p (Ts T ) V o This can be recast into a more convenient form by substituting qo h(Ts T ) and 2 1 into the above, and rearranging: o 2 Cf V h V cp Cf 2 The terms on the left side can also be written in terms of the Reynolds, Nusselt, and Prandtl numbers Stx Nu x Rex Pr Cf (8.90) 2 where St x is called the Stanton number Equation (8.90) is commonly referred to as the Reynolds Analogy; it can also be derived for laminar flow over a flat plate for Pr =1. The Reynolds Analogy is limited to Pr = 1 fluids; it is appropriate for gases, but not for most liquids. (ii) Prandtl-Taylor Analogy The Reynolds analogy doesn’t account for the varying intensity of molecular and turbulent diffusion in the boundary layer Very close to the wall, molecular forces are expected to dominate: M , and H (8.91) Further away from the wall, turbulent effects dominate: M , and H Neither condition above restricts us to Pr = 1 fluids. (8.92) Prandtl and Taylor independently divided the boundary layer into two regions: A viscous sublayer where molecular effects (8.91) dominate A turbulent outer layer, where (8.92) is assumed to hold In order for an analogy to exist, the momentum and boundary layer equations, and their boundary conditions, must be identical in both regions, so: The viscous sublayer is defined as the portion of the boundary layer beneath y = y1, where y1 is some threshold value, with boundary conditions: u (0) T (0) Ts , 0, u ( y1 ) u1 , T ( y1 ) T1 The following normalized variables make the boundary conditions and equation (8.82) identical: u , V u1 U v , u1 T Ts , X T1 Ts x and Y y1 y . y1 For the viscous sublayer, the ratio of the apparent heat flux and apparent shear stress (Eqn. 8.86) leads to the following: Ts T1 where qapp / app = qo / o qo Pr u1 o cp (8.93) = constant The outer layer closely resembles the Reynolds Analogy, with M H (or Prt 1 ), but this time we assume that the turbulent effects outweigh the molecular effects; this region has boundary conditions: u ( y1 ) u(y u1 , T ( y1 ) T1 , ) V , T (y ) T The following normalizing variables make the analogy valid in this region: U u u1 , V V u1 v - u1 , V u1 T T T1 , X T1 x , and Y L y L For the outer region, the ratio of the apparent heat flux and apparent shear stress (equation 8.86) leads to: T1 T The ratio qapp / app qo (V c o p u1 ) (8.94) is constant, so we have chosen the value at y=y1, which can be represented by qo / o . Adding (8.93) and (8.94) yields: qo V o cp Ts T Substituting o 1C 2 f u1 Pr 1 V 1 V 2 into the above yields Cf /2 qo V c p (Ts T ) St u1 ( Pr 1) 1 V The velocity at the edge of the viscous sublayer, u1 , is estimated using the universal velocity profile (Fig. 8.13); a value of u y 5 is chosen to approximate the edge of the viscous sublayer. From the definition of u : u 5 u1 V 5 u1 V 2 , or Cf Cf (8.95) 2 Thus the Prandtl-Taylor analogy is: Stx Cf /2 Nu x Rex Pr 5 Cf 2 (8.96) ( Pr 1) 1 (iii) von Kármán Analogy Theodore von Kármán extended the Reynolds analogy even further to include a third layer – a buffer layer – between the viscous sublayer and outer layer: Stx Cf /2 Nu x Rex Pr 1 5 Cf 2 (8.97) 5Pr 1 ( Pr 1) ln 6 Note: Refer to Appendix D for development. (iv) Colburn Analogy Colburn proposed a purely empirical modification to the Reynolds analogy that accounts for fluids with varying Prandtl number, using an empirical fit of available experimental data: Stx Pr 2 / 3 The exponent (2/3) is entirely empirical Cf 2 (8.98) The Colburn Analogy yields acceptable results for Rex 107 (including the laminar flow regime) and Prandtl number ranging from about 0.5 to 60. An example, Example 8.3, is presented in which the average Nusselt number for heat transfer along a flat plate of length L with constant surface temperature is determined, using White’s model for turbulent friction factor and assuming the flow over the plate has an initial laminar region. Equations developed within the example: The average heat transfer coefficient from Equation (2.50) is split into laminar and turbulent regions: xc hL 1 L L hlam ( x)dx 0 (8.99) hturb ( x)dx xc After several manipulations, substitutions and assumptions, the equation for the average Nusselt number for heat transfer along a flat plate of length L with constant surface temperature and initial laminar region is: Nu L 0.0158ReL6 / 7 739 Pr1/ 3 (8.100) If the laminar region had been neglected, the above would be: Nu L 0.0158ReL6 / 7 Pr1/ 3 (8.101) 8.5.2. Validity of Analogies Momentum-heat transfer analogies are frequently used to develop heat transfer models for many types of flows and geometries. Although derived for a flat plate, these analogies are considered generally valid for slender bodies, where pressure gradient does not vary greatly from zero. They are approximately valid for internal flows in circular pipes, although other analogies have been developed specifically for internal flow (See Chapter 9). Although they are derived assuming constant wall temperature, the above correlations work reasonably well even for constant heat flux. The large temperature variation near the wall means that the assumption of uniform properties, particularly for Pr, becomes a weakness; this is overcome by evaluating properties at the film temperature: Tf Ts T 2 (8.102) The analogies were derived assuming that the turbulent Prandtl number is equal to unity. Experimentally-measured values of Prt are as high as 3 very near the wall, though outside the viscous sublayer the values range from around 1 to 0.7. The turbulent Prandtl number seems to be affected slightly by pressure gradient, though largely unaffected by surface roughness or the presence of boundary layer suction or blowing. A value of Prt 0.85 is considered reasonable for most flows, so the analogies should be approximately valid for real flows. Though only an empirical correlation, the Colburn analogy was shown to represent experimental data well over a variety of fluids. However, it has been demonstrated that the Colburn analogy under-predicts the Nusselt number by 30-40% for fluids with Prandtl numbers greater than 7. 8.5.3. Universal Turbulent Temperature Profile Development of a universal temperature profile in turbulent flow provides physical insight. An approximate temperature profile can be used in an integral approach to solve for the heat transfer. (i) Near Wall Profile Beginning with the turbulent energy equation, assume that, near the wall, the velocity component v ~ 0, as is the temperature gradient T / x . The left-hand side of (8.39) approaches 0: H y T ~ 0 , near wall y This implies that the apparent heat flux is approximately constant with respect to y: qapp H cp T ~ constant y (8.103) Solve the above relation for the temperature profile: Since qapp / c p is constant throughout this region, replace qapp with qo and substitute wall coordinates into (8.103) and rearrange so both sides are dimensionless: T c p u* y qo (8.104) H (8.104) suggests a definition for a temperature wall coordinate: T (Ts T ) The above is cast into a simpler form: c p u* qo (8.105) T (8.106) y H Though the above can now be integrated, H is unknown, but substituting M into (8.106) by using the definition of the turbulent Prandtl number (8.81) yields: y dy T (8.107) H 0 This is a general expression for the temperature profile in wall coordinates. The above is evaluated by dividing it into two regions, as was done with the universal velocity profile. (ii) Conduction Sublayer Very close to the wall, molecular effects are expected to dominate the heat transfer, so H . By evoking this, (8.107) reduces to: T Pr dy Pr y C The constant of integration, C, is found by applying the boundary condition that T ( y 0) 0 , resulting in C 0 and a temperature profile in the conduction sublayer of: T Pr y , ( y y1 ) (8.108) y1 is the dividing point between the conduction and outer layers (iii) Fully Turbulent Region Outside the conduction-dominated region close to the wall, turbulent effects dominate, so H The temperature profile (8.104) becomes: y T T1 dy (8.109) H y1 must be evaluated to evaluate (8.109); this is done by relating it to the momentum eddy diffusivity: H H Prt M Recall (8.44), the model for the eddy diffusivity from Prandtl’s mixing length theory: u y 2 2 y M (8.44) This is written in terms of wall coordinates: 2 M ( y )2 u (8.110) y The partial derivative u / y can be found from the Law of the Wall Substituting the above and (8.58) into (8.109) yields: y Prt T (8.111) dy y y1 Assume Prt and are constants, so (8.111) becomes: T Prt y ln y1 , y y1 (8.112) The temperature profile defined by Equations (8.108) and (8.112) depends on the fluid (Pr), as well as the parameters Prt and . Kays et al. assumed Prt = 0.85 and = 0.41, but found that the thickness of the conduction sublayer ( y1 ) varies by fluid White reports a correlation that can be used for any fluid with Pr 0.7: T Prt 13 Pr 2 / 3 7 ln y (8.113) In this model, Prt is assumed to be approximately 0.9 or 1.0 Figure 8.18 is a plot of this model, along with viscous sublayer, for various values of Pr Note that the temperature profile increases with increasing Prandtl number. (iv) A 1/7th Law for Temperature A simpler 1/7th power law relation is sometimes used for the temperature profile: T T Ts Ts y 1/ 7 (8.114) t 8.5.4. Algebraic Method for Heat Transfer Coefficient The existence of a universal temperature and velocity profile makes for a fairly simple method to estimate the heat transfer: The definition of the Nusselt number, expressed using Newton’s law of cooling is: hx k Nu x qo x (Ts T )k (8.115) The universal temperature profile T is to be invoked, so using the definition of T , equation (8.105), the free stream temperature is defined: T c p u* (Ts T ) (Ts T ) qo c pV Cf /2 qo (8.116) where (8.47) was substituted for the friction velocity u * Substituting (8.116) into (8.115) for (Ts T ) and rearranging yields: c pV Nu x Cf /2 x T k Then, multiplying the numerator and denominator by yields: Re x Pr C f / 2 Nu x (8.117) T The universal temperature profile, Equation (8.113),is used to evaluate T : T Prt 13Pr 2 / 3 7 ln y (8.118) A precise value for y is not easy to determine, but a clever substitution is made by using the Law of the Wall velocity profile. Equation (8.58) is evaluated in the free stream as: u 1 ln y (8.119) B Substituting (8.119) into (8.118) for ln y , the Nusselt number relation then becomes Nu x Re x Pr C f / 2 Prt (u B ) 13Pr 2 / 3 7 The above is simplified using the definition of Stanton number, St Nu x /( Rex Pr ) , selecting B = 5.0 and Prt = 0.9, and noting that the definition of u leads to u 2/Cf Stx 0.9 13 Pr Cf 2/3 /2 0.88 C f / 2 (8.120) Note how similar this result is to the more advanced momentum-heat transfer analogies, particularly those by Prandtl and Taylor (8.96) and von Kármán (8.97). 8.5.5. Integral Methods for Heat Transfer Coefficient The universal temperature profile allows us to model heat transfer using the integral energy equation. A case of turbulent flow over a flat plate where a portion of the leading surface is unheated is examined, illustrated in Figure 8.19. The simplest solution is to assume the 1/7th power law for both the velocity and temperature profiles This is mathematically cumbersome, and is developed in Appendix E. The result of the analysis is: Stx Nu x Rex Pr Cf 1 2 xo x 9 / 10 1/ 9 (8.121) where xo is the unheated starting length Note that Equation (8.121) reduced to the Reynolds Analogy when xo = 0, (the Prandtl number in that derivation was assumed to be 1). The model has been used to approximate heat transfer for other fluids: Equation (8.121) is expressed as: Nu xo Nu x 1 where Nu xo 0 xo / x 0 9 / 10 1 / 9 (8.122) represents the heat transfer in the limit of zero insulated starting length In this form, other models for heat transfer, like von Kármán’s analogy, could be used to approximate Nu xo 0 for Pr ≠ 1 fluids. 8.5.6. Effect of Surface Roughness on Heat Transfer Surface roughness is important in such applications as turbomachinery, where rough surfaces can be used to enhance heat transfer in the cooling of turbine blades. Figure 8.16 shows that the viscous sublayer diminishes and disappears as roughness increases, implying that the turbulent fluid elements exchange momentum with the surface directly (i.e. profile or pressure drag) and the role of molecular diffusion (i.e. skin friction) is diminished. Heat transfer relies on molecular conduction at the surface, no matter how rough the surface, or how turbulent the flow. Fluid in the spaces between roughness elements is largely stagnant, and transfers heat entirely by molecular conduction. The conduction sublayer can be viewed as the average height of the roughness elements. The major resistance to heat transfer is formed by the stagnant regions between roughness elements Roughness cannot improve heat transfer as much as it increases friction. This means that we cannot predict the heat transfer by using a friction factor for rough plates along with one of the momentum-heat transfer analogies In developing a model for heat transfer on a rough surface, the following are expected: Roughness size has no influence until it extends beyond the viscous and conduction sublayers. The influence of roughness reaches a maximum beyond some roughness size (the fully rough limit). The Prandtl number should be present in any model, and fluids with higher Prandtl number (lower conductivity) would be affected more by roughness In these fluids, lower-conductivity fluid trapped between the roughness elements will have a higher resistance to heat transfer. The conduction sublayer is shorter for these fluids, so roughness elements penetrate relatively further into the thermal boundary layer A correlation for a rough plate was developed by Kays, et al. [29]: St Cf 2 Prt C (ks )0.2 Pr 0.44 C f / 2 1 (8.123) where k s k s u * / is based on the equivalent sand-grain roughness, k s , and C is a constant that depends on roughness geometry This model displays the expected behavior. For high-Prandtl-number fluids the second term in the parentheses dominates For sufficiently low-Pr fluids the second term diminishes, in spite of roughness size Bogard, et al. showed a 50% increase in heat transfer on rough turbine blades compared to heat transfer in smooth turbine blades, and that increasing roughness beyond some value showed little increase in the heat transfer, which is consistent with the above expectations. Chapter 9: Convection in Turbulent Channel Flow 9.1 Introduction Laminar channel flow was discussed in Chapter 6; many features of turbulent flow are similar Chapter begins with the criteria for fully developed velocity and temperature profiles Chapter Focus: Analysis of fully developed flows Analysis is limited to the following general boundary conditions: (i) uniform surface temperature (ii) uniform surface heat flux 9.2 Entry Length Criteria for entry length was discussed in Chapter 6 As a rule of thumb, fully developed velocity and temperature profiles exist for Lh De Lt De 10 (6.7) where D e is the hydraulic or equivalent diameter De 4Af P where A f is the flow area and P is the wetted perimeter Eq. (6.7) is recommended for Pr = 1 fluids More elaborate correlations exist, especially for hydrodymanic entry length; the following approximations are recommended: From White Lh De 4.4 Re1D/e6 (9.1) Lh De 0.623Re1D/e4 (9.2) From Latzko Thermal entry length doesn’t lend itself to a simple, universally-applicable equation since the flow is influenced by fluid properties and boundary conditions Hydrodynamic entry length is much shorter for turbulent flow than for laminar, so much so that sometimes it’s neglected from analysis Thermal entry length is often important Analysis of heat transfer in the thermal entry length is complicated and is not covered in the text 9.3 Governing Equations Figure 9.1 shows a circular pipe with the velocity in the x-direction is labeled as u Assumptions: Two-dimensional Axisymmetric Incompressible flow 9.3.1 Conservation Equations After Reynolds-averaging, conservation of mass reduces to: u x 1 (rv r ) 0 r r (9.3) Using the same conditions, the Reynolds-averaged x-momentum equation reduces to: u u x vr vr r T x vr T r 1 p x 1 r( r r M ) u r (9.4) Conservation of energy becomes: u 1 r( r r H ) T r (9.5) 9.3.2 Apparent Shear Stress and Heat Flux The apparent shear stress and heat flux are defined similarly to that of the flat plate development: app q app cp ( M ( u r ) H ) (9.6) T r 9.3.3 Mean Velocity and Temperature Mean velocity and bulk, or mean, temperature are used in correlations for predicting friction and heat transfer in duct flow Mean velocity is calculated by evaluating the mass flow rate in the duct (9.7) ro m um A u (2 r )dr 0 Assuming constant density, this becomes ro 1 um ro u (2 r )dr ro2 0 2 ro2 u rdr (9.8) 0 The mean temperature in the duct is evaluated by integrating the total energy of the flow: ro mc p Tm c p T u (2 r )dr 0 Substituting (9.8) for the mass flow rate, and assuming constant specific heat, ro T u rdr Tm 0 ro u rdr 0 This can be simplified by substituting the mean velocity, equation (9.8): ro Tm 9.4 2 u m ro2 T u r dr (9.9) 0 Universal Velocity Profile 9.4.1 Results from Flat Plate Flow It was shown that universal velocity profile in a pipe is very similar to that over a flat plate (see Fig. 8.13), especially when the plate is exposed to a zero or favorable pressure gradient A pipe flow friction factor model was used to analyze flow over a flat plate using the momentum integral method (see Section 8.4.3) The characteristics of the flow near the wall of a pipe are not influenced greatly by the curvature of the wall of the radius of the pipe, so, invoking the two-layer model that we used to model flow over a flat plate: Viscous sublayer u (8.54) y Law of the Wall 1 u ln y (8.58) B Continuous wall law models by Spalding and Reichardt were applied to pipe flow and discussed in Section 8.4.2 For pipe flow, the wall coordinates are a little different than for flat plate flow, so the ycoordinate is y ro r (9.10) This yields the y+ wall coordinate: y r0 (ro r r )u * (9.11) The velocity wall coordinate is the same as before: u u (8.49) u* The friction velocity is the same u* o / (8.46) The friction factor is based on the mean flow velocity instead of the free stream velocity: Cf o (1 / 2) u m2 (9.12) Therefore, the friction velocity can be expressed as: u* um C f / 2 9.4.2 Development in Cylindrical Coordinates Since the velocity profile data for pipe flow matches that of flat plate flow, it allowed us to develop expressions for universal velocity profiles solely from flat plate (Cartesian) coordinates Developing expressions for universal velocity profiles using cylindrical coordinates is developed after revealing important issues and insights Assuming fully-developed flow, the left side of the x-momentum equation (9.4) goes to zero, leaving 1 r r r 1 p x (9.13) Rearranging and integrating yields an expression for shear stress anywhere in the flow (r ) r p 2 x (9.14) C The constant C is zero since the velocity gradient (and hence the shear stress) is expected to go to zero at r = 0 Evaluating (9.14) at r and ro and taking the ratio of the two gives (r) r ro o (9.15) Equation (9.15) shows that the local shear is a linear function of radial location and raises the following important issue: Equation (9.15) is a linear shear profile depicted in Fig. 9.2 This expression contradicts how the fluid is expected to behave near the wall (recall the Couette Flow assumption led us to the idea that τ is approximately constant in the direction normal to the wall for the flat plate) Yet, in Section 8.4 that the universal velocity profile that resulted from this assumption works well for flat plate flow as well as pipe flow This is reconciled as follows The near-wall region over which we make the Couette flow assumption covers a very small distance Assume that, in that small region vary close to the wall of the pipe, the shear is nearly constant: 0 The Couette assumption approximates the behavior near the pipe wall as ( M ) u r o constant (9.16) Experimental data shows that near the wall, velocity profiles for flat plate and pipe flow are essentially the same and suggests that the near-wall behavior is not influenced by the outer flow, or even the curvature of the wall (see Fig. 8.13) 9.4.3 Velocity Profile for the Entire Pipe From the previous development the velocity gradient (and the shear stress) is expected to be zero at the centerline of the pipe; none of the universal velocity profiles developed thus far behave this way Reichardt attempted to account for the entire region of the pipe by suggesting the following model for eddy viscosity: y 6 M r 1 ro r 1 2 ro 2 (9.17) This leads to the following expression for the velocity profile 1 u where Reichardt used ln y 0.40 and B The slope of this equation is zero at r of the pipe 1.5 1 r / ro 1 2(r / ro ) 2 B (9.18) 5.5 0 , so the behavior matches that at the core The viscous sublayer is not accounted for As r 9.5 r0 , equation (9.18) reduces to the Law of the Wall form, eq. (8.56) Friction Factor for Pipe Flow 9.5.1 Blasius Correlation for Smooth Pipe Blasius developed a purely empirical correlation for flow through a smooth pipe using dimensional analysis and experimental data: 0.0791 Re D1 / 4 , Cf 4000 Re D where the friction factor is based on the mean flow velocity C f 105 o (9.19) /(1 / 2) u m2 Though less accurate and versatile than later correlations, this lead to the 1/7th Power Law velocity profile 9.5.2 The 1/7th Power Law Velocity Profile A crude but simple approximation for the velocity profile in a circular pipe was discovered by Prandtl and von Kármán, leading from the Blasius correlation Formulation Recasting the Blasius correlation in terms of wall shear stress o 1 2 u m2 0.0791 2ro u m 1/ 4 Then rearranging: o 0.03326 u m7 / 4 ro 1 / 4 1/ 4 Assume a power law can be used to approximate the velocity profile: (a) u u CL q y ro (b) with uCL representing the centerline velocity Assume that the mean velocity in the flow can be related to the centerline velocity as u CL (c) (const)u m Substituting (b) and (c) for the mean velocity in (a) yields y u ro o (const) o (const) u 7 / 4 y ( 1/ q 7 / 4 ro 1/ 4 1/ 4 This is simplified: 7 / 4 q ) ( 7 / 4 q 1 / 4) ro 1/ 4 (d) Prandtl and von Kármán argued that the wall shear stress is not a function of the size of the pipe, so the exponent on r0 should be zero Setting the exponent to zero, the value of q must be equal to 1/7, leading to the classic 1/7th power law velocity profile: u uCL y ro 1/ 7 (9.20) Experimental data shows that this profile adequately models velocity profile through a large portion of the pipe and is frequently used in models for momentum and heat transfer (recall Section 8.4.3) Limitations Accurate for a narrow range of Reynolds numbers: (104 to 106) Yields an infinite velocity gradient at the wall Does not yield a gradient of zero at the centerline Nikuradse (a student of Prandtl’s) measured velocity profiles in smooth pipe over a wide range of Reynolds numbers, and reported that the exponent varied with Reynolds number: u u CL y ro n (9.21) Table 9.1 lists Nikuradse’s measurements, including measurement of pipe friction factor of the form Cf C Re1D/ m (9.22) Nikuradse’s results show that the velocity profile becomes fuller as the mean velocity increases 9.5.3 Prandtl’s Law for Smooth Pipe A more theoretical model for friction factor is developed by employing the universal velocity profile Formulation Beginning with the Law of the Wall, equation (8.58), substituting the wall um C f / 2 coordinates u and y , as well as the friction velocity u * o / yields u um 2 Cf 1 ln yu m Cf B 2 (9.23) Assume that the equation holds at any value of y; evaluate the expression at the centerline of the duct, y ro D / 2 , where u uCL , by substituting these and the Reynolds number: u CL um 2 Cf 1 ln Cf Re D 2 B 2 (9.24) By looking at the Law of the Wall, one can obtain a functional relationship for the friction factor, though unfortunately a relationship for uCL / u m is unknown To evaluate the mean velocity, u m , substitute the velocity profile (8.58) into an expression for the mean velocity, equation (9.10): ro 1 um ro 2 u (2 r )dr ro2 ro2 0 u (ro y )dy (9.25) 0 where y ro r Performing the integration, the mean velocity becomes: um u * ln ro u * Cf 1 1 B 3 2 (9.26) Making substitutions again um um 2 ln Re D 2 Cf 2 B 3 2 (9.27) u m cancels out of the equation; however uCL does not appear either, so the above expression can be used directly to find an expression for C f by rearranging and substituting 0.40 and B 5.0 1 Cf /2 2.44 ln Re D C f / 2 0.349 The expression is not yet complete, as it was assumed that the Law of the Wall is accurate everywhere and ignores the presence of a viscous sublayer or wake region The constants are adjusted to fit the experimental data, yielding 1 2.46 ln C f / 2 D C f / 2 Cf /2 0.29, Re D 4000 (9.28) This is Prandtl’s universal law of friction for smooth pipes; it is also known as the Kármán-Nikuradse equation Despite the empiricism of using a curve fit to obtain the constants in (9.28), using a more theoretical basis to develop the function has given the result a wider range of applicability than Blasius’s correlation Equation (9.28) must be solved iteratively for C f A simpler, empirical relation that closely matches Prandtl’s is Cf 2 0.023Re D1 / 5 , 3 104 Re D 106 (9.29) This equation is also suitable for non-circular ducts with the Reynolds number calculated using the hydraulic diameter 9.5.4 Effect of Surface Roughness Roughness shifts the universal velocity profile downward (see Fig. 8.16 and Section 8.4.4) The velocity profile in the logarithmic layer can be written as 1 u ln y B B where B is the shift in the curve, which increases with wall roughness k The behavior also depends on the type of roughness, which ranges from uniform geometries like rivets to random structures like sandblasted metal Equivalent sand-based roughness model: 1 f 1/ 2 2.0 log10 Re D f 1 / 2 1 0.1(k / D) Re D f 1 / 2 0.8 (9.30) where it is common to use the Darcy friction factor f (9.31) 4C f If the relative roughness k / D is low enough, it doesn’t have much of an effect on the equation Scaling shows that roughness is not important if (k / D) Re D 10 If (k / D)Re D 1000 , the roughness term dominates in the denominator, and the Reynolds number cancels Friction is no longer dependent on the Re D Colebrook and White developed the following formula for commercial pipes: 1 f 1/ 2 2.0 log10 k/D 3.7 2.51 Re D f 1 / 2 (9.32) Representative roughness values presented in Table 9.2 This function appears in the classic Moody chart (Fig. 9.3) 9.6 Momentum-Heat Transfer Analogies Analogy method is applied to pipe flow to the case of constant heat flux boundary condition Though an analogy cannot be made for the case of a constant surface temperature, resulting models approximately hold for this case as well Formulation For hydrodynamically fully developed flow the x-momentum equation (9.4) becomes 1 dp dx 1 r( r r M 1 r( r r H) ) u r (9.33a) The energy equation reduces to: u T x T r (9.33b) Recall that an analogy is possible if the momentum and energy equations are identical, so Note that in pipe flow the pressure gradient is non-zero, although constant with respect to x. To ensure an analogy, the left side of (9.33b) must then be constant For thermally fully developed flow and a constant heat flux at the wall, the shape of the temperature profile is constant with respect to x, leading to T x constant Analogy appears to be possible; however, boundary conditions must also match Boundary conditions are: At r At r 0: du (0) dr 0, T (0) r u (ro ) 0, T (ro ) Ts ( x) (9.34a) 0 r0 : du (ro ) dr T (ro ) r , k o (9.34b) (9.34c) qo where qo is assumed to be into the flow (in the negative r-direction) By normalizing the variables as: U u , um T Ts , X Tm Ts x , and R L r , ro it can be shown that both governing equations and boundary conditions are identical in form 9.6.1 Reynolds Analogy for Pipe Flow Assuming that (Pr = 1) and H M ( Prt = 1), the same assumptions used to develop Reynold’s analogy for a flat plate, then the governing equations (9.33a) and (9.33b) are identical Following the same procedure as in the original derivation, the Reynolds analogy is essentially identical for pipe flow: St D qo u m c p (Ts Cf Tm ) 2 , Pr 1 or St D Nu D Re D Pr Cf 2 (9.35) Note that in this case the Stanton number is defined in terms of the mean velocity and bulk temperature, as is the wall shear stress: o 12 C f u m2 9.6.2 Adapting Flat Plate Analogies to Pipe Flow Other flat plate analogies can be adapted to pipe flow, with modifications Example: von Kármán analogy For pipe flow, conditions at the edge of the boundary layer are approximated by conditions at the centerline of the pipe: V u CL and T TCL These substitutions affect the friction factor, which translates to: Cf o 1 2 2 u CL Following the development exactly as before, the result is almost identical: qo uCL c p (Ts Cf /2 TCL ) 1 5 Cf (9.36) 5Pr 1 6 ( Pr 1) ln 2 The left side of (9.36) and the friction factor are expressed in terms of centerline variables instead of the more common and convenient mean quantities u m and Tm This is corrected: qo u m c p (Ts u m (Ts Tm ) u CL (Ts Tm ) TCL ) C f / 2 u m / u CL 2 u 1 5 m u CL Cf 2 ( Pr 1) ln 5Pr 1 6 where C f is defined in terms of the mean velocity C f o 2 / 12 u m The terms q o / u m c p (Ts Tm ) collectively are the Stanton number for pipe flow Simplifying yields von Kármán Analogy for pipe flow: St D T s Tm Ts TCL C f / 2 u m / u CL 1 5 um u CL Estimates for the ratios u m / uCL and Ts Tm / Ts definition of mean temperature, equation (9.9): Cf 2 ( Pr 1) ln (9.37) 5 Pr 1 6 TCL are found using the th u m and Tm are estimated using the 1/7 Law profiles, which for a circular pipe are: y ro u u CL 1/ 7 (9.20) Similar to (8.111) for a flat plate: T Ts TCL Ts y ro 1/ 7 (9.38) Substituting these models into (9.8) and (9.9), it can be shown that um u CL 0.817 Tm Ts TCL Ts (9.39) 0.833 (9.40) 9.6.3 Other Analogy-Based Correlations A simple correlation for turbulent flow in a duct is based on the Colburn analogy: Starting equation (8.96), and using equation (9.27) for the friction factor, one finds St D Nu D 0.023Re D1/ 5 Pr 2/3 0.023Re D4 / 5 Pr 1 / 3 (9.41) The Dittus-Boelter correlation is an empirical correlation based on the Colburn analogy: Nu D where n = 0.4 for heating ( Ts 0.023Re D4 / 5 Pr n (9.42) Tm ) and n = 0.3 for cooling Simplicity and the fact that this analogy compares well with experimental data make it popular In recent years its accuracy, along with that of the Colburn analogy, have been challenged Models by Petukhov and the Gnielinski correlation (see Section 9.8) are preferred for their improved accuracy and range of applicability Analogies remain a common way to model heat transfer in pipes; there are models developed specifically for pipe flows 9.7 Algebraic Method Using Uniform Temperature Profile As was done for the flat plate, the universal temperature and velocity profiles can be used to estimate the heat transfer in a circular duct Formulation The Nusselt number for flow in a duct is defined as hD k Nu D qo D (Ts Tm )k (9.43) To invoke the universal temperature profile, the definition of T , equation 8.102, is used to define the mean temperature as Tm (Ts c p u* Tm ) qo (Ts Tm ) c pum C f / 2 qo (9.44) For duct flow, the friction velocity u * is defined in terms of the mean velocity, so substituting the above into (9.43) for q o and invoking the definitions of the Reynolds and Prandtl numbers: Nu D Re D Pr C f / 2 (9.45) Tm There are several ways to proceed from here: One approach: evaluate Tm using a dimensionless version of (9.33): ro Tm 2 u m ro 2 T u (ro y ) dy (9.46) 0 Substituting appropriate universal temperature and velocity profiles into (9.46) and integrating must be done numerically A second approach yields a simpler closed-form solution Rewrite the original Nusselt number relation (9.43) as follows: Nu D qo D (Ts (Ts Tm )k (Ts TCL ) TCL ) where TCL is the centerline temperature Substituting the definition of T denominator: Nu D for the centerline temperature in the Re D Pr C f / 2 (Ts TCL ) (Ts Tm ) TCL (9.47) The universal temperature profile, equation (8.118), is used to evaluate TCL : TCL Prt ln yCL 13Pr 2 / 3 7 (9.48) As was done for the flat plate, substitute the Law of the Wall velocity profile (8.59) for ln y CL : u CL 1 ln y CL (9.49) B Substituting these into the Nusselt number relation: Re D Pr C f / 2 Nu D Prt (u CL B) 13Pr Expressions are needed for u CL and (Ts 2/3 TCL ) /(Ts For the centerline velocity, use the definition of u u CL u CL u* u CL um (Ts TCL ) 7 (Ts Tm ) (9.50) Tm ) : for pipe flow 2 Cf (9.51) Mean velocity and temperature are needed, so, the 1/7th power law is used to avoid the complexity of the logarithmic velocity and temperature profiles From the previous section, the 1/7th power law yields: um u CL 0.817 and Tm Ts TCL Ts 0.833 Using the definition of Stanton number, St D NuD /( Re D Pr) , and selecting Prt = 0.9 and B = 5.0, equation (9.50) is rearranged: St D Cf /2 0.92 10.8 Pr 2/3 0.89 C f / 2 (9.52) A first approximation suggests this model be limited to ReD < 1 x 105 The ultimate test is to compare this model to experimental data 9.8 Other Correlations for Smooth Pipes Petukhov evoked Reichardt’s model for eddy diffusivity and velocity profile (9.15, 9.16) to obtain: St D Cf /2 1.07 12.7 Pr 2/3 1 C f / 2 , 0.5 10 4 Pr ReD 2000 5 106 (9.53) This compares well to experimental data over a wide range of Prandtl and Reynolds numbers The following model was used for the friction factor: Cf 2 (2.236 ln Re D 4.639) 2 (9.54) Note the similarity between Petukhov’s relation (9.53) and the algebraic result, equation (9.52) In 1976, Gnielinski modified Petukhov’s model slightly, extending the model to include lower Reynolds numbers: Nu D ( Re D 1000) PrC f / 2 1 12.7 Pr 2/3 1 C f / 2 , 0.5 3 3 10 Pr ReD 2000 5 106 (9.55) Pethkhov’s friction model can be used in (9.55) for the friction factor For the above models, properties should be evaluated at the film temperature These correlations are reasonable for channels with constant surface temperature as well as constant heat flux; the flows are relatively insensitive to boundary conditions 9.9 Heat Transfer in Rough Pipes The effects of roughness on the heat transfer from flat plates was discussed in Section 8.5.6, and much of the same physical intuition applies to flow in channels Norris presents the following empirical correlation for flow through circular tubes: Nu Nu smooth Cf C f ,smooth n , Cf C f ,smooth 4 (9.56) where n 0.68Pr 0.215 A correlation like Colebrook’s (9.30) could be used to determine the rough-pipe friction factor The behavior of this relation reflects what is expected physically: The Prandtl number influences the effect of roughness For very low Prandtl fluids the roughness plays little role in heat transfer The influence of roughness size is limited The effect of increasing roughness vanishes beyond (C f / C f ,smooth) 4 , hence a maximum is reached Though roughness enhances heat transfer, it increases friction, which increases pumping costs, though the increase of friction due to roughness also reaches a limiting value The application of roughness to increase heat transfer requires benefits to be weighed against increasing costs CHAPTER 10 CORRELATION EQUATIONS: FORCED AND FREE CONVECTION 10.1 Introduction A key factor in convection is the heat the heat transfer coefficient h. Instead of determining h we determine the Nusselt number Nu, which a dimensionless heat transfer coefficient. Whenever it is difficult or not possible to determine the Nusselt number analytically, we search for a correlation equation which gives the Nusselt number. Correlation equations are usually based on experimental data. This chapter gives correlation equations for: (1) External forced convection over plates, cylinders, and spheres. (2) Internal forced convection through channels. (3) External free convection over plates, cylinders and spheres. (4) free convection in enclosures. 10.2 Experimental Determination of Heat Transfer Coefficient h Newton's law of cooling is used to determine h experimentally: h qs Ts T (10.1) Measure surface temperature Ts , surface heat flux q s , and free stream temperature T , and use (10.1) to determine h. Example: Determining q s . In Fig. 10.1 the cylinder is heated electrically, current and voltage give power (heat) dissipated, heat flux q s is power divided by surface area qs ΔV Ts T i V Fig. 10.1 Experimental data is correlated in terms of dimensionless variables and parameters. Example: Forced convection for constant properties and no dissipation, local Nusselt number is correlated as Nu x = f ( x* ; Re, Pr ) 10.3 Limitations and Accuracy of Correlation Equations (2.52) 2 All correlation equations have LIMITATIONS. They MUST be carefully noted. Examples of limitations: Geometry: An equation for each configuration. Range of parameters, such as the Reynolds, Prandtl and Grashof numbers, for which a correlation equation is valid, are determined by the availability of data and/or the extent to which an equation correlates the data. 10.4 Procedure for Selecting and Applying Correlation Equations There are many, many correlation equations. Each is for a specific application and is valid under specified conditions. Presenting all correlation equations and discussing their applications and limitations is not the most effective and efficient approach to studying and learning the material. Instead, we will describe a systematic procedure for searching, identifying and selecting correlation equations. Some of the common applications will be presented as examples. Selecting correlation equations for applications not discussed in this chapter follow the same procedure described below. (1) Identify the geometry under consideration. Is it flow over a flat plate, over a cylinder, through a tube, or through a channel? (2) Identify the classification of the heat transfer process. Is it forced convection, free convection, external flow, internal flow, entrance region, fully developed region, boiling, condensation, micro-gravity? (3) Determine if the objective is finding the local heat transfer coefficient (local Nusselt number) or average heat transfer coefficient (average Nusselt number). (4) Check the Reynolds number in forced convection. Is the flow laminar, turbulent or mixed? (5) Identify surface boundary condition. Is it uniform temperature or uniform flux? (6) Examine the limitations on the correlation equation to be used. Does your problem satisfy the stated conditions? (7) Establish the temperature at which properties are to be determined. For external flow properties are usually determined at the film temperature T f Tf (Ts T )/2 (10.2) and for internal flow at the mean temperature Tm . However, there are exceptions that should be noted. (8) Use a consistent set of units in carrying out computations. (9) Compare calculated values of h with those listed in Table 1.1. Large deviations from the range of h in Table 1.1 may mean that an error has been made. 3 10.5 External Forced Convection Correlations 10.5.1 Uniform Flow over a Flat Plate: Transition to Turbulent Flow Boundary layer flow over a semi-infinite flat plate. Mixed flow: laminar and turbulent. Transition or critical Reynolds number Re xt : V xt Rext 5 105 (10.3) V ,T xt (1) Plate at Uniform Surface Temperature. The local heat transfer coefficient is determined from the local Nusselt number. In the turbulent region, x Valid for: Fig.10.2 8.2 Fig. xt hx k Nu x x turbulent transition laminar 0.0296 ( Rex ) 4 / 5 ( Pr )1 / 3 (10.4a) flat plate, constant Ts 5 105 Rex 107 0.6 Pr 60 properties at Tf (10.4b) Average heat transfer coefficient for mixed flow: consider both laminar and turbulent: xt L h 1 L 1 L h( x)dx 0 L hL ( x)dx h t ( x)dx (10.5) xt 0 Use (4.72) for hL (x) and (10.4a) for ht (x ) , integrate h k 0.664 ( Rext )1 / 2 L 0.037 ( Re L ) 4 / 5 ( Rext ) 4/5 ( Pr )1 / 3 (10.7a) Limitations: see limitations on the respective correlations for the Nusselt numbers. Expressed in terms of the average Nusselt number Nu L : hL k Nu L 0.664 ( Re xt )1 / 2 0.037 ( Re L ) 4 / 5 ( Re xt ) 4 / 5 ( Pr )1 / 3 (10.7b) (2) Plate at Uniform Surface Temperature with an Insulated Leading Section. Turbulent flow Nu x hx k 0.0296( Re x ) 1 ( xo /x ) V 4/5 1/3 ( Pr ) 9 / 10 1 / 9 (10.8) T t insulation xt xo Ts Fig. 10.3 x 4 Limitations: see (10.4b) and review limitations on (4.72) (3) Plate with Uniform Surface Flux. Nu x hx k 0.030( Re x ) 4/5 Pr 1 / 3 (10.9) Properties T f (Ts T ) / 2, where Ts is the average surface temperature. 10.5.2 External Flow Normal to a Cylinder T V Average Nusselt number Fig. 10.5 N uD hD k 0.3 0.62Re1D/ 2 Pr 1 / 3 1 0.4 Pr 2 / 3 1/ 4 1 ReD 282,000 5/8 4/5 (10.10a) Valid for: flo w n o rm a ltoc y lin d e r PeRe Pr0 .2 D p ro p e rtie sa tT f (10.10b) For Pe < 0.2, use NuD hD k 1 0.8237 0.5 ln Pe (10.11a) Valid for: flo w n o rm a ltoc y lin d e r PeRe Pr0 .2 D p ro p e rtie sa tT f (10.11b) The above gives examples of correlation equations and their limitations. Cor r elation equations for other configur ations will be listed without details. Limitations and conditions on their use should be noted. 10.5.3 External Flow over a Sphere 10.6 Internal Forced Convection Correlations 5 10.6.1 Entrance Region: Laminar Flow through Tubes at Uniform Surface Temperature (1) Fully Developed Velocity, Developing Temperature: Laminar Flow. (2) Developing Velocity and Temperature: Laminar Flow. this case is given by [5, 7] 10.6.2 Fully Developed Velocity and Temperature in Tubes: Turbulent Flow (1) The Colburn Equation (2) The Gnielinski Equation 10.6.3 Non-circular Channels: Turbulent Flow 10.7 Free Convection Correlations 10.7.1 External Free Convection Correlations (1) Vertical Plate: Laminar Flow, Uniform Surface Temperature. (2) Vertical Plates: Laminar and Turbulent, Uniform Surface Temperature. (3) Vertical Plates: Laminar Flow, Uniform Surface Heat Flux. (4) Inclined Plates: Laminar Flow, Uniform Surface Temperature. (i) Heated upper surface or cooled lower surface (ii) Heated lower surface or cooled upper surface (6) Vertical Cylinders. (7) Horizontal Cylinders. (8) Spheres. 10.7.2 Free Convection in Enclosures (1) Vertical Rectangular Enclosures. (2) Horizontal Rectangular Enclosures. (3) Inclined Rectangular Enclosures. 10.8 Other Correlations Keep in mind that this chapter presents correlation equations for very limited processes and configurations. There are many other correlation equations for topics such as: Condensation Boiling High speed flow Jet impingement Dissipation Liquid metals Heat transfer enhancements Finned geometries Irregular geometries Micro-gravity Non-Newtonian fluids Etc. Consult textbooks, handbooks and journals. CHAPTER 11 CONVECTION IN MICROCHANNELS 11.1 Introduction 11.1.1 Continuum and Thermodynamic Hypothesis. Previous chapters are based on two fundamental assumptions: (1) Continuum: Navier-Stokes equations, and the energy equation are applicable (2) Thermodynamic equilibrium: No-velocity slip and no-temperature jump at boundaries. Validity criterion: The Knudsen number: Kn (1.2) De is the mean free path. Continuum: valid for: Kn 0.1 (1.3a) 0.001 (1.3b) No-slip, no-temperature jump: Kn 11.1.2 Surface Forces Surface area to volume ratio increases as channel size is decreases. Surface forces become more important as channel size is reduced. 11.1.2 Chapter Scope Classification Gases vs. liquids Rarefaction Compressibility Velocity slip and temperature jump Analytic solutions: Couette and Poiseuille flows Table 11.1 gas 11.2 Basic Considerations 11.2.1 Mean Free Path Ideal gas: (11.2) RT p 2 Properties if various gases; Table 11.1 7 R Air Helium J/kg K kg/m3 287.0 2077.1 Hydrogen 4124.3 Nitrogen 296.8 Oxygen 259.8 1.1614 0.1625 0.08078 1.1233 1.2840 10 kg/s m m 184.6 199.0 89.6 178.2 207.2 0.067 0.1943 0.1233 0.06577 0.07155 2 11.2.2 Why Microchannels? The heat transfer coefficient increases as channel size is decreased. Examine fully developed flow through tubes and note the effect of diameter h 3.657 k D (11.3) 11.2.3 Classification. Based on Knudsen number Kn 0.001 continuum, no slip flow (11.4) 0.001 Kn 0.1 continuum, slip flow 0.1 Kn 10 transition flow 10 Kn free molecular flow 11.2.4 Macro and Microchannels Macrochannels Continuum and thermodynamic equilibrium model applies. No-velocity slip and no-temperature jump. Microchannels Failure of macrochannel theory and correlation. Distinguishing factors: two and three dimensional effects, axial conduction, dissipation, temperature dependent properties, velocity slip and temperature jump at the boundaries and the increasing dominant role of surface forces. 11.2.5 Gases vs. Liquids Mean free paths of liquids are much smaller than those of gases. Onset of failure of thermodynamic equilibrium and continuum is not well defined for liquids. Surface forces for liquids become more important. Liquids are almost incompressible while gases are compressible. 11.3 General Features Rarefaction: Knudsen number effect. Compressibility: large channel pressure drop, changes in density (compressibility). Dissipation: Increased viscous effects. 11.3.1 Flow Rate No-velocity slip: Fig. 11.3a Velocity slip: Fig. 11.3b Flow rate Q : Macrochannel theory underestimates flow rate: (a) (b) Fig. 11.3 3 Qe Qt 1 (11.5) w (4.37a) 11.3.2 Friction Factor Friction coefficient C f Cf (1 / 2) u m2 Friction factor f f 1D 2 L p (11.6) u m2 Fully developed laminar flow in macrochannels: f Re (11.7) Po Po is known as the Poiseuille number Macrochannel theory does not predict Po. The following ratio is a measure of prediction error Po Po e C* (11.8) t Po appears to depend on the Reynolds number. Both increase and a decrease in C are reported. 11.3.3 Transition to Turbulent flow Macrochannels Ret uD 2300 (6.1) Microchannels: reported transition Reynolds numbers ranged from 300 to 16,000 11.3.4 Nusselt number Macrochannels: Fully developed laminar flow: constant Nusselt number, independent of Reynolds number. Microchannels: Macrochannel theory does not predict Nu. The following ratio is a measure of reported departure from macrochannel prediction 0.21 ( Nu ) e ( Nu ) t 100 (11.9) 11.4 Governing Equations In the slip-flow domain, 0.001 Kn 0.1 , the continuity, Navier Stokes equations, and energy equation are valid. Important effects: Compressibility, axial conduction, and dissipation. 4 11.4.1 Compressibility Compressibility affects pressure drop, Poiseuille number and Nusselt number. 11.4.2 Axial Conduction Axial conduction is neglected in macrochannels for Peclet numbers greater than 100. Microchannels typically operate at low Peclet numbers. Axial conduction may be important. Axial conduction increases the Nusselt number in the velocity-slip domain. 11.4.3. Dissipation Dissipation becomes important when the Mach number is close to unity or larger. 11.5 Velocity Slip and Temperature Jump Boundary Conditions In microchannels fluid velocity is not the same as surface velocity. The velocity slip condition is 2 u ( x,0) u u ( x,0) u s (11.10) n u u(x,0) = fluid axial velocity at surface u s surface axial velocity x = axial coordinate n = normal coordinate measured from the surface u = tangential momentum accommodating coefficient Gas temperature at a surface differs from surface temperature: T ( x,0) Ts 2 T T 2 1 T ( x,0) Pr n (11.11) T(x,0) = fluid temperature at the boundary Ts = surface temperature c p / cv , specific heat ratio T = energy accommodating coefficient u and T are assume equal to unity. (11.10) and (11.11) are valid for gases. 11.6 Analytic Solutions: Slip Flows Consider Couette and Poiseuille flows. Applications: MEMS. Thermal boundary conditions: Uniform surface temperature and uniform surface heat flux. Examine the effects of rarefaction and compressibility. 5 11.6.1 Assumptions (1) Steady state (2) Laminar flow (3) Two-dimensional (4) Slip flow regime (0.001 < Kn < 0.1) (5) Ideal gas (6) Constant viscosity, conductivity, and specific heats (7) Negligible lateral variation of density and pressure (8) Negligible dissipation (unless otherwise stated) (9) Negligible gravity (10) The accommodation coefficients are assumed equal to unity, u T 1 .0 . 11.6.2 Couette Flow with Viscous Dissipation: Parallel Plates with Surface Convection Stationary lower plate, moving upper plate. Insulated lower plate, convection at the upper plate. y ho T us Determine: (1) Velocity distribution (2) Mass flow rate (3) Nusselt number x u H Fig. 11.6 Flow Field x-component of the Navier-Stokes equations for compressible, constant viscosity flow (2.9), simplifies to d 2u (11.12) 0 dy 2 Boundary conditions: apply (11.10) du( x,0) dy u( x,0) u ( x, H ) du( x, H ) dy us (g) (h) Solution u us 1 (y 1 2 Kn H Kn) (11.14) Mass Flow Rate. The flow rate, m, for a channel of width W is H m W u dy 0 (11.15) 6 (11.14) into (11.15) m us 2 WH (11.16) Macrochannels flow m o mo WH us 2 (11.17) Thus m mo (11.18) 1 Nusselt Number. Defined as 2 Hh k Nu (l) Heat transfer coefficient h: T (H ) y Tm Ts k h Substitute into (l) Nu T (H ) y 2H Tm Ts (11.19) k thermal conductivity of fluid T fluid temperature function (variable) Tm fluid mean temperature Ts plate temperature NOTE: Fluid temperature at the surface, T ( x, H ), is not equal to surface temperature Ts . Surface temperature is unknown in this example Relation between T ( x, H ) and T s is given by the temperature jump condition: Ts 2 1 T ( x, H ) T ( x, H ) Pr y (11.20) uT dy (11.22) Mean temperature Tm Tm H 2 us H 0 Temperature distribution: Energy equation simplifies to 2 k T y2 0 (11.23) 7 Dissipation function : 2 u y (11.24) (11.24) into (11.23) d 2T du k dy dy 2 2 (11.25) Boundary conditions dT (0) dy k dT ( H ) dy (m) 0 ho (Ts T ) Use (11.20) k dT ( H ) dy ho T ( x, H ) 2 1 T ( x, H ) T Pr n (n) Solution: Use (11.14) for u, substitute into (11.25), solve and use boundary conditions (m) and (n) kH H2 2 Kn 2 T y2 H T (11.26) 2 ho 2 1 Pr us k H (1 2 Kn ) 2 Nusselt number: Use (11.26) to formulate Ts , (11.19) Nu (p) dT ( H ) and Tm , substitute into dy 8(1 2 Kn ) 8 8 (1 2 Kn ) Kn 1 Kn 3 1 Pr (11.27) NOTE: The Nusselt number is independent of Biot number. The Nusselt number is independent of the Reynolds number. This is also the case with macrochannel flows. The Nusselt number depends on the fluid (Pr and ). Nusselt number for macrochannel flow, Nu o : set Kn Nu o Thus 8 0 in (11.27) (11.28) 8 Nu Nu o 1 2 Kn 8 (1 2 Kn ) Kn 1 Pr 8 1 Kn 3 (11.29) 11.6.3 Fully Developed Poiseuille Channel Flow: Uniform Surface Flux Inlet and outlet pressures are p i and p o Surface heat flux: q s Determine: qs y H/2 (1) Velocity distribution (2) Pressure distribution (3) Mass flow rate (4) Nusselt number H/2 x qs Poiseuille flow in microchannels differs from macrochannels as follows: Fig. 11.7 Streamlines are not parallel. Lateral velocity component v does not vanish. Axial velocity changes with axial distance. Axial pressure gradient is not linear. Compressibility and rarefaction are important. Assumptions. See Section 11.6.1. Additional assumptions: (11) Isothermal flow. (12) Negligible inertia forces. 2 (13) The dominant viscous force is u y2 . Flow Field. Determine the axial velocity distribution. Axial component of the Navier-Stokes equations p x 2 u 0 y2 (c) Boundary conditions u ( x,0) y 0 (e) u ( x, H / 2) y (f) H 2 dp y2 1 4 Kn ( p ) 4 2 8 dx H (11.30) u ( x, H / 2) Solution u Must determine pressure distribution and lateral velocity v. Continuity for compressible flow: 9 u x v y (h) 0 Use ideal gas law in (h) pv y (i) pu x (11.30) into (i) H2 dp y2 p (1 4 Kn ( p ) 4 2 ) 8 x dx H ( p v) y (j) Boundary conditions on v v(x,0) (k) 0 v( x, H / 2) (l) 0 Multiply (j) by dy, integrate y H2 dp p 8 x dx d ( p v) 0 y (1 4 Kn ( p) 4 0 y2 H2 ) dy (m) 0 (n) Evaluate the integrals, solve for v, and use (l) x p 4 y3 3 H3 dp y 1 4 Kn( p) dx H y H /2 Introduce Knudsen number Kn Evaluate (n) at y H H 2 1 p RT (11.33) H / 2, substitute (11.33) into (n) and integrate 1 2 p 6 H 2 RT p Cx (o) D where C and D are constants of integration. The solution to this quadratic equation is 2 p( x) 3 2 RT H 18 RT H2 6Cx (p) 6D Boundary conditions on p p (0) pi , p ( L ) po (q) Use (q) to find C and D, substitute into (p) and use the definition of Knudsen number p ( x) po 6 Kn o 6 Kn o pi po 2 (1 pi2 2 po ) 12 Kn o (1 Mass Flow Rate. The flow rate m for a channel of width W is pi x ) po L (11.35) 10 H/ 2 m 2W (s) u dy 0 Use (11.30), (11.35) and the ideal gas law WH 3 p 6 12 RT H m 2 RT dp dx (11.38) Using (11.35) to formulate the pressure gradient, substituting into (11.38), assuming constant temperature ( T To ), and rearranging, gives m 1 W H 3 po2 24 LRTo pi2 1 12 Kn o ( 2 po pi po 1) (11.39) For macrochannel mo 1 W H 3 po2 12 LRTo pi po 1 (11.40) 1 12 Kno (11.41) Taking ratio m mo 1 2 pi po NOTE: m in microchannels is very sensitive to channel height H. (11.39) shows the effect of rarefaction and compressibility. Nusselt Number. Follow Section 11.6.2 Nu 2H qs k (Ts Tm ) (v) T s is given by (11.11) Ts T ( x, H / 2) 2 1 T ( x, H / 2) Pr y (11.42) Tm is given by H /2 uT dy Tm 0 H /2 udy 0 Temperature distribution. Solve the energy equation. Additional assumptions: (11.43) 11 (14) Axial velocity distribution is approximated by the solution to the isothermal case. (15) Negligible dissipation, 0 2 T / y2 (16) Negligible axial conduction, 2T / x 2 (17) Negligible effect of compressibility on the energy equation, u / x (18) Nearly parallel flow, v 0 Energy equation: (2.15) simplifies to T c pu x 2 k T v/ y 0 (11.44) y2 Boundary conditions T ( x,0) y (w) 0 and k T ( x, H / 2) y (x) qs Assume: (19) Fully developed temperature. Define T ( x, H / 2) T ( x, y ) T ( x, H / 2) Tm ( x) (11.45) (y) (11.46) 0 (11.47) Fully developed temperature: Thus x Equations (11.45) and (11.46) give dT ( x, H / 2) dx T x ( y) dT ( x, H / 2) dx dTm ( x) dx 0 (11.48) The heat transfer coefficient h, is given by T ( x, H / 2) y Tm ( x) Ts ( x) k h (y) Use (11.42) and (11.45) into (y) h k[T ( x, H / 2) Tm ( x)] d ( H / 2) Ts ( x) Tm ( x) dy Newton’s law of cooling: h Equating the above with (11.49) qs Ts ( x) Tm ( x) (11.49) 12 qs d ( H / 2) dy T ( x, H / 2) Tm ( x) (11.50) constant Combining this with (11.48), gives dTm ( x) dx dT ( x, H / 2) dx T x (11.51) Conservation of energy for the element in Fig. 11.8 gives 2q sWdx mc pTm mc p Tm qs dTm dx dx Simplify and eliminate m dTm dx m 2q s = constant c pum H Tm Tm (11.52) dx dTm dx dx qs (11.52) into (11.51) dTm ( x) dx dT ( x, H / 2) dx T x Fig. 11.8 2q s c pum H (11.53) into (11.44) 2 2q s u kH u m T y2 (11.54) where u m is given by um 2 H H /2 (cc) udy 0 (11.30) into (cc) um H 2 dp 1 6 Kn 12 dx (11.55) (11.30) and (11.55) u um 6 1 1 6 Kn 4 Kn y2 (11.56) H2 (11.56) into (11.54) 2 T y2 qs 1 12 1 6 Kn kH 4 Kn y2 (11.57) H2 Integrating twice and use (w) T ( x, y ) 12q s 1 1 ( (1 6 Kn )kH 2 4 Kn) y 2 y4 12H 2 g ( x) (11.58) 13 To determine g(x), find Tm using two methods. First method: Integrate (11.52) Tm x 2q s c pum H dTm Tmi Tm (0) dx 0 Tmi (11.59) 2q s x Tmi c pum H (11.60) Evaluating the integrals Tm ( x) Second method: use definition in (11.43). (11.30) and (11.58) into (11.43) Tm ( x) 3q s H k (1 6 Kn) 2 13 Kn 40 ( Kn) 2 13 560 (11.61) g ( x) (11.60) and (11.61) give g(x) g ( x) Tmi 2q s x c pum H 3q s H k (1 6 Kn ) Surface temperature Ts ( x, H / 2) : 3q s H 1 Ts ( x ) Kn k (1 6 Kn ) 2 2 5 48 ( Kn ) 2 2 13 Kn 40 qs H Kn 1 kPr 13 560 (11.62) (11.63) g ( x) Nusselt number: (11.61) and (11.63) into (v) Nu 2 3 1 Kn (1 6 Kn) 2 5 48 1 ( Kn) 2 (1 6 Kn) (11.64) 13 Kn 40 (i) The Nusselt number is an implicit function of x since Kn is a function p which is a function of x. (iii) The effect of temperature jump on the Nusselt number is represented by the last term in the denominator of (11.64). (iv) The Nusselt for no-slip, Nu o , is determined by setting Kn 0 in (11.64) 2 1 Kn 1 Pr 9 NOTE: (ii) Unlike macrochannels, the Nusselt number depends on the fluid, as indicated by Pr and in (11.64). 13 560 8 7 Nu 6 5 4 0.04 0 0.08 0.12 Kn Fig. 11.9 Nusselt number for air flow between parallel plates at unifrorm surface heat flux for air, = 1.4, Pr = 0.7, u T 1 14 Nu o 140 17 (11.65) 8.235 (v) Rarefaction and compressibility have the effect of decreasing the Nusselt number. 11.6.4 Fully Developed Poiseuille Channel Flow: Uniform Surface Temperature Assumptions: same as the uniform flux case. The velocity, pressure, and mass flow rate, are the same as for uniform flux. Surface boundary condition is different. Must determine temperature distribution Ts y H/2 x H/2 Fig. 11.10 Temperature Distribution and Nusselt Number. Newton’s law of cooling the Nusselt number for this case is given by Nu 2 Hh k 2H T ( x, H / 2) Tm ( x) Ts y Ts (11.66a) Energy equation: Include axial conduction c pu 2 T x k( T 2 2 y2 x T ) (11.67a) Boundary conditions: T ( x,0) y 2 T ( x, H / 2) Ts (11.68a) 0 H T ( x, H / 2) Kn 1 Pr y T (0, y ) (11.69a) Ti (11.70a) T ( , y ) Ts (11.71a) Axial velocity is given by (11.56) u um 6 1 1 6 Kn 4 Kn y2 (11.56) H2 Dimensionless variables T Ts , Ti Ts y , Re H x , H RePr 2 um H , Pe RePr (11.72) Use (11.56) and (11.72), into (11.66a)-(11.71a) Nu 2 m ( , / 2) (11.66) 15 6 (1 1 6 Kn 4 Kn 2 ( Pe) 2 ( ,0) ( ,1 / 2) 2 2 1 ) 2 2 (11.67) 2 (11.68) 0 1 Kn 1 Pr ( ,1 / 2) (11.69) (0, ) 1 (11.70) ( , ) 0 (11.71) Solution: method of separation of variables 8 Pe = 0 1 5 8 Result: Fig. 11.11. NOTE: 7 Nu The Nusselt number decreases as the Knudsen number is increased. Axial conduction number. increases the 6 Nusselt 5 No-slip (Kn = 0) and negligible axial conduction (Pe ): Nu o 7.5407 0 0.08 0.04 0.12 Kn Fig. 11.11 Nusselt number for flow between parallel plates at uniform surface temperature for air, Pr = 0.7, , 1 .4 , u T 1 [14] (11.73) 11.6.5 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Flux This problem is identical to Poiseuille flow between parallel plates at uniform flux presented in Section 11.6.3. qs r z Determine the following: (1) Velocity distribution (2) Nusselt number qs Fig. 11.12 Assumptions. See Section 11.6.3. Flow Field. Following the analysis of Section 11.6.3. Use cylindrical coordinates. Results: r ro 16 ro2 dp 1 4 Kn 4 dz vz vz v zm p( z ) po 8Kn o 2 ro4 po2 16 LRTo m 2 (11.77) pi2 (1 pi z ) po L ) 16 Kn o (1 2 po pi2 pi 1 16 Kn ( 1) o po po2 ro4 p o2 8 LRT mo (11.74) ro2 1 4Kn (r / ro ) 2 1 8Kn pi po 8Kn o r2 ( pi po (11.78) (11.79a) 1) (11.79b) Nusselt Number. Define 2ro h k Nu Nu (d) 2 ro q s k (Ts Tm ) (e) Results T (r , z ) Tm g ( z) Tmi q s ro 2 14 Kn 3 16Kn 2 1 r4 4 ro2 7 24 k (1 8Kn) 2q s q s ro z 16Kn 2 2 c p ro v z m k (1 8Kn ) Ts (ro , z ) Nu qs (1 4 Kn ) r 2 (1 8 Kn ) k ro 4q s ro 3 Kn k (1 8Kn ) 16 4 (11.92) g ( z) (11.95) g ( z) 14 Kn 3 7 24 (11.96) q s ro Kn g ( z ) 1 kPr (11.97) 2 4 ( Kn 3 ) (1 8Kn ) 16 1 (1 8Kn ) 2 16Kn 2 14 Kn 3 7 24 Nusselt number variation with Knudsen number for air, with is plotted in Fig. 11.14. No-slip Nusselt number, Nu o , is obtained by setting Kn Nu o 48 11 4.364 4 1 Kn 1 Pr (11.98) 1.4 and Pr 0.7, 0 in (11.98) (11.99) 18 11.6.6 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Temperature The uniform surface flux of Section 11.6.5 is repeated with the tube maintained at uniform surface temperature Ts . r r z Temperature Distribution and Nusselt Number Fig. 11.15 Ts Same flow field as the uniform surface flux case of Section 11.6.5 Follow the analysis of Section 11.6.4. and use the flow field of Section 11.6.5. Dimensionless variables T Ts , Ti Ts z , R 2ro RePr 2 um ro r , Re ro , Pe RePr (11.106) Nusselt number, energy equation, and boundary conditions in dimensionless form Nu 2 m 1 4 Kn R 2 2(2 16Kn) (0. ) R (1, ) 2 0 Kn 1 Pr (1, ) R 1 (R ) R R R (11.100) (2 Pe) 2 (11.101) 2 4.5 (11.102) (1, ) R 2 1 Pe = 0 1 5 4.0 (11.103) 3.5 (R,0) 1 (11.104) (R, ) 0 (11.105) Nu 3.0 Solution: By separation of variables. 2.5 Results: Fig. 11.16. 2.0 0 0.12 0.08 Kn Fig. 11.16 Nusselt number for flow through tubes at uniform surface temperature for air, Pr = 0.7, 1.4, u T 1 , [14] 0.04 ro