* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Convection Principles
Survey
Document related concepts
Renormalization group wikipedia , lookup
Center of mass wikipedia , lookup
Flow conditioning wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Classical central-force problem wikipedia , lookup
Routhian mechanics wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Heat transfer physics wikipedia , lookup
Equations of motion wikipedia , lookup
Fluid dynamics wikipedia , lookup
Reynolds number wikipedia , lookup
Transcript
1 7. PRINCIPLES OF CONVECTION Objective is to determine h values for the combinations of: › Different geometric configurations » internal flow » external flow (flat plates, cylindrical & spherical bundles) › Types of fluid flow » laminar flow » turbulent flow › Types of convection » forced convection » natural convection 1. Introduction - Definitions q" h(Ts T ) where, h = LOCAL heat transfer coefficient › Again, we have used an average h defined as: h 1 h dAs As A s › Thus, the heat flux varies locally and the total heat flux is: q q" dAs h(Ts T ) dAs h As (Ts T ) As As 2. Boundary Layer Concept 2 - Velocity Boundary Layer › Shear stress causes the retardation of flow from the wall to some point where velocity is almost that of free stream (u) du dy At the plate surface, us = 0 (i.e., nonslip surface) › Velocity BL thickness, The distance from the wall to a point where u( y) 0.99u - Thermal Boundary Layer › The thermal BL develops because of the temperature difference between the free stream and the plate wall › Thermal BL thickness, t The distance from the wall to a point where T ( y) 0.99T › At the surface, heat flux must obey Fourier’s Law: q"s k dT dy y 0 Coupling with convection rate equation q"s h(Ts T ) , h can be calculated by: k dT h (Ts T ) dy y 0 3 - Laminar vs Turbulent Flow › For a flat plate: u x u x Critical Reynold Number: Re c c ~ 5 x105 › For a tube: Reynold Number: Re u D 2000 to 4000 3. Boundary Layer Equations - Continuity Equation (Mass Conservation) Re c 4 Mass conservation: m in m out m st Then, (u )dy (v)dx u (u )dx dy v (v)dy dx (dxdy) x y t After simplification, (u ) (v) x y t - Momentum Equation Goal is to apply Newton’s second law: Fx d (mV ) x dt Need to consider: › Body forces: gravitational, magnetic, electrical, etc. › Surface forces: normal and shear Basic Assumptions › Steady-state, incompressible fluid › = const. › P 0 and 0 y x 5 Inflow of momentum = (udy)u (vdx)u Outflow of momentum = [(u )u ] [(v)u ] (u )u x dxdy (v)u y dy dx Net pressure force = Pdy Pdy P dxdy x Net viscous force = u dx u u dy dx y y y x Substitute into Newton’s law and simplify 2 u u v u dP u2 y dx x y Net momentum normal force - Energy Equation Basic Assumptions › Steady-state, incompressible fluid › Constant , k, cp shear force 6 Energy balance equation is E k , x E k , x dx E k , y E k , y dy E h, x E h, x dx E h, y E h, y dy W E g Note that the work done by viscous force, W › At the bottom: › At the top: u dx u y (u u / ydy) u dx u dy y y Substitute into the energy balance equation and simplify after neglecting the second-order differentials: 2 2 2 T T k d T k d T u q u v x y c p dy 2 c p dx 2 c p y 2 2 k d T d T Note that and c p dy 2 dx 2 The above equation becomes 2 u T v T d T2 q x y dy For the cases with no heat generation 2 u T v T d T2 x y dy - Mass Equation Assumptions: › SS › No reactions By analogy, we can write: C A C A d 2C A u v D AB x y dy 2 7 4. Solutions to BL Equations Mass: Momentum: u v 0 x y 2 p u u v u u y x x y 2 2 u T v T d T x y dy 2 Energy: - Analytical Solution › Assumptions » Laminar flow over a flat plate » Steady-state, incompressible fluid » Constant fluid properties » Negligible viscous dissipation › Blasius Solution to Momentum Equation » Noting the momentum equation is independent of temperature, we can define a stream velocity function : and v u y x to satisfy the continuity equation. » Boundary layer similarity gives: u ( y ) u » Order-of-magnitude analysis leads to: ~ x u x 8 » Define the similarity variable y u / x Holding x constant, udy u () dy d u x ()d u x f () d » Write u, v and their derivatives f u f u u x u y y x 2 Similarly, we can write v, u , u , u x y y 2 » Substitute equation: these d2 f 2 3 f 2 d d d3 f expressions into 0 with boundary conditions: u ( x,0) v( x,0) 0 , and in terms of : f and f (0) 0 0 » Velocity BL solution f f’ u( x, ) u f 1 f” 0 0.4 0 0.027 0 0.133 0.332 0.331 4.8 5.2 3.085 3.482 0.988 0.994 0.039 0.022 » BL thickness At f ' u / u 0.99 , 5.0 yu 0.99u 5.0 5.0 x u / x Re x (Eqs. 7.13 to 16) the momentum 9 » Shear stress at the surface s u y y 0 u u 2 f x 2 0.332u 0 u x The local friction coefficient is C f ,x s 0.664 Re x u 2 / 2 › Pohlhausen Solution to Energy Equation » Define the dimensionless parameters T Ts T Ts » Assume the similarity function as () » Substitute into the energy equation and simplify d 2 Pr f d 0 d 2 2 d with boundary conditions: and (0) 0 () 1 where: Pr / (control the relation between the velocity and temperature distribution) Solve numerically (Pr 0.6) / 0 0.332(Pr)1 / 3 » Local heat transfer coefficient, hx hx k dT (Ts T ) dy y 0 k (T Ts ) dθ dη u k dθ (Ts T ) dη dy 0 x dη 0 Nu x hx x 0.332(Re)1/ 2 (Pr)1/ 3 k where: Nux = Nusselt No. Pr 0.6 10 › Average friction and heat transfer coefficients, ( C f , L , hL ) » Average friction coefficient C f , L 1 0L C f , x dx 1.328 L Re L » Average friction coefficient hx L L 1 L 1/ 2 1/ 3 0 hx dx 0.664(Re L ) (Pr) k kL Pr 0.6 - Mass Convection in Laminar Flow › Local Mass Transfer Coefficient By analogy Shx where: hm x 0.332(Re)1 / 2 ( Sc)1 / 3 D AB Shx = Sherwood No. Sc = Schmidt No. ( ) D AB › Average Mass Transfer Coefficient ShL hm L 0.664(Re L )1 / 2 ( Sc)1 / 3 D AB - Dimensional Analysis Assume the dimensionless variables: y L x* x L y* u* u V T Ts T* T Ts v* v V C *A C A C A, s C A, C A, s p* p V 2 11 Substitute into the BL equations and simplify (see Table 6.1): u * v * 0 Mass: x * y * Momentum: * * 2 * p * * u u 1 u u v * * * Re L *2 x y x y Energy: * * 2 * u * T * v * T * 1 d T2 Re L Pr * x y dy * Concentration: u * C *A x * v * C *A y * 1 d 2 C *A Re L Sc dy * 2 › Velocity From the above dimensionless momentum equation * * p * u f1 x , y , Re L , * x * The shear stress at the surface: s u y y 0 V u * L y * y* 0 From the coefficient of friction, we can write: s 2 u * Cf V 2 / 2 Re L y * y* 0 * p * f 2 x , Re L , * x Since p*(x*) is related to the surface geometry, it can be written as: C f f 2 x * , Re L Integrating over the length L, x*/L = 1. Thus, C f f 2 Re L Friction coefficient is only a function of Reynolds Number in laminar flow conditions! 12 › Heat and Mass Transfer Coefficients (See Table 6.3) - Reynolds Analogy › dp*/dx* = 0 and Pr = Sc = 1 Basic BL equations along with their boundary conditions are of exactly same form. So are their solutions. Re L Cf Nu Sh 2 Define the Stanton Numbers are: k Nu Heat transfer: St h hL u c p k u L c p Re Pr hm hm L DAB Sh u DAB u L Re Sc The above equation can be written as: Mass transfer: Stm 13 Cf St St m 2 › dp*/dx* = 0 only Use the laminar flow on a flat plate as an example & recall: C f 0.664 Re 1 / 2 Nu 0.332(Re)1 / 2 (Pr)1 / 3 Combine these two equations: Cf (Pr)2 / 3 1 1 / 3 2(Re) (Pr) 2 Nu Re Pr Cf St Pr 2 / 3 jH 2 By analogy, the Reynolds analogy for mass transfer is: Cf Stm Sc 2 / 3 jm 2 where: jH and jm = Colburn j factors for heat and mass 5. Turbulent Flow - Turbulent Boundary Layer P P P' where: P u, v, T , C A › Steady-state is defined as P is independent of time › Due to their random fluctuating nature, it is impossible to predict the exact variation with time 14 - Turbulent Equations Due to the fluctuating nature, BL equations must be modified! Momentum: u u v u dP u u 'v' y dx y y x Energy: u T v T 1 d k T c p v'T ' x y c p dy y Concentration: u C A C C A v A d D AB v' C A ' x y dy y - Definition of Eddy Diffusivity Momentum: u ' v' M u Energy: Concentration: y v'T ' H T y C A v' C A ' m y Thus, Total shear stress: Total heat transfer: Total mass transfer: tot (v M ) u y q"tot c p ( H ) T y C N "A, tot ( D AB m ) A y Note: Eddy diffusivities in turbulent core as much as 100 times larger than molecular counterparts