华中科技大学：传热学（二）：英文版教案

Heat Transfer by J P Holman To Research on the Heat Diffusing , Transporting and Radiating in the Material Systems Chapter One Introduction ? Heat:Heat is a kind of energy, Which occurs in the transfer process. (the transferred energy contains the orderly work and disorderly heat) ? Energy: Concept of energy is used to specify the state of a system in thermodynamics ? Heat Transfer:A kind of application science, is concerned with analysis of the rate of heat transfer taking place in system, or with the research on the heat-transfer regularity. 1 Some Basic Concepts and Definitions ? System research region or scope, a physical space ? Internal energy measure of motion of microscopic particles ? Temperature measure molecular mean kinetic energy ? Enthalpy internal energy plus displacement work ? Work orderly energy, force multiplies displacement in the force direction ? Heat flow transferred heat quantity per unit time 1 Some Basic Concepts and Definitions ? Heat flux transferred heat quantity per unit time and per unit area perpendicular to the direction of the heat flow ? Heat source equipment or region to produce or provide heat flow for system ? Heat transfer elements thermal resistance, thermal capacity, thermal inductance ? Change in phase substance state is changed ? Continuity Supposition the macroscopic continuity but the microscopic discontinuity 2 Process of heat transfer ? heat transfer process occurs when the internal energy(or enthalpy) in a system is not well- distribution (has n’t been even), or when the internal energy of system is different from that of its surrounding. ? The energy transport by heat flow can’t be measured directly. The concept has physical meaning, because it is related to the measurable temperature. 2 Process of heat transfer ? Substantially, the heat flow amounts to disorderly variation of internal energy with time ? The temperature difference takes place the heat transfer. 3 Applications of heat transfer ? The problem of determining the temperature distribution and the heat flow is of interest in many branches of science and technology (or engineering). ? In power engineering heat exchangers, boilers condensers, burners, nuclear reactor cores, radiators ? In aerospace technology temp. and heat flow,high velocity,weight, safety ? In heating and air-conditioning building structure, estimate insulation, prevent excessive heat losses ? In electronic and electrical engineering heat distribution, heat stress, dissipation 4 Study method ? Classical macroscopic study method (macroscopic phenomenology Method) ? Phenomenon ? Abstract ? Mathematical description(mould) ? Solution (Analysis, Numerical, Experimental) ? Building mould Mass, momentum or energy conservation law plus the mass,fluid viscous or heat conductive basic law ?differential equation 5 Learning heat transfer ? Have a profound grasp of the essential of heat transfer construction, research method, general regularity, mathematical and physical moulds. ? Know well the basic definitions, concepts, laws, calculation methods. ? Normal learning listen to teaching, write note, discuss questions, accomplish fixed exercises, make experiments. 1-1 Three Basic Modes of heat Transfer ? Three modes: conduction the aid of the motion of the microscopic particles convection the aid of the motion of the macroscopic fluids Radiation the aid of the propagation of electromagnetic waves The temperature field is controlled by combined effects, it is not possible to isolate entirely one mode. For simplicity, one mode can consider separately. Give a brief qualitative description of the three modes. 1 Thermal Conduction ? Mechanism The achievement of heat transfer must have the aid of the motion of the microscopic particles. the microscopic particles have: the kinetic motion or impact of molecules, as in the fluids at rest; the drift of electrons, as in the case of metals;the vibration of the lattice, as in the case of solids. ? Basic heat conduction law—Fourier law The heat conduction regularity is result of Biot’s experimental observation and Fourier’s analysis theory of heat ? The law states that: heat flow rate by conduction in a given direction is proportional to the area normal to the direction, to the difference of temperature in the direction, and is inversely proportional to the distance in the direction. For example,large plate(slab) 1 Thermal Conduction t 1 t 2 0 x Q A x t A q A tt q ? ? ?=? ? = κ δ κ where,κ proportionality constant is called thermal conductivity of materials ? Thermal conductivity There is a wide difference in conductivity of various materials. metals > alloys > nonmetal solids > nonmetal fluids > insulating materials >gases (at atmospheric pressure) ? The conductivity of materials is a physical property quantity,which is related to the state parameters of system. 1 Thermal Conduction 2 Thermal convection ? The achievement of heat transfer has the aid of the macroscopic displacement of the fluids, and is accompanied with the conduction. ? Convective heat transfer —the fluid flows over a solid body or inside channel, while temperature of fluid and solid surface are different,the heat transfer between the fluid and the solid surface takes place as a consequence of the motion of fluid relative to the surface. ? Forced convection fluid motion is artificially induced. ? Free(natural) convection fluid motion is set up by buoyancy effect, resulting from density diff., caused by temp. diff.. ? The convective heat transfer is important to the engineering application ? The h can be computed analytically for laminar flow over body having simple geometries,but an experimental approach is only means to determine it for flow over body having complex configuration. ? Typical values of heat transfer coefficient various factors as velocity, geometric size,flow type, fluid sort, fluid temp., flow state, change in phase. 2 Thermal convection )( ∞ ?= TThAq w ?In order to simplicity the heat transfer calculation,a heat transfer coefficient h is defined as 3 Thermal radiation ? When two bodies at different temperature are separated by perfect vacuum, heat transfer between them by conduction or convection is not possible, but it takes place(occurs) by thermal radiation. ? The radiation energy emitted by a body because of its temperature. ? The radiation energy is transmitted in the space in the form of electromagnetic waves or in the form of discrete photons. ? Maxwell’s theory and Planck’s hypothesis ? Surface process—thermal radiation is absorbed or emitted by surface of body for solids or fluids; ? Bulk process—thermal radiation is absorbed or emitted by the bulk of gases. ? Stefan-Boltzman law—radiation energy emitted by a body is proportional to the fourth power of absolute temperature for black body,as 3 Thermal radiation 4 TAq emitted σ= )( 4 2 4 1 TTAq ?= σ )( 4 2 4 1 TTAq n ?= σε ?Net exchanged energy of radiation for black body ?Net exchanged energy of radiation for practical body 1-2 Heat transfer process and overall coefficient of heat transfer ? The heat flow transfers from a hot fluid to a cool fluid through a solid wall ? For example: a plane wall exposed at a hot fluid A on one side and a cooler fluid B on the other side. In steady state the heat transfer is expressed by 热流体 冷流体 t f1 q t w2 t w1 α 2 α 1 t f2 图 8-2通过平壁的传热过程 ? Where U called overall heat transfer coefficient,and 1-2 Heat transfer process and overall coefficient of heat transfer )( 11 21 2211 21 ff w ff TTUA AhAAh TT q ?= ++ ? = κ δ 2211 111 hAAhAAU w ++= κ δ R TT RRR TT AU TT q ff w ffff 21 21 2121 )/(1 ? = ++ ? = ? = ?Lead into a concept of thermal resistance Chapter 2 Steady state heat conduction ? 2-1 Basic law of heat conduction ? 2-2 Differential equation of heat conduction ? 2-3 One-Dimensional steady-state conduction ? 2-4 Conduction through the fin ?Contain parts of 1-1 and 2-1~11 2-1 Basic law of heat conduction ? 1 Temperature field and its gradient ? Temperature field—A physical quantity field is a distribution in space and time, or the physical quantity varies with space and time. ? Mathematical expression of temperature field as T=f(x,y,z,τ ) ? Species(types):steady state 3-D T=f(x,y,z), 2-D T=f(x,y) 1-D T=f(x);unsteady state 3-D T=f(x,y,z,τ ) , 2-DT T=f(x,y,τ ), 1-D T=f(x,τ ),0-D T=f(τ ). ? Isotheral face(visible presentation of temp. field) ? Characteristics of isothermal face are that: (1)any two iso-faces don’t intersect; (2)they can enclose each other or disappear in the system boundary. ? Temperature gradient along the iso-face the temp. variation does not be well, and along the non-iso-face the temp. variation with the direction can be measured. ? Exist maximal value Δ T/Δ n in the direction normal to the iso-face. Taking the limit of Δ T/Δ n obtain: ? the temperature gradient ? written as gradT a vector. 2-1 Basic law of heat conduction n n T n T n v ? ? = ? ? →? lim 0 2-1 Basic law of heat conduction ? 2 Basic conduction law ? Boit’s experimental relation ? General situation ? This is general expression of Fourier’s law Δ A A δ q T 2 T 1 Δ n n Δ T Δ q x T Aq ? ? = κ gradT n T A q q nA κκ ?= ? ? ? ? ? ? ? ? = ? ? = ′′ →?→? limlim 00 2-1 Basic law of heat conduction ? It makes know that the heat flux vector is proportional to the temp. gradient in a given point of system,and the direction is converse counter . ? For a continual derivable temp. field a continual heat flux field exists certainly. ? Thermal conductivity ? Physical property quantity, its value is relative to material, temp., pressure, without Some adiabatic materials (fibrous,porous materials). n q p gradt t+Δt t t-Δt ? From the Fourier law heat flux can be computed if the temp. gradient is known, and the gradient can be determined by the temp. distribution. The temp. distribution can be determined from the solution of the field equation of temp. field (conduction equation), subject to appropriate boundary and initial conditions. ? 1 differential equation of heat conduction ? Consider a differential volume element dxdydz, the energy balance equation (thermodynamics first law) ? Δ q c +Δ q v =Δ E ? The Δ q c 2-2 Differential equation of heat conduction ? x-direction flow in , flow out 2-2 Differential equation of heat conduction dxxqq xx )( ??+ dxxq x )( ??? dyyq y )( ??? dzzq z )( ??? dxdydz z T zy T yx T x q c ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? =? κκκ ?Net rate of heat flow in x-direction ?Similarly in y-direction and in z-direction ?Summing up the three components and using the Fourier’s law, we have ? Δ q v there are distributed energy sources in the volume, gererating heat at a rate of per unit time and per volume .then we obtain 2-2 Differential equation of heat conduction ),,( zyxq v & dxdydzzyxqq vv ),,( & =? dxdydz T cE p τ ρ ? ? =? ?Δ E the rate of increase of internal energ is reflected in the rate of energy storege in the volume,and is given by 2-2 Differential equation of heat conduction ?Substituting above equations into energy balance equation, we obtain: ?For constant conductivity κ τα / 1 2 2 2 2 2 2 v q z T y T x TT & + ? ? + ? ? + ? ? = ? ? vp q x T xx T xx T x T c & + ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? = ? ? κκκ τ ρ ? ?α is thermal fiffusibility ? 2 Thermal diffusivity ? The physical significance of α =κ /(ρ c) is associated with the speed of propagation of heat into the medium during change of temp. with time. It is a physical property to characterize the comprehensive effect of heat conduction and heat capacity.(dynamical characteristic of conduction) ? 3 The conditions for determining the solution of the conduction equation . ? Geometrical conditions,physical conditions, initial condition, and boundary conditions. 2-2 Differential equation of heat conduction ? Initial condition—the temp. distributionis specified at initial time。 ? Boundary conditions(three kinds) ? First kind:the temp. of boundary is given; ? Second kind:the heat flow is given at the boundary; ? Third kind:first kind plus second kind,ex. Heat exchange of the boundary by convection with a fluid at a prescribed temperature 2-2 Differential equation of heat conduction ? 1 the slab(plane wall) ? Governing equation is 2-3 One-dimensional steady state heat conduction ?Simplified form conduction equation. ?Subject to appropriate boundaries for solution. ?The heat flux is 0 x δ 0)( =+ v q dx dT k dx d dx dT q κ?= ′′ ? Example 1 ? uniform thermal conductivity and no heat source, boundary surfaces: at x=0 and x=δ are keep at uniform temp.s T 1 and T 2 respectively. 2-3 One-dimensional steady state heat conduction ?Math. Mould ?Integrating ?Substituting boundary conditions δ δ == == ≤≤= xatTT xatTT xat dx Td 2 1 2 2 0 00 21 cxcT += δ x TT TT = ? ? 12 1 ?T varies linearly with the direction x 0 xδ T 2 T 1 ? Example 2 uniform conductivity and uniform heat source ? Boundary conditions at x=0,T=T 1 ;a at x=δ ,T=T 2 . ? The governing equation 2-3 One-dimensional steady state heat conduction ?The heat flux δ κκ 21 TT dx dT q ? =?= ′′ δ δ κ == == ≤≤=+ xatTT xatTT xat q dx Td v 2 1 2 2 0 00 ?First integrating: 1 c xq dx dT v +?= κ ?Second integrating: 21 2 2 cxc xq T v ++= κ 2-3 One-dimensional steady state heat conduction ? Substituting the boundary conditions,w we obtain 0 x δ T 2 T 1 ?When the T 1 =T 2 , ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?=? 2 2 1 2 δδκ δ xxq TT v () 12 2 2 12 1 2 TT xxqx TT TT v ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?+= ? ? δδκ δ δ 0 x δ T 2 T 1 ?When the x=δ /2, T=T c (central temp.) κ δ 8 2 1 v c q TT =? ?then ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?= ? ? 2 1 1 4 δδ xx TT TT c ? Example 3 with a variable conductivity and no heat source Math. mould 2-3 One-dimensional steady state heat conduction δ κ == == = ? ? ? ? ? ? xatTT xatTT dx dT dx d 2 1 0 0 Integration have 1 c dx dT =κ dx dT κ Separating variables dxcdTT 1 )( =κ Integrating the equation,we obtain ∫ ∫ ∫ ∫ = + + δ βκ βκ 0 1 0 1 0 0 2 1 1 )1( )1( dxc dxc dTT dTT x T T T T and δβ β x TTTT TTTT = ?+? ?+? 2/)()( 2/)()( 2 1 2 212 2 1 2 1 ? The heat flux ? conductivity corresponding to arithmetic mean value of temp.at the two surfaces 2-3 One-dimensional steady state heat conduction δ κ δ βκ δ βκ )( )](2/)(1[]2/)([ 21 21120 2 1 2 2120 TT qor TTTTTTTT q m ? = ′′ ?++ = ?+? ?= ′′ ]2/)(1[ 120 TTwhere m ++= βκκ ? 2 the cylinder(hollow or solid) ? In a cylinder temperature is assumed to be a function of the r variable only. In the cylindrical coordinate the conduction equation simplifies to ? Subjected to appropriate boundary conditions the temp. distribution is obtained and heat flow is got. 2-3 One-dimensional steady state heat conduction 0 1 =+ ? ? ? ? ? ? κ v q dr dT r dr d r ? Example 1 hollow cylinder with a source and const. conductivity ? The mathematic mould has a form as 2-3 One-dimensional steady state heat conduction 0 1 =+ ? ? ? ? ? ? κ v q dr dT r dr d r 22 11 rratTT rratTT == == ? The integration yields 21 2 4 cnrcr q T v ++= l κ ? Substituting the boundary conditions ,we obtain the integration constants r T 1 r 2 r 1 T 2 0 The temp. field The radial heat flow 2-3 One-dimensional steady state heat conduction ? ? ? ? ? ? ?+??+= )( 4 )( 4 2 1 2 212 2 1 12 rr q TT rq Tc vv κκ 1 2 2 1 2 2121 1 )( 4 )( r r n rr q TTc v l ? ? ? ? ? ? ?+?= κ 1 2 2 1 2 2122 )( 4 )(2 2 2 )( r r n rr q TTL rLq rq v v l ? ? ? ? ? ? ?+? ?= κ κπ π ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? += ? ? 1 1 )(4 )( 2 1 2 2 1 1 2 1 12 2 1 2 2 1 2 1 12 1 r r r r r r n r r n TT rrq r r n r r n TT TT v l l l l κ ? For the case of no heat source by setting q v =0 ? The temp. distribution and radial heat flow are 2-3 One-dimensional steady state heat conduction )/( )/( 12 1 12 1 rrn rrn TT TT l l = ? ? ? Example 2 ? solid cylinder with const. conductivity r T c T 0 0 0 1 =+ ? ? ? ? ? ? κ v q dr dT r dr d r 00 00 rratTT rat dr dT == == )/( 2 1 )( 12 12 rrn L TT rq l πκ ? = First integrate Second integrate 2-3 One-dimensional steady state heat conduction The temp. distribution 00 2 1 0 1 ==+?= = cthen dr dT for r c r q dr dT r v κ 2 2 002 2 0 4 4 r q Tcthen TTrrforcr q T v v κ κ += ==+?= The central temp. [ ] 2 0 2 0 )/(1 4 0 rrr q TT v ?=? κ 2 0 0 4 r q TT v c κ =? The heat flow r q dr dT q v 2 =?= ′′ κ Surface heat flow 0 2 0 r q dr dT q v rr =?= ′′ = κ ? 3 composite wall ? A composite structure consisting of parallel plates, coaxial cylinders, or concentric hollow spheres 2-3 One-dimensional steady state heat conduction ? Parallel slabs ? The steady state ? Heat flow is the same through each layer,then there is x T 1 T 4 T 3 T 2 )/()/()/( 333222111 41 κδκδκδ AAA TT q ++ ? = )/()/()/( 333 43 222 32 111 21 κδκδκδ A TT A TT A TT q ? = ? = ? = ? The thermal resistance form 2-3 One-dimensional steady state heat conduction total R TT RRR TT q 41 321 41 ? = ++ ? = ? The interface temperature is calculated with 1 41 21 R R TT TT total ? =? ? Coaxial hollow cylinders ? In the steady state heat flow is the same through each layer,then there is 3 4 3 3 2 3 2 2 1 2 1 1 3 43 2 32 1 21 2 1 ; 2 1 ; 2 1 r r n L R r r n L R r r n L R where R TT R TT R TT q lll πκπκπκ === ? = ? = ? = ? The interface temp. is obtained as ? For third kind of B. C. there is 1 41 21 R R TT TT total ? =? 22 2 11 1 23211 21 2 1 ; 2 1 Lhr R Lhr R where RRRRR TT q cc cc ff ππ == ++++ ? = ? The R c1 , or R c2 is convective thermal resistance 2-3 One-dimensional steady state heat conduction 2-3 One-dimensional steady state heat conduction ? Thermal contact resistance ? Two plates with different temperature brought into contact under steady-state condition. ? Experience shows that the actual temperature profile through the two materials varies approximately as shown in Fig. The temperature drop at the contact plane between the two materials, is said to be the result of a thermal contact resistance. t 1 t 2 Δ t t 0 Illustrations of thermal contact resistance effect x 2-3 One-dimensional steady state heat conduction ? Performing an energy balance on the two materials, we obtain Where the quantity 1/(h c A) is called the thermal contact resistance and h c is called the contact coefficient. ? The physical mechanism of contact resistance may be better understood by examining a joint in more detail. The actual surface roughness is believed to play a central role in determining the contact resistance. There are two principal contributions to the heat transfer at the joint: 1.the solid-to-solid conduction at the spots of contact; 2.the conduction through entrapped gases in the void spaces created by the contact. )/()/(1)/( )/(1 21 21 AkxAhAkx TT q or x TT Ak Ah TT x TT Akq BBcAA B B B c BA A A A ?++? ? = ? ? = ? = ? ? = 2-4 Heat conduction through the fin ? Heat transfer of convection can be increased by attaching thin strips of metal(called fin) to the surface. ? Fin is used on the surface where the heat transfer is low ? 1 One dimensional heat-balance equation for fin ? Fin of variable cross section ? Assume :temp. at any cross section of the fin is uniform, so that T is a function of x only. (really,L>>δ and κ /δ >>h , namely hδ / κ <<1 0 x dx Q x Q x+dx Q c W δ L ? Consider a small volume element of thickness dx, the energy balance equation has a form as (net rate by conduction in x into volume)+(net rate of heat by convection through lateral surface into volume)=0 The math. Form is we obtain the differential equation for a fin with variable section 2-4 Heat conduction through the fin 0)( =?+ ∞ p x dATThdx dx dq Substituting ? For uniform cross section the A c and p are constant values, then the form of fin equation is rewritten as pdxdAandA dx dT q pcx =?= κ 0)( =?+ ? ? ? ? ? ? ∞ pc dATT hp A dx dT dx d κ ? 2 Temp. profile and heat flow for uniform section ? Fin base x=0 T=T 0 , different physical situations are possible at the fin top x=L . ? For negligible heat flow at the fin top ( insulation of top) 2-4 Heat conduction through the fin ∞ ?= ? ? ? ? ? ? ? ? ==? TT A hp mwherem dx d c θ κ θ θ 2/1 2 2 2 0 Lxat dx d xat m dx d == == =? 0 0 0 0 2 2 2 θ θθ θ θ ? The general solution is ? Complete solution becomes )(/0 )]([)]([ 012 21 mLchcc xLmshcxLmchc θ θ == ?+?= Temp. profile Temp. of fin top Heat flow through the fin )()()( )( 1 )( )]([ 2/1 00 0 0 mLthhpAmLmthAq mLch mLch xLmch cc L κθθκ θ θ θ θ == = ? = 2-4 Heat conduction through the fin ? 3 fin efficiency ? There is a value for evaluating thermal performance of fin. ? Fin efficiency is defined as Ideal heat transfer through fin, if entire fin surface were at fin-base temperature. For uniform section the ideal heat flow is 2-4 Heat conduction through the fin ideal actual q q finthroughtransferheatideal finthroughtransferheatactual = mL mLth then mLthhpAqfrompLhq f cactualideal )( )()( 2/1 00 = == η κθθ ? Discussion for mL ? Straight fin of rectangular profile ? Multiplying numerator and denominator by L 1/2 gives 2-4 Heat conduction through the fin L h L z zh L A hp mL c κδδκ δ κ 2)22( ≈ + == Straight pin fin L d h L d dLh L A hp mL c κκπ π κ 4 4/ )( 2 === . , 22 2/32/3 fintheofareaprofileisALwhere L A h L L h mL m m = == δ κδκ ? Fin effectiveness ? In some case a valid method of evaluating fin performance is to compare the heat transfer with the fin to that which would be obtained without the fin. The ratio of these quantities is ? This is sometime called the fin effectiveness 2-4 Heat conduction through the fin )]/()[( )( 0 0 phL mLth hL pL q q f finwithout finwith κδδθ θη == Chapter 3 unsteady state heat conduction ? 3-1 Concepts of transient heat conduction ? Two types: ①periodical process;②non-periodical process( heating or cooling). The latter is discussed now. T 0 T ∞ τ 2 δ For example, a slab is set in surrounding at temp. T ∞ . Interior temp. and heat flow will vary with time.there are T=f(x,τ ) and q=f(x,τ ). Two stages:①initial stage—initial temp. T 0 has an effect on the slab temp. distribution ;②regular regime stage —the rule of temp. variation with time is same at any position of slab. The temp. variation with time is simplified. ? A discussion of the boundary conditions ?①When δ /κ >>1/h (Bi=h δ /κ >>1), there is temp. variation only at solid side. ? (regarded as first kind of boundary condition) ?②When δ /κ ～ 1/h (Bi=h δ /κ ～ 1), there is temp. variation at fluid and solid sides. ? (normal third kind of boundary condition) ?③When δ /κ <<1/h (Bi=h δ /κ <<1), there is temp. variation only at fluid side. ? (the solid can be regard as lumped parameter system) Chapter 3 unsteady state heat conduction T ∞ ① 2 δ ② ③ ? 3-2 Analysis solution of 1-D transient heat conduction ? Let θ =T-T ∞ (excess temp.) we can obtain homogeneous equation and boundary conditions Chapter 3 unsteady state heat conduction 1 Transient heat conduction in a slab Governing equation and Initial and boundary conditions are given as δκ τα τ =?= ? ? ?== ? ? == ? ? = ? ? ∞ xatTTh x T xat x T whenTT x TT )(;00 ;0; 0 2 2 T ∞ x 2 δ T 0 0 h The equation becomes Chapter 3 unsteady state heat conduction δθ θ κ θ τθθ θ α τ θ ==+ ? ? == ? ? == ? ? = ? ? xath x xat x when x 0;00 ;0; 0 2 2 )()(),( ττθ Γ?Χ= xx let 2 2 11 x? Χ? Χ = ? Γ? Γ τα Two sides of the equation must be equal to some constant which is equal to a negative value (-β 2 ) in order to ensure that excess temp. will decay with time. We get two ordinal differential equations and their initial and boundary conditions. δκβ ταβ τ ==Χ+ ? Χ? == Χ =Χ+ ? Χ? Χ >=Γ+ ? Γ? Γ xath x xat dx d x for 0;000 1 ;00 1 2 2 2 2 ? The general solutions are get separately as ? The form of general solution is Chapter 3 unsteady state heat conduction ταβ ββ 2 321 )sin()cos( ? =Γ+=Χ ecandxcxc )]sin()cos([ 2 xBxAe ββθ ταβ +=ΓΧ= ? Using the boundary conditions has )()/()( )00()cos( 2 δκβδ κβ βδ βθ ταβ =Χ+ Χ == == Χ = ? xath dx d Bi h tg xat dx d xAe The latter is a transcendental equation, the cos(β x) is eigenfunction. The solution of the equation is called eigenvalue β n , which is a series for n=1,2,3,…,n. ? The basic solution has following form ? Using the initial condition has Chapter 3 unsteady state heat conduction )cos( 2 xeA nnn n βθ ταβ? = Because only one of basic solution is not satisfied the initial condition θ =θ 0 at τ =0, the formal solution becomes linear sum of all individual solutions as ∑ ∞ = ? = 1 )cos( 2 n nn xeA n βθ ταβ ∑ ∞ = = 1 0 )cos( n nn xA βθ For determining A n the both sides of above equation are multiplied by cos(β m x) and are integrated, which is as According to the orthogonal property of eigenfunction there is Integrating above equation we obtain Chapter 3 unsteady state heat conduction dxxxAdxx n n mnm )cos()cos()cos( 1 00 0 βββθ δδ ∑ ∫∫ ∞ = = ? ? ? ≠= =≠ ∑ ∫ ∞ = )(0 )(0 )cos()cos( 1 0 nm nm dxxxA n n mn ββ δ )cos()sin( )sin(2 0 δβδβδβ δβθ nnn n n A + = Substituting the A n into the formal solution we derive final solution as Fo n nnn nn n nnn nn nn e x e x 22 11 0 cossin )/cos(sin 2 )cos()sin( )cos()sin( 2 μταβ μμμ δμμ δβδβδβ βδβ θ θ ? ∞ = ? ∞ = ∑∑ + = + = ? where Fo=ατ / δ is called Fourier number, μ n =β n δ is a function of Bi . Therefore the temp. distribution can be written as θ /θ 0 =f(Fo,Bi.x/δ ). ? 2 Transient temp. charts ? For simple geometries such as slab, cylinder ,or sphere, the temp.distribution result is plotted as a function of time and position available in the form of charts Chapter 3 unsteady state heat conduction For a slab Let θ m =T m -T ∞ ,θ 0 =T 0 - T ∞ ,Fo=ατ /δ 2 ,Bi=hδ /κ ,then θ / θ 0 =f(Fo,Bi,x/δ )=(θ m / θ 0 )(θ / θ m )=f 1 (Fo,Bi)f 2 (Bi,x/δ ) Calculating the temperature Known the Bi and Fo get (θ m / θ 0 ), known the Bi and x/δ get (θ / θ m ), the temp. is obtained θ / θ 0 =(θ / θ 0 )(θ / θ ) 2 δ 0 x T m T 0 T ∞ h Chapter 3 unsteady state heat conduction Calculating the process time ? Known the θ / θ 0 =(θ m / θ 0 )(θ / θ m ), from Bi and x/δ get (θ / θ m ), then get (θ m / θ 0 ), the Fo. is obtained, and time is calculated. θ m / θ 0 θ m / θ 0 0 FoFo 1/Bi θ / θ m x/ δ 10 1/Bi ? For long cylinder ? the math. Mould is ? For sphere the math. Mould is Chapter 3 unsteady state heat conduction T ∞ T m 0 r 2R h () κατ θ θ τ θ α θ //)/,,( . , 11 2 0 2 2 hRBiRFowhereRrBiFof isfieldtemptheand r rr === ? ? = ? ? () κατ θ θ τ θ α θ //)/,,( . , 11 2 0 2 2 hRBiRFowhereRrBiFof isfieldtemptheand r rr === ? ? = ? ? T ∞ T m 0 r2R h ? The dimensionless heat losses for the slab, cylinder, and sphere are given in figs.,where Q 0 =ρ cVθ 0 represents the initial energy content of the body in reference to the environment temperature. Q is the actual heat loss by the body in time τ . Chapter 3 unsteady state heat conduction Q/Q 0 Bi FoBi 2 ? The physical significances for Fo and Bi Chapter 3 unsteady state heat conduction )//()/()//(/ 3222 τδρδδκαδτδατ cFo === )/1/()/(/ hhBi κδκδ == The Bi is ratio of the convective capacity of solid surface to the conduction capacity of solid over the characteristic dimension, which has on effect on the temperature profile of solid interior The Fo is a measure of the rate of heat conduction in comparison with the rate of heat storage in a given volume(or system). ? 3 The discussion of regular regime stage ? regular regime stage—rate of the temperature variation with time is same at any position of solid. The effect of initial temperature has disappeared(when the Fo is more than 0.2). The infinite series of the solution has left over first term, and other terms rapidly tend to zero. Conducting logarithm operation of above equation we obtain Chapter 3 unsteady state heat conduction 2 11 111 1 0 111 11 0 )cos( )cos()sin( )sin(2 )cos()sin( )cos()sin( 2 2 1 αββ δβδβδβ δβ θ θ δβδβδβ βδβ θ θ τ ταβ == + = = + = ? ? mxUA whereAUe e x m ? Derivation of above relation with respect to time yields ? It is shown that the relative rate of variation of the excess temp. with time is a constant at any position of solid in the regular regime stage. The m is called heating or cooling rate Chapter 3 unsteady state heat conduction τ θ θ mAUnn ?= )( 0 ll τ τ θ θ m?= ? ?1 θ / θ 0 X/δ =1 surface X/δ =0 center plane Τ (Fo) ? 3-3 Lumped parameter analysis for the transient conduction ? When the Bi<<1, the system (solid) temp. is uniform only function with time. This system is of responding with time,which is called quasi-zero-dimensional system or thermal thin body. ? A solid is suddenly immersed at the time τ =0 in a well-stirred fluid which is kept at a uniform temperature T ∞ . Chapter 3 unsteady state heat conduction T ∞ T ρκ c p A V h solid fluid The energy balance relation of the system is {rate of heat flow into solid of volume V through boundary surface A} ={rate of increase of internal energy of solid of volume V} , that is V d dT cTThA p τ ρ=? ∞ )( ? Introduced excess temp. θ =T-T ∞ and given τ =0 θ = θ 0 , the equation and initial condition become as ? The general solution is Chapter 3 unsteady state heat conduction )/(0 0 0 VchAmwhereat m d d p ρτθθ θ τ θ === =+ The significance of exponent mτ τ θ m ce ? = Substitution of initial condition into the general solution has τ θθ m e ? = 0 / AVLwhere BiFo Lc hL Vc hA m s sp s p /= === ρ κτ κρ τ τ characteristic size of system ? 1/m=ρ c p V/(hA) is called time constant(relaxation time). When the τ =1/m ,the temp. valueθ /θ 0 =e -1 =0.368. ? The heat flow at any instant q=hA(T ∞ -T), from θ /θ 0 = (T-T ∞ )/(T 0 -T ∞ )=e -mτ , can be writed q=hA(T ∞ -T 0 ) e -mτ . Chapter 3 unsteady state heat conduction The total heat quantity )1()( 0 0 τ τ ρτ m p eVcTTqdQ ? ∞ ??== ∫ Note:The heat balance equation of lumped parameter system is different from above situation for another surrounding condition. mτ θ / θ 0 The time constant is a characteristic parameter,which is determined by geometrical parameters,physical parameters, and convective situation(surrounding condition). The less 1/m,the response faster is. ? 3-4 Analysis of transient conduction for semi- infinite solid ? Transient heat conduction—a boundary of solid is suddenly contact with a surrounding at a uniform temperature and the initial condition is maintained in the direction normal to the boundary. ? The math. Mould can be written as Chapter 3 unsteady state heat conduction T ∞ T 0 x 0 ∞∞ ?=?==∞→ = ? ? =+ ? ? ==> == ? ? = ? ? TTTTwherex c x or h x orx x 000 0 2 2 0000 0 1 θθθθ θ θ κ θ θτ θθτ θ τ θ α ? The temp. can be shown as a function of single variable for above situation. ? Substituting above variables into the differential equation, we obtain Chapter 3 unsteady state heat conduction ατ ηη ατ ατ θ θ 2 )(: 2 0 2 0 x TT TT YwherefYisfunctionthe x for x f = ? ? == ? ? ? ? ? ? ? ? ? ? ? ? = ∞ ∞ 1)(;0)0(:..,02:. 2 2 =∞==+ YYcband d dY d Yd eq η η η Integrating the equation, we obtain . 21 2 cdecY += ∫ ? η η ? This solution can be written as ? If given Chapter 3 unsteady state heat conduction By substituting the error function and boundary conditions, where ? ? ? ? ? ? = ? ? ?= ∞ ατ η 2 )( x erf TT TT erfY ∫ ? = φ η η π φ 0 2 2 )( deerf 00 =+ ? ? = θ κ θ h x x The solution is ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? ? ++? ? ? ? ? ? = ατ κ ατ κ ατ κ ατ θ θ 2 1exp 2 2 2 0 xh erf hhxx erf The problem of semi-infinite solid can be used in the initial stage. So the solution of semi-infinite solid is available. When the η =2,the θ /θ 0 =0.995, we can think the temp. at η is equal to initial temp.. When are geometric position x>4(ατ ) 1/2 and the time τ <x/(16 α ), the temp. is not variable. ? The heat flux (at the position x and the surface) The total heat quantity from 0 to τ Chapter 3 unsteady state heat conduction πατ κ πατ κηκ θ κ ατ 0 )4/( 0 0 2 )]([)( TT q e TT erf x TT x q w w x w wx ? = ′′ ? = ? ? ??= ? ? ?= ′′ ? )( )(/2 0 0 tcoefficiencendothermiheatofabsorptioniscwhere TTcAdqAQ p wpw κρ κρπττ τ ?= ′′ = ∫ ? 3-5 2-D or 3-D transient heat conduction ? The solution for 2-d or 3-D can be obtained in a result of the solution of 1-D for some simple geometric forms. ? It is to prove that the 2-D solution can be represented by product of two 1-D solutions. ? An infinite long rectangular prism, whose mathematical mould and initial and boundary conditions are the formula (1) Chapter 3 unsteady state heat conduction 00),,( 00),,( 01 0 2 0 1 2 2 2 2 2 1 = ? Θ? = ? Θ? +Θ = ? Θ? = ? Θ? +Θ ==Θ ? ? ? ? ? ? ? ? ? Θ? + ? Θ? = ? Θ? == == yy xx yyh x xxh y at yx δ δ κ τδ κ τδ τα τ ∞ ∞ ? ? =Θ TT TT where 0 x y 2δ 1 2 δ 2 ? Assume that Θ =Θ x Θ y (2) ? Θ x is satisfied by (3) : Substituting (2) into (1) yields (6): Chapter 3 unsteady state heat conduction 0001 0 2 2 1 1 = ? Θ? = ? Θ? +Θ==Θ ? Θ? = ? Θ? == = x x x x x xx xx xxh at x δ δ κ τα τ ? ? ? ? ? ? ? Θ? ? ? Θ? Θ= ? ? ? ? ? ? ? ? ? Θ? ? ? Θ? Θ 2 2 2 2 yx yy x xx y τ α τ From eqs.(3) (4) in the eq (6) the two sides are equal to zero respectively. Θ y is satisfied by (4) : 0001 0 2 2 2 2 = ? Θ? = ? Θ? +Θ==Θ ? Θ? = ? Θ? == = y y y y y yy yy yyh at y δ δ κ τ τ ? The result shows that the Θ x and Θ y is is satisfied with the equation (1). ? Similarly boundary conditions are satisfied as ? We can infer that solution form is established for 3-D problems Chapter 3 unsteady state heat conduction 00 1 1 = ? Θ? Θ+ΘΘ?= ? Θ? +Θ = = δ δ κκ x x yyx x xhxh Chapter 4 Numerical method of heat conduction ? 4-1 Numerical analysis of heat conduction ? Solution methods of differential equation ? (1) analysis solution—simple problems, the results are complex series; ? (2)numerical solution—complex problems,the solution is get with the aid of the computer, the method has three steps: ? a. divide the given research region into many small volume controlled by nodal point; ? b. establish the nodal point equation,based on some rules (energy balance),the methods have finite difference approximate, finite element method etc. ; ? c. solve the algebraic equation set. ? 4-2 Finite differential approximate of 2-D heat conduction equation ? A 2-D problem is shown in figure. ? The math. mould is that ? Subject to a set of initial and boundary conditions Chapter 4 Numerical method of heat conduction W E NN SS P S P WE N P h T ∞ 绝热 换热 τακ ? ? =+ ? ? + ? ? Tq x T x T v 1 2 2 2 2 ? The continuous region is divided into discrete differential element volumes, which are called control volumes. every one control volume is governed by a representative point called nodal point. We can establish the nodal point equations by having the aid of the energy balance ? Derivation of interior node equation ? A interior node is bordered on four neighboring points. The heat balance relation can be written as ? {rate of heat entering volume through the boundaries}+{rate of heat generated in the volume}={rate of increase of internal energy of volume} Chapter 4 Numerical method of heat conduction 0=? ?? ? ′′ ? =? ?? ? ? ′ =? ?=??+? ? ? +? ? ? +? ? ? +? ? ? ?=++++ Eor yx TT cEor yx TT cEwhere Eyxqx y TT x y TT y x TT y x TT Eqqqqq PP p PP p v PNPSPEPW vNSEW τ ρ τ ρ κκκκ T P ’ representing the temp. after a interval time Δτ T P ” representing the temp. before a interval time Δτ When the Δ E=0 and Δ x=Δ y the above equation is simplified as The former is an explicit form of finite difference equation, where the future temp.of point P is predicted by the temp. of the neighboring points. Chapter 4 Numerical method of heat conduction When the Δ E≠0 and Δ x=Δ y the above equation becomes 0/4 2 =?+?+++ κxqTTTTT vPNSEW for first situation(T P ’ is temp. at future time) for first situation(T P ” is temp. at past time) )/()41()( 22 pvPNSEWP cqT x TTTT x T ρτ τατα ?+ ? ? ?++++ ? ? = ′ ? ? ? ? ? ? ?++++ ? ? + ′′ ? ? + = )/()( )41( 1 2 2 pvNSEWPP cqTTTT x T x T ρτ τα τα ? Derivation of equation of node P on convective boundary ? The heat balance equation is Chapter 4 Numerical method of heat conduction The latter is one of implicit form of finite difference equation. It use the past time temp.of point P and the current temps of the neighboring points to express the point temp. now. 2 )( 222 yxT cyTTh yx q x y TTx y TT y x TT Eqqqqq pPv PNPSPW vcSNW ?? ? ? =??+ ?? + ? ? ? + ? ? ? +? ? ? ?=++++ ∞ τ ρκκκ ? For steady state ? For unsteady state the explicit form 0/)24(22 2 =?+ ? +? ? +++ ∞ κ κκ xqT xh T xh TTT vPNSW )/(2412)2( 2222 pvPNSWP cqT x xh x T x xh TTT x T ρτ τα κ τατα κ τα ?+ ? ? ? ? ? ? ? ?? ? ? ? ?+ ? ?? +++ ? ? = ′ ∞ ? Derivation of equation of node P on an adiabatic boundary ? The heat balance Chapter 4 Numerical method of heat conduction ? For unsteady state the implicit form ? ? ? ? ? ? ?+ ? ?? +++ ? ? + ′′ ? ?? + ? ? + = ∞ )/(2)2( 241 1 22 22 pvNSWPP cqT x xh TTT x T x xh x T ρτ τα κ τα τα κ τα 2222 yxT c yx q x y TTx y TT y x TT Eqqqq pv PNPSPE vSNE ?? ? ? = ?? + ? ? ? + ? ? ? +? ? ? ?=+++ τ ρκκκ ? For steady state ? For unsteady state the explicit form 4 /2 2 κxqTTT T vNSE P ?+++ = Chapter 4 Numerical method of heat conduction ? The physical significances of dimensionless numbers ? αΔτ /Δ x 2 —element volume Fourier’s number Fo Δ x reflects the dynamic characteristic of element volume. ? hΔ x/κ —element volume Biot’s number Bi Δ x reflects the relation of the convection with the conduction at boundary node. ? q v Δ x 2 /κ or q v Δτ /(ρ c p ) —characterize that the thermal source conducts the variation of temperature for a element volume. )/(41)2( 22 pvPNSEP cqT x TTT x T ρτ τατα ?+ ? ? ? ? ? ? ? ? ?+++ ? ? = ′ ? ? ? ? ? ? ?+++ ? ? + ′′ ? ? + = )/()2( 41 1 2 2 pvNSEPP cqTTT x T x T ρτ τα τα ? For unsteady state the implicit form Chapter 4 Numerical method of heat conduction ? 4-3 The method of solution for set of algebraic equations ? The heat conduction equation is replaced by a set of simultaneous algebraic equations for temperatures at the nodal points of a network constructed over the region.? The Gauss elimination method ? The equations have a form of matrix as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??? n n nn nnnn c c T T a aa aaaaa aaaaa aaaa aaa 1 1 , ,11,1 6,45,44,43,42,4 5,34,33,32,31,3 4,23,22,21,2 3,12,11,1 0 0 0 0 00 0 00 00 Chapter 4 Numerical method of heat conduction ? The set of algebraic equations have a form of a banded matrix, which can be solved efficiently with a digital computer by using the Gauss elimination process. The above matrix can become following form (through the operation of matrix): ? The last equation immediately gives T n , knowing the T n , the temp. T n-1 is determined from the (n-1)th equation and the calculations are carried out until T 1 is determined from first equation. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??? * * 1 * 2 * 1 1 2 1 * , * ,1 * 1,1 * 6,4 * 5,4 * 4,4 * 5,3 * 4,3 * 3,3 * 4,2 * 3,2 * 2,2 * 3,1 * 2,1 * 1,1 0 0 0 0 00 0 00 00 n n n n nn nnnn c c c c T T T T a aa aaa aaa aaa aaa Chapter 4 Numerical method of heat conduction ? The iterative method ? When the number of equation os very large, the matrix is not sparse, and the computer storage is critical, an iterative method is frequently used to solve the equations. ? The iterative begins with an arbitrary initial approximation to the solution and then successively modifies the approximation according to some rule. ? The Gauss-Seidel iterative method, for example, is one method frequently used for iterative solution, which is available to accelerate the convergence of the iteration process. ? A iterative method of solution of a system of coupled algebraic equations is discussed, as following ? ? ? ? ? ? ? ? ? =++++ =++++ =++++ nnnnnnn ininiii nn bTaTaTaTa bTaTaTaTa bTaTaTaTa K LLL K LLL K 332211 332211 11313212111 Chapter 4 Numerical method of heat conduction ? When the a ii ≠0 the equations can be written iterative forms ? The general expression is ? ? ? ? ? ? ? ? ? ++?= ????= ????= ? ?? nnnnnnnnn iininiiii nn aTaTaTabT aTaTaTabT aTaTaTabT /)( /)( /)( 1 112211 3 2211 11131321211 K LLL K LLL K ii n ijj jijii aTabT /)( )(1 ∑ ≠= ?= ? And the Gauss-Seidel iterative general expression is ii n ij k jij i j k jijii aTaTabT /)( 1 )( 1 1 )1( ∑∑ += ? = + ??= Chapter 4 Numerical method of heat conduction ? The iterative steps: ? A initial set of values for the T i is assumed ? Next, now value of the nodal temp. T i is calculated according to above equation, always using the most resent values of the temp. of the surrounding nodal point. ? The process is repeated until successive calculation defer by a sufficiently small amount in terms of a computer program. This means that a test will be inserted to stop the calculation, when|T i k+1 -T i k |≤δ for all T i . Alternately,a non-dimensional test may be selected such that ε ≥|(T i k+1 -T i k )/T i k |. Chapter 5 convective heat transfer ? 4-1 Concepts of convective heat transfer ? 1 The physical description of heat transfer process ? When the fluid flows over a solid body or inside a channel with a temperature difference between the fluid and the solid wall, the heat transfer process takes place. ? The mechanism of hat transfer: thermal convection plus thermal conduction. ? The typical physical model is shown as T ∞ u ∞ T w T m T w u m fluid solid fluid wall T u ∞ T ∞ T w u q w (Contain main materials of the chapters 5, 6, and 7) ? The classification for convective heat transfer ? According to the cause of fluid motion: ? Two main sorts: forced convection and free convection, for the former the fluid motion is induced by external forces; for the latter fluid motion is set up by buoyancy effects resulting from density difference caused by the temperature difference in the fluid. ? According to the mode of contacting the fluid with the solid surface: ? Internal flow convection—fluid flows through the tube or channel; ? external flow convection—fluid flows over the a plate or across bluff body. ? According to fluid flow state: the laminar flow and turbulent flow. ? According to fluid change in phase: single phase and multi-phase flow. ? Therefore determination of temp. distribution and heat flow is a complicated matter. Chapter 5 convective heat transfer ? 2 convective(or surface) heat transfer coefficient ? In order to simplify the heat transfer calculation, a heat transfer coefficient is defined as ? Assuming the main stream is x direction,and y axis is normal to the wall surface, the heat flux in the wall surface in the fluid side is Chapter 5 convective heat transfer 0= ? ? ?= ′′ y y T q κ )/()( ∞∞ ? ′′ =??= ′′ TTqhTThq ww as a result of assuming that the fluid does not slip at the wall. Therefore we obtain ∞ = ?=? ? ? ? ?= TTTwhere y T T h w y 0 κ and the solving the convection becomes determination of the heat transfer coefficient, which concentrates the various effects of convective heat transfer ? The following factors have an effect on the h : ? The fluid velocity, the physical properties of fluid, the surface geometric form, the flow state of fluid, buoyancy force of fluid, and the change of phase of fluid. So the function form of h is h=f(u, ρ, μ, κ, β?T, L, T). ? The heat transfer coefficient can be evaluated by the three ways: a). The analytical method—solving governing equations b). The experimental method —measuring the fluid flow field c). The numerical method —calculating set of algebraic equations conducted from the governing equations. ? We lay emphasis on three points: ? 1). Know well grasp of the rules of convective heat transfer; ? 2). Distinguish the primary and the secondary; ? 3). Be good at calculating the criterion relations. Chapter 5 convective heat transfer ? 5-2 The mathematical description for convective heat transfer ? The CHT is complicated by the fact that the motion of the fluid plays an important part in the heat transfer. ? For determining the temperature field of fluid the velocity field is essential. Then the continuity, momentum and energy equations are derived by the mass balance, momentum balance and energy balance (conservation). ? The 2-D fluid flow is considered in order to emphasis on the significance. ? 1 The continuity equation ? It is derived by a mass conservation in a differential element volume dxdy·1 in a flow field. ? The mass balance is in the flow field: ? Net rate of mass flow entering the element volume=rate of increase of mass in the element volume. Chapter 5 convective heat transfer ? The entering mass flow rate in x direction Chapter 5 convective heat transfer dx dy 1?udyρ 1)( ? ? ? + dxdy y v vρ 1)( ? ? ? + dydx x u uρ 1?vdxρ 1? ? ? dxdy τ ρ 1? ? ? ? ? ? ? ? ? ? ? + ? ? dxdy y v x u ρρ － The leaving mass flow rate in x direction The entering mass flow rate in y direction The leaving mass flow rate in y direction The net rate of mass flow entering the dxdy.1 The rate of increase of mass with time in the dxdy.1 According to the mass balance in the dxdy.1, we can obtain the continuity equation 0= ? ? + ? ? + ? ? y v x u ρρ τ ρ 0= ? ? + ? ? y v x u ρρ 0= ? ? + ? ? y v x u For general form For steady state For incompressible fluid ? 2 The momentum equation ? According to the momentum conservation law the momentum balance in a element volume is ? (rate of increase of momentum with time) = (momentum variation resulting from the fluid flow) + (the external forces acting on the element) in a given direction ? The momentum rate of entering mass from x direction Chapter 5 convective heat transfer uudy ??1ρ The momentum rate of leaving mass from x direction ? ? ? ? ? ? ? ? +?? ? ? ? ? ? ? ? ? + dx x u udydx x u u 1 ρ ρ The momentum rate of entering mass from y direction The momentum rate of leaving mass from y direction uvdx ??1ρ ? ? ? ? ? ? ? ? ? ? +?? ? ? ? ? ? ? ? ? ? ? + dy y u udxdy y v v 1 ρ ρ The momentum variation rate in x direction : uudy??1ρ ? ? ? ? ? ? ? ? +??? ? ? ? ? ? ? ? + dx x u udydx x u u 1 ρ ρ uvdx ??1ρ ? ? ? ? ? ? ? ? ? ? +?? ? ? ? ? ? ? ? ? ? ? + dy y u udxdy y v v 1 ρ ρ (1) momentum variation caused by the fluid flow The momentum change rate in the x direction can be got by using continuity equation and omitting the smaller order amount, as ? (2)body forces ? The body force is resulting from the gravitational, electric, or magnetic field. ? We let Chapter 5 convective heat transfer Similarly, the momentum change rate in the y direction can be got as 1? ? ? ? ? ? ? ? ? ? ? + ? ? ? dxdy y u v x u uρ (1a) 1? ? ? ? ? ? ? ? ? ? ? + ? ? ? dxdy y v v x v uρ (1b) Body force in the x direction Body force in the y direction 1?dxdyF y 1?dxdyF x (2a); （ 2b ） ? (3)surface forces ? Force acting per unit area is called stress. ? Normal stress—the force is vertical to the acted surface; ? shear stress—the force is parallel to the acted surface. ? The net surface forces in the x direction and the net surface forces in the y direction are respectively Chapter 5 convective heat transfer dy dx 1? ? ? ? ? ? ? ? ? ? ? + dydx x xy xy τ τ ;1? ? ? ? ? ? ? ? ? ? ? + dxdy y yx yx τ τ 1? ? ? ? ? ? ? ? ? ? ? + dxdy y y y σ σ 1?dx y σ 1?dy x σ ;1?dy xy τ 1? ? ? ? ? ? ? ? ? + dydx x x x σ σ 1?dx yx τ 11 ? ? ? +? ? ? dxdy x dxdy y x yx σ τ 11 ? ? ? +? ? ? dxdy y dxdy x yxy στ σ x —the x direction normal stress σ y —the y direction normal stress τ yx —the x direction shear stress in the y surface τ xy —the y direction shear stress in the x surface ? For 2-D impressible constant property flow field, the stress related to the velocity components by: ? Substituting the stress relations into above surface force relations we obtain: Chapter 5 convective heat transfer ? ? ? ? ? ? ? ? ? ? + ? ? == x v y u yxxy μττ x u p x ? ? +?= μσ 2 y v p y ? ? +?= μσ 2 1 2 2 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? dxdy y u x u x p μ 1 2 2 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? dxdy y v x v y p μ (3a) (3b) in y direction in x direction ? (4)the momentum increase rate with time ? For incompressible fluid we can write ? For the x direction Chapter 5 convective heat transfer (5)the momentum equations the momentum equations are obtained by substituting the relations (1), (2), (3), and (4) into the momentum balance relation as 1? ? ? dxdy u τ ρ 1? ? ? dxdy v τ ρ (4a); (4b). For x direction For y direction For the y direction ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? 2 2 2 2 y u x u x p F y u v x u u u x μ τ ρ ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? 2 2 2 2 y v x v y p F y v v x v u v y μ τ ρ The right term characterize inertia force, the left first term is body force, the left second term is pressure force ,the left third term is viscous force. ? 3 The energy equation ? There is energy balance relation in the element volume as ? (net heat flow by conduction)+(net heat flow by convection)+(heat flow by viscous energy dissipation) =(enthalpy increase rate with time) ? Net heat flow by conduction Chapter 5 convective heat transfer Quoting Fourier law we obtain dy dx 1? ? ? ? ? ? ? ? ? ? ? + dxdy y Q Q y y 1?? ? ? ? ? ? ? ? + dydx x Q Q x x 1?dxQ y 1?dyQ x dy y Q dx x Q Q y x ? ? ? ? ? ?=? 1 111 2 2 2 2 ? ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? = dxdy y T x T dydx y T y dxdy x T x Q x κκκ Net heat flow by convection dy dx 1?utdyc p ρ 1?? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? + dydx x t tdx x u uc p ρ 1?vtdxc p ρ 1? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? + dxdy y t tdy y v vc p ρ The x direction: entering heat flow rate 1?uTdyc p ρ 1? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? + dydx x T Tdx x u uc p ρ leaving heat flow rate Chapter 5 convective heat transfer The net heat flow rate 1 2 ? ? ? ? ? ? ? ? ? ? ? + ? ? ?=? dxdy y T v x T ucQ p ρ The internal energy or enthalpy increase rate 1? ? ? ? dxdy T cE p τ ρ＝ Φ+ ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? μλ τ ρ 2 2 2 2 y T x T y T v x T u T c p Φ+ ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? μλρ 2 2 2 2 y T x T y T v x T uc p 1?vTdxc p ρ leaving heat flow rate 1? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? + dxdy y T Tdy y v vc p ρ The y direction: entering heat flow rate 12 22 2 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? =? dxdy x v y u y v x u Q μμ The viscous dissipation changes the heat flow The energy equation can be obtained by substituting the above various terms into the energy balance relation,as ? The first is unsteady state energy equation and second is steady state energy equation. For the low velocity flow of the low viscous fluid the viscous dissipation term can be omitted. For the conduction the fluid velocity become to zero and the equation degenerates differential equation of heat conduction. ? The set of the differential equations for convection heat transfer can be written: Chapter 5 convective heat transfer ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? 2 2 2 2 y u x u x p F y u v x u u u x μ τ ρ ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? 2 2 2 2 y v x v y p F y v v x v u v y μ τ ρ 0= ? ? + ? ? y v x u Φ+ ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? μλ τ ρ 2 2 2 2 y T x T y T v x T u T c p the energy differential equation 22 2 2 ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? =Φ x v y u y v x u the viscous dissipation function the continuity equation the x direction momentum eq. the y direction momentum eq ? 5-3 The experimental investigation method of convective heat transfer ? 1 non-dimensional groups for convection ? The complex momentum equations and energy equation to solve them is extremely difficult. The experimental research is adopted for solving problems of convection heat transfer. The non-dimensional groups involved by the empirical relation are used often. ? Two different methods are used to determine non-dimensional groups: ? (1)dimensional analysis—the produce is straight for ward, but the analysis may load to incorrect result if one or more of the variable are omitted. ? (2) similar analysis—the governing equations change the non-dimensional forms . ? We adopt the second method. For example a 2-D steady incompressible constant property fluid flow is considered. The main flow is in the x direction,and the body force is neglected for simplicity equation ? The governing equations are given as Chapter 5 convective heat transfer ? Select the characteristic parameters: ? Size L, velocity u ∞ , temp. diff. ?T=T w - T ∞ , pressure diff.?p=p in -p out ? The non-dimensional variables Chapter 5 convective heat transfer . 0 2 2 2 2 2 2 2 2 2 2 2 2 ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ?= ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? + ? ? y t x t y t v x t uc y v x v y p y v v x v u y u x u x p y u v x u u y v x u p λρ μρ μρ ； ； ； ? ? ? ? ? ? ? ? ? Θ? + ? Θ? ? Θ? ? Θ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ? ? ? ? 2 2 2 2 2 2 2 2 2 2 2 2 PrRe 1 Re 1 Re 1 0 YXY V X U Y V X V Y P Eu Y V V X V U Y U X U X P Eu Y U V X U U Y V X U ＝＋ ＝－＋ ＝－＋ ＝＋ )/()(;/;/;//;/ ∞∞∞∞ ??=Θ?===== ttttppPLyYLxXuvVuuU w u ∞ T ∞ L p in p out T w Introducing the above variables into the set of equations,we obtain: ? The non-dimensional groups(criteria) are defined as Chapter 5 convective heat transfer Ectert number )/( 2 p TcuEc ?= ∞ Ratio of dynamic temp. rise due to fluid motion to parameter temp. diff. for a given flow field Eular number ? ? ? ? ? ? ?= ∞ 2 2 1 upc D ρ )/( 2 ∞ ?= upEu ρ Ratio of pressure force to inertial force, or resistance coefficient for a given flow field Reynolds number νμρ LuLu ∞∞ ==Re Ratio of inertial force to viscous force for a given flow field Peclet number PrRe)/)(( ?==== ∞∞∞ aLuaLuLucPe p ννλρ Ratio of thermal convection to thermal conduction for a given flow and heat transfer process Prandel number aν=Pr Ratio of diffusivity of momentum to diffusivity of heat for a given flow field ? Similarly the convective heat transfer equation can be non-dimensionalized as Chapter 5 convective heat transfer 0 0 = = ? Θ? ?=? ? ? ? Y y Y Nu y T t κ α＝－ κhLNu= is called Nusselt number,which characterizes the ratio of convective capability of fluid to conductive capability of fliud 流体侧 固体侧 L s L f λ f λ s Nu Bi Θ The Bi characterizes the ratio of the convective capability of a fluid surrounding a solid wall to the conductive capability of the solid. Note the Nu is differnet with the Bi ? 2 the solution forms of the non-dimensional equatios ? For Above non-dimensional equations the solution forms are: ? the velocity field The derivation of the temp.field with respect to Y, at Y=0, yields Chapter 5 convective heat transfer ( ) ( )YXPEufVYXPEufU vu ,,,Re,,,,,Re, == ( ) ( )Re,,, ep fEuYXEufP =＝ ( )YXVUf ,,,Pr,Re, θ ＝Θ the temperature field can be represented as ( ),,Pr,Re, YXf θ ＝Θ () x Y NuXf Y ?= ′ = ? Θ? = Pr,Re, 0 θ Taking the average value of the temp.field with respect to Y, at Y=0, yields ( )PrRe, θ fNu x ′ = the pressure field the temperature field ? According to this relation for geometrical similar surfaces the heat transfer coefficient for forced convection at moderate flow velocity can be correlated in terms of two dimensionless groups instead of several variables the t enter the problem. This is important in the experimental investigation. 3 the experimental method for determining criterion relationships Consider a convective heat transfer that the fluid flows over a plate,which has some sizes shown in the figure. When the electrical current I and voltage V of heating the plate are measured, the convective coefficient h can be obtained Chapter 5 convective heat transfer ( )[ ]LBttIVh w ∞ ?= N uh uh uh uh N ∞ ∞ ∞ ∞ ? ? ? ? L 3 2 1 3 2 1 ν ν ν ν LukLhNu LukLhNu LukLhNu LukLhNu N NNN ∞ ∞ ∞ ∞ =?= =?= =?= =?= Re Re Re Re 3 2 1 333 222 111 L t w q L B the h is corresponding with the velocity ,size, flow state, heating state.when the velocity is changed we can obtain a set of correlative data Non-dimensionalized ? Assume the criterion relation had a following form Chapter 5 convective heat transfer ),,3,2,1( kinxay ii L=+= ′ () 2 1 ∑ = ? ′ = k i ii yyW 0,0 = ? ? = ? ? a W n W ? ? ? ? ? ? ? =+ =+ ∑∑ ∑∑∑ == == k i k i ii k i k i k i iiii yakxn yxxaxn 11 111 2 n cNu Re 1 = Introducing to the logarithm operation we can get nxaygngcgNu +=?+= Re 1 lll using the method of least square for evaluating the values of n and c 1 . Test at k times of measurement obtains relations as Square difference sum up In order to minimizing the W let the we obtain ? Solving above equations yields the values of n and a . Then the relation is got as ? Notes: ? L—characteristic size: along the stream wise for external flow; perpendicular to the stream wise for internal flow. ? u—characteristic velocity: coming velocity for over plate or bluff-body; mean velocity for flow inside pipe or channel; maximum velocity for flow over tube banks. ? T—temperature for determining fluid properties: average fluid temperature T f =(T f1 +T f2 )/2 for flow inside pipe; T m =(T ∞ +T w )/2 for external flow. Chapter 5 convective heat transfer n cNu Re 1 = nxay += The criterion relationship is determined Using some method we obtain mn cNu PrRe= ? 5-4 Boundary layer concept and boundary layer equations ? 1 Boundary layer—proposed by Prandtl successful simplifying momentum and energy equations ? using the boundary layer concept the fluid flow over a body is divided into 2 regions: ? One is boundary layer (thin layer). The gradients of velocity and temperature are steep. ? Another is potenial flow or external flow. The gradients of velocity and temperature are small. ? For example: fluid flow over a flat plate ? Edge of the velocity boundary layer is appointed as u=0.99u ∞ at y=δ. ? Edge of the thermal (temp.) boundary layer is appointed as θ=0.99 θ w θ w =T w -T ∞ at y=δ t . ? So the definition of the boundary-layer thickness may be arbitrary. δ/δ t =f(Pr),they are same order of magnitude. Chapter 5 convective heat transfer t w δ t δ T ∞ u ∞ edge x0 ? The laminar and turbulent boundary layers ? The boundary layer is developed and can be turbulent flow, when the Re= u ∞ x/ν=5×10 5 , which is called turbulent boundary layer. ? The turbulent boundary layer can be divided into three layers,which are : ? Viscous sub-layer—molacular diffusion ? Buffer sub-layer —molacular diffusion + turbulent lump diffusion ? Turbulent sub-layer —turbulent lump diffusion ? 2 Boundary layer equations ? According to the boundary layer concept the 2-D equations can be simplified by on order-magnitude comparison Chapter 5 convective heat transfer Laminar boundary layer turbulent boundary layer Critical distance X c is specified in term of Re ? Example: the fluid flow along a plate ? When the δ/x<<1 and δ t /x<<1 are assumed, the order-magnitude x/L～1, and y/L～? can be determined, then the governing equations have following order-magnitudes: ? In order to the (1/1 2 +1/? 2 )/Re～1,the Re must have order-magnitude 1/? 2 . Then the term of (1/Re)(1/1 2 )～?, can be negated(omitted). The x direction momentum equation becomes Chapter 5 convective heat transfer ? ? ? ? ? ? ? ? ? ? ? ? ? ? += ? ?+? ? ? ? ? ? ? ? ? ? Θ? + ? Θ? ? Θ? ? Θ? ? ? + ? ? + ? = ? ? ?+ ? ? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ? ++= ? ?+? ? ? ? ? ? ? ? ? ? ? + ? ? + ? ? ? ? ? ? ?≈?= ? ? +? ? ? ? ? ) 1 1 1 ( PrRe 11 1 1 1 PrRe 1 )( Re 11 1 1 Re 1 ) 1 1 1 ( Re 1 1 11 1 1 1 Re 1 0 1 1 0 222 2 2 2 222 2 2 2 222 2 2 2 YXY V X U Y V X V Y P Eu Y V V X V U Y U X U X P Eu Y U V X U U V Y V X U ＝＋ ＝－＋ ＝－＋ ＝＋ 2 2 Re 1 Y U X P Eu Y U V X U U ? ? + ? ? ? ? ? ? ＝－＋ ? and the y direction momentum equation becomes through the order- magnitude comparison ? Assuming the RePr～1/? 2 we can obtain the energy equation of boundary layer as Chapter 5 convective heat transfer 0= ? ? ? Y P The momentum equation becomes of the boundary layer is got 2 2 Re 1 Y U dX dP Eu Y U V X U U ? ? + ? ? ? ? ＝－＋ 2 2 PrRe 1 YY V X U ? Θ? ? Θ? ? Θ? ＝＋ The set of boundary layer equation changes to dimensional form as ； ； ； 2 2 2 2 0 yy v x uc y u dx dp y u v x u u y v x u p ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? ? ? +?= ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? + ? ? θ κ θθ ρ μρ ? Removal of the x direction diffusion terms of U and Θ implies that U and Θ is bounded on the upper stream side, and the down stream side does not effect on the upper stream side. There is a single tunnel problem. We must take note of the condition of the formation, Re～1/? 2 and RePr～1/? 2 ? 3 the solutions of the boundary layer equations ? The boundary layer equations can be analyzed for some simple flow problems. ? for example fluid flow along a flat plate is a typical boundary layer flow, the momentum equation was solved by Blasius and the energy equation is solved by Bohlhausen. The results are: ? δ/x=4.96Re -1/2 and c f =0.664Re -1/2 , where Re=u ∞ x/ν; ? δ t /x=4.51Re -1/2 Pr -1/3 and δ t /δ=0.975Pr -1/2 . ? The criterion relations for the convective heat transfer of the flow over the plate Chapter 5 convective heat transfer 3/12/1 PrRe332.0 xx Nu = (for local nusselt number) 3/12/1 PrRe664.0 xx Nu = (for average nusselt number) ? 5-4 analogies between heat transfer and momentum transfer in turbulent flow ? 1 the turbulent flow and heat transfer ? The boundary layer can change form the laminar state to the turbulent state. ? The mechanism of momentum and heat transfer in the turbulent flow—the crosswise diffusion and mixing of macroscopic lumps of fluid. We can define the eddy viscosity and eddy thermal diffusivity,which may be 10 times as large as the molecular diffusivity of momentum and energy. ? Description of the turbulent flow of fluid ? Instantaneous velocity is equal to the sum of mean velocity up fluctuation velocity, as Chapter 5 convective heat transfer . . ∫ ?+ = ′ += ′ += ′ += ′ += ττ τ τφφφφφ dwhereformgeneral TTTvvvSimilarlyuuu ii iii ? The ?τ—a small time interval is large enough for turbulent fluctuation time. ? The rules: ? Explain: a turb. Lump moves across a plane p-p. the mass flow is ρv′, and lead to a change in x direction component velocity -u′. There is a force action,which is a shear stress Chapter 5 convective heat transfer 000 2 ≠ ′′ ≠ ′ = ′ φφφφ uv t ′′ ?= ρτ A eddy viscosity or eddy momentum diffusivity is defined ityviseddyiswhere y u uv mmt cosερερτ ? ? = ′′ ?= Similarly: ρv′ associated with this mass transport is a change in the temp. fluctuation T′. The result in turb. Heat transfer is Tvcq pt ′′ = ρ ? We can define a eddy heat diffusivity ε h that is contained by Chapter 5 convective heat transfer y T cTvcq hppt ? ? ?= ′′ = ερρ We can define a turb. Prandtl number Pr t = ε m /ε h The total transfer performances in the turb. Flow are y T cqqq hptl ? ? +?=+= )( εαρ y u mtl ? ? +=+= )( ενρτττ In the turb. Boundary layer corelayerturbulentforqq sublayerviscousforqq tthm llhm ==>>>> ==<<<< τταενε τταενε ? 2 Reynolds analogy ? Assumed that turb. Boundary layer consist of a single zone of high turb. Region for a flow over a plate. (neglect viscous sub-layer and buffer sub-layer). There are ? Integrating above equation from T=T w at u=0 to T=T ∞ at u=u ∞ , as Chapter 5 convective heat transfer du c q dTand dy dT c q dy du then p hm h p mhm τ εεε ε ρ ε ρ τ αενε ?=?== ?==>>>> numbertonScalled Nu Stwhere c uc h StyieldWe equationabovetoucandTThqngsubstituti u c q TTobtainwedu c q dT f p fw p u p T T w tan PrRe2 2/)( 2 0 === =?= ′′ =??= ∞ ∞∞ ∞ ∫∫ ∞∞ ρ ρτ ττ ? For fluid flow inside tube the shear stress at wall of tube can be written as ? In engineering application the colbuen analogy has a form as Chapter 5 convective heat transfer Integrating the dT=-qdu/(τc p ) form T=T w at u=0 to T=T m at u=u m , where T and u are mean bulk stream temp. and velocity respectively, we obtain: 2 2 ∞ = u f ρ τ 8 f uc h St p == ∞ ρ 2 Pr 3/2 3/2 f p c uc h St = ? ? ? ? ? ? = ∞ α ν ρ For flow over a plane plate the local skin friction coeff. The average skin friction coeff. 2.0 2.0 Re074.0 Re059.0 ? ? = = xfx xfx c c ? Applying above relation to Colburn analogy we get ? Local convective coeff. and average convective coeff. ? Prandtl analogy Chapter 5 convective heat transfer 3/18.05 3/18.0 3/18.0 Pr)850Re037.0(105Re PrRe037.0 PrRe0296.0 ?=×= = = xxc xx xx uNthenfor uN Nu u ∞ T ∞ u L T L T W Viscous sub-layerε m << νε h <<α T W -T L =Prq ” u L /(τc p )(1) Turb. Core ε m >> νε h >>α T L -T ∞ =q ” u ∞ /(τc p )[1-(Pr-1)u L /u ∞ (2) (1)+(2) St=c f /2{1/[1+(Pr-1)u L /u ∞ ]} St=c f /2{1/[1+5(Pr-1)(c f /5) 1/2 ]} ? 5-6 Empirical and practical relations for forced convective heat transfer ? 1 the characteristics of flow and heat transfer for forced convection in a tube Chapter 5 convective heat transfer Laminar and turbulent flows Mean velocity and bulk temperature .)( 1 )/( constfor A V u udA A AmuAum m A mm == ==?= ∫ ρ ρ ρ ρρ & && Developing region and fully developed region The properties effect on the heat transfer ? The effect on the entrance region ? Entry length—developing region ? Laminar velocity entry length ? Laminar thermal entry length ? Turbulent entry length Chapter 5 convective heat transfer V TudA Tthenconstcandfor udAc TudAc T A p p p f & ∫ ∫ ∫ === .ρ ρ ρ To take into account the property variations, we introduce the correction terms as (μ f /μ w ) n or (Pr f /Pr w ) n in the relations for convective heat transfer 50/Re623.0/ .)(07.0/ .)(055.0/ Re06.0/ 4/1 == = ′′ = == = dLordL constqPedL constTPedL dL wf wt f ? 2 The empirical relations for turbulent flow in a tube ? For fully developed turbulent flow in smooth tube ? Recommending: Nu=0.023Re 0.8 Pr n , ? where n=0.4 for heating fluid or n=0.3 for cooling fluid. ? The equation can be used Re=10 4 ~1.2 ? 10 5 , Pr=0.6~100 and moderate temperature difference:?T<50K for gases; ?T<30K for water; ?T<10K for oils. ? For Pr=0.6~700;Re=10 4 ~1.7 ? 510 6 we must take into account the variable properties,then the relation become ? Nu=0.027Re 0.8 Pr 1/3 (μ f /μ w ) 0.14 . ? In the entrance region Nu=0.036Re 0.8 Pr 1/3 (d/L) 0.055 , for L/d=10~400 Chapter 5 convective heat transfer ? Some discussions: ? 1. non-circular cross of tube or channel hydraulic diameter d e =4A/P; ? 2. Entry effect correction ε L =1+(d/L) 0.7 ; ? 3.bent tube correction (because second recirculating flow) ? ε R =1+1.77(d/L) for gases; ? ε R =1+10.3(d/L) 3 for liquids; ? 4.Characteristic size, Characteristic velocity and temperature for determining the properties Chapter 5 convective heat transfer ? In the transition flow ? For gases Nu=0.0214(Re 0.8 -100)Pr 0.4 [1+(d/L) 2/3 ](T f /T w ) 0.45 , there Re=2200~10000 and Pr=0.6~1.5; ? For liquids Nu=0.012(Re 0.87 -280)Pr 0.4 [1+(d/L) 2/3 ](Pr f /Pr w ) 0.11 , there Re=2200~10000 and Pr=0.6~1.5. Chapter 5 convective heat transfer ? 3 Empirical relations for laminar and transition flows ? For laminar flow Nu=1.86(RePrd/L) 1/3 (μ f /μ w ) 0.14 , when RePrd/L>10; ? Nu=0.46Re 0.5 Pr 0.42 (Pr f /Pr w ) 0.25 (d/L) 0.4 , for application region Re=186~4530, Pr f /Pr w =1.43~18.2. ? 5-7 Empirical relations for flow across cylinder or its banks Chapter 5 convective heat transfer 分离流动速度分布 边界层速度分布 1 flow acros a cylinder Flow characteristics Physical analysis: p ↓, u ↑ along the front side of cylinder and p ↑, u ↓ along the back side of cylinder. on the wall of cylinder fluid velocity is equal to zero u=0, no inertial force may be to transform the pressure force, but the pressure outside can transfer into the wall. So there is a negative pressure gradient in the close layer,which can lead to reverse flow or separation flow. Mathematical analysis: ρ dp/dρ =-d(u 2 /2)=-udu=-u[(?u/?x)dx+(?u/?y)dy]. There is a point of dp=0 and where ?u/?x=0, so must ?u/?y=0 Chapter 5 convective heat transfer ? Drag coefficient ? F=c D Aρ u 2 /2 ? F—drag force(frictional resistance and form or pressure drag); ? c D —drag coefficient; A —frontal area; ρ u 2 /2—inertial force. ? When Re=ρ u ∞ d/μ <1 only friction force; Re=10~1000 friction+form drag; Re>1000 form drag; Re>10 5 laminar flow transit into turbulent flow. Heat transfer characteristics Heat transfer calculation in engineering Average heat transfer coefficient Nu=cRe n Pr m A table of c and n Chart. Size; chart.velocity; temp. for determining physical properties Chapter 5 convective heat transfer ? 2 Flow across thetube banks ? Two typical sorts:Staggered tubes and in-line tubes d d s 1 s 1 s 2 s 2 u max u ∞ u ∞ t ∞ t ∞ 图 5-24 流体绕流管束时的流动特征及几何尺寸示意图 (1)叉排管束 (2)顺排管束 The flow is different from that of a tube except first and second rows Chapter 5 convective heat transfer ? The calculation relation for air Nu=cRe n , where the c and n can be given in a table, ? and Re=u max d/ν , where u max is maximum velocity through the minimum flow area, d is external diameter, ν is fluid viscosity determined under the film temperature T m =(T w +T ∞ )/2. ? The relation is used in tube banks having 10 or more rows in the flow direction. ? The heat transfer performance in first or second row is different to that of subsequent rows. Thus influence is remained till to 10 rows. The correction coefficients are given in a table. Chapter 5 convective heat transfer ? 4-7 Natural convective heat transfer ? 1 Natural convection ? Natural convection resulting of the motion of the fluid due to density change araising from the heating or cooling process ? 2Natural convective heat transfer on a vertical flat plate 紊流流动状态 层流流动状态 边界层速度分布曲线 边界层温度分布曲线 t w t ∞ x 0 y x 0 y （ 1）速度分布和温度分布 （ 2）自然对流边界层的发展 图 5-26 竖直平板在空气中冷却过程 Chapter 5 convective heat transfer ? This is a type of boundary layer flow, but unlike the boundary layer flow of forced convection. ? Mathematical mould: ； ； ； 2 2 2 2 0 yy v x uc y u dx dp F y u v x u u y v x u p x ? ? = ? ? ? ? ? ? ? ? ? ? + ? ? ? ? +?= ? ? ? ? ? ? ? ? ? ? + ? ? = ? ? + ? ? θ λ θθ ρ μρ gF x ρ?=For momentum equation of natural convection the and beside the boundary layer g dx dp ∞ ?= ρ Chapter 5 convective heat transfer ? Substituting above relations into the momentum equation we can obtain ? The density difference maybe expressed in terms of the volume coefficient of expansion,defined by ρθ ρρρρ ρ β ? = ? ? ? ? ? ? ? ? ? ? ≈ ? ? ? ? ? ? ? ? = ∞ ∞ ∞ TTT v v p 11 () 2 2 y u g y u v x u u ? ? +?= ? ? ? ? ? ? ? ? ? ? + ? ? ∞ μρρρ so that T1＝β () 2 2 y u TTg y u v x u u ? ? +?= ? ? ? ? ? ? ? ? ? ? + ? ? ∞ μβρρ For ideal gases The equations can be solve simultaneously for obtaining the velocity profile and temperature profile Chapter 5 convective heat transfer ? The experimental solution ? The equations after giving the reference variables,as ? Can be non-dimensionalized as 。 ； ； 2 2 2 2 2 0 YLu a Y V X U y u Luu Lg Y U V X U U Y V X U a aa w ? Θ? = ? Θ? + ? Θ? ? ? +Θ= ? ? + ? ? = ? ? + ? ? νβθ waa u v V u u U L y Y L x X θ θ =Θ==== ,,,, ∞ ?= ttθ ∞ ?= tt ww θ where When the reference velocity is defined as Lgu wa βθ= The equation be written as Chapter 5 convective heat transfer where called Grashof number,whose physical significance is 。 ； ； 2 2 2 2 2 Pr 1 1 0 YGrY V X U y u GrY U V X U U Y V X U ? Θ? = ? Θ? + ? Θ? ? ? +Θ= ? ? + ? ? = ? ? + ? ? 2 3 ν βθ Lg Gr w = forceviscous buoyancetodueforceinertial forceviscous forcebuoyance Lu LuTg Ge a a ?= ?? = 22 2 )/( / μ ρβρ The solution form has be obtained by analyzing as Nu=f(Gr, Pr) Chapter 5 convective heat transfer ? 3 empirical relations ? The general form as Nu=c(GrPr) n , where c and n are determined by the experimental data in the basis of different convection types. ? For vertical plate(or cylinder) ? Gr=gβΔ TL 3 /ν 2 , where L is plate height ? Temp. of determining physical properties T m =(T w +T ∞ )/2 ? When GrPr=10 4 ~10 9 , c=0.59, n=1/4 ? When GrPr=10 9 ~10 13 , c=0.10, n=1/3 Chapter 5 convective heat transfer ? For horizontal cylinder ? Gr=gβΔ Td 3 /ν 2 , where d is external diameter ? Temp. of determining physical properties T m =(T w +T ∞ )/2 ? When GrPr=10 4 ~10 9 , c=0.54, n=1/4 ? When GrPr=10 9 ~10 13 , c=0.13, n=1/3For horizontal plate Gr=gβΔ TL 3 /ν 2 , where d is characteristic size (square, side; rectangle, short side; disk, 0.9diameter) Temp. of determining physical properties T m =(T w +T ∞ )/2 Upper surface of heated plate or lower surface of cooled plate, when GrPr=2*10 4 ~8*10 6 , c=0.54, n=1/4, and when GrPr=8*10 6 ~10 11 , c=0.16, n=1/3; Upper surface of cooled plate or lower surface of heated plate, when GrPr=10 5 ~8*10 11 , c=0.55, n=1/4. pg transfer ? Heat transfer process is associated with a change of fluid phase in a research system ? 6-1 condensation heat transfer phenomena ?Consider a vertical plate exposed to a condensable vapor. If the temperature of the plate is below the saturation temperature of the vapor, condensate will form on the surface and the action of gravity will flow down the plate. If the liquid wets the surface, a smooth film is formed ,and the process is called film condensation. If the liquid does not wet the surface, droplets are formed which fall down the surface in some random fashion. This process is called drop-wise condensation. ?It is extremely difficult to maintain the drop-wise condensation. The convective coefficient α film <<α drop . ? 6-2 Analytical solution of film condensation of pure vapor transfer ? 1. Following simplifying assumptions: ? Plate uniform temperature , and T w <T s ; ? Vapor is saturate and exerts no drag on the motion of condensate; ? Condensate flow is laminar; ? Fluid acceleration may be neglected; ? Constant properties; ? Condensate layer heat transfer is pure conduction. 0 x dx x y δ ?film condensation on a vertical plate may be analyzed in a manner first proposed by Nusselt in 1916. Chapter 6 Condensation & boiling heat transfer 2. Determining velocity distribution of condensate film We consider a differential element volume of condensate. The force balance relation is given as: dx dy du dxyg lvl μδρρ =?? )()( Giving the boundary condition u=0 at y=0, we can integrate above equation to obtain the velocity profile as ) 2 ( )( 2 y yu l vl ? ? = δ μ ρρ 3. Determining the mass flow rate of condensate film Chapter 6 Condensation & boiling heat transfer 3 0 2 3 )( 2 )( δ μ ρρρ δ μ ρρ ρ δ gdy y ym l vll l vl l ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ∫ & Substituting u(y) into above integration relation, and performing the integration yields ∫ = δ ρ 0 udym lx where m x is mass flow rate per unit width of the plate. When the condensate film flows from x to x+dx, the film thickness grows from δ to δ +dδ . The amount of condensate added between x and x+dx is l vll l vll dg dx g dx d md μ δδρρρ μ δρρρ 23 )( 3 )( ? = ? ? ? ? ? ? ? = & 4. Determining the relation of mass rate with heat transfer Chapter 6 Condensation & boiling heat transfer According to the energy balance the rate of heat released during the condensation of mass dm must be transferred by conduction across the condensate layer to the plate. Then we get dx tt k dg md ws l l vll δ γ μ δδρρρ γ ?? = ＝ 2 )( & Separating the variables and performing integration of above equation, we yield the condensate thickness expression as 41 )( )(4 ? ? ? ? ? ? ? ? = vll wsll g ttxk ρργρ μ δ Chapter 6 Condensation & boiling heat transfer 5. Determining the heat transfer coefficients of condensation The local heat transfer coefficient: dx tt kdxtthdq ws lwsx δ )( )( ? =?= δ lx kh = we obtain Substituting thickness relation(δ ) yields 4 1 3 )(4 )( ? ? ? ? ? ? ? ? = wsl lvll x ttx kg h μ γρρρ The average coefficient(from 0 to L) can be obtained Lx L xm hdxh L h = == ∫ 3 41 0 4 1 3 )( )( 943.0 ? ? ? ? ? ? ? ? = wsl lvll m ttL kg h μ γρρρ or From Chapter 6 Condensation & boiling heat transfer 4 1 3 )( )( 943.0 ? ? ? ? ? ? ? ? == wsll vll l ttk Lg k hL Nu μ γρρρ Expressed in dimensionless form in terms of the Nusselt number, this is Above relations can be used to the vertical tube when the tube radius is large compared with the thickness of condensate film. 6.The heat transfer calculation for other condensate manner The condensation on horizontal tube The condensation on inclined plate 4 1 3 )( )( 725.0 ? ? ? ? ? ? ? ? = wsl lvll ttd kg h μ γρρρ 4 1 3 )( sin)( 943.0 ? ? ? ? ? ? ? ? = wsll vll ttk Lg Nu μ ?γρρρ g φ Chapter 6 Condensation & boiling heat transfer 7. Determining the Reynolds number for condensate flow Turbulent flow may be start at the lower position of plate. To establish a criterion for transition from laminar flow to turbulent flow, Reynolds number Re of condensate flow is defined as: Re=D e u m ρ l /μ l where u m -average velocity of condensate film; D e - hydraulic diameter for condensate flow D e =4A/P 4A-cross section area for condensate flow; P-wetted perimeter. For vertical plate of width b, P=bδ , then De=4δ . We can obtain Re=4δ u m ρ l /μ l = 4M e /μ l , where M h =δ u m ρ l –mass flow rate of condensate width of plate. For vertical tube of out side diameter D, We can obtain Re=4W/[μ l (π D)], where W= π ( δ D+ δ 2 )u m ρ l –mass flow rate at bottom of vertical tube. Chapter 6 Condensation & boiling heat transfer For horizontal tube of out side diameter D, We can obtain Re=4M h /μ l , where M h =δ u m ρ l –mass flow rate of condensation for unit length at bottom of horizontal tube . Usable form of Re For vertical tube the condensate heat flow is h m (T s -T w )L=γ W, then the mass rate W=h m (T s -T w )π DL/ γ . The Reynolds number can be written as Re=4h m (T s -T w )L/(μ l γ ). For vertical plate the condensate heat flow is h m (T s -T w )Lb=γ M h b, then the mass rate M h =h m (T s -T w )L/ γ . The Reynolds number can be written as Re=4h m (T s -T w )L/(μ l γ ). Chapter 6 Condensation & boiling heat transfer For horizontal tube the condensate heat flow is h m (T s -T w )π DL=γ ML, then the mass rate M=h m (T s -T w )π D/ γ . The Reynolds number can be written as Re=4h m (T s -T w ) π D /(μ l γ ). When the Re>1800 the turbulent flow occurs for vertical plate or tube When the Re>3300 the turbulent flow occurs for horizontal tube 8. Criterion relations for condensation heat transfer The average heat transfer coefficient is generally rearranged as a function of the Reynolds number. for vertical plate: (h m /k l )[μ l 2 /(ρ l 2 g)] 1/3 =1.74(4M/ μ l ) -1/3 or Co=1.74Re -1/3 . or experimental relation Co=1.76Re -1/3 Chapter 6 Condensation & boiling heat transfer for horizontal tube: (h m /k l )[μ l 2 /(ρ l 2 g)] 1/3 =1.51(4M/ μ l ) -1/3 or Co=1.51Re -1/3 ; for turbulent flow when the Re>1800 the experimental relation of vertical plate is: (h m /k l )[μ l 2 /(ρ l 2 g)] 1/3 =0.0076(4M/ μ l ) 0.4 or Co=0.007651Re 0.4 . The total heat transfer coefficient h=h l x c /L+h t (1-x c /L) 6-3 A discussion of some factors effecting on the film condensation 1. Non condensate gases as air: very small content of air can effect heavily the vapor to condensate. Chapter 6 Condensation & boiling heat transfer a.The partial pressure of vapor decreases and the partial pressure of air increases in the condensation process. On the film surface the air layer concentrates which obstructs the vapor to condensate in the wall. b.the vapor pressure drop can result in the saturation temperature decrease, then the temperature difference of heat transfer must decrease. So air content of weight 1% results to decrease of convective heat transfer coefficient h about 60%. 2. Vapor velocity a. Downward flow of vapor—can decrease the film thickness, hence the heat transfer coefficient h can increase ; b. Upward flow of vapor —by causing the thickening condensate layer the heat transfer coefficient h is to decrease; c. Higher upward flow can break through the film and causes the increase of the heat transfer coefficient Chapter 6 Condensation & boiling heat transfer 3. Hot vapor The vapor is superheated, so there is a enthalpy difference between the hot vapor and the condensate film. The latent heat γ '=i v -i l . 4. Cool condensate film The film is excess cool, the latent heat γ ‘ = γ +0.68c p (T s -T w ) Chapter 6 Condensation & boiling heat transfer When the surface is exposed to liquid and is maintained at a temperature avove the saturation temperature of liquid, boiling may occur, and the heat flux will depend on the difference temperature between the surface and saturation temperatures. Classifying: a.pool boiling and tube boiling; b. Sub-cooled (local) boiling and saturated (bulk) boiling. 1. The heat flux cure of boiling 6-3 Boiling heat transfer B C E A D Δ t s q " 0 Fig.6-2 pool boiling curve Pure free convection Interface evaporation Nucleate boiling Sub-cooled boiling(bubbles condense in heated liquid) and saturate boiling(bubbles form more rapidly and rise up the surface of liquid) Transition from nucleate boiling to film boiling A abyupt loss in the heat flux Stable film boiling heat flux drops to a minimum point and the heated surface is blamketd with a stable film of vapor Peak heat flux q c " Chapter 6 Condensation & boiling heat transfer There is a critical point or a peak heat flux which obtainable with a temperature difference of less than 55.6° C when the wall temperature is controllable. If the same heat flux is to be obtained in the film boiling , it requires a large temperature difference, will above the melting point of materials. So the maximum heat flux point in nucleate boiling is called the burnout point. (when the heat flux is controllable) 2. The bubble analysis In the nucleate boiling bubble are created by the expansion of entrapped gases or vapor at small cavities. The bubbles may collapse,expand in liquid or detach from the surface, which is controlled by the force equilibrium and heat balance. Chapter 6 Condensation & boiling heat transfer σ p v R σσ p l liquid interface Surface tension pressure The pressure forces of liquid and vapor must be balanced by the surface tension force at the vapor-liquid interface. For a spherical bubble we can obtain σππ RppR lv 2)( 2 ≥? R pp lv σ2 ≥? The force balance condition is In order to the bubbles to grow and escape to the surface of liquid, they must receive heat from the liquid. So that T l ≥T v . The heat balance condition is T l =T v . Assume that T v =T vs (corresponding to the p v ), if T l =T ls (corresponding to the p l ), because p l <p v inside the bubble,consequently heat must conduct out the bubble, the vapor inside the bubble must condense, and the bubble must collapse. So the liquid must be in a superheated condition. There is ?T s =T l -T ls =T v -T ls . Chapter 6 Condensation & boiling heat transfer The minimum radius of bubble being There has a thermodynamic relation at the phase interface as () slv TdT dp ρρ γ 11 ? = which for ρ v << ρ l the difference form is s v sv lv TTT pp γρ = ? ? Substituting the equation into force balance equation yields )( 2 svv s TT T R ? ≥ γρ σ Near heated surface ?T s =T v -T ls =T l -T ls =T w -T s , this maximum superheated temperature difference, the minimum radius can be obtained as () max min 2 sv s t T R ?γρ σ ＝ Chapter 6 Condensation & boiling heat transfer Analysis : When T w ↑, the R min ↓, the local point where the bubble can create increases; When R ↑, the p v ↓, then T vs ↓,which result in heat flux(from L to V )↑, and the bubble grows, causing the R ↑. When R ↓, the p v ↑, then T vs ↑,which result in heat flux(from L to V ) ↓, and the bubble grows, causing the R ↓.when the T vs ↑, the heat transfer (from V to L) may occur, the bubble may collapse rapidly. Chapter 6 Condensation & boiling heat transfer 6-5 The calculation relationships for boiling heat transfer 1. Nucleate boiling Analysis process is extremely difficult. Exact mechanism of bubble formation and motion are not get fully understood. a considerable amount of experimental date is available on boiling heat flux as a function of the temperature difference ?T s =T w -T s for a variety of liquid and over a wide range of pressure level. Rohsennow relation: n l vll wl swpl g q C TTc Pr )( )( 3 1 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ρρ σ γμγ Chapter 6 Condensation & boiling heat transfer Chapter 6 Condensation & boiling heat transfer where： c pl specific heat of saturation liquid,J/(kg℃ )； γ latent heat of vaporization at t s ,J/kg； μ l viscosity of saturation liquid， kg/(ms)； ρ l 、 ρ v densities of liquid and vapor at saturation state， kg/m 3 , Pr l Prandtl number of saturation liquid； σ surface tension of liquid-vapor interface， N/m， g gravitation acceleration， m/s 2 ； C wl constant to be determined from experimental data depending on heating surface-fluid combination. 3. Stable film boiling Chapter 6 Condensation & boiling heat transfer 2. Peak heat flux in nucleate boiling This equation is found applicable for boiling of single-component liquid on the clean surface. for dirty or contaminated surface the exponent n of Pr l has been found to vary between 0.8 and 2.0， instead of 1.7， for water n=1.0。 The spread of experimental data is approximately ± 20% . 2 1 4 1 2 1 1 )( 24 ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? = l v v v vc g rpq ρ ρ ρ ρρσπ () () 4 1 3 62.0 ? ? ? ? ? ? ? ? = swv vvlv c TTd g h μ λρργρ Chapter 6 Condensation & boiling heat transfer σππ RppR lv 2)( 2 ≥? lv pp R ? ≥ σ2 () slv TdT dp ρρ γ 11 ? = s v sv lv TTT pp γρ = ? ? )( 2 svv s TT T R ? ≥ γρ σ vl tt ≥ ( ) sws ttt ?=? max () max max 2 sv s t T R ?γρ σ ＝ )( 2 svv s TT T R ? ≥ γρ σ Chapter 6 Condensation & boiling heat transfer Chapter 7 Radiation Heat Transfer ?77-1 Basic Concepts of Thermal Radiation thermal radiation--a physical process in which the body emits the radiation energy because of their own temperature. Other types of radiation include, for example, x-rays, gamma rays, cosmic rays, and electromagnetic. 无线电波 红外线 可 见 光 紫外线 X射线 γ射线 10 -8 10 -6 10 -4 10 -2 10 -1 10 0 10 1 10 2 10 4 10 6 10 8 图 7-1电磁辐射波谱分布图 Actual mechanism: electromagnetic waves according to properties of propagation; photon quantum according to properties of energy. propagation speed =speed of light :c=λν where, c--light speed, λ--wavelength, ν--frequency of radiation. 2 Incident radiation and Body Chapter 7 Radiation Heat Transfer Q Q r Q t Q a 图 7-2物体对辐射的吸收、反射和穿透 the average radiation properties over the entire wavelength: Q i = Q a +Q r +Q t , and we can obtain 1=α+ρ+τ, where α-- absorptivity, ρ-- reflectivity, τ--transmissivity. For spectral radiation: 1=α λ +ρ λ +τ λ , where α λ -spectral absorptivity, ρ λ - spectral reflectivity,τ λ -spectra transmissivity. when the radiation is incident on a semitransparent body, part of incident radiation is absorbed, part of incident radiation is reflected, and remainder radiation is transmitted through the body. Chapter 7 Radiation Heat Transfer For opaque surface α+ρ =1(as liquids and solids). For gases the radiation is a bulk process. Then the reflection is neglected, and we can obtained α+τ =1 . Some idealized bodies are: Black body(perfect absorber) α = 1; Perfect transparent body τ =1 Perfect reflected body ρ =1 (white body or specular body) specular reflection -- incident and reflected rays lie symmetrically with respect to the normal at the point of incident; Diffuse reflection-- intensity of reflected radiation is constant for all angles of reflection Real surface neither diffuse nor specular φ φ n n 图 7-4物体表面对热射线的反射特征 (a)镜反射 (b)漫反射 Chapter 7 Radiation Heat Transfer Black surface Isothermal surface Incident radiation Manual black cavity dA dA dA df dφ dQ dQ （ a）微元表面总辐射情况 （ b）微元表面单色辐射情况 （ c）微元表面方向辐射情况 图 7-5物体表面向半球空间的辐射示意图 r n n n 7-2 Black body radiation Black body is a perfect absorber. Natural state: hole, pupil Manual state: black cavity 1 Radiation energy flux and intensity Chapter 7 Radiation Heat Transfer 1.1 emissive power E=dQ/dA It is used to characterize the amount of energy per unit time per unit radiative area emitted over an entire hemisphere space and the entire wavelength (spectrum from λ =0 to λ =∞). 1.2 spectral emissive power E λ =dQ/(dAd λ) It is used to characterize the amount of energy per unit time per unit radiative area emitted over an entire hemisphere space and per unit wavelength about the λ. The E is related to the E λ by )/( 0 λλ λλ ddEEordEE == ∫ ∞ Chapter 7 Radiation Heat Transfer 1.3 Directional emissive power E ? =dQ/(dAdω)=dE/dω It is used to characterize the amount of energy per unit time per unit radiative area emitted over a solid angle about the direction ? and the entire wavelength (spectrum from λ =0 to λ =∞). solid angle is defined as ω=f/r 2 , then the differential solid angle is dω=df/r 2 , for the spherical coordinate df=rsin?dθrd? . Substituting above relation into defined relation of solid angle have dω=sin?dθd? .then dθ dφ df r φ θ n ∫∫∫ == π ? ? ? π θ??? 2 00 2 sin ddEdEE The concept of solid angle Chapter 7 Radiation Heat Transfer 1.4 spectral Directional emissive power E λ? =dQ/(dAdω d λ) It is used to characterize the amount of energy per unit time per unit radiative area emitted over a solid angle about the direction ? and per unit wavelength about the λ.. The E ? is related to the E λ? by )/( 0 λλ ?λ?λ?? ddEEordEE == ∫ ∞ 1.5 Directional radiation intensity I ? =dQ/(dAcos?dω)=dE ? / cos? It is used to characterize the amount of energy per unit time per unit radiative area normal to the direction ? emitted over a solid angle about the direction ? and the entire wavelength (spectrum from λ =0 to λ =∞). Chapter 7 Radiation Heat Transfer 1.5 Spectral directional radiation intensity I λ? =dQ/(dAcos?dωd λ)=dI ? /d λ=dE λ? /cos? It is used to characterize the amount of energy per unit time per unit radiative area normal to the direction ? emitted over a solid angle about the direction ? and per unit wavelength about the λ.. dAcosφ dA φ n dω ∫∫∫ ∞ = 0 2 00 2 sincos π λ? π λθ??? dddIE The E is related to the E λ by Chapter 7 Radiation Heat Transfer 2 Basic laws of black body radiation The basic radiation laws for black body are Prank distribution law, Wein displacement law , Stefen-Boltzmann law and Lanbert law. Chapter 7 Radiation Heat Transfer )/( 0 λλ ?λ?λ?? ddEEordEE == ∫ ∞ E bλ 0 λ 2 λ 1 λ 图7-10 一定波长范围内的黑体辐射力 E bλ 黑体 实际物体 E bλ E λ λ 0 Chapter 7 Radiation Heat Transfer Chapter 7 Radiation Heat Transfer