Presentation on theme: "Chapter 2 Introduction to Heat Transfer"— Presentation transcript:
1Chapter 2 Introduction to Heat Transfer 2.1 Basic Concepts2.1.1 Conduction, convection, and radiationHeat transfer is the transfer of heat due to a temperature difference with differentmechanisms: conduction, convection, and radiation. Conduction refers to heattransfer that occurs across a stationary solid or fluid in which a temperature gradientexists. Convection refers to the heat transfer that occurs across a moving fluid inwhich a temperature gradient exists. Radiation refers to the heat transfer betweentwo surfaces at different temperatures separated by a medium transparent to theelectromagnetic waves emitted by the surfaces.
22.1.2 Fourier’s law of conduction One-dimensionalConsider the conduction of heat through a slab of thickness L, as shown inFig The lower and upper surfaces are kept at a constant temperature T1 andT2, respectively. A steady-state temperature profile T(y) is established in the slab.Consider two surface in slab separated with a infinitesimal distance dy, as shownin Fig due to temperature gradient generated in the slab, heat flow from thesurface y to the surface y+dy. A heat flux is defined as the amount of heat transferredper unit area per unit time, and can be expressed as[2.1-1]where k is the thermal conductivity of the medium. This equation is Fourier’s lawof conduction for one-dimensional heat conduction in the y-direction. The mks units ofthe heat flux and the thermal conductivity are W/m2 and Wm-1K-1, respectively.
3The thermal conductivities of some common materials are given in Fig and TableThree-dimensionalFor heat transfer in a three-dimensional medium, the Fourier’s law can beexpressed for each of the three coordinate directions[2.1-2][2.1-3][2.1-4]And can be expressed in a three-dimensional form of Fourier’s law of conduction.The Thermal DiffusivityThe thermal diffusivity, α , is defined asWhere r and Cv are the density and specific heat of the material, respectively.
42.1.3 Thermal boundary layer Consider a fluid of uniform temperature T∞ approaching a flat plate of constanttemperature Ts in the direction parallel to the plate. At the solid/liquid interface thefluid temperature is Ts since the local fluid particles achieve thermal equilibrium atthe interface. The fluid temperature T in the region near the plate is affected by theplate, varying from Ts at the surface to T∞ in the main stream. This region is calledthe thermal boundary layer.
5oppositive sides approach the centerline. This occurs at Definition of thermal thickness: The thickness of thermal boundary layer δTis taken as the distance from the plate surface at which the dimensionlesstemperature (T-TS)/(T∞-TS) reaches In practice it is usually specifiedthat T=T∞ and at y=δT.The effect of conduction is significant only in the boundary layer. Beyond itthe temperature is uniform and the effect of conduction is no longer significant.A fluid of uniform temperature T entering a circular tube of inner diameter D anduniform wall temperature TS, as illustrated in Fig A thermal boundary layerbegins to develop at the entrance , gradually expanding until the layers fromoppositive sides approach the centerline. This occurs at[2.1-7]whereInertial forceViscous diffusivity[2.1-8]Viscous forcethermal diffusivity(Reynolds number) (Prandtl number)
6Define average temperature [2.1-10]The thermally fully developed temperature profile in a tube is one with adimensionless temperature (Ts-T)/(Ts-Tav) independent of the axial position,that is[2.1-12]2.1.4 Heat transfer coefficientConsider the thermal boundary layer. At the solid/liquid interface heat transferoccurs only by conduction since there is no fluid motion. Therefore, the heat fluxacross the solid/liquid interface is[2.1-13]This equation cannot be used to calculate the heat flux when the temperaturegradient is an unknown. A convenient way to avoid this program is to introducea heat transfer coefficient, defined as follows:[2.1-14]
7From this and Eq.[2.1-12], we have The absolute values are used to keep h always positive. From Eq. [2.1-14][2.1-15]The equation is called Newton’s law of cooling. For fluid flow through a tube ofan inner radius R and wall temperature TS, a similar equation can be used:[2.1-16]Where Tav is the average fluid temperature over the cross-sectional area pR2.Consider the thermally fully developed region shown in Fig In the case of aconstant heat flux, the heat transfer coefficient h is constant in the thermally fullydeveloped region. From Eq. [2.1-16] we see that (TS-Tav) is also constant.From this and Eq.[2.1-12], we have(constant )[2.1-17]Since TS and Tav are independent of r, is also independent of r, Let usconsider the case of a constant wall temperature TS. Eq. [2.1-12] can be expandedand solved for to give
8(constant TS)[2.1-18]Since T is dependent on r, is also dependent on r.2.2 Overall energy-balance equation2.2.1 DerivationConsider a control volume Ω bounded by control surface A through which amoving fluid is flowing. As defined in previous chapter, the control surface iscomposed by Ain , Aout, and Awall. Consider an infinitesimal area dA in vector formis ndA, the inward and outward heat transfer rate through area dA is -q． ndAand q． ndA, respectively.The energy conservation law (first law of thermodynamics) written for an opensystem under unsteady-state condition is[2.2-3]
9Term 1: Rate of energy accumulation The thermal, kinetic, and potential energy per unit mass of the fluid areCvT, v2/2, and ψ, respectively, where Cv, T, v are the specific heat, temperature,and velocity of the fluid. The total energy per unit mass of the fluid[2.2-4]The total energy in the differential volume element dΩis dEt=ρetdΩ. dm =ρdΩThis can be integrated over Ω to obtain the total energy in the control volume EtEt (overall) = (integral)And the rate of energy change in Ω isTerms 2 & 3: Rate of energy in by mass inflowThe inward energy flow rate is energy per unit mass, et, times inward mass flowrate through dA and can be expressed asSince v=0 at the wall, above term can be expressed as
10Term 4: Rate of heat transfer The heat transfer by conduction dA isTerm 5: The rate of work done by the fluid in the C.V. on the surroundings, includes:(1) The rate of shaft work doneThe rate of shaft work done by the fluid in the C.V. on the surroundings,that is, through a turbine or compressor, is Ws(2) The rate of pressure work doneTo leave the C.V. through dA, the fluid has to work against the pressure ofthe surrounding fluid. Since the pressure force is pndA, the rate of pressurework required is dWp= pv‧ndA. Therefore, the rate of pressure work the fluidhas to do to go through the C.V. is(3) The rate of viscous work doneTo overcome the viscous force t‧ndA, the rate of viscous work required isdWv= (t‧n)‧vdA. The rate of viscous work the fluid has to do is
11Term 6: Heat generation rate The heat generation rate per unit volume, such as that due to Joule heating,phase transformation, or chemical reaction. The rate of heat generation in thedifferential volume element dΩ is dS=s dΩ. The heat generation in the controlvolume is S,Substituting the integral form of terms(1) through (6) into Eq.[2.2.3]=[2.2-5]In most problem, including those in materials processing, the kinetic andpotential energies are neglegible as compared to the thermal energy. Furthermore,the pressure, viscous and shaft work are usually negligible or even absent. As such,Eq.[2.2-5] reduces to[2.2-6]
12According to [2.2-3], Eq. [2.2-6] can be written as [2.2-7]Where ET is the thermal energy in the control volume, substituting Eq. [2.1-11](definition of Tav) into this equation and assuming constant Cv, we obtain[2.2-8]2.2.2 Bernoulli’s EquationConsider the steady-state isothermal flow of an inviscid incompressible fluidwithout heat generation, heat conduction, shaft work, and viscous work. SubstitutingEq. [2.2-4] into [2.2-5] and assuming uniform properties over the cross-sectional areaA, we have
13[2.2-9]Since T1=T2 and (rvA)1=(rvA)2, Eq.[2.2-9] reduces toIf the z direction is taken vertically upward, f =gz, where g is the gravitationalacceleration. As such, Eq/.[2.2-10], on multiplying by r, becomesor simplyWhich is the Bernoulli equation.
14Example 2.2.2 Conduction through cylindrical composite wall
15Example 2.2.3 Heat transfer in fluid flow through a pipe Based on the C.V.selected in the figureBased on the definition of overallHeat transfer coefficient
16Example 2.2.4 Counterflow heat exchanger Given: Hot stream Th1, Th2, mhcold stream inlet(Tc2), outlet(Tc1),and mass flow rate mcOverall heat transfer coefficient U, turbulentFind Qe (steady state heat exchange rate Qein terms of U, Th and Tc).For hot streamFor cold streamFromviewofoverallThereforeand[2.2-47][2.2-48]
17From view of C.V.Because we want to express Qe in terms of U,therefore, considering the C.V. in the inner pipe.Similarly, for the C.V. in the outer pipeandSubtracting Eq.[2.2-52] from Eq. [2.2-50], we have
18Integrating from z=0 to z=, and substituting [2. 2-47] and [2 Integrating from z=0 to z=, and substituting [2.2-47] and [2.2-48] to [2.2-53]
19Example 2.2.5 Heat transfer in laminar flow over a flat plate Given: Steady state, constant physical properties,no heat generationFind: dT and heat transfer coefficientApproach:Construct C.V.Find the mass flow rate into the C.V. from surface 4Consider the energy balanceSubstituting m4 into Eq. [2.2-57], andaccording to Fourier’s law of conduction
20we have5. Assumeand6. AssumePlease see the derivation in other pages
217. By the definition of heat transfer coefficient From Eq.[1.4-62]From Eqs. [2.2-65], [2.2-68], and [2.2-69]We have[2.2-65][2.2-69]