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c06 supl.qxd1/24/065:35 PMPage W-21W-216S.1Derivation of the Convection TransferEquationsIn Chapter 2 we considered a stationary substance in which heat is transferred byconduction and developed means for determining the temperature distributionwithin the substance. We did so by applying conservation of energy to a differentialcontrol volume (Figure 2.11) and deriving a differential equation that was termedthe heat equation. For a prescribed geometry and boundary conditions, the equationmay be solved to determine the corresponding temperature distribution.If the substance is not stationary, conditions become more complex. For example, if conservation of energy is applied to a differential control volume in a movingfluid, the effects of fluid motion (advection) on energy transfer across the surfacesof the control volume must be considered, along with those of conduction. Theresulting differential equation, which provides the basis for predicting the temperature distribution, now requires knowledge of the velocity field. This field must, inturn, be determined by solving additional differential equations derived by applyingconservation of mass and Newton’s second law of motion to a differential controlvolume.In this supplemental material we consider conditions involving flow of a viscous fluid in which there is concurrent heat and mass transfer. Our objective is todevelop differential equations that may be used to predict velocity, temperature, andspecies concentration fields within the fluid, and we do so by applying Newton’ssecond law of motion and conservation of mass, energy, and species to a differentialcontrol volume. To simplify this development, we restrict our attention to steady,two-dimensional flow in the x and y directions of a Cartesian coordinate system. Aunit depth may therefore be assigned to the z direction, thereby providing a differential control volume of extent (dx dy 1).6S.1.1Conservation of MassOne conservation law that is pertinent to the flow of a viscous fluid is that mattermay neither be created nor destroyed. Stated in the context of the differential control volume of Figure 6S.1, this law requires that, for steady flow, the net rate atwhich mass enters the control volume (inflow outflow) must equal zero. Massenters and leaves the control volume exclusively through gross fluid motion. Transport due to such motion is often referred to as advection. If one corner of the controlvolume is located at (x, y), the rate at which mass enters the control volume throughthe surface perpendicular to x may be expressed as ( u) dy, where is the total massdensity ( A B) and u is the x component of the mass average velocity. Thecontrol volume is of unit depth in the z direction. Since and u may vary with x, the

c06 supl.qxd1/24/06W-225:35 PMPage W-226S.1Derivation of the Convection Transfer Equations ρ v ] ( ρ v) dy]ydyyxρ u ] ( ρ u) dx]yρuz(x, y)dxρvFIGURE 6S.1Differential control volume(dx dy 1) for massconservation in two-dimensionalflow of a viscous fluid.rate at which mass leaves the surface at x dx may be expressed by a Taylor seriesexpansion of the formdx dy ( u) ( u) xUsing a similar result for the y direction, the conservation of mass requirementbecomes ( u) dy ( v) dx u ( u) ( v)dx dy v dy dx 0 x yCanceling terms and dividing by dx dy, we obtain ( u) ( v) 0 x y(6S.1)Equation 6S.1, the continuity equation, is a general expression of the overallmass conservation requirement, and it must be satisfied at every point in the fluid.The equation applies for a single species fluid, as well as for mixtures in whichspecies diffusion and chemical reactions may be occurring. If the fluid is incompressible, the density is a constant, and the continuity equation reduces to u v 0 x y6S.1.2(6S.2)Newton’s Second Law of MotionThe second fundamental law that is pertinent to the flow of a viscous fluid is Newton’ssecond law of motion. For a differential control volume in the fluid, this requirementstates that the sum of all forces acting on the control volume must equal the net rate atwhich momentum leaves the control volume (outflow inflow).Two kinds of forces may act on the fluid: body forces, which are proportionalto the volume, and surface forces, which are proportional to area. Gravitational, centrifugal, magnetic, and/or electric fields may contribute to the total body force, and wedesignate the x and y components of this force per unit volume of fluid as X and Y,respectively. The surface forces Fs are due to the fluid static pressure as well as toviscous stresses. At any point in the fluid, the viscous stress (a force per unit area)

c06 supl.qxd1/24/065:35 PMPage W-236S.1 Derivation of the Convection Transfer EquationsW-23σyy ] (σ yy)dy]yτyx ] (τ yx)dy]yτxy ] (τ xy)dx]xyxzσxxσxx ] (σxx)dx]xτxy(x, y)dydxτyxσyyFIGURE 6S.2Normal and shear viscousstresses for a differential controlvolume (dx dy 1) in twodimensional flow of a viscousfluid.may be resolved into two perpendicular components, which include a normal stress ii and a shear stress ij (Figure 6S.2).A double subscript notation is used to specify the stress components. The firstsubscript indicates the surface orientation by providing the direction of its outwardnormal, and the second subscript indicates the direction of the force component.Accordingly, for the x surface of Figure 6S.2, the normal stress xx corresponds to aforce component normal to the surface, and the shear stress xy corresponds to aforce in the y direction along the surface. All the stress components shown are positive in the sense that both the surface normal and the force component are in thesame direction. That is, they are both in either the positive coordinate direction orthe negative coordinate direction. By this convention the normal viscous stressesare tensile stresses. In contrast the static pressure originates from an external forceacting on the fluid in the control volume and is therefore a compressive stress.Several features of the viscous stress should be noted. The associated force isbetween adjoining fluid elements and is a natural consequence of the fluid motionand viscosity. The surface forces of Figure 6S.2 are therefore presumed to act on thefluid within the control volume and are attributed to its interaction with the surrounding fluid. These stresses would vanish if the fluid velocity, or the velocity gradient, went to zero. In this respect the normal viscous stresses ( xx and yy) must notbe confused with the static pressure, which does not vanish for zero velocity.Each of the stresses may change continuously in each of the coordinate directions. Using a Taylor series expansion for the stresses, the net surface force for eachof the two directions may be expressed as yxFs,x x p x y dx dyFs,y x y p y dx dyxx xy yy(6S.3)(6S.4)To use Newton’s second law, the fluid momentum fluxes for the control volume must also be evaluated. If we focus on the x-direction, the relevant fluxes are asshown in Figure 6S.3. A contribution to the total x-momentum flux is made by themass flow in each of the two directions. For example, the mass flux through the xsurface (in the y-z plane) is ( u), the corresponding x-momentum flux is ( u)u. Similarly, the x-momentum flux due to mass flow through the y surface (in the x-z

c06 supl.qxd1/24/06W-245:35 PMPage W-246S.1 Derivation of the Convection Transfer Equations( ρv)u ] [(ρ v)u]dy]yy, vdyx, uz(ρ u)u ] [(ρ u)u]dx]x(ρ u)ux, y( ρv)udxFIGURE 6S.3Momentum fluxes for a differential control volume (dx dy 1)in two-dimensional flow of a viscous fluid.plane) is ( v)u. These fluxes may change in each of the coordinate directions, andthe net rate at which x momentum leaves the control volume is [( u)u] [( v)u]dx (dy) dy (dx) x yEquating the rate of change in the x momentum of the fluid to the sum of theforces in the x direction, we then obtain [( u)u] [( v)u] xx p yx X x y x x y(6S.5)This expression may be put in a more convenient form by expanding the derivativeson the left-hand side and substituting from the continuity equation, Equation 6S.1,giving u yx u u v ( xx p) X x y x y (6S.6)A similar expression may be obtained for the y direction and is of the form u xy v v v ( yy p) Y x y x y (6S.7)We should not lose sight of the physics represented by Equations 6S.6 and6S.7. The two terms on the left-hand side of each equation represent the net rate ofmomentum flow from the control volume. The terms on the right-hand side accountfor the net viscous and pressure forces, as well as the body force. These equationsmust be satisfied at each point in the fluid, and with Equation 6S.1 they may besolved for the velocity field.Before a solution to the foregoing equations can be obtained, it is necessary torelate the viscous stresses to other flow variables. These stresses are associated withthe deformation of the fluid and are a function of the fluid viscosity and velocity gradients. From Figure 6S.4 it is evident that a normal stress must produce a lineardeformation of the fluid, whereas a shear stress produces an angular deformation.Moreover, the magnitude of a stress is proportional to the rate at which the deformation occurs. The deformation rate is, in turn, related to the fluid viscosity and to thevelocity gradients in the flow. For a Newtonian fluid1 the stresses are proportional to1A Newtonian fluid is one for which the shear stress is linearly proportional to the rate of angulardeformation. All fluids of interest in the text are Newtonian.

c06 supl.qxd1/24/065:35 PMPage W-256S.1 W-25Derivation of the Convection Transfer Equationsτ yxτ xyσxxτ xyσxxτ yx(a)(b)FIGURE 6S.4Deformations of a fluid element due to viscous stresses.(a) Linear deformation due to a normal stress. (b) Angulardeformation due to shear stresses.the velocity gradients, where the proportionality constant is the fluid viscosity.Because of its complexity, however, development of the specific relations is left tothe literature [1], and we limit ourselves to a presentation of the results. In particular,it has been shown that xx 2 u 2 u v x 3 x y (6S.8) yy 2 v 2 u v y 3 x y (6S.9) xy yx u y v x (6S.10)Substituting Equations 6S.8 through 6S.10 into Equations 6S.6 and 6S.7, thex- and y-momentum equations become u u p u u u u v v 2 2 x y x x x 3 x y u v y y x X(6S.11) p v v v u v v 2 2 x y y y y 3 x y u v x y x Y (6S.12)Equations 6S.1, 6S.11, and 6S.12 provide a complete representation of conditions ina two-dimensional viscous flow, and the corresponding velocity field may be determined by solving the equations. Once the velocity field is known, it is a simple matter to obtain the wall shear stress s from Equation 6.2.Equations 6S.11 and 6S.12 may be simplified for an incompressible fluid ofconstant viscosity. Rearranging the right-hand side of each expression and substituting from Equation 6S.2, the x- and y-momentum equations become u p u u 2u 2u v X x y x x2 y2(6S.13)

c06 supl.qxd1/24/06W-265:35 PMPage W-266S.1 Derivation of the Convection Transfer Equations u6S.1.3 p v v 2v 2v v Y x y y x2 y2(6S.14)Conservation of EnergyTo apply the energy conservation requirement (Equation 1.11c) to a differentialcontrol volume in a viscous fluid with heat transfer (Figure 6S.5), it is necessary tofirst delineate the relevant physical processes. If potential energy effects are treatedas work done by the body forces, the energy per unit mass of the fluid includes thethermal internal energy e and the kinetic energy V2/2, where V 2 u2 v2. Accordingly, thermal and kinetic energy are advected with the bulk fluid motion across thecontrol surfaces, and for the x-direction, the net rate at which this energy enters thecontrol volume is V u e dx dy x2V u e dx dy x2Ėadv,x Ėadv,x dx u e V2V2dy u e 2222(6S.15)Energy is also transferred across the control surface by molecular processes. Theremay be two contributions: that due to conduction and energy transfer due to the diffusion of species A and B. However, it is only in chemically reacting flows thatspecies diffusion strongly influences thermal conditions. Hence the effect isneglected in this development. For the conduction process, the net transfer ofenergy into the control volume is T x dy k T x x k T x dx dy T k dx dy x xĖcond,x Ėcond,x dx k(6S.16)Energy may also be transferred to and from the fluid in the control volume bywork interactions involving the body and surface forces. The net rate at which workis done on the fluid by forces in the x-direction may be expressed as E cond, y dyE adv, y dy Wydy xz E cond, xE cond, x dx Eg E adv, x E adv, x dxx, ydx E cond, y E adv, yFIGURE 6S.5Differential control volume(dx dy 1) for energyconservation in two-dimensionalflow of a viscous fluid with heattransfer.

c06 supl.qxd1/24/065:35 PMPage W-276S.1 W-27Derivation of the Convection Transfer Equations [( xx p) u] dx dy ( yx u) dx dy(6S.17) x yThe first term on the right-hand side of Equation 6S.17 represents the work done bythe body force, and the remaining terms account for the net work done by the pressure and viscous forces.Using Equations 6S.15 through 6S.17, as well as analogous equations for they-direction, the energy conservation requirement (Equation 1.11c) may be expressed asẆnet,x (Xu) dx dy y v e V2 T T k k (Xu Yv) (pu) (pv) x x y y x y V2 u e x2 ( u xy v) ( yx u yy v) q̇ 0 x xx y2(6S.18)where q is the rate at which thermal energy is generated per unit volume. Thisexpression provides a general form of the energy conservation requirement for flowof a viscous fluid with heat transfer.Because Equation 6S.18 represents conservation of kinetic