c06_supl.qxd 1/24/06 5:35 PM Page W-21 W-21 6S.1 Derivation of the Convection Transfer Equations In Chapter 2 we considered a stationary substance in which heat is transferred by conduction and developed means for determining the temperature distribution within the substance. We did so by applying conservation of energy to a differential control volume (Figure 2.11) and deriving a differential equation that was termed the heat equation. For a prescribed geometry and boundary conditions, the equation may 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 moving fluid, the effects of fluid motion (advection) on energy transfer across the surfaces of the control volume must be considered, along with those of conduction. The resulting differential equation, which provides the basis for predicting the temperature distribution, now requires knowledge of the velocity field. This field must, in turn, be determined by solving additional differential equations derived by applying conservation of mass and Newton’s second law of motion to a differential control volume. 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 to develop differential equations that may be used to predict velocity, temperature, and species concentration fields within the fluid, and we do so by applying Newton’s second law of motion and conservation of mass, energy, and species to a differential control 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. A unit depth may therefore be assigned to the z direction, thereby providing a differential control volume of extent (dx ⅐ dy ⅐ 1). 6S.1.1 Conservation of Mass One conservation law that is pertinent to the flow of a viscous fluid is that matter may 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 at which mass enters the control volume (inflow Ϫ outflow) must equal zero. Mass enters 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 control volume is located at (x, y), the rate at which mass enters the control volume through the surface perpendicular to x may be expressed as (u) dy, where is the total mass density ( ϭ A ϩ B) and u is the x component of the mass average velocity. The control volume is of unit depth in the z direction. Since and u may vary with x, the which are proportional to the volume.1. respectively. y) dy ρu ρ u + ] ( ρ u) dx ]y dx ρv FIGURE 6S. and the continuity equation reduces to Ѩu Ѩv ϩ ϭ0 Ѩx Ѩy (6S. as well as for mixtures in which species diffusion and chemical reactions may be occurring.1. Two kinds of forces may act on the fluid: body forces. For a differential control volume in the fluid. If the fluid is incompressible. The equation applies for a single species fluid. The surface forces Fs are due to the fluid static pressure as well as to viscous stresses.qxd 1/24/06 5:35 PM Page W-22 W-22 6S. and we designate the x and y components of this force per unit volume of fluid as X and Y.1 Derivation of the Convection Transfer Equations ρ v + ] ( ρ v) dy ]y y x z (x. centrifugal. the conservation of mass requirement becomes (u) dy ϩ (v) dx Ϫ u ϩ ΄ Ѩ(u) Ѩ(v) dx dy Ϫ v ϩ dy dx ϭ 0 Ѩx Ѩy ΅ ΄ ΅ Canceling terms and dividing by dx dy. this requirement states that the sum of all forces acting on the control volume must equal the net rate at which momentum leaves the control volume (outflow Ϫ inflow). the viscous stress (a force per unit area) . and surface forces. is a general expression of the overall mass conservation requirement.c06_supl. we obtain Ѩ(u) Ѩ(v) ϩ ϭ0 Ѩx Ѩy (6S.1) Equation 6S. At any point in the fluid. which are proportional to area.2 Newton’s Second Law of Motion The second fundamental law that is pertinent to the flow of a viscous fluid is Newton’s second law of motion. the continuity equation. rate at which mass leaves the surface at x + dx may be expressed by a Taylor series expansion of the form ΄(u) ϩ Ѩ(Ѩxu) dx΅ dy Using a similar result for the y direction. and it must be satisfied at every point in the fluid. Gravitational.1 Differential control volume (dx ⅐ dy ⅐ 1) for mass conservation in two-dimensional flow of a viscous fluid.2) 6S. magnetic. and/or electric fields may contribute to the total body force. the density is a constant. The associated force is between adjoining fluid elements and is a natural consequence of the fluid motion and viscosity. The first subscript indicates the surface orientation by providing the direction of its outward normal. and the second subscript indicates the direction of the force component. the mass flux through the x surface (in the y-z plane) is (u). the relevant fluxes are as shown in Figure 6S. Using a Taylor series expansion for the stresses.4) To use Newton’s second law. may be resolved into two perpendicular components. y) σxx + ] (σxx)dx ]x dy dx τyx σyy FIGURE 6S. If we focus on the x-direction. which does not vanish for zero velocity. A contribution to the total x-momentum flux is made by the mass flow in each of the two directions.3. Several features of the viscous stress should be noted. x ϭ Fs. for the x surface of Figure 6S. and the shear stress xy corresponds to a force in the y direction along the surface.1 Derivation of the Convection Transfer Equations W-23 σyy + ] (σ yy)dy ]y τyx + ] (τ yx)dy ]y τxy + ] (τ xy)dx ]x x z y σxx τxy (x. which include a normal stress ii and a shear stress ij (Figure 6S. These stresses would vanish if the fluid velocity. the fluid momentum fluxes for the control volume must also be evaluated.2 are therefore presumed to act on the fluid within the control volume and are attributed to its interaction with the surrounding fluid. went to zero. That is. The surface forces of Figure 6S.2 Normal and shear viscous stresses for a differential control volume (dx ⅐ dy ⅐ 1) in twodimensional flow of a viscous fluid. In contrast the static pressure originates from an external force acting on the fluid in the control volume and is therefore a compressive stress.qxd 1/24/06 5:35 PM Page W-23 6S. the x-momentum flux due to mass flow through the y surface (in the x-z .2. In this respect the normal viscous stresses (xx and yy) must not be confused with the static pressure. For example.c06_supl. A double subscript notation is used to specify the stress components. or the velocity gradient. y ϭ Ѩp Ϫ ϩ dx dy Ѩ Ѩx Ѩx Ѩy xx Ѩyx (6S. All the stress components shown are positive in the sense that both the surface normal and the force component are in the same direction. the normal stress xx corresponds to a force component normal to the surface. Similarly. the net surface force for each of the two directions may be expressed as Fs.2). By this convention the normal viscous stresses are tensile stresses. Each of the stresses may change continuously in each of the coordinate directions. the corresponding x-momentum flux is (u)u. Accordingly.3) p dx dy Ѩx ϩ Ѩy Ϫ Ѩ Ѩy Ѩxy Ѩyy (6S. they are both in either the positive coordinate direction or the negative coordinate direction. These fluxes may change in each of the coordinate directions. These stresses are associated with the deformation of the fluid and are a function of the fluid viscosity and velocity gradients. Before a solution to the foregoing equations can be obtained.4 it is evident that a normal stress must produce a linear deformation of the fluid. whereas a shear stress produces an angular deformation.1. From Figure 6S. v x. plane) is (v)u. For a Newtonian fluid1 the stresses are proportional to 1 A Newtonian fluid is one for which the shear stress is linearly proportional to the rate of angular deformation. These equations must be satisfied at each point in the fluid.1 Derivation of the Convection Transfer Equations y. and the net rate at which x momentum leaves the control volume is Ѩ[(u)u] Ѩ[(v)u] dx (dy) ϩ dy (dx) Ѩx Ѩy Equating the rate of change in the x momentum of the fluid to the sum of the forces in the x direction. it is necessary to relate the viscous stresses to other flow variables. u z (ρ u)u ( ρv)u + ] [(ρ v)u]dy ]y dy (ρ u)u + ] [(ρ u)u]dx ]x x. Equation 6S. and with Equation 6S.qxd 1/24/06 5:35 PM Page W-24 W-24 6S.c06_supl. Moreover.5) This expression may be put in a more convenient form by expanding the derivatives on the left-hand side and substituting from the continuity equation.6) A similar expression may be obtained for the y direction and is of the form u Ѩxy Ѩv Ѩv Ѩ ϩv ϭ ϩ (yy Ϫ p) ϩ Y Ѩx Ѩy Ѩx Ѩy (6S.6 and 6S. y ( ρv)u dx FIGURE 6S. . in turn. the magnitude of a stress is proportional to the rate at which the deformation occurs.7) We should not lose sight of the physics represented by Equations 6S. as well as the body force. we then obtain Ѩ[(u)u] Ѩ[(v)u] Ѩxx Ѩp Ѩyx ϩ ϭ Ϫ ϩ ϩX Ѩx Ѩy Ѩx Ѩx Ѩy (6S. All fluids of interest in the text are Newtonian. related to the fluid viscosity and to the velocity gradients in the flow. giving u Ѩyx Ѩu Ѩu Ѩ ϩv ϭ (xx Ϫ p) ϩ ϩX Ѩx Ѩy Ѩx Ѩy (6S.1 they may be solved for the velocity field.7.3 Momentum fluxes for a differential control volume (dx ⅐ dy ⅐ 1) in two-dimensional flow of a viscous fluid. The deformation rate is. The two terms on the left-hand side of each equation represent the net rate of momentum flow from the control volume. The terms on the right-hand side account for the net viscous and pressure forces. In particular. and the corresponding velocity field may be determined by solving the equations.7.10 into Equations 6S.12) Equations 6S. the x.12 provide a complete representation of conditions in a two-dimensional viscous flow.qxd 1/24/06 5:35 PM Page W-25 6S. Equations 6S. it is a simple matter to obtain the wall shear stress s from Equation 6. however.1. development of the specific relations is left to the literature [1]. the velocity gradients. Once the velocity field is known. Because of its complexity.c06_supl. (b) Angular deformation due to shear stresses. the x.11 and 6S.and y-momentum equations become u Ѩp Ѩu Ѩu Ѩu Ѩu Ѩv Ѩ ϩv ϭϪ ϩ 2 Ϫ2 ϩ Ѩx Ѩy Ѩx Ѩx Ѩx 3 Ѩx Ѩy ϩ Ѩu Ѩv Ѩ ϩ Ѩy Ѩy Ѩx ΄ ΄ Ά΄ Ά΄ ΅ ϩ X ΅· (6S.10) Substituting Equations 6S. (a) Linear deformation due to a normal stress.9) xy ϭ yx ϭ (6S.8) (6S. and 6S. where the proportionality constant is the fluid viscosity. it has been shown that xx ϭ 2 yy ϭ 2 Ѩu 2 Ѩu Ѩv Ϫ ϩ Ѩx 3 Ѩx Ѩy Ѩv 2 Ѩu Ѩv Ϫ ϩ Ѩy 3 Ѩx Ѩy u Ѩv ϩ Ѩ Ѩy Ѩx (6S. Rearranging the right-hand side of each expression and substituting from Equation 6S. 6S.6 and 6S.4 Deformations of a fluid element due to viscous stresses.1 Derivation of the Convection Transfer Equations W-25 τ yx τ xy σxx σxx τ yx (a) (b) τ xy FIGURE 6S.13) .and y-momentum equations become u Ѩp Ѩu Ѩu Ѩ2u Ѩ2u ϩv ϭϪ ϩ ϩ ϩX Ѩx Ѩy Ѩx Ѩx2 Ѩy2 (6S.8 through 6S.11.2.11) u Ѩp Ѩv Ѩv Ѩv Ѩu Ѩv Ѩ ϩv ϭϪ ϩ 2 Ϫ2 ϩ Ѩx Ѩy Ѩy Ѩy Ѩy 3 Ѩx Ѩy ϩ Ѩu Ѩv Ѩ ϩ Ѩx Ѩy Ѩx ΅ ϩ Y ΅· (6S.12 may be simplified for an incompressible fluid of constant viscosity.2. and we limit ourselves to a presentation of the results. x + dx E adv. Hence the effect is neglected in this development.5). There may be two contributions: that due to conduction and energy transfer due to the diffusion of species A and B.5 Differential control volume (dx ⅐ dy ⅐ 1) for energy conservation in two-dimensional flow of a viscous fluid with heat transfer. x E adv. x Ϫ Econd.c06_supl. x Ϫ Eadv. y + dy W y x z E cond. If potential energy effects are treated as work done by the body forces. it is only in chemically reacting flows that species diffusion strongly influences thermal conditions.14) 6S. y + dy • E cond. y • • • • • dy Eg • E cond.16) ΄ Ѩ ѨT ϭ k dx dy Ѩx Ѩx ΅ Energy may also be transferred to and from the fluid in the control volume by work interactions involving the body and surface forces. y E cond. The net rate at which work is done on the fluid by forces in the x-direction may be expressed as E adv.3 Conservation of Energy To apply the energy conservation requirement (Equation 1. where V 2 ϵ u2 ϩ v2. y FIGURE 6S. it is necessary to first delineate the relevant physical processes. the net rate at which this energy enters the control volume is V ˙ ˙ E adv. thermal and kinetic energy are advected with the bulk fluid motion across the control surfaces. . x ϩ dx ϵ u e ϩ 2 dy Ϫ Άu e ϩ V2 V Ѩ ϩ ΄u e ϩ ΅dx· dy Ѩx 2 V Ѩ ϭ Ϫ ΄u e ϩ ΅ dx dy Ѩx 2 2 2 2 2 (6S. and for the x-direction. x + dx dx • • • E adv. x x. Accordingly.1 Derivation of the Convection Transfer Equations u Ѩp Ѩv Ѩv Ѩ2v Ѩ2v ϩv ϭϪ ϩ ϩ ϩY Ѩx Ѩy Ѩy Ѩx2 Ѩy2 (6S. xϩdx ϭ Ϫ k Ѩx Ѩx Ѩx Ѩx (6S.11c) to a differential control volume in a viscous fluid with heat transfer (Figure 6S. For the conduction process. the net transfer of energy into the control volume is ѨT ѨT Ѩ ѨT ˙ ˙ dy Ϫ Ϫ k Ϫ k dx dy E cond.1. the energy per unit mass of the fluid includes the thermal internal energy e and the kinetic energy V2/2.qxd 1/24/06 5:35 PM Page W-26 W-26 6S.15) Energy is also transferred across the control surface by molecular processes. However. which is termed the thermal energy equation. the viscous dissipation.17. x ϭ (Xu) dx dy ϩ Ѩx Ѩy The first term on the right-hand side of Equation 6S. and subtracting the results from Equation 6S.17) net. and ⌽. the thermal energy equation may then be expressed as cp u where ⌽ ϭ Ѩv u Ѩv Ѩu ϩ ϩ 2΄ ϩ ΅· ΆѨ Ѩy Ѩx Ѩx Ѩy 2 2 2 ѨT ѨT Ѩ ѨT Ѩ ѨT ˙ ϩv ϭ k ϩ k ϩ ⌽ ϩ q Ѩx Ѩy Ѩx Ѩx Ѩy Ѩy (6S.1 Derivation of the Convection Transfer Equations W-27 Ѩ Ѩ ˙ W [(xx Ϫ p) u] dx dy ϩ (yx u) dx dy (6S.2. it is rarely used in solving heat transfer problems.22) . the terms account for the rate at which mechanical work is irreversibly converted to thermal energy due to viscous effects in the fluid.6 and 6S.19 and 6S. respectively. and the remaining terms arise from the viscous normal stresses.7 by u and v.19) where the term p(Ѩu/Ѩx ϩ Ѩv/Ѩy) represents a reversible conversion between mechanical work and thermal energy. Using Equations 6S.20 may be simplified by substituting Equation 6S. Moreover. a more convenient form.c06_supl. it follows that [2] u Ѩe Ѩe Ѩu Ѩv Ѩ ѨT Ѩ ѨT ⅐ ϩ v ϭ k ϩ k Ϫp ϩ ϩ ⌽ ϩ q Ѩx Ѩy Ѩx Ѩx Ѩy Ѩy Ѩx Ѩy 2 2 (6S. is defined as ⌽ ϵ Ά Ѩu Ѩv ϩ Ѩy Ѩx ΄ ΅ ϩ2 Ѩu Ѩx ϩ Ѩv Ѩy 2 Ѩu Ѩv Ϫ2 ϩ 3 Ѩx Ѩy · 2 (6S.18. Equations 6S. the energy conservation requirement (Equation 1.17 represents the work done by the body force. and the remaining terms account for the net work done by the pressure and viscous forces. as well as analogous equations for the y-direction. This expression provides a general form of the energy conservation requirement for flow of a viscous fluid with heat transfer.qxd 1/24/06 5:35 PM Page W-27 6S. Collectively. If the fluid is incompressible.15 through 6S. After considerable manipulation. with de ϭ cvdT and cv ϭ cp for an incompressible fluid.18) ⅐ where q is the rate at which thermal energy is generated per unit volume. Because Equation 6S.20 originates from the viscous shear stresses. is obtained by multiplying Equations 6S.20) The first term on the right-hand side of Equation 6S.11c) may be expressed as Ϫ ΄ ΅ Ϫ ѨѨy ΄ve ϩ V2 ΅ Ѩ ѨT Ѩ ѨT Ѩ Ѩ ϩ k ϩ k ϩ (Xu ϩ Yv) Ϫ (pu) Ϫ (pv) Ѩx Ѩx Ѩy Ѩy Ѩx Ѩy V2 Ѩ u e ϩ Ѩx 2 2 ϩ Ѩ Ѩ ˙ ϭ0 ( u ϩ xy v) ϩ (yx u ϩ yy v) ϩ q Ѩx xx Ѩy (6S.18 represents conservation of kinetic and thermal internal energy. Instead.21) (6S. qxd 1/24/06 5:35 PM Page W-28 W-28 6S. dif.23) and using Equation 6S. Introducing the definition of the enthalpy. adv. dif. dif. The concentration may also be affected by chemical reactions. y + dy dy MA. x + dx MA. y • • • • • MA.24) If the fluid may be approximated as a perfect gas. Equation 6S. g • MA. y + dy MA. y FIGURE 6S.6.4 Conservation of Species If the viscous fluid consists of a binary mixture in which there are species concentration gradients (Figure 6.1 to replace the third term on the right-hand side of Equation 6S.6 Differential control volume (dx ⅐ dy ⅐ 1) for species conservation in two-dimensional flow of a viscous fluid with mass transfer. p iϭeϩ (6S.19 by spatial derivatives of p and (p/).25) 6S. x MA. adv. y • MA. x x. and we designate the rate at which the mass of ˙ A.24 becomes cp u Ѩp Ѩp T ѨT ѨT ѨT Ѩ Ѩ ˙ ϩ v ϭ k ϩ k ϩ u ϩ v ϩ ⌽ ϩ q Ѩ Ѩx Ѩy Ѩx Ѩx Ѩy Ѩy Ѩx Ѩy (6S. the energy equation may be expressed as [2] u Ѩp Ѩp Ѩi Ѩi Ѩ ѨT Ѩ ѨT ˙ ϩ v ϭ k ϩ k ϩ u ϩv ϩ ⌽ ϩ q Ѩx Ѩy Ѩx Ѩx Ѩy Ѩy Ѩx Ѩy (6S. di = cpdT.1. instead of its internal energy e. adv. dif.9). Consider the control volume of Figure 6S. x + dx dx • • MA.c06_supl. there will be relative transport of the species. adv.1 Derivation of the Convection Transfer Equations The thermal energy equation may also be cast in terms of the fluid enthalpy i. species A is generated per unit volume due to such reactions as n The net rate at which species A enters the control volume due to advection in the x-direction is MA. . and species conservation must be satisfied at each point in the fluid. The pertinent form of the conservation equation may be obtained by identifying the processes that affect the transport and generation of species A for a differential control volume in the fluid. Species A may be transported by advection (with the mean velocity of the mixture) and by diffusion (relative to the mean motion) in each of the coordinate directions. adv.qxd 1/24/06 5:35 PM Page W-29 6S.27.y Ϫ MA. multiplying both sides of Fick’s law (Equation 6.1 Derivation of the Convection Transfer Equations W-29 Ѩ(Au) ˙ ˙ M dx dy A.adv. Equation 6S.1 One of the few situations for which exact solutions to the convection transfer equations may be obtained involves what is termed parallel flow. for example.26 and 6S. What is the appropriate form of the continuity equation (Equation D.x A. determine the velocity distribution between the plates. x Ϫ MA.2).dif. 1.29 then reduces to u or in molar form u ѨCA ѨC ѨCA ѨCA Ѩ Ѩ ˙ ϩ Aϭ D ϩ D ϩN A Ѩx Ѩy Ѩx AB Ѩx Ѩy AB Ѩy ѨA ѨA ѨA ѨA Ѩ Ѩ ˙A ϩ ϭ D ϩ D ϩn Ѩx Ѩy Ѩx AB Ѩx Ѩy AB Ѩy (6S.yϩdy ˙ ˙ ˙ ˙ ˙ ϩM A. the species conservation requirement is ˙ ˙ ˙ ˙ M ϪM ϩM ϪM A.6) by the molecular weight ᏹA (kg/kmol) of species A to evaluate the diffusion flux.27 may be formulated for the y-direction. Referring to Figure 6S.27) ѨA ѨA Ѩ Ѩ ϪDAB dx dy ϭ D dx dy Ѩx Ѩx Ѩx AB Ѩx ΅ Expressions similar to Equations 6S.dif.31) EXAMPLE 6S.adv. with the intervening space filled by an incompressible fluid.xϩdx A. Consider a special case of parallel flow involving stationary and moving plates of infinite extent separated by a distance L.x Ϫ MA.adv. This situation is referred to as Couette flow and occurs.yϩdy ϩ MA. it follows that Ѩ(Au) Ѩ(Av) ѨA ѨA Ѩ Ѩ ˙A ϩ ϭ D ϩ D ϩn Ѩx Ѩy Ѩx AB Ѩx Ѩy AB Ѩy (6S.28) Substituting from Equations 6S. the net rate at which species A enters the control volume due to diffusion in the x-direction is ѨA ˙ ˙ M dy Ϫ A.adv.y A.dif.dif.29) A more useful form of this equation may be obtained by expanding the terms on the left-hand side and substituting from the overall continuity equation for an incompressible fluid.26) Ѩx Similarly. Beginning with the momentum equation (Equation D.xϩdx ϩ MA.x Ϫ MA.30) (6S. In this case fluid motion is only in one direction.6. xϩdx ϭ (Au) dy Ϫ (Au) ϩ Ѩx Ѩ(Au) dx dy ϭϪ (6S. in a journal bearing.1)? 2.g ϭ 0 (6S.dif.26 and 6S.c06_supl.adv.xϩdx ϭ ϪDAB Ѩx ϩ ΄ ΅ ΄ϪD AB ѨA Ѩx (6S. as well as from similar forms for the y-direction.dif. . steady-state conditions with v ϭ 0. v x.145 W/m ⅐ K. determine the temperature distribution between the plates.8. Two-dimensional flow (no variations in z). Equation D. 4.c06_supl. The speed of the moving plate is U ϭ 10 m/s. respectively. For an incompressible fluid (constant ) and parallel flow (v ϭ 0). 3. 4. and the temperatures of the stationary and moving plates are T0 ϭ 10°C and TL ϭ 30°C. Properties: Table A. ϭ ϭ 0. Surface heat fluxes and maximum temperature for prescribed conditions. SOLUTION Known: Couette flow with heat transfer.qxd 1/24/06 5:35 PM Page W-30 W-30 6S. 5. No internal energy generation. ϭ 900 ϫ 10Ϫ6 m2/s. the x velocity component u is independent of x. Find: 1. 2. 4. u Assumptions: 1. Velocity distribution. No body forces. It may then be said that the velocity field is fully developed. 2. Form of the continuity equation.2 kg/m3.2 reduces to 0ϭϪ Ѩp Ѩ2u ϩ Ѩx Ѩy2 . Analysis: 1.799 N ⅐ s/m2. (Ѩu/Ѩx) ϭ 0. 2. Steady-state conditions. Equation D. Schematic: Moving plate TL = 30°C T0 = 10°C U = 10 m/s Engine oil L = 3 mm Stationary plate y. Consider conditions for which the fluid is engine oil with L ϭ 3 mm. Temperature distribution. k ϭ 0.1 Derivation of the Convection Transfer Equations 3. engine oil (20°C): ϭ 888. Incompressible fluid with constant properties. although depending on y.1 reduces to Ѩu ϭ0 Ѩx ᭠ The important implication of this result is that. 3. and X ϭ 0.4). Beginning with the energy equation (Equation D. Calculate the heat flux to each of the plates and determine the maximum temperature in the oil. For two-dimensional. Accordingly. the temperature field must also be fully developed. it follows that particular. The appropriate form of the energy equation is then 0ϭk Ѩu Ѩ2T ϩ Ѩy Ѩy2 2 The desired temperature distribution may be obtained by solving this equation. but by an external force that provides for motion of the top plate relative to the bottom plate. The velocity distribution is then u(y) ϭ ᭠ 3. In ˙ ϭ 0. Applying the boundary conditions u(0) ϭ 0 u(L) ϭ U y U L it follows that C2 ϭ 0 and C1 ϭ U/L. we obtain u(y) ϭ C1y ϩ C2 where C1 and C2 are the constants of integration. we obtain T(y) ϭ Ϫ U 2 2 y ϩ C3y ϩ C4 2k L T(L) ϭ TL TL Ϫ T0 U 2 ϩ L 2k L The constants of integration may be obtained from the boundary conditions T(0) ϭ T0 in which case C4 ϭ T0 and T(y) ϭ T0 ϩ y 2 y U Ϫ L L 2k y ΄ ΅ ϩ (T Ϫ T ) L 2 L 0 and C3 ϭ ᭠ . Ѩp/Ѩx.4) may be simplified for the prescribed conditions. with v ϭ 0. Hence (Ѩp/Ѩx) ϭ 0. (Ѩu/Ѩx) ϭ 0. motion of the fluid is not sustained by the pressure gradient. The energy equation (D.qxd 1/24/06 5:35 PM Page W-31 6S. Rearranging and substituting for the velocity distribution.1 Derivation of the Convection Transfer Equations W-31 However. k d2T du ϭϪ 2 dy dy ϭ Ϫ U L 2 2 Integrating twice. in Couette flow. in which case (ѨT/Ѩx) ϭ 0. Integrating twice. because the top and bottom plates are at uniform temperatures. and q cpu Ѩu ѨT Ѩ2T Ѩ2T ϭ k 2 ϩ k 2 ϩ Ѩx Ѩy Ѩx Ѩy 2 However. the x-momentum equation reduces to Ѩ2u ϭ0 Ѩy2 The desired velocity distribution may be obtained by solving this equation.c06_supl. Tmax ϭ 89. it follows that qЉ 0ϭϪ U 2 k Ϫ (TL Ϫ T0) L 2L and qЉ Lϭϩ U 2 k Ϫ (TL Ϫ T0) L 2L Hence. The temperature distribution is a function of the velocity of the moving plate. it follows that ymax ϭ k (T Ϫ T ) ϩ 1΅L ΄U 2 2 L 0 or for the prescribed conditions ymax ϭ W/m ⅐ K (30 Ϫ 10)ЊC ϩ 1΅ L ϭ0.qxd 1/24/06 5:35 PM Page W-32 W-32 6S.799 N ⅐ s/m2 ϫ 100 m2/s2 0.799 0.315 W/m Ϫ 967 W/m ϭ 12.c06_supl.2ЊC ᭠ Comments: 1. respectively.315 W/m Ϫ 967 W/m ϭ Ϫ14.145 W/m ⅐ K Ϫ (30 Ϫ 10)ЊC 2 ϫ 3 ϫ 10Ϫ3m 3 ϫ 10Ϫ3m ᭠ ᭠ 2 2 2 2 2 2 qЉ 0 ϭ Ϫ13. Hence qЉ y ϭϪk 2 1 2y TL Ϫ T0 dT ϭϪk U Ϫ 2 ϩ L L L dy 2k ΄ ΅ At the bottom and top surfaces. the maximum temperature occurs in the oil and there is heat transfer to the hot. for the prescribed numerical values.536L ΄0. and the effect is shown schematically below.145 2 N ⅐ s/m ϫ 100 m /s 2 2 2 Substituting the value of ymax into the expression for T( y). the surface heat fluxes may be obtained by applying Fourier’s law. Given the strong effect of viscous dissipation for the prescribed conditions.1 Derivation of the Convection Transfer Equations 4. Knowing the temperature distribution. as well as to the cold.3 kW/m qЉ L ϭ ϩ13. plate.3 kW/m The location of the maximum temperature in the oil may be found from the requirement that TL Ϫ T0 dT 2 1 2y ϭ U Ϫ ϩ ϭ0 L L2 L dy 2k Solving for y. y L U= 0 U1 U2 T0 TL T(y) . qЉ 0 ϭϪ 0. New York. one for which the plates are separated by water and the other for which the plates are separated by air. (c) Derive the energy equation and simplify it as much as possible. Wiley. 6S. 5 mm apart. while the other plate is moving at a speed of 200 m/s. (a) For each of the two fluids. 1979. T Ϸ 55°C). and ϭ const. where the Prandtl number Pr ϭ cp/k and the Eckert number Ec ϭ U2/cp ⌬T are dimensionless groups.2 Consider a lightly loaded journal bearing using oil having the constant properties ϭ 10Ϫ2 kg/s ⅐ m and k ϭ 0. W. Stewart. B. 7th ed. and k ϭ 0.3 Consider a lightly loaded journal bearing using oil having the constant properties ϭ 800 kg/m3. The characteristic velocity and temperature difference of the problem are designated as U and ⌬T.5 A judgment concerning the influence of viscous dissipation in forced convection heat transfer may be made by calculating the quantity Pr ⅐ Ec.1 Consider the control volume shown for the special case of steady-state conditions with v ϭ 0. 2. E. Consider two cases. respectively.. which is not a good measure of the average oil temperature. Boundary Layer Theory. Problems Conservation Equations and Solutions 6S. References 1. and the temperature distribution is linear. v p x. ϭ 10Ϫ5 m2/s. water. Schlichting.g. If the journal and the bearing are each maintained at a temperature of 40°C. and how much power is needed to rotate the journal? 6S. (b) Derive the x-momentum equation and simplify it as much as possible. New York. Consider Couette flow for which one plate moves at 10 m/s and a temperature difference of 25°C is maintained between the plates. and the bearing operates at 3600 rpm. For more precise calculations. the properties should be evaluated at a more appropriate value of the average temperature (e. what is the force per unit surface area required to maintain the above condition? What is the corresponding power requirement? (b) What is the viscous dissipation associated with each of the two fluids? (c) What is the maximum temperature in each of the two fluids? 6S. and E.qxd 1/24/06 5:35 PM Page W-33 Problems W-33 For velocities less than U1 the maximum temperature corresponds to that of the hot plate. 2. τ + ]τ dy ]y y. Both plates are maintained at 27°C. . R. determine the value of Pr ⅐ Ec for air. Bird. the clearance is 0. dissipation effects may be neglected. 1966. For U ϭ 0 there is no viscous dissipation. Lightfoot.15 W/m ⅐ K.c06_supl. and engine oil. If Pr ⅐ Ec Ӷ 1. what is the maximum temperature in the oil when the journal is rotating at 10 m/s? 6S. McGrawHill. N. u dy dx τ p+ ]p dx ]x (b) What is the rate of heat transfer from the bearing. and the calculations should be repeated. Recognize that the properties were evaluated at T ϭ (TL ϩ T0)/2 ϭ 20°C...25 mm. The journal diameter is 75 mm. T ϭ T( y). H.13 W/m ⅐ K. One plate is stationary.4 Consider two large (infinite) parallel plates.. Transport Phenomena. (a) Determine the temperature distribution in the oil film assuming that there is no heat transfer into the journal and that the bearing surface is maintained at 75°C. Evaluating properties at 27°C. What is the value of Pr ⅐ Ec for air if the plate is moving at the sonic velocity? (a) Prove that u ϭ u(y) if v ϭ 0 everywhere. fluid motion between the plates is now sustained by a finite pressure gradient.7 Consider Couette flow with heat transfer for which the lower plate (mp) moves with a speed of U ϭ 5 m/s and is perfectly insulated. kb dp __ < 0 dx T2 Tb Lubricant x Ts Shaft Lubricant Shaft 100 mm diameter 200 mm Twc = 30°C Water-cooled surface.and y-momentum equations take? What is the form of the velocity profile? Note that. Its outer surface is maintained at Tsp ϭ 40°C. Determine the temperature of the insulated plate.6 Consider Couette flow for which the moving plate is maintained at a uniform temperature and the stationary plate is insulated.1 Derivation of the Convection Transfer Equations 6S. (a) What is the form of the continuity equation for this case? In what way is the flow fully developed? (b) What forms do the x. Lo). (b) Determine the rate of heat transfer (W) from the lubricant. 6S.qxd 1/24/06 5:35 PM Page W-34 W-34 6S. y(mm) 1 Bearing. determine the temperatures of the bearing and shaft. assuming that no heat is lost through the shaft. (b) Derive an expression that prescribes the conditions under which there will be no heat transfer to the upper plate.15 W/m ⅐ K. in terms of the plate speed U. 6S. expressing your result in terms of fluid properties and the temperature and speed of the moving plate. 1. y Lsp Stationary plate. The dimensionless groups are the Prandtl number Pr ϭ cp/k and the Eckert number Ec ϭ U2/cp(TL Ϫ T0). while the bearing material has a thermal conductivity of kb ϭ 45 W/m ⅐ K.145 W/m ⅐ K. The upper plate (sp) is stationary and is made of a material with thermal conductivity ksp ϭ 1. T(0) ϭ To. infinite parallel plates maintained at different temperatures. 10. ko.c06_supl. Obtain an expression for the heat flux at the moving plate. 4. u 0 Infinite parallel plates (a) On T( y)–y coordinates. (c) If the bearing housing is water-cooled.5 W/m ⅐ K and thickness Lsp ϭ 3 mm. Tb and Ts. ksp. 2. in the lubricant. sketch the temperature distribution in the oil film and the moving plate. The lubricant properties are ϭ 0.8 A shaft with a diameter of 100 mm rotates at 9000 rpm in a journal bearing that is 70 mm long.799 N ⅐ s/m2 and thermal conductivity ko ϭ 0.03 N ⅐ s/m2 and k ϭ 0. insulated U (c) Derive an expression for the heat transfer rate to the lower plate for the conditions identified in part (b). µ 0 Moving plate. (b) Obtain an expression for the temperature at the lower surface of the oil film. ⌽ (W/m3). incompressible laminar flow between two stationary. ksp (a) Determine the viscous dissipation. v x. How is this pressure gradient related to the maximum fluid velocity? . Tsp Lo Oil (o) ko. Lsp) and the oil parameters (. but the y and z components (v and w) are zero.10 Consider the problem of steady. the stationary plate parameters (Tsp. 6S. 6S. A uniform lubricant gap of 1 mm separates the shaft and the bearing. (a) Rearrange the temperature distribution to obtain the dimensionless form () ϭ [1 ϩ 1 2 Pr Ec(1 Ϫ )] where ϵ [T( y) Ϫ T0]/[TL Ϫ T0] and ϭ y/L. unlike Couette flow.9 Consider Couette flow with heat transfer as described in Example 6S. The plates are separated by a distance Lo ϭ 5 mm.1. Referred to as Poiseuille flow with heat transfer. (d) Generate a plot of versus for 0 Յ Յ 1 and values of Pr Ec ϭ 0. kb 0 Bearing. Calculate this temperature for the prescribed conditions. Explain key features of the temperature distributions. which is filled with an engine oil of viscosity ϭ 0. this special case of parallel flow is one for which the x velocity component is finite. T1 L y. such that the outer surface of the bearing is maintained at 30°C. CA. What is the mass rate of pure water production per unit surface area? Express your results in terms of CA.31). Express your result in terms of CA. and the thermal conductivity k. If it is further assumed that the transport of species A across the gas–liquid interface does not penetrate very far into the film. determine an expression for A(x. T0. What is the heat flux at the upper ( y ϭ L) surface? Species Conservation Equation and Solution 6S. such that the x velocity component is finite while the y and z components are zero. respectively. . what is the appropriate form of the species A continuity equation? Obtain expressions for the species concentration distribution and the species flux at the upper surface. u ( y). hƒg (the latent heat of vaporization of water). Express your result in terms of ␦. which is maintained at an elevated temperature T0. The species concentration and temperature may be assumed to be independent of x and z. (a) Obtain an expression for the distribution of the water vapor molar concentration CA( y) in the air.L. laminar airflow exists between the plates. L.0. 6S.14 The falling film is widely used in chemical processing for the removal of gaseous species. The film is in fully developed laminar flow over the entire plate. g. identify the conditions under which they have the same form. y) that applies in the film.0.13 Consider the conservation equations (6S. 6S. Comment on the existence of a heat and mass transfer analogy. incompressible. Write an expression for the maximum velocity umax. 2 at the top and bottom surfaces.24) and (6S.c06_supl. Solve this equation for the distribution of the x velocity component. CA. for all practical purposes. (b) Obtain an expression for the rate at which heat must be supplied per unit area to maintain the lower surface at T0.L. The corresponding molar concentrations of water vapor at the lower and upper surfaces are designated as CA.29 and 6. (b) Obtain an appropriate form of the A species conservation equation for conditions within the film. when the fluid is a binary mixture with different molar concentrations CA. (c) Assuming viscous dissipation to be significant and recognizing that conditions must be thermally fully developed. while condensation occurs at the upper surface.0 and CA. . The mass density of A at y ϭ 0 in the liquid is a constant A. what is the appropriate form of the energy equation? Solve this equation for the temperature distribution. Hint: This problem is analogous to conduction in a semi-infinite medium with a sudden change in surface temperature.10. o Condensate Air flow δ uid film (B) TL T0 L x= Liq φ y x Thin film of salt water A slow. (a) Describe the physical significance of each term. TL. and the vapor–air diffusion coefficient DAB. 1 and CA. be viewed as y ϭ ϱ. and the liquid properties and . The flow is sustained by gravity. x Gas (A) y ρ A. respectively. L.30). the position y ϭ ␦ may. which is maintained at a reduced temperature TL. (a) Write the appropriate form of the x-momentum equation for the film.qxd 1/24/06 5:35 PM Page W-35 Problems W-35 6S.11 Consider Problem 6S. such that its velocity components in the y and z directions are zero. Evaporation occurs from the liquid film on the lower surface. Comparing these equations.12 A simple scheme for desalination involves maintaining a thin film of saltwater on the lower surface of two large (infinite) parallel plates that are slightly inclined and separated by a distance L. in the film.L. This condition implies that to a good approximation. (b) Identify the approximations and special conditions needed to reduce these expressions to the boundary layer equations (6. DAB. For the region between the plates. o independent of x. u ϭ umax in the region of penetration. It involves the flow of a liquid along a surface that may be inclined at some angle Ն 0. Subject to these assumptions. and the gas species A outside the film is absorbed at the liquid–gas interface. x hm.c06_supl. x ϵ A. such that the mass density of NH3 at the gas–liquid interface (but in the liquid) is 25 kg/m3. o where n A. Љ x is the local mass flux at the gas–liquid interface.qxd 1/24/06 5:35 PM Page W-36 W-36 6S. develop a suitable correlation for Shx as a function of Rex and Sc. (d) Develop an expression for the total gas absorption rate per unit width for a film of length L (kg/s ⅐ m). A dilute solution of ammonia in water is formed. An airstream containing ammonia (NH3) moves through the tube.1 Derivation of the Convection Transfer Equations (c) If a local mass transfer convection coefficient is defined as nЉ A. What is the mass rate of NH3 removal by absorption? . (e) A water film that is 1 mm thick runs down the inside surface of a vertical tube that is 2 m long and has an inside diameter of 50 mm. and the diffusion coefficient is 2 ϫ 10Ϫ9 m2/s.
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