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1 1.1 Conservation Equations Conservation of mass Mass conservation states that the net rate of change of the mass of a control volume (CV) is equal to the net rate of transport across the boundary of the CV. In vector form, Z Z d ρ dV + n̂ · ρu dA = 0 (1) dt V A in which n̂ is the outward pointing normal along the surface of the control volume. The quantity ρu is the mass flux vector, and since the normal is defined as pointing out, the second term will be positive/negative if a net amount of mass is leaving/entering the CV. The divergence theorem of vector calculus transforms the surface integral into a volume integral, and the order of differentiation and integration in the first term can be switched. This gives Z ∂ρ + ∇ · (ρu) dV = 0 (2) ∂t V in which ∇· is the divergence operator; the specific form of this function will depend on the coordinate system. The above relation must hold for any sub–volume of the CV, and thus the integrand must be zero: ∂ρ + ∇ · (ρu) = 0 ∂t (3) This is the divergent form of the continuity equation. The chain rule of calculus can be applied to the second term, ∇ · (ρu) = ρ ∇ · u + u · ∇ρ (4) Note that the density gradient, ∇ρ, is a vector, and u · ∇ρ gives a scalar. When replaced into Eq. (3) the result is Dρ + ρ∇ · u = 0 (5) Dt in which the material derivative is defined as D ∂ = +u·∇ Dt ∂t (6) Equation (5) is the conservative form of the continuity equation. The divergent and conservative forms are completely equivalent; their distinction becomes more relevant when developing numerical methods for computing flowfields. 1.2 Momentum conservation Newton’s law states that the rate change of linear momentum of a system is equal to the net force acting on a system. When applied to a CV, the rate of change of the CV momentum is given by the 1) the instantaneous time derivative of the CV momentum, plus 2) the net rate of transport of momentum across the CV boundaries. Note that these two contributions are analogous to the two sources of mass change in a CV, i.e., the ”rate of storage” and the ”rate of transport” terms. Unlike mass, however, momentum can be created or destroyed by application of forces to the CV. The forces arise from two contributions: 1) surface forces which act on the surface of the control volume, and 2) body forces which act directly on the mass within the CV. In vector form, the linear momentum balance on the CV appears as Z Z Z Z d ρ u dV + n̂ · ρu u dA = n̂ · σ dA + ρ f dV (7) dt V A A V Note that this is a vector equation; it can be resolved into the mass flux vector, and the dyadic tensor ux ux ρu u = ρui uj = ρ uy ux uz ux 1 three orthogonal components. As before, ρu is ux uy uy uy uz uy ux uz uy uz uz uz (8) denotes the flux of j–directed momentum in the ith direction. The quantity σ is the stess tensor in the fluid. This is a second–order tensor which has the stucture σxx σxy σxz σ = σij = σyx σyy σyz (9) σzx σzy σzz The stress tensor σij appears as a tensor (i.e., a 3× 3 matrix) because two distinct directions are involved in specifying the stress (which is force/unit area) acting on a surface: the directional component of the force (the second subscript) and the surface normal direction (the first subscript). Consequently, the force vector acting on a surface element dA that has a normal in the x–direction would be x̂ · σ dA. The stress tensor is often resolved into an isotropic part (independent of direction) and a deviatoric part via σ = −P I + τ (10) in which P is the thermodynamic pressure, I is the identity matrix, and τ denotes the shear stress tensor. Strictly speaking, the diagonal elements of the shear stress tensor (i.e., τxx , etc.) do not denote shear stresses – rather, they describe normal stresses – yet for the situations we encounter the only significant normal stress will be due to the pressure, and the diagonal components of τ will be zero. The shear stress tensor will be related to the rate of fluid deformation (i.e., velocity gradients) via constitutive relations; Newton’s viscosity law is a simple example. Such relations will be deferred for later. The quantity f denotes the body force per unit mass and has units of acceleration. Indeed, the most typical body force component is gravity, for which f would be the gravitational acceleration times the downwards unit vector. Other mechanisms can result in body forces, such as electric currents, magnetic fields, and coriolis effects. Application of the divergence theorem to Eq. (7) yields the differential form of linear momentum conservation: ∂ (ρ u) + ∇ · (ρu u) = ∇ · σ + ρ f (11) ∂t and combination of this with the continuity equation in Eq. (3) leads to ρ Du = ∇ · σ + ρf Dt (12) This is the most general form of the momentum equation. Again, a constitutive law is needed to connect σ to the velocity gradients and thereby ”close” the equation. 1.3 Energy conservation Let e denote the internal energy of the fluid per unit mass. The total energy, per unit mass, of the fluid is internal plus kinetic energy, or e + u2 /2 where u2 = u · u is the square magnitude of the velocity. From the first law, the rate of storage plus rate of transport of total energy in the CV must be balanced by the rate of work and rate of heat transfer to the CV. Work is force over a displacement, and rate of work will be force times a rate of displacement, or force times velocity. As was the case with the momentum equation, the force components include surface stresses and body forces. Putting it all together, the first law for the CV has Z Z u2 u2 d ρ e+ dV + n̂ · ρu e + dA dt V 2 2 A Z Z Z Z 00 = n̂ · σ · u dA + ρ f · u dV − n̂ · q dA + q 000 dV (13) A V A V In the above, q00 denotes the heat flux, and the surface integral containing this term is negative if there is a net transfer of heat from the CV; hence the negative sign. The quantity q 000 denotes a volumetric source of heat (W/m3 ) within the fluid; this term would arise from chemical reactions, Joule heating from an electric current – but not from fluid friction: this effect will be accounted by the stress tensor work term. It is left as an exercise to show that the differential form of the energy equation becomes ρ Di DP = + τ : ∇u − ∇ · q00 + q 000 Dt Dt 2 (14) in which i = e + P/ρ is the enthalpy per unit mass of the fluid (yes, the symbol h is commonly used in thermodynamics to denote enthalpy, but h has been reserved for the heat transfer coefficient). Note that the body force term has disappeared – this results from combining the momentum equation with the energy equation. What is left, with regard to work, is the τ : ∇u dissipation term. The operator : denotes a contraction between the two tensors τ and ∇u and this produces a scalar. In tensor notation, this operation is written ∂ τ : ∇u = σij uj (15) ∂xi In the tensor notation used above, summation over repeated subscripts is implied. That is, the operation sums over i = 1, 2, 3 and over j = 1, 2, 3. Equation (14) is a general form of the energy equation for a moving fluid. To close the equation, constitutive relations are needed to related τ to velocity gradients (Newton’s laws) and q00 to temperature gradients (Fourier’s law). We also need an equation of state to connect enthalpy i to temperature. All such details will be addressed when we examine specific flows. Also note that no explicit reference is made to a coordinate system. The differential relations for mass, momentum, and energy conservation are presented in a general, vector calculus form. Of course, the specific forms of the gradient (i.e., ∇ρ), the vector gradient (∇u), and the divergence (∇ · q00 ) would be needed to apply the formulas to a given problem with a given coordinate system. Such information can be found in the text, or on Mathematica. Exercises 1. Fill in the details in the derivation of Eq. (12) from Eq. (11). This will take only line or two of formulas. 2. Derive Eq. (14) beginning with Eq. (13). To do this, you will need to (among other things) (a) Apply the divergence theorem per the previous derivations. (b) Use the vector identity ∇ · (σ · u) = (∇ · σ) · u + σ : ∇u (16) (c) Replace e with i − P/ρ. (d) Dot the momentum equation into u, use the fact that Du2 /2 Du ·u= Dt Dt (17) (note that the resulting equation represents a balance of mechanical energy in the CV), and use this equation to eliminate the kinetic energy terms in the energy equation. (e) Use the stress tensor relation in Eq. (10) to split σ into the normal and shear parts. Equation (15) will give you a hint at what results from I : ∇u – note that I, in tensor notation, is the Kronecker delta function δij = 1 if i = j and 0 if i 6= j. 3. Consider a sphere, of radius R, that is immersed in a stationary fluid. The surface temperature of the sphere is maintained at TS , and the temperature far from the sphere in the fluid is T∞ . Using the steady 1–D conduction equation for the fluid temperature, 1 ∂ 2 2 ∂T ∇ T = 2 r =0 (18) r ∂r ∂r determine the formula for the Nusselt number of the sphere, which is defined as NuD = hD 2hR = kf kf (19) You will need to solve the conduction equation for this problem and use the solution to determine the heat flux at the surface of the sphere. Your result, in the end, will be a pretty simple formula. 3