Viscous Fluids Amanda Meier December 14th, 2011 Abstract Fluids are represented by continuous media described by mass den- sity, velocity and pressure. An Eulerian description of uids focuses on the transport of a small xed volume. Real uids have the ability to resist stress and are known as viscous uids. Motion of these uids is governed by the Euler equation, where the Navier-Stokes equation is a simplica- tion of the Euler equation due to imcompressibilty. These equations are derived from conservation laws of mass, momentum and energy. 1 1 Introduction Viscosity is a measure of the resistance of a uid to a stress and diers among materials that may have similar properties such as density and weight. For example, water ows very dierently than oil. In order to analyze this, we can think of placing a material between two plates, where the bottom plate is xed and the top plate is allowed to move. If we have a uid between the plates and we apply a pressure to the upper plate, the plate will move continuously with a certain velocity. The uid in contact with the upper plate moves with a specic velocity, but the uid in contact with the lower plate has a zero velocity due to the solid boundary condition on the lower plate. This tells us that the uid between the plates developes a velocity gradient. The viscosity is the uid property that relates the shearing stress, or pressure per area, to the velocity gradient. If the shearing stress is linearly related to the velocity gradient, then the uid is known as Newtonian, whereas if the shearing stress is not linearly related to the velocity gradient, the uid is Non-Newtonian. In order to under- stand how the velocity varies, we need to be able to examine an innitesimal volume within the ow, which we do through dierential analysis. The equa- tions of motion for uids are governed by conservation of mass, momentum and energy. For incompressible, Newtonian uids, the Navier-Stokes equation is the dierential equation of motion. What follows is the derivation of these equations and their use in a simple example example describing uid ow in a pipe. 2 Derivations 2.1 Conservation Laws In order to look at what happens to a particular part of a uid as it moves through a system we need to look at the changes of an extensive property of the uid over a control volume, which is done using the Reynolds Transport theorem. The theorem states that these changes are equal to what is lost (or gained) through the boundary of the volume and what is created (or consumed) by the sources or sinks in the control volume. This is shown in the following integral where L is an extensive property of the uid, Ω is the control volume, v is the velocity and Q is the sink or source in the uid. d L dV = − Lv · n dA − Q dV (1) dt ˆΩ ˆ@Ω ˆΩ The divergence theorem can then be applied to change the surface integral,@Ω, to a volume integral,Ω. d L dV = − r · (Lv) dV − Q dV (2) dt ˆΩ ˆΩ ˆΩ Applying Leibniz's rule for dierentiation under an integral we get the following. 2 @L dV = − r · (Lv) dV − Q dV (3) ˆΩ @t ˆΩ ˆΩ And then combining the integrals we get, @L + r · (Lv) + Q dV = 0 (4) ˆΩ @t where the integrand is equally zero. @L + r · (Lv) + Q = 0: (5) @t This relationship gives us a general continuity equation in which we can then examine the conservation of mass, momentum and energy. 2.1.1 Conservation of Mass The continuity equation that describes liquids arises from looking at the con- servation of mass. Assuming Q = 0 or no sources or sinks of mass and looking at the mass density, ρ is the extensive property, the general continuity equation becomes @ρ + r · (ρv) = 0 (6) @t For an incompressible uid, ρ is constant and this equation simplies to r · v = 0 (7) which is a statement of conservation of volume. 2.1.2 Conservation of Momentum Conservation of momentum leads to discussion of a stress tensor and euler's equations of motion for the uid. Applying the continuity equation (Eq. 5) to momentum, ρv, we get the following equation. @ (ρv) + r · (ρvv) + Q = 0 (8) @t Expanding the derivatives and taking b to be a body force as a sink or source of momentum, @ρ @v v + ρ + r(ρv) · v + ρvr · v = b (9) @t @t we have expanded the divergence of a second rank tensor which gives a rst rank tensor @ρ @v v + ρ + r(ρ)v · v + ρr(v) · v + vρr · v = b (10) @t @t 3 if we rearrange this equation to be of the form @ρ @v v + ρ + vv · rρ + ρv · rv + ρvr · v = b (11) @t @t and pulling out v and ρ, @ρ @v v + v · rρ + ρr · v + ρ + v · rv = b (12) @t @t we can then recognize v · rρ + ρr · v = r · (ρv). @ρ @v v + r · (ρv) + ρ + v · rv = b (13) @t @t The left most term in Eq. 13 is the conservation of mass we saw earlier that is equal to zero (Eq. 6,7). Therefore, we are left with @v Dv ρ + v · rv = b ) ρ = b (14) @t Dt where D is the convective derivative. This equation is just an expression for Dt Newton's second law where we have body forces instead of point forces. And Euler's equations of motion follow directly from this, where b can be broken up into two terms, one describing forces resulting from stresses (σ is stress tensor) and the other describing forces such as gravity (f). Dv ρ = r · σ + f (15) Dt 2.1.3 Conservation of Energy Conservation of energy will lead to information pertaining to the thermody- namics of the system, which I am choosing not to focus on in this paper. For those interested, the extensive property energy density which has two parts, the kinetic energy density, 1 2, associated with the center-of-mass motion, and the 2 ρv internal energy density, ρ, where is the internal energy per unit mass will be used as L in Eq. 5. 2.2 Viscous Stress Tensor If we have a uid in a steady shear ow, where vx increases linearly with y, v = vx(y)^x (16) we have incompressible ow because the divergence is zero. The drag force acting on uid lying below area dA is then @v F = η x dA (17) x @y 4 where η is the viscosity. Refer to Fig. 1 for a schematic on viscous drag force. Figure 1: Viscous drag force exerted by uid above surface y=y0on uid below for linear shear ow. Since viscosity contributes to the surface stress, it is added in to the stress tensor. Newton's second law for a xed volume, subject to an external force f is as follows 3 @ (ρvk) 3 X 3 d x = − TkldA + ρf d x (18) ˆ @t ˆ ˆ k V l=1 A V And in dierential form this is exactly what we got out of applying the continuity equation to momentum (Eq. 15). 3 @ (ρvk) X @Tkl = − + ρfk (19) @t @xl l=1 The symmetric stress tensor for nonviscous or ideal uids has the form ideal Tkl = ρvkvl + pδkl (20) Where δkl is the Kronecker delta and we just need to add the part that describes the viscous forces. Since the uid is stationary, i.e. isotropic with no preferred spatial direction, the only terms that remain in the stress tensor are 5 @v @v k + l (21) @xl @xk 3 X @vj δkl = δkl(r:v) (22) @x j=1 j We assume the viscous stress tensor is a linear combination of these two. v @vk @vl 2δklr:v Tkl = −η + − − ζδkl(r:v) (23) @xl @xk 3 Where η is the viscosity and ζ is the bulk viscosity. This simplies for incom- pressible uids where the divergence of the velocity is zero meaning the bulk viscosity no longer contributes. v @vk @vl Tkl = −η + (24) @xl @xk So for our example shown in Figure 1, the viscous stress tensor becomes the following. @v T v = T v = −η x (25) xy yx @y The total stress tensor is then the sum of the ideal and viscous stress tensors. ideal v Tij = Tij + Tij (26) @vi @vj 2δijr:v Tij = ρvivj + pδij − η + − − ζδij(r:v) (27) @xj @xi 3 Using the expression for Newton's second law (Eq. 18) for a particular stress tensor. 3 3 2 3 2 X @ (ρvivj) @ (ρvi) η X @ vj X @ vi @p + = ζ + + η + ρf − (28) @x @t 3 @x @x @x 2 i @x j=1 j j=1 i j j=1 j i The left side simplies with the continuity equation (Eq. 5) 3 X @ (ρvj) @ρ + = 0 (29) @x @t j=1 j to give the following result. 3 3 2 3 2 X @vi @vi 1 η X @ vj η X @ vi 1 @p v + = ζ + + + f − (30) j @x @t ρ 3 @x @x ρ @x 2 i ρ @x j=1 j j=1 i j j=1 j i 6 And in vector form we get the generlized Euler equation for a vicous uid. @v 1 η η 1 (v:r)v + = ζ + r(r:v) + r2v + f − rp (31) @t ρ 3 ρ ρ With the simplication for incompressible ow, we get the Navier-Stokes equa- tion.