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 simpli ca- 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 speci c 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 in nitesimal 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 = − ∇ · (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 = − ∇ · (Lv) dV − Q dV (3) ˆΩ ∂t ˆΩ ˆΩ And then combining the integrals we get,
∂L + ∇ · (Lv) + Q dV = 0 (4) ˆΩ ∂t where the integrand is equally zero.
∂L + ∇ · (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 + ∇ · (ρv) = 0 (6) ∂t For an incompressible uid, ρ is constant and this equation simpli es to
∇ · 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) + ∇ · (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 + ∇(ρv) · v + ρv∇ · 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 + ∇(ρ)v · v + ρ∇(v) · v + vρ∇ · v = b (10) ∂t ∂t
3 if we rearrange this equation to be of the form
∂ρ ∂v v + ρ + vv · ∇ρ + ρv · ∇v + ρv∇ · v = b (11) ∂t ∂t and pulling out v and ρ,
∂ρ ∂v v + v · ∇ρ + ρ∇ · v + ρ + v · ∇v = b (12) ∂t ∂t we can then recognize v · ∇ρ + ρ∇ · v = ∇ · (ρv).
∂ρ ∂v v + ∇ · (ρv) + ρ + v · ∇v = 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 · ∇v = 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 = ∇ · 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)ˆx (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(∇.v) (22) ∂x j=1 j We assume the viscous stress tensor is a linear combination of these two. v ∂vk ∂vl 2δkl∇.v Tkl = −η + − − ζδkl(∇.v) (23) ∂xl ∂xk 3 Where η is the viscosity and ζ is the bulk viscosity. This simpli es 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 2δij∇.v Tij = ρvivj + pδij − η + − − ζδij(∇.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 simpli es 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.∇)v + = ζ + ∇(∇.v) + ∇2v + f − ∇p (31) ∂t ρ 3 ρ ρ With the simpli cation for incompressible ow, we get the Navier-Stokes equa- tion.
∂v η 1 (v.∇)v + = ∇2v + f − ∇p (32) ∂t ρ ρ Where we de ne the kinematic viscosity, υ, as the ratio of viscosity to mass density. η ν = (33) ρ 3 Important Quantities in Characterizing Flow 3.1 Reynolds number The Reynolds number is important in characterizing the relative eect of inertial forces to viscous forces, so the dimensionless parameter is de ned as the ow velocity, v, times the characteristic length scale, L, divided by the kinematic viscosity, η . ν = ρ vL Re = (34) ν A Reynolds number much less than 2100 describes completely laminar or smooth ow. A Reynolds number around 2100 is where vortices begin to appear. And a Reynolds number much greater than 2100 describes a regime of turbulence.
3.2 Mach number Another dimensionless parameter that is important in characterizing the uid ow is the mach number which is ratio of the speed of an object moving through a uid to the speed of sound in that uid for its particular physical conditions. Speci cally it is de ned as the ow velocity, , divided by the speed of sound √ v cp cs = kRT ,where k= is the ratio of speci c heat at constant pressure to cv speci c heat at constant volume, R is the universal gas constant, and T is the temperature. v Ma = (35) cs The uid is considered compressible for a Mach number greater than 0.3 and incompressible for a Mach number of less than 0.3. Since we are already assum- ing incompressible ow in deriving the equations we do not need to look further at this parameter.
7 3.3 Knudsen number The Knudsen number is helpful in classifying how rare ed a ow is or how far apart the molecules of the ow are relative to the scale of the ow. It is de ned as the mean free path of the atom, λ = √ 1 , where d is the diameter of the 2πd2n gas molecule and n is the density of the gas, divided by the characteristic length scale of the ow, L. λ Kn = (36) L A Knudsen number of less than 10−3 indicates no-slip boundary conditions in the Navier-Stokes Equations, between 10−3 and 10−1 indicates slip boundary conditions, between 10−1 and 101 indicates slip transitional ow, and greater than 101 indicates free molecular ow. The Knudsen number can be related to the Reynolds and Mach number by the following, where we once again have the ratio of speci c heats included . r Ma kπ Kn = . (37) Re 2 4 Boundary Conditions
Very important in understanding a physical situation are the boundary condi- tions of the system. Some examples are as follows:
1. For slow motion, a viscous uid in contact with a moving surface moves with the velocity of the surface, whereas the velocity is zero for a viscous uid in contact with a xed surface. 2. On a surface separating two uids the velocities have to be the same in order to maintain continuity of the velocity and its derivatives.
3. Continuity of force requires Eq. 38 to be continuous across any surface, including the one separating two uids.
3 X v (38) (pδij + Tij)ˆn j=1
4. At a free surface adjoining a vacuum, the tangential component of velocity is not restricted and the nonviscous or ideal condition of zero pressure in the comoving frame is given by the following.
3 X v (39) (pδij + Tij)ˆn = 0 j=1
8 5 Example: Fluid ow in channel or pipe 5.1 Analytic Solution From Navier-Stokes Equation
Figure 2: Steady viscous ow in a channel of width d.
Following the steady ow in a channel or pipe example in Fetter and Walecka, an incompressible uid owing in the z-direction has the following ow velocity.
v(r, t) = vz(r)ˆz (40) The assumption that ow is invariant in the z-direction is in order to have steady ow.
∂v z = 0 (41) ∂t ∂v z = 0 (42) ∂z Where the Navier-Stokes equations for incompressible ow reduce down to two terms.
1 ∂p ∇2v = (43) z η ∂z Assuming a uniform pressure gradient, we get the total change in pressure over length, where the minus sign comes from a decreasd pressure along the direction of ow.
9 ∂p ∆p = − (44) ∂z L Then the velocity satis es the 2-D Poisson equation.
1 ∆p ∇2v = − (45) z η L
With the boundary condition that vz must be zero at the walls of the channel, the solution to the dierential equation is given by,
∆px2 v (x) = − + c (46) z 2ηL and by solving forc at the boundaries, refer to Fig. 2,
d v ± = 0 (47) z 2 The result is a parabolic pro le that is a maximum at the center and decreases to zero at the walls,
∆p d2 v (x) = − x2 (48) z 2ηL 4 which the drag force follows from.
3 d X F = − TijdA = − dydz Tix| − Tix| (49) i ˆ ˆ x=d/2 x=−d/2 j=1 A The pressure contributions cancel and leave only the viscous stress.
∂vz ∆px Tzx = −η = (50) ∂x L So the total drag force is simpli ed to the following.
A d − − d ∆p F d = − 2 2 = −∆pY d (51) z L And the total mass ow through the channel per unit length in the y-direction is as so.
d 2 3 p 0 d ∆ M = ρ vz(x) dx = (52) ˆ d 12Lv − 2 In order to avoid edge corrections is it common to examine a long pipe with a circular cross section of radius R where the velocity if axisymmetric. The equations for velocity pro le, drag force and mass owrate are then as follows.
∆p v (r) = R2 − r2 (53) z 4ηL
10 d p 2 (54) Fz = −∆ πR
πR4∆p M 0 = (55) 8Lv Eq. 55 is the Poiseuille formula, that gives the pressure drop in the uid ow- ing through a long cylindrical pipe and is only valid for viscous, laminar, and incompressible ow. One simpli cation of the above velocity pro le is to look at the centerline velocity, de ned as the change in pressure times the diameter of the pipe squared, divided by the dynamic viscosity times the characteristic length of ow.
∆pD2 V (∆P, µ, l) = (56) c 16µL And the average velocity is the owrate divided by the cross-sectional area.
4 p πR ∆ ∆pD2 V V = 8Lv = = c (57) πR2 32µL 2 5.2 Numerical Solution In most physical situations, there are not as many simpli cations and we are not able to solve the Navier-Stokes equation analytically. Numerical analysis in software like COMSOL ( nite element analysis software) is able to solve the Navier-Stokes equation with an input set of boundary conditions de ned by the given parameters above (the Mach, Knudsen, and Reynold numbers).
5.2.1 Speci c Boundary Conditions In COMSOL, the ow is de ned to be laminar and incompressible, with no- slip boundary conditions (zero velocity at xed walls of channel) to model get the correct form of the Navier Stokes equation the program will solve based on the ow parameters of our system (usually de ned by the Reynolds, Mach and Knudsen numbers). We then have to draw the channel geometry we want and give input parameters, such as the speci c uid we are owing down the channel and in ow and out ow velocities or pressures it has. A mesh or grid is de ned over the entire geometry that examines the velocity in dierent volumes speci ed by the mesh size ( ne, coarse, etc.). If the mesh is Physics Controlled, there is usually a smaller grid size closer to the boundaries of the system, so the walls of the channel in our system. The solver is then run, and the results show a velocity pro le in the speci ed geometry. Referring to Fig. 3, we can see the parabolic velocity pro le in the channel that matches the analytic description we found above.
11 Figure 3: Velocity magnitude and eld pro le of ow through channel.
More complicated systems can then be input to COMSOL and convergence studies run to make sure the equations and boundary conditions make physical sense. Having an analytic check with the pipe example helps in understanding what the nite element analysis is doing with the speci ed input and boundary conditions.
6 Conclusion
Derivation and examinination of the Euler and Navier-Stokes equations leads to understanding of the continuous motion of viscous uids. Once equipped with the equations of motion, it is important to know the ow regime and physical boundaries of the system in order to simplify the equations of motion as much as possible. When we simpli ed our ow to be viscous, laminar, and incompressible, we were able to get an analytic description of the velocity, drag force and mass ow of the uid. Comparison with numerical analysis veri ed our analytic solution. Now that the most simple case of viscous uid ow is understood, we can analyze more complicated uid systems which are more prevalent in the real world.
12 References
[1] Alexander L. Fetter and John Dirk Walecka. Theoretical Mechanics of Particles and Continua. Dover Publications, 2003.
[2] Nam-Trung Nguyen and Steven T. Wereley. Fundamentals and Ap- plications fo Micro uidics, 2nd Edition. Artech House, 2006. [3] Bruce R. Munson, Donald F. Young and Theodore H. Okiishi. Fundamentals of Fluid Mechanics, 4th Edition. John Wiley & Sons, 2002.
[4] Ain A. Sonin Equations of Motions for Viscous Fluids, 8th Edition. Massachusetts Institute of Technology, 2001. [5] COMSOL Multiphysics, registered trademark of COMSOL AB, Stockholm, Sweden
[6] Derivation of the Navier-Stokes Equations http://en.wikipedia.org/wiki/Derivation_of_the_Navier%E2%80%93Stokes_equations Date accessed: December 2011
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