Lai Ch8

Lai Chapter 8: Non-Newtonian Fluids PhL 9.23.12 Part A: Linear ViscoElastic Fluid (444) 2 8.1 Linear Maxwell Fluid (444) 2 8.2 Fluid with multiple components (450). [ generalized linear Maxwell discrete] 7 8.3 Green's Function solution to our ODE's (451) 9 8.4 Continuous spectrum version (450). 11 8.5 Computation of H(λ) from φ from G* (454). 12 Part B: Nonlinear Viscoelastic Fluid (456) 13 8.6 Current configuration as the reference configuration (456) 13 8.7 The Relative Deformation Gradient (457) 15 8.8 The Relative "other tensors" (459) 15 8.9 The Relative C tensor in various coordinates (460) 16 8.10 "Deformation History" and the Rivlin-Ericksen Tensors (463) 16 8.11 Relating A1 to D; a recursion formula for the AN (468) 17 8.12 Some tensor relations involving Ft and (v) and the polar decomposition (471) 20 8.13 Change of Frame Stuff (471) 23 8.14 Change of Frame for the A tensors (474) 23 8.15 Incompressible Simple Fluid (474) 23 8.16 Two specific models which reduce to Maxwell for small deformation (475) 24 8.17 Single-integral non-linear constitutive equations (478) 26 8.18 Differential-type constitutive equations for incompressible fluid (481) 28 8.19 Objective Rate of Stress (483) 29 8.20 Rate-type constitutive equations (487) 30 Part C: Viscometric flow for incompressible simple fluid (491) 30 8.21 Viscometric Flow (491) 30 8.22 Form of Tij for Viscometric Flow (493) 31 8.23 Channel Flow (495) 31 8.24 Couette Flow (497) 32 Appendix 8.1. 33 Water does behave as a Newtonian fluid as discussed in Chapter 6, but paint and polymer fluids don't. In this chapter, which has Part A,B,C, we are going to examine other Fluid Models. In Ch 6 we found this rule for an isotropic Newtonian fluid Tij = -p δij + T'ij T'ij = λ Δ δij + 2μDij Δ = tr(D) D = (∂ivj + ∂jvi)/2 where μ was the constant viscosity. For incompressible, tr(D) = ∂ivi = div(v) = 0 dρ/dt = 0, so the incompressible equation is simpler; D is the "rate of deformation tensor" : Tij = -p δij + T'ij T'ij = 2μDij D = (∂ivj + ∂jvi)/2 Part A: Linear ViscoElastic Fluid (444) 8.1 Linear Maxwell Fluid (444) Now suddenly, without any comment, the piece T' is renamed S, the "extra stress" so for a Newtonian fluid we would say Tij = -p δij + Sij Sij = 2μDij D = (∂ivj + ∂jvi)/2 Finally, then, we are able to see the new feature involved with the "linear Maxwell fluid" Tij = -p δij + Sij Sij = 2μDij - λ∂Sij/∂t D = (∂ivj + ∂jvi)/2 where we now have some "drag" or "velocity dependent force" element added in, though it is not yet clear why that name is appropriate. First, recall that any equation like F = -kx which relates "stress" to "strain" is called a constitutive relation/equation, and this describes the nature of the medium and is separate from the equations of motion and the various other "conservation of" principles. I think this is a "linear" Maxwell fluid because the term ∂Sij/∂t = ∂tSij is linear, and we end up with a system of linear ODE's. Why Maxwell? I just read the entire wiki site on Maxwell, Scottish. He did so much stuff that this little fluid thing is not even mentioned: E&M, thermo, kinetic theory of gases, color photos, rings of Saturn as particles, on and on. But in 1867 Maxwell managed to work in the above little model for visco-elastic materials. Kelvin was a contemporary and did similar stuff. Example 8.1.1. As an elementary model of this Maxwell fluid, we study the "spring plus dashpot" situation of 8.1.3. You apply some force Lai calls S on the left. Each of the two series elements has its own displacement which here are called εe (spring, elastic) and εη (viscous), and the total displacement is ε = εe+ εη. Like components in an electric circuit, each element here has a governing equation: S = Gεe which says F = kx // G and η are assumed constants S = η dεη/dt which says dashpot represents a frictional force proportional to velocity So G is the spring constant, and η is the viscosity (the dashpot drag coefficient F = η * velocity ) The two S's are the same because these circuit elements are in series (see comment below!). If we then differentiate the geometric fact that ε = εe+ εη we get, dε/dt = dεe/dt + dεη/dt = (1/G) dS/dt + (1/η) S or S + (η/G)dS/dt = η (dε/dt) //which is 8.1.6 Remember that S = force ~ "stress". If we define λ = (η/G) = drag/spring, this equation says S + λ dS/dt = η (dε/dt) // λ has dimensions of time // 8.1.6 which we compare to our Maxwell fluid model for the extra stress tensor S Sij + λ (∂Sij/∂t) = 2μDij = μ (∂ivj + ∂jvi) // partial derivative on S !! In this fairly close analogy, we can see that μ ~ η = viscosity λ = (η/G) = (viscosity/elasticity) ratio = relaxation time Somehow the RHS drag term η (dε/dt) gets replaced by μ (∂ivj + ∂jvi). In a simple layer situation, adjacent layers have a relative velocity between them which causes drag: Δv ~ (∂vx/∂y) Δy, so we expect the drag force to be η Δv = η (∂vx/∂y) Δy. So when you make the adjustment to "per unit area" and all that, you end up with a RHS as shown. So for S we end up with a first order matrix ODE which is "driven" by a RHS inhomogeneous term which is the drag. But the dS/dt term feels the effect of both drag and elasticity through the coefficient λ. Comment : The RHS of 8.1.6 has sort of η * velocity which we know is the usual frictional force, so the role of the η (dε/dt) term in 8.1.6 seems very clear (think S = force), and this is the viscous effect. But where is the "spring effect" coming from? If the dashpot is very viscous, η >> 1, then in effect the entire system is just the spring. In this limit 8.1.6 says (η/G)dS/dt = η (dε/dt) which you can integrate in effect to get S = G Δε and that is your F = kΔx, and there is your spring effect. But in general the two effects are very well mixed together, and the term viscoelastic fluid seems very appropriate. Comment on massless spring. If you apply F1 and F2 to the ends of a spring, it stretches Δx = (F1+F2)/k and it does this instantly. Both forces might be time dependent, so Δx(t) = [F1(t)+F2(t)]/k and things are still instantaneous. If the spring had distributed mass, you might have some inertial ma force at different points along the spring, but it is massless so inertial force is zero, even if the spring is in the act of stretching. Therefore, at any instant of time you have F1(t) = -F2(t). If you have both forces pulling to the left, then we negate F2 and we then have F1(t) = F2(t). What this says is that the spring "transmits force through itself instantly. This then means that, even in a dynamic situation, both the spring and the dashpot in our little model experience the same force. Comment on electrical analogy: In this series circuit problem, the two voltages which you add are the two displacements, so you then have ε = εe + εη . Since the spring transmits force through itself instantly, the force S across the spring is the same as the force S across the dashpot, so S is like the current in a series circuit. Then Ohm's Law for the spring says V = IR or εe = S (1/G) and the resistance is R = 1/G = 1/k, so G is more like a conductance. The dashpot is like a capacitor Q = CV and I = C dV/dt so get S = η deη/dt and then you can identify the capacitance with η. A stiff dashpot means large C = η, so it is hard to change the voltage across a large capacitor, meaning here that εη does not change, and you have just the spring. A stiff spring on the other hand means large G = k and small R and then εe does not change and then you have just a dashpot. Now we are going to set up some "experiments" similar to those at the start of the elastic solid chapter which serve to clarify the meaning of these model constants. We first apply these experiments to our spring and dashpot, then later we will show how they apply to stuff like S and D for the medium. Creep Experiment. Just apply constant force S0 starting at t = 0 and what happens? From 8.1.6 S0 = η (dε/dt) => ε(t) = (S0/η) t + ε(0) At t = 0 when S0 is applied, the spring element instantly stretches so εo = S0/G, whereas the dashpot just sits there with εη = 0. How do we know these things. The first is from F = kx. The second is from S0 = η dεη/dt => εη(t) = (S0/η) t which is 0 at t = 0+. So it takes the dashpot "time" to do something (capacitor), and at t = 0+ we have εη(0+) = 0. Therefore, only the spring contributes to ε(0) and we get ε(t) = (S0/η) t + S0/G So from such a creep experiment, we could in theory measure G and η for our Maxwell fluid: G coming from the jump just after t = 0, and η from the rate of movement. The spring here instantly reaches its amount of stretch εe = S/G and that then never changes. The dashpot then just keeps moving at a constant velocity and ε grows linearly. The dashpot creeps along at constant v = (S0/η) . Notice that this creep experiment does not tell you the value of μ. Relaxation Experiment. Apply an impulse displacement ε0 at time 0 and watch what happens. The spring uses up all this displacement, so we get εe(t=0+) = ε0 and εη(t=0+) = 0, because the dashpot has no time to get rolling. Then S(0+) = Gε0 and we have to solve our ODE which says (ε = ε0 = constant) S + λ ∂S/∂t = 0 S(0+) = Gε0 => S(t) = (Gε0) e-t/λ εe(t) = S/G = εo e-t/λ starts at ε0 and decays to no stretch eη(t) = ε - εe = ε0[ 1 - e-t/λ] starts at 0 and eventually uses all the ε0 So this gives some intuitive feel for meaning of parameter λ. But we have NOT yet connected these experiments to experiments we might do with fluids. Hopefully the examples will do that. Notice that this experiment lets you measure only parameter λ. Key idea: the λ parameter which is the elasticity/viscosity ratio has units of time and should be thought of as a characteristic time for the material, like τ = RC for a resistor capacitor circuit. I could imagine another book calling this τ, or some other symbol that more looks like a time. (RC = (1/G)η = η/G = λ). Example 8.1.2. Simple shear flow of a Maxwell fluid (steady state flow). The stress, whatever it is, should be constant, so ∂S/∂t = 0, so then Sij = 2μDij or S = 2μD. Same as Newtonian fluid. Implication: You can measure parameter μ, the viscosity of a fancy Maxwell fluid, by doing some static flow experiment, just as you would do with a Newtonian fluid. For example you could set up a simple layer flow between two parallel plates and then measure the force of the liquid on either plate. Lai has failed to make this point clearly. So our "new" task for a Maxwell fluid is to find an experiment by which we can measure that ratio parameter λ. ( Sij + λ ∂Sij/∂t = 2μDij ), and of course this has to be some kind of non-static experiment. You might say that μ is the "static flow viscosity". We have seen how to measure λ in the model, but we need soon to do it with the real fluid. Example 8.1.3. Here we shall get the relaxation idea going with a fluid. Assume a step function here u1 = εoH(t) x2 so this is a transverse (shear as in half river)) step in displacement in a 3D liquid, so we are talking shear. This is like the relaxation experiment above. Maybe you could physically do this with a fluid between two plates and you suddenly slide one of the plates so u1 = εoh at the upper moved plate. Probably this creates the above displacement pattern in the fluid at time t=0+. We can compute Dij by first computing v1 = ε0x2δ(t) and then D12 = (∂v1/∂x2)/2 = εoδ(t)/2 and then we have (notice that D12 is not a function of transverse x2 and is thus the same at all x2) S12 + λ ∂S12/∂t = 2μD12 = 2μεo δ(t)/2 = μεo δ(t) Integrate in time range (-e,e) where e is small to get 2eS12 + λ [S12(e) - S12(-e)] = με0 or λ [S12(0+) - S12(0-)] = με0 But before our step application nothing was happening so there was not strain, so S12(0-) = 0 and then S12(0+) = με0/λ Then for t > 0 we solve S12 + λ dS12/dt = μεo δ(t) = 0 => S12(t) = S12(0+)e-t/λ = (με0/λ) e-t/λ and this shows that the shear strain S12 (same at all x2) jumps to a constant value then decays away, just like the dashpot spring system where recall S(t) = (Gε0) e-t/λ . So the analogy says G = μ/λ = the springiness of the medium, since G was the spring constant. This is our first interpretation of the μ and λ parameters in the actual medium model. This is the "relaxation experiment" done for the actual medium. The stress S12 starts out and then decays away as the dashpot element flows. Measure initial S12 and that is a measure of μ/λ. Then measure decay rate and that tells you λ and then you can have μ as well. Example 8.1.4. (448) Two infinite plates separated by h. The lower plate is fixed, the upper one does shear oscillation, so we have various BC's on the u displacement (no slip condition) as shown in (i) and (ii). We try out ux(y,t) = u0(y/h)eiωt as a solution everywhere between the plates. Then vx = iω u0(y/h)eiωt and then D12 = (1/2)∂vx/∂y = iω (u0/2h)eiωt. This rate of deformation thing is independent of y, it is the same at every level. Next, we try out S12 = S0eiωt as a solution everywhere in the gap, also independent of y. We plug this form into S12 + λ dS12/dt = 2μD12 and we find that it works just fine as long as we select S0 properly: S0 = (u0/h)G* where G* = as shown 8.123 through 8.125 so the solution is then S12(t) = (u0/h)G*eiωt and G* is complex indicating phase stuff. Eventually we take the real part and get S12(t) = (u0/h)|G*| cos(ωt+δ) tanδ = 1/(λω) λ = drag/spring Not surprisingly , since D12 was the same at every level (indep of y), so is S12. So what do we learn from this example? We learn that the shear "extra stress" S12 oscillates along with that top plate, and is the same at all levels y, but has a certain phase δ relative to the driving oscillation displacement u. If we remember that λ is the relaxation time, we see that there will be no phase if λ = 0 (no drag), but as λ increases, we have δ increasing at a given ω and reaching a max of π/2 when λ = ∞ (no spring). The phase delay is the same at all y. So really this example is an experiment to measure the λ parameter. Note added: Let's examine this plane oscillation experiment again. We assume that the shear displacement ux at the driving plate has zero phase in (i), and then we find that the shear stress is S12 = G* (u0/h) in 8.1.22. In practice, you drive the plate with S12 at zero phase, and then the displacement will be out of phase, think either way. But in practice that second way allows you to measure the displacement including its phase. The model says that uo = h (S0/G*) that is uo is the shear displacement just below the driving plate If G* is real, then displacement and force are in phase, your medium just vibrates like a purely elastic substance work goes into stored spring energy. But if G* is imaginary, then π/2 phase shift, and dashpot dominates things and you have a heat loss going on. So the real thing G' is called storage modulus since energy is stored in the springs, while G" is called the loss modulus. My main point is that this kind of experiment let's you measure G*, and later Lai will assume that one knows G* experimentally and wants to do other things with that data. 8.2 Fluid with multiple components (450). [ generalized linear Maxwell discrete] Something like paint is a mixture of fluids, so we need to get that into the model. So imagine that n = 1 to N lists off different "species" of materials in the fluid. This mix idea is called "generalized". Sn + λn∂tSn = 2μnD S = Σn=1N Sn So here each species has its own private "extra stress" tensor Sn. The claim is that you have N dashpots in parallel, each one with force Sn and total force S. Each has its own spring and viscosity constant. But all the dashpots see the same total fluid velocity v so they have the same Dij tensor. Lai and I make no pretense toward explaining exactly how the mixture behaves like parallel dashpots. OK for now. I verify below that (8.2.3) is valid for any N mixture. I know now that the an and bn exist because I know exactly how to compute them. ____________________________________________________________________________________ Digression: Maple Lai_p450.mws. Now I am going to solve the problem he poses in the text and solve it with Maple. If we apply ∂t multiple times to the above equation, we have a whole set of equations of the form ∂tiSn = λn∂ti+1Sn = 2μn∂tiD i = 0,1,2... This is a recursion relation and on scratch I solved it to find this solution: ∂tiSn = (2μn) Σk=1i(-1)k-1 cnk (∂ti-kD) + (-1)i cni Sn cn ≡ 1/λn which expresses any derivative of Sn in terms of Sn and derivatives of D. In Maple, I make these definitions DS(i,n) ≡ ∂tiSn D[r] ≡ ∂trD S[n] = Sn Then the above equation reads DS(i,n) = (2μn) Σk=1i(-1)k-1 cnk D[i-k] + (-1)i cni S[n] cn ≡ 1/λn Our task is to find coefficients an and bn such that (8.2.3) is valid for some given N. Σr=1NSr + Σn=1N an [Σr=1N∂tnSr] = ΣN=0N-1bn∂tnD or Σr=1NSr + Σr=1N Σn=1N an DS(n,r) = ΣN=0N-1bnD[n] or Σr=1N[ Sr + Σn=1N an DS(n,r) ] = ΣN=0N-1bnD[n] The problem is to find an and bn such that Q = 0. For N=3 we get for Q this expression which you see has terms for S1, S2,S3 and for D0, D1, D2. Since these are all independent functions, their coefficients have to separately be zero. So then we can get the coefficients this way: You can see that the Sn coefficients define an NxN linear problem which can be solved for the an. Then those an are inserted into the first set of equations to get the bn. I first solve the NxN problem this way: where the assign makes the an accessible in Maple. I then obtain the bn values as B[n] this way where you can see that I cancel off the bn's appearing in the first set of equations above. Then I print out the solution an and bn and this agrees with the result quoted on p 451. Here are my results for N = 2, which agrees with (8.26). Just for fun here are the N=4 results ____________________________________________________________________________________ 8.3 Green's Function solution to our ODE's (451) Now forget the mixture stuff of the last section, and go back to the original single-component ODE which is the constitutive relation for a linear Maxwell fluid: S + λ∂tS = 2μD We are now going to solve this to get S as a function of D using the Green's Function method which I have spent a year studying lately! The associated Green's Function problem is this (1 + λ∂t)G(t,0) = δ(t-0) Claim that the solution is G(t,0) = (1/λ) e-t/λ H(t-0) so G(t,t') = (1/λ) e-(t-t')/λ H(t-t') // the propagator Proof: ∂tG(t,0) = (-1/λ) G +(1/λ) e-t/λ δ(t-0) = -G/λ + δ(t)/λ Then (1 + λ∂t)G(t,0) = G + λ [-G/λ + δ(t)/λ] = δ(t) so yes, is solution Then the inhomo solution is then given by integrating the Green's against the RHS, S(t) = !Syntax Error, Idt' G(t,t') [2μD(t')] = !Syntax Error, Idt' [(1/λ) e-(t-t')/λ H(t-t')] [2μD(t')] = !Syntax Error, I dt' [(1/λ) e-(t-t')/λ [2μD(t')] = (2μ/λ) !Syntax Error, I dt' e-(t-t')/λ D(t') = 2 !Syntax Error, I dt' [ (μ/λ)e-(t-t')/λ] D(t') = 2 !Syntax Error, I dt' [ φ(t-t')] D(t') = (8.3.1) This is the solution !! I also did this the other way Lai shows, as a sort of verification. ( ignoring BC's ! ) SO, this looks more like a standard-issue "constitutive relation" where stress S is on the left side and some function of strain/deformation is on the right side (in this case, rate of deformation D). Written in this integral form, we don't have a ∂tS  term sitting in a constitutive relation. the "coefficient" that you might find in a usual stress/strain constitutive relation is now that function φ, and you see that, unlike the simpler cases, the stress at current time t gets a contribution from D at each point in the past time (history), and the coefficient is then a function of both times as t-t'. This coefficient φ is called the stress relaxation function. The larger the difference in the times t - t', the less influence of a past point in time. Each past point in time "emits" a propagator to the present, and older propagators die out while more recent once are still active. So φ is a history weight function, and it could have been called a memory function for that reason, but that term is saved for below when E enters the scene. Now having done all this for a one-component fluid, we add back in the "mixture" idea. We have, Sn + λn∂tSn = 2μnD Then solution is Sn = 2 !Syntax Error, I dt' [ (μn/λn)e-(t-t')/λn] D(t') and then sum to get S = 2 !Syntax Error, I dt' Σn[ (μn/λn)e-(t-t')/λn] D(t') = 2 !Syntax Error, I dt' Σn[φn(t-t')] D(t') (*) with φ(t) as shown in 8.3.4. So now φ has a contribution for each component of the mixture. 8.4 Continuous spectrum version (450). In the discrete mix problem we had φ(t) = Σn=1N (μn/λn) e-t/λn = the "stress relaxation function" Looking just above, you see that φ(t-t') is a convolution form propagator! The driver is the rate of deformation tensor D, and the output is the stress S. Equation (*) is similar to earlier constitutive relations, but here we are not relating stress to deformation, but rather to rate of deformation! The propagator is obtained in the mixed medium by simple addition over the components. I think Stakgold would refer to this propagator as a "fundy solution". This is always the Green's function with no boundary conditions, the free-space propagator. Stak never did this with such a simple differential equation. Imagine a large number of components in the mixture so this becomes a continuum, and then the propagator is an integral. Each component in the mix has some value of λ. φ(t) = !Syntax Error, I[μ(λ)/λ] e-t/λdλ μ(λ) he calls H(λ) [ H not Heaviside as in 8.1.17] Seems a bad choice for μ(λ) but OK. Notice that λ must have dimensions of time! We need to know the ratio (μn/λn) for each component, and this is H(λ)/λ, the "relaxation spectrum". One idea now is to assume small displacement so you can replace D with ∂E/∂t as page 205. D is the rate of deformation tensor, E is the infinitesimal strain tensor that we used a lot in elastic solids work. If we replace D in this way, we get various results quote p 452-453. We start with S = 2∫φ D were φ is a convolution kernel. Parts gets this to be S = -2 ∫ds E φ' where now E is the kernel and φ' = f. Remember that φ is called the relaxation function, and its derivative f is called the memory function. I derived everything. Comment: Look at this form Sij(t) = 2!Syntax Error, I φ'(t-t') Eij(t')dt' f(t-t') ≡ φ'(t-t') = memory function What you see here is that the stress at current time t is determined by the strain E at all earlier times! This is just coming from the infinitesimal relation D = ∂tE and parts integration. The weight ascribed to various times in the history is provided by f = φ' . The current stress "remembers" the entire history of the strain E, and the weight function as a function of historical time t is called "the memory function". It all makes perfect sense. Dimensions of things: D = v/L = sec-1; S = stress = force/area = ma/area = kg/(m-sec2) = nt/m2. μD = stress = nt/m2. Look at p 452 to see that S = φ D dt = φ, so φ must also be nt/m2. Then 8.4.1 says that φ = H/λ e-t/λdλ = H in units, so H is also stress. μ = stress/D = stress * time. = nt/m2 * sec so I think Hdλ has the same units as μ which is stress*time. G* from 8.1.21 is stress as well, which is also clear from 8.1.28, so here is a little summary of units: S = μD = H = φ = G* = stress units = nt/m2 λ = sec D = sec-1 μ = stress * sec Example 8.4.1. Here we are back to the oscillating plate, but now instead of using the ODE 8.1.2 to get S, we use 8.4.12's integral form instead. Either way D12 is exactly the same. We now find that G* is given as the integral 8.4.13 in terms of the stress relaxation function φ(t), but then we write that as its integral over H(λ) which is the μ(λ) object from p 452 (not sure how to interpret this thing H), so then we end up with 8.4.14 where G' and G" are given as integrals of H, reminds me of dispersion relations. Remember that G* = (h/uo) S12 from the oscillating plate experiment and it has phase. 8.5 Computation of H(λ) from φ from G* (454). We assume we have done some oscillating plate experiment and we have data for G*(ω), and we have at our disposal 8.4.15 and 16. I bet there is a nice way of inverting those equations, but instead Lai describes a numerical method for finding a reasonable H(λ) that makes those integrals work. λ is a time variable. It is an iterative method with a 1968 paper reference of Tanner. Again, we have here a method for measuring the relaxation spectrum H(λ)/λ for any complicated mixed fluid having a continuum of components. This method in theory should work for a finite number of components, and delta functions will then appear for H. Remember that we are constantly looking for experiments that can measure model parameters. Above we found μ, then we found λ for single-component fluid, and now H(λ) for a fluid with many components. Example 8.5.1 (455) We are given some G* curves for synovial fluid of young, old, and arthritic as plots page 455. Both storage and loss generally seem to increase with ω in the 10 Hz range as shown. We are supposed to do the numerical method just described and obtain plots of H(τ) and then φ(τ) from 8.4.1. Only the latter plots are shown on page 456. I really don't know how to interpret the conclusions because I have only just learned about φ(τ). Part B: Nonlinear Viscoelastic Fluid (456) This section has no opening comments at all and just dives into the reference version of things, so we are not told what the term "non-linear" refers to. We have seen that the Maxwell type fluids always have a constitutive equation of the form S = KD where K is a linear operator as for example in 8.3.1 p 451. In Hooke's law we thing of F = kx as being "linear", stress is linear function of the strain. I guess in fluids, the "strain" thing is the rate of deformation tensor D, so if S = KD, then your fluid model is linear. The very first "model" discussed in Part B is way down at page 474, where S is said to be a functional of Ct. Then finally on page 475 we see the first ACTUAL model which has the [ Ct-I ] form. Is this non-linear because it is not linear in D or E (except for small deformations)? Or is it non-linear because linear would mean linear in Ft but Ct is quadratic in Ft? I think Ct is the "acting" strain object, so I think it is the fact that the -I is in there in [ Ct-I ] that makes it non-linear. Question: How are Ft and D related? Answer: On page 469 we see how D and A1 are related, and we know A1 = Ct' page 463, so roughly this says that 2D = A1 = Ct' where Ct = FtTFt so we then have 2D = (dt[FtTFt]) in effect. This then is how the "rate of deformation tensor" is related to the "relative deformation tensor". We don't for example have the result D = dtFt . 8.6 Current configuration as the reference configuration (456) When I first started here, I was very confused. Having written Appendix J, it all seems very simple, but I will leave my past scribblings here between marks. ___________________________________________________________________________________ Suddenly we have a whole new Lagrange/Euler functional form world to ponder. Recall that long ago Lai talked about (see Ch 3 notes) x = x(X,t) X = x(X,t0) t0 = the reference time Lagrangian v = (X,t) Eulerian v = (x,t) Now we have some all new stuff: x = x't(x,t) x' = x't(x,τ) x = where particle is at time t t = the reference time x' = where particle is at time τ I think the analogy is this reference time and position x = x(X,t) = xt0(X,t) X = x(X,t0) x' = x't(x,τ) x = x't(x,t) Lagrangian v = (X,t) So we seem to have a mapping of all the names: reference time t0 → t reference position X → x non-reference time t → τ non-reference position x → x't current time t = t Consider a particle which has position x at time t. Where was or will be that same particle at time τ? The answer is that it is at x' = x't(x,τ). So this x' is position away from time t and position x. Consider a particle which has position X at time t0. Where was or will be that same particle at time t? The answer is that it is at x = xto(X,t). So this x is position away from time t0 and position X. So yes, I think it is all just a name change. Before we had dx = F dX ( p 86) where F was the deformation gradient. The dumbbell dX moved to dx. Now we are going to have dx' = Ft dx, it is all the same thing with the names changed. The subscript t on Ft reminds you of the reference time when you have dumbbell dx. And this will be the "relative" deformation gradient. I think it helps a little to think of τ as being in the future relative to t if we are talking forward flow. When we said x = x(X,t), this thing shows where the particle actually goes in space, so as time goes on, this is going to map out a pathline, not a streamline. In the new notation, x' = x't(x,τ) is again the pathline for the particle that starts at position x at time t. The Eulerian velocity would at some future point x' and time τ, v = (x',τ) As a derivative, this would be v = (x',τ) = dx'/dτ // he writes dxt'/dτ Example 8.6.1. (457). We have a steady state velocity field at time t. Thus, v(x') = v(x). OK, I think I am making a mountain out of a molehill here, but Lai is so silent. The example seems pretty reasonable. The idea is that "now" always means τ = t. Sometime in future or past has some other τ. ______________________________________________________________________________ Here is the picture in Appendix J and a few comments: dx = F(x,t) dX dxi = FijdXj // Lai p 105 (3.18.13) Fij = (∂xi/∂Xj) or F = (x) // Lai p 105 (3.18.4) F* = Q(t) F or F*(x*,t) = Q(t) F(x,t) // Lai p 336 (5.56.21) dx' = Ft(x,τ) dx // compare to dx = F(x,t) dX from the lower left (Ft)ij = (∂x'i/∂xj) or Ft = (x') My Appendix explains all the Subscript t relative objects, but it does not (at this time) say why this is of any practical use. 8.7 The Relative Deformation Gradient (457) Fine, we just replace dx = F dX which was referenced to time t0 and X with dx' = Ft dx which has reference t and x. So we immediately translate earlier results for Cartesian in both systems, Cylindrical in both systems, and Spherical in both systems, all from my huge effort with section 3.29 long ago. The translation rule for something like (8.7.8) is s→s' and s0 → s compared with the Section 3.29 notation. Notice that, in this entire section and the next. we never see the arguments of Ft which are Ft(x,τ). 8.8 The Relative "other tensors" (459) The relative polar decomposition is stated with its Ft = RtUt = VtRt and various relations. Then we get the left and right Cauchy-Greens which are now Bt and Ct . Remember that F-1 just means you do that s↔s' swap for all the coordinates. Then it is easy to write Ct-1 (Finger) and Bt-1(Piola). All the time, the t label indicates the reference time. If you actually GO to that time, then all tensors are unity since there is no movement from reference time to current time! [all this is in Appendix J ] Example 8.8.1. This reprises results found earlier for dumbbell dot products. Text following this example are more reprises. 8.9 The Relative C tensor in various coordinates (460) In this section Lai does some actual examples of the relative tensors, some in curvilinear coordinates, which is all based on Section 3.29 which, luckily, I did completely. A. Both x and x' Cartesian coordinates, we just translate some section 3.29 results Example 8.9.1. For a specific simple "pathline equation" (which means x' = xt'(x,τ) ) we compute first Ft and Ct, and then inverting the equation to get x = xτ(x', t) we can get Ft-1 and Ct-1. In this example, we have a simple shear velocity field v1(x2) with all else 0. We earlier found the pathline for this example, and so that pathline is then used to find Ft where k = ∂2v . This example will be used later in Example 8.16.1. B. Both x and x' Cylindricals, compute a few Cij elements, more Section 3.29 translation. C. Both x and x' Sphericals, compute a few Cij elements, more Section 3.29 translation. 8.10 "Deformation History" and the Rivlin-Ericksen Tensors (463) Here for the first time we see some relative tensor arguments, and we see Ct(x,τ) and of course the arguments are the same for ALL the subscript-t deformation related tensors. Fancy words for simple object. 1. Any of our tensors, including Ct(x,τ), can be tracked backwards in time toward τ = -∞, and thus any tensor gives a "history of the deformation". Recall that F, C and B are all called "deformation tensors", with adjective nothing, right Cauchy-Green and left Cauchy-Green. So this notion of "history" seems pretty obvious. 2. If you Taylor expand C(τ) around C(t), you get derivatives as coefficients and these get the fancy name Rivlin-Ericksen tensors and are called An. This derivative idea seems very simple. The examples are drills in computing actual tensors in different situations. No new theory here. Example 8.10.1 . Compute C and An for the little v1 = v(x2) we have had going this chapter. Example 8.10.2. For a pipe flow in cylindricals with vz = v(r), compute pathline, C and An. Example 8.10.3. For Couette flow in cylindricals with vθ = v(r), compute pathline, C and An. Example 8.10.4. For a sink flow in sphericals vr = =-a/r2, compute pathline, C and An. These are all just exercises in turning the crank, I do appreciate their presence in the book. 8.11 Relating A1 to D; a recursion formula for the AN (468) Note Added: I have done all this now in full detail in "RE tensors An.doc". Leaving notes below. There is some subtle notation in use here after equation (8.11.4). Remember that our Euler functions now all look like ct(x,τ). I think what Lai has in mind is setting x to some fixed value, call it x1. Consider then a function α(x,τ) . Suppose we set x = x1 = a constant vector, and then define a function only of τ, g(τ) ≡ α(x1,τ) => ∂τg = dτg (1) We can certainly write [∂τα(x,τ)]x=x1 = ∂τα(x1,τ) (2) because that partial derivative only hits the τ argument and ignores the first argument, so we can set x to x1 either before or after doing the derivative. Therefore, from (1) and (2) we get dτg = dτα(x1,τ) = ∂τα(x1,τ) = [∂τα(x,τ)]x=x1 (3) Setting α = (ds'2) this becomes, // note that ds'2 really means (ds')2 dτ[(ds'2)(x1,τ)] = ∂τ{(ds'2)(x1,τ)} = [∂τ{(ds'2)(x,τ)}]x=x1 (4) Meanwhile, we have (ds'2)(x1,τ) = dx Ct(x1,τ) dx = a function of τ, where dx = some constant dumbbell value where the tail end of dx lies at location x1 dτ[(ds'2)(x1,τ)] = dx [dτCt(x1,τ)] dx (5) So now write LHS(5) = RHS(4) = RHS(5) to get dτ[(ds'2)(x1,τ)] = [∂τ{(ds'2)(x,τ)}]x=x1 = dx [dτCt(x1,τ)] dx (6) and suppressing arguments and setting d = D, this becomes Dτ(ds'2) = [∂τ(ds'2)]x=x1 = dx [dτCt] dx (7) and this is p 468 equation A. The entire argument above can be repeated with ∂t → ∂tN and also with dt→ dtN. The result is then DτN(ds'2) = [∂τN(ds'2)]x=x1 = dx [dτNCt] dx (8) = // (8.11.5) where Lai does not use my value x1 but just says "xi-fixed" which means exactly the same thing. When we set τ = t (t = reference time), then (ds'2)(x1,t) = dx Ct(x1,t) dx = dx I dx = dx dx = (ds)2 So now we arrive at the last line on page 468 which is not clear to me. First, it has three typos which I have fixed. But how now do be bring in the object dt(ds)2 ? In what sense is (ds)2 a function of t ? Using the above development, you would say (ds)2 = dx dx = a constant. He wants to claim this. Start with (8) above and set τ = t wherever τ appears. So let's put the arguments back into (8) so we can see them, DτN(ds'2(x1,τ)) = [∂τN(ds'2(x,τ))]x=x1 = dx [dτNCt(x1,τ)] dx Now do an honest setting of τ = t, {DτN(ds'2(x1,τ))}τ=t = [∂τN(ds'2(x,τ))]x=x1τ=t = dx [dτNCt(x1,τ)]τ=t dx In the last term, Ct(x1,τ) is just some h(τ), so we can replace dτ by ∂τ there. Then (8.10.2) says Then using (8.10.2) this says {DτN(ds'2(x1,τ))}τ=t = [∂τN(ds'2(x,τ))]x=x1τ=t = dx AN dx But now Lai wants to claim that the LHS's of this equation are in fact DtN (ds2). That is the part I am unable to show. I agree that ds'2(x1,τ=t) = ds2. Well, consider the middle term above, [∂τN(ds'2(x,τ))]x=x1τ=t = ( ∂τN(ds'2(x1,τ))] )τ=t = ( ∂rN(ds'2(x1,r))] )r=t where I have set the dummy variable to r. Now consider from regular calculus, [drf(r)]r=t = f'(r)r=t = f'(t) = dtf(t) So I guess we then can write the above line with an extra = at the end [∂τN(ds'2(x,τ))]x=x1τ=t = ( ∂τN(ds'2(x1,τ))] )τ=t = ( ∂rN(ds'2(x1,r))] )r=t = ∂tN(ds'2(x1,t)) But again, x1 is fixed, so ds'2(x1,t) is a function only of t, so could add another = dtN(ds'2(x1,t)) . So at this point we have [∂τN(ds'2(x,τ))]x=x1τ=t = dtN(ds'2(x1,t)) = dtN (ds2) Now it is true that ds'2(x1,t) = ds2 as shown above which is dx dx, so add extra = above. Now I think we have to change our footing in order to tie back to earlier in the book. In the new footing, x = a reference point in space. But in the old footing, X = reference point and x moves in time. In this old footing, we had dx dx being a function of time, so we have ds2(t) and this derivative dtN (ds2) ≠ 0. This footing change is indeed confusing, I will just have to "buy it" for the moment. My counter argument is that in the new footing, ds2 = constant because dx = constant. But this tensor D really is an animal from the old footing world so maybe that makes it OK. Status: OK, I will accept 8.11.6 for now. Yes, when this is combined with the much earlier result relating ds2/dt to the D tensor, yes, we get 8.11.8 for A1 . Let's review the path Ft → Ct = FtTFt ∂tCt = A1 D = (v)S = rate of deformation. result is A1 = 2 D Both objects being related involve the time change for deformation. Then if you want to get into the acceleration of deformation, so to speak, you can relate A2 to dtD as in 8.11.9. Example 8.11.1. We take our spherical sink flow example, we compute v and then (v)S = D and from it A1 as in 8.11.12. Notice there is no explicit time dependence in A1 so that ∂tA1 = 0. We are then going to compute A2 using the formula 8.11.2, and then DtA1 = (A1) v if I have written it right (see top line of page 470). This is the first time in this book we have encountered an object like (A1) which is the gradient of a rank-2 tensor. This is not an object I treated in tensor doc. Let's go back to DAij/Dt = ∂Aij/∂t + ∂Aij/∂xk * ∂xk/dt = ∂Aij/∂t + [object]ijk* vk so this is the thing we need to compute [object]ijk = [A]ijk // = ∂kAij = Aij,k in Cartesian coordinates only so we have this same "reversal" of the last index, I could of course (and may) power up my tensor doc machinery to compute this object. Maybe I would treat ∂nAijk... as a general case. There is nothing mysterious about this, it is just turn the crank. Lai in fact turns this crank in his usual cases of Polar, Cylindrical and Spherical coordinates in his Appendix 8.1, which is the concluding tour de force section of the book. He also does divA at the same time, but I have already done that. Recall that Aij,k is the form of the result in Cartesian coordinates only. Looking at tensor doc, I would start with this from my examples section Tab;α ≡ ∂α Tab + Γaαn Tnb + Γbαn Tan . // and Γ = 0 for Cartesians T'ab;α ≡ ∂'α T'ab + Γ'aαn T'nb + Γ'bαn T'an . // prime means curvilinear coordinates and that is basically it! Most of my tensor doc work expresses Γ in terms of the R type object. There is of course one detail here: the above T'ab is a true contravariant tensor, whereas in Lai we always use "unit vector tensors". I fiddled on this for a while and managed to confuse myself so I moved those confused notes into a separate doc because I want to stay on message right now regarding Chapter 8. So Lai uses his appendix results to go ahead and compute (A1) and thereby to compute DtA1 and thereby to compute A2 as shown in (8.11.12). But we just computed these same A2 in Example 8.10.4 by computing Ct and then differentiating it wrt τ, and his answers found here agree with those results. Comments: This section 8.11 was a bit "rocky" for me. I get the main point, but I had trouble with the fine details. At some point I will need to clean this up. 8.12 Some tensor relations involving Ft and (v) and the polar decomposition (471) We did a lot with the various "deformation tensors", but we never did time derivatives of them. This is the famous math that makes me very uncomfortable, so I will do all the lines. (a) Pondering equations (8.12.1) We start with dx'(τ) = x't(x+dx,τ) - x't(x,τ) Even this simple equation is confusing, even after 470 pages of reading Lai, so I have to laboriously wrote out all the words. A Particle starts out at location x at time t. The position of that particle is x't(x,t) = x. Another particle is nearby at the same time t, and it has position x't(x+dx,t) = x+dx. The separation of these two particles at time t is x+dx – x = dx. Now we look at some time τ > t in the future, perhaps far in the future. The particle what was at position x is now at position x't(x,τ), wherever that might be. And the particle that was at x + dx is now at x't(x+dx,τ). Thus, at time τ, the distance between these two particles is x't(x+dx,τ) - x't(x,τ). We might call this distance dx'(x,τ). So we then have  dx't(x,τ) = x't(x+dx,τ) - x't(x,τ) so I disagree with the first equation in (8.12.1) where we see dx'(τ) because that does not show the dependence on parameters x and t. Now the general dumbbell rule goes on to say x't(x+dx,τ) - x't(x,τ) = Ft(x,τ) dx. So here then is my reading of (8.12.1) dx't(x,τ) = x't(x+dx,τ) - x't(x,τ) = Ft(x,τ) dx // (8.21.1) (b) Pondering the three equations p471 A We are supposed to apply dτ to the above line and add more stuff. Question: is there any τ dependence "through" the x argument of dx't(x,τ)? That's like asking in the old world if there was any t dependence of X. We would have said then "no, because X is a fixed material location which is our reference. ". So we must say the same thing here, "no, because x is a fixed material location which is our reference.". Therefore we can say dτ [dx't(x,τ)] = dτ [x't(x+dx,τ)] - dτ [x't(x,τ)] = dτ [Ft(x,τ) dx] (1) Now we pause to ask about dτ [x't(x,τ)] . Since x is again a reference, this is dτ [x't(x,τ)] =[ x't(x,τ+dτ) - x't(x,τ)]/dτ = v'(x,τ) (2) Notice in passing the convention that Lai only puts that t subscript on the position vector, not other vectors like velocity. I am not sure why that is. But OK, and then using (2) twice in (1) we get dτ [dx't(x,τ)] = v'(x+dx,τ) - v'(x,τ) = (dτFt(x,τ)) dx (3) So far then we have three pieces of equations A. The missing piece is this dv'i ≡ v'i(x+dx,τ) - v'i(x,τ) = ∂v'i(x,τ)/∂xk * dxk = ∂v'i/∂xk * dxk (4) This dv'i is only non-zero due to the first arguments of the two terms being different, and that is what justifies the expression on the right. Now we need to think of ∂v'i/∂xk as a matrix (v')ik and then we are allowed to write the above as dv'i ≡ v'i(x+dx,τ) - v'i(x,τ) = ∂v'i(x,τ)/∂xk * dxk = ∂v'i/∂xk * dxk = (xv')ik dxk = [(xv') dx]i Therefore removing component index i we get dτ [dx't(x,τ)] ≡ v'(x+dx,τ) - v'(x,τ) = [(xv') dx] = (dτFt(x,τ)) dx // p 471 A The last equality above says matrix = matrix, or, re-displaying the arguments of v'. dτFt(x,τ) = x[v'(x,τ)] // (8.12.2) Now I arrive at another "standard point of confusion". Suppose we evaluate both sides of the above equation at τ = t. We must then write [dτFt(x,τ)] τ=t = { x[v'(x,τ)] }τ=t On the RHS, we can set τ = t either before or after applying x, so we can write RHS = x[v'(x,t)] . But since we are at the reference place and time, we want to say that v'(x,t) = v(x,t) where we have a convention I think of representing any object at x and t without a prime. We do this for x, for example. This may be something I missed. In this case, the above says [dτFt(x,τ)] τ=t = x[v(x,t)] Now we examine the LHS. I had this situation earlier, and I said that [ f'(x)]x=x1 = f '(x1) in our usual notation of single variable calculus. So [f '(τ)]τ=t = f '(t). This is the derivative evaluated at t. Now I guess you might write this as [ df(τ)/dτ]τ=t. But you could also write it as f '(t) = df/dt and be done with it. But this would be dtf. Therefore, thinking of the LHS Ft as f, we have dtFt and we are done with it. Then we have dtFt(x,t) = (x[v(x,t)]) = (xv) // (8.12.3) I guess any time you have a dτ acting on something and then you set τ = t, you can do what I just did. Comment: We now have in 8.12.3 some information on the total time derivative of one of the deformation tensors, namely Ft. We are able to relate it to the (xv) tensor. Ft has no dimensions, so both sides of the above are sec-1. Let's go back now to an earlier equation which I am happy now to write as dx'(τ) = x't(x+dx,τ) - x't(x,τ) = (xx't(x,τ)) dx = Ft(x,τ) dx // (8.7.1,2) so that Ft(x,τ) = (xx't(x,τ)) Now what happens here if we evaluate at τ = t? We get Ft(x,t) = (xx't(x,t)) but since we are at reference x and t, the LHS is just the unity matrix. And RHS = (xx) = unity as well. So we don't have anything "interesting" at this no-derivative level. But we could write this pair Ft(x,τ) = (xx't(x,τ)) dτFt(x,τ) = (x[v'(x,τ)]) or Ft = (xx't) at x,τ dτFt = (xv') so there is some parallel structure here. (c) the symmetry question. I am happy with (8.12.6), but how does Lai arrive at his symmetry claims here? As I review polar decomposition.doc in the Lai support folder, I see that in that decomposition, U is always symmetric! I forgot that fact. So consider Qab = dtUab = dtUba = Qba so of course dtUab is symmetric. now if you write for matrices that A = B + C and you know that C is symmetric, I think you then know that B is antisymmetric. Proof: You know you can decompose A = As+ Aa in a unique manner: Asab = (Aab + Aba)/2 etc Therefore if A = B + C and C is symmetric, it must be that C = As and therefore B = Aa QED. (d) finish I have derived (8.12.5) and I want to continue down this page 471. We now think of (8.12.3)'s LHS in the sense I had it above. LHS = [dτFt(x,τ)] τ=t The object here is (8.12.5) if you evaluate each term at τ = t, and THAT is where (8.12.6) comes from. Note that U and R become unity when τ = t, so this (8.12.6) has a lot in it. From (c) above I now agree with the symmetry comments . But I do know that D + W = v because we have used that a zillion times. And so finally I have 8.12.7 derived. 8.13 Change of Frame Stuff (471) I now have to go reread previous sections on this subject starting on p 334 in Chapter 5. A thorough review is essential here! // I have now done this, with lots of notes taken in "essay on covariance of...". So I am now ready to start reading this section of Lai. We are going to see the same discussion but reframed in terms of the various "relative" tensors. With Appendix J I think I am now happy on this change of frame stuff, BUT I have to redo things because things are still not optimally presented. 8.14 Change of Frame for the A tensors (474) This tiny section just shows that the A tensors are objective relative to Q(t), nothing more or less. 8.15 Incompressible Simple Fluid (474) OK, we are starting to make sense. From Appendix J, we found that Ct was objective (and Bt was not). Therefore, a covariant model for some fluid might be (8.15.2) which is somewhat similar to our Maxwell fluid model (constitutive relation), S(t) = 2 !Syntax Error, I dt' [ (μ/λ)e-(t-t')/λ] D(t') where all of history plays a role. Both S and D are objective, though I am not sure Lai has said so, but that does make the above equation covariant. Now in this section, we could imagine a similar thing S(t) = 2 !Syntax Error, I dτ [some kernel] Ct(τ) where the weight function or kernel I guess should be a rotational scalar. We know that Ct is objective and so is S, so we have covariance for a model of this form. But Lai wants to be even more general and in (8.15.2) he just says that S is a "functional of the history of Ct" which is fine by me. I think he means a linear functional, but maybe not. In Stak you could have either kind of functional. The word "simple" means that the functional is a functional of Ct which means of Ft in effect, and not of the gradient of Ft or multiple gradients thereof! He requires that the functional of Ct be objective in (8.15.5). Just as we did with f = f* in our isotropic example in App J, here we assume H = H* and again we are isotropic. So in our model with isotropic, we have to make sure that (8.15.5) is satisfied, which is again similar to our App J thing, so I expect soon to see lower powers of Ct appearing somewhere. It is obvious that our linear Maxwell fluid fits into this general model, as does Newtonian. At this point come some comments I will need to return to later. // Comments now make sense. 8.16 Two specific models which reduce to Maxwell for small deformation (475) Here we go back to our solution form for S + λ∂tS = 2μD which was S = 2!Syntax Error, Ids f(s) E(t-s) in the case of infinitesimal displacement, and this is now (8.16.1). He shows (and I follow) that E can be related to our new objective friend Ct as shown, so we get two "models" on page 475. They are both the same in the case that Ct is such that E is small. Remember that f(s) = φ' was called a "memory function". Now, why are these two new models non-linear? Well consider Y = CX // linear Y = CX + D // non-linear so I guess it is the presence of two terms that makes these models non-linear, the way a Galilean transformation is non-linear. By the way, here Lai uses "frame indifferent" exactly to mean covariant, so he is applying this to an equation, not to a tensor, so I need to repair things. // Repair done. Example 8.16.1. We have our usual super-simple shear velocity field v1 = kx2. This fits into the more general form of Example 8.9.1 (p 461) and so we already know Ft and Ct and Ct-1. We then plug Ct into our first "model" (8.16.3) and we thereby obtain constitutive relations for S in terms of an integral of the memory weight function f(s) and no other unknowns. For example, we find that S12 = -k !Syntax Error, Ids f(s) s ≡ τ(k) = shear stress function (makes sense) Some of the other elements of Sij are also computed on page 476. The differences of diagonal elements are given special σ names. Notice that for some f(s), the various ds integrals are just numbers. Back in our Newtonian flow with this same example, we found T12 = μ k where μ was the viscosity. Here in our fancier flow, we are finding that T12 = S12 = [!Syntax Error, Ids f(s) s] k . So comparison shows that the integral over ds, a number, is in fact the viscosity in this fancier model, called the apparent viscosity. In this particular model, τ(k) is linear in k, but the viscosity is still a constant, just as it was a constant in the Newtonian fluid. He then asks us in a problem to repeat the above work for the other of the two models. The results are exactly the same with f1→ f2. Now recall from our discrete mixture Maxwell fluid we had φ(t) = Σn=1N (μn/λn) e-t/λn => f(t) = – Σn=1N (μn/λn2) e-t/λn and for single component then f(t) = - (μ/λ2)e-t/λ. What happens if we use that for f(s) here in our two new models? We are cribbing this form of the memory function from the Maxwell linear fluid model and applying it here to our new Ct models. Well, Tanner and Simmons (I think 1967 from below) have a certain "network model" using this f(s), but they cut it off in some way I don't care about right now. So basically this section has introduced two non-linear "new models" both of which reduce to the linear Maxwell fluid model for small displacements, and both of which are "covariant". [ seems similar to my covariant method of finding differential operators.] The idea is that maybe these models might describe some actual fluids in the real world. Tanner and Simmons made use of one of these models. I would certainly call this "an advanced topic", but for me it is another example of constructing a covariant model! I like how it reduces to a known simple model in a limit. Sort of like relativity reduces to Newtonian in a limit. Notice that these two new models are examples of functionals of Ct, but they are not the form I conjectured in my Section 8.15 notes above. Example 8.16.2. A routine exercise I already did Example 8.16.3. In App J I review the case of T = f(B) where B was objective, and I showed that if f is polynomial in B, then it has to be at most second degree with objective scalar coefficients. Here that derivation is repeated for a generic F(A) which is some polynomial in some symmetric matrix A. If F is an isotropic function, then we know that the coefficients are scalars. He then says we can apply this in the case F(Ct) since Ct is symmetric. Again, any function of Ct which can be fitted by a polynomial and which is isotropic can only be a quadratic in Ct with scalar coefficients which are the invariants of Ct. Doubtless he will be using this fact soon (as in, next section). Notice that he uses the form I, A, A-1 as the most general form, but when he applies this to A = Ct, he omits the I term with no comment! Suppose an I term were present? That would imply an "extra stress S" that had the form scalar * δij and such a term would by like Tij = scalar function * δij and I guess the idea is that we would just absorb such a term into the pressure term Tij = -p δij + Sij. I don't think he has ever noted that pressure p is a scalar, I guess that is obvious. So that then is the reason for just the two terms appearing in 8.16.20. 8.17 Single-integral non-linear constitutive equations (478) For models in this class, the specific form of the "simple" model is shown in (8.17.1). Why is this model non-linear? Answer : Well, the model first term is S = FCt where F is an integral operator. The first term in 8.17.1 is linear because then, for example, F(αCt) = αCt . The second term even by itself is non-linear since S = G Ct-1, so both terms present is non-linear. In the analogy to Hooke's law F = kx being linear, the object playing the role of x is Ct and the object playing the role of F is S. The BKZ Fluid (1963). We start with the general form 8.17.1 whose generalness was explained two paragraphs above in these notes. The two scalar coefficients are then written as derivatives of a potential U (similar I think to the Airy functions idea). We then have a specific expression for U which includes these ingredients: scalar invariants of Ct, c(s), β(s) where these two add up to φ(s)/s where φ(s) is the kernel function which appears in the S = φD relation 8.3.1, and later in the S = 2φE relation. Lai now considers in this BKZ context the usual simple shear flow situation v1= kx2 for which we computed Ct and Ct-1 earlier in the book, so this lets us then compute the invariants I1 and I2. What happened to I3? For incompressible, Lai claims det(Ct) = 1 and this follows from det(Ft) = 0 and this goes back to the idea that det(F) is the Jacobian of our transformation, and if the volume does not change, det(F) = 1, as on page 130 and as in my tensor doc. That is why only I1 and I2 appear in the general formulas. OK, now look at the form 8.17.3 for U. Right at that point we can compute the two derivatives to get f1 and f2 but for some reason Lai does not do that at that point. So I have Maple do these right here: where I leave the dots off b and c. He has computed I1+I2+3 and I1+15. I have verified his results and he lost a factor of 2 in the last term as you see above. I go on to compute so he has quietly assumed that t = 0 in 8.17.1 so we are computing S12(t=0) which is fine. In the last expression above, the terms with the 48 cancel, and the b terms add, so the result according to me is S12 = 18bks/(9+2k2s2) + 2cks = 2k [ 9bs/(9+2k2s2) + cs ] which agrees except I get a minus sign overall which arises from Ct(t-s) = Ct(-s) = +ks for the C12. Next, we have μ = S12/k as on page 476 and then I agree with his μapp except for sign. But 8.17.10 which I agree with says that μ is + integral of c(s) and probably c(s) is positive. So I will have a complicated errata report here. Review of BKZ: For some unknown reason, the f1,2 memory functions are derived from some kind of potential U which has a strange form shown. If you then "do the math" for the usual simple shear flow situation, you end up with S12 and μapp as shown in 8.17.9 and μ∞ as he shows. The key idea here is that for the first time, we have μapp = μapp(k), so viscosity varies with the shear speed parameter k. Example 8.17.1. We are now going to apply the BKZ theory to synovial fluid. The claim is that at high shear rates (large k), synovial fluid basically has no viscosity at all, so from 8.17.10 we set c = 0 and get corresponding μ∞ = 0. This my sign error noted plays not role in this term at least. Next, he writes the function μapp(k) as an integral of H(τ) using the definition of φ, and I agree with all this part (a) stuff. Now earlier we obtained H(τ) for our three age people in a table on page 456. So we can insert H(τ) into our equation (ii) here and numerically compute μapp. Thus, the BKZ model gives the curves shown for the apparent viscosity of synovial fluid of people of three ages. For the young at heart, you have more viscosity at low shear rates than for the old. Lai wrote a paper on this subject, reference on page 455. So to recap, we have a general-form covariant equation 8.17.1 for an incompressible fluid. It is claimed that this is the most general possible "simple", "single integral" and incompressible fluid equation. Then the BKZ model somehow comes up with some f1 and f2 functions using theory that Lai does not present, and the result is an expression for apparent viscosity and it is found to be a function of shear rate k. I thihk Lai has done a nice job here, one page on the BKZ theory, and one page applying it to synovial fluid. 8.18 Differential-type constitutive equations for incompressible fluid (481) Instead of saying that S = integral(Ct and Ct-1) as in 8.17.1, we want now to say S = f(Ai). The idea is that maybe we can approximate Ct for times close to time t using the first N derivatives. Well, my first question is: do these derivatives have information about Ct-1 as well as Ct? CtCt-1 = 1 => dt(CtCt-1) = 0 => (dtCt) Ct-1 + Ct(dtCt-1) = 0 => (dtCt-1) = – (dtCt)/Ct2 // for first derivative Then what about the next derivative? (dtCt) Ct-1 + Ct(dtCt-1) = 0 => (d2tCt) Ct-1 + (dtCt) (dtCt-1) + (dtCt) (dtCt-1) + Ct(d2tCt-1) = 0 => (d2tCt) Ct-1 + 2(dtCt) (dtCt-1) + Ct(d2tCt-1) = 0 So OK, then we can relate d2tCt-1 to all the other stuff. So the answer is this: knowledge of Ct expanded near τ = t does in fact include knowledge of Ct-1expanded about the same point. Lai might have commented on this but kept quiet. The details are a bit messy, I could do something if I wanted. So in theory if functions are smooth, then S = f(Ai) with all derivatives is as good as 8.17.1. How now does "incompressible" enter this model? Before it just means I3 = 1 so we could forget about I3 in 8.17.1. Well he showed on page 469 that A1 = 2D, so incompressible means Δ = tr(D) = 0, so here it means that tr(A1) = 0, fine. We then get a little cookbook of models: (1) Keep the first N of the Ai, that is Rivlin-Ericksen fluid of complexity N (2) Keep A1, A12 and A2 , that is a second order fluid. Newtonian has only A1= 2D for incompressible, so this is apparently "the next step up". Not obvious why these two extra terms are used, but I guess they both have the same "smallness" in ε = (τ-t). Example 8.18.1. Back to our usual shearing example. He computes A1, A12 and A2. He gets the corresponding expression for Tij. We see here normal stress adjustments away from just -p. He then quotes the two σi objects. You can do an experiment and measure things and deduce the three constants which are μ1, μ2 and μ3. Since T12 = μ1k = S12, this second order model has constant viscosity μ1 so that is one of the three constants. Recall 2μD in the Newtonian equation. Example 8.18.2. Again the simple shearing flow. Lai shows that Ai for i ≥ 3 are all zero in this simple situation, though I would have to check why that is. He then computes all the lower Ai and their powers, and allows for cross terms like A2A12. So here he is saying that you might think of 8.18.3 as being a polynomial in the Ai. He does a term A13 so it is not true that powers only up to 2. I guess basically he computes all possible non-zero Ai poly-like terms and then gives each one a coefficient. There are then eight different constants μi. Certain things are even and odd in k. 8.19 Objective Rate of Stress (483) OK, I am familiar with this section from appendix J. The first observation is that dtT is not covariant, if T transforms as a rank 2 tensor when the transforming rotations Q(t) are time dependent. Recall that you added a note showing that v = dx/dt is not a vector if R= R(t) in tensor doc. So you cannot have dtT sitting in any covariant equation. We are led then to the Jaumann and two Oldroyd alternates to T, call them T', so that then dtT' is in fact covariant and transforms as a rank-2 tensor. Example 8.19.1. Here we start with a simple stress T11 = constant, other = 0. We then observe this from a rotating frame Q(t) = Rz(ωt) and we get T* as shown. He then computes dtT* and we of course find that in the * frame, dtT* ≠ 0, whereas in the original frame dtT = 0. So the point here is that the rate of stress change differs when measured in a fixed frame or in a rotating frame, the quantity dtT cannot be a ran-2 tensor. I think now he wants to show that the "corotational stress" is the same in these two frames. He is going to sue 8.19.10 to get this corotational stress, but that requires knowledge of the spin tensor W. He uses a strange equation we had earlier in 5.56.20 which allows him to compute (*v*) in the starred frame. I think in the rest frame since we have a static situation with uniaxial stress, v = 0 and of course v = 0 everywhere. Then once you compute (*v*), you know W from the antisym part, as in (v). Then we compute that the Jaumann tensor is in fact 0 in the rotating * frame. Of course the Q(t) we use there is the same as the Q(t) relating the frames, so we are just undoing the rotation, it seems a bit circular. But interesting to see how it works going through the v pathway with W. We know the Jaumann stress is a rank-2 tensor, so if it is 0 in one frame, it has to be zero in the other frame. 8.20 Rate-type constitutive equations (487) The game here is to play with all our covariant derivatives and use them as building blocks in fancier equations. The "general idea" is 8.20.2 where the overstar means any of the three covariant derivative types. Two stars means a second time derivative. These extra terms "could be present" from a covariant point of view for some fluid. A. The convected Maxwell fluid. This is a sort of "correction" to our original Maxwell fluid which we recall had a dS/dt term in the equation, which we now realize is non-covariant. So maybe that theory only works for slow-flowing fluids or some such. We correct it by using the Jaumann time derivative of S as shown in 8.20.4. In Example 8.20.1 we do our usual simple shear flow and we compute the Jaumann thing as in 8.20.7. The big deal is that if we then compute S12 we get a viscosity that varies with k (I think we saw this only once before). For small (kλ) we get the result of the regular Maxwell fluid where μ is the constant viscosity. B. The Jeffrey Fluid. This is same as Maxwell, but we add in a time derivative on the RHS, so the old λ is now λ1 and the new term has λ2. In Example 8.20.2, since we already know W, we compute and as shown. We then use the right equation in 8.20.12 to solve for S including S12, and once again of course we have μapparent being a function of k. Simple Maxwell is recovered now when kλ1 is small and when λ1= λ2. So this model allows some "extra normal stress". C. Oldroyd 3-constant fluid (fluid A). Exact same as Jeffrey, but we use Oldroyd upper instead of Jaumann So think of the triangle base as lower when down, and upper when up. It turns out that this fluid A thing gives a constant μ. He should have made this a problem perhaps. D. Oldroyd 4-constant fluid. We stick with Oldroyd's upper convected stuff, we add that extra term with λ2 as was done in Jeffrey above, and we add a totally new term μ0(trS)D. This is patently covariant, and seems to add a viscosity which increases as S increases, ie, when there is more S and thus more deformation. In this model μ approaches a constant at both low k and high k. This one is made a problem. Part C: Viscometric flow for incompressible simple fluid (491) 8.21 Viscometric Flow (491) I suspect that for even very exotic fluids, the "space" is filled all the time with fluid, so to first order approximation the fluid is incompressible. That is why almost all the fancy models we are examining in this chapter are incompressible models. They are of course simpler models for that reason. So what is viscometric flow? First ingredient is that we keep only the A1 and A2 terms in the Ct expansion as in 8.21.1. The second item seems very strange to me. There must be some frame in which the A1 and A2 have the very simple forms shown. This frame is allowed to change with position. It is easy to show, given the simple N matrix shown, that 8.21.3 is true. At first blush, it seems hard to imagine that you would ever get this simple form, but now he is going to show that it happens all the time. Example 8.21.1. Not the simple shear flow, but a generalization of it called unidirectional flow, v1= v(x2). However, back on page 464 we already computed A1 and A2 for this kind of flow, and they have exactly the form required to be called "viscometric flow". So you might say this is just "unidirectional flow". Those basis vectors ni are just the ei and are the same everywhere. Example 8.21.2. Cylindricals, and flow is vz = v(r), so this is similar to saying v1 = v(x2) but of course is not the same. Now we have to compute A1,2 in cylindricals. But this was done as an exercise on page 465 which is replicated into 8.21.7, where k(r) = dv(r)/dr. . These Ai do not have the right form in the er, eθ, ez basis in which they are computed, but a trivial reordering so that 1,2,3 = z,r,θ gets it into standard form. Therefore our assumed velocity field is "in voscometric form". It happens now that the matrix elements are functions of position. Example 8.21.3. This is the Couette concentric cylinders situation where vθ = v(r). We compute the Ai for this case as well on page 466 where now k(r) = dv(r)/dr - v/r . This time we do 1,2,3 = θ,r,z and we are then in standard viscometric form as in 8.21.13. So three very basic flows all fit into the viscometric pigeonhole. 8.22 Form of Tij for Viscometric Flow (493) Lai first restates things, then uses the phrase "the objectivity condition" for "the covariance condition". Lai next defines Q to be a simple z reflection matrix. I think he just shows that our special T is invariant under Q. The result is that for this viscometric flow, a lot of the Tij are 0. With our usual σ definitions as in 8.22.15 one can summarize the Tij of viscometric flow as in 8.22.16. The problem in this case is reduced to the three functions τ(k), σ1,2(k). So you do one experiment, find these functions, and then you know them for any other viscometric flow. τ is odd in k, the σi are even. Newtonian has τ = μk and the others 0 (that is, the Tii are not altered). You can then try power series in small k for the three functions to model a fluid. 8.23 Channel Flow (495) Scene is v1= v(x2) between static parallel plates which we know is viscometric. This flow between such parallel plates (separation is h) is called "channel flow". A new k is defined here, namely k = dv(x2)/dx2 so whatever v is, this is the slope in the x2 direction. v = kx2 would be a special case with a constant k. I think Lai makes an ansatz here which is that the Sij are functions only of this k thing. That was the case for the v = kx2 flow. So we know the general viscometric forms in 8.23.3. Next, we seek a static solution for this problem, which means EOM says ∂jTij= 0. Note k = k(x2). We get some simple ODE's for p, τ, S22. It is trivial then to obtain 8.23.5 which then says that -∂1p = f, a constant. This is the constant pressure gradient in anywhere between the plates, so the pressure is dropping off linearly as we move to the right in our little flat pipe. I have verified all equations on page 496-7 top. The problem is to concoct an experiment that lets you measure τ(k) for the fluid. We find that τ(k) = -f x2 so k = τ-1(-fx2) which we know is odd, so k = - τ-1(fx2) = - γ(fx2) where γ is a new name for function τ-1. The plan is first to find the total flow Q (per unit cross section I guess) and we get 8.23.14. Fiddling gives 8.23.17 and this lays out the required experiment. For a set of pressure gradient values fi measure Q. Then using this data, plot f2Q versus f. Differentiate this plot to get a new plot which is ∂f(f2Q) for different values fi. Then you get points for function γ(fh/2) asd shown in 8.23.17. So, if we set up this channel flow experiment, we have a way to measure function γ(S) = τ-1(S) and from that we deduce τ(k). This is one of the three viscometric functions of interest. Example 8.23.1. We want to use the model just described and reduce the results to the Newtonian fluid limit and see if we replicate earlier results of page 370. We start with knowledge that τ(k) = μk and at once we know that γ(S) = S/μ. So we compute Q from 8.23.14 and lo and behold the result agrees with that found on page 370. This is just a good check of signs and factors of 2, etc. 8.24 Couette Flow (497) The channel flow is perhaps impractical since you have to make it very wide. We should be able to construct a concentric cylinders Couette experiment as an alternate method of finding γ(S), and that is the goal of this section, where things are going to be more complicated. Right off the bat we know the A1,2 matrices from p 493 (we just did this). Lai replaces vθ(r) by rω(r) so ω(r) is our function of interest. We showed Couette is viscometric. The general form 8.22.16 is supposedly a covariant equation, so we can write it in terms of the cylindrical unit vectors to get 8.24.4 based on 8.22.14, I agree. Again, we just want to find our three functions σ1,2 and τ. As before, we write the equations of motion, but we do this in cylindricals as shown on page 170, a long time ago! For example, ∂θTrθ = 0 because Trθ = τ(k(r)) and does not depend on θ. The middle of our three equations tells us that Trθ = constant/r2 and constant = M/2π where M is the torque per unit height applied to the fluid from the end caps of the cylinder. We apply a constant torque onto this Couette apparatus to make it go. It just turns out that M is the same at all r. The equations fly and we end up with 8.24.16 and 17. Lai notes that if we knew γ(S) ahead of time, we could use these equations to find ω(r) assuming we are driving R1 inner at some fixed Ω1, and then of course we know ω(R2) which is Ω2. But we want to work backwards somehow to deduce γ(S). But first, a little digression. We now use the first of the three EOM as shown in 8.24.18. Notice how σ1 naturally appears due to the form of this EOM, now maybe we know why σ1 is a quantity of interest. Lai then shows the strange fact that, since σ1 is positive in all known materials, the second term in 8.24.20 can outweigh the first, causing the radial force to be larger on the inner cylinder, whereas you would think centrifugal would make the opposite be true. Now on page 500 we resume the goal of deducing γ(S) from this experiment. The first task is to plot Γ(S1) versus S1 , where S1 = M/(2πR12) and Γ is the difference of γ's shown. We have to do this plot using 8.24.23. This means that you use lots of different M values, and for each you measure ΔΩ. You have a meter that measures M. In practice you apply some M, and this gives some Ω2 hence ΔΩ. It seems reasonable that this gives you a plot of Γ(S1) and you don't have to do anything unpleasant like vary Ri. Lai then shows that γ(S) is given by 8.24.26 so you can use your Γ(S) data to get γ(S). S detail here is that it is assumed that γ(0) = 0, and I think that follows from τ(0) = 0 since there is no stress when there is no shear action at some radius. So that is the answer to how you do an experiment to get γ. If you use a very small gap so ΔR << R, we get a simpler result 8.24.28. Again, you apply M, measure ΔΩ and work out points on this graph. Here is a commercial Couette machine. 1. Purpose & characteristics Rigchina produces a range of true Couette coaxial cylinder rotational viscometers. The test fluid is contained in the annular space or shear gap between the cylinders. Rotation of the outer cylinder at known velocities is accomplished through precision gearing. The viscous drag exerted by the fluid creates a torque on the inner cylinder or bob. This torque is transmitted to a precision spring where its deflection is measured and then related to the test conditions and instrument constants. This system permits the true simulation of most significant flow process conditions encountered in industrial processing. Direct Indicating Viscometers combine accuracy with simplicity of design, and are recommended for evaluating materials that are Bingham plastics. These instruments are equipped with factory installed R1 Rotor Sleeve, B1 Bob, F1 Torsion Spring, and a stainless steel sample cup for testing according to American Petroleum Institute Specification RP 13B. Other rotor-bob combinations and/or torsion springs can be substituted to extend the torque measuring range or to increase the sensitivity of the torque measurement. Shear stress is read directly from a calibrated scale. Plastic viscosity and yield point of a fluid can be determined easily by making two simple subtractions from the observed data when the instrument is used with the R1-B1 combination and the standard F1 torsion spring. It applies to viscosity measurement of drilling fluid in all the big oil fields, scientific research institutions and laboratories and it also applies to geology, chemical industry, coal, wine making and etc. Appendix 8.1. A. Polar coordinates. B. Cylindrical and Sphericals. Finally Lai is forced to use the affine connection Γ, but he calls is Γijk with all indices down. The rank 2 tensor of interest is T. He does things "his way" as I think I outlined in one place in tensor doc (in fact, right in the v section!). The solution for cylindricals is stated by Lai in the table on page 504. As shown in separate notes, I think ΓLai might not be the same as the real affine connection because he is dealing with unit basis vectors. But OK, eventually I could verify everything in this appendix, but for now I will just let it be. It's not like Lai is likely to apply this to something critically important before the book ends. Task: I could try to work up (T) as an extension to my tensor doc work, of course valid in all coordinate systems, then I would have to verify this against the massive Lai results on page 504. This could be a big job, but maybe not. I would rather do meta notes for Chapter 8 instead.