Macromolecular Hydrodynamics Quiz Solutions. (i) To start, we recognize the following relationships on the stress and strain

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1 Question 1 i To start, we recognize the following relationships on the stress and strain γ = γ k + γ 2 1 τ = G k γ k + μ k γ k = μ 2 γ 2 Therefore, the following relationships are also true γ = γ k + γ 2 2 τ = G k γ k + μ k γ k = μ 2 γ 2 By substituting the results for strain in Eq. 1, γ k = γ γ 2 into our stress equation, we get τ = G k γ γ 2+μ k γ γ 2 3 Furthermore, by substituting the result γ 2 = τ/μ 2 we have τ τ = G k γ + μ k γ τ 4 μ2 μ s We can rearrange this result and we obtain τ + 1+ μ 2 λkτ = μ 2 γ + μ 2 γ 5 where μ k /G k. The Jeffreys model is given by μ k τ + τ =μ m + μ s γ + μ s γ 6 So we can map the parameters from the new model to the Jeffreys model parameters by = 1+ μ 2 μ k λk μ m + μ s = μ 7 2 μ s = μ 2 which also gives where G m μ m /. μ 2 = μ m + μ s μ k = μ s 1+ μs μ m μ = λ s m μ m+μ s G k = G m 1+ μs μ m 2 8 ii We can integrate the Jeffreys model by using an integrating factor as we discussed in class. Hence we take Eq. 6 and replace the left hand side by } { } t t {τ exp = exp μ m + μ s γ + μ s γ 9 t Integrating this result from the appropriate limits, we have 1 Solution by TJO & VS

2 So that we have t τt exp = 1 t { } t exp μ m + μ s γt + μ s γ t dt 10 1 t τt = λm exp t { } t μ m + μ s γt + μ s γ t dt 11 λ m Next, we can eliminate the terms with γ by using integration by parts on the term with γ. 1 t τt = λm exp t t μ m +μ s γt dt +exp t t μ s γt which, provided γ is finite at t =, reduces to 1 t τt = λm exp t =t t = 1 t exp t t μsγ t dt 12 t t μmγ t dt + μ s γ t 13 For a cessation of steady shear test, we have γ t =γ 0 1 Ht = γ0 H t, where Ht is the Heaviside function. We take Eq. 13 and substitute in the strain rate function to obtain τt = 1 t exp t t μmγ 0H t dt + μ s γ 0H t τt = 1 0 exp t t μ m γ 0 dt + μ s γ 0 H t 14 τt = exp t t μ m γ 0 + μ s γ 0H t In dimensionless form, this result is τt = exp μ m γ 0 t t + βh t 15 λ where β μ s /μ m. So we see that the initial stress is τt =0 =μ m + μ s γ, and that the stress instantaneously drops to τt =0 + =μ m γ, since the solvent no longer contributes to the stress once the shearing is stopped. iii To determine the complex modulus G ω it is convenient to use complex variables expiωt = cosωt + i sinωt, where i 1. Note, that ultimately, we are concerned only with the real part of this complex variable. So we have γt =Re { } iγ 0 expiωt γ t =Re { } { γ 0 ω expiωt 16 γ t =Re iγ 0 ω 2 } expiωt { We also assume that the stress will take the form τt =Re G ω+ig } ω iγ 0 expiωt = Re { ig ω+g ω γ 0 expiωt }, or equivalently τt =G ωγ 0 sinωt+g ωγ 0 cosωt, when only the real part is considered. When we substitute these results in to Eq. 6 and ignore the Real operator Re {} which can be taken as implicit, we have 0 Integration by parts is uv x=b = x=a m R b u dv dx + R b du vdx. a dx a dx 2 Solution by TJO & VS

3 τ t μm γ β =0.1 β =0.05 β = t/ ig ω+g ωγ 0 expiωt+iλω ig ω+g ωγ 0 expiωt =μ m +μ s γ 0 ω expiωt+iλμsγ 0 ω expiωt 17 where we are writing λ = for simplicity. Simplifying, we have which is ig ω+g ω 1 + iλω =μ m + μ s ω + iλμ s ω 2 18 μ m + μ 2 ig s ω + iλμ s ω ω+g ω = 1+iλω To obtain a real number in the denominator, we multiply the denominator and the numerator by the complex conjugate of the denominator 1 iλω Simplifying, we have ig ω+g ω = μ m + μsω1 iλω+iλμ s ω 2 1 iλω 1+λω 2 20 μ ig ω+g ω = m λλω 2 i 1+λω 2 + μ mλλω + μ sω 21 1+λω 2 Recognizing that μ m = G m,wehave λω G 2 m 1+λω λω 2 G ω = G λω 22 ω =G m μ Again, there is a viscous and a viscoelastic contribution to G only. In dimensionless form these results are G ω G m = λω2 1+λω 2 G ω + 1+λω 2 s ω G m = λω 23 βλω 3 Solution by TJO & VS

4 The complex modulus is defined as G ω ig ω+g ω = G 2 ω+g 2 ω, so we have the following plots. See overleaf. iv For a creep compliance test, we instantly apply a constant stress of τ 0 for all time t>0. Since we now wish to solve for the strain as a function of time, neither the differential form of the governing equation for the mechanical model given in Eq. 6 nor the integral form given in Eq. 13 is very convenient. Instead, it is useful to reconsider the original equations for stress and strain given in Eq. 1 for our new model, because strains add linearly for elements in series. We can solve for the two strains independently and then add them together to determine the total strain in time. The initial condition for the solvent strain is γ 2 0=0. Solving for the solvent, we have Hence γ 2 = τ 0 μ t. s For the viscoelastic element, we have τt =τ 0 Ht =μ 2 γ 2 24 τt =τ 0 Ht =G k γ k + μ k γ k 25 This is a first order, ordinary differential equation in γ with a constant on the left side for all t>0. We can solve this equation to obtain 1 } { τ γ k t = 0 G k 1 exp t 26 This result is consistent with the initial condition at t = 0 that γ k = 0 and γ k = τ 0 /μ k. The total strain is therefore τ0 γ t = t + τ { } 0 t 1 exp 27 μs G k The compliance is defined Jt γt/τ 0, and for a creep test with the Jeffreys model we obtain Jt = 1 t + 1 } t {1 exp μ s G k The normalized compliance is defined as Jt γt/τ 0, and for the creep test we can thus write { } 1 t t JtG k = + 1 exp 29 β 28 v To solve for the start up of this uniaxial extensional flow, we use Eq. 13 in tensorial form to obtain three evolution equations for the three normal stresses. τ xx t =τ yy t = 1 t exp t t μ m ε 0 Ht dt μ s εht τ zz t = 1 t exp 30 t t λ 2μ m ε 0Ht dt +2μ s ε 0Ht For times t>0, we have 1 Eq. 25 can be solved with an integrating factor by t γk exp ` t = 1 R t exp ` t τ0 Ht dt. G k m 4 Solution by TJO & VS

5 G ω Gk β =0.1 β =0.05 β = λω G ω Gk β =0.1 β =0.05 β = λω G ω Gk 10 1 β =0.1 β =0.05 β = λω Solution by TJO & VS

6 J tgk β =0.1 β =0.05 β = t/λ τ xx t =τ yy t = 1 exp t μ m ε 0 μ s ε 0 31 τ zz t = 1 exp t 2μ m ε 0 +2μsε 0 Hence the stress in the viscous solvent adapts immediately to the imposed extensional flow, just like a Newtonian fluid, while the stress in the viscoelastic component of the model grows toward steady state with an exponential decay on the time scale. The extensional viscosity is given by λm τ η + E zz τ xx = 1 exp ε 0 t 3μ m +3μ s 32 So at long times the Trouton ratio is always Tr μ +μ = 3, and so this model does not give rise to extensional strain-hardening. To generate m s strain-hardening, we need to consider the convected derivative discussed in class. η E 6 Solution by TJO & VS

7 Question 2 This problem was based on problem 9.15 on page 408 in Rubinstein & Colby. We are given DNA with molar mass M = g/mol corresponding to n = base pairs. i For a base pair length l =0.34 nm, the contour length is R max = nl = nm = nm=55.8 μm 33 Equivalently, we have R max = Nb, where N is the number of Kuhn segments and b is the Kuhn segment length. We are given the Kuhn segment length of b = 300 nm, hence the number of Kuhn segments is N = R max /b = nm/300 nm = We can also calculate the root mean square size of the molecule which is R = Nb =1.27 μm, so it is clearly possible to see a molecule of DNA under a microscope, especially if using fluorescence microscopy. Overlap occurs when the volume of each polymer 4 3 π R 2 3/2 is roughly equal to the total system volume divided by the number of polymers M/c N A, where N A is Avagadro s number. c M 1 = M 1 N A 4 3 π R2 3 2 N 4 = mg/ml 35 A 3 π Nb2 3/2 Some students could have also used c φ M 0 b 3 = M 1 =1.52 N A N A Nb 2 3/ mg/ml 36 where M 0 = M/N is the monomer molecular weight and we use the scaling φ N 1/2. Note that these results are just approximate and that we could have used the Fox-Flory equation given by Eq in Rubinstein & Colby, in which case we have c 1 [η] M =6.42 Φ R 2 3/ mg/ml 37 where Φ = 0.425N A = mol 1 is a universal constant. ii Given that the entanglement concentration is c e =10c = mg/ml, where we have used our first result for c above, we note that for the solution concentration of c =0.5 mg/ml the system is in the entangled regime. To determine the contour length of an entanglement strand, we need the volume fraction at the entanglement concentration. To find this quantity, we must first find the volume fraction at the overlap concentration, φ. We can use the expression in Eq. 36 b 3 N φ c A = M 0 7 Solution by TJO & VS

8 The volume fraction at the entanglement concentration is φ e =10φ =0.17. Note that a number of students simply used the formula φ N 1/2 =0.073, but this approach gives an erroneous result since it is just a scaling relationship. The conversion factor A for c = Aφ from Eq. 38 that applies for a dilute solution is A = b 3 N A /M 0 = mg/ml, which is small indicating that a small mass of polymer coil occupies a large volume. On the other hand, in the melt, where φ =1,A = ρ or the density of the polymer. For intermediate concentrations, mg/ml <A<10 3 mg/ml, so without knowing the exact value of the prefactor A, we end up using one of these limits. As a consequence, we will notice below that our estimates for modulus, viscosity and relaxation time for semi-dilute solutions of DNA appear to be physically unreasonable in parts ii and iii. However the results for parts iv and v are reasonable. An example study DNA solutions is Mason, Dhople & Wirtz, Macromolecules, 31, With this in mind, we proceed as follows. We calculate N e 1 using Eq in Rubinstein & Colby for a θ-solvent φ e φ = c 3 e c =10=N e N 1 4 hence N e = 123. Therefore the contour length of an entanglement strand is N e R e = R max N = nm = nm=36.9 μm Likewise, the molar mass of an entanglement strand is given by M e = M N e N = g/mol = g/mol Finally, the plateau modulus at this concentration of 0.5 mg/ml and 30 o C can be calculated using G e φ =G e 1 7 φ 3 where we calculate the entanglement modulus first = ρrt M e φ g/m J/mol.K303 K G e 1 = =34.65 Pa g/mol Now we can obtain the value of G e φ by rewriting the equation as φ 7 3 G e φ =G e 1 φ 7 c 7 3 φ 3 = Ge 1 φ mg/ml 3 7 c 3 = Pa =14.7 kpa mg/ml 44 This number is very large for reasons quoted above. iii The specific viscosity can be estimated by Eq in Rubinstein & Colby, namely 7 8 Solution by TJO & VS

9 η sp = 14 φ φ 3 N 2 3 [N e 1] 2 = 14 c 3 N 2 3 c [N e 1] 2 = So to calculate the solution viscosity, we have η sp 0.5 mg/ml mg/ml = η ηs η = η s η s and hence the solution viscosity is η = η s = Pa.s, which is again quite large. The relaxation time is given by τ rep φ η/g e φ = Pa.s/14700 Pa = 4770 s. iv The relaxation time of an entangled melt is given in Eq. 9.8 in Rubinstein & Colby, ζb 2 τ rep kt N 3 47 N e Here we have to estimate the drag coefficient in the melt ζ, however the background viscosity is unknown. In the limit of concentrated solution with φ approaching unity, we can estimate ζ η s b. Hence, the relaxation time is τ rep η sb 3 N 3 = 338 s 48 kt N e 1 In the case of the shortened DNA chains, the molecular weight is less than the entanglement molecular weight calculated above in part ii. Hence the Rouse relaxation time is relevant in this regime. From Eq in Rubinstein & Colby, we can determine the Rouse relaxation for the original chain, N τ rep =6τ R 49 So the Rouse time of the original chain is τ R =37.3 s. Now, the determine the Rouse time of the shortened DNA chains τ R,wehave Accordingly, we have τ R = τ R N N τ R τ R = N e N 2 N 50 2 = 37.3 s = 3.73 ms The result is roughly five orders of magnitude smaller than the relaxation time of the melt. v For a dilute solution of DNA, a CaBER test measures the Zimm relaxation time. Zimm time is given by Eq from Rubinstein & Colby The 9 Solution by TJO & VS

10 η τ Z =0.163 s R 3 kt η =0.163 s Nb 3 52 kt We note from part i, that for a molecular weight of M = g/mol, we have N = 186 Kuhn segments. So we estimate that the Zimm relaxation time is τ Z =0.163 η sr 3 kt = Pa.s m 3 =0.014 s J/K303 K Solution by TJO & VS

11 MIT OpenCourseWare J / J Macromolecular Hydrodynamics Spring 2016 For information about citing these materials or our Terms of Use, visit: