HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Σχετικά έγγραφα
Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Homework 3 Solutions

Section 8.3 Trigonometric Equations

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Matrices and Determinants

2 Composition. Invertible Mappings

Areas and Lengths in Polar Coordinates

Areas and Lengths in Polar Coordinates

EE512: Error Control Coding

Section 7.6 Double and Half Angle Formulas

derivation of the Laplacian from rectangular to spherical coordinates

C.S. 430 Assignment 6, Sample Solutions

Homework 8 Model Solution Section

Solutions to Exercise Sheet 5

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Risk! " #$%&'() *!'+,'''## -. / # $

Problem Set 3: Solutions

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

4.6 Autoregressive Moving Average Model ARMA(1,1)

Math221: HW# 1 solutions

( ) 2 and compare to M.

Concrete Mathematics Exercises from 30 September 2016

Example Sheet 3 Solutions

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

1) Formulation of the Problem as a Linear Programming Model

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

ST5224: Advanced Statistical Theory II

( y) Partial Differential Equations

Second Order Partial Differential Equations

1 String with massive end-points

Section 9.2 Polar Equations and Graphs

Mean bond enthalpy Standard enthalpy of formation Bond N H N N N N H O O O

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Finite Field Problems: Solutions

[1] P Q. Fig. 3.1

Partial Differential Equations in Biology The boundary element method. March 26, 2013

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Quadratic Expressions

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

Συστήματα Διαχείρισης Βάσεων Δεδομένων

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Exercises to Statistics of Material Fatigue No. 5

Overview. Transition Semantics. Configurations and the transition relation. Executions and computation

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

CRASH COURSE IN PRECALCULUS

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Notes on the Open Economy

Numerical Analysis FMN011

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

TMA4115 Matematikk 3

Forced Pendulum Numerical approach

w o = R 1 p. (1) R = p =. = 1

The challenges of non-stable predicates

6.3 Forecasting ARMA processes

PARTIAL NOTES for 6.1 Trigonometric Identities

Parametrized Surfaces

Second Order RLC Filters

Approximation of distance between locations on earth given by latitude and longitude

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

Lecture 2. Soundness and completeness of propositional logic

Srednicki Chapter 55

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

Part III - Pricing A Down-And-Out Call Option

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Inverse trigonometric functions & General Solution of Trigonometric Equations

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

Example of the Baum-Welch Algorithm

Strain gauge and rosettes

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

The Simply Typed Lambda Calculus

Fractional Colorings and Zykov Products of graphs

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Capacitors - Capacitance, Charge and Potential Difference

Statistical Inference I Locally most powerful tests

Variational Wavefunction for the Helium Atom

Orbital angular momentum and the spherical harmonics

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

10.7 Performance of Second-Order System (Unit Step Response)

Every set of first-order formulas is equivalent to an independent set

the total number of electrons passing through the lamp.

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

ENGR 691/692 Section 66 (Fall 06): Machine Learning Assigned: August 30 Homework 1: Bayesian Decision Theory (solutions) Due: September 13

EE101: Resonance in RLC circuits

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

Lecture 15 - Root System Axiomatics

Written Examination. Antennas and Propagation (AA ) April 26, 2017.

Tridiagonal matrices. Gérard MEURANT. October, 2008

On a four-dimensional hyperbolic manifold with finite volume

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

If we restrict the domain of y = sin x to [ π, π ], the restrict function. y = sin x, π 2 x π 2

Similarly, we may define hyperbolic functions cosh α and sinh α from the unit hyperbola

Assalamu `alaikum wr. wb.

Tutorial problem set 6,

ΚΥΠΡΙΑΚΟΣ ΣΥΝΔΕΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY 21 ος ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ Δεύτερος Γύρος - 30 Μαρτίου 2011

Mock Exam 7. 1 Hong Kong Educational Publishing Company. Section A 1. Reference: HKDSE Math M Q2 (a) (1 + kx) n 1M + 1A = (1) =

If we restrict the domain of y = sin x to [ π 2, π 2

Partial Trace and Partial Transpose

Transcript:

HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying the chain rule: P zz = ψ λ z = ψ λ z = ψ = ψ λ. The elastic energy per unit volume can be written as: ψ = [ λ z + ] λ z. Hence: ψ = λ z λ z.4 Finally, the nominal stress P zz can be written as: P zz = λ z λ z = λ z λ z.5 In order to plot the stress versus the stretch we define a normalized stretch: P zz = P zz = λz λ z.6 In the figure bellow we can see how as we increment the compression on the hydrogel, the stress also increments.

Figure : Normalized nominal stress P zz versus λ z b In the slow loading case, we know that the elastic energy per unit volume can be expressed as: f gel = λ z + λ z + f mix.7 Because the loading is very slow, the system is always in an equilibrium state. Therefore: f gel =.8 f gel = λ z f mix + f mix =.9 And rearranging the terms we obtain: λ z λ z = f mix + f mix = f mix + f mix. Remind that the osmotic pressure can be computed as: Π = f f + f.

And for long chains N the mixing energy per unit volume is: k B T ν [ ln + χ ]. From the previous expression it s easy to see that f =, so we can conclude: λ z = Π. c First of all we need to derive an expression for Π: Π = f + f = k BT [ ln + χ ] + k BT [ ln + χ χ ] ν ν = k BT [ ln + + χ ].4 ν Which means that the relationship λ z can be expressed as: Or = k BT [ ln + + χ ].5 λ z ν ν + [ ln + + χ ] =.6 λ z For a given value of χ, using any solver we can find the pairs λ z where λ z = λz ν. As expected, when λ z tends to zero, the volume fraction tends to one, because the polymer is so compressed that there is almost no solvent in it. Then, λ x is directly found thanks to the following relation: λ x = λ z = λ z ν.7 Since we know the values of the pairs λ z and we can directly plot the pairs λ x λ z where: λ x = λ x ν.8

Figure : versus the normalized λ z for three different values of χ Figure : Normalized λ x versus versus normalized λ z, computed for a value of χ = d As before we know that P zz = ψ.9 4

For slow loading the elastic free energy density can be written as: ψ = λ z + λ x = λ z + λ z. And the derivative then becomes: ψ = λ z λ z. So the nominal stress becomes: P zz = λ z λ z. The term <. Hence comparing this equation to.6 we can clearly see that the stress is much less in this case. The more solvent that leaves, the less stress we have. Finally, in order to plot the stress we define: P zz = λ z λ z. Figure 4: Normalized nominal stress P zz versus λ z 5

Problem In order to prove that the gel shrinks, we first compute the volume fraction of the polymer that we added. Considering that we added a total amount of polymer V p we get: sol = V p V V = V p V Vg = sol V sol V p. And if we take the derivative of sol respect to we obtain: = V sol V p V sol sol V sol V p = V p V. Since this derivative is negative, it means that the slope of the curve sol is negative and therefore, whenever sol decreases, the gel block must shrink increases. Which is due to the fact that when we add polymer the mixing energy on the gel is reduced some solvent wants to mix with the new polymer instead., sol V g, V g, f el = trc = [ ] / V Vg f mix sol = sol f mix sol sol Figure 5: Scheme of the full process carried out in this experiment Next we want to find the equilibrium state after introducing the new polymer. First we write the total energy in the two states. On the first state, since there is no polymer in the solution, the total energy is the elastic energy of the gel plus its mixing energy: U tot = V g f sol + f el. On the second state we added some polymer to the solution which affected the gel, that changed its volume to V g. This polymer, however, does not have elastic energy because the chains are not cross-linked. Hence, we can write the total energy on state as the elastic energy on the gel plus the mixing energy on both gel and solution: U tot = V g f sol + f el + V V g f sol sol.4 Which means that the difference in energy between the two states is: U tot = f el f el + V g f sol V g f sol + V V g f sol sol.5 6

Then, we consider the total energy per unit reference volume of gel: f tot = U tot = f el f el + V g f sol V g f sol + V V g f sol sol.6 Because the amount of polymer in the block is the same during all the process we can establish the following relationship: V g = V g =.7 So we obtain: f tot = f el f el + f sol + V f sol + f sol sol.8 The system will find the minimum energy state, which means that the variation of energy will be minimum too, hence: f tot = = f el + f sol V f sol sol.9 The elastic energy of a system which shrinks λ in all directions can be written as: f el = trc = λ = [ ]. Which means that: f el =. Then, going back to.9 we obtain: = f sol + f sol + V f sol sol + f sol sol. And rearranging the terms we can write: = f sol + f sol } {{ } Π V +f sol f sol sol sol +. For the term in parenthesis, we can write: V Then, using the chain rule we can express: = V.4 7

f sol sol = f sol sol sol sol.5 And from equation. it s easy to show that sol = sol V sol = sol V V.6 Therefore, we obtain = Π + f sol sol V sol f sol sol V.7 sol = Π + f sol sol sol f sol sol sol }{{} Π sol.8 So we finally obtain: = Π Πsol.9 8