ASTR 320: Solutions to Problem Set 2

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1 ASTR 320: Solutions to Problem Set 2 Problem 1: Streamlines A streamline is defined as a curve that is instantaneously tangent to the velocity vector of a flow. Streamlines show the direction a massless fluid element will travel at any point in time (see Chapter 1 of Lecture Notes). Consider a two-dimensional fluid with velocity field u = (ax, by), where a, b are constants and we adopt Cartesian coordinates. Assume that at time t = 0 the density everywhere is equal to ρ 0. a) Draw streamlines for the following two cases: (i) a = b > 0, and (ii) a = b > 0. SOLUTION: We can characterize the streamlines by an implicit function of the form f(x, y) = 0. We know that f u = 0, which expresses that the gradient in f is perpendicular to u (i.e., streamlines are everywhere tangent to the streaming motion). Hence, we have that ( ) ( ) f f u x + u y = 0 x y If we parameterize the location along the streamline by the parameter λ, then we have that df/dλ = 0 (i.e., moving along the streamline does not change the value of f). Using that f = f(x, y) = f(x(λ), y(λ)) we can write this as df dλ = f x dx dλ + f y dy dλ = 0 By comparing the above two equations, we see that u x = dx dλ u y = dy dλ which we can write as a relation between y and x as follows: dy u y = dx u x 1

2 For case (i), this becomes dy/y = dx/x, which can be solved by integration: dy y = dx x the solution of which is ln y = lnx + C, where C is an integration constant. We can rewrite this solution as y = k x, where k = e C is a parameter that characterizes the streamline. It is now straightforward to draw the streamlines (see left-hand panel of Fig. 1). In fact, in this particular case, one can immediately see what the streamlines should look like by simply using that u = (ax, ay) = a(x, y) = a r: the velocity vector points in the same direction as the radial vector, and hence the streamlines must be along the direction of r. In case (ii), our relation between y and x becomes dy y = dx x the solution of which is lny = ln x + C, and thus y = k/x. The corresponding streamlines are depicted in the right-hand panel of Fig. 1, where the direction follows from the fact that a > 0 upon inspection. b) Under what condition(s) is the flow incompressible? Under what condition(s) is it curl-free SOLUTION: A flow is incompressible if u = 0. Using that u = u x x + u y y = a + b we see that the flow is incompressible if a + b = 0. Similarly, a flow is curl-free if u = 0. Using that ( uy u = x u ) x e z = (0 0) e z = 0 y Hence, we see that this flow is always curl-free, independent of the values of a or b. 2

3 c) Consider the location A = (2, 2). How does the density at A change as a function of time in cases (i) and (ii) above? SOLUTION: The density is related to the divergence of the velocity field via the continuity equation ρ t + ρ u + u ρ = 0 Using that u = (a + b), and using that initially ρ = 0, we have that which has as solution ρ t = (a + b)ρ ρ(t) = ρ 0 e (a+b)t where we have used the boundary condition ρ(t = 0) = ρ 0. Note that ρ(t) is independent of position, which implies that ρ will remain zero at all times. Hence, the density at location A is described by the above equation. Now consider a two-dimensional velocity field u = ( ay, bx). Again, assume that at time t = 0 the density everywhere is equal to ρ 0. d) Draw streamlines for the following two cases: (i) a = b > 0, and (ii) a = b > 0 SOLUTION: Using the same method as under (a), we find that the streamlines in case (i) are characterized by u x = ay and u y = +ax. Hence, we obtain the relation y dy = x dx with solution y 2 /2 = x 2 /2 +C or x 2 +y 2 = 2C. Such curves correspond to circles of radius 2C (see left-hand panel of Fig. 2). The directionality of the streamlines follows from the fact that a > 0 upon inspection. 3

4 Figure 1: Solution to Problem 1a. In case (ii) we have that u x = by and u y = bx. Hence, we obtain the relation y dy = x dx with solution y 2 /2 = x 2 /2 + C or y 2 x 2 = 2C. Such curves correspond to hyperbolae (see right-hand panel of Fig. 2). Once again, the directionality of the streamlines follows from the fact that a > 0 upon inspection. e) Under what condition(s) is the flow incompressible? Under what condition(s) is it curl-free SOLUTION: A flow is incompressible if u = 0. Using that u = u x x + u y y = = 0 Hence, this flow is always incompressible, independent of the values of a or b. Similarly, a flow is curl-free if u = 0. Using that ( uy u = x u ) x e z = (b a) e z = (a + b) e z y 4

5 Figure 2: Solution to Problem 1d. Hence, we see that this flow is curl-free if a + b = 0. f) How does the density at A change as function of time at A in cases (i) and (ii)? SOLUTION: Since we have shown that the flow is incompressible, we have that ρ/ t = 0 (this follows from continuity equation and the fact the the flow is divergence free). Hence, ρ A (t) = ρ 0 at all t. Problem 2: Navier-Stokes; from index form to vector form The Navier-Stoker equation in index form, as derived in class, is given by ρ du i dt = P + [ ( ui µ + u j 2 )] x i x j x j x i 3 δ u k ij + ( η u ) k ρ Φ x k x i x k x i Show clearly, step-by-step and using text were needed, that this can be written in vector form as ρ d u dt = P + µ 2 u + (η + 1 ) 3 µ ( u) ρ Φ 5

6 You may assume that the kinetic and bulk viscocity are constant throughout space. SOLUTION: Let us start with the trivial terms: and ρ du i dt ρd u dt P x i P ρ Φ x i ρ Φ In addition, using the notion that if an index appears twice in the same term then this implies Einstein summation convention, we have that u k / x k u. Hence, using that the viscosity is constant, we have that x i ( η u k x k ) = η x i ( u) η ( u) Finally, using the same concept, the second term on the l.h.s. of the equation in index form reduces to µ [ ui + u j 2 ] x j x j x i 3 δ ij u which can be written as [ 2 u i µ + 2 u j 2 ] (δ ij u) x j x j x j x i 3 x j Let s first examine the first term, where we have to realize once again that if an index appears twice it implies summation over that index; 2 u i 2 u i = + 2 u i + 2 u i x j x j x 1 x 1 x 2 x 2 x 3 x 3 ( ) 2 = u x 2 1 x 2 2 x 2 i 3 = 2 u i 2 u 6

7 Similarly, for the second term we have 2 u j = ( ) uj = ( u) ( u) x j x i x i x j x i And finally, for the third term we have 2 (δ ij u) = 2 ( u) 2 ( u) 3 x j 3 x i 3 Combining all of the above, our equation becomes ρ d u dt = P + µ 2 u + µ ( u) 2 µ ( u) + η ( u) ρ Φ 3 = P + µ 2 u + (η + 1 µ) ( u) ρ Φ 3 Problem 3: The Stress Tensor Consider a fluid in a 2-dimensional, Cartesian coordinate system (x 1, x 2 ), with stress tensor ( ) σ11 σ σ ij = 12 σ 21 σ 22 and let ˆn and ˆt be the unit normal and unit tangent vectors of a surface S, for which ˆn is rotated by angle θ with respect to the x 1 axis. a) Express ˆn and ˆt in terms of θ, i.e., what are n 1, n 2, t 1 and t 2 in ˆn = (n 1, n 2 ) and ˆt = (t 1, t 2 )? SOLUTION: Simple geometry shows that ˆn = (cosθ, sin θ) and ˆt = ( sin θ, cosθ). Note that ˆn ˆt = 0, as required for two vectors that are perpendicular to each other. Note also, one is free to choose ˆt in the opposite direction, i.e., with ˆt = (sin θ, cos θ). In that case, the tangential stress will have the opposite sign as in the case used here. b) Show that the normal stress, Σ n, can be written as Σ n = σ 11 cos 2 θ + σ 22 sin 2 θ + σ 12 sin 2θ 7

8 and derive a similar expression for the shear stress, Σ t. SOLUTION: The normal stress is the stress along the normal. Hence, Σ n = Σ ˆn = Σ i n i = σ ij n j n i = σ 11 n 1 n 1 + σ 12 n 2 n 1 + σ 21 n 1 n 2 + σ 22 n 2 n 2 = σ 11 cos 2 θ + σ 12 sin θ cosθ + σ 21 sin θ cosθ + σ 22 sin 2 θ = σ 11 cos 2 θ + σ 22 sin 2 θ + σ 12 sin 2θ where, in the last step, we have used that the stress tensor is symmetric, so that σ 12 = σ 21. Similarly, for the shear stress, we have that Σ t = Σ ˆt = Σ i t i = σ ij n j t i = σ 11 n 1 t 1 + σ 12 n 2 t 1 + σ 21 n 1 t 2 + σ 22 n 2 t 2 = σ 11 sin θ cos θ σ 12 sin 2 θ + σ 21 cos 2 θ + σ 22 sin θ cosθ ( ) σ22 σ 11 = sin 2θ + σ 21 cos 2 θ σ 12 sin 2 θ 2 ( ) σ22 σ 11 = sin 2θ + σ 21 cos 2θ 2 where in the last step we have used once again that σ 12 = σ 21. c) Consider the case σ 11 = σ 12 = σ 21 = 0. Under what angle θ is the shear stress on S maximal? For what angle θ does the shear on S vanish? What are the normal stresses in both cases? SOLUTION: Using the expression for the shear stress derived under b), we see that, when σ 11 = σ 12 = σ 21 = 0 we have that Σ t = 1 2 σ 22 sin 2θ. This will be maximal when sin 2θ = ±1, which occurs for θ {π/4; 3π/4; 5π/4; 7π/4}. The corresponding normal stress is Σ n = σ 22 sin 2 θ = σ 22 /2. NOTE: the shear stress is maximal when Σ t is maximal; the direction of the shear is irrelevant!! 8

9 Similarly, the shear stress vanishes when sin 2θ = 0, which occurs for θ {0, π/2, π, 3π/2}. The corresponding normal stresses are Σ n {0; σ 22 ; 0; σ 22 }. d) Answer the same questions as under c), but now for the case with σ 11 = σ 22 = 0. SOLUTION: In this case, Σ t = σ 12 cos 2θ and Σ n = σ 12 sin 2θ. The shear stress is maximal whenever cos 2θ = ±1, which corresponds to θ {0, π/2, π, 3π/2}. The corresponding normal stresses vanish (i.e., Σ n = 0). The shear stress will vanish whenever cos 2θ = 0, which happens whenever θ {π/4; 3π/4; 5π/5; 7π/4}. The corresponding normal stresses are Σ n = {σ 12 ; σ 12 ; σ 12 ; σ 12 } e) Under what condition are both the normal and shear stress independent of θ, and what are their values in that case? SOLUTION: Using the derivations for Σ n and Σ t under b), it is fairly easy to see that the normal and shear stresses are identical whenever σ ij = σ 0 δ ij. In words, we have that the normal stress is equal to the shear stress whenever the matrix representing the stress tensor is diagonal, with identical diagonal elements. Substituting these values for σ ij in the expressions for Σ n and Σ t we see that the shear stress vanishes everywhere, while the normal stress is equal to σ 0. 9

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