Assignment 1

ME651 – Problem Set 1 1. Cartesian tensor notation problems (a) Show δij xj = xi (show explicitly, expanding the summations) (b) Show (c) Show

∂xi ∂xj ∂r ∂xi

= δij =

xi r ,

where r = (x · x)1/2 (Use Cartesian tensor Notation)

(d) Use the relations ωi = −ijk Wjk and ijk lmk = δil δjm − δim δjl to show that Wij = − 12 ijk ωk . Here Wij = −Wji . 2. Starting with the Navier Stokes equation for u(x, t):   ∂u + u · ∇u = −∇P + µ∇2 u ρ ∂t

(1)

where ρ(x, t) is the density field, P (x, t) is the pressure field and µ is the dynamic viscosity of the fluid (a) Use Cartesian tensor notation to show that the advection term is given as u · ∇u = 21 ∇u · u + ω × u, where ω = ∇ × u is the vorticity field (b) Use Cartesian tensor notation to take the Curl of equation (1), and show that, for constant density, the equation for ω = ∇ × u is given by: ∂ω + u · ∇ω = ω · ∇u + ν∇2 ω ∂t where ν = µ/ρ is the kinematic viscosity.

(2)

3. An open cylindrical container, with weight W , is immersed in a rigid closed cylindrical container as shown in Fig 1. Air is trapped at the top of the smaller container, because of which it floats above the surface of the surrounding fluid (gravity is acting downwards as usual). If the height of the closed container, H, is reduced, then does the open container rise, sink, or stay at the same level with respect to the bottom surface of the closed container ? Assume that water is incompressible. (Hint: Take into account the compressibility of the air. Relate pressure inside and outside the open container by (a) drawing the free body diagram of the open container, and (b) via the hydrostatic pressure change. Then obtain a constraint for the difference in height between the liquid surfaces inside and outside the open container, and deduce the answer.) 1

Figure 1:

4. Consider a rigid object with some arbitrary shape, represented by the volume V , completely immersed in a quiescent incompressible fluid of density ρ. If the pressure field in the fluid is P (x), then the force acting on the object can R be given as F = − ∂V P ndS, where n is the normal vector at any point on the surface ∂V , and dS is the differential surface element. In Cartesian tensor notation, we can write P ni = P δij nj (a) From hydrostatics, show that P = P0 − ρgxj δj2 (x is the position vector, g is acceleration due to gravity acting in the j direction, ρ is density and P0 is a constant). (b) Now use the divergence theorem for second order tensors to prove Archimedes principle: ”A body totally immersed in a fluid is subject to an upward force equal in magnitude to the weight of fluid it displaces”. 5. The velocity components in an unsteady plane flow are given by u = x/(1 + t) and v = 2y/(2 + t). Determine equations for the streamlines and pathlines, subject to x = x0 at t = 0. 6. For the flow field u(x) = U + Ω × x, where U and Ω are constant linear and angular velocity vectors, use Cartesian coordinates to show that (a) strain rate ∂u ∂ui − ∂xji tensor Sij is zero and (b) determine rotation tensor Rij = ∂x j

2

7. For a smooth single valued function F (x) that depends only on space, and an arbitrary shaped control volume that moves with a velocity b(t), where t R is time, show, using Reynolds transport theorem, that d/dt V ∗ (t) F (x)dV = R b · ( V ∗ (t) ∇F (x)dV ) 8. Show that, for a velocity field u(x), ∇u can be shown to be a second order tensor (hint: start by writing coordinate transformation relations for u and x) 9. A small thermocouple probe floats over water, in which the velocity and temperature fields are given by: u = (2x, y + 3t2 y, 2) and T = 2x2 + yz + t. Find the rate of change of temperature w.r.t. time recorded by the thermocouple when it is at a position x = (3, 1, 3) and t = 1 10. In the class, we showed that, under coordinate transformation, the unit vectors along the axes, ei , rotate to e0i , and that the stress tensor components in the 0 = σ C C , where C = e · e0 . For a new coordinate frame are given as σij ij i lm li mj j 2D stress tensor, show that, for a coordinate frame rotation of θ, we can say: 0 σ12 = σ12 cos 2θ −

σ11 − σ22 sin 2θ 2

(3)

0 is maximized ? What are What are the angles for which the magnitude of σ12 0 is equal to zero ? the angles for which σ12

3