SOLUTION TO TEST – 1 01-04-2020 The exam is open book and notes 50 Marks 1. Given two vectors a = −2δ1 +3δ2 +δ3 and b = −4δ1 +δ2 +δ3, and the components of the orthogonal tensor Q = P3i=1 P3j=1 Qij δiδj given by the following matrix representation, 2 −2 1 Qij=131 2 2 Show that: which demonstrates that orthogonal matrices preserve the dot product. Solution: 2 1 −2 a · b = (Q · a) · (Q · b) CHE 3167 —Transport Phenomena and Numerical Methods 2. Prove the identity, where v is a vector and ε is the third rank permutation tensor, ε = P3i=1 P3j=1 P3k=1 εijk δiδjδk, −∇ × (∇ · vv) = ε . {∇v · ∇v} − v · ∇ (∇ × v) with εijk being the permutation symbol. Solution: CHE 3167 —Transport Phenomena and Numerical Methods 3. The equation that governs the time evolution of the entropy per unit mass, Sˆ, in a flowing fluid is given by the expression: ∂Sˆ ! 1 1 ρ ∂t+v·∇Sˆ =−T(∇·q)−T(τ.∇v) where ρ is the density, T is the temperature, q is the heat flux vector with components (qx,qy,qz), v is the velocity field with components (vx,vy,vz), and τ is the stress tensor with the appropri- ate components in Cartesian coordinates. Determine the simplified form of the equation for a Newtonian fluid, for which the stress tensor τ is related to the rate of deformation by the relation, τ =−μ?∇v+∇vT? (where μ is the viscosity), in a shear flow. Note that in a shear flow, the components of the velocity field are, vx = γ ̇ y, vy = 0, vz = 0, where, γ ̇ is the shear rate. In matrix form, this implies that the tensor ∇v can be written as, 000 ∇v=γ ̇ 0 0 000 From symmetry arguments, one can also show that q = (qx,qy,0). Solution: CHE 3167 —Transport Phenomena and Numerical Methods 4. An incompressible Newtonian fluid (i.e., one with constant density ρ), approaches a stationary sphere of radius R, with a uniform, steady velocity u∞ in the upward z direction. For this creeping flow the following expressions for the velocity components in the vicinity of the sphere have been derived: ur =u∞ uθ =u∞ ” ” 3 ?R? 1 ?R?3# 1−2 r +2 r cosθ 3 ?R? 1 ?R?3# −1+4 r +4 r sinθ uφ = 0 Verify that these equations satisfy the equation of continuity, ∇·u = 0. The equation of continuity for a compressible fluid is given in Table 1. Table 1: Continuity equation: ∂ρ/∂t + ∇ · ρu = 0 Cartesian coordinates (x, y, z) ∂ρ + ∂ (ρux)+ ∂ (ρuy)+ ∂ (ρuz)=0 ∂t∂x ∂y ∂z Cylindrical coordinates (r, θ, z) ∂ρ + 1 ∂ (ρrur)+ 1 ∂ (ρuθ)+ ∂ (ρuz)=0 ∂tr∂r r∂θ ∂z Spherical coordinates (r, θ, φ) ∂ρ + 1 ∂ (ρr2ur)+ 1 ∂ (ρuθsinθ)+ 1 ∂ (ρuφ)=0 ∂t r2 ∂r rsinθ ∂θ rsinθ ∂φ Solution:
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