CHE 3167 —Transport Phenomena and Numerical Methods
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:
