Exercise 3: Conservation Equations and Stress Tensor Example 1

Home , Shear stress, Viscous stress tensor

Fluid Mechanics, SG2214, HT2013 September 11, 2013

Exercise 3: Conservation Equations and Stress Tensor

Example 1: conservative form of continuity equation (also in lecture notes)

a) Consider a ﬁxed, closed surface S in a ﬂuid. Show that conservation of mass implies

∂ρ ∂ + (ρui) = 0 . ∂t ∂xi

The change of mass in the volume is a balance between the mass ﬂux in an out the domain and the change in density. Conservation means that this should be zero Z ∂ρ I dV + ρui ni dS = 0 . V ∂t S Using the Gauss Theorem we can rewrite the surface integral as volume integrals, I Z ∂ ρui ni dS = (ρui) dV . S V ∂xi We then have Z ∂ρ ∂ { + (ρui)} dV = 0 . V ∂t ∂xi This must be true for an arbitrary volume leading to the diﬀerential form ∂ρ ∂ + (ρui) = 0 . ∂t ∂xi

b) Show that this can be written as Dρ ∂u + ρ i = 0 . Dt ∂xi

∂ ∂ρ ∂ui ∂ρ ∂ui (ρui) = ui + ρ = ui + ρ ∂xi ∂xi ∂xi ∂xi ∂xi This gives ∂ρ ∂ρ ∂ui Dρ ∂ui + ui +ρ = + ρ . ∂t ∂xi ∂xi Dt ∂xi | {z } Dρ Dt

c) Explain what the following means

∂u Dρ i = 0 ⇒ = 0 . ∂xi Dt

1 Remember the decomposition of ∂ui in its invariant parts ∂xj ∂ui 1 ∂ui ∂uj 1 ∂uk 1 ∂ui ∂uj 1 ∂uk = + − δij + − + δij . ∂xj 2 ∂xj ∂xi 3 ∂xk 2 ∂xj ∂xi 3 ∂xk | {z } | {z } | {z } ¯ deformation e¯ij rotation ξij volume change e¯ij

The volume change e¯ij = 0 due to the divergence-free condition. Therefore, if we are following a ﬂuid particle, the density and volume are constant. But at a ﬁxed position, the density can change as ∂ρ ∂ρ = −ui . ∂t ∂xi Example 2: Reynolds Transport Theorem (also in lecture notes)

a) Use the Reynolds Transport Theorem to provide an alternative derivation of the conservation of mass equation ∂ρ ∂ + (ρui) = 0. ∂t ∂xi Reynolds Transport Theorem:

Z Z D ∂Tijk ∂ Tijk dV = + (ulTijk) dV Dt V (t) V (t) ∂t ∂xl

Put Tijk = ρ, D Z Z ∂ρ ∂ ρ dV = + (ρui) dV Dt V (t) V (t) ∂t ∂xi No ﬂow through the surface, so the integrals are zero and valid for all V (t)

∂ρ ∂ + (ρui) = 0 . ∂t ∂xi

b) Use this result to show Z Z D DFijk ρFijk dV = ρ dV . Dt V (t) V (t) Dt

Put Tijk = ρFijk in the Reynolds Transport Theorem

D Z Z ∂ ∂ ρFijk dV = (ρFijk) + (ulρFijk) dV = Dt V (t) V (t) ∂t ∂xl Z ∂Fijk ∂ρ ∂ ∂Fijk ρ + Fijk + Fijk (ρul) + ρul dV = V (t) ∂t ∂t ∂xl ∂xl Z ∂ρ ∂ ∂Fijk ∂Fijk Fijk + (ρul) + ρ + ul dV = V (t) ∂t ∂xl ∂t ∂xl | {z } | {z } DF =0 ijk = Dt Z DF ρ ijk dV . V (t) Dt

2 The stress tensor Tij

The equation for the stress tensor was deduced by Stokes in 1845 from elementary hypotheses. Writing the stress tensor in the form

Tij = −pδij + τij ,

the following statements should be true for the viscous stress tensor τij in a Newtonian fluid: (i) τij should vanish if the flow involves no deformation of fluid elements: Tij = −pδij. ∂ui (ii) τij is a function of the velocity gradients . This is dependence is assumed “local” and “at present ∂xj time”, that is the stresses depends on the local deformation without any effect from past deformations. ∂ui (iii) the relationship between τij and should be isotropic as the physical properties of the fluid ∂xj are assumed to show no preferred direction. (iv) τij should be frame indifferent, so independent of the antisymmetric part ξij

∂ui Finally linear dependence is assumed between τij and . ∂xj

Tij = −pδij + 2 µeij . This proofed to be good!

Example 3: Derivation of the viscous stress tensor

If τij is a linear function of the components of eij it can then be written as

τij = cijkl ekl + dij ,

where dij = 0 from the ﬁrst assumption above. It can be shown that the most general fourth-order isotropic tensor is of the form cijkl = Aδijδkl + Bδikδjl + Cδilδjk where A, B and C are scalars. a) Use this to show that τij = λekkδij + 2µeij where λ and µ are scalars.

τij = cijkl ekl = (Aδijδkl + Bδikδjl + Cδilδjk)ekl = Aδijekk + Beij + Ceji

Say that A = λ and that B = C = µ, then since eij = eji we have

τij = λekkδij + 2µeij

1 b) Show that if we have a Newtonian ﬂuid where p = − T then 3 ii 2 ∂uk ∂ui ∂uj Tij = − p + µ δij + µ + 3 ∂xk ∂xj ∂xi

We have Tij = −pδij + τij = −pδij + λekkδij + 2µeij Then Tii = −3p + 3λeii + 2µeii = −3p + (3λ + 2µ)eii

3 2 This gives λ = − 3 µ, and thus 2 T = −p − µe δ + 2µe ij 3 kk ij ij or 2 ∂uk ∂ui ∂uj Tij = − p + µ δij + µ + . 3 ∂xk ∂xj ∂xi

2 λ = − 3 µ is assumed and shown to be valid for monoatomic gases. This implies that the mean pressure (normal stress) is the thermodynamic pressure. As consequence, there is no irreversible viscous loss in the case of ﬂuid expansion or compression (Work for expansion/compression is only provided by the pressure, not by viscosity)

Note that for incompressible ﬂuid ∇ · u¯ = 0:

Tij = −pδij + 2µeij .

Example 4

Calculate the stress vector at a surface through the origin of example 3, recitation 2 for a Newtonian ﬂuid if the pressure at the origin is p0, and the unit normal is • a) in the x-direction

• b) in the y-direction • c) at 45o in the x − y plane Also, for each case identify the normal stress and the shear stress, the viscous stress and the isotropic part of the stress. Solution: Given the velocity the ﬁeld u = αx, v = −αy,(w = 0) and the deﬁnition of the stress tensor Tij = −p0δij + 2µeij, the strain rate tensor and the stress tensor become respectively  ∂u 1 ∂u ∂v  ( + )    ∂x 2 ∂y ∂x  α 0   eij =   =   1 ∂u ∂v ∂v  0 −α ( + ) 2 ∂y ∂x ∂y   −p0 + 2µα 0 Tij =   0 −p0 − 2µα

a) Unit normal n = ex and tangent t = ey, or equivalently

1 0 n =   ; t =   ; 0 1

Stress vector Ri = Tijnj       −p0 + 2µα 0 1 −p0 + 2µα Ri =     =   0 −p0 − 2µα 0 0

Normal stress Rini 1 −p0 + 2µα 0   = −p0 + 2µα |{z} |{z} 0 isotropic part viscous stress

4 Shear stress Riti 0 −p0 + 2µα 0   = 0 1

b) Unit normal n = ey and tangent t = ex, or equivalently

0 1 n =   ; t =   ; 1 0

Stress vector Ri = Tijnj       −p0 + 2µα 0 0 0 Ri =     =   0 −p0 − 2µα 1 −p0 − 2µα

Normal stress Rini 0 0 −p0 − 2µα   = −p0 − 2µα |{z} |{z} 1 isotropic part viscous stress

Shear stress Riti 1 0 −p0 − 2µα   = 0 0 c) 1  1  1 1 n = √   ; t = √   ; 2 1 2 −1

Stress vector Ri = Tijnj

−p + 2µα 0  1 −p + 2µα 0 1 1 0 Ri =   √   = √   2 2 0 −p0 − 2µα 1 −p0 − 2µα

Normal stress Rini 1 1 1 √ −p0 + 2µα −p0 − 2µα √   = −p0 2 2 |{z} 1 isotropic part of stress

Shear stress Riti  1  1 1 √ −p0 + 2µα −p0 − 2µα √   = 2µα 2 2 |{z} −1 viscous stress

Example 5

Consider a control volume x ∈ [0,L], y ∈ [0,L] for the ﬂow in the example 3 of recitation 2 and calculate the net force resulting from the isotropic stress tensor on the control volume using the momentum theorem. Momentum theorem in the absence of external body forces d Z Z Z ρudV + ρu(u · n)dS = RdS dt V ∂V ∂V

5 x-component for the steady ﬂow u = αx, v = −αy,(w = 0) Z Z Z Z ρu(u · n)dS = RxdS = −pnxdS + τxjnjdS ∂V ∂V ∂V ∂V | {z } | {z } x-comp from isotropic stress tensor x-comp from viscous stress tensor y-component Z Z Z Z ρv(u · n)dS = RydS = −pnydS + τyjnjdS ∂V ∂V ∂V ∂V | {z } | {z } y-comp from isotropic stress tensor y-comp from viscous stress tensor

Given the viscous stress tensor 2µα 0  τij =   0 −2µα we integrate long the boundary of our domain. Along x = L the normal is njdS = (ex)jdy and thus

2µα 0  1 2µα τijnj =     =   . 0 −2µα 0 0

Along the side x = 0 the normal is in the opposite direction njdS = −(ex)jdy and

2µα 0  −1 −2µα τijnj =     =   . 0 −2µα 0 0

Hence we get no net contribution from the two surfaces of S at x = 0, and x = L       Z L Z L 2µα −2µα 0 τijnj(for x=0 and x=L) =   +   dy =   0 0 0 0 0

We now consider the side along y = L the normal is njdS = (ey)jdx and thus

2µα 0  0  0  τijnj =     =   . 0 −2µα 1 −2µα

Along the side y = 0 the normal is in the opposite direction njdS = −(ey)jdx and

2µα 0   0   0  τijnj =     =   . 0 −2µα −1 2µα

Hence we get no net contribution from the two surfaces of S at y = 0, and y = L       Z L Z L 0 0 0 τijnj(for y=0 and y=L) =   +   dx =   0 0 2µα −2µα 0

Therefore we have for the x-component Z Z ρu(u · n)dS = −pnxdS = ∂V ∂V | {z } x-comp from isotropic stress tensor

6 Z L Z L Z L Z L α2L3 α2L3 ρu(−u)dy + ρu(u)dy + ρu(−v)dx + ρu(v)dx = α2L3 − = 0 0 0 0 2 2 | {z } | {z } | {z } | {z } =0 at x=0 =α2L2 at x=L =0 at y=0 =αx(−αL) at y=L and the y-component Z Z ρv(u · n)dS = −pnydS = ∂V ∂V | {z } x-comp from isotropic stress tensor Z L Z L Z L Z L α2L3 α2L3 ρv(−u)dy + ρv(u)dy + ρv(−v)dx + ρv(v)dx = − + α2L3 = 0 0 0 0 2 2 | {z } | {z } | {z } | {z } =0 at x=0 =−αy(αL) at x=L =0 at y=0 =α2L2 at y=L Note that you should obtain the same result using Bornoulllis theorem for irrotational ﬂow and integrating the pressure over the surface of the control volume.

Example 6

Show that the stress tensor Tij = −pδij for the ﬂow u¯ =ω ¯ × x,¯ where ω¯ is constant.

Tensor notation: ui = εiklωkxl The stress tensor then becomes: ∂ui ∂uj Tij = −pδij + µ + = ∂xj ∂xi ∂ ∂ −pδij + µ (εiklωkxl) + (εjklωkxl) = ∂xj ∂xi ∂xl ∂xl −pδij + µωk εikl + εjkl = ∂xj ∂xi

−pδij + µωk(εiklδlj + εjklδli) =

−pδij + µωk(εikj + εjki) =

−pδij + µωk(−εijk + εijk) = −pδij