Math 575-Lecture 14 1 Viscous Dissipation of Energy

Math 575-Lecture 14 We continue to study some properties of N-S equations and look at some examples of viscous ﬂows.

1 Viscous dissipation of energy

Consider the kinetic energy 1 Z E = ρ|u|2dV 2 Rt where Rt is a material volume of ﬂuid. Then, by the second convection theorem d 1 Z D Z D Z E = ρ |u|2dV = ρu · udV = u · (−∇p + µ∆u + ρb)dV dt 2 Rt Dt Rt Dt Rt Z Z Z Z Z Z = −∇·(up)dV +µ u·∆udV + ρb·udV = u·(−pn+µn·∇u)dS−µ |∇u|2dV + ρb·udV Rt Rt Rt S Rt Rt In the last step, we did the following: Z Z Z Z Z 2 uj∂i ujdV = ∂i(uj∂iuj)dV − ∂iuj∂iujdV = niuj∂iuj − ∂iuj∂iujdV Rt Rt S This is some energy estimate that may be useful if u = 0 on S, but it does not have enough physical meaning. Now, consider the fact Z Z Z Z T 0 = µu · ∇ · (∇u )dV = µui∂j∂iujdV = µuinj∂iujdS − ∂ui∂iujdV Rt Rt S Rt Adding this to the equation, we have d Z Z Z E = u · (−pn + n · 2µE)dS − µ ∂iuj(∂iuj + ∂jui)dV + ρb · udV dt S Rt Rt Z Z Z = u · (n · σ)dS − 2µ ∇u : EdV + ρb · udV S Rt Rt Note ∇u : E = E : E and thuswe have d Z Z Z E = u · (n · σ)dS − 2µ |E|2dV + ρb · udV dt S Rt Rt The ﬁrst term is the work done by hydrodynamics force; the second term represents the energy dissipated and therefore it is the dissipation rate; the last term means the work done by body force.

1 2 The scaling invariance

d α+1 If we take the scaling for Ω = R as x = λx¯, t = λ t¯, then, by the physical meaning dx 1 u = = u ⇒ u = λαu(λx,¯ λα+1t¯). (2.1) dt λα λ λ

2α α+1 With the scaling pλ = λ p(λx,¯ λ t¯), we ﬁnd

2α+1 ∂t¯uλ + uλ · ∇x¯uλ + ∇x¯pλ = λ (∂tu + u · ∇u + ∇p)

The viscosity term is scaled as

α+2 ∆x¯uλ = λ ∆u

Hence, the equation is scaling-invariant if

α = 1. (2.2)

d 2 2 2 Theorem 1. Let Ω = R . If (u, p) is a solution, then (λu(λx, λ t), λ p(λx, λ t)) is also a solution.

A self-similar solution will be of the form (by choosing λ = t−1/2) 1 x u(x, t) = √ U(√ ) t t

3 Stress in other coordinates

The stress vector is

t = n · σ = −pn + 2µn · E = −pn + µ[2(n · ∇)u + n × (∇ × u)].

One may use this to compute the stress vector. Alternatively, one may compute E and then compute the stress. Example. Consider the ﬂow outside a rotation circular cylinder of the form

u = v(r)θˆ

Find the stress vector acting on the ﬂuid at the surface of the cylinder. Direct computation shows

∂ θˆ ∂v(r) 1 v(r) ∇u = (ˆr + ∂ +z∂ ˆ )[v(r)θˆ] =r ˆ ⊗ θˆ + θˆ ⊗ v(r)∂ θˆ = v0(r)ˆr ⊗ θˆ − θˆ ⊗ r.ˆ ∂r r θ z ∂r r θ r

2 Hence, v(r) 2E = ∇u + ∇uT = v0(r) − (ˆr ⊗ θˆ + θˆ ⊗ rˆ). r At the surface, the normal of the surface is n = −rˆ. Then, v(r) d v t = prˆ − µ v0(r) − θˆ = prˆ − µr θ.ˆ r dr r

4 Some examples of incompressible viscous ﬂows

In this section, we look at some examples. Below, we consider constant density so that we introduce ν = µ/ρ.

4.1 The Couette ﬂow Two rigid planes at y = 0,H. y = 0 plane is stationary while y = H plane is moving with constant velocity Uxˆ. Solve the N-S equation in between these two planes. Due to symmetry, all quantities depend on y only. We assume

u = hu(y), 0i.

By the equation 0 = u · ∇u = −∇p + µ∆u We ﬁnd py = 0, 0 = µ∆u(y) Uy so p is a constant and uyy = 0 and u(y) = H . We have shear ﬂow. The viscosity stress τ is nonzero.

4.2 Rayleigh problem Consider the unsteady problem: the ﬂuid is in the region y > 0. y = 0 is moving in x direction with velocity U(t) = U0 cos ωt. By symmetry, one may assume

u = hu(y, t), 0i

Similar as above, we ﬁnd p = const, ut − νuyy=0 Now, the equation for u is linear and we can assume (Fourier transform)

u(y, t) = Re(f(y)eiωt)

3 Then, iωf − νfyy = 0, f(0) = U0 Hence, r ω √ u(y, t) = U cos ωt − y e−y ω/(2ν) 0 2ν The velocity dies away quickly. To be more precise, after a layer of thickness O(ν1/2) = O(Re−1/2), the eﬀect of the moving plate dies away.

4.3 Poiseuille flow A Newtonian viscous fluid is flowing inside a cylinder tube with radius R in steady motion. Let the axis be the z-axis. We then by symmetry have

u = u(r)ˆz, p = p(r, z)

The equations u · ∇u = −∇p + µ∆u read in cylindrical coordinates as 1 1 0 = −(ˆr∂ p +z∂ ˆ p) + µ( ∂ (r∂ (u(r)ˆz)) + ∂2(u(r)ˆz) + ∂2(u(r)ˆz)) r z r r r r2 θ z

Hence, we must have ∂rp = 0 and p = p(z). Also 1 0 = −∂ p + µ (ru0)0 z r

Separation of variables, we have ∂zp = −G = const and 1 G u00 + u0 = − r µ

The general solution is given by (the homogeneous equation is the Euler ODE)

G u(r) = C log r + C − r2 1 2 4µ

G 2 Clearly, C1 = 0 and C2 = 4µ R by the no-slip boundary condition. The importance of Poiseuille ﬂow is that it can be used to measure the viscosity by measuring the pressure needed and the ﬂux.

4 4.4 Stagnation Point ﬂow Consider a 2D steady viscous ﬂow in y > 0 plane whose streamfunction is given by

ψ(x, y) = UL−1xF (y)

The vorticity is given by ω = −∆ψ = −UL−1xF 00(y) The N-S equations in vorticity form read

u · ω − ν∆ω = 0 where ν = µ/ρ. Then, F 0F 00 − FF 000 − Re−1F 0000 = 0 where Re = UL/ν. 0 The boundary condition: u = 0, y = 0. Hence, ψy = 0 or F (0) = 0 and ψx = 0 or F (0) = 0. If one aims to obtain a stagnation point ﬂow so that ψ ∼ UL−1xy at inﬁnity, then, F (y) ∼ y as y → ∞. Then, as y → ∞, F 0(y) → 1 and F 00(y) → 0,F 000(y) → 0. Hence, integrating once, we have (F 0)2 − FF 00 − Re−1F 000 = 1 Doing the scaling η = Re1/2y and F (y) = Re−1/2f(η), one have

(f 0)2 − ff 00 − f 000 = 0 for f(η). f(η) satisﬁes f(0) = f 0(0) = 0 and f 0(∞) = 1. Then, after η = 1, f 0 ∼ 1. This means the velocity transits from 0 to 1 . This layer is of thickness O(Re−1/2), which is the typical length of boundary layer.