Boussinesq Arguments About the Energy Equation

Lecture 9 9.1 Administration • None. 9.2 Continuation of energy equation From the last time, we ended with an expression for the energy equation
 1 D e+ u ρ = ρg · u + ∇ · (τ · u) − ∇ · q Dt 2 (9.1) Where ρg · u ≡ changes in potential energy −∇ · q ≡ changes due to heat flux ∇ · (τ · u) ≡ a kinetic energy part and an internal heating part Clearly, the stress term is the painful one: ∇ · (τ · u) = u · (∇ · τ ) + τ : ∇u (9.2) The first term, u · (∇ · τ ), is the rate of work done by the surface forces. Imbalances in stress accelerate the fluid blob and change its kinetic energy. The second term, τ : ∇u, is the work done by deformation. Not all stress moves the blob – some of it deforms and causes heat change. Note the double contraction, A : B ⇒ scalar. The u · (∇ · τ ) terms is the easiest of the two since we came up with an expression for ∇ · τ when we looked at momentum conservation.
 2 ∇ · τ = ∇ · 2µe − p − µ∇ · u 3 µ = −∇p + µ∇2 u + ∇(∇ · u) 3   µ 2 (9.3) ⇒ u · (∇ · τ ) = u · −∇p + µ∇ u + ∇(∇ · u) 3 We’re advecting τ through our fixed blob of fluid. It is bringing in kinetic energy associated with the stress on the blob. (Jim – You have a note on the side here saying “think about this more”; I am not sure if this is directed at you or the students. In the least, “advecting τ ” is a bit confusing.) Now the second term, τ : ∇u double dot product 1 Notice, there is no · between the ∇ and the u. ∇u is the gradient of the velocity... The velocity gradient tensor! ∇u = G (9.4) τ :G (9.5) Thus, this term looks like But G = e + 12 r,
τ:  1 e+ r 2 (9.6) Recall that r is an antisymmetric tensor, and τ is a symmetric tensor. The double dot product of asymmetric and antisymmetric matrix is zero  Aij Bij (9.7) A:B = i j In 2D  τ11 τ21 τ12 τ22   0 : 1 2ω − 12 ω 0  = τ11 (0) + τ12 ω 2 ω ω = τ12 − τ21 2 2 = 0  ω + τ21 − + τ22 (0) 2 (9.8) This last line is due to the fact that since τ is symmetric, τ12 = τ21 . Thus, you can get rid of the 1 2 r term and end up with τ :e Substitute in our expression for τ ,
   2 2µe − p + µ∇ · u I : e 3 2 2µe : e − pI : e − µ∇ · uI : e 3 (9.9) (9.10) Lets start with an easy one pI : e ∂v ∂w ∂u +p +p ∂x ∂y ∂z = p∇ · u = p (9.11) That is, change in energy due to compression or expansion. The last term is also pretty easy 
2 2 ∂u ∂v ∂w µ∇ · uI : e = µ∇ · u + + 3 3 ∂x ∂y ∂z 2 = µ∇ · u(∇ · u) 3 2 (9.12) = µ(∇ · u)2 3 2 Figure 9.1: (fig:Lec9MuParameter) µ parameterizes work associated in pulling apart molecules in a blob of fluid. Finally, the first term 2µe : e e:e ⎡ ⎤ ⎡ e11 e12 e13 e11 = ⎣ e21 e22 e23 ⎦ : ⎣ e21 e31 e32 e33 e31 e11 e11 + e12 e21 + e13 e31 + = e21 e12 + e22 e22 + e23 e32 + e31 e13 + e32 e23 + e33 e33 Note e12 e22 e32 ⎤ e13 e23 ⎦ e33 2 ∂u ∂x 2 
1 ∂u ∂v + 2 ∂y ∂x
2
2 ∂u 1 ∂u ∂v ∂v +2 + 4 ∂y ∂y ∂x ∂x
2
2 1 ∂u ∂v 1 ∂u 1 ∂v + + 4 ∂y 2 ∂y ∂x 4 ∂x (9.13)
e11 e11 = e12 e21 = = = (9.14) (9.15) etc. Consider just the e11 e11 term. Remember, 2µe : e ⇒ 2µ
∂u ∂y 2 = 
∂u ∂u 2 µ ∂y ∂y (9.16) ∂u The µ ∂u ∂y bit is associated with normal stress, and ∂y with the velocity gradient. Thus, this term has something to do with the work it takes to pull apart molecules in a blob of fluid (see figure ∂u ∂v 2 9.1). The other terms, such as ( ∂u ∂y ) and ∂y ∂x , say something similar about shear and twisting. Clearly this is a bit of a mess. Thankfully, what is typically done is to sweep it all under the rug and define
 2 φ = 2µe : e − µ(∇ · u)2 (9.17) 3 ≡ viscous component of the deformation work rate This allows us to write τ : ∇u = −p(∇ · u) + φ Tracking this all
D ρ e+ Dt (9.18) the way back to the energy equation gives:    1 µ u = ρg · u + u · −∇p + µ∇2 u + ∇(∇ · u) − p(∇ · u) + φ − ∇ · q (9.19) 2 3 3 9.3 Thermal energy (or heat) equation What a mess. What games can we play to make this more comprehensible? Let’s go back to a version of the momentum equation: ρ Du Dt = ρg + u · (∇ · τ ) (9.20) Multiply (dot) through by u ρ D Dt
1 2 u 2  = ρu · g + u · (∇ · τ ) (9.21) This looks a little familiar. It is the mechanical energy equation. Subtract this from the total energy equation
 D 1 ρ e+ u = ρg · u + u · (∇ · τ ) − p(∇ · u) + φ − ∇ · q Dt 2
 D 1 2 = ρu · g + u · (∇ · τ ) u − ρ Dt 2 and we get the thermal energy (or heat) equation: ⇒ OR De Dt DT ρCv Dt ρ = −p(∇ · u) + φ − ∇ · q (9.22) = −p(∇ · u) + φ − ∇ · q (9.23) The second equation is obtained using de = Cv dT . Here −p(∇ · u) ≡ change in thermal energy due to compression φ ≡ change in thermal energy due to viscosity −∇ · q ≡ change in thermal energy due to heat transfer 9.4 Approximations Almost always it is safe to say that φ  1, and it is ignored (Jim – Comparing φ to what other term that is O(1)?). ρCv DT + p(∇ · u) Dt = −∇ · q (9.24) Here, ρCv DT Dt is temperature changes at constant volume, and p(∇ · u) is temperature changes due to changing volume. Remember the Boussinesq approximation? I stated it as ρ = ρo + ρ
(x, y, z, t) (9.25) And it allowed us to simplify the continuity equation to ∇·u = 0 (9.26) It would be tempting to simply remove the −p(∇ · u) term, leaving ρCv DT Dt = −∇ · q 4 (9.27) But this isn’t correct, because here ∇ · u is similar in scale to ρCv DT Dt , so we’ll keep it. If we can’t say ∇ · u = 0, let’s substitute in the continuity equation instead Dρ + ρ∇ · u = 0 Dt ⇒ ∇·u=− 1 Dρ ρ Dt (9.28) Substitution yields ρCv Rewrite as DT p Dρ − Dt ρ Dt = −∇ · q 
DT 1 Dρ ρCv + ρp − 2 Dt ρ Dt Using the fact that − ρ12 Dρ Dt = = −∇ · q (9.29) (9.30) D 1 Dt ρ ρCv DT Dα + ρp Dt Dt = −∇ · q (9.31) where α = ρ1 ≡ specific volume. The equation of state for a perfect gas is p = ρRT or pα = RT (9.32) Differentiate and rearrange: p ⇒ Dα Dp +α Dt Dt Dα p Dt DT Dt DT Dp = R −α Dt Dt = R (9.33) (9.34) Substitute: ρCv DT DT Dp + ρR − ρα Dt Dt Dt DT Dp − ρ(Cv + R) Dt Dt DT Dp ρCp − Dt Dt = −∇ · q = −∇ · q = −∇ · q (9.35) Where Cv + R = Cp . Here, ρCp DT Dt is temperature change from heating at constant pressure, and Dp is the correction term for constant pressure. Thus the approximations made are φ small and Dt an ideal gas (though structurally similar to case with any equation of state). The Boussinesq approximation says that Dp Dt is small. Hence ρCp DT Dt = −∇ · q (9.36) Further, using Fourier’s Law of conduction q = −k∇T (9.37) which comes from measuring the heat flow through a block with different temperatures on each side, we get the following result ρCp DT Dt = ∇ · (k∇T ) 5 (9.38) Here k ≡ thermal conductivity, and if constant we have ρCp DT Dt = k∇2 T (9.39) (Jim – Here we have used an atmospheric approximation, ideal gas, and an oceanic one, Boussinesq; perhaps less than ideal). This assumption that Dp Dt is small is okay for the ocean, but poor for the atmosphere. What can we do about that? Go back to ρCp DT Dp − Dt Dt = −∇ · q (9.40) ∇·q ρ (9.41) Divide through by ρ DT 1 Dp − Dt ρ Dt Cp = − The source on the RHS has gone from a per volume to a per mass. Lets give it a new name, Q̇. DT 1 Dp − Dt ρ Dt = Q̇ Cp 1 DT 1 1 Dp − T Dt ρ T Dt = Q̇ T (9.43) Cp 1 DT 1 Dp −R T Dt p Dt = Q̇ T (9.44) Cp (9.42) Divide through by T Noting that p = ρRT ⇒ Since 1 dx x dt = d dt 1 pT = R ρ ln x Cp D D Q̇ Dη ln T − R ln p = ≡ Dt Dt T Dt (9.45) Where η ≡ entropy per mass, and δη ≡ δQ T where Q ≡ heat per mass. How on earth could that possibly be useful? Let me introduce a new variable
θ=T ps p R/Cp ≡ potential temperature (9.46) Here ps ≡ surface pressure. The potential temperature is the temperature a parcel would have if you moved it adiabatically (no heat enters or leaves the parcel) from whatever height it sits at, down to the surface. Take the ln
 R ps ln θ = ln T + ln Cp p Cp ln θ = Cp ln T − R ln p + R ln ps (9.47) Take the time derivative Cp D ln θ Dt = Cp D D ln T − R ln p Dt Dt 6 (9.48) Compare with our expression for rate of change of entropy: Dη Dt = Cp D D ln T − R ln p Dt Dt (9.49) So, Cp D ln θ Dt Dη Dt = (9.50) This tells us that a parcel that conserves entropy travels along a surface of constant θ. Why do we care? In a compressible atmosphere it isn’t the density gradient with height that determines vertical stability, it is the entropy gradient that determine stability. The potential temperature analog in the ocean is potential density, defined as the density obtained if a parcel is taken to a reference pressure at constant salinity and adiabatically. Back to the heat equation Cp D Dη Q̇ ln θ = = Dt Dt T Cp Dθ θ Dt Dθ Cp Dt No assumptions here except ideal gas and φ small. 9.5 Reading for class 10 None assigned in class. 7 −∇ · q ρT −∇ · q = ρT θ = − ∇·q ρT = Dp Dt retained. (9.51)