From the last time, we ended with an expression for the energy equation. u = ρg u + (τ u) q (9.1)

Transcription

1 Lecture 9 9. Administration None. 9. Continuation of energy equation From the last time, we ended with an expression for the energy equation ρ D (e + ) u = ρg u + (τ u) q (9.) 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.) he 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. he 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. he u ( τ) terms is the easiest of the two since we came up with an expression for τ when we looked at momentum conservation. τ = (µe p 3 ) µ u = p + µ u + µ 3 ( u) u ( τ) = u ( p + µ u + µ ) 3 ( u) (9.3) 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

2 Notice, there is no between the and the u. u is the gradient of the velocity... he velocity gradient tensor! hus, this term looks like u = G (9.4) τ : G (9.5) But G = e + r, τ : (e + ) r (9.6) Recall that r is an antisymmetric tensor, and τ is a symmetric tensor. he double dot product of asymmetric and antisymmetric matrix is zero A : B = A ij B ij (9.7) i j In D [ ] τ τ : τ τ [ 0 ω 0 ω ] ( ω ) ( = τ (0) + τ + τ ω ) + τ (0) ω = τ τ ω = 0 (9.8) his last line is due to the fact that since τ is symmetric, τ = τ. hus, you can get rid of the r term and end up with τ : e (9.9) Substitute in our expression for τ, ( µe [p + 3 ] ) µ u I : e Lets start with an easy one µe : e pi : e µ ui : e (9.0) 3 pi : e = p u x + p v + p w z = p u (9.) hat is, change in energy due to compression or expansion. he last term is also pretty easy 3 µ ui : e = ( u 3 µ u x + v + w ) z = 3 µ u( u) = 3 µ( u) (9.)

3 Figure 9.: (fig:lec9muparameter) µ parameterizes work associated in pulling apart molecules inabloboffluid. Finally, the first term µe : e e : e = e e e 3 e e e 3 : e e e 3 e e e 3 e 3 e 3 e 33 e 3 e 3 e 33 e e + e e + e 3 e 3 + = e e + e e + e 3 e 3 + (9.3) e 3 e 3 + e 3 e 3 + e 33 e 33 Note e e = e e = = 4 ( ) u (9.4) x [ ( u + v )] x ( ( u ) + u ( ) ) v v x + x = 4 etc. ( u ) + u v x + 4 ( v x ) (9.5) Consider just the e e term. Remember, ( ) u µe : e µ = ( µ u ) u (9.6) he µ u u bit is associated with normal stress, and with the velocity gradient. hus, this term has something to do with the work it takes to pull apart molecules in a blob of fluid (see figure 9.). he other terms, such as ( u ) and u v x, say something similar about shear and twisting. Clearly this is a bit of a mess. hankfully, what is typically done is to sweep it all under the rug and define φ = (µe : e 3 ) µ( u) (9.7) his allows us to write viscous component of the deformation work rate τ : u = p( u) +φ (9.8) racking this all the way back to the energy equation gives: ρ D (e + ) u = ρg u + u [ p + µ u + µ ] 3 ( u) p( u)+φ q (9.9) 3

4 9.3 hermal 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: Multiply (dot) through by u ρ D ρ Du ( ) u = ρg + u ( τ) (9.0) = ρu g + u ( τ) (9.) his looks a little familiar. It is the mechanical energy equation. Subtract this from the total energy equation ρ D (e + ) u = ρg u + u ( τ) p( u)+φ q ρ D ( ) u = ρu g + u ( τ) and we get the thermal energy (or heat) equation: ρ De = p( u)+φ q (9.) OR = p( u)+φ q (9.3) he second equation is obtained using de = C v d.here 9.4 Approximations 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 Almost always it is safe to say that φ, and it is ignored (Jim Comparing φ to what other term that is O()?). + p( u) (9.4) Here, is temperature changes at constant volume, and p( u) is temperature changes due to changing volume. Remember the Boussinesq approximation? I stated it as And it allowed us to simplify the continuity equation to ρ = ρ o + ρ (x, y, z, t) (9.5) It would be tempting to simply remove the p( u) term, leaving u = 0 (9.6) (9.7) 4

5 But this isn t correct, because here u is similar in scale to, so we ll keep it. If we can t say u = 0, let s substitute in the continuity equation instead Dρ + ρ u =0 u = Dρ (9.8) ρ Substitution yields p Dρ (9.9) ρ Rewrite as ( + ρp ) Dρ ρ (9.30) Using the fact that ρ Dρ = D ρ + ρpdα where α = ρ specific volume. he equation of state for a perfect gas is (9.3) p = ρr or pα = R (9.3) Differentiate and rearrange: p Dα + αdp p Dα = R = R αdp (9.33) (9.34) Substitute: + ρr ραdp ρ(c v + R) Dp ρ Dp (9.35) Where C v + R =. Here, ρ is temperature change from heating at constant pressure, and Dp is the correction term for constant pressure. hus the approximations made are φ small and an ideal gas (though structurally similar to case with any equation of state). he Boussinesq approximation says that Dp is small. Hence ρ (9.36) Further, using Fourier s Law of conduction q = k (9.37) which comes from measuring the heat flow through a block with different temperatures on each side, we get the following result ρ = (k ) (9.38) 5

6 Here k thermal conductivity, and if constant we have ρ = k (9.39) (Jim Here we have used an atmospheric approximation, ideal gas, and an oceanic one, Boussinesq; perhaps less than ideal). his assumption that Dp is small is okay for the ocean, but poor for the atmosphere. What can we do about that? Go back to ρ Dp (9.40) Divide through by ρ Dp ρ = q ρ (9.4) hesourceontherhshasgonefromapervolumetoapermass.letsgiveitanewname, Q. Dp ρ = Q (9.4) Divide through by ρ Dp = Q (9.43) Noting that p = ρr p = R ρ R Dp p = Q (9.44) Since dx x dt = d dt ln x D ln R D Q ln p = Dη (9.45) Where η entropy per mass, and δη δq where Q heat per mass. How on earth could that possibly be useful? Let me introduce a new variable θ = ( ) R/Cp ps potential temperature (9.46) p Here p s surface pressure. he 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. ake the ln ln θ = ln + R ( ) ps ln p ln θ = ln R ln p + R ln p s (9.47) ake the time derivative D ln θ = C D p ln R D ln p (9.48) 6

7 Compare with our expression for rate of change of entropy: So, Dη D = ln R D ln p (9.49) D Dη ln θ = (9.50) his 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. he 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 D Dη ln θ = = Q θ Dθ Dθ ρ ρ = θ q (9.5) ρ No assumptions here except ideal gas and φ small. Dp retained. 9.5 Reading for class 0 Noneassignedinclass. 7