R g. o p2. Lecture 2: Buoyancy, stability, convection and gravity waves

Transcription

1 Lecture : Clarifications of lecture 1: Hydrostatic balance: Under static conditions, only gravity will work on the fluid. Why doesn't all the fluid contract to the ground? Pressure builds u and resists the comression. The balance between this ressure force and gravity yields the hydrostatic balance relation. What actually causes the resistance is the vertical gradient in ressure the ressure acts equally in all directions but it decreases with height, so at height z it is a certain value and at height z+δz it is slightly lower, which will cause the fluid to want to move uwards. But gravity will ull it down and balance will ensue: = ρg. z Hysometric relation: relates a layer thickness to the vertically integrated (over logressure) temerature. Can be in terms of: Height: z ( ) z( ) = R g Geootential height: zφ ( ) zφ ( ) = Geootential: φ ( z ) φ( z1) = R R g o Lecture : Buoyancy, stability, convection and gravity waves T1-radiative equilibrium and observed temerature rofile: The jum at the surface, and as we will see later also the troosheric lase rate are unstable and will lead to convection. T-schematic radiative equilibrium and observed temerature rofile. The trooshere is in Radiative-convective equilibrium. T3-schematic of Radiative-convective equilibrium. Understanding this convection is the toic of this class. In atmoshere we have convection in moist comressible fluid. Much more comlicated than a an of water. We will start from simle to more comlex. Why does convection occur? Horizontally uniform heating of water from below will force a satially uniform temerature decreasing with height. Even though light fluid is on bottom, the gravitational force can be balanced by ressure gradient, according to hydrostatic balance law. Zero net force. So why does convection develo? How come it is satially varying even if balanced state is horizontally homogeneous?

2 Instability: Lets look at the following examle of a ball on a hill-valley: Comonent of gravitational acceleration along the sloe: -gdh/dx Ball at oint A, where dh/dx=0 is in equilibrium stable. What if dislaced an increment δx? Sloe at x A +δx: dh/dx(x A +δx) dh/dx(x A )+(d h/dx ) A δx=(d h/dx ) A δx The equation of motion at δx: d (δx)/dt =-gdh/dx=-g(d h/dx ) A δx σ + t σ t Solution: δ x ( d h t) = C+ e + C e σ = ± ± g dx At the eak (d h/dx ) A<0 so the solution is exonentially growing- a small erturbation will grow. Instability, convection. At the trough (d h/dx ) B>0 so the solution is oscillatory. Energetics: At the trough, the erturbed ball moves uhill- we need to do work since its otential energy increases, and it won't haen. In the absence of an external energy source. At the ridge, the ball moves downhill, decreasing otential energy, and increasing kinetic energy. Motion will continue. Convection in water: Buoyancy: A light object in water will bounce back u. What about a fluid arcel? Fluid arcel: small but finite size, which is thermally isolated from environment and always at same ressure as environment. Assume ρ=ρ(t).

3 Above oint A the fluid is the same in 1, and (arcel). Hence ressure at A1, A, A is the same. Above oint B fluid column is lighter, and ressure is lower. The forces acting on the arcel: (subscrit E for environment and for arcel) Gravity deends on the arcel mass: F g =-gρ δaδz Pressure gradient force- deends on the environmental ressure: F t + F b =-δ E δa= gρ E δa δz Net force on arcel: F g +F t + F b = g(ρ E ρ ) δa δz Divide by the arcel mass ρ δa δz to get the acceleration, which is a reduced form of gravity: Buoyancy: b F tot /m = -g(ρ ρ E )/ ρ If ρ <ρ E, the arcel is ositively buoyant: b>0, and it will accelerate uwards. If ρ >ρ E, the arcel is negatively buoyant: b<0, and it will accelerate downwards. Lets look at a secific density rofile, which we assume deends only on temerature (incomressible water): ρ ref=ρ(1- α[t-t ref ]) Assume that a arcel which is similar to the environment at z1 starts rising, to height z=z1+ Δz, and when it rises it does not exchange energy with the surroundings, thus its temerature remains constant. At z1: ρ = ρ E (z1), T=T E (z1) At z: ρ = ρ E (z1), T=T E (z1) but ρ E = ρ E (z)= ρ E (z1)+ Δz dρ E /dz The arcel's buoyancy is: b(z)=-g(ρ ρ E )/ρ =g/ρ dρ E /dz Δz =g/ρ E(z1) dρ E /dz Δz The buoyancy is ostivie/negative/0 if dρ E /dz is ostivie/negative/0, and the arcel will rise more/sink back/remain still. For the equation of state above, b>0,=0,<0 if dt/dz <0, =0, >0. In the absence of heating and daming, a arcel will be unstable if the density increases with height. Mathematically: the equation of motion for the arcel's dislacement Δz: d (Δz)/dt =b= g/ρ E(z1) dρ E /dz Δz -N Δz N -b/δz =-g/ρ E(z1) dρ E /dz is the buoyancy frequency which we will return to later. if dρ E /dz>0, N <0 and the solution is exonential (instability): N t N t Δ z = Δ e + Δ e 1 If dρ E /dz<0, N >0 and the solution is oscillatory: Δ z = Δ1 cos( Nt) + Δ sin( Nt) We will discuss this case in a bit.

4 Energetics view: The energetics argument of the ball on the hill can't be alied to the arcel, because we are dealing with a continuous fluid, and when the arcel rises, another arcel has to go back down. Thus, we need to examine the initial and final states of the entire fluid, and see what haened to the otential energy: Lets assume the erturbations occurs with a arcel at z1 rising to z, while a arcel at z descends to z1 and besides that all the fluid remains unchanged. The initial otential energy: PE initial =g(ρ 1 z1+ ρ z). The final otential energy: PE final =g(ρ 1 z+ ρ z1) Their difference: PE final - PE initial =-g(ρ -ρ 1 )(z-z1) -g dρ E /dz (z-z1) This yields the same condition if dρ E /dz>0 the arcel's otential energy will be reduced, and its kinetic energy will increase, meaning it will continue moving. If dρ E /dz<0 the arcel's otential energy will increase, on exense of its kinetic energy, and the arcel will sto moving. Note: energetics considerations only show if instability is lausible, but we need an analysis like the one above to determine if instability will arise or not. Water heated from below: We assume a constant known heating rate er meter squared of H (W/m ). This heating will be carried uwards by convection, at a rate H. The heat content of a unit volume of fluid is: ρct where C is the secific heat of water. The heat flux (amount of heat transorted across a unit area) is then ρctw, where w is vertical velocity. W is highly variable in convection, hence we need to integrate over time (a few convective cycles) and sace (over a few convective cells). Assume that over the domain and some time, 1/ the fluid is moving uward with temerature T+ΔT, and half the fluid is moving downward with temerature T. The net integrated flux is: <H> = 1/ ρc(t+δt)w- 1/ ρctw=1/ ρcδt. How do we get w? from energy considerations: Assume all kinetic energy comes from release of otential energy: ΔKE = ΔPE= g Δρ Δz ΔKE= ρ / (u + v + w ) 3/ ρw w /3 g Δρ/ ρ Δz For the equation of state above: ρ ref =ρ(1- α[t-t ref ]), Δρ=ρ ref αδt Plugging in w, and then in <H> we get: <H>=1/ ρ ref C(/3 g αδz) 1/ ΔT 3/ Dry convection in a comressible atmoshere: We need to account for comression due to ressure changes as the arcels rise/descend. Assume that a arcel at z1 has the environmental ressure and temerature: = E (z1), T1=T E (z1), ρ1=/(rt1) The arcel rises to z adiabatically. Its ressure decreases like the environment. What haens to T? Consider a arcel of unit mass: ρv=1. Its exchange of energy with the surroundings follows the law:

5 δq= δu+ δw where δq is the heating from outside, δu is the added internal energy and δw is the work done by the arcel on its surroundings. δq= CvdT+dV where Cv is the secific heat er unit volume (choose this since if dv=0, all δq goes to a change in T. Want to find dt/, so change from dv to using the ideal gas law: dv=d(1/ρ)=-1/ρ dρ, dρ=/(rt)-/(rt )dt dv= -/(ρ RT) + /(ρ RT ) dt =RdT-/ρ δq= (Cv+R)dT-/ρ = CdT - /ρ Note: C>Cv, meaning temerature will change less for a given heating when ressure, rather than volume is ket constant. This is because some of the energy goes towards work due to volume changes. For adiabatic dislacements, δq=0, so CdT = /ρ Where the variables are taken to be the arcel's. From the hydrostatic relation, =-ρ E gdz, so dt/dz= - ρ E /ρ g/c For small dislacements, ρ ρ E, so that their ratio can be taken to be 1, and we get the dry adiabatic lase rate: Γd=-dT/dz=g/C For C=1005J/Kg/K, Γd 10K/km.