Post on 17-Dec-2015
AOSS 321, Winter 2009 Earth System Dynamics
Lecture 102/10/2008
Christiane Jablonowski Eric Hetland
cjablono@umich.edu ehetland@umich.edu
734-763-6238 734-615-3177
Today’s class
• Conservation principles:– Conservation of mass: Continuity equation– Thermodynamic equation– Conservation of total energy
• The equations of motion for the atmosphereLeads to
So far: Equations of motion
€
Du
Dt−
uvtan(φ)
a+
uw
a= −
1
ρ
∂p
∂x+ 2Ωvsin(φ) − 2Ωw cos(φ) + ν∇ 2(u)
Dv
Dt+
u2tan(φ)
a+
vw
a= −
1
ρ
∂p
∂y− 2Ωusin(φ) + ν∇ 2(v)
Dw
Dt−
u2 + v2
a= −
1
ρ
∂p
∂z− g + 2Ωucos(φ) + ν∇ 2(w)
p = ρRT and α =1
ρ 6 unknowns: u, v, w, , p, T4 equations
We need two more equations!
Recall: Eulerian Point of View
• This point of view, where is observer sits at a point and watches the fluid go by, is known as the Eulerian point of view.
– Useful for developing theory
– Looks at the fluid as a field.
– Requires considering only one coordinate system for all parcels
– A value for each point in the field – no gaps or bundles of “information.”
Recall: Lagrangian point of view is that the parcel is moving.
x
y
z
and it changes shape!
Parcel of air
The Eulerian point of view: our parcel is a fixed volume and the fluid flows through it.
x
y
z
≡ density =mass per unit volume (V)
V = xyz
m = xyz
p ≡ pressure = force per unit area acting on the particle of atmosphere
The Eulerian point of view: our parcel is a fixed volume and the fluid flows through it.
x
y
z
Introduce mass flux, ρu
x
y
z
ρu = mass flux at (x, y, z) in the x direction.
.
(x, y, z)
x axis
Mass flux out = ρu + (∂ρu/∂x)x/2 + higher order terms
Introduce mass flux, ρu
x
y
z
ρu = mass flux at (x, y, z) in the x direction.
.
(x, y, z)
x axis
Mass flux in = ρu - (∂ρu/∂x)x/2 + higher order terms
What is mass into and out of the fixed volume?
• Mass flux times the area of the face of the box which it is flowing through.
zyx
ux
u
zyx
ux
u
∂∂
−
∂∂
+
)2
)((
)2
)((
Mass out right (downstream) face
Mass in left (upstream) face
Conservation of mass
• The change of mass in the box is equal to the mass that flows into the box minus the mass that flows out of the box.
zyxux
zyx
ux
uzyx
ux
u
∂∂
−=
∂∂
+−
∂∂
−
)(
)2
)(()2
)((
Extend to 3-Dimensions
€
=−∂∂x
(ρu) +∂
∂y(ρv) +
∂
∂z(ρw)
⎛
⎝ ⎜
⎞
⎠ ⎟ΔxΔyΔz
• The change of mass in the box is equal to the mass that flows into the box minus the mass that flows out of the box.
t∂∂
=
This is equal to the local rate of change
Mass flux per unit volume:
€
−∂∂x
(ρu) +∂
∂y(ρv) +
∂
∂z(ρw)
⎛
⎝ ⎜
⎞
⎠ ⎟
Conservation of Mass: The continuity equation
€
∂∂t
= −(∂
∂x(ρu) +
∂
∂y(ρv) +
∂
∂z(ρw))
∂ρ
∂t= −∇ • (ρ
r v )
Conservation of Mass: The continuity equation
€
dρ
dt= −ρ∇ •
r v
Can you show that the continuity equation can also be written as
So far: Equations of motion
€
Du
Dt−
uvtan(φ)
a+
uw
a= −
1
ρ
∂p
∂x+ 2Ωvsin(φ) − 2Ωw cos(φ) + ν∇ 2(u)
Dv
Dt+
u2tan(φ)
a+
vw
a= −
1
ρ
∂p
∂y− 2Ωusin(φ) + ν∇ 2(v)
Dw
Dt−
u2 + v2
a= −
1
ρ
∂p
∂z− g + 2Ωucos(φ) + ν∇ 2(w)
Dρ
Dt= −ρ∇ • u
p = ρRdT and α =1
ρ
6 unknowns: u, v, w, , p, T5 equationsAlmost closed system!One more equation is missing.
Conservation of thermodynamic energy: The thermodynamic equation
• Change in internal energy is equal to the difference between the heat added to the system and the work done by the system.
• Internal energy is due to the kinetic energy of the molecules.
• Total thermodynamic energy is the internal energy plus the energy due to the parcel moving.
Thermodynamic Equation
€
cv
DT
Dt+ p
Dα
Dt= J
J is the source or sink of heating which are – radiation, latent heat release, thermal conductivity, frictional heating
= 1/ is the specific volumecv = 717 J K-1 kg-1 is the specific heat of dry air at constant volume
Another form of the Thermodynamic Equation
cp = 1004 J K-1 kg-1 is the specific heat of dry air at constant pressure
Note: cp = cv + Rd = 1004 J K-1 kg-1 Rd = 287 J K-1 kg-1 gas constant for dry air
€
c p
DT
Dt−α
Dp
Dt= J
Can be derived via the ideal gas law (show):
Full equations of motion(Navier-Stokes equations)
€
Du
Dt−
uvtan(φ)
a+
uw
a= −
1
ρ
∂p
∂x+ 2Ωvsin(φ) − 2Ωw cos(φ) + ν∇ 2(u)
Dv
Dt+
u2tan(φ)
a+
vw
a= −
1
ρ
∂p
∂y− 2Ωusin(φ) + ν∇ 2(v)
Dw
Dt−
u2 + v2
a= −
1
ρ
∂p
∂z− g + 2Ωucos(φ) + ν∇ 2(w)
Dρ
Dt= −ρ∇ • u
cv
DT
Dt+ p
D
Dt(1
ρ) = J
p = ρRT and α =1
ρ
6 unknowns: u, v, w, , p, T6 equationsClosed system!These are our equations of motions in the atmosphere!
Conservation of total energy• The law of conservation of energy states that the sum of
all energies in the universes is constant.
• There are many forms of energy in the atmosphere, e.g.: kinetic energy, potential energy, latent heat energy, radiant energy, …
• Radiant energy from the Sun is the source of nearly all of the total energy in the atmosphere/ocean system.
• When solar energy is absorbed at the Earth’s surface it appears as internal energy (noticeable as temperature changes).
• One of the major challenges in atmospheric science is determining how this internal energy is converted into the other forms of energy.
Derivation of the total energy equation
• Let’s gain some insights into the nature of the energies in the atmosphere.
• Start by taking the dot (scalar) product of the acceleration vector with the velocity vector
• Equivalent to multiplying the components of the momentum equations with their respective component velocities (u,v,w).
€
dr v dt
€
rv
Derivation of the total energy equation
• Multiplication leads to
• Sum all three expressions:€
1
2
Du2
Dt−
u2vtan(φ)
a+
u2w
a= −
u
ρ
∂p
∂x+ 2Ωuvsin(φ) − 2Ωuwcos(φ) + uFx
1
2
Dv2
Dt+
u2vtan(φ)
a+
v2w
a= −
v
ρ
∂p
∂y− 2Ωuvsin(φ) + vFy
1
2
Dw2
Dt−
w u2 + v2( )
a= −
w
ρ
∂p
∂z− wg + 2Ωuwcos(φ) + wFz
€
D
Dt
u2 + v 2 + w2
2
⎛
⎝ ⎜
⎞
⎠ ⎟= −
1
ρ
r v • ∇p − gw +
r v •
r F
€
Fx,Fy,Fz are the components of the friction
Metric terms and Coriolis forces cancel!
Derivation of the total energy equation
• Leads to
€
D
Dt
u2 + v 2 + w2
2
⎛
⎝ ⎜
⎞
⎠ ⎟= −
1
ρ
r v • ∇p − gw +
r v •
r F
Rate of change of kinetic energy perunit mass
Pressure advectiondivided by density
€
−gw = −gDz
Dt= −
DΦ
Dt
€
D
Dt
u2 + v 2 + w2
2+ Φ
⎛
⎝ ⎜
⎞
⎠ ⎟= −
1
ρ
r v • ∇p +
r v •
r F
Measure of thework required toraise a unit massa distance z
Sum of the kinetic and potential energiesper unit mass of an atmospheric parcel
frictionaleffects
Derivation of the total energy equation
• Also referred to as mechanical energy equation:
• Adding the first law of thermodynamics
leads to
• Use:
€
D
Dt
u2 + v 2 + w2
2+ Φ
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
ρ
r v • ∇p −
r v •
r F = 0
€
cv
DT
Dt+ p
Dα
Dt= J
€
cv
DT
Dt+ p
Dα
Dt+
D
Dt
u2 + v 2 + w2
2+ Φ
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
ρ
r v • ∇p −
r v •
r F = J
€
1
ρ
r v • ∇p = α
Dp
Dt− α
∂p
∂t
€
pDα
Dt=
D
Dtαp( ) − α
Dp
Dt
Total Energy Equation
• Rewriting leads to the energy equation:
• Relationship implies that if the flow is frictionless , adiabatic (J=0) and steady-state
, then the quantity
€
D
Dt
u2 + v 2 + w2
2+ Φ + cvT + pα
⎛
⎝ ⎜
⎞
⎠ ⎟− α
∂p
∂t−
r v •
r F = J
€
∂p ∂t = 0( )
€
rF = 0( )
€
u2 + v 2 + w2
2+ Φ + cvT + pα =
u2 + v 2 + w2
2+ Φ + cvT + RdT =
u2 + v 2 + w2
2+ Φ + c pT
is constant (conserved).