1

Energy Transfers Link Ch. 1-4

•  Ch. 1 – Atmospheric Composition–  “dry air” (N2, O2, Ar, CO2), Water, (particles)

•  Ch. 2 – First and Second Laws– Energy Balance (System/Environment)

•  Ch. 3 – Stefan-Boltzmann Equation– Kirchoff’s Law

•  Ch. 4 – Water and Phase Changes– Latent Heat of Water

What did we learn in Ch. 1?•  What P, T, U are for a fluid•  What an ideal gas is•  How P, T, v relate for an ideal gas (and we call

this relationship an equation of state)•  What chemical components constitute the

atmosphere (for homosphere <110 km)•  What the hydrostatic balance is•  How p, T vary with z for observed, “standard,”

isopycnic, isothermal, constant lapse-rate atmospheres

Key Combined 1st+2nd Law Results•  1st Law: du=dq+dw; u is exact Eq. 2.8

•  du=dqrev-pdv (reversible, expansion only) p. 56•  Define Enthalpy: H=U+PV Eq. 2.12

•  dh=du+pdv+vdp (also Eq. 2.32: dh=vdp-Tdη)•  2nd Law: [dqrev/T]int.cycle=0 Eq. 2.27•  Define Entropy: dη=dqrev/T Eq. 2.25a

•  Tdη=dqrev

•  du=Tdη-pdv•  Define Gibbs: G=H-Tη Eq. 2.33

•  dg=dh-Tdη-ηdT=(du+pdv+vdp)-Tdη-ηdT•  dg=du-(Tdη-pdv)+vdp-ηdT=vdp-ηdT p. 58

•  (δp/δt)g=η/v Eq. 2.40

What did we learn in Ch. 3?•  Radiative transfer definitions

–  Diffuse vs. direct–  From all directions (irradiance F) or one (radiance I)–  Absorption coefficient and optical thickness–  Blackbody radiation

•  Radiative transfer equations–  Kirchoff’s law (gaseous molecules): Eλ = Aλ–  Planck’s radiation law: F = fcn(λ, T)–  Wien’s displacement law: λ ~ 3000/T–  Stefan-Boltzmann law (black body: Fbb = σT4

What did we learn in Ch. 4?•  Phase equilibrium definitions

–  Criteria of phase equilibria (thermal, mechanical, chemical)

–  Degrees of freedom reduced by phases–  Phase diagram of (pure) water

•  Clausius-Clapeyron equation (dp/dT=Llvp/RvT2)–  Strong dependence of esat on temperature (and Llv)

•  Doubles every 10°C–  There are two ways to saturate, i.e. H=e/esat=1

•  Increase water vapor in parcel (e)•  Decrease temperature (and hence esat)

What did we learn in Ch. 4?•  Phase equilibrium definitions

–  Criteria of phase equilibria (thermal, mechanical, chemical)

–  Degrees of freedom reduced by phases–  Phase diagram of (pure) water

•  Clausius-Clapeyron equations: Will H2O condense?–  Strong dependence of esat on temperature (and Llv)

•  Doubles every 10C–  There are two ways to saturate, i.e. H=e/esat=1

•  Increase water vapor in parcel (e)•  Decrease temperature (and hence esat)

2

Reversible-Adiabatic-Work

Adiabatic

First Law

Reversible

Internal Energy

Ideal Gasp1v1T1

= R = p2v2T2

Δu = cvdT

€

dw = −pdv

Δu =Q +W

Q = 0

€

T2T1

=P2P1

"

# $

%

& '

Rcp

thick walls

Low P, High T

Frictionless

Reversible, Adiabatic

(mass is conserved)

€

−pdv = cvdT

−RTv

#

$ %

&

' ( dv = cvdT

− R1

2

∫ dvv

= cv1

2

∫ dTT

Reversible, Adiabatic

Expansion of an Ideal Gas

€

−pdv = cvdT

−RTv

#

$ %

&

' ( dv = cvdT

− R1

2

∫ dvv

= cv1

2

∫ dTT

−R lnv2 − lnv1( ) = cv lnT2 − lnT1( )

ln v2v1

#

$ %

&

' (

−R

= ln T2T1

#

$ %

&

' (

cv

T2T1

#

$ %

&

' ( =

v2v1

#

$ %

&

' (

−R cv

T2T1

#

$ %

&

' ( =

RT2p2

#

$ %

&

' (

RT1p1

#

$ %

&

' (

#

$

% % % %

&

'

( ( ( (

−R cv

=T2T1

#

$ %

&

' (

−R cv p2p1

#

$ %

&

' (

Rcv

T2T1

#

$ %

&

' (

1+R cv=

p2p1

#

$ %

&

' (

Rcv

T2T1

#

$ %

&

' ( =

p2p1

#

$ %

&

' (

RR +cv

=p2p1

#

$ %

&

' (

Rcp

Hurricane as Carnot Cycle

1:Add Heat (isothermally)

2:Adiabatic

3:Lose Heat (isothermally)

4:Adiabatic

Hurricane as Carnot Cycle •  Step 1 (A-B): air spirals in, acquiring heat

isothermally expanding •  Step 2 (B-C): air ascends and expands adiabatically

in eye •  Step 3 (C-D): air loses heat and begins to compress

isothermally •  Step 4 (D-A): air compresses adiabatically and

returns to surface

•  Caveats –  Not open: rain, mixing

Potential Temperature

3

Hydrostatic Balance

•  Applicable to most atmospheric situations (except fast accelerations in thunderstorms)

€

g = −1ρ∂p∂z

∂p = −pgRdT

∂z

Curry and Webster, Ch. 1

Special Cases of Hydrostatic Equilibrium

•  1. ρ=constant (homogeneous)– H=8 km =RT/g=scale height Eqn. 1.39

•  2. constant lapse rate (e.g. if hydrostatic, homogeneous, and ideal gas)–  -dT/dz=constant=-g/R=-34/deg/km

•  3. isothermal T=constant (and ideal gas)–  p=p_0*exp(-z/H)

Homogeneous Atmosphere

•  Density is constant •  Surface pressure is finite•  Scale height H gives where pressure=0

€

p0 = ρgH

H =pρg

=RdT0g

€

g = −1ρ∂p∂z

dp = −ρgdz

dpp0

0∫ = − ρgdz

0

H∫

0 − p0 = − ρgH − 0( )

Curry and Webster, Ch. 1

Hydrostatic + Ideal Gas + Homogeneous

•  Evaluate lapse rate by differentiating ideal gas law

€

p = ρRdT∂p∂z

= ρRd∂T∂z

−1ρ∂p∂z

%

& '

(

) * = Rd −

∂T∂z

%

& '

(

) *

g = −1ρ∂p∂z

€

Γ = −∂T∂z

=gRd

= 34.1oC/km

Density constant

Ideal gas

Hydrostatic

Curry and Webster, Ch. 1

Hydrostatic Equilibrium Example���(Constant Lapse Rate) Terminology Review

•  Isotropic–  Same in all directions, such that F=πI

•  Reflection–  Change in direction but not energy or wavelength

•  Isentropic –  Adiabatic+reversible

•  For adiabatic, ideal: –  p determines T and vice versa

•  Potential temperature–  temperature that air would have if raised/lowered to a

reference pressure adiabatically and reversibly.

4

Summary: Solar Radiation•  Luminosity of the sun L0 ~ 3.9x1026 W (p. 331)•  Irradiance F=Luminosity/Area=L0/(4πr2)= 6.44x107 W/m2

–  rsun=6.96x108 m [p. 437]•  Estimate blackbody radiation TBB,sun=(F/σ)0.25~ 5800K

–  σ=5.67x10-8 W m-2 K-4 [p. 437]

•  Use Wien’s law to evaluate λsun ~ 0.5 µm (visible)

•  Similarly, λearth ~ 10 µm (infrared) for Tearth ~ 300K

Comparison -Earth & Sun Radiation•  Sun – more energy & shorter wavelength

•  Earth-lower energy and longer wavelength

FL Emitted from sphere

surface 4πr2 FS

Incident on projected disc πr2

FS = FL

Simplified Climate Model

•  Incoming shortwave = Outgoing longwave•  Energy absorbed = Energy emitted

FS = 0.25*S0(1- αp) FL = σTe4

Add an Atmosphere!•  Atmosphere is transparent to non-reflected portion

of the solar beam•  Atmosphere in radiative equilibrium with surface•  Atmosphere absorbs all the IR emission (A=1)

TOA: FS = Fatm

0.25*S0(1- αp) = σTatm4

Tatm = 255K

Fsurf

FS

Fatm

Fatm Atmos: Fsurf = 2Fatm

σTsurf4

= 2σTatm4

Tsurf = 303K

Absorptivity Example

•  Solar heating of the atmosphere– What is the absorptivity?

•  Eqn. 3.23–  kabs=3x10-5m2/kg– ρ=1 kg/m3–  cp=103 J/kgK

• A =0.3 (30%)

10 km

Io=340 W/m2

Heating Rate Example•  What is the temperature change due to absorbed ΔI?

–  Start with Eqn. 3.1, neglect p change: –  Multiply by density: ρcpdT/dt=(1/V)dQ/dt–  p. 83 (def’n): AdF=dQ/dt –  Write for one θ: AdI=dQ/dt–  Divide by V: ΔI/Δz=(1/V)dQ/dt–  Eqn. 3.22: dI=-A I=-Idτ–  Eqn. 3:26: I=Ioexp(-τsecθ)–  Evaluate: dI=-[Ioexp(-τsecθ)]dτ

•  dI/dz=-Ioexp(-τsecθ)(dτ/dz)– ρcpdT/dt=ΔI/Δz~dI/dz=-Ioexp(-τsecθ)(dτ/dz)

Δz

Io

Io-ΔI

θ

5

Water Saturation Pressures

es doubles with every 10C!

T(C) eS (hPa)

10 12.3

20 23.4

30 42.4

40 73.8

(this is one consequence of Clausius-Clapeyron’s equation)

Water Vapor Metrics

Mixing ratio Specific humidity Relative humidity

Water vapor by mass

Water vapor by partial pressure

Water saturation

Virtual temperature

Virtual potential temperature

€

qv =mv

md +mv

=wv1+ wv

€

wv =mv

md

=ρvρd

€

qv = 0.622 ep − 1− 0.622( )e

#

$ %

&

' (

€

ws = 0.622 esp − es

#

$ %

&

' (

€

H ≈wv

ws

€

H =ees

€

θv = T 1+ 0.608qv( ) p0p

#

$ %

&

' (

Rdc pd

€

Tv = T 1+ 0.608qv( )€

wv = 0.622 ep − e

#

$ %

&

' (

€

H = 1

€

qv = 0.622 esp − 1 − 0.622( )es

#

$ %

&

' (

€

MvMd

" # $ %

& ' = Rd

Rv" # $ %

& ' = 0.622

€

RvRd

" # $ %

& ' − 1 = 0.608

Virtual Temperature Simplified Climate Model:���First 2 Assumptions

•  Atmosphere described as one layer–  Albedo αp~0.31: reflectance by surface, clouds,

aerosols, gases –  Shortwave flux absorbed at surface

•  Earth behaves as a black body–  Temperature Te: equivalent black-body temperature of

earth–  Longwave flux emitted from surface

FS=0.25*S0(1- αp)

FL=σTe4

Simplified Climate Model•  At thermal equilibrium (what happens if not?)

•  Observed surface temperature T = 288K•  What’s missing?

FS = FL 0.25*S0(1- αp) = σTe

4

Te = [0.25*S0(1- αp)/σ]0.25

Te ~ 255K

Sensitivity to Albedo•  What if albedo changes?

•  1% decrease in albedo warms temperature 1K•  1% increase in albedo cools temperature 1K

Te = [0.25*S0(1- αp)/σ]0.25

αp=0.31, Te ~ 255K αp=0.30, Te ~ ?

6

Add an Atmosphere!•  Atmosphere is transparent to non-reflected portion

of the solar beam•  Atmosphere in radiative equilibrium with surface•  Atmosphere absorbs all the IR emission (A=1)

TOA: FS = Fatm

0.25*S0(1- αp) = σTatm4

Tatm = 255K

Fsurf

FS

Fatm

Fatm Atmos: Fsurf = 2Fatm

σTsurf4

= 2σTatm4

Tsurf = 303K

Solar Constant•  Luminosity of the sun •  Irradiance at earth S0 = L0/(4πd2) = 1.4x103 W/m2

€

d =1.5 ×1011m (p.437)

€

Area = 4πd 2L0 ~ 3.9x1026 W (p. 331)

Textbook “Greenhouse Effect”

FS + Fghg= σTsurf4

Tsurf≈ 15°C or 59°F

•  How do we know this?–  The Greenhouse Effect of

CO2•  Joseph Fourier 1824•  John Tyndall 1858•  Svante Arrhenius 1896

Tyndall measured how increases with CO (and H2O).

CO2+ + Aerosols

FS

FL

FghgTsurf

Review Problems

•  Hydrostatic equilibrium– Special cases– Pressure altitude dependence

•  More Midterm Review problems– Terminology review

Hydrostatic Equilibrium ExampleConsider a planet with an atmosphere in hydrostatic equilibrium. Assume that theatmosphere is an ideal gas. Also assume that the temperature is a maximum at thesurface of the planet, and, as height increases, the temperature in the atmospheredecreases linearly (in other words, temperature decreases with height at a constant rate).Derive a formula for atmospheric density as a function of height in this atmosphere.

From the hydrostatic equation for an ideal gas (Eqn. 1.42)

€

∂p = −pgRdT

∂z

and a constant lapse rate

€

Γ = −dTdz

we get

€

dp = −pgRdT

dz−dT dzΓ

%

&

' '

(

)

* *

= −pgΓRd

dTT

dpp

=gΓRd

%

& '

(

) * dTT

dpp

=g

ΓRd

%

& '

(

) * dTT∫∫

ln pp0

=gΓRd

%

& '

(

) * ln

TT0

p = p0TT0

%

& '

(

) *

gRdΓ

% & ' (

) *

which is Eqn 1.48. Then dividing both sides by RT and noting that for an ideal gas

€

ρ =pRT

, we get

€

pRT

= ρ =p0

RTTT0

%

& '

(

) *

gRdΓ

=p0

R T0 −Γz( )T0 −ΓzT0

%

& '

(

) *

gRdΓ

Degrees of Freedom Example

7

Clausius Clapeyron ExampleThe saturation vapor pressure at a temperature of 30°C is 42.4 hPa. The gas constant fordry air is 287 J K-1 kg-1. The gas constant for water vapor is 461 J K-1 kg-1.

In addition to the constants given above, here is one more: the saturation vapor pressureat a temperature of 40°C is 73.8 hPa. Assuming that the latent heat of vaporization isconstant, use this information to calculate the numerical value for this latent heat.

The Clausius Clapeyron equation can be integrated if L is assumed constant, and theresult is Eqn. 4.23. Using 30°C=303K and 40°C=313K, and knowing saturationvapor pressure values for each, the only unknown is L. Solving Eqn. 4.23,

€

e2 = e1 exp − LlvRv

1T2

−1T1

#

$ %

&

' (

)

* +

,

- .

Llv = −RvT1T2

T1 −T2

#

$ %

&

' ( ln

e2

e1

#

$ %

&

' ( = −461 303× 313

303− 313#

$ %

&

' ( ln

73.842.4

#

$ %

&

' ( = 2.4 ×106

L=2.4x106 J/kg.

Definition ExampleDefine the following terms, briefly and clearly, in light of their use in the kinetic theoryof gases and the first and second laws of thermodynamics:

a) an ideal gas: vapor whose molecules have collisions with perfect elasticity,typical at low pressures (for air ≤1 atm) and high temperatures (typically ≥300K);vapor that satisfies pv=RT and has the properties that dh=cpdT, du=cvdT, cp-cv=R(p. 44).

b) temperature: the intensive property describing the internal energy of a gas, whichfor an ideal gas depends only on the average speed of the molecules.

c) entropy: a state property whose differential describes the amount of energy that isnot available for doing work for a reversible process (in which maximum work isdone) and satisfies the criteria of an exact differential; it can be evaluated fromEqn. 2.25a and 2.26b:

€

dη =dQT

#

$ %

&

' ( rev

= cpd(lnT) − Rd(ln p) .

d) exact differential: a function ξ for which dξ has the properties (1) for any closedpath

€

dξ = 0∫ , and (2) for ξ(x,y) where x and y are independent, then

€

dξ =∂ξ∂x#

$ %

&

' ( y

dx +∂ξ∂y#

$ %

&

' ( x

dy ≡ Mdx + Ndy⇒ ∂M∂y

#

$ %

&

' ( x

=∂N∂x#

$ %

&

' ( y

.

e) enthalpy: a state property whose differential describes the change in heat for aconstant-pressure process and satisfies the criteria of an exact differential; it isdefined as H=U+pV (Eqn. 2.12).

Water Vapor Quantities ExampleConsider moist air at a temperature of 30°C, a pressure of 1,000 hPa, and a relativehumidity of 50%. Find the values of the following quantities:

a) vapor pressureb) mixing ratioc) specific humidityd) specific heat at constant pressuree) virtual temperature

The saturation vapor pressure at a temperature of 30°C is 42.4 hPa. The gas constant fordry air is 287 J K-1 kg-1. The gas constant for water vapor is 461 J K-1 kg-1.

a) vapor pressure: e=H*es = 0.50*42.4 = 21.2 hPa [Eqn. 4.34a]b) mixing ratio: wv=mv/md=(Mv/Md)*(e/(p-e))=0.0135 [Eqn. 4.36]c) specific humidity: qv=mv/(mv+md)=wv/(wv+1)=0.0133 [Eqn. 1.20]d) specific heat at constant pressure: cp~(7R/2)(1+0.87qv)=1016 J/K/kg [Eqn. 2.65]e) virtual temperature: Tv=(1+0.608qv)*T=305.6 K [Eqn. 1.25]

Radiation Balance ExampleIt has been estimated from satellite observations that variations in solar radiance duringthe last 20 years amounted to ≤0.2 W m-2, or less than 0.1% of the incoming shortwaveradiation. Calculate the approximate change in the temperature at the Earth’s surface fora 0.1% decrease in solar luminosity for a simplified climate model. State allassumptions, simplifications, and equations used. Values of constants that you may needare Earth’s albedo 0.31, solar luminosity 3.92x1026 W, Earth-sun distance 1.50x1011 m, Stefan-Boltzmann constant 5.67x10-8 W m-2 K-4.

Assume that: (1) the earth behaves as a blackbody, (2) atmosphere is transparent tonon-reflected portion of the solar beam; (3) atmosphere in radiative equilibrium withsurface; (4) atmosphere absorbs all the infrared emission. Then, at equilibrium, theincoming shortwave flux and outgoing longwave flux are equal (i.e. there is noaccumulation) so for the normal solar luminosity we can write:FL= σTsurf

4 (assumption 1; Eqn. 3.20)FS = FL(assumption 2-4)0.25*S0(1- αp) = σTsurf

4 (Eqn. 12.1b)Tsurf = [0.25*S0(1- αp)/σ]0.25

where S0 = L0/(4πd2) = 1.3938x103 W m-2 (p. 332), αp=0.31, σ=5.67x10-8 W m-2 K-4

Tsurf = 254.84KReducing L to 99.9% of its value, we get S0’ = 0.999*1.3938x103 = 1.3924x103 W m-2

Tsurf = 254.78KThe resulting cooling of 0.06K associated with a 0.1% reduction in solar radiation isnegligible for this simplified climate model.