Ch.  3  Atmospheric  thermodynamics  

Chapter  3  in  the  text  book  References:  Thermodynamics  of  the  atmosphere  and  ocean  by  Curry  and  

Webster      What  controls  air  density  varia?ons  and  changes?  

Outline:  •  Dry  gas  laws  (The  text  3.1-­‐3.4,  p63-­‐79)  •  Moist  air  (The  text,  3.5-­‐3.7,  p79-­‐101)  

How  do  we  define  dry  and  moist  air?    

–  Dry  air:  can  include  water  vapor  but  does  NOT  involve  water  phase  change,  i.e.,  condensa?on/freeze,  melt/evapora?on.  

–  Moist  air:  air  involves  water  phase  change.      

3.1  Dry  Gas  Laws  

•  The  idea  gas  law  •  Virtual  temperature  •  The  hydrosta?c  equa?on:  geopoten?al  height  and  thickness,  the  scale  height,  

•  The  first  law  of  the  thermdynamics  •  Adiaba?c  processes  and  poten?al  temperature    

The  idea  Gas  Law:  

•  The  the  atmosphere,  whether  considered  individually  or  as  a  mixture,  follows  a  state  equa?on  that  describes  the    rela?onship  between  air  pressure,  volume  and  temperature.  

•  Various  forms  of  the  idea  gas  law:  –  PV=mRT        

•  P:  pressure  (Pa),  V:  volume  (m3),  m:  mass  (kg),  T:  absolute  temperature  (K),  R:  the  gas  constant  for  1  kg  of  gas.  

–  P=ρRT    or  αP=RT  •  Where  ρ=m/V,  air  density,  α=1/ρ:  the  specific  volume  

–  PV=nR*T      If  we  let  PV=m(kg)RT=(1000m(g)/M)(MR)T=nR*T        •  Where  n=m/M,  M:  molecular  weight  of  a  mole  of  gas  (6.022X1023  molecules),  n:  

the  number  of  moles  in  mass  m.  R*=MR/1000:  gas  constant  for  1  mol  of  gas,  =8.3145  J  K-­‐1  mol-­‐1,  or  universal  gas  constant  

The  idea  Gas  Law-­‐2:  

•  Various  forms  of  the  idea  gas  law:  –  P=nokT      for  a  gas  containing  no  melecules  

•  k=R*/NA:  Boltzmann  constant,  no:  number  of  moleucles  per  unit  volume,  NA:  number  of  molecules  per  mole.  

–  Pdαd=RdT    for  dry  air  •  “  d”:  dry  air  •  Molecular  weight  of  dry  atmosphere,  Md:  28.97~29,    •  Gas  constant  for  1  kg  of  dry  atmosphere,  Rd=1000xR*/Md  

=1000*8.3145/28.97=287.0  J  K-­‐1kg-­‐1.  

–  evαv=Rv*T    for  water  vapor  •  ev:  vapor  pressure,  “  v”:  water  vapor  •  Water  vapor  gas  constant:  Rv=1000xR*/Md=1000*8.3145/18.016  =461.51  J  K-­‐1kg-­‐1.    

–  Rd/Rv=Mv/Md=18.016/28.97=0.622    Ques?on:  can  we  apply  the  idea  gas  law  to  a  mixture  of  dry  air  and  water  vapor?  

The  virtual  temperature,  Tv:  

•  Atmosphere  is  a  mixture  of  dry  air  and  water  vapor.  To  determine  the  influence  of  air  humidity  on  air  density,  the  meteorologists  introduce  vritural  temperature,  to  translate  the  influence  of  atmospheric  water  vapor  on  air  density  ,  into  something  comparable  to  the  influence  of  temperature  on  air  density.      

•  Virtual  temperature  is  yhe  temperature  of  a  dry  air  mass    that  has  the  same  air  density  of  the  mixture  of  dry  air  and  water  vapor  at  the  same  pressure.  

The  virtual  temperature,  Tv-­‐2:  

•  What  does  Tv  mean?  –  The  temperature  of  dry  air  that  has  the  same  air  density  at  

the  same  pressure  as  the  mixture  of  dry  air  and  water  vapor.  

•  Why  is  Tv  warmer  than  T?  –  Because  water  vapor  has  lighter  molecular  weight  (18)  than  

that  of  dry  air  (~29),  Tv  is  higher  than  T  so  it’s  corresponding  dry  air  density  would  be  lowed  to  that  of  the  mixture  of  dry  air  and  water  vapor.  

•  What  controls  the  devia?on  of  Tv  from  T?  –  The  high  humidity  is,  the  higher  Tv  is  rela?ve  to  T.  

€

Tv = 1

1− ep

(1+ε ) T

The  virtual  temperature,  Tv-­‐2:  

•  What  does  Tv  mean?  –  The  temperature  of  dry  air  that  has  the  same  air  density  at  

the  same  pressure  as  the  mixture  of  dry  air  and  water  vapor.  

•  Why  is  Tv  warmer  than  T?  –  Because  water  vapor  has  lighter  molecular  weight  (18)  than  

that  of  dry  air  (~29),  Tv  is  higher  than  T  so  it’s  corresponding  dry  air  density  would  be  lowed  to  that  of  the  mixture  of  dry  air  and  water  vapor.  

•  What  controls  the  devia?on  of  Tv  from  T?  –  The  high  humidity  is,  the  higher  Tv  is  rela?ve  to  T.  

€

Tv = 1

1− ep

(1+ε ) T

Example:  

•  Compare  Tv  for  surface  air  consists  of  4%  of  water  vapor  to  that  consists  of  1%  of  water  vapor.    Assuming  that  the  surface  pressure  is  1000  hPa  and  T  is  25C.    

€

Tv = 1

1− ep

(1−ε) T

Example:  

•  Compare  Tv  for  surface  air  consists  of  4%  of  water  vapor  to  that  consists  of  1%  of  water  vapor.    Assuming  that  the  surface  pressure  is  1000  hPa  and  T  is  25C.  

€

At the surface, p = 1000 hpa, for air with 4% water vapor, e = 0.04X1000 hpa = 40 hPa

Tv =T

1− ep

(1−ε)=

(25+ 273)K

1− 40hpa1000hPa

(1− 0.622)=

298K0.985

= 302.5K = 29.5C

For air with 1% water vapor, e = 0.01X1000 hpa = 10 hPa

Tv =T

1− ep

(1−ε)=

(25+ 273)K

1− 10hpa1000hPa

(1− 0.622)=

298K0.996

= 299.2K = 26.2K

3% of extra water vapor in the atmosphere has the same impact onair density as 3.3K temperature increase for atmosphere with 1% water vapor.

Exercise:  

•  Compare  Tv  between  the  surface  air  and  air  at  500  hPa.    In  both  case,  water  vapor  is  4%  of  the  total  air  mass.    Assuming  that  the  surface  pressure  is  1000  hPa  and  T  is  25C,  and  T  at  500  hPa  is  -­‐8C.  

•  Compare  the  difference  between  T  and  Tv  at  both  the  surface  and  500  hPa.    Explain  what  cause  the  difference.  

Exercise:  

•  Compare  Tv  between  the  surface  air  and  air  at  500  hPa.    In  both  case,  water  vapor  is  4%  of  the  total  air  mass.    Assuming  that  the  surface  pressure  is  1000  hPa  and  T  is  25C,  and  T  at  500  hPa  is  -­‐8C.  

•  Compare  the  difference  between  T  and  Tv  at  both  the  surface  and  500  hPa.    Explain  what  cause  the  difference.  

€

At the surface, p = 1000 hpa, 4% water vapor, e = 40 hPa

Tv =T

1− ep

(1−ε)=

(25+ 273)K1− 0.04(1− 0.622)

=298K0.985

= 302.5K = 29.5C

At 500 hP, T = (-8 + 273)K = 265K, 4% water vapor, e = 0.04X500 hpa = 20 hPa

Tv =T

1− ep

(1−ε)=

265K1− 0.04(1− 0.622)

=265K0.985

= 269.0K = 26C

The difference between T and Tv is smaller because of lower temperature at 500 hPa.

• The  hydrosta?c  equa?on:  geopoten?al  height  and  thickness,  the  scale  height,  

• The  first  law  of  the  thermodynamics  

• Adiaba?c  processes  and  poten?al  temperature    

Two  physical  principles  and  applica1ons  for  dry  air  thermodynamics:  

3.2  The  hydrosta/c  equa/on  

•  Air  pressures  at  any  given  height  in  the  atmosphere  is  due  to  gravita?onal  force  of  the  air  mass  above  that  height,  

•  Thus,  change  of  surface  pressure  at  a  height  is  due  to  change  of  air  mass  above  that  height.  –  dp(z)=-­‐ρgdz    or  dp/dz=-­‐ρg            The  hydrosta?c  equa?on  

 •  It  is  a  good  approxima?on  for  large-­‐scale  

atmospheric  and  climate  dynamic  processes,  but  not  a  good  approxima?on  for  fast  mesoscale  convec?ve  processes,  such  as  supercell,  tornadoes.      Why?  

€

p(z) = ρ(z)gdzz

∞

∫

Why  cannot  we  use  hydrosta?c  equa?on  in  case  of  rainfall  and  thunderstorms?  

€

ρ ⋅az = Fz∑

z : vertical direction, az : vertical acceleration, Fz : forces in vertical direction, ρ : air density

ρdwdt

= −gρ − dpdz

− ρa friction

g : gravitational acceleration, p : pressure, a friction : deceleration by friction, w : vertical velocity

Away from convection, dwdt

~ 0, friction of the atmosphere is small, thus a friction ~ 0

Thus − gρ − dpdz

~ 0 1ρdpdz

= - g hydrostatic equation

3.2.1  Geopoten/al  

•  The  work  to  against  gravity  for  an  rising  air  mass,  or  poten?al  energy  of  air  at  a  height  z,  

•  Geopoten?al  height:    

   •  Because  gravita?onal  accelera?on  decreases  with  

height,  the  geometric  height  does  not  exactly  represent  the  poten?al  energy  of  the  air  at  high  al?tude.    Thus,  we  use  geopoten?al  height  to  represent  the  TRUE  value  of  the  poten?al  energy  of  the  atmosphere  at  a  given  height.  

–  In  the  lower  troposphere,  gègo,  Z=z  –  In  the  upper  troposphere,  Z  starts  to  deviate  from  z  

€

dΦ = gdz = −1ρdp = −αdp = −RT dp

p= RTd ln p

Φ(z) = gdz =0

z∫ − αdp

p(z)

ps∫

€

Z ≡Φ(z)go

= −1go

gdzz

∞

∫g:  gravita?onal  accelera?on  at  al?tude  z  Go=9.81  m/s2,  :  global  mean  g  at  surface  

Table  3.1:  Values  of  the  geopoten?al  height  (Z),  and  accelera?on  due  to  gravity  (g),  at  40◦  la?tude  for  geometric  height  (z)  

     z(km)    Z(km)    g(m  s−2)          0              0            9.81          1                              1.00          9.80        10            9.99        9.77    100                            98.47          9.50    500            463.6          8.43  

3.2.2  Geopoten/al  height:  

•  Geopoten?al  height  is  more  commonly  used  in  pressure  coordinate,  in  which  it  represents  the  poten?al  height  of  the  air  at  a  given  pressure  level:    

•  Geopoten?al  height  at  the  given  pressure  level  is  determined  by  T  of  the  air  from  the  surface  to  that  level,  which  in  turn  determines  the  height  of  the  pressure  level.  €

Z ≡Φ(z)go

= −1go

αdp = −1go

RTd ln pp

ps∫p

ps∫

z  

Z500  hPa  

colder  warmer  

3.2.3    Geopoten/al  thickness:  

•  Geopoten?al  height  difference  between  the  boqom  and  top  of  an  atmospheric  layer:  

•  Ques?on:  What  determine  the  thickness  of  an  atmospheric  layer?  €

dZ ≡dΦ(z)go

For dry air :

Z2 − Z1 =1go

dΦz1

z2∫ = −1go

RTd ln pp2

p1∫ = RT lnp2

p1

for dry air with vapor :

Z2 − Z1 =1go

dΦz1

z2∫ = −1go

RTvd ln pp2

p1∫ = RTv ln p2

p1Hypsometric  equa/on  

Scale  height:  

•  Scale  height:  the  geopoten?al  height  of  the  atmospheric  layer  if  it  were  an  isothermal  layer.    It  depends  on  the  temperature  and  gas  constant  of  the  air.    It  is  used  as  a  scaling  factor  in  determining  the  geopoten?al  height  of  a  given  pressure  level.  

€

Z2 − Z1 =Rgo

Tvdppp2

p1∫For an isothermal atmosphere layer, if the virtual effect is neglected

Z2 − Z1 =RTgo

ln p2

p1

= Hn p2

p1

where H ≡RTgo

is the scale height

•  Different  gases  have  different  scale  height  for  the  same  T  •  Scale  height  increase  with  T,  the  isothermal  gas  column  expands  ver?cally  with  T.  •  Why  use  scale  height?  

•  Easy  to  convert  between  Z  and  P  •  An  important  characteriza?on  of  a  fluid  or  gas  layer.  

•  Scale  height  of  the  earth’s  atmosphere  under  current  insola?on:  

€

For the earth's atmosphere, Rd = 287.0 J K-1 kg-1, go = 9.81ms−1

H =Rd

goT =

287.0 J K-1 kg-1

9.81ms−1 T = 29.3T m K-1

For T = 255K, the approximate mean T of the atmosphere (Te)H = 7.47 ×103m = 7.47km The earth's atmosphere scale height is about 7.5 km averaged globally

 In  class  exercise:    What  would  be  the  scale  height  of  the  earth’s  atmosphere,    •  A)  if  the  solar  radia?on  were  decreased  by  1%  rela?ve  to  the  current  solar  

radia?on  intensity  and  lead  to  an  1k  decrease  of  earth’s  effec?ve  temperature,  •  B)  if  the  atmosphere  consisted  100%  water  vapor?    Use  T=255K  

 Rv=461.51  JK-­‐1kg-­‐1,    •  C)  if  we  need  to  consider  the  virtual  effect  of  the  atmosphere  with  1%  water  vapor    

What  would  be  the  scale  height  of  the  earth’s  atmosphere,    •  A)  if  the  solar  radia?on  were  decreased  by  1%  rela?ve  to  the  current  solar  

radia?on  intensity  and  lead  to  an  1k  decrease  of  earth’s  effec?ve  temperature,  •  B)  if  the  atmosphere  consisted  100%  water  vapor?    Use  T=255K  

 Rv=461.51  JK-­‐1kg-­‐1,    •  C)  if  we  need  to  consider  the  virtual  effect  of  the  atmosphere  with  1%  water  vapor    

€

a)Te = 254K

H = 29.3X254 m K-1 = 7.44 ×103m = 7.44km The earth' s atmosphere scale height would be 7.44 km averaged globally, about 300 m lower than that of the current climate.b)

Rv = 461.51JK −1kg−1

H =RvTgo

=461.51JK −1kg−1

9.81ms−2 255K = 11,996 m = 12.00 km

c)

H = RdTvgo

=RdT

go 1− ep

(1− 0.622)$

% &

'

( )

=287.0JK −1kg−1 × 255K

9.81ms−2 1− 0.01* (1− 0.622)( )= 7489m = 7.49km

•  Thickness  of  the  atmosphere  between  two  pressure  levels:  

€

ΔZ = Z2 − Z1 =RTvgo

ln p2

p1

= 29.3Tv ln p2

p1

if we want to consider Tv

Between 1000 hPa to 700 hPa, ΔZ = 29.3Tv ln1000hpa700hPa

=10.45Tv m/K

For Tv = 280K, ΔZ1000-700hPa = 2926mFor Tv = 282K, ΔZ1000-700hPa = 2947m

Warm  core  system   Cold  core  system  

Summary:  

•  The  atmosphere  is  governed  by  ideal  gas  law,  p=ρRT  •  To  simplify  the  ideal  gas  law  for  mixed  dry  air  and  water  

vapor,  we  introduce  virtual  temperature,  

•  To  describe  atmospheric  layer  thickness  change  with  temperature,  we  introduce  geopoten?al  height,  Z,  and  scale  height,  H.    Both  increase  with  an  increase  of  T.    

€

Tv = 1

1− ep

(1−ε) T

€

Z ≡Φ(z)go

= −1go

αdp = −1go

RTd ln pp

ps∫p

ps∫

4.3  The  first  law  of  thermodynamics  

•  Energy  conserva?on  in  a  closed  system,  energy  can  only  be    transformed  from  one  form  to  another,  cannot  be  generated  or  destroyed.  

Figure  3.4  Representa?on  of  the  state  of  a  working  substance  in  a  cylinder  on  a  p—V  diagram.  The  work  done  by  the  working  substance  in  passing  from  P  to  Q  is  p  dV  ,  which  is  equal  to  the  blue  shaded  area.  

P:  pressure  

A:  cross  sec?on  area  

€

dq− dW = du (Eq. 1)where dq : heat added to the systemdW : external work done by the systemdu : change of the internal energy of the systemWork done to against external forcing :

dW = p(V )dVV1

V 2∫ = (pAΔx =)pΔV If dV > 0, dW > 0

dq - pdα = du (Eq. 2)

•  Cp=1004  JK-­‐1,  Cv=717  JK-­‐1,  R=287  JK-­‐1kg-­‐1    for  dry  air  •  Thus,  Cp:Cv:R=7:5:2  €

Because d(pα) = pdα +αdP, the Eq. 2 becomes dq - d(pα) +αdp = cvdTdq +αdp = cvdT +RdT = cpdT where cp = cv +R is the specific heat for a constant pressuredq = cpdT −αdp (Eq. 3)

€

dq = du = cvdT =T1

T2∫ cvΔT If the volume of the system stay constant

where cv : specific heat for constant volumeThus, Eq. 2 becomes : Δq - pΔV = cvΔTFor Δ→ 0 and a unit mass, we have dq - pdα = cvdT (Eq. 3)

If  we  add  heat  to  a  system  and  keep  the  volume  the  same,  we  have  

3.3.3  Enthalpy  and  dry  sta/c  energy  

•  Based  on  the  first  law  of  thermodynamics,  the  change  of  enthalpy  of  the  air  dh=CvdT+pdα=CpdT,  due  to  change  of  the    internal  energy  and  work  related  to  expansion  

•  For  a  dry  mo?onless  air  mass  above  the  earth’s  surface,  its  total  energy  S=h+Φ,      –  Where  Φ:  geopoten?al,  or  poten?al  energy  

•  For  an  air  parcel  that  raises  or  sinks  slowly  (neglect  kine?c  energy  associated  with  the  ver?cal  mo?on),  its  first  law  of  thermodynamics  for  atmosphere  becomes            

 dq=d(h+Φ)=cpdT+gdz    

€

h = cpdT0

T∫ = cpT

4.4  Adiaba/c  process:  

Adiaba?c  process:  If  no  heat  being  added  or  removed,  dq=0.  •  dq=d(h+Φ)=cpdT+gdz=0 •  Γd=-dT/dz=g/cp

•  Γd=g/cp=9.8 ms-2/1004JK-1kg-1≈ 9.8K/km –  T in a dry air rising adiabatically decreases at a constant rate of

9.8 K per km.

•  The potential temperature, θ=Τ(po/p)R/cp: –  The temperature of a dry air parcel at a given altitude would be,

if it moves adiabatically to the sea level (1000 hPa). A conserved value for a dry adiabatic process.

Poten?al  temperature  distribu?on  and  atmospheric  flow:  

• Without  diaba?c  hea?ng,  atmospheric  flows  (e.g.,  associate  atmospheric  waves)  along  isentropic  surface  (constant  θ)  because  it  is  neutral  for  dry  adiaba?c  process.    This  type  of  flow  is  referred  to  as  the  isentropic  flow.  Diaba?c  hea?ng  (e.g.,  rainfall,  clouds)  drives  cross-­‐isentropic  surface  flow.    •   In  some  case,  for  example,  mid-­‐la?tude  planetary  waves  or  stratosphere  circula?on,  it  is  more  convenient  to  use  θ, instead of p, as the vertical coordinate.  

Ver?cal  gradient  of  poten?al  temperature  and  sta?c  stability    

dθ/dz:  sta?c  stability  for  dry  convec?on  

Unstable,  ascending  mo?on  

Stable,  atmospheric  waves  

Strongly  stable  

•  Why  use  poten?al  temperature?  

Skewed  T-­‐lnp  chart  

More  informa?on  at  hqp://airsnrt.jpl.nasa.gov/SkewT_info.html  

 To    enlarge  the  small    devia?on  from  the  constant  θ  lines  for  the  atmospheric  temperature  profiles,  we  use  slated  temperature  lines.  

Determine  stability  of  the  atmosphere  for  dry  convec?on:  

Summary:  •  Thermodynamics  of  dry  atmosphere  is  governed  by  the  idea  gas  law,  the  hydrosta?c  equa?on  and  the  first  law  of  the  thermodynamics;  

•  A  set  of  variables,  including  Tv,  ϕ,  Z,  ΔZ,  H,  θ,  and  dry  sta?c  energy,  are  defined  to  effec?vely  apply  these  laws  to  meteorological  and  clima?c  applica?ons.  

€

p = ρRT , dp/dz = -ρg, dq = -αdp +CpdT = pdα +CvdT