Ch.  3  Atmospheric  thermodynamics  

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

Webster  

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.,  condensaSon/freeze,  melt/evaporaSon.  

–  Moist  air:  air  involves  water  phase  change.      

3.1  Dry  Gas  Laws  

•  The  idea  gas  law  •  Virtual  temperature  •  The  hydrostaSc  equaSon:  geopotenSal  height  and  thickness,  the  scale  height,  

•  The  first  law  of  the  thermdynamics  •  AdiabaSc  processes  and  potenSal  temperature    

The  idea  Gas  Law:  

•  The  the  atmosphere,  whether  considered  individually  or  as  a  mixture,  follows  a  state  equaSon  that  describes  the    relaSonship  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    QuesSon:  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  ρ =

md +mv

V= ρd + ρv

dry air and water vapor each follow the idea gas law:pd = ρdRdT and ev = ρvRvTthe total air pressure (dry +vapor), p=pd + e Dalton's law (eq. 1)

ρ=ρd + ρv =p− eRdT

+eRvT

=pRdT

1− ep+eRdpRv

"

#$

%

&'=

pRdT

1− ep

(1+ε)"

#$

%

&' (eq. 2)

where ε= RdRv

= 0.622

If we insist on applying the idea gas law for the mixed dry air and vaporand treat it as if it were dry air:b p=ρRdT', where T' has to be slightly different warmer than T (why?) and substitute eqs. 1 and 2: we would have

p= pRdT

1− ep

(1+ε)"

#$

%

&'RdT' T'= 1

1− ep

(1+ε) T

Tv =T'= 1

1− ep

(1+ε) T defined as the virtual temperature

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  deviaSon  of  Tv  from  T?  –  The  high  humidity  is,  the  higher  Tv  is  relaSve  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  hydrostaSc  equaSon:  geopotenSal  height  and  thickness,  the  scale  height,  

• The  first  law  of  the  thermodynamics  

• AdiabaSc  processes  and  potenSal  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  gravitaSonal  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  hydrostaSc  equaSon  

 •  It  is  a  good  approximaSon  for  large-­‐scale  

atmospheric  and  climate  dynamic  processes,  but  not  a  good  approximaSon  for  fast  mesoscale  convecSve  processes,  such  as  supercell,  tornadoes.      Why?  

€

p(z) = ρ(z)gdzz

∞

∫

Why  cannot  we  use  hydrostaSc  equaSon  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  potenSal  energy  of  air  at  a  height  z,  

•  GeopotenSal  height:    

   •  Because  gravitaSonal  acceleraSon  decreases  with  

height,  the  geometric  height  does  not  exactly  represent  the  potenSal  energy  of  the  air  at  high  alStude.    Thus,  we  use  geopotenSal  height  to  represent  the  TRUE  value  of  the  potenSal  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:  gravitaSonal  acceleraSon  at  alStude  z  Go=9.81  m/s2,  :  global  mean  g  at  surface  

Table  3.1:  Values  of  the  geopotenSal  height  (Z),  and  acceleraSon  due  to  gravity  (g),  at  40◦  laStude  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:  

•  GeopotenSal  height  is  more  commonly  used  in  pressure  coordinate,  in  which  it  represents  the  potenSal  height  of  the  air  at  a  given  pressure  level:    

•  GeopotenSal  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:  

•  GeopotenSal  height  difference  between  the  bopom  and  top  of  an  atmospheric  layer:  

•  QuesSon:  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  geopotenSal  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  geopotenSal  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  verScally  with  T.  •  Why  use  scale  height?  

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

•  Scale  height  of  the  earth’s  atmosphere  under  current  insolaSon:  

€

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  radiaSon  were  decreased  by  1%  relaSve  to  the  current  solar  

radiaSon  intensity  and  lead  to  an  1k  decrease  of  earth’s  effecSve  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  radiaSon  were  decreased  by  1%  relaSve  to  the  current  solar  

radiaSon  intensity  and  lead  to  an  1k  decrease  of  earth’s  effecSve  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  geopotenSal  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∫