EAS 6140 - Thermodynamics of Atmospheres and Oceans
Saturated adiabatic processes
1. The saturated adiabatic lapse rate is (less than, greater than, equal to) the dry adiabatic
lapse rate
less than
2. Why are moist adiabats steeper in slope than dry adiabats?
The temperature decrease is slower because moist air has a larger heat capacity
than dry air
3. Why does the saturated adiabat finally (at cold temperatures) become parallel to the
dry adiabat? At low temperatures the air can hold very little moisture so the adiabat
becomes similar in behavior
4. The adiabatic enthalpy equation can be written as
0 = c p dT – Llv dwl  v dp
Derive an expression for dwl that depends on d, s. This is the adiabatic liquid water
content.
0  c p dT  Llv dwl  vdp
sub
dp
  g
dz
c p dT  Llv dwl  v (  gdz )
c p  dT g 
  dz

Llv  dz c p 
c
dwl  p   d   s  dz
Llv
dwl 
sub  d 
g
cp
and
s 
dT
dz
5. The variable equivalent potential temperature, e, is not conservative for the
following:
a) freezing of drops
b) precipitation
c) water clouds in super- or subsaturated conditions
d) ice clouds in ice super- or subsaturated conditions
e) mixed phase clouds away from the triple point
f) dry adiabatic processes
g) condensation and evaporation in adiabatic ascent/descent
6. The variable ice-liquid water potential temperature, il is not conservative for the
following:
a) freezing of drops
b) precipitation
c) water clouds in super- or subsaturated conditions
d) ice clouds in ice super- or subsaturated conditions
e) mixed phase clouds away from the triple point
f) dry adiabatic processes
g) condensation and evaporation in adiabatic ascent/descent
7. The variable entropy potential temperature,  is not conservative for the following:
a) freezing of drops
b) precipitation
c) water clouds in super- or subsaturated conditions
d) ice clouds in ice super- or subsaturated conditions
e) mixed phase clouds away from the triple point
f) dry adiabatic processes
g) condensation and evaporation in adiabatic ascent/descent
Saturated adiabatic ascent
In the following, use the Stuve or Skew-T diagram to consider a parcel of air with T =
25°C and TD = 20°C at 1000 mb that is lifted mechanically to a height of 500 mb.
1. What is the mixing ratio of this parcel? 15g/kg
2. On the thermodynamic diagram (NOTE: ignore the plotted sounding, just consider the
parcel), determine the lifting condensation level (LCL), which is where w=ws in adiabatic
ascent. Above the LCL, continued ascent follows the saturated adiabat. To determine
the LCL on the thermodynamic diagram, lift a parcel from the surface dry adiabatically
until the temperature (and associated value of ws) equals the value of w of the parcel
(represented by the w value at the surface; recall that that the value of w remains constant
in the moist adabatic ascent). Graphically, the LCL is at the intersection of the  value
(dry adiabat) associated with the surface temperature) and the mixing ratio line associated
with the surface mixing ratio). The LCL is _~930mb___
3. As the lifting continues above the LCL, the temperature cools following the saturated
adiabat. The relative humidity of the parcel remains at 100%. At 500 mb, what is the
temperature and mixing ratio at 500 mb? -5C and 5.5g/kg
4. The adiabatic liquid water content is the amount of water condensed in adiabatic
ascent, with no precipitation or evaporation. The adiabatic liquid water content at a
certain level is difference of the original mixing ratio of the parcel, minus the saturated
mixing ratio of the parcel at that level. At 500 mb, what is the adiabatic liquid water
content of the parcel? 15g/kg - 5g/kg = 10g/kg
5. On the thermodynamic diagram, determine the equivalent temperature and equivalent
potential temperature Te and e for a parcel of air at the surface. The equivalent
temperature is defined as the temperate the air would be if all of the water vapor were
condensed (i.e. all of the latent heat is realized from the water vapor). This can be
determined on the thermodynamic diagram by lifting the parcel dry adiabatically to the
LCL, the lifting the parcel along the saturated adiabat to 200 mb or so, where the
saturated adiabat becomes parallel to the dry adiabat (implying that there is no more
latent heat to be realized by further condensation of vapor). Then follow this dry adiabat
down to the original pressure level, which gives Te. e is determined by continuing down
the dry adiabat to 1000 mb. Te = 65C e = 65C
6. On the thermodynamic diagram, determine TW and W for a parcel of air at the
surface. The adiabatic web bulb temperature is defined as the temperate the air would be
if all of the water vapor were condensed (i.e. all of the latent heat is realized from the
water vapor). This can be determined on the thermodynamic diagram by lifting the
parcel dry adiabatically to the LCL, then follow this moist adiabat down to the original
pressure level, which gives Tw. w is determined by continuing down the moist adiabat to
1000 mb.
TW = 21C W = 21C
7. Write the following temperatures for a given air parcel with temperature T and water
vapor mixing ration w in ascending order (that is, in order of the lowest to the highest
temperature): T, Tv, Tw, Te, TD.
TD <Tw <T <Tv <Te