EAS 6140 – Thermodynamics of Atmospheres and Oceans
Practice Exam II
A parcel of air consisting of dry air, water vapor, and condensed liquid water is rising, undergoing to
reversible saturated adiabatic expansion. The air parcel is a closed thermodynamic system (i.e. no
mass enters or leaves the system). As the parcel rises from p1 to p2 (p2<p1), compare the
thermodynamic state of the parcel at p2 with its state at p1 (i.e., state whether the following
thermodynamic variables increase, decrease, or remain the same as a result of the sinking):
1. T (temperature)
decreases
2. wv (water vapor mixing ratio)
decrease
3. wl (liquid water mixing ratio)
increase
4. H (relative humidity)
same
5. e (vapor pressure)
decrease
6. TD (dewpoint temperature)
decrease
7.  (potential temperature)
decreases
8. e (equivalent potential temperature)
same
A cloud is observed to have an amount of condensed liquid water corresponding to the adiabatic
liquid water content. State how the liquid water content would change (increase, decrease,
remain the same) if each of the following processes were included:
9. lateral entrainment of environmental air
decrease
10. precipitation
decrease
11. radiative cooling increase
12. glaciation
decrease
13-14. Which variables are conserved when unsaturated air is lifted (circle all that apply)
a) Potential temperature
b) Mixing ratio
c) Saturation mixing ratio
d) Equivalent potential temperature
14-15. Which variables are conserved when saturated air is lifted (circle all that apply)
a) Potential temperature
b) Mixing ratio
c) Saturation mixing ratio
d) Equivalent potential temperature
16-17. Wet bulb temperatures are often measured by wrapping the bulb of a thermometer with an
absorbent material (e.g. cotton cloth) thoroughly misted with the clean water. The lowest
temperature recorded while the cloth is still wet is the wet bulb water. Supposed the water is
salty (not pure). What would happen to the measured web bulb temperature? Would it increase,
decrease, remain the same? Explain.
Tw would be higher (ie closer to the dry bulb temperature). Salt water is harder to evaporate than
fresh water (See section 4.3). At relative humidities below 100%, water evaporates from the
bulb which cools the bulb below the dry bulb/ambient temperature. If one bulb is 100%
pure water and the other 99% pure water the rate of evaporation rate would be less. This
is the same with the case for salty water.
The number of CCN per unit volume of air that have critical supersaturation values less
than (S–1) for a maritime airmass might be NCCN=50 (S-1)0.4 whereas a typical
continental (non-urban) spectrum might be NCCN =4000 (S-1)0.9.
See section 5.2 through 5.4
Recall N = ∫ n(r) dr ; so that the size distribution determines N.
For a given updraft speed, which spectrum (maritime or continental) will result in:
18. the larger cloud drop concentration
continental
19. the larger maximum supersaturation
maritime
20. the smaller mean drop size
continental
21. the larger dispersion (spread) of drop sizes
continental
In the following diagram, assume that a parcel of air is lifted to point A mechanically and
reversibly (i.e., no precipitation or mixing with the environment). In the diagram below,
state at each level indicated whether the acceleration (du/dt) and vertical velocity (u) due
to buoyancy (not the mechanical lifting) are equal to, greater, or less than zero.) ------represents the temperature profile of the environment, while ______ represents the path
of a parcel.
u
du/dt
inside LNB 22. less than zero
23. negative
INSIDE
CAPE
24. greater than zero
25. positive
@LFC
26. equal to zero
27. zero
Between LFC 28. less than zero
29. negative
& LCL
* Recall to solve this problem look at temperature values if parcel is warmer than the
environment there will be upward motion due to buoyant forces if parcel is colder than
environment there will be downward motion due to buoyant forces if parcel is equal to
environment there should be no buoyant force.
30-33. The expression for liquid water path is given by
zt

 a wl dz
Wl =
zb
Derive an expression for the liquid water path between p1 and p2 for a situation in which the
liquid water mixing ratio increases linearly with height between these two levels, wl=0 at zb and
wl=W at zt.
See Worksheet 16 and change accordingly writing liquid water path in terms of dp instead
of dz use hydrostatic approximation to substitute for dz.
34-37. The Kelvin equation for the homogeneous ice nucleation for a liquid drop is written as
r 
2 il
i RvT ln Si 
where Si is the saturation ratio with respect to ice. Show that an alternative form of the Kelvin
equation may be written as
r* =
2 il Ttr
 i L il T tr  T
r* 
2 il

i RvT ln( Si )
2 il
e 
i RvT ln  s 
 esi 
 es  Lil  1 1 
2 il Ttr
     substituting gives, r* =
 i L il T tr  T
 esi  Rv  Ttr T 
Where ln 
38-41. Starting from the first law of thermodynamics for a saturated adiabatic process,
derive an expression for the adiabatic liquid water content as a function of height above
cloud base.
42-46. If the production of saturation occurs via isobaric cooling rather than by adiabatic cooling,
then
dS = A dT  A dwl
3
4
dt
dt
dt
(5.39)
where dT/dt is the isobaric cooling rate resulting from radiative cooling or other isobaric
processes. Determine expressions for the terms A3 and A4 .
Assuming ascent without condensation, term A3 can be derived using the following
relation, see pg 146-147
dS
dT
 A3
dt
dt
A3  
Llv
RvT 2
and
A4 
p
es
dp L lv p
=
dT Rv T2
Wv =

0
e 2 = e 1 exp –
v dz
e s r,n solt
= 1 + ar  b3
es
r
L lv 1
– 1
R v T2 T1
r* =
dr
r  =
dt
S1
2
L lv  l
 R vT
dwl
dS
= A1 u z  A2
dt
dt
2  lv
l R v T ln S
2
+
1
= S
K+D
l R v T
es T D
L

Tw = T  lv ws Tw  wv
cp
dR  aE

wl uT (R)
dt
4 l
uT = k 1 r2
 = m d c pd + mv cpv + m l c l dT + Llv dmv – V dp – v dm v – l dm l
Td
L w
 = c pd d(lnT)  R d d(lnp) + d lv s
T

s = d
L w

 e =  exp lv s

c pdT
cp
dwl,ad =
   s dz
L lv d
* =
 ext

re =
ext dz

r3 n r dr
0

r2 n r dr
0
g
N2 = 0

ext =

L lv ws
RdT
 L 2lvws
1+
2
c pd R d T
1+
d 
d
dz – dz
n r  r2 Qext x dr
0
 3Wl
* =
 ext
2l re