Transcript Document 7801088

Cumulus Convection
Eric A. Pani
The University of Louisiana at Monroe
Overview



Have taken simplified view of convection so
far
Only force considered has been buoyancy
Are there other forces involved? If so, what
are they?
Entrainment
Mathematical Development
Consider a sample of saturated cloud mass, m, ascending through a distance
dz and experiencing a pressure change dp. As the sample rises, it entrains a
mass dm of unsaturated environmental air. Model the rising as a saturated
adiabatic cooling of m, an isobaric cooling of m and warming of dm during
mixing, and the evaporation of part of the liquid water content of m into dm to
produce a final mixture which is saturated. We will assume that the cloud is
horizontally uniform in temperature, water vapor, and liquid water and that
there are no sources or sinks of heat.
Heat required to warm entrained air from T  to cloud temp (T ) is
dQ1  c p (T  T )dm
Heat needed to evaporate some liquid water from cloud to saturate entrained air is
dQ2  L( ws  w)dm where ws is the saturation mixing ratio in cloud and w is the mixing
ratio of entrained air
As air ascends,water vap or is condensed(dws ) and heat is released
dQ3   mLdw s
First Law of Thermodynamics for cloud mass becomes
dp
dQ  dQ1  dQ2  dQ3  m(c p dT  Rd T )
p
assuming heat changes of vapor and liquid water are small compared to that of dry air
Substituting,
 c p (T  T )dm  L( ws  w)dm  mLdw s  m(c p dT  Rd T
But ws 
es
p
dp
)
p
 ln( ws )  ln( )  ln( es )  ln( p ) so
dws des dp
dw
de
pg
g

 . But dp   gdz  
dz, so s  s 
dz
ws
es
p
Rd T
ws
es
Rd T
Therefore, dws  ws
des ws g
pg

dz and, since dp   gdz  
dz
es
Rd T
Rd T
 c p (T  T )dm  L( ws  w)dm  mL ( ws
des ws g
dp

dz)  m(c p dT  Rd T )
es
Rd T
p
 c p (T  T )dm  L( ws  w)dm  mL ( ws
des ws g

dz)  m(c p dT  gdz)
es
Rd T
Dividing by mdz ,
 c p (T  T )

w de
wg
1 dm
dT
[c p (T  T )  L( ws  w)]  L( s s  s )  c p
g
m dz
es dz Rd T
dz
But

w de
wg
1 dm
1 dm
dT
 L( ws  w)
 L( s s  s )  c p
g
m dz
m dz
es dz Rd T
dz
des des dT
w de dT ws g
1 dm
dT

so 
[c p (T  T )  L( ws  w)]  L( s s

)  cp
g
dz
dT dz
m dz
es dT dz Rd T
dz
Lw s g
1 dm
dT Lw s des dT
[c p (T  T )  L( ws  w)] 
 g  cp

m dz
Rd T
dz
es dT dz
dT
from the right side
dz
c p dm
Lw s
L
dT Lw s des

[(T  T )  ( ws  w)]  g (1 
)  cp
[
 1]
m dz
cp
Rd T
dz c p es dT
Factor c p from 1st term on LHS of the equation and c p
dT
, the parcel' s lapse rate
dz
Lw s
1 dm
L
g
dT Lw s des



[(T  T )  ( ws  w )]  (1 
)
[
 1]
m dz
cp
cp
Rd T
dz c p es dT
Divide by c p and solve for -
Lw s
g
1 dm
L
(1 
)
[(T  T )  ( ws  w)]
Rd T
m dz
cp
dT c p


Lw s des
dz
1
c p es dT
By Clausius - Clapeyron Equation,
1 de s
L

so
2
es dT Rd T

dT

dz
But

Lw s
g
1 dm
L
(1 
)
[(T  T )  ( ws  w)]
cp
Rd T
m dz
cp
1
L2 ws
c p Rd T 2
g
  d so (finally)
cp
dT

dz
 d (1 
Lw s
1 dm
L
)
[(T  T )  ( ws  w)]
Rd T
m dz
cp
1
L2 ws
c p Rd T 2
Lw s
)
Rd T
1 dm
dT
dT
Note that if
 0, 



and,
if
w

0
,

d
s
s
2
m dz
dz
dz
L ws
1
c p Rd T 2
 d (1 
Results from




dT

dz
 d (1 
Lw s
1 dm
L
)
[(T  T )  ( ws  w)]
Rd T
m dz
cp
1
L2 ws
c p Rd T 2
1 dm
If entrainment rate ( m dz ) > 0 and cloud is
buoyant, then cloud cools faster than moist
adiabatic
Example: at 700 mb, -1°C, and 67% environmental
RH, if cloud is at 0°C, then moist adiabatic lapse
rate is 5.8°C/km compared to cloud lapse rate of
6.6°C/km if entrainment rate is 0.25/km
Typical entrainment rates are 1-2/km (i.e., cloud
doubles or triples its mass with each km of ascent)
Where does entrained air go?
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Large entrainment rates indicate cloud mass is
increasing quickly
Would expect width of cloud to increase with height,
but observations do not support this
Some air must move downward in downdrafts not
modeled by theory