Numerical Weather Prediction
Parametrization of diabatic processes
Introduction to Moist Processes
ADRIAN
Adrian Tompkins
tompkins@ecmwf.int
Clouds
Water in the atmosphere can be present in all three phases: vapour,
liquid and solid
Clouds, water in
the form of liquid
or ice, play an
crucial role in the
earth’s climate
Clouds and the
precipitation formed in
them are important
forecast products
Clouds - why are they important?
Clouds are a common atmospheric feature and cover large parts of the globe
What is the
mean
coverage?
Latent heat release or consumption occur either directly in clouds or in the
precipitation produced in them.
Clouds - Why are they important?
The entire model hydrological cycle will depend
on the representation of clouds and their
microphysical processes
Cloud radiative effects (“forcing”)
TOA- ERBE - JJA 87
Clouds - Why are they
important?
They strongly affect the radiative
fluxes throughout the
atmosphere
Sherwood et al JGR 1994, show
that the tropical circulation is
strongly influenced by the
interaction between radiation
and clouds
LWCRF  L  Lclr
SWCRF  S  Sclr
CRF  LWCRF  SWCRF
Clouds and Convection in GCMs What are the problems ?
Many clouds and especially the processes within them are of subgrid-scale size (both
horizontally and vertically) and thus must be parameterized.
This means a mathematical model is constructed that attempts to assess their effects in
terms of the Large-scale model resolved quantities.
The aim of this
course
component is to
introduce a few
of the
approaches used
for this important
task
GCM Grid cell 20-400km
T,q,u,v,w...
Clouds in GCMs - What are the
problems ?
stratocumulus
There is a huge variety of cloud types
cirrus
cumulus/cumulonimbus
http://australiasevereweather.com/photography
In this course, we will consider:
composite infra-red image
Warm and ice phase microphysics
This schematic of a front illustrates the interactions involved
In this course, we will consider:
Stratocumulus
Why are they dark?
Clouds in GCMs - What are the
problems ?
Stratocumulus
Driven by turbulence in the
boundary layer
Very common, particularly over the
oceans near the west Coast of the
major continents
Difficult to model: formation, regime
(cloud cover), decoupling, break
up… one of the common
deficiencies of climate and forecast
models
Schematic: StCu to Cu Transition
destabilization:
zi
• surface buoyancy flux
• atmos. radiative cooling
• uplift (cooling, moistening)
• convergence
• shear
zcb
zi
dry BL
Stratocumulus
Shallow cumulus Deep cumulus
Figure from Martin Koehler
Cloud Cover
JJA
Total Cloud
Cover 90°W
ehh2 June
2001 nmonth=3
Glob
Mean: 90°E
59.2 50N-S
Mean: 56.7
135°W
45°W
0°
45°E
135°E
DJF
Total Cloud
Cover ehh2
December
2000 nmonth=3
Glob
Mean: 62.2
50N-S Mean:
59.3
135°W
90°W
45°W
0°
45°E
90°E
135°E
[percent]
99.80
60°N
60°N
99.45
60°N
60°N
65
35
95
30°N
30°N
65
80
95
30°N
65
0°
65
65
65
65
0°
0°
65
35
65
50
30°S
35
65
60°S
90°W
45°W
0°
45°E
90°E
30°S
65
20
60°S
60°S
135°W
90°W
45°W
0°
45°E
90°E
135°E
Total Cloud
Cover 90°W
ISCCP D245°WDecember
2000 nmonth=3
50N-S
Mean:
62.9
135°W
0°
45°E
90°E
135°E
[percent]
[percent]
99.47
60°N
60°N
65
65
97.41
60°N
60°N
95
95
95
30°N
30°N
35
80
30°N
30°N
65
65
80
65
65
65
65
0°
0°
65
0°
30°S
65
0°
65
65
50
30°S
35
65
65
35
65
35
65
65
50
65
30°S
30°S
65
65
65
35
35
95
60°S
60°S
20
60°S
60°S
5
135°W
90°W
45°W
0°
45°E
90°E
135°W
90°W
45°W
0°
45°E
90°E
135°W
90°W
45°W
0°
45°E
90°E
135°E
Difference ehh2 - ISCCP 50N-S Mean err -3.64 50N-S rms 12.9
135°E
135°W
90°W
45°W
0°
45°E
90°E
135°E
[percent]
[percent]
43.03
60°N
20
9.090
135°E
Difference ehh2 - ISCCP 50N-S Mean err -4.74 50N-S rms 14
Difference
20
5.751
135°E
Total Cloud
Cover
ISCCP45°WD2 June0° 2001 nmonth=3
50N-S Mean:
61.4
135°W
90°W
45°E
90°E
135°E
ISCCP D2
35
65
5
95
135°W
30°S
35
60°S
50
35
65
30°S
65
35
65
35
80
65
35
65
0°
30°N
65
65
35
ECMWF
[percent]
60°N
42.42
60°N
60°N
40
40
-20
30
30°N
30°N
30
30°N
30°N
20
10
-20
20
10
0°
0°
10
-10
10
0°
0°
-20
30°S
30°S
-20
30°S
30°S
-30
-30
-40
10
60°S
60°S
-40
60°S
60°S
-50
-50
-60
135°W
90°W
45°W
0°
45°E
90°E
135°E
-10
-60
135°W
90°W
45°W
0°
45°E
90°E
135°E
In this course, we will consider:
Cirrus
Clouds in GCMs - What are the
problems ?
Cirrus
Ice cloud found in the
upper troposphere.
e.g: fronts, convective
anvils.
Important for the
radiative impact
Microphysics of the ice
phase even less well
understood than liquid
phase
In this course, we will consider:
Deep and Shallow (un)organised convection
Convection
In atmospheric science
generally restrict the term
“convection” to refer to
fluid motions resulting
from the action of gravity
on unstable vertical mass
distribution.
Encompasses a huge range of scales of organisation:
• Turbulent boundary Layers
• Deep moist convection - Thunder storms
• Squall Lines and Hurricanes
Current models do not take organised systems into account
Complex interaction of a host of
processes
Overview of Clouds/Convection
• Introduction
– Motivation, moist thermodynamics
• Parametrization of moist convection (4 lectures, Peter),
– Theory of moist convection
– Common approaches to parametrization including the ECMWF scheme
• Cloud Resolving Models (1 lecture - Adrian)
– their use as parametrization tools
• Parametrization of clouds (3 lectures - Adrian)
– Basic microphysics of clouds
– The problem of representing cloud cover
– The ECMWF cloud scheme (and issues concerning validation)
• Exercise Classes (1 afternoon, Peter and Adrian)
Moist Thermodynamics
Brief Overview of what we presume you
are familiar with, and an introduction to
thermodynamic charts
For simplified Overview:
Rogers and Yau (1989) “A short course
in cloud physics”
For rigorous definitions:
K. A. Emanuel (1994) “Atmospheric
Convection”
How many have
used tephigrams?
Moist Thermodynamics
• Assume that “moist air” can be treated as mixture of two
ideal gases: “dry air” + vapour
Density=1/specific volume
Dry air equation
of state:
(   1 )
pd  d RdT
Pressure
Water Vapour
equation of
state:
Vapour
pressure
Gas Constant for
dry air = 287 J Kg-1 K-1
e  v RvT  v
vapour
density
Temperature
Rd
Gas constant for
Vapour = 461 J Kg-1 K-1

T
0.622
First law of thermodynamics
Heat is a form of energy, and energy is conserved
dq  du  dw
Heat supplied by
diabatic process
(lowercase=per unit mass)
Change in internal
Energy
Can write as
dq  Cv dT  pd
Work done
by Gas
Cv  ( )
Specific Heat at
constant volume
Cp  ( )
Specific Heat at
constant pressure
dq
dT 
dq
dT p
First law of thermodynamics: form
Differentiating the equation of state and using
C p  Cv  Rd
dq  C p dT  dp
Special processes: “Adiabatic Process” dq=0
C p dT  dp
Special significance since many atmospheric
motions can be approximated as adiabatic
Conserved Variables
Using equation of
state and
integrating, obtain
Poisson’s
equation
 p 
T

 
T0  p0 
Rd Cp
Setting reference pressure to 1000hPa
gives the definition of potential
temperature (Note: for dry air)
 p0 

  T 
 p 
Rd C p
Meteorological energy diagrams
Total heat added
in cyclic process:
 dq   C T 
p

Rd dp
Cp p
  C  Td (ln  )
p
rotate to have pressure (almost) horizontal
Thus diagram with ordinates T versus
ln  will have the properties of
“equal areas”=“equal energy”
Called a TEPHIGRAM
dT
T
The Clausius-Clapeyron equation
air+water vapour
water
Consider this closed system in equilibrium:
T equal for water & air, no net evaporation
or condensation
Air is said to be saturated
• For the phase change between water and water
vapour the equilibrium pressure (often called
saturation water vapour pressure) is a function of
temperature only
• with v >> w, and the ideal gas law v=RvT/es
des 1
Lv

dT T ( v   w )
des
Lv es

dT
RvT 2
The Clausius-Clapeyron equation Integration
• The problem of integrating
the Clausius-Clapeyron
equation lies in the
temperature dependence
of Lv.
• Fortunately this
dependence is only weak,
so that approximate
formulae can be derived.
 es  L  1 1 
ln      
 es 0  Rv  T0 T 
es0 = 6.11 hPa at T0=273 K
Nonlinearity has
consequences for
mixing in convection
Humidity variables
There are a number of common ways to describe vapour
content :
1. Vapour Pressure, e
Pa
2. Absolute humidity kg m 3
3. Specific humidity kg kg
1
Mass of water vapour per moist air mass
4. Mixing ratio
5. Relative humidity
kg kg 1
v
v
e
e
q



pe
p
v
e
e
r


d
p  (1   )e
p
e
r
RH  (sometimes )
es
rs
Tephigram (II)

T
Saturation
specific humidity
r
es
qs  
p  es
Saturation
mixing ratio
es
rs  
p  (1   )es
pressure
Function of temperature and
pressure only – tephigrams
have isopleths of rs
Using a
Tephigram
At a pressure of
950 hPa
Measure
T=20 oC
r=10 gkg-1
plot a atmospheric
sounding
Virtual temperature Tv
Another way to describe the
vapour content is the virtual
temperature , an artificial
temperature.
It describes the temperature
dry air would have to have in
order to have the same
density as a sample of moist
air
By extension, we define the
virtual potential temperature,
which is a conserved variable
in unsaturated ascent, and
related to density
1 r 
v

p  pd  e  Rd T (  d  )  Rd T 
 1 r 


1 r 



Tv  T
 T (1  0.608r )
 1 r 


 p0 
 v  Tv  
 p 
Rd C p

Water variables
• absolute liquid water content
l
• liquid water mixing ratio
l
rl 
d
kg kg 1
• total water mixing ratio
rt  rv  rl
kg kg 1
• specific liquid water content
l
ql 

kg kg 1
Analogous to Tv can define the density
temperature T, which is the T dry air would
have to have equal density to moist cloudy air
kg m 3
1 r 

T  T 
 1  rt 


How can we saturate a
parcel of air?
The description of water content in its liquid (and/or ice) state
Ways of reaching saturation
Several ways to reach saturation:
All of these are
important cloud
processes!!!
• Diabatic Cooling (e.g. Radiation)
• Evaporation (e.g. of precipitation)
• Expansion (e.g. ascent/descent)
Cooling: Dew point temperature Td
Temperature to which air must be
cooled to reach saturation, with p
and r held constant
Evaporation: Wet-bulb temperature Tw
C pmdT 
 Lv dr
 Lv dr
 C p dT 
  Lv dr
1 r
(1  r )(1  0.9r )
Temperature to which air
may be cooled by
evaporation of water into
it until saturation is
reached, at constant p
Will show how to determine graphically from tephigram
Ways of reaching saturation:
Expansion: (Pseudo) Adiabatic Processes
As (unsaturated) moist air expands (e.g.
through vertical motion), cools
adiabatically conserving .
Eventually saturation pressure is reached,
T,p are known as the “isentropic
condensation temperature and pressure”,
respectively.
If expansion continues, condensation will
occur (assuming that liquid water
condenses efficiently and no super
saturation can persist), thus the
temperature will decrease at a slower rate.
Ways of reaching saturation:
Expansion: (Pseudo) Adiabatic Processes
Have to make a decision concerning the
condensed water.
• Does it falls out instantly or does it remain in the
parcel? If it remains, the heat capacity should be
accounted for, and it will have an effect on parcel
buoyancy
• Once the freezing point is reached, are ice
processes taken into account? (complex)
These are issues concerning microphysics, and
dynamics. The air parcel history will depend on
the situation. We take the simplest case: all
condensate instantly lost as precipitation, known
as “Pseudo adiabatic process”
Pseudo
adiabatic
process
Tephigram (III)

T
For a pseudoadiabatic
process:
r
0  dq  C p dT  dp  Lv drs
gives:
dp Lv
dT  kT 
drs
p Cp
pressure
Pseudoadiabat
Remember: Involves
an arbitrary “cloud
model”
(or moist adiabat)
parcel mixing ratio=5g/kg
Cooling:
(Isobaric
process)
gives dew
point
temperature
Expansion,
(adiabatic
process) gives
condensation
temperature
Wet Bulb
Temperature
Raise parcel
pseudoadiabatically
until all humidity
condenses and then
descend dry
adiabatically to
reference pressure
Equivalent Potential
temperature
 e  e
 Lv rv

 c pT





conserved in adiabatic motions
Parcel at 850 hPa,
T=12.5oC
r=6 g/kg
Te
e
(=315K)
Summary: Conserved Variables
(approximately, r.e. Emanuel 1994)
Dry adiabatic processes
Moist adiabatic processes
potential temperature
equivalent potential temperature
 p0 
  T  
 p
Rd
cp
dry static energy
s  c pT  gz
water vapour mixing ratio
rv
 e  e
 Lv rv

 c pT





moist static energy
h  c pT  gz  Lv rv
total water mixing ratio
rt  rv  rl
Practice!
1. Plot an air parcel at 750 hPa
which has
T=10C
q=2 g kg-1
State the dew-point temperature
2. If it is subjected to forced
ascent,
State the pressure at which it
saturates
3. The parcel continues to rise
pseudoadiabatically to
200hPa, what is its
temperature and humidity (r)
there?
4. The parcel then descends back
to 750hPa, what is the final
temperature and r?
Practice!
Results 1 - 10 of about 6,060,000 for
practice makes perfect [definition].
(0.05 seconds)
1. The environment at 750 hPa
has T=10C and r=2 g kg-1.
A saturated thermal
arrives at this level with
the temperature T=5C and
entrains the environmental
air with equal ratios (1-1),
what is the T/r of the
mixed parcel?
2. Then precipitation falling
through the layer
evaporates and brings this
new parcel to saturation,
what is the temperature?
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Finally: A few Unofficial Social Tips…
Answer Sheet for Lesson 1
Sheet 1
•
Td = -10oC
•
550 hPa
•
T= -78oC
For the humidity it is not possible to read from this tephigram. Therefore
you have to use the Clausius Clapeyron equation and the definition of
mixing ratio to find q=6x10-6 kg kg-1 using T=-78oC and p=200 hPa
4. T=15oC, so the parcel has gained 5oC from the latent heating released by
the condensation process.
Since humidity is conserved then q =6x10-6 kg kg-1
Sheet 2
1. T=12.5oC and q=4.7 g kg-1. Note that the humidity scale is not linear!
2. T=4oC