Global budgets and transports
1
Global budgets and transport
Introduction
On a large scale the atmospheric circulations show a regular structure which depends
primarily on latitude. Using climate data on a three-dimensional (3-D) grid (latitude ,
longitude  and pressure p) we are able to calculate transports and budgets in order to
understand the large-scale circulation. How to calculate these quantities is explained in
this note.
Average values
Vertical averaging
Most atmospheric quantities depend on location (,) and on height (or pressure p). We
may define the so-called mass-weighted vertical average of a quantity A by:


0
0
A   A dz /   dz
(1)
If A is given on pressure surfaces this integral is simplified because density () is no
longer needed:
0
A 
 A dp /
p0 ,
(2)
p0
where p0 is the surface pressure.
Vertical summation
Usually one is not really interested in the vertical average but in the total amount of A in
a vertical column stretching from the surface to the top of the atmosphere ACOL To this
end we should multiply with the mass of air per m2 (mCOL) in the same column:
mCOL 
p0
,
g
(3)
hence
ACOL  mCOL
1
A 
g
0
 A dp .
(4)
p0
For instance, if A is specific humidity q (in kgH2O/kg), then qCOL is the Total amount of
water vapour in a column from the surface to the top of the atmosphere and is given in
kgH2O/m2.
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Meridional transports
In climate usually the meridional transports (i.e. the north-south transports) are the most
interesting. Usually the total transport over a latitude circle is considered. In this case
and for a given situation (day and time) averages should be taken over (1) pressure
(from the surface to the top of the atmosphere) and (2) the latitude circle. We will carry
out this calculation for water vapour transport in two steps. For transports of other
quantities (heat, CO2, momentum etc.) the calculation is similar.
The meridional transport of water vapour in a specific point in 3-D space is defined by
the product the specific humidity q (in kgH2O/kg) and the meridional component of the
wind speed v (in m/s) hence by q.v (in m/s kgH2O/kg). Figure 1 presents an example of
this transport for the January climatology (monthly averaged values of v and q and
averaged over many years). The positive (negative) areas denote a northward
(southward) transport over a latitude circle. It appears that positive areas are found
mainly over the oceans (especially on the eastern side). It also appears that most
transport is taking place in the lower troposphere because of the high values of specific
humidity there. However, from Figure 1 no net transport of water vapour over a latitude
circle can be determined.
Figure 1 The northward transport of water vapour v.q across 40ºN in January
in m s-1 kgH2O kg-1 (NCEP/NCAR Reanalysis climatology).
Global budgets and transports
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The first step is calculating a vertical average:
0
qv 
 qv  dp /
p0 .
(5)
p0
To calculate the total meridional transport through a vertical column from the surface to
the top of the atmosphere (Q) we must multiply with the mass of air in the column (p0/g).
This yields:
Q  qv
p0 1

g
g
0
 qv  dp ,
(6)
p0
with Q in kgH2O m-1 s-1. It represents the amount of water vapour that is transported per
second and per m latitude circle to the north if Q > 0 or to the south if Q < 0 (Figure 2).
Figure 2 Vertically integrated northward transport of water vapour Q40 (in kgH2O
m-1 s-1) across 40ºN in January (NCEP/NCAR Reanalysis Climatology).
Figure 2 gives this vertically integrated transport, and the positive and negative areas of
Figure 1 are now clearly visible.
The last step is to calculate the transport across the total length of the latitude circle
Q,TOT by (1) averaging Q over the latitude circle which yields [Q]:
Global budgets and transports
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2
Q   21  Q


d ,
0
And multiplying by the length of the latitude circle (2 Ra cos):
 
2
Q ,TOT  2Ra cos  Q  Ra cos   Q d ,
(7)
0
In Equation (7) Ra is the radius of the Earth (6370 km).
We now have Q,TOT in kgH2O s-1 and it is a measure of the total amount of water vapour
which crosses the latitude circle through the entire depth of the atmosphere either
northward (positive value) or southward (negative value).
If v and q are hourly, daily, weekly, monthly, seasonal or yearly averages, then Q,TOT
represents the same average and we might want to multiply with the amount of seconds
in e.g. a month in order to express Q,TOT in kgH2O month-1.
Application: the global hydrological cycle
For a basin the following balance equation holds for water:
S  P  E  R0  RU ,
(8)
where
S = storage term of water
P = precipitation both liquid and solid
E = evaporation including the evapotranspiration and sublimation
R0= surface runoff
Ru= soil runoff.
For the average over a large area we find that Ru is usually small. We are left with:
S P  E R ,
0
(9)
where { } indicates a spatial average over an area with size A. If we also average over
time (indicated by and an overbar) the storage term may be neglected as well. The
precipitation P is measured at many locations and R0 is measured in the flow of water in
rivers and streams. Measurements of evaporation, however, are difficult and therefore
not always available.
For the atmospheric branch of the hydrological cycle we define the amount of
precipitable water by:
0
W (  ,  ,t ) 
1
q dp ,
g p0
(10)
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where q is specific humidity (Equation 4). W is expressed in kg m-2 (may be transformed
into mm of water) and it represents the amount of water vapour in a column from the
surface to the top of the atmosphere. The horizontal water transport through the
atmosphere (Q) is:
Q ( ,  , t ) 
1
g
0
 qv dp ,
(11)
p0
Q is a vector, with the zonal and meridional components van Q given by, respectively:
0
Q 
0
1
qu dp
g p0
Q 
and
1
qvdp .
g p0
(12)
Integration from the surface to the top of the atmosphere yields the conservation of the
amount of precipitable water:
Wc
W
 div Q 
 div Qc  P  E ,
t
t
(13)
where Wc and Qc are analogous to W and Q but for clouds: liquid water and ice. If we
average Equation (13) over a long period of time then the third and fourth term will
disappear. If furthermore we average over a basin then we are left with:

 

 W 

  div Q  E  P .
 t 
(14)
This equation represents the atmospheric branch of the hydrological cycle. Except in
heavy storms and during short periods of time the storage term W / t is very small
and Equation (14) indicates that convergence (or divergence) of water vapour can be
found in areas with more (less) precipitation than evaporation. The right hand side of
Equation (14) is also present in the terrestrial branch (9) and is responsible for coupling
these two branches. Elimination of the right hand side of Equation (14) using Equation
(9) results in:
R  S  div Q  Wt  .
0


(15)
For the calculations we restrict ourselves to latitude belts. Table 1 (next page) gives the
zonally- and yearly averaged values of precipitation and evaporation for 10º latitude
belts. The runoff may then easily be calculated by the difference in precipitation and
evaporation.
Using this table and the NCEP climatological atmospheric data we may be able to
estimate global transports and budgets of water or water vapour in the hydrological
cycle.
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Table 1 Estimated yearly sums of precipitation (P), evaporation (E), runoff (P-E), the
evaporation ratio (drought-index) (E/P) and runoff ratio ((P-E)/P).
Practical calculations
Vertical averages and summations
The calculation of a vertical average is less obvious than it seems from the simple
Equations (1), (2) etc. We will explain this with the aid of Figure 3 (next page).
Usually atmospheric quantities are only available on pressure levels 1,2,3,... with
pressures p=1000, 925, 850, 700,... hPa). However, pressure intervals may not be
constant. We assume then that the value of a quantity e.g. specific humidity (q), on a
given pressure level is representative for the layer which is located from this level to
halfway up and down the next pressure levels. For example (Figure 3) q at 925 hPa is
representative for the layer from 962.5 hPa (= (1000-925)/2) to 887.5 hPa (=(925850)/2). In this way the atmosphere can be divided into layers and levels. The thickness
of the bottom layer is determined by the sea level pressure (pmsl = mean sea level
pressure), assuming that there are no mountains.
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Figure 3. Definition of pressure levels (solid lines) on which quantities are known (black
circles) and the boundaries between the layers (dashed lines).
Because the pressure intervals are not constant, the thickness of each layer (expressed
in hPa) will be different from the next. E.g. in Figure 3 q(2) is representative for a layer
with thickness of 962,5-887,5 = 75 hPa, while q(3) is representative for a layer with
thickness 775 - 887,5 = 112,5 hPa. If we want to average or integrate we must account
for these differences in thickness, i.e. we need a weighted average. If we indicate the
pressure at the very top by ptop (275 hPa in Figure 3), the thickness of layer i by pi and
the value of the quantity with qi then for the integral in Equation (1) we have:
q 
1
p msl  ptop
n
 q p
i 1
i
i
,
(16)
where n is the number of layers in the vertical direction.
In GrADS the function VINT performs vertical integration. It is a mass-weighted
integration as defined in Equation (4). However, the result is multiplied by 100 to account
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for the difference between hPa and Pa. In this way if you use VINT the dimension of a
quantity M with unit X is changed into unit X kg m-2.
Examples
- if M is specific humidity q (in kgH2O/kg), then the result of VINT is the Total amount of
water vapour in the vertical column (in kgH2O m-2)
- if M is the northward flux of water vapour vq (in m/s kgH2O/kg), then the result of VINT is
the vertically integrated water vapour flux Q (in kgH2O m-1s-1).
Determining the water budget of a latitude belt
The determination of the vertically averaged meridional transport Q,TOT across a latitude
circle enables us to calculate the water budget for a latitude belt (Figure 4).
Figure 4. Scheme for the determination of the
water budget of a latitude belt.
Figure 4 shows the atmospheric branch of the hydrologic cycle. If we determine the
transport Q,TOT across two latitudes (say 40ºN and 50ºN), hence: Q40,TOT and Q50,TOT,
then the difference Q = Q50,TOT - Q40,TOT is a measure for the divergence of the water
vapour transport (div Q). The divergence may be positive or negative. With positive
divergence water vapour will disappear by atmospheric flows from the column above the
latitude belt; according to Equation (14) this is only possible for a prolonged period of
time if we have more evaporation than precipitation in the latitude belt (P < E, or P-E<0).
The water for evaporation must then be supplied through the soil or more likely by rivers
and/or ocean currents (terrestrial branch of the hydrological cycle). With negative
divergence we have the opposite situation, the amount of water vapour cannot increase
continuously hence we have more precipitation than evaporation (P > E, or P-E>0).
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Figure 5. The northward water vapour transport averaged over a year and over the
winter (December, January and February: DJF) and over the summer (June, July and
August: JJA) [From: Peixoto and Oort, 1996].
Figure 5 presents the northward water vapour transport for seasonally and yearly
averaged situations. The data in Figure 5 across latitude belts of 10º must be similar to
the data in Table 1 when we apply equation (14) with W / t = 0.