MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
Lecture W4
Thermal advection
W4 1 - The concept of advection:
The temperature at an observing station may change in two ways:

The air parcel which is being sampled might change its thermodynamic state. For
example, sunlight might increase its internal energy, and hence its temperature will
rise.

The air parcel might be replaced by a different parcel with a different thermodynamic
state as the wind blows past the station. This process is called advection.
In practise, both processes will operate. However, on the synoptic scale, temperature
changes on timescales less than a few days are dominated by advection effects.
W4 2 - Lagrangian & Eulerian rates of change:
The diagram shows the variation of temperature in the vicinity of an observing station at
A. Suppose that the temperature observed is T0 .
A short distance  x upstream, at point B, the temperature will be:
 T
  x.
T  T0  
  x
But after a short time  t , the air parcel at point A will have been replaced by the air
parcel originally at point B, a distance  x  U t upstream of point A. The rate of change
of temperature at point A is therefore:
Which simplifies to:
 T  x  x
T
T

 U
.
t
t
x
The Norwegian frontal model has
already emphasized this, with the
concept of air mass, and the
changes of air mass associated with
the passage of a depression
system.
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
In this case, we assumed that the properties of the individual air parcels were constant.
Now suppose some process was changing the temperature of the air parcels themselves.
We denote the rate of change of temperature of an air parcel by DT Dt . This is
sometimes called the “Lagrangian rate of change” of T, i.e., the rate of change following
an individual air parcel and is related to the rate of change at the observing station by:
 T DT
T
.

U
 t Dt
x
In general, the wind will not blow precisely parallel to the x-axis. The expression is easily
generalized to give:
 T DT
T
T
T
.

u
v
w
 t Dt
x
y
z
Other quantities than temperature can be advected in the same way. Similar expressions
hold for them, simply replace T by the quantity concerned.
W4 3 - Thermal advection and the thermal wind:
Suppose that geostrophic balance holds. Then the low level wind blows parallel to the
height contours at the lower level. If the temperature gradient is constant with height,
then isotherms are parallel to lines of constant thickness. There will be thermal advection
if the lines of constant height make an angle with lines of constant thickness. If they are
parallel, then no thermal advection is taking place.
The “thermal wind”, i.e., the vector difference between the wind at 100 kPa and 50 kPa
was given by:
VT 
g  ( Z2  Z1 )
;
f
y
the thermal wind blows parallel to the lines of constant thickness. This enables us to
recognize three different situations:
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
1. No thermal advection: the thermal wind is parallel to the thickness contours. Hence,
in this case, the geostrophic winds at upper and lower levels are parallel.
Cold
V1
VT
V2
Warm
2. Cold advection. The thermal wind points to the left of the low level wind. Hence, the
geostrophic wind must back with height. The wind vector points from cold towards
warm air at all levels.
Cold
V2
VT
V1
Warm
3. Warm advection. In this case, the thermal wind vector points to the right of the low
level wind vector. The geostrophic wind points from warm air toward cold air at all
levels, and the geostrophic wind veers with height.
Cold
V1
VT
V2
W4.4 Estimating thermal advection from charts
Thermal advection is important when contours of height and thickness cross at a
significant angle. From a thickness chart we can estimate the magnitude of this thermal
advection. We can therefore identify likely regions of vertical motion by examining the
thermal advection across the chart.
Warm
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
Suppose both height and thickness contours are plotted, both with the same contour
interval Z. Denote the horizontal distance between height contours Z and the horizontal
distance between thickness contours T.
The magnitude of the geostrophic wind is therefore:
Ug 
T
g Z
f Z
Z

The temperature gradient can be estimated from the spacing of the thickness contours
using the thickness equation:
T
g
Z

s R ln  p1 p2  T
If the angle between the thickness and height contours is , then the magnitude of the
thermal advection is:
But the area of the parallelogram formed by the height and thickness contours is:
Ug
T
g 2 Z 2
sin 
sin  


s
fR
A ln  pT 1 Zp2  T  Z
sin 
and so we have a very simple formula for the magnitude of the thermal advection:
v.T 

A
where  
g 2 Z 2
fR ln  p1 p2 
For standard contour intervals of 60 m,
pressures
of
p1= 100 kPa
and
p2 = 50 kPa and at a latitude of 51N, 
is 1.54x107K m2 s-1.
MT11B Weather Systems Analysis
Eleanor Highwood, December 2003
Worked Example: Examining some charts, you find that at a station at 50N, the
surface geostrophic wind is 15 m s-1, bearing 240, and the 100-50 kPa thickness
decreases from south to north at 60 m in 200 km. What is the wind at 50 kPa? Assuming
these conditions do not change, estimate the change of temperature at the station in the
next 12 hours.
Step 1: The two components of the surface geostrophic wind are ug  15 cos( 30 )  13 m s-1
and vg  15 sin ( 30 )  7.5 m s-1.
Step 2: The thermal wind has magnitude
VT 
g  ( Z2  Z1 )
 26. 3 m s-1
f
x
and points from west to east.
Step 3: Hence the components of the geostrophic wind at 50 kPa are (39.3, 7.5) m s-1.
Step 4: The magnitude of the wind is 40 m s-1 and its bearing is 270   tan 1 v g / u g ,


which is 259.
Step 5: Draw it.
Step 6: We have a warm advection configuration. To calculate the temperature rise, we
note that the increase due to thermal advection is v g  T /  y .
Step 7: From the thickness equation:
T
g
 ( Z2  Z1 )

 y R ln( p1 / p2 )
y
Hence,  T /  y is 1.4810-5K m-1.
Step 8: The temperature increase due to thermal advection is therefore 1.1110-4 K s-1
or 4.8 K in 12 hours.
In fact, you will find that this type of calculation always over-estimates the observed
temperature change, by a factor typically around 2. Why do you think this might be?