Okay, so we were looking at this long problem
where there is a wind right. There is this
wind which is coming on one side. It is carrying
lot of moisture. It’s having some particular
temperature and pressure. Then that air parcel
is rising. It rises up to the LCL. Then it
rises beyond. Then it becomes supersaturated.
So it sheds some of its moisture as rain on
the windward side of the mountain.
It further climbs and then comes on to the
other side of the mountain. Why is it moving
because of the wind. It’s moving to the
other side of the mountain. Then when it is
coming it is following what is called a dry
adiabat. When it’s falling in dry adiabat
we see that the temperature is increased alright.
So we’ll now plot it on this Skew-T ln P
chart and show it and then if the internet
works, if the internet works after this okay
if the internet works we’ll also look at
what the internet is saying about this phenomenon.
That is some wind is coming on the windward
side, shedding its moisture and going to the
right side. They have some phenomenon like
this in Germany also, some particular type
of wind alright. So this was problem number
38. So we will. The first 5 minutes I will
just quickly run uh jot down only the highlights,
the important variables so that it helps me
when I sit down and plot it on the Skew-T
ln P chart.
That will be a revision for you and also for
those people who missed the class and it will
be very useful for the people who are taking
the course or want to watch the course on
nptel alright. So for the benefit of the people
who did not come to the last class. The problem
runs like this. An air parcel at 1000 hPa
has a temperature of 30 degree Celcius and
a mixing ratio 14 grams per kg correct. Is
this correct.
An air parcel at 1000 hPa has a temperature
of 30 degree Celsius and a mixing ratio 14
grams per kilogram. What is its wet bulb temperature,
first question. Then the air parcel is lifted
to the 700 hPa level by passing over a mountain
and then 80% of the water vapour that is condensed
out by the ascent is removed by precipitation.
Determine the temperature, potential temperature,
mixing ratio, wet bulb potential temperature
of air parcel after it has descended to the
1000 hPa level, same 1000 hPa level but now
it is on the other side of the mountain okay.
So we are looking at the story of the air
parcel as it moves from left to right or right
to left whatever and mountain is in between.
So the first thing was saturation mixing ratio
28 right. How much did you say 30 is it, 28
okay. T d is, 18. What is the theta? How do
you get theta? 30 I thought, 30, dry adiabat
okay.
840 okay. Then. Now follow the saturated adiabat
till it reaches, intersection gives, very
good wet bulb. Now we’ll stop and then do
it on the chart okay so that it is helpful.
Okay what are the initial condition, 30 degrees
okay. See today, today it’s giving trouble
okay. Then the dry adiabat is, where is the
saturation mixing, saturation mixing ratio.
Where is the saturation mixing ratio, 14 here
is it no. Saturation mixing ratio is the dashed
line? Where are the values for that? Just
above, oh this 0.4, 1, 2, 5, 10 ah I got it
I got it. Omega s is 14? They are intersecting.
Now I follow the saturation adiabat. So it
cuts the 1000 somewhere. Now Karthik I want
to clean this up. I want fresh sheet. Okay
very good.
So we got. Let me write whether, I want to
check whether the line is following 30. Problem
number 10? Okay, it’s alright. We will proceed.
What is the wet bulb potential temperature?
In this case, the wet bulb potential temperature
is same as T w because the original pressure
of the air parcel was only, you write that.
We had a rule right. The wet bulb temperature
is the arithmetic mean of the actual temperature
and the dry bulb temperature.
Actual temperature is 30, sorry actual temperature
dew point temperature. Actual temperature
is 30. Dew point temperature is 18. Average
is 24 but wet bulb was 22. But not bad, it
is close. It is alright okay. We will proceed
now.
So when it is lifted, when it is lifted up
to 700 hPa which line the red line, it follows
some red line or whatever that theta is a
constant line theta equal to 30 degree line
is what it follows up to the LCL then correct?
Then it follows the saturated adiabat up to
so it went here okay. LCL you can call. Did
it go here. Now what is omega s at 3. But
omega of parcel is 14. It cannot load so much
okay. So the excess is 4 grams per kg. 80%
of that becomes rain okay that’s what is
in the problem.
Here I am saying remaining moisture the correct
English will be moisture remaining in the
parcel whatever remaining moisture in the
parcel. So 20% of that 4 grams per kg plus
10 which is the saturation corresponding to
that so it is or you can take it as 14 minus
3.2. It is having 14, 3.2 became rain. So
10.8 is left. Either way it is correct. It
is the same. Now what you have to do is, again
I will erase. You please note the, you noted
point 3 no? Um very good.
So it was 700 and 710. That’s correct oh
very good. Let me fix the point. Now, attention
please. Please follow the saturated adiabat
until the omega s becomes 10.8 okay. So that
is a very fine this thing. So the are you
getting my point? The saturation mixing ratio
is 10 and saturation mixing ratio is 20. So
I look at saturation mixing ratio which is
like 10.8 or something. Okay so just wait.
That point is correct right okay. Come again
710?
Hey you said isotherm 10 what did you, dashed
line 10 okay very good 10 okay. Wait wait.
This one? Here fine right. Then, I’ll change
the colour. Marius what happened? You are
thinking about the holidays. That is dead
and gone in time. We live in the present.
Okay now what do we do. Saturation mixing
ratio. Now is this okay? 10 and 20? I want
10.8 so I’ll take it as 11 okay. Alright,
I got that. So it will come down up to that
alright. Then now it is fully saturated.
After that it is going to the other side of
the mountain. It is coming down. Not shedding
any moisture. As it comes down it becomes
drier and drier. So it will follow the dry
adiabat okay. Where is the dry adiabat, okay.
So I will like to call it as so T 5 equal
to 38. What did I ask? Okay when it cuts what
is the T 4, what is T 3, 8 very good. T 4,
I got 12; 10 okay so 
what is the omega s at what is omega at what
will be omega of the air parcel now okay.
So we started with 14 and we are running with
10.8. What happened to the 3.2. It went out
as rain. But we started with 30 degree centigrade
and we came out to 38 degree centigrade. Please
evaluate the relative humidity at 0.5. What
is the RH at 0.5? How to do that? You know
the omega. You have to get the omega s right?
Which one? What is that 41, 42? Omega s. Where
did you go all the way? Which one okay. Let’s
say 45. How much is it? Did you see that?
Air at 30 degree centigrade and 50% relative
humidity which was very comfortable became
air at 38 degree centigrade and 24% RH. It
becomes dry and hot. It became dry and hot.
Reasonably a moist and comfortable air became
so this is let us work it out let us write
all this on the board okay. So remaining moisture
okay. So T 3 is please tell me. T 3 is 8,
T 3 is 80% okay.
P at this level 4 is 720. So something like
this you can expect in the quiz. It can keep
you occupied for half an hour. You have to
be systematic. No surprises but you have to
do you have to be very meticulous. I’ll
be able to test whether you know everything;
wet bulb potential, dry adiabat, this thing
and all these things. All the concepts which
we have studied in thermodynamics will be
put to test okay.
Omega, omega s okay. Such a wind is called
a Chinook wind okay.
See that is the Chinook wind alright. Adiabatic
warming of downward moving air produces the
warm Chinook wind. So it is rising here. Then
it is dissipating its moisture. Then there
is adiabatic heating and you are getting a
clear and dry air. So this is Chinook wind.
Chinook winds are simply Chinooks or how do
you pronounce this Marius foehn winds which
are where the Canadian Prairies and Great
Plains meet okay.
It is in mythology it is known to be in popular
etymology it is known to be a snow eater because
it can eat the snow on the other side. It
is a snow eater okay. “Professor - student
conversation starts” You have foehn in Germany
also right. Ya foehn. Ya you have experienced
that. Have you experienced. Ha let’s say
uh ya. Okay. “Professor - student conversation
ends”. So right so a strong Chinook wind
can make one foot can make snow one foot deep
almost vanishing one day.
So it can remove snow right. The snow melts
and partly evaporates in the dry wind. Okay
and in Canada it takes place in these regions
and then Calgary and so on. So these are the
typical the cloud pattern when the Chinook
wind happens okay. In Alberta in Canada. How
Chinook occurs is basically rain shadow results
and all that we have seen. We have seen the
complete thermodynamics okay.
It is also in Chinook winds are sometimes
cause sharp increase in number of migraine
headaches suffered by locals and are often
called Chinook headaches. They have conducted
studies in department of clinical neurosciences
and they they support the belief that this
can cause headaches, irritability, sleeplessness
and all that right okay. Do you have something
Chinook wind in India, what is it? Your Andhi
is not this right. That is not this, ya okay.
So Chinooks also occur in Denver. Denver is
very notorious place in Colorado where always
because rocky mountains are there. I have
experienced this myself and missed flights
and all that. That story I will tell separately.
So Denver is a very dangerous region in the,
it’s a big airport DIA the Denver International
Airport but till you climb 20, 25000 feet
is all it’s all dizzy. So alright. This
is the Chinook wind. So let us go to Chinook
wind and see some images.
See this is good one. Get this 10 degrees,
moisture lost, heat added, minus 12 degrees,
Chinook wall cloud, Chinook okay. Any doubts.
So you’ll be able to solve problems of this
nature right. I asked some more things, wet
bulb, potential temperature, what did I ask?
Determine the temperature, potential temperature,
mixing ratio, wet bulb potential temperature.
What is the wet bulb potential temperature?
No change. Wet bulb potential temperature
no change okay.
Theta w is 22 is it? Was it 22? Okay. Let’s
move on to the next topic. We have to look
at a stability of an air parcel. If there
is an air parcel which is rising, you give
it some force. Will the air parcel continue
to rise or will the air parcel return to its
equilibrium okay. If it returns to its equilibrium
we again give a force and if starts oscillating
then is there a natural frequency of the system.
So we have to get deep into the maths.
Then you have to do F equal to ma then a equal
to d square z by d t square and we’ll identify
all the forces and then we get a characteristic
frequency of this air parcel which is called
the Brunt-Vaisala frequency. Don’t worry
too much about it. We’ll derive it in tomorrow’s
class. There is something called the Brunt-Vaisala
frequency, name of the 2 scientists who figured
out, the air parcel we need to oscillate.
What is the natural frequency depending on
whether this frequency is small, high and
this thing we can look at the stability and
all that. But before that first we have to
understand what we mean by stability. Stability,
we have to see with respect to unsaturated
air and saturated air. So we’ll start with
considerations of static stability with unsaturated
air okay. Let’s start so this will take
a couple of classes.
First we are going to look at unsaturated
air. Please note that it is slightly funny
curve because height height is always our
y coordinate in atmosphere. Temperature is
not the y coordinate. So I am doing temperature
versus height. How do you get the temperature
versus height in the atmosphere. If you want
how will you obtain? Radiosonde. You can put
a balloon and get the values and then with
that you have to find out how the temperature
changes with z.
That is called the gamma lapse rate which
is like 5 to 6 Kelvin per kilometer correct?
I have drawn 2 lines, 2 curves. What do you
think they are? One is for one is for the
air, one is for the parcel. Please note the
parcel pressure is the same as the air pressure
but the parcel temperature need not be the
same. If everything is the same, no activity
we have to, atmospheric science we have to
close the show.
There is some difference in the temperature,
there is buoyancy and then upward and downward
force, some activity is there in the atmosphere.
So I want to call this which one the steep
one is the surrounding air. This one is the
air parcel. If it is an unsaturated air, the
air parcel will follow what? Gamma, what is
its dharma. What will it follow? Gamma of
dry air. It will follow gamma d. But what
you get from Radiosonde measurements, if you
put a balloon and the Radiosonde instrument
you launch the balloon now at 12 o’clock,
it goes up.
It will not detect air parcel. It will just
detect, it will find out what is the temperature
in the atmosphere okay. So what will be that,
gamma okay. So this is from Radiosonde. Now,
O is the initial equilibrium point. Now tell
me gamma equal to gamma d, gamma greater than
gamma d, gamma less than gamma d. No no I
am giving you the slope, how can it be equal.
Gamma greater than gamma d surrender how?
This is height versus temperature not temperature
versus height.
Don’t go, don’t do too much research.
Look at the 2 graphs. Which is steeper? Which
is steeper yellow or pink? Gamma is greater
than gamma d? What is the lapse rate? It is
d z by d T or d T by d z. Then which pink
is steeper or yellow is steeper. Yellow is
steeper. Yellow is steeper from the point
of view of lapse rate. Basic funda isn’t
it. So what do you want to write here? Gamma
okay. Clear. If you keep on look if you keep
on looking at it for 5 minutes you’ll get
confused.
Just accept it okay. Now we are pushing the
air parcel from O okay. now give a gentle
upward push to the air parcel so that it comes
to a new level. At this level, the temperature
of the air parcel is A. The temperature of
the air is B. I am pushing it up okay. Come
down, call it T A. Come down, call it T B.
T A is 
but once you push, within a very small amount
of time the air parcel gets, the air parcel
gets its pressure adjusted to the pressure
of the surrounding. That is the assumption.
That is the assumption. So don’t worry about
the pressure remains the same but the temperature
is different. So what will happen to the density?
Very good. Rho A is A is the parcel and B
is the air. If it is denser and I push it
up what will happen? It will come back.
So is it a stable situation or unstable situation?
Stable situation. So this is called, this
situation is called positive static stability
okay. This is called . I’ll give all of
you 2 minutes. Please consider a situation
where the parcel is pushed downwards, draw
one more picture, the parcel is pushed downwards,
take the 2 new points as T c and T d, argue
out whether rho c, rho d is greater or equal
and then find out whether the parcel will
go down further or or the parcel will come
back to O.
Therefore, once you have done that analysis
in the next few minutes can you conclude with
reasonable confidence that if gamma is less
than gamma d irrespective of whether you push
it up or push you or you push the air parcel
down it will always come back to O. Please
get yourself convinced. Please complete that
exercise.
If it is pushed down correct? Therefore, the
parcel will come back to O. Tomorrow’s class
what we will do is we’ll take up the situation
where gamma greater than gamma d and prove
that it is unstable or negative static stability
okay. After that we’ll get an expression
for the acceleration of the air parcel and
then we’ll get an once it starts oscillating
what will be the characteristic frequency.
That is the Brunt-Vaisala frequency.
After that what you have to do is for saturated
air it becomes more difficult. Then quiz 2
it will be too late for quiz 2. In the final
exam, I’ll give you a Radiosonde reading.
I’ll give you the pressure and temperature
at various heights. You have to plot it on
the Skew-T ln P chart, find out the LCL whatever
and then you’ll have to see ab, bc, cd,
at all the levels where the parcel is stable,
unstable, convectively stable all that.
So that’s going to be, that’s going to
be the tough question in the exam. That means
you are becoming a meteorologist alright.
So we are getting deeper into this thermodynamics.
I thought this is an interesting part. So
I am spending more time. After this chapter
is over, we’ll look at radiation and climate
change. Dynamics I will just touch upon 3,
4 classes. I’ll just give geostrophic approximation
and the cyclostrophic approximation is relevant
to us.
You can, I want you to be able to calculate
the maximum wind in a cyclone okay; maximum
wind speed in a cyclone based on pressure
differences and all.
