SSI ＆ Potential Temperature

SSI is "Showalter Stability Index" here.
SSI is used for the following etc.
Does a thundercloud develop?
Does heavy rain fall on short time?
However, it is calculated by only 850 and 500 hPa.
It is impossible to compare two indices and to get to know the whole atmosphere.
As an index, it is insufficient.
Probably, you who know SSI have the following dissatisfaction.
Isn't there better method anything?

A potential temperature emagram will cancel your dissatisfaction.

0.　 Two definitional equations in potential temperature
There are the following two definitional equations in potential temperature.

θ1=T(z) (1000/P (Z))^(R/Cp)          (0-１)

θ2=T(z)+(mg/Cp) (z-z1000)    　　  (0-2)

(0-1) is the definitional equation which you know.
It is the definitional equation that entropy does not change.
(0-2) is calculated from a dry adiabat lapse rate.
It is the definitional equation that total energy does not change.
θ2 is the same as θ1.
We discuss using the θ2.
You who read the dry adiabat lapse rate and the potential temperature emagram will welcome θ2.
Supposing you do not welcome θ2, let's perform it as follows.

We define SSI new from now on.

1. Potential Temperature Emagram and SSI  
 SSI that is Showalter Stability Indexare defined as follow.
SSI= T500-T850'　　 (1-1)
T500 is temperature at 500 hPa.
T850' is the temperature related to 850 hPa.
T850' is explained later.
First, the relation between SSI and θe*500-θe850 is shown.

θe*500 is 500hPa saturation equivalent-potential temperature(SET), and θe850 is 850hPa equivalent-potential temperature.

SET is the equivalent-potential temperature which assumed humidity to be 100%.

The next is this relation.
If θe*500-θe850<0 then SSI<0.  
If θe*500-θe850>0 then SSI>0.  
It is supposed that a thundercloud progresses easily if SSI<0, and SSI>0 does not progress easily.
The next figure is a potential temperature emagram of 09:00 Tateno on August 04, 2008.
Data is using preliminary figures.

Tateno is an upper-air observation point nearest to Tokyo.

θ*500-θ850 of a graph is an error of θe*500-θe850.
Altitude of 850 hPa is 1495 m, the temperature 19 ℃, and the dew point temperature is 11 ℃.
Altitude of 500 hPa is 5890m,,temperature is -4.7 ℃.
It was calculated with SSI=4.3 at T850'=-9.0 ℃.
SSI said.
"The state of atmosphere is stable. Please judge that a thundercloud does not progress easily. "
The potential temperature emagram said.
“Equivalent-potential temperature is high at 1000 m or less.
Energy has accumulated.
CAPE seems to be very large.
If a convection takes place, cloud top height will exceed 8000 m.
You could not judge the state of atmosphere is stable.
Be careful. “
The rain of 79 mm/hour was observed in Otsuki City, Yamanashi at 19:40 on August 4.
Yamanashi is a prefecture contiguous to Tokyo.

Potential temperature emagram was right.

Although it must be careful, even when the state of the atmosphere is the same as this, it may not rain in your country USA.
Mt. Fuji sees from Tokyo.
Almost no place whose horizon can be seen is located in Japan.

The difference in geographical feature makes a phenomenon the different thing.

2. Comparison of Potential Temperature and SSI
A comparison is troublesome.
We compare θe*500-θe850 with SSI.
θe*500 is 500hPa saturation equivalent potential temperature.

Saturation equivalent potential temperature can be defined as follows.

Cpθe^＊500＝CpT500＋mgh500＋L（E500/(P500－E500)）　　　（2-1）

Saturation equivalent potential temperature is assumed to be 100% of humidity.
E500 is saturation water vapor pressure.
The right-hand side of (1-2) is the enthalpy, potential energy, and latent heat of vapor.

If you cannot convince (2-1), please read Potential Temperature Emagram.

850-hPa equivalent-potential temperature defines θe850 as follows.

Cpθe850＝CpT850＋mgh850＋L（E850/(P850-E850)）　　　　（2-2）

T500, h500, and E500 are temperature, height, and saturation water vapor pressure at 500hPa.
T850, h850, and E850 are temperature, height, and s water vapor pressure at 850hPa.

Let's put 850-hPa air into a balloon and raise it.

A balloon is raised from h850 to h500 so that the value of θe850 may not be changed.

The energy sources to be used are CpT850 and L (E850/(P850-E850)).
Since it raises in the beginning using CpT, temperature falls.
Relative humidity will be 100% if temperature falls.
Latent heat is also made into potential energy when it becomes 100% of relative humidity.

Latent heat is calculated that it becomes ice from vapor.

If a balloon is lifted, T850, E850, and P850 will change.
Let's set the values to T850', E850', and P850'.
P850' is raised to the height h500 up to 500 hPa.
Although it is P850'!=P500, P850' approximates with P500.
(2-2) is as follows.

Cpθe850＝CpT850´＋mgh500＋L（E850´／(P500－E850´）)（2-3）

T850' can be found if troublesome calculation is done.
A program does troublesome　calculation.

You can get a weather function by　Weather Function Excel VBA　.

The definitional equation of SSI is as follows.
SSI= T500-T850' (1-1)  
T850' presupposes that it is 100% of humidity.
And it can say as follows.
 
T850' is the temperature which lifted the balloon to 500 hPa without changing equivalent-potential temperature.
Also as follows, it can say.

T850' is temperature when a balloon is lifted to 500 hPa, without changing entropy and total energy.

If T850' is larger than T500, the mass density in a balloon will become small and it will acquire lift.
On the other hand, it will become energy if Cp is applied to θe*500-θe850.

It is as follows.

Cpθe^＊500－Cpθe_８５０＝Cp T500－CpT850´

　　　　　　　＋L(E500/(P500ーE500)) －L(E850´／(P500－E850´)）

θe^＊500－θe850＝

　　T500－T850´＋（L／Cp）｛E500/(P500－E500) －E850´／(P500－E850´）｝

　　(　P500－E500≒P500、 P500 －E_８５０’≒P500　)

　　　　　　＝T500－T850´＋（L／Cp）（ E500－E850´）／P500

θe^＊500－θe 850＝T500ーT850´＋（L／Cp）（ E500ーE850´）／P500 

　　　　　　（・・・SSI＝ T500－T850´…）

θe^＊500－θe 850＝SSI＋（L／Cp）（ E500－E850´）／P500　　（2-4）

If temperature is decided on 100% of relative humidity, water vapor pressure will be decided.
If temperature is high, water vapor pressure will also become high.
If temperature is low, water vapor pressure will also become low.

There is the following relation to SSI=T500-T850' and E500-E850'.

If T500-T850'<0 then E500-E850'<0
If T500-T850'>0 then E500-E850'>0

These are as follows.
If T500－T850´＜0 then θe＊500－θe850＜0
If T500－T850´＞0 then θe＊500－θe850＞0
T500-T850' is SSI=T500-T850' from a definition.

If SSI＜0 then θe＊500－θe850＜0
If SSI＞0 then θe＊500－θe850＞0

Potential temperature emagram tells us where it is that energy buildup.
Also, tells the energy can be accepted.
Equivalent-potential temperature is energy and saturation equivalent-potential temperature can be said to be energy capacity.

Clouds develop when the lower energy (=θe)  is greater than upper capacity (=θe*).

unstable atmosphere

3. Heavy Rain by Unstable atmosphere (August 4～9, 2008)

Workers were passed within the manhole and died five persons in Tokyo at August 5.

A lower graph is the potential temperature emagram, and Pressure/Water Vapor Pressure of 09:00 Tateno on August 4.

I say as follows experientially.
Clouds are beginning to be made to an altitude of 1000 m from moisture.

The saturation equivalent-potential temperature of 8000～12000m is in 344k.
On the other hand,  equivalent-potential temperature of 0～1000m is more than 349k.
The 8000～12000m  said.
Superfluous energy had accumulated in 1000 m from surface.

I already explained the next.

The rain of 79 mm/hour was observed in Otsuki City, Yamanashi .

The potential temperature emagram said, "Cloud top may be set to 12000 m or more."

Observed time is 09 in the morning.
Earth surface is heated by solar radiation.
Solar radiation brings about the next.
In a mountain, the value of potential temperature becomes the same easily. ^＊
Energy crawls on a surface of a mountain, and goes up.
The weather survey is performed in Mt. Fuji.

Next Graf is August 4, and the potential temperature and equivalent-potential temperature of Mt. Fuji.

＊I Think.

We should use not upwelling current but potential temperature which is easy calculation and easy observation.

Vapor will be conveyed if the value of potential temperature becomes the same.

Please refer to Gibbs' paradox 

Blue is potential temperature and red is equivalent-potential temperature.

I think as follows.

A cloud top will be set to 12000 m or more if equivalent-potential temperature becomes more than 345k.

The potential temperature emagram of Tateno of August 4 12Z.

21時 is 12Z

The unstable still continues.
An altitude of 850 hPa is about 1500 m, and 500 hPa is about 5500 m.
Is SSI minus or plus?  
If you know SIN and CAPE, SIN is small and will judge that CAPE is large.

Since it will be 12Z at 21:00, the energy of solar radiation is not added.

The potential temperature emagram of Tateno of August 5 00Z.

It is possible that 0～2000m air goes up to 12000 m.

I can think that I am as follows.
Development of the thundercloud has already started.
All energy by solar radiation willl be released high up in the sky.
It is so large that CAPE is also fearful.
It is very dangerous.
You should make work in a manhole stopped.
The rain of 72 mm/hour was observed in Narita-shi, Chiba at 11:50.

You may know Narita Airport near Tokyo.

The potential temperature emagram of Tateno of August 5 12Z.

The rain of 71.5 mm/hour was observed in Osaka at 17:40 on August 6.

You may know Osaka in Japan.

August 6 12Z Shiono-misaki

自由対流高度 is  level of free convection (LFC)

The Sion‐omisaki is an upper-air observation point nearest to Osaka.

A level of free convective is about 1000m.
You may think that it is high.
The potential temperature of this height is 308k.
If set to 308k at 0～1000m, vapor will be conveyed to a level of free convective.
 And a thundercloud will be progresses.
Potential temperature will be set to 308k if ground temperature will be 35 ℃.
It is not new that it will be 35 ℃ in Osaka in August.

4. Potential Temperature and Energy

It is important to consider that equivalent-potential temperature is energy.

It can be considered that saturation equivalent-potential temperature is energy capacity.

In marine or a plain, the energy accumulated becomes large.

It is because energy cannot escape easily high up in the sky there.

I think as follows.

A tornado becomes strong and a typhoon grows large.

By the way, is global warming considered using energy?

Even the energy which the air of unit volume has may not be calculated.

In the near future, you may understand the following thing.

Anticipation of global warming is carried out now using the wrong model.

******

This is only story.

This story includes a different occurrence from the fact. 

 

Story of No.3　Plant‘s　melt down.

Next knowledge is required to understand the melt-through of No. 3.

A. Structure which cools a pressure vessel. & B. Vent

A. Structure which cools a pressure vessel.

The upper figure is a key map of RCIC (Reactor Core Isolation Cooling) and HPCI（High-Pressure Core Injection system）.

A pump is turned by the pressure difference of a pressure vessel and a pressure suspension chamber.

If pressure difference is lost, circulation of water will be lost, and fuel will cause meltdown, and also it will start melt-through.

If a RCIC and a HPCI are used, pressure difference will be lost gradually and will be stopped automatically.

The vent needed to be carried out before stopping automatically.

A:pressure vessel

B:pressure suspension chamber

B. Vent

The following procedures are required in order to prevent meltdown.

1. Open a vent valve.

2. Open a safety valve and raise the pressure of a container.

3. A rupture disk is destroyed by high pressure and pressure turns into atmospheric pressure.

4. Pour water into fuel and it continues cooling.

1 to 3 is the work called a vent.

The valve on the outside of a container was closed. 

The valve which is inside a container has been then open. 

Story

RCIC has already stopped. 

HPCI stopped automatically around 2:42 on the 13th.

However, they have changed that they stopped manually.

And the pressure vessel lost circulation of water.

The work of the vent was long overdue although stopping automatically was clear.

The vent work is as follows about.

1． 04:50 on March 13

   Opening the vent valve of a pressure suspension chamber went wrong.

2. 05:23

   In order that there may be no compressed air, cylinder exchange judges it as necessity.

3．Unknown Time

It exchanged cylinders.

4．08:41

   Operation which a valve opens was carried out. And it checks (but the valve was not opened).

5. 09:08

   Rapid decompression of the reactor pressure by safety valves opened.

6. 09:24

   It was judged that vent enforcement was carried out.

7. 11:17

   The compressed air of a cylinder is lost and a valve closes.

8. 12:30

   Cylinder exchange and a valve opened. A valve is opened, it is going to act as Locke, but it will be at failure.

9. 06:10 on March 14

   A valvula is opened. Also after that, it repeats several times.

HPCI has stopped around 2:42 on the 13th.

9:08 are about 6 hours and half after HPCI stops.

Melt-through has actually occurred.

It is said that according to theinterim report the S/C Vent valve of the pressure suspension chamber was fully closed, and the pressure of the cylinder was 0. （on P 201）

The safety valve was opened.

And the worker heard sound.

It is the sound of steam blows to a pressure suspension chamber from pressure vessel.

I consider the next act is misjudgment.

A safety valve is opened with a vent valve closed. This action has the risk of breaking the containment vessel.  

This judgment is irrational.

I think as follows.

I think 5. and 6. are the following errors.

5．　 09時24分。

　Rapid decompression of the reactor pressure by melt-through happened.

6．　 09時24分。

　　It was judged that melt-through happened.

And also in this way, I think.

They had forgot for a HPCI to stop automatically.