2014年10月11日土曜日

Do you know greenhouse effect?

Do you know greenhouse effect?

Do you know greenhouse effect?
The Japanese meteorologist said.” Greenhouse effect is a scientific fact.”
I do not know what kind of basis he has.

I guess you regard him as right.

A scientific basis is not in greenhouse effect.
Why?
You need Radiative Equilibrium for the Earth.

1Radiative Equilibrium for the Earth
greenhouse effect is protected by Radiative Equilibrium.
Do you know next
Earth's average surface is 32°C warmer than it would be if it had no atmosphere. A planet the size of earth at earth's distance from the sun, and in thermodynamic equilibrium with solar energy (sunlight), would have an average surface temperature of -18°C. 

He is wrong.

Let's calculate.

Radiation input is
S0(1-A)πe2
S0S0 solar constant
AAlbedo
e: The radius of the earth

Radiation output is
e2 σTe4  

σ:Stefan-Boltzmann constant
e: The radius of the earth
Te Radiation equilibrium Temperature

He calculated as follows.
S0(1-A)πe2e2 σTe4  
Te4=(1/4)S0(1-A)  (A)
Te=(273-18) k=-18℃

Please look (A)
Te  has not e.
 Derivative of Te by re is 0.
Therefore Te is constant.
The temperature of Everest and the sea becomes the same.

Te is not surface temperature but Radiation equilibrium Temperature.

Do you know surface temperature of Gas planet like Jupiter.

I think 500hPa has Radiation equilibrium Temperature.


2013年2月21日木曜日

What is proof of global warming?


It is not scientifically right although investigating with mean temperature is common.
In 2 air, if wet-bulb temperature is 0 , snowflakes do not melt.
This is an observation fact.
It is not appropriate that the following value discusses warming.
Average value of the temperature of many different places.
I think that we should use the energy density of air.
It is rare to treat energy density in meteorology.

1.Energy which melts 1 g of ice
It is review of fundamental physics here.
It is said that the heat of fusion of water is 335 (kJ/kg).
335kJ of energy is necessary to ice into water of 0 1 kg.
I think our A and B boxes of 1 cubic meter volume, and two ice 1g of 0 .
Air temperature Ta 1mol contains the A.
Air temperature Tb contains 0.5mol The B.
I put ice in the box.
The air and the water I became 0 after a while, all the ice melts into water.
I'll try to calculate the value Ta, of Tb.
Main component of the air is nitrogen and oxygen.
Degrees of freedom of nitrogen and oxygen are 5.


Specific heat at constant volume of 1 mole of air is Cv = (5/2) R=.2.5 ×8.3143 20.8 (J / mol)
Energy that flows into the ice from the air is as follows: A.
Cv × (Ta-0) = 20.8 × Ta (J)
Energy that flows into the ice from the air of B is as follows.
0.5 × Cv × Tb-0) 8 = 10.4 × Tb (J)
The energy which melts 1 g of ice is 335J.
Ta and Tb are as follows.
20.8xTa (J) =335 (J)
10.4xTb (J) =335 (J)
Ta16.1 () =289.3 (K)
Tb32.2 () =304.4 (K)
Although A becomes higher 16 than B, the quantity of the melting ice is the same.

Thus, the average value of the temperature of A and B does not have a scientific meaning.
 
2. The 0 atmosphere and the 20 atmosphere
Let’s compare the atmosphere of the next two.
The atmosphere whose ground pressure and temperature is 1000 hPa at 0 .
The atmosphere whose ground pressure and temperature is 1000 hPa at 20 .
.
The atmospheric pressure of height Z is as follows.
P (Z) = P0exp-∫ (mg / RT (Z ')) dZ'
Integration range is from 0 to Z.
 0 of atmospheric pressure is as follows.

20 of atmospheric pressure is as follows.

 It can hardly distinguish.
The energy of 1 mol of air which it has is as follows.
mgZ+CpT=Const
mgZ is potential energy, and CpT is enthalpy.
The equation of state of 1 mol of ideal gas is as follows.
PV=RT
Energy density is as follows.

(P/RT)mgZ+Cp(P/R)

The former is potential energy density and the latter is enthalpy density.
Enthalpy density comparison is the same as comparing the atmospheric pressure P.

The next is the graph which pulled the pressure of 20 from the pressure of 0 .



It can be judged from a graph that the enthalpy density of 20 is large.
By warming, the height B1 to which an atmospheric pressure difference becomes large exists.
Pressure difference that became the largest is 30.7hPa of 8250m.
Let’s compare 0 and 20 mass density.



A1 was 16500 m, and -0.025kg/.
B1 is 8250 m.
There is height which guesses warming also about potential energy density.



The energy which carried out warming is saved up also as potential energy of an air parcel.

Let’s compare 0 and 1



The 1.5hPa pressure difference appeared among 5000 to 10000 m.




Let’s compare 0 and 01



A1 16500m -0.025kg/
A2 15950m -0.0013 kg/
A3 15950m -0.00013 kg/

B1 8250m -30.7hPa
B2 8000m   -1.6hPa
B3 8000m  -0.16hPa
0 of water was 0 of ice, and air also became 0 .
The temperature before putting in ice is as following.
A was 16.1 .
B was 32.2 .


********
Nuclear power generation and Greenhouse effect
Greenhouse effect has infringed the 2nd law.
And the principle of conservation of energy is also infringed.
For example, suppose that a marine temperature went up by warming by 1.
Greenhouse effect cannot explain the immense energy which raises 1 of marine temperature.
Scholars ignore such a basic law and assert Greenhouse effect for nuclear power plant promotion.
They have not calculated the energy density of air.
The definition of their potential temperature is wrong.
Even the scientific proof of warming is not acquired.
They did not inquire, either but have only said "Propel nuclear power plants."
Where on earth did their curiosity go?
And we seem to be the selfish children who think as follows.
Warming is not carried out even if it consumes energy by Nuclear power generation.


The mystery in which greenhouse effect is allowed
Let's check that it is fundamental about temperature.
The temperature of air is the average of the kinetic energy of air particles.
Air particles have potential energy and kinetic energy.
The next can be said from a principle of conservation of energy.
Kinetic energy will become small if potential energy becomes large.
If it goes high up in the sky, temperature will fall.
The temperature of air is dependent on gravity.
The air of the earth also has the latent heat by vapor.
This complicates structure of the atmosphere of the earth.
On the other hand, Venus does not have most water.
It can approximate by kinetic energy and potential energy.
If it approaches 100 m on the surface of Venus, it is a rate which 1 temperature goes up.
Those with about 50 km and 500 difference in temperature are required between clouds and the surface.
One hot in the surface of Venus is for this.
Now, the basis of greenhouse effect is as follows.
The thing whose temperature of surface of the earth is higher than radiation equilibrium temperature is for greenhouse effect.
However, there is no basis which radiation equilibrium temperature makes the temperature of earth surface.
Please consider the surface of a gas planet to be somewhere.
We believe such unscientific explanation.
The surface is about 6500 km from the center of the earth.
Radiation equilibrium temperature is not related to the distance from the center of the earth.
It will be set to 0 if radiation equilibrium temperature is differentiated from the distance R from the center.
Radiation equilibrium temperature is a value fixed regardless of R.
The high school student who studied differentiation and physics can understand this.
Radiation equilibrium temperature has disregarded the influence of gravity.
Although it is not believed, this is a fundamental physics level of meteorology.
Greenhouse effect is a trick for promoting a nuclear power plant.
Is it necessary why to propel a nuclear power plant?
It is for processing plutonium and uranium of nuclear weapons curtailed tens of thousands of shots.
It is made to process in the nuclear power plant in Asia.
The name of the military operation is nuclear peaceful use.
Therefore, Greenhouse effect is required.
A method passes a research cost to the scholar who studies Greenhouse effect, and gives scientific authority to Greenhouse effect.
Scholars get social standing and pretend that there is a scientific basis.
Or since a research cost can be got, it believes firmly.
A nuclear power plant group of promoters uses greenhouse effect.
The United States is not building the nuclear power plant for 30 years or more.
It is because it turns out that it fails.
It is clear that they have technical capabilities.
However, it is Japan and South Korea that make a nuclear power plant to Asia.
When a breakdown becomes actual, it becomes Japan and South Korea to have promoted atomic power.

It seems that we cannot escape the influence of the military operation of the name called nuclear peaceful use.


2013年2月14日木曜日

Internal energy and enthalpy of ideal gas


This is the image which I have.
When I was a freshman at college, and studied entropy.
However, I was not able to understand.
Words did not become a question easily.
I think that there are people with the same question.
And I even forget having had a question.
The next is a clear thing.
I do not understand entropy.

1. Internal Energy of Ideal Gas
The internal energy of n mol ideal gas is expressed as follows.
U=nCvT
Experiment person put n mole ideal gas into the box.
He took care that the size of a box did not change.
He found that energy required to warm 1 was nCv.
This is known as experimental fact.
The theorist explained as follows.
It will be set to nCvT if n mole kinetic energy is all added.

All kinetic energy may be set to 0.
Temperature was set to 0k when kinetic energy was 0.
This is the absolute temperature.
Internal energy is likely to have such a meaning.

Particles in a box collide and rebound upon a stopping wall.
Energy is not given by the wall although particles were perfectly elastic collision.
A box is changed into a piston and compressed.
Particles said.
"Piston’s side is approaching“
Particles get energy from the piston’s side.

The following is an expression of that energy.
nCvdT =-PdV  (1-1)
dV is dV = Sdx.
S is the piston’s area.
dx is the distance.
I think P does not change even if you press only dx.
(1-1) is perhaps the following meanings.
When you press the piston only dx, the temperature was raised only dT.
I think so・・・.

I did the addition of energy in the piston, the kinetic energy of the particles increases our.
I expressed as follows: U internal energy of an ideal gas is increased only dU.
dU = nCvdT =-PdV
I may not know.
What is the necessity to think of P does not change?
Cv is a proportionality constant that does not change the volume.
dV is small, the volume is changed.
I would be using the Cv?
Do not know if I will always think of them.

I cannot be convinced to (1-1).
However, I decided to be convinced that it is right.
(1-1). may be called quasistatic process.
In Japan, it is a freshman at college and learns.
That I learned in first grade college is not always easy.
I was explained as follows.
Quasistatic process takes infinite time.
I thought as follows.
May I be based on such a thing?
Moreover, there are also the following talks.
Quasistatic process does not need time so much.

2Entropy was not changing.

CvT=-PV (1-1)
This is the energy conservation law (1-1).
We have asked the definition of potential temperature equation from this equation.
Japanese meteorologist said.
 "A potential temperature is defined from the law of conservation of energy nCvdT =PdV"
I feel this lacking something.
Consider the ideal gas (T2, P2, V2) and (T1, P1, V1).
Two gases are very similar.
I want the difference between the two gas was like following.
nCv (T1T2) = (P1P2) (V1V2)
If this equation holds, the entropy of (T2, P2, V2) is the same as that (T1, P1, V1).
Someone said I think it reached the definition of entropy.
“On quasistatic process, Entropy does not change.”
(It may be my mistake.)

If nCvdT =PdV, entropy is not changing.
As good as saying that entropy is nCvdT =PdV..
We have described the entropy in entropy.

Entropy is it correct. And Memorize entropy!

I do not know what it is (1-1).
However, nCvdT =-PdV What is the special law of conservation of energy.
Even if it heats the inside of a box with a heater, a principle of conservation of energy is realized (δQ).
It is realized even if it turns a fan in a box (δW).
The principle of conservation of energy is fulfilled also by the following formula.
dU = nCvdT = δQ + δW

nCvdT =PdV is the law of conservation of energy ,and entropy is not changing.
3. Dry Aiabatic lapse rate.
We find the dry adiabatic lapse rate.
Please see the Dry Adiabatic Lapse Rate for more information.
If premised on Hydrostatic equilibrium, there is the following relation.

dP (Z) / P (Z) = (mg / RT (Z)) dz  (3-1)
I do the following with ease.
dP / P = - (mg / RT) dz
Consider the ideal gas of n mole gravitational field.
The state equation is as follows.
P (Z) V (Z) = nRT (Z)
It is as follows if this is differentiated from Z.
dP / P + dV / V = ​​dT / T
The left hand side of (3-1) is now modified as follows.
dP / P = dT / T dV / V
= (dT / T)-PdV / PV
= dT / T-PdV / nRT
Here, We use the (1-1) nCvdT =PdV .

 P/P    =(dT/T)+CvT/RT)  (32)

(3-1) was the following.
 dP/P      -(mg/RT)dz  (31)

It is as follows from (3-1) and (3-2).

(dT/T)+CvT/RT-(mg/RT)dz

We multiplied by RT to both sides.
RTCvT=-mgdZ  (33)
It is Cp = R + Cv from Cp-Cv = R equation of Meyer.
(33) is as follows.
CpT=-mgdZ 

We divide both sides by Cpdz.

T/Z=-mg/Cp (34)

(34) is the dry adiabatic lapse rate.
We will be integral as follows.
CpT+mgZConst (35)

(35) is the law of conservation of energy in something.
Let's define a θ as follows.

CpθCpT+mgZConst  (36)

θ is the value does not change by the Z height.

3Familiar formula of entropy
I asked for the definition of potential temperature equation from the equation of Hydrostatic equilibrium
Hydrostatic equilibrium was that of the following.
dP (Z) / P (Z) = (mg / RT (Z)) dz (3-1)
We integrate both sides.
The integration range of the left side is P (Z) from P0.
P/PlnP(Z)lnP0
From (3-4), the right hand side is the Cp / RdT / T.
The integration range is from T0 to T (Z).
Cp/RT(Z))dz=Cp/R)(lnT(Z)lnT0)

Because it is left-hand side = right side, it looks like this.

lnP(Z)lnP0 Cp/R)(lnT(Z)lnT0)) 
lnP(Z)-(Cp/RlnT(Z)lnP0-(Cp/RlnT0 (42)

Even if Z is any value, lnP(Z)-(Cp/RlnT(Z) is the same value as lnP0-(Cp/RlnT0 which is an initial value.

lnP(Z)-(Cp/RlnT(Z)lnP0-(Cp/RlnT0Const   (43)

An amount that does not change is the entropy of the (43)
.(43) is as follows.
lnP(Z)/P0)=(C/RlnT(Z)/T0) (44)

P0 is set to 1000 and T0 is set to θ.

lnP(Z)/1000)=(C/RlnT(Z)/θ) (45)

(45) defines the potential temperature θ which you know.
This definitional equation has a serious problem.
Please confirm the serious problem by Definition of potential temperature.

5Enthalpy
nCvdT =-PdV (1-1)
I have studied (1-1) in first grade college.
I do not know what that is (1-1) today.
However, it was found that the energy conservation law is entropy does not change (1-1).
The ideal gas, there is a law of conservation of energy that does not change entropy another.
It is the law of conservation of energy using the enthalpy.

It is as follows if the equation of state of ideal gas is differentiated from Z.

VdPPdVnRdT
PVVPnRdT
(31) is as follows.

nCvdT=VPnRdT
nCvdT+nRdT=VdP

Mayer's expression of relations R=Cp-Cv is used.

nCpdT=VdP (51)

(51) was calculated from the following three relations.
They are the Ideal gas law, Mayer's equation, and a principle of conservation of energy (1-1) which does not change entropy.
(51) can be said to be the principle of conservation of energy which does not change entropy.
We can have an image which pushes a piston by (1-1).
However, we cannot have even an image by (5-1).