4.7.The2 nd$law$of$thermodynamics$and$Clausius9 … · 2015. 3. 31. · q s ≈ 0.622e s p e s =e o...

4.7. The 2 nd law of thermodynamics and Clausius Clapeyron equa<on • The ClausiusClapeyron equa<on describes the change of satura<on vapor pressure on T. • What does the ClausiusClapeyron equa<on mean physically? Envision energy conserva<on and the Carnot cycle. Work added to the system adiabically Work removed from the system From A to B : heat and work in this case, due to vapor condensation and air volume change, are added to the system to increase T 2 to T 1 . Based on the Carnot cycle model : Q 1 T 1 = Q 2 T 2 = Q 1 − Q 2 T 1 − T 2 Q 1 − Q 2 = α 2 − α 1 ( ) de s ~ dp dT = T 1 -T 2 , Q 1 = L v , T 1 = T, Thus , α 2 − α 1 ( ) de s dT = L v T ⇒ de s dT = L v T α 2 − α 1 ( ) ≈ L v Tα 2 because α 2 >> α 1 de s dT = L v TR v T e s = L v T 2 R v e s ⇒ 1 e s de s = L v T 2 R v dT e e = e s,o ⋅ exp − L v R v ( 1 T − 1 T o ) ' ( ) * + , T o = 273 K, e s,o = 6.11mb e e = 6.11 mb⋅ exp − L v R v ( 1 T − 1 273 ) ' ( ) * + , Heat, Q1 extracted from the system Heat, Q2 added to the system

Upload
others
Category

Documents
view
0
download
0

Embed Size (px):

Transcript of 4.7.The2 nd$law$of$thermodynamics$and$Clausius9 … · 2015. 3. 31. · q s ≈ 0.622e s p e s =e o...

4.7. The 2nd law of thermodynamics and Clausius-‐Clapeyron equa
€

qs ≈0.622es

p

es = eo ⋅ expLRv⋅

1To−

1T

%

& '

(

) *

%

& '

(

) *

where qs : saturation mixing ratio; p : air pressurees : saturation vapor pressureeo : saturation vapor pressure at a reference temperature, To,often To = 273.15K = 0

oCL : 2.5 ×106 J deg-1, latent heat of vaporation, Rv = 461 Jdeg

−1 kg−1 : gas constant for water vaporT : temperature;

What control atmospheric water vapor: •  The satura
Discussion:

•  Why do storm intensity and extreme rainfall tend to increase, esp. in tropics, under a warmer climate?
Summary: •  Most of
Exercises-‐Moist thermodynamics

•  We oOen experience a cool downdraO air during a summer thunderstorm. –  What causes a cool downdraO in the thunderstorm? –  Compare air temperatures at 1000 hPa for these two cases. In both cases, air

temperature was 0C at 700 hPa. In the first case, a dry air subsides to 1000 hPa. In the second case: 5 g/kg liquid water evaporated in the air as it subsides to 1000 hPa.

2. Air at 1000 hPa and 25C has a wet bulb temperature at 20C. Using the in

SkewT-‐lnp chart: –  Find the dew-‐point, Td, –  If this air were expanded un
3. Water vapor feedback is the strongest atmospheric feedback to climate change. •  Assuming that the rela
4. Plot the temperature profile and use the Te (ambient temperature) and Td (dew point) at the surface to determine the pressure level of free convec
Te values are measured at various pressure levels. We need to compute Tc for convec
Solu%on for 4b: To compute CAPE, we use Te and Tc values in each atmospheric layer between

the LFC and LOC given in the above discussed tables and the formula

•  Calculate CAPE uses formulas (2) for Td=20C and 15C, respec
Materials covered by Quiz-‐2 (April 7th):

•  All materials in the thermodynamics of dry and moist air