1. Thermodynamics 200713680 Lee, Jinuk 200930386 Park, sang ah 201013768 Yuk, kyeong min Group 6

2. I. Latent Heats II. Saturated Adiabatic and Pseudoadiabatic Processes III. The Saturated Adiabatic Lapse Rate IV. Equivalent Potential Temperature and Web-bulb Potential Temperature V. Normand’s Rule VI. Net Effects of Ascent Followed by Descent contents

3. The amount of energy released or absorbed by a chemical substance during a change of state that occurs without changing its temperature. Latent Heat

4. As a liquid is received the heat. then, molecular configurations are dispersed. So the heat appears to be latent In this phase, temperature is not change

5. Vapor Solid Liquid Deposition Sublimation Evaporation Condensation Freezing Melting

6. Solid Liquid Freezing Melting Latent heat of melting is 3.34*105 J kg -1. Melting Point is 0 ℃ at 1atm.

7. Vapor Liquid Evaporation Condensation Latent heat of evaporation is 2.25*106 J kg -1. Boiling Point is 100 ℃ at 1atm.

8. Deposition Sublimation Solid Vapor Latent heat of Deposition is 2.85x106 J kg-1

9. When air parcel rises, its temperature decreases with altitude at the dry adiabatic lapse rate until it becomes saturated. Then, they releases latent heat. Consequently, the rate of decrease in the temperature of the rising parcel is reduce. Saturated Adiabatic and Pseudoadiabatic Processes

10. Saturated adiabatic processSaturated adiabatic process - If all of the condensation products remain in the rising parcel, the process may still be considered to be adiabatic. Pseudoadiabatic processPseudoadiabatic process - If all of the condensation products immediately fall out of the air parcel, the process is irreversible, and not strictly adiabatic.

11. But, the amount of heat carried by condensation products is small compared to that carried by the air itself. Therefore, the saturated-adiabatic and the pseudoadiabatic lapse rates are virtually identical.

12. The Saturated Adiabatic Lapse Rate About the saturated adiabatic lapse rate - Change with pressure & temperature. - Actual range : ⅰ ) about 4K/km ⅱ ) about 6~7K/km

13. Γ d < Γ s It because water vapor condenses when a saturated air parcel rises. But, Γ s is only slightly less than Γ d because the saturation vapor pressure of the air is so small that the effect of condensation is negligible.

14. Saturated adiabats Lines that show the rate of decrease in temperature with height of a parcel of air that is rising or sinking in the atmosphere under saturated adiabatic conditions are called saturated adiabats (or pseudoadiabats).

15. Equivalent potential temperature & Wet-bulb potential temperature Now derive equation that describes how temperature varies with pressure under conditions of saturated adiabatic ascent or descent.

16. = – dq T C p dT T R T dp T P P ⍺ = RT ⍺ = RT P dq = C p dT – dP RT P ⍺ = – dq T C p dT T Rdp P Ideal gas equation First law of thermodynami cs Dividing T

17. dΘ = dT - RdP Θ T c p P cpcp cpcp cpcp dΘ = dT - RdP Θ T P cpcp cpcp Θ = T ( ) P0PP0P RcpRcp From Poison’s equation Multiplied by c p lnΘ = lnT - lnP + const RcpRcp Logarithm

18. dΘ = dT - RdP Θ T P dq = dT - RdP T T P cpcp cpcp cpcp dΘ ΘdΘ Θ =cpcp dq = -L v dW s dq T -L v dW s L v is the latent heat of vaporization First law of thermodynamics First law of thermodynamics Poisson’s equation Poisson’s equation

19. dΘ ΘdΘ Θ = L v dw s T ㅡ cpcp cpcp cpcp dΘ ΘdΘ Θ = T cpcp ㅡ cpcp T ≃ d ( ) LvwsLvws cpcp T dΘ ΘdΘ Θ - d LvwsLvws cpcp T ≃ Dividing Cp

20. dΘ ΘdΘ Θ - d ( ) LvwsLvws cpcp T ≃ lnΘ + C ≃ LvwsLvws cpcp T ㅡ LvwsLvws cpcp T ≃ ㅡ ( ) ln Θ Θe Θ Θe ≃ Θ exp ΘeΘe LvWsLvWs CpTCpT ( ) Integrated Equivalent potential temperature → Θe C= -ln Θe

21. 1000 p T θ Td θsθs Tw θwθw DRY ADIABAT θeθe LC L All the water vapor has condensed

22. Potential energy Dry static energy When height, rather than pressure, is used as the independent variable, the conserved Quantity during adiabatic or pseudoadiabatic ascent or descent with water undergo transitions between Liquid and vapor phases is Moist static energy(MSE) MSE = C p T + Φ + L v q The temperature of air parcel The geopotential The specific humidity enthalpy The latent heat constant

23. When air is lifted dry adiabatically : - Enthalpy is converted into potential energy. - The latent heat content remains unchanged. In saturated adiabatic ascent : - Energy is changed among all three terms. - Potential energy increases. - Enthalpy and latent heat content decrease. The sum of the three terms remains constant.

24. Normand’s Rule It is helpful in many computation involving the Skew T –ln P chart L ifting C ondensation L evel is located at the intersection of the potential temperature line, the equivalent potential temperature line and the saturation mixing ratio line Using LCL, we guess a certain value So, this rule is illustrated with temperature T, pressure P, dew point Td, and wet-bulb temperature Tw For example, if we know T, Td and P, we will find Tw using Normand’s rule

25. LC L θ T Tw Td p Ws θeθe

26. The effects of the saturated ascent coupled with the adiabatic descent are : I.net increases in the temperature and potential temperature of the parcel II.A decrease in moisture content III. No change in the equivalent potential temperature or wet-bulb potential temperature, which are conserved quantities for air parcels undergoing both dry and saturated processes Net Effects of Ascent Followed by Descent

27. Exercise 3.10 [1] Air parcel 950hPa has a temperature of 14 ℃ and mixing ratio of 8g/kg 1. What is the wet-bulb potential temperature of the air?

28. Θe=14 ℃ 14 ℃ LCL=890hPa 950hPa Ws=8g/kg Tw=14 ℃ 1000hPa θ

29. The air parcel is lifted to the 700hPa level by passing over a mountain, and 70% of the water vapor that is condensed out by the ascent is removed by precipitation 2. Determine the temperature, potential temperature, mixing ratio, and wet-bulb potential temperature of the air parcel after it has descended to the 950hPa level on the other side of the mountain Exercise 3.10 [2]

30. θ e=14 ℃ LCL1=890hP a 950hPa 700hPa 760hPa Ws=5.7g/kg LCL2=760hPa Ws=4.7g/kg 700hPa level – mixing ratio is 4.7g/kg 8 - 4.7 = 3.3g/kg 3.3 * 0.7 = 2.3g/kg (precipitated out!) Water vapor remains in the air 4.7+1.0=5.7g/kg (=mixing ratio) 20 ℃ T = 20 ℃ Mixing ratio(W) = 5.7g/kg Wet-bulb temperature(Θw) = constant Potential temperature(Θ) = 25 ℃

31. Thank you