1. * Reading Assignments: 2.2 2.2.1 2.2.2 2.3 2.4 2.4.1 2.4.2 2.5

2. 3. The First Law 3.1 Internal Energy Observations show that expansion work performed on an adiabatic system only depends on the initial and final states. So it behaves as a state variable. Further, changes in expansion work are reflected by changes internal to the system, i.e., the internal energy (u). Or for infinitesimal change,

3. 3.2 The First Law (Diabatic change) If heat is exchanged between the system and the environment, but So, the difference betweenandis called the heat transfer into the system,

4. * For an incremental process, This is the mathematical form of the first law. It describes that the change of internal energy between two states is the net effect between the heat transfer into the system and the work performed by the system on surroundings. Four physical points can be deduced from the first law. (3.1)

5. The net work performed by the system is balanced by the net heat absorbed by the system during a cyclic process. Then, 2) Since u is a state variable, for a cyclic process. 1) For an adiabatic process, For a process where no work is done, Because du is an exact differential and depends only on the Initial and final states, two different processes can change the internal energy of a system by the same amount.

6. 3) A closed system performs work through a conversion of heat absorbed by it -- heat engine. 4) A closed system rejects heat through a conversion of work performed on it -- refrigerator. Example: Troposphere Example: Stratosphere

7. * Another state variable, the specific enthalpy, can be defined for the convenience, From this, Then, For constant pressure processes, the first law becomes The heat transferred into the system can be measured by the change in enthalpy. Use the first law, (3.2)

8. 3.3 Heat Capacity * How to determine the heat absorbed or released in thermodynamic processes? 1)For a constant volume process, a homogeneous system, no physical change of state and no chemical reaction, The heat absorbed is proportional to its mass and temperature variation, or The constant of proportionality is called the specific heat capacity at constant volume, and is called the heat capacity.

9. To relate to state variables, consider Use this, the first law can be written as When the volume is held constant

10. 2) Similarly, at constant pressure is the specific heat capacity at constant pressure. For we can relate to the enthalpy when p is held constant, So, Changes of heat appear as changes of temperature for constant pressure situations.

11. * What did the Joule experiment demonstrate? Why is this result important? 1) Consider is constant, Experiment shows

12. Therefore, the internal energy of an ideal gas is a function of temperature only, 2) Consider Since can not depend on pressure. An easy extension can be made to enthalpy, only.

13. * If we define u=0 and h=0 at T=0, integrate We get or Based on simplified kinetic theory, in general, where j is the number of degree of freedom, R is the specific gas constant. So,

14. For a diatomic gas, j=5, translation + rotation + vibration. In dry air which mainly includes For a monatomic gas, j=3, translation only. we would expect j=5

15. * We will use the following forms of the first law frequently: (3.3) (3.4)

17. 3.4 Adiabatic Processes * If the other terms must balance in the first law. Use So Integrate Finally, (3.5) 3.4.1 Poisson’s Equations

18. * Similarly, use We have and * To get a relationship in p & v, use log derivatives on idea gas law So, From, (3.6)

19. Integrate to get or (3.7) Since We can get Poisson’s equations from (3.5)-(3.7), (3.8)

21. 3.4.2 Potential Temperature From the relationship between pressure and temperature for an adiabatic process, we can derive another state variable, i.e., potential temperature. Consider Integrate it from to, We have

22. Finally, The potential temperature of an air parcel is the temperature that it would take if we compressed or expanded it adiabatically to a reference pressure of 1000 mb. is conserved during an adiabatic process. (3.9)

27. 3.4.3 Vertical Motions and Adiabatic Constraints To describe the change of temperature with height under adiabatic condition, we need to combine two equations: primes are value for an air parcel for the environment Note that the parcel pressure also obeys 2) since horizontal pressure difference between the parcel and environment adjust at sonic speeds. So Use this in 1)

28. Rearrange it, we have Since Finally is called the dry adiabatic lapse rate.

30. 3.5 Diabatic Processes We can intuitively predict that is not constant when significant diabatic processes occur. How do changes in relate to changes in ? Use log derivatives on We have (3.10)

31. We can also put the first law in log derivative form, (3.11) Compare (3.10) and (3.11), we have (3.12) This is an alternative form of the first law. It shows that the heat transferred into the system is directly related to the change in potential temperature.

33. Homework (2) 1.A sample of dry air has an initial pressure p 1 =1000 hPa and temperature T 1 =300K. It undergoes a process that takes it to a new pressure p 2 =500hPa with unchanged temperature T 2 =T 1. Compute the mechanical work per unit mass performed by the sample under the following scenarios: a) Isochoric pressure reduction to p 2 followed by isobaric expansion to final state. b) Isobaric expansion to final specific volume v 2 followed by isochoric pressure reduction to final state. c) Isothermal expansion to final state. 2. A unit mass of dry air goes through the following steps: a) adiabatic compression from 60 kPa and 0 o C to a temperature of 25 o C, and b) isothermal expansion to a pressure of 70 kPa. Calculate the work done by the air during each step. 3. Through sloping convection, dry air initially at 20 o C ascends from sea level to 700 mb. Calculate a) its initial and final specific volume, b) its final temperature, c) the specific work performed, and d) changes in its specific energy and enthalpy. 4. A plume of heated air leaves the cooling tower of a power plant at 1000 mb with a temperature of 30 o C. If the air may be treated as dry, to what level the plume will ascend if the ambient temperature varies with altitude as: a) T(z)=20-8z ( o C) and b) T(z)=20+z ( o C), with z in kilometers.