Atmospheric Science 4310 / 7310

1. Atmospheric Science 4310 / 7310 Atmospheric Thermodynamics - II By Anthony R. Lupo

2. Day 1 • The work done by an expanding gas • Let’s draw a piston:

3. Day 1 •  Consider a mass of gas at Pressure P in a cylinder of Cross section A • Now, Recall from Calc III or Physics: • Work = force x distance or Work = Force dot distance • So only forces parallel to the distance travelled do work!

4. Day 1 • Then,

5. Day 1 • But, we know that: • Pressure = Force / Unit Area • So then, • Force = P x Area

6. Day 1 • Total work increment now: • Well, • Area x length = Volume • soo………………… A * ds = dVol

7. Day 1 • Then we get the result: •  Work : • Let’s “work” with Work per unit mass: •  Thus, we can start out with volume of only one 1 kg of gas!!!

8. Day 1 •  And we know that: • a = Vol/m • then…….. • Note the “heavy D”

9. Day 1 • So…. • Very Important!!! This is not path independent, it’s not an exact differential! •  Thus is is easy to see that if a parcel undergoes expansion, or a2 > a1

10. Day 1 • Parcel does work expanding against environment! •  But if the parcel undergoes contraction (a2 < a1) • Environment is doing work on the parcel!

11. Day 1/2 • A derivation more relevant to an air parcel •  Given a spherical parcel of air with radius R, Surface Area = 4pr2 and the volume is 4/3pr3.

12. Day 2 •  We calculate dW in expanding the parcel from r to r + dr in an environment where the pressure is p! • Then, • Work = F x dist = Pressure x Area x distance

13. Day 2 • Well, • Dist = dr • And; • Area = A + dA

14. Day 2 • As with the “expanding piston” • DW = p*A*dr = p * dV •  Recall from Calc II (or is it physics?) • dV = 4p r2 dr

15. Day 2 • Then we’ll deevide by the unit mass again to get: • dw = p da • Now the short comings in this problem are obvious: • 1. (Surf)Area = A + dA = 4 p r2 + 8p r dr

16. Day 2 • 2. The parcel expands and p decreases; • so: p = p + dp • So….. the REAL problem is: • dW = (p+dp)(A+dA) dr

17. Day 2 •  Which “foils” out to, • DW = pAdr + dpAdr + pdAdr + dpDAdr • then assume; • Dw = p da + dp da + p dA dr + dp dA dr

18. Day 2 • Which then reduces to (by scale analysis): • Dw = p da

19. Day 2 • The First Law of Thermodynamics: A derivation. •  The First Law of Thermodynamics, is really just a statement of the principle of the conservation of energy. •  This law can be used as a predictive tool: ie, we can calculate changes in T, a (r), or pressure for a parcel.

20. Day 2 •  The first law of thermodynamics is associated with changes in these quantities whereas the equation of state relates the quantities themselves, at a given place at a given time (or at a given place for a steady state). • Forms of energy relevant to our treatment of the 1st Law • “Energy” is (simply) the capacity to do work! • for our purposes, a system is an air parcel.

21. Day 2 • KE (kinetic energy)  associated with motion, energy of motion • (Definition: “Kinetic” means the study of motion w/o regard to the forces that cause it)

22. Day 2 • PE (potential energy)  is the energy of position relevant to some reference level, or within some potential (e.g., gravitational) field.

23. Day 2 • **The location of Z = 0 is arbitrary, but it makes good sense to use sea level. • Internal energy (IE)  Is the energy (KE and PE) associated with the individual molecules in the parcel. • **So IE is temperature dependent (use Absolute T)!!!

24. Day 2 • Deriving the First Law: (Hess pp 25 – 26) • Statement of Conservation of Energy: “Energy of all sorts can neither be created nor destroyed” (Of course we ignore relativistic theory E = mc2) • Written in incremental form (here it is!):

25. Day 2/3 • The “book definition”: • “Any increment of energy added to a system is equal to the algebraic sum of the increments in organized KE, PE, IE (thermal), work done by the system on it’s surroundings, and whatever forms of electrical and magnetic energy may appear” • Dheat = DKE + DPE + DIE + DWork + DE&M

26. Day 2/3 •  We can neglect electrical and magnetic for atmospheric motions we consider (scale analysis). • ** Recall: our system is the air parcel! •  Consider the LHS; “increment of energy added to a system” can be broken down into “heat added to the system” + work done on the system by the surroundings”. • We want to simplify, if possible to quantify our physical principle (the conservation of energy).

27. Day 3 •  Kinetic Energy: We first note that increments of Kinetic energy are small compared to changes in internal energy and increments of work done. For example, consider a parcel of air changing speeds from 10 m/s to 50 m/s • DKE = ½ (50 m/s)2 – ½ (10 m/s)2 or 1.2 x 103 m2 sec -2

28. Day 3 • Next, let us consider a change of 20 C for a parcel of air at a constant volume, a change of internal energy: • DIE = Cv DT = 717 J K-1 kg-1 x 20 K • or 1.4 x 104 m2 sec-2

29. Day 3 • Then, let’s consider the change in PE at, oh, say 500 hPa, • So, • DPE = g*Dz = 10 m s-2 * 60 m = 600 m2 s-2

30. Day 3 • Pressure work; • We have dw = p da and pa = RT • so dW = (RT/a) da = RT d[ln(a)] • Since T = Const, we can write: • = ln 0.1 * 287.04 J/kg K * 280 K • W = 8.47 x 103 m2 s-2

31. Day 3 • So, we’ve shown that: • Dh/dt = IE + PE + KE - Work done on the system by the surroundings + work done by the system on the surroundings+ EE + ME •  We neglected EE and ME by scale analysis, and we can neglect KE and PE by scale analysis (or do we)?

32. Day 3 • Well, • DKE = DKE(horiz) + DKE (vert) • ***KE vertical is very tiny, by several orders of magnitude (at least 4)! • Work done on the system by surroundings (W): • -(Work horiz + Wvert)

33. Day 3 •  The Horizontal pressure gradient is responsible for changes in wind speed. • The pressure gradient is in DKE and Wh (which is horizontal wind times distance covered) • Then, DKE must cancel with Wh! • How about Wv?

34. Day 3 •  Well, DPE = gdz • and, • Wv = vert. PGF x distance = - (1/r) (dp/dz) (dz) • If hydrostatic balance is assumed to hold then g = -(1/r) (dp/dz).

35. Day 3 • Thus, DPE + Wv = gdz – g dz = 0 (and these cancel)! •  So in the incremental form of the first law we are left with: • Dh/Dt = IE + work done by the system on the surroundings (1) • **(note, the “heavy D”)

36. Day 3/4/5 • Then if we • define heat added as dh/dt (dq/dt) or just dh or dq • Internal energy as “du”, • and work done by the system as “dw”, then: • We can quantify first law (this is it)!  ? • Dh = du + Dw or du = dh - dw

37. Day 5 • The First Law of Thermodynamics and Entropy. • What happens when heat is added to parcel? • 1st Law: dh = du + dw •  Heat can be added (or subtracted, of course)! How? • Conduction, convection, radiation, or phase change (All “Diabatic” heating)

38. Day 5 • Diabatic processes: • Sensible heating  warm to cold by contact • Radiative heating  “Short wave in” or “long wave out” • Latent Heating  phase changes of water mass

39. Day 5 •  So we can say formally “if a small quantity of heat* is added to the system (parcel) some goes into increasing the int. energy of the parcel, and some gets expended as the parcel does work (dw) on the surroundings”. • *(Note: heat is NOT temperature!) •  We know the 1st law, however, how do we express in terms of state thermodynamic variables (T, a, or P?) so we can obtain useful expressions for atmospheric calculations?

40. Day 5 •  Unfortunately, there are no formal, or analytical, mathematical expressions for dh (especially for atmospheric processes – these typically “parameterized” (which is an empirical “fudge factor”)). •  Thus, we are reduced to solving as a residual, and in the 1st Law leave as dh/dt or dq/dt (Q-dot). • Well, we showed that the work done by an expanding gas could be derived by considering a piston (closed controlled system), or an air parcel.,

41. Day 5 Thus, dw = p da. So the 1st Law is now:

42. Day 5 • Internal Energy •  With the pressure work term expressed in terms of state variables p and a, we must now consider that internal energy may be dependent on (T and a) or (T and p) (since p and a can be related) • So: Consider that u = u(T,a) or u = u(T,p)

43. Day 5 • The expression

44. Day 5 • Joule’s Law; and his experiment conceptualize this term • Consider an insulated container with two chambers, one filled with Gas at pressure P = Po, and the other a vaccuum at P = 0. (connected by valve) • Insulated chamber, thus Q dot = 0. (No surroundings also, so dw = 0).

45. Day 5 • Draw:

46. Day 5 • Then the valve is opened and gas expands into vacuum chamber (let ‘er rip!) P = 0. (No temp change occurs), then from first law: • Hmm, Dq = du + dw • Well: • 1. Dq = 0 (no heat added) • 2. dw = 0 (since chamber cannot expand)

47. Day 5 • Thus, du must be 0!!  • So,

48. Day 5 • Thus: • Buuut, Joules experiment also showed that dT = 0 soooo,

49. Day 5 •  This (celebrated) result is known as Joule’s law, which is strictly true for an Ideal gas. • Thus,

50. Day 5/6 •  Thus, u = u(T) internal energy is a function of T only!!! • Q: So what did Joule’s experiment show? • A: Joule’s experiment, separate from Joules’ law showed that heat and mechanical energy are two forms of the same thing.