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Atms 4310 / 7310

Atms 4310 / 7310. Anthony R. Lupo Test 3 material. Day 1. Then this vapor equation is: e s a v =R v T or e s = r v R v T Saturation or Equilibrium Vapor Pressure (e s )  “e s ” is a function of temperature only and not dependent on the pressure of the other gasses present. Day 1.

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Atms 4310 / 7310

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  1. Atms 4310 / 7310 Anthony R. Lupo Test 3 material

  2. Day 1 • Then this vapor equation is: • esav=RvT or es = rvRvT • Saturation or Equilibrium Vapor Pressure (es) •  “es” is a function of temperature only and not dependent on the pressure of the other gasses present

  3. Day 1 • The concept of equilibrium vapor pressure over a plane of pure water (does the atmosphere “hold” water vapor?):

  4. Day 1 • The Variation of es (es over water and es over ice – or “on the rocks”) with temperature:

  5. Day 1 • Graph here:

  6. Day 1 • So, some summary points: • 1) es is the maximum possible vapor pressure for a particular temp. (e most often less) • 2) es is dependent on temperature only (highly non-linear – exponential) • 3) Note that a 10o C increase (decrease) in temperature yields a doubling (halving) of es

  7. Day 1 • 4) The vapor pressure on a water surface equals es which depends only on water temperature • 5) The actual vapor pressure in the air may range from 0 mb to es hPa  0 < e < es • 6) Thus by definition RH = e/es or e actual = RH * es • 7) Water vapor will diffuse from regions of higher e values toward lower (evaporation).

  8. Day 1 • Now, you can convert mixing ratio (g/kg) to e (vapor pressure) on a thermodynamic diagram. •  Follow Temperature line (go straight up) to roughly 620 – 630 hPa and the mixing ratio here will roughly equal e in hPa!!!

  9. Day 1 • Changes of phase of water mass and associated latent heats • ** Recall that as we add heat (specific heat), we raise the temperature until we reach the melting point! •  then all heat at 0 C (273.15K) goes into changing the phase

  10. Day 1 •  once all the ice changes to water, then we raise temperature again, •  until vaporization, then there is a phase change first before we can raise temperature again (process also works in reverse) • ** Recall 1 cal = 4.186 J

  11. Day 1 • ** Recall 1 cal = 4.186 J • Raise the temperature of ice or steam: 0.5 Cal / kg (2.09 J / kg) • Raise the temperature of water: 1.0 Cal / kg

  12. Day 1/2 • Liquid to gas phase transformation: • Latent heat of vaporization condensation • L = 2.500 x 106 J/kg = 5.972 x 105 Cal/kg • Liquid to solid phase transformation: • Latent heat of fusion  melting • Lf = 3.34 x 105 J/kg = 7.98 x 104 Cal /kg

  13. Day 2 • Solid to gas phase transformation: • Latent heat of ablation  sublimation • La = Lf + L = 2.834 x 106 J/kg = 6.770 x 105 cal/kg

  14. Day 2 • The Phase Change Diagram for a Water Substance • First thing to note, that at Terrestrial pressures and temperatures water exists in all three phases (liquid, solid, and gas) at the same time.

  15. Day 2 • This has tremendous implications for earth’s weather (clouds, etc) and climate (recall other parts of the climate system, the oceans, ice sphere (cryosphere)) and the interactions between various components of the climate system. • (Hand out phase diagram, in 2-D and 3-D). • Phase diagram  Describes the state of a system in physical space. In our case (specific volume, vapor pressure, and Temperature)

  16. Day 2 • Water…

  17. Day 2 • Carbon

  18. Day 2 •  Water behaves as an ideal gas so the isotherms are hyperbolas in the phase plane (provided we’re far from phase changes). • Consider these important points: • At point A: all water is in the form of vapor, if we increase the pressure (with temperature constant), then volume shrinks roughly in accord w/ ideal gas law (Boyle’s Law).

  19. Day 2 • Eventually you can reach a point B: where increases in the pressure will force some water vapor to liquefy. • If we reach this point, the a small increase in pressure forces all vapor to condense out, then as we move from B to C with little change in pressure and constant temperature (we change from a gas to liquid with a huge decrease in volume)

  20. Day 2 • This constant P is called saturation vapor pressure for that particular temperature (That’s the straight lines across the parabola). • At point C: all sample is liquid which is for all practical purposes incompressible (constant a).

  21. Day 2 • 1) To the right of B (in A) water mass is all vapor. • 2) B to C vapor and water co-exhist. • 3) To the left of C water mass is all liquid. • If we follow triple state isotherm  at the triple point moisture (again fixed temperature and pressure) condenses out as liquid and solid

  22. Day 2/3 • Equilibrium triple state (6.11 hPa and 0.0o C). If pressure and temperature fall below triple state, then ice and gas will equilibrate • Two Exceptions: • 1) Super cooled water • 2) Another special case Follow the point D isotherm. This is a critical point.

  23. Day 3 • At the critical point, the distiction between vapor and liquid disappears (surface tension is 0) and there is no more interface. • This occurrs at 211 atmospheres (bars) or 21100 Kpa 211,000 hPa, and 374o C. • Defn: Critical point – above this value it is impossible to liquefy a gas by compression (above pc), or cooling (Tc).

  24. Day 3 • Aha! This explains why N2 O2 CO2 and Ar do not condense in our atmosphere. • Their critical points are WAY below terrestrial T and P’s! • The CO2 critical point is 31o C and 74000 hPA, thus condensation of CO2 could take place (under special conditions) if it were in sufficient quantities.

  25. Day 3 • CO2 behaves in Venus mid-upper atmosphere and below like water vapor does here. • The variation of the latent heat of vaporization (Ll –v) or condensation (L(v-l)) with temperature. • Let us integrate the First Law of Thermodynamics during a change of phase from 1 to 2 • (recall ds = dq/T) and p = es

  26. Day 3 • Thus; • And • The definition of latent heat for the phase transformation L 1 to 2 (where 1 = liquid (l) and 2 = gas (v))

  27. Day 2 • Latent heat • Recall that during a change of phase, T and es are constant!, so;

  28. Day 3 • L(1 – 2) is defined as the latent heat of vaporization. • Thus our equation can be rewritten as: • Density of air is = 1 kg /m3

  29. Day 3 • Then, if the mixing ratio is = 10 g/kg • rv must be: 0.01 kg/m3 • Density of liquid water is: 1 gm cm-3 or 1000 kg / m3 • Flip each to get specific volume, but look….. there’s 5 orders of magnitude between the two.

  30. Day 3 • Thus av is 100,000 times larger than al, so the equation becomes • Use the equation of state: esav = Rv T • L(l-v) = RvT + uv – ul

  31. Day 3/4 • This equation is good for any change from liquid to vapor regardless of initial and final values of internal energy (exact differential!) • Although changes occur at constant temperature, we can look at how L will vary with changes in temperature. • How? • dL/dt = RvdT/dt + duv/dt – dul/dt

  32. Day 4 • Recall: Defn of specific heat cl = dul / dT • and Cvv = duv / DT • So now we rewrite as: • dL/dt = RvdT/dt + CvvdT/dt – CldT/dt • -or- • (by chain rule) dL/dT = Rv + Cvv – Cl

  33. Day 4 • and, of course, Rv + Cvv = Cpv • So DL/DT = Cp – Cl = -2369 J/kg K • (aha! – slope of L versus Temperature, so Latent heat is temp dependent!)

  34. Day 4 • We can then integrate above expression from T = To = 0 degrees C, and L(l – v) = Lo (at T=0) to an arbitrary temperature T: • After applying the snake: • L (l – v) = Lo + (Cp – Cl) (T – To)

  35. Day 4 • The product term on the RHS is small for T – To less than 40 C, thus L(l – v) is approximately Lo for typical weather situations, and is taken to be 2.5 x 106 J/kg • Thus, L is not strictly a constant, and in Latent heat release (cloud and precipitation) schemes (advanced ones) this fact is taken into account. • For water: • Cpv = 1811 J/kg • Cv = 1350 J/kg • Rv = 451 J/kg • Cl = 4186 J/kg

  36. Day 4 • Over typical ranges of T here value of L varies 6%

  37. Day 4 • So, in the range of 20 to –20 C there is only 2% error in using Lo, thus to within 98% accuracy L = Lo. This is good enough to win $50.00 at the bar this weekend. • The variation of es with temperature (the Clausius Clapeyron equation) • If as before L(l – v) is the latent heat associated with a change in phase: • Dq = Tds = du + es da

  38. Day 4 • Since during a phase change T and es are constant then; • where;

  39. Day 4 • then, • L(1-2) = T(S2 – S1) = (u2 – u1) + es(a2 – a1) • or rearrange the above to isolate each state, state 2 and state 1 • TS2 – u2 – esa2 = TS1 – u1 – esa1

  40. Day 4 • During the phase change: • T S – u – esa = Constant • or – • u + esa - T S = Constant = G • (J. Willard Gibbs potential)

  41. Day 4 • Gibbs function this is a “fundamental” Thermodynamic (Gibbs) function for simple compressible systems (such as an air parcel) of fixed chemical composition, and using the concept of an exact differential. • A thermodynamic function provides a complete description of the thermodynamic state of a system. In principle, all properties of interest (v,T,P) can be determined from the function given a suitable set of boundary or initial conditions.

  42. Day 4 • In plain English: the phase diagram represents all possible states of system. • G is constant during a phase change (T and es constant) it has values in accord with T and es depending on the T and es at which the phase change takes place. G(T,es) (recall diagrams?)

  43. Day 4/5 • So, let us look at the variation of G; • Take the derivative with respect to time (remember to use product rule!) • DG/Dt = du/dt + es da/dt + a des/dt – T dS/dt – S dT/dt

  44. Day 5 • Rearrange: • dG/dt = du/dt + es da/dt – T dS/dt + a des/dt – S dT/dt • On the RHS, the first three terms are the 1st Law of Thermodynamics!

  45. Day 5 • If T dS/dt = du/dt + es da/dt • Then dG = a des/dt – S dT/dt • And if G = a constant during phase change at T and es; • And if G + dg is constant at a phase change for T + dT, es + des

  46. Day 5 • Then dG must also be a constant! • If dG is constant; • Then; • a2 des/dt – S2 dT/dt = a1 des/dt – S1 dT/dt

  47. Day 5 • so; • des/dt (a2 – a1) = dT(S2 – S1) • Or • des/dT = (S2 – S1) / (a2 – a1)

  48. Day 5 • But, • So: • This is it! (Make no mistake)! • Generalized Clausius Clapeyron equation for any phase change from phase 1 to 2.

  49. Day 5 • Saturation Vapor Pressure over Water • Conditions governing saturation or equilibrium vapor pressure over a plane of pure water surface (es), which involves the phase changes from liquid to vapor (evaporation) = phase changes from vapor to liquid (condensation) • Evaporation = condensation

  50. Day 5 • In the case of the CC equation: • L(1-2) is L (l-v) • a2 = av and a1 = al • es = esw

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