GEF2200 Atmosfærefysikk 2012

Transcription

1 GEF2200 Atmosfærefysikk 2012 Løsningsforslag til oppgavesett 4 WH (WH 2.49) The air parcel has the properties p = 1000hPa, T = 15 C and T d = 4 C. b Lifting the air parcel to p 2 = 900hPa, T 2 we still have unsaturated air, and we also have a new dew point temperature T 2d. This means that the mixing ratio is still the same, w(t 2 ) = w(t 2d ) = w(t d ) = 5.1g/kg. By reading off the diagram, the saturation mixing ratio at T 2 is w s (T 2 ) = 6.8g/kg, giving RH = 5.1/6.8 = 0.73 = 73%. The potential temperature is given by decending along the dry adiabat, giving θ = 15 C, so it is conserved. Repeating the procedure for wetbulb temperature, but stopping at p 2, we have T w = 4.7 C. Following the moist adiabat all the way to 1000hPa, we get θ w = 9.3 C, which is conserved. c Lifting the air parcel to p 3 = 800hPa, we pass the line for constant mixing ratio (5.1g/kg), which is at the LCL, so we now have saturated air. The parcel then follows the moist adiabat in acent. At p 3 we get the new values w = w s = 4.4g/kg, that is RH = 100%. For saturated air we have T = T w = T d and here T = T w = T d = 1 C. Following the moist adiabat downwards, we see that θ w = 9.3 C is still conserved, but when following the dry adiabat we get θ = 17 C, meaning that potential temperature is not conserved for moist adiabatic processes. d As already noted, LCL = 845hPa.

2 WH r (WH m) When unsaturated air is lifted, the potential temperature, the equivalent potential temperature and the mixing ratio are conserved quantities. The saturation mixing ratio on the other hand is not. Consequently the first pair of quantities are conserved. WH s (WH n) When saturated air is lifted, the only conserved quantity out of the four is the equivalent potential temperature. The potential temperature, the mixing ratio and the saturation mixing ratio are not conserved quantities when lifting saturated air. WH (WH ) The initial dew point temperature of the air is located on the skew T lnpby cooling the air parcel along the 1000 hpa isobar from the initial temperature of the air parcel to the intersection between the 1000 hpa isobar and the w s = 10 g/kg contour. The initial dew point temperature is: T d,1000 hpa = 14 C To find the final temperature of the air parcel, follow the dry adiabat from the initial location of the air parcel (the intersection between the 1000 hpa isobar and the 15 C isotherm) to the intersection between the dry adiabat and the w s = 10 g/kg contour at the lifting condensation level (LCL). As the air parcel is now saturated, further lifting must be along the saturated adiabat until the air parcel reaches the 700 hpa isobar at the intersection between the 700 hpa isobar and the w s = 6 g/kg. Itisassumedthatonly80%ofthecondensedwaterintheairparcelfalls out of the air parcel as precipitation during the lift. As the air parcel was lifted from the LCL to the 700 hpa isobar, the saturated mixing ratio decreased by 4 g/kg from 10 g/kg to 6 g/kg. 80% of the 4 g/kg, corresponding to a mixing ratio of 3.2 g/kg, falls out of the air parcel, and 20%, corresponding to a mixing ratio of.8 g/kg, remains in the air parcel.

3 Figure 1: Plot of the radiosonde values of Exercise As the air parcel starts to sink on the other side of the mountain, the saturation mixing ratio is 6 g/kg. As the air parcel sinks, the remaining water in the air parcel evaporates until there is no water left at which point the saturation mixing ratio of the air parcel is 6.8 g/kg. Up to this point, the air parcel is saturated and sinks along the saturated adiabat. Having passed this point, the air parcel will no longer be saturated and sinks along the dry adiabat until it reaches the 900 hpa isobar. The temperature of the air parcel after descending to the 900 hpa isobar is: T 900 hpa = 19.5 C WH (WH ) a Solvetheproblemusingtheskew T lnp. First,plotthesoundingonthe skew T lnp(showninfigure1). Then, toconsiderthestaticstabilityof the layers remember that for dry/unsaturated air the following is true: Γ > Γ d : the layer is unstable, Γ = Γ d : the layer is neutral, and Γ < Γ d : the layer is stable.

4 Similarly, for saturated air the following is true: Γ > Γ s : the layer is unstable, Γ = Γ s : the layer is neutral, and Γ < Γ s : the layer is stable. AB: The layer is not saturated as the dew point temperature is smaller than the temperature. The temperature decrease following the dry adiabat from p A to p B is smaller than the actual temperature decrease between A and B. Consequently: Γ > Γ d The layer is unstable BC: The layer is not saturated, as also here the dew point temperature is smaller htan the temperature of the layer. Now, the temperature decrease following the dry adiabat is equal to the temperature decrease of the layer. Consequently: Γ = Γ d The layer is neutrally stable CD: The layer is saturated as the dew point temperature equals the temperature throughout the layer. The temperature decrease in the layer equals the temperature decrease found following the saturated adiabat. Consequently: Γ = Γ s The layer is neutrally stable DE: The layer is unsaturated as the dew point temperature is smaller the temperature throughout the layer. This layer is special compared to the other layers as the temperature indreases throughout the layer yeilding a negative lapse rate Γ. Increasing temperature with height is called an inversion. Following the dry adiabat from p D to p E the temperature is found to decrease. Consequently: Γ < Γ d The layer is stable EF: The layer is again unsaturated as the dew point temperature is smaller the temperature throughout the layer. Following the dry adiabatfromp E top F yeildsatemperaturedecreasewhichislargerthanthe temperature decrease in the layer. Consequently: Γ < Γ d

5 Obs p(hpa) T(C) T d (C) θ(k) dθ/dz w s θ e (K) dθ e /dz A < 0 < 0 B < 0 C (> 0) 0 (or > 0) D > 0 < 0 E > 0 < 0 F > 0 < 0 G Table 1: Souning of WH The layer is stable FG: The layer is again unsaturated as the dew point temperature is smaller the temperature throughout the layer. Following the dry adiabat from p F to p G again yeilds a temperature decrease which is larger than the temperature decrease in the layer. Consequently: Γ < Γ d The layer is stable Alternatively, potential temperature may be calculated from the well known equation: ( ) 1000hPa R/cp θ = T (1) p and the stability of the various layer can be found by considering the sign of θ θ z. θ values and the sign of z are given in 1. Remember that according to equation (3.77) on page 91 in Wallace & Hobbs (2006), the layer is stable when θ θ z > 0, neutral when z = 0 and unstable when θ z < 0 Thus, for dry air, layer AB is unstable, BC is neutral and the rest are stable. However, if the layer is saturated we have to consider the saturated adiabat, not the dry adiabat! This is the case for layer CD. Consequently, the stability of layer CD cannot be estimated by considering θ, we must consider the equivalent potential temperature θ e.

6 There are two ways to find θ e, but we need the skew T lnpanyway: 1. To find the θ e value of say an air parcel at point A, initially locatedattheintersectionbetweenthe1000hpaisobarandthe30.0 isotherm, follow the dry adiabat until the air parcel reaches saturation at the LCL. Subsequently, follow the saturated adiabat until all water vapor in the air parcel has condensed (that is, until the dry adiabats and saturated adiabats are parallel). Then, follow the dry adiabat back to 1000 hpa. 2. We calculate it from equation (3.71) in Wallace & Hobbs (2006): ( ) Lv w s θ e = θexp (2) c p T To do this, we read off the saturation mixing ratio of water vapor (w s )forthetemperaturet ateachpoint(w s (T),notw(T) = w s (T d ), you should know the difference). L v = Jkg 1, and c p = 1004Jkg 1 K 1. The θ e -values are given in table 1. Both point C and D are saturated (T = T d ), but the equivalent potential temperature increases with height (or is almost unchanged). Therefore, this layer is approximately neutral. b Consideringconvectiveinstability,thecriterionforinstabilityisdθ e /dz < 0. Consider the stability of each layer by finding the θ e value at each pointandconsiderthesignofdθ e /dz withinthelayer. Usingtheskew T ln pyeilds: A:θ e = 80,B:θ e = 72.5,C:θ e = 70,D:θ e = 70,E:θ e = 65,F:θ e = 58, and G: θ e = 50. As the height increases within a layer, it is the change in θ e which determines the sign of dθ e /dz < 0. In layer AB dθ e /dz < 0 decreases, consequently dθ e /dz < 0 and the layer is convectively unstable. The same is true for layers BC, DE, EF and FG. In layer CD on the other hand, θ e does not change and dθ e /dz = 0. Thus layer CD is not convectively unstable.

7 Alternatively, use the θ e values in table 1. WH (WH ) Will find the entropy for converting m = 2 g = kg of ice at T 1 = 10 C to vapor at T 1 = 100 C by heating. The entopy is given by ds = dq rev T soweneedtofindexpressionsfordq rev. Notethatwedonotusespecific valuesforq rev ands here. Theheatsuppliedtochangethetemperature is given by the first law of thermodynamics: dq temp = mc px dt wherec px isthespecificheatatconstantpressureforeithericec pi orliquid water c pv. But we also have additional heating due to phase change at temperature T phase : Q phase = ml x where L x is latent heat, which has to be considered for melting and evaporation. Integrating Equation (3) is then done in four steps: S = = = S2 S 1 ds = [ T2 T2 dq temp T 1 T 273K 263K 373K + = m 273K dq rev T 1 T ] + Q phase T phase dt mc p T +ml melt 273K mc p dt T +m L vap 373K [ c pi ln L melt 273 +c pv ln L vap 373 The constants are (given at the start of the book): c pi = 2106 Jkg 1 K 1 c pv = 4218 Jkg 1 K 1 L melt = Jkg 1 L vap = Jkg 1 ] (3)

8 giving the answer S = JK 1 ( ) = 17.3 JK 1 Clearly, the phase changes are important, and as expected, since gases are more disordered than liquids or ice, the change in entropy is much larger for liquid-gas change than for ice-liquid change.