GEF2200 Atmosfærefysikk 2017 Løsningsforslag til sett 3 Oppgaver hentet fra boka Wallace and Hobbs (2006) er merket WH06 WH06 3.18r Unsaturated air is lifted (adiabatically): The rst pair of quantities are conserved: Potential temperature and mixing ratio. For the second pair, only equvivalent potential temperature is conserved, but saturation mixing ratio changes. WH06 3.18s Saturated air is lifted (pseudoadiabatically): Only equvivalent potential temperature is conserved. Potential temperature, mixing ratio and saturation mixing ratio change and are not conserved. WH06 3.18t The frost point is the temperature at which water vapor becomes saturated with respect to ice. Considering Figure 3.9 in WH, we see that the ice saturation pressure is lower than the water saturation pressure, for a given temperature. But since the saturation pressures increase with temperature, this means that for a given ambient vapor pressure, we will reach saturation w.r.t. ice at a higher temperature than saturation w.r.t. water, which means that the frost point temperature is higher than the dew point temperature. See my drawing in Figure??. This is why we get frost and not dew on the ground when the temperature drops to below 0 C. WH06 3.18v The refridgerator is driven by a reversed heat engine transferring heat from the cold reservoir (within it) to the warm reservoir (the kitchen). As we have learned in chapter 3.7, such an installation needs external work to be transferred into heat. This means that the total system of the kitchen plus 1
Obs p(hpa) T(C) T d (C) θ(k) dθ/dz w s θ e (K) dθ e /dz A 1000 30 21.5 303.15 16.5 347.15 < 0 < 0 B 970 25 21 300.75 16.4 344.91 0 < 0 C 900 18.5 18.5 300.57 15.1 341.92 (> 0) 0 (or > 0) D 850 16.5 16.5 303.42 14.0 342.23 E 800 20 5 312.46 6.9 331.32 F 700 11-4 314.65 4.0 325.87 G 500-13 -20 317.16 1.6 322.05 Table 1: Souning of WH06 3.53. refridgerator is heated. As the door is left open, the heat engine works harder to maintain the chill of the refridgerator, so that more work is transferred into heat. If the heat engine were to cool the kitchen, we would have to use something other than the kitchen as the warm reservoir, for example the outdoor air. This is what is done in air-condition systems. WH06 3.53 a We calculate the potential temperature and equlivalent potential temperature and see how they change with height (given in Table 1). We know the following about the static stability of dry/unsaturated air: dθ/dz < 0: the layer is unstable. dθ/dz = 0 the layer is neutral. dθ/dz > 0 the layer is stable. Potential temperature may be calculated from the well known equation ( ) 1000hPa R/cp θ = T (1) p 2
or you can read the values o the sonde diagram. 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 moist adiabat, not the dry adiabat! This is done by looking at the equivalent potential temperature. There are two ways to nd θ e, but we need the sonde diagram anyway: 1. We can read it o the sonde diagram, by lifting the air in each point (along the dry adiabat) to saturation, and then upwards along the pseudo adiabat, until all moisture has been removed. We then follow the air downwards along the dry adiabat and read o the temperature at 1000hPa. 2. We calculate it from Equation (3.71) ( ) Lv w s θ e = θ exp c p T (2) To do this, we read o the saturation mixing ratio of water vapor (w s ) for the temperature T at each point (w s (T ), not w(t ) = w s (T d ), you should know the dierence). L v = 2.5 10 5 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 stable (or neutral). Alternative solution by use of skew-t ln p chart Alternatively, just consider the lapse rate compared to the dry and moist adiabates in Figure (1). b Considering convective instability, the criterion for instability is dθ e /dz < 0, so we see that all layers are convetively unstable, except layer CD which we have already noted is stable (or neutral). 3
Figure 1: Plot of the radiosonde values of Exercise 3.53. WH06 3.58 From (3.83): T1 Q1 = Q2 T2 (3) and the work done in one cycle is: W = Q1 Q2 (4) So the work done in 10 cycles is: W10 = 10 W = 10 (Q1 Q2 ) = 10 (20 14.64)J = 146.4J (5) A.20.T a Potential temperature (θ): The temperature an air parcel will have if it is compressed (lowered) or expanded (lifted) from its pressure to the standard pressure p0. We use p0 = 1000hPa. θ=t p0 p R/cp (6) The layer is stable when the lapse rate of the surroundings (Γ) is less than the dry adiabatic lapse rate (Γd ), where air lifted will be colder than its 4
surroundings. Dierentiating the logartithm of the equation above, ln θ = ln T + R c p (ln p 0 ln p) (7) θ dz T dz R 1 dp c p p dz (8) Inserting the hydrostatic equation for dp/dz and using the ideal gas equation (p = ϱrt ): θ dz T dz Rgϱ 1 c p p T dz g (9) T c p where Γ d = g/c p and Γ = dt/dz. = 1 T ( Γ + Γ d) = 1 T (Γ d Γ) (10) We then have the criterion for stable air (Γ < Γ d ) given by potential temperature: θ dz = 1 T (Γ d Γ) > 0 (11) b Since θ increase with height, the layer is stable. We can also see that Γ < Γ d. 5