GEF2200 atmospheric physics 208 Solutions: thermodynamics 3 Oppgaver hentet fra boka Wallace and Hobbs (2006) er merket WH06 WH06 3.8r Unsaturated air is lifted (adiabatically): The first 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 3.8s 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 3.8t 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. This is why we get frost and not dew on the ground when the temperature drops to below 0 C. WH06 3.8v 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 refridgerator is heated. As the door is left open, the heat engine works harder
Obs p(hpa) T(C) T d (C) θ(k) dθ/dz w s θ e (K) dθ e /dz A 000 30 2.5 303.5 6.5 347.5 < 0 < 0 B 970 25 2 300.75 6.4 344.9 0 < 0 C 900 8.5 8.5 300.57 5. 34.92 (> 0) 0 (or > 0) D 850 6.5 6.5 303.42 4.0 342.23 > 0 < 0 E 800 20 5 32.46 6.9 33.32 > 0 < 0 F 700-4 34.65 4.0 325.87 > 0 < 0 G 500-3 -20 37.6.6 322.05 Table : Souning of WH06 3.53. 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 ). 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 ( ) 000hPa R/cp θ = T () p or you can read the values off the sonde diagram. Thus, for dry air, layer AB is unstable, BC is neutral and the rest are stable. 2
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 find θ e, but we need the sonde diagram anyway:. We can read it off 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 off the temperature at 000hPa. 2. We calculate it from Equation (3.7) ( ) Lv w s θ e = θ exp c p T (2) To do this, we read off 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 difference). L v = 2.5 0 5 Jkg, and c p = 004Jkg K. The θ e -values are given in Table. 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 (). 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 : Plot of the radiosonde values of Exercise 3.53. WH06 3.58 From (3.83): Q Q2 = T T2 (3) T2 T 273.5K = 20J = 4.6J 00 + 273.5K Q2 = Q (4) Q2 (5) and the work done in one cycle is: W = Q Q2 (6) So the work done in 0 cycles is: W0 = 0 W = 0 (Q Q2 ) = 0 (20 4.64)J = 54J (7) The engine thus rejects 4.64J 0 = 46.4J. The efficiency of the maschine is the work done divided by the heat absorbed: η= Q Q2 54J = = 0.27 Q 0 20J 4 (8)
WH06 3.64 We use that at the boiling temperature, T B, the saturation vapor pressure of water is equal to the atmospheric pressure: e s (T B ) = p atmos. (9) We now use the Clausius Clapeyron equation (eq 3.92) pluss the expression above to get the expression: or dp atmos dt B = L v T B (α 2 α ) (0) dt B = T B(α 2 α ) () dp atmos L v Now, notice that the spesific volume of liquid water α is very much smaller than that of gas α 2, so we can ignore this term. Also, we use the ideal gas law to express α 2 = T BR v p atmos. This gives us: dt B = T B( TBRv dp atmos L v (2) dt B R v = dp atmos L v p atmos (3) dt B = R v dp atmos L v p atmos (4) T 2 B TB (600hPa) T 2 B T B (000hPa) TB 2 T B + dt B = p atmos ) 600hPa 000hPa R v dp atmos (5) L v p atmos 373.5K = R v ln 600hPa (6) L v 000hPa ( T B (600hPa) = 373.5K R v ln 600hPa ) (7) L v 000hPa T B (600hPa) = 360K = 87.5 C (8) So the boiling point is reduced to 87.5 C at 600hPa. A.20.T 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 p 0. We use p 0 = 000hPa. ( ) R/cp p0 θ = T (9) p 5
Static stability: The layer is stable when the lapse rate of the surroundings (Γ) is less than the dry adiabatic lapse rate (Γ d ), where air lifted will be colder than its surroundings. In other words: Γ < Γ d (20) dt g < T dz c p T (2) d ln T g < dz c p T (22) (23) Now we can express ln T in terms of potential temperature θ. ( p0 ) R/cp θ = T (24) p ( p0 ) R/cp T = θ (25) p ln T = ln θ + R ( p ) ln (26) c p p 0 Putting this into the expression above: ( d ln θ dz d ln T g < dz c p T + R dp c p p dz ) < g c p T (27) (28) (29) (30) Now we can use the ideal gas law and substitute p (p = ρrt ), and the hydrostatic equation, dp dz = ρg: ( d ln θ dz + R c p (ρrt ) ( ρg)) < g c p T d ln θ dz (3) + g c p T < g c p T (32) d ln θ dz > 0 (33) (34) In other words: if potential temperature increases with height, the layer is stable. If potential height decreases with height, the layer is unstable. b) Since θ increase with height, the layer is stable. We can also see that Γ < Γ d. 6
Figure 2: Exercise A.20.T d) c) The frequency is called the Brunt-Väisälä frequency [ g N = T (Γ d Γ)] (35) d) To find the mixing ratio at A and B, we use the definition of the relative humidity: RH = w w s 00% (36) which gives: w A = RH w s = 7.5g/kg 0.95 = 7.g/kg 00% (37) w B = RH w s = 5.g/kg 0.80 = 4.0g/kg 00% (38) We can now mark these saturation vapor pressure lines the skew-t lnp chart, as in figure 2. We noe raise a theoretical air parcel from point A: First we follow the dry adiabat because the parcel is not yet saturated, but when we reach the level/temperature where 7.g/kg is the saturation mixing ratio, we know that we have reached saturation and LCL. From this level we follow the pseudoadiabat. We reach the level of free convection when the temperature of the theoretical air parcel is warmer than its surroundings. In this case, this is approximately at 420hPa. When we do the same for point B, we get that as soon as the air parcel reaches saturation it becomes warmer than its surroundings, i.e. LCL is equal to LFC. Seeing as the height air at A has to be lifted in order to become buoyant is so high compared to B, and that the negative buoyancy it has to 7
Figure 3: Temperatur T (heltrukken linje) og duggpunkt T d (stiplet) for en sonde. overcome is so large, it seems more likely that air at B will release the potential instability. A.74.T Eksamen GEF2200 08- See figure 4. Part of a skew-t log p chart is shown in figure 3, in which the temperature T and the dew point temperature T d are given for a layer A-B. a. How is the stability of layer A-B? Layer A to B is stable. We can see this because the environmental sonding crosses the adiabats such that potential temperature increses with height. b. The layer A-B is now lifted by 00hPa. Is the layer more or less stable now? We lift the end points as an approximation. We know that the air at point A is saturated because the temperature is equal to the dew point temperature. We thus follow the pseudoadiabat. B is not saturated so we follow the dry adiabat to B. The layer is less stable. c. The layer is lifted another 00hPa. What is the stability now? the airparcel A is still saturated so we keep following the pseudoadiabat. For 8
Figure 4: Temperatur T (heltrukken linje) og duggpunkt T d (stiplet) for en sonde. the air parcel at B we are now approaching saturation and follow the pseudoadiabat from the LCL. The layer is now conditionally unstable! 9