GEF2200 Atmosfærefysikk 2017

Transcription

1 GEF2200 Atmosfærefysikk 2017 Løsningsforslag til sett 2 Oppgaver hentet fra boka Wallace and Hobbs (2006) er merket WH06 I ere av oppgavene vil du få bruk for sondediagrammet (skew T- lnp-chart) som kan lastes ned fra kursets semesterside E.2.T An air parcel containing 1 g of water vapor and 400 g of dry air has temperature 5.0 C and is located at 900 hpa. a. To nd the mixing ratio, we use the denition: w = m v m d = b. The specic humidity is dened: q = 1g = 2.5g/kg (1) 400g m v m d + m v (2) Since the vapor content is so small in this case, the specic humidity is very close to the mixing ratio. q = 2.5 g/kg. c. We nd the saturation mixing by inspecting the skew T-lnp chart. We rst locate the 900 hpa isobar (a black horizontal line near the bottom of the gure). We then nd the isotherm (red line) marked 5 C. The intersection between these two lines represent the thermodynamic state of our air parcel. Now we consider the saturation mixing ratio line passing through this point on the diagram (the dashed green lines mark costant saturation mixing ratio). We see that the line marked 6 g/kg passes exactly through the 900hPa-5 C intersection. We thus conclude that our air parcel has a saturation mixing ratio of 6 g/kg. 1

2 Now we may use the denition of relative humidity: RH = w w s 100% = 2.5g/kg 6g/kg 100% = 41.7% (3) d. The dew point temperature is the temperature to which our air parcel needs to be cooled at constant pressure to be saturated. Since at saturation, the mixing ratio should be equal to the saturation mixing ratio, we thus need to make the saturation mixing ratio equal 2.5 g/kg. We therefore try to locate the 2.5 g/kg line in the diagram. But this is not drawn. However, the 2 g/kg and 3 g/kg lines are, and we know the 2.5 g/kg line should lie about halfway between them. We draw this line and nd its intersection with 900 hpa. Considering the (red) isotherms, we conclude the temperature to be about 7 C at this point. So the dew point is 7 C. e. To nd the virtual temperature in this case, we use the formula (3.60). This is reasonable since the mixing ratio is so small, w 2 w. T v = T ( w) (4) = K ( ) = K 5.4 C f. The lifting condensation level is the pressure to which the air parcel must be lifted adiabitically to reach saturation. In the skew T-lnp chart, an adiabatic process is represented by following the dry adiabat lines (isentrops, yellow lines). Such a line does not pass through the 900hPa-5 C intersection, but we are able to approximate the direction from the surrounding lines. We follow this direction upwards and to the left until we reach the 2.5 g/kg line. This is at about 750 hpa. Thus, the lifting condensation level is at 750 hpa. g. We nd the potential temperature by adiabatic compression to 1000 hpa. Following the adiabat downwards and to the right, we nd its intersection with the 1000hPa line. Considering the isotherms, the temperature at this point is about 14 C. So the potential temperature is θ = 14 C. h. To nd the wet bulb temperature, we use Normand's rule. The pseudoadiabat (solid green line) passing through the lifting condensation level leads to the wet bulb temperature at its intersection with the isobar of our parcel (900 hpa). We thus nd the wet bulb temperature T w = 0 C. i. We read the equivalent potential temperature o the pseudoadiabat line intersecting our lifting condensation level. We nd θ e = 293K. 2

3 E.3.T a. Air parcels that are saturated with respect to water vapor will get supersaturated (w s > w) as they are cooled further. As a consequence, some of the vapor condensates, forming liquid water droplets. This process releases latent heat, thus having a warming eect on the air. The cooling is therefore less rapid for saturated parcels than for unsaturated parcels, from which no condesation occurs. b. In an adiabatic process, the system (air parcel) does not exchange heat with the surroundings. If, in addition, no condensation or evaporation occurs, we have a dry adiabatic process. A saturated adiabatic process is an adiabatic process where the air is saturated with respect to water vapor. As evaporation or condensation occurs, latent heat is taken/released within the parcel, but no heat is exchanged with the surroundings. The water droplets are therefore not allowed to leave the air parcel in a saturated adiabatic process, since that would represent exchanging heat with the surroundings. A pseudoadiabatic process is like the saturated adiabatic process, except that the liquid water is assumed to be removed immediately from the air after condensation. This makes the process irreversible and not adiabatic. WH Considering the moist adiabats (solid green lines) on the skew T-lnp chart, we notice they diverge going up. If we start at 25 C and 26 C at 1000 hpa and follow the moist adiabats up to 250 hpa, the temperature dierence will grow to about 3 C. WH We have an air parcel at pressure p = 1000hPa and temperature T = 25 C, and a wet-bulb temperature of T w = 20 C. a. The dew point is found by following the dry adiabat upwards until lifting condensation level is reached (where it itersects the moist adiabat from the wet-bulb temperature, 900hPa), where we can read o the mixing ratio of water vapor ( 12g/kg), and then following the line for this mixing ratio back to the initial level, which in this case is p = 1000hPa. Doing that, we nd the dew point temperature T d = 17.5 C. 3

4 b Lifting the parcel until all moisture is condensed and removed, and then lowering the parcel back to 1000hPa, means that we lift the parcel from the LCL upwards along the moist adiabat until the moist and dry adiabats converge, and then downwards along the dry adiabat. c. This is the equivalent potential temperature, and from the sonde diagram we get θ e = 62 C. WH On the skew-t ln p chart, we locate the w = 10g/kg stippled line, and at 1000hPa we nd the temperature T = 20 C. We see that the saturation mixing ratio at T = 20 C, p = 1000hPa is w s = 15g/kg. The dew point is found by compressing air at constant pressure until saturation. In the process of compression the air cools, and will be saturated when the saturated mixing ratio is w = 10g/kg. Following the 1000hPa line from T = 20 C until we hit the stippled mixing ratio w = 10g/kg, we can read o the dew point temperature, T d = 14.5 C. The parcel is then lifted to 700 hpa. We rst follow the dry adiabat up to the lifting condensation level (LCL). Above the LCL the parcel is saturated, so we must follow the moist adiabat. Where this line reaches 700 hpa, we can read o the saturation mixing ratio 6 g/kg. This is the amount of water vapor remaining in the air parcel. The remaining 4 g/kg has condensated onto water drops. We are then told that 80 % of the condensed water falls out. That is 4 g/kg 80 % = 3.2 g/kg. So the total water content of the air parcel has been reduced to 10 g/kg g/kg = 6.8 g/kg. Now the air parcel descends again. Until it reaches the 6.8 g/kg 'line' (located betwen the 6 and 8 g/kg lines) the parcel contains liquid water which evaporates into the air. We therefore follow the moist adiabat line. However, when we reach the 6.8 g/kg line, the air parcel becomes unsaturated, and we must then follow the dry adiabat line for the rest of the descent. We reach 900 hpa and read o the temperature 19.5 C. WH We consider the water vapor contained in our air parcel. From the ideal gas 4

5 equation (3.1) and the denition of partial pressure (page 66), we get: ev = m v R v T (5) m v = ev R v T (6) where m v and e is the mass and partial pressure of water vapor in the parcel, respectively, and V is the total volume of the air parcel. We may use (6) both for the state before and after compression. The dierence in vapor mass is due to the mass m c of condensed water. Thus, m c = m 1 m 2 = e 1V 1 e 2 V 2 R v T where subscripts '1' and '2' indicates the state before and after compression, respectively. Since the compression is isothermal, T is the same for the two states. Since condensation has occured after the compression, we know the parcel is saturated at state '2'. Thus, e 2 = e s (T ). We are further told that the initial relative humidity was 60 %. Using (3.64), we conclude e 1 = RHe s (T ). Inserting this, we nally have all we need to calculate the mass of condensate: m c = e s(t )(RH V 1 V 2 ) R v T = J/m 3 ( m m 3 ) 461.5J/Kkg K = kg (10) So the mass of condensed water is g. A.20.T 1. 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 = 1000hPa. ( ) R/cp p0 θ = T (11) p 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. Dierentiating the logartithm of the equation above, (7) (8) (9) ln θ = ln T + R c p (ln p 0 ln p) (12) 1 dθ θ dz = 1 dt T dz R 1 dp c p p dz 5 (13)

6 Inserting the hydrostatic equation for dp/dz and using the ideal gas equation (p = ϱrt ): 1 dθ θ dz = 1 dt T dz Rgϱ 1 c p p = 1 dt T dz g (14) T c p where Γ d = g/c p and Γ = dt/dz. = 1 T ( Γ + Γ d) = 1 T (Γ d Γ) (15) We then have the criterion for stable air (Γ < Γ d ) given by potential temperature: 1 dθ θ dz = 1 T (Γ d Γ) > 0 (16) 2. Since θ increase with height, the layer is stable. We can also see that Γ < Γ d. 3. The frequency is called the Brunt-Väisälä frequency [ g N = T (Γ d Γ)] (17) E.4.T When the two air parcels mix, heat will be exchanged from the warmer to the colder air parcel. Thus the warmer parcel will cool and the colder parcel will warm. Consider the saturation vapor pressure as a function of temperature (gure 3.9). When the warm parcel is cooled, it can hold less vapor. Since it was already saturated, some of the vapor can no longer be held by it. On the other hand, the cold parcel is warmed, and can hold more vapor than previously. As we can see from the gure, the change of saturation pressure will be larger for the warm parcel, since e s (T ) has a steeper slope for larger T. Thus, e s for the cold parcel will increase less than e s for the warm parcel decreases. Thus, in total, the air can hold less vapor after the mixing, and some water will therefore condensate. 6