Solutions to questions from chapter 5 (and part of ch.6.) in GEF4310 - Cloud Physics i.h.h.karset@geo.uio.no Problem 1 a) When figuring out if an atmospheric layer is stable when lifting/lowering an unsaturated air parcel, we can use these criteria: dt v > Γ d, or dθ v > 0 (stable) dt v = Γ d, or dθ v = 0 (neutral) dt v < Γ d, or dθ v < 0 (unstable) Explain why we have these criteria (draw plots of Γ d and dtv system to explain). in the T-z-coordinate Figur 1: Stability criteria for an unsaturated atmosphere 1
Answer: It s easy to mix up and get confused by the signs when looking at stability criteria. We will look at an example: The dry adiabatic lapse rate is approximately 10 K/km (which means that Γ d = 10 K/km). Lets look at an atmosphere where the virtual temperature decreases with only 8 K/km ( dtv = 8 K/km). This is the red graph in Figure 1. An air parcel that s not saturated, will follow the dry adiabat, Γ d, when lifted/lowered up/down in the atmosphere. We can see that lifting an air parcel from A to B in this (red) atmosphere will make the parcel at point B colder and heavier than the surrounding air. The parcel will sink. If we lower the air parcel from A to C in this (red) atmosphere, the parcel at point C will be warmer and less dense than the surrounding air, and the parcel will rise. Since the parcel will get an acceleration and move in the opposite direction of the displacement, we say that the (red) atmosphere is stable. By looking at how large dtv (-8) is compared to Γ d (-10), we see that dtv > Γ d (since 8 > 10), which matches with the criteria for a stable atmosphere given in the book. The green graph in Figure 1 is showing the oppisite case where we have an unstable atmosphere. We are using T v since the atmosphere can contain water vapor, and the amount of water vapor can change with height. For the parcel, the amount if water vapor is kept constant during lifting/lowering since it s happening adiabatically and the air is not getting saturated. The criteria with θ v can be explained in the same way as for T v, but remember that this is potential virtual temperature, so we need to follow the dry adiabat down to the reference pressure before comparing the values. b) When finding out if an atmospheric layer is stable when lifting/lowering a saturated air parcel, we use these criteria: dθ v < 0 (absolutely unstable) dθ es dθ es < 0 < dθ v (conditionally unstable) > 0 (absolutely stable) Explain the difference between these criteria and the criteria for unsaturated air. Answer: We will just look at lifting, but the same thinking can be used for lowering: When we have a saturated air parcel, or the air parcel gets saturated on its way when being lifted from A to B in Figure 2, condensation happens and energy is released. This causes the temperature to not decrease as much as the dry adiabat, but somewhere between the dry adiabat and the moist adiabat (depending on when on its way between A and B the condensation starts). The max possible temperature the air parcel can get at B is given by the moist adiabat. If the temperature of the atmosphere decreases less with height than this value, the air parcel will sink down again bacause it will be colder and denser than the suroundings. we say that the atmospheric layer is absolutely stable (red line). The reason why this is given as dθ es < 0 is because θ es is the max possible potential temperature an air parcel can get (this happens when it is completely saturated in A), and if the this value increases 2
with height, it means that the air parcel never will get a temperature higher than the surroundings at B, even though we have condensation all the way up from A to B. When the virtual temperature of the air decreases more with height than the dry adiabat (meaning that the virtual potential temperature is decreasing with height), the air parcel will always be warmer than the surroundings when being lifted up from A to B, so it will continue to rise. We say that we have a absolutely unstable atmospheric layer (green line). When the temperature of the atmospheric layer is changing with height somewhere between the moist adiabat and the dry adiabat, we say that it is conditionally stable, because it depends on when the condensation of the air parcel on its way from A to B starts. If the condensation start right above A, the final temperature at B will be almost as high as when following the moist adiabat, which means that the condition for stability is that the atmospheric lapse rate must be almost the moist adiabatic value. In the case of conditionally stability, the atmosphere is stable with respect to dry air parcels, but if the potential temperature of the surrounding atmosphere is decreasing more with height than that of the wet adiabat (along Γ s, we have that dθes = 0), meaning that we can say this about the temperature of the surrounding atmosphere: dθes < 0, this atmosphere will have a lower temperature than a wet parcel i B (or at least the parcels that gets wet early on its way from A to B), and hence it will be unstabel for these parcels. Problem 2 What is... a) LFC? Answer: Level of free convection. This is the height an air parcel needs to be lifted adiabatically to make the temperature of the parcel higher than the environment. If it reaches this level, it will continue to rise buoyantly by its own. c) LNB? Answer: Level of neutral buoyancy. This is the height where the temperature of the air parcel that is rising buoyntly by its own following the moist adiabat from LFC and up will reach before it s getting lower than the environment. This is where we go from unstability to stability. Problem 3 - About mixing a) How will drier surrounding air affect the equivalent potential temperature of the final mixture? Answer: When the air is drier, θ e will be lower because there is less energy available to be released through condensation. b) How will mixing with drier surrounding air affect LCL and LFC? Answer: when drier air is mixed with the air parcel, it has to reach higher up in the atmosphere before it gets saturated (reach LCL). The air parcel doesn t start to follow the moist adiabat before it is saturated. When it follows the moist adiabat, the temperature decrease with height is low, so it s easier to reach LFC (temperature higher than 3
Figur 2: Stability criteria for a saturated atmosphere suroundings) than when following the dry adiabat. Thus, if it isn t saturated before high up in the atmosphere, LFC will also be higher up. Problem 4 Use Figure 3 to answer the questions. a) What does the figure shows? Answer: It shows two parcels with different temperatures and vapor pressures. If these parcels are mixed isobarically, the final mixed parcel will end up somewhere on the straight line since the quantities T and e is conserved. The curved line shows how the equlibrium saturation vapor pressure varies with temperature according to Clausius-Clapeyron. Since this line isn t straight, but exponential, it can happen that two unsaturated parcels is mixed and the result is a saturated parcel. This will happen if the mixed parcel ends up on the mixing line somewhere over the grey shaded area. b) We get cloud formation right behind an aircraft. What is happening? Explain it by using Figure 3. Answer: Warm vapor rich air from the engine of the aircraft (Parcel 4
Figur 3: Figure 6.1 from the book (some info is erased) 1 in the Figure 3) is mixed with cold, dry, surrounding air (Parcel 2 in Figure 3). The mixed parcel ends up at the mixing line over the gray shaded area. Problem 6 from the midterm exam in 2015 Figur 4: Figure from Problem 6 in the midterm exam from 2015. It shows an atmospheric sounding from the south central United States on 5 June 2001. a) Explain the meaning of the two concepts CAPE and CIN, and derive equations expressing them. Answer: CAPE is convective available potential energy, the total energy available from the environment between the LFC (level of free convection) 5
and the LNB (level of neutral buoyancy). We find this by integrating the buoyancy between LFC and LNB: LNB LNB T vp T v CAP E = B = g. LF C T v LF C (Remember that CAPE is energy. When integrating the buouyancy force over distance, we get energy). CIN is convective inhibition, the work required to lift the parcel up to LFC. We find this by integrating the negative buoyancy between the ground and LFC: LF C CIN = B 0 b) How would you determine CAPE and CIN from the diagrams in Figure 2? Answer: CAPE is the shaded area between the moist adiabat and the temperature profile of the surroundings from LFC to LNB. CIN is the area between the dry adiabat from the ground, the wet adiabat from LCL and the temperature profile of the surroundings from the ground and up to LFC. c) Is there a negative CIN in this case? Explain your answer. Answer: It s hard to say by looking at the figure, but it doesn t look like we have any CIN here. Usually the temperature profile is to the right of the adiabates that the air parcel follows all the way up to LFC, meaning that the temperature of the surroundings is lager than the temperature of the air parcel. This doesn t seems to be the case here. This is a real situation, so it s more complicated. If we look at the figure below (in the book) we can see that LCL is higher up than what we would expect from the sonding-figure, so it wil probably not follw the wet adiabat from the level right below 850 hpa (as we though in the lecture). d) Is CAPE small or large in this case? Explain your answer. Answer: CAPE is large in this case. From the figure text in the book, you can see that this is taken from a tornado storm event, where CAPE values are high. Probem 7 from the midterm exam in 2015 (related to figure 5.4 in the book) a) Set up equations defining the quantities θ e and r t that are used in Figure 5.4 in the book. θ e = χθ e,p + (1 χ)θ e,s r t = χr t,p + (1 χ)r t,s Answer: The equations tells us how the equivalent potential temperature, θ e, and the total water mass mixing ratio, r t is calculated when we have mixing between an air parcel (subscript p) and the surrounding air (subscript s). χ is the mass mixing fraction, the mass-fraction of the parcel in the final mixture (χ = m parcel m mixture ). 6
b) Explain the physical meaning and logic of the black line and the four arrows in the figure. Answer: When we have mixing between two air masses with r t and θ e given by the two dots in the figure, θ e and r t to the final mixed parcel will be given by a point on the black line since these quantities are conserved. Other physical processes in the cloud can move it away from the line in the directions given by the arrows. Warming/cooling will not affect the total water mass mixing ratio, but it will affect the equivalent potential temperature. Precipitation in or out will increas/decrease r t, but it won t affect θ e (increased amount of water vapor will increase θ e, but when we have preciptation, it s the liquid water that increases, not the vapor). 7