EART 121 PROBLEM SET #5 SOLUTIONS

Transcription

1 ERT 121 PROLEM SET #5 SOLUTIONS 1. Condensation is not sufficient to make rain Consider 1 mg of air with T = 15 C and RH = 80%. If this small parcel of air gets lofted very high (say, in a cumulonimbus cloud), then let s assume that all the water vapor that was initially present condenses (equivalent to assuming that the SVP at low temperatures is negligible). If the condensation occurs onto 200 particles in this parcel of air, compute the size of each drop. How does this compare with the size of a typical rain drop? Solution: Strategy for the problem: figure out how much water vapor is present in the initial parcel. This is then how much water condenses. Divide that water into 200 drops to calculate its size. Step 1: mass of water initially present. y calculation (or by table), it is easy to show that SVP at 15 C is 1.63 kpa. Thus the partial pressure of water vapor is 1.63 kpa * 0.8 = 1.30 kpa This implies that the *mole* fraction of water is 1.30 kpa / 101 kpa = (assuming the small air parcel initially started at sea level pressure) The mass fraction, then, must be (18 g/mol * mol) / ( 29 g/mol * 1 mol total) = Thus, our initial 1 mg of air contains 8.0 µg of water vapor. Each of the 200 drops, then, must have a mass of 8.0 µg / 200 = µg To calculate the drop diameter, use m = ρv where V = (π/6)* d 3 Thus, d = (6m/ρπ) 1/3 = [ ( 6 * g 10-3 kg/g) / (10 3 kg/m 3 * π) ] 1/3 = m = 42.4 µm This is something like 100 times smaller than the size of a typical raindrop (which might be like 4 mm in diameter), which is equivalent to a million times less massive. Since this problem represents just about the maimum size drop that condensation can produce (since we assume that all the water vapor turns into liquid, which is an etreme assumption), it implies that collisional growth MUST occur to make rain. It can NOT just come from condensation alone.

2 2. Tule fog [25 pts] Tule fog forms when radiative cooling of the ground causes the ground temperature to drop below the dewpoint temperature. We will assume that the ground gains and loses energy through radiation. For the IR absorbed by the Earth, use the solution from Question #1 on Problem Set #3 (where we find that T GHG = 242 K and f = 0.77). We start at sunset (so there s no incoming sunlight). t this moment (t = 0), the ground temperature T g and the air temperature T a are the same, 15 C. The RH is 50%. We will assume that radiative cooling cools the first 5 cm of the ground, which has a heat capacity of c g = 2 kj/(kg K) and a density of ρ g = 1000 kg/m 3. (a) Find an epression for the temperature of the top 5 cm of the ground (T g ) as a function of time t. Hint: you need to do an energy balance, and then integrate this epression. Plot this function over the course of the night (say 12 hrs). (b) Calculate what the ground temperature must be in order for a fog to form. (c) Using the answer to (a) and (b) above, determine how long after sunset we would see the first signs of fog. Do this analytically rather than from the graph. Solution: See attached. In the solutions, (a) to (e) correspond to (a) in our problem; (f) actually corresponds to (b); and (g) corresponds to (c).

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7 3. Seeing your breath [30 pts] miing cloud can form when two parcels of air combine (both can even be subsaturated!). Under certain conditions, the resulting combined parcel can be supersaturated and thus cause condensation. Let s find the conditions under which the ambient air will cause us to be able to see our own breath. The miing looks like this: m, T, p, RH + m, T, p, RH m, T, p, RH m = total mass of the parcel T = temperature of the parcel p = partial pressure of water vapor in the parcel RH = relative humidity of the parcel subscript = ambient air subscript = breath subscript = mied parcel Define the mass fractions and, which denote the relative amounts (by mass) of air being combined where = m /m. and similarly for. Note that + = 1. We will assume that our breath is at body temperature (37 C) and 100% RH. Our variables are:, RH and T. (a) Using conservation of energy, show that T = T + T if you assume the latent heat contribution to the temperature change is negligible. Let s define energy E as relative to some reference temperature, T ref. I m guessing all of you will choose T ref = 0 (either 0 C or 0 K), but I ll just leave it as T ref and show you that the choice of reference doesn t matter at all (which it shouldn t). This means that the sensible heat energy possessed by parcel is given by: E = m C p (T T ref ) and similarly for parcels and. Now, using conservation of energy, we know that E + E = E. Therefore, m C p (T T ref ) + m C p (T T ref ) = m C p (T T ref ) Cancel Cp and collecting terms: m T + m T (m + m )T ref = m T m T ref

8 ut we also know from conservation of mass that m + m = m, which means that the terms with T ref cancel each other out. This means we re left with m T + m T = m T which leads to the final equation! Divide both sides by m : T = (m /m )T + (m /m )T = T + T (b) Using conservation of water vapor, show that M p = p + p if you assume M M where M is the molecular weight of the air parcels. Denote the mass of water vapor in the parcel with the symbol y. So conservation of water vapor states: y + y = y Using the ideal gas law for strictly the water vapor, we can write: n w = pv / RT y = pvm w / RT where M w is the molecular weight of water (remember we ve defined p is the partial pressure of water vapor) Therefore, we can re-write conservation of mass as: pv M RT w = pv M RT w + pvm RT w Canceling M w (didn t cancel R for a reason you ll see in a bit), pv RT p V RT = + () p V RT Now we have some pesky volumes, but we really want mass relationships. So let s use the ideal gas law again, this time for the entire volume (i.e. N2, O2, water vapor, etc): m = PVM / RT where now m is total mass and P is total pressure. Rearranging, V/RT = m/pm Therefore, () becomes: m m p = p + PM PM p m PM Cancel P in all terms. lso cancel all M s by the given assumption Now we re left with: M M M.

9 p m = m p + m p Divide both sides by m and use definition of and yields the desired relationship: p = p + p (c) Find an equation describing the relationship among these three variables such that a cloud can be seen. Note that this relation will be an inequality! [No need to make this equation pretty.] The condition for a cloud is that p must be greater than or equal to the SVP at T : p p T sat ( ) Substituting the results from (a) and (b) yields: p + p C ep[ D /( T + T where C and D are constants given from class: C = kpa; D = 5223 K. To eliminate p as a variable, we p = RH p sat (T ) RH C ep( D / T ) + p C ep[ D /( T + T )] (d) During the evening in Santa Cruz, the RH typically reaches about 80% (i.e. RH = 80%). Given this, plot the equation from (c) and shade in the area where a cloud would be seen. Unfortunately, this equation is not easily separable and is non-linear. To complete this, then, use a spreadsheet program (Ecel or OpenOffice Calc) and use Goal Seek to calculate pairs of values ( a, T a ) for a = 0.50, 0.55, 0.60, [This is better than Wolfram/lpha sometimes since you can store you answer and also change things quickly.] To solve, the first thing you want to do is to normalize the equation. We ll do that by rearranging the equation from (c) to be: )] RH C ep( D / T) + p C ep[ D /( T + T )] 1 () We ll use Goal Seek to solve Eq. (). For a given RH and, we ll find T such that Eq. is satisfied. See the attached spreadsheet (Ecel format)

10 Note that for clarity in the spreadsheet, I ve actually written out the terms. I ve used: p p T ) where p = p + p sat ( (e) How does this function agree with your own observation? [Words only] The temperatures seem a bit high (seeing one s breath at 20 C outside temperature seems unrealistic). I suspect the problem is in assuming that our breath is saturated at 37 C.

11 C = 1.22E+08 kpa D = 5223 K T = 310 K p = kpa <-- using the SVP equation RH = 0.8 Change: T (K) T (K) p (kpa) psat, (kpa) Eqn: For Goal Seek, assign it to change the "T" column such that the "Eqn" column equals 1. You have to do it for each cell, unfortunately.

12 In this region, one can NOT see one's breath. In this region, one can see one's breath. If T is less than the value calculated, then one's breath can be seen - as we know from eperience, colder temperatures make it easier to see one's breath! Mass fraction of ambient air Temperature (K)