Liquid water static energy page 1/8
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1 Liquid water static energy age 1/8 1) Thermodynamics It s a good idea to work with thermodynamic variables that are conserved under a known set of conditions, since they can act as assive tracers and rovide information about the hysical rocesses acting on the domain. In articular, we have available: 1. Conservation of mass 2. Conservation of momentum (newton s 2nd law) 3. Conservation of energy (first law) 4. Conservation of entroy (second law) The conditions we re interested in are horizontal and vertical dislacement on time scales that are too short to be affected by radiation. 1.a) Conservation of mass Starting with conservation of mass, we assume that in the absence of reci/evaoration the total water mixing ratio, defined as q T = kg (H 2 O)/kg (moist air), is conserved as a arcel moves, because while the volume, ressure and temerature all change, the mass remains constant (in the absence of mixing in regions with varying r v ), so: dq T = 0 (1) if mass is conserved and there is no condensation evaoration. 1.b) Conservation of energy What about a conserved energy variable? A good choice is the dry static energy, h d, which is conserved for adiabatic motion in a hydrostatic atmoshere. To see why this is true, recall the definition of the first law: du = q w (2) where u is the secific internal energy ( J kg 1 ) and q and w ( J kg 1 s 1 ) are resectively the rates at which the arcel is heated and work is done by the arcel. In all the roblems in this course, we will only be concerned with the ressure-volume working rate, which is defined by: w = dα (3) where α ( m 3 kg ), the secific volume, is 1/density = 1/ρ. It s not often that we get to work with constant volume situations in the atmoshere. More useful than the internal energy in atmosheric thermodynamics is the secific enthaly, h ( J kg 1 ), defined as: h = u + α (4)
2 Liquid water static energy age 2/8 Note that the α term in (4) looks like ressure-volume work, and can be interreted as the work required at constant ressure to make room for the system with internal energy u 1. Using h we can rewrite the first law (du/ = q dα/) as = q + αd (5) Taking d/ = 0 in (5), we can say that the heating rate q is equal to the time rate of change of enthaly at constant ressure. Since evaoration and condensation occur at constant ressure we will find h useful when we deal with hase change. Now exand h in, T using the chain rule: and substitute (29) into (5) to get: = h dt T + h d (6) q = h dt T + ( h ) d α If the ressure is constant, then d/=0, and (7) roduces the definition of the heat caacity at constant ressure: (7) c = h T T q = c t (8) at constant ressure (9) For dry air c d = 1004 J kg 1 K 1. Also, thermodynamics tells us that over the range of atmosheric temeratures and ressure h/ 0, so using (7) and and (5) we can get: = c dt (10) since c deends only very weakly on ressure. Now what about d/ in (5) for the case of vertical motion? For the boundary layer, it is a very good aroximation to assume hydrostatic balance, which is the statement that an the vertical ressure differential across a layer rovides exactly the amount of force necessary to balance gravity: 1 Schroeder, D. V., An Introduction to Thermal Physics, Addison-Wesley,. 149
3 Liquid water static energy age 3/8 0 d 0 ρg Figure 1: hydrostatic balance for a 1 m 3 layer In symbols, the balance shown by Figure 1 imlies that: d = ρg (11) Question: is this the same ressure as the local ressure given by the equation of state: = ρr d T? Substituting (11) into (5) and recognizing that αρ=1, the first law in the case of hydrostatic balance becomes: = q g In articular, if the heating rate q=0, then: (12) + g = d = 0 (13) defines a conservation equation for the dry static energy h d, where: h d = h + gz (14) This conservation makes h d a useful tracer, since changes to h d have to come from some diabatic rocess (radiation, evaoration), and can t come from simly lifting or sinking air. Note also that gz is just the otential energy er kilogram. So as the arcel decreases or increases its altitude, energy is moving between otential and internal, while h d is staying the same. We can also take (13) aart to get an equation for the dry adiabatic lase rate: d so dt = c dt g = 0 (15a) = Γ d = g c 9.8 K km 1 (15b) which is the rate at which internal energy is converted to otential energy during adiabatic ascent.
4 Liquid water static energy age 4/8 1.c) Conservation of entroy First remember that the 2nd law of thermodynamics relates the entroy s to the heating rate q and temerature T : ds q T where the > sign holds when the rocess is irreversible. Also note that the equation of state for dry air is given by: (16) = ρr d T (17) where R d (287 J kg 1 K 1 ) is the gas constant for dry air. So divide (5) by T and rewrite the first law as: q T = 1 ( ) T αd = c dt T R dt d T = c dt T R d d ds where I ve also included invoked the second law to relate this to s. Again, if the rocess is adiabatic then q = 0, (18) 1 dt c = R d d (19a) T c d log T = R d d log (19b) Integrating both sides of (19b) from the surface, with (temerature, ressure) given by (θ, 0 ) to a lower ressure and lower (why?) temerature T gives Poisson s equation: ( ) ( ) c T log = log (20) R d θ Equation (20) defines the otential temerature, θ which is conserved for adiabatic ascent/descent (BLM 1.5.1c): θ = T ( ) Rd /c 0 = T 0 ( ) (c cv)/c 0 = T ( ) γ 1 0 γ where R d /c = 287/1004 = and γ = c /c v. Like h d, θ is constant for a dry adiabatic rocess. To see this first differentiate (21) to get: (21) Then use (18) to show that: c θ dθ = c dt T R d d (22) c dθ θ = ds q T and if q=0 and the rocess is reversible, both θ and s are constant. (23)
5 Liquid water static energy age 5/8 This means that there s a one to one relationshi between θ and the entroy s: or ds = c dθ θ (24) s c log θ (25) 2) Liquid water static energy Once we introduce water vaor and liquid water, we re dealing with a new mixture that can t be described by the ideal gas equation. It does have an enthaly H (J): H = m d h d + m v h v + m l h l (26) where m stands for mass, H is the total enthaly (J) and h x ( J kg 1 ) is the secific enthaly where subscrits x = (d, v, l) resectively denote dry air, water vaor and liquid water. We will assume that the mass of dry air stays fixed, and write the secific enthaly of the mixture as: Note that the total entroy still obeys the first law: h = H/m d J kg 1 (27) But what is? Calculus says: q = αd (28) dh = H T And by the definition of the heat caacity: H H dt + d + dm v + H dm l (29) m v m l H T = m dc d + m v c v + m l c l (30) Which, after dividing through by m d, roduces the heat caacity of the mixture, c ( J kg 1 K 1 ): The second term in (29), H and looking at (26) it s aarent that H m v h T = c d + r v c v + r l c l = c (31) = 0 to an excellent aroximation (remember that H = U + P V ), = h v and H m l = h l, yielding a new version of (29): = c dt + h v dr v + h l dr l (32) We ll start with the simlest scenario: the total water mass is conserved and the mixture is at its saturation mixing ratio r sat = ɛe s (T )/( e s (T )). Then the total water mass m t = m v + m l
6 Liquid water static energy age 6/8 satisfies dm t = 0 = dm v + dm l = dm s + dm l = dr sat + dr l, so that dr v = dr sat = dr l and (32) becomes: = c dt + (h v h l )dr sat = c dt + l v dr sat (33) where l v ( J kg 1 ) = h v h l is called the enthaly of evaoration or the latent heat. It reresents the energy needed to break a liquid water molecule away from its neighbors and send it into the vaor hase. It takes about J to evaorate a kg of water at 0 C. Note that because the mass of water is conserved, I could just as well write (33) as: = c dt l v dr l (34) I could also go through the stes above for a more general mixture of drolets, rain and ice and get the exression for enthaly used in Khairoutdinov and Randall: = c dt + c dt l v (dr l + dr r ) l s (dr i + dr s + dr g ) (35) Since is still the enthaly, it has to obey the first law: or if we have hydrostatic balance: = q αd (36) + g = q (37) Note that (37) can be rearranged to give the saturated moist adiabatic lase rate with q=0: ds v = + g = c dt + l v dr sat + g = 0 (38) Equation (37b) defines the moist static energy, s v, which is conserved for adiabatic ascent. Again, since total water is conserved, we are also free to define the liquid water static energy, s l : Integrating (39) gives ds l = + g = c dt l v dr l + g = 0 (39) s l = c T l v r l + gz (40) How much does condensation heating change the dry adiabatic lase rate? We can calculate this: + g = c dt + l v dr sat + g = 0 (41) = c dt + l dr sat v = c dt + l dr sat dt v dt = g (42) = g (43)
7 Liquid water static energy age 7/8 ( ) ( ) dt g = 1 c dr 1 + l sat v dt For tyical midlatitude conditions, the evaoration and condensation of water reduces the magnitude of dt/ from about -10 K/km to -6 K/km. 3) Moist entroy We can divide (28) by T and substitute (32) for to get the entroy of the mixture: ds = q T (44) = c T dt + l v T dr sat R d d (45) Note that we have made the aroximation that α, the secific density, is about the same for moist and dry air. You ll integrate (45) numerically using ython, but for now we can make some rogress by using the aroximation that ( ) l v T dr lv r sat sat d (46) T (note that this is not a good aroximation (errors of about 30% at high temerature, but fortuitously there s cancellation in the neglected terms that makes it nearly exact, we ll derive the full version later in the course) We ll also need the result from the entroy notes that: dθ c θ = c dt T R d d Substituting (47) and (46) into (45), and taking the adiabatic case (q = 0) gives a a differential equation for the otential temerature under an adiabatic transformation: ( ) dθ c θ = l v T dr lv r sat sat d (48) T As usual, I can fatten and thin (48) to get an equation for the change in θ wrt, say, height, for an adiabatic arcel: d log θ = l v dr sat c T Convince yourself that (49) imlies that θ increases with height along a moist adiabat. We can integrate (49) to give θ as a function of height (or equivalently, ressure or time). Since dr l = dr sat, (49) is equivalent to: d log θ = l v dr l c T There are quite a few different ways to label (49) or (50) but they all rovide different names for the same adiabat. In this course we ll use either the equivalent otential temerature θ e, the (47) (49) (50)
8 Liquid water static energy age 8/8 liquid water otential temerature θ l or the wet bulb otential temerature θ w. To find the θ e label use (46) and integrate the aroximate equation from a height where the temerature is so cold that effectively r sat = 0, down to the height of interest, where r sat = r w : θ θ e d log θ = rsat 0 ( d l ) vr sat c T Since both sides are erfect differentials we can immediately write down (51) as: ( ) θ log = l vr sat θ e c T or: ( ) lv r sat θ e = θ ex c T I ll leave it to you to show by the same aroach that (50) can be integrated to give: ( θ l = θ ex l ) vr l c T One esecially useful feature of θ e is that it is aroximately constant for reciitation arcels, because r l doesn t aear in (45), excet as a small term in the mixture heat caacity c. In contrast θ l assumes imlicitly that there is no reciitation along the vertical integration, since we assume dr l = dr sat everywhere. The advantage of θ l is that it reduces to θ when liquid water is zero (say below or outside of cloud). (51) (52) (53) (54)
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