GEF2200 vår 2017 Løsningsforslag sett 1

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1 GEF2200 vår 2017 Løsningsforslag sett 1 A.1.T R is the universal gas constant, with value JK 1 mol 1. R is the gas constant for a secic gas, given by R R M (1) where M is the molecular weight of the gas (usually given in g/mol). For the equation of state: V nr T m M R T mrt (2) It is imortant to note that we usually use mass m in units of [kg], which requires that the units of R is changed accordingly (if given in [J/gK], it must be multilied by 1000 [g/kg]). A.2.T Aarent molecular weight is the average molecular weight for a mixture of gases. We introduce it to calculate a gas constant R R /M d for the mixture, where M d is the aarent molecular weight of i dierent gases given by Equation (3.10): M d mi mi ni M i mi M i n n n i n M i (3) In meteorology the most common aarent molecular weight is the one of air. WH Concentrations by volume (See exercise A.8.T): v N2 V N 2 V v CO2 V CO 2 V (4) (5) Assuming ideal gas, we have total volume V V N2 + V CO2 at a given temerature (T ) and ressure (). This means volume concentration is equal to molecular fraction: v i V i V n ir T/ nr T/ n i n (6) The aearent molecular mass is the molecular mass a mixture of n moles of gass and a total mass m aears to have: ni M i M ve m n mi n n n i n M i v i M i (7) Inserting values for CO 2 and N 2 : M ve 0.95 ( ) g/mol (2 14) g/mol 43.2g/mol (8) This is quite higher than for our atmoshere, and gives the gas constant R ve R M ve JK 1 mol gmol Jg 1 K Jkg 1 K 1 (9) which is lower than for the Earth's atmoshere. A.4.T We have that R R (10) M To nd if R increases when adding water vaor (or any gas) to dry air, will be the same as nding that the mixture becomes lighter (molecular mass decreases). Looking at dry air: mi M d mi m d (11) M i n d 1

2 Adding mass m x with molecular weight M x gives mi + m x M new mi m d + m x M i + mx m d M x M d + mx M x M d m d + m x m d + m x M d M x (12) If M x < M d, we see that M new < M d (lighter than dry air), which means that R new > R. If M x > M d, then M new > M d (heavier than dry air) and R new < R. A.5.T Because R is deendent on humidity, we move this deendency from R to T : RT R d T v (13) so that we can use the gas constant for dry air in calculations. A.6.T When we look at a single gas we use the gas constant for that gas. For water vaor we use R v 461JK 1 kg 1. We therefore do not use virtual temerature. WH Given 1% of water vaor in the air, what is T v T? The mole fraction of water vaor is equal to the ressure fraction exerted by the water vaor (see A.7.T). Thus e/ Using a tyical temerature T 288 K, equation (3.16) gives: T v T T 1 e T (1 ɛ) 288K ( ) 288K 1.1K A.7.T At sea level the ressure is about 1013 hpa. The ressure fraction exerted by water vaor is therefore e 18hPa (14) 1013hPa The ideal gas law as stated in (3.6) is V nr T (15) The artial ressure exerted by water vaor is the ressure it would exert had it alone occuied the same volume as the total gas mixture. We may therefore use (3.6) again on water vaor alone: ev n v R T (16) where n v is the number of moles of water vaor contained in the volume V. Combining the two equations, we nd that n v n e (17) (From this deduction, we see that the mole fraction of any gas comonent in a mixture equals the ressure fraction exerted by that comonent.) So the mole fraction of water vaor in this case is 1.78 %. A.9.T The hydrostatic equation gives the relationshi between gravity and ressure for an air mass at rest, a relationshi between a change in ressure d and a change in height dz (see Figure 1). Since the ressure at a given height is due to the weight of the above air, we exect the ressure to decrease with height, meaning that a decrease in ressure equals an increase in height. We need to consider the forces acting uon the layer, and to do so, we look at the edges of the layer (height z and z+dz. Gravity works downwards, and the force on the to of the layer is F g (z + dz). Furthermore, the layer acts on the air below with a force F g (z). Since the layer is at rest, we have from Newton's thrid law that 2

3 A. ϱgdz d 0 (20) It is very common to do such calculations er area, leaving out A from the start. Figure 1: The forces on a layer of air. the layer exerience a similar force in the oosite direction ( F g (z)). For a body with mass m, the gravity is mg ϱv g, where V is the volume. Given that the layer has a thickness dz and area A, and therefore a volume dv Adz, we have (assuming ositive axis uwards): F g F g (z + dz) F g (z) ϱga(z + dz z) ϱgadz (18) Forces acting uwards: Pressure (force er area). The uward force at the layer to is oosed by a similar force acting on the layer ( F ( + d)). F F () F ( + d) [ ( + d)] A Ad (19) Where F () > F ( + d), that is, d is negative. F is then value of the force acting uwards. To nd the equilibrium between gravity and ressure forces, we set F + F g 0, and cancel When using vector calculation (k-vector): F g k ϱga(z + dz)k ( ϱgaz)k ϱgadzk (21) F k [ ( + d)] Ak Adk (22) In any case, when the sum of forces is zero, we get d dz ϱg (23) A.10.T The geootential at a given location is the energy required to lift 1 kg from sea level and u to that location. φ(z) z 0 gdz (24) Dividing by the average surface acceleration of gravity, we get the geootential height. Z 1 z gdz (25) Using the hydrostatic equation, the equation of state and the denition of virtual temerature: 0 d dz ϱg RT g g (26) R d T v Using the denition of geootential height gdz dz (27) we have d dz (28) R d T v Which we integrate from 1, Z 1 to 2, Z 2 : Z2 dz R dt v Z 1 dz 2 1 R d d R dt v 2 1 d T v d (29) (30) 3

4 By introducing the weighted mean virtual temerature T v 2 d 1 T v 2 d 1 (31) and inserting into Equation (30), we may write Z 2 Z 1 R 2 d T v R d T v ln d ) 1 ( 2 1 (32) (33) This means that the thickness is linearly roortional to the mean temerature of the layer. A horizontal gradient in this temerature will then mean a horizontal gradient in layer thickness. This is an imortant feature of the atmoshere, esecially when it comes to storm systems and low/high ressures. WH Meteorological station 50m below sea surface. The temerature at the surface is T vs 15 C 288K, and the mean temerature in the layer between 1000 and 500hPa is T v 0 C 273K. Z s 50m s 1020hPa 9.81ms 2 R 287Jkg 1 K 1 To nd the height of the 500hPa level, we use the hysometric equation. You have to assume that the temerature from the surface u to 1000hPa is constant. (If the layer between 1000hPa and 500hPa was isothermal, the would be hysically correct to calculate a mean temerature for the layer between 1020hPa and 1000hPa, to avoid temerature jums.) Assuming the mean temerature in the layer between s and hPa to be T vs, we integrate the hysometric equation Z500 dz R 500 d Z s Z 500 Z s R d Z 500 Z s R d Inserting the values: s [ 1000 T v d s T vs d d T v 1000 [ T vs ln 1000 s + T v ln (34) ] (35) ] (36) Z km (37) WH Ignoring the virtual temerature correction, we may use the hysometric equation (3.29) on the location at the storm centre Z Z sea level R dt A and on the surrounding region: Z Z sea level R dt B ln 940hPa ln 1010hPa (38) (39) T A is the (unknown) mean temerature between the sea surface and 200 hpa at the storm centre, and T B 3 C K is the mean temerature in the surrounding region. Since the 200 hpa surface is erfectly at, and the ocean surface is also at, the left-hand sides of the two above equations are the same. Therefore: R d T A ln 940hPa R dt B ln 1010hPa (40) ln 1010hPa T A T B ln 940hPa (41) 1010hPa ln 270.2K ln 940hPa (42) 282.7K 9.5 C (43) 4

5 The air column above the storm centre has in average 12.5 C higher temerature than the surroundings. A.11.T dq is the heat added to a system. dw is the work done by the system. du is the change in internal energy. Dierent kinds of work include exansion, surface exansion, extension and electrical work. Exansion work is the imortant one in thermodynamics of the atmoshere, so we can assume dw dα. WH j ressure. From the rst law of thermodynamics, we have dq du + dw c v dt + dα 0 (44) In the exansion, the secic volume (volume er mass) increases (dα > 0), so that the temerature change is negative: dt c v dα < 0 (45) Since the air exands, the work done on the surroundings is ositive, equals to dw dα. The decrease in temerature means that the internal energy (du c v dt ) also decreases. This is what haens when an air arcel rise in the atmoshere; it exands and cools. This is also imortant for the heat transort from the equator to the oles. A.12.T An adiabatic rocess is a rocess where no heat exchange between the system and the surroundings is allowed. That means dq 0.../FIGURES/tire-es-converted-to.df Figure 2: An air arcel exands when leaving a tire: T 1 T, 1 >. We follow an air arcel leaving the tire, to get out into the surroundings. The ressure inside the tire is higher than the ressure of the surroundings ( 1 > ), but the temerature is the same (T 1 T ). The air arcel will then exand adiabatically, since it is entering air with lower 3.18h When a low-ressure is colder than its surroundings, the isobars are closer together in the low-ressure than in the surroundings. Going uwards (towards lower ressure), the height of a given isobar will get increasingly lower in the low-ressure centre comared to in the surroundings (see gure 3.3b in WH06). 3.18k In our study of thermodynamics, the only work taken into account is the work done in exanding the volume of the system. We then have: dw dα (46) Here, is the ressure at the 'edge' of our system, where the environment is 'ushed' away as the system exands by dα. To get zero work we can either let the volume remain 5

6 constant (dα 0) or let the system exand into a vacuum ( 0, there are no surroundings to do work on). WH (WH ) We have isothermal comression of 2kg dry air at T 15 C, to one tenth of its initial volume. Work done on the air arcel work done by the air arcel. The only work done by the arcel is exansion work: W V2 We have the ideal gas equation dv (47) mrt V (48) Inserting and integrating gives, when remembering that T is constant, W V2 dv mrt ln V 2 (49) 2kg 287Jkg 1 K 1 288K ln J This is the work done by the arcel, so the work done on the arcel when comressing it is W J. WH (WH ) a We have an ideal gas, undergoing an adiabatic transformation, and we will rove that V γ constant for this rocess, where γ c /c v. The rst law of thermodynamics (TD1) is given by Eqn. (WH ): dq c v dt + dα (50) where dq 0 for an adiabatic rocess. Dierentiating the ideal gas equation and inserting (50) for dt, we get α RT (51) αd + dα RdT (52) R dα (53) c v ) αd (1 + Rcv dα (54) Using Eqn. (WH ), we have that and we get 1 + R c v 1 c v (c v + R) c c v γ (55) Re-arranging: αd (1 + Rcv ) dα γdα (56) d γ dα α Now er integrate from 0, α 0 to, α 0 d ln 0 0 and because we have α (57) γ dα (58) α 0 α γ ln α ( ) α γ ln (59) α 0 α 0 ( ) α γ (60) α 0 α 1/ϱ ϱ 0 α 0 1/ϱ 0 ϱ m/v 0 m/v V (61) V 0 ( ) V γ 0 V 0 V γ 0 V γ 0 (62) b For an isothermal transformation we have from the equation of state 1 2 V 2 (63) 6

7 So for an isothermal comression from ressure hPa and volume 7.5cm 3 to the volume V 2 2.5cm 3, we nd the ressure hPa 7.5 V hPa To nd the ressure after adiabatic exansion until the original volume V 3, we use the exression from Equation (62) 3 V γ 3 2 V γ ( V2 3000hPa ) γ (64) ( ) 2.5 γ 643hPa 7.5 The temerature after exansion is found using the equation of state 3 V 3 T 3 2V 2 T 2 (65) T 3 3V 3 2 V 2 T 2 (66) where we can insert the ressure 3 from Equation (64) (or the value calculated). We also note that T 2 T 1, since the rst transformation was an isothermal transformation (and still V 3 ): ( ) V γ 3 V2 T 3 T 2 2 (67) 2 V 2 V 3 ( ) 1 ( ) γ V2 V2 T 1 T 1 ( V2 290K ) γ 1 ( ) 2.5 γ K 7.5 Comment on the hysical side: The result that V γ is constant for an adiabatic rocess means that the volume changes dierently from an isothermal rocess (where V is constant). In the case of exansion from one ressure to another (), the volum of the adiabatic system (V a ) after transformation is given by Va γ C a V a ( Ca ) 1/γ (68) where C a is the constant value. The corresonding volume change for the isothermal system (V i ) is given by V i C i V i C i (69) where C i is the constant value. Because γ c /c v > 1, isothermal exansion gives a faster/larger increase in volume than does the adiabatic exansion (Equation (68) vs (69)). The hysical reason is that the adiabatic system cools during the exansion, whereas the isothermal system is ket at constant temerature by adding heat to the system. The increased temerature increases the volume or the ressure. In this exercise, the volume was the same for the two rocesses, but the ressure was not: The adiabatic exansion resulted in a lower ressure (643hPa vs 1000hPa), and a lower temerature (187K vs 290K). Clearly, adding heat to kee the rocess isothermal means heating the air by 103K, which indeed should increase the ressure (remember that the volume is constant). In the case of comression, Equation (68) and (69) show that the decrease in volume is smaller for the isothermal rocess than for the adiabatic rocess. The argument follows the same lines as for exansion; since an adiabatic comression means that the temerature increases, whereas heat is removed from the isothermal 7

8 system to kee it at constant temerature, the isothermal volume (or ressure) will be smaller. A.13.T Secic heat: The amount of heat you have to suly to a body of unit mass to raise its temerature by 1 degree. dq dt c (70) So that if dt 1, we have dq c. c deends on whether the system is allowed to change its volume because in that case, work will be done either by or on the system. This will also contribute to changing its temerature. To nd the relationshi c v + R c, we start with constant volume: ( ) ( ) dq du c v (71) dt αconst dt αconst and because of Joule's law for an ideal gas, du is only deendent on temerature c v du dt (72) Then we have from the TD1, and by rewriting dα dq c v dt + dα (73) c v dt + d(α) αd (74) c v dt + d(rt ) αd (75) (c v + R)dT αd (76) So we see that for constant ressure, we have d 0 and ( ) dq c c v + R (77) dt const E.1.T a When the air is contained in a closed and rigid box, its volume is constant. We therefore use the secic heat at constant volume: dq dt c v (78) q c v (T 2 T 1 ) (79) Since we have 2 kg of air, the heat that must be added is: Q mq (80) mc v (T 2 T 1 ) (81) 2kg 717J/Kkg 10K (82) 14.3kJ (83) b Now ressure is constant. We do the same as in (a), just using c 1004J/Kkg instead of c v. We then get Q 20.1kJ. c CANCELLED d For an adiabatic rocess we have T 2 T 1 ( 2 1 ) R/c (84) (Eq in WH06). We solve for the ressure: 2 1 ( T2 T 1 ) c/r ) (85) ( 283.2K 1000hPa 273.2K (86) 1134hPa (87) We nd the secic volume from the ideal gas equation: α 2 R dt 2 (88) J/Kkg 283.2K J/m 3 (89) 0.717m 3 /kg (90) WH To nd the height of a ressure level, given an 8

9 isotherm atmoshere and surface ressure hPa, we use the hysometric equation, and assume T T v 240K. Z 1 Z 0 R dt v ln 0 1 (91) and for Z 0 0 at 0 we get Z J kgk 240K 9.81ms 2 ln 0 1 (92) 7.021km ln 0 1 (93) For ressure levels 1 100, 10 and 1hPa, we get Z km (94) Z km (95) Z km (96) This is the geootential height, which dier slightly from the geometrical height. The assumtion of isothermal atmoshere is also not very correct, but ok for this exercise. 9