39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 3 / Solution. Solution

Transcription

1 Solution. For an altitude change dz, the atmospheric pressure change is : dp = ρgdz () where g is the acceleration of gravity, considered constant, ρ is the specific mass of air, which is considered as an ideal gas: Put this expression in () : m pμ ρ = = V R dp = dz p R.. If the air temperature is uniform and equals, then dp p = dz R After integration, we have :.2. If then z R p e p z = (2) z = z (3) dp = dz p R z (4).2.. Knowing that : dz d z = = ln z z by integrating both members of (4), we obtain : g z p z z μ ln = ln = ln p R R ( z) z = p p z R (5)

2 .2.2. he free convection occurs if: ρ ρ ( z) > he ratio of specific masses can be expressed as follows: ρ z p z z = = ρ p z R he last term is larger than unity if its exponent is negative: < R hen : μ g K > = = 34. R 83. m 2. In vertical motion, the pressure of the always equals that of the surrounding air, the latter depends on the altitude. he temperature depends on the pressure. 2.. We can write: d d dp = dz dp dz p is simultaneously the pressure of air in the and that of the surrounding air. Expression for d dp By using the equation for adiabatic processes pv γ = const and equation of state, we can deduce the equation giving the change of pressure and temperature in a quasi-equilibrium adiabatic process of an air : p γ γ = const (6) 2

3 where cp γ = is the ratio of isobaric and isochoric thermal capacities of air. By c V logarithmic differentiation of the two members of (6), we have: d γ dp + = γ p Or d = (7) p γ γ dp Note: we can use the first law of thermodynamic to calculate the heat received by the m in an elementary process: dq = cv d + pdv, this heat equals zero in an μ adiabatic process. Furthermore, using the equation of state for air in the m pv = R we can derive (6) μ dp Expression for dz From () we can deduce: where dp pgμ = ρg = dz R is the temperature of the surrounding air. On the basis of these two expressions, we derive the expression for d dz : / d γ = = G (8) dz γ R In general, G is not a constant If at any altitude, =, then instead of G in (8), we have : or γ = = const (9) γ R 3

4 = (9 ) c p Numerical value: K 2 K = = m m hus, the expression for the temperature at the altitude z in this special atmosphere (called adiabatic atmosphere) is : z = z () 2.3. Search for the expression of ( z ) Substitute in (7) by its expression given in (3), we have: d γ dz = γ R z Integration gives: z γ z ln = ln γ R Finally, we obtain: z ( z) = 2.4. From () we obtain z ( z) = z <<, then by putting x =, we obtain z If ( z) = + x x z z z e z () 4

5 hence, z z (2) 3. Atmospheric stability In order to know the stability of atmosphere, we can study the stability of the equilibrium of an air in this atmosphere. At the altitude z ( z ) ( z ), where =, the air is in equilibrium. Indeed, in this case the specific mass ρ of air in the equals ρ '- that of the surrounding air in the atmosphere. herefore, the buoyant force of the surrounding air on the equals the weight of the. he resultant of these two forces is zero. Remember that the temperature of the air z is given by (7), in which we can assume approximately G = at any altitude z near z = z. Now, consider the stability of the air equilibrium: Suppose that the air is lifted into a higher position, at the altitude z (with d>), ( z + d) = ( z ) d and z + d = z d. In the case the atmosphere has temperature lapse rate > + d, we have z + d > z + d, then ρ < ρ '. he buoyant force is then larger than the air weight, their resultant is oriented upward and tends to push the away from the equilibrium position. Conversely, if the air is lowered to the altitude z d < z d and then ρ > ρ '. z d (d>), he buoyant force is then smaller than the air weight; their resultant is oriented downward and tends to push the away from the equilibrium position (see Figure ) So the equilibrium of the is unstable, and we found that: An atmosphere with a temperature lapse rate >is unstable. In an atmosphere with temperature lapse rate <, if the air is lifted to a higher position, at altitude z + d (with d>), ( z + d) < ( z + d), then 5

6 ρ > ρ '. he buoyant force is then smaller than the air weight, their resultant is oriented downward and tends to push the back to the equilibrium position. Conversely, if the air is lowered to altitude z d (d > ), z d > z d and then ρ < ρ '. he buoyant force is then larger than the air weight, their resultant is oriented upward and tends to push the also back to the equilibrium position (see Figure 2). So the equilibrium of the is stable, and we found that: An atmosphere with a temperature lapse rate <is stable. z > z +d z z -d > ρ < ρ < ρ up > ρ down unstable ( z ) Figure z < z +d z z -d < ρ > ρ > ρ down < ρ up stable ( z ) Figure 2 6

7 In an atmosphere with lapse rate =, if the is brought from equilibrium position and put in any other position, it will stay there, the equilibrium is indifferent. An atmosphere with a temperature lapse rate =is neutral 3.2. In a stable atmosphere, with <, a, which on ground has temperature > and pressure p equal to that of the atmosphere, can rise and reach a maximal altitude h, where h = h. In vertical motion from the ground to the altitude h, the air realizes an adiabatic quasi-static process, in which its temperature changes from h = h. Using (), we can write: h = = h h h = ( ) to - - h = h= ( - - ) = So that the maximal altitude h has the following expression: h= ( ) ( ) (3) 7

8 4. Using data from the able, we obtain the plot of z versus shown in Figure 3. 3 Altitude [m] 2 D(2.6 o C;42 m) C(2.8 o C; 9 m) B(2. o C; 96 m) A( 22 o C; m) emperature [ o C] Figure We can divide the atmosphere under 2m into three layers, corresponding to the following altitudes: ) < z < 96 m, K = = m 2) 96 m < z < 9 m, 2 =, isothermal layer K 3) 9 m < z < 25 m, 3 = = m In the layer ), the temperature can be calculated by using () 96 m = K 294. K that is 2. o C In the layer 2), the temperature can be calculated by using its expression in z = exp isothermal atmosphere ( z). 8

9 he altitude 96 m is used as origin, corresponding to m. he altitude 9 m corresponds to 23 m. We obtain the following value for temperature: h And 9 m = K that is 2.8 o C 4.2. In the layer 3), starting from 9 m, by using (3) we find the maximal elevation = 23 m, and the corresponding temperature K (or 2.6 o C). Finally, the mixing height is H = = 42 m. 42 m = K that is 2.6 o C From this relation, we can find 9 m K and h = 23 m. Note: By using approximate expression (2) we can easily find z = 294 K and K at elevations 96 m and 9 m, respectively. At 9 m elevation, the difference between and surrounding air temperatures is.7 K (= ), so that the maximal distance the will travel in the third layer is.7/( 3 ) =.7/.3 = 23 m. 5. Consider a volume of atmosphere of Hanoi metropolitan area being a parallelepiped with height H, base sides L and W. he emission rate of CO gas by motorbikes from 7: am to 8: am M = /36 = 3 3 g/s he CO concentration in air is uniform at all points in the parallelepiped and denoted by C( t ). 5.. After an elementary interval of time dt, due to the emission of the motorbikes, the mass of CO gas in the box increases by Mdt. he wind blows parallel to the short sides W, bringing away an amount of CO gas with mass LHC ( t) udt. he remaining part raises the CO concentration by a quantity or Mdt LHC ( t) udt = LWHdC dc in all over the box. herefore: 9

10 dc u M + C() t = (4) dt W LWH 5.2. he general solution of (4) is : ut M C() t = Kexp + W LHu (5) From the initial condition C =, we can deduce : M ut C() t = exp LHu W (6) 5.3. aking as origin of time the moment 7: am, then 8: am corresponds to t =36 s. Putting the given data in (5), we obtain : 3 ( 36 s) (. 64) 2. 3 mg/m C = =