Fill dynamics and sample mixing in the AirCore 1

6 The AirCore is a long coiled tube that acts as a “tape recorder” of the composition of air as it is 7 slowly filled or flushed. When launched by balloon with one end of the tube open and the other 8 closed, the initial fill air flows out during ascent as the outside air pressure drops. During descent 9 atmospheric air flows back in. I describe how we can associate the position of an air parcel in the 10 tube with the altitude it came from by modeling the dynamics of the fill process. The conditions 11 that need to be satisfied for the model to be accurate are derived. The extent of mixing of air 12 parcels that enter at different times is calculated, so that we know how many independent 13 samples are in the tube upon landing, and later when the AirCore is analyzed. 14

in which Qm is mass flow (kg s −1 ), Qn = Qm/M is amount flow (mol s -1 ) with M molecular weight 106 of dry air (0.02896 kg mol -1 ), ρ is gas density (kg m −3 ), ρn is amount density (ρ/M in mol m -3 ), η 107 is viscosity (kg m −1 s −1 ), r is tube radius (m), P is pressure in Pascal (kg m −1 s −2 ), and z is 108 distance along the tube (m). Pressure is given by the ideal gas law as P = (n/V) RT, with n/V = ρn 109 the number density in mol m −3 , T is temperature in degrees Kelvin (K), and R the universal gas 110 constant, 8.3144 J mol -1 K -1 . The flow velocity is parabolic as a function of radius, zero at the 111 wall, and maximum in the center where the speed is twice the average speed. 112 The viscosity (η) depends on temperature, but it is very nearly independent of pressure in our 113 range of interest. The latter is of primary importance to the fill process. A simple approximate 114 molecular expression for viscosity is η ≅ (1/3) ρ c λ, in which c is the average molecular speed 115 and λ is the mean free path between collisions which is inversely proportional to ρ (Jeans, 1952) 116 so that it cancels the factor "ρ" in η ≈ (1/3) ρ c λ. Since the volume flow (m 3 s -1 ) is Qv = Qm/ρ, 117 Eq. (1) states that the volume flow depends on viscosity, but not on gas density. It takes the same 118 amount of force (pressure difference) to push a volume flow irrespective of the density of air in 119 that volume. During steady flow through any tube the flow needs to speed up at the low pressure 120 end to conserve mass, so that the pressure gradient always steepens at the low pressure end.

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The z-coordinate is for position along the length of the tube. The pressure change at any point in 122 a small section of the tube with length dz can be due to temperature change or to more amount 123 flow coming in from z than leaving from z+dz. The latter term is 124 dρ n dt = − 1 πr 2 dQ n dz , so that 125 dP dt = ρ n R dT dt + RT dρ n dt = P T dT dt − RT πr 2 dQ n dz Eq.
(2) 126 Because we assumed that the tube cross section is round (not elliptical for example) the amount 127 flow Qn is given by Poiseuille's equation, and Eq. (2) can be represented numerically in a very 128 efficient manner. In that case the flow is in effect solved as a succession of steady state flows 129 that evolve slowly in time and along the length of the tube. In the rest of this section we will 130 discuss a number of assumptions we are making for our "succession of steady state flows" 131 approximation to Eq.

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The first one is that inertial effects, i.e. accelerations, die out very rapidly. Suppose we suddenly 133 set the pressure gradient that is driving the flow to zero. What is the time scale for the flow to die For a tube with a radius of 3 mm and ρ corresponding to 1 bar and 285 K, τ ≅ 0.07 s. At an 142 altitude where the density is 10 times lower (~18 km), τ ≅ 0.007 s. Recently NOAA GML has 143 been flying AirCores with r ≅ 1.46 mm, for which the adjustment time at 1 bar and 285 K is τ ≅ 144 0.017 s. A succession of steady state flows is indeed a very close approximation. 145 Next we assume that the temperature of the gas is the same as that of the wall. How rapidly does 146 the temperature of the gas equilibrate with the wall of the tube? The heat capacity of a volume of 147 air is cp ρn ≅ (7/2) R * P/RT in which cp is the molar heat capacity at constant pressure and ρn is 148 the number density (mol m -3 ) of the gas, so that cp ρn has units of J m -3 K -1 . The heat 149 conductivity of gas is κ ≅ (1/3) cv ρn c λ (Jeans, 1952) in which cv is the molar heat capacity at 150 constant volume, c is the average speed of individual molecules and λ the mean free path. It has 151 units of (J/s) m -2 (K/m) -1 , the heat flow per area per temperature gradient. As in the previous 152 paragraph we divide the heat energy change corresponding to ΔT in a volume of gas residing in a 153 length Δz by the heat flow from the wall assuming the temperature gradient is close to which has units of seconds. For r = 3 mm and λ corresponding to 1 bar, and 285 K, the parameters. If we were to wind our coil much tighter, say with r/R of 1/20, then the maximum 175 relative flow correction during a flight would be +2 10 −4 for the same Reynolds number.

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Therefore we can neglect the corrections for the tube coil curvature.

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If the tube is elliptical (as a result of bending, for example) instead of circular, we can use a good 178 approximation for the change in flow resistance. Following Lekner (2019), Eq.1 can be written 179 for volume flow as (η Qv) / (dP/dz) = π r 4 / 8, neglecting the sign. Note that π r 4 / 8 equals A 3 / (2 180 P 2 ) for a circular cross section, with A the cross sectional area, and P the perimeter of the tube. 181 Lekner shows that A 3 / (2 P 2 ) applies quite generally for many cross sectional shapes. So if the 182 tube is somewhat squashed into an ellipse with major axis 1.05 times the original radius, and a 183 minor axis slightly smaller (in order to keep the perimeter the same) than 0.95 times radius, the 184 term A 3 / (2 P 2 ) has become ~1% smaller. This correction is not major, but easy to apply if 185 needed. 186 We assumed the ideal gas law. Non-ideality is often described by the virial expansion relating 187 pressure and density, PV/nRT = 1 + B(n/V) + C(n/V) 2 +….. Note that n/V is called ρn above.

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Taking only the second (and largest) virial coefficient B (m 3 mol -1 ) into account we can 189 approximate the number density ρn as (P/RT)(1-BP/RT). The relative change of number density 190 is thus BP/RT which has dimension one. At 300 K and 1 bar, B is -7.3 10 −6 m 3 /mol 191 (Sevast'yanov, 1986) which leads to a relative density increase of 2.9 10 −4 . B increases to -18.9 192 10 −6 and -37.8 10 −6 m 3 mol -1 at 250 K and 200 K respectively, but at the higher altitudes the 193 density is lower so that the largest non-ideality effect occurs near the ground. Therefore the 194 fractional density increase relative to ideal gas during a flight remains well below 0.001

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When the mean free path increases at lower pressures there could be "wall-slip", non-zero 196 velocity at the wall which can be modeled as an effective decrease in viscosity increasing the 197 volume flow. Berg (2005) gives an approximate expression for the factor by which the flow quantitatively investigating molecular flow effects, although this limit depends also on the 211 sampling accuracy we require.
The above expressions for viscosity, η ≅ (1/3) ρ c λ, and heat conductivity, κ ≅ (1/3) cv ρn c λ, 213 and similar for diffusivity, D ≅ (1/3) c λ are approximate. More precise forms of these equations 214 vary depending on the treatment of intermolecular forces. Instead, we use a curve fit to empirical 215 data for viscosity in dry air as a function of temperature, as presented by Kadoya (1985). The 216 empirical data show, as expected, that there is no dependence on pressure in our range of 217 interest.

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For diffusivity of trace gases in air as a function of temperature and pressure we use the is warmer, and thus less dense, than outside air. In the tropopause the tube continues to cool so 252 that the "deficit" becomes smaller, but at higher altitudes, around ~25 km the amount by which 253 the pressure in the tube is higher than outside becomes substantial relative to the low outside 254 pressure -as a result the ratio at ~34 km altitude becomes a bit larger than 1. Then, during 255 descent the outside pressure increases rapidly and the inflow cannot keep up because the 256 viscosity of air at low pressure is the same as at 1 bar (see section 3). Back in the troposphere the 257 tube warms up, but much more slowly than outside air. When the tube hits the ground, it is 258 colder than ambient air temperature so that the ratio is greater than 1. If one wants to sample still higher into the stratosphere the diameter of the first 10 to 20 m at the 320 open end needs to be widened further than 6 mm diameter ( Table 1). All of this is consistent with  fill starts at ambient pressure of 4.7hPa. We also note that in this case the pressure drop inside 329 the two valves and the dryer is a large part of the overall pressure drop across the entire tube, an 330 effect that becomes more pronounced as the tube diameter gets larger.

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In these calculations I have experimented with another strategy to fill the AirCore. One could 332 launch it with both valves open, but the one in the back is closed as soon as the descent starts.

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That would decrease the amount of fill air that remains in the back. However, the difference from 334 having the back valve closed during the entire flight is negligible.

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So far the treatment of valves and the dryer has been missing from this description. As a first 338 approximation we could treat the valves as short pieces of tubing with reasonably "average" 339 internal diameter and length such that their internal volume is correct. This does not provide 340 enough flow resistance, when we compare it to differential pressure measurements made during 341 flights between the closed end of the AirCore and the outside ambient air (Fig. 8). (Eq. 5b), which is obtained from the previous expression by 373 substituting XTPR (a constant) for X. In these expressions we prefer to express the flow, instead 374 of in standard L/min as in the Swagelok brochure, as 0.04403 mol/min. This is the same, when 375 using the molecular weight of dry air (28.97 g/mol), as a mass flow of 1.276 g/min.

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In Fig. 8C we optimized both Cvand XTPR to get the best match for the calculated pressure  spreading of a packet calculated at each time step "k" is decreased as the increasing pressure 411 compresses the packet further. So the contribution of each step to the final spreading at valve closure is calculated by dividing the density during that time step by the final density in the tube. 413 We are thus accumulating the "2Dt" term of Xrms = (2Dt) 0.5 , with Taylor diffusion added: For an AirCore with (almost) uniform diameter we get mixing as in Fig. 10    For an AirCore with two sections of different diameter we see an interesting effect (Fig.10 B). 427 The air that comes in at high altitudes and ends up in the back of the tube, has to go through 428 the1/4" section first. When a packet enters the 1/8" section, its spread becomes approximately 429 four times larger, while its 2Dt accumulation term stays the same. Approximately, because the 430 inner diameters (ID) matters, not the outer (OD). To correct for the jump we add another factor 431 to Eq. 6, and we will call this corrected rms diffusion distance: In Eq. 7 dvol/dx is the increment in volume per increment in length of the tube, while (dvol/dx)ref 434 is the total volume divided by the total length, both in units of m 2 . This prevents a jump at the 30 435 m position, but more importantly, what matters for mixing is the spread relative to total mass in 436 the tube, not whether it is in the 1/4 or 1/8" section. From now on we call this configuration "1/4 437 -1/8". Fig. 10 B shows that air closer to the back has been in the 1/4" section for a shorter time, 438 and thus experienced less mixing relative to mass. When plotting mixing not as a function of position, but as a function of cumulative mass in the tube, Fig. 10 C also shows that the 1/8" mixing than air ending up at the 0.57 point, the first transition between 1/4 and 1/8". 459 We will now express the amount of spreading (in both directions -twice the rms distance) of 460 each equal-mass "packet" of air as a fraction of the total mass of air in the tube, assuming that 461 the temperature inside the tube has become uniform. If that fraction were 0.01 everywhere in the 462 tube there would be slightly less than 100 independent samples in the AirCore. Slightly less 463 because the remaining fill air in the back takes up space. Fig. 12 shows a more realistic situation.  Let us assume that after the valve has been closed there has been a half hour delay before 500 analysis starts. Therefore, additional diffusion has taken place, as shown in Fig. 14 for the case 501 1/4 -1/8 -1/4 (Fig. 11B). The 2Dt term has been increased by an amount dependent on the 502 diameter of the tube, normalized as in Eq. 7. In the upper right (panel B) the square root of the 503 sum has been taken, and then transformed into the spreading width relative to total mass in the  Often the AirCore is analyzed significantly later than 30 min. after valve closure, and the 520 measurement process itself may take half an hour. In Fig. 15 the state of mixing four hours after 521 valve closure has been calculated, and two AirCore configurations are compared. The spreading width of air "packets" near the closed end is nearly twice as large for the 1/4 -1/8 -528 1/4 case as for the 1/4 -1/8 case, and the initial fill air penetrates almost 50% further into the 529 tube. It would in most cases not be a good idea to have a wide bore section at the closed end. If 530 one waits 24 hours (6 times longer) before starting the analysis, the spreading width near the 531 closed end, centered at x = 0.9470, is 2.32 times larger than after 4 hours, not quite √6 because 532 after 4 hours the spreading that occurred during the descent still makes a small, but still 533 noticeable, contribution.
Qj is centered in the middle of segment dzj. The first factor in Qj is the average amount density 570 (ρj). The pressure change at the boundary between segments dzj-1 and dzj caused by the 571 imbalance of the flows Qj-1 and Qj is equal to that imbalance divided by the volume between the 572 mid points of dzj-1 and dzj. Adding in the pressure change due to temperature (Eq. 2) we get for 573 the change at boundary j: The first term (P/T)(dT/dt) is handled separately from the two other terms describing the amount 585 change. We write the latter two with the time step going from n to n+1 (superscript): 586 P j n+1 − P j n = � 2T j n �P j+1 n + P j n � T j+1 n + T j n r j 4 η j P j+1 n+1 − P j n+1 dz j 587 − 2T j n �P j−1 n + P j n � T j−1 n + T j n r j−1 4 η j−1 P j n+1 − P j−1 n+1 dz j−1 � t n+1 − t n 8�dz j−1 r j−1 2 + dz j r j 2 � Eq. 9

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On the right hand side we have defined the pressure differences at the end of the time step. The 589 reason is to make the solution of the matrix equation described below unconditionally stable. 590 This method has been described as "fully implicit" or "backward time" (Press, 1992). We leave 591 the pressure and temperature averages as defined at the start of the time step. They determine the 592 average amount density of the air and do not create any numerical instability. Eq. 9 can be 593 further re-arranged, for j =1 to k-1, as 594 P j n = − t n+1 − t n 8�dz j−1 r j−1 2 + dz j r j 2 � � 2T j n �P j+1 n + P j n � T j+1 n + T j n r j 4 η j dz j � P j+1 n+1 + �1 + t n+1 − t n 8�dz j−1 r j−1 2 + dz j r j 2 � � � 2T j n �P j+1 n + P j n � T j+1 n + T j n r j 4 η j dz j + 2T j n �P j−1 n + P j n � T j−1 n + T j n r j−1 4 η j−1 dz j−1 � P j n+1 596 − t n+1 − t n 8�dz j−1 r j−1 2 + dz j r j 2 � � 2T j n �P j−1 n + P j n � Eq. 10 597 This is a tridiagonal matrix equation, A•P n+1 = P n , linking the k+1 dimensional pressure vector 598 P n+1 at the end of the time step to the pressure vector P n at the start of the time step. The solution 599 is P n+1 = A -1 •P n , in which A -1 is the inverse matrix calculated by the subroutine TRISOL which is 600 the IDL version of TRIDAG described by Press et al (1992). If the tube is closed at z = 0, then in (permeability is more important than porosity) could be helpful for further improving dynamics 647 code as described in this paper, and will be especially helpful for potential revisions of sample 648 altitude assignments of older flights.

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The precision of the sample mixing estimates could be improved by laboratory measurements of 650 the pulse response of analyzers, especially when an AirCore is analyzed quickly in the field 651 because very little mixing has yet occurred for the air that came in last.

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In addition to measuring the pressure inside the tube during a flight at the closed end, one could  In cases where people want to fly AirCores without a dryer it could be helpful to study wall 657 effects. Water vapor tends to adhere tightly to many surfaces, and as anyone experienced with 658 vacuums knows, it can take a long time to pry it off the walls. One possible experiment would be 659 to inject a short pulse of wet air at one end of a dry tube and register what comes out at the other