Solution to gas weighing challenge
- PDF / 122,912 Bytes
- 3 Pages / 595.276 x 790.866 pts Page_size
- 83 Downloads / 221 Views
ANALYTICAL CHALLENGE
Solution to gas weighing challenge Lukas Flierl 1 & Olaf Rienitz 1 & Axel Pramann 1 Published online: 26 June 2020 # Springer-Verlag GmbH Germany, part of Springer Nature 2020
This Analytical Challenge sets out to develop an equation to obtain mass of gases added to cylinders while accounting for buoyancy corrections [1]. The gas containers have tare masses, e.g., ms, encasing the inner volume Vgas. This volume contains an amount of gas, e.g., mA. The total mass of the container and the gas ms + A is: 1−
ρair;1 ρcal
0 0 msþA ¼ ms þ mA ¼ K sþA msþA ¼ ρ msþA air;1 ρsþA
ð1Þ
where Ks + A is the buoyancy correction factor, m′sþA is the balance reading, ρair, 1 is the density of air at the time t1 when the mass is recorded, ρcal is the density of the calibration mass, and ρs + A is the density of the container filled with gas. The latter can be expressed as follows: ρsþA ¼
ms þ mA V s þ V gas
ð2Þ
The numerator is the sum of the mass of the container ms and the mass of the contained gas mA. This is divided by the sum of the inner volume, Vgas, which is the volume occupied by the gas, and the outer volume, Vs, which is the volume occupied by the container walls (and valves). If mA becomes zero (the container is evacuated), Eq. 2 can be rewritten to yield the density ρs of the empty container ρs ¼
ms V s þ V gas
ð3Þ
This article is the solution to the Analytical Challenge to be found at https://doi.org/10.1007/s00216-019-02168-4 * Olaf Rienitz [email protected] 1
Physikalisch-Technische Bundesanstalt, Bundesallee 100, 38116 Braunschweig, Germany
The inverse of ρs can be written as: V s þ V gas V gas 1 1 ¼ ¼ þ ρs ρm ms ms
ð4Þ
Here, ρm is the combined density of all metallic parts the container is made from (the walls, the valve etc.). Equation 4 and Equation 1 can be used to find the mass of the evacuated container at time t0 (mA = 0): ρair;0 1− ρ 0 cal ms;0 ms ¼ V gas 1 1−ρair;0 þ ρm ms
ð5Þ
where m′s, 0 is the balance reading of the empty container “s.” We assume that the densities of the two evacuated containers are the same within the limits of their uncertainties. We also assume that the ambient conditions (air temperature, humidity, and pressure) do not change significantly during weighing the two containers (for a better understanding of the applied weighing procedure, please see the figure 1 in the challenge [1]). At the time t0 (before container “s” is filled with gas), both containers “s” and “r” are evacuated and weighed which provide the following equations: 0 ms ¼ K s;0 ms;0
ð6Þ
mr ¼ K r;0 mr;0
ð7Þ
0
The masses ms and mr are the (tare) masses of the to-be filled container “s” and the empty reference container “r,” respectively, whereas m′r, 0 and m′s, 0 are the balance readings. After filling the container “s” with gas A and weighing it (at the time t1), the total mass of the container “s” is the sum of ms and mA: msþA ¼ ms þ mA
ð8Þ
The tare mass ms at time point t1 is 0 ms ¼ K s;1 ms;1
ð9Þ
3958
Flierl L. et al.
Also, container “r” is weighed at
Data Loading...
