Use an Austrian instrument for carbon monoxide level analysis

2CO + O2 = 2CO2 CH4 + 2O2 = CO2 + 2H2O. From these equations, it can be seen that 2 volumes of CO react with 1 volume of O2 to produce 2 volumes of CO2; the total volume after the reaction is 2/3 of the total volume before the reaction. 1 volume of CH4 reacts with 2 volumes of O2 to produce 1 volume of CO2 and 0 volume of water. The total volume after the reaction is 1/3 of the total volume before the reaction. It is actually calculated based on the change in volume before and after the reaction. I’m not sure if the above analysis is helpful to you.

The calculation principle was already explained upstairs; I just don’t know what the specific process used by the original poster is How much sample is needed? The process for conducting a complete analysis of semi-water gas is as follows: 1. Take 100 ml of the sample gas. 2. Absorb CO2 using a KOH solution; volume measured as V1. 3. Absorb O2 using a potash solution; volume measured as V2. 4. Absorb CO using an ammoniacal copper(II) chloride solution, and then remove any volatile NH3 with H2SO4; volume measured as V3. 5. Take 25 ml of the remaining gas and mix it with 75 ml of air to initiate an explosion; volume measured as V4. 6. After the explosion, absorb the remaining CO2; volume measured as V5. The calculation formulas are as follows: CO2%: V1; O2%: V2 – V1; CO%: V3 – V2; CH4%: (V5 – V4) * (100 – V3) / 25; H2%: 2/3 * (100 – V3) / 25; N2%: 100 – (sum of the previous values). This post was last edited by zjs0536 on 2009-3-23 at 12:22

The above operations are fine. The combustible gases that should remain after the explosion are H2 and CH4. The volume changes of hydrogen are as follows: 2H2 (2 volumes) + O2 (1 volume) = 2H2O (0 volumes). All of the CO2 absorbed after the explosion is produced by CH4. The amount of CH4 can be calculated using the volume reduction formula. Then, the amount of H2 is calculated based on the reduced total volume and the reduced methane volume.

Thank you all for your participation. In the past, I only focused on the analysis results and never considered the analysis process; this time, I paid special attention to it, and the details are as follows: 1. Take 100 ml of the transformed gas, feed it into a NaOH absorption bottle to allow full absorption, then measure the remaining volume and denote it as V1. 2. Take 25 ml from the remaining gas and add 75 ml of air to it; feed this mixture into an explosion chamber, and after the explosion, measure the remaining volume and denote it as V2. 3. Feed the gas back into the NaOH absorption bottle for full absorption again, then measure the remaining volume and denote it as V3. The concentration of CH4 is analyzed from time to time, and it is currently 0.6%. The main components of the gas entering the explosion chamber are as follows: H-explosion% = H% * 25/V1, CH4-explosion% = CH4% * 25/V1, CO-explosion% = CO% * 25/V1. The main reaction equations for explosions are: CO + 1/2 O2 = CO2, CH4 + 2 O2 = CO2 + 2 H2O, H2 + 1/2 O2 = H2O. During the first absorption step, CO2 from the transformed gas is absorbed, so CO2% = 100 – V1. During the second absorption step, CO and CH4 are absorbed, along with the CO2 generated in the explosion chamber. From the equations, it can be seen that one volume of CO produces one volume of CO2, and the same applies to CH4. Therefore, V2 – V3 = CO-explosion% + CH4-explosion% = CO% * 25/V1 + CH4% * 25/V1. Hence, CO% = (V2 – V3) * V1/25 – CH4%. From the reaction equations, it is clear that when carbon monoxide explodes, it produces one volume of CO2 while consuming 1/2 volume of O2. When methane explodes, it also produces one volume of CO2 but consumes 2 volumes of O2. When hydrogen explodes, it consumes 1 volume of H2 and 1/2 volume of O2. Thus, 100 – V2 = 1/2 * CO% * 25/V1 + 2 * CH4% * 25/V1 + 3/2 * H2% * 25/V1. By substituting the value of CO% obtained earlier, we get: H2% = 1/3 (200 + V3 – 3V2) * V1/25 – CH4%. The remaining gases are denoted as N2 + Ar

The above calculation methods have been compared with those of online analyzers on multiple occasions, and they are generally accurate; however, there are many sources of error in this analysis process. 1. The methane content remains relatively stable, but it is by no means a constant value. 2. There are deviations inherent in the Oxymeter itself. 3. There are errors in the readings taken by the analyst. 4. Carbon dioxide in the air is not taken into account in the calculations; its concentration is only 0.03, so it is likely to have little impact on the results. 5. The absorption reaction and the combustion reaction do not necessarily proceed to completion. Nevertheless, the analysis results are still fairly accurate when used to reflect changes in the composition of the gases.

Our sampling process is the same as that on the 3rd floor. If you are also using this type of Oxygen analyzer, and if gases can be absorbed in various solutions, then the volume fraction of the gas should equal the volume decrease as the gas passes through the corresponding solution!

1. Deviations in the instrument and in the readings are inevitable, as the scale is very fine, and it is necessary to wait for the liquid level to stay steady for more than half a minute before taking a reading! 2. As for the impact of CO2 in the air, it can be calculated at 0.03%, so the impact is minimal! 3. Judgment of complete absorption: If the readings after two absorptions are identical, it indicates that absorption is complete; otherwise, continue absorbing! As for whether the explosion was complete, it is generally believed that a complete reaction occurred, which indicates that there was sufficient oxygen available; if there are doubts, two more explosions can be attempted! (I haven’t tried it myself)