Calculation of standard gas volume----Discussion is welcome; it remains valid

I think option 2 is right; solid particles cannot be included in the conversion using the standard volume formula, and can be ignored.

The problem lies with solid particles and water vapor. Solid particles can be specified by their dust content. Standard volume actually refers to conditions of fixed temperature and pressure. It is water under standard conditions, and it shouldn’t be taken into account. In fact, posing the question in this way is unscientific. As we all know, matter exists in three states: gas, liquid, and solid. Has anyone ever asked about the standard volume of iron? Because as we all know, it is in a solid state under standard conditions. Therefore, when discussing water vapor, standard volume should not be considered. Conversely, only substances that are gaseous under standard conditions can have their standard volume determined.

Agree with the second floor’s opinion; choose answer 2. Regarding the comments on the 3rd floor, I would like to share my personal view. Based on the properties of substances, solids can also be transformed into gases; this is determined by the vapor pressure of the substance. When calculating the amount of gas, we take into account the fact that the gas state is greatly influenced by temperature and pressure. To facilitate unified calculations and comparisons, data are provided under standard conditions. Similarly, when designing steam transmission systems in factories, standards for volume flow rates are used as a basis. So it is not only substances that are gaseous under standard conditions for which we consider their standard volume. Haha, let’s talk about the comments from the 5th floor: For gases, what is the difference between their flow rate in NM3/h and Kg/h? Looking at the formula: (standard cubic meters/h) ÷ 22.4 standard cubic meters/Kmol × molecular weight in kg/Kmol = kg/h. In fact, it’s just a matter of a conversion factor. This post was last edited by zzlxxm on 2008-4-10 at 13:04

Disagree with the explanation on floor 4. When discussing vapor flow rate properly, the unit should be kg/h. The fact that many people, including some experts, talk about the standard volume of steam shows a lack of clarity in understanding this concept. This post was last edited by Huahua Jianghu on 2008-4-10 11:51.]

The discussion was really lively! I’m joining in the fun too! 1. If it is purely a conceptual calculation problem, that is, to find the volume of a substance under standard conditions, I agree with the opinion from the third floor; solids and water vapor should indeed not be included. 2. If this calculation is intended to be used as design parameters for a cyclone separator in engineering design, I recommend adopting the suggestions for floor 2 and floor 4, including the amount of water vapor. After all, almost no cyclone separator operates under standard conditions. :D

I don’t know where the friend on the fourth floor went to school. Under standard conditions, water vapor is already water; so where does the idea that 1 mole of water occupies 22.4 liters come from? 1 mole being 22.4 L applies only to ideal gases. I expressed my opinion because I had participated in discussions with professionals in the fields of calculation and instrumentation regarding control valves on wet chlorine pipelines. High-temperature wet chlorine gas contains a high amount of water vapor. For the instrumentation field, when dealing with ordinary gases, the flow rate of the valve is calculated by converting it to standard conditions. Steam uses another algorithm, but many people are not very clear about how to calculate it for mixed gases. Regrettably, any formula has conditions under which it can be applied. If it’s already in a liquid state under those conditions, isn’t it too rigid to still use the ideal gas equation? Furthermore, I also disagree with the compromise approach proposed on floor 6. In fact, there’s no need to convert everything to standard conditions at all. It is sufficient to clearly indicate the working status. Why use the standard state? It is to make it convenient for manufacturers to fill in the data on the samples.

Without a doubt, the answer should be 3. As for the opinion from the 4th floor, I can’t agree with it: funk:

I think there might be a problem with the question posed by the original poster – under standard conditions, water vapor? However, considering what the original poster meant, I think 3 should be chosen. I believe that both water vapor and solids are included in the volume of air.

I just saw this post, so let me share some of my thoughts: Conditions like these are commonly encountered in some dehumidification processes, in the coolers located after air compressors. The issue raised by the original poster isn’t that designers insist on converting actual operating conditions to standard conditions; rather, in most cases, the gas flow rates provided on-site are based on the values indicated on the compressor’s nameplate, and those values represent flow rates under standard conditions. Therefore, during the design of actual equipment, it’s necessary to convert these standard-condition flow rates into those corresponding to the actual operating conditions. This can also be considered an answer to the discussions by those who posted above. Now, let me share my opinion on the original poster’s question: First of all, I wouldn’t choose any of the three options suggested by the poster. I think this issue should be analyzed as follows: Solid particles are relatively very small compared to the overall volume of gas, so their volumetric flow rate can be ignored. Both water vapor and air are gases under these conditions, so both must be included in the total flow rate. However, it’s not possible to simply multiply the values from the options given by the poster by “0.0224 Nm3” – after all, water vapor is present, so it can’t be treated entirely as if it were air. In my designs, I usually follow this approach as a reference: I determine the corresponding partial pressures for water vapor and air based on their proportional composition, then use the equation of state to convert these values into their respective flow rates under standard conditions, and finally add them together.