Urgent! ! Urgent! ! ! Issues regarding the reduction in ammonia consumption during urea production! ! !

We craftsmen never explain the specifics of the problem we’re seeking help with; it’s as if asking questions on the forum is the same as asking the technicians in your workshop. :Q: Different processes result in different exhaust gas compositions; for example, there is no CO2 in the exhaust gas from medium-pressure venting during ammonia stripping, while CO2 is present in the exhaust gas from medium-pressure venting during carbon dioxide stripping. For better discussion, be more detailed when posting a topic thread: handshake :loveliness:

The moderator is right; different processes have different components. I can’t explain it specifically. I wonder what kind of process you use. This post was last edited by lxq700918 on 2009-3-21 19:11]

The amount of 100% CO2 required per ton of urea is 733.3 kg; the purity of this CO2 is 98.5%. After adding air, the oxygen content in the CO2 becomes 0.5%. By reducing the ammonia content in the exhaust gases from 10% to 5%, how many kilograms of ammonia can be saved per ton of urea? ! ! Our company uses the ammonia stripping method! !

I’ll treat this as a chemistry problem for calculation, without considering the manufacturing process or various operating conditions. With an oxygen content of 0.5%, the inert gases (such as nitrogen in air) account for 2%. 733.3 kilograms of carbon dioxide correspond to 373 standard cubic meters of carbon dioxide; approximately 7.6 standard cubic meters of inert gas need to be added to this amount. Considering only the inert gas and ammonia, with the ammonia ratio given by the moderator on a molar basis, a loss of 0.84 m³ of ammonia occurs at 10% and a loss of 0.4 m³ at 5%. This results in a savings of 0.44 m³ of ammonia, thereby reducing the ammonia consumption per ton of urea by 0.334 kilograms. (Since the amount of gas released per ton of urea produced is much greater than the 7.6 m³ of inert gas calculated, the reduction in ammonia consumption is also greater than 0.334 KG.) This post was last edited by lxq700918 on 2009-3-21 at 19:12.]

{(733.3/98.5%*1.5%)+(733.3/98.5%*0.5%/21%)}/5%=1.444Kg Hehe, it’s just a rough calculation – I’m not sure if it’s correct. Please feel free to correct me, fellow sailors; let’s learn together*. This post was last edited by Fan Zhou Wu Hu on 2009-3-21 at 18:00.]

Throughout the entire production process, oxygen, nitrogen, and other inert gases do not participate in the reaction; in other words, the same amount remains as it entered. Based on this, the calculations are as follows: 1. Oxygen addition amount: 733.3 kilograms of carbon dioxide correspond to 373 Nm³; 373×0.5÷98.5 = 1.89 Nm³ of oxygen. 2. Oxygen makes up 20.95% of air, so to maintain an oxygen content of 0.5%, 1.89÷20.95% = 9.02 Nm³ of air needs to be added. 3. When all this added air is expelled, and the ammonia content is 10%, the amount of gaseous ammonia lost per ton of urea is: 9.02×10÷90 = 1.0 Nm³ of gaseous ammonia, which is equivalent to 1.0×17÷22.4 = 0.76 kg. 4. When the ammonia content is 5%, the amount of gaseous ammonia lost per ton of urea is: 9.02×5÷95 = 0.47 Nm³ of gaseous ammonia, which is equivalent to 0.47×17÷22.4 = 0.36 kg. 5. Therefore, 0.76 – 0.36 = 0.4 kg of ammonia can be saved per ton of urea. Is this the correct way of calculation? I welcome any feedback from fellow experts.

Hehe, I learned it! Yes, it should be calculated based on volume, not mass. The percentages given are generally volume ratios. But there are two questions: 1 The total amount of inert gas released as mentioned seems to need to include the 1.5% of inert gas present in CO2 as well ； 2 The temperature and pressure of the medium-pressure vent are not at standard conditions; can 22.4 be removed directly? Please ask the sailors for help in resolving this!

Thank you for your answers! ! Thank you so much! ! !

A rough estimate is around 0.35. 5th and 7th floors are correct; on the 6th floor, oxygen was treated as a mass concentration