How to calculate the methanol off-gas volume!
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Calculate the release amount of methanol off-gas! Our facility produces 200,000 tons of methanol per year from coke oven gas. The gas composition is as follows! Composition of raw coke oven gas: Technical specifications for raw materialsSerial No. | Name | Specification
1 | Coke oven gas | Composition: V%
H2: 58.0
CO: 6.2
CO2: 2.2
CH4: 26.0
N2: 4.5
CmHn: 2.5
O2: 0.6
H2S: 50 mg/Nm3
Organic sulfur: 250 mg/Nm3
1. Fresh gas: 40°C, 2.1 MPa(g), 93,509.43 Nm3/h
Components (Nm3/h): VOLUME/%
H2: 67,028.14 → 71.68%
CO: 15,389.12 → 16.46%
CO2: 7,265.89 → 7.77%
CH4: 7,740.3 → 0.83%
N2: 27,249.6 → 2.91%
H2O: 3,272.9 → 0.35%
Total S: ≤0.1 ppm; Total volume: 935,509.43 Nm3/h → 100.00%
2. Recycled gas (at the outlet of the separator): 40°C, 5.5 MPa(g), 515,203.48 Nm3/h
Components (Nm3/h): VOLUME/%
H2: 39,895.29 → 77.44%
CO: 29,150.24 → 5.66%
CO2: 19,418.08 → 3.77%
CH4: 14,028.44 → 2.72%
N2: 50,398.15 → 9.78%
H2O: 2,466.9 → 0.05%
CH3OH: 3,004.59 → 0.58%
Total S: ≤0.1 ppm; Total volume: 515,203.48 Nm3/h → 100.00%
3. Gas entering the tower (fresh gas + recycled gas): 220°C, 5.9 MPa(g), 608,588.07 Nm3/h
Components (Nm3/h): VOLUME/%
H2: 465,985.38 → 76.57%
CO: 44,538.02 → 7.32%
CO2: 26,677.03 → 4.38%
CH4: 14,801.98 → 2.43%
N2: 53,122.93 → 8.73%
H2O: 4,581.3 → 0.06%
CH3OH: 3,004.60 → 0.49%
Total S: ≤0.1 ppm; Total volume: 608,588.07 Nm3/h → 100.00%
4. Gas exiting the tower: 225°C, 5.7 MPa(g), 569,023.21 Nm3/h
Components (Nm3/h): VOLUME/%
H2: 420,444.86 → 73.89%
CO: 30,731.25 → 5.40%
CO2: 20,701.37 → 3.64%
CH4: 14,801.98 → 2.60%
N2: 53,122.93 → 9.34%
H2O: 6,433.79 → 1.13%
CH3OH: 22,787.03 → 4.00%
Total S: ≤0.1 ppm; Total volume: 569,023.21 Nm3/h → 100.00%
I think nitrogen should be used for the calculations. The nitrogen entering the tower equals the circulating nitrogen plus the nitrogen that is vented. The amount of nitrogen vented is equal to the vent volume multiplied by the nitrogen concentration in the circulating gas. The calculated vent volume is 29,173 Nm3/h; of course, this does not take into account the amount of nitrogen dissolved in the crude methanol. At 50°C and 5 MPa, the value is 9 cubic centimeters per gram of methanol
I think it should be calculated this way: as the inert gases in the synthesis loop, they should be methane + nitrogen; the total amount of inert gases in the off-gases should equal the total amount of inert gases in the fresh gas. The ratio of the amount of purge gas to the amount of fresh gas can be calculated from the test data. (0.83+2.91)/(2.72+9.78) = 29.992% By treating the synthesis system as a stable system with an equilibrium between inflow and outflow, it is possible to calculate this value: the total amount of inert gas introduced with the fresh gas equals the total amount of inert gas removed through the off-gas. That’s incorrect; please correct me.
The amount of inert gas that enters is equal to the amount that exits; for now, consider the system to be in equilibrium. What enters is not only fresh gas but also the hydrogen-rich gas recovered from behind the membrane; the gas that exits consists of the gas going through the membrane, the flashed fuel gas, and the vent gas sent to the flare system. It is calculated based on the equilibrium of nitrogen, methane, and argon in the system.
Ha’s consignment point card welder’s Republican party has received the goods