The feed rate and sulfur content are known; now the sulfur content needs to be changed. What is the calculation process?
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A reforming unit has a reforming feed rate of 75 t/h, with a sulfur content of 0.05 wtppm. It is now required that the sulfur content in the feed be 0.25 wtppm. Calculate how many milliliters (ml) of DMDS need to be injected per minute Answer: 0.345 milliliters per minute
This post was last edited by ylb913 on 2015-10-24 at 19:32. 1. DMDS is dimethyl disulfide; its molecular formula is C2H6S2, with a molecular weight of 94 (24+6+64). The sulfur content constitutes 64/94 = 68.1% (by mass). Its specific gravity is 1.065 grams per milliliter (at 20 degrees). The sulfur content in 1 milliliter of DMDS = 1.065*0.681 = 0.725 grams per milliliter. 2. “The feed rate for reprocessing is 75 t/h, with a sulfur content of 0.05 wtppm; now the desired sulfur content in the feed is 0.25 wtppm.” This increase in sulfur content is achieved by adding DMDS, so the increase in sulfur content is 0.25 – 0.05 = 0.2 ppm (by mass). 3. 75 t/h = 75/60 = 1.25 t/minute. 4. 1 ppm is one part in 10 to the power of 6; 1 t = 1000 kg = 1,000,000 g = 10 to the power of 6 grams ； Kilograms correspond to ‘ rise’, while milliliters correspond to grams – unit conversion is quite complicated. ppm (by mass) corresponds to ‘grams per ton’. 5. The mass of sulfur that needs to be added is 1.25 t/min * 0.2 ppm = 0.25 g/min. 6. 0.25 g/minute divided by 0.725 g/milliliter = 0.345 milliliters/minute.