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High fructose corn syrup (HFCS) is a sweetener made from corn syrup (100% Glucose). In a particular process that is considered here, the feed (streaml)

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High fructose corn syrup (HFCS) is a sweetener made from corn syrup (100% Glucose). In a particular process that is considered here, the feed (streaml) containing 100% Glucose (G) is mixed with a recycle stream (8) and fed to an isomerization reactor where the glucose (G) is converted to fructose (F). The reaction can be simplified as Glucose + Fructose The product of the isomerization reaction (Stream 3) contains 42% Fructose. This product stream is split into two, namely streams 4 and 7. Stream 4 is sent to a chromatographic separation unit that results in two product streams (5 and 8). Stream 5 contains 90% Fructose, while stream 8 contains 10% Fructose. Stream 5 is blended with the bypass stream 7 in a mixing tank to provide the product HFCS-55 (Stream 6) which contains 55% Fructose. Stream 8 from the chromatographic column is recycled to the isomerization n reactor. A particular plant has 100 kg/h of mixture flowing into the chromatographic unit as shown in the Table. In this problem the percentages refer to wt %, and we assume that there are just two components, i.e., Fructose and Glucose. Hence a stream containing 30% Glucose implies that the other 70% is Fructose. Questions: 1. Show the appropriate mass balances (for each mass balance, provide the envelope and the equations) and complete the stream table. Showing the envelope around which, a balance is drawn, clearly indicating the species for which a balance is drawn and showing the numerical values for each step is essential. [25 marks) 2. What is the per-pass conversion of glucose for the isomerization reactor? [5 marks) 3. What is the overall conversion of glucose for the entire process? [5 marks] 4. What is the extent of reaction in the isomerization reactor? [5 marks] 5. Let us say you can directly produce a 55% fructose stream from the separation unit, i.e., you can operate the process in such a way that stream 5 can directly produce 55% fructose. Such an operation will eliminate the need for the bypass stream (stream 7). Which operation mode will you prefer? Provide a short rationale. [5 marks]

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