Reference no: EM133025065
EXAMPLES OF ENERGY BALANCE
Question 1. In the production of soya bean oil, the oil is extracted from the original beans using hexane as solvent. The soya beans contain 12% w/w oil, the rest being insoluble bean waste. 500 kg min-1 of soya beans are to be extracted. The bean waste leaving the process (known as raffinate) contains 0.5% w/w oil and 2% w/w hexane. The extract is 22% w/w oil and the rest is hexane. No bean waste leaves with the extract.
The extract is then passed to a distillation column which separates the mix into pure hexane, which is recycled back to the extractor for reuse, and 98.5% oil as final product. Hexane losses are made up by continuous make up from the hexane storage system.
(a) Draw a block diagram of the process incorporating all theinformation given in the question.
(b) Calculate (in kg min-1):
(i) the amount of extract leaving the extractor
(ii) the amount of hexane required in the extraction stage
(iii) the amount of both streams leaving the distillation stage
(iv) the amount of hexane make-up required to ensure the process is steady state.
Check your answer by carrying out an additional balance not already used in your calculation.
Question 2. 180 kmol h-1 of a three component mixture is made up of 60 mol % 'A', 25 mol % 'B' and 15 mol % 'C'. Component 'A' is the most volatile and 'C' the least volatile. The mixture is to be separated by continuous distillation in two columns so that 80% of 'C' in the feed is removed as bottom product in the first column as a 90 mol % solution with B.
The top product from the first column will be separated in the second column to give a top product here of 95 mol % 'A' and a bottom product of 85 mol % 'B'.
Determine the flows from the two columns and the mol % 'A' in bottom product from column 2.
(Assume there is to be no 'A' in the bottom product of column 1 and no 'C' in the top product of column 2.)
Question 3. The product from a reactor contains 60% w/w A, 25% w/w B and 15% w/w C. The final saleable product is required to contain 90% A.
The mixture can be purified in a distillation column but under the conditions possible in the column the product produced is 100% A with the bottom product containing no A. It is possible to mix the reactor products and distillation product.
Calculate:
(i) the amount of reactor product that could by-pass the distillation system to give the maximum amount of saleable product at minimum cost
(ii) the amount and composition of 'waste' material.
Question 4. 500 kg h-1 of a 10% w/w solution of a compound X(OH)2 is to be treated using a 5% excess of 20% w/w solution of sodium carbonate (Na2CO3) to precipitate the X as its carbonate according to the balanced reaction equation:
X OH( )2(aq) + Na CO2 3(aq) → XCO3( )s + 2NaOH(aq)
All the solid is filtered off but some solution remains with the solid such that the solution forms 4% w/w of the total mass removed.
The filtered liquid product is then neutralised with 2 molar hydrochloric acid according to the reactions:
Na CO2 3(aq) +2HCl(aq) → 2NaCl(aq) + H O2 ( )l + CO2( )g
NaOH(aq) + HCl(aq) → NaCl(aq) + H O2 ( )l
Calculate:
(i) the amount (kg h-1) of sodium carbonate solution required
(ii) the amount (kg h-1) and composition (% w/w) of the liquid product after filtration
(iii) the amount (kg h-1) and composition (% w/w) of the solid product after filtration
(iv) the amount of acid (in litres per hour) required for neutralisation.
Data: Relative atomic mass of X = 40, O = 16, H = 1, Na = 23, C = 12, Cl = 35.5.
Question 5. A chemical process is represented by the equation and diagram below:
A + 3B 2C
The feed to the process (F) contains 'A' and 'B' in stoichiometric proportions. It also contains 0.5 mol % of impurity 'I'.
The mixture (X) entering the reactor (i.e. feed and recycle) contains 'A' and 'B' in stoichiometric proportion but the level of impurity must not exceed 4 mol%.
In the reactor, the reaction goes to 60% completion. Pure product (C) is removed completely in the separator. The remainder of the reaction products (i.e. unconverted 'A' and 'B' together with any impurity) is recycled. In order not to exceed the maximum quantity of impurity in the reactor feed, some of the recycle is purged (P), the remainder (R) going back to the reactor.
On the basis of 100 kmol of 'A' plus 'B' entering the reactor at (X), carry out mass balances to find the flows of:
(i) feed (F)
(ii) recycle (R) (iii) purge (P).
Attachment:- EXAMPLES OF ENERGY BALANCE.rar