Example Bean Oil In the production of bean oil, beans containing 13.0 wt% oil and 87.0% solids are ground and fed to a stirred tank (the extractor) along with a recycled stream of liquid n-hexane. The feed ratio is 3 kg hexane/kg beans. The ground beans are suspended in the liquid, and essentially all of the oil in the beans is extracted into the hexane. The extractor effluent passes to a filter. The filter cake contains 75 wt% bean solids and the balance bean oil and hexane, the latter two in the same ratio in which they emerge from the extractor. The filter cake is discarded and the liquid filtrate is fed to a heated evaporator in which the hexane is vaporized and the oil remains as a liquid. The oil is stored in drums and shipped. The hexane vapour is subsequently cooled and condensed, and the liquid hexane condensate is recycled to the extractor. (a) Draw and label a flowchart of the process, do the degree-of-freedom analysis, and write in an efficient order the equations you would solve to determine all unknown stream variables, circling the variables for which you would solve.
Strategy In this system, there are three process units and a mixing point where the recycle stream joins the feed. Balances can be written around the overall system and each of these subsystems. Why isn't a balance around the condenser helpful?
Following is an outline of how degree-of-freedom analyses might be used to plan the solution for all unknown system variables. (Fill in the missing information.) DEGREE-OF-FREEDOM ANALYSIS: OVERALL UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION
+ 4 unknowns - 3 balances 1 DOF
_ , _ , _ , _ __ 5 _____ ) _____
Don't start with overall balances DEGREE-OF-FREEDOM ANALYSIS: EXTRACTOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION
-
balances 0 DOF
Start here. In subsequent analyses, treat m2, x2, and y2 as known
+
DEGREE-OF-FREEDOM ANALYSIS: FILTER UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION unknowns
-
balances
-1
Do this system next, m3, y3, m4, and y4 are then known.
0 DOF
+
DEGREE-OF-FREEDOM ANALYSIS: EVAPORATOR UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION unknowns
-
balances 0 DOF
+
DEGREE-OF-FREEDOM ANALYSIS: MIXING POINT UNKNOWNS AND INFORMATION JUSTIFICATION/CONCLUSION unknowns
balances 0 DOF
Solution Here are the equations with the variables to be determined circled. Fill in the blanks, and notice how the equations are written to minimise the number that must be solved simultaneously.
Mass balance on extractor: (300 + 87 + 13) kg = m2 S balance on extractor: 87 kg S = m2 x2 Oil balance on extractor: Oil/hexane ratio equality (filter cake & filter inlet):
Write down mass balance for the below components and solve them simultaneously: Mass balance on filter: S balance on filter: Oil balance on filter:
Hx balance on evaporator: Oil balance on evaporator:
Mass balance on mix point: We now have a complete and efficient set of equations for solving the flowchart. Do the algebra to determine the values of the stream variables listed below.
m2 =
kg
x2 =
kg S/kg y2 =
m3 =
kg
y3 =
kg oil/kg
m4=
kg
y4 =
kg oil/kg
m5 =
kg
m6 =
kg
m1 =
kg
kg oil/kg
(b) Calculate the yield of bean oil product (kg oil/kg beans fed), the required fresh hexane feed (kg C6H14/kg beans fed), and the recycle to fresh feed ratio (kg hexane recycled/kg fresh feed).
Solution The answer to Part (b) is trivial since all the stream variables are known. Remember that the total bean feed is 87 + 13 = 100 kg beans. Using the values you determined above, calculate the ratios requested and record them in the following box.
These ratios deserve a few words. Virtually all processes are characterised by those who work on them by using a few key parameters—often yields, feed ratios, and recycle ratios. Since we cannot possibly keep all of the various stream flows and mass fractions in our heads, the use of a few key indicators of process stability and productivity are used. In chemical plants that produce hundreds of millions of kg of product per year, fractions of a percentage improvement in product yields can result in millions of pounds in increased annual income. Likewise, tiny reductions in the amount of solvent used to process a pound of product or small reductions in the amount of material that must be recycled can have huge environmental and financial consequences due to the shear tonnage of material being handled in the plant. The repeated requests for such calculations in the text are true to form and very realistic. Practice calculating and understanding process ratios; before long, you may be asked to "think" in terms of such key indicators.