cbe 40 f14 midterm solutions

8/10/2019 CBE 40 F14 Midterm Solutions

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4. 50% loss of water by transpiration. (5 pts.)

Note that water is now the limiting reactant, and CO 2 is in excess. (100% excess, sincewe have twice as much as we need to match the stoichiometry of water)

If we only have 50% inputs, we would only have 54 g H2O reacting.

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Problem 3 (30 pts. total)

Notice that the stoichiometry dictates what comes out of the reactor.

A. (8 pts.) Drawing a system boundary around the entire system, we find:

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Products C and D are created in equal molar amounts. The total amount of C leaving thesystem must equal the total amount of D leaving the system. A and B must also be

present in equal molar amounts throughout the system. Then we find:

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Note that this is 16.7% A, 16.7% B, 67% C.

Consider reaction conversion:

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This can also be interpreted as:

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Thus we can related the flow of these components in the specified streams.

B. (12 pts.) Note that all the D that comes out of the separator must come from stream 3.

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Since there was 50% conversion of the reactor (the moles of unreacted A and B areequal to the moles of A and B that reacted to form C and D):

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The mole fractions in streams 3, 5, 6, and 7 must be the same. Based on thecomposition from stream 6,

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C. (5 pts.) Stoichiometry and the overall mass balance require that the component flowrates in the purge do not change. If the reactor has 70% conversion but still produces80 mol/min of D, less A & B are required in Stream 2 and less A & B exit the reactorin Stream 3. The amounts of A & B that exit the system (in the purge) do not change,so the amounts of A & B in the recycle are smaller than in the system with 50%reactor conversion. The ratio of C to A (or C to B) is the same in Streams 3, 5, 6, and7, and is also the same as in the system with 50% reactor conversion. Since the flowrates of A & B in the recycle have been reduced, the flow rate of C in the recycle willalso be reduced compared to the system with 50% reactor conversion. In summary,the flow rates of the components in the purge will not change, but the flow rates ofthe components in the recycle will decrease, when the reactor conversion is changedfrom 50% to 70%.

D. (5 pts.) With 50% reactor conversion, it is not possible to produce 100 mol/min of Din Stream 4 without adding additional units. If 100 mol/min of D were to exit thesystem in Stream 4, then Stream 6 would have 100 mol/min of C and 0 mol/min of A& B. Since the recycle has the same composition as the purge, no A or B would getrecycled to the reactor. The reactor would only receive A & B from Stream 1, at 100mol/min each. With 50% conversion in the reactor, 100 mol/min of D could not beachieved with A & B inputs of only 100 mol/min each.


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Problem 4 (20 pts)

A detailed, workable design is shown on the following page. This design washes thestalks to remove dirt prior to crushing the stalks, so that sugar is not lost in the dirty watereffluent from the washer. The design also includes two approaches to recycle energy (forextra credit): (1) transfer of heat energy from the combustion of bagasse to theevaporator, and (2) a heat exchanger to cool/condense the steam from the evaporator and

pre-heat the input to the evaporator. Points were deducted for the following commonmistakes:

If the stalks were crushed prior to washing, some of the sugar will be lost in thewashers dirty water effluent because sugar dissolves in water.

If steam from the evaporator was not recycled (also requires a condenser). If the molasses and sugar crystals were not separated using a centrifuge. A

centrifuge is used because the density difference between molasses and sugarcrystals is small.

If the mud was not removed with a settler. A settler is used because the densitydifference between mud and sugar water is large.

If essential unit operations or streams were missing. If extra unit operations made the design unnecessarily inefficient. If units or streams were not properly labeled. If flow chart conventions were not followed.


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cbe 40 f14 midterm solutions

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