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# Purdue - CHE 20500 - Class Notes

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Purdue - CHE 20500 - Class Notes

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ChE 20500 Exam 1 – Fall 2015

Problem 1
B = benzene, C
6 H 6   T = toluene, C 7 H 8   X = xylene, C 8 H 10     Points for part a (5 pts total):
1 pt each for correct mix point, DC #1, DC #2, split point (4 pts)
1 pt for correct symbology (y for vapor, x for liquid,
𝒏̇ for flow).
Part B (Degrees of Freedom Analysis Around DC #1)
5
Unknowns ( 𝑛̇ 3 , 𝑛̇ 4 , 𝑦 4𝐵 , 𝑦 4𝑇 , 𝑦 4𝑥 1 pt  -3   Balances (Pick 3 of 4: X, B, T, total) 1 pt  -2  Extra Equations (90% X recovery in bottoms product #1, sum of mole fractions is 1 in
stream 4) 2 pts  (1 for each)
= 0 D.O.F.

4 pts total for part (b)

Part C (Molar Flow Rates out of DC #1)

Start from the extra condition of 90% xylene recovered:
𝑛̇ 3 𝑥 3𝑋 = 0.9𝑛̇ 2 𝑥 2𝑋   (5 pts 𝑛̇ 3 = 0.9𝑛̇ 2 𝑥 2𝑋 𝑥 3𝑋 = 𝟕𝟑. 𝟓 𝒎𝒐𝒍 𝒉 −𝟏   (1 pt)
Then, a total mole balance on DC #1 can be written:
𝑛̇ 2 = 𝑛̇ 3 + 𝑛̇ 4   (5 pts 𝑛̇ 4 = 𝑛̇ 2 − 𝑛̇ 3 = 𝟏𝟑𝟗. 𝟓 𝒎𝒐𝒍 𝒉 −𝟏   (1 pt)    12 pts possible for part (c)

Part D (Calculation of Recycle Flow Rate, Bottoms #2 Composition)

It is now necessary to calculate mole fractions in stream 4 to solve the remainder of the problem.

Benzene Balance around DC #1:
𝑛̇ 2 𝑥 2𝐵 = 𝑛̇ 4 𝑦 4𝐵   𝑦 4𝐵 = 𝑛̇ 2 𝑥 2𝐵 𝑛̇ 4 = 0.876 𝑚𝑜𝑙 𝐵/𝑚𝑜𝑙
Xylene Balance around DC #1:
𝑛̇ 2 𝑥 2𝑋 = 𝑛̇ 3 𝑥 3𝑋 + 𝑛̇ 4 𝑦 4𝑋   𝑦 4𝑋 = 𝑛̇ 2 𝑥 2𝑋 − 𝑛̇ 3 𝑥 3𝑋 𝑛̇ 4 = 0.053 𝑚𝑜𝑙 𝑋/𝑚𝑜𝑙
Sum of mole fractions in stream 4 is 1:
𝑦 4𝑇 = 1 − 𝑦 4𝑋 − 𝑦 4𝐵 = 0.071 𝑚𝑜𝑙 𝑇/𝑚𝑜𝑙
5 pts awarded for using component balances to obtain the mole fractions of stream 4.

Do a benzene balance on DC #2; remember that mole fractions in streams 6, 7, and 8 are equal:
𝑛̇ 4 𝑦 4𝐵 = 𝑛̇ 6 𝑦 6𝐵   𝑛̇ 6 = 𝑛̇ 4 𝑦 4𝐵 𝑦 6𝐵 = 130.1 𝑚𝑜𝑙 ℎ −1
5 pts awarded for obtaining the flowrate of stream 6.

Now, perform a total mole balance at the split point:
𝑛̇ 6 = 𝑛̇ 7 + 𝑛̇ 8     𝑛̇ 8 = 𝑛̇ 6 − 𝑛̇ 7 = 𝟏𝟎𝟎. 𝟏 𝒎𝒐𝒍 𝒉 −𝟏   (2 pts)
To calculate the mole fractions in the bottom stream, first perform a total mole balance on DC#2:
𝑛̇ 4 = 𝑛̇ 5 + 𝑛̇ 6   𝑛̇ 5 = 𝑛̇ 4 − 𝑛̇ 6 = 9.4 𝑚𝑜𝑙 ℎ −1
Now, perform a xylene balance on DC#2:
𝑛̇ 4 𝑦 4𝑋 = 𝑛̇ 5 𝑥 5𝑋     𝑥 5𝑋 = 𝑛̇ 4 𝑦 4𝑋 𝑛̇ 5 = 𝟎. 𝟕𝟖𝟏 𝒎𝒐𝒍 𝑿/𝒎𝒐𝒍  (1 pt)  𝑥 5𝑇 = 1 − 𝑥 5𝑋 = 𝟎. 𝟐𝟏𝟗 𝒎𝒐𝒍 𝑻/𝒎𝒐𝒍  (2 pts)
15 pts possible for part (d)

Part E (Composition and Flow Rate of the Fresh Feed)

𝑛̇ 1 + 𝑛̇ 8 = 𝑛̇ 2   (5 pts 𝑛̇ 1 = 𝑛̇ 2 − 𝑛̇ 8 = 𝟏𝟏𝟐. 𝟗 𝒎𝒐𝒍 𝒉 −𝟏   (1 pt)
Perform a xylene balance around the mixing point:
𝑛̇ 1 𝑥 1𝑋 = 𝑛̇ 2 𝑥 2𝑋     𝑥 1𝑋 = 𝑛̇ 2 𝑥 2𝑋 𝑛̇ 1 = 𝟎. 𝟔𝟓 𝒎𝒐𝒍 𝑿/𝒎𝒐𝒍  (1 pt)
Perform a benzene balance around the mixing point:
𝑛̇ 1 𝑥 1𝐵 + 𝑛̇ 8 𝑦 8𝐵 = 𝑛̇ 2 𝑥 2𝐵     𝑥 1𝐵 = 𝑛̇ 2 𝑥 2𝐵 − 𝑛̇ 8 𝑦 8𝐵 𝑛̇ 1 = 𝟎. 𝟐𝟓 𝒎𝒐𝒍 𝑩/𝒎𝒐𝒍  (1 pt)
Then, the sum of mole fractions is equal to 1:
𝑥 1𝑇 = 1 − 𝑥 1𝐵 − 𝑥 1𝑋 = 𝟎. 𝟏𝟎 𝒎𝒐𝒍 𝑻/𝒎𝒐𝒍  (1 pt)
5 pts awarded for using component mole balances to get the mole fractions.

14 pts possible for part (e)

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##### Description: ChE 20500 Exam 1 – Fall 2015 Problem 1 B = benzene, C H 6 6 T = toluene, C 7 8 X = xylene, C H 8 10 Points for part a (5 pts total): 1 pt each for correct mix point, DC #1, DC #2, split point (4 pts) 1 pt for correct symbology (y for vapor, x for liquid, ???? for flow)
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