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Chemical Engineering 140 November 21, 2005 Midterm #
Exam 3
Points Possible: 170
Avg = 82
StdDev = 29.
High = 151
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Chemical Engineering 140 November 21, 2005 Midterm #
(170) Researchers at the University of Wisconsin-Madison have recently developed an alternative process for producing H 2 for the hydrogen economy^1. This approach uses renewable resources such as biomass as opposed to nonrenewable resources such as natural gas and petroleum. In this process, glucose (obtained from starch) is first converted to sorbitol, which is then reformed over a platinum catalyst to form hydrogen.
Sorbitol is reformed over a Pt/Al 2 O 3 catalyst in a tubular reactor at 498K and 29 bar according to the following stoichiometry.
C 6 O 6 H14(liq) +6H 2 O(liq) → 13H2(gas) + 6CO2(gas)
where the subscripts “(liq)” and “(gas)” denote liquid and gas states. Sorbitol is the limiting reactant. The reaction is run with 10% excess water. Water is fed in 10% excess to sorbitol. Conversion of sorbitol in the tubular reactor is 60%.
As a side reaction, carbon dioxide and hydrogen readily react to form methane and water.
CO2(gas) + 4H2(gas) → CH4(gas) + 2H 2 O(gas)
The effluent of the reactor is sent to a flash tank where a vapor phase and liquid phase exit in equilibrium. Upon analysis of the vapor stream exiting the flash tank, only 40% of the H 2 that would be produced under complete conversion with only the primary reaction is present. Approximate that H 2 , CH 4 , and CO 2 are not present in the liquid phase and that the vapor pressure of sorbitol is negligible.
(^1) Cortright, R. D., Davda, R. R. & Dumesic, J. A. Hydrogen from catalytic reforming of
biomass-derived hydrocarbons in liquid water. Nature 418 964-967 (2002).
50 a) Determine the overall composition of the species leaving the reactor. What is the conversion of the side reaction? Define conversion of the side reaction as:
2, reaction 2
2 H 2
mol H reacted in reaction 2 X = mol H produced in reaction 1
F = Basis mols/time Sorbitol in feed G = 6*(1.1F) = 6.6F = mols/time water in feed
Splitting the first process unit into two reactors; one for the primary reaction and
one for the secondary, the conversion of the second reactor is
Pvap 2181
(+10 for vapor pressure overall if they use only one lookup)
Antoine eqn method: 19 total (lose 1 b/c out of T range for constants in book) Antoine eqn: (+7) filled: (+10) Answer: (+2)
Water is the only material which will be split between the liquid and vapor streams leaving the tank.
y=
n (^) vap
5.23 F+ 2.96 F+ 0.644 F+nvap
n (^) vap
8.83 F+nvap
x=
nliq
0.4 F+nliq
4.29 F n (^) vap
0.4 F+ 4.29 F n (^) vap
4.29 F n (^) vap
4.69 F n (^) vap P y=Pvap x 10 y=^ 25.46^ x
(x and y: +8 each)
(Raoult's law filled in: +5. Not filled in -> +0)
x=
4.29 F n (^) vap
4.69 F n (^) vap
y=
nvap 8.83 F+nvap 10 y= 25.46 x 10 nvap
8.83 F+n (^) vap
25.46 4.29 F n (^) vap
4.69 F nvap
0.393 n (^) vap
8.83 F+nvap
4.29 F nvap
4.69 F nvap
(Quadratic setup: +5)
0.393 n 4.69 F n 4.29 F n 8.83 F+n
1.84 F n 0.393 n^2 37.88 F^2 8.83 F n+ 4.29 F n n^2
0 37.88 F^2 6.38 F n 0.607 n^2
(Solve quadratic: +5)
Roots: 4.23F and -14.75F
- negative 14.75F is obviously incorrect
- 4.23F is the number of mols of water in the vapor phase. (+7)
- 0.06F is the number of mols of water in the liquid phase. (+7)
K method: get Ki via simple method: (+7) acknowledge that Ki via simple method not applicable (polar compound under pressure): (+7) mass balance: (+8 w/o #s, +8 more with) Apply to get x and y: (+15)
Explaination instead of application: (up to 15 pts)
20 c) Endothermic reactions in tubular reactors are often heated by condensing steam on the exterior side of the tubes at about 10°C above the reaction temperature. If we wish to heat the reactor using this method, what pressure steam (in bar) should be used?
Reaction temperature=498K; Temperature wanted=508K (+2)
The minimum pressure of steam that could be used for this task is the weighted average of 2637 kPa (@500) and 3163 kPa (@510). (+4, +4)
Wtd average calculation:
2637
The result is 3058 kPa (or 30.58 bar) (+5)
No obvious use of weighted average/chose to use only one value: Loose interpolation resulting in 3060: (+10 total)
Simply takes value at 510K, producing 3160: (+10 total)
“About 30 bar” from steam table lookup of 507.05K: (+10 total)
Mention of 29 bar if implied lookup of pressures at 500 and 510 and use of a non-weighted average: (+9)
