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Midterm Exam Answers - Chemical Engineering Material/Energy Balances | CHEN 2120, Exams of Chemistry

Material Type: Exam; Professor: De Grazia; Class: Chemical Engineering Material and Energy Balances; Subject: Chemical Engineering; University: University of Colorado - Boulder; Term: Fall 2009;

Typology: Exams

2009/2010

Uploaded on 11/07/2010

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1. Carbon monoxide and hydrogen react to form methanol. A fresh feed containing the reactants joins a
recycle stream and the combined stream flows to the reactor. The reactor outlet stream is 350 mol/min
with wt %’s of 10.6 H2, 64 CO and 25.4 CH3OH. This stream enters a condenser where most of the
methanol is condensed and withdrawn as a product stream. The gas stream leaving the condenser,
which contains CO, H2 and 0.40 mole % methanol vapor, makes up the recycle stream. (30 pts)
Find the following USING ATOMIC SPECIES BALANCES:
a) molar flow rates of CO and H2 in the fresh feed
b) production rate of liquid methanol
c) single-pass and overall conversion of CO
n6 (CH3OH), n5 (CO), n4 (H2)
350 mol/min
n1(CO), n2 (H2)
.095 CH3OH np (CH3OH)
.274 CO
.631 H2
Find mole %’s : 6 pts
DOF analysis:
Overall: 3 unknowns, only 2 independent atomic balances
Condenser: 4 unknowns, 3 species, 1 piece of info (mole fraction of CH3OH)
Around condenser:
n4 = .631* 350 = 222 (2 pts)
n5 = .274 * 350 = 94.5 (2 pts)
.004 = n6/(222 + 94.5 + n6) n6 = 1.28 (3 pts)
np = 350 – 222 – 94.5 – 1.28 = 32.2 moles (2 pts)
Overall atomic species balance to find n1 and n2:
C: n1 = np = 32.2 (2 pts)
H: 2n2 = 4nPn2 = 64.4 (2 pts)
Mixing point balance to find CO to reactor: n1 + n5 = 126.7 (3 pts)
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  1. Carbon monoxide and hydrogen react to form methanol. A fresh feed containing the reactants joins a recycle stream and the combined stream flows to the reactor. The reactor outlet stream is 350 mol/min with wt %’s of 10.6 H 2 , 64 CO and 25.4 CH 3 OH. This stream enters a condenser where most of the methanol is condensed and withdrawn as a product stream. The gas stream leaving the condenser, which contains CO, H 2 and 0.40 mole % methanol vapor, makes up the recycle stream. (30 pts) Find the following USING ATOMIC SPECIES BALANCES: a) molar flow rates of CO and H 2 in the fresh feed b) production rate of liquid methanol c) single-pass and overall conversion of CO n 6 (CH 3 OH), n 5 (CO), n 4 (H 2 ) 350 mol/min n 1 (CO), n 2 (H 2 ) .095 CH 3 OH np (CH 3 OH) .274 CO .631 H 2 Find mole %’s : 6 pts DOF analysis: Overall: 3 unknowns, only 2 independent atomic balances Condenser: 4 unknowns, 3 species, 1 piece of info (mole fraction of CH 3 OH) Around condenser: n 4 = .631* 350 = 222 (2 pts) n 5 = .274 * 350 = 94.5 (2 pts) .004 = n 6 /(222 + 94.5 + n 6 ) n 6 = 1.28 (3 pts) np = 350 – 222 – 94.5 – 1.28 = 32.2 moles (2 pts) Overall atomic species balance to find n 1 and n 2 : C: n 1 = np = 32.2 (2 pts) H: 2n 2 = 4nP n 2 = 64.4 (2 pts) Mixing point balance to find CO to reactor: n 1 + n 5 = 126.7 (3 pts)

Overall conversion: 100 % (3 pts) Single pass: 126.7 – 94.5/126.7 = 25.4 % (3 pts)

  1. A vapor mixture of n- butane and n -hexane contains 50 mole % butane at 120 degrees C and 1 atm. A stream of this mixture flowing at a rate of 150.0 L/s is cooled and compressed, causing some of the vapor to condense. Liquid and vapor product streams emerge from the process at 57 degrees C and some pressure. The vapor product contains 60.0 mole % butane. (15 pts) Find: a. the final pressure b. the composition of the liquid product c. the molar flow rates of the liquid and vapor products. Find moles: no = 4.65 mol/s (2 pts) 4 unknowns, n 1 , n 2 , x 2 , P 2 Raoult’s, 2 species .6P = x 2 pB(57O) = x 2 (4417) 3 pts .4P =1- x 2 pT(57O) = (1-x 2 )(517.6) 3 pts P = 1100.6 mm Hg (1 pt) xB = .15 (1 pt) xH = .85 (1 pt) Doing a material balance: 4.65 = n 1 + n 2 (1 pt) .5(4.65) = .6n 1 + .15n 2 (1 pts) n 2 = 1.03 (1 pt) n 1 = 3.62 (1 pt)
  1. A gas stream containing 40 mole % hydrogen gas, 35 % carbon monoxide 20 % CO 2 and the rest methane is cooled from 100 oC to 10o^ C at a constant absolute pressure of 35 atm. Gas enters the condenser at 120 m^3 /min and upon leaving is fed to an absorber, where it is contacted with liquid methanol. The methanol is fed to the absorber at a molar flow rate 1.2 times that of the inlet gas and absorbs all the CO 2 , 98 % of the methane and nothing else. The gas leaving the absorber is saturated with methanol at -12o^ C. Find the volumetric flow rate of methanol entering the absorber (m^3 /min) and the molar flow rate of methanol in the gas leaving the absorber. RT/P = 0.87 < 20 not ideal (1 pt) H 2 : Tc corrected: 41.3 Pc corrected: 20. CO 133 34. CO 2 304.2 72. CH 4 190.7 45. 1 pts for H 2 , ½ pt each for the others (2.5 total) Use Kay’s rule: T’c: 133.4 K (2 pts) p’c: 37.3 atm (2 pts) Tr = 2.8 (1 pt) Pr = 0.94 (1 pt) z = about .98 (1/2 pt) n = PV/zRT = 140 kmol/min of gas (2 pts) nmeth = 1401.2 = 168 kmol/min of MeOH(1 pt) * 32 kg/kmol = 5376 kg/min (1 pt) /792 kg/m^3 = 6.78 m^3 / min (1 pt) at outlet: yP = p(-12oC) (1 pt) p(-12oC) = 13 mm Hg (Antoine’s (2 pts) y = 13/(35760) = 4.9 X 10-4^ (2 pts)
  2. Three moles of benzene and seven moles of toluene are placed in a closed cylinder which is immersed in boiling water that maintains the temperature at 100o^ C. Find the composition of the liquid and the vapor at 1000 mm Hg (abs). (5 pts) no vapor; liquid is .3 B and .7 T