o (4) A gas stream containing 40.0 hydrogen, 35 carbon monoxide,20.0% of 35 diox
ID: 523812 • Letter: O
Question
o (4) A gas stream containing 40.0 hydrogen, 35 carbon monoxide,20.0% of 35 dioxide, and 5.00% methane is cooled from 1000°C to 10°C at a constant pressure atm (absolute). Gas enters the cooler at 120 m3/min and upon leaving the cooler is fed to an absorber where it is contacted with refrigerated liquid methanol to strip out the carbon dioxide. The methanol is fed to the absorber at a molar flow rate 1.2 times that of the inlet gas and absorbs essentially all of the CO2, 98% of the methane, and none of the other components of the feed gas. The gas leaving the absorber is at -12 Cand is fed into a cross-country pipeline. Calculate the volumetric flow rate of the gas entering the absorber and the molar flow rate of the methanol leaving the absorber.Explanation / Answer
Flow rate of gas entering the cooler= 120 m3/min =120*1000 L/min.
Temperature= 1000 deg.c= 1000+273= 1273K
Pressure = 35 atm. Molar flow rate of gas= PV/RT= 35*120*1000/(0.0821* 1273)= 40186.27 moles/hr
The gases leave the tower at 10 deg.c. Assuming pressure remains constant
V1/T1= V2/T2, V2= V1*T2/T1= 120* (10+273)/1273=26.7 m3/hr.
CO2 in the feed= 35%, molar flow rate of CO2 in the feed= 40186.27*35/100 =14065 moles/hr. Molar flow rate of methane= 40186.27*5/100 = 2009 moles/hr.
Liquid methanol removes all the CO2 and 98% of methane.
Hence methane not removed= 2009*2/100 = 40.18 moles/hr
Rest of the molar flow rate of gases= 40186.27 -2009-14065= 24112. 27 moles/hr
Gases leaving the absorber contains 40.18 moles/hr CH4 +24112.27 moles/hr other gases =24152.45 moles/hr.
Molar flow rate of methanol =1.2* 40186.27= 48223.52 moles/hr. Methanol carries with it 40186.26-24152.45 moles/hr of CO2 and CH4 removed= 16033.8 moles/hr of CO2 and CH4.
So molar flow rate of methanol leaving the tower= 48223.52+ 16033.80 =64257.32 moles/hr
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