Q3: Methane is completely burned with air. The outlet gases from the burner, whi

Question

Q3: Methane is completely burned with air. The outlet gases from the burner, which contain no oxygen and methane, are passed through an absorber where some of the water is removed by condensation. The gases leaving the absorber have a nitrogen mole fraction of 0.8335. If the exit gases from the absorber are at 130 °F and 20 psia: a) To what temperature must this gas be cooled at constant pressure in order to start condensation of water? b) To what pressure must this gas be compressed at constant temperature before condensation will occur?

Explanation / Answer

to answer

at what temp. must this gas be colled at constant P in order to start condensation of water .

first we will calculate composition of this gas

it contains

nitrogen =83.35%

note that some water get condensed in absorber

after combustion

CH4+2O2----->CO2+2H2O

take basis of 100 moles of CH4

gas must have contain nitrogen ,carbon dioxide & water vapours

100 moles CH4 will require 200 moles O2

as we use air for burning

nitrogen =200*79/21 = 752.38 moles

combution will form 100 moles CO2 & 200 moles H2O as per stoichiometry

so gas in absorber will contain 752.38 moles of N2 & 100 moles of CO2 abasorbed & out of 200 moles of H2O vapor some will get condensed

mole fraction of N2 in exit gas from absorber =0.8335=moles of N2/total moles of exit gas

0.8335= 752.38/G

G=902.67

out of which 752.38 is of N2 & 100 of CO2

so water vapor in exit gas = 902.67-752.38-100=50.29 moles

now we have to find out at what temp vapor will start to condense at constant P.

If we consider gas mixture as ideal gas

partial pressure of water vapors =50.29/902.67*20 psia= 1.1142psia=7.68213858 kPa

use steam stable to find saturation temp. at this P from table you see that its around 314 K

b) the temp is 130 F =327.594 K

its around 15.3 psia . to get more accuracy you can use interpolation.

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