A combustion gas has the following composition (by volume): 75% N2, 5.5% O2, 11% CO2, 8.4%...

A combustion gas has the following composition (by volume): 75% N2, 5.5% O2, 11% CO2,
8.4% H2O, and 0.1% CO.
a. What is the CO concentration in mg/dscm (dry standard cubic meter)?
b. What is the average molecular weight of the gas?
c. What is the density of the gas at T = 350 °F and P = 755 mmHg?
d. If the measured flowrate was 1000 L/s at the temperature and pressure stated in
part (c), what would the volumetric flowrate be at STP?

Expert Solution

a)ppm of CO = 0.1/100= 0.1*10⁴/10⁴*100 = 1000ppm

mg/dscm= ppm*Molecular weight/22.4L/gmole= 1000*28/22.4=1250

b) Molecular weight= mole fraction of nitrogen* molecular weight of nitrogen + mole fractions of oxygen* molecular weight + mole fraction of CO2* Molecular weight+ mole fraction water* molecular weight of water+ mole fraction CO* molecular weight of CO= 0.75*28+0.055*32+0.11*44+0.084*18+ 0.001*28=27.82 g/mole

c) 1 mole of any gas at standard conditions occupies 22.4 Liters

hence density at STP= mass/ Volume = 27.82/22.4=1.242 gm/L

T= 3500F correcponds to (350-32)/1.8 deg.c =177deg.c= 177+273.15=450.15K and pressure = 755/760 atm=0.99atm

since the temperature is raised from 273.15 (STP) to 450.15, the volume increases and hence density decreases. Since pressur is reduced from 1 atm to 0.98 atm. volume increases and density decreases. Hence

density= 1.242*(273.15/450.15)* 0.99/1=0.746 g/L

d) V1= 1000L/S T1= 450.15K P1= 0.99atm T2= 273.15K P2= 1atm V2=

From P1V1/T1= P2V2/T2

V2= P1V1*T2/P2T1= 0.99*1000*273.15/1*450.15)=600.643 L/s

queen_honey_blossom answered 2 years ago

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