Question

A fuel gas produced by gasifying coal is to be burned with 20% excess air. The gas contains 50.0 mole % nitrogen and the balance carbon monoxide and hydrogen. A sample of the gas is passed through an infrared spectrometer, which registers a signal R that depends on the mole fraction of carbon monoxide in the sample, and a reading R 38:3 is recorded. Analyzer calibration data are as follows: x (mol CO/mol) 0.05 0.10 0.40 0.80 1.00 R 10.0 17.0 49.4 73.6 99.7 A power law (x = aRb ) should be suitable for fitting the calibration data. Derive the equation relating x and R (use a graphical method), and then calculate the molar flow rate of air required for a fuel feed rate of 175 kmol/h, assuming that CO and H2 are oxidized but N2 is not.

Answer 1

Power law , x= a.R^b

taking ln, ln x⁼ lna +blnR

So a plot of lnx vs ln R needs to be drawn.

The plot is shown below

lnX= 1.334lnR*- 6.073

lna= -6.073

a= 0.0023, b= 1.334

, ln X⁼ lna +blnR

The equation of best fit becomes   x= 0.0023 R ^1.334

At R= 37.3, x= 0.0023*(37.3)^1.334=0.29

Mole fraction of CO= 0.29, mole fraction of N2= 0.5, mole fraction of H2= 1-0.5-0.29=0.21

Flow rate of gas = 175 kmole/hr

Flow rate of CO= 175*0.29=50.75 kmo/hr, H2= 175*0.21= 36.75 Kmol/hr

Combustion reactions are CO+0.5O2àCO2, mole of oxygen = 50.75/2=25.375 kmol/hr, H2+0.5O2-à H2O, moles of O2= 36.75/2= 18.375 kmol/hr

Total moles of oxygen required = 23.375+18.375=43.75 kmole/hr

Air contains 21% O2 and 79%N2, moles of air required = 43.75/0.21=208.33kmol/hr

Air is supplied 20% in excess, molar flow rate of air = 1.2*208.33 = 250 kmole/hr