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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3012/CH14/EX14.4/Ex14_4.sce | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '3012/CH14/EX14.4/Ex14_4.sce')
| -rwxr-xr-x | 3012/CH14/EX14.4/Ex14_4.sce | 24 |
1 files changed, 24 insertions, 0 deletions
diff --git a/3012/CH14/EX14.4/Ex14_4.sce b/3012/CH14/EX14.4/Ex14_4.sce new file mode 100755 index 000000000..71190fea4 --- /dev/null +++ b/3012/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,24 @@ + +// Given:- +// For a complete reaction of CO with the theoretical amount of air +// CO + .5 O2 + 1.88N2 ---- CO2 + 1.88N2 +// Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is +// CO + .5O2 + 1.88N2 -- zCO + z/2 O2 + (1-z)CO2 + 1.88N2 + +K = 0.0363 // equilibrium constant the solution to Example 14.2 +p =1.0 // in atm +pref = 1.0 // in atm + +// Calculations +// Solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives +z = 0.175 +yCO = 2.0*z/(5.76 + z) +yO2 = z/(5.76 + z) +yCO2 = 2.0*(1.0-z)/(5.76 + z) +yN2 = 3.76/(5.76 + z) + +// Results +printf( ' The mole fraction of CO is: %.3f ',yCO) +printf( ' The mole fraction of O2 is: %.3f ',yO2) +printf( ' The mole fraction of CO2 is: %.3f',yCO2) +printf( ' The mole fraction of N2 is: %.3f',yN2) |