diff options
author | priyanka | 2015-06-24 15:03:17 +0530 |
---|---|---|
committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /593/CH4/EX4.3 | |
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 '593/CH4/EX4.3')
-rwxr-xr-x | 593/CH4/EX4.3/ex4_3.sce | 25 |
1 files changed, 25 insertions, 0 deletions
diff --git a/593/CH4/EX4.3/ex4_3.sce b/593/CH4/EX4.3/ex4_3.sce new file mode 100755 index 000000000..da95ca411 --- /dev/null +++ b/593/CH4/EX4.3/ex4_3.sce @@ -0,0 +1,25 @@ +clear;
+//clc();
+
+// Example 4.3
+// Page: 80
+printf("Example-4.3 Page no.-80\n\n");
+
+//***Data***//
+// We have the system which consists of isobutane and normal butane and isomerisaation is taking place between them
+// The equilibrium constant for this reaction is given by
+// K = (mole fraction of isobutane)/(mole fraction of n-butane) = x_iso/x_normal
+
+// For this reaction, at 25C,
+K = 4.52;
+
+// and
+// x_iso + x_normal = 1
+// so
+// K = x_iso/(1-x_iso)
+
+// solving for x_iso
+deff('[y]=f(x_iso)','y = x_iso/(1-x_iso)-K');
+x_iso = fsolve(0,f);
+
+printf(" Mole fraction of isobutane isomer in equilibrium is %0.2f",x_iso);
|