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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /409/CH8/EX8.3/Example8_3.sce | |
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Diffstat (limited to '409/CH8/EX8.3/Example8_3.sce')
| -rwxr-xr-x | 409/CH8/EX8.3/Example8_3.sce | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/409/CH8/EX8.3/Example8_3.sce b/409/CH8/EX8.3/Example8_3.sce new file mode 100755 index 000000000..5e37c0351 --- /dev/null +++ b/409/CH8/EX8.3/Example8_3.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+// Example 8.3
+printf('Example 8.3\n\n');
+// Page no. 202
+// Solution
+
+// Basis : 1 hr so F = 1000 kg
+F = 1000 ;// feed rate-[kg/hr]
+P = F/10 ;// product mass flow rate -[kg/hr]
+
+n_un = 9 ;// Number of unknowns in the given problem
+n_ie = 9 ;// Number of independent equations
+d_o_f = n_un-n_ie ;// Number of degree of freedom
+printf('Number of degree of freedom for the given system is %i .\n',d_o_f);
+
+// Overall mass balance: F = P+B
+B = F-P ;// bottom mass flow rate -[kg/hr]
+printf('\n Bottom mass flow rate - %.1f kg \n',B);
+
+// Composition of bottoms by material balances
+m_EtOH = 0.1*F-0.6*P ;// By EtOH balance-[kg]
+m_H2O = 0.9*F - 0.4*P ;// By H2O balance-[kg]
+total = m_EtOH+m_H2O ;//[kg]
+f_EtOH = m_EtOH/total ;// Mass fraction of EtOH
+f_H2O = m_H2O/total ;// Mass fraction of H2O
+
+printf(' Mass of EtOH in bottom - %.1f kg \n',m_EtOH);
+printf(' Mass of H2O in bottom - %.1f kg \n',m_H2O);
+printf(' Mass fraction of EtOH in bottom - %.3f \n',f_EtOH);
+printf(' Mass fraction of H2O in bottom - %.3f \n',f_H2O);
\ No newline at end of file |