clear ; clc; // Example 26.1 printf('Example 26.1\n\n'); //page no. 804 // Solution printf('Table to carry out degree of freedom analysis:\n'); // Number of variables involved printf('\nI. Number of variables involved.\n'); printf(' Species in F1 1 \n'); printf(' Species in F2 2\n'); printf(' Specie in F3 5\n'); printf(' Total stream flows 3\n'); printf(' Stream temperatures 3\n'); printf(' Stream pressures 3 \n'); printf(' Q 1 \n'); printf(' Extent of reactions 2\n'); printf('\__________________________________________________________________________________________\n'); printf(' Total 20\n'); printf('\n\nII. Number of equations and specifications.\n'); printf(' Independent species material balances 6\n'); printf(' Sum of species in each of the two streams 2 \n'); printf(' Energy balance 1\n'); printf(' Total stream flows 2\n'); printf(' Species values(CO) 1\n'); printf(' Pressures 3 \n'); printf(' Temperatures 2 \n'); printf(' O2 to N2 ratio specified in F2 1 \n'); printf(' Complete reaction, hence the extent of reaction is implied to both reactions 2\n'); printf('\___________________________________________________________________________________________\n'); printf(' Total 20\n'); printf('\n Therefore, by analysing the above table it is clear that degrees of freedom of system is (20 - 20) = 0 \n');