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+clear;

+clc;

+printf("\t\t\tProblem Number 6.34\n\n\n");

+// Chapter 6: The Ideal Gas

+// Problem 6.34 (page no. 284) 

+// Solution

+

+//From the table at 1000 R:                     //From the table at 500 R:

+h2=240.98;                                       h1=119.48; 

+//Btu/lbm //enthalpy                             //Btu/lbm //enthalpy

+u2=172.43;                                       u1=85.20; 

+//Btu/lbm //internal energy                      //Btu/lbm //internal energy 

+fy2=0.75042;                                     fy1=0.58233; 

+//Btu/lbm*R                                      //Btu/lbm*R

+

+//The change in enthalpy is

+deltah=h2-h1; //Btu/lbm

+//The change in internal energy is

+deltau=u2-u1; //Btu/lbm

+printf("The change in enthalpy is %f Btu/lbm & the change in internal energy is %f Btu/lbm\n",deltah,deltau);

+//Because in the constant-pressure process -R*log(p2/p1) is zero,

+deltas=fy2-fy1; //Btu/lbm*R //The entropy when air is heated at constant pressure

+printf("The entropy when air is heated at constant pressure is %f Btu/lbm/R\n",deltas);