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+clear;

+clc;

+printf("\t\t\tProblem Number 6.17\n\n\n");

+// Chapter 6: The Ideal Gas

+// Problem 6.17 (page no. 261) 

+// Solution

+

+//data of problem6.16

+cp=0.24; //Specific heat at constant pressure //Btu/lbm*R

+p2=15; //psia //final pressure

+p1=100; //psia //initial pressure

+T2=460+0;  //absolute  final temperature //unit:R

+T1=460+100; //absolute initial temperature //unit:R

+J=778; //conversion factor

+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality

+//Because cp and R are given,let us first solve for cv,

+//cp=(R*k)/(J*(k-1))

+k=(cp*J)/((cp*J)-R); //k=cp/cv //ratio of specific heats

+printf("Ratio of specific heats k is %f\n",k);

+//k=cp/cv

+cv=cp/k; //Specific heat at constant volume //Btu/lbm*R

+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);

+//Now, deltas=(cv*log(p2/p1))+(cp*log(v2/v1));

+//But, v2/v1=(T2*p1)/(T1*p2)

+v2byv1=(T2*p1)/(T1*p2); // v2/v1 //unitless

+deltas=(cv*log(p2/p1))+(cp*log(v2byv1)); //The change in enthalpy  //unit:Btu/lbm*R

+printf("The change in enthalpy is %f Btu/lbm*R\n",deltas);

+//The agreement is very good.