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+//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.12\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.12  (page no. 334) 
+// Solution
+
+//The change in entropy of the mixture is the sum of the changes in entropy of each component.
+//Given in problem 7.9,: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500F and nitrogen  is at 200 F. //cp=specific heat at constant pressure 
+//In 7.9,for the oxygen,the temperature starts at 500F(960 R) and decreases to 328.7 F.For the nitrogen,the temperature starts at 200F(660 R) and increase to 328.7 F.
+//deltas = (cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy
+
+//For the oxygen,
+cp=0.23; //specific heat at constant pressure //Unit:Btu/lbm*R
+T2=328.7+460; //Unit:R //final temperature
+T1=500+460; //Unit:R //starting temperature
+deltas=(cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy for oxygen
+DeltaS=160*deltas; //Btu/R //The total change in entropy of the oxygen
+printf("The total change in entropy of the oxygen is %f Btu/R\n",DeltaS);
+
+//For the nitrogen,
+cp=0.25; //specific heat at constant pressure //Unit:Btu/lbm*R
+T2=328.7+460; //Unit:R //final temperature
+T1=200+460; //Unit:R //starting temperature
+deltas=(cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy for nitrogen
+deltaS=196*deltas; //Btu/R //The total change in entropy of the nitrogen
+printf("The total change in entropy of the nitrogen is %f Btu/R\n",deltaS);
+deltaS=deltaS+DeltaS; //the total change in entropy for the mixture //Btu/lbm*R
+printf("The total change in entropy for the mixture is %f Btu/R\n",deltaS);
+
+//Per pound of mixture,
+deltasm=deltaS/(196+160); //increase in entropy per pound mass of mixture
+printf("Increase in entropy per pound mass of mixture is %f Btu/lbm*R\n\n",deltasm);
+
+
+printf("An alternative solution:\n");
+//As an alternative solution,assume an arbitrary datum of 0 F(460 R).
+cp=0.23; //specific heat at constant pressure //Unit:Btu/lbm*R
+//For initial entropy of oxygen,
+T2=500+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+deltas=cp*log(T2/T1); //the initial change in entropy for oxygen // Btu/lbm*R
+printf("The initial change in entropy for oxygen is %f Btu/lbm*R\n",deltas);
+//For final entropy of oxygen,
+T2=328.7+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+Deltas=cp*log(T2/T1); //the final change in entropy for oxygen // Btu/lbm*R
+printf("The final change in entropy for oxygen is %f Btu/lbm*R\n",Deltas);
+deltaS=Deltas-deltas; //The entropy change of the oxygen //Btu/lbm*R
+printf("The entropy change of the oxygen is %f Btu/lbm*R\n",deltaS);
+
+//For nitrogen,
+cp=0.25; //specific heat at constant pressure //Unit:Btu/lbm*R
+//For initial entropy of nitrogen,
+T2=200+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+deltas=cp*log(T2/T1); //the initial change in entropy for nitrogen // Btu/lbm*R
+printf("The initial change in entropy for nitrogen is %f Btu/lbm*R\n",deltas);
+//For final entropy of nitrogen,
+T2=328.7+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+Deltas=cp*log(T2/T1); //the final change in entropy for nitrogen // Btu/lbm*R
+printf("The final change in entropy for nitrogen is %f Btu/lbm*R\n",Deltas);
+deltaS=Deltas-deltas; //The entropy change of the nitrogen //Btu/lbm*R
+printf("The entropy change of the nitrogen is %f Btu/lbm*R\n",deltaS);
+
+//The remainder of the problem is as before.The advantage of using this alternative method is the negative logarithms are avoided by choosing a reference          temperature lower than any other temperature in the system
+