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+//ques6
+//isothermal steady state processes
+clear
+clc
+//from table A.2
+P1=8;//pressure at state 1 in MPa
+P2=0.5;//pressure at state 2 in MPa
+T1=150;//Temperature at state 1 in K
+Pr1=P1/3.39;//Reduced pressure at state 1
+Pr2=P2/3.39;//Reduced pressure at state 2
+Tr1=T1/126.2;//Reduced temperature
+T2=125;//temperature at state 2
+//from compressibility chart h1*-h1=2.1*R*Tc
+//from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)
+//h2*-h2=0.5*R*Tc
+//this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc
+R=0.2968;//gas constant for given substance
+Tc=126.2;//K, Constant temperature
+Cp=1.0416;//heat enthalpy at constant pressure in kJ/kg
+dh=(2.35)*R*Tc+Cp*(T2-T1);//
+printf('Enthalpy change = %.2f kJ/kg \n',dh);
+//change in entropy
+//ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)
+//s1*-s1=1.6*R
+//s2*-s2=0.1*R
+//s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)
+//so
+ds=1.6*R-0.1*R+Cp*log(T2/T1)-R*log(P2/P1);
+printf(' Entropy Change = %.4f kJ/kg.K ',ds); \ No newline at end of file