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+// Example 10_9
+clc;funcprot(0);
+// Given data
+T=400;// °C
+P=1;// MPa
+v=0.3066;// m^3/kg
+
+// Calculation
+ds=7.619-7.302;// kJ/kg.K
+dT=450-350;// K
+c_p=(T+273)*(ds/dT);// kJ/kg.K
+dv=0.3304-0.2825;// m^3/kg
+mu_j=(1/(c_p*10^3))*[((T+273)*(dv/dT))-v];// K/Pa
+printf("\nThe Joule thomson coefficient,mu_j=%1.2e K/Pa",mu_j);
+dT=403.7-396.2;// K
+dP=(1.5-0.5)*10^6;// Pa
+mu_j=dT/dP;// K/Pa
+printf("\nThe Joule thomson coefficient,mu_j=%1.2e K/Pa",mu_j);