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+// Implementation of example 5.2
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+//Work done W, Velocity V, Pressure p, Specific volume v, Height Z
+W = 135;//kJ/kg
+Q = - 9;//kJ/kg
+v1 = 0.37; //m^3/kg
+p1 = 600;//kPa
+V1 = 16;//m/s
+Z1 = 32;//m
+v2 = 0.62;
+p2 = 100;
+V2 = 270;
+Z2 = 0;
+g = 9.81;//m/s^2
+//First law: 
+//u1 + p1v1 + V1^2/2 + Z1g + dQ/dm = u2+ p2v2 + V2^2/2 + Z2g + dW/dm
+//Change in specific internal energy 'dU'
+dU = (p2*v2 - p1*v1) + (V2^2 - V1^2)*10^(-3)/2 + (Z2 - Z1)*g*10^(-3) + W - Q;
+if(dU>0)
+  printf('Specific internal energy decreases by %0.3f kJ',dU);
+else
+  printf('Specific internal energy increases by %0.3f kJ',-dU);
+end
+// end
+\ No newline at end of file