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+clear;
+clc;
+disp('Example 4.13');
+
+//  aim : To determine 
+//  the dryness fraction of steam 
+//  and the internal diameter of the pipe
+
+//  Given values
+
+//  steam1
+P1 = 2;// pressure before throttling, [MN/m^2]
+t = 300;//  temperature,[C]
+ms1_dot = 2;// steam flow rate, [kg/s]
+P2 = 800;//  pressure after throttling, [kN/m^2]
+
+//  steam2
+P = 800;// pressure, [N/m^2]
+x2 = .9;//  dryness fraction
+ms2_dot = 5; //  [kg/s]
+
+//  solution
+//  (a)
+//  from steam table specific enthalpy of steam1 before throttling is
+hf1 = 3025;//  [kJ/kg]
+//  for throttling process specific enthalpy will same so final specific enthalpy of steam1 is
+hf2 = hf1;
+// hence
+h1 = ms1_dot*hf2;// [kJ/s]
+
+//  calculation of specific enthalpy of steam2
+hf2 = 720.9;//  [kJ/kg]
+hfg2 = 2046.5;//  [kJ/kg]
+//  hence
+h2 = hf2+x2*hfg2;//  specific enthalpy, [kJ/kg]
+h2 = ms2_dot*h2;//  total enthalpy, [kJ/s]
+
+//  after mixing
+m_dot = ms1_dot+ms2_dot;//  total mass of mixture,[kg/s]
+h = h1+h2;//  Total enthalpy of the mixture,[kJ/s]
+h = h/7;//  [kJ/kg]
+
+//  At pressure 800 N/m^2 
+hf = 720.9;//  [kJ/kg]
+hfg = 2046.5;//  [kJ/kg]
+//  so total enthalpy is,hf+x*hfg, where x is dryness fraction of mixture and which is equal to h
+//  hence
+x = (h-hf)/hfg;// dryness fraction after mixing
+mprintf('\n (a) The condition of the resulting mixture is dry with dryness fraction  =  %f \n',x);
+
+//  (b)
+// Given
+C = 15;//  velocity, [m/s]
+//  from steam table
+v = .1255;//  [m^/kg]
+A = ms1_dot*v/C;//  area, [m^2]
+//  using ms1_dot = A*C/v, where A is cross section area in m^2 and
+//  A = %pi*d^2/4, where d is diameter of the pipe 
+
+//  calculation of d
+d = sqrt(4*A/%pi); // diameter, [m]
+
+mprintf('\n (b) The internal diameter of the pipe is  =  %f mm \n',d*1000);
+
+//  End