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+clear;

+clc;

+disp('Example 4.4');

+

+//  aim : To determine

+//  the specific enthalpy of steam

+

+//  Given values

+P = 2.5;//  pressure, [MN/m^2]

+t = 320; //  temperature, [C]

+

+//  solution

+//  from steam table at given condition the saturation temperature of steam is 223.9 C, therefore steam is superheated

+tf = 223.9;//  [C]

+

+//  first let's calculate estimated enthalpy

+//  again from steam table 

+

+hg = 2800.9;//  enthalpy at saturation temp, [kJ/kg]

+cp =2.0934;//  specific heat capacity of steam,[kJ/kg K]

+

+//  so enthalpy at given condition is

+h = hg+cp*(t-tf);//  [kJ/kg]

+mprintf('\n The estimated specific enthalpy is  =  %f kJ/kg \n',h);

+

+//  calculation of accurate specific enthalpy

+//  we need double interpolation for this

+

+//  first interpolation w.r.t. to temperature

+//  At 2 MN/m^2

+Table_t_h = [[325,300];[3083,3025]];// where, t in [C] and h in [kJ/kg]

+h1 = interpln(Table_t_h,320); //  [kJ/kg]

+

+//  at 4 MN/m^2

+Table_t_h = [[325,300];[3031,2962]]; //  t in [C] and h in [kJ/kg]

+h2 = interpln(Table_t_h,320); //  [kJ/kg]

+

+//  now interpolation w.r.t. pressure

+Table_P_h = [[4,2];[h2,h1]]; //  where P in NM/m^2 and h1,h2 in kJ/kg

+h = interpln(Table_P_h,2.5);//  [kJ/kg]

+mprintf('\n The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C is  =   %f kJ/kg \n',h);

+

+//  End