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clear;
clc;
disp('Example 4.3');
// aim : To determine
// the specific enthalpy
// given values
P = 2; // pressure ,[MN/m^2]
t = 250; // Temperature, [C]
cp = 2.0934; // average value of specific heat capacity, [kJ/kg K]
// solution
// looking up steam table it shows that at given pressure saturation temperature is 212.4 C,so
tf = 212.4; // [C]
// hence,
Degree_of_superheat = t-tf;// [C]
// from table at given temperature 250 C
h = 2902; // specific enthalpy of steam at 250 C ,[kJ/kg]
mprintf('\nThe specific enthalpy of steam at 2 MN/m^2 with temperature 250 C is = %f kJ/kg \n',h);
// Also from steam table enthalpy at saturation temperature is
hf = 2797.2 ;// [kJ/kg]
// so enthalpy at given temperature is
h = hf+cp*(t-tf);// [kJ/kg]
mprintf('\n The specific enthalpy at given T and P by alternative path is = %f kJ/kg \n',h);
// End
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