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clear;
clc;
disp('Example 4.16');
// aim : To determine
// the minimum dryness fraction of steam
// Given values
P1 = 1.8;// testing pressure,[MN/m^2]
P2 = .11;// pressure after throttling,[MN/m^2]
// solution
// from steam table
// at .11 MN/m^2 steam is completely dry and specific enthalpy is
hg = 2680;// [kJ/kg]
// before throttling steam is wet, so specific enthalpy is=hf+x*hfg, where x is dryness fraction
// from steam table
hf = 885;// [kJ/kg]
hfg = 1912;// [kJ/kg]
// now for throttling process,specific enthalpy will same before and after
// hence
x = (hg-hf)/hfg;
mprintf('\n The minimum dryness fraction of steam is x = %f \n',x);
// End
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