Question

What is the specific enthalpy for steam at 17.5 bar and 443 °C? show the steps for double interpolation to arrive at your answer

Answer 1

h_* at 17.5 bar and 443⁰

By linear interpolation

$h_2=h_1+\frac{h_3-h_1}{T'''-T'}(T''-T')=3253.5+\frac{3470.6-3253.5}{500-400}(443-400)=3345.465kJ/kg$

$h_5=h_4+\frac{h_6-h_4}{T'''-T'}(T''-T')=3251.9+\frac{3469.5-3251.9}{500-400}(443-400)=3345.468kJ/kg$

$h_*=h_2+\frac{h_5-h_2}{P_3-P_1}(P_2-P_1)=3346.85+\frac{3345-3346.85}{18-17}(17.5-17)=3346.15kJ/kg$