Wet Steam Enthalpy from h_f and h_g

Same mixture, written from the other pair of columns. Most tables print $h_f$, $h_{fg}$ and $h_g$ side by side, and since $h_g = h_f + h_{fg}$ you can blend the two end states directly: a fraction x of the mass arrives as dry steam at $h_g$ and the remaining (1 − x) as water at $h_f$. At 1 MPa, with $h_f = 762.5$ and $h_g = 2{,}777.1$ kJ/kg, steam that is 95% dry carries 0.05 × 762.5 + 0.95 × 2,777.1 = 2,676.4 kJ/kg. Run it the other way and it is the lever rule from school: x = (h − $h_f$)/($h_g$ − $h_f$), the distance you have travelled along the horizontal tie line divided by its full length.

Which form you use is purely about which columns are in front of you, and the answers agree to the last digit. The reason to know both is that the same weighting works on every other property in the table, not just enthalpy. Specific volume is the one that bites: at 1 MPa water occupies 0.001127 m³/kg and dry steam 0.19444, a ratio of 173, so a mixture at x = 0.95 has almost exactly 95% of the dry-steam volume and a pipe sized on the mixture is a pipe sized on the vapour. Entropy blends the same way, which is how the exhaust point of a turbine expansion gets pinned down. One thing the lever rule cannot do is average temperatures, because in the wet region there is only one temperature and both ends of the lever are already sitting at it.