## Specific Enthalpy of Wet Steam

**Specific Enthalpy of Wet Steam**

The **specific enthalpy of saturated liquid water** (x=0) and **dry steam** (x=1) can be picked from steam tables. In case of **wet steam**, the actual enthalpy can be calculated with the vapor quality, *x,* and the specific enthalpies of saturated liquid water and dry steam:

*h*_{wet}* = h*_{s}* x + (1 – x ) h*_{l}* ** *

*where*

*h*_{wet}* = enthalpy of wet steam (J/kg)*

*h*_{s}* = enthalpy of “dry” steam (J/kg)*

*h*_{l}* = enthalpy of saturated liquid water (J/kg)*

As can be seen, wet steam will always have lower enthalpy than dry steam.

**Example:**

A high-pressure stage of steam turbine operates at steady state with inlet conditions of 6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186°C and x = 0.87 (point D). Calculate the enthalpy difference between these two states.

The enthalpy for the state C can be picked directly from steam tables, whereas the enthalpy for the state D must be calculated using vapor quality:

*h*_{1, wet}* = ***2785 kJ/kg**

*h*_{2, wet}* = h*_{2,s}* x + (1 – x ) h***_{2,l}** = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 =

**2523 kJ/kg**

**Δh = 262 kJ/kg**

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