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). Determine the vapor quality of the steam when throttled from 1.15 MPa to 1.0 MPa. Assume the process is adiabatic and no work is done by the system.
See also: Steam Tables
The enthalpy for the state D must be calculated using vapor quality:
hD, wet = hD,vapor x + (1 – x ) hD,liquid = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg
Since it is an isenthalpic process, we know the enthalpy for point T. From steam tables we have to find the vapor quality using the same equation and solving the equation for vapor quality, x:
hT, wet = hT,vapor x + (1 – x ) hT,liquid
x = (hT, wet – hT, liquid) / (hT, vapor – hT, liquid) = (2523 – 762) / (2777 – 762) = 0.874 = 87.4%
In this case of the throttling process (1.15MPa to 1MPa) the vapor quality increases from 87% to 87.4% and the temperature decreases from 186°C to 179.9°C.