Determine the specific volume of superheated water vapor at 15 mpa and 350°c, using a. the ideal-gas equation 0.01917 m3 /kg b. the generalized compressibility chart 0.01246 m3 /kg c. the steam tables 0.01148 m3 /kg

specific volume by ideal gas equation = 0.01917 m³/kg

specific volume by compressibility chart = 0.01246 m³/kg

specif volume by super heated stream table is 0.0114810 m³/kg

Explanation:

given data

temperature  T = 350°C = 623 K

pressure P = 15 MPa = 15000 kPa

to find out

specific volume by  ideal-gas equation ,generalized compressibility chart and steam tables

solution

we will apply here ideal gas equation that is

specific volume =   ..............1

here P is pressure and T is temperature and R is gas constant i.e 0.4615 kJ/kg-K

specific volume =  

specific volume = 0.01917 m³/kg

and

by the compressibility chart

critical pressure of water Pcr = 22.06 Mpa

and critical temperature of water Tcr = 647.1 K

so

reduced pressure will be =

reduced pressure = = 0.68 Mpa

and

reduced temperature will be =

reduced pressure = = 0.963 K

so by compressibility chart pressure 0.68 Mpa and temperature 0.963 K

compressibility factor Z is 0.65

so

specific volume = compressibility factor Z × ideal specific volume

specific volume = 0.65 × 0.01917

specific volume = 0.01246 m³/kg

and

by the steam table

use here super heated stream table for

pressure = 15 Mpa

ans temperature = 350°C

so

specif volume by super heated stream table is 0.0114810 m³/kg

efficiency of reversible power cycles operating between a hot re

explanation:

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: