Question

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

Answer 1

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

Step-by-step 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 =
`(R*T)/(P)` ..............1

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

specific volume =
`(0.4615*623)/(15000)`

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 =
`(P)/(Pcr)`

reduced pressure =
`(15)/(22.06)` = 0.68 Mpa

and

reduced temperature will be =
`(T)/(Tcr)`

reduced pressure =
`(623)/(647.1)` = 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