# Calculation of specific volumes

1. Apr 26, 2017

### mech-eng

1. The problem statement, all variables and given/known data
300 kPa pressure and v=0.5 m3/kg.

The state is fixed. I should find the temperature and quality if defined.

Table values for temperature

T P vf vfg vg
130C 270.1kPa 0.001070 0.66744 0.66850
135C 313 kPa 0.001075 0.58110 0.58217

I do not understand why table values seems bad when thread is created?

2. Relevant equations

Because the temperature for 300 kPa is not given it should be found by interpolation.
3. The attempt at a solution
$\frac {300-270.1}{T-130}$ = $\frac {313.-270.1}{135-130}$

But something is confusing for me in the pic. If the picture is not good I write here the related part of it: "Note that if we did not have table B.1.2 (as would be the case with other substances listed in Appendix B), we could have interpolated in Table B.1.1 between the 130 and 135 C entries to get the vf and vg values for 300 kPa. " This explanation is confusing for me because I understand it to be that if we calculate the temperature by interpolation then it is like vf and vg values will appear automatically without doing interpolation. So is an interpolation is required for finding them?
Thank you.

Source: Introduction to Engineering Thermodynamics by Sonntag/Borgnakke

Last edited: Apr 26, 2017
2. Apr 26, 2017

### Staff: Mentor

It might help if you could tell us what is in these two tables that are referred to, in terms of the column and row headings.