How to find specific volume given temperature?

[EastWindBreaks]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't understand how the solution got its values, not from the tables right? because I couldn't find such value from all of the tables. Since the pressure is less than the saturated pressure at given temperature, it is in super heated phase. but 90.4 kpa equals 13.1 psia, which is not in the super heated table. can we calculate specific volume without the tables? can we find its volume by finding the density at given temperature and pressure since the mass is known? but which would also require a table to find such density...

 

[Chestermiller]

Your book (Moran et al) also has a table in metric units, doesn't it?

 

[BvU]

Else try http://www.egr.msu.edu/classes/me417/somerton/R134a Tables.pdf

 

[EastWindBreaks]

Else try http://www.egr.msu.edu/classes/me417/somerton/R134a Tables.pdf

Your book (Moran et al) also has a table in metric units, doesn't it?

Thank you, I found the SI system tables, but i still don't see 0.2302, it is in this region correct?

where P= 0.10 MPa

 

[Chestermiller]

Thank you, I found the SI system tables, but i still don't see 0.2302, it is in this region correct? View attachment 216340
where P= 0.10 MPa

Yes. You have to interpolate in the table. Since the behavior is close to ideal gas behavior (with Pv nearly constant), for good accuracy, you should interpolate on (Pv) rather than simply on v.

 

[EastWindBreaks]

Yes. You have to interpolate in the table. Since the behavior is close to ideal gas behavior (with Pv nearly constant), for good accuracy, you should interpolate on (Pv) rather than simply on v.

Else try http://www.egr.msu.edu/classes/me417/somerton/R134a Tables.pdf

got it, so we have to do linear interpolate (assume the trend is linear?) for vapors when the table doesn't have the value we want, and we use quality(

)to find the properties for saturated state where the temperature and pressure are dependent on each other. Do we use

to find properties of sub cooled liquid as well? since I did not find a table for sub-cooled liquid.

 

[Chestermiller]

got it, so we have to do linear interpolate (assume the trend is linear?) for vapors when the table doesn't have the value we want, and we use quality(View attachment 216469 )to find the properties for saturated state where the temperature and pressure are dependent on each other. Do we use View attachment 216470 to find properties of sub cooled liquid as well? since I did not find a table for sub-cooled liquid.

You should be able to find the tabulated values of sub cooled liquid in your tables (sometimes called pressurized liquids).

I also point out again that, unless you interpolate on Pv rather than v, you will not get an accurate answer for the interpolated v. I guarantee that you linear interpolation will not deliver 0.2303.

 

[EastWindBreaks]

You should be able to find the tabulated values of sub cooled liquid in your tables (sometimes called pressurized liquids).

I also point out again that, unless you interpolate on Pv rather than v, you will not get an accurate answer for the interpolated v. I guarantee that you linear interpolation will not deliver 0.2303.

Thank you, now I understand what you said about Pv being nearly constant. I just want to confirm, you said interpolate on Pv, you mean like this?:

I got v= 0.2326, a little off from 0.2302.

 

[Chestermiller]

At P= 60000 Pa, v = 0.35048 m^3/kg, and Pv = 21029 J/kg

At P = 100000 Pa, v = 0.20743 m^3/kg, and Pv = 20743 J/kg

So, at P = 90400 Pa, $$Pv=21029+\frac{(90400-60000)}{(100000-60000}(20743-21029)=20812$$ And, $v=20812/90400=0.2302$ m^3/kg.

 

[EastWindBreaks]

At P= 60000 Pa, v = 0.35048 m^3/kg, and Pv = 21029 J/kg

At P = 100000 Pa, v = 0.20743 m^3/kg, and Pv = 20743 J/kg

So, at P = 90400 Pa, $$Pv=21029+\frac{(90400-60000)}{(100000-60000}(20743-21029)=20812$$ And, $v=20812/90400=0.2302$ m^3/kg.

Thanks a lot! I am so careless, I made a mistake by using P2= 0.14 Mpa instead of 0.1 Mpa.