2
$\begingroup$

Typically in an exam or problem set we have a problem that's like this: Determine the specified property at the indicated state of water. Locate the state on a sketch of the T-v and P-v diagram.

p= 3 bar, T= 240°C. Find v, in $m^3/kg$.

I believe I can figure out $v,u,h$ when it ask to find this information, however I'm having trouble figuring out what's the best way to look at the steam table and figure out how to draw the correct information. For reference I've provided what the answer to this problem is.

Here is the solution:

Problem Set Answer

Thanks,

$\endgroup$

1 Answer 1

2
$\begingroup$

I'm going to start with a very short answer, but will be happy to elaborate on any point you find confusing.

Steam tables for superheated steam are organized by pressure and temperature. In most examples I've seen, you look up the pressure first and then scan that row, column, or page for the relevant temperature. The entry for your temperature and pressure combination should give you things like specific volume, specific enthalpy, etc.

The curve on the P-V and T-V diagrams in the solution is the saturation dome for water. Below it, water is a liquid; above it, water is a vapor (steam). On the saturation dome water is a saturated vapor.

The way to sketch a particular point on a P-V or T-V diagram if you don't have a numerical steam table handy and are given the "wrong" properties is to identify the correct isotherm on a P-V diagram (you can see a small part of it drawn in your example P-V) or the correct isobar on a T-V diagram (also shown in the example).

I've included a figure of isobars on a temperature-entropy (T-S) diagram (sorry I didn't have a T-V diagram handy) and isotherms on a P-V diagram that I've used when teaching this material. In general, instructors do not give impossible problems on this topic. They either provide diagrams like the ones below (except labeled!) when asking for a sketch so that you can find the correct isotherm (-bar,-chor, etc.) or they'll accept any sketch that is close to correct.

TS-PV

$\endgroup$
1
  • $\begingroup$ Thanks a lot, the thorough explanation, makes sense, I also noticed I was thinking that the diagrams were different because of the way the answer keys would draw the curves, when there was really only one curve for PV and then TS. $\endgroup$
    – eWizardII
    Commented Sep 28, 2011 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.