# Pressure, temp, volume change of water vapor, and steam table clarity

1. Sep 15, 2014

### wmgabe74

1. The problem statement, all variables and given/known data

Two pounds of water vapor at 30 psia fill the $4ft^3$ left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is $40^°F.$

2. Relevant equations
The change in internal energy is heat minus work, $ΔU=Q-W$ where work done by the system is positive.
Steam tables
$water = 0.0397 lb/mol$

3. The attempt at a solution
I begin with my known properties of state: for state one (before partition removal), I have the following:
m = 2 lbs water
P = 30 psia
V = 4ft^3
T = ?
U = ?

State two:
m = 2 lbs water
P = ?
V = 12ft^3 (the original space plus eight more cubic feet)
T = 40 °F.
U = ?

I want the initial temperature, so using the steam tables and interpolation, I find 30 psia falling between values as follows: 29.823 psia at 250 °F and 35.422 psia at 260 °F. My calculation results in 250.316 °F, which looks reasonable.

My questions begin......

My understanding of the steam table values is lacking clarity I believe. Here is what I think: When I look at these values of pressure and temperature for Saturated water, I see specific volumes between 13 and 11 cubic feet per molar pound, between 29.8 and 35.4 psia. With the problem giving me 2 lbs. of water, it seems to me I need nearly 26 cubic feet of volume to achieve the stated temperature of 250.316 °F. Pushing the vapor into a smaller area, 4 cubic feet, should require work done on the system, adding to the internal energy. This internal energy should be seen as increased temperature, if the pressure is to be held at the 30 psia stated. I need to look at the superheated vapor tables.

Assuming this is all correct, I don't think I understand how to read the superheated table.
two sections are headed as follows: 20psia(227.96 °F) and 40psia(267.26 °F). The first two columns are Temp (°F) and specific volume V (cubic feet per molar pound).
Do I use these section headings and my 30psia to interpolate my initial temperature?
This would give me an initial temperature of 247.61 °F.

How does the vapor being in 4 cubic feet play into my initial state?

State two pressure should be less, with a larger volume, and the cooling to the final temperature, but I am unclear regarding moving from state one to state two.
40 °F has a specific pressure of 0.122psia in the saturated water table....and a much higher specific volume rating, 2445.1 for saturated vapor, suggesting a higher pressure for the temp to be held constant at the 12 cubic feet of volume allowed.

I can determine my internal energy, U of state one, find the difference between that and the U at state two, but this comes after the above.

I am not looking for a specific answer to the problem here, but help understanding where I am going awry would be appreciated.

2. Sep 15, 2014

### BiGyElLoWhAt

Im honestly not sure what youre talking about. When you calculated the initial temperature, what were you using? You have enough information to figure it out, and it appears as though youve don that. If you did, then why are you trying to compress it more?

With the initial temperature known, you know the initial state of your water.

You now only need to solve for the one unknown variable in the second state. Since you seem to be looking at energy, does it matter how i add energy to a system? (Meaning chronologically) if the water moves over to the other chamber, what happens to the energy of the water as a whole? Does it matter if i heat it before or after my "change in pressure energy exchange"?

State 1 is state 1, state 2 is state 2. Does it matter how it got there?

Last edit.
What kind of gas are you treating water as?

3. Sep 15, 2014

### wmgabe74

"What kind of gas are you treating water as?"
The closer to liquid a gas gets, the higher the error when using the ideal gas laws. One of the assumptions in using an ideal gas law such as $PV=nRT$ is that there are no molecular interactions (low density). This becomes questionable with water vapor. In short, I cannot use the ideal gas equations of state for this, (also a requirement of my class, but there is a certain sense to this, even without the professor's decree). This is why the steam tables are in play for this question.

It is correct that for the answer to the problem I only need to solve for the final pressure. I am seeking greater understanding however, in looking at the energy. I am aware that when the energy is used is not relevant, this is inherent in the concept of first and second states. With a better understanding in interpreting the steam tables, I can find the internal energy, and subsequently the change in heat.

I am not trying to compress the vapor more, the problem states a static first state of 4 cubic feet for a volume; pressure is force applied divided by an area - if there is no volume at all (free space), there would be no pressure, or perhaps ambient pressure, if we are talking open space. This indicates there is a pressure associated with volume;

This leads me to consider that the specific volumes in the steam tables only work for the pressures and temperatures they are associated with. If my given volume of water vapor is different, I think either my temperature or my pressure must change. Is this not correct? From that if my pressure is stated in the problem, I must change the temperature.

My question is, if this is incorrect, why? or where is my reasoning inaccurate?

4. Sep 16, 2014

### BiGyElLoWhAt

Sorry to leave you hanging for so long. But, in short, yes, that works logically. I'm not sure what you were saying about energy however,
You have an initial state, of which you know certain conditions. I'm not familiar with these steam table things that you're talking about (is this engineering?), but regardless, ideal gas law is a good first order approximation. You can't use IGL, which is fine, but if you look at the relationship between Pressure Volume and Temperature you can see that you have the right idea, and you can see how they (approximately) vary with respect to each other.

5. Sep 18, 2014

### wmgabe74

The end result of this problem is that I was way overthinking the problem. In the steam tables, the pressure for saturated liquid water at the final temperature of 40 degrees F was given, and that was it.

Thanks for your looking at it, BiG

6. Sep 19, 2014

### BiGyElLoWhAt

Anytime. Glad you got it figured out.