Water in a piston-cylinder device losing heat

[ShizukaSm]

Homework Statement

The Attempt at a Solution

Let's divide it in three states.
E_1(\text{Initial state}) \\P = 3.5 \text{MPa} \\T = T_\text{sat@P} + 5 = 242.56 + 5 = 247.56\text{ C} \\ h_v = \text{Known value (table)}\\ v_v = \text{Known value (table)}

E_2(\text{When the piston hit the stops.}) \\P = P_\text{@ state 1} = 3.5 \text{MPa}(\text{Because the piston is still providing pressure}) \\T = 242.56\text{ C} \\ h_f = \text{Known value (table)}\\ v_f = \text{Known value (table)}\\ x = 0

E_3(\text{Final state}) \\P = ?? \\T = 200 \text{ C}

That were as far as I could get, which allowed me to solve a and b. I can't understand however, how to go from state 2 to state 3. The book solution affirms that specific volume is constant(s2>s3), and as such I should use it to calculate the final value of specific volume and find the data in "Saturated water" table.

I don't understand why, though. Apparently the water became steam again? How could I determine that? What is there preventing the water from simply cooling in a compressed liquid state and decreasing its specific volume?

 

[.Scott]

To answer your last question, if the liquid water does not occupy the full volume, the remaining space will contain water vapor. It wouldn't contain a vacuum because that would be below the vapor pressure of the water.

 

[Chestermiller]

I think the problem statement intended to say that, just as the piston hits the stops, the first drop of liquid water is just beginning to condense out. After this happens, the volume of the contents is no longer decreasing, so that the weighted average specific volume of the vapor and the liquid in the tank no longer changes. But, there is liquid water and water vapor in the tank thereafter. These are at equilibrium in state 3. So what is the pressure at 200 C? At these conditions, what is the specific volume of the liquid and what is the specific volume of the vapor?
What fraction of the mass is liquid water, and what fraction is vapor?

 

[ShizukaSm]

I think the problem statement intended to say that, just as the piston hits the stops, the first drop of liquid water is just beginning to condense out. After this happens, the volume of the contents is no longer decreasing, so that the weighted average specific volume of the vapor and the liquid in the tank no longer changes. But, there is liquid water and water vapor in the tank thereafter. These are at equilibrium in state 3. So what is the pressure at 200 C? At these conditions, what is the specific volume of the liquid and what is the specific volume of the vapor?
What fraction of the mass is liquid water, and what fraction is vapor?

I think the problem wasn't very well written. In state 2 all water is already condensed (that's what I thought initially, though your theory was also plausible in my mind, and later I confirmed that in the solution manual). I suppose the only way to figure out that the third state was saturated was if the problem informed me about that, right?

Supposing that there's 100% liquid water at state 2, I can't simply discover if it will go into a liquid-vapor mixture or if it will simply decrease its specific volume, right?

 

[Chestermiller]

I think the problem wasn't very well written. In state 2 all water is already condensed (that's what I thought initially, though your theory was also plausible in my mind, and later I confirmed that in the solution manual). I suppose the only way to figure out that the third state was saturated was if the problem informed me about that, right?

Supposing that there's 100% liquid water at state 2, I can't simply discover if it will go into a liquid-vapor mixture or if it will simply decrease its specific volume, right?

Well, if there were 100% liquid water in state 2, and the water were cooled, the liquid density would decrease, and this would leave a little room at the top for vapor to form (at the equilibrium vapor pressure). This would mean that the third state is saturated.