How Does Entropy Change in Evaporation Processes?

[AriAstronomer]

Homework Statement

"You allow 1kg of saturated liquid water at 100 degrees C to evaporate in a room with a constant volume. The final pressure is 10kPa. The initial and final temperature are equal, and the walls of the room are maintained at 100C. Use steam tables unless otherwise indicated."

Now there is a part a and b and c that I've done, here's the part I'm stuck on:

d) Now let us track the generation of entropy. Assume a two step process. In the first step, the room expands slowly (cylinder-piston) so that the vapour remains saturated. What is the entropy change during this first step?

e) Then the vapour is allowed to spread freely to the final volume. By how much does the entropy change of the steam during this second step?

Homework Equations

Steam tables

The Attempt at a Solution

d) I'm a bit confused conceptually what they're saying. If the volume changes so that the vapour remains saturated, then at the end of the expansion, isn't there still going to be water present (by definition of saturated steam, there is an equilibrium of water/steam based on temp/pressure)? I'm asking this because the answer is ds = s_final - s_initial = s_g - s_f = s_fg = 6.0480. But this implies the state is all gas afterward, and all liquid beforehand? I thought saturated steam by definition is an equilibrium of water/gas? There is something conceptual I'm missing here, if someone could help me out I'd really appreciate it. Our textbook doesn't cover things like sfg, sg, sf, quality factor, etc. at all, if there are any resources that explains this well I'd love to read it.

On a side note for 100 degrees C for saturated steam I find v = .001044m^3/kg. But I thought our piston was changing the volume? Should I simply be thinking, as long as 'they' are 'somehow finding a way' to keep it saturated at T=100, just forget about this volume term?

e) So at this point I'm unsure if I'm dealing with all steam or an equilibrium of steam/liquid. I feel like I'm dealing with all steam at this point, but again, I don't see how it can still be called 'saturated' then (what's then the difference between saturated and superheated?). The final state however is superheated vapour at 100Celsius, 10kPa, s = 8.4479. I guess before letting our saturated steam expand the entropy was s_g = 7.3549, so ds = 8.4479 - 7.3549 = 1.0930kJ/kgK. This answer is right (we have answers).

Help would be appreciated.
Ari

 

[mark726]

ana, I can definitely help you with this problem. First, let's clarify some concepts about saturated steam and entropy.

Saturated steam is a thermodynamic state where the steam is in equilibrium with liquid water at a given temperature and pressure. This means that if you have a closed container with saturated steam at a certain temperature and pressure, there will be a balance between the amount of water and steam present. This does not mean that the steam is all gas or all liquid, but rather a mixture of both.

Entropy is a measure of the disorder or randomness of a system. In the case of steam, it is a measure of the amount of energy that is unavailable for work. When steam is at its saturated state, it has a certain entropy value, let's call it s_1. When it is allowed to expand and reach a new state, it will have a different entropy value, s_2. The change in entropy, Δs, is equal to s_2 - s_1.

Now, let's apply this to the problem at hand. In part d), we are considering a two-step process. In the first step, the room expands slowly while the steam remains saturated. This means that the temperature and pressure remain constant, but the volume increases. In this case, the entropy change is simply the difference between the final and initial entropy values, s_2 - s_1 = s_g - s_f = s_fg, where s_g is the entropy of the saturated gas and s_f is the entropy of the saturated liquid. This is because the steam remains in its saturated state throughout the process.

In part e), the steam is allowed to spread freely to the final volume. This means that the temperature and pressure are no longer constant, and the steam will reach a superheated state at the final volume. In this case, the entropy change is the difference between the final entropy value of the superheated steam, s_2, and the initial entropy value of the saturated steam, s_1. This is because the steam has now reached a different state with a different entropy value.

I hope this clarifies things for you. For more information on steam tables and thermodynamics, I recommend checking out some online resources or textbooks on the subject. Best of luck!