Heated piston & cylinder device with saturated water and vapor states

[EastWindBreaks]

[Mentor note: Thread title changed to describe actual problem being presented]

1. Homework Statement

Homework Equations

The Attempt at a Solution

I understand you have to interpolate temperature and pressure of the saturated vapor from the table, since there is no matched final specific volume value on the saturated table.

But I don't understand why my first approach is wrong: Since the pressure is constant, I can find pressure using initial properties ( h=u+Pv, where h is the enthalpy of saturated liquid and u is the internal energy of saturated liquid) and then find T using ideal gas equation.

[/B]
the pressure is not correct either, why can I not use h=u+Pv in this case?
does ideal gas equation only works near standard pressure and temperature condition?

 

[stockzahn]

But I don't understand why my first approach is wrong: Since the pressure is constant, I can find pressure using initial properties ( h=u+Pv, where h is the enthalpy of saturated liquid and u is the internal energy of saturated liquid) and then find T using ideal gas equation.
the pressure is not correct either, why can I not use h=u+Pv in this case?
does ideal gas equation only works near standard pressure and temperature condition?

Why do you think the pressure is constant? And no, you can't use the ideal gas equation. An easy way to estimate, if the use of the ideal gas equation is valid, is to have a look at the T,s diagramm (e.g. https://commons.wikimedia.org/wiki/File:T-s_diagram.svg). If the isenthalps and the isotherms are parallel (i.e. if the isenthalps are horizontal) you can use the ideal gas equation for a good estimation. In the diagram of the link you can see that this isn't the case. The correlation $h=u+pv$ should be valid though anyhow.

 

[EastWindBreaks]

Why do you think the pressure is constant? And no, you can't use the ideal gas equation. An easy way to estimate, if the use of the ideal gas equation is valid, is to have a look at the T,s diagramm (e.g. https://commons.wikimedia.org/wiki/File:T-s_diagram.svg). If the isenthalps and the isotherms are parallel (i.e. if the isenthalps are horizontal) you can use the ideal gas equation for a good estimation. In the diagram of the link you can see that this isn't the case. The correlation $h=u+pV$ should be valid though anyhow.

thank you, because I assumed a piston-cylinder device has a constant pressure from state 1 to state 2, since the piston is free to move. I guess the pressure is not constant then? why is it not constant though?
so in the case of exams where we are not given the T,S diagram, how do we know the ideal gas equation will work?

 

[stockzahn]

thank you, because I assumed a piston-cylinder device has a constant pressure from state 1 to state 2, since the piston is free to move. I guess the pressure is not constant then? why is it not constant though?
so in the case of exams where we are not given the T,S diagram, how do we know the ideal gas equation will work?

The final pressure is obtained by two information:

1) The volume quadruples (hence you know the volume)
2) The entire water becomes saturated vapour (you are on the right arm of the saturation line)

Therefore, you have all data you need to calculate the thermodynamic state (there is only one point in the T, s diagram satisfying these two conditions). Pressure and temperate then must be results, otherwise you have an over-determined system. However, if you are close or even in the two-phase region, you can confidentially assume that the use of the ideal gas equation is not valid.

 

[EastWindBreaks]

The final pressure is obtained by two information:

1) The volume quadruples (hence you know the volume)
2) The entire water becomes saturated vapour (you are on the right arm of the saturation line)

Therefore, you have all data you need to calculate the thermodynamic state (there is only one point in the T, s diagram satisfying these two conditions). Pressure and temperate then must be results, otherwise you have an over-determined system. However, if you are close or even in the two-phase region, you can confidentially assume that the use of the ideal gas equation is not valid.

ok, but why is the final pressure not equal to initial pressure in this problem?

 

[stockzahn]

ok, but why is the final pressure not equal to initial pressure in this problem?

Well, there is no statement about it. But think about a combustion engine. If you supply energy/heat, the pressure in the cylinder rises, pushing the piston to move a car or run a generator. In fact, a piston free to move is a nice idea to approach to some basic ideas of thermodynamics, but in reality it always should transfer energy/work to get something run. So a free moving piston is of theoretical interest, but not in engineering or "applied" physics.

 

[EastWindBreaks]

Well, there is no statement about it. But think about a combustion engine. If you supply energy/heat, the pressure in the cylinder rises, pushing the piston to move a car or run a generator. In fact, a piston free to move is a nice idea to approach to some basic ideas of thermodynamics, but in reality it always should transfer energy/work to get something run. So a free moving piston is of theoretical interest, but not in engineering or "applied" physics.

but in this case, the piston is not connected to anything right?

I got another question,for part b) i keep getting 370.697 for the final T and 21.223 MPa for the final pressure, a little off from the solution. I have a feeling that the solution did not simply interpolate on v, but I failed to find how it got those values.

 

[stockzahn]

but in this case, the piston is not connected to anything right?

No, but again: You know the volume and the quality after heating. Therefore you know the state, no more information necessary. Doesn't matter what's above the piston.

I got another question,for part b) i keep getting 370.697 for the final T and 21.223 MPa for the final pressure, a little off from the solution. I have a feeling that the solution did not simply interpolate on v, but I failed to find how it got those values.

Your calculations seem to be correct.Of course there will be a mistake, if you interpolate linearly, since the saturation line is curved. If I connect the critical Point (you used as one limit) and the point at the temperature of 370°C by a straight line in the T,s diagram, the obtained values by the interpolation are lower. Although I didn't check the numbers, your result is realistic. If you have only the tables, I'd say that's as close as you can get without guessing.

 

[EastWindBreaks]

No, but again: You know the volume and the quality after heating. Therefore you know the state, no more information necessary. Doesn't matter what's above the piston.
Your calculations seem to be correct.Of course there will be a mistake, if you interpolate linearly, since the saturation line is curved. If I connect the critical Point (you used as one limit) and the point at the temperature of 370°C by a straight line in the T,s diagram, the obtained values by the interpolation are lower. Although I didn't check the numbers, your result is realistic. If you have only the tables, I'd say that's as close as you can get without guessing.

Thanks a lot. so my initial pressure was correct (using h=u+Pv) ? I hope I am not asking too many questions...I am a bit confused on how temperature and pressure are independent in super-heated region, if you increase the temperature, the vapor pressure will increase since the molecules are traveling at faster velocities right? for example, if volume is held constant for a rigid container, increasing gas's temperature will increase the pressure correct?

 

[stockzahn]

Thanks a lot. so my initial pressure was correct (using h=u+Pv) ?

The idea was correct, the execution not, since you used the inner energy and the enthalpy in kJ/kg (instead J/kg) and you mixed specific values (for $h$ and $u$) and absolute values (for the volume $V$). However if you have liquid water of 200°C, you know the pressure must be much higher than atmospheric pressure. Always check the plausibility of your results. But try it again with the correct units, my calculation (based on your values for $h$ and $u$) yielded a pressure of 15.56 bar.

I am a bit confused on how temperature and pressure are independent in super-heated region, if you increase the temperature, the vapor pressure will increase since the molecules are traveling at faster velocities right? for example, if volume is held constant for a rigid container, increasing gas's temperature will increase the pressure correct?

I hope this is not a misunderstanding due to language difficulties: The vapour pressure increases with temperature, but that's not what you are interested now, I think. The vapour pressure is the maximal possible partial pressure of a substance in the gas phase.

Please let me know, if I'm mistaken, but you are asking about the correlation between pressure and temperature of a single-phase substance in the gaseous state (superheated)? In this case, I think, your idea is correct. An increased temperature (= increased velocity of the molecules) causes stronger collisions (= increased pressure).

 

[EastWindBreaks]

I hope this is not a misunderstanding due to language difficulties: The vapour pressure increases with temperature, but that's not what you are interested now, I think. The vapour pressure is the maximal possible partial pressure of a substance in the gas phase.

Please let me know, if I'm mistaken, but you are asking about the correlation between pressure and temperature of a single-phase substance in the gaseous state (superheated)? In this case, I think, your idea is correct. An increased temperature (= increased velocity of the molecules) causes stronger collisions (= increased pressure).

My professor said that temperature and pressure are independent in super-heated region, I searched for independent definition, it says that if two properties are independent, i can vary one while holding one at constant . So its like saying I can increase or decrease pressure while holding temperature constant, which is confusing to me, since like you said, increase temperature will increase the pressure.

 

[Chestermiller]

You can figure out whether the ideal gas law is going to apply or not by calculating the reduced temperature and the reduced pressure, and then going to a compressibility z plot to see how close z is to unity. In your system, you are very close to the critical point, so the ideal gas law is going to be inaccurate. Why don't you calculate the value of z from your steam table results and see what value you get?

 

[EastWindBreaks]

You can figure out whether the ideal gas law is going to apply or not by calculating the reduced temperature and the reduced pressure, and then going to a compressibility z plot to see how close z is to unity. In your system, you are very close to the critical point, so the ideal gas law is going to be inaccurate. Why don't you calculate the value of z from your steam table results and see what value you get?

oops, I forgot to convert the temperature to Kelvin, so my Z should be about 0.458, which is far from unity.

 

[Chestermiller]

View attachment 216726
oops, I forgot to convert the temperature to Kelvin, so my Z should be about 0.458, which is far from unity.

That’s at the initial condition. Try the final condition.

 

[EastWindBreaks]

That’s at the initial condition. Try the final condition.

ok, this time I got 18.46 which is very far from unity. but finding Z requires final conditions, and the question ask for final conditions...i feel like its too late to use Z to determine whether not i should use ideal gas equation, since I already have final conditions.

 

[Chestermiller]

ok, this time I got 18.46 which is very far from unity. but finding Z requires final conditions, and the question ask for final conditions...i feel like its too late to use Z to determine whether not i should use ideal gas equation, since I already have final conditions.

It should be about o.35

 

[EastWindBreaks]

It should be about o.35

what did I do wrong?

 

[Chestermiller]

You forgot to multiply by 18 and divide by 1000

 

[EastWindBreaks]

You forgot to multiply by 18 and divide by 1000

oh right, 18 is the MW. Thank You!

 

[stockzahn]

My professor said that temperature and pressure are independent in super-heated region, I searched for independent definition, it says that if two properties are independent, i can vary one while holding one at constant . So its like saying I can increase or decrease pressure while holding temperature constant, which is confusing to me, since like you said, increase temperature will increase the pressure.

In the two-phase region (under the "dome"), pressure and temperature are dependent on each other. That means, if you know one of them, you know the other. Just have a look at your steam tables. You will see that every temperature corresponds to exactly one pressure and vice versa.

In the single-phase (over-heated) region, there are three state variables (temperature, pressure, density). If you keep one of them constant you can investigate the correlation of the two others. In the example you've described in a previous post, the gas was in a rigid container, therefore you held the density constant. The ideal gas equation (without the gas constant) reads

$\frac{p}{\rho}\propto T$

In your example, if the density is a constant, you can neglect it and the equation reads

$p \propto T$.

Since the volume can't change, the derived correlation (higher temperature$\rightarrow$higher pressure) is correct. This is called a isochore change of state.

Now another example: Heating a house in the winter. Since a house is not air tight, the density can change. But the pressure in the house will correspond to the pressure of the atmosphere, therefore $p=const.$. Now the equation reads

$\frac{1}{\rho}\propto T$.

If you are heating at constant pressure and the temperature rises, the density must decrease (by molecules exiting the house to the atmosphere). This called an isobar change of state.

Finally, a change of state where the temperature is constant (isothermal). Isothermal changes of state are not common, it requires very slow changes with a good heat transfer. However, you can again use the adapted ideal gas equation, neglecting the constant temperature:

$p \propto \rho$.

If the temperature doesn't change, pressure and density are directly proportional.

 

[EastWindBreaks]

In the two-phase region (under the "dome"), pressure and temperature are dependent on each other. That means, if you know one of them, you know the other. Just have a look at your steam tables. You will see that every temperature corresponds to exactly one pressure and vice versa.

In the single-phase (over-heated) region, there are three state variables (temperature, pressure, density). If you keep one of them constant you can investigate the correlation of the two others. In the example you've described in a previous post, the gas was in a rigid container, therefore you held the density constant. The ideal gas equation (without the gas constant) reads

$\frac{p}{\rho}\propto T$

In your example, if the density is a constant, you can neglect it and the equation reads

$p \propto T$.

Since the volume can't change, the derived correlation (higher temperature$\rightarrow$higher pressure) is correct. This is called a isochore change of state.

Now another example: Heating a house in the winter. Since a house is not air tight, the density can change. But the pressure in the house will correspond to the pressure of the atmosphere, therefore $p=const.$. Now the equation reads

$\frac{1}{\rho}\propto T$.

If you are heating at constant pressure and the temperature rises, the density must decrease (by molecules exiting the house to the atmosphere). This called an isobar change of state.

Finally, a change of state where the temperature is constant (isothermal). Isothermal changes of state are not common, it requires very slow changes with a good heat transfer. However, you can again use the adapted ideal gas equation, neglecting the constant temperature:

$p \propto \rho$.

If the temperature doesn't change, pressure and density are directly proportional.

Thank you! really appreciated!