Thermodynamics problem -- Steam cooled until it is just dry saturated

[shobaky]

"HOMEWORK" PROBLEM ORIGINALLY POSTED IN WRONG FORUM, SO NO TEMPLATE

a rigid vessel of volume 1m^3 contains steam at 25 bar and 450c the vessel is cooled until the steam is just dry saturated calculate the mass in the vessel the final pressure and the heat rejected during the process?
i know i have to use the temperature table i tried to solve it i just want to make sure that i clearly understand .. thanks in advance

 

[Chestermiller]

Is this a homework problem?

 

[shobaky]

no that's a problem my professor solved it in today's lecture but i didn't clearly understand so i tried to solve it again but my fault is that i didn't write the final answers :/

 

[Chestermiller]

no that's a problem my professor solved it in today's lecture but i didn't clearly understand so i tried to solve it again but my fault is that i didn't write the final answers :/

This is still a homework-like problem, so I am moving it to the homework forums (Engineering and Computer Science).

In the initial state, from your steam tables, what is the specific volume of your superheated steam?

 

[shobaky]

it's .13

 

[Chestermiller]

it's .13

So, based on this, what is the mass of water in the vessel?

 

[shobaky]

i think it's V/v so it's 7.69 kg i could calculate the mass and the final pressure , i amn't sure though, but my main problem is with the heat rejected

 

[Chestermiller]

i think it's V/v so it's 7.69 kg i could calculate the mass and the final pressure , i amn't sure though, but my main problem is with the heat rejected

How did you determine the final pressure? What is the specific volume in the final saturated state?

 

[shobaky]

first of all thanks of your attention and for your time...
the specific volume is constant right? so the pressure can be determined from the table as v(double dash) = .13?

 

[Chestermiller]

first of all thanks of your attention and for your time...
the specific volume is constant right? so the pressure can be determined from the table as v(double dash) = .13?

Excellent. So you know the final state. The key to getting the rejected heat is the word "rigid" in the problem statement. How much work does the steam do on the rigid vessel in this problem? From the first law, what does that tell you about the relationship between the change in internal energy and the heat?

 

[shobaky]

do you mean it's a closed system so i have to use the equation " delta h = delta u + delta P * v" ?? then i have to determine delta h from the table and i know the value of specific volume and of course i can determine delta p .. am i right? :/ .. thanks in advance

 

[Chestermiller]

do you mean it's a closed system so i have to use the equation " delta h = delta u + delta P * v" ?? then i have to determine delta h from the table and i know the value of specific volume and of course i can determine delta p .. am i right? :/ .. thanks in advance

No. I mean that $$\Delta U=Q$$

 

[shobaky]

yes but how can i calculate the delta U :/ ?

 

[Chestermiller]

yes but how can i calculate the delta U :/ ?

You know everything about the initial state, so you know the internal energy for that.

For the final state, you know that it is dry saturated, and you know its specific volume, so you should be able to precisely pinpoint this state in your steam tables. You can then determine its internal energy.

 

[shobaky]

You know everything about the initial state, so you know the internal energy for that.

For the final state, you know that it is dry saturated, and you know its specific volume, so you should be able to precisely pinpoint this state in your steam tables. You can then determine its internal energy.

oh i got your point .. Thanks a lot i really appreciate your valuable time