Non-flow process with steam (Constant Pressure)

Click For Summary

Homework Help Overview

The discussion revolves around two questions involving steam processes in closed systems. The first question concerns wet steam at a pressure of 5 bar and a dryness fraction of 0.2, which is heated at constant pressure until a specified volume is reached. The second question involves superheated steam at 10 bar cooling at constant pressure. Participants are exploring the implications of heating wet steam and the conditions under which it may remain wet versus becoming dry or superheated.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which wet steam can remain wet after heating and question how to determine the state of steam when not explicitly stated in the problem. They explore the relationship between temperature, pressure, and steam state, including the implications of constant pressure heating.

Discussion Status

The discussion is active, with participants providing insights into the behavior of steam under various conditions. Some guidance has been offered regarding the relationship between steam state and temperature, but there is no explicit consensus on how to approach the problems or determine the steam state without additional information.

Contextual Notes

Participants note the lack of explicit hints in the problems regarding the steam state and express confusion about the implications of heating wet steam at constant pressure. There is also mention of the need for steam tables to determine specific properties, which may not be readily available to all participants.

freshbox
Messages
290
Reaction score
0
Question 1
In a closed system, 0.6kg of wet steam at a pressure of 5 bar and a dryness fraction of 0.2 is heated at constant pressure until the volume occupied is 0.18m3.
(A)Calculate the change in internal energy, the work energy and the heat energy transferred during the process.

Question 2
In a closed system, 0.25m3 superheated steam at a pressure of 10 bar and a temperature of 300°C cools at a constant pressure until the volume becomes 0.15m3. Determine:
(A)The mass of the steam in the system
(B)The dryness fraction of the steam after cooling.

For question 2, the question says that find the dryness fraction of the steam after cooling, this shows that the steam is wet steam.

But for question 1, the question did not give me a hint to find dryness fraction of the steam. How do I know that I need to find the dryness fraction in order to solve the question?

And what confuses me is that how did the wet steam from question 1 remains as wet steam after heating? I thought it will be either dry saturated steam or superheated steam.Thanks..
 
Physics news on Phys.org
What happens to the wet steam depends on how much heat goes into it. It may not have received enough to become dry. That's what you need to find out.
 
Do you mean that after heating the wet steam may still become a wet steam because it has not received enough to become dry?
 
Exactly.
 
Woah that solve part of the mystery hehe... Ok if you look at question 2, i know it is a wet steam through "determine The dryness fraction of the steam after cooling."

For question 1, how do i determine what kind of steam it is because it did not give me any hint at all. Thanks.
 
If a wet steam is heated under constant pressure, what happens to its temperature and volume (before it gets completely dry)?
 
temperature rises and volume getting lesser?
 
The wet steam is in water-vapor equilibrium. At constant pressure, can the equilibrium temperature change?
 
I'm sorry I don't understand. I only know in a constant pressure process, p1=p2=p.
 
Last edited:
  • #10
When you boil water in your kitchen, you add heat to the water-vapor system held at a constant (atmospheric) pressure. Does its temperature change while it is boiling?
 
  • #11
No the temperature will not change because it reaches it's boiling point already. So what are you trying to tell me?

Am I right to say:
Compressed Water/Saturated Water/Wet Steam: Have Volume
Dry Saturated Steam/Superheated Steam: No Volume
 
  • #12
Wet vapor, when held at a constant pressure, won't change in temperature when heated: the heat will be use to "boil" water in it.

I am not sure what you are trying to say about "having volume". Everything has some volume.
 
  • #13
So do you mean to find out whether stage 1 steam is the same as stage 2 steam, i have to work out the temperature first?

So for wet steam when heated at a constant pressure remain as it is. Does this applies to wet steam only or how about the other steam like example: dry saturated.Thanks
 
Last edited:
  • #14
Wet steam, when heated at constant pressure, does not remain as it is. It becomes drier, which means there is more steam in the water-steam mixture. And if you have more steam, with pressure and temperature constant, what happens to its volume?
 
  • #15
with more steam, i think the volume will increase.

Since you say wet steam when heated at constant pressure, it becomes drier and does not remain as it is but how come for question 1 it remains as wet steam? And you say that because it has not received enough to become dry.

So how do i determine whether it has changed into a different steam or remain the same steam?

Sorry I'm very confused :confused:
 
  • #16
Start from the beginning. 0.6kg of wet steam at a pressure of 5 bar. What is its temperature? Remember, the water in this mixture must be at the boiling point for the pressure given.
 
  • #17
Ok if according to the steam table @ 5bar, the temperature should be 151.8°C
 

Attachments

  • for.jpg
    for.jpg
    28.4 KB · Views: 636
  • steamtable.jpg
    steamtable.jpg
    62.5 KB · Views: 751
  • #18
Very well. Now that you know the temperature and pressure, what is the volume? Compute separately the volume of the saturated steam component, and the volume of the water component.
 
  • #19
Stage 1 working:

Given from question: mass =0.6kg, P=5 Bar, x=0.2


vx=xvg
vg@5bar=0.3748 from chart.
0.2=vx/vg
0.2=vx/0.3748
vx=0.07496m3/kg

v=V/m
0.07496=V/0.6
V=0.04496m3 (Ans)
 
  • #20
That is not exactly what I asked, but that is a start. Now, assume that all the water in the wet steam becomes steam, so you have saturated steam. What volume would that be? Compare with the volume given. What is your conclusion?
 
  • #21
This is what i know:

compressed water > saturated water > wet steam > dry saturated steam > superheated steam

"so you have saturated steam." what steam are you talking about here, dry saturated?


Thanks.
 

Attachments

  • 20121120_172911.jpg
    20121120_172911.jpg
    22.2 KB · Views: 555
  • #22
if the water in the wet steam becomes steam, i have to recalculate my specific volume.

v=V/m
vg@5bar=0.3748 from chart.
0.3748=V/0.6
V=0.22488m3

My conclusion is the volume with pure steam is heavier than wet steam.
 
  • #23
So if 0.6 kg of wet steam becomes dry saturated stream at constant 5 bar, the volume becomes 0.22 m3. However, the problem states that the end volume was 0.18 m3. Knowing this, answer this question: has all of the wet steam become dry saturated steam?
 
  • #24
No it did not changed to dry saturated steam because for it to become dry saturated it has to be 0.22m3
 
  • #25
Correct. So what you need to find out is how much water became steam, and, knowing the pressure and temperature, you can answer all the questions.
 
  • #26
In a closed system, 0.05.kg of dry saturated steam is heated at a constant pressure of 8 bar until the volume is occupied is 0.018m3
(A)Calculate the change in internal energy, the work energy transferred and the heat energy transferred during the process.

For this question how do I determine? dry saturated steam and superheated are both steam, no water components in it.
 
  • #27
Dry steam is basically an ideal gas, forget about phase transitions and use the ideal gas laws.
 
  • #28
What do you mean by ideal gas laws?
 
  • #29
"perfect" gas as per your formula sheet.
 
  • #30
I thought this question is on non flow process with steam?

I am not given R/Cp/Cv/Gamma, how do I use U2-U1=mCv(T2-T1)?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K