Thermodynamics: Isobaric process finding volume of steam.

In summary, the equations state that v=vf+x(vf-vg), where x is the dryness fraction. If the water is a pure liquid, then v=vf. If the water is a pure vapor, then v=vfg. When water changes from liquid to vapor, there is a change in pressure. When calculating heat supplied, Q is used to find the difference in temperatures between two points. For this problem, U1 was found to be 1670.4 kilojoules, and U2 was found to be 2734 kilojoules.
  • #1
craglyboy
8
0

Homework Statement


http://img140.imageshack.us/img140/2963/isobaricsteam.jpg

P1 and P2 are 1 bar. V2 4.812 and V2 specific volume is 2.406

X1=0.6


Homework Equations



v=vf+x(vf-vg)

W=-P(v2-v1)

Q=(u2+pv2)-(u1+pv1)



The Attempt at a Solution



Hi guys first timer here just looking where to start off, I guess I need to find out the specific volume to find out V1 but I am stumped at how to use the dryness fraction to do that any help would be greatly appreciated
 
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  • #2
The dryness fraction is used to compute the specific volume before heating.

v=vf+x(vf-vg)

Is the above correct? Looks to me that you'll get a negative value for v because vg>vf.

You can compute v at the end of the process.

Apply first law.
 
  • #3
thanks yeah that was just a typo supposed to be (Vg-Vf).

The thing is I don't know values Vg and Vf are I have looked up in my steam tables and it says Vg is 1.694 at 1 bar but what is Vf?
 
  • #4
Is Vf zero?
 
  • #5
No, vf is not zero but it is quite a bit smaller than vg. My steam tables have 3 columns. One is vf, another is vfg, and the third is vg. vfg = vg-vf. It's the change in specific volume at a specified pressure when water changes from liquid to vapor.
 
  • #6
yeah Vf is tiny now I have found it got work done to -277.9KJ.

for U1 I got 1670.4KJ from using the formula u=uf+x(ug-uf) and for u2 i thought because i know the specific volume look it up in the steam tables it comes to 2734KJ but I don't get the right answer for heat supplied.

Because it is a isobaric process I used Q=(u2+pv2)-(u1+pv1) but getting no where near the answer so I must of done something wrong.
 
Last edited:
  • #7
OK, I'll check values from my columns and supply them to you. Then you can decide for yourself whether using vf is worthwhile.

My system is in SAE units.

at 14.696 psi vf= 0.01672; vfg=26.811; vg=26.828

Units are ft^3/lb

So forgetting about vf only creates a very small error.
 
  • #8
craglyboy said:
yeah Vf is tiny now I have found it got work done to -277.9KJ.

for U1 I got 1670.4KJ from using the formula u=uf+x(ug-uf) and for u2 i thought because i know the specific volume look it up in the steam tables it comes to 2734KJ but I don't get the right answer for heat supplied.

Because it is a isobaric process I used Q=(u2+pv2)-(u1+pv1) but getting no where near the answer so I must of done something wrong.

thanks man can you see where i am being an idiot here, is it just for U2 I can't use that value because its not a pure dry mixture.
 
  • #9
You can determine whether it is superheated by computing v2. Recall, it was .6 quality before heating at constant pressure. You know the volume, the pressure, and how many kg you have. See where that value of v falls in the steam tables. From that value, you can interpolate to get u2.
 
  • #10
I got the values of U right I was just being an idiot and forgot the values were specific values and forgot factor in that I had 2kg of mass not 1. So I have got the right answer finally!

Thankyou Lawrence C you have been extremely helpful!
 

Related to Thermodynamics: Isobaric process finding volume of steam.

1. What is an isobaric process in thermodynamics?

An isobaric process is a thermodynamic process in which the pressure remains constant. In other words, the system undergoes a change in volume while the pressure remains the same.

2. How is the volume of steam calculated in an isobaric process?

The volume of steam in an isobaric process can be calculated using the ideal gas law, which states that the product of pressure and volume is equal to the product of the number of moles, the gas constant, and the temperature. So, if the pressure and temperature are known, the volume can be calculated.

3. What is the significance of an isobaric process in thermodynamics?

An isobaric process is important because it allows us to study the behavior of gases under constant pressure, which is a common scenario in many real-world situations. It also helps in understanding the relationship between pressure, volume, and temperature in a system.

4. How does an isobaric process differ from an isothermal process?

An isobaric process is different from an isothermal process in that the temperature remains constant in an isothermal process, while the pressure remains constant in an isobaric process. In other words, an isothermal process involves a change in volume while keeping the temperature constant, while an isobaric process involves a change in volume while keeping the pressure constant.

5. Can an isobaric process be reversible?

Yes, an isobaric process can be reversible, meaning that the system can be taken back to its original state without any change in entropy. This is possible if the process is carried out slowly enough, allowing the system to reach equilibrium at each step and minimizing any energy losses.

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