Thermodynamics: Liquid-Vapour Mixture Internal Energy

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Discussion Overview

The discussion revolves around calculating the dryness fraction of wet steam at 11 bar, given that 75% of the total internal energy is contained in the vapor portion. Participants explore the relationship between internal energy, dryness fraction, and the definitions of saturated liquid and vapor states, engaging in a homework-related problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that the equation for internal energy is u = (1-x)uf + x*ug, where u is the total internal energy, uf is the internal energy of the saturated liquid, and ug is that of the saturated vapor.
  • One participant suggests that if 75% of the total internal energy is in the vapor, then the dryness fraction could be interpreted directly as 0.75.
  • Another participant proposes that the total internal energy should be expressed as 0.25*uf + 0.75*ug to reflect the contributions from both phases.
  • There is a discussion about the correct values for uf and ug, with some participants noting discrepancies in steam tables.
  • One participant expresses confusion regarding the treatment of internal energy as an absolute quantity, raising concerns about the reference point for internal energy values.
  • A later reply emphasizes the need to translate the problem statement into algebraic formulas to solve for the dryness fraction accurately.
  • Another participant arrives at a dryness fraction of 0.148, questioning its implication regarding the proportions of vapor and liquid in the mixture.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the dryness fraction, with multiple competing views and interpretations of the problem statement remaining unresolved.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about internal energy and the reference states used in calculations. Participants also note potential errors in the values of uf and ug from different steam tables.

Sirsh
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Homework Statement


In a sample of wet steam at 11 bar, 75% of the total internal energy is contained in the vapour portion. Calculate the dryness fraction of the steam to 3 decimal places.

Homework Equations


u(kJ/kg) = (1-x)uf + x*ug

x - dryness fraction
f - saturated liquid state
g - saturated vapour state

The Attempt at a Solution



I have done the following to no avail:

At 11 bar using steam tables uf = 780 kJ/kg and ug = 2568 kj/kg

Assuming u = 75% of ug.

u = (1-x)uf + x*ug
0.75*ug = (1-x)uf + x*ug
0.75*2568 = (1-x)*(780) + x*(2568)
x = 0.642

Therefore, dryness fraction is equal to 0.642.

However this is incorrect and I'm unsure what I'm doing wrong. Possibly interpreting the 75% of total internal energy wrong.

Could it be that I need to use u = 0.25*uf + 0.75*ug?

Thanks for your help, it's appreciated.
 
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Sirsh said:

Homework Statement


In a sample of wet steam at 11 bar, 75% of the total internal energy is contained in the vapour portion. Calculate the dryness fraction of the steam to 3 decimal places.

Homework Equations


u(kJ/kg) = (1-x)uf + x*ug

x - dryness fraction
f - saturated liquid state
g - saturated vapour state

The Attempt at a Solution



I have done the following to no avail:

At 11 bar using steam tables uf = 780 kJ/kg and ug = 2568 kj/kg

Assuming u = 75% of ug.

u = (1-x)uf + x*ug
0.75*ug = (1-x)uf + x*ug
0.75*2568 = (1-x)*(780) + x*(2568)
x = 0.642

Therefore, dryness fraction is equal to 0.642.

However this is incorrect and I'm unsure what I'm doing wrong. Possibly interpreting the 75% of total internal energy wrong.

Could it be that I need to use u = 0.25*uf + 0.75*ug?

Thanks for your help, it's appreciated.
Nice work, but unnecessary.

Review the definition of vapor quality:

https://en.wikipedia.org/wiki/Vapor_quality

Now can you tell what the dryness fraction of the steam is if 75% of the internal energy is in the vapor?
 
SteamKing said:
Nice work, but unnecessary.

Review the definition of vapor quality:

https://en.wikipedia.org/wiki/Vapor_quality

Now can you tell what the dryness fraction of the steam is if 75% of the internal energy is in the vapor?

I read the article but am still a bit confused.

Im thinking that the dryness fraction should be 0.75 as if it was 100% of internal energy in the vapor it'd be 1.0 and if 0% in the vapor it would be 0.0, is this thinking correct?

Thank you for your response.
 
Sirsh said:
I read the article but am still a bit confused.

Im thinking that the dryness fraction should be 0.75 as if it was 100% of internal energy in the vapor it'd be 1.0 and if 0% in the vapor it would be 0.0, is this thinking correct?

Thank you for your response.
The dryness fraction represents that fraction of the steam which is vapor. If you have steam which is all vapor, then the dryness fraction is 1.0 by definition.
 
SteamKing said:
The dryness fraction represents that fraction of the steam which is vapor. If you have steam which is all vapor, then the dryness fraction is 1.0 by definition.

X = internal energy of vapor / total internal energy

X = (0.75*ug)/(uf+ug)
X = (0.75*2568)/(780+2568)
X = 0.58

Some reason I'm not confident with this, though it would be reasonable given a mass instead of internal energy..
 
Sirsh said:
X = internal energy of vapor / total internal energy

X = (0.75*ug)/(uf+ug)
X = (0.75*2568)/(780+2568)
X = 0.58

Some reason I'm not confident with this, though it would be reasonable given a mass instead of internal energy..
The OP stated that 75% of the total internal energy was from the vapor, not 75% of the internal energy of the vapor, which is what 0.75*2568 represents...

The total internal energy of 1 kg of steam cannot be (780 + 2586) kJ/kg, since that would mean this sample is composed of 100% liquid and 100% vapor, a physical impossibility.

BTW, my steam tables say that ug = 2586 kJ/kg
 
I have a real issue with this problem statement. In particular, I have a problem with treating the internal energy as if it has an absolute quantity. The internal energy in the steam tables is specified relative to liquid water at 0 C, and is thus a relative quantity. If we adopt a different reference temperature for zero internal energy, the answer to this problem changes.

Chet
 
SteamKing said:
The OP stated that 75% of the total internal energy was from the vapor, not 75% of the internal energy of the vapor, which is what 0.75*2568 represents...

The total internal energy of 1 kg of steam cannot be (780 + 2586) kJ/kg, since that would mean this sample is composed of 100% liquid and 100% vapor, a physical impossibility.

BTW, my steam tables say that ug = 2586 kJ/kg

So would the total internal energy be equal to: 0.25*uf + 0.75*ug? This would represent 75% of the internal energy from the vapor and 25% of the internal energy from the liquid.

If that's correct, then would the mass of the total vapor be 0.75*ug so;

X = (0.75*ug)/(0.75*ug+0.25*uf)
X = (0.75*2568) / (0.75*2568+0.25*780)
X = 0.91
 
Sirsh said:
So would the total internal energy be equal to: 0.25*uf + 0.75*ug? This would represent 75% of the internal energy from the vapor and 25% of the internal energy from the liquid.

If that's correct, then would the mass of the total vapor be 0.75*ug so;

X = (0.75*ug)/(0.75*ug+0.25*uf)
X = (0.75*2568) / (0.75*2568+0.25*780)
X = 0.91
Let's start over with the analysis of what the problem statement says:

Sirsh said:

Homework Statement


In a sample of wet steam at 11 bar, 75% of the total internal energy is contained in the vapour portion. Calculate the dryness fraction of the steam to 3 decimal places.

Homework Equations


u(kJ/kg) = (1-x)uf + x*ug

x - dryness fraction
f - saturated liquid state
g - saturated vapour state

The Attempt at a Solution



I have done the following to no avail:

At 11 bar using steam tables uf = 780 kJ/kg and ug = 2568 kj/kg

I agree that uf = 780 kJ/kg; but my tables show ug = 2586 kJ/kg, rather than 2568 kJ/kg (I think the last two digits got transposed somehow)

Assuming u = 75% of ug.

u = (1-x)uf + x*ug

I agree up to here.

0.75*ug = (1-x)uf + x*ug

The problem wants 75% of the total internal energy to come from the internal energy of the vapor portion.

So what you must do is:
1. Translate "75% of the total internal energy" into an algebraic formula.
2. Translate "internal energy of the vapor portion" into an algebraic formula.
3. Solve the two equations for the dryness faction X which makes both statements true.
 
  • #10
SteamKing said:
Let's start over with the analysis of what the problem statement says:
I agree that uf = 780 kJ/kg; but my tables show ug = 2586 kJ/kg, rather than 2568 kJ/kg (I think the last two digits got transposed somehow)
I agree up to here.
The problem wants 75% of the total internal energy to come from the internal energy of the vapor portion.

So what you must do is:
1. Translate "75% of the total internal energy" into an algebraic formula.
2. Translate "internal energy of the vapor portion" into an algebraic formula.
3. Solve the two equations for the dryness faction X which makes both statements true.

If 75% of the total internal energy is from the vapor then the rest of the internal energy (25%) comes from the liquid.

Can we not just do:

u(total) = 0.25*uf + 0.75*ug

If u(total) = (1-x)*uf + x*ug,

then x is just 0.75?

Thanks for your help
 
  • #11
Sirsh said:
If 75% of the total internal energy is from the vapor then the rest of the internal energy (25%) comes from the liquid.

Can we not just do:

u(total) = 0.25*uf + 0.75*ug

If u(total) = (1-x)*uf + x*ug,

then x is just 0.75?

Thanks for your help
No. $$\frac{xu_g}{(1-x)u_f+xu_g}=0.75$$
Does that make sense?

Chet
 
  • #12
Chestermiller said:
No. $$\frac{xu_g}{(1-x)u_f+xu_g}=0.75$$
Does that make sense?

Chet

That makes complete sense, thank you!

@SteamKing thank you for all your help it's very appreciated.
 
  • #13
I end up with a dryness fraction, x, of 0.148.

Doesn't this signify that the mixture is 14.8% vapour and 85.15% liquid?
 
  • #14
Sirsh said:
I end up with a dryness fraction, x, of 0.148.

Doesn't this signify that the mixture is 14.8% vapour and 85.15% liquid?
I think you've messed up your arithmetic. Did you plug your value of X back into the formula in Post #11?
 
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  • #15
SteamKing said:
I think you've messed up your arithmetic. Did you plug your value of X back into the formula in Post #11?

Ah yes, I had a negative where a positive should have been. I solved it and got the dryness fraction to be equal to 0.475.

Thank you very much.
 

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