# Thermodynamics: Liquid-Vapour Mixture Internal Energy

## 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.

SteamKing
Staff Emeritus
Homework Helper

## 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?

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?

SteamKing
Staff Emeritus
Homework Helper
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?

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.

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..

SteamKing
Staff Emeritus
Homework Helper
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

Chestermiller
Mentor
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

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

SteamKing
Staff Emeritus
Homework Helper
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:

## 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.

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?

Chestermiller
Mentor
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?

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

Chet

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.

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?

SteamKing
Staff Emeritus
Homework Helper
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?

Sirsh
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.