Liquidification of Helium under constant Enthelphy Condition

[Marcus95]

Homework Statement

a)Helium enters a closed system as a flow of compressed gas at a temperature
of 14 K and at pressure p (enthalpy conserved). A fraction α emerges as liquid and the rest as gas at 14 K, both at atmospheric pressure p0. Find an expression for α in terms of the enthalpy H(p) of He gas and the enthalpy Hℓ of the liquid. [6]

b)From the data below, find the maximum value of α and the initial pressure p for
which this occurs. [4]

p/[atm] | H/[$10^4$J/kg] at 14 K

0 | 8.74

10| 7.85

20 | 7.31

30 | 7.18

40 | 7.26

[The enthalpy Hℓ of liquid He at p0 is $1.01 × 10^4$ J kg−1.]

Homework Equations

Enthalpy: $H = U + pV$

The Attempt at a Solution

a) In a closed flow process, enthalpy is conserved. Hence we must have:

$$ H(p) = \alpha H_l + (a-\alpha) H(p_0) $$

which simplifies to:
$$ \alpha = \frac{H(p) - H(p_0)}{H_l - H(p_0)} (\star) $$

to write this as a function of $H_l$ and $H(p)$ only, we somehow need to express $H(p_0)$ in the other functions.

The definition of enthalpy: I thought that if we take the helium to be an ideal monoatomic gas we simply have: $$H = \frac{3}{2} nRT + nRT = \frac{5}{2} nRT $$ or $$H = \frac{5}{2} RT/M $$ per kg (as that seems to be the prefferred unit), M being the molar mass.

However, the problem is that $T = T_0 = 14 K$, hence the expression above would suggest that $H(p_0) = H(p)$ and $\alpha = 0$. This must clearly be wrong and the helium cannot be modeled as an ideal gas.

Does anybody know how $(\star)$ can be simplified? Considering that 60% of the marks (old exam question) are awarded for this part there must be something more to it than this.

b) Nevertheless, I have tried to progress with the question here by simply interpolating H(p_0) linearly from the data given. It leaves me with:¨
$$\alpha = \frac{8.651 - H(p)}{7.641} $$
and this is a linear function without a maximum...

 

[Chestermiller]

Why are you assuming ideal gas behavior? It’s obviously not. Also, shouldn’t your equation have $1-\alpha$ ?

 

[Marcus95]

Why are you assuming ideal gas behavior? It’s obviously not. Also, shouldn’t your equation have $1-\alpha$ ?

As mentioned in the post, I was assuming ideal gas behaviour but realized it didn't work. This is the reason I am asking for help, to find alternatives.

Yes, the equation is a typo. It should indeed read $1-\alpha$

 

[Chestermiller]

As mentioned in the post, I was assuming ideal gas behaviour but realized it didn't work. This is the reason I am asking for help, to find alternatives.

Yes, the equation is a typo. It should indeed read $1-\alpha$

So, what do you get for alpha as a function of p?

 

[Marcus95]

So, what do you get for alpha as a function of p?

Well the enthalphy is defined as $H = U + pV$, or as a differential: $dH = TdS + V dp$ which means that it is indeed a function of p, but what kind of function I don't know. Should I assume a van der Walls gas instead of an ideal one? Would that help?

 

[Chestermiller]

Well the enthalphy is defined as $H = U + pV$, or as a differential: $dH = TdS + V dp$ which means that it is indeed a function of p, but what kind of function I don't know. Should I assume a van der Walls gas instead of an ideal one? Would that help?

Using the table, what is the value of alpha at each pressure?

 

[Marcus95]

Using the table, what is the value of alpha at each pressure?

Alright, so using the table I can see that by choosing the value at 30 atm, ie the approximate minimum of the H function, I get the maximum alpha. This solves the second part of the question.

Now only the first part remains, ie how to express $\alpha$ in $H(p)$ and $H_l$ only. Any hint on this?

 

[Chestermiller]

Alright, so using the table I can see that by choosing the value at 30 atm, ie the approximate minimum of the H function, I get the maximum alpha. This solves the second part of the question.

Now only the first part remains, ie how to express $\alpha$ in $H(p)$ and $H_l$ only. Any hint on this?

$$\alpha=\frac{H(p)-H(p_0)}{H(p)-H_l(p_0)}$$

 

[Marcus95]

$$\alpha=\frac{H(p)-H(p_0)}{H(p)-H_l(p_0)}$$

well yes, that was my initial answer! Does that mean that it cannot be further simplified?

 

[Chestermiller]

well yes, that was my initial answer! Does that mean that it cannot be further simplified?

That is all you need to quantify the relationship between alpha and P (using the data that has been supplied to you in this problem.)

 

[Marcus95]

That is all you need to quantify the relationship between alpha and P (using the data that has been supplied to you in this problem.)

Are you hinting on that I could calculate alpha for each given value, assuming a linear interpolation for $H(P_0)$?
Or that I should leave it as it is considering the questions says: " Find an expression for α in terms of the enthalpy H(p) of He gas and the enthalpy Hℓ of the liquid. "? and the overall H(p) function is not nicely linear?

 

[Chestermiller]

Are you hinting on that I could calculate alpha for each given value, assuming a linear interpolation for $H(P_0)$?
Or that I should leave it as it is considering the questions says: " Find an expression for α in terms of the enthalpy H(p) of He gas and the enthalpy Hℓ of the liquid. "? and the overall H(p) function is not nicely linear?

Part (b) requires you to actually calculate alpha. How you do the interpolation to get H(p0) is up to you.