0
$\begingroup$

I'd appreciate some help in squaring this thermodynamic circle (which may exist only in my mind, but I digress ...)

Let's consider pure water in a two-phase(liquid/vapour) state. Within the vapour dome, the density of water can be determined from mass and volumetric balances according to

$$\rho = \Bigg[\frac{x}{\rho_V(P)}+\frac{1-x}{\rho_L(P)}\Bigg]^{-1} $$ where $x$ is the vapour mass (not volume) fraction (which we knuckle-dragging engineers call the "steam quality"), $P$ is the pressure, and where $\rho_L(P)$ and $\rho_V(P)$ are the saturated liquid and saturated vapour densities to the left and right of the critical point (apex of the vapour dome) respectively. Both $\rho_L$ and $\rho_V$ are functions of pressure $P$, alone. The classical definition of the compressibility (really the isenthalpic compressibility within the vapour dome since the vapour mass fraction is a proxy for enthalpy here) is

$$ \beta=-\frac{1}{V}\frac{\partial V}{\partial P}=\rho\frac{\partial \rho}{\partial P} $$

Now along the saturated liquid curve $x=0$ (to the left of the vapour dome apex), the density drops with pressure (because the dominant effect is that of the increase in the boiling point with an increase in pressure), so that for a saturated liquid, the compressibility is

$$ \beta_{x=0}=\rho_L\frac{\partial \rho_L}{\partial P} $$

which would be negative. This defies not only the physics, but is also inconsistent with the (positive) compressibility that would be evaluated form a simple finite difference between a point on the saturated liquid line and a point at the same pressure in the subcooled region just above and to the left of the vapour dome boundary. I suspect that I'm missing some key concept here.

Any advice?

Thanks

$\endgroup$
  • $\begingroup$ I agree that by, increasing pressure, the boiling temperature increases. But why does density decrease? Would it increase as well? $\endgroup$ – user115350 Oct 29 '17 at 1:32
  • $\begingroup$ The density decrease is simply due to the particular behavior of water at saturated liquid conditions. If you look at the tabulated values of specific volume (reciprocal of density) you will notice that the saturated liquid specific volume increases with pressure which implies a decrease in density with an increase in pressure up to the top of the vapor dome. Continuing along the vapor dome boundary from left to right past the apex results in a decrease in pressure as well as density which is what one would expect. My issue really has to do with the evaluation of the compressibility. $\endgroup$ – Sharat V Chandrasekhar Oct 29 '17 at 3:05
0
$\begingroup$

I just realised the fallacy in my assertion that the mass fraction is a proxy for enthalpy. This is only true along an isobar within vapour dome. Since the partial derivative in question is w.r.t pressure at constant enthalpy, an evaluation along the vapour dome boundary is meaningless. Very stupid of me not to have realised this earlier.

I have to define the partial derivative at constant enthalpy, not mass fraction. When I evaluate it with a backward finite difference between the density at a point on the saturated liquid line and a point vertically below it within the vapour dome on a Pressure-Enthalpy plot, It all works out as one would expect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.