Question

If given pressure is less than saturated pressure for any given temperature what would be the phage of state ?

If the tank full of methane, contain 50% liquid and 50% vapor, by volume at 100 kPa. What is the dryness fraction?

Answer 1

(a) Answer: Superheated Phase.

Higher the pressure of a liquid, the higher the temperature required to reach its saturation state. Hence having actual pressure () less than saturation pressure ( $P_{sat@T}$ ) means, that the saturation temperature ( $T_{sat@P}$ ) corresponding to the actual pressure () of the liquid will be less than the actual temperature (). This would imply that the phase is superheated.

Simpler way to remember it is, if the actual pressure is more than the saturated pressure, then the phase is compressed (because it has more pressure) and if the actual pressure is less than the saturation pressure, then the phase is superheated (since the actual pressure is low, the saturation temperature corresponding to the actual pressure will be low).

____________________________________________________________________

(b)

From 'Properties of Saturated Methane' Table, at   ;

$\nu_g=0.55665\;\;m^3/kg$

$\nu_f=0.002366\;\;m^3/kg$

Hence,

Mass of fluid will be;

$m_f=\frac{0.5V}{\nu_f}$

Mass of vapor will be;

$m_g=\frac{0.5V}{\nu_g}$

Total mass will be;

$m = m_f+m_g=\frac{0.5V}{\nu_f}+\frac{0.5V}{\nu_g}$

Dryness fraction will be;

$x = \frac{m_g}{m} = \frac{m_g}{m_f+m_g}=\frac{\frac{0.5V}{\nu_g}}{\frac{0.5V}{\nu_f}+\frac{0.5V}{\nu_g}}$

$\Rightarrow x = \frac{\nu_f\nu_g}{\nu_g(\nu_f+\nu_g)}$

$\Rightarrow x = \frac{(0.002366)(0.55665)}{(0.55665)(0.002366+0.55665)}$

$\Rightarrow x =0.00423$