Chapter 6, Problem 6.49P

Interpretation Introduction

Interpretation:

The fraction of propane gas liquified in the given processneeds to be calculated.

Concept Introduction:

The equation to calculate the reduced vapor pressure as a function of reduced temperature is:

  $P_r^{\text{sat}}=\exp \left(\frac{A\tau +B{\tau }^{1.5}+C{\tau }^3+D{\tau }^6}{1-\tau }\right)$ ...... (1)

Here,

  $\begin{array}{c}\tau =1-T_r\\ T_r=\frac{T}{T_c}\\ P_r=\frac{P}{P_c}\end{array}$

The equation used to calculate the molar volume of vapor is:

  $V_{vap}=\frac{ZRT}{P}$ ...... (2)

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature of the system, and $R$ is the universal gas constant. $Z$ is the compressibility factor and is defined as:

  $Z=1+\left(B^0+\omega B^1\right)\frac{P_r}{T_r}$ ...... (3)

Here, $\omega$ is an acentric factor given in table B.1 and $B^0\text{ and }{B}^1$ are calculated as:

  $\begin{array}{c}{B}^0=0.083-\frac{0.422}{T_r^{1.6}}\\ B^1=0.139-\frac{0.172}{T_r^{4.2}}\end{array}$ ...... (4)

To calculate molar volume of saturated liquid, the formula to be used is:

  $V_{liq}=V_c{Z}_c^{{\left(1-T_r\right)}^{2/7}}$ ...... (5)

The formula to calculate enthalpy of phase transition from liquid to vapor $\left(\Delta H^{lv}\right)$ is:

  $\Delta H^{lv}=\left(V_{vap}-V_{liq}\right)T\frac{d{P}^{\text{sat}}}{dT}$ ...... (6)

The formula to calculate enthalpy of vapor $\left(\Delta H^{ig}\right)$ is:

  $\Delta H^{ig}={\displaystyle {\int }_{T_1}^{T_2}{C}_p^{ig}dT}$ ...... (7)

The value of $C_p^{ig}$ is taken from table C.1.

The equation to calculate the value of $H^R$ is defined as:

  $H^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1$ ...... (8)

Here, ${\left(H^R\right)}^0\text{ and }{\left(H^R\right)}^1$ are constant values.