Question

Consider a mixture of hydrocarbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. the mixture pressure before and after the separation is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture.

Answer 1

Answer:

$\Delta P_m=6\text{kPa}\\\Delta P_e=3\text{kPa}\\\Delta P_p=-9\text{kPa}$

Explanation:

mole fraction of propane after passing through the separator is $\beta_p$

$\frac{\beta}{0.6+0.3+\beta}=0.01$

$\beta =9.09\times 10^-^3$

mole fractions of ethane $\beta _e$ and methane $\beta_m$ after passing through separator are:

$\beta_e =\frac{0.3}{0.3+0.6+0.00909}=0.66\\\beta_m=\frac{0.6}{0.3+0.6+0.00909}=0.33$

Change in partial pressures then can be written as:

$\Delta P=(y_2-y_1)\cdot P$  where $y_2$ and $y_1$ are mole fractions after and before passing through the separator

Hence,

$\Delta P_m=(0.66-0.6)\cdot 100\text{k}=6\text{kPa}\\\Delta P_e=(0.33-0.3)\cdot 100\text{k}=3\text{kPa}\\\Delta P_p=(0.01-0.1)\cdot 100\text{k}=-9\text{kPa}$