Documentation:CHBE Exam Wiki/Midterm Exam 2 2016W/MT2 Question 2

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Chemical and Biological Engineering
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Past Exams
Final Exam 2016W
Midterm Exam 1 2016W
Midterm Exam 2 2016W
Problem Sets
Module 1 - Process Basics
Module 2 - Reactors
Module 3 - Separations 1
Module 4 - Separations 2
Module 5 - Non-reactive Energy Balances
Module 6 - Reactive Energy Balances

Oil extracted from a well may contain a significant amount of water. This water is typically removed in a 3 phase separator which splits the entering stream into 2 liquid streams and a gas stream. To simplify this system, we will assume the entering stream contains only water (H₂O), n-butane (C₄H₁₀) and n-octane (C₈H₁₈) and that the streams exit the separator at equilibrium.Total pressure in the separator is 6.382 bar. You may also assume that water and octane are not miscible. Make sure to list any other assumptions that you make.
The following information is given:

• Unit operates at 60°C
• MW_water = 18 g/mol ; MW_butane = 58 g/mol ; MW_octane = 114 g/mol
  
• p*_water = 149.61 mmHg ; p*_butane = 6.2 bar
  
• H_butane(at 60°C) = 0.000403 mol_butane/(kg_water·bar)
  
• K_c(mass fractions at 60°C) = ${\frac {(w_{butane})_{octanephase}}{(w_{butane})_{waterphase}}}$
  
  

The Antoine equation is: $log_{10}p^{*}(bar)=A-{\frac {B}{T(K)+C}}$

Table 1: Antoine equation coefficients for octane, respectively

Compound	A	B	C
Octane	4.049	1355	-63.63

Question 2a [5 points]

What is the vapour pressure of octane?

Hint
How can the Antoine's equation be applied in this situation?

Solution
Using the Antoine's equation and the given constants, ${\begin{aligned}log(p*_{octane})&=4.049-{\frac {1355}{(273.15+60)-63.63}}\\&=-0.9784\\p*_{octane}&=0.1051bar\\\end{aligned}}$

Question 2b [10 points]

What is mass fraction of all components in the liquid water phase?

Hint
Step 1 How can Henry's law be applied in this situation? What simplyfing assumption needs to be made? Step 2 Set a specific mass of water as the basis and determine the mass fraction of the components

Solution

Step 1 Assume that octane is immiscible, so the liquid phase only contains only butane and water. Using Henry's law, ${\begin{aligned}x_{butane}&=p_{butane}^{*}\times H_{butane}\\&=6.382bar\cdot 0.000403mol_{butane}/(kg_{water}\cdot bar)\\&=0.002569mol_{butane}/kg_{water}\\\end{aligned}}$ Step 2 Using 1 kg of water as the basis, ${\begin{aligned}m_{butane}&=0.002569mol_{butane}\times 58g/mol_{butane}\times {\frac {1kg}{1000g}}\\&=0.000149kg\\w_{butane}&={\frac {0.000149kg}{0.000149kg+1kg}}\\&=0.0015\\w_{water}&=1-w_{butane}\\&=0.9985\\\end{aligned}}$

Question 2c [5 points]

What is the mass fraction of all components the octane rich liquid?

Hint
How can the value of Kc be used in this situation?

Solution
Assume that there is no water as it is immiscible. ${\begin{aligned}(w_{butane})_{octane}&=85.7\cdot (w_{butane})_{water}\\&=85.7\cdot 0.000149\\&=0.01277\\w_{octane}&=0.9872\\\end{aligned}}$