Documentation:CHBE Exam Wiki/Midterm Exam 2 2016W/MT2 Question 2
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Midterm Exam 2 2016W  
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Module 1  Process Basics  
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Module 4  Separations 2  
Module 5  Nonreactive 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 (H_{2}O), nbutane (C_{4}H_{10}) and noctane (C_{8}H_{18}) 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) =
The Antoine equation is:
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,

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.
Step 2
Using 1 kg of water as the basis, 
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.

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