Documentation:CHBE Exam Wiki/Final Exam 2016W/Question 2

CHBE 241 Exam resources wiki
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

A liquid mixture of 70 mol% n-hexane (C₆H₁₄) and 30 mol% n-heptane (C₇H₁₆) at 40°C is fed at a rate of 100 mol/s into a separating tank, known as a flash drum. In the flash drum, the temperature of the liquid is brought to 80°C and leads to the creation of a vapour phase. The vapour and liquid product streams can be assumed to be in equilibrium and are removed continuously at 80°C. The entire process operates at a uniform 1 atm pressure. Data for this system are provided below:

The Antoine equation is: $log_{10}P^{sat}(mmHg)=A-{\frac {B}{T(^{\circ }C)+C}}$

Table 1: Antoine equation coefficients for n-hexane and n-heptane, respectively

Compound	A	B	C
n-hexane	6.87601	1171.17	224.41
n-heptane	6.89677	1264.90	216.54

Table 2: Thermodynamic properties

Compound	Cp [J/(mol K)]	Ĥvap[kJ/mol)]
n-hexane	150	28.85
n-heptane	180	31.69

Note that Ĥ^vap can be assumed to be independent of temperature and pressure

Question 2a [25 points]

Calculate the n-hexane and n-heptane molar compositions in the vapour and liquid product streams as well as the product vapor and liquid stream molar flow rates.

Hint

Step 1 Determine the components in Raoult's law which can be determined using the given operationing condition (P,T), before solving, do you have the information you need to solve this problem? Step 2 How can Raoult's Law be applied in this situation, while taking into account the relationship between mole fractions of each species? Use the relationship between mole fractions (vapor and liquid) to reduce the amount of equations to solve

Solution

Step 1 Using the Antoine equation, determine the vapor pressure of n-hexane(a) and n-heptane(b) ${\begin{aligned}P_{a}^{sat}&=10^{(6.87601-{\frac {1171.17}{80+224.41}})}\\&=1068mmHg\times {\frac {1atm}{760mmHg}}\\&=1.406atm\\\end{aligned}}$ ${\begin{aligned}P_{b}^{sat}&=10^{(6.89677-{\frac {1264.90}{80+216.54}})}\\&=427.8mmHg\times {\frac {1atm}{760mmHg}}\\&=0.563atm\\\end{aligned}}$ Step 2 Using Raoult's law, $P=y_{a}\cdot P+y_{b}\cdot P=x_{a}\cdot P_{a}^{sat}+x_{b}\cdot P_{b}^{sat}$ and $x_{a}+x_{b}=1$ Liquid composition: ${\begin{aligned}x_{a}&=x_{a}\cdot P_{a}^{sat}+(1-x_{a})\cdot P_{b}^{sat}\\&={\frac {P-P_{b}^{sat}}{P_{a}^{sat}-P_{b}^{sat}}}\\&={\frac {1atm-0.563atm}{1.406atm-0.563atm}}\\&=0.518\\\end{aligned}}$ ${\begin{aligned}x_{b}&=1-x_{a}\\&=0.482\\\end{aligned}}$ Vapor composition: ${\begin{aligned}y_{a}&={\frac {x_{a}\cdot P_{a}^{sat}}{P}}\\&={\frac {0.518\cdot 1.106}{1}}\\&=0.728\\\end{aligned}}$ ${\begin{aligned}y_{b}&=1-y_{a}\\&=0.272\\\end{aligned}}$ Step 3 Using mole balance, ${\dot {n}}_{1}={\dot {n}}_{2}+{\dot {n}}_{3}$ ${\dot {n}}_{1}\cdot x_{1,a}={\dot {n}}_{2}\cdot y_{2,a}+{\dot {n}}_{3}\cdot x_{3,a}$ Rearranging the equations, ${\begin{aligned}{\dot {n}}_{3}&={\frac {{\dot {n}}_{1}\cdot x_{1,a}-{\dot {n}}_{1}\cdot y_{2,a}}{x_{3,a}-y_{2,a}}}\\&={\frac {100mol/s\cdot (0.7-0.728)}{0.518-0.728}}\\&=13.3mol/s(liquid)\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{2}&={\dot {n}}_{1}-{\dot {n}}_{3}\\&=100-13.3\\&=86.7mol/s(vapor)\end{aligned}}$

Question 2b [15 points]

Is energy being transferred to or from the separation tank and what is the amount of energy transferred in kW?

Hints
Step 1 What are the enthalpy values for each components in the liquid and vapor streams leaving the flash drum? Step 2 How does energy balance come into play in this situation?

Solution

Step 1 Determine the enthalpy values for all the streams, $H=\int _{T_{1}}^{T_{2}}\mathrm {C_{p}} \,\mathrm {d} T$ Since $C_{p}$ is independent of temperature and pressure, $H=C_{p}\cdot (T_{2}-T_{1})$ For the liquid streams: ${\begin{aligned}{\hat {H}}_{l,a}&=0.15kJ/molK\cdot (80-40)K\\&=6kJ/mol\\\end{aligned}}$ ${\begin{aligned}{\hat {H}}_{l,b}&=0.18kJ/molK\cdot (80-40)K\\&=7.2kJ/mol\\\end{aligned}}$ For the vapor streams: ${\hat {H}}_{v}={\hat {H}}_{l}+{\hat {H}}^{vap}$ ${\begin{aligned}{\hat {H}}_{v,a}&={\hat {H}}_{l,a}+{\hat {H}}_{a}^{vap}\\&=6kJ/mol+28.85kJ/mol\\&=34.85kJ/mol\end{aligned}}$ ${\begin{aligned}{\hat {H}}_{v,b}&={\hat {H}}_{l,b}+{\hat {H}}_{b}^{vap}\\&=7.2kJ/mol+31.69kJ/mol\\&=38.89kJ/mol\end{aligned}}$ Step 2 Using energy balance ${\begin{aligned}Q&=\Delta H\\&=\sum _{i}{\dot {m}}_{out,i}\cdot h_{out,i}-\sum _{i}{\dot {m}}_{in,i}\cdot h_{in,i}\\&=x_{a}\cdot n_{3}{\hat {H}}_{l,a}+x_{b}\cdot n_{3}{\hat {H}}_{l,b}+y_{a}\cdot n_{2}{\hat {H}}_{v,a}+y_{b}\cdot n_{2}{\hat {H}}_{v,b}\\&=0.519\cdot (13.3mol/s)\cdot (6kJ/mol)+0.482\cdot (13.3mol/s)\cdot (7.2kJ/mol)+0.728\cdot (87.7mol/s)\cdot (34.85kJ/mol)+0.272\cdot (87.7mol/s)\cdot (38.89kJ/mol)\\&=3188kW\end{aligned}}$