Documentation:CHBE Exam Wiki/Final Exam 2016W/Question 1

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

The water gas shift reaction is commonly used for producing hydrogen and follows the reaction given below:

CO + H₂O ↔ CO₂ + H₂

Say we have a water gas shift reactor with 3 entering streams. The first stream (#1) contains 10 weight% CO2, 30 weight% CO and the remainder as nitrogen (N₂) and enters at 100 kg/h. The second stream (#2) contains 40 mol% hydrogen and the remainder as CO. The third entering stream (#3) contains only water. Only one product stream exits and it contains a mixture of all the entering species. The reactor achieves an 80% conversion of the entering CO. The reactants and products enter and leave at 300 °C and 1 atm and the reactor exchanges energy by giving off or taking in heat to achieve this. The exiting stream is not under equilibrium.

Physical Data
MW CO₂ : 44 g/mol
MW CO : 28 g/mol
MW N₂ : 28 g/mol

Question 1a [5 points]

Draw a flowchart for the process labeling all streams and components.

Hint
Consider each stream individually and write the mass/molar flow rate and mass/mole fraction of the components in the stream? Consider if energy streams are present.

Solution

Question 1b [5 points]

Indicate the mole fractions of all components and the total molar flow in kmol/h for stream #1.

Hint
Use the relationship between the molar and mass flow rates of individual species involving molar mass. Find the ratio of the individual molar flow rates of each species to the total molar flow rate

Solution
${\dot {n}}_{1,CO_{2}}={\frac {10kg/h}{44kg/kmol}}=0.227kmol/h$ ${\dot {n}}_{1,CO}={\frac {30kg/h}{28kg/kmol}}=1.07kmol/h$ ${\dot {n}}_{1,N_{2}}={\frac {60kg/h}{28kg/kmol}}=2.14kmol/h$ ${\dot {n}}_{1,tot}=3.437kmol/h$ $y_{1,CO_{2}}={\frac {0.227kmol/h}{3.437kmol/h}}=0.066$ $y_{1,CO}={\frac {1.07kmol/h}{3.437kmol/h}}=0.311$ $y_{1,N_{2}}={\frac {2.14kmol/h}{3.437kmol/h}}=0.623$

Question 1c [5 points]

Do a degrees of freedom analysis on the reactor. Is the problem over-specified, under-specified or adequately specified?

Hint
How many components are in each stream and how many flows are known? Don't forget to include an energy balance.

Solution
Parameter Value Unknowns 8 Atomic balances 3 (C,O,H) Molecular balances 1 (N2) Energy balance 1 Additional equations 1 (conversion) Degree(s) of Freedom 2 Since the degrees of freedom is positive, it is underspecified.

Question 1d [5 points]

The pressure gauge for the reactor was found to not be accurate and instead a manometer was installed with one end attached to the reactor and then other end open to the atmosphere. A liquid with a specific gravity of 5.27 (relative to water at 4°C) was used in the manometer. The atmospheric pressure was found to be 100 kPa. The fluid was higher on the side open to the atmosphere by 50 cm. What is the absolute pressure inside the reactor in kPa?

Hint
If the fluid is higher on the side that is open to the atmosphere, is the absolute pressure higher than atmospheric pressure? What is the relationship between gauge and absolute pressure?

Solution
$SG={\frac {\rho }{\rho _{ref}}}$ ${\begin{aligned}\rho &=5.27\times 1g/cm^{3}\\&=5.27g/cm^{3}\\&=5270kg/m^{3}\\\end{aligned}}$ ${\begin{aligned}P_{reactor}&=P_{o}+\rho \cdot g\cdot (h_{o}-h)\\&=100kPa+5270kg/m^{3}\cdot 9.81m/s^{2}\cdot 0.5m\\&=100kPa+25.85kPa\\&=125.85kPa\\\end{aligned}}$