Documentation:CHBE Exam Wiki/Final Exam 2016W/Question 3

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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

A furnace is used in a plant to heat a stream of 1 tonne per hour of pure H₂O. The H₂O enters as a mixture of water (liquid) and steam (vapour) at 5 bar to produce superheated steam at 300°C and 5 bar. The heat to do this is provided by the complete combustion of 1 kmol/h propane (C₃H₈) with 100% excess air at 1 atm. The air consists of 79 mol% nitrogen (N₂) and 21 mol% oxygen (O₂) and enters the furnace with the propane at 200°C. The fractional conversion of propane is 1.00 and the combustion products leave the furnace at 400°C. The heat of propane combustion forming gaseous water is -2220 kJ/(mol C₃H₈) at 400 °C. The steam and combustion streams are not in direct contact, but energy is transferred between the two streams and you can assume the furnace to be adiabatic with negligible energy contributions from shaft work, kinetic energy, or potential energy.

Compound	Cp[J/mol K]
C3H8	68
N2	29
H2O	33.5
O2	29
CO2	36

Note : enthalpy and internal energy values below are in kJ/kg and volumes are in m³/kg

Question 3a [5 points]

Write out the equation for the combustion reaction. What is the molar flow of oxygen and nitrogen into the process?

Hints
Write out a balanced equation for the reaction. Start by balancing the atoms in this order: C, H, O If air is fed in excess, what is the ratio between propane and oxygen in air? What does 100% excess mean?

Solution

Combustion reaction ${C_{3}H_{8}}+5{O_{2}}\rightarrow 3{CO_{2}}+4{H_{2}O}$ Using stoichiometry, the ratio of propane to oxygen is 1:5. Since air is fed with 100% excess to the furnace, the ratio is doubled. Hence, ${\begin{aligned}{\dot {n}}_{O_{2}}&=10\cdot {\dot {n}}_{C_{3}H_{8}}\\&=10kmol/h\\\end{aligned}}$ The composition of nitrogen and oxygen in air are 79 mol% and 21 mol% respectively. Hence, ${\begin{aligned}{\dot {n}}_{N_{2}}&={\frac {0.79}{0.21}}\cdot {\dot {n}}_{O_{2}}\\&=37.62kmol/h\\\end{aligned}}$

Question 3b [10 points]

What are the molar fractions of all species in the gas exiting the furnace?

Hints
Determine the molar flow rates of the each species exiting the furnace by taking into account the extent of reaction (conversion)

Solution

Step 1 Reactant species: Propane and oxygen Unreacting species: Nitrogen Product species: Carbon dioxide and water Conversion of propane is 1.00. Hence $\epsilon =1kmol/h$ Determine the molar flow rates of each species ${\begin{aligned}{\dot {n}}_{out,C_{3}H_{8}}&={\dot {n}}_{in,C_{3}H_{8}}-\epsilon \\&=0\\\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{out,O_{2}}&={\dot {n}}_{in,O_{2}}-5\cdot \epsilon \\&=10kmol/h-5\cdot (1kmol/h)\\&=5kmol/h\\\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{out,N_{2}}&={\dot {n}}_{in,N_{2}}\\&=37.62kmol/h\\\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{out,H_{2}O}&=4\cdot \epsilon \\&=4kmol/h\\\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{out,CO_{2}}&=3\cdot \epsilon \\&=3kmol/h\\\end{aligned}}$ ${\begin{aligned}{\dot {n}}_{out}&={\dot {n}}_{out,C_{3}H_{8}}+{\dot {n}}_{out,O_{2}}+{\dot {n}}_{out,N_{2}}+{\dot {n}}_{out,H_{2}O}+{\dot {n}}_{out,CO_{2}}\\&=49.62kmol/h\\\end{aligned}}$ Step 2 The molar fractions of the species in the exit stream of furnace are: $y_{out,O_{2}}=0.101$ $y_{out,N_{2}}=0.758$ $y_{out,CO_{2}}=0.060$ $y_{out,H_{2}O}=0.081$

Question 3c [15 points]

What is the amount of energy transferred due to the combustion reaction in kJ/h?

Hints
Step 1 How does the temperature change in the streams relate to the change in enthalpy? Step 2 How does energy balance come into play in this situation?

Solution

Step 1 Take the reference temperature, to be $T_{1}=400^{\circ }C$ Assuming that the $C_{p}$ value is independent of temperature and pressure, the molar enthalpy is given by the following formula ${\hat {H}}=C_{p}\cdot (T_{2}-T_{1})$ Determine the molar enthalpy for all the species ${\begin{aligned}{\hat {H}}_{C_{3}H_{8}}&=68kJ/kmolK\cdot (200-400)K\\&=-13600kJ/kmol\\\end{aligned}}$ ${\begin{aligned}{\hat {H}}_{O_{2}}&=29kJ/kmolK\cdot (200-400)K\\&=-5800kJ/kmol\\\end{aligned}}$ ${\begin{aligned}{\hat {H}}_{N_{2}}&=29kJ/kmolK\cdot (200-400)K\\&=-5800kJ/kmol\\\end{aligned}}$ Step 2 Determine the energy transfer using energy balance ${\begin{aligned}Q&=\Delta H\\&=\sum H_{rxn}+\sum _{i}{\dot {n}}_{out,i}\cdot {\hat {H}}_{out,i}-\sum _{i}{\dot {n}}_{in,i}\cdot {\hat {H}}_{in,i}\\&=1kmol/h\cdot (-2220kJ/mol)\cdot {\frac {1000mol}{1kmol}}-(-13600kJ/kmol\cdot 1kmol/h)-(-5800kJ/kmol\cdot 10kmol/h)-(-5800kJ/kmol\cdot 37.62kmol/h)\\&=-1.93\times 10^{6}kJ/h\\\end{aligned}}$

Question 3d [10 points]

What are the weight fractions of water and steam in the entering pure H₂O?

Hints
Step 1 Look for the appropriate values in the steam table based on the stream conditions Step 2 How can the energy balance involving specific enthapy be applied in this situation?

Solution

Step 1 From the steam table, the following values are obtained $h_{in,l}=640.1kJ/kg$ $h_{in,v}=2747.5kJ/kg$ $h_{out}=3065kJ/kg$ Let the fraction of liquid water be $w_{l}={\frac {\text{mass of liquid water}}{\text{total mass of water}}}$ Step 2 Using energy balance ${\begin{aligned}Q&=\Delta H\\&=\sum _{i}{\dot {m}}_{out,i}\cdot h_{out,i}-\sum _{i}{\dot {m}}_{in,i}\cdot h_{in,i}\\&=m_{total}\cdot h_{out}-m_{in,l}\cdot h_{in,l}-m_{in,v}\cdot h_{in,v}\\&=h_{out}-w_{l}\cdot h_{in,l}-(1-w_{l})\cdot h_{in,v}\\&={\frac {-1.93\times 10^{6}kJ/h}{m_{total}}}{\text{ (From Question 3c)}}\\\end{aligned}}$ Rearranging the expression and solving for $w_{l}$ $w_{l}=0.765$