Chemical Processes. What is Engineering? July 25, 2007. Chemical Processes Outline. Motivations Reactions Separations Calculations using Conservation of Mass and Energy Distillation. Chemists Design reaction pathways to produce a chemical from raw materials
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What is Engineering?
July 25, 2007
Design reaction pathways to produce a chemical from raw materials
Work in the laboratory setting to produce material on the gram to kilogram scale
Design a process to scale the chemist’s process to mass produce the product
Work in a chemical plant to produce material in the ton and beyond rangeChemists vs Chemical Engineers
Chemicals Are All Around
Hydrogen Flouride (HF)
Polyvinyl Chloride (-CH2-CHCl-)n
Energy Produced by reaction is proportional to reactor volume L3
Energy Removed is proportional to surface area L2
Possible Scale up Problem
Affinity to a stationary phase
Selective affinity to solid particles
Exploits Differences of Material Properties
Input + generation – Output = Accumulation
Mass and Energy Balances reduce to
Input = Output
x % C2H5OH
x % C2H5OH
Conservation of total Moles 100 – (V+80) = 0
Conservation of moles of C2H5OH 100*.1 – (.4*V+x*80) = 0
x = 2.5%
Separates liquids based on differences in volatility!
Consider a liquid mixture of A and B:
Boiling point of A: 70 C
Boiling point of B: 100 C
As mixture begins to boil, the vapor phase becomes richer in A than the liquid phase!
Condense vapor phase to get a mixture with a higher concentration of A!
As temperature increases, the concentration of B in the vapor phase increases.
What would be the composition of the vapor phase if the entire liquid mixture vaporized?
A) Phases are brought into close contact
B) Components redistribute between phases to equilibrium concentrations
C) Phases are separated carrying new component concentrations
D) Analysis based on mass balance
Represent vapor liquid equilibrium data for more volatile component in an x-vs-y graph
Pressure constant, but temperature is changing!
Calculation of theoretical number of equilibrium stages
Applicable for many liquid systems
Technology is well developed
Due to molecular interactions. Composition of vapor equal to composition of liquid mixture.
Batch distillation apparatus – only one equilibrium stage!
Want efficient transfer and conversion of energy ($$)
In lab, will be examining energy transfer in the form of heat: warming a pot of water with a hot plate – what is the efficiency of energy transport from electricity to the water?
**Methanol and 2-propanol are poisons! Wear safety goggles, do not ingest or inhale and rinse skin immediately if spilled.
**Ethanol is a poison! Wear safety goggles, do not ingest or inhale and rinse skin immediately if spilled.
xA = mass fraction of A in stream L
yA = mass fraction of A in stream V
(e.g., L0 xA0 = mass of component A in stream L0 )
Component mass balance (mass/time):
L0 xA0 + V2 yA2 = L1 xA1 + V1 yA1 = M xAM
L0 xC0 + V2 yC2 = L1 xC1 + V1 yC1 = M xCM
(equation for B not necessary because xA + xB + xC = 1)
Suppose the following: V is oil (B) contaminated with dye (A). L is water (C) which is used to extract the dye from the oil. When V comes in contact with L, the dye redistributes itself between the V and L. L and V are immiscible (i.e., two distinct liquid phases).
Water flow = L(1 - xA) = L’ = constant
Then, for mass balance of the A component:
Another assumption: dye concentrations yA1, xA1 come into equilibrium according to Henry’s Law: yA1 = H xA1 , where H depends on the substances A, B, C.
Specific problem: 100kg/hr of dye-contaminated oil (1% by weight) is mixed with 100 kg/hr of water to reduce the dye concentration in the oil. What is the resulting dye concentration in oil after passing through the mixing stage if dye equilibrium is attained and Henry’s constant H = 4?
L’ = 100kg/hr V’ = 100 ( 1 - .01) = 99 kg/hr
xA0 = 0 (no dye in incoming water)
yA2 = .01 (initial contamination in oil)
yA1 = 4 xA1 (equilibrium concentration of dye between oil and water)
* weight) is mixed with 100 kg/hr of water to reduce the dye concentration in the oil. What is the resulting dye concentration in oil after passing through the mixing stage if dye equilibrium is attained and Henry’s constant H = 4?
Rectifying operating line
y-int ~ 0.36
Stripping operating line
Nideal = 6 2/3