Syllabus
for ChE 301
INTRODUCTION TO CHEMICAL ENGINEERING THERMODYNAMICS
Fall,
2011, 2:10 – 3:00p in Carpenter 101
Prerequisites: ChE 201, ChE or BE
major
Instructor: Bernard J. Van Wie
Office: EE/ME B57 Phone: 509-335-4103 Office Hours: Tu 11a – 12 noon, Th. 1 – 2p, appointment, or by e-mail: bvanwie@che.wsu.edu
Teaching Assistant: Kat Tran
Phone: 617-412-5599; Time & Location: M, W 6 – 8p; F
3:15 – 5p in Dana 215, or by e-mail: giao.tran@email.wsu.edu
Text: Introduction to Chemical Engineering
Thermodynamics by Smith, Van Ness and Abbott (McGraw-Hill, 7th
ed.; 6th ed. will work)
Yes
Picture:
Topical Outline: Clicking on the “Chapters” will lead you to
supplementary notes developed by Prof. Emeritus Reid Miller and Prof. Bernie
Van Wie. Found going to the Voiland
School Website http://www.voiland.wsu.edu/,
clicking on Faculty & Staff, clicking on Bernard Van Wie, and then on ChE 301 – Introduction to Chemical Engineering
Thermodynamics.
Chap. 1 & 2 Introduction and 1st Law - Aug. 22-Sept. 2
Chap. 3
& 4
Volumetric Properties and Heat Effects – Sept. 2 – 12
Help Session: Th., Sept. 22, 7p
Exam 1 - Sept. 23 Practice_Exam_I_2008_2006_2005.pdf
Practice_Exam_Ib_2008_2003_1999.pdf
Chap.
6 & 7 Thermodynamic
Properties and Flow Processes - Sept. 26-Oct. 10
Chap. 10 -
12 Phase
Equilibrium & Solution Thermodynamics - Oct. 10-28
Help Session: Sun., Oct. 30, 7p
Exam
2 – Oct. 31 Practice_Exam_II_2008_2005_2003.pdf
Practice_ExamII_2008_2005_2003_No_Solutions.doc
Stagewise
Processing Stagewise
Processes - Nov. 2-18
·
Introduction to Pro II: Pro
II Basics Tutorial; Introduction
to Pro II
·
Pro II
Flash, Distillation, Gibbs Reactor Modules
·
Distillation Tutorial in HYSYS
Chap. 13 Chemical-Reaction
Equilibrium – Nov. 28 – Dec. 7
· Pro II Flash, Distillation, Gibbs Reactor Modules
·
Tutorial
on Equilibrium Reactors in HYSYS
Possible Help Session: Th. Dec. 15, 7p
Final Exam – Fri. Dec. 16 (7:00 - 10:00 a.m.)
Links
to the following:
Course
Objectives:
7.
Students can assess their team problem-solving processes to improve these
processes. (evaluation)
Special
Features:
Active
Learning – Teams
of four students will be formed, self-selected but subject to veto by the
professor. Roles of team leader, recorder, technical specialist, and
reflector should rotated by the group during the semester. These groups
will be used in three ways. First, in-class discussions will be by group.
Second, most homework assignments are group; for
individual assignments you may talk with your group, though you must turn in your own individual work.
Third, a team design project will be assigned during the
semester. Information on organization and operation of teams is
available through this link.
Relationships
to Chemical Engineering Practice – Every
attempt will be made to relate the information developed in class to the real
world of engineering design operations. Students will be asked to
identify different applications of thermodynamics that appear to be important
in specific chemical processing operations.
Learning
Styles and Levels – Lesson
plans will be developed to incorporate different learning styles and different
levels of learning (see information on Bloom's Taxonomy).
Course objectives have been developed to involve a range of learning levels, as
indicated in parentheses above.
Policies
and Procedures:
30%
Final Exam
25%
Homework & Design Projects
4.
Homework and Design Projects
A problem
statement with a diagram (0.5 pts)
Approaches
and equations used for the solution (varies)
Values
with units used for the solution, e.g., P = 10 atm (1 pt)
The
answer clearly indicated – underlined or boxed with appropriate units (0.5 pts)
Separate
group work from individual assignments
Students
with Disabilities:
Reasonable
accommodations are available for students with a documented disability. If you have
a disability and may need accommodations to fully participate in this class,
please visit the Disability Resource Center (DRC). All accommodations MUST be
approved through the DRC (Washington Building, Room 217). Please stop by or
call 509-335-3417 to make an appointment with a disability specialist.
ChE
301 – Homework Assignments:
Assignment
-- Problems -- Due Date; (I) indicates individual solutions required
- Chapter 1 &
2 - 1.5, SP1, 12(I), 20 & 2.1, 27, 28, 34(I) – Sept. 2
- Chapter
3 & 4 – 3.4, 19, 33, 38(m)(I),
4.3, 4.10(e), 4.21(z), 4.49(I) – Sept. 12
- Chapter
5 – 5.4, 7(I), 19, 21, 26, SP2
– Sept. 21
- Chapter
6 & 7 – 6.8, 23, 36, 45(I),
87(j); 7.4, 19(e), 35(e)(I), SP
3, SP4(I) – Oct. 5
- Chapter
10 & 11 – 10.1 b,c,d,f, 10.2a(I)
- 5pt EC by Mathcad, 7, 17b; 26a(I),
SP5, SP6(I), 11.17, 19, 28, 27(I) 10pt EC (only) by Mathcad –
Oct. 17
- Chapter
12 – 12.3a(I), 15(methanol
+ benzene), 30, 45a,b(I), SP5, SP6(I), SP8 – Oct. 28
- Design Project: Simplified VLE
Calculations – Due Nov. 18
- Stagewise
Processes – Dec. 2
- Chapter 13.1c, 13(I) – Pro II Gibbs Reactor, and by hand, Sp11 – Dec. 9
Other
Interesting Web Sites
History of Chemical Engineering
Some Basic
Definitions
System - that portion of the
universe set aside for study
Surroundings - the environment - the rest
of the universe
Boundaries - walls - separate system
from surroundings
Closed
system - constant mass -
impermeable boundaries - energy can cross boundaries
Open
system - mass and energy
can cross boundaries - some permeable or semi-permeable boundaries
Isolated
system - constant mass
and energy - impermeable, rigid, adiabatic boundaries
Adiabatic
walls - prevent thermal
equilibrium - no heat
Diathermal
walls - instant thermal
equilibrium
Units - scales used to quantify
dimensions (e.g., g, lb, ft, s, K, etc.)
Extensive
properties - depend on
the extent of the system (volume, mass, internal energy, etc.)
State
(of a system) -
specified by a unique set of intensive properties
Cycle - series of processes leading
back to the initial state of the system
Temperature - the property which tells us
whether systems are in thermal equilibrium
Heat - energy in transition across
the boundaries of a system due to a temperature difference
Special Problems
Related to Bioengineering
SP 1 – Due date: Sept. 2
It
is common to state that the energy content of human food is only the calories
consumed, but considerable energy is involved in getting the food from its
source to the eater.
It is reported1
that while the world’s human population has grown from 107 to 7x109
over the last 10,000 years, the energy used to feed the world population has
grown by a factor of 5x103 from 8x103 kcal per person
each day. How many calories are now used
daily to feed each person?
Comment on the
causes of the added calories since the actual intake is about 2500 kcal per
day.
1Clark
ME, “Adriane’s Thread”, St. Martin’s Press, New York, 1989, p 102, as quoted at
http://telstar.ote.cmu.edu/environ/m3/s3/all_ene_sys.htm
SP 2 – Due date: Sept. 21
The
core temperature of the body should be maintained relatively constant at 37oC
regardless of the environment. Exposure to cold for
extended periods will require extra energy intake and the heat transfer will
generate extra entropy.
Consider treating
severely burned humans by cooling their body in a air or water bath at 15oC for 12-20
hours. Since more heat will be lost to
the surroundings than under normal conditions, there will need to be
augmentation of the patient’s energy intake.
Given the data
below, estimate:
a) The extra
required rate of energy input, 
, in kcal/h
b) The rate of
extra entropy generation, 
,
in kcal K-1 h-1, for such a patient, regardless of the energy source.
c) Suggest ideas
for putting in this energy.
Note:
Consider the
typical patient to be 70 kg mass and 1.75 m in height, so the empirical formula
for surface area can be used: 
The rates of heat
transferred from the blood system to the skin and from the skin to the bath can
be calculated by heat transfer formulae.
An estimate of the rate in such a bath is 
with a skin temperature of 25oC
while the normal rate 
when the skin temperature is 34oC.
SP 3 – Due date: Sept. 30
Living
systems use energy for many functions. The heart of humans does work to
circulate blood. This problem explores using easily measured quantities to
obtain the fraction of total body energy converted in circulating blood at
various exercise conditions
Consider the work
that the human heart does when pumping blood and its relation to oxygen
consumption at rest and when exercising.
The heart consists of two chambers, the right ventricle that sends the
blood to the lungs (pulmonary) and the left ventricle which sends the blood to
the rest of the body (systemic). The
blood pressure of the left ventricle is denoted PS while that of the right ventricle is denoted PP. The volume of blood ejected at each beat is
the difference of the volume of blood taken in by the ventricles, or the stroke
volume, S. The rate of work, or power, for the heart to
circulate blood is the frequency of heart beats, f, times the sum of PV
work by the ventricles for each beat, W,
i.e., power = 
. It has been found that under a variety of conditions,
. The
origin of the body’s energy is oxidation of carbohydrates, fats, and proteins
by oxygen taken into the body. The
energy released per liter of oxygen is identified as E. The part of this energy
consumed by the heart to circulate blood is identified as η. The total rate of
oxygen consumption in the body is Ω.
Show that the
relationship for the rate of cardiac output of blood, Q, in terms of the stroke volume of the heart and the quantities
given above is 
where Ω
is the total rate of oxygen consumption by the body.
SP 4 – Due date: Sept. 30
Thermodynamic
property relations, especially partial derivatives, hold for all systems whose
properties can be defined. Biochemicals are often polymeric, and even
elastomeric, and the relevant state properties are not all the same as those
for normal fluids. This exercise explores the different properties for
elastomers, and demonstrates how some experimental observations can be
leveraged to predict behaviors under other conditions
Consider a rubber
band. The variables of its state are
length, L, stress, t, and temperature, T. Thermodynamics can take some state
information about a system and predict other behaviors, principally through
signs on partial derivatives. A series
of experiments and analysis with a large rubber band can illustrate the
procedure.
First, our
knowledge of the chemical structure of elastomers is that there are chains with
crosslinks at intervals so that the chains between the crosslinks are able to
take on many conformations as in the left figure.
When the length, L,
is changed, say at constant temperature, T, the chains are stretched, so the
number of conformations is reduced, lowering the entropy, S, as in the right
figure. The partial derivative
characterizing this process is 
.
Next, do an
experiment. Slowly stretch the rubber
band; increasing L at fixed T takes increased stress, t. The partial derivative characterizing this
process is 
.
Another experiment
is, with the band pressed against your upper lip, quickly pull on the
band. It should feel warm. The result of this adiabatic (~isentropic)
process gives 
.
From these
relationships and some mathematics, certain predictions of behavior can be made
for situations.
a) Stretch the
rubber band to a fixed L hold it while heating. What happens to the stress?
b) Put a weight on
the band and heat. What happens to the
weight?
SP 5 – Due date: Oct. 17
Living
systems function with semipermeable membranes that allow some components to
pass through, but not others. This problem explores the thermodynamics and
modeling via the osmotic virial equations for blood
hemoglobin and cell walls
Consider the
situation of red blood cells containing hemoglobin (MW ~ 68000) at a
concentration of 0.3 M. Water passes
through the cell walls, but the hemoglobin cannot.
a) Ignoring the
presence of other substances inside & outside of the cell membrane, prove
that there would be a flow of water into the cell at ambient conditions.
b) To stop the flow
of water, show that the fundamental equilibrium relations would require
increased pressure on the inside of the cell.
c) Write a relation
for the pressure difference across the cell membrane, 
,
and estimate the pressure assuming a binary ideal solution inside the cell and
pure water outside the cell.
d) Real binary
solutions are often described by the second osmotic virial
equation
where
is the molar concentration of hemoglobin and 
is the second osmotic virial coefficient,
which is a measure of the interaction between pairs of solute particles in the
solvent medium. (Recall that the 2nd virial coefficient of the virial
equation of state measures the pair interactions in vacuum.) If the osmotic pressure, 
,
at body temperature of 37oC is found to be 0.83 MPa, obtain the
value of .
e) Red blood cells
will rupture under such an osmotic pressure, so Nature arranges things to
obviate this by changing the solution outside the cell by addition of a
different impermeant.
Suppose the added solute is glucose (MW ~ 180). If the measured osmotic pressure for 0.7 M
glucose is 20 bar, estimate the glucose concentration
outside a red blood cell to balance the pressure from 0.3M hemoglobin inside
the cell.
SP 6 – Due date: Oct. 17
Protein molecular
structure has two general forms, native (folded) and unfolded. In solution,
these two forms will be in reaction equilibrium, with equal activities of the
species at the melting temperature, Tm. This extensive problem applied the
thermodynamics of reactions to develop relations and obtain calculations of
thermodynamic properties as functions of several variables
Protein denaturation can be considered as a
reaction between the native species and the unfolded species,
. The “melting” point, 
, is given as the
temperature when equimolar concentrations are found at equilibrium. This temperature can be changed by pH,
pressure, added denaturant, or added salt1. For lysozyme (MW = 14.313 kDa),
calorimetric measurements have been reported2:
1. The heat effect to fully unfold the
protein in 2 mL of a pH = 4 solution at a concentration of 1 
at = 78 oC
is 0.0755 J.
2. The difference in partial molar heat
capacities of the species, independent of T, is2
Using these data and thermodynamic relations,
solve the following problems.
a) What is the Gibbs energy difference of the
system with native and unfolded species
at78oC?
b) Show that at 78oC, the molar
enthalpy and entropy differences of the species are 
, and, 
.
c) Show that the Gibbs energy and enthalpy of
unfolding as functions of T are 
and 
where 
is a temperature at which the properties of
unfolding are known. Suggest why might be a convenient .
d) Using the relations of part c, calculate
the molar Gibbs energy, enthalpy and entropy differences at pH = 4 and T = 23 oC. Is energy or
entropy “driving” the protein to fold at this temperature?
e) Prove that the ratio of unfolded to folded
species concentrations is 
.
f) The relationship Δ
from 
has been found2 when pH is used to change . From this, what is the relationship for 
and ?
g) Using the relations and data above, plot
the ratio of unfolded to folded species concentrations as a function of (1/T)
from 20oC to 100oC when pH = 4.
h) Add to your plot for part g the ratio of
species concentrations when the pH has been adjusted to pH = 1, giving = 43oC.
i) Lysozyme has four disulfide bonds that
constrain the structure. Cooper et al.2
have modified the wild protein by removing one of the S-S bonds and measuring
the change in . They find that the relationship of part f
above, for Δ
with when varied by pH, still holds, even though can be as low as 23oC at pH 2.5 for
the mutant. Prove this indicates that
the reduction of stability by disulfide bond removal is purely entropic.
j) The volume change of denaturating
lysozyme has been obtained3
Δ
. Is
its sign reasonable for the different structures? If this protein is the same as that for parts
a - h, at what pressure would the protein melt when T = 23oC?
1. Haynie DT,
“Biological Thermodynamics, 2nd Ed.” Cambridge University Press, Cambridge, UK,
2008, chapter 2.
2. Cooper A, Eyles
SJ, Radford SE, Dobson CM, J. Mol. Biol. 1992 225 939-943.
3. Sasahara K,
Sakurai M, Nitta K, Proteins 2001 44 180–187.
SP 7 – Due date: Oct. 28
A
statement of the Second Law of Thermodynamics is that doing something in
opposition to Nature takes work. The reversible work is the minimum and its
amount depends on the conditions. Separation and purification of fluid
mixtures, such as for chemicals, do not happen spontaneously, so work must be
put in and there will be an associated heat effect.
a) Derive the expression
for the minimum reversible isothermal-isobaric work to completely purify 1 mol of a solute from a solvent in an ideal solution.
b) Calculate the
required work for mole fractions from 10-1, 10-2, . . .
to 10-9 at 300 K.
c) Derive the
relation for the heat effect and calculate the amount for the separation
process of parts a) and b).
d) Comment on how
solution non-idealities would affect the work and heat effects.
e) Since such
separations are done for chemicals from pharmaceuticals to sewage, comment on
the economics for recovery of dilute products and the wisdom of dilution before
concentration.
SP 8 – Due date: Oct. 28
Purification
of biochemicals is often done through dissolution in solvents followed by precipitation.
Effective solvent selection is important for such processing. This problem
examines thermodynamic property relations and modeling for estimating
biochemical solubility in solvents using hydrocortisone as an example
Pharmaceuticals and
other biochemicals are often processed by dissolution in solvents and then
precipitated for purification. Toxicity,
waste disposal, and economics have a significant impact on solvent selection. Further, choosing a solvent by exhaustive
experimental search is inefficient, so thermodynamic models have been developed
to leverage a minimum of data for finding new and alternative solvents for
biochemical processing.
a) Derive the
following expression for the solubility of a solute, i, that is pure in the
solid form, in a solvent, j:
where
is the enthalpy of melting the solute at the
melting temperature, 
,
and 
is the solute activity coefficient in the
solvent. Assume that the heat capacities
of the solid and liquid solute are equal.
For compounds sparingly soluble in a solvent k, this may often be
taken as the value at infinite dilution,
.
b) Using the
following data, estimate the solubility of hydrocortisone (2) in hexane (1) and
n-butyl acetate (1’) from ideal solution and with the solubility parameter
method for estimating activity coefficients.
Compare these estimates with the experimental values.
|
|
|
|
|
|
|
34.87 |
485.5 |
25.37 |
0.293 |
-5.43 |
|
Solvent |
|
|
|
|
Hexane (1) |
14.8 |
131.6 |
-18 |
|
n-butyl acetate (2) |
16.5 |
132.5 |
-7.4 |
c) An alternative approach
is to use a measured solubility of the solute in a reference solvent, j, and
ratios of activity coefficients to estimate the solubility in another solvent,
k. Derive the fundamental relation for
this method
Note that the pure
solute properties are not needed here.
Reference
: Poling BE, Prausnitz JM, O’Connell JP, Properties of Gases and
Liquids, 5th ed., McGraw-Hill, 2000
SP 9 – Due date: Dec. 9
Knowing
the preference for biocompounds to dissolve or partition
in liquid phases is important in bioprocessing. A biothermodynamic
approach to estimating relative solubilities of amino
acids in aqueous solutions and organic solvents is explored in this problem
When placed in
systems containing immiscible, or partially miscible, liquids such as water
with organic solvents, biochemicals will partition between the two liquid
phases. The partition coefficient,
, is defined as the ratio of
concentrations of a solute i in an
organic solvent, 
,
to that in an aqueous phase, 
,
. This concept can be used even if the organic
and water are not immiscible. In all
cases, the value of 
would also be a measure of the relative
solubility of a solid solute in the solvent to that in water since the chemical
potential of the solute is determined by the solid form and the difference in
activity coefficients of the solutes in the solvents will cause different
solution concentrations.
A further useful
way to determine the relative partitioning of biocompounds,
such as amino acids, is the ratio of the partition coefficient of the acid of
interest to that of glycine, which is the same as the selectivity relative to
glycine, 
. This quantity can be related to the
differences in Gibbs energy of transfer of the solute, 
,
from water to an organic solvent,
.
Damordaran and Song1
give values for 
at 37oC for many amino acids to
different organics (kcal mol-1) while Hammes2 gives
values for transfer from water to aqueous urea at 25oC; some are
|
Amino Acid |
Side Chain |
N-methylacetamide |
ethanol |
hexane |
8M Urea |
|
Glycine |
H |
0 |
0 |
0 |
0 |
|
Alanine |
CH3 |
-0.67 |
-0.80 |
0.44 |
-0.07 |
|
Leucine |
(CH3)2CHCH2 |
-2.55 |
-2.15 |
-2.80 |
-0.38 |
|
Phenylalanine |
C6H5CH2 |
-3.02 |
-2.88 |
-2.59 |
-0.70 |
|
Tyrosine |
HO-pC6H4CH2 |
-3.58 |
-2.67 |
- |
-0.73 |
Using these data,
order the solubilities of the amino acids in the different
solvents and discuss the impact of the side chain structures on the values.
1Damodaran
S, Song, KB, J Biol Chem,
1986, 261, 7220-7222.
2Hammes,
GG, “Thermodynamics and Kinetics for the Biological Sciences”, Wiley-Interscience, 2000, p 57.
SP 10 – Due date: Dec. 9
Certain
proteins can bind to DNA chromosomes for the purpose of regulating functioning.
Using the thermodynamics of binding two different proteins, this problem
illustrates how properties can suggest differences in ligand function
Certain proteins
can bind to DNA chromosomes for the purpose of regulating functioning. A protein, identified as WT1, has a domain
containing zinc that binds to DNA in order to suppress a particular type of
kidney tumor. A protein of very similar
structure, identified as EGR1, also binds to the same DNA site and is involved
in regulating cell proliferation.
Research suggests that there may be an important regulatory link between
these two proteins.
The amounts of
bound protein have been studied for these proteins as a function of
temperature. Some results are given
below. The uncertainties in ln K are ± 0.1.
Determine the
standard state enthalpies and entropies for binding and comment on what these
values suggest about the molecular binding processes for WT1 and EGR1.
|
Protein |
WT1 |
EGR1 |
|
ln K @ 4oC |
19.6 |
20.4 |
|
ln K @ 11oC |
20 |
20 |
|
ln K @ 27oC |
20.9 |
19 |
SP 11 – Due date: Dec. 9
Living
systems couple unfavorable reactions to favorable reactions to make biochemicals
that would not normally form. This problem treats the unfavorable
phosphorylation of glucose by coupling to the favorable ATP hydrolysis to form
ADP
An important biochemical reaction is the
phosphorylation of glucose
C6H12O6
+ H3PO4 
C6H13O9P + H2O (1)
a) Show that the
standard state Gibbs energy and enthalpy of this reaction are 12 and 1 kJ mol-1
respectively. Does this suggest that the
reaction is favorable or not?
b) The
concentrations of the species are actually not 1 M. They are given below. Under such conditions, show that the Gibbs
energy change for the reaction is unfavorable.
c) Living systems
overcome this barrier by coupling this reaction with a reaction involving
dissociation of adenosine triphosphate (ATP) to adenosine diphosphate
(ADP) that is coupled to the phosphorylation reaction by removing the H2O
and providing the phosphoric acid. Using
the tabulated values compute the standard state and actual state property
changes for the reaction
ATP + H2O ADP + H3PO4 (2)
Is this reaction
favorable?
d) If the two
reactions occur in the same system,
C6H12O6
+ ATP C6H13O9P
+ ADP (3)
what
are the standard state and solution Gibbs energy changes and the standard state
enthalpy change?
The standard state
properties (1 M, 1 atm, 298 K, pH 7.0, pMg 3.0, ionic
strength 0.25 M) of the aqueous species are1
|
Species |
C6H12O6 |
H3PO4 |
C6H13O9P |
H2O |
ATP |
ADP |
|
|
-427 |
-1060 |
-1319 |
-156 |
-2098 |
-1230 |
|
|
-1267 |
-1299 |
-2279 |
-286 |
-2996 |
-2006 |
The
concentrations of species in erythrocytes are1
|
Species |
C6H12O6 |
H3PO4 |
C6H13O9P |
H2O |
ATP |
ADP |
|
Concentration, M |
0.005 |
0.001 |
0.000083 |
56 |
.00185 |
.00014 |
1Hammes GG, Thermodynamics and Kinetics for the Biological
Sciences, Wiley-Interscience, 2000, chapter 3,
Appendix 1.