SOLUTIONS MANUAL
Basic Principles and
Calculations in
Chemical Engineering
Eighth Edition
David M. Himmelblau
James B. Riggs
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This work is protected by United States copyright laws and is provided solely for the use of
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needs of other instructors who rely on these materials.
ISBN-10: 0-13-288551-4
ISBN-13: 978-0-13-288551-5
ACKNOWLEDGMENTS
We want to thank Christine Bailor for preparing this Solutions Manual,
and for the many students and graders who have contributed to
the solutions it contains.
David M. Himmelblau
James B. Riggs
iii
TABLE OF CONTENTS
Page
1.
2.
3.
4.
5.
6.
7.
8.
9.
To the Instructor ............................................................................................
Example Course Syllabus .............................................................................
Course Objectives .........................................................................................
Exam and Recitation Section Schedules .......................................................
Suggestions for Taking Exams .....................................................................
What You Should Know About This Course ...............................................
Standards for Chemical Engineering Homework .........................................
Typical Assignments for One Semester ........................................................
Typical Examinations for a One Semester Course ........................................
iv
v
vi
viii
ix
x
xi
xii
xiv
xvii
To the Instructor
This Solutions Manual accompanies the book Basic Principles and Calculations in
Chemical Engineering, Eighth Edition, published by Prentice Hall. In addition to the
detailed, worked-out solutions for all the problems that follow each chapter in the
textbook and answers to the thought problems, you will find in what follows a number of
useful components of a syllabus for students, information that usually are handed out
during the first day of class:
1.
2.
3.
4.
5.
Class grading policies, homework and reading assignments, and examination
information.
Class objectives.
Schedule of topics covered.
Suggestions for taking examinations.
Format standards for submitting homework.
Suggested Content for the Introductory Course in Chemical Engineering
The introductory course in chemical engineering is usually taught over an interval of one
or two semesters, or one or three quarters. The textbook contains more material than can
be successfully presented in one quarter and probably in one semester (depending on the
background and previous coursework of students). Although an instructor would like to
assume that a student has learned all of the material covered in earlier courses in
chemistry and physics, it takes just one time in teaching the introductory course to
abandon that expectation. The textbook is organized into four parts comprised of 11
chapters plus 6 additional chapters on the accompanying CD that treat material usually
not included in a one semester course. The following list suggests the chapters to include
in courses of various duration:
One quarter
One semester
Two quarters
Two semesters
1–6, 8, 9–10
1–11
1–7 followed by 8 and 11
1–11 followed by 12–17
v
Example Course Syllabus
Information for ChE 317
Introduction to Chemical Engineering
Instructor:
D.M. Himmelblau
Office hours: M-F 10-11 a.m.
1.
Office: CPE 5.410
GENERAL
a. The prerequisites for ChE 317 are Ch 302 and Math 808. If you have not completed
these two courses, you will have to drop ChE 317 and should do so at once.
b. Class conduct is informal. Feel free to raise your hand at any time to ask a question or
for an explanation.
2.
EXAMINATIONS
a. Five two-hour examinations plus a final exam will be held at specified announced
dates as shown on the assignment sheets. The last examination will be scheduled
during the final exam period (refer to the course schedule for details). The lowest
exam of the first 5 (excluding the final exam) will be omitted in calculating your final
grade. You must take the final. If you will miss an exam, notify me prior to the exam,
not afterwards, to arrange for a makeup exam.
3.
GRADING
a. The grading is based on scores on the examinations, each of which is weighted
equally (90%), plus class discussion and homework (10%). The grades are assigned
on an absolute basis, not a curve:
A
B
C
D
F
> 82
71-82
61-70
51-60
< 51
hence there is no penalty for working together and helping each other.
b. You will have a grader assigned to this course whose name is ___________, office
number is Room ______, and office hours are __________.
c. The recitation session assistant is ___________________, office number is Room
_____, and office hours are __________.
d. If you disagree with the grader’s method of grading and with the total points he has
given you on a particular problem, discuss it with the grader first, but if you cannot
reach a decision, I will be the referee. Bring exam grade questions to me.
e. Prepare a grade sheet on which you can keep account of your homework and exam
grades so that you will be able to compute your status at any time you wish.
vi
f. A grade of at least a C is required in this course for subsequent courses in chemical
engineering.
4.
HOMEWORK PROBLEMS
a. CHEMICAL ENGINEERING STANDARDS WILL BE REQUIRED AND
ENFORCED. (Capital letters intended.)
b. Problems are due at the beginning of each class according to the assignment. No late
problems can be accepted.
c. Turn in a much of a problem as you can get. It is better to get a low grade than a
“miss.”
d. Working together is an important part of professional practice. After the second week
of class, students will be assigned to work on homework (not exams!) in pairs. During
the first two weeks of class look for a possible compatible partner. You will receive a
list of all of the class members with their phone numbers to help in the selection.
Exceptions can be made for individuals who insist on working alone.
e. After each scheduled homework assignment has been turned in, the solution(s) will
be placed in a file located in the ChE stockroom that may be checked out for 2 hours
at a time.
5.
If you have difficulty in the early part of the course, confer with me before you get into
trouble.
vii
Course Objectives
The objectives for Chemical Engineering 317 are as follows:
1.
To introduce you to the principles and calculation techniques used in the field of
chemical engineering.
2.
To acquaint you with the fundamentals of material and energy balances as applied to
chemical engineering.
3.
To acquaint you with efficient methods of problem solving so that you can effectively
solve problems you will encounter after leaving school.
4.
To offer practice in defining problems, collecting data, analyzing the data, and breaking it
down into basic patterns, and selection of pertinent information for application.
5.
To review certain principles of applied physical chemistry.
6.
To help you decide you have chosen the right field.
Contributions to Program Outcomes
By graduation a chemical engineering student should have achieved certain knowledge, skills,
and abilities known as Program Outcomes. ChE 317 contributes to five significant outcomes,
namely an ability to:
1.
Apply knowledge of mathematics, chemistry, physics computing, safety, ethical practice,
and technology to solve engineering problems.
2.
Apply and integrate elements of chemical engineering to solve problems in design,
operation, and control of processes.
3.
Participate in team activity effectively and demonstrate leadership.
4.
Communicate effectively via oral, written, and graphic means.
5.
Appreciate the societal and economic impact of engineering decisions locally and
globally.
viii
Fall Semester Exam Schedule
The first 5 exams are evening exams on Thursday, open book, of 2 hours duration, specific times
to be arranged (such as 5-7, 6-8, 7-9, etc.):
Exam 1
Exam 2
Exam 3
Exam 4
Exam 5
September 14
October 1
October 16
November 1
November 15
The final exam is listed in the University final exam schedule (that will appear about
December 1).
Fall Semester Schedule for the Recitation Section
The recitation section will meet on Thursday, 2–3:30 p.m. in CPE 2.220. The objective of the
recitation section is to let you ask questions and provide assistance in problem solving for old or
new ChE 317 homework and exam problems on a one-to-one basis.
Attendance is not required, but you will miss the unique opportunity to get personal
attention if you want it. You will also miss questions asked by other students that you have not
considered. You will also have the chance to meet other students in the class, and discuss
anything (!) with them.
No assignments are made, no grades given, no lectures presented, and no formal structure
exists for the recitation section. It’s up to you to make use of it.
ix
Suggestions for Taking Exams
1.
Bring what you want to the exams—they are open book. Be sure to have adequate
pencils, batteries, etc.
2.
Read the entire examination through quickly before starting to work any one problem.
Then work first on those problems which seem the simplest or about which you are most
confident in solving.
3.
Be sure to allot your working times to the questions roughly according to the grade value
of each. If a problem is not completed in the time allotted, it is usually better to
discontinue work on it and spend time on the other problems. Be sure to spend at least
some time on each problem. Partial solutions to all problems usually result in a higher
overall grade than complete solutions to only a small portion of the problems (provided
you do enough work on a problem to indicate that the correct method of attack is being
used).
4.
When starting work on a problem read it through carefully and be certain you understand
it. Spend a short time thinking about the method of solution instead of writing down what
first comes to mind.
5.
When writing down the solution, organize your work in a neat and logical manner in spite
of the time constraints. This step not only impresses the grader but also permits him or
her to follow the work closely enough so that if a mistake is made he or she can still
evaluate the succeeding work. Neatness and organization also permit you to check your
work more easily and to find quickly information needed later in the problem.
6.
In answering a question write enough so that the grader does not have to guess what you
had in mind. For example, when using equations, write down the equation first and then
substitute numbers. A group of numbers alone may confer little information to the grader,
especially if they are the wrong numbers. When using data obtained from tables or charts,
state the source—and in some cases the method of using the source. Draw pictures, and
separate subproblems from each other.
7.
If it is obvious that you are not going to finish a problem, carefully outline the remainder
of the solution by numbered steps, and include sufficient details, such as pertinent
equations and methods of solving them, sources for remaining necessary data, etc.
8.
If you start to get rattled, slow down a bit— perhaps even think of something besides the
examination for a minute or two. Remember that this one examination is not going to
make or break you whatever success you have on it. View the problem bothering you as
you would a bridge hand, crossword puzzle, or other game that involves solving a
problem based on a given set of facts with available information.
9.
Sample old exams are located in the ChE Stockroom, and can be taken out and copied.
Practice solving old exams two or three days in advance of each exam to isolate your
weaknesses in subject material and exam taking skills.
x
What you should know about this course at the beginning
that will be clear by final exam time
1.
You no longer are a freshman so that the material covered proceeds at a rapid pace.
2.
Your notions of teaching and learning will require substantial adjustment. Our goal is not
for you to reproduce what was told to you in the classroom or you read in the text. Your
study habits probably must change.
3.
Lecture time is at a premium and must be used efficiently. Listening is not learning any
more than lecturing is teaching. You are responsible for learning the material, a phase
that will occur primarily outside the classroom. The instructor cannot “teach” all the
skills you need in the short time of a class. It will take you two or three hours on the
average per hour of class time to become proficient.
4.
The instructor’s job is to provide a framework of the topic along with demonstrations to
guide you in your learning of concepts, methods, and efficient problem solving skills. It
is not to imprint you with isolated facts and problem types.
5.
If you read the material in the assigned section for the next period before coming to class,
the lecture will make more sense, and you can ask questions to clarify any uncertain
issues.
xi
Standards for Chemical Engineering Homework Assignments
1.
Engineering paper must be used (paper ruled on the back with a grid).
2.
Use the unruled side of the sheet for the calculations, and the back side for drawings
(rarely required).
3.
Use the sheet with the holes to the left.
4.
Turn in your work with the paper folded vertically.
5.
Write neatly. Make the text in your calculations in letters 0.20 inches high—these match
the horizontal grid spacing on the back of the paper. Leave 0.20 inches (one grid interval)
between lines. The idea is to be professional in presenting your work.
6.
Use engineering/scientific notation for numbers such as 0.341 and 1.453 x 105. Use
judgment as to how many zeros you put after the decimal point or before the first
significant figure. Note: always put a zero before the decimal point for a number less than
unity.
7.
Indicate multiplication and division using units as seen below. (Note: Use vertical
and horizontal rules as necessary.)
3.45 lb NaCl 4 gal soln 1 ft 3
=
7.48 gal
1 ft 3 soln
As you gain experience, you can suppress the units for simple problems and show
multiplication and division thus (use parenthesis rather than centered dots as the dots get
confused with periods, dust specks, etc.)
1 $
(3.45)( 4) !#" 7.48
&% =
8.
Use a solid line across the page between the vertical rules on the engineering paper to
demark the end of a part of a problem with multiple parts. Denote the end of the entire
problem by a line across the page from the left hand rule to the far right edge of the
paper.
9.
Always show the units of your answer, underline the numbers and units, and draw an
arrow from the right edge of the paper to the answer so that the answer is easy to pick out
on the page thus
8.5 lbH 2 1 lb mol H 2 1 lb mol Zn 4.21 lb mol Zn
=
lb F
lb F
2.02 lb H 1 lb mol H 2
!"
"
part (a) of the problem
10.
Indicate a new problem by placing the problem number in the left hand margin.
11.
Always show the basis of your calculations thus
Basis: 100 lb feed
xii
12.
On each submission place your class number, the date, the assignment number, your
name, and the page numbering at the top of each and every page, even if you staple the
pages together, thus
317
↑
Class
Sept. 10,
2003
↑
Date
Assignment No. 5
Jones, Robert
2/3
↑
Assignment Identification
↑
Your name
↑
Page 2 of 3
pages
submitted
xiii
TYPICAL ASSIGNMENTS FOR ONE SEMESTER
Topic and Problem Assignments Due
All assignments are in the 8th
edition. Study:
1.
First Class meeting. No assignments due
2.
UNITS, DIMENSIONS, UNIT CONVERSION
2.1.1, 2.2.2, 2.2.6
Chapter 2
3.
DIMENSIONAL CONSISTENCY, SIGNIFICANT
FIGURES, VALIDATION, MOLES
2.3.1, 2.3.3, 2.3.8
Chapter 2
4.
METHODS OF ANALYSIS AND MEASUREMENT
2.6.1a, 2.6.4a, 2.9.1
Chapter 2
5.
BASIS, TEMPERATURE, PRESSURE
2.7.1a,b,c; 2.11.2, 2.11.5
Chapter 2
6.
PRESSURE MEASUREMENT
2.11.9
Chapter 2
7.
INTRODUCTION TO MATERIAL BALANCES
3.1.11, 3.1.19, 3.1.16, 3.1.7, 3.1.8
Chapter 3
8.
STRATEGY FOR SOLVING MATERIAL
BALANCES
3.2.2, 3.2.4, 3.2.5, 3.2.9, 3.2.14
Chapter 3
9.
No class meeting. Exam No. 1 in the evening.
10.
MATERIAL BALANCES WITHOUT REACTION—
SINGLE UNITS
4.1.7, 4.1.8, 4.1.10, 4.1.12
Chapter 4
11.
MATERIAL BALANCES (CONTINUED)
4.1.18, 4.1.20, 4.1.23, 4.1.25
Chapter 4
12.
STOICHIOMETRY
5.1.2a,e; 5.1.5; 5.2.14; 5.2.15
Chapter 5
13.
MATERIAL BALANCES WITH REACTION—
SINGLE UNITS
5.3.1, 5.3.2, 5.3.6, 5.3.7
Chapter 5
14.
MATERIAL BALANCES WITH REACTION—
SINGLE UNITS (CONTINUED)
5.5.5, 5.5.7, 5.5.10, 5.5.13
Chapter 5
xiv
Topic and Problem Assignments Due
All assignments are in the 8th
edition. Study:
15.
Review for Exam No. 2.
Chapters 4–5
16.
No class meeting. Exam No. 2 in the evening.
17.
MATERIAL BALANCE PROBLEMS WITH
MULTIPLE UNITS
6.1.2, 6.1.5, 6.2.1, 6.2.7
Chapter 6
18.
MATERIAL BALANCE PROBLEMS WITH
RECYCLE (NO REACTION)
6.3.1b,d; 6.3.2; 6.3.16, 6.3.17
Chapter 6
19.
MATERIAL BALANCE PROBLEMS WITH
RECYCLE (WITH REACTION)
6.3.8, 6.3.13, 6.3.21
Chapter 6
20.
IDEAL GAS AND PARTIAL PRESSURE
7.1.1, 7.1.5, 7.1.12c, 7.1.22, 7.1.31
Chapter 7
21.
MATERIAL BALANCES WITH IDEAL GASES
7.1.52, 7.1.44, 7.1.55
Chapter 7
22.
REAL GASES—COMPRESSIBILITY
7.3.1, 7.3.3, 7.3.13
Chapter 7
23.
REAL GASES—EQUATIONS OF STATE
7.2.8, 7.2.5, 7.2.17, 7.2.13
Chapter 7
24.
Review for Exam No. 3.
Chapters 6–7
25.
No class meeting. Exam No. 3 in the evening.
26.
SINGLE COMPONENT-TWO PHASE SYSTEMS
(VAPOR PRESSURE)
8.2.1, 8.2.9, 8.3.2 a to e, 8.3.4, 8.3.5b, 8.3.19, 8.3.20
Chapter 8
27.
TWO PHASE GAS-LIQUID SYSTEMS
8.3.17, 8.3.22a, 8.4.2, 8.4.4, 8.4.7
Chapter 8
28.
TWO PHASE GAS-LIQUID SYSTEMS
(CONTINUED)
8.4.14, 8.4.16, 8.4.18, 8.4.27
Chapter 8
xv
Topic and Problem Assignments Due
All assignments are in the 8th
edition. Study:
29.
VAPOR-LIQUID EQUILIBRIA AND THE PHASE
RULE
8.2.4, 8.2.5, 8.5.3, 8.5.6, 8.5.14, 8.5.18
Chapter 8
30.
Review for Exam No. 4
Chapter 8
31.
No class meeting. Exam No. 3 in the evening.
32.
ENERGY: TERMINOLOGY, CONCEPTS, AND
UNITS
9.1.1, 9.1.5, 9.1.7, 9.1.11, 9.2.4, 9.2.21, 9.2.22
Chapter 9
33.
ENERGY BALANCES—CLOSED SYSTEMS
9.3.31a,b; 9.3.9; 9.3.25; 9.3.45
Chapter 9
34.
ENERGY BALANCES—OPEN SYSTEM
9.3.31, 9.3.32, 9.3.17a, 9.3.33
Chapter 9
35.
CALCULATING ENTHALPY CHANGES
9.2.26, 9.2.37, 9.2.46
Chapter 9
36.
APPLICATIONS OF ENERGY BALANCES
WITHOUT REACTION-CLOSED SYSTEMS
9.3.19, 9.3.39, 9.3.40
Chapter 9
37.
APPLICATIONS OF ENERGY BALANCES
WITHOUT REACTION-OPEN SYSTEMS
9.3.21, 9.3.41
Chapter 9
38.
Review for Exam No. 5
Chapter 9
39.
No class meeting. Exam No. 5 in the evening.
40.
ENERGY BALANCES WITH REACTION
10.1.6; 10.1.8a,b; 10.2.1a; 10.2.3; 10.2.4
Chapter 10
41.
CONTINUED
10.2.17, 10.2.6, 10.2.10, 10.4.6
Chapter 10
42.
PARTIAL SATURATION AND HUMIDITY
11.1.1
Chapter 11
43.
HUMIDITY CHARTS
11.2.1, 11.3.1, 11.3.7
Chapter 11
44.
Exam No. 6 is the final exam held on the scheduled
final exam period (3 hours).
xvi
TYPICAL EXAMS FOR A ONE SEMESTER COURSE
(scheduled in the evening to avoid time constraints on students
Exam No. 1
(Open Book, 1 1/2 hours)
PROBLEM 1 (5%)
Hydrogen can be separated from natural gas by diffusion through a round tube. The rate
of separation is given by
N = 2!D"R
where
N = rate of transport of H2 from the tube, g moles/(sec)(cm of length of tube)
D = diffusion coefficient
! = molar density of H2, g moles/cm3
R
=
!r $
log mean radius of tube, r2 –r1 / l n # 2 & , with r in cm.
" r1 %
What are the units of D?
PROBLEM 2 (10%)
A pallet of boxes weighing 10 tons is dropped from a lift truck from a height of 10 feet.
The maximum velocity the pallet attains before hitting the ground is 6 ft/sec. How much
kinetic energy does the pallet have in (ft)(lbf) at this velocity?
PROBLEM 3 (5%)
The specific gravity of a fuel oil is 0.82. What is the density of the oil in lb/ft3? Show all
units.
PROBLEM 4 (10%)
Sulfur trioxide (SO3) can be absorbed in sulfuric acid solution to form more concentrated
sulfuric acid. If the gas to be absorbed contains 55% SO3, 41% N2, 3% SO2, and 1% O2,
how many parts per million of O2 are there in the gas? (b) What is the composition of the
gas on a N2 free basis?
PROBLEM 5 (15%)
You have 100 kilograms of gas of the following composition:
CH4
H2
N2
30%
10%
60%
What is the average molecular weight of this gas?
xvii
PROBLEM 6 (15%)
If the heat capacity of a substance is 5.32 J/(g)(°C) and its molecular weight is 37.4, what
is its heat capacity in
(a)
(b)
(c)
J/(g)(°F)
J/(lb)(°R)
J/(gmol)(K)
PROBLEM 7 (20%)
A rock containing 100% BaSO4 is burned with coke (94%C, 6% ash) and the
composition of the product is BaSO4 (11.1%) BaS (72.9%), C (13.9%, ash (2.2%). The
reaction is
BaSO 4 + 4C ! BaS + 4CO
Calculate the percent excess reactant, and the degree of completion of the reaction.
PROBLEM 8 (20%)
A gas cylinder to which is attached a Bourden gauge appears to be at a pressure of 27.38
in. Hg at 70°F. the barometer needs 101.8 kPa. A student claims that the pressure in the
tank is 1.3 psia, but another student points out that this is impossible—the pressure is
really 28.2 psia. Can 1.3 psia be correct? Explain and show calculations to back up your
explanation.
xviii
EXAM NO. 2
(Open Book, 2 hours)
PROBLEM 1 (25%)
A chemist attempts to prepare some very pure crystals of Na2SO4‑10H2O by dissolving
200 g of Na2SO4 (MolWt=142.05) in 400 g of boiling water. He then carefully cools the
solution slowly until some Na2S04·10H20 crystallizes out. Calculate the g of
Na2SO4·10H2O recovered in the crystals per 100 g of initial solution, if the residual
solution after the crystals are removed contains 28% Na2SO4.
Right answer but:
–10 if answer is in g of
Na2SO4 and not g of
Na2SO4·10H2O
–10 if answer not
per 100 g of initial
soln
PROBLEM 2 (25%)
Water pollution in the Hudson River has claimed considerable recent attention, especially
pollution from sewage outlets and industrial wastes. To determine accurately how much
effluent enters the river is quite difficult because to catch and weigh the material is
impossible, weirs are hard to construct, etc. One suggestion which had been offered is to
add a trace Br– ion to a given sewage stream, let it mix well, and sample the sewage
stream after it mixes well.
On one test of the proposal you add ten pounds of NaBr per hour for 24 hours to a
sewage stream with essentially no Br– in it. Somewhat downstream of the introduction
point a sampling of the sewage stream shows 0.012% NaBr. The sewage density is 60.3
lb/ft3 and river water density is 62.4 lb/ft3.
What is the flow rate of the sewage in lb/min?
–10 if answer based on
0.012 fractin and not
0.00012.
–15 if 24 hr basis
was used and then
not converted back
to per hour basis.
PROBLEM 3 (25%)
In preparing 5.00 moles of a mixture of three gases (SO2, H2S, and CS2), gases from
three tanks are combined into a fourth tank. The tanks have the following compositions
(mole fractions):
Gas
SO2
H2 S
CS2
Tank 1
0.10
0.40
0.50
Tank2
0.20
0.20
0.60
Tank3
0.25
0.25
0.50
Tank 4
0.20
0.26
0.54
How much of Tanks 1, 2, and 3 must be mixed to give a product with composition of
Tank4?
–10 for correct answer but wrong
basis
–15 A lot of people said no soln.
They used wrong basis, etc. No
soln but correct mat’l balance
xix
PROBLEM 4 (25%)
10%
a)
For the given distillation process, calculate the composition of the bottoms
stream.
15%
b)
If steam leaked into the column at 1000 mole/sec and all else was
constant, what would the new bottoms composition be? –5 (should be gmole), if assumed to k-mole and not stated.
xx
Exam No. 3
(Open Book Exam, 2 hours)
PROBLEM 1 (35%)
A company burns an intermediate product gas having the composition 4.3% CO2, 27%
CO, 10% H2, 1.0% CH4, and the residual N2 together with a waste oil having the
composition 87% C, 13% H2. Analysis of the stack gas gives an Orsat analysis of 14.6%
CO2, 0.76% CO, and 7.65 O2 and the rest N2. Calculate the fraction of the total carbon
burned that comes from the product gas.
PROBLEM 2 (35%)
Benzene, toluene and other aromatic compounds can be recovered by solvent extraction
with sulfur dioxide. As an example, a catalytic reformate stream containing 70% by
weight benzene and 30% non-benzene material is passed through the counter-current
extractive recovery scheme shown in the diagram below. One thousand pounds of the
reformate stream and 3000 pounds of sulfur dioxide are fed to the system per hour. The
benzene product stream (the extract) contains 0.15 pound of sulfur dioxide per pound of
benzene. The raffinate stream contains all the initially charged non-benzene material as
well as 0.25 pound of benzene per pound of the non-benzene material. The remaining
component in the raffinate stream is the sulfur dioxide.
(a)
How many pounds of benzene are extracted per hour, i.e. are in the extract?
(b)
If 800 pounds of benzene containing in addition 0.25 pound of the non-benzene
material per pound of benzene are flowing per hour at point A and 700 pounds of
benzene containing 0.07 pound of the non-benzene material per pound of benzene are
flowing at point B, how many pounds (exclusive of the sulfur dioxide) are flowing at
points C and D?
xxi
PROBLEM 3 (30%)
Reactant A is polymerized as shown in the figure. It is mixed with fresh catalyst and
recycled catalyst. Conversion of A is 40% on one pass through the reactor. Fresh catalyst
(G) enters at the rate of 0.40 lb G per lb of A in stream H. The separator removes 90% of
the catalyst and recycles it as well as recycling unreacted A. Nevertheless, the product
stream P constraints 15% of the unreacted A and 10% of the catalyst exiting in stream K
as well as the polymer product. Determine the ratio of stream R to G.
Note: catalyst does not react in the process!
xxii
Exam No. 4
(Open Book, 2 hours)
PROBLEM 1 (25%)
In the vapor-recompression evaporator (not insulated) shown in the figure below, the
vapors produced on evaporation are compressed to a higher pressure and passed through
the heating coil to provide the energy for evaporation. The steam entering the compressor
is 98% quality at 10 psia, the steam leaving the compressor is at 50 psia and 400°F, and 6
Btu of heat are lost from the compressor per pound of steam throughput. The condensate
leaving the heating coil is at 50 psia, 200°F. The replacement liquid is at the temperature
of the liquid inside the evaporator.
Computer:
(a)
the work in Btu needed for compression per pound of H2O going through the
compressor.
(b)
the Btu of heat transferred from the heating coil to the liquid in the evaporator per
pound of H2O through the coil.
(c)
Bonus of 5 points for correct answer to the question: What is the total heat gained
or lost by the entire system?
PROBLEM 2 (25%)
An insulated, sealed tank that is 2 ft3 in volume holds 8 lb of water at 100°F. A 1/4 hp
stirrer mixes the water for 1 hour. What is the fraction vapor at the end of the hour?
Assume all the energy from the motor enters the tank.
For this problem you do not have to get a numerical solution. Instead list the following in
this order:
1.
2.
3.
4.
5.
6.
State what the system you select is.
Specify open or closed.
Draw a picture.
Put all the known or calculated data on the picture in the proper place.
Write down the energy balance (use the symbols in the text) and simplify it as
much as possible. List each assumption in so doing.
Calculate W.
xxiii
7.
8.
Lists the equations with data introduced that you would use to solve the problem.
Explain step by step how to solve the problem (but do not do so).
PROBLEM 3 (15%)
What is the enthalpy change in Btu when 1 pound mole of air is cooled from 600°F to
100°F at atmospheric pressure.
Compute your answer by two ways:
1)
Use the tables of the combustion gases.
2)
Use the heat capacity equation for air.
PROBLEM 4 (10%)
Answer the following questions by placing T fro true and F for false on your answer
page. Grading: +2 if correct, 0 if blank, -1 if wrong.
(a)
(b)
Heat and thermal energy are synonymous terms used to express one type of
energy.
You can find the enthalpy change at constant pressure of a Substance such as CO2
from the solid to the gaseous state by integrating
T2
" C dT
!
from T1 (solid temperature) to T2 (gas temperature) for a
constant pressure path.
T1
(c)
(d
(e)
The enthalpy change of a substance can never be negative.
Heat and work are the only methods of energy transfer in a non-flow process.
Both Q and ∆H can be classed as state functions (variables)
PROBLEM 5 (25%)
Hot reaction products (assume they have the same properties as air) at 1000°F leave a
reactor. In order to prevent further reaction, the process is designed to reduce the
temperature of the products to 400°F by immediately spraying liquid water into the gas
stream.
400°F?
How many lb of water at 70°F are required per 100 lb of products leaving at
xxiv
For this problem you do not have to get a numerical solution. Instead list the
following in this order.
1.
2.
3.
4.
5.
6.
State what the system you select is.
Specify open or closed.
Draw a picture.
Put all the known or calculated data on the picture in the proper places.
Write down the material and energy balances (use the symbols in the text) and
simplify them as much as possible, list each assumption in so doing.
Insert the known data into the simplified equation(s) you would use to solve the
problem.
xxv
Exam No. 5
(Open Book, 2 hours)
PROBLEM 1 (10%)
Answer the following questions briefly (no more than 3 sentences);
a.
Does the addition of an inert dilutent to the reactants entering an exothermic
process increase, decrease, or make no change in the heat transfer to or from the
process?
b.
If the reaction in a process is incomplete, what is the effect on the value of the
standard heat of reaction? Does it go up, down, or remain the same?
c.
How many properties are needed to fix the state of a gas so that all of the other
properties can be determined?
1
Consider the reaction H 2 ( g ) + O 2 ( g ) ! H 2O ( g ) .
2
Is the heat of reaction with the reactants entering and the products leaving at 500K
higher, lower, or the same as the standard heat of reaction?
d.
PROBLEM 2 (10%)
Explain how you would calculate the adiabatic reaction temperature for Problem 5 below
if the outlet temperature is not specified. List each step. (You can cite some of the steps
listed in Problem 5 if you list them by number in Problem 5.)
PROBLEM 3 (20%)
A flue gas at 750°F and 1 atm of composition 14.0% CO2, 1.0% CO, 6.4% O2, and the
balance N2 is the product of combustion with excess air at 750°F and 1 atm that is used to
burn coke (C).
a.
What is the volume in ft3 of the flue gas leaving the furnace per pound of carbon
burned?
b.
What is the volume of air in ft3 entering the furnace per pound of C burned?
PROBLEM 4 (20%)
Seven pounds of N2 are stored in a cylinder 0.75 ft3 volume at 120°F. Calculate the
pressure in the cylinder in atmosphere:
a.
Assuming N2 to be an ideal gas.
b.
Assuming N2 is a real gas and using compressibility factors.
PROBLEM 5 (40%—one half each for the material and energy balances)
Pyrites (FeS2) is converted to sulfur dioxide (SO2) gas according to the reaction
4FeS2 ( s ) + 11 O 2 ( g ) ! 2 Fe 2O3 ( s ) + 8 SO 2 ( g )
xxvi
The air, which is 35% excess (based on the above reaction) for combustion, enters at 27°C,
the ore at 18°C, and the products leave at 900K. Because of equipment degradation,
unburned FeS2 exits from the process. In one hour 8000 kg of pyrites are fed to the process,
and 2000 kg of Fe2O3 are produced. What is the heat added or removed from the process?
Data: For FeS2, C p = 44.77 + 5.590 ! 10 –2 T where T is in Kelvin and Cp is in J/(g mol)(K).
For F2O3, C p = 103.4 + 6.71! 10"2 T with the same units.
xxvii
Exam No. 6
(Open Book, 2 hours)
PROBLEM 1 (20%)
A high pressure line carries natural gas (all methane) at 10,000 kPa and 40°C. How
would you calculate the volume of the gas under these conditions that is equivalent to
0.03 m3 of CH4 at standard conditions using an equation of state? Select one equation
other than van der Waal's equation, and list it on your solution page. Give a list of steps
to complete the calculations. Include all the proper equations, and include a list of data
involved, but you do not have to obtain a solution for the volume.
PROBLEM 2 (20%)
From the following data estimate the vapor pressure of sulfur dioxide at 100°C.
Temperature (°C)
Vapor pressure (atm)
–10
1
6.3
2
32.1
5
55.5
10
PROBLEM 3 (20%)
Dry atmospheric air at the ambient
conditions of 90°F and 29.42 in. Hg
absolute passes through a small blower
and is bubbled up through water so that
the air leaving the water is saturated. The
temperature of the water is constant at
80°F, and because of the back pressure in
the system, the pressure in the vapor space
in the top of the bottle is 2.7 in.
H2O greater than atmospheric pressure. The bottle is weighted after the air is blown for 2
hours, 13 minutes, 47 seconds, and the decrease in weight was found to be 8.73 lb. What
was the hourly rate of flow of air at ambient conditions in ft3?
PROBLEM 4 (20%)
A vessel with a volume of 2.83 m3 contains a mixture of nitrogen and acetone at 44.0°C
and 100.0 kPa. The dew point of the mixture is 20.0°C and the relative saturation of the
acetone in the mixture is 58.39%. The vapor pressure of acetone at 44.0°C is 65.35 kPa
and it is 24.62 kPa at 20.0°C.
a.
What is the partial pressure of acetone vapor in the original mixture, in kPa?
b.
How many kg moles of acetone does the original mixture contain?
c.
If the nitrogen-acetone mixture is cooled with the volume remaining at 2.83 m3
constant so that 27.0 percent of the acetone condenses, what is the final
temperature of the mixture in °C?
xxviii
PROBLEM 5 (20%)
A wet sludge contains 50 percent water by weight. This sludge is first centrifuged, and
0.1 kilograms of water are removed per kilogram of wet sludge feed. The sludge is dried
further using air so that the final product contains 10 percent by weight water. The air for
drying is heated, passed into an oven drier, and vented back into the atmosphere. On a
day when atmospheric pressure is 760mm Hg, the temperature is 70°F and the relative
humidity is 50%, calculate the cubic meters of wet air required to dry one kilogram of
sludge fed to the process. The air vented from the oven drier is at 100°F and 780mm Hg.
It has a dew point of 94°F.
xxix
Solutions Chapter 2
2.1.1
2.2.1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
a.
N/mm or nm (nanometer)
°C/M/s
100 kPa
273.15 K
1.50m, 45 kg
250°C
J/s
250 N
Basis: 1 mi3
1mi3 ⎛ 5280 ft ⎞ 3 ⎛ 12 in ⎞ 3 ⎛ 2. 54 cm ⎞ 3 ⎛ 1 m ⎞ 3
⎝ 1 mi ⎠ ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠
9
3
= 4.17 × 10 m
b.
Basis: 1 ft3/s
a.
0.04 g 60 min (12 in )3 1 lb m
lb m
= 9.14
3
3
(min )( m ) 1 hr 1 ft 454g
(hr ) (ft 3 )
b.
2L 3600 s 24 hr 1 ft
ft 3
= 6.1× 10 3
s 1 hr 1 day 28.32L
day
2.2.2
1 ft 3 60 s 7.48 gal
= 449 gal / min
1 s 1 min 1 ft 3
3
c.
1 ft 2
(1 in)2
1 yr 1 day 1 hr 2.2 lbm
6 (in.)(cm 2 )
2
2
2
(yr)(s)(lbm )(ft ) (12 in) (2.54 cm) 365 days 24 hr 3600s 1 kg
m
1 ft 0.3048 m
–11
= 1.14 × 10
12 in
1 ft
( kg) (s2 )
2–1
Solutions Chapter 2
2.2.3
a. Basis: 60.0 mile/hr
60.0 mile 5280 ft 1 hr
ft
= 88
hr
1 mile 3600 sec
sec
b. Basis: 50.0 lbm/(in)2
50.0/lbm 454 g 1 kg
1(in)2 (100 cm)2
kg
= 3.52 × 104 2
2
2
2
(in)
1 lb 1000 g (2.54 cm)
(1 m)
m
c. Basis: 6.20 cm/(hr)2
6.20 cm 1 m 109 nm 1(hr) 2
nm
= 4.79 2
2
2
(hr) 100 cm 1 m (3600 sec)
sec
2.2.4
20 hp 0.7457 kW
= 14.91 kW
1 hp
No , not enough power even at 100% efficiency; 68 kW = 91.2 hp.
2.2.5
1 hr 2200 gal 1000 mile
= 4190.5 gal
525 mile 1 hr
1 hr 2000 gal 1000 mile 4210 gal
=
475 mile 1 hr
(20 gal)
None: 20 gal more are needed.
2.2.6
Let tA be the time for A to paint one house; tB for B
A does a house in 5 hours, or 1 house/5 hr. B does one house
in 3 hours, or 1 house/3 hr.
1 house tA hr 1 house tB hr
+
= 1 house
5 hr
3 hr
Also tA = tB so that
tA =
3
5
8
tA +
tA = 1 or
tA =1
15
15
15
15
hr = tB = 1.875 hr or 112.5 min
8
2–2
Solutions Chapter 2
2.2.7
2.2.8
(a) mass, because masses are balanced
(b) weight, because the force exerted on the mass pushes a spring
2
1(hr)2
20.0g 1 lbm 0.3048m 3600s 1 (lbf )(s )
1 hr 32.174(lbm )(ft) (3600)2s 2
(m)(s) 453.6 g 1 ft
= 1.16 × 10−7
2.2.9
(lbf )(hr)
ft 2
1.0 Btu
24 hrs 1 ft 2
1 in 2
(100 cm)2 1.8o F 2.54 cm
⎛ o F ⎞ 1 day (12 in)2 (2.54 cm 2 ) 1 m 2
in
1oC
(hr)(ft 2 ) ⎜ ⎟
⎝ ft ⎠
kJ
252 cal 12 in 4.184 J 1 kJ
= 1.49 × 104
2
1 Btu 1 ft 1 cal 1000 J
(day)(m )(°C / cm)
2.2.10
Basis: 1 lb H2O
3
a.
1 2 1 1 lb m ⎛ 3 ft ⎞
(s 2 )(lb f )
KE= mv =
= 0.14(ft)(lbf )
⎜
⎟
2
2
⎝ s ⎠ 32.174(ft)(lb m )
b.
Let A = area of the pipe and v = water velocity. The flow rate is
⎛ πD 2 ⎞
(v)
q = Av = ⎜
⎝ 4 ⎟⎠
π (2 in)2 (1 ft)2 3 ft 60 s 7.48 gal
= 29.37 gal/min
4
(12 in)2 S 1 min 1 ft 3
2.2.11
PE =
75 gal 8.33 lbm 32.2 ft 100 60 s s 2 -lbf
Btu
= 4818 Btu/hr
2
min
ft hr 32.2 lbm -ft 778 ft-lbf
gal
sec
2 hp 2545 Btu
= 5090 Btu/hr
hp-hr
Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr
Pump Work =
2–3
Solutions Chapter 2
2.2.12
The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the
force used to support the mass.
2.2.13
In American Engineering System
Power =
=
FV
800 lbf 300 ft
1
min
=
2.4 ×105
(lbf )(ft)
or 7.27 hp
min
In SI
Power =
4000 N 1.5 m 1 (watt)(s)
1
s
1(N)(m)
=
6000 watts
=
1
m v2
2
2.2.14
KE
=
1 2300 kg 1 lb m ⎛ 10.0 ft ⎞ 2
1
1 Btu
2
0. 454 kg ⎝ s ⎠ 32. 2 lb m ft 778.2(ft )(lb f )
2
lb f
sec
= 10.11 Btu
2.2.15
Basis: 10 tons at 6 ft/s
2
KE =
2.2.16
1
1 20,000 lbm ⎛ 6 ft ⎞ 1(s 2 )(lbf )
2
mv =
= 11,200(ft)(lbf )
⎜
⎟
2
2
⎝ s ⎠ 32.2(ft)(lbm )
Basis: 1 mRNA
1 nRNA 3x ribonucleotides 1 active ribosome
nRNA
1200 amino acids
protein
264 ribonucleotides min-active ribosome x amino acids
13.6 protein molecules formed per min per nRNA
2–4
=
Solutions Chapter 2
2.3.1
2.3.2
A
is
in
g/cm3
B
is
in
g/(cm3)(°C)
Since the exponent of e must be dimensionless
C
is
in
atm-1
a 1.
a 2.
a 3.
b1
A=
b2
B=
b3
C=
lb
1.096 g (30.48)3cm3 1 lb m
= 68.4 m3
cm3
ft 3
454g
ft
lb
0.00086 g (30.48)3cm3 1 lbm 1oC
= 0.0298 3 mo
3 o
3
o
(cm )( C)
ft
454 g 1.8 R
(ft )( R)
0.000953 1 atm
1
= 0.0000648
2
atm 14.7 lbf /in
lb f /in 2
Introduce the units. The net units are the same on both sides of the equation.
⎛ ft ⎞
(ft)(ft) ⎜ 2 ⎟
⎝s ⎠
1.5
1/2
=
ft 3
s
1/ 2
2.3.3
No.
⎡
⎤
⎢
⎥
-3
3
1 ⎥
2 ⎢ (2) 9.8m 10 m 50×10 (kg)(m)
q = 0.6(2m ) ⎢
2 ⎥
s2
kg
(s 2 )(m 2 )
⎛ 2 ⎞
⎢
1- ⎜ ⎟ ⎥
⎢
⎝ 5 ⎠ ⎥⎦
⎣
The net units on the right hand side of the equation are
1/2
⎡ m3 ⎤
m ⎢ 4 ⎥
⎣ s ⎦
2
≠
m3
s
Consequently, the formula will not yield 80.8 m3/s, presumably in the formula the g
should be gc for use in the AE system.
2–5
Solutions Chapter 2
1.19
2.3.4
Q = 0.61S
(2Δp)/ρ
assume hole is open to atmosphere
⎡
lbf ⎤
14.7
⎢
⎥
73 in gas. 0.703 H 2O 1 ft
144 in ⎢ 23 lbf
in 2 ⎥
+
Δp =
1 gas 12 in 33.91 ft H 2O ⎥
1 ft 2 ⎢ in 2
⎢
⎥
⎣
⎦
2
= 3,579 lbf /ft
2
62.4
lb
ρ=
0.703ft3
3
1 lb/ft H 2O
lbH 2O
ft 3H 2O
=43.87
lbm
ft 3
2
⎛ 1 ⎞
S = π⎜
= 3.41× 10-4 ft 2
⎟
⎝ (4)12 ⎠
Q = (3600)(0.61)(3.41× 10−4 ) (2)(3579)g c / 43.87 = 54
2–6
ft 3
hr
Solutions Chapter 2
2.3.5
a.
b.
Z
= 1 + ρB + ρ2C + ρ3D
B
C
D
Units
cm3 / g mol
(cm3/ g mol)2
(cm3/ g mol)3
Z
= 1 + ρ* B* +(ρ* )2C* +(ρ* )3D*
B*
C*
D*
Units
ft3 / lbm
(ft3/ lbm)2
(ft3/ lbm)3
If B is the original coefficient, B* is obtained by multiplying B by conversion factors.
Let MW is the molecular weight of the compound.
ft 3
B cm3
B
=
lb m
g mol
*
2
3
0.016
⎛ 1 ft ⎞ 1 g mol 454 g
= B
⎜
⎟
MW
⎝ 30.48 cm ⎠ MW g 1 lb m
2
6
2
2
⎛ ft 3 ⎞
⎛ cm3 ⎞ ⎛ 1 ft ⎞ ⎛ 1 g mol ⎞ ⎛ 454g ⎞
2.57 × 10-4
C ⎜
⎟ = C ⎜
⎟ = C
⎟ ⎜
⎟ ⎜
⎟ ⎜
MW 2
⎝ g mol ⎠ ⎝ 30.48 cm ⎠ ⎝ MWg ⎠ ⎝ 1 lbm ⎠
⎝ lb m ⎠
*
3
⎛ ft 3 ⎞
4.096×10-6
D ⎜
⎟ = D
MW 3
⎝ lb m ⎠
*
2–7
Solutions Chapter 2
2.3.6
⎡m ⎤
u ⎢ ⎥ = k[1]
⎣s⎦
⎡⎛ τ ⎞ 1/2 ⎛ N ⎞ 1/2 ⎛ m 3 ⎞ 1/2 ⎤
⎢⎜ ⎟ ⎜ 2 ⎟ ⎜
⎥
⎢⎝ ρ ⎠ ⎝ m ⎠ ⎝ kg ⎟⎠ ⎥
⎣
⎦
To get u in ft/s, substitute for τ and for ρ , and multiply both sides of the equation by
3.281 ft/1 m (k is dimensionless).
2
τ
τ′lbf ⎛ 3.281 ft ⎞
N
1N
=
2
2 ⎜
⎟
m
ft ⎝ 1 m ⎠ 0.2248 lbf
3
ρ′ lbm ⎛ 3.281 ft ⎞
kg
1 kg
ρ 3 =
m
ft 3 ⎜⎝ 1 m ⎟⎠ 0.454 lbm
⎛ τ′ ⎞
u ′ = 2.57 k ⎜ ⎟
⎝ ρ′ ⎠
2.3.7
Place units for the symbols in the given equation, and equate the units on the left and
right hand sides of the equation by assigning appropriate units to the coefficient 0.943.
LHS
RHS
3
⎡⎛
⎤
⎞⎟ ⎛ lb m ⎞ 2 ft Btu (hr)(ft )
Btu
Btu
⎜
?
(hr ) ft 2 (Δ ° F) ⎢⎣⎝ (hr )(ft )(Δ ° F)⎠ ⎝ ft 3 ⎠ (hr)2 lb m ft lb m Δ ° F ⎥⎦
1
4
( )
The units are the same on the right and left so that 0.943 has no units associated with
it.
2.3.8
η
=
numerator
denominator
Numerator
=
Yxc/ s γ b ΔHbc/ e−
−
⎛ mol cell C ⎞ ⎛ e equiv. ⎞⎛ energy ⎞
⎟⎜
⎟
⎟ ⎜
−
⎝ mole substrate C ⎠ ⎝ mol cell C ⎠ ⎝ mol e equiv. ⎠
= ⎜
=
energy
mol substrate C
(1)
2–8
Solutions Chapter 2
Denominator
η
= ΔH ccat
=
energy
mol substrate C
=
(1)
energy
=
(2)
energy
(2)
There is no missing conversion factor
What the author claimed about the units is correct.
For dimensional consistency:
B – absolute temperature in either ºR or K
C – absolute temperature in either ºR or K
A – dimensionless
2.3.9
In most cases the argument of a logarithm function should be dimensionless, but in
this case it will not be. Therefore, the numerical values of A, B, and C will depend
upon the units used for temperature and the units used for p*.
2.3.10
The equation is
Δp = 4fρ ⎡⎣(v 2 / 2g) (L/D) ⎤⎦
The units on the right hand side (with f dimensionless) in SI are
ρ
⎞
kg ⎛ m 2
⎜
m ⎟⎟
1
m 3 ⎜ s2
→ 2
⎜ (kg)(m) m ⎟
m
⎟
⎜ s2
⎠
⎝
hence the equation is not dimensionally consistent because Δp has the units of N/m2.
If g is replaced with g, the units would be correct.
2.4.1
Two because any numbers added to the right hand side of the decimal point in 10 are
irrelevant.
2.4.2
The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an
implied decimal point, the answer should be 1287 (no decimal point).
2–9
Solutions Chapter 2
2.4.3
The number of significant figures to the right of the decimal point is 1 (from 210.0m),
hence the sum of 215.110 m should be rounded to 215.1 m.
2.4.4
A calculator gives 569.8269 000, but you should truncate to 4 significant figures, or
569.8 cm2.
2.4.5
Two significant figures (based on 6.3). Use 4.8 × 103.
2.4.6
Step 1: The product 1.3824 is rounded off to 1.4
Step 2: Calculate errors.
For absolute error, the product 1.4 means 1.4 + 0.05
Thus
0.05
× 100% = 3.6% error
1.4
Similarly 3.84 has
0.005
× 100% = 0.13% error
3.84
and 0.36 has
0.005
× 100% = 1.4% error
0.36
Total 2.7% error
2.6.1
a)
4 g mol mg Cl 2 (95.23)g MgCl 2
= 380.9 g
gmol MgCl 2
b)
2 lb mol C3H8 (44.09 )lb C 3H8 454g C3 H8
4
= 4 × 10 g C3 H8
lb mol C3 H8 1 lb C3 H8
c)
(d)
16 g N2
gmol N 2 1 lb mol N 2
= 1.26 × 10−3 lb mol N 2
(28.02)g N2 454 gmol N 2
3 lb C2 H6O 1 lb mol C2 H6O 454 g mol
= 29.56 g mol C2 H6O
(46.07) lb C2 H6O 1 lb mol
2–10
Solutions Chapter 2
2.6.2
2.6.3
(a)
16.1 lb mol HCl 36.5 lb HCl
= 588 lb HCl
1 lb mol HCl
(b)
19.4 lb mol KCl 74.55 lb KCl
= 1466 lb KCl
1 lb mol KCl
(c)
11.9 g mol NaNO3
(d)
164 g mol SiO2 60.1 g SiO2 2.20 ×10-3lb
= 21.7 lb SiO2
1 g mol SiO2
1g
85 g NaNO3 2.20 ×10-3lb
= 2.23 lb NaNO3
1 g mol NaNO3
1g
Basis: 100 g of the compound
comp.
C
O
H
g
42.11
51.46
6.43
MW
12
16
1.008
g mol
3.51
3.22
6.38
13.11
Ratio of Atoms
1.09
1
2
Multiply by 11 to convert the ratios into integers
The formula becomes C12O11H22
Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)
2–11
Solutions Chapter 2
2.6.4
Vitamin A, C20O H30, Mol Wt.: 286
Vitamin C, C6 H8O6, mol. wt: 176
Vitamin A:
a.
Vitamin A =
2.00 g mol 286 g 1 lb
= 1.26 lb
1 g mol 454 g
16 g 1 lb
= 0.0352 lb
454 g
Vitamin C =
2.00 g mol 176 g 1 lb
= 0.775 lb
g mol 454 g
16 g 1 lb
= 0.0352 lb
454 g
b.
Vitamin A =
1.00 lb mol 286 lb 454 g
= 130,000 g
1 lb mol 1 lb
Vitamin C =
1.00 lb mol 176 lb 454 g
= 79,900 g
1 lb mol 1 lb
For both
12 lb 454 g
= 5450 g
1 lb
2.6.5
1 kg 1160 m 3
= 223.1 kg/kg mol
5.2 m 3 kg mol
2.6.6
Mass fraction to mole fraction:
ω1
MW1
x1 =
ω1
(1− ω1 )
+
MW2
MW1
Mole fraction to mass fraction
ω1 =
x1MW1
(x1 )(MW1 ) + (1− x1 )(MW2 )
2–12
Solutions Chapter 2
2.6.7
Basis: 100 kg mol gas
Comp.
CH4
H2
N2
Total
Mol % = mol
30
10
60
100
MW
16
2
28
kg
480
20
1680
2180
21.8 kg
kg mol
Basis: 100 g mol gas
2.6.8
Comp.
CO2
N2
O2
H 2O
mol = %
19.3
72.1
6.5
2.1
100.0
Avg. mol. wt =
MW
44
28
32
18
g
849.2
2018.8
208.0
37.8
3113.8
3113.8
= 31.138
100.0
Basis: 100 lb mol
2.6.9
Comp.
CO2
CO
N2
% = mol
16
10
30
MW
44
28
28
lb
2640
280
840
3760
Avg. MW = 37.6 lb/lb mol
2.7.1
(a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if use SI
units).
(b) A gas requires a convenient basis of 1 or 100 lb moles (if use AE units).
(c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass
is a convenient basis.
(d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas.
(e) Same answer as (e).
2–13
Solutions Chapter 2
2.7.2
Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.
2.7.3
Pick one day as a basis that is equivalent to what is given—two numbers:
(a) 134.2 lb C1
2.8.1
(b) 10.7 × 106 gal water.
Basis: 1000 lb oil
lb oil
lb H2 O
62.4
3
ft
ft 3 = 57.78 lb oil
lb H2 O
ft 3 oil
1.00
ft 3
0.926
1000 lb oil
2.8.2
1 ft 3 oil 7.48 gal
57.78 lb oil 1 ft 3 = 129.5 gal
Basis: 10,010 lb
10,010 lb 1 gal 0.134 ft3
=
3
8.80 lb
gal
152 ft
Basis: 1 g mol each compound
2.8.3
g mol
mw
g
V (cm3)
Pb
1
207.21
207.21
11.33
18.3
Zn
1
65.38
65.38
7.14
9.16
C
1
12.01
12.01
2.26
5.31
V̂ =
2.9.1
ρ (g/cm3)
Component
Na
C1
O
g mol
1
1
3
5
mass (g)
density (g/cm 3 )
mol fraction
0.20
0.20
0.60
2–14
Set 2 is the correct one
MW
23
35.45
16
g
23
35.45
48
106.45
mass fraction
0.22
0.33
0.45
1.00
Solutions Chapter 2
2.9.2
Basis: 1 gal of solution.
Mass of solution:
1.0824 lb soln
lb H O
62.4 3 2
3
ft H 2O
ft H2 O 1 ft 3 1 gal
= 9.03 lb soln .
lb H2 O
7.
481
gal
1.00 3
ft H 2O
Mass fraction KOH =
0.813 lb
9.03 lb
Mass fraction H2O = 1 – 0.09
0.09
0.91
1.00 Total
Basis: 30 lb gas
2.9.3
Comp.
CO2
N2
2.9.4
=
=
lb
20
10
30
a)
1000
b)
No
MW
44
28
lb mol
0.455
0.357
0.812
mol fr.
0.56
0.44
1.00
c)
Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases
the ratio is in moles.
2–15
Solutions Chapter 2
2.9.5
On a paper free basis the total ppm are:
Brand A: 6060 ppm
Brand B: 405 ppm
The respective mass fractions are:
The other entries are similar
Brand A:
Fe
Cu
1310
= 0.216
6060
2000
= 0.330
6060
350
= 0.864
405
50
= 0.123
405
2750
= 0.454
6060
Brand B:
5
= 0.0123
405
Vs =
5 m 30.2 m 200 mm
1m
2 sides
3
= 60.4 m
1000 mm
5 m 27.2 m 200 mm
1m
2 ends
= 54.4 m3
1000 mm
Ends
Ve =
2–16
Pb
Solutions Chapter 2
Floor
Vf =
27.4 m 30.4 m 200 mm
1m
1000 mm
= 166.592 m
Total volume = 281.392 m3
Mass of concrete =
281.392 m3 2080 kg
= 585,295
m3
Volume of displaced water required to float:
586,543.36 kg 1 m 3 H2 O
= 586.543 m 3
1000 kg H2 O
V = LWh → h =
h=
2.9.6
V
LW
586.54 m 3
27.4m 30.4 m
Basis: 190,000 ppm
190,000 g PCB
× 100 = 19%
106
2–17
= 0.703 m
3
Solutions Chapter 2
Basis: 100 g of the sample
2.9.7
The biomass sample is
g = % dry weight of cells
C
O
N
H
P
50.2
20.1
14.0
8.2
3.0
95.5
4.5
100.0
other
Total
10.5 g cells 50.2 g C 1 g mol C
= 0.439 g mol C/g mol ATP
g mol ATP 100 g cells 12 g C
2.9.8
MMM(s) → NN(s) + 3CO2 (g)
a.
b.
2.9.9
2×107 disintegrations 1 min
min
60
1 curie
106µ curie
3×1010 disintegrations/s
1 curie
= 11µ curie
2 ×107 disin tegrations 0.80
= 1.6 ×107 cpm
min
The relation to use is
t1/ 2 = l n(2) /(k)(OH − )
with (OH − ) = 1.5 ×106
k
t1/ 2 (sec onds)
Methanol
Ethanol
MTBE
30.8 ×105
4.6 ×105
7.7 ×105
0.15 ×10−12
1×10−12
0.60 ×10−12
The order is in increasing persistence ethanol, MTBE, and methanol.
2–18
Solutions Chapter 2
2.9.10
Yes.
Bases are first entries.
4800 g mol CC14 103 g mol air kg mol air 154 g CC14 103 mg CC14
=
109 g mol air
kg mol air 22.8 m3 g mol CC14
g CC14
32.4 mg CC14 /m3 which exceeds the NIOSH standards
2.9.11
a.
25,600 ton P 2.6 yr
1
1 gal 2000 lb 454 g 106 µ g
yr
1.2 × 1014 gal 3.785 L 1 ton 1 lb
1 g
= 133.1µ g / L
b.
19,090 lb(P )municipal
= 63. 4%
30,100 lb(P)total
c.
19,090 lb(P )municipal 0.70 lb(P)det. 100 lb(P )tota l
= 44.4%
30,100 lb(P)total 1 lb(P)municipal 30,100 lb(P)tota l
For d. and e. assume that (P)outflow remains unchanged.
d.
P retained/yr = 2,240 + 6,740 + 0.7
× 19,090 – 4,500 = 17,843 ton/yr.
P conc. in ppb =
17,843 ton 2.6 yr
1
2000 lb 1 g / cm 3
109 g
= 92.6 ppb
yr
1.2 × 1014 gal ton 8.345 lb / gal 1billion g
This is greater than 10 ppb. Eutrophication will not be reduced.
e.
P retained/yr = 2,240 + 6,740 + 0.3
P conc. in ppb =
× 19,090 – 4,500 = 10,207
ton
yr
10,207 ton 2.6 yr
2000 lb 1 g / cc
109 g
= 53.2 ppb
1.2 × 1014 gal ton 8.345 lb / gal 1 billion g
yr
This will not help.
2–19
Solutions Chapter 2
Basis: 106 g mol gas
2.9.12
350 g mol H 2S 1 g mol gas 34g H 2S 1 g mol CO2
106g mol gas 1 g mol CO2 1 g mol H 2O 44 g CO2
=
270 g H 2S
270
g H 2S
= 6
6
10 g CO2
10 g total liquid
mass fraction H2S = 2.70 ×10−4
2.9.13
(a)
1 mol O2
104 mol O2
⇒
or 104 ppm
6
100 mol gas
10 mol gas
(b)
Basis: 100 mol gas
Comp.
SO3
SO2
O2
Total
2.9.14
MW
answer
% = mol
55
3
1
59
CaCO3:
Ca
Mg
C
O
mol fr.
0.932
0.051
0.017
1.000
or mol %
93.2
5.1
1.7
100.0
100.06
40.05
24.3
12.01
16.00
100.06 g CaCO3 1 g mol CaCO3 1 g mol Ca
g CaCO3
= 2.50
g mol Ca CO3
1 g mol Ca 40.05 g Ca
g Ca
Similarly
g CaCO3
g CaCO3
= 4.118
g Mg
g Mg
Total alkalinity = 2.50 (56.4) + 4.118 (8.8) = 177
2–20
mg CaCO3
L
Solutions Chapter 2
1 molecule in 1023 or more is not 13-20 ppb
2.9.15
No.
2.9.16
On a mol basis, the carbon dioxide concentration in air is about 350 parts per million
(ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration
of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to
353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration
causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is
less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at
present.
2.10.1
TK = –10 + 273 = 263K
T°F = –10 (1.8) + 32 = 14°F
T°R = 14 + 460 = 474°R
2.10.2
Yes, if the temperature scale is a linear relative one (°C, °F), or a logarithmic scale
(ln 1° is zero). No, if the scale is absolute, but read J. Wisniak, J. Chem. Educ., 77,
518-522 (2000) for a different conclusion.
2.10.3
C p = 8.41 +
2.4346 × 10-5TK

⎛
⎞
J
⎜ 2.4346×10-5
⎟ (T )
⎜⎝
(gmol)(K)2 ⎟⎠ K
Substitute
Cp
=
=
1.8 TK = T°R
8.41 + 2.4346 × 10œ5
J
⎛ T°R ⎞
2 ⎝
(gmol)(K) 1.8 ⎠
8.41 + 1.353 ×10−5 To R
2–21
Solutions Chapter 2
a)
10° C 1.8° F
1.0° C + 32 = 50°F
b)
10°C 1.8°F
1.0°C + 32 + 460°R = 510°R
2.10.4
c)
d)
−25o F – 32o F 1.0oC
+ 273K = 241.3K
1.8
1.8o F
150K 1.8° R
1.0K = 270°R
2.10.5 First multiply the RHS of the equation so that
Btu
(lbmol)(°F)
= 4.182
1054.8J
1 Btu
1 lb mol
454 gmol
1.8°F
1.0°C
1°C
1K
J
(gmol)(°K)
and substitute T°F = 1.8T°C + 32
C p = [8.488 + 0.5757 × 10−2 (1.8T°C + 32) − 0.2159 × 10−5 (1.8T°C + 32)2 +
J
0.3059 × 10−9 (1.8T°C + 32)3 ]4.182
(gmol)(°K)
Simplifying,
Cp = 36.05 + 0.0447T – 0.2874
2.10.6
× 10–4 T2 + 0.7424 × 10–8 T3
The instrument does not contain mercury, but has to contain a fluid that responds
at –76°C and can be calibrated to measure temperature.
2–22
Solutions Chapter 2
Basis: 15cm3 water
2.11.1
1000 kg 10 m 10 m 0.15 m
= 15,000 kg
m3
a.
m = ρV=
b.
F
m g 15,000 kg 9.80 m
1 N 1 Pa
=
=
= 1470 Pa
2
A
A
s
10 m 10 m 1 (kg)(m) N
s2
m2
= 1.47 kPa
1.470 kPa
or
2.11.2
1 atm 14.7 psi
= 0.21 psi
101.3 kPa atm
15 cm H2O
1 in. 1 ft 14.696 psi
= 0.21
2.54 cm 12 in. 33.91 ft H2 O
ρconcrete = 2080 kg/m 3
ρwater = 1000 kg/m 3
Volume of concrete:
Sides
Vs =
5 m 30.2 m 200 mm
1m
2 sides
3
= 60.4 m
1000 mm
5 m 27.2 m 200 mm
1m
2 ends
= 54.4 m3
1000 mm
Ends
Ve =
Floor
Vf =
27.4 m 30.4 m 200 mm
1m
1000 mm
2–23
= 166.592 m
3
Solutions Chapter 2
Total volume = 281.392 m3
Mass of concrete =
281.392 m3 2080 kg
= 585,295
m3
Volume of displaced water required to float:
586,543.36 kg 1 m 3 H2 O
= 586.543 m 3
1000 kg H2 O
V = LWh → h =
h=
2.11.3
V
LW
586.54 m 3
27.4m 30.4 m
= 0.703 m
The pressure is a gauge pressure.
60°F
Basis: 50.0 psig
a.
50.0 psig 33.91 ft H2 O
= 115 ft
14.7 psia
(difference)
b.
No. Insufficient height.
Alternate solutions can be applying ∆p = ρΔhg
2.11.4
lbf in.2
= lbf → lbm in the AE system
in.2
so the procedure is ok, although the unit conversion is ignored.
2–24
148
ft
Solutions Chapter 2
2.11.5
Basis: Dept = 1000 m
p = p o + ρgh
p = 1 atm +
1000 m 1.024 g 9.8 m 1 kg ⎛ 100 cm ⎞
cm 3
s2 103 g ⎝ 1 m ⎠
=
3
1 N
1 kPa
1 atm
(kg)(m ) 1N 1.013 × 10 5 N / m 2
(kg)(m )
s2
2
s
100.1 atm 101.3 kPa
4
= 1.014 × 10 kPa
1 atm
Alternative solution:
101.3 kPa +
1000 m sea H 2 O 1.024 g H 2 O 3.28 ft
1.00 g sea H 2 O 1 m
101.3 kPa
33.91 ft H 2 O
= 1.003 × 104
+ 0.01× 104
= 1.013 × 104
2.11.6
(a) p ABS = pGauge + patm
⎛ 22.4 lbf 144 in.2 ⎞ ⎛ 28.6 in. Hg 14.7 psia 144 in.2 ⎞
2
⎟ = 5250 lbf /ft
⎜
⎟ + ⎜
2
2
2
1 ft ⎠ ⎝
1
29.92 in. Hg 1 ft ⎠
⎝ 1 in.
⎛ 22.4 psig 29.92 in. Hg ⎞
+ 28.6 in. Hg = 74.2 in. Hg
14.7 psia ⎟⎠
(b) ⎜
⎝
⎛ 22.4 psig 1.013 × 105 N/m 2 ⎞ ⎛ 28.6 in. Hg 1.013 × 105 N/m 2 ⎞
+⎜
= 2.51× 105 N/m 2
⎟
⎟
14.7
psia
27.92
in.
Hg
1
1
⎝
⎠ ⎝
⎠
(c) ⎜
⎛ 22.4 psig 33.91 ft H2 O ⎞ ⎛ 28.6 in. Hg 33.91 ft H 2O ⎞
⎟ + ⎜
⎟ = 84.1 ft H 2 O
1
14.7 psia ⎠ ⎝
1
29.92 in. Hg ⎠
⎝
(d) ⎜
2.11.7
Neither John is necessarily right. The pressure at the top of Pikes Peak is
continually changing.
2–25
Solutions Chapter 2
2.11.8
Δp = ρgh
a.
=
1 g 9.8 m 13.1 m 1 kg (100 cm )3 Pa
1 kPa
1 kg 1000 Pa
cm 3 s2
1000 g
m3
m s2
= 128. 4 kPa
b.
mg
ρVg
ρ(S.A.)(t )g
F
=
=
=
A A Bottom A Bottom
ABottom
A Bottom
2
π 2 π (30.5 m )
2
= D =
= 730.62 m
4
4
2
ATop = 730.62 m
Aside = πDh = π(30.5 m)(13.1m) = 1255.2 m
2
S.A.= ATop + A Bottom + ASide
2
= (730.62 + 730.62 +1255.22 ) m = 2716.46 m
2
ρ(S.A.)(t )g
=
A
(
)
3
7.86 g 100 cm 2,716.46 m 2 9.35 × 10 œ3 m
9.8 m
3
3
2
cm
s2
m
730.6 m
1 kg
1 Pa
1 kPa
=
2
1000 g 1 kg / ms 1000 Pa
2–26
2.68 kPa
Solutions Chapter 2
2.11.9
p A + ρ1gh1 + ρ 2gΔh = PD
p B + ρ1gh1 + ρ 2gΔh = PD
pA -pB = ρ1gh1 +ρ2gΔh −
ρ 1gh1ρ− 1gΔ h
= ρ2gΔh1 − ρ1gΔh = gΔh(ρ2 − ρ1 )
=
9.8 m 0.78 in 1 m 13.546 œ0.91g 1 kg
s2
3.28 ft
cm 2
1000 g
1 N
1 ft ⎛ 100 cm ⎞ 3 m 2 1 Pa 760 mm H2
12 in ⎝ 1 m ⎠ 1 kg 1 N 101.3 × 10 3 Pa
(m )(s)2 m 2
=18.4 mm Hg
The pressure at A is higher than the pressure at B.
Alternate solution:
Δp =
(0.78 in.) (13.546 − 0.91) 760 mm Hg
= 18.5 mm Hg
13.546
29.92 in. Hg
2–27
Solutions Chapter 3
3.1.1
Boundary:
Around both pumps and include the soil at the end of the pipes.
It’s an open system.
It’s at steady-state except at startup (note system boundary limits fluid), or
if fluid enters, unsteady state.
3.1.2
Either open (flow) or closed (batch) is acceptable if explanation is given
(1)
flow – material comes in and out continuously over a suitably long
period of time
(2)
batch – material is injected into the system, and then in a short period
of time, a reaction occurs with the system valves closed.
(a)
open if you have to replace water, and water evaporates; otherwise closed
(b)
open
3.1.3
3.1.4
If the overall flows in and out over a time period for several batches are
considered, and the local batches ignore, the process can be treated as continuous.
3.1.5
a) closed
b)open
c)open
d)closed
3–1
Solutions Chapter 3
3.1.6
The system is the radiator.
Open System
Closed System
Steady-State
Unsteady State
a)
X
X
b)
X
X
c)
X (Before filling)
d)
X (After filling)
X (After filling)
X
X
X (Before Filling)
3.1.7
system
contains
ice + H2O
Ice
boundary
boundary
(c)
Unsteady state for any assumptions
b ⎫
⎧open
⎬ assume water (melted ice) leaves the system ⎨
a ⎭
⎩flow
b ⎫
⎧closed
⎬ assume water stays in sytem because the ice and water ⎨
a ⎭
⎩batch
remain on melting
3.1.8
P
F1
F
2
F
unsteady state
open
P
unsteady state
open
steady state
open
(a)
(b)
(c)
3–2
Solutions Chapter 3
3.1.9
(a)
(b)
(c)
(d)
(e)
(f)
(g)
1.
x
x
2.
x
x
____________________________________________________
3.
x
x
4.
Depends on the time period considered
____________________________________________________
5.
x
x
6.
x
x
x
____________________________________________________
7.
x
x
3.1.10
If the overall flows in and out over a time period for several batches are
considered, and the local batches ignore, the process can be treated as continuous.
3.1.11
Basis: 1 minute
Accumulation (kg)
mfinal − 0
In (kg)
= 300 + 100
Out (kg)
−
380
n final = 20 kg
20 kg 60 min
= 1200 kg
min 1 hr
3.1.12
Basis: 600 kg solution
Accumulation (kg)
n final − 0
In (kg)
= 100 + 500
Out (kg)
−
300
n final = 300 kg if no water evaporates in which case less than 300 kg
would remain
3–3
Solutions Chapter 3
3.1.13
(c)
27.0 g
3.1.14
Based on the process measurements, there are 5,000 lb/h more flow for the
process leaving the heat exchanger than the feed rate to the heat exchanger; therefore, the
material balance for the process fluid does not close. The reason for this discrepancy
could be faulty flow sensor readings or possibly a leak of the condensate or steam into the
process stream.
3.1.15
Basis: Data shown on flowsheet units are MTA
In
2.5 × 106
Out
404 ×103
228 ×103
152 ×103
101 ×103
67 ×103
36 ×103
100 ×103
1190 ×103
2,278 × 103
mass in does not equal to mass out
Reasons:
(1) Some of the material was burned as fuel
(2) Some of the material formed gases that were exhausted to atmosphere
(such as H2O, CO2).
(3) errors in measurement.
(4) Some process streams are not shown.
3–4
Solutions Chapter 3
3.1.16
In = 250,000 ton/yr.
In (ton/yr)
250,000
Out = 244,500 ton/yr.
Out (ton/yr)
Combustibles
Combustibles
Polyethylene
Polystyrene
PVC
Acrylonitorile
DB
Phenol
Acetone
Rubber
LPG
Aromatics
Total
3,800
39,500
30,000
5,000
40,000
20,000
8,000
10,000
5,750
10,000
24,000
48,000
244,050
Balance is not exact but very good for an operating plant.
3.1.17
Basis: 1 week
in (tons) = 920 + 0.6 = 920.6
out (tons) = 3.8 + 620 + 0.01 + 0.01 + 0.08 + 1.1 + 275 + 20 = 920
Yes
3–5
Solutions Chapter 3
3.1.18
The density of the crystalline silicon in the cylinder is 2.4 g/cm3.
Basis: 62 kg silicon
The system is the melt, and there is no generation or consumption. Let Δmt be the
accumulation.
Accumulation
Δmt
=
Input
0
-
Output
0.5(62 kg)
Δm t = − 31 kg
Let t be the time in minutes to remove one-half of the silicon
2.4 g π(17.5cm)2 0.3 cm t min 1
= (62,000 g)
cm3
4
min
2
t = 179min
3.1.19
Basis: 1 hr
Overall material balance (kg): 13,500 + 26,300 ? 39,800
Yes. The balance is satisfied
NaC1 balance: 0.25 (13,500) + 0.05 (26,300) ? 0.118 (39,800)
4690 ≅ 4696
The balance is closely satisfied but not exactly.
The closure is good for industrial practice.
3–6
Solutions Chapter 3
3.1.20
Basis: 1 hour
The overall material balance is
In (lb)
106,000
?
Out (lb)
74,000 + 34,000 = 108,000
The error in the overall material balance is 2000 lb/h or 1.9%; therefore, the overall
material balance is within the expected error for industrial flow sensors.
Propylene balance:
0.7 × 106,000 ? (0.997) (74,000) + (0.1) (34,000)
74,200 ? 73,778 + 3,400 = 77,178 (3.8% error)
Propane balance:
0.4 × 106,000 ? (0.003) (74,000) + (0.9) (34,000)
31,800 ? 222 + 30,600 = 30,622 (3.8% error)
Note that the relative error for the component balances are twice as large as the relative
error for the overall material balance, indicating that there is additional error in the
composition measurements used for the component balances.
3.1.21
Basis: 100 kg wet sludge
The system is the thickener (an open system). No accumulation, generation, or
consumption occur. The total mass balance is
In
Out
=
100 kg
70 kg + kg of water
Consequently, the water amounts to 30 kg.
3.2.1
(a)
(b)
(c)
(d)
water and air.
Insulation, air and what is in the atmosphere.
Yes (cold water in, hot water out)
Yes
3–7
Solutions Chapter 3
3.2.2
Four balances are possible, 3 components plus 1 total.
0.10F1 + 0.50F 2 + 0.20F 3 = 0.35P
0.20F1 + 0F2
+ 0.30F 3 = 0.10P
0.70F1 + 0.50F 2 + 0.50F 3 = 0.55P
Total balance F1 + F2 +F3 = P
Only 3 of the equations are independent.
3.2.3
If you specify F, P, W, you can calculate all of the stream variables.
a)
Unknown: three stream values F, P, W (plus two compositions if you take into
account all of the variables).
b)
The two known compositions are not given but may be calculated from Σxi = 1,
c)
Two components exist; hence two independent material balances can be written.
The problem cannot be solved unless one stream value is specified.
3.2.4
(a)
No. The equations have no solution – they are parallel lines.
The rank of the coefficient matrix is only 1 because the
⎡1 2⎤
det ⎢
=0
⎣1 2⎥⎦
The rank of the augmented matrix is 2
⎡1 2 1⎤
Largest non zero det. of ⎢
is of order 2
⎣1 2 3⎥⎦
Thus, although the 2 equations are independent and the number of variables is 2
(the necessary conditions), the sufficient conditions are not met.
3–8
Solutions Chapter 3
x2
x +2x =1
1
2
x + 2x = 3
1
2
x1
(b)
x2
•
(x1 − 1)2 + (x2 − 1)2 = 0
No; 2 solutions
•
The equations are independent
x1
x1 + x 2 = 1
Two solutions exist as can be seen from the plot hence no unique solution exists.
3.2.5
The number of independent equations is just 3. The number of unknown
quantities is 3, hence a unique solution is possible.
3.2.6
a)
No for both
⎡1 1 1 ⎤
⎢1 2 3 ⎥ rank of coefficient matrix is2.
⎢⎣3 5 7 ⎥⎦ rank of augmented matrix is2.
r = 2, n = 3
b)
multiple solutions exist
⎡1 0 1⎤
⎢5 4 9⎥ 3d column issum of1st two columns,so rank is2.
⎢⎣2 4 6⎥⎦ rank of augmented matrix is2, also.
r = 2, n = 3
multiple solutions exist
3–9
Solutions Chapter 3
3.2.7
(a) F; (b) F; (c) F (the maximum can be more than the number of independent
equations).
3.2.8
P = 16 kg
F = 10 kg
ωF1= 0.10
ωF2= ?
ωF3= 0 (assumed)
ωP1= 0.175
ωP2= ?
ωP3= ?
A = 6 kg
ωA1= 0.30
ωA2= ?
ωA3= 0.20
Unknowns (4): ω2F ,ω2A ,ω2P ,ω3P
Equations:
Mass balances:
(1):
F(0.10) + A(0.30) = P(0.175)
or
10(0.10) + 6(0.30) = 16(0.175)
(2):
F (ω 2 F ) + A (ω 2 A ) = P (ω 2 P )
or
10 (ω 2 F ) +6 (ω 2 A ) = 16 (ω 2 P )
(3):
F(0)
+A(0.20) = P (ω 2 P )
or
10(0)
+ 6(0.20) = 16 (ω 2 P )
redundant
ω 2 F ,ω 2 A ,ω 2 P ,ω3P
10
6 -16 0
0
0 -0 16
coeff. matrix
Two independent material balances exist (the rank of the coefficient matrix is 2).
In addition three sum of mole fraction equations exist. The total number of equations is
5, hence the degrees of freedom = -1. The problem is overspecified, and will require at
least squares solution.
3–10
Solutions Chapter 3
3.2.9
Examine the row of C3H8. None of the concentrations are greater than the desired
50% so 50% is not achievable by any combination of A, B, or C. Or look at the CH4 row.
3.2.10
A
x1
Basis: D =100 lb
B
x2
C
x3
D
0 M 1.4 ⎤
⎡ 5.0 0
⎢90.0 10.0 0 M 31.2 ⎥
⎢
⎥
Coefficient matrix is ⎢ 5.0 85.0 8.0 M 53.4 ⎥
⎢
⎥
⎢ 0 5.0 80.0 M 12.6 ⎥
⎢⎣ 0
0 12.0 M 14 ⎥⎦
No unique solution exists with 5 equations and 4 variables. A least squares solution
could be determined, or the 4 equations with the most accurate data solved.
The rank of the coefficient matrix is 3
The rank of the augmented matrix is 4
Hence no unique solution exists.
3.2.11
You can see by inspection that no combination of tanks 1,2 and 3 will give a mole
fraction of 0.52 for the mixture.
Basis: 2.50 mol of tank 4
Let xi = be the total moles of tank i
The balances are
0.23x1 + 0.20x2 + 0.54x3 = 0.25(2.5) = 0.625
0.36x1 + 0.33x2 + 0.27x3 = 0.23(2.5) = 0.575
0.41x1 + 0.47x2 + 0.19x3 = 0.52(2.5) = 1.300
The coefficient matrix has a rank of 3 as does the augmented matrix so the set of
equations has a solution. However, the solution is
x1 = −4.00
x 2 = 6.36
x 3 = − 0.327
The values of x1and x3 are not physically realizable!
3–11
Solutions Chapter 3
3.2.12
Number of unknowns:
F, W, P and 9 compositions:
Equations
W
Specifications: x CF
=0
Sum of mole fractions = 1
Material balances (3 species)
12
1
3
3
Degrees of freedom =
7
5
You can make any set of measurements that results in independent equations (assuming
equal accuracy). For example, you cannot use all the flows, or all the compositions in
one stream, as the resulting set of material balances will not consist of 5 independent
balances, but a lesser number.
3.2.13
The inerts are treated as a compound.
Unknowns: 5 compounds × 5 streams plus 5 streams =
30
Equations: Specifications:
Concentrations not shown in diagram assumed
to be zero
8
Concentrations with % given
7
E = 11 kg
1
Material balances: 5
5
Implicit equations (∑ ωi = 1)
5
26
a.
Degrees of freedom (additional specifications) =
4
You must make 4 measurements that result in independent equations
b.
No.
3–12
Solutions Chapter 3
3.2.14a
a.
b.
The remaining gas is 100% minus the N2, and was put on the figure as R.
c.
Basis: 100 mol A
d.
Unknowns: A, B, C
Equations: N2 and R material balances
Basis = A = 100 mol
Note: You could treat the values of R as unknowns in each stream,
and then there would be 3 more unknowns and 3 more independent
equations ( ∑ x i = 1 or ∑ mi = total mass flow i )
100(0.90) + B(0.30) = C(0.65)
e. & f. N2:
R:
100(0.10) + B(0.70) = C(0.35)
Total:
100 + B
=C
Two of the above equations are independent
g.
Solution:
B = 71.4 mol
C = 171.4 mol
A
100
=
= 1.40
B
71.4
3–13
Solutions Chapter 3
3.2.14b
a.
DT = dry timber
b.
The DT is the balance of each stream.
c.
Basis: F = 100 kg
d.
Unknowns: F, W, P
OR
Equations:
Basis: F = 100 kg
Material Balances: Water, DT
(total); 2 independent
Degrees of freedom = 0
F, W, P, DTF, DTW, DTP, H2OF,
H2 OP , H2 OW
Equations:
F = 100 kg
Material Balances: Water, DT,
(total); 2 independent
Specifications: 3 for water
Implicit equations: 3 of ∑ ω i = 1
Degrees of freedom = 0
e.&f. Introducing the specifications and basis into the material balances:
Water: (0.201)100 = (1)W + (0.086)P
DT:
(0.80) 100 = 0
+ (0.914)P a tie element
Total:
100 = W + P
g.
W = 12.5 kg
P = 87.5 kg
W 12.5 kg
=
= 0.125kg/kg
F
100 kg
3–14
Solutions Chapter 3
3.2.14c
a. & b.
A N2 1.00
mol. fr.
N2 0.70
CH4 0.30
1.00
P
F
d.
CH4 n CH 4
C 2H 6 n C 2 H 6
N2
n
N2
B
c.
or
N2
C 2H 6
0.90
0.10
1.00
∑=P
∑
x CH 4
x C2 H 6
x N2
=1.00
Basis: B = 100 mol
Unknowns:
Assume all of the variables except those that are specified as zero are
included in the analysis.
F
A
P
B
stream variable
1
1
1
1
4
+
component variables
2
1
3
2
8
=
12
Equations:
Basis:
Material balances:
N2, CH4, C2H6
Specified ratio: n CH4 /n C2H6 = 1.3
Specified values of n: 2 + 1 + 0 + 2 =
Implicit equations (∑ xi = 1 in P)
(Note: the implicit equations of streams
F, A, and B are redundant with the
specifications)
Total
Degrees of freedom =
1
3
1
5
1
11
1
More information is needed to solve the problem uniquely.
3–15
Solutions Chapter 3
3.2.15
Here are some possibilities. Consult the tables at the end of the Chapter for more
suggestions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Rephrase the problem to make sure you understand it?
Draw a simple diagram of what was happening?
Think about what was going into the tank and what was coming out?
Imagine yourself inside the tank, and ask what was going on around you?
Ask whether there were any physical laws to consider (such as conservation
of matter or energy)?
Try to imagine the answer as a number, graph, table, or whatever?
Try to identify essential variables?
Choose a notation?
Look for a ready-made formula for the answer?
Look for simplifying assumptions?
Try to find an easier version of the problem?
Look for bounds (simple models that would definitely underestimate
or overestimate the answer)?
3–16
Solutions Chapter 4
4.1.1
Basis: 100 kg cucumbers
P − 100 = evaporated
Cucumber
Initial
1 kg
Final (P)
2%
Water
99 kg
98%
100%
100 kg
100%
100%
Cucumber balance:
Evaporated
−
0.02P − 0.01(100) = 0
P = 50 kg
4.1.2
Answer is: Yes
Initial
Final
plant 5 ppm
soil 6 ppm
Plant 755 ppm
Soil 5 ppm
Final − Initial
Final − Initial
Arsenic balance: S (5) − S(6) =
750P = S or
P(755) − P(5)
S
= 750
P
4.1.3
Step 5: Basis is 1 ton (2000 lb) sludge
Steps 1, 2, 3, 4:
Sludge
0.70 H 2 O
0.30 Solids
1.00
F
Drier
W
H2 O
1.00
4–1
P
Dried Sludge
0.25 H 2O
0.75 Solids
1.00
Solutions Chapter 4
Steps 6 and 7:
Basis: 2000 lb F
Balances: H2O
Solids
Unknowns: P, W
Steps 8 and 9:
Total: 2000 = P + W
Solids: 2000 (0.30) = P (0.75)
P = 800 lb
W = 2000 − 800 = 1200 lb
4.1.4
Step 5:
Basis: 1 min
Steps 2, 3, 4:
D
220 mL
V 215 mL
B
Urea 2.30 mg/mL
H2O 220 mL
P
1.70 mg/mL
H2O 215 mL
Steps 6 and 7: Unknowns: mP urea and mP H2O with two equations: urea and H2O. Degrees
of freedom = 0.
Steps 8 and 9:
(a)
H2O balance: 220 mL − 215 mL = 5 mL
Urea balance:
(b)
2.30 mg 220 mL
1.70 mg 215 mL
−
= 141 mg
mL
mL
Dialisate: 1500 + 5 = 1505 mL
Urea: 141 mg
Concentration: 141/1505 = 0.0934 mg/mL
4–2
Solutions Chapter 4
4.1.5
Step 5:
Basis: 1 day
Steps 2, 3, 4:
W ton
1.00 H2O
F=2 ton
P (ton) NaOH
mass fr
NaOH 0.03
H2O 0.97
1.00
H2O
Step 6: Unknowns, 2: P, W
Step 7: Balances 2: NaOH, H2O, total (2 of the 3)
Steps 8 & 9:
Total: 2 = W + P
NaOH: 2 (.03) = P(.18)
P = 1/3 ton
(666 lb)
4.1.6
W = 1 2/3 ton
(3334 lb)
Steps 2, 3, 4:
500 lb 10% solution
A
20% solution
3000 lb
13% polymer
pure solvent
S
Step 5:
Basis: 3000 lb of final product
Steps 6 and 7: Unknowns: A and S; balances: total and polymer
4–3
Mass fr.
0.18
0.82
1.00
Solutions Chapter 4
Steps 8 and 9:
Total wt
A + S + 500 = 3000
A + S = 2500
Polymer
0.2A + 50 = 390
A = 5 (340) =
1700
S = 2500 – 1700 = 800
4.1.7
Steps 1, 2, 3 and 4:
N = nitrocellulose
S – solvent
Treat the problem as a steady state flow
problem, or as an unsteady state batch
system. As a flow system:
M (lb) N 100%
F (lb)
N
S
0.055
0.945
1.000
Step 5:
Plant
P 1000 lb
Mass fr.
N 0.08
S 0.92
1.00
Basis: 1000 lb P
Steps 6 and 7: Two unknowns, F and M. Two balances can be made, N and S.
Steps 8 and 9:
Solve to get
N:
F ( 0. 055) + M (1. 00) = 1000 (0. 008)
S:
F (9. 045) + M ( 0) = 1000 ( 0. 92)
4–4
⎫ F = 973. 5 lb
⎪
⎬
M = 26. 5 lb
⎪
⎭
P = 1000.0 lb
Solutions Chapter 4
4.1.8
Steps 1, 2, 3 and 4:
F = 100 kg mol/min
CH4
He
Mol fr.
0.80
0.20
1.00
W (kg mol) CH
4
He
System
P (kg mol)
CH4
He
Mol.
n CH
4
n He
Mol fr.
0.50
0.50
1.00
Step 5:
Basis
1 min → 100 kg mol F
Step 4:
To get P, calculate first the average mol. wt. of F and P, or transform the
mole fractions to mass fractions.
Step 5:
Basis: 1.00 mol F
Mol
CH4
He
0.80
0.20
1.00
Basis : 1.00 mol P
Mol
CH4
He
0.50
0.50
1.00
100 kg mol F
MW
kg
16.03
4
12.8
0.8
13.6
MW
kg
16.03
4
8.01
2.0
10.01
13.6 kgF 0.20 kgP 1kgmol P
1.00kg mol F 1kgF 10.01kgP
P = 27.2 kg mol
Steps 6, 7, 8 and 9:
100(0.80) = 27.2(0.50) + n CH 4 ⎫
⎬
100(0.20) = 27.2(0.50) + nHe ⎭
n CH 4
nHe
Mol
66.4
6.4
72.8
4–5
In W
Mol.fr.
0.912
0.088
1.00
Solutions Chapter 4
4.1.9
Fermenter
40% void
60% cell
volume
F1
Separator
discharge
W1
F2
wet cell
75% H2 O
25 % dry cell
100% H2 O
Basis: 1000 cm3 F1
Drycells =
1000cm 3 in fermenter
3
60cm cell
1.1gcell 25gsolids
3
100 cm in fermenter 1cm 3 cell 100g cell
= 16.5gDrycell / 1000cm 3 in fermenter
4.1.10
a. Basis: 1 kg mole of mixture
Three components exist: (CH4)x, (C2H6)x, (C3H8)x
Let A, B, C respectively represent kg mol of each mixture; these are the unknowns.
Equations:
0.25A + 0.35B + 0.55C = 0.30
(CH4)x balance
0.35A + 0.20B + 0.40C = 0.30
(C2H6)x balance
0.40A + 0.45B + 0.05C = 0.40
(C3H8)x balance
There is a unique solution to the set of equations (in kg mol)
The solution is
A = 0.600
B = 0.350
C = 0.05
b. It is proposed to prepare the final mixture by blending four different compounds (A,
B, C, D); there will still be three equations, but now there will be four unknowns. Since
the rank is now less than n, there will be an infinite number of possible blends of the four
mixtures. Not required: (An optimization of a revenue function subject to the equations
is needed.)
4–6
Solutions Chapter 4
4.1.11
a.
A
100% CO2 100 lb = 2.27 lb mol
CH4 100%
(MW = 16)
P (lb) = ?
mol fr.
CH4 0.9796
CO2 0.0204 (average)
1.000
F=?
(lb)
100 lb 1 lb mol
= 2.27 lb mol
44 lb
Steps 2, 3, 4:
Step 5: Basis 1 min
Step 6: Unknowns: F, P
Step 7: Balances: CH4, CO2
Steps 8 and 9:
Balances in moles
CH4: F (1.00) + A(0) = P (0.9796)
CO2: F (0) + 2.27 = P (0.0204)
F=
b.
P = 111.41 lb mol
111.27 lb mol 0.9796 lb mol CH4 16 lb CH4
= 1746 lb/min
1 lb mol P
1 lb mol CH 4
(a)
Redo the problem with a new composition for F:
CH4
CO2
0.99
0.01
1.00
CH 4 : F (0.99) + A(0) = P (0.9796) ⎫
⎬ F = 3520 lb/min
CO2 : F (0.01) + 2.72 = P (0.0204) ⎭
error
4–7
3520-1744
100 = 50% (b)
3520
Solutions Chapter 4
4.1.12
Steps 1, 2, 3, and 4:
NH 3
F(kg)
Pipeline
72.3 kg/min for 12 min
P (kg)
Mass fr.
NH 3
0
gas
1.00
1.00
NH 3
gas
Mass fr.
0.00382
0.99618
1.00
Step 5: Basis: 1 min
Steps 6 and 7:
Two unknown, F and P. Two balances can be made, NH3 and gas.
You can use the total balance as a substitute.
Steps 8 and 9:
Total
F + 72.3 = P
NH3
F (0) + 72.3 = P (0.00382)
P =18,900 kg / min
Step 10:
F = 18,900kg/min or 1.14 × 106 kg/hr
Check using the gas balance
4.1.13
Steps 1, 2, 3, and 4:
10 lb/hr
Mass fr.
H2 O 1- 0.00012
0.00012
P Br
1.00
F
Mass fr.
H 2 O 1.00
Br
0
1.00
Step 5: Basis: 1 hr
Step 6: Unknowns:
F, P
4–8
Solutions Chapter 4
Step 7: Balances:
Steps 8 and 9:
Total:
Br:
H2O:
H2O, Br
F + 10 =
F (0) + 10
F (1.00) + 0
=
=
P
P (0.00012) ; P = 8.33 × 104 lb/hr
P (1-.00012)
4
4
F = 8.33 × 10 − 10 = 8.33 ×10 lb / hr
4.1.14
Step 5: Basis: 1 year
Steps 2, 3, 4:
P = 15 x106 lb
F = ? lb
PET
PVC
mass fr.
0.98
0.02
1.00
W
PVC 100%
Steps 6 and 7:
Two unknowns: F, W
Two balances: PET, PVC
Steps 8 and 9:
PET balance:
PVC balance:
F (.98) = P (.999990) ≅ 15 × 106
F (.02) = W (1.00) + P (0.000010)
= W + (15 × 106 (10-5)
Total balance could be used in lieu of one of the above
F = 15 × 106 + W
From PET: F ≅ 15.3 × 106 lb
W = (15.3 × 106) (0.02) – 15 × 106 (10-5) = 0.31 × 106 lb
4–9
mass fr.
PET 0.999990
PVC 0.000010
(10 ppm)
Solutions Chapter 4
4.1.15
Basis. 300 g initial solution
(a)
In
Na2SO4
H 2O
g
100
200
MW Na 2 SO4 ⋅10 H2O is 322.2
MW
142.05
18.016
g mol
0.704
11.10
Out
Na 2SO4 ⋅10H2O
Material balances in g mol (in = out):
g mol
Na2SO4: 0.704 = 0.310 + n Na 2SO4
n Na 2SO4 = 0.394
H2O:
n H2O
11.10 = 0.310 (10) + n H 2O
= 8.0
Total
Mother liquor:
(b)
g
100
8.394
g mol
0.310
g
%
55.97
28
144.1
72
200.07
100
Na 2SO4 is 28% and H2 O is 72%
100 g crystals 100 g solution
= 33.3 g crystals/100 g initial soln.
300 g solution
4.1.16
Assume steady state flow problem (alternate is unsteady state batch problem).
Steps 1-4:
200 g H2O
1.00
100 g
Na2B4O7
1.00
No rxn.
No accum.
P
(g)
Na2B4O7 10H2O
F (g)
Final solution
mol fr.
Na2B4O7 0.124
H2O
0.876
1.000
Calculate composition of P: Basis: 100 mol Na2B4O7.10 H2O
4–10
Solutions Chapter 4
Na2B407
H2 O
mol
1
10
Step 5:
Basis: 200 g H2O + 100 g Na2 B4O7
Step 6:
Unknowns: F, B
Step 7:
balances: Na2B4O7, H2O
Step 8:
Na 2 B4O7 :
Step 9:
MW
201.27
18
g
201.27
180.0
381.27
mol fr.
0.528
0.472
1.00
H 2O:
100 = P (0.528) + F (0.124) ⎫
⎬ 2 indept.
200 = P (0.472) + F (0.876) ⎭
Total:
100 + 200 = P + F
Use H2O and Total to get P = 155.5g and F = 144.5 g
ratio:
155.5
(100) = 51.8 g Na 2 B4O7 /100 g H 2O
300
Check use Na2B4O7
100.0 = 100.0 ok
Step 10:
4.1.17
Assume the process is a steady state one without reactions.
Steps 1, 2, 3, 4:
F1
.
F2
.
FeC13 H2O
MW
Fe C13
Fe C13.6H2O
Fe C13.H2O
Fe C13.2.5 H2O
H 2O
.
FeC13 6H2O
1000 kg
P
162.22
270.32
180.24
207.26
18.02
4–11
FeC13 2.5 H2O
Solutions Chapter 4
Calculate the compositions for just one iron compound. FeC13 is the simplest to use.
For F1:
Basis: 1 kg mol
For F2:
FeC13 ⋅ 6H 2 O
FeC13 ⋅1H 2 O
kg mol
1
6
FeC13
H 2O
Basis: 1 kg mol
MW
162.22
18.03
kg
162.22
108.18
270.40
FeC13
H 2O
kg mol
1
1
MW
162.22
18.03
For P Basis: 1 kg mol
kg mol
1
2.5
FeC13
H 2O
MW
162.22
18.03
Step 5: Basis: 1000 kg FeC13.6H2O
kg
162.22
45.08
207.30
Step 6: Unknowns P, F
Step 7, 8, 9: Balances (kg)
IN
OUT
⎛ 162.22 kg FeC13 ⎞
⎛ 162.22 ⎞
⎛ 162.22 ⎞
⎟ + F2 ⎜
⎟ = P ⎜
⎟
kg tot ⎠
⎝ 270.40 ⎠
⎝ 180.25 ⎠
⎝ 207.30
FeC13: 1000 ⎜
Total: 1000 + F2 = P
soln
Step 10:
F2 = 1555.7 kg
P = 2555.7 kg
Check: F2 + F1 = 1554.5 + 1000 = 2554.5
4–12
ok
kg
162.22
18.03
180.25
Solutions Chapter 4
4.1.18
Steps 2, 3, 4:
The process can be viewed as an unsteady process without reaction, or as a flow process.
Step 5:
Take as a basis 100 g of Ba(NO3)2.
Step 4:
The maximum solubility of Ba(NO3)2 in H2O at 100°C is a saturated solution, 34 g/100 g of
H2O. Thus the amount of water required at 100°C is
100 g H2 O
100 g Ba(NO3 )2
34 g Ba(NO3 )2
= 294.1 g H2 O
If the 100°C solution is cooled to 0°C, the Ba(NO3)2 solution will still be saturated so that
the composition of the final solution is
mass fr.
Ba(NO3)2:
5
= 0.0476
100 + 5
H2O:
100
= 0.9524
100 + 5
The composition of the crystals is
Ba(NO3)2:
100
= 0.9615
100 + 4
4–13
Solutions Chapter 4
4
= 0.0385
100 + 4
H2O:
The composition of the original solution is
BaNO3
H2 O
mass fr.
0.254
0.746
100g
294.1g
Steps 6 and 7:
We have two unknowns, F and C, and can make two independent mass balances so that the
problem has a unique solution.
Steps 8 and 9:
Balance
Final solution
Ba(NO3)2:
H2O:
Total:
F(0.0476)
F(0.9524)
F
Transport through
boundary (out)
Initial solution
100
294.1
- (100 + 294.1)
=
=
=
-C(0.9615)
-C(0.0385)
-C
Solve the Ba(NO3)2 and total balance to get
F = 305.2 g
Step 10:
C = 88.89 g
Check using the water balance
305.2(0.9524) – 294.1 ? -88.89(0.0385)
- 3.42 = -3.42
The Ba(NO3)2 that precipitates out on a dry basis is
88.89 g C 0.9615 g Ba(NO3 ) 2
= 85.5 g Ba(NO3 ) 2
1gC
4–14
Solutions Chapter 4
4.1.19
M.W. Na2 S2 O2 = 142
M.W. Na 2 S2 O2⋅ 5H2 0 = 232
feed = F
60% Na2S2 O2
1% impurity
39% H 2O
H2 O
W
0
H2 O
Wi
S
I
10°C
C
Na 2 S2 O2⋅ 5H2 O
crystals
0.06 lb satd. soln/lb crystals
II
Na 2 S2 O2⋅ 5H2 O
0.1 % impurity
0% H 2O as H 2 O
D
Process II Compositions:
a)
satd soln
1.4 lb Na 2 S2 O2⋅ 5H2 O
1 lb H 2O
? lb impurity
dried cyrstals
Basis: 1 lb mol Na2S2O2 ⋅ 5 H2O, impurity free
Stream D
Comp. lb mol
Na2S2O2
H2O
Total
1
5
6
mol wt
lb
142
18
wt fr
142
90
232
0.612
0.388
1.000
Basis: 100 lb D (Na2S2O2⋅ 5 H2O = 99.9 lb)
Comp.
Na2S2O2
H2 O
impurity
Total
Stream C
99(0.612)
99(0.388)
lb
=
=
61.1
38.8
0.1
100.0
Let y = lb impurity/100 lb free water in saturated solution
Basis: 100 lb dry crystals, impurity free
lb from dry
Comp.
crystals
+
lb from adhering soln
from calculation below
4–15
= Total
Solutions Chapter 4
Na2S2O2
61.2
⎛ 85.7 ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
⎛ 85.7 ⎞
⎟
61.2 + 6 ⎜
⎝ 240 + y ⎠
H2O
38.8
⎛ 154.3 ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
⎛ 154.3 ⎞
⎟
38.8 + ⎜
⎝ 240 + y ⎠
impurity
-----
⎛ y ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
Totals
100.0
Stream S
Comp.
⎛⎜ y ⎞
⎟
⎝ 240 + y ⎠
6
106
Basis: 100 lb free water in satd. soln., impurity free
lb from salt lb from free water
Na2S2O2
1.4(61.2) = 85.68
H2O
1.4(38.8) = 54.32
Total
100
wt. fr.
wt.fr.total
85.7
0.357
85.7
(240 + y)
154.3
0.643
154.3
(240 + y)
__
140.00
100
240.0
1.000
y
240 + y
impurity =
Total
Basis: 100 lb F
(Rather than use the impurity balance, use the total balance instead if you want)
Water in, Wi, comes from balances on Unit I.
Total:
100 + Wi = S + C
4–16
Solutions Chapter 4
⎛ 85.7 ⎞
⎟S+
60= ⎜
⎝ 240 + y ⎠
Na 2 S2 O2
H 2O
⎛ 154.3 ⎞
⎟S +
39 + W i = ⎜
⎝ 240 + y ⎠
⎛ 85.7 ⎞
⎟
61.2 + 6⎜
⎝ 240 + y ⎠
106
⎛ 154.3 ⎞
⎟
38.8+ 6⎜
⎝ 240 + y ⎠
106
⎫
⎪ Unknowns: W ,S,C, y
i
C⎪
⎬
Total:4
⎪
C⎪
⎭
Balances on Unit II needed as well because 4 unknowns exist.
Total:
C = W0 + D
Na 2 S2 O2
H 2O
⎛ 85.7 ⎞
⎫
⎟C
61.2 + 6⎜
⎪ Unknowns:W o ,D
⎝ 240 + y ⎠
= 0.611D
⎪
106
⎬
Total:2
⎛⎜ 154.3 ⎞
⎟C
⎪
38.8+ 6
⎝ 240 + y ⎠
= W o + 0.388D⎪
⎭
106
6 equations and 6 unknowns
Note that the complex term involving y can be eliminated to solve for D, W0, Wi, and S,
i.e., in effect making total Na2S2O2 and H2O balances overall the process.
The solution is
4.1.20
(a)Wi = 23.34 lb
(b)66.5%
Basis: 100 lb pulp as received
Comp
H2 O
lb = %
22
Pulp
78
Total
100
Freight cost =
$1.00 2000 lb
100 lb 1 ton
Assume air dried pulp means the 12% moisture pulp.
Allowed water =
78 lb pulp 12 lb H 2 O
88 lb pulp
= 10.65 lb H 2 O
4–17
= $20.00/ton
Solutions Chapter 4
Pulp on contract basis = 78 + 10.65 = 88.65 lb
Basis: 1 ton pulp as received
Cost =
4.1.21
88.65 lb 12% pulp rec'd 1 ton shipped $60.00
100 lb shipped
1 ton
$53.19/ton
You pay for soap plus transportation.
Basis: 100 kg soap with 30% water
Convert soap in wet soap to soap in dry soap. Data:
W (wet soap)
H2O 30 kg
soap
D(dry soap)
5%
70 kg
95%
100 kg
100
Soap balance: 0.70W = 0.95D
or 0.70(100) = 0.95D
D = 73.68 kg of which 70 kg is soap
$6.05
= $36.05
100
Cost of W at your site (containing 70 kg soap):
100 ($0.30 +
Cost of D at your site (containing 70 kg soap):
⎛ x$ $6.05 ⎞
73.68 ⎜
+
⎟ = $36.05
⎝ kg 100 ⎠
x = $ 0.43/kg
4–18
Solutions Chapter 4
4.1.22
The problem here is to decide on the balances to make. Not all will be indept balances.
Steps 1, 2, 3, and 4:
F = 100 lb
H2 O
VM
C
Ash
H(lb)
Mixer
Mass fr.
0.124
0.166
0.575
0.135
1.000
(lb)
P
Mass fr.
H2 O
VM
C
Ash
0.10
ω VM
ωC
0.10
1.00
Mass fr.
VM 0.082
C
0.887
H2O 0.031
1.000
Mass
mH O
2
m VM
mC
m Ash
P
Step 5:
Basis: F = 100 lb
Step 6:
Unknowns:
m H 2 O , mVM, mC, mAsh, P, H (6)
Step 7:
Balances:
H2O, VM, C, Ash, Σmi = P, m H 2 O /P = 0.10
mAsh/P = 0.10: (7), and presumably 6 are independent
Steps 8 and 9:
H2O (lb):
In
100 (0.124) + H (0.082)
=
Out
m H 2O
VM :(lb)
100 (0.166) + H (0.082)
=
mVM
C (lb):
100 (0.575) + H (0.887)
=
mC
Ash (lb)
100 (0.135) + H (0)
=
mAsh
=
P
Total
100 + H
4–19
Solutions Chapter 4
Solution:
mAsh = 13.5 lb
m H 2 O = 13.51
Not all of the equations have to be written down as above. Note that ash is a tie element to
P, so that the ash balance gives
100(0.135) + H(0) = P(0.10)
P = 135 lb
and mAsh = 13.5 lb
From a total balance
H + 100 = P
H = 35 lb
Check via H2O balance
Step 10:
12.4 + 35(0.031) = 13.5 ⎫
⎬ ok
13.49
=13.5⎭
4.1.23
Step 5:
Basis: 1 day
Steps 2, 3, and 4:
%
C3
C3
i − C4
n − C4
C5 +
%
1.9
51.6
46.0
0.5
100.0
D
F=5804
i − C4
n − C4
kg mol/day
B
Steps 6 and 7:
i−C
n- C 4
C 5+
3.4
95.7
0.9
100.0
%
1.1
97.6
1.3
100.0
There are two unknowns, and we have 4 independent equations, but the precision of
measurement is not the same in each equation, hence different results are obtained
depending on the equations used. Choose the most accurate to use.
4–20
Solutions Chapter 4
C3
(0.019)(5804) =
i-C4
n-C4
C5+
Total:
In
110
2992
2667
37
5804
=
=
=
=
=
0.034D
0.057D + 0.011B
0.009D + 0.976B
0.013B
D+B
Steps 8 and 9:
Use total + i-C4 balances:
kg mol/day
B = 2563
D = 3240
2992 = 0.975(5804-B) + 0.011 B
Use total + n-C4 balances:
B = 2704
D = 3100
2667 = 0.009 (5804 - B) + 0.976B
The other two balances give
C3 as a tie component:
kgmol D
5804 kgmol F 0.019kg mol C3
1kg mol D
= 3243
day
1.00 kgmol F 0.034 kg mol C3
1day
C5+ as a tie element:
5804 kgmol F 0.006kg mol C5 +
1kgmol B
= 2679 kgmol B / day
1.00 kg mol F
1day
0.013 kgmol C5 +
4–21
Out
Solutions Chapter 4
4.1.24
Steps 1, 2, 3, and 4:
Assume the other components have the same density as water in all flows.
3.785 × 106 L/day
150 mg/L BOD
a.
B
5 mg/L BOD
0.3 m 3 /s
A
C
BOD = ? = x
Step 5: Basis: 1 day
0.3m 3 3600s 24hr 1000 L
7
= 2.592 ×10 L / day
3
s
hr day m
A=
Steps 6 and 7:
Unknowns: C, x
Balances:
H2O, BOD
Steps 8 and 9:
Total mass balance A + B = C
1day A L 1kg 1day B L 1kgB 1day C L 1kg C
+
=
day L
day L
day L
BOD mass balance
−3
−3
1day A L 5×10 gBOD 1day B L 150 ×10 gBOD 1day C L x g BOD
+
=
day
L
day
day
day
L
2.592 × 107 kg + 3.785 × 106 kg = C = 2.9705 × 107 kg
5 × 10-3 A + 150 × 10-3 B = xC
5× 10 −3A +150 ×10 −3 B
x=
C
4–22
Solutions Chapter 4
5 × 10−3(2.592 × 10 7 ) + 150 × 10 −3(3.785 × 106 )
=
2.9705 × 107
−3
= 23.475 × 10 g / L
b.
g/L
N BOD = ?
B
530 x 106 L/day
3x10-3 g/L BOD
Total mass balance
= 15.8x106 L/day
C
A
5x10-3 g/L BOD
A+B=C
1day A L 1kg 1day B L 1kg 1day C L 1kg
+
=
day L
day L
day L
530 × 106kg + 15.8 × 106 kg = C = 5.458 × 108 kg
BOD mass Balance
−3
−3
−3
1day A L 3× 10 gBOD 1day B L N ×10 g 1day C L 5×10 g
+
=
day
L
day
L
day
L
3 × 10-3A + N × 10-3 B = 5 × 10-3C
5 × 10 −3 C − 3 × 10 −3 A
N=
B
(5 × 10−3 )(5.458 × 10 8 ) − (3 × 10−3 )(5.30 × 108 )
≅ 72.1x10 −3 g / L
=
7
1.58 × 10
4–23
ok
Solutions Chapter 4
4.1.25
The extractor is assumed to be a steady state process.
Steps 1, 2, 3, and 4:
All the known data have been placed in the figure.
9.7 L/min,
Solvent (organic stream)
Aqueous
Stream
F = 100 L/min
Waste (aqueous stream)
W(L/min) Enzyme C W
E
Water, etc.
P (L/min)
Enzyme C PE
Solvent
P
CE
=18.5
CW
E
Step 5: Basis: 1 min (F = 100L)
Step 6: Unknowns are P and W and their concentrations.
Steps 7 and 8: We can make enzyme, solvent, and aqueous balances, plus we have
C PE =18.5C W
E
The solvent and aqueous phases are tie components.
Balances
In
Out
P
C
g
(W L) C W
100 L 10.2 g (P L) E
E g
+
=
L
L
L
Enzyme (g)
Phase relation (g/L)
(D = 18.5)
Balances
P
Out
9.7L ρ Sg
(P L ) ρ P g
=
L
L
100 L ρ F g W L ρ W g
=
ρ F ≅ ρ W and ρ=
g ρ
L
L
Solvent
Aqueous
P = 9.7L
1020 = 97 C
W
C E = 18.5C E
In
W = 100 L
P
E
+ 100 C
W
E
= 9.7 (18.5) C PE + 100 C W
E
g
CW
E = 3.65
L
C PE = 67.53
4–24
P
Solutions Chapter 4
Fraction recovery =
67.53 g 9.7 L
L
10.2g 100L
L
= 0.642(1020) = 655 g/min
4.1.26
Assume the process is in the steady state, and no reaction occurs. Solution assumes no S is
in stream A and no N2 in stream B.
Steps 1, 2, 3 and 4:
mass fr.
A
ω NH
3
NH 3
ω HA2
1.00
N2
mass fr.
0.0633
0.9367
1.00
mol fr.
A
x NH
3
NH 3
x NA2
S
1.00
mass fr.
ω BNH 3
ω SB
1.00
Gas
1000 kg
mol fr.
NH 3
0.10
N2
0.90
1.00
Step 5: Basis 1.0 hr
A
, ω NA2 , ω BNH 3 , ω SB , a total of 6.
Step 6: The unknowns are A, B, ω NH
3
Steps 7 and 8: Three compound balances can be made, N2, NH3, and S. One relation is
given:
A
B
ωNH 3 = 2 ωNH 3
4–25
Solutions Chapter 4
Two summations exist:
∑ω
A
i
= 1 and
∑ω
B
i
=1
hence, we have an adequate number of balances (unless some are not independent) to solve
the problem.
Material balance equations in grams
Total:
F+S= A+B
1000 + 1000 = A + B
NH3:
A
1000(0.0633) + 0 = Aω NH
+ Bω BNH 3
3
N2 :
A
1000(0.9367) + 0 = Aω NH
+0
2
S:
0 + 1000 = 0 + Bω SB
Constraints
A
ω NH
+ ω NA2 = 1
3
ω BNH 3 + ω SB = 1
A
ω NH
= 2ω BNH 3
3
Solve in Polymath
Note: The equations can be converted to linear equations by letting
A
A
m NH
= ω NH
A, m NA2 = ω NA2 A, etc. The set can be solved in
3
3
A
.
Polymath or reduced to a quadratic equation in m BNH 3 or m NH
3
4.1.27
Basis: 12 hours
The process is illustrated in the figure
MTBE
H2O
mass fr. = 1.00
MTBE
H2O
4–26
H2O
MTBE
Solutions Chapter 4
The MTBE entering the pond in 12 hours is
25 boats 0.5 L gasoline 1000 cm3 0.72 g 0.10 g MTBE
= 900 g MTBE
1 boat
IL
1 cm3 1 g gasoline
The pond holds (ignoring the MTBE in the pond which is negligible)
3000 m 1,000 m 3m 1000 kg H 2O
= 9 × 109 kg H 2O
1 m 3 H 2O
The increase in the concentration of MTBE is
900 g MTBE 1 kg
= 1× 10-10 g MTBE/g H 2O
9
1000
g
9 × 10 kg H 2O
4–27
Solutions Chapter 5
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
5.1.1
Mol. wt.: 208.3
142.05
233.4
58.45
a. Basis: 5.0 g Na2SO4
Example:
5 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2 208.3 g BaCl 2
=
142.05 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2
7.33 g BaCl 2
Problem
b.
c.
d.
e.
f.
g.
h.
i.
Basis
5 g BaSo4
5 g NaCl
5 g BaCl2
5 g BaSO4
5 lb NaCl
5 lb BaCl2
5 lb Na2SO4
5 lb NaCl
Answer
4.47 g BaCl2
8.91 g BaCl2
3.41 g Na2SO4
3.04 g Na2SO4
6.08 lb Na2SO4
5.59 lb BaSO4
8.21 lb BaSO4
9.98 lb BaSO4
AgNO3 + NaCl → AgCl + NaNO3
5.1.2
Mol. wt.: 169.89
58.45
143.3
85.01
a. Basis: 5.0 g NaCl
1 g mol NaCl 1 g mol AgNO3 169.89 g AgNO3
Example: 5.0 NaCl
=
58.45 g NaCl 1 g mol NaCl 1 g mol AgNO3
14.53 g AgNO3
Problem
b.
c.
d.
e.
f.
g.
h.
i.
Basis
5 g AgCl
5 g NaNO3
5 g AgNO3
5 g AgCl
5 lb NaNO3
5 lb AgNO3
5 lb NaCl
5 lb AgNO3
5–1
Answer
5.92 g AgNO3
10.00 g AgNO3
1.721 g NaCl
2.04 g NaCl
3.44 lb NaCl
4.22 lb AgCl
12.27 lb AgCl
4.22 lb AgCl
Solutions Chapter 5
5.1.3
a.
The balanced equation is:
3 As2S3 + 4 H2O + 28 HNO3 = 28NO + 6 H3AsO4 + 9 H2SO4
and the reaction is unique within relative proportions.
b.
2 KC1O3 + 4 H C1 = 2 KC1 + 2 C1O2 + C12 + 2 H2O
or
KC1O3
or
3 KC1O3 + 10 HC1 = 3 KC1 + 2 C1O2 + 4 C12 + 5 H2O
+ 6 H C1 = KC1
+ 3 C12
+ 3 H2 O
We classify these reactions as non-unique since they are not simply proportional
equations, but are linearly independent, and infinitely many equations can be obtained by
linear combination.
5.1.4
Basis: 2 g mol
C: 6 × 12 = 72 H: 8 × 1 = 8
mol. wt. = 72 + 8 + 96 = 176
2 g mol 176 g 1 lb
= 0.775 lb
1 g mol 454 g
5–2
O: 6 × 16 = 96
Solutions Chapter 5
5.1.5
a)
C
H
Zn
O
LHS
RHS
4
10
1
14
4
10
1
14
Yes, the equation is balanced.
b) Basis: 1.5 kg ZnO
1.5 kg ZnO 1000 g ZnO 1 gmol ZnO 1 gmol DEZ 123.4 gDEZ
1.0 kg ZnO 81.40 g ZnO 1 gmol ZnO 1 gmol DEZ
×
c)
1.0 kg DEZ
= 2.27 kg DEZ
1000 g DEZ
20 cm 3 H2 O 1.0 g H2 O 1.0 gmol H2 O 1 gmol DEZ 123.4 g DEZ
1 cm 3 H2 O 18.0 g H2 O 5 gmol H 2O 1 gmol DEZ
= 27.4 g DEZ
5.1.6
Basis: Data given in problem statement
55.847 gFe 1 gmol Fe 1 gmol X2 O3 2 gmol X
= 1 gmol X
1
55.847 gFe 2 gmol Fe 1 gmol X2 O3
55.847 gFe 1 gmol Fe 1 gmol X 2 O3 3 gmol O
16 g O
= 240 g O
1
55.847 gFe 2 gmol Fe 1 gmol X2 O3 1 gmol O
50.982 g X 2 O3 − 24.0 g O = 26.982 g X
26.982 g X
g
= 26.982
hence X =A1
1 gmol X
gmol
5–3
Solutions Chapter 5
5.1.7
Basis: Data given in problem statement
1.603g CO2 1 gmol CO2 1 gmol C 12.00g C
= 0.437 g C
1
44.01g CO2 1 gmol CO2 1 gmol C
0.2810gH2 O 1 gmol H2 O 2 gmol H 1.008g H
= 0.03144 g H
1
18.02g H2 O 1 gmol H2 O 1 gmol H
0.6349 g CxHyOz – 0.437g C – 0.03144g H = 0.1661g O
Based on O:
0.1661g O 1 gmol O 0.01038 gmol O
=
= 1 gmol O
1
16.00g O
0.01038
0.437 g C 1 gmol C
0.03642 gmol C
=
= 3.5 gmol C
1
12.00 g C
0.01038
0.03144 g H 1 gmol H
0.03119 gmol H
=
= 3 gmol H
1
1.008 g H
0.01038
The formula in integers is
C 7 H6 O2
5.1.8
Basis: Data given in the problem
Mass of hydrate
Mass of dry sample
Mass of water
10.407 g
-9.520 g
0.887 g
9.520 g BaI2 1 g mol BaI2
= 0.0243 g mol BaI2
391 g BaI2
0.887 g H2 O 1 g mol H2 O
= 0.0493 g mol H2 O
18.02 g H2 O
0.0243
= 0.49, or 2 H 2 O for 1 BaI2 . Thus the hydrate is BaI2 ⋅ 2H 2 O.
0.0493
5–4
Solutions Chapter 5
5.1.9
Basis: 2000 tons 93.2% H2SO4 (1 day)
a.
2000 tons soln 0.932 ton H 2SO4 1 ton mole H 2SO4
1 mol S
32 ton S
1 ton soln
98 ton H 2SO4 1 mol H 2SO4 1 ton mol S
= 609 ton S
1.5 mol O2 32 ton O2
b. 609 ton S 1 ton mol S
= 913.5 ton O2
32 ton S 1 mol S 1 ton mol O2
1 mole H 2 O 18 ton H 2 O
= 342.6 ton H2O
c. 609 ton S 1 ton mole S
1 mol S 1 ton mol H 2 O
32 ton S
5.1.10
Basis: 1 lb Br2
2Br − → 1Br2
MW:
2Br– + Cl2 → 2Cl– + Br2
70.9
159.8
MW:
Br2 + C2H4 → C2H4Br2
28
187.9
6
a.
0.27 lb acid 1×10 lb seawater = 2.08 lb 98% H SO / lb Br
2
4
2
2000 lb seawater
65 lb Br 2
b.
1 lb Br 2 mole Br 2 1 mole Cl 2 70.9 lb Cl 2
= 0.445 lb Cl 2
159.9 lb Br 2 1 mole Br 2 mole Cl 2
1 lb Br 2 1×10 6 lb sea water
c.
= 15,400 lb seawater
65 lb Br 2
d.
1 lb Br2
mol Br2 1 mol C2 H 4 Br2 187.9 lb C2H 4Br2
= 1.176 lb C2 H 4 Br2
159.9 lb Br2 1 mol Br2
1 mol C2 H 4 Br2
5–5
Solutions Chapter 5
5.1.11
FeSO4 ⋅7H2O
MW
277.9
FeSO4 ⋅H 2O
MW
169.9
FeSO4 ⋅4H2O
223.9
FeSO4
151.9
It is best to evaluate the three types of ferrous sulfate in terms of the cost/ton of FeSO4
a.
FeSO4 ⋅7H2O:
Basis: 475 ton material
$83,766.25
= $176.35/ton FeSO4 ⋅ 7H2 O , which is equivalent to
475 ton FeSO4 ⋅ 7H 2 O
$176.35
277.9 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅7H2 O
1 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅ 7H2 O
1 ton mol FeSO4
1 ton mol FeSO4
= $323/ton FeSO4
151.9 ton FeSO4
b.
FeSO4 ⋅4H2O:
$323 151.9 ton FeSO4
1 ton mol FeSO4
1 ton mol FeSO4 ⋅ 4H2 O
ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ 4H2 O 223.9 ton FeSO 4 ⋅ 4H 2O
= $219 / ton FeSO4 ⋅ 4H2O
c.
FeSO4 ⋅H 2O:
$323 151.9 ton FeSO4
1 ton mol FeSO4
1 ton mol FeSO4 ⋅H 2O
ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ H2 O 169.9 ton FeSO4 ⋅ H2 O
= $289 / ton FeSO4 ⋅ H2O
5–6
Solutions Chapter 5
5.1.12
The reaction is C 4H10 + 6 12 O2 → 4CO2 + 5H2O
Basis: 1 mol C4H10
⎛ mol O2 for complete combustion ⎞
Minimum O2 ≅ (LFL) ⎜
⎟
mol C4 H10
⎝
⎠
≅ (1.9%) (6.5/1) = 12.4%
5.1.13
Na2CO3 + Ca(OH2 ) → 2NaOH + CaCO3
Basis: 1 ton of soda ash (Na 2 CO3 )
1 ton Na 2 CO3 1 ton mol Na2 CO3 2 ton mol NaOH 40.0 ton NaOH
=
106 ton Na 2 CO3 1 ton mol Na 2 CO3 1 ton mol NaOH
0.755 ton NaOH
$130
1 ton Na 2 CO3
= $172 / ton Na 2CO3
1 ton Na 2CO3 0.755 NaOH
The above result is for the case of free Ca(OH2). Otherwise, the value must be reduced to
compensate for the cost of the CA(OH2)—which might come from the CaCO3 produced,
and possibly be cheaper than Ca(OH2) purchased directly.
5–7
Solutions Chapter 5
5.1.14
Basis: 1 ton dolomite
Assume complete
reactions
CaCO3 + H2SO4 → CaSO4 + H2O + CO2
MgCO3 + H2SO4 → MgSO4 + H2O + CO2
Comp.
%
lb
CaCO3
MgCO3
SiO2
68
30
2
1360
600
40
100
2000
Total
a.
mol.wt.
100.0
84.3
60.0
lb mol
that react
13.60
7.13
20.73
20.73 lb mol react 1 mol CO2 produced
44 lb
1 mol react
ton dolomite
1 lb mol CO2 = 912 lb CO2
1 mol H 2 SO4 reqd 98 lb H 2 SO4 100 lb acid
= 2160 lb acid
b. 20.73 lb mol react
1 mole H 2 SO4 94 lb H 2 SO4
1 mol react
ton dolomite
5–8
Solutions Chapter 5
5.1.15
Basis: 1 kg dichlorobenzene (DCB)
The balanced stoichiometric reaction for dichlorobenzene is shown below:
C6 H 4 C12 + 6.5 O2 → 6 CO2 + H 2 O + 2 HC1
Therefore, for 1 kg of dichlorobenzene, the following moles of HC1 are produced:
1 kg DCB 1 kg mol DCB 2 kg mol HC1
= 0.0136 kg mol HC1
147 kg DCB 1 kg mol DCB
The balanced stoichiometric reaction for tetrachlorobiphenyl is as shown below:
C12 H6 C14 + 12.5 O 2 → 12 CO 2 + H 2O + 4 HC1
Therefore, for 1 lb of tetrachlorobiphenyl (TCB), the following moles of HC1 are
produced:
Basis: 1 kg TCB
1 kg TCB 1 kg mol TCB 4 kg mol HC1
= 0.0138 kg mol HC1
290 kg TCB 1 kg mol TCB
Thus, the amount of acid produced does not change significantly (≈ 1.5%) , and neither will
the amount of base required for neutralization.
5.2.1
Basis: 1 L solution = 1000 g H2O
50 g H2 S 1 g mol H2 S 1 g mol HOCl 52.45g HOCl 100 g HOCl soln
10 6 g soln 34 g H2 S 1 g mol H2 S 1 g mol HOCl
5 g HOCl
1000 g soln 2
= 3.08 g HOCl solution
1 L soln
You can use the density of H2O for the density of the solution as the H2 S content has
negligible effect on the density.
5–9
Solutions Chapter 5
5.2.2
Basis: 10 lb moles phosgene
For phosgene:
ξ=
5.2.3
n f − n i 10 − 0
=
= 10 reacting moles
1
1
n CO,i = n CO,f -ν COξ = 7 − (−1)(10)
=
17 lb mol
n C12 ,i = n C12 ,f -νC12 ,iξ = 3 − (−1)10
=
13 lb mol
Basis: 135 mol CH4
ξ=
0 − 135
= 22.5 reacting mols
−6
5.2.4
2FeS + 3O2 → 2FeO + 2SO2
MW
87.85
32
71.85
64
Basis: Use 100 g slag
Component
FeO
Fes
For the FeO:
ξ=
g=%
80
20
100
MW
71.85
87.85
g mol
1.113
0.228
nf − ni
1.113-0
=
= 0.557 reacting moles
ν
2
For the FeS:
ξ=
0.228 − n i
nf − ni
=
= 0.557
ν
-2
ninitial FeS = 1.341 g mol
5–10
Solutions Chapter 5
5.2.5
Basis: 1080 lb bauxite
A12 O3 +3H 2SO4 → A12 (SO4 )3 + 3H 2 O
1080 lb bauxite 55.4 lb A12 O3 1 lb mol A12 O3
= 5.87 lb mol A12 O3
100 lb bauxite 101.9 lb A12 O3
ξ=
0 − 5.87
= 5.87 reacting moles
−1
Basis: 2510 lb H2SO4 (77%)
2510 lb H2SO4 (77%) 0.77 lb H 2SO4 1 lb mol H 2SO 4
= 19.70 lb mol H 2SO4
1 lb H2SO4 (77%) 98.1 lb H 2SO4
ξ=
0 − 19.70
= 6.57 reacting moles
−3
(a)
A12O3 is the limiting reactant and H2SO4 is the excess reactant.
(b)
%xs H2SO4:
(19.70)(1/ 3) − 5.87
6.57 − 5.87
(100) =
(100) = 11.9%
5.87
5.87
or H2SO4 (77%) required:
%xs =
(5.87)(3)(98.1)
= 2244 lb
0.77
2510 − 2244
= 12%
2244
Basis: 2000 lb A12(SO4)3
2000 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3 1 lb mol A12 O3
342.1 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3
101.9 lb A12 O3 1 lb bauxite
=1075 lb bauxite
1 lb mol A12 O3 0.554 lb A12 O3
1075
× 100 = 99.5%
1080
5–11
Solutions Chapter 5
5.2.6
Basis: 100 kg of fusion mass
BaSO 4
+
4C → BaS
+
4CO
Comp.
Wt.% = kg
MW
kg mol
BaSO4
BaS
C
Ash
11.1
72.8
13.9
2.2
100.0
233.3
169.3
12.0
____
0.0476
0.430
1.16
____
BaSO4 orig. = 0.0476 + 0.430 = 0.4776 mol
Corig. = 1.16 + 4 (0.430) = 2.88 mol
Carbon is the limiting reactant.
Needed C for complete reaction = 4 (0.4776) = 1.9104 mol
% excess =
2.88 − 1.91
× 100 = 50.8% excess C
1.91
Degree of completion =
0.4776 − 0.0476
= 0.900
0.4776
You could calculate the extent of reaction from the BaSO4, and use it to get the original
amounts of reactants.
5–12
Solutions Chapter 5
5.2.7
CO + C12 → COC12
Basis: 20 kg products
Comp.
C12
COC12
CO
kg
3.00
10.00
7.00
20.00
kg
mol
0.0423
0.1011
0.250
0.393
mol.wt.
70.91
98.91
28.01
kg mol
C12 in
0.0423
0.1011
0.143
kg mol
CO in
0.1011
0.250
0.351
The extent of reaction is
CO:
0 − 0.351
= 0.351
−1
C12:
0 − .143
= 0.143 (smallest ξ )
−1
C12 is the limiting reactant; CO is the excess reactant
a.
0.351 − 0.143
(100) = 145% excess CO
0.143
b.
0.1011
(100) = 70.7% C12 converted
0.143
c.
0.1011 kg mol COC12
= 0.205
(0.143 + 0.351) kg mol reactants
5.2.8
Amount of o-n hydrolized:
Total protein:
0.10 m mol 1000 µ mol
= 100 µ mol
1 m mol
1.00 mg 0.10 mL
= 0.100 mg
1 mL
Specific activity =
100 µ mol
= 200 µ mol/(min)(mg protein)
(5 min)(0.10 mg protein)
5–13
Solutions Chapter 5
5.2.9
Basis: 100 kg mol A
100 kg mol C 6H 5NO2 86.9 kg mol C 6H 8NOCl 0.95 kg mol C 6H 7NO 3 kg mol C 8H 9NO2
100 kg mol C 6H 5NO2 kg mol C 6H 8NOCl 4 kg mol C 6H 7NO
→ 0.619 fraction overall conversion.
5–14
Solutions Chapter 5
5.2.10
C5 H7 O2 N
C5
H7
O2
N
60
7
32
14
113 kg/k mol
NH +4
N
14
4
H4
18
20,000 kg C 5 kg NH +4
= 1000 kg NH +4
100 kg C
1000 kg NH +4 kg mol NH 4+
= 55.556 k mol NH +4
+
18 kg NH 4
55.556 kg mol NH +4 1 kg mol C5H 2O2 N 95 kg mol NH +4 113 kg C5H 2O2 N
55 kg mol NH +4 100 kg mol fed kg mol C5H 7O 2 N
= 108.43 kg C5H7O2N
5.2.11
2C2 H4 + O2 → 2C2 H4O
For 100% conversion of C2H4 according to reaction (a):
For reaction b3:
2 mol C2 H 4O / time
=1
2 mol C2 H 4 / time
1 mol C2 H 4 O/time
= 0.75
1.33 mol C2 H 4 /time
5–15
Solutions Chapter 5
5.2.12
Step 4:
The molecular weights needed to solve the problem and the gram moles forming the basis
are:
Component
Sb2S3
Fe
Sb
FeS
kg
0.600
0.250
0.200
Mol. wt.
339.7
55.85
121.8
87.91
g mol
1.766
4.476
1.642
The process is illustrated in Fig. P9.19
1.766 g mol Sb2S3
Reactor
4.476 g mol Fe
Fe S
1.642 g mol Sb
Figure P9.19
Step 5:
Basis: Data given in problem statement
Steps 6-9:
(a)
To find the limiting reactant, we examine the chemical reaction equation. The
ratio of Sb2S3 to Fe in the equation is 1/3 = 0.33. In the actual reaction the
corresponding ratio is 1.766/4.476 = 0.395, hence Sb2S3 is the excess reactant and
Fe is the limiting reactant. The Sb2S3 required to react with the limiting reactant is
4.476/3 = 1.492 g mol.
(b)
The percentage of excess reactant is
% excess =
(c)
1.766 − 1.492
(100) = 18.4% excess Sb 2S3
1.492
Although Fe is the limiting reactant, not all the limiting reactant reacts. We can
compute from the 1.64 g mol of Sb how much Fe actually does react:
5–16
Solutions Chapter 5
1.64 g mol Sb 3 g mol Fe
= 2.46 g mol Fe
2 g mol Sb
If by the fractional degree of completion is meant the fraction conversion of Fe to
products, then
Fractional degree of completion =
(d)
2.46
= 0.55
4.48
Let us assume that the percent conversion refers to the Sb2S3 since the reference
compound is not specified in the question posed.
1.64 g mol Sb 1g mol Sb 2S3
= 0.82 g mol Sb 2S3
2 g mol Sb
% conversion of Sb 2S3 to Sb =
(e)
0.82
(100) = 46.3%
1.77
The yield will be stated as kilograms of Sb formed per kilogram of Sb2S3 that was
fed to the reaction
Yield =
0.200 kg Sb
0.33 kg Sb
=
0.600 kg Sb2S3
1 kg Sb2S3
5–17
Solutions Chapter 5
5.2.13
Reaction on which to base excess is
Fe2O3 + 3C → 2Fe + 3CO
Basis: 1 ton Fe2O3
a. Theoretical C =
1 ton Fe 2O3 2000 lb Fe 2O3 1 lb mol Fe 2O3
3 lb mol C 12 lb C
= 451 lb
1 ton Fe 2O3 159.7 lb Fe 2O3 1 lb mol Fe 2O3 lb mol C
600 œ451
× 100 = 33% excess C
451
b.
1 ton Fe 2O3 1200 lb Fe lb mol Fe 1 lb mol Fe 2O3 159.7 lb Fe 2O3
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol Fe 2O3
= 1720 lb Fe2O3 reacts
1720 × 100 = 86.0% Fe O
2 3
2000
c. (i)
1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol CO 28 lb CO
= 35.62 lb CO
1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO 1 lb mol CO
1 ton Fe 2O3 1200 lb Fe 1 lb mol Fe 3 lb mol CO 28 lb CO
= 902.4 lb CO
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol CO
Total (35.62 lb CO + 902.4 lb CO) = 938 lb CO
(ii)
1 ton Fe 2O3 1200 lb Fe lb mol Fe 3 lb mol C 12 lb C
= 387.1 lb C
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe lb mol C
1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol C 12 lb C
= 15.3 lb C
1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO lb mol C
387.1 + 15.3 = 402.4 lb C used / 1ton Fe2O3
5–18
Solutions Chapter 5
or use
938 lb CO 1 lb mol CO 1 lb mol C 12 lb C
28 lb CO 1 lb mol CO 1 lb mol C
Selectivity: Mass basis:
d.
1200 lb Fe formed
= 6.56
183 lb FeO formed
1200 lb Fe lb mol Fe
= 21.49 lb mol Fe
55.85 lb Fe
183 lb FeO ×
Mole Basis:
lb mol FeO
= 2.55 lb mol FeO
71.85 lb FeO
21.49 lb mol Fe
= 8.44
2.55 lb mol FeO
5.2.14
Cl2 + 2NaOH → NaCl + NaOCl + H2O
MW: 71.0 40.01
58.5
74.5
18.0
Basis: 1145 lbm NaOH + 851 lb Cl2 (→ 618 lbm NaOCl)
a. Determine limiting reactant
Assume all Cl2 reacts, calculate lb of NaOH required
NaOH required =
2 lb mol NaOH lb mol Cl 2 851 lb Cl 2 40.01 lb NaOH
71 lb Cl 2
lb mol NaOH
lb mol Cl 2
= 959.11 lb NaOH required
C12 is limiting reactant
or use extent of reaction
NaOH:
C12:
1145
= 28.63 lb mol
40
851
= 11.99 lb mol
71.0
(0-28.63)/-2 = 14.3
(0-11.99)/ − 1 = 12.0 (minimum)
5–19
Solutions Chapter 5
b. % excess NaOH =
=
lb mol NaOH in excess
(100)
lb mol NaOH for rxn
1145 lb m/NaOH
lb mol NaOH
1
œ 959.11
40.01 lb NaOH
40.01
(100)
1
959.11
40.01
= 19.4% excess NaOH
618
lb mol C12 that reacted
c. Degree of completion = =
= 74.5 = 0.692
lb mol C12 available to react 851
71.0
ξ
8.30
or
=
=0.69
ξ min
12.8
lb
d. Yield of NaOCl = 618 lb NaOCl = 0.726
851 lb Cl 2
lb
e.
618
= 8.30 lb mol NaOC1
745
ξ=
8.30 − 0
= 8.30 reacting moles
1
5–20
Solutions Chapter 5
5.2.15
4HC1+ O 2 → 2C12 + 2H 2O
MW
35.45 32
71.0
18
Basis: 100 mol gas
Exit comp.
Entering moles
HC1
O2
% mol
HC1
4.4
C12
19.8
H2 O
19.8
4.4
-
(2)(19.8)
39.6
-
(1/2)(19.8)
-
9.9
O2
4.0
-
4.0
N2
52.0
-
-
44.0
13.9
100.0
(The N2 balance gives the O2 as 13.8).
Calculate the extent of reaction for complete reaction
For HC1:
0 − 44.0
= 11.0 (the minimum)
−4
For O2:
0 − 13.9
= 13.9
−1
(a)
HC1 is the limiting reactant
44.0
4 100 = 26.36%
44.0
4
139 −
(b)
% excess =
(c)
degree of completion =
44 − 4.4
= 0.90
44
(or 9.9 from (d)/11.0) = 0.90
(d)
extent of reaction ξ =
19.8 − 0
= reacting moles
2
5–21
Solutions Chapter 5
5.2.16
Basis: 30 mol CO, 12 mol CO2, 35 mol H2O feed = 1 hr
CO
CO2
H2O
mol
30
12
35
mol
F
CO + H2O
MW:
a)
CO
CO2
H2
18
H2O
P
28
18
CO2 + H2
fed
in eqn
CO 1
CO 30
=
= 0.86 <
= =1
H2 O 1
H2 O 35
44
2
CO is limiting reactant
b)
H 2O is the excess reactant
c)
18
= 0.514
35
d)
18
= 0.60
30
e)
18 kg mol H2 2.016 kg H 2
1 kg mol H 2O
1 kg mol H2 35 kg mol H2 O 18 kg H2 O
(for each mol of H2 produced, 1 mol CO reacts)
= 0.0576 kg H2
f)
1 mol of CO2 is produced for each mol of H 2
mol CO 2 produced
18
= 0.60
30
mol CO fed
(g)
ξ=
18 − 0
= 18 reacting moles using H 2
1
5–22
Solutions Chapter 5
5.2.17
Basis: 100 lb mesitylene (C9H12)
The reactions are
C9H12
MW
120.1
C9 H12 + H 2 → C8 H+10 CH 4
C8H10 106.08
C8 H10 + H 2 → C7 H+8 CH 4
C7 H8
92.06
The initial selectivity is: C8 H10 /C7 H8 = 0.7 and 0.8
The number of moles of C9H12 in the basis is
100 lb C9 H12 1 lb mol C9 H12
= 0.833 lb mol
120.1 lb C9 H12
For the selectivity of 0.7 (the equations are in lb mol)
(1) C8 H10 + (1)C7 H8 = 0.833
C8 H10 / C7 H8 = 0.7
with the results:
(lb mol)
(in lb)
C7 H8 = 0.490
C8 H10 = 0.343
45.1
36.4
For the selectivity of 0.8 the results are
C7 H8 = 0.463
C8 H10 = 0.370
42.6
39.3
The catalyst used in the respective cases is
Selectivity of 0.7
Selectivity of 0.8:
0.490 lb mol C7 H8 92.06 lb C7 H8
1 lb catalyst
500 lb C7 H8
1 lb mol C7 H8
0.463 lb mol C7 H8 92.06 lb C7 H8
1 lb catalyst
500 lb C7 H8
The economic analysis of the change is
5–23
1 lb mol C7 H8
= 0.090 lb
= 0.085 lb
Solutions Chapter 5
Income change
Increase in C8H10:
Decrease in C7H8:
$
(39.3 − 36.4)($0.65) = +1.89
(45.1 – 42.6) ($22) =
− 0.55
Expense change (decrease)
Change in catalyst used: (0.090-0.085) ($25) = + 0.13
Net change
5.3.1
$1.47
Basis: 1 mol A
A → 3B
50 percent conversion gives as the product:
A
B
mol
0.5
1.5
2.0
(a)
Mole fraction A = 0.5/2.0 = 0.25
(b)
Extent of reaction ξ =
0.5 − 1.0
= 0.5 using A
−1
5–24
Solutions Chapter 5
5.3.2
Step 5:
2 FeS2 + 121 O2 → Fe2 O3 + 4 SO2
Basis: 100 lb pyrites in
Steps 2, 3, and 4:
SO3
Y
Feed
100 lb pyrite
32% S
10 lb S
Z
X
O2
N2
mol fr
0.21
0.79
Gas
mole %
SO 2
13.4
O2
2.7
N2
83.9
100.0
Air
Cinder : Fe2 O3 , inert
1.00
Let X = mol air in, Y = mol SO3 out, Z = mol flue gas out
Steps 6 and 7:
Using a reduced set of variables:
Unknowns (4): X1 Y1 Z, cinder
Balances (4): O, N, S, cinder
Steps 8 and 9:
1.
Sulfur Balance:
(32 + 10)/32 = Y + 0.134Z
2.
2O Balance:
0.21X = 1.5Y + (0.134 + 0.027) Z
+
3.
2N Balance:
1mol S mol Fe 1mol Fe 2 O3 1.5mol O2
2mol S 2 mol Fe 1mol Fe 2 O3
0.79 X = 0.839Z
Solve (1), (2), (3): Y = 0.118 mol, Z = 8.9 mol
Step 10:
Check: mol S = 0.118 + 0.134(8.9) = 1.31
% conversion of S to SO3 = (11.8)/(1.315) = 8.98%
5–25
Solutions Chapter 5
5.3.3
Steps 1, 2, 3 and 4:
mole %
Product
Methanol F
CH3 OH 100%
Reacts
P
A
Air
O2
N2
mol frac
0.21
0.79
1.00
Step 5:
Basis: 100 mol P
Step 6: Unknown
F, A
Step 7: Balances
N, H, O, C
H2 O
5.9
O2
13.4
N2
62.6
CH2 O
4.6
CH3 OH
12.3
HCOOH
1.2
100.0
More balances than unknowns. Are they independent? Consistent?
Steps 8 and 9:
Balances in moles
Check by first solving C, N (tie elements)
C:
F (1) = 4.6 + 12.3 + 1.2 = 18.1
F = 18.1 mol
2N:
A (0.79) = 62.6
A = 79.24 mol
F
A
Check via O: (2) (79.24)(.21) + 18.1
51.38
?
=
13.4(2) + 5.9+4.6+12.3+1.2(2)
?
=
52.0
Close but not exactly the same, more oxygen out than in. Might be caused by round off.
Check via H:
4(18.1)
?
=
5.9(2) + 4.6(2) + 12.3(4) + 1.2(2)
72.40
?
=
72.6
Not exactly the same; more hydrogen out than in, but again may be caused by round off.
The conclusion is that probably no error exists in the measurements.
5–26
Solutions Chapter 5
5.3.4
a.
Additional information was the molecular weights of the compounds and the
chemical reaction equations.
MW CaCO3 = 100.0, MW MgCO3 = 84.3, MW CO2 = 44.
b.
Assume a batch process
Final
Initial
F
CaCO3
CaO
MgCO3
MgO
P
G
CO2
c.
Basis: 1 g sample
d.
F = 1.000 g; P is not explicitly given but can be calculated: P = 0.550 g;
G is not explicitly given and can be calculated, G = 0.450 g
e.
Mass fraction of CaO in P and F, mass fraction MgO in P and F
(or mass of each)
f.
Species balances
g.
It appears to be a carbon balance
h.
0 degrees of freedom
i.
Yes
5–27
Solutions Chapter 5
5.3.5
Assume a steady state flow process with reaction.
Step 5: Basis: 1 hr (6.22 kg mol = F)
Steps 1, 2, 3 and 4:
H2 (g)+SiHC13 (g) → Si(s)+3HC1(g)
F = 6.22 kg mol
Mol fr.
0.580 H 2
Reactor
Mol fr. mol
W(kg mol) HCl x HCl nHCl
SiHCl3 x SiHCl 3 n SiHCl 3
H2
P
0.420 SiHCl 3
1.00
Si
0.223 nH 2
Mol fr.
1.00
MW Si = 28.086
n H 2 /W = 0.223
Step 6:
Unknowns:
W, P, 3 n i
Step 7:
Balances:
element balances are H, Si, Cl, ∑ ni = W, n H 2 /W = 0.223
Or, using the extent of reaction
Step 6:
Unknowns = 6:
W, P, 3ni ,ξ
Step 7:
Balances = 6
Species balances: H2, SiHC13, HC1, Si
∑ ni = W
n H2 /W = 0.223
Steps 8 and 9:
Balances are in kg mol.
5–28
Solutions Chapter 5
Cl (kg mol):
Out
In
=
6. 22(0. 420)(3) nHCl + n SiHCl3 (3)
Si (kg mol):
6.22(04.20)(1) = P(1) + n SiHCl3
H (kg mol):
6.22[( 0.580 )( 2) + 0.420(1)] = n HCl + nSiHCl3 + nH 2 (2 )
Balances using ξ :
H2 : ⎡⎣n H2 − (0.580)(6.22) ⎤⎦ = ( 1)(ξ )
SiHC13 : ⎡⎣nSiHC13 − (0.420)(6.22) ⎤⎦ = ( 1)(ξ )
HC1:
−
−
[n HC1 − 0] = (+3)(ξ )
[nSi − 0] = (+1)(ξ )
Si :
ξ = 1.83
Solution:
W = 8.05
P=1.83
n H 2 = 1.78
n HCl = 5.49
n SiHCl3 = 0.7824
1.83kgmol Si 1hr 20 min 28.086 kgSi
= 17.13kgSi
1hr
60min
1kgmol Si
1.46 kg Si initial
18.59 kgfinal Si
5–29
Solutions Chapter 5
5.3.6
Steps 1, 2, 3, and 4:
G(kg mol)
Ore
Cus
Cus*
inert
SO3
P(kg mol)
Reactor
n CuS (kg mol)
Y (kg mol)
CuO
A(kg mol)
Air
O2
N2
*
100%
SO2
mol fr.
0.072
O2
0.081
N2
0.847
1.000
unreacted CuS
inert
mol fr.
0.21
0.79
1.00
This CuS doesn’t react.
This is a steady state open process with reaction.
Step 5: Basis: 100 kg mol P
Steps 6 and 7: Unknowns: n CuS ,A, G, Y
Element balances: S, O, N, Cu
Steps 8 and 9: Balances are in kg mol on the elements
S:
n CuS (1) = G(1) + 0.072 (100)
2O:
A(0.21) = G(1.5) + 100(0.072) + 100(0.081) + Y( )
2N:
Cu:
A(0.79) = 100(0.847)
n Cus (1) = Y(1)
1
2
5–30
Solutions Chapter 5
Solution yields (in kg mol): G = 1.81, nCuS = Y = 9.01
1.81 mol SO3 1 mol S
1 mol SO3
= 0.20 of CuS that reacts goes to SO3 20%
9.01 mol CuS that reacts 1 mol S
1 mol CuS
If you use the extent of reaction, you need to write down the balanced reactions
CuS + O2 → CuO + SO2
SO2 +
1
2
(1)
(2)
O2 → SO3
Steps 6 and 7:
The unknowns: 4 above plus ξ1 and ξ2
Balance: 6 species balances
Steps 8 and 9:
CuO:
CuS
Y
0
=
=
0 + (1) (ξ1 )
n CuS + (−1)(ξ1 )
O2 :
P(0.081)
=
A(0.21) + (−1)(
+ξ1−) (
N2
SO2
SO3:
P(0.847)
P(0.072)
G
ξ1 = 9.01
ξ2 = 1.81
=
=
=
A(0.79)
0 + (1)(ξ1 ) + ( 1)(ξ2−)
0 + (1)(ξ2 )
5–31
1
2
)(ξ 2 )
Solutions Chapter 5
5.3.7
(1)
6HF(g) + SiO2 (s) → H2SiF6 (l ) + 2H2O(l )
(2)
H2SiF6 (g) → SiF4 (g) + 2HF(g)
The net reaction is
4HF + SiO2 → SiF4 + 2H2O
Steps 2, 3, and 4:
SYSTEM
mol fr.
0.50 HF
0.50 N2
1.00
F
SiO2
on wafers
P
The process is an unsteady state, open process
Step 5:
Basis: 1 mol F entering
Steps 6 and 7:
Unknowns:
4 ni , P1 ξ, loss SiO2 from wafer (L)
Balances:
HF, N2 , SiF4 , H2O, SiO2 (5 species)
Other equations: ∑ ni = P; 10% HF reacts
Steps 8 and 9:
SiO2:
HF:
n HF = 0.50 + (−4)ξ = 0.50(.90) = 0.45
ξ = 1.25 ×10−2 reacting moles
SiF4:
n SiF4 = 0 + (1)(1.25×10-2 ) = 1.25 × 10-2 mol
H2O:
N2 :
n H2O = 0 + (2)(1.25 × 10-2 ) = 2.50 × 10-2 mol
n N2 = 0.50 mol
n SiO2 ,f − n SiO2 ,i =(−1)(ξ ) = −1.25 × 10-2 mol
5–32
nHF
nN2
nSiFH
nH2O
Solutions Chapter 5
Composition
HF
SiF4
H2 O
N2
mol
0.45
1.25 10-2
2.50 10-2
0.50
0.9875
mol fr.
0.456
0.013
0.025
0.506
1.000
If you use element balances:
Step 6 and 7:
The unknowns: 4ni, P, L
The balances: H, F, Si, O, N (presumably 4 are independent)
Other equations: ∑ n i = P, 10% HF reacts
Steps 8 and 9:
H:
F:
Si:
O:
N:
0.50(1) = n HF + 2n H O
0.50(1) = n HF + 4nSiF
(nSiO ,f -nSiO ,i ) = nSiF
2(nSiO ,f -nSiO ,i ) = n H O
2(0.50) = 2n N
2
4
2
2
2
4
2
2
2
n HF = 0.90(0.50)=0.45
5–33
Solutions Chapter 5
5.3.8
We will view this process as a steady state process in an open system with flow in and out, and a
change in the material in the vessel from the initial to the final states. For component i, Equation
(10.1) becomes Equation (10.6)
R
n iout = n ini + ∑ νijξ j
j=1
The bioorganisms do not have to be included in the solution of the problem.
Steps 1, 2, 3, and 4
Figure P10.9 is a sketch of the process.
CO2 (g)
H2O
C6H12O6
H2O (88%)
F
C6H12O6 (12%)
C2H5OH
C2H3CO2H
Figure P10.9
Step 5 Basis: 3500 kg F
Step 4
Convert the 3500 kg into moles of H2O and C6H12O6.
n initial
=
H O
3500(0.88)
2
n initial
=
C6 H12 O6
18.02
= 170.9
3500(0.12)
180.1
= 2.332
Step 6 and 7
The degree of freedom analysis is as follows:
Number of variables: 9
5–34
Solutions Chapter 5
Out
Out
Out
Out
n InH2O , n CIn6H12O6 , n Out
H2O , n C6 H12O6 , n C2 H5OH , n C2 H2CO2 H , n CO2 , ξ1 , ξ2
Number of equations: 9
Basis: F = 3500 kg of initial solution (equivalent to initial moles of
H2O plus moles of sucrose)
Species material balances: 5
H2O, C2 H12O6 , C2 H5OH, C2 H3CO2 H, CO2
Specifications: 4 (3 independent)
n InH2O = 170.4 or n CIn6H12O6 = 2.332 (one is independent, the sum is F in mol)
n Out
C6 H12 O6 =
90
= 0.500
180.1
n Out
CO2 =
120
= 2.727
44
The degrees of freedom are zero.
Step 8
The material balance equations, after introducing the known values for the variables are:
H 2O :
n Out
H2O = 170.9 + (0)ξ1 + (2)
C2 H12O6 :
0.500 = 2.332 + (−1)ξ1 + (−1)ξ2
(b)
C2 H5OH :
n Out
ξ
C2 H5OH 0 + 2ξ1 + (0)
(c)
2
C2 H3CO2 H : n Out
ξ
C2 H3CO2 H + 0 + (0)ξ1 + (2)
CO2
2
2.727 = 0 + (2)ξ1 + (0)ξ
2
2
ξ
(a)
(d)
(e)
If you do not use a computer to solve the equations, the sequence you should use to solve
them would be
(e), plus (b), (a), (c), (d)
Step 9
The solution of Equations (a) – (e) is
ξ1 = 1.364 kg moles reacting
ξ 2 = 0.469 kg moles reacting
5–35
Solutions Chapter 5
Results
Conversion to mass percent
Species
kg moles
MW
kg
mass %
H 2O
171.8
18.01
3094.1
89.0
C2H5OH
2.727
46.05
125.6
3.6
C2H3CO2H
0.939
72.03
67.5
1.9
CO2
2.272
44.0
100.0
2.9
C2H12O5
0.500
180.1
90.1
3477.3
2.6
1.00
Step 10
The total mass of 3477 kg is close enough to 3500 kg of feed to validate the results of the
calculations.
5.3.9
Steps 1, 2, 3 and 4:
SiH4 +O2 → SiO2 +2H2
4PH3 + 5O2 → 2P2O+5 6H2
100% O2
100% SiH4
100% PH
B
A
D
C
3
SiO2
H2
100%
E
P2 O5
Step 5: Basis:
1 g mol PH3
Step 6: No. Unknowns:
E
A, B, D, E, n P 2 O 5
5
Step 7: Element balances:
Si, H, P, O, ωEP2O5
5
Note: n EP 2 O 5 + n ESiO2 = E
Steps 8 and 9:
5–36
Solutions Chapter 5
P Bal (in grams):
in
Out
⎫
1 g mol PH3 2 g mol P2O5 2 g mol P
31gP
= 31g P ⎪
4 g mol PH3 1 g mol P2O5 1 mol P
⎪
⎬ in P2O5
1 g mol PH3 2 g mol P2O5 5 g mol O
16 g O
⎪
= 40 g O ⎪
4 g mol PH3 1 g mol P2O5 1 g mol O
⎭
Total:
or since all the P is in stream E:
Si Bal:
________
71 g P2O5
1 g mol PH3 2 g mol P2O5 142 g P2O5
= 71g P2O5
4 g mol PH3 1 g mol P2O5
in
Out
B g mol SiH4 1g mol SiO2 1g molSi 28.1g Si
= 28.1B g Si
1g mol SiH4 1g mol SiO2 1g mol Si
B g mol SiH4 1g mol SiO2 2 g mol O 16 g O
= 32 B g O
1g mol SiH4 1g mol SiO2 1g mol O
Total:
71
= 0. 05
71 + 60.1B
1 g mol PH 3
3.55 + 3.0058B = 71
34 g PH 3
= 34 g PH 3
1g mol PH 3
23.5
. g mol SiO2 32.1g SiH 4
= 7.54 g SiO2 O2
1g mol SiH4
754
34
= 22.2
gSiO 2
gPH
5–37
60.1 B g SiO2
B = 23.5 g mol
Solutions Chapter 5
5.3.10
Steps 1, 2, 3 and 4:
Cu
Cu(NH 3 ) 4 Cl 2
NH 4 OH
A
B
C
D
E
Cu(NH 3 ) 4 Cl
H2 O
F
Cu
Step 5:
Basis: 1 board
3
4in 8in 0.03in (2.54cm) 8.96g l gmol Cu
= 2.219g mol Cu
in 3
cm 3 63.55g Cu
A:
Cu foil
F:
Cu remaining
1 − 0.75 2.219 gmol Cu
Steps 6 and 7: Unknowns:
= 0.555g mol Cu
B, C, D, E
Balances: Cu, N, Cl, H, O should be enough if one is redundant
Steps 8 and 9:
Cu Balance:
2.219g mol Cu+
(Balances to be solved:)
Bg mol Cu(NH3 )4 Cl 2
= 1.664 gmol Cu +
D gmol Cu (NH3 ) 4 Cl
1gmol Cu
1g mol Cu(NH3 )4 Cl 2
1gmol Cu
1gmol Cu (NH3 )4 Cl
5–38
Solutions Chapter 5
Cl Balance:
Bg mol Cu(NH3 )4 Cl 2
D g mol Cu(NH3 )4 Cl
2 gmol Cl
1gmol Cl
=
gmol Cu(NH3 )4 Cl 2
gmol Cu(NH3 ) 4 Cl
N Balance:
Bg mol Cu(NH3 )4 Cl 2
=
D gmol Cu(NH3 ) 4 Cl
Solution:
Cu overall:
Cg mol NH4 OH 1gmol N
4 gmol N
+
gmol Cu(NH3 )4 Cl 2
gmol NH4 OH
4gmol N
g mol Cu(NH4 )Cl
A + B = F + D →A + B = F + (2B)
2.219 + B = 0.555 + (2B)
B = 2.219 - 0.555 = 1.664 g mol Cu(NH3)4Cl2
Cl:
D = 2B= 2(1.664) = 3.328 g mol Cu(NH3)4Cl
N:
4B + C = 4D
C = 4(D-B) = 4(3.328 - 1.664) = 6.656 g mol NH4OH
MW of NH4OH
N
14
H4
4
O
16
H
1
35
Cu(NH3)4CL2
Cu
N
H
Cl2
35gNH4 OH 6.656gmol
= 232.0 g NH4 OH
gmol
63.55
56.00
12.00
70.90
202.45
202.45 (g/gmol) (1.664 g mol) = 336.88 g Ca ( NH 3 ) 4 Cl 2
5–39
Solutions Chapter 5
5.3.11
F
Feed
Mixer
contaminated
sand
( 16,000 lb )
F'
Incinerator
60%
combustible
sand plus
combustion
products
H
O2 enriched air
x O 2 = 0.4
100% excess
Basis: 8 tons of contaminated sand containing 30% PCB
hexane
a)
4, 4' dichlorobiphenyl ⇒
Cl
Cl
C12H8Cl2, MW = 222
8tons sand + PCB 0.3ton PCB 2000lb PCB
= 4800 lbPCB
tonsand + PCB ton PCB
If we want the net feed to be 60% combustible, we see that the sand is a tie component:
A sand balance around the feed mixer: 16000(0.7) = F'(0.4) so F' =28,000 lb.
A total balance around the feed mixer.: F + H = F' ⇒ H = 12, 000lb m hexane
b)
C1 2H8Cl 2 + 13.5O2 → 12CO2 + 2HCl + 3H 2 O
C 6 H1 4 + 9.5O2 → 6CO2 + 7H2 O
Even with total combustion, we will produce undesirable HCl, which will adversely
affect the environment if it is simply vented. Therefore, it would be wise to condense
the HCl and H2O by cooling. Perhaps the acid solution might be sold to help pay for the
process.
(c)
21.62 moles PCB
P
139.5 moles nC 6
16.54% CO2
12.20% O2
H2 O = ?
HCl = ?
N2 = ?
A
enriched air
(40% O2, 60% N 2 )
5–40
Solutions Chapter 5
By a tie component for CO2 , we know:
12(21.62) + 139.5 (6) = 0.1654 P hence P = 6629 mol
By an O tie: 0.40(2)A = 0.1654(2)(6629) + 0.1220(2)(6629) + x H 2 O (1) (6629)
x H 2 O (6629) =
21.62 PCBReacted 3H2 O 139.6 C6 Reacted 7 H2 O
=
= 1042
1 PCB
1 C6
so x H 2 O = 0.1572
From O tie balance: A = 6065.4 mol of 40% O2 air
Thus: (N2) in = (N2) out = 0.6(6065.4)=3639.2 mol N2
xN2 =
3639.2
= 0.549
6629.0
We can get HCl by a Cl balance:
21.6PCBReacted 2 HCl
= 43.2mol H Clin product
1PCB
xHCl= 0.0065
Component
CO2
O2
H 2O
HCl
N2
xi
0.1654
0.1220
0.1572
0.0065
0.5490
We know % excess O2 used is : % excess =
100 [O2 fed − O2 req' d ]
(O2 req' d)
O2 fed = 0.4 (air fed) = 0.4 (6065.4) = 2426.2 mol O2
Also, since we have total combustion,
(O2 req'd) = (O2 consumed) = (O2 fed - O2 out) = 2426.2 - 0.1220(6629)
= 1617.5 moles req'd
⎛ 2426.2 − 1617.5⎞
50%excessO 2 used
Thus: % O2 = 100 ⎝
⎠ =
1617.5
5–41
Solutions Chapter 5
5.3.12 & 5.4.1
Steps 1, 2, 3, and 4: To solve this problem you must recall that CaCO3 reacts with
H2SO4 to form CaSO4 and that CaCO3 when heated yields CO2 and CaO whereas when
CaSO4 is heated at the same time it does not decompose. Although the problem seems to
be underspecified, when you draw a diagram of the process and place the known data on
it, the situation becomes clearer. Assume the process is a steady state open one with
reaction. Then the element balances are just in = out. If you use the extent of reaction,
you make species balances.
Note that we have designated the composition of P and A in terms of the mass of
each component, m Pi , rather than the mass fraction ω Pi , because this choice makes the
element or species balances linear (avoids products such as ωP).
The CO2 liberated from the sludge is equivalent to
1lbCO2 1lb mol CO2 1lb mol CaCO 3 100.09lb CaCO3
10 lbP 44lb CO2 1lb mol CO2 1lb mol CaCO3
= 0.227 lb CaCO3/lb P
A (lb) waste acid
H2 SO4
H 2O
lb
m H SO
A
2
m
Y (lb mol)
4
A
H 2O
CO2 gas
1.000 = mol fr.
∑ =A lb
Feed F (lb)
Reaction
Vessel
P (lb)
dry
sludge
mass fr.
CaCO 3
inert
0.95
0.05
1.00
Water
W (lb)
H2 O mass fr. = 1.00
Step 5:
lb
CaCO 3
m PCaCO
CaSO 4
m CaSO
inert
m inert
3
P
4
P
∑ = P (lb)
also
m PCaCO
P
3
= 0.227
Basis: F = 100 lb
A
Step 6:
The variables whose values are unknown are A, m AH 2 SO 4 ,m H 2 O ,m PCaCO3
m PCaSO 4 , m Pinert , P, W, and Y, a total of 9.
5–42
Solutions Chapter 5
Step 7:
The element balances that can be made are Ca, C, O, S, H, inert (6 total)
and we know Σ m Ai = A , Σ m Pi = P, and m PCaCO3 / P = 0.227 for a total of 9. If the
equations are independent, we can find a unique solution.
(
)
Steps 8 and 9: The equations (in = out) are (except for the inert balance which is in lb)
in moles
1 lb mo lCa
95 lb CaCO3 1 lb mol CaCO3
100.09 lb CaCO3 1lb mol CaCO3
Ca:
m PCaCO3 lbCaCO3 1lb mol CaCO3
1lb mol Ca
=
100.09lb CaCO3 1lb mol CaCO3
m PCaCO3 lbCaSO4 1lb mol CaSO4
1lb mol Ca
+
136.15lb CaSO4 1lb mo l CaSO4
1 lb mol C
0.95 lb CaCO 3 1 lb mol CaCO3
Y lb mol CO2 1 lb mol C
=
100.09 lb CaCO3 1lb mol CaCO 3
1 lb mol CO2
C:
m PCaCO3 lbCaCO3 1 lb mol CaCO3
1 lb mol C
+
100.09 lb CaCO3 1 lb mol CaCO 3
A
m H 2 SO 4 lb H2 SO4 1 lb mol H2 SO4
1 lb molS
S:
98.08 lb H2 SO4 1 lb mol H2 SO4
m PCaSO4 lb CaSO4 1lb mol CaSO4
1lb molS
=
136.25lbCaSO4 1lb mol CaSO4
m AH 2 SO 4 lb H2 SO 4 1 lb mol H2 SO 4 1 lb mol H 2
2H :
98. 08 lb H 2 SO4 1 lb mol H2 SO4
+
m AH 2 O lb H2 O 1 lb mol H2 O 1 lb mol H 2
18. 016 lb H2 O 1 lb mol H 2 O
=
W lb H2 O 1 lb mo l H2 O 1 lb mol H2
18. 016 lb H2 O 1 lb mol H2 O
O:
+
95 lb mol CaCO 3 1 lb mo l CaCO3
3 lb mol O
100.09 lb CaCO3 1 lb mol CaCO 3
m AH 2 SO4 lb H2 SO4 1 lb mol H2 SO4
4 lb mol O
98.08 lb H2 SO4 1 lb mol H2 SO4
5–43
Solutions Chapter 5
A
m lb H2 O 1lb mol H2 O 1lb mol O
+ H 2O
18.016lb H2 O 1lb mol H 2 O
=
Y lb mol CO2 2lb mol O
1mol CO2
+
m PCaCO3 lbCaCO3 1lb mol CaCO3
3lb mol O
100.09lb CaCO3 1lb mol CaCO 3
m PCaSO4 lb CaSO4 1lb mol CaCO 3
4lb mol O
+
136.15CaCO 4 1lb mol CaSO4
+
W lb H2 O lb mol H 2 O 1lb mol O
18.016lb H2 O l lb mol H 2O
Inert: 5 lb inert = m Pinert lb inert
The solution of these equations using a computer code would give all of the values of the
unknown quantities. However, a review of the set of equations, after having gone into
great detail about each equation, indicates that only 4 of the equations have to be solved
to get the desired answer, namely the Ca balance, the Σm Pi = P, and the inert balance plus
m PCaCO3 = 0.227 :
Σm Pi = P:
P
m PCaCO3 + m PCaSO4 + 5 = P = m CaCO
/ 0.227
3
Ca:
P
P
0.949 = 0.010 m CaCO3 + 0.00734 m CaSO4
m PCaCO3 = 28.55 lb and 100
28.55
= 30%unreacted .
95
5–44
Solutions Chapter 5
5.5.1
Basis: 100 kg mol
synthesis
gas
Burner
flue gas
air
One composition is unknown, the flue gas. After selection of a basis, the air flow is
known. Hence, this problem can be solved by direct addition and subtraction
Unknowns:
(CO2, H2O, N2,O2) = 4
Balances:
(C, H, O, N) = 4
Comp.
%=mol
CO2
O2
CO
H2
N2
6.4
-------------0.2
----------(0.2)
40.0 CO+1/2 O2→CO2
20.0
50.8 H2+1/2 O2→H2O
25.4
2.6
-------------XS O2: (0.40)(45.2) =
18.08
Total O2 in
63.28
N2 in with O2 (63.28)(79/21) =
Totals
Comp.
mol
Reaction
%
CO2
46.4
13.0
H 2O
50.8
14.3
240.65
67.6
N2
O2
Totals
reqd.O2
18.1
5.1
356.0
100.0
5–45
CO2
H2 O
N2
O2
6.4
40.0
50.8
46.4
50.8
2.6
238.05
240.65
18.08
18.1
Solutions Chapter 5
5.5.2
Steps 1, 2, 3, and 4:
H2 O
W
C
H
mass fr.
0.80
0.20
F
30 lb
mol
n CO
n CO2
nO2
coal
A
1.00
O2
N2
nN2
600 lb
mol fr.
0.21
0.79
∑ =F
1.00
Steps 5 and 6: With a basis of 30 lb coal, only n N 2 ,n O2 ,n CO2 ,n CO ,F and W are unknown.
Total 6
Step 7: Can make C, H, O, N element balances.
plus
n CO2
nCO
=
3
and ∑ n i = F
2
Do calculations in moles, but you don't need to make all of the element balances
Comp.
C
2H
lb
mol wt
lb mol
24
12
2
6
2
3
30
3.5lb mol O2 1.00lb mole air
= 16.67lb mol air reqd.
0.21lb mol O2
or
4.35− 3.5
(100)
3.5
= 24.3%
5–46
Rxn
C+O2→CO2
H2+1/2O2→H2O
reqd O2
2.0
1.5
3.5
Solutions Chapter 5
5.5.3
Steps 1,2 3 and 4:
H2 O 100 %
mol %
W
P
mol frac F (mol) = ?
CH 4 x
N2
y
A
1.00
O2
N2
CO
1.0
C O2
8.7
O2
3.8
N2
86.5
air
100.0
mol frac
0.21
0.79
1.00
Step 5: Basis: 100 mol P
Step 6: Unknowns:
x, y, A, W, F Total 5
Step 7: Element Balances:
Equations:
C, H, N, O
a)
If x and y are moles, not mole fractions,
add: x + y = F
b)
If x, y are mole fr., add x + y = 1.00
Steps 8 and 9: Use x and y as mol:
C balance:
x = 8.7 + 1.0 = 9.7
2N balance:
y + .79A = 86.5
2H balance:
2x = W = 2(9.7) = 19.4
2O balance:
0.21A = (1/2) (W + 8.7) + 1.0 + 3.8
Use 2O bal:
A=
A and P will be mol
9.7 + 8.7 + 0.5 + 3.8
= 108
0.21
5–47
Total 5
Solutions Chapter 5
Use 2N bal:
y = 1.10
required O2: CH4 +2O2→ CO2+2H2O
(1)
(9.7)(2) = 19.4 mol
total O2 in: (0.21) (108) = 22.7
% xs air = % xs O2 =
3.3
(100)
19.4
= 17%
(a)
(b)
excess O2= 22.7 - 19.4 = 3.3 mol
Percentage of CH4, N2: CH4 =
10.2%
9.7
(100) = 89.8%
9.7 + 1.1
N2 =
Alternate solution:
Use extent of reaction and species balances.
Another equation needed for CO:
(2)
CH 4 + 1.5O2 → CO + 2H 2 O
Species balances to get ξ1 and ξ 2 :
F
CH 4 : O = n CH
+ ( −1)+ξ1 − ( 1)ξ 2
4
CO2 : 8.7 = 0 + (1)ξ1
CO: 1.0 = 0 + (1)ξ 2
ξ1 = 8.7 reacting moles
ξ 2 = 1.0 reacting moles
F
Thus n CH
= 8.7 + 1.0 = 9.7 mol and other compounds follow.
4
5–48
1.1
(100)=
9.7+1.1
Solutions Chapter 5
5.5.4
Steps 1, 2, 3 and 4:
H2 O
100%
W
CH4 100%
P
F
A
CO2 20.2
O2
4.1
N2 75.7
100.0
enriched O2 40
N2 60
air
100
Step 5:
Basis 100 mol P
Step 6:
Unknowns: F, A, W (all compositions known)
Step 7:
Element balances: C, H, O, N (1 extra balance)
Steps 8 and 9:
C balance: 20.2 = F
2N balance: 75.7 = 0.6A, A = 126.2
2H balance: 2F = W, W = 40.4
2O balance: 0.4A = 1/2 W + 20.2 + 4.1, 50.5 ≠ 20.2 + 20.2 + 4.1 = 44.5
Analysis isnot correct.
5–49
Solutions Chapter 5
5.5.5
Steps 2, 3, and 4:
H2 O 100%
W
F
mass frac
C
0.90
H
0.06
inert
0.04
P
Furnace
A
1.00
O2
N2
mole %
R
mol frac
0.21
0.79
mass frac
0.10
C
inert 0.90
1.00
Step 5: Basis: 100 kg mol P
13.9
C
13.9
CO
0.8
0.8
O2
4.3
N2
81.0
CO2
100.0
14.7
1.00
Steps 6 and 7:
Unknowns: F, R, A, W
4
Probably ok but check for redundancy
Balances:
C, H, O, inert, N2
5
Steps 8 and 9:
Balances (in moles):
C:
H:
F(0.90)
R(0.10)
= P(0.139+0.008) +
+ W(0)
12
12
F (0.06)
= P (0) + R(0) + W (2)
1.005
2N:
A (0.79) = P(0.81) + R (0) + W (0)
2O:
A (0.21) = P (0.139 +
0.008
W
+ 0.043) +
2
2
Inert (mass): F (0.04) = R (0.90)
5–50
Solutions Chapter 5
From 2O,
100(.81)
= 102.53kgmol
0.79
W = 5.863 kg mol
From H,
F = 196.98 kg
From inert,
R = 8.7548 kg
A=
From 2N,
196.98(0.06)
= 5.94
2
5.94
O2 in:
= 2.96
2
196.98(0.90)
C in =
= 14.77
12
H in:
Step 10:
Check via C: 196.98 (0.90) /12 = 100 (0.147) + (8.7548 (0.10)/12)
ok
Required O2: 14.7 + (5.863/2) = 17.63 kg mol Total O2in = 102.53 (0.21) = 21.53
Excess (21.53 - 17.63)/17.63 = 0.221 or 22%
Alternate solution using extent of reaction:
C+O2 → CO2
1
C+ O2 → CO
(1)
(2)
2
1
2H+ O 2 → H 2 O
(3)
2
CO 2 :
P
F
n CO
=13.9 = n CO
+(1) ξ1
2
2
ξ1 = 13.9
CO:
P
F
nCO
=0.8 = nCO
+(1) ξ2
ξ 2 = 0.8
N2 :
n PN2 = 81.0 = n AN2
A = 102.5 mol
A
O2
n = 0.21(102.5) = 21.5 mol
O2:
1
1
2
2
4.3 = 21.5 + (-1) ( ξ1 ) + (- )( ξ 2 ) + (- ) ( ξ 3 )
C:
0 = nCF + (-1)(13.9) + (-1)(0.8)
H:
0 = nFH + (-2) (2.90)
nCF = 14.7 mol
nHF = 5.80 mol (some round off)
5–51
ξ 3 = 2.90
Solutions Chapter 5
5.5.6
Basis: 100 mol F
P
F = 100 mol
Moles = %
CH4 : 60
C 2 H6 : 20
CO:
5
O2 :
5
N 2:
A
0.21 O2
0.79 N
1.00 2
10
100
50% xs air:
CH4 + 2O2
n CO2
Moles
281.25
1058
or
x CO2
n H 2O
x H 2O
n
O2
xO2
nN2
x n2
∑ =P
∑ =1.00
Calculate O2 and N2 in with the air in
reqd O2 (mols)
60 (2) = 120
→ CO2 + 2H2 O
C 2 H6 + 312 O2 → 2CO2 + 3H 2O
20(3.5) = 70
2.5
⎛1
5⎝ ⎞⎠ =
192.5
2
1
CO + O 2 → CO 2
2
Less O2 in gas
5.00
Net required O2
xs O2: 187.5 (.5) =
Total
N2 in with O2
187.5
93.75
281.25
281.25 0.79
= 1058
0.21
Material balances (elemental):
in (F)
+
in (A)
=
out (P)
C:
60 + 2 (20) +5 +
0
=
n CO2
H:
60 (4) + 20 (6) +
0
=
n H 2 O (2)
5
+5
2
+
281.25 =
O as O2:
5–52
n CO2 +
n H 2O
2
+ nO 2
Solutions Chapter 5
N as N2:
10
+
n O 2 = 288.75 - 105 -
nN2
=
180
= 93.75
2
moles
composition of stack in % moles
n CO2 = 105
7.26
12.44
73.82
6.48
n H 2 O = 180
n N 2 = 1068
n O 2 = 93.75
Total moles =
out
1058
1446.75
100.00
5.5.7
Steps 1, 2, 3, 4:
F = 100 (basis)
mol = %
CO 13.54
CO2 15.22
H2 15.01
CH4 3.20
N2 53.03
100.00
P
CO2
O2
N2
H2O
O2 reqd
6.77
-7.505
6.40
--20.675
A
mol fr.
mol fr.
x
y
z
1-x-y-z
1.00
0.143
0.037
0.724
0.096
1.00
0.21 O2
0.79 N2
1.00
Step 5: Basis: 100 mol F
Step 6: Unknowns: P, x, y, z
Step 7: Balances
C, H, O, N
ok
Steps 8 and 9:
Air in is based on complete combustion
CO +
1
2
O2 → CO2 and CH 4 + 2O2 → CO2 +2H 2 O
5–53
mol
calculated
31.96
8.27
161.92
21.41
Solutions Chapter 5
xs O2 = 0.40 (20.675) = 8.27
reqd O2 =
20.675
total O2
28.945
N2 in with air
28.945
28.945
⎛ .79 ⎞
⎜
⎟ = 108.89 mol
⎝ .21 ⎠
Air = 137.84 mol
Balance
2N
C
2H
N2 out = 108.89 + 53.03 = 161.92 mol
CO2 out = 13.54 + 15.22 + 3.20 = 31.96 mol
H2O out = 15.01 + 2(3.20) = 21.41 mol
2O
O2 out = 28.945 +
(as O2)
13.54
21.41
+ 15.22 − 31.96 −
= 8.27 mol
2
2
Alternate solution: Use the extent of reaction and species balances
5.5.8
Steps 2, 3, and 4:
7.9%N2
21%O2
Step 5:
Step 4:
A
F = 100 mol
P
CS2 40.0
SO2 10.0
H2O 50.0
xCO
2
xSO
2
xO
2
xN
W
100% H2O
Basis: 100 mols feed; assume complete combustion
Equations:
CS2 + 3O2 → CO2 + 2SO2
O2 required:
% excess:
40 mol CS2 3 mol O2
= 120.0 mol O2
1
1 mol CS2
O 2 excess
× 100 =% excess
O 2 required
Steps 6, 7, 8 and 9:
5 unknowns and 5 equations → unique solution
5–54
2
Solutions Chapter 5
Element balances
Component
C
Mol in
40.0
Mol out
Px CO
2
2S
1
2O
10.0 + (50.0) + .21A
2
1
(10.0) + 40.0
1
2N
2H
.79A
50.0
(0.02)P
2
1
(x C O2 + 0.02 + x O2 )P+ W
2
2
(1-x O2 − 0.02 − x CO2 )P
W
2(45.0)
= 4500
.02
x CO2 = 0.0089
P=
% Excess =
0.1829(4500 molO2 )
(100) = 686%
120 mol
Alternate solution: Use extent of reaction and species balances.
5.5.9
Steps 2, 3 and 4:
W
Fuel
F
mass fr.
C 0.74
H 0.14
ash 0.12
1.00
Flue gas
A
air O2 0.21
N2 0.79
1.00
Step 5:
H2O 1.00
R
ash
1.00
Basis: P = 100 mol
Step 6: Unknowns:
F,A,R,W ⎫⎪ a discrepancy unless one
⎬
Step 7: Balances: C,H,O,N,ash ⎪⎭ balance is not independent
Steps 8 and 9:
all balances in moles
5–55
P
CO2
CO
O2
N2
mol
12.4
1.2
5.7
80.7
100.
Solutions Chapter 5
C balance:
F(0.74)
= 12.4 + 1.2 = 13.6
12
F=
(13.6)(12)
= 220.5 lb
0.74
2N balance:
A(0.79) = 80.7
A = 102.2 mol
2H balance:
F(0.14)
= W;
2
W = 15.44 mol
2O balance:
A (.21) = W
(1.00)
1.2
+ 12.4 +
+ 5.7
2
2
21.46 = 26.42
not equal hence
Some discrepancy exists
A total balance can be used, but must be in mass, and the mass of P calculated.
5.5.10
Basis: 100 moles exhaust gas
Comp
CO2
O2
N2
Total
exhaust gas mol
16.2
tie element
exit gas
%
13.1
4.8 ⎫
⎬
79.0⎭ 83.8
100.0
16.2 mol CO2
100.0 mol exit gas
= 123.6mol exit gas
100.0 mol exhaust gas 13.1mol CO2
exhaust gas = 100.0 mol
air leaked in = 23.6 mol
23.6
= 0.236 mol air / mol exhaust gas
100.0
5–56
mol
16. 2
Solutions Chapter 5
5.5.11
Steps 2, 3, and 4:
H2 O 100%
W mol
Sludge
S mass
mass frac
C
S
H2
O2
0.40
0.32
0.04
0.24
mol
1.00
O2
N2
mole %
P mol
Furnace
A
F mass
mol frac
Fuel oil lb
0.21
0.79
mC
mH
1.00
Step 5:
Basis: 100 mol P
CO
2.02
CO2
10.14
O2
4.65
N2
81.67
SO 2
1.52
100.00
∑ F (lb)
Steps 6 and 7:
Unknowns:
S, W, F (or mH),A, mC
5 Element Balances: S, C, H, O, N
Element Balances (moles)
S:
2N:
C:
2O:
0.32S
=
32
1.52
A(.79) = 81.67
S(0.40) m C
+
=10.14+2.02
12
12
S = 152 lb
A= 103.38 lb mol (2998.01 lb)
mC = 85.12 lb
W
2.02
0.24(152)
+ A (0.21) =
+ 1.52 + 10.14 + 4.65 +
2
2
32
W = 11.04 lb mol (198.71 lb)
5–57
Solutions Chapter 5
0.04(152) m H
+
=W
2.03
2.03
2H:
mH = 16.32 lb
Check via a total mass balance
152 + 16.32 + 85.12 + 2998.01
3251.45
?
=
?
=
198.71 + 3035.56
3234.27
close enough
lb
85.12
16.32
101.44
C
H
152 lb
S
=
=1.50
F 101.44 lb
5.5.12
mass fr
0.84
0.16
1.00
B
Basis: 1 min
[CaO]
=
theor.
ft 3
1000 gal
min
1.05×62.4 lb soln. 0.02lb H 2SO4 mol H 2SO4
7.48 gal
56.1 lb CaO
mol CaO
ft
3
lb soln.
mol CaO
98 lb H 2SO4 mol H 2SO4
=100.3 lb CaO/min
Feed rate of CaO = 1.2 [CaO ]theor . = 120.4 lb CaO/min
CaSO4 production rate =
×
100.3 lb CaO
min
60 min 24 hr 365 d
hr
d
yr
mol CaO
mol CaSO4 136.1 lb CaSO4
56.1 lb CaO mol CaO
= 63,947 ton/yr
5–58
mol CaSO4
ton
2000 lb
Solutions Chapter 5
5.5.13
Basis: 100 mol of product gas
NH3 + 2O2 → HNO3 + H2O
The product gas from such a reactor has the following composition (water free basis):
0.8% NH3
9.5% HNO3
3.8% O2
85.9% N2
Determine the percent conversion of NH3 and the percent excess air used.
in
n out
NH3 = n NH3 +
NH3
(9.5)
υ
0.8 = n inNH3 + ( 1)(9.5)
−
F = ninNH3 = 0.8 + 9.5 = 10.3 mol
n out − n in
=ξ
υ
For HNO3 :
9.5 − 0
= 9.5 =
1
ξ
Percent conversion f = (100%) ξ /F = 92.2%
Calculate the entering oxygen using a N2 balance:
N2: 0.79A = 85.9; therefore, A = 108.73 mol
Calculate the theoretical O2 needed for the reaction
(O2)theor = 2 × 10.3 = 20.6 mol
Feed rate of O2 = 0.21A = 22.83 mol
% Excess O2 = [(22.83-20.6)/20.6](100) = 10.84%
5–59
Solutions Chapter 5
5.5.14
Basis: 100 mol of product gas (20.5%, C2H4O; 72.7 N2; 2.3 O2; and 4.5%CO2
C2 H 4 +
1
O2 → C2 H 4O
2
C2H4 + 3O2 → 2CO2 + 2H2O
Calculate ξ1 :
n iout − n ini
ξ1 =
υi
Use C2H4O:
Calculate ξ2 :
Use CO2:
ξ2 =
ξ1 =
20.5 − 0
= 20.5
1
4.5 − 0
= 2.25
2
Calculate the entering air using a N2 balance:
N2: 0.79A = 72.7; therefore, A = 92.03 mol
Calculate the moles of C2H4 entering (F)
in
n out
CH4 − n CH4 = υ1ξ1 + υ2 ξ2
0 − F = −ξ1 − ξ2 hence F = 22.75
O2 feed rate = 0.21A = 19.33 mol
(O2)theor = F/2 = 11.375 mol
% Excess O2 = [(19.33-11.375)/11.375](100) = 70.0%
To get the ethylene feed:
22.75 C2 H4 100,000 ton C2 H4O 2000 lb C2 H4O mol C2 H 4O
20.5 mol C2 H4O
yr
ton C2 H4O 44 lb C2 H 4O
×
yr
d 28 lb C2 H 4
= 16,123 lb C2 H 4 /h
365 d 24 h mol C2 H 4
5–60
Solutions Chapter 5
5.5.15
mole fr
mol fr
CH4
0.70
C3 H8
0.05
CO
0.15
O2
0.05
N2
0.05
Air
1.00
mol frs
0.21
0.79
Fuel
Flue gas
Flare
F mol
0.0773
H2 O
0.1235
O2
P mol
O2
N2
CO2
N2
A mol
xO2
xN2
1.00
1.00
Unknowns A, F, P, x O 2 , x N 2 , Pick F = 100 as basis
4
Balances: C, H, O, N (is one redundant ?) plus
0.0773+0.1235+ x O 2 + x N 2 = 1
Balances:
C: 100 (0.70) + 100 (0.05) (3) + 100 (0.15)
N2:
100 (0.05) + 0.79A
H2:
100 (0.70) (2) + 100 (0.05) 4
O2:
100 (0.15)
1
2
= P (0.0773)
= P xN2
= P (0.1235) (redundant with C)
+ 100 (.05) + A (0.21)
Σxi
Solve C (or H2) balance for P;
Solve N2, O2 and Σxi = 1
5
0.7992
P = 1294
0.1235
⎡
⎤
= P ⎢(0.0773) +
+ x O2 ⎥
2
⎣
⎦
= xO2 + xN2
simultaneously for A, x O 2 , x N 2
A = 1203 moles
x O 2 = 0.0648 and x N 2 = 0.7344 (not needed)
Calculation of the required O2
Required O2
C H4 + 2O 2 → CO2 + 2H2 O
70(2)
5–61
= 140
Solutions Chapter 5
C3 H8 + 5O2 → CO 2 + 4H 2 O
CO+ 12 O2
5(5)
→CO2
15(
1
2
O2 present in F
A(0.21) =
less
= 25
) = 7.5
-5.0
167.5 reqd
252.54 O2 in
167.5 reqd O2
85.04 xs O2
85.04 (100) = 51 %
167.5
5.5.16
F = 100 kg mol
CO2
N2
C
1.00
O2 0.21
N2 0.79
C + O2 → CO2
Basis: 100 kg mol coke
Step 5:
a.
O2 in = 100 kg mol
100 kg mol C 1 mol O 2 79 mol N 2
= 376 kg mol N 2
1 mol C 21 mol O2
Steps 6 and 7:
Unknowns: moles (ni) CO2, O2, and N2 in P, and P
Balances: elements C, O, N
Equations:
P
i
∑n
=P
or add ξ and have C, CO2, N2, O2 species balances
5–62
in
Solutions Chapter 5
Steps 8 and 9:
Element balances
C:
2O:
2N:
in
100
100
376
out
n CO
n CO +n O
nN
=
=
=
2
2
n O2 = 0
2
2
Component in P
N2
O2
CO2
kg mol
376
0
100
476
mol %
79
0
21
100
With 50% xs air, xsO2, is 50 mol
The total O2 in is 150 kg mol
The N2 in is
150 0.79
= 564
0.21
The N and C balance are the same but the 2O balance is
In
150
out
−
n O2
reacts
=
100
Component in P
N2
O2
CO2
n O2 = 50
kg mol
564
50
100
714
mol %
79
7
14
100
c. Same 150 mol O2 enter and 564 mol N2 exit but O2 is different
in
Oxygen balance:
150
mol CO2 used
–
90
5–63
–
Used CO
5 = 55 kg mol O2 lb fg
Solutions Chapter 5
Component
CO2
CO
O2
N2
Total
kg mol
90
mol%
12.51
10
1.39
55
564
719
7.65
78.45
100.00
5.5.17
Steps 1, 2, 3 and 4:
B
CH3CHO 1.00
P = 100 mol (Basis)
F
CH3CH2OH 1.00
CO2
O2
CO
H2
CH4
N2
A
mol. wt.
CH3CHO
44
CH3CH2OH 46
W H2O
1.00
Air .21 O2
.79 N2
1.00
Step 5: Basis: 100 mol P
Step 6: Unknowns: F, A, B, W
Step 7: Balances:
C, O, H, N
ok
Steps 8 and 9:
Via algebra
2N balance: A(.79) = 85.2
A = 107.85 mol
C balance:
F(2) = B(2) + 5.6
2H balance: F(3) = B(2) + W(1) + 12.3
2O balance:
1
1
1
2
2
2
F( ) + (.21)A = B ( ) + W ( ) + 3.95
5–64
%
0.7
2.1
2.3
7.1
2.6
85.2
100.0
C
0.7
2.3
O2
0.7
2.1
2.3/2
7.1
5.2
2.6
5.6
H2
3.95
12.3
Solutions Chapter 5
Solution:
B = 44.1 mol
F = 46.9 mol
W = 40.2 mol
44.1 mol acet. 44 kg acet.
100 mol P 1 kg mol EtOH
= 0.899
100 mol P 1 kg mol acet. 46.9 mol EtOH
46 kg
kg acetaldehyde
kg ethanol
Alternate solution: Use extent of reaction and species balances, but it is more
complicated.
5.5.18
Basis: F = 100 lb given in Example 10.9
This basis gives P = 50 lb moles. Note: If the problem calculations were redone because
of the stated additional reactions, the value of P = 50 for two significant figures would not
change. Similarly, the other values of the moles in P would not change. Thus, the SO2 + CO2 is
0.154 (50) = 7.7 lb mol. The values of the pertinent components (for NOx use NO1.5) for this
problem are
lb mol MW kg
ELU/kg ELU
NOx: (0.0024) (80.6)
=
0.193
2.2
4.25
0.22
0.93
CO:
(0.0018) (7.7)
=
0.090
28
2.52
0.27
0.68
SO2:
(0.014) (7.7)
=
0.002
64
0.128
0.09
0.012
CO2:
(0.986) (7.7)
=
7.59
48
364
0.10
36.4
Total
the total lb mol of nitrogen (N) entering is 2(50) (0.806) + 0.001 = 80.6
5–65
38.0
Solutions Chapter 5
5.5.19
Assume ammonia and glucose (MW = 180.1) are fed in stoichiometric proportions.
The reaction equation has to be set up and balanced to use the value given of 60% (mole
assumed) conversion of glucose.
a C6 H12O6 + b NH3 → c CH1.8O0.5 N0.2 + d C2 H6O + e CO2 + f H2O
Pick a basis of 4.6 kg of EtOH (C2H6O, MW = 46.07)
4600 g EtOH 1 g mol EtOH
= 100 g mol EtOH produced
46.07 g EtOH
To balance the chemical reaction equation use element balances. Take a basis of a = 1. Assume
60% conversion of the glucose means that 1 mole of glucose produces 0.6 mol of ethanol, not that
one mole of glucose produces 3 mol of ethanol.
C:
H:
O:
N:
6 = c + 2d + e
12 + 3b = 1.8c + 6d + 2f
6 = 0.5c + d + 2e + f
b = 0.2c
Specifications 0.60 = d
Results
a = 1 (the basis)
b = 0.8
c=4
d = 0.6
e = 0.8
f = 1.8
167 g mol C6 H12O .08 g mol NH 3
= 133.6 g mol NH 3
1 g mol C6 H12O
167 g mol C6 H12O6 180.1 g mol C6 H12O6
= 30,076 g or 30.1 kg C6 H12O6
1 g mol C6 H12O6
133.6 g mol NH 3 17.03 g NH 3
= 2,275 g or 2.28 kg NH 3
1 g mol NH 3
5–66
Solutions Chapter 6
6.1.1
6.1.2
2 (The two overall balances on A and B sum up to the overall balance on
the system)
4 (such as 2 component balances for each unit)
6.1.3
Unit I involves A, B, C
Unit II involves D, E
Unit III involves A, B, C, D
Components
3
2
4
Number of independent balances
9
If A and B are always combined in the same ratio, then you have to reduce the
independent balances by 1 for Unit I and 1 for Unit III, a total of 2.
6.1.4
Six independent eqautions if all compositions are known. The sum of the
components (in moles) would equal the total flow, hence not all of the equations
that could be written would be independent, only 2 per subsystem.
Total balances:
Condenser
F2 = F 4 + F5
(F3 does not mix, hence cancels out or can be omitted)
Column
F1 + F5 + F9 = F 2 + F8
Reboiler
F8 =F9 + F7 (F6 does not mix, hence cancels out or can be omitted)
Component balances:
In addition, (2) component balances could be written for each component for each
subsystem cited above. Multiply Fi by the composition xij.
6–1
Solutions Chapter 6
6.1.5
Unknowns:
stream flows including F:
Mass fractions (7 plus 3 in F) 10
Independent material balances:
Species in 1 : 3
2:2
3:3
6
16
8
Other independent equations:
∑ ω i = 1 in each stream including F 8
6.1.6
Basis: F = 100 kg
mass fr.
A 0.60
B 0.20
C 0.20
1.00
m = mass fraction
mass fr.
A 0.10
B 0.85
P2 C 0.05
1.00
P1
(1)
% = kg
A 15
B 30
C 55
100
(2)
F = 100 kg
(a) SEPARATE ANALYSIS
16
W2=20 kg
W1
mass fr.
A m1A
B m1B
C m1C
1.00
Column (1) balances
⎤
= W1 (m1A ) + P1 (0.60) ⎤
⎥ any 3 ⎥
1
(2) B: 0.30(100) = W1 (m B ) + P1 (0.20) ⎥
⎥ 4 indept. eqns.
are indept.⎥
⎥
1
(3) C: 0.55(100) = W1 (m C ) + P1 (0.20) ⎥
eqns. ⎥
⎥
⎥⎦
100 = W1
+ P1
⎥
(4)
m1A + m1B + m1C = 1
⎥
⎦
(1) A: 015(100)
Unknowns: W1 ,P1 ,m1A ,m1B ,m1C
5 – 4 = 1 degree of freedom
5 unknowns
6–2
A 0.002
B m2B
C m2C
1.00
Solutions Chapter 6
Column (2) balances
(5) A: m1A W1
⎤
= 0.10 P2 + 0.002 (20) ⎤
⎥ any 3 ⎥
2
(6) B: m W1 = 0.85P2 + m B (20)
⎥ are indept.⎥ 5 indept. eqns.
⎥
⎥
1
2
(7) C: m C W1 = 0.05P2 + m C (20)
⎥ eqns. ⎥
⎥
⎥⎦
W1 = P2 + 20
⎥
m1A + m1B + m1C = 1
⎥
⎦
2
2
0.002 + m B + mC = 1
1
B
Unknowns: W1 , P2 , m1A , m1B , m1C , m 2B , m C2 > unknowns
7 – 5 = 2 degrees of freedom
JOINT ANALYSIS
Adding the 2 units to make a joint system
No. of
Unknowns
• From the separate calculations start with:
• Delete the 4 common variables
W1 , m1A , m1B , m1C
that are treated as unknowns
in both subsystems leaving:
• Delete the equation
m1A + m1B + m1C
that will no longer be independent in
the joint system (it appears in both
subsystems) once leaving:
No. of
Degrees
indept. Eqns. of freedom
12
9
3
8
9
-1
8
8
0
ANALYSIS OF OVERALL SYSTEM
System overall balances
(1) A: 100 (0.15)
= P1 (0.60) + P2 (0.10) + 20 (0.002) ⎤
⎥
(2) B: 100 (0.30) = P1 (0.20) + P2 (0.85) + 20 (m 2B ) ⎥
4 indept.
(3) C: 100 (0.55) = P1 (0.20) + P2 (0.05) + 20 (mC2 ) ⎥
⎥ eqns.
⎥
⎥
0.002 + m2B + mC2 = 1
⎥⎦
2
2
Unknowns:
4 unknowns
P1 , P2 , m B , m C
4 - 4 = 0 degrees of freedom as expected
6–3
Solutions Chapter 6
6.2.1
The results show that the system of equations is very sensitive to small perturbations in
the coefficients (measurements).
6–4
Solutions Chapter 6
6.2.2
Basis: 1 hr (1000 kg)
a. Overall balances:
Total: 1000 = W + D
Salt: 1000 (0.0345) = 0 + 0.069D
W= 500 kg
D = 500 kg
b. Salt balance on the freezer:
1000 (0.0345) = 0 + 0.048B
B = 718.75 kg
Total balance on the freezer:
1000 = A + B
A = 281.25 kg
Total balance around filter:
B = 718.75 = C + D
6.2.3
C =218.75 kg
Basis: 220g IgG from the reactor
Fraction recovered =
140
= 0.64
220
6–5
Solutions Chapter 6
6.2.4
Two subsystems exist, hence 4 component balances can be written. No reaction
occurs and the process is assumed to be in the steady state. Steps are omitted here. The
balances are
System : Splitter
System : Stock Chest
Total: R = E + P
Fiber : 2.34 = xE+xP
Water: 7452 = 4161+3291
Total: P +N = L
Fiber: xP+xN =103.26
Water: 3291 + 18 = 3309
(a)
Also N(0.15) = 18 N = 120
0.85 (N) =xN = 0.85 (120)
Overall balances
Total: R+ N = E + L
Fiber: 2.34 + xN = xE + 103.26
Water: 7452 + 18 = 4161 + 3309
(b)
(c)
Substitute (b) into (c), solve (c) for xE , and then solve (a) for xP.
x N =102
x P =1.26
x E =1.08
all in kg
6.2.5
Step 5:
Basis: 100 kg mol = F
Steps 1, 2, 3, and 4:
H2 O 100%
W
F (kg mol)
E (kg mol)
P (kg mol)
Furnace
CH4
H2
C2 H 6
mol frac
0.70
0.20
0.10
1.00
Duct
mole frac.
mol
-
n ECO
CO 2
A (kg mol)
O2
N2
mol frac
0.21
0.79
O2
H2 O
N2
0.02
n
mole frac.
Air ?
2
E
O2
-
n HE O
-
nN
1.00
E
B (kg mol)
2
E
1.00
2
O2
N2
mol frac
0.21
0.79
CO 2
-
mol
n PCO
O2
0.06
n
N2
-
nPN
1.00
P
2
P
O2
2
1.00
Assume an air leak occurs first and use material balances to calculate the amount. This is
a steady state flow process without reaction.
6–6
Solutions Chapter 6
Overall system
Step 6:
Unknowns are:
A, B, W, P, x N 2 (or n N2 ), x CO2 (or n CO2 )
Step 7:
Balances are:
C,H, O, N, Σ x Pi = 1 (o r ∑ n Pi = P)
Steps 8 and 9:
In
C(kg mol):
Out
0.70(100) + 0.10 (100) 2
=
n CO2
=
W (2)
n CO2 = 90 kg mol
H (kg mol):
0.70 (100)(4) + 0.20 (100)(2)
+ 0.10 (100)(6)
W
=
190 kg mol
190(1.00)
+ 90 + 0.06P
2
2O (kg mol): 0.21A + 0.21B
=
2N (kg mol): 0.79A + 0.79B
=
n PN 2
=
P
90 + 0.06 P + n PN 2
A balance about the furnace or about the duct is needed; or these two alone would have
been sufficient, omitting the overall balance.
System: Furnace (or Duct)
E
Step 6: Unknowns:
A, n ECO2 ,n EO 2 , nH 2 O , n EN2
Step 7: Balances:
C, H, O, N2, ∑ nEi = E
Steps 8 and 9:
C (kg mol):
H (kg mol):
0.70 (100) + 0.10 (100)(2)
n ECO2 = 90 kg mol
=n ECO2
E
0.70 (100)(4) + 0.20 (100)(2) + 0.10 (100)(6) = n H 2 O
E
n H 2 O = 190
190
2O (kg mol): 0.21A
= n EO 2 + 90 +
2
6–7
Solutions Chapter 6
E
= nN2
2N (kg mol): 0.79A
0.02
= n EO 2 / E
90 + 190 + n EO 2 + n EN 2 = E
Solution:
kg mol in E
n ECO2
kg mol
90
A
982
n H 2O
E
190
E
1077
n EO 2
21.5
n EN 2
776
1077.5
From the overall N2 balance:
O2 balance:
0.79 (982) + 0.79B = n PN 2
(1)
0.21 (982) + 0.21B = 185 + 0.06P
(2)
(1)
22.2 + 0.21B = 0.06P
B = 207 kg mol
(3)
90 + (0.79)(982) + 0.79B = 0.94P
P = 1095 kg mol
It is an air leak.
207 kgmol / 100 kgmol F
6–8
Solutions Chapter 6
6.2.6
Steps: 1,2,3, and 4: This is a steady state process with no reaction taking place. All the
compositions are known except for stream C. We can pick the overall system first to get
D, and then make balances on the first (or second) units to get C and its composition.
Step 5:
Basis:
F = 290 kg (equivalent to 1 second)
Step 6:
Unknowns:
A, B, C, D and E and the composition of C
Step 7:
Balances:
NaCl, HCl, H2SO4, H2O, inert solid
Step 8:
For the overall system there are 4 unknowns (C is excluded) and 5 species
balances:
Overall balances (kg)
In
Out
NaCl:
HCl:
H2SO4:
H2O:
A(0.040)
A(0.050)
A(0.040)
A(0.870)+B(0.910)
=
=
=
=
290(0.0138)
290(0.0255)+D(0.020)+E(0.015)
290(0.0221)+D(0.020)+E(0.015)
290(0.9232)+D(0.960)+E(0.970)
Inerts:
Totals:
B(0.09)
A+B
=
=
290(0.0155)
D + E +290
Steps 6 and 7:
Two of the equations are not independent: HC1 and H2SO4.
HC1: D(0.020) + E (0.015) = 7.40 – 5.00 = 2.40
H2SO4: D(0.020) + E (0.015) = 6.41 – 4.00 = 2.41
Thus, overall the degrees of freedom are 0.
Steps 8 and 9:
The solution of the equations is (in kg/s)
A = 100 ,
B = 50 ,
D = 60 and E = 80
from which C = 150 via a total balance about the first unit.
6–9
Solutions Chapter 6
6.2.7
Notation: Subscripts Benzene = bz, Xylene = xy, Solids = s
W = mass fraction, Fi = liquid stream, Si = solids stream
0
ωS bz = 1.0
0
ωS xy = 0
2
1
ωS xy
1
ωS xy
0
2
2
ωS bz = (1-ωSxy)
1
1
ωS bz = (1-ωSxy)
2
ωS s = 0
1.0
ωS s = 0
1.0
ωS s = 0
1.0
S0
S1
S2
1
2
F1
F2
2
2
1
1
ωF bz = (1-ωFxy)
F0
ωF bz = (1-ωFxy)
2
1
ωF xy
ωF xy
2
ωF s = 0.9
1.0
0
ωF bz = 0
0
ωF xy=0.1
1
ωF s = 0.9
1.0
0
ωF s = 0.9
1.0
Step 5: Basis is 1 hr (F0 = 2000 kg and S0 = 1000 kg)
Steps 6 and 7:
2
1
1
Unknowns:
Unit 1: ωFxy , ωSxy , ωFxy , S1 , F1 , F2 ⎫⎪
⎬ Net = 8
S1
F1
S2
1
1
2
Unit 2: ωxy , xy
ω, xy
ω, S , F , S ⎪⎭
Material Balances:
Unit 1: bz, xy, s
Unit 2: bz, xy, s
Additional Equations: ωFxy = ωSxy and ωFxy = Sxy
ω (the related equations for benzene are
redundant). Total equations = 8.
2
1
1
2
The solution is simplified if two balances are made first (not an essential step):
All material balances are in kg.
Solids balance on Unit 2 and also Unit 1:
2000 (0.9) = F1 (0.9)
2000 (0.9) = F2 (0.9)
F1 = 2000
F2 = 2000
6–10
Solutions Chapter 6
Total balance on Unit 1 and also Unit 2:
2000 + S1 = 2000 + S2
1000 + 2000 = 2000 + S1
S1 = S 2
S1 = 1000
S2 = 1000
The other balances are
Unit 2:
1
1
2
Benzene:
1000(1 − ωSxy ) = 2000(1 − ω Fxy ) + 1000(1 − ωSxy )
Xylene:
2000(0.1) + 1000ωSxy = 2000 Fxy
ω + 1000 Sxy
ω
1
1
1
2
2
10ωFxy = Sxy
ω
Unit 1:
Benzene:
1.0(1000) + 2000(1 − ωF ) = 2000(1 − ωFxy ) + 1000(1 −ωSxy )
Xylene:
0 + 2000ωFxy = 2000ωFxy + 1000(ωSxy
1
1
2
2
2
1
1
)
1
10ωFxy = Sxy
ω
Solve the equations to get the compositions of the streams:
b)
Stream
Component
wt fr
Stream
Component
wt fr
S0
Bz 1.0
Xy 0.0
F0
Bz 0.9
Xy 0.1
S1
Bz 0.97
Xy 0.03
F1
Bz 0.082
Xy 0.018
S2
Bz 0.82
Xy 0.18
F2
Bz 0.097
Xy 0.003
Xylene in Feed
= 2000 × 0.1 = 200 kg
Xylene in Product
= 1000 × 0.18 = 180 kg
% Recovery =
180
×100 = 90%
200
6–11
Solutions Chapter 6
6.2.8
Step 5: Basis: F = 100 kg and D3 = 10 kg
Steps 2, 3, and 4:
A
0.50
B
D1
C
0.23
D2
0.27
A
0.17
B
0.10
C
0.73
D
3
1.00
1.00
D3
A
mA
B
mB
C
0
D3
10 kg
F = 100 kg
P1
1
A
0.50
B
0.20
C
0.30
P2
2
3
P3
A
A
B
C
B
0
C
A
0.70
B
0.30
C
0
1.00
1.00
1.00
E
A
B
C
0
1.00
Additional information:
P3 = 3D3 = 30 kg
P2 = D 2
m PA2 = 4m PB2
Begin with overall balances.
Steps 6 and 7:
Unknowns 4: D1 , D 2 , m AD3 , m BD3
Balances (3) : A, B, C (plus total)
Implicit equation (1): m DA3 + m DB3 = 10 kg
⎫
⎪
⎬ Degrees of freedom = 0
⎪
⎭
Steps 8 and 9:
Overall balances
Total :
100
= D1 + D2 + 10 + 30
6–12
or 60 =D1 + D2
Solutions Chapter 6
C:
100 (.30) = 0.27D1 + 0.73 D2 +0
Solve these two equations to get D1 = 30 and D2 = 30
Get m DA 3 and m DB3 from A balance and m DA 3 + m DB 3 = 1
A:
100(.50) = 30(.50) + 30(.17) + 10 m DA 3 + 30 (.70) so, m DA 3 = 0.89
m DA 3 + m DB 3 = 1 so
D
m B 3 = 0.11
Step10:
Note - Can check using the B balance (not independent)
Balances on Unit No. 3
P
A
m A2
B
m PB2
C
0
1.00
A
m
E
A
B
m
E
B
C
0
D
3
A
0.89
B
0.11
C
0
1.00
P2 = 30
E
1.00
P3
3
A
0.70
B
0.30
C
0
1.00
(D3 = 10; P3 = 30)
Steps 6 and 7:
Unknowns (5): E, mPA2 , mPB2 , mAE , mBE
Balances (2): A, B
Implicit equations (2): mPA2 + mPB2 = 30, mAE + mBE = E
Other equations: mPA2 = 4mBP2
Steps 8 and 9:
Total:30 + E = D3 + P3 = 10 + 30 = 40
so
One component balance:
6–13
E = 10 kg
Solutions Chapter 6
A: 30 m PA2 + 10 m EA = 10 (.89 ) + 30 (. 70) = 29. 9⎫
⎬ not independent equations
P
E
B: 30 m B2 + 10 m B = 10 (.11) + 30 (. 30) = 10.1 ⎭
∑ m i 2 = m A2 + m B2 =1⎫
m PA2 = 0.80
P
⎬
m B2 = 0.20
m PA2 = 4m PB2
⎭
30 (.80) + 10 m EA = 29.9 so that m EA = 0.59
P
P
P
E
E
so that
or 59%
E
m A + m B =1
or 41%
m B = 0.41
_____
1.00
6.2.9
Steps 1, 2, 3 and 4:
mol %
64.29H2
14.29SiCl 4
21.42 H2 SiCl 2
HCl (g)
A
100% Si
B
Step 5: Basis:
C
Reactor
D
Separator
H 2 ,SiCl 4 ,HSiCl 3
H 2 SiCl 2
100%
E
HSiCl 3
100 kg B
100 kg Si 1kg mol Si
= 3.560 kg mol Si
28.09 kgSi
Steps 6 and 7: Unknowns:
A, D, E
Balances:
Steps 8 and 9: Balances to be solved:
System: overall process
Si overall mole balance
kg mol Si = 3.560
6–14
H, Cl, Si
Solutions Chapter 6
= D kg mol gas
+
⎡ 0.1429 kg mol SiCl 4 1 kg mol Si
0. 2142 kg mol H 2 SiCl2 1 kg mol Si ⎤
+
⎢⎣
kg mol gas
kg mol SiCl 4
kg mol gas
kg mol H 2Si Cl 2 ⎥⎦
E kg mol HSiCl3
1 kg mol Si
= 0. 3571D + E
kg mol HSiCl3
Cl overall mol balance
⎡ 0.1429kg mol SiCl 4 4 kg mol Cl
A kgmol HCl 1kg mol Cl
= D kg mol gas
⎢⎣
kgmol SiCl 4
kg mol gas
kg mol HCl
+
0.2142 kgmol H2 SiCl 2 2 kgmol Cl ⎤ E kg mol HSiCl3 3kg mol Cl
+
kg mol gas
kgmol H2 SiCl2 ⎥⎦
kg mol HSiCl 3
A=
D + 3E
H overall mol balance
⎡ 0.6429kg mol H2 2 kgmol H
A kgmol HCl 1kgmol H
= D kg mol gas
⎢⎣
kg mol H2
kgmol gas
kg mol HCl
+
0.2142 kgmol H2 SiCl 2 2 kgmol H ⎤ E kg mol HSiCl3 1kg mol H
+
kg mol gas
kgmol H2 SiCl2 ⎥⎦
kg mol HSiCl 3
A=
Solution:
2
1
3
1.71421D + E
3.560 = 0.3571 1D + E
1
A
= D + 3E
2
A
= 1.7142D + E
3
2E
= 0.7142 1D
E
= 0.3571 1D
4 → 3.56
4
= 0.3571D + 0.3571 D
6–15
Solutions Chapter 6
H
Si
Cl3
D
= 4.98
E
= 1.78
A
= 10.32
MW
1.01
28.09
106.35
135.45 kg/mol
135.45 kg HSiCl 3 1.78 kg mol
= 241 kg HSiC13
kg mol HSiCl 3
6.2.10
Step 5: Basis: 100 kg mol = F
Steps 1, 2, 3, 4:
W
F (kg mol)
E (kg mol)
Furnace
mol fr.
CH4 0.70
H2 0.20
C2H6 0.10
1.00
A (kg mol)
mol fr.
0.21 O2
0.79 N2
1.00
100%
H2O
Duct
mol fr. mol
CO2
nECO2
O2 0.02 nEO2
H2O
nEH 2O
N2
nN2
1.00
E
P (kg mol)
mol fr. mol
B (kg mol)
Air ?
mol fr.
0.21 O2
0.79 H2
1.00
nPCO2
CO2
O2 0.06 nPO 2
nPN2
N2
1.00
P
Assume first an air leak occurs and use material balances to calculate the amount. This is
a steady state flow process without reaction.
Overall system
Step 6: Unknowns are: A, B, W, P, x N2 (or n N2 ), x CO2 (or n CO2 )
6–16
Solutions Chapter 6
Step 7: Balances are:
Step 8:
C, H, O, N,
∑x
P
i
=1 (or ∑ niP = P)
In
0.70(100) + 0.10(100)2
C (kg mol):
Out
= n CO2
n CO2 = 90 kg mol
0.70(100)(4) + 0.20(100)(2) + 0.10(100)(6) = W(2)
W = 190 kg mol
190(1.00)
0.21A + 0.21 B
=
+ 90 + 0.06P
2
0.79A + 0.79 B
= n PN2
H (kg mol):
2O (kg mol):
2N (kg mol):
90 + 0.06P + n PN2
=P
A balance about the furnace or about the duct is needed, or those two alone would have
been sufficient, omitting the overall balance.
System: Duct
E
Step 6: Unknowns: A, E, n CO
, n OE 2 , n EH2O
2
Step 7: Balances:
C, H, O, N2 , ∑ niE = E
Steps 8 and 9:
C (kg mol):
0.70(100) + 0.10(100) (2) =
E
n CO
2
E
= 90 kg mol
n CO
2
H (kg mol):
0.70(100)(4) + 0.20 (100) (2) + 0.10(100) (6) = n EH2O
n EH2O = 190
2O (kg mol):
0.21A = n OE 2 + 90 +
2N (kg mol):
0.79A = n EN2
190
2
90 + 190 + n OE 2 + n EN2 = E
0.02 =
Solution
n
E
CO2
n EH2O
n OE 2
n EN2
n O2
E
kg mol in E
90
190
A
kg mol
982
E
1077
21.5
776
6–17
Solutions Chapter 6
1077.5
From the overall 2N balance
(1)
0.79(982) + 0.79B = n PN2
2O balance
(1') 22.2 + 0.21B = 0.06P
0.21(982) + 0.21B = 185 + 0.06P
(2)
⎫ B = 207 kg mol
⎬
(3) 90 + (0.79)(982) + 0.79B = 0.94P ⎭ P = 1095 kg mol 207 kg mol / 100 kg mol F
6.2.11
Assume steady state process
P = 100 mol
100% H2 O
W
gas
F2
F
1
mol fr.
C 0.357
H 0.643
1.00
oil
Aair
A1
A2
Basis: 100 mol natural gas
Comp.
mol=% atoms C
atoms O
atoms H
CH4
C2H2
CO2
96
2
2
96
4
2
--4
384
4
--
Total
100
102
4
388
Basis: 100 mol oil
Comp.
atoms C
atoms H
(CH1.8)n
100
180
Basis: 100 mol flue gas (dry)
6–18
Solutions Chapter 6
Comp. mol%
mol C mol O
atoms H
mol N2
CO2
CO
O2
N2
Total
10.0 10.00 20.00
-0.63 0.63 0.63
-4.55 -9.10
84.82 --100.00 10.63
29.73
Note:
If have separate air streams, we have 5 unknowns and can't solve.
----
84.82
84.82
Basis: 100 mol natural gas (or use 100 mol dry gas product.)
Let F2
= mol oil
F1 or P
= mol dry flue gas
A
= mol air to natural gas fed boiler plus oil fed boiler
W
= mol water associated with the dry flue gas
Four balances: In = Out
C balance
102 + F2
= 0.1063 P
(1)
= 0.8482 P
(2)
N2 balance
0.79A
H balance
388 + 1.80 F2 = 2 W
(3)
O balance
4 + 0.42A
= 0.2973P +W
(4)
P
= 1729 mol dry flue gas
W
= 268 mol H2O
F2
= 82 mol oil and 1 mol C = 1 mol oil so
C in oil = F2
= 82 mol
6–19
Solutions Chapter 6
Total C = 102 + 82
= 184 mol
% C burned from oil = (0.82)(100) = 44.5%
184
6.2.12
Steady-state, reaction take place
mA=. 30
mB = .70
1.00
NaC1 A
mANaC1 salt
mAH2O H2O
B
H2O
C12 H2
D
E
G
F
C
30%
Overall balances
Steps 6 and 7:
5 unknowns: A, B, D, E, G
4 balances: Na, C1, H, O
other equations: A/B = 30/70
Steps 8 and 9:
(a)
Percent conversion of salt to sodium hydroxide.
Basis: 1 lb product = H
1b mol of NaOH:
1 lb 0.5 lb NaOH lb mol NaOH
= 1.25 ×10-2lb mol NaOH
lb
40 lb NaOH
lb mol NaC1:
1 lb 0.07 lb NaC1 lb mol NaC1
= 1.20 ×10-3 lb mol NaC1
lb
58.45 lb NaC1
1.25 ×10−2
Conversion =
(100) = 91.2%
1.25 ×10−2 + 1.20 ×10−3
6–20
H
50 % NaOH
7% NaC1
43% H2O
Solutions Chapter 6
(b)
How much chlorine gas is produced per lb of product?
1.25×10-2 lb mol NaOH 0.5 lb mol C12 70.9 lb C12
= 0.44 lb C12 /lb product
lb mol NaOH lb mol C12
(c)
⎛ 58.45 lb ⎞
A = (1.25×10-2 lb mol + 1.20 × 10-3 lb mol ) ⎜
⎟
⎝ lb mol ⎠
= 0.80 lb
Using salt as a tie component:
C=
A
1
=
0.8 lb = 2.67 lb
0.30 0.30
B = C – A = 2.67 – 0.8 = 1.87 lb
Balance on oxygen:
⎛ 1.87 lb mol 0.43 lb mol 0.50 lb mol ⎞ ⎛ 18 lb ⎞
G = ⎜
−
−
⎟⎟ ⋅ ⎜
⎟ = 1.22 lb
⎜
18
lb
18
lb
40
lb
mol
⎝
⎠
⎝
⎠
6.2.13
Step 5: Basis = 100 lb A
Steps 2, 3, and 4:
Fe added: 36(100) = 3600 lb
MW TiO2 = 79.9 lb/lb mol
MW Fe = 55.8 lb/lb mol
MW H2SO4 = 98.1 lb/lb mol
MW TiOSO4 = 159.7 lb/lb mol
6–21
Solutions Chapter 6
For Part (a), the system boundary has been drawn on a light solid line.
Steps 6 and 7:
F
Unknowns (9): B, C, J, K, H, F, ξ1 , ξ 2 , m Fe
Species balances (7): TiO2 , Fe, H2SO4 , H2O, TiOSO4 , O2 , inert
Other equations (2): Reactions 1 and 2 are complete
Degrees of freedom = 0
Steps 8 and 9
a. lb H2O removed in evaporator (J):
Mol TiO2 in slag:
Mol Fe in Slag:
(0.70)(100 lb) lb mol
= 0.876 lb mol
79.9 lb
(0.80)(100 lb) lb mol
= 0.143 lb mol
58.8 lb
Amount of H2SO4 based on theoretical requirements of Equations (1) and (2)
0.876 + 0.143 = 1.02 lb mol H2SO4
6–22
Solutions Chapter 6
1.02 lb mol H 2SO 4 98.1 lb
= 100 lb H 2SO 4
lb mol
100 lb
B=
= 149.2 lb Water in stream B = 49.2 lb
0.67
O2 in:
0.143 lb mol Fe 0.5 lb mol O2 32 lb O2
= 2.29 lb O2 =C (0.0715 lb mol)
1 lb mol Fe 1 lb mol O2
Using TiO2:
ξ1 =
0 − 0.876
= 0.876 moles reacting
( − 1)
Using O2:
ξ2 =
0 − 0.0715
= 0.143 moles reacting
(0.5)
TiOSO4 in stream K
K
n TiOSO
= 0 + (1) (0.876) = 0.876 lb mol TiOSO4
4
K=
0.876 lb mol 159.7 lb 1
= 170.6 lb
1 lb mol 0.82
H2O in K = 170.6 (0.18) = 30.7 lb
Water formed in the reactions (use ξ1 and ξ2 ) :
n out,rxn
= (1)(0.876) + (1)(0.143) = 1.019 lb mol (or 8.36 lb H2O)
H2 O
Water exiting from the evaporator J (lb):
49.2 + 18.36 – 30.7 = 36.9 lb
b.
Exit H2O from dryer:
inlet air =
18 lb mol 29 lb
= 522 lb dry air
lb mol
inlet water =
0.036 mol H 2 O 18 mol air 18 lb
= 11.7 lb H 2O
mol air
lb mol
water added to air = (0.18) 170.9 lb – (0.876 lb mol)
6–23
18 lb
= 15.0 lb
lb mol
Solutions Chapter 6
Humidity = (15.0 lb H2O + 11.7 lb H2O)/522 lb air
= 0.051 lb H2O/lb air
c.
The pounds of TiO2 produced (P):
By reaction (3), 0.876 lb mol of TiOSO4 goes to TiO2.
(0.876) (79.9) = 70 lb TiO2
6.3.1
a. 1; b. 3; c. 0; d. 7
a.
b.
6–24
Solutions Chapter 6
c.
d.
6–25
Solutions Chapter 6
6.3.2
Step 5: Basis: 60 kg W
Pick the overall system
Steps 6 and 7:
Unknowns: F, P
Balances: A, B
Steps 8 and 9:
Total F= 60 + P ⎫ F = 380 kg
⎬
B: 0.80F = 0 + 0.95P ⎭ P = 320 kg
Pick mixing point as the system
Steps 6 and 7:
Unknowns: G, R
Balances: A, B (or total as alternate)
Steps 8 and 9:
Total: 380 + R = G
⎫
⎬ R = 126.7
B: 0.80(380) + R(0) = 0.60G ⎭
R 125.7
=
= 0.33 kg R/kg F
F
380
6.3.3
Basis: 100 kg of fresh feed
Overall balance around junction
100 + R = F
KC1 balance around mixing point
6–26
Solutions Chapter 6
20 + 0.6 R = 0.4 F = 0.4 (100 + R)
20 + 0.6 R = 40 + 0.4 R
R = 100 kg R/100 kg fresh feed
6.3.4
Steps 2, 3, and 4:
B1 0.75
B2 0.25
S 0.00
1.00
B1 = butene
B2 = butadiene
S = solvent
A
5,000 lb/hr
I
C (lb) B1
B2
S
E (lb) = 10,000
II
B1 0.0
B2 0.01
S 0.99
B (lb) B1 1.00
B2 0
S 0
D (lb) B1 0.05
B2 0.95
S 0
Step 5: Basic: 1 hr (A = 5000 lb)
Select whole process as the system.
Steps 6 and 7:
Unknowns: B, D
Balances: Total, B1, B2, S (not all independent)
Steps 8 and 9:
Total balance: 5000 = B + D
B1 balance:
5000 (0.75) = B(1) + D(0.05)
D = 1316 lb
B = 3684
Apparently the calculated values are not correct. (The value for C can be obtained from
balances on Unit I or II).
6–27
Solutions Chapter 6
6.3.5
Basis: 1 hour
An overall balance shows that Rout from the adsorber must equal Rin to the adsorber if a
steady state exists. Let Ri = R.
Adsorber balance of U (units are U):
600 mL 1.37U
R mL 5.2U
600 mL 0.08U
R mL 19.3U
+
=
+
1 mL
1 mL
1 mL
1 mL
600 (1.37-0.08) = R(19.3-5.2)
R = 55 mL/hr
6.3.6
Step 5: Basis: 1000 kg = W ≡ 1 hour
Steps 1-4:
Revised compositions are in mass fractions.
Note: compositions identical
W (mass) = 1000 kg
E (kg)
Wet cereal
Exit air
H2O
air
mass fr.
H2O 0.200
cereal 0.800
1.00
mol fr.
0.263
0.737
1.000
Recycle
R (kg)
D (mass)
Dried cereal
mass fr.
0.050
H2O
Cereal 0.950
1.00
G (kg)
mol fr.
H2O 0.066
air 0.934
1.00
mass fr.
0.042
0.958
1.000
mol fr.
H2O 0.0132
0.9868
air
1.000
6–28
F (kg)
mass fr.
0.0082
0.992
1.000
Fresh Air
Solutions Chapter 6
Conversion of mole fractions to mass fractions is not required, but since both mass
fractions and mol fractions are listed as data, convert all to mass fractions (can’t convert
cereal to mol) to avoid confusion in making the balances.
Fresh air, Basis 100 mol
mol MW mass(kg)
air
98.68 29
2861.7
H2 O
1.32 18
23.76
100.00
2885.48
Exit air, Basis 1.00 mol
mol MW mass mass fr.
0.737 29
21.37 0.819
0.263 18
4.73 0.181
1.00
26.11 1.000
mass fr.
0.992
0.0082
1.000
Air entering drier, basis 1.00 mol
mol MW mass mass fr.
air
0.934 29
27.09 0.958
H2O 0.066 18
1.19 0.042
28.27 1.000
Overall balances
(unknowns D, E, F; balances H2O, cereal, air)
Total: can be made in mass:
1000 + F = E + D
in
cereal (dry): 1000 (0.80) + F (0)
= E (0) + D (0.950) D = 842 kg
1000 (0.20) + F (0.0082) = E (0.181) + D (0.050)
H2O:
air:
1000 (0)
+ F (0.992) = E (0.819) + D (0)
3 are independent eqns. check by 4th eqn.
a.
D = 842 kg
E = 906 kg
F = 748 kg/hr
Balance on mixing point (to get R)
Total: R + F = G
air : R (0.819) + F (0.992) = G (0.958)
H2O: R (0.181) + F (0.0082) = G (0.042)
2 equations are independent, check via 3d equation
b.
R = 183 kg/hr
6–29
Solutions Chapter 6
6.3.7
Steps 1, 2, 3 and 4:
H 2 O 100%
W
KNO3 Solution
P
Evaporator
mass frac.
H2 O
0.50
KNO3
0.50
1.00
mass frac.
10,000 kg/hr 20%
mH O
m KNO
F
H2 O
0.80
KNO3
0.20
Feed
1.00
2
300°F
M
50%
KNO3
3
∑P
R
Recycle 100
°F
Saturated Solution
Crystallizer
0.6 kg KNO 3
mass frac.
H2O
0.625
HNO3
0.375
kg H 2O
1.000
Boundary line
for overall balance
C
mass frac.
H2 O
0.04
KNO3
Boundary line for
balance around
crystallizer
0.96
1.00
Step 5: Basis: 1 hr = 10,000 kg KNO3 solution
Step 4 cont’d:
Compute the weight fraction composition of R. On the basis of 1 kg of water, the
saturated recycle steam contains (1.0 kg of H2O + 0.6 kg of KNO3) = 1.6 kg total. The
recycle steam composition is
0.6 kg KNO3
1 kg H2 O
= 0.375 kg KNO3 / kg solution
1 kg H2 O 1.6 kg solution
or 37.5% KNO3 and 62.5% H2O (which has been added to the figure).
Analysis of complete process ( 6 streams):
Unknowns:
W, M, C, R, m PH 2 O m PKNO 3
=6
Balances:
(2 units + 1 mix point) × 2 components
=6
6–30
Solutions Chapter 6
Other compositions are (in %)
M
H2O 50
KNO3 50
F
80
20
W
100
0
C
4
96
R
62.5
37.5
Start with overall balances as substitute for unit balances
Steps 6 and 7:
Unknowns:
W, C,
Balances:
2 components
Overall balance to calculate C:
Total:
10,000
=W+C
KNO3: 10,000 (0.20) = W (0) + C (0.96)
10, 000 kgF 0.20 kgKNO3 1kgcrystals
= 2083 kg / hr crystals = C
1kg F
l hr
0.96 kgKNO3
W = 10,000 - 2083 = 7917 kg
To determine the recycle stream R, we need to make a balance that involves the stream R.
Either (a) a balance around the evaporator or (b) a balance around the crystallizer will do.
The latter is easier since only three rather than four (unknown) streams are involved.
Total balances on crystallizer:
M=C+R
a
M = 2083 + R
Component (KNO3) balance on crystallizer:
M ω M = Cω C + Rω R
0.5M = 0.96C + R (0.375)
Solving equations a and
b
b we obtain
0.5 (2083 + R) = 0.375R + 2000
R = 7670 kg/hr
6–31
Solutions Chapter 6
6.3.8
Single pass conversion is:
a.
fsp =
−υLR ξ
feed
n reactor
LR
Single pass conversion based on H2 as the limiting reactant:
1.979 − 3.96
= 0.99
−2
0 − 3.96
ξ max =
= 1.98
−2
ξ=
SP conversion =
−(−2)(0.99)
= 0.50
3.96
Use Eq. (12.1) and Eq. 12.2 with H2 the limiting reactant
b.
0.99 =
fsp =
c.
− υLR ξ
−( − 2) ξ
=
ξ = 0.980
fresh feed
n LR
1.98
−(−2)(0.98)
= 0.986
1.989
Overall conversion of H2
f overall of H 2 =
d.
−(−2)(0.99)
= 1.0
1.98
Overall conversion of CO
f overall of CO =
−(−1)(.99)
= 0.99
1
6–32
Solutions Chapter 6
6.3.9
Steps 2, 3, and 4:
Recycle R(mol)
CO, H2O
F(mol)
Reactor
mol
%
52 CO nCO
48 H2O nH2O
L(mol)
P(mol)
H2: 3 mol %
Step 5:
CO2
H2
CO
Basis: P = 100 mol
Pick the overall process as the system.
Step 6:
F
Unknowns: n CO
, n FH O
Step 7:
Balances:
2
C, H, O
Steps 8 and 9: Element balances
C (mol): n CO = 48 + 4 = 52
Step 10:
⎫⎪
⎬ total 100
H (mol): n H2O (2) = 48 (2)
n H2O = 48⎪⎭
O (mol): n H2O (1) + n CO (1) = 48(2) + 4 check is ok.
a.
Composition of fresh feed
40% H 2 O and 52% CO
Alternate solution: Use extent of reaction
Based on
Then
48 − 0
= 48
1
F
nCO
= 4 + ξ = 4 + 48 = 52 mol
F
n H2O = 0 + ξ = 48 mol
CO2:
ξ=
To get the recycle, make a balance about the separator (no reaction occurs)
6–33
mol % = mol
Separator
48
48
4
100
Solutions Chapter 6
H 2 balance (mol):
Total (mol):
L(0.03) = 48⎫ L=1600 mol
⎬
L=100 + R ⎭ R=1500 mol
1500 mol R
= 31.3
48 mol H 2
b.
6.3.10
Steps 2, 3, and 4:
MW Ca (Ac)2 = 158.1
MW HAc = 60
100 kg Ca(Ac)2 1 kg mol Ca(Ac)2
= 0.633 kg mol Ca(Ac)2
158.1 kg Ca(Ac)2
Step 5:
Basis: 1000 kg of Ca(Ac)2 feed
Inspection of the diagram shows that the overall conversion of Ca(Ac)2 is 100%
(none leaves the process). Thus, fOA = 1. You are given fSP = 0.90. Then
R = 0.703 kg mol
a.
fSP
0.90
6.33
=
=
f OA
1
6.33 + R
b.
The single pass conversion for Ca(AC)2 is 0.90.
0.90 =
− υ LR ξ
−( − 1)ξ
=
feed
n LR
6.33
or 111 kg
ξ = 5.697
Overall HAc balance:
6.33 kg mol Ca (Ac) 2
2 kg mol HAc
= 12.66 kg mol HAc or 760kg
1 kg mol Ca(Ac) 2
6–34
Solutions Chapter 6
6.3.11
Steps 2, 3, 4:
MW C2 H2 = 26.02
MW Zn = 65.37
MW C2H2Br4 = 346
a. C2 H5 produced per hour
Step 5: Basis: 1 hr ≡ 1000 kg C2 H 2 Br4 (2.890 kg mol)
Make overall balances:
Get the extent of reaction using C2 H 2 Br4
n out − n in = υξ
0 – 2.890 = (-1) ( ξ )
ξ = 2.890 reacting moles
C2 H2 balance: n C2H2 = 0 + (1)(2.890) = 2.890 kg mol
2.890 (26.02) = 75.2 kg
ZnBr2 balance: n ZnBr2 = 0 + 2(2.890) =5 .78 kg mol
Alternate solution
100 kg C2 H2 Br4 1 kg mol C2 H2 Br4 1 kg mol C2 H2
26 kg C2 H2
= 75.2
1
346 kg C2 H2 Br4 1 kg mol C2 H2 Br4 1 kg mol C2 H2
b.
Recycle
Make balance on the mixing point. First, get the feed of C2H2Br4 to the reactor
0.80 =
−υξ -(-1)(2.896)
=
n feed
n feed
LR
LR
C2 H 2 Br4 balance:
n feed
LR = 3.61 kg mol
2.890 + R = 3.61
R = 0.72 kg mol
0.72 (346) = 249 kg
Alternate solution (in kg). Balance over separator.
(1000 kg + R) (0.20) = R
R = 250 kg
6–35
Solutions Chapter 6
c.
Feed rate required for 20% excess Zn in reactor:
1000 kg C2 H 2 Br4 1 kg mol C2 H 2 Br4 2 kg mol Zn 65.37 kg Zn 1.2 kg Zn in feed
= 454 kg
1
346 kg C2 H 2 Br4 1 mol C2 H 2 Br4 1 kg mol Zn 1 kg Zn required
d.
Mole ratio of ZnBr2 to C2H2 in final products:
5.78
= 2
2.89
2:1 (as per reaction)
6.3.12
Step 5:
Basis: 100 kg F1 ≡ 1 hr
Overall balance
Step 6 and 7:
Unknowns: F2, P1, P2
Element balances: Ca, Na, C1, CO3 (enough, not all independent)
CO3 balance:
900 kg CaCO3 1 kg mol CaCO3 1 kg mol CO3
100 kg CaCO3 1 kg mol CaCO3
= 9.00 kg mol CO3
100 kg Na 2 CO3 1 kg mol Na 2 CO 3 1 kg mol CO 3
0.94 kg mol CO3
=
106 kg Na 2 CO3 1 kg mol Na 2 CO 3
9.94 kg mol CO3
a.
9.94 kg mol CO3 1 kg mol Na 2 CO3 106 kg Na 2 CO3
= 1054 kg Na 2 CO3
1 kg CO3
1 kg mol Na 2 CO3
6–36
Solutions Chapter 6
Species balance about the reactor plus the separator on CaCO3:
IN


OUT
[1000(0.90) + R] (.24) = R



UNREACTED
R = 284 kg
Alternate solution:
Use the extent of reaction to get the same results.
6.3.13
This is steady state process with reaction and recycle. Pick the overall process as the
system.
Step 5: Basis: 100 kg mol CH4
Steps 1, 2, 3, and 4:
The system is as shown in the diagram with recycle added.
0.21
0.79
Air
kgmol
O2 230
N2 865.24
CH4
CO2
O2
100 kg mol
H 2O
NO
N2
R
Calculate the moles of each component entering.
C H4 (g) + 2O2 (g)→C O2 (g)+ 2 H2 O (g)
Calculation of the required O2 and accompanying N2 in moles
req'd O2
100(2) = 200
xs
200 (0.15)
Total O2
= 30
230
6–37
Solutions Chapter 6
⎛ 230 .79 ⎞
⎟ =
N2 in ⎜
.21 ⎠
⎝
865.24
The exit concentrations (via stoichiometry) are
CO2
O2
H2 O
NO
N2
Mol
100
30
200
415 × 10-6
865.24
1195.24
Next, consider adding recycle as shown in the diagram. Recycle is not involved in the
overall balance, hence the concentration of NO will not be affected because the extent of
reaction with and without recycle remains the same. The recycle does reduce the
combustion temperature, which in turn will reduce the exit concentration of NO.
6.3.14
Basis: 1 hr ≡ 1L = F
all concentrations are g/L.
W
D
F=1L
Zn 0.10
Ni 1.00
H2O
Zn 100
Ni 10.0
P0
Zn 190.1
Ni 17.02
Figure P12.9
Pick the overall process as the system.
Steps 6 and 7:
Balances:
3
H2O, Zn, Ni
Unknowns:
3
W, D, P
6–38
(all in L)
Solutions Chapter 6
Steps 8 and 9:
D(L) 0.10g
+ P 0 (190.1g / L)
L
Zn:
100 (1) + 0
=
Ni:
10(1) +0
= D (1.00) + P0 (17.02)
Solve to get P0 = 0.525L
D = 1.056L
Pick Unit 3 as the system:
Steps 6 and 7:
Steps 8 and 9:
Balances:
Unknowns:
2
2
Zn, Ni
R32, P2 (all in L)
Zn:
P2 (3.50) + 0 = 0.10(1.056) + R32 (4.35)
Ni:
P2 (2.19) + 0 = 1.00(1.056) + R32 (2.36)
Solve to get R 3 2 =2.75L/hr
6.3.15
H
Mass frac.
P
oil + dirt ω o+d
P
H2 O
ω HO
ω
H
ωH O
2
2
1.00
P
H
o+d
D
mass fr.
0.229
0.771
H2 O
0.9927
0.771
1.000
Filter
R
G
o+d
ω
ω HGO
Mass frac.
oil + dirt 0.229
H2 O
0.771
1.000
2
oil + dirt 0.0073
oil + dirt 0.229
H2 O
2910 gal/day
mass fr.
mass fr.
1.000
G
90 gal/day
F
3000 gal/day
1.000
6–39
Solutions Chapter 6
Pick the total process as the system
Step 5:
Basis: 1 day (equivalent to D = 90 gal, F = 3000 gal, P = 2910 gal)
Steps 6 and 7:
Unknowns:
P
ωo+d
,
Balances:
oil + dirt, H2O
Steps 8 and 9:
Overall oil + dirt balance:
ω
P
H2 O
In
Out
P
3,000(0.0073) = 2,910(ωo+d ) + 90(0.229)
P
ωo+d
=
21.9 − 20.61
= 4.43 10−4 ×
2,910
(a)
To solve for R, make balances on the mixing point, the filter, and the splitter. Not all of
the balances are independent:
Total
oil + dirt
Splitter:
H = 90 + R
0.229H = 0.229(90) + 0.229(R)
Filter:
G = 2910 + H
ωGo+d G = 2910 (4.43 × 10-4) + H(0.229)
Mixing point: 3000 + R = G
0.0073 (3000) + 0.229R = ωGo+d G
Unknowns: H, R, G, ω Go+d
Balances:
6 (4 independent)
Steps 8 and 9:
The solution is
R = 57.2 gal/day
6–40
ωGo+d = 0.01145
Solutions Chapter 6
6.3.16
Start with the overall system
mass fr.
0.95
rice
mol fr.
H2 O
0.0473
gas
0.9527
1.0000
0.05
H2 O
1.00
P (lb)
S (lb mol)
Rice Product
Dryer
mass fr.
W (lb mol)
F (lb)
mol fr.
rice
0.75
x
H2 O
0.25
x
w
0.0931
H2 O
w
0.9069
gas
1.00
Steps 1, 2, 3 and 4:
1.0000
In the diagram use mol and mass percent for
compositions and mol and mass for flow as specified.
Step 5:
Basis
P = 100 lb
Step 6:
Unknowns:
F, S, W
Step 7:
Can make total and 3 component balances: rice, H2O, dry gas (G)
Steps 8 and 9:
lb
Rice:
F(0.75) = P (0.95) = 100 (0.95)
lb mol
Dry gas:
S (0.9527) = W (0.9069)
lb mol H2O
H2O:
S (0.0473) +
P (0.05)
F (0.25)
= W (0.0931) +
18.02
18.02
F = 126.67 lb
S = 27.35 lb mol
W = 28.75 lb mol
6–41
Solutions Chapter 6
R
S
Make balances on the mixing point (easiest)
lb mol gas:
Dryer
F1
(27.35)(0.9527) + R (0.9069) = F1 x FDG1
F
lb mol H2O:
(27.35)(0.0473) + R (0.0931) = F1 x H12 O
lb mol total:
27.35 + R = F1
x FH12O = 0.0520
1
x FDG
+ x HF12O =1
R = 3.12 lb mol / 100 lb P
0.0931 H 2O
R
or, make balances on separation point
H2O:
total
(or G):
P
1 0.0931 H 2O
P1 (0.0931) = R (0.0931) + 30.768 (0.0931)
(only 1 independent equation)
0.0473 S + 0.0931 R = F1 (0.052)
27.35 + R = F1
1.385 + 0.0931 R = 0.052 (27.35 + R) = 1.523 + 0.052R
0.041 R = 0.138
R = 3.37lb mol
6–42
30.78 lb mol
0.0931 H2 O
Solutions Chapter 6
6.3.17
F°
Organic Solvent
100%
I
Aqueous Phase
FA = 10 L/min
A
C0 =
F°
FA = 10 L/min
100 g/L
A
O
C1 = 0.01g / L
Fermentation product balance:
100
g 10L / min
10 L / min 0.1g / L
= F°(0.01) +
L
F° =
1000 − 1 = 99, 900 L
min
0.01
B:
F2 °
F1 °
10L/min
I
II
A
C1
FA = 10 L/min
CA = 100 g/L
o
C1 = 0.1 g/L
O
A
C1 = 0.1C1
Fermentation
balance:
F3 °
10L/min
III
A
C2
O
A
C 2 = 0.1C 2
A
C3 =0.1 g/L
O
A
C3 = 0.1C 3
Assuming F1° = F2° = F3° = F°/3
I:
100
A
II:
C1
g
L
F°
A
L 10 min = C A ⋅10 L
1
mn + 3 × 0.1C (1)
g
L
F°
A
L 10 min = C A ⋅10 L
2
min + 3 × 0.1C (2)
6–43
F =10L
A
Solutions Chapter 6
A
III:
C2
g
L
F°
A
L 10 min = C A ⋅10 L
3
min + 3 × 0.1C3 (3)
Solving (1), (2) and (3) Simultaneously
C 1 = 10.0 g L
C 2 = 1.0 g L
A
C 3 = 0.1 g L
A
A
F = 2702 L min
C:
Fermentation Product Balance:
I:
100 g L 10 L min
A
A
A
+ 0.1 C 2 ⋅ F° = 0.1 C1 ⋅ F° + C1 ⋅ 10
II:
C1 ⋅10 + 0.1(0.1)F° = 0.1C 2 ⋅ F° + C2 ⋅10
III:
C 2 ⋅10 + 0 = 0.1(0.1) ⋅ F° + 0.1 (10)
A
A
A
A
Solving (1), (2) and (3) Simultaneously
A
C1 = 10.36g / L
F = 960 L min
A
C 2 = 1.06 g / L
A
C3 = 0.10 g / L
This one is the least
6.3.18
Steps 1, 2, 3, and 4:
Mol. Wt: Na2S, 78; CaCO3, 100; Na2 CO3, 106; CaS, 72
Step 5: Basis 1 hr
6–44
Solutions Chapter 6
Steps 6, 7, 8, and 9:
System: Overall (System A)
Na Balance:
=
b.
(1000)(0.45)lb Na 2S 1 lb mol Na 2S 2 lb mol Na
78 lb Na 2S 1 lb mol Na 2S
m Na 2CO3 lb 1 lb mol Na 2CO3
2 lb mol Na 100 lb Na 2CO3 soln
106 lb Na 2CO3 1 lb mol Na 2CO3
80 lb Na 2CO3
mNa2CO3 = 764 lb/hr
System: Reactor plus separator (System B)
in
out
gen. Consumption accum
Na 2S balance: ⎡100(0.45)
R ⎤ R
⎡100(.45) R ⎤
+
−
+ 0 − ⎢
+ ⎥ 0.90=0
⎢
⎥
(lb mol) ⎣ 78
78 ⎦ 78
78 ⎦
⎣ 78
[1000(0.45) + R] (0.10) = R
a.
R = 50 lb/hr
Alternate solution
reacts.
Na2S is the limiting reactant (LR). Mole Na2S = (0.45)(1000)/78 = 5.77. All of it
1.00 =
( − 1)( − 1)ξ
,
5.77
ξ = 5.77
Overall fraction conversion of Na2S is 100%; fOA = 1.
fSP
n freshfeed
0.90
5.77
LR
=
= freshfeed
=
recycle
f OA
1.00 n LR +n LR
5.77+R
5.19 + 0.9R = 5.77
R = 0.64 lb mol (or 50 lb)
6–45
Solutions Chapter 6
6.3.19
A basis of 1 hr requires the listing and solution of many simultaneous equations.
A basis of 100 mol of toluene in G is more convenient
Makeup
M
H2
F
Feed
1
total
2
3450 lb/hr
C7 H
8
Gas Recycle
P Pure
RG
mol
1
CH4
Mixer
100% toluene
100% H2
n wH
only
2
n
W
w
CH 4
One system
Gross
Feed
G
4H 2
4CH 4
1toluene
Reactor
B
Separator
Benzene
100%
<-- 4H2 to 1 toluene
Liquid Recycle
D
Diphenyl
100%
RL
100% toluene
Basis: 100 mol toluene in stream G (gross feed)
System: Reactor plus separator
Species balances
(1) toluene:
(2) H2:
In
Out
Gen
Cons.
100
-RL
+0
-100 (.80 +.08)
− nHW2
+0
⎛1
-100 (4) (.80 +.08 ⎝ ⎞⎠ ) = 0
2
W
− nCH
4
+ 100(4)(.80 + .08)
100 (4)
(3) CH4:
100 (4)
Solve above for
RL = 12 mol
= 64 mol⎫
⎬ Total = 816 mol
nW
=
752
mol
CH 4
⎭
n
W
H2
6–46
-0
=0
=0
Solutions Chapter 6
System : Mixer plus makeup point
F
(Species balances but no reaction so In = Out)
RG
M
(1) Toluene:
F (1.0) +
0
+
(2) H2:
0
+
M(1.0)
+
(3)CH4:
0
+
0
+
0
G
R
L
+
12
=
100
⎛ 64 ⎞
R G ⎜
⎟ +
⎝ 816 ⎠
⎛ 752 ⎞
R G ⎜
⎟ +
⎝ 816 ⎠
0
=
400
0
=
400
Solve (3) for RG = 434 mol
Solve (1) for F = 88 mol
M = 366
Change basis to 1 hour
3450 lb 1 lb mol
= 37.5 lb mol of toluene in F
92 lb
100 lb mol tol inG 37.5 lb mol F 12 lb mol R L
= 5.11 lb mol R L /hr
88 lb mol F
1 hr
100 lb mol tol in G
100 37.5 434 lb mol R G
= 185 lb mol R G /hr
88 1 100 lb mol tol in G
If F = 37.5 lb mol of toluene is selected as the basis, you have to make the same
balances as above plus benzene and diphenyl balances on the reactor plus separator
because F is a known but G becomes an unknown.
The unknowns would be
Alternate solution
G, RL, RG, D, B, M n PH 2 ,n PCH4
Calculate the extent of reaction for each reaction, and use them as shown in the
book to get the outputs of the reactor plus separator system instead of using generation
and consumption terms.
6–47
Solutions Chapter 6
6.3.20
Basis: 1 hr = 50 kg mol F5
Unknowns:
F
F
F
F
F
F1, F4, F6, n C23 H 8 , n C23 H 6 , n C33 H 8 ,n C33 H 6 ,n H32
8
Balances:
Mixing Point:
Reactor:
Abs. /Distil.:
⎫⎪
C3 H8 ,C 3 H6 ,H 2 ⎬
C H ,C H ,H ⎪⎭
C3 H8 ,C 3 H6
3
8
3
6
8
2
Use overall balances as substitute for some of the above
Overall balances (element balances because of the reaction)
In
Out
C: F1 (3) = F 5 (3)
H: F1 (8) = F 4 (2) + F5 (6)⎫
⎬
8(50) = 2F 4 + 50(6) ⎭
F1 = 50 kg mol
F4 = 50 kg mol
Mixing point balances (no rxn):
In
Out
F
C3 H8 : F1 (1.0) + F6 (0.8) = n C23 H 8 = 50 + 0.8F6
C3 H6 : F1 (0) + F6 (0.2) = n FC23 H 6 = 0.2F 6
6–48
Solutions Chapter 6
Reactor plus absorber/distillation balances:
In
Out
n FC23 H 8 -
C3 H8 :
Generated
F 6 (0.80)
(recycle)
+0
Consumed
-
0.4n FC23 H 8
(50 + 0.8F6) - 0.80 F6 - 0.4 (50 + 0.8 F6) =0
n FC23 H 8 = 75 + 50 = 125 kgmol
F6 (.20) - 50 + 0.40 (50+ 0.8F6)
-0
F2
n C3 H 6 = 18.75 kg mol
Recycle:
F6 = 93.75 kg mol
C3 H6 :
n FC23 H 6 -
=0
=0
Note: since F2 = F1 + F6 = 50 + 93.75 = 143.75 kg mol (easier calculated this way)
n FC23 H 6 = 0 . 2, F 6 = 18 . 75 kg mol
F2 = 18.75 + 125 = 143.75 kg mol
Omit H2 balance (Use C3 H6 )
Absorption/distillation tower
C3 H6 :
n FC33 H 6 = F5 = 50 kgmol
H 2:
n FH32 = F4 = 50 kgmol
total:
F 3 = F4 + F5 + F 6 = 50 + 50 + 93.75 = 193.75kg mol
n FC33 H 8 = 193.75 − 100 = 93.75 = F6
Summary (all kg mol):
(a)
F1 = 50
F 4 = 50
F 2 = 143.75
F 5 = 50
F 3 = 193.75
F6 = 93.75
(b) 100%(no.C3 H8 exists)
Alternate solution using extent of reaction
f OA = 1 =
−1( − 1)ξ
50
ξ = 50
6–49
Solutions Chapter 6
fSP = 0.4 =
−1( − 1)ξ
feed
n Creactor
3 H8
=
50
n CF23H8
n CF23H8 = 125 kg mol
0.40
50
=
1
50 + n CF63H8
Also
n CF63H8 = 75 kg mol
n CF63H6 =
n CF23H8 = 125 = 50 + 0.8 F6
0.20
(75) =18.75 kg mol
0.80
F6 = 93.75
n CF23H6 = n CF63H6 + 0 = 18.75
F2 = 18.75 + 125 = 143.75
Similar calculations to these will yield the same results as in the original solution.
6.3.21
This is a steady state problem with reaction and recycle.
Step 5:
Basis: 100 mol F
Steps 1, 2, 3 and 4:
Make the overall balances first
1
SO2 + O2 ↔ SO3
2
SO2
O2
N2
kg mol
10.0
9.0
81.0
F (mol)
Overall
Process
P(mol)
Calcd.!!!!!!!Calcd.
Use stochiometry mol
mol fr
SO3 0.95 (10) = 9.50 0.100
SO2 0.05 (10) = 0.50 0.00525
O2 9-4.75
= 4.25 0.0446
N2
81.0 0.850
!
95.25 1.000
Step 6:
The unknowns are the 4 exit compositions plus the extent of reaction if it
is to be used.
Steps 7, 8, and 9: We have 3 element balances plus the fraction conversion of the SO2 to
SO3, or 4 species balances (SO3, SO2, N2,O2). We can use a mixture of element and
species balances.
6–50
Solutions Chapter 6
Element S:
10
= 0.95 (10) + 0.05 (10)
Compound
N2:
81.0 = 81.0
check is ok
Select unit 1 as the system
mol
SO 2
10.0
O2
9.0
N2
81.0
calcd. mol
SO 3 7.5
fract.
conv.
0.75
1
0.75
SO 2
2.5
O2
5.25
N2
81.00
SO 3 : 7.5
S : 10
SO 2 : 2.5
0.25
= 1.5 (7.5) + 2.5 + n O 2 = 11.25 + 2.5 + n O 2
Element 2O: 9.0 + 10.0 = 19
n O 2 = 5.25
Balance around converter 2 plus separator: Note we need H = P + R, and observe that the
composition of H, P and R is the same.
kg mol
SO 3
mol
7.5
SO 2
2.5
fract.
conv.
O2
5.25
0.65
N2
81.00
P
H
2
R
System
mol fr.
SO 3
9.50
0.100
SO 2
0.50
0.00525
O2
4.25
0.0446
N2
81.00
0.85
1.000
same composition as P
R
Make a species balance on SO2
In
Out
Consumption
[2.5 + R(0.00525)] - [0.50 + R(0.00525)] +0
R = 111 kg mol
6–51
Generation
-[2.5 + R(0.00525)]0.65 = 0
Solutions Chapter 6
6.3.22
Steps 2, 3, and 4:
(Non Bz is non-benzene)
Step 5: Basis: 1 hr
Process is steady state, no reaction
Steps 2, 3 and 4:
Compositions:
at P
SO2
Bz
0.15 kg SO2
= 0.130
1+ 0.15
1 − 0.130
=
mWBz
at W
W
NonBz
m
mWSO2
0.870
1.00
Pick the overall process as the system
Steps 6 and 7:
6–52
⎫
⎪
⎬ ∑ =W
⎪
⎭
Solutions Chapter 6
Unknowns: P, 3miW (4)
Balances:
Steps 8 and 9:
Bz, SO2, non Bz,
Balances are in kg
Total
3000 + 1000 = P + W
Bz:
1000 (0.70) = P ⎛⎜
1.00 ⎞
W
⎟ + m Bz
⎝ 1.15 ⎠
Non Bz: 1000 (0.30) = P (0) + mW
N-Bz
SO2:
W
W
mBz
/mnonBz
= 0.25 (4)
0.15
⎞
W
3000 = P ⎛⎜
⎟ + n SO
⎝ 1.15 ⎠
W
mnonBz
= 300 kg
W
mBz
= 300 (0.25) =75 kg
2
a. Solution: P = 719 kg
W
mSO
= 2906 kg
2
W = 3281 kg
System: Unit 1
Steps 6 and 7:
D
Unknowns: m DBz , m DnonBz , mSO
Balances:
2
Bz, nonBz, SO2
Steps 8 and 9: Balances are in kg
b.
Bz + nonBz: Dʹ′ = A + 375
Dʹ′ = 1000 + 375 = 1375 kg
System: Unit 2
6–53
Solutions Chapter 6
Steps 6 and 7:
Steps 8 and 9: Balances are in kg
Bz + nonBz: Cʹ′ = 749 + G ʹ′
System: mixing point
D' = 1375 kg
G'
F = 1000 kg
G ʹ′ = 1375 – 1000 = 375 kg
c.
Cʹ′ = 749 + 375 = 1124 kg
6.3.23
Steady state process with reaction
Step 1, 2, 3, and 4:
Get the amount of {reqd., excess} HNO3 in G.
Step 5:
Basis: 1 hr
Step 4:
1 ×103 kg Glycerine 1 kg mol Gly
= 10. 86 kg mol Gly
92.11 kg Gly
Glycerine
C3 H8 O3 + 3HNO3 →C3 H5 O3 (NO2 )3 + 3H2 O
Required:
HNO3 is 3 (10.86)
6–54
= 32.58 kg mol
Solutions Chapter 6
Excess:
Steps 6 and 7:
HNO3 is 32.56 (.20) = 6.516
Total:
39.10 kg mol in G
System is overall
W
Unknown are F, P,m W
H 2 SO 4 ,and m H 2 O
(Σmi=W)
Equations are: C, H, S, O, N
of reaction)
C:
(or you can use stoichiometry and make species balances, or use the extent
10.86 = 0.9650 P
P = 11.25 kg mol or 11.25 (227.09)
= 2556kg
(a)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - -- - - - - - - - - - - - - System is the mixing point
5
Unknowns:
F, m H 2 SO 4 , m H 2 O ,R, G
Balances:
H 2SO4 , H 2O , HNO3 and m H 2 O + m H 2 SO 4 + 39.10 =G
(a) Glycerine Feed
=
Nitroglycerine Produced =
1000
= 10.86 kg mol
92.11
1086 kg mol 227.98 kg
= 2466 kg
lb mol
0.9650 P = 2465
(a)
P = 2555 kg
(b) Mol HNO3 req'd
10.86 mols Glyc. 3mol HNO3
mol Glyc
= 32.58 mol
=
Actual HNO3 Req'd
= 1.200 (32.58) = 39.10 mol
HNO3 in R
= 39.10 - 32.58 = 6.52 mol
= 6.52 (63.01) = 411 kg
HNO3 is 70.00% R, 0.7000R = 411
6–55
4
Solutions Chapter 6
R = 587lb
(c)
(b)
HNO3 in G = 39.10 mol = 2464 kg
HNO3 in R = 411 kg
HNO3 Balance:
F+R=G
F + 411 = 2464
F = 2053 lb HNO3
F is 43.00% HNO3 so that 0.4300 F = 2053
F = 4774lb
(d)
H 2SO4 in F
(c)
= 0.5000 (4774) = 2387 kg
= H 2SO4 in W
H2O in F
H2O Generated by reaction
= 0.0700 (4774) = 334 kg
= 10.86 Mole Glyc × 3
= (10.86) (3) (18.02) = 587 kg
H2O in Stream P
= 0.0350 (2555) = 89 kg
H2O in W
= H2O in F + H2O generated - H2O in P
= 334 + 587 - 89 = 832 kg
Component
kg
wt%
H 2SO4
2387
74.15
H 2O
832
3219
25.85
100.00
Stream W = 3219 kg/ hr
H2 SO 4
= 74.15 %
H2 O
= 25. 85%
(d)
Step 10:
Do numbers check?
Glycerine + F
1000 + 4774
5774
=
=
=
P+W
2555 + 3219
5774
6–56
Solutions Chapter 6
6.3.24
a.
No. They all contain AN and/or NH3.
b.
The bottom stream from the distillation column and the wastewater stream from
the condenser are candidates. The bottoms stream from the distillation column
contains no NH3 and has the highest mass fraction of AN of any waste stream so
that the entire stream could be fed to the scrubber.
c.
The change in the scrubber feed will not affect any of the stream flows or
compositions upstream of the scrubber that are connected to the scrubber, namely
those associated with the reactor and the subsequent condenser, nor any
downstream flows not connected to the scrubber. A sequential set of material
balances can be used to get the flows and concentrations in the rest of the process.
Basis: 1 second
(1)
The water flow rates will not change except for the stream going to treatment,
which will be (10.1 – 0.7) = 9.4 kg/s.
(2)
The AN clues in the streams can be determined from the following balances.
Let x be the kg of AN entering or leaving a particular unit.
Scrubber balance
In
In
x Scrubber
+ 4.6 = x Decanter
Decanter balance
In
x InDecanter = x Distill
+ 5.5(0.073)
Distillation column plus two condenser plus stream jet plus product stream balance
Out
In
x InDistill + 0 = x Out
Jet + x Pr oduct + x Scrubber
Distribution of AN entering the distillation column
x Out
0.2
Jet
=
In
x Distill 4.2
x Out
3.9
Pr oduct
=
In
x Distill 4.2
In
x Scrub
0.1
=
In
x Distill 4.2
One of these relations is redundant with the distillation column balance.
The solution of these equations is in kg (note the changes are quite small, hence
the number of significant figures is exaggerated):
6–57
Solutions Chapter 6
In
In
Out
x InDistill = 4.302 x Scrub
= 0.102 x Decant
= x Out
Jet = 0.205 x Pr od = 3.995
Because the changes are so small, the NH3 concentration changes are negligible.
6.4.1
No purge stream exists out of the separator so that CO will build up.
6.4.2
By inspection you can see that the flow of C12 and H2 in the separator is going the
wrong way, hence any calculations make no sense.
6.4.3
Step 1, 2, 3 and 4:
CO2
Makeup
3 : 1 H2 /CO 2
6
O2
7
5
4
100%
9
Condenser
10
Methanol
CH 4 feed
11
Reactor
Reformer
steam
12
CO conv.
3
217.8 kg mol
13
2
CO
99 kg mol CH 4
Pure 5% N 2
8
3 : 1 H 2 /CO
Converter
1 kg mol N 2
Recycle
conversion
100% CH4
90% CO2
yield
55%
conversion
Methanol
solution
Chemical Reactions
a)
CH4 + 2H2 O → CO2 + 4H2
(main reformer rxn)
b)
CH4 + H2 → CO + 3H2
(reformer side rxn)
c)
2CO + O2 → 2CO2
(CO converter rxn)
d)
CO2 + 3H 2 → CH 3 OH + H 2O
(methanol rxn)
CH4 feed is 1% N2 or 1 kg mol N2
steam feed is 10% excess based on reaction (a).
6–58
Solutions Chapter 6
⎛ kgmol H 2O ⎞
⎟ = 198kgmol steam
99 kg mol CH4 ⎜ 2
⎝ kgmol CH4 ⎠
1.1 (198) = 217.8 kg mol steam
Step 5:
Basis: 100 kg mol CH4 in feed
Steps 6 and 7:
Unknowns:
6-CO2 makeup
10-reactor product
3-reformer product
7-3:1 H2/CO2
11-Methanol solution
4-O2 feed, stoichiometric
8-recycle, H2/CO2=3
12-condenser tops
5-CO conv. products
9-reactor feed, H2/CO2=3
13-purge, 5% N2
Balances:
Reformer balance
Condenser balance
CO conv. balance
purge/recycle balance
CO2 makeup balance
Feed/recycle balance
Methanol reactor balance
Steps 8 and 9:
Solve balances serially.
Reformer balance gives stream 3
CO2 =
99 kgmol CH4 conv. 0.9conv by(a) 1kgmol CO2
= 89.1kgmol CO2
1kgmol CH 4
1conv.
CO = 99(0.1) = 9.9 kg mol CO
H 2O reacted =
+
2 kgmol H 2 O 89.1kg mol CO2
1kgmol CO2
1kg mol H2 O 9.9kg mol CO
= 188.1kgmol H2 O
1kgmol CO
H 2O remaining = 217.8 - 188.1 = 29.7 kg mol H 2O
H2 = 4(89.1) + 3(9.9) = 386.1 kg mol H2
6–59
Solutions Chapter 6
N2 = 1 kg mol
CO conv. balance gives streams 4 & 5:
Stream 4:
O2 =
9.9 kgmol CO (1 / 2) kgmol O2
= 4.95kgmol O 2
1kgmol CO
Stream 5:
CO2 = 89.1 + 9.9 = 99 kg mol CO2
H2O = 29.7 kg mol H2O
H2 = 386.1 kg mol H2
N2 = 1 kg mol N2
CO2 makeup gives streams 6 and 7:
stream 7 is 3:1 H2/CO2
CO2 = 386.1/3 = 128.7 kg mol CO2 needed
Stream 6:
CO2= 128.7 - 99 = 29.7kg mol CO2
b.
purge/recycle gives stream 13:
N2 is inert species:
stream 13 =
1.0 kgmol N2
= 20 kgmol instream 13
0.05kg mol N2 / kg mol stream13
Stream 13:
H2/CO2 = 3
Let x = mol frac. of CO2 in stream 13
1 = 0.05 + 3x + 1x
4x = 0.95, x= 0.2375
N2 = 1 kg mol N2
H2 = 20 (3) (0.2375) = 14.25kg mol H2
6–60
a.
Solutions Chapter 6
CO2 = 20 (0.2375) = 4.75 kg mol CO2
Special balance gives stream 11:
128.7 CO2
29.7 H2O
386.1 H2
1.0 N 2
1 rxn occurs
13
7
11
4.75 CO2
14.25 H 2
1.00 N2
CO2 reacted = 128.7 - 4.75 = 123.95 kg mol CO2 reacted
H2 reacted = 386.1 - 14.25 = 371.85 kg mol H2 reacted
CH3OH produced = 123.95 kg mol CH3OH
H2O produced = 123.95 kg mol H2O
Stream 11:
H2O = 29.7 + 123.95 = 153.65 kg mol H2O
CH3OH = 123.95 kg mol CH3OH
mass stream 11 = 153.65 (18) + 123.95 (32) = 6732.1kg
123.95 (32)
wt. % CH3OH =
= 58.9% CH3 OH wt. %
6732.1
d.
Methanol reactor balance for 55% conversion
From special balance, each pass uses 123.95 kg mol CO2
123.95 = 0.55 (CO2)in so (CO2)in = 225.36 kg mol CO2
Stream 8 = stream 9 - stream 7
Stream 8: CO2 = 225.36 - 128.7 = 95.66
So
recycle
95.66
=
= 20.35
purge
4.75
c.
6–61
Solutions Chapter 7
7.1.1
14.7 psia
100 ft 3
psia ft 3
(296)(1.8)o R
10.73
o
(lb mol) ( R)
n=
pV 15.5 mm Hg
=
RT
760 mm Hg
n=
(15.5)(14.7)(100)
= 0.00524 lb mol
(760)(10.73)(533)
lb H 2 O = (0.00524)(18) = 0.0944 lb H 2O
Basis: 1 L gas at 780 mm Hg and T
7.1.2
1L 780 mm Hg
= 1.026 L
760 mm Hg
7.1.3
(pV)1 = (pV)2
1×1 = p×1.2
p=
7.1.4
1
= 0.83 atm
1.2
Specific volume:
V̂ =
MW = molecular weight
(R)(T) (R/MW)T (1545.3/28.97) (78 + 460)
=
=
= 13.56 ft 3 /lb m
p
p
(14.7)(144)
Molal specific volume:
V̂m =
(R)(T) (1545.3) (78 + 460)
=
= 392.8 ft 3 /lb mol
p
(14.7)(144)
Note that
V̂ =
V̂m
392.8
=
= 13.56 ft 3 /lbm
MW 28.97
7–1
Solutions Chapter 7
7.1.5
Assume ρH2O = 62.4 lb/ft3 , pbar = 14.696 psia
p=
500 f tH 2O 14.696 psia
+ 14.696 psia = 231.387 psia
33.91 ft H2 O
ft 3
359 ft3 14.696 45 + 460
Vˆ =
= 23.4
lb mol
lb mol 231.387 32 + 460
7.1.6
pV = nRT
T=
3
pV
121 kPa 25 L 1 m
(kg mol)(K)
=
3
nR 0.0011kg mol
1000 L 8.314 (kPa)(m )
= 330.8 → 331K
7.1.7
You must first convert the temperatures and pressures into absolute units:
460 + 70 = 530°R
460 + 75 = 535°R
atmospheric pressure = 29.99 in. Hg = 14.73 psia
final pressure = 29.99 in. Hg +
4 in. H 2 O 29.92 in. Hg
12 in H 2 O 33.91 ft H 2 O
ft H 2 O
= 29.99 + 0.29 = 30.28 in. Hg absolute
The simplest way to proceed, now that the data are in good order, is to apply the ideal gas
law. Take as a basis, 16.01 ft3 (do not forget to include the volume of the O2 tank in your
system) of O2 at 75°F and 30.28 in. Hg. Determine the initial pressure in the O2 tank
alone.
7–2
Solutions Chapter 7
⎛ V ⎞⎛ n ⎞⎛ T ⎞
p1 = p2 ⎜ 2 ⎟⎜ 1 ⎟⎜ 1 ⎟
⎝ V1 ⎠⎝ n2 ⎠⎝ T2 ⎠
⎛ 16.01 ft 3 ⎞ ⎛ 530°R ⎞
p1 = 30.28 in. Hg ⎜
⎟ ⎜
⎟ = 480 in. Hg absolute
3
⎝ 1 ft ⎠ ⎝ 535°R ⎠
In gauge pressure,
p1 =
(480 − 29.99) in. Hg 14.696 psia
= 221 psig
29.92 in. Hg
7.1.8
Basis: 5 L at 1 atm and T
Assume T is constant throughout the dive.
At the end of the dive the pressure is p = ρgh , or easier let x = depth in m.
x m 1 atm
+ 1 = p atm
10.34
The pressure for 1 L is obtained from p1V1 = p2 V2
⎛ V ⎞
⎛ 5 L ⎞
p2 = p1 ⎜ 1 ⎟ = 1 atm ⎜
⎟ = 5 atm
⎝ 1 L ⎠
⎝ V2 ⎠
x
=4
10.34
hence x = 41.4 m
7.1.9
Basis: air at 30 psi and 75°F in volume of tire
(1)
Initial State
(2)
Final State
p1 = 44.7 psia
p2 = ?
T1 = 535°R
T2 = 600°R
V1 = ?
V2 =
7–3
Solutions Chapter 7
N1 = ?
n2 = ?
⎛ n ⎞ R ⎛ T ⎞
p1V1
= ⎜ 1 ⎟ ⎜ 1 ⎟
p2 V2 ⎝ n 2 ⎠ R ⎝ T2 ⎠
Assume volume is the same
⎛ T2 ⎞
⎫
⎛ 600 ⎞
⎬ p2 = p1 ⎜ ⎟ = (44.7) ⎜
⎟ = 50.2 psia
The number of mols are the same ⎭
⎝ 535 ⎠
⎝ T1 ⎠
50.2 – 14.7 = 35.5 psia
7.1.10
a)
just over the limit
Basis: 1 hr
Q= ν A
3
11.3 ft 3600 s (18.0 in)2 1 ft 2
5 ft
Q=
2.875
×
10
π
=
hr
s
1 hr
(12 in)2
5 3
In 1.0 hr 2.875 × 10 ft
b)
m = ρν s
m = 0.0796
In 1 day:
lb m
lbm 11.3 ft π (18.0 )2 in 2 1 ft 2
3
2 = 6.358
s
ft
s
(12 in )
6.358 lb m 3600s 24 hr
5
= 5.49 × 10 lb m
s
1 hr
7–4
Solutions Chapter 7
7.1.11
Apply pV = nRT twice or use it once with R
V (0.7302)(560)
=
n
1.54
Basis: 1 pound mole fg
(1) 100°F (560°R) and 1.54 atm
(2) 32°F (492°R) and 1 atm (SC)
⎛ T ⎞⎛ p ⎞
p1V1
T
= 1 so V1 = V2 ⎜ 1 ⎟⎜ 2 ⎟
p2 V2
T2
⎝ T2 ⎠⎝ p1 ⎠
=
359 ft 3 560o R 1 atm
265 ft 3
=
1 lb mol 492o R 1.54 atm
lb mol
7.1.12
(a) 1.987 cal/(g mol)(K);
(b) 1.987 Btu/(lb mol)(°R);
14.7 psia 359 ft 3
3
(c )
= 10.73 ( psia ) ft / (°R )(lb mol )
492°R 1 lb mol
( )
5
2
1.013 × 10 N/m 22.4 m 3
1J
1 kg mol
(d)
3
273 K
1 kg mol 1 (N) (m)
10 g mol
= 8.314 J/(g mol) (K)
(e)
1 atm 22,400 cm3
3
= 82.06 cm (atm) /( K)( g mol)
273 K 1 g mol
( )
1 atm 359 ft 3
3
(f)
= 0.7302 ft ( atm)/ (lb mol )(°R )
492°R 1 lb mol
( )
7–5
Solutions Chapter 7
7.1.13
Basis: 1 ft3 O2 at 100°F, 740 mm Hg
n=
(740)(14.7)(1)
= 0.00238 lb mol
(760)(10.73)(560o R)
Density at 100°F, 740 mm Hg
ρ=
(0.00238)(32) 0.0761 lb/ft 3
=
1
(1.22 g/liter)
7.1.14
Basis: 1 kg mol gas at 200 kPa and 40°C
(a)
R = 8.31
3
(kPa) (m )
(K) (kg mol)
T = 40 + 273 = 313 K
MW = 44 kg/kg mol
p = 200 kPa
Density = p (MW)/RT
=
200 kPa 44 kg
313 K 1 kg mol
(b)
(kPa )(m )
( kg mol)( K)
3
8.31
= 3.38 kg/m
3
Specific gravity here is assumed to be density of propane at SC/density of
air at SC.
Sp. gr. =
(
pC3 H 8 n MWC3 H 8
RTC3 H 8
)
p air n(MW )air
RTair
=
MWC 3 H 8
7–6
MWair
=
44
= 1.52
29
Solutions Chapter 7
7.1.15
Basis: 1 lb mol gas at 760 mm Hg and 60°F
Sp. gr. =
7.1.16
(
p C3H8 MWC3H8
)
RTC3H8
p air (MW) air
RTair
=
800(44) 520 1
=1.483
560 760 29
Basis: 1 m3 H2 at 5°C and 110 kPa
a.
1 m 3 273 110 kPa 1 kg mol 2 kg
3
= 0.0952 kg/m
3
278 101.3 kPa 22.4 m 1 kg mol
b.
2 kg H 2 1 kg mol air
= 0.069 = specific gravity
1 kg mol H 2 29 kg air
7.1.17
Basis: 2m3 at 200 kPa and 25°C
pCO2 = 200(0.8) = 160 kPa
7.1.18
Basis: Gas at 30 psig and 20°C
Assume the barometric pressure = 14.7 psia
14.7 + 30.0 = 44.7 psia
pO2 = (44.7) (0.01) = 0.447 psia @ 20oC
7–7
Solutions Chapter 7
7.1.19
Basis: 1 liter final volume
Assumptions:
(1)
Temperature is constant
(2)
Ideal gas law applies
pT = pO2 + p N2 = (760 + 760) = 1,520 mm Hg
7.1.20
a. F
b. F
c. T
7.1.21
V=?
90°F
4 in. H 2O gauge
1 ft 3
70°F
215 psia
460 + 70 = 530°R
460 + 90 = 550°R
atmospheric pressure = 29.92 in. Hg = std atm = 14.7 psia
initial pressure =
200 psig + 14.7 psia 29.92 in. Hg
= 437 in. Hg
14.7 psia
7–8
Solutions Chapter 7
final pressure = 29.92 in. Hg +
4 in. H2 O 29.92 in. Hg
12 in. H 2 O 33.91 ft H2 O
ft H2 O
= 29.92 + 0.29 = 30.21 in. Hg
Basis: 1 ft3 of oxygen at 70°F and 200 psig
final volume =
3
1.00 ft 550°R 437 in. Hg
530°R 30.21 in. Hg
3
= 15.0 ft at 90°F and 4in. H2 O gauge
Formally, the same calculation can be made using
p T
V2 = V1 1 2
since n 1 = n 2
p 2 T1
7.1.22
?
20 ft 3
10 lb
CO2
30°C
10 lb CO2
known vol.
at S.C.
Solution
We can write (the subscript 1 stands for standard conditions, 2 for the conditions
in the tank)
V T
p2 = p1 1 2
V2 T 1
14.7 psia 10 lb CO 2 1 lb mol CO2 359 ft 3
303K
= p2 = 66.6
3
44 lb CO 2 1 lb mol 20 ft 
273K


psia


  T
V2
V1
2
T1
Hence the gauge on the tank will read (assuming that it reads gauge pressure and that the
barometer reads 14.7 psia) 66.6 – 14.7 = 51.9 psig
7–9
Solutions Chapter 7
7.1.23
Basis: Data in the diagram
State 1 is before corking; state 2 is after equilibrium is reached after filling. Assume pA =
pZ1 = 29.92 in. Hg abs. pZ2 = pA + ρ Hg gh = 29.92 + h where pZ2 is in inches Hg.
Use the ideal gas law: pZ2 VZ2 = pZ1VZ1
pZ2 = 29.92 (8/6) = 39.89 in. Hg
39.89 = 29.92 + h
or
h = 9.97 in. Hg
9.97 + 14 = 24 in. Hg
7.1.24
Basis: fixed amount of air in manometer
h
A = area of manometer
L
736
h = height of air, mm
L = length of manometer, mm
?
748
p1 = 755
p2 = 740 p3 = 760
Vair = (h)(A)
p 1 V2 h 2
=
=
p 2 V1 h 1
p1 = 755 – 748 = 7 mm Hg
L = 748 + h1
(1)
p2 = 740 – 736 = 4 mm Hg
L = 736 + h2
__________
12 + h1 = h2
(2)
7Ah1 = 4Ah2
(3)
(h2/h1) = 7/4 (4)
p3 = 760 – (L – h3)
Solving (3) and (4): h1 = 16 mm
L = 764 mm
7(A)(h1) = [760 – (L h3)] (A) (h3)
(5)
7–10
Solutions Chapter 7
Substituting values of h1, L in (5), one obtains:
h3 = 18 mm
The height of barometer = L – h3 = 751 mm Hg
7.1.25
1.
Basis: Flue gas at 1800°F, constant pressure, and given volume
Calculate the inlet cross-sectional area A:
Ai = π[(Di )2 ]/4 = π(42 )/4 = 12.57 ft 2
2.
Calculate the inlet volumetric flow rate Q:
Q = (velocity)×(cross-sectional area) = 25(12.57) = 314.16 ft 3/s
3.
Calculate the outlet volumetric flow rate using the ideal gas law:
Qo = Qi (To /Ti ) = 314.16(460 + 550)/(460 + 1800) = 140.40 ft 3/s
4.
Calculate the outlet cross-sectional area:
Ao = Qo /νo = 140.40/20 = 7.02 ft 2
5.
Calculate the outlet duct diameter:
(Do )2 = 4(A0 )/π = 4(7.02)/π
Do = (4(7.02)/π)0.5 = 2.99 ft
7.1.26
Basis: 10 kg FeS
(MW = 87.9)
FeS + 2HC1 → H2S + FeC13
10 kg FeS 1 kg mol FeS
= 0.114 kg mol FeS
87.9 kg FeS
Assume pressure most likely is gage. Absolute pressure = 15.71 + 76.0 = 91.71 cm Hg =
122.2 kPa
V=
nRT 0.114 kg mol FeS 8.314(kPa)(m3 ) (30 + 273)K
=
p
(kg mol)(K) 122.2 kPa
= 2.35 m3 @ 30o C and 15.71 cm Hg gauge
7–11
Solutions Chapter 7
7.1.27
Basis: 1 hr = 500 lb waste (W)
Get μg/ft3 at SC
1 μg/mL = 0.001g/L; ignore HC in totaling W
500 lb W 454 g 0.01 g HC
= 5.32×10-3g HC/ft 3SC
3
1 lb
1 g W 427,000 ft SC
The minimum volume of flue gas is
1 ft 3SC
1 g 10 µg 25 mL
= 0.047 ft 3SC
-3
6
5.32 × 10 g HC 10 µg mL
a.
0.047 ft 3SC
b.
7.1.28
1 hr
= 1.1 × 10-7 hr
3
427,000 ft SC
Basis: 1 min = 10 g VC
air
F
T = 25°C
p = 101.3 kPa
P
air
T = 25°C
VC = 1 ρρm p = 101.3 kPa
10g VC
Apply pV = nRT to get the volume of VC at SC and then use the mole fraction
information to get the volume of air (the addition of VC to air has negligible effect on the
volume)
MW
VVC
VC = 78
( )
3
n VC RT 0.0101 kg VC 1 kg mol VC 8.314 (kPa ) m
298K
=
=
(kg mol)(K) 101.3 kPa
p
78 kg VC
–3
3
= 3.01 × 10 m VC at SC
Volume fraction is the same as mole fraction here.
3.01 × 10–3 m3 VC at SC 106 m 3 air
3 3
= 3.01 × 10 m /min an absurd number!
3
1 m VC
(1.06 × 106 ) ft 3/min
7–12
Solutions Chapter 7
If a hood is used:
area =
(30)(25) in 2 1 ft 2 100 ft ⎛ 0.3048 m ⎞ 3
144 in
2
s
⎝
60s
⎠ × min
1 ft
3
= 882 m /min a smaller amount of air.
Besides inconvenience, the discharge concentration from the hood vent may be
unacceptable.
7.1.29
Basis: 100 ppm of TCE (MW 131.5) in air
ρ=
p MW
where MW is the average MW. At 100 ppm, MW =
RT
(100 ×10−6 )(131.5) + (1 − 100 ×10−6 )(29) 29.013
=
The density is essentially the same as that of air.
7.1.30
Basis: 1 min
T = (68 + 460) / 1.8 = 293K
MW = benzene = 78.11 g/g mol
ρ Bz liq = 0.879 g/cm 3
pV = nRT
R = 0.08206 (L)(atm)/(g mol)(K)
n=
2.5 cm3 liq 0.879 g Bz liq 1 g mol Bz
g mol
= 0.281
3
min
min
1 cm
78.11 g Bz
V=
0.0281 g mol 0.0206 (L )(atm )
L
293K
= 0.694
min
(g mol)(K) (740/760 ) atm
min
To dilute to 1.0 ppm, mulitiply by 106 or use 6.94 ×105 L/min (695 m 3 /min or
24,500 ft 3 /min ).
7–13
Solutions Chapter 7
7.1.31
Basis: 1 m3 gas passing at 10°F and 1 atm. vs 60°F and 1 atm.
(assume p1 = p2), T1 = 520°R; T2 = 470°R
1 m3 520o R
=1.106 m3
o
470 R
Since the moles, and mass, are directly proportional to volume at constant pressure, %
increase = 10.6%
7.1.32
Basis: CO2 in cylinder
Volume of cylinder
2
⎛ 0.75 ⎞
3
∏ r h = ∏ ⎜
⎟ (4.33) = 1.91 ft
⎝ 4 ⎠
2
Assume 100% CO2 in cylinder at 0°C
1.91 ft 3 82.6 psia 530o R
= 11.55 ft 3 at 14.7 psia, 70o F
o
14.7 psia 492 R
If the machine has a 4 gal capacity
4 gal
1 ft 3
= 0.535 ft 3 at 14.7 psia and 70oF
7.48 gal
You have more than enough CO2 to fill the machine, if it operates under atmospheric
pressure.
7–14
Solutions Chapter 7
7.1.33
Basis: 100 kg mol gas.
a)
Total
b)
Mol
87
12
1
100
CH4
C2 H6
C3H8
MW
16
30
44
kg
1392
360
44
1796
composition in vol. percent = mol percent
CH4 87%
C2H6 12%
C3H8 1%
c)
MW of gas =
kg
1796 kg
= 17.96
kg mol
100 kg mol
R = 8.314 (kPa)(m3)/(kg mol)(K)
no. of moles of gas =
T = 9°C = 282.15K
80 kg
= 4.45 kg mol
17.96 kg kg mol
( 3 ) 282.15 K
nRT 4.45 kg mol 8.314(kPa) m
V=
=
p
(kg mol )(K)
d)
e)
600 kPa
= 17.4m3
17.96 kg
1 kg mol
density =
= 0.801 kg/m3
3
1 kg mol 22.415 m at SC
Specific gravity =
density of gas at 9°C and 600kPa
density of gas at SC
600
1 17.96
8.314 282
_______________
101.3
29
8.314 273
= 3.55
kg/m3gas at 9oC, 600 kPa
kg/m3 gas at SC
7–15
wt%
77.5
20.0
2.5
100.0
Solutions Chapter 7
7.1.34
Basis: 100 mol mixture
a.
Comp
Mol
Mol. Wt.
Air
99
29
Br2
1
159.8
100
b.
Avg. Mol. Wt. =
c.
sp.gr. at SC =
d.
sp.gr. =
e.
(30.3)
(359)
sp.gr. =
(29)
(359)
3030
lb
= 30.3
100
lb mol
30.30
= 1.045
29
30.30
= 0.1895
159.8
(492) (114.7)
(560) (14.7)
= 7.54
492
30
520 29.92
7–16
lb
Wt %
2870
94.72
160
5.28
3030
100.00
Solutions Chapter 7
7.1.35
Basis: 1 lb mol of gas mixture
At 20°C
pCO2 = pT yCO2 = 0.20(740) = 148 mm Hg
pO2 = pT yO2 = 0.60(740) = 444 mm Hg
p N2 = pT y N2 = 0.20(740) = 148 mm Hg
At 40°C the pressure is
740 mm Hg 313 K
= 790.5 mm Hg
293 K
(pT )yCO2 = pCO2 = (0.20)(790.5) = 158.1 mm Hg
(pT )yO2 = pO2 = (0.60)(790.5) = 474.3 mm Hg
(pT )yN2 = pN2 = (0.20)(790.5) = 158.1 mm Hg
7.1.36
Steps 1, 2, 3:
CH4 + 2 O2 → CO2 + 2 H2O
Step 4: Unknowns:
Step 5: Balances:
CH4 + 3/2 O2 → CO + 2 H2O
P and all the compositions of P (5 unknowns)
C, H, O, N + fact that 30% of C → CO
7–17
Solutions Chapter 7
Step 6: Basis:
100 mol F
Steps 7, 8:
C balance: 100 mol in CO = 30 mol ⎫
⎪
CO 2 = 70 mol ⎬ out
100 mol ⎪⎭
2N balance:
Reqd. O2:
200 mol
XS O2: 0.20 (200) =
Total O2 in
total N2 in =
40
240 mol
240 O2 0.79 N2
= 903 mol = n N2 out
0.21 O2
2H balance: mol H 2O out = mol H 2 in =
400
= 200 mol
2
2O balance: O2 in – O2 in CO, CO2, H2O = O2 out
240 –
30
200
+ 70 +
= 55 mol
2
2
alternate: XS O2 + O2 not used for CO = 40 +
pCO = pTyCO
= (740)
30
1258
= 17.7 mm Hg
7–18
30
= 55 mol
2
Solutions Chapter 7
7.1.37
Steps 2, 3, 4:
Step 5: Basis: Gas listed in the figure
Steps 6 and 7:
Unknowns p (final), V (final); Equations: ideal gas laws
Vfinal = 1m3
n1 + n2 = nfinal
Steps 8 and 9:
n1 =
p1V1
pV
pV
n2 = 2 2 nf = f f
RT1
RT2
RTf
(600 kPa)(0.5m3 ) (150 kPa)(0.5m3 ) pf (kPa)(1.0m3 )
+
=
293K
303K
288K
pfinal = 366 kPa
7–19
Solutions Chapter 7
7.1.38
Steps 2, 3, and 4: p1 = 55 + 14.7 = 69.7 psia
Step 5: Basis: Gas with given data
Steps 6 and 7: Unknowns: T1 , T2 , Tf , pf (Vf = 400 + 50 = 450 ft 3 )
Equations:
p1V1 = n1 RT1
T1 = T2 = Tf
p2V2 = n 2 RT2
n1 + n 2 = nf
pf Vf = nf RTf
n1 =
p1V1
400(69.7)
=
RT
RT
n2 =
(50)(14.7)
RT
⎛ 50(14.7) + 400(69.7) ⎞
pf Vf = ⎜
⎟ RT
RT
⎝
⎠
pf =
50(14.7) + 400(69.7)
28635
=
= 63.8 psia = 48.9 psig
450
450
7.1.39
Steps 2, 3, and 4:
100 psig = 114.7 psia
MW =
28 lb
1 lb mole
60°F = 520°R
Step 5: Basis: 100 ft3 N2 @ 100 psig and 60°F final moles
Final Moles:
7–20
Solutions Chapter 7
100 ft 3 114.7 psia 492°R 1 lb mol
= 2.056 lb mol
14.7 psia 520°R 359 ft 3
Initial moles: 2.056 +
Initial pressure:
1
= 2.092 lb mol
28
p=
nRT
2.092 lb mol (80 + 460)o R 0.7302(ft 3 )(atm)
=
V
100 ft 3
(lb mol)(o R)
= 8.24 atm abs (121.2 psia)
(106.5 psig)
7.1.40
Step 5: Basis: 1 min
Steps 2, 3, and 4:
28.8 m3 at SC
1 kg mol
= 1.29 kg mol
22.4 m3 at SC
Step 6:
Unknowns: P, F
Step 7:
Indept. balances: air, SF6
Steps 8-9:
Balances are in moles
SF6 : F(0) + 1.29 = P(4.15×10-6 ) hence P = 3.098 × 105 kg mol
3.098 × 105 kg mol 22.4 m3 at SC 60 + 273
= 8.46×106 m3 /min
1 kg mol
273
7–21
Solutions Chapter 7
Alternate solution:
28.8 m3 at SC 333K
1
= 8.46 ×106
6
273K 4.15 ×10
7.1.41
Step 5: Basis: 15 lb CO2 added ≡ 30 min
Steps 2, 3, and 4:
Step 6:
Unknowns: F, P
Step 7:
Balances: Gas, CO2
Steps 8 and 9: Balance overall on process in moles
Gas:
0.988 (F) = P(0.966)
CO2:
0.012 (F) + 0.341 = P(0.034)
Total: F + 0.341 = P
F = 14.79 lb mol
F=
P = 15.13 lb mol
14.79 lb mol 21.9(in Hg)(ft 3 ) (60 + 460)o R
= 182 ft 3 /min
o
30 min 31.2 in Hg (lb mol)( R)
at 60oF and 31.2 in Hg
7–22
Solutions Chapter 7
7.1.42
Steps 2, 3, and 4:
Step 5: Basis: 2 min
Steps 6, 7, 8 and 9:
mol of gas out =
30,000ft 3 720 mm Hg 492o R 1 mole
= 75 mol
760 mm Hg 520o R 359 ft 3
60 mol gas 0.162 mol CO2
= 9.7 mol CO2 in
1 mol gas
75 mol gas 0.131 mol CO2
= 9.7 mol CO2 out
1 mol gas
mol of gas in =
47,800ft 3 740 mm Hg 492o R 1 mole
= 60 mol
760 mm Hg 1060o R 359 ft 3
Thus since there are more moles out than in, one may assume the system does contain a
leak. Also, assuming a leak would bring no additional CO2 into the system, the above
CO2 balance calculations indicate the given analysis is probably correct.
7–23
Solutions Chapter 7
7.1.43
Steps 2, 3 and 4:
Step 5: Basis: 1 day
Step 4:
mol in
pV = nRT
n=
pV 689 kPa 3,000 m 3 (kg mol)(K)
=
= 845.635 kg mol/day
RT
8.314(kPa ) m 3 294K
y=
p
pT
( )
yB =
mol C 4H 1 0
586 kPa
= 0.8505
mol feed
689 kPa
Steps 6-9:
mol C 4 H10 out
L=
845.635 kg mol F 0.8505 mol C 4 H10 0.8 mol C4 H10
=
mol feed
mol C4 H10 feed
day
575.35
overall mol balance
(no reaction)
F=L+V
7–24
mol C4 H10
day
Solutions Chapter 7
V = F – L = 845.635 – 575.35 = 270.26 kg mol
pV = nRT
( )
3
nRT 270.26 kg mol 8.314 ( kPa ) m 311K
3
V=
=
= 1,270.57 m
(kg mol) (K)
550 kPa
p
7.1.44
Steps 1, 2, 3, 4: Steady state problem with a reaction.
Step 5:
Basis: 1 kg mol n-butane
Step 6:
Unknowns are the moles in P (4) plus P.
Step 7:
The element balances are C, H, O, N, and we can use Σni = P so we can
get a unique solution. You can use element balances or compound
balances. The former is easier.
Step 4:
Calculate A given the 40.0% excess air.
C 4 H10 + 6 1 2 O2 → 4 CO2 + 5 H2 O
1.00 kg mol C4 H10 6 1 2 kg mol O 2
1 kg mol C4 H10
7–25
In
= 6.5 kg mol O 2
Solutions Chapter 7
= 2.60 kg mol O2 xs
= 9.10 kg mol O2 total
6.5(0.40)
9.10
(0.79 0.21)
= 34.23 kg mol N2
Steps 7, 8, and 9:
Balances (kg mol)
In
C:
1(4)
H:
Out
=
n CO2
n CO2 = 4
1(10) =
n H 2 O (2)
n H2O = 5
O2:
9.10
n CO2 +n O2 +
N2
34.23 =
=
n H2O
n O2 = 2.6
2
n N2 = 34.23
n N2
P = 4 + 5 + 2.6 + 34.23 = 45.83 kg mol
( 3) 533.1K
nRT 45.83 kg mol 8.314(kPa) m
V=
=
p
100.0 kPa
(kg mol)(K )
= 2.03x103m3
a.
flue gas analysis kg mol
mol%
CO 2
4
9
H 2O
5
11
6
O2
N2
Total
2.6
34.23
45.83
7–26
75
100
Solutions Chapter 7
7.1.45
H2 100%
F1
F2
HSiC3 100%
Reactor
1000°C
HC 
100%
W
P
Si 100%
P = (D)(V) = D
π
2
2
L (Do – DI ) =
4
2
2
2. 33 g π 100 cm (10 cm) – (1 cm)
cm
3
644.95 mol
4
g mol
= 644. 95 g mol Si product
28. 086 g
Si balance
F 2 g mol HSiCl 3
1 g mol Si
= P g mol Si, hence F 2 = P
1 g mol HSiCl 3
H g mol balance
F1 g mol H 2 2 g mol H F2 g mol HSiCl 3 1 g mol H
W g mol HCl 1 g mol H
+
=
1 g mol H 2
1 g mol HSiCl3
1 g mol HCl
Cl mol balance
F 1(0) +
F 2 g mol HSiCl 3
Solution:
3 g mol Cl
W g mol HCl 1 g mol Cl
=
1 g mol HSiCl 3
1 g mol HCl
F2 = P = 1 g mol HSiCl3
3F2 = W = 3 g mol HCl
2F1 + F2 = W
F1 =
3 g mol – 1 g mol
= 1 g mol H 2
2
7–27
Solutions Chapter 7
1 g mol H 2 644.95 g mol Si
= 644.95 g mol H 2
g mol Si
pV = nRT
V=
nRT
p
( )
3
644.95 g mol 82.06 cm ( atm) 1273 K
1L
=
( g mol)( K) 1 atm 1000 cm3
= 67,373 L
7.1.46
Basis: 1 hr
a.
2.5×10-7 g mol O2 2.9×106 cells 104 mL 103 mg mol
725 mg mol
=
3
(10 cells)(hr)
mL
1 g mol
hr
b.
45 L gas 0.40 L O2 110 kPa 1 m3
(kg mol)(K)
3
h
1 L gas
10 L 298 K 8.314(kPa)(m3 )
= 7.99 × 10-4 kg mol/hr
c.
or
799 mg mol/hr
Increase
7.1.47
Basis: 106m3 at SC of CH4 burned completely
Ignore the oxidation of S, N, and C going to CO in the material balance to simplify the
calculations. Including those products will have negligible effect on the N2 and CO2 produced.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H 2O(g)
Calculate the kg mol of CH4 burned:
106 m3at SC
kg mol
= 44,613 kg mol
22.415 m3
7–28
Solutions Chapter 7
Assume 10% excess air is used. Base the CO2 on the emission factor value.
Required O2 entering: 44,613(2) =
Excess O2 (0.1)(89,226)
Total O2:
N2 entering: 98,149(79/21) =
Exit gas
CO2 produced:
N2 exiting:
H2O exiting: (2)(44,613) =
O2 exiting: (0.10)(89,226) =
SO2: (9.6)/64 =
kg mol
89,226
8,923
98,149
369,227
(kg mol) Emission factor
44,613
369,227
89,226
8,923
43,812
0.15
0.086
0.722
0.174
0.017
2.9 × 10-7
⎧3040 / 46 =
⎪
NO 2 : ⎨1600 / 46 =
⎪1500 / 46 =
⎩
66
1.3 ×10−7
35
6.8 × 10−5
33
6.4 ×10−5
⎧1344 / 28 =
CO ⎨
⎩640 / 28 =
48
23
9.3 ×10−5
⎯⎯
→
⎯⎯
→
⎯⎯
→
Total (about)
On a dry basis:
4.5 ×10−5
511,200
1.00
mol fr.
SO2
NO2
CO
3.5 ×10−7
1.6×10-4 , 8.2×10-5 , and 7.8×10-5 respectively
1.1×10-4 and 5.5×10-5 , respectively
7.1.48
Basis: 103 L of No. 6 fuel oil
103L oil 0.86 g 1000 cm3
= 8.6×105 g or 8.6×102 kg of oil
3
1 cm
1L
To get the moles of pollution components alone in the gas produced, the kg have to be converted
using MW (ignore the particulate matter)
SO2
SO3
NO2
kg
MW
g mol
19 (0.84) = 15.96
0.69 (0.84) = 0.580
8
64
80
46
249
7.3
174
7–29
m3 at SC on the basis
of 103/kg
6.5
0.18
4.5
Solutions Chapter 7
CO
CO2
0.6
3025
28
44
21
6.875 × 104
0.55
1,790
Example of calculating the m3 at SC on the basis of 1 metric ton (103 kg) of No. 6 fuel oil burned:
SO2:
0.249 kg mol 22.415 m3 at SC 103 kg oil
= 6.5 m3 at SC
2
8.6×10 kg oil
1 kg mol
7.1.49
Basis: 103 L of No. 6 fuel oil
103 L oil 0.86 g 1000 cm3
= 8.6×105 g or 8.6×102 kg oil
1 cm3
1L
Calculate the kg mol of each compound in the oil:
C
H
O
N
S
ash
Mass fraction
0.8726
0.1049
0.64 ×10−2
0.28 ×10−2
0.84 ×10−2
0.04 ×10−2
kg
750.4
90.21
5.5
2.4
7.2
0.3
MW
12
1.008
16
14
32
-
kg mol
62.53
89.49
0.34
0.17
0.23
-
MW
44
kg
2751
46
7.8
Calculate the moles of product gas
CO2:
C + O2 → CO2
kg mol
62.53
N:
N + O2 → NO2
0.17
S:
The SO2 and SO3 are from the EPA data and Problem 13.55
kg
0.58
15.96
SO3:
SO2
kg mol
0.0073
0.249
If the S + O2 → SO2 is used as the product from the S in the oil, the value of 0.23 can be
compared with 0.25 (ignoring the SO3).
7–30
Solutions Chapter 7
7.1.50
7–31
Solutions Chapter 7
7.1.51
Basis: 1 mol of NH3 fed
NH3 balance:
(1 + R) (0.2) = R
0.8 R = 0.2
⎛ mol NH3 recycled ⎞
R = 0.25 ⎜
⎟
⎝ mol NH3 fed ⎠
1 m3 (100o C, 150 kPa)NH3 fed 0.25 mol recycled (150 + 273)K
1 mol NH3 fed (100 + 273)K
⎛ m3 recycled at 150o C, 150 kPa ⎞
= 0.284 ⎜ 3
⎟
o
⎝ m NH3 fed at 100 C, 150 kPa ⎠
7.1.52
Process: Steady state with reaction and recycle
Step 5:
Basis: 1 min
Steps 2, 3, and 4:
260L C6H6
950L H2
C6 H 6 + 3 H 2 → C6 H12
Covert L → mol
7–32
Solutions Chapter 7
pV
RT
n=
feed
p = 150 kPa
n H 2 = 45.96 g mol
T = 100°C + 273 = 373K
(kPa ) m3
R = 8.314
(kg mol)(K)
( )
n C 6 H 6 = 12.58 g mol
Steps 6, 7, 8 and 9:
Select the overall system for the first balances.
H2 :
In –
45.96 –
C 6 H6 :
12.58 –
C6H12:
0
–
Out
n °H 2
n °C 6 H 6
n °C 6 H12
+ generation
+
0
– Consumption = 0
– 0.75(45.96) = 0
+
–
0
+ 0.75(45.96)
1
3
0.75(45.96)
–
0
1
3
= 0
= 0
from 1: n °H 2 = 11.5 g mol
from 2: n °C 6 H 6 = 1.09 g mol
from 3: n °C 6 H12 = 11.5 g mol
b.
Volumetric flow rates:
( n ) (RT) =
=
o
VH2
H2
p
T = 200°C = 473K
P = 100 kPa = 0.987 atm
452 liters/min
VC6H6 = 42.8 liters/min
C6 H12 = 452 liters/min
c.
H2 balance on Reactor + Separator
H2: (45.95 + 0.90R) – (0.90R +11.5 ) + 0 – 0.48 (45.95 + 0.90R) = 0
Substitute for n °H 2 and solve to get R = 28.7 kg mol
Volumetric flow rate (R) = 890 liters/min
7–33
Solutions Chapter 7
7.1.53
Basis: 1 day = 104 kg C2H2O
If the overall system is chosen, the unknowns are F1 and ni, and three independent
equations can be written from C, H, and O balances plus nCO2 = nH2O.
Pick the system of the reactor plus separator, and use the data for the conversion.
Step 5: Basis
10, 000 kg 1.00 kg mol
= 227.3 kg mol C 2H 4O/day
day
44 kg
Steps 6, 7, 8 and 9:
Mixing point: F1 + R = F2 ⇒ 0.40F1 + R = n FC22 H 4 , 0.60F1 = nFO22 . Unknowns:
F
F
F1 ,F2 ,R,n O 2 , nC 2 H 4
Reactor plus separator:
C2H4 balance:
consumed
⎡ in
⎤ out 
⎢⎣0.4(F1 )+ R ⎥⎦ – R – [0.4(F1 ) + R]0.5 = 0
generation


(
)
(
)
(
)=0
F
0
.
4
+
R
0
.
50
0
.
70
]
C2H4O balance: 0–227.3+[ 1
F1 = 811.79 kg mol/day
R = 324.7 kg mol/day
7–34
Solutions Chapter 7
F1 + R = F2 = 1136.49
1136.5 kg mol 22.40 m3 at SC 1 day
3
(a)
= 1059 m at SC/hr
day
1 kg mol
24 hr
adjust recycle


324.71 kg mol of C2 H4 in R
22.40 m at SC 283K 101.3 kPa
3
(
)( )
(b) 811.79 0.40 kg mol of C 2 H 4 in F1 22.40 m at SC 273K 100 kPa = 1.05
3
Overall balances (kg mol): unknowns n1 tr 3
F1
C:
811.79 (.40) (2) = 227.3 (2) + nCO2
Balances 3
nCO2 = 194.83
H:
811.79 (.40) (4) = 227.3 (4) + nW (2) nW
O:
811.79 (.60) (2) = 227.3 (1) + 194.83 (2)
+ 194.83 (1) + nO2 (2)
nO2 = 81.18
G = 194.83 + 194.83 + 81.18 = 470.84
470.84 kg mol P 2 22.4 m 3 at SC 353 101.3
day
273 100
kg mol
(c)
4
3
= 1.38 × 10 m gases/day at 80°C and 100 kPa
7–35
= 194.83
Solutions Chapter 7
7.1.54
Y H2 O (100%)
From
incineration
4.0%
2.60%
2.0%
16.0%
52.0%
@ 60°F and 30 in Hg abs
P=?
CO2 12.2%
CO 0.7%
O2
2.4%
3
N
84.7%
2
W = 1200 ft @ 60°F and
30 in Hg abs.
80.5% CH4
17.8% C 2 H6
1.7% N 2
CO2
CO
CH4
H2
N2
Step 1, 2, 3, and 4:
Steady state process with reaction.
Step 5: Basis: 1 min.
Step 6, 7, 8, 9:
Element balance: have C, H, O, N, and have 4 unknowns P, Y, F, A.
3
Basis: 1200 ft of W at 60°F and 20 in. Hg
C:
N2:
H2:
O:
Fin
(0.04 + 0.26 + 0.02) F
(0.52) F
(0.04+0.16) F
(0.08+0.26)F
Ain
Win
+
0
+ (0.805 + 2×0.178)W
+ 0.79A +
0.017W
+
0
+
(1.61+0.534)W
+ 0.42A +
0
–
–
–
–
Yout
Pout
0 –
(0.122 + 0.007)P
0 –
(0.847) P
Y –
0
Y – (0.244+0.007+0.048)P
Introduce W = 1200 ft3
C:
N2:
H2:
O:
0.32F
0.52F
0.20F
0.34F
+
+
+
+
0
0.79A
0
0.42A
+
+
+
+
1393.2 ft3
20.4 ft3
2572.8 ft3
0
–
–
–
–
0
0
Y
Y
–
–
–
0.129P
0.847P
0
0.299P
=0
=0
=0
=0
3
Solve: F = 3,975 ft
3
A = 19,510 ft
3
Y = 3,370 ft
3
P = 20,660 ft
Air in =
19,510 ft 3 @ 60°F & 30 inches Hg 540°R 30 inches Hg
3
= 20,535 ft
520°R 29.6 inches Hg
7–36
=0
=0
=0
=0
Solutions Chapter 7
7.1.55
Basis: 1 lb mol of 50% H2 and 50% C2H4O
p1V1 = n1RT1
p1 n1
=
p2 n2
p2V2 = n2RT2
T1 = T 2
760 mm Hg 1
hence n 2 = 0.922 lb mol
=
700 mm Hg n 2
Let y = mol of C2H6O formed
(0.5 – y) + (0.5 – y) + y = 0.922
from which y = 0.079 lb mol
Degree of completion =
C2H6O formed
0.079
×100 = 15.8%
0.50
7.1.56
C6 H12O6 is glucose and C3H8O3 is glycerol
The reaction equation is
a CH1.8O0.5N 0.5 + b NH 3 + c C6 H12O6
→ d C3H8O3 + e CO2 + f N2 + g H2 O
The amount of CO2 measured (52.4L at 95 kPa and 300 K) is
52.4 L CO2 1 m3 95 kPa
(kg mol)(K)
= 2 ×10−3 kg mol or 2.00 g mol
3
1000 L
300 K 8.314 (kPa)(m )
Use element balances and specifications to get the coefficients in the reaction equation. Let c = 1
(divide both sides by c) be the basis for obtaining the coefficients: then there are 6 unknowns, 4
equations, and two specifications:
C:
a + 6c = 3d + e
H:
1.8a + 3b + 12 = 8d + 2g
7–37
Solutions Chapter 7
O:
0.5a + 6 = 3d + 2.00(2) + g
N:
0.5a + b = 2b
Specifications:
f
=1
b
and
e
=2
c
The solution of the equations is:
a = 7.27,
b = 3.64,
d = 1.09,
f = 3.64,
g = 1.64
The glycerol produced was 1.09 g mol , and
The biomass reacted was 7.27 g mol
7.2.1
1.a
Immediate but approximate
2.e
Exact but requires access to a handbook
3.b
Good accuracy, needs calculation of z
4.c
Good accuracy, more complicated calculation than b even in a program
with a data base
5.e
May require a long search to filter out a good value
For other answers, look at the answer to P15.1.
7.2.2
Will get the most accurate value, and be quick, assuming you have the handbook that
contains the required data. For other answers, look at the answer to P15.1.
7–38
Solutions Chapter 7
7.2.3
Some valid answers are
(1)
continuous analytical functions, and these functions can be differentiated
and integrated
(2)
higher accuracy (with complex equations)
7.2.4
B
C
D
+ 2 + 3
V
V
V
ˆ
pV
The results for
= z are shown in the Excel table below. You can decide how much
RT
z=1+
accuracy is needed.
p(atm)
4 terms
2 terms
1 term
% difference
2 from 4
1 from 4
0
10
50
100
200
500
1000
5000
1.00
0.95
0.88
0.77
0.80
0.90
0.95
0.99
1.00
0.95
0.88
0.76
0.79
0.90
0.95
0.99
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0
0
0
-0.8
1.6
0
0
0
7.2.5
⎛ L ⎞
a ⇒ (atm)⎜
⎝ g mol⎠
2
b ⇒ L/g mol
α ⇒ dimensionless
7–39
0
-4.9
-13.3
-29.8
-25.1
-10.7
-3.6
-1.1
Solutions Chapter 7
7.2.6
Redlich-Kwong
p=
RT
a
− 1/ 2
ˆ
ˆ
ˆ + b)
(V − b) T V(V
1
2
(
3
a = 86,870 psia (°R) ft /lb mol
)
2
b = 0.3536 ft 3 /lb mol
n=
63.9
= 1.997 lb mol
32
V
6.7
3
Vˆ = =
= 3.355 ft /lb mol
n 1.997
Solving, p = 1335 psia
Therefore, the reading on the pressure gauge is incorrect.
7.2.7
Basis: 1 min
(2)
CO2
(1)
CO2
2000 m3/min
20oC
500 kPa
110oC
4800 kPa
Calculate the moles entering the compressor:
500
atm
p r1 = 101.3
= 0.068
72.9 atm
4800
atm
p r2 = 101.3
= 0.650
72.9 atm
Tr1 =
(20 + 273)K
= 0.96
304.2 K
Tr2 =
(110 + 273)K
= 1.26
304.2 K
2
⎛ cm3 ⎞
⎛ m3 ⎞
a = 3.6 × 106 atm ⎜
=
364.68
kPa
⎟
⎜
⎟
⎝ g mol ⎠
⎝ kg mol ⎠
7–40
2
Solutions Chapter 7
b = 42.8
cm3
m3
(kPa)(m3 )
= 0.0428
, R = 8.314
g mol
kg mol
(kg mol)(K)
Use Van der Waals’ equation:
⎛
n 2 a ⎞
p
+
⎜
⎟ (V − nb) = nRT
V2 ⎠
⎝
As the initial guess for the number of moles in State 1 use the ideal gas law
n1 =
p1V1
(500)(2000)
=
= 410.5 kg mol
RT1
(8.314)(293)
The number of moles at condition (1) is obtained from
⎛
n12 (364.68) ⎞
=
⎜ 500 +
⎟ (2000 − 0.0428 n1 ) (8.314)(293)n
1
20002 ⎠
⎝
From Polymath (on the CD)
n1 = 420 kg mol
Calculate the m3 exiting the compressor:
Insert into Vander Waals’ equation new parameters:
p = 4800 kPa
T = 383 K
n = 420 kg mol
Solve for V
V = 246 m3 at 383K and 4800 kPa
7–41
Solutions Chapter 7
7.2.8
Basis: 460 kg CO2
a. RK Equation:
460 kg CO2 1 kg mol
= 10.455 kg mol CO2
44 kg
Tc = 304.2K, pc = 7385 kPa
a = 6458 (kPa) (m6) (K0.5)/(kg mol)2
b = 0.0297 m3/kg mol
(kPa )( m3 )
R = 8.314
(kg mol)(K)
V
10.4 m3
m3
V̂ =
=
= 0.995
n
10.455 kg mol
kg mol
p=
(8.314)(290)
6458
−
0.995 − 0.0297 0.995(0.995 + 0.0297)(290)1/ 2
= 2.126 ×103 kPa
b. SRK Equation:
T/Tc = 290/304.2 = 0.953
Data as above:
aʹ′ =
0.42748(8.314)2 (304.2)2
= 379.2(kPa)(m6 )/(kg mol)2
7358
b=
0.08664(8.314)(304.2)
= 0.0297 m3 /kg mol
7358 × 103
p=
(8.314)(290)
(379.2)(1.040)
−
0.995 − 0.0297 0.995(0.995 + 0.0297)
2
κ = 0.480 + (1.574)(0.225) − 0.176(0.225)
=
0.843
1
2
λ = [1 + 0.843(1 − 0.953
= )]2 1.040
No significant difference.
p = 2.111×103 kPa
7–42
Solutions Chapter 7
7.2.9
Basis: 100 ft3 of NH3 at 20 atm and 400ºF
Part I. Calculate the moles of NH3:
Data:
Tc = 405.3 K
→
729.5º R
Pc = 111.3 atm → 1636 psia
= 0.250
400º F → 860º R
R = 10.73 (psia) (ft3) / (lb mol) (ºR) 20 atm = 293.9 psia
Tr = T/Tc = 860/729.5 = 1.179
p=
RT
aα
−
ˆ − b V(V
ˆ ˆ + b) + b(V
ˆ b) −
V
α = [1 + κ(1 − T
=r1/ 2 )]2
0.9890
κ = 0.37464 + 1.54226ω − 0.26992 ω2= 0.1276
⎛ R 2Tc2 ⎞
a = 0.45724 ⎜
⎟ = 17,124
p
⎝ c ⎠
⎛ RT ⎞
b = 0.07780 ⎜ c ⎟ = 0.3722
⎝ pc ⎠
From Polymath:
V̂ = 29.9 ft 3 / lb mol
100
= 3.34 lb mol NH3 → 56.8 lb NH3
29.9
7–43
Solutions Chapter 7
Part II: Calculate the final pressure
5 ft 3
Final V̂ =
= 1.497 ft 3 /lb mol
3.34 lb mol
350ºF
810ºR
Introduce V̂ into the PR equation. From Polymath
p=
(10.73)(810)
17,124(0.984)
−
1.497 − 0.3722 1.497(1.497 + 0.3722) + 0.3722(1.497 0.3722)
−
= 2462 psia
7.2.10
Basis: CO2 at 81 psig and 25oC
Pressure in can (81.0 psig + 14.7 psia) (1 atm/14.7 psia) = 6.51 atm
Temperature 25°C ⇒ 298.15K
Tc = 304.2 K
V = π (4.05 cm )2 (17.0 cm) = 876 cm 3
pc = 72.9 atm
n = ? CO2
ω = 0.225
MW= 84.00
C6 H8 O7 + NaHCO 3 →
MW =44.01
CO2
+ H2 O + NaC 6 H7O 7
Using Peng-Robinson equation
(
)
(
) ( )
n 3 baα – b 2 RT – b3 p – n2 (V) aα – 2bRT – 3b 2 p – n V2 (bp – RT) – V3 p = 0
Equations ⇒
( )
f ( n) = n3 q – n2 (V)(z ) – n V2 ( bp – RT ) – V3 p = 0
q = baα – b 2 RT – b3 p
z = aα – 2bRT – 2b2 p
⎛ R 2Tc 2 ⎞
6
a = 0.45724 ⎜
⎟ = 3.908 10
⎝ pc ⎠
×
7–44
Solutions Chapter 7
b = 0.07780(RTc / pc ) = 26.64
1/ 2
α = ⎡1 + κ (1 − (T
= / Tc ) ⎤
⎣
⎦
Tr = 298.15/304.2 = 0.980
(Tr)1/2 = 0.990
2
1.007
2
κ = 0.37464 + 1.54226ϖ – 0.26992ϖ = 0.7080
The ideal gas law gives 0.233 mol CO2
R = 82.06 (cm3)(atm)(g mol)(K)
T = 298.18K
p = 6.51 atm
Introduce the values into the PR equation to get n = 0.2420 mol CO2
0.2420 mol CO2 1 mol NaHCO3 84.00 g NaHCO3
= 20.3 g NaHCO 3
1 mol CO2
1 mol Na HCO3
For propane: Tc = 369.9K, pc = 42.0 atm, ω = 0.152
7.2.11
2
κ = 0.37464 + 1.5422(0.152) − 0.26992(0.152)
=
2
α = [1 + 0.60283(1 − 1.00686)]
=
Propane Coefficients
RK
PR
0.60283
0.992
a
b
180.5 ×106 (atm)(K1/2 )(cm6 )/(g mol)2
10.04 ×106 (atm)(cm6 ) /(g mol)2
Computer results:
Redlich-Kwong: 1193.14 cm3 /g mol
Peng-Robinson:
1168.91 cm3 /g mol
Determine the volume at condition of state 2 (383 K, 4800 kPa)
⎛
(419.9)2 (364.68) ⎞
=
⎜ 4800 +
⎟ (V2 − (0.0428)(419.9)) (419.9)(8.314)(383)
V22
⎝
⎠
V = 246 m3 at state 2
7–45
62.7 cm3 /g mol
56.3 cm3/g mol
Solutions Chapter 7
7.2.12
For CO2, ω = 0.225 and Tc = 304.K, pc = 72.9 atm or 7385 kPa
R = 8.314 (kPa) (m3)/(kg mol) (K)
a = 6458 (kPa) (m6) (K0.5)/(kg mol)2
b = 0.0297 m3/(kg mol)
1200 =
(8.314)(290)
6458
−
−
2
ˆ − 2.97 ×10 ) V(V
ˆ ˆ + 2.97 × 10−2 )(290)1/ 2
(V
V̂ = 1.876 m3 /kg mol from Polymath
n CO2 =
V 10.4
=
= 5.54 kg mol
ˆ 1.876
V
n CO2 =
5.54 kg mol
48 kg
= 266 kg CO2
1 kg mol
7.2.13
nRT
n2 a
p=
–
V − nb V2
T = 492.0°R
nRT ⎞ ⎛ V2 ⎞
⎛
⎜–
⎟
a = ⎝p –
V – nb ⎠ ⎝ n2 ⎠
R = 0.7302
( ft 3)(atm)
(lb mol )(°R )
Basis: 1.0 lb mol
1.0 lb mol 0.7302
200 atm =
200 =
(ft 3)(atm )
( lb mol)(°R)
492.0°R
(1.806 ft3 – (1.0 lb mol) b)
( )
359.3 ft 3 ( atm)
(1.860 ft 3 – b)
– 0.2891 a
similarly
7–46
–
(1.0 1b mol)2 a
(1.86ft 3 )
2
Solutions Chapter 7
1000 =
( )
359.3 ft 3 (atm)
( 0.741 ft 3 – b )
–1.82a
359.3 ⎞ ⎛ 1 ⎞
⎛
–
a = ⎝ 1000 –
0.741 – b ⎠ ⎝ 1.82 ⎠
200 =
359.3 ⎞
359.3
⎛ 1 ⎞⎛
1000 –
– (0.289)⎝ –
⎠
⎝
0.741 – b ⎠
(1.860 – b )
1.82
200 =
57.07
359.3
+ 158.8 –
(1.860 – b)
(0.741 – b )
41.20 =
57.07
359.3
–
(1.860 – b) (0.741 – b)
b = 0.4808
ft 3
lb mol
⎛ ft 3 ⎞
a = 209.3(atm) ⎜
⎟
⎝ lb mol ⎠
2
7.2.14
a. Solution by compressibility factor method:
Basis: 80 lb water at 900°K
V̂ =
10 ft 3 18 lb
1cm3 /g mol
= 140.8 cm3 /g mol
3
80 lb 1 lb mol 0.016 ft /lb mol
Tc = 647.4K;
pc = 218.3 atm from Appendix D1 (217.6 atm for the CD)
RTc
82.06(cm3 )(atm) 647.4 K
Vci =
=
= 243 cm3 /g mol
o
pc
( K)(g mol) 218.3 atm
Vrʹ′ =
V̂
140.8
T
900
=
= 0.579; Tr =
=
= 1.39
Vci
243
Tc
647.4
From the compressibility charts, pr = 1.9
7–47
Solutions Chapter 7
p = pr pc = (1.9)(218.3) = 415 atm
b. solution using the Redlich-Kwong equation of state:
p=
R 2 Tc
RT
a
where a = 0.4278
− 1/2
ˆ
ˆ ˆ
pc
(V-b)
T V(V+b)
RTc
b = 0.0867
pc
2.5
2
⎡
(cm3 )(atm) ⎤
0.4278 ⎢82.06
(647.4 K)2.5
⎥
(K)(g mol) ⎦
(cm3 ) 2 (atm)(K)1/2
⎣
a=
= 1.407 × 108
218.3 atm
(g mol)
0.0867 82.06(cm3 )(atm) 647.4 K
cm3
b=
= 21.1
(K)(g mol)
218.3 atm
g mol
p=
(82.06)(900)
1.407 × 108
−
= (617 206) atm
−
(140.8 − 21.1) (900)1/ 2 (140.8)(140.8 + 21.1)
= 411 atm
7.2.15
For ethane: Tc = 516.3 and pc = 63.0 atm
a. Use Van der Waals’ equation.
Basis: Ethane at 100oF and 2000 psig (MW = 30)
p = 2014.7 psia
V = 1.0 ft3
T = 560ºR
2
2
⎛
⎛ cm3 ⎞ ⎞
⎛ ft 3 ⎞
6
-3
4
a = ⎜ 5.50 × 10 atm ⎜
⎟ ⎟ (3.776 × 10 ) = 2.07 ×10 psia ⎜
⎟
⎜
g mol ⎠ ⎟
⎝
⎝ lb mol ⎠
⎝
⎠
⎛
cm3 ⎞
ft 3
-2
b = ⎜ 65.1
1.60
10
=1.04
×
)
⎟ (
g mol ⎠
lb mol
⎝
⎛
an 2 ⎞ ⎛ V
⎞
p
+
− b ⎟ = RT
⎜
2 ⎟ ⎜
V ⎠ ⎝ n
⎠
⎝
⎛ ba ⎞
⎛ a ⎞
n 3 ⎜ 2 ⎟ − n 2 ⎜ ⎟ + n(pb + RT) − pV= 0
⎝ V ⎠
⎝ V ⎠
7–48
Solutions Chapter 7
n3 − 0.962n 2 + 0.376n − 0.0936 = 0
Iterate to find value of n.
Use Excel to solve this cubic equation to obtain a positive, real answer.
n = 0.594 lb mol or 17.8 lb ethane
b. Use the compressibility factor method.
pc
Tc
pr
Tr
709 psia
549ºR
2.84
1.02
mass =
1(ft)3 2014.7 psia (oR)(lb mol) 30 lb
= 23.4 lb ethane
560oR
0.43
10.73(psia)(ft)3 mol
7.2.16
a.
Van der Waals equation
2
⎛ 27 ⎞ R Tc
a = ⎜ ⎟
⎝ 64 ⎠ pc
⎛ 1 ⎞ RT
b = ⎜ ⎟ c
⎝ 8 ⎠ pc
Methane
Propane
Tc = 190.7K
Tc = 369.9K
pc = 45.8 atm
pc = 42.0 atm
Methane:
190.7
= 4.16
45.8
Propane:
369.9
= 8.81
42.0
Propane will be higher
for both coefficients
b.
z from the compressibility charts
SRK equation:
a ʹ′ =
0.42748 R 2 Tc
pc
b=
0.08664 RTc
pc
Answer is same as for (a); propane
will be higher for both coefficients
7–49
0.43
Solutions Chapter 7
7.2.17
pabs = pg + barometric pressure
T = 273 – 50 = 223 K
pabs = 39 + 1 = 40 atm
Basis: 1 g mol H2
H2 Coefficients
a
b
Van der Waals
0.246 ×106 (atm)(cm)6 /(g mol)2
26.6(cm)3 /g mol
Redlich-Kwong
1.439 ×106 (atm)(K1/2 )(cm)6 /(g mol)2
18.5(cm)3 /g mol
Computer results
Van der Waals:
Redlich-Kwong
471.78 cm3 /g mol
471.26 cm3 /g mol
5L 1000 cm3 1 g mol
= 10.60 g mol(van der Waals)
L
471.78 cm
5 (1000)/471.26 = 10.61 g mol (Redlich − Kwong)
7.2.18
For CO2, ω = 0.225 and Tc = 304.2 K, pc = 72.9 atm
a = 62.75 ×106 (atm)(K1/2 )(cm6 )/(g mol)2
b = 29.9 cm3 /g mol
Computer results:
3
Redlich-Kwong = 1559.6 cm /g mol
6250 cm3
3
= 1562.5 cm /g mol
The experimental molar volume =
4 g mol
7–50
Solutions Chapter 7
7.3.1
a.
b.
Basis: 7 lb N2 or 7/28 = 0.25 lb mol N2
nRT 0.25 lb mol (120 + 460)o R 0.732(atm)(ft 3 )
=
V
0.75 ft 3
(lb mol)(o R)
= 141 atm
p=
Tc = 126.3K
pc = 33.54 atm
580
= 2.55
(126.3)(1.8)
Tr =
0.75 ft 3
V̂
0.25 lb mol
Vrʹ′ =
=
= 0.607
RTc
(atm)(ft 3 )
(1.314)
(126.3K)
pc
(lb mol)(K)
33.54 atm
From the compressibility charts
pr = 4.3
so that
p = pr pc = (4.3) (33.54) = 144 atm
Note: You calculate z from Tr and Vrʹ′ , and then use p =
znRT
V
7.3.2
Basis: 2 g mol C2H4
pC = 50.5 atm
TC = 283.1K; p = ?
T = 95 + 273.1 = 368.1 K
RTc
82.06(cm3 )(atm) 283.1 K
V̂Ci =
=
= 460 cm3 /g mol
pc
(g mol)(K)
50.5 atm
418 cm3
V̂=
= 209 cm3 /g mol
2
V̂ ˆ
=Vri = 0.45
V̂ci
Tr =
368.1
= 1.30
283.1
From chart, pr = 2, hence p = 101 atm.
Use of the Pitzer ascentric factor would require a trial and error solution.
7–51
Solutions Chapter 7
7.3.3
Basis: 3 lb mol of gas at 252°C and 463 psia.
Volume = 50 ft3
R=
(1 atm)(359 ft 3 )
(atm)(ft 3 )
= 1.315
(1 lb mol) (273 K)
(lb mol)(K)
⎛
⎞
⎜ 463 psia ⎟
50 ft 3 )
⎜
psia ⎟ (
⎜ 14.7
⎟
atm
⎝
⎠
z=
= 0.76
⎛
(atm)(ft 3 ) ⎞
o
(3 lb mol) ⎜1.315
⎟ (273 + 252 C)
(lb mol)(K) ⎠
⎝
The Pitzer ascentric factor is not as convenient to use as the compressibility charts.
From compressibility charts (see the CD for better precision)
z = 0.76
Tr =
525
= 1.05
500
pr = 0.7
Thus
463 psia
= 0.7
pc
pc ≅ 661 psia (45.0 atm)
7–52
Solutions Chapter 7
7.3.4
Basis: 1 lb n-octane
Tc = 1025°R
pr =
pc = 24.6 atm
MW = 114
27
= 1.10
24.6
To get the temperature of the gas
1 lb 1 lb mol 359 ft 3SC 1 atm T K z
114 lb
27 atm 273 K
Tz = 845
= 0.20 ft 3
zTR = 845/1025 = 0.825
From the compressibility Chart Fig. 14.4b (expanded in the CD)
read Tr = 1.16, so that T = (1.16)(1025) = 1189o R (729°F)
Alternate solution: Calculate z = 0.71 and use pV = znRT.
7.3.5
Basis: 50 lb CO2 (50/44) = 1.138 lb mol CO2
V = 5.0 ft3
p = 1600 + 14.7 = 1614.7 psia
Pc = 1070 psia
Tc = 547.5°R
pr =
1614.7
= 1.51
1070
V̂ =
5.0
= 4.40 ft 3 /lb mol
1.138
Vci = 5.50 ft 3 /lb mol so that Vri =
4.40
= 0.800
5.50
Tr from Figure 14.4b is (expanded in the CD) is 1.44
T = (1.44) (547.5) = 788o R (328°F)
The Pitzer ascentric factor is less convenient to use for this problem.
7–53
Solutions Chapter 7
7.3.6
Basis: 10 kg CH4
MW CH4 = 16.03
10 kg 1 kg mol
= 0.624 kg mol
16.03 kg
Tc = 305.4K and pc = 4883 kPa
V = 0.0250 m3
pr =
p = 14,000 kPa gauge = 14,015 kPa absolute
14, 015
= 2.87
4883
V̂ci =
RTc
(8.314)(305.4)
=
= 0.520 m3 /kg mol
pc
4883
Vr =
0.0250
= 0.048
0.520
From the compressibility chart, Tr ≅ 1.05 hence T = (1.05)(305.4) = 321 K
7.3.7
Basis: 1ft3 CH4
pV = nRTz
R=
p = 214.7 psia
10.73(psia)(ft 3 )
(lb mol)(oR)
Tc = 191°K → 344°R
pc = 45.79 atm → 672.9 psia
Tr = 540/344 = 1.57
pr =
214.3
= 1.0
214.7
z = 0.975
n=
(214.3)(1)
= 0.038 lb mol
(10.73)(540)(0.975)
weight (mass) = (0.038) (16.03) = 0.608 lb
By use of the Pitzer ascentric factor (ω = 0.008 for CH4)
z = z0 + z1ω = 0.976 + 0.024 (0.008) = 0.976
7–54
Solutions Chapter 7
7.3.8
Basis: 25L of CO2 at 25°C and 200 kPa
For CO2:
pc = 7385 kPa
Tc = 304.2 K
pr =
p
200 kPa
=
= 0.0271
pc
7385 kPa
Tr =
T
(25 + 273) K
=
= 0.98
Tc
304.2o K
From the compressibility chart, at these coordinates, z ≅ 0.99
pV = nzRT
n=
200 kPa
1 atm 25L
(g mol)(K)
101.3 kPa
0.08206(L)(atm) 298 K 0.99
= 2.04 g mol
m = (2.04) (44) = 87.9 g
or 0.0879 kg
7.3.9
Basis: 106 ft3 at 60°F and 14.7 psia
pr Vr = zr n r R r Tr
at the reservoir
pV = znRT
at SC
Tr =
T
120 + 460 580
=
=
= 1.69
Tc
191(1.8)
344
pr =
p
1000
1000
=
=
= 1.48
pc
(45.79)(14.7)
674
zr = 0.95
⎛ p ⎞ ⎛ T ⎞⎛ z ⎞
⎛ 14.7 ⎞⎛ 120 + 460 ⎞⎛ 0.95 ⎞
3
Vr = V ⎜ ⎟⎜ r ⎟⎜ r ⎟ =106 ⎜
⎟⎜
⎟⎜
⎟ = 15,600 ft
⎝ 1000 ⎠⎝ 60 + 460 ⎠⎝ 1 ⎠
⎝ pr ⎠ ⎝ T ⎠⎝ z ⎠
7–55
Solutions Chapter 7
7.3.10
Basis: 1 kg propane
For propane: Tc = 370 K
pc = 4255 kPa
and R = 0.18855 (kPa)(m3 )/(kg)(K)
Reduced temperature = Tr =
T
230 + 273.2
=
= 1.36
Tc
370
Reduced pressure
p
6000
=
= 1.408
pc
4255
= pr =
From a generalized compressibility chart
at pr = 1.408 and Tr = 1.36, z ≈ 0.80
Since pV = znRT,
V̂=
V̂ =
zRT
p
(0.80)(0.18855)(503.2)
= 0.0127 m3/kg
(6000)
7.3.11
a)
Yes
g)
No
b)
No
h)
No
c)
Yes
d)
No
e)
Yes
f)
No
7–56
Solutions Chapter 7
7.3.12
Basis: 1 g mol C6H5C1
pc = 4518 kPa
pr =
230
= 0.051
4518
(MW = 112.56)
Tc = 632.4 K
Tr =
380
= 0.601
632.4
These values fall below the gaseous region in the compressibility chart here C6H5C1 is
not a gas. It is a liquid (sp.gr = 1.107), so that the volume is (1)(112.56)/1.107 =
102 cm3
7.3.13
The answer is it depends on the gas temperature and the pressure. If pV = znRT applies,
the ideal gas law is represented by z = 1, and for a non-ideal gas, z can be greater or less
znRT
than 1. The pressure is p =
. So preal = pideal z, for fixed V and for a fixed number
V
of moles and temperature (the volume of cylinder is fixed). The pressure prediction by
the ideal gas law will be conservative (higher than the true pressure) for z < 1, and lower
than the true pressure for z > 1.
7.3.14
Basis: Data on gases cited in problem.
a.
The cubic feet associated with “165 ft3” and “240 ft3” probably means that the
values are the respective volumes of an ideal gas at standard conditions, because the
volumes calculated for the gases in part b. below are quite different than the stated
values.
b.
To find the amount of gas in the cylinder, first find the approximate volume in the
cylinder.
Vcyl =
π d2h
4
(3.14)(9)2 (52)
=
= 1.914 ft 3
(4)(1728)
Use the gas law as ratios:
p 2 V2
z RT
= 2 2
p1V1
z1RT1
7–57
Solutions Chapter 7
⎛ p ⎞⎛ z T ⎞
V2 = V1 ⎜ 1 ⎟⎜ 2 2 ⎟
⎝ p2 ⎠⎝ z1T1 ⎠
For C2H4:
Tc = 283 K, pc = 50.5 atm
Assume T1 = 80°F = 540°R = 300°K
300
⎫
= 1.06
⎪⎪
283
⎬ z1 = 0.35
1515
p r1 =
= 2.05⎪
(14.7)(50.5)
⎪⎭
Tr =
(At standard conditions: z2 = 1.00)
⎛ 1515 ⎞⎛ 1 ⎞⎛ 492 ⎞
3
V2 = (1.914) ⎜
⎟⎜
⎟⎜
⎟ = 514 ft
14.7
0.35
540
⎝
⎠⎝
⎠⎝
⎠
For CH4:
Tc = 191 K, pc = 45.8 atm
Tr =
300
= 1.57
191
pr2 =
2015
= 2.98
(14.7)(45.8)
z2 = 0.84
⎛ 2015 ⎞⎛ 1 ⎞⎛ 492 ⎞
3
V2 = (1914) ⎜
⎟⎜
⎟⎜
⎟ = 285 ft
14.7
0.84
540
⎝
⎠⎝
⎠⎝
⎠
MW of C2H4 = 28
MW of CH4 = 16
wt of C2H4 =
(514)
wt of CH4 =
(285)
(359)
16
(359)
Therefore
c.
28
= 40.1
= 12.7
wt C2 H4 > wt CH 4
For C2H4:
pV = znRT
= z
m
RT
MW
7–58
Solutions Chapter 7
m=
(1515)(1.91)(28)
pV(MW)
=
(0.35)(10.73)(590)
zRT
= 40 lb C2 H4
For CH4:
m=
(2015)(1.91)(16)
= 12.7 lb CH 4
(0.84)(10.73)(540)
Check with cylinder:
163 – 90 = 123 lb cylinder
105 – 9.5 = 135 lb cylinder
7.3.15
a)
p1V1 = z,n,RT
R = constant
pa V1 = z 2 n 2 RT1
V1 = constant, the vol. of cylinder
zn
p1
= 1 1
p2 z 2 n 2
Tc = 133.0 K
Tr = T T =
24.44 + 273
= 2.24
133
c
pr1 =
pc = 34.5 atm
2000 psig + 14.7
= 3.97
14.7 psig/atm 34.5 atm
pr2 = 3.80
z1 = 1.0 ⎫
⎬ Ideal gas behavior
z 2 = 1.0 ⎭
p1
p2
⎛ n ⎞ ⎛ 1.0 ⎞ n
= ⎜ 1 ⎟⎜ ⎟ = 1
⎝ n 2 ⎠ ⎝ 1.0 ⎠ n 2
7–59
Solutions Chapter 7
n CO =
pV 14.7 psia
=
RT
536°R
Vcylinder =
10.73
175 ft 3
(psia ) ft 3
( )
= 0.447 lb mol
( lb mol)(°R)
nCO RT1
p1
( )
3
0.447 lb mol 10.73(psia ) ft 536°R
Vc =
= 1.276 ft 3
2014.7psia (lb mol)(°R)
n final,CO =
p2 Vc
=
RT
1924.7 psia
1.276 ft 3
= 0.427 lb mol
( psia ) ft 3 536°R
10.73
(lb mol )(°R )
( )
n CO, lost = (0.447 – 0.427)lb mol ≅ 0.020 lb mol
rate of loss =
b)
n air
0.020 lb mol
48 hr
pV 14.7 psia
=
=
RT 536.0°R
(
100 ppm = 100 10
–6
= 4.167 ×10−4
lb mol
hr
1600 ft 3
= 4.09 lb mol air
( psia ) ft 3
10.73
(lb mol)(°R)
( )
)(4.09 lb mol) = 4.09 × 10 –4 lb mol
4.09 × 10-4
t=
= 0.982 hr
-4 lb mol
4.167 × 10
hr
c)
t = 67 hr
CCO =
d)
n CO =
4.167 × 10-4 67
= 0.0279 lb mol
0.0279 lb mol
lb mol
= 1.75×10-5
3
1600 ft
ft 3
The CO alarms would go on to alert people in the building.
7–60
Solutions Chapter 7
7.3.16
3
The density of the gas must be 2.0 g/ m . MW is Si is 28.086 so
= (2/28.086) = 0.0712 g mol/cm3. Let T - 298K. Also z = 1 at Tr = 1.98
V̂
and Vri very large.
a)
zRT 1 8.314(Pa)(m3 )
=
p=
ˆ
(g mol)(K)
V
3
⎛ 100 cm ⎞ 298 K
= 2.48 ×109 Pa
⎜
⎟
⎝ 1 m ⎠
b)
For a lower pressure, use a less dense gas (lower MW).
c)
At too high a temperature, you cannot get adequate density.
7.3.17
3
Basis: 240 ft methane at 80°F and 1964.7 psia. Assume the number of moles does not
change.
p1V1 nRT1z1
=
p2 V2 nRT2 z 2
p1 T1 z1
=
p2 T2 z2
T2 =
T1z1 p2
z 2 p1
Calculation of z1
Methane ⇒ Tc = 190.7K
pc = 45.8 atm
T1 = 80°F => 299.82K
Tr1 =
299.82K
= 1.572
190.7K
pr1 =
133.7 atm
= 2.919
45.8 atm
∴z1 = 0.83
7–61
Solutions Chapter 7
Calculation of z 2
(
)
= (3000 psig + 14.7 psia )(1 atm 14.7 psia ) = 205.08 atm
p1 = (1950 psig + 14.7 psia ) 1 atm 14.7 psia = 133.7 atm
p2
pr =
205.08 atm
= 4.478
45.8 atm
(
)(
V = 240 ft 3 2.832 × 10–2 m 3 /1 ft3 100 cm3 /1 m3
6
= 6.797 × 10 cm
)
3
)
( )
3
RTc 82.06 cm (atm) /( g mol)( K) (190.7K )
ˆ
Vc i =
=
pc
45.8 atm
= 341.68
cm3
g mol
(
(45.8 atm) 6.797 × 10 6 cm3
p1 V1
n=
=
(299.82K)(82.06)(0.83)
z1 RT1
)
= 15244.5 g mol
V 6.797 × 10 6 cm3
cm 3
ˆ
∴V = =
= 445.87
n
15244.5 g mol
g mol
ˆ
V
445.87
∴Vri = ˆ =
= 1.305
Vc i 341.68
∴z 2 = 1.06
T2 =
T1z1 p2 540°R( 0.83)(3014.7 psia )
=
z 2 p1
1.06(1964.7 psia)
T2 = 649o R
(189°F)
7–62
Solutions Chapter 7
7.4.1
T = 180°F (640°R)P =
2400 + 14.7 = 2415 psia
Basis: 33.6 lb gas
pV = znRT
Basis: 100 mol gas
mol
mol.wt.
lb
CO2
10
44
440
CH4
40
16
640
C 2 H4
100
50
28
2560
n=
avg. mol wt. =
1400
2480
= 24.8
100
33.6
= 1.35 lb mol
24.8
Tcʹ′ = 0.10 (304.2) + 0.40 (190.7) + 0.50 (283.1) = 248.25 K
pcʹ′ = 0.10 (72.9) + 0.40 (45.8) + 0.50 (50.5) = 50.86 atm
p′r =
p
2415
= 3.23
′ =
p c (50.86 )(14.7)
Tr′ =
T
640
= 1.43
′ =
Tc (248.25)(1.8)
From Fig. in text, z ≅ .75
( 3)
znRT 0.75 1.35 lb mol 10.73(psia ) ft 640°R
V=
=
(lb mol)(°R)
p
2415 psia
3
= 2.88 ft at 180 ° F, 2415 psia of gas is cylinder volume
7–63
Solutions Chapter 7
7.4.2
0.20 EtOH
0.80 CO2
1.00
T = 500K
V = 180 cm3 /g mol
p=?
Basis: gas as above
pcʹ′ = (0.20) (63.0) + (0.80) (72.9) = 70.92 atm
Tcʹ′ = (0.20) (516.3) + (0.80) (304.2) = 346.6 K
p r′ = ? =
T r′ =
500
= 1.44
346.6
Vci′ =
′
Vri =
p
p
=
p c 70.92
RTc′
p′c
=
(1 atm )( 22.4 L) 1000 cm3 346.6 K
= 401 cm3 /g mol
(273 K )(1 g mol) 1 L 70.92 atm
180
= 0.45
401
From the compressibility chart, pr ≅ 2.5 , p ≅ 177 atm
7–64
Solutions Chapter 7
7.4.3
Basis: 540 kg gas @ 393 K, 3500 kPa absolute
Component
kg
kg mol
mole fraction
pc, atm
Tc , K
CH4
C2 H6
C3H8
N2
100
240
150
50
6.25
8.00
3.41
1.79
0.321
0.412
0.175
0.092
45.8
48.2
42.1
33.5
191
305
370
126
Total
540
19.45
1.000
Tcʹ′ = (0.32l) (191) + (0.412) (305) + (0.175) (370) + (0.092) (126) = 262.8 K
pcʹ′ = (0.321) (45.8)+(0.412) (48.2)+(0.175) (42.1)+(0.092) (33.5) = 44.96 atm
T r′ =
3.93
= 1.50
262.8
p r′ =
3500 1 atm
= 0.77
44.96 101.3
From Nelson and Obert Chart, zʹ′ = 0.935
MW (average) = 540/19.45 = 27.8 kg/kg mol
ρ=
=
3500 kPa
27.8 kg
393 K 0.935
(K) (kg mol)
3
kg mol 8.31 (kPa) (m )
31.9 kg
m at 393K and 3500 kPa
3
7–65
Solutions Chapter 7
7.4.4
Use Kay’s Method
pcʹ′ = Σnipci ; Tcʹ′ = ΣniTci
Component
C2H4
A
He
n
pc
0.57
0.40
0.03
50.5
48.0
10.26
npc
28.8
19.2
0.308
Σnipci= 48.31
p r′ =
120
= 2.49
48.3
V=
( 0.70)(82.05)(298)
3
= 142.8 cm /g mol
(120)
nTc
283
150.7
13.19
61.2
60.2
0.395
∑ n i Tci =
298
= 1.342
222
T r′ =
Tc
∴
221.8
z = 0.70
⎛ 142.8 –140 ⎞
2.00%
% diff = ⎝
⎠ 100 =
140
7.4.5
Basis: 30ft3 mixture of 100 atm and 300°F
Comp
Mol. frac
Tc
pc
Tcʹ′
pʹ′c
C2H4
0.60
283
50.5
170
30.3
A
0.40
150.7
48
60.2
19.2
230.2
49.5
Trʹ′ = T/Tcʹ′ =
pʹ′r =
760
= 1.835
(230.2)(1.8)
p
100
=
= 2.02
pʹ′c
49.5
z = 0.95
7–66
Solutions Chapter 7
Avg. Mol. wt. = (0.60)(28) + (0.40)(40) = 32.8 lb/lb mol
pV = znRT
n=
pv
(100)(14.7)(300)
=
= 57.0 lb mol
zRT (0.95)(10.71)(760)
Assume capacities are given at storage conditions. Then type IA is selected because it is
the cheapest.
Number of tanks required =
(57.0)(32.8)
= 30
62
7.4.6
Steps 2, 3, and 4:
Assume a continuous process although a batch process would be acceptable and give the
same results.
F=?
mol fr.
0.20 C2H4
0.80 N2
1.00
Step 5: Basis:
P=?
Mixer
G 100% C2H4 @ 1450 psig and 70oF
G at given conditions
Steps 6 and 7:
Unknowns:
Balances:
F, P
⎫
⎪
C2 H 4 , N 2 , (or total) ⎬ degrees of freedom = 0
⎪
⎭
Steps 8 and 9:
Calculate G:
Tr =
mol fr.
C2H4 0.50
N2 0.50
1.00
70 + 460
= 1.044
508.3
7–67
MW
28
28
Solutions Chapter 7
1450 + 14.7
= 1.992
735
pr =
z = 0.34
V=
(359)(n)(14.7)(530)(0.34) (2640)
=
(1465)(492)
(1728)
n=
(2640)(1465)(492)
= 1.157 lb mol
(1728)(359)(14.7)(530)(0.34)
Balances:
C2H4
F(0.20) + 1.157 = P(0.50)
Total
F + 1.157 = P
F = 1.928 lb mol
7.4.7
P = 3.085 lb mol
Basis: Gas at 50 atm and 600°R
From the compressibility charts:
pr =
50
= 3.5
14.3
Tr =
600
= 1.2
40.0 + 460
z = 0.58
100,000 std ft 3 1 lb mol 359(0.58) act ft 3 ⎛ 1 ⎞ 600
3
Q=
= 1415 actual ft /hr
3
⎝
⎠
hr
1 lb mol
359 std ft
50 492
or use pV = znRT twice
p2 V2 z 2 n2 RT2
=
p1V1
z1n1 RT1
V2
⎛ T ⎞⎛ p ⎞⎛ z ⎞
= V1 ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 2 ⎟
⎝ T1 ⎠ ⎝ p 2 ⎠ ⎝ z1 ⎠
7–68
Solutions Chapter 7
⎛ 600 ⎞ ⎛ 1 ⎞ ⎛ 0.58⎞
= 100,000⎝
492 ⎠ ⎝ 50 ⎠ ⎝ 1.00 ⎠
= 1415 actual ft 3 /hr
7.4.8
Basis: initial gas at 200 psig and final gas at 1000 psig
MW C2H4 = 28
initial
214.7/14.7
pr1=
= 0.290
50.5
final
1015/14.7
pr2=
= 1.37
50.5
Tr1 = 1.05
Tr2 = 1.05
z1 = 0.90
z2
= 0.35
Use pV = znRT
n 2 p 2 z2
=
n 1 p 1 z1
–1
1015 0.35
=
215 0.90
–1
= 12.14
Alternate way to get z is to use the Pitzer ascentric factor.
z = z0 + z1 ω
z1 = 0.910 + (-0.023)(0.085) = 0.91
z2 = 0.37 + (0.1059)(0.085) = 0.38
Material balance
W1 + Wc = 222
W2 + Wc = 250
W2 – W1 = 28 lb
Solve (I) and (II) to yield n1 = 0.089
and n2 = 1.089
7–69
n2 – n1 = 28/28 = 1
(II)
Solutions Chapter 7
a.
Mass of gas initially = (0.089) (28) = 2.49 lb
(28 lb) ($0.41/lb) = $11.48 billing
b.
Wt. of cylinder = (222 – 2.49) lb = 219.5 lb
c.
V=
n1z1 RT 0.089 0.9 10.73 298(1.8)
3
=
= 2.17 ft
p1
214.7
7.4.9
Step 5:
Basis: 1 hr.
Steps 2, 3, and 4: Steady state, open system
liquid
30,000 = F(kg)
mass fr.
Bz 0.50
Tol 0.30
Xy 0.20
1.00
mole fr.
Bz 0.912
Tol 0.072
Xy 0.016
1.000
Separator
vapor
P1(kg)
MW
78.11
92.13
106.16
kg
71.24
6.63
1.70
79.57
liquid
P3(kg)
mass fr.
Bz 0.06
Tol 0.09
Xy 0.85
1.00
liquid
P2 = 9,800 (kg)
mass fr.
0.895
0.084
0.021
1.00
mBz
mTol
mXy
Σ = 9,800
mass fr.
Convert F to moles and then to mass:
Basis: 100 kg = F
Component
kg
MW
kg mol
mol fr.
Tc(K)
pc(atm)
Bz
50
78.11
0.640
0.592
562.6
48.6
Tol
30
92.13
0.324
0.299
593.9
40.3
Xy
20
106.16
0.118
0.109
619
34.6
1.082
1.000
100
7–70
Solutions Chapter 7
Calculate the kg of F; get z:
Tcʹ′ = 562.6 (0.592) + 593.9 (0.299) + 619 (0.109) = 578.1
pʹ′c = 48.6 (0.592) + 40.3 (0.299) + 34.6 (0.109) = 44.6
Tr =
607K
=1.05
578.1K
pr =
26.8 atm
= 0.60
44.6 atm
z = 0.80
Together mass of feed:
pV
26.8 atm 483 m3
101.3 kPa (kg mol) (K)
n=
=
zRT
0.80 1 atm 8.314(kPa)(m3 ) 607 K
= 324.7 kg mol
F = (324.7) (100)/(1.082) = 30,017 kg, say 30,000 kg
The material balances are in kg.
Step 6:
Unknowns: P1 , P3 , m Bz , m Tol , m Xy
Step 7:
Balances:
Bz
Tol
mBz 3
= = 1.5
mxy 2
Xy
∑ mi = 9,800
Step 8:
The balances in kg are (only 3 are independent mass balances)
(1)
Bz
30,000 (0.50) = P1 (0.895) + m Bz + P3 (0.06)
(2)
Tol:
30,000 (0.30) = P1 (0.084) + mTol + P3 (0.09)
(3)
Xy:
30,000 (0.20) = P1 (0.021) + mXy + P3 (0.85)
(4)
Total 30,000
= P1
+ 9,800 + P3
7–71
Solutions Chapter 7
(5)
m Bz + mTol + m Xy = 9800
(6)
m Bz = 1.5m Xy
Step 9:
Solution: Using (1), (2), (3), (5), and (6) via Polymath
kg
mass fr.
P3 = 5500 kg/hr
mBz = 1518
0.156
P1 = 14,700 kg/hr
mtol = 7270
0.745
0.103
mXy = 1012
9800
7–72
1.00
Solutions Chapter 8
8.2.1
a. (1); b. (2); c. (3); d. (4); e. (3); f. (1)
8.2.2
8.2.3
F=2–P+C=2–2+2= 2
8.2.4
F=2–P+C
(a)
(b)
F=2–2+1= 1
F=2–3+2= 1
8.2.5
C=1
P=2
F=2–2+1=1
8.2.6
C = 3 (O2, N2, H2O)
P=2
F=2–2+3= 3
8–1
Solutions Chapter 8
8.2.7
The number of components is 3, but one independent reaction exists among them so that
C=2
NH 4 C1(s) Ä NH3 (g) + HCl(g)
P=2
F=2–2+2= 2
8.2.8
(1) F = 2 – P + C
The rank of
Ca
C
O
CaCO3(s)
1
1
3
CaO(s)
1
0
1
⎡1 0 9 ⎤
⎢1 1 0⎥ is only 2 so C = 2
⎢
⎥
⎢⎣3 2 0 ⎥⎦
P = 3 hence F = 2 – 3 + 2 = 1
8.2.9
(a) 40ºC; (b) 190ºC; (c) 60ºC; (d) Compound B
8.2.10
(e)
8–2
CO2(g)
0
1
2
Solutions Chapter 8
8.2.11
a.
F=2–P+C
C=3
P = 2 (assume that p is not unreasonably high)
F=2–2+3= 3
b.
Possible variables are: pressure
mol or mass fraction of
H2O
⎫
⎪
acetic acid ⎬ only 2 are independent
ethyl alcohol⎪⎭
8.2.12
(a)
F = 2 –P + C = 2 – P + 2
if F = 0
P=4–F
Max P = 4
(b)
F=2–P+C
3 = 2 – 2 + C C=3
8.3.1
All true
8.3.2
(a) higher; (b) lower; (c) higher; (d) lower; (e) no change; (f) lower;
(g) higher; (h) lower; (i) higher; (j) lower; (k) lower; (l) higher
8.3.3
From the p-T diagram for water you can see that raising the pressure by a large amount
on ice at constant temperature causes the water to go from the solid phase to the liquid
phase.
8–3
Solutions Chapter 8
8.3.4
The vapor pressure of methanol is less than that of gasoline so that at lower temperatures
the fuel-to-air mixture is insufficient for combustion.
Automotive parts can be made of alcohol tolerant materials such as stainless steel or other
corrosion resistant materials and additives to the methanol, such as 15% gasoline, can
help solve the problem of difficult cold-weather engine starts.
8.3.5
Appendix G (T is in K)
(a)
CD (T is in ºC)
Acetone at 0ºC
ln (p*) = 16.6513 −
2940.46
T − 35.93
log10 (p*) = 7.2316 −
p* = 70.51 mm Hg
(b)
p* = 70.55 mm Hg
Benzene at 80ºF
ln (p*) = 15.9008 −
2788.51
T − 52.36
log10 (p*) = 6.90565 −
p* = 102.73 mm Hg
(c)
1277.03
T + 237.23
1211.033
T + 220.79
p* = 102.74 mm Hg
Carbon tetrachloride at 300 K
ln (p*) = 15.8742 −
2808.19
T − 45.99
log10 (p*) = 6.8941 −
p* = 123.81
1219.58
T + 227.17
p* = 122.89 mm Hg
8–4
Solutions Chapter 8
8.3.6
Fit the Antoine Equation using the three data points to get A, B, and C. In mm Hg
kPa
2.53
15.0
58.9
1n(18.98) = A −
mm Hg
18.98
112.54
441.90
B
= 2.9434
C + (−40+ 273)
1n(112.54) = A −
B
= 4.7233
C + (−10+ 273)
1n(441.90) = A −
B
= 6.0911
C + (−20+ 273)
The solution from Polymath is
A = 16.5386
B = 2707.43
At 40ºC, ln (p*) = 16.538 −
C = -33.8541
2707.43
−33.85 + (40 + 273)
p* = 939 mm Hg (125 kPa)
8.3.7
At the triple point, all the phases (solid, liquid and vapor) exist in equilibrium. Therefore,
the vapor pressure of liquid ammonia = the vapor pressure of solid ammonia
15.16 – 3063/T = 18.7 – 3754/T
T = 195 K
8.3.8
No. It is a typo. Probably the number was 98.8ºC because if you equate the pressures
from the two equations, T = 371.75K, or 98.7oC.
8–5
Solutions Chapter 8
8.3.9
For benzene (p* is in mm Hg and T in K)
1n(p*) = 15.9008 −
2788.51
T − 52.36
p* = 760 mm Hg
ln(p*) = 6.6331
T = 353K agrees with data base on CD
For toluene
1n(p*) = 16.0137 −
3096.52
T − 53.67
p* = 760 mm Hg
T = 383.7 agrees with data base on CD
8.3.10
Merge the two equations (A1 MW = 27)
1/2
W = 10-4 = 5.83 × 10-2
(p v )(27)
T1/2
where p v =
[10
1.594×104
)
T
(8.79 −
]
1/ 2
T
The solution from Polymath is
T = 1492K
8.3.11
ˆ =x V
ˆ
ˆ
All of the values of specific volume are obtained using V
with
vapor vapor + (1-x vapor )Vliquid
V̂ data from the steam tables.
(a)
V̂ = 0.5 (1.673) + 0.5 (0.001043) = 0.837 m3/kg
(b)
V̂ = 0.5 (0.6058) + 0.5 (0.001073) = 0.303 m3/kg
(c)
V̂ = 0.3 (350.8) + 0.7 (0.01613) = 105.25 ft3/lb
(d)
V̂ = 0.7 (1.8431) + 0.3 (0.0187) = 1.296 ft3/lb
8–6
Solutions Chapter 8
8.3.12
(a) T; (b) F; (c) T; (d) T; (e) F; (f) T
8.3.13
Basis: 2.10 lb water
Specific volume:
ˆ = (1 − x) V
ˆ +xV
ˆ
V
l
g
0.35
Specific volume of water = Vˆl =
= 0.175m3 /kg
2
From the steam tables:
Specific volume of saturated liquid,
Vˆl = 0.001 088 m3 /kg
specific volume of saturated vapor =
Vˆg = 0.414 m3 /kg
Then, 0.175 = (1 – x) 0.001 088 + x (0.414)
The quality = x = 0.42
8.3.14
Since the two phases coexist in equilibrium, we have a saturated mixture, and the
pressure must be the saturation pressure at the given temperature:
p = p* @ 90ºC = 70.14 kPa
ˆ and V
ˆ values are V = 0.001036 m3 /kg and V
ˆ = 2.361 m3 /kg
At 90ºC, V
l
l
g
g
Add the volume occupied by each phase:
ˆ +m V
ˆ
V= Vl + Vg = mV
l
g g
= (8 kg)(0.001 m3/kg) + (2 kg)(2.361 m3/kg) = 4.73 m3
8–7
Solutions Chapter 8
8.3.15
v=
ˆ
Q mV
=
A
A
where
v = velocity, ft/s
V̂ = specific volume of the fluid, ft3/lb
A = pipe cross sectional area, ft2
V̂ can be obtained from the steam tables: V̂ = 0.9633 ft3/lb
2
⎛ 2.9 ⎞
2
The area is A = ∏ d / 4 = ⎜
⎟ = 0.046 ft
4 ⎝ 12 ⎠
2
v=
∏
25,000 lb 0.9633 ft 3
ft
= 5.24 ×105
or 145 ft/s
2
hr
lb
0.046 ft
hr
8.3.16
The hot oil vaporized the water, and the increase in volume in the system caused the
damage.
8.3.17
Use data from the steam tables at the initial condition of p = 101.33 kPa and saturated
water (i.e. two phases):
V̂liq = 1.044 cm3/g and V̂vap = 1673.0 cm3/g.
Then the mass of both phases in the container can be calculated:
m liq = (0.03 m3) (100 cm/m)3 / 1.044 cm3/g = 28,376g
m vap = 2.97 m3 (100 cm/m)3 / 1673.0 cm3/g = 1,775 g
mt otal
= 30,511 g
8–8
Solutions Chapter 8
At the final conditions all of the liquid has been evaporated, and only saturated vapor will
exist having a specific volume of:
Vv ap = (3.00 m3) (100 cm/m)3 / 30,511 g = 98.3 cm3/g
Interpolating in the steam tables
(212-T) ºC/(212-214) ºC = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g
T = 212oC
(1985.2-p) kPa/(1985.2-2065.1) kPa = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g
p = 2002 kPa
8.3.18
Basis: 10 lb of water
Specific volume:
V̂ =
10.0
= 4.975 ft 3 /lb
2.01
From the paper steam tables (in ft3/lb) at 80 psia:
V̂liq = 0.0176
V̂vapor = 5.476
Mass balance:
m L + m V = 2.01
Volume balance:
10.0 = m L (0.0176) + m V (5.476)
Mass of liquid is 0.184 lb
Mass of vapor is 1.826 lb
The volume of liquid is
Vliq = m liq V̂liq = (0.184)(0.0176) = 3.28 × 10−3 ft 3
The volume of vapor is
8–9
Solutions Chapter 8
Vvapor = m vap V̂vapor = (10.0 − 3.28 × 10−3 ) = 9.997 ft 3
Quality is 1.826 / 2.01 = 0.908
8.3.19
Log (T)
(log (T))
5.6101
5.7038
5.7991
5.8861
5.9402
5.9915
6.0403
6.0868
6.1527
6.2146
31.4727
32.5331
33.6295
34.6462
35.2856
35.8976
36.4847
37.0488
37.8561
38.6214
DATA:
(log (T))3
2
vp
176.5329
185.5619
195.0204
203.9313
209.6027
215.0795
220.3767
225.5079
232.9186
240.0166
0.61130
3.53600
17.21000
62.15000
128.80000
245.60000
437.00000
733.20000
1454.00000
2637.00000
1n p* = a+b ln T + c(ln T)2
+ d(ln T)3
THE COEFFICIENTS ARE
-1199.2548
532.2715
a
b
-79.0261
3.9638
c
d
8.3.20
Based on data from the steam tables:
State 1: liquid
State 3: solid
State 5: liquid (saturated)
State 2: superheated vapor
State 4: saturated vapor
State 6: vapor (saturated)
You can calculate the properties of a mixture of vapor and liquid in equilibrium (for a
single component) from the individual properties of the saturated vapor and saturated
liquid by averaging the properties of the two saturated phases. The weights are the
respective amounts of each phase.
8–10
Solutions Chapter 8
8.3.21
Based on data from the steam tables:
State 1: saturated vapor
State 3: two phase (saturated liquid and vapor)
State 2: saturated liquid
State 4: superheated vapor
8.3.22
Prepare a Cox chart as described in the text. Use semi-log paper with the horizontal axis
log10. Obtain vertical scale rules by use of steam properties at even integers for the
temperature. Or you can use Antoine Equation, Appendix G, to get 2 points and draw a
line between them.
Tcritical (o F)
(a) Acetone
(b) Heptane
(c) Ammonia
(d) Ethane
454
512
270
89.7
p critical
cox chart (psia)
670
700
1500
750
8–11
pcritical (psia)
691
397
1636
708
Solutions Chapter 8
8.3.23
Steps for preparing a Cox Chart (using water as the reference substance):
1.
On logarithmic paper, set up the pressure scale on one axis. The other axis will be
a nonlinear temperature scale, which will be set up by the following method.
2.
Draw a diagonal line from the origin to the opposite corner of the graph.
3.
From the steam tables, obtain values for the vapor pressure of water at evenly
spaced temperature increments.
4.
Plot these vapor pressures along the diagonal line.
5.
Now, beginning with the first point on the diagonal, draw a line extending from
that point to the temperature axis and label the intersection as the corresponding
temperature for that vapor pressure.
6.
Repeat step five for every point on the diagonal line. This process will establish
the temperature scale.
7.
Using the established temperature scale, plot the data for benzene and draw a line
through the points.
8.
From this line, obtain the vapor pressure for benzene at 125ºF.
The vapor pressure for benzene at 125ºC (257oF) is estimated to be 3.3 atm.
8–12
Solutions Chapter 8
8.3.24
The procedure is the same as outlined for P16.28. The result from the Cox chart is that
the vapor pressure for aniline at 350o is 22 atm .
8.3.25
From the CD that accompanies the text the respective vapor pressures at 300K are in
increasing value
Ethanol
Methanol
MTBE
p* in mm Hg
PEL (ppm)
65.8
139.7
294
1000
200
100
The relative values of p* are about the same as the inverse of the relative values of the PEL.
8.4.1
(a) Volume increases
(b) Pressure increases
8–13
Solutions Chapter 8
8.4.2
Basis: Data given in problem statement
dry N2
H 2O

27 C
p* = 3.536kPa p = 101.3kPa
27 C
(a) pT = p N 2 + pH 2 O = (101.3 + 3.536) kPa = 104.8kPa
(b)
n H 2O
n N2
=
p H 2O
p N2
=
3.536 kPa
= 0.0349
101.3 kPa
8.4.3
T= − 20oC
p* =14.1 mm Hg
O2
p t =760 mm Hg
N2
C 6 H1 4
pair =760 − 14.1= 745.9 mm Hg
pC6H14 = pC6H14 =14.1 mm Hg
Basis: 760 mol saturated gas
(or use 100 mol)
p ⇒ moles
O2
N2
C6H14
mol
= 156.6
= 589.3
= 14.1
760
.21 (745.9)
.79 (745.9)
For complete combustion
C6H14 + 9 1/2 O2 → 6 CO2 + 7 H2O
14.1 mol C6H14 required 14.1 (9.5) = 134 mol O2 required
156.6 – 134 = 22.60 mol O2 excess
22.60
× 100 = 17 %
134
8–14
Solutions Chapter 8
8.4.4
mol. wt. CCl3NO2 = 164.5
Basis: 1 mol saurated gas
mol fr.
0.02
0.98
1.00
CCl3NO2
air
100 kPa 0.02 mol CCl 3NO3 760 mm Hg
= 15.0 mm Hg
1 mol gas
101.3 kPa
(a)
Using linear interpolation, which may introduce a slight error, 15.0 mm Hg
corresponds to
⎛ 15.0 − 13.8 ⎞
o
⎜
⎟ (5) = 1.3 + 15 = 16 C
⎝ 18.3 − 13.8 ⎠
(289.5K)
Basis: 100 m3 at 100 kPa and 16.5°C
(a)
3
100 m air 100 kPa 273 K
101.3 kPa 289.5 K
1 kg mol air
3
22.4 m air at SC
0.02 mol CCl3 NO2 164.5kgCCl 3 NO2
= 13.95 kg
0.98mol air
1kgmol CCl 3 NO2
8.4.5
Basis: 1 mol air
y = 0.12
pH2O = pT (0.12) =101.3 kPa (0.12)
= 12.2 kPa
The dewpoint is the temperature at which p*H O =12.2 kPa , or from the steam tables 323K
2
or 50 C
o
8–15
Solutions Chapter 8
8.4.6
Assume the tank with air fills with the hazardous liquid vapor at 80ºF. The pressure in
the tank with air will become after the transfer (10 + 14.7) + 13 = 37.7 psia hence the
seal will possibly rupture during the transfer
8.4.7
At 60ºF, p*H O =0.52 in Hg; at 75ºF, p* = 0.87 in Hg
2
Basis: 12,000 ft3 air at 75ºF and 29.7 in Hg absolute
12,000 ft 3
0.52 in Hg H 2 O 492oR 18 lb H 2 O
1 lb mol
= 9.62 lb H 2 O
359 ft 3 at SC 29.92 in Hg total 535oR lb mol
8.4.8
Basis: 1 gal benzene (sp. gr. 0.879, MW 78.1) at 750 mm Hg, 70ºF
Moles of air =
(3600 ft 3 ) (750) (14.7)
= 9.18 lb mol
(10.73) (760) (530)
Moles of benzene =
(1) (1ft 3 )(0.879)(62.4)
= 0.094 mol
(7.48)(78)
nTotal = 9.18+0.094 = 9.274 mol
yBz =
0.094
= 0.0101
9.274
mol % = 1.01% therefore beneath explosive limit
8–16
Solutions Chapter 8
8.4.9
Basis: 350 ft3 C2H2 – O2 mixture at 25ºC, 745 mm Hg
n1 =
pV
(745)(14.7)(350)
=
= 0.874 lb mol
RT
(760)(10.73)(298)(1.8)
⎛ 745 ⎞⎛ 14.7 ⎞⎛ 300 ⎞⎛ 1 ⎞
n 2 = ⎜
⎟⎜
⎟⎜
⎟⎜
⎟ = 0.671 lb mol
⎝ 760 ⎠⎝ 10.73 ⎠⎝ 333 ⎠⎝ 1.8 ⎠
p*H2O at 60ºC = 149.4 mm Hg
149.4
y H2O at2 =
= 0.201
745
n H2O = (0.201)(0.671) = 0.135 lb mol
C2 H 2 +
5
O 2 → 2CO 2 + H 2 O
2
You get 1 mol H2O/mol C2H2 on reaction
Therefore moles C2H2 = 0.135
mol O2 = 0.874 – 0.135 = 0.739 lb mol
y O2 at 1 =
0.739
= 0.845
0.874
VO2 =(0.845)(350) = 296 ft 3 at 745 mm and 25ºC
VC2H2 = 350 − 276 = 54 ft 3 at 745 mm and 25ºC
8.4.10
Elephant seals have an average body temperature several degrees higher than ours. That
means their breath can hold a bit more moisture than ours, and on that day, that extra
moisture could have been just enough for condensation to occur.
But there are other reasons the elephant seals’ breath may have been easier to see. It
could be that it was a few degrees colder on the beach than where the observer was sited.
And, as you know if you have ever experimented with seeing your breath, the bigger the
volume of air, the more moisture available to condense out and so the condensed
moisture was more visible.
8–17
Solutions Chapter 8
8.4.11
(a) p*Bz exists if saturation occurs. Then, with ptot = 100 kPa
pBz = 100 (0.014) = 1.4 kPa, which
corresponds to
−11oC from Perry
or use Antoine Eqn. and solve for T ( p*Bz = 10.5 mm Hg)
ln (10.5) = 15.9008 -
b)
(-15oC from the CD)
2788.51
T − 52.36
Repeat for p*Bz = 8.0 kPa (60 mm Hg) corresponding to 15.4oC
8.4.12
Assume equilibrium.
At 75˚F, from the Antoine equation, the vapor pressure of benzene is 90 mmHg.
Assume the barometer is 760 mm Hg abs.
p∗B
90
mol Bz
mol Bz
=
=
= 0.13
∗
mol air
mol air pt − pB 760 − 90
a) The OSHA limit is 1 ppm over 8 hours. The above greatly exceeds this limit.
b) No, because the garage is probably not well ventilated, and also if a water heater is in
the garage, the LEL (Lower Explosive Limit) may be exceeded.
8.4.13
Basis: 50 ft3 air at 29.92 in. Hg and 70ºF
p*H2O at 50ºF = 0.36 in Hg
The amount of water per hour is
1.68 lb H 2O
50 ft 3 60 min 1 lb mol 429oR 0.36 in. Hg of H 2O 18 lb
=
3
o
min 1 hr 359 ft 530 R 29.92 in. Hg total lb mol
hr
Alternate solution: make air and water balances.
8–18
Solutions Chapter 8
8.4.14
Basis: 1 lb air
If the air is saturated, p*C8 = 2.36 in Hg and pair = 29.66 - 2.36 = 27.30 in Hg.
p*C8
pair
=
2.36
mol C8
= 0.0864
27.30
mol air
(a)
lb air
lb air 27.30 mol Air 29 lb air 1 lb mol C 8
= 2.94
=
l bC 8
lb C8 2.36 mol C 8 1 lb mol air 114.2 lbC 8
(b)
nc
2.36 mol C 8
= 8 = 0.080 which is also the fraction of the volume.
(27.30 + 2.36) mol total n tot
2.94lb air 1 lb mol air 359 ft 3 at SC 580 R 29.92in Hg
lb C8
29 lb air 1lb mol air 492 R 29.66in Hg
(c)
= 43.3 ft 3 at 120 ˚F and 29.6 in. Hg/lb octane
8.4.15
At equilibrium, the pressure of Hg is its vapor pressure
−4
p* H g 1.729× x10 g mol Hg
=
=
=
pt
nt
pair
99.5g mol air
pHg
nHg
Note: pt = pair + pHg = pair for all purposes so
1.729×10 −4 gmol Hg
99.5gmol total gas
Basis: 1.729 × 10-4 g mol Hg
−4
1.729×10 gmol Hg 200.59g 1000mg
= 38.14 mgHg
1gmol Hg 1g
nRT
=
p
3
99.5gmol air 293.15K 8.314(kPa)(m ) 1kg
3

= 2.44m at 20 C and 99.5kPa
99.5 kPa (kg mol)(K) 1000g
V=
8–19
Solutions Chapter 8
15.65
mg
at 20˚C and 99.5 kPa.
m3
Level is not acceptable.
A mercury spill can be cleaned up reasonably effectively by dusting with sulfur, then
vacuuming with a vacuum cleaner specifically designed for mercury pick-up, which
prevents the escape of mercury vapor.
The instructor should point out to the student that the assumption of the “no
ventilation” is not a very good approximation unless the surface of the mercury is rather
large or there is truly no ventilation in the storeroom.
8.4.16
Step 5: Basis: 1 min
Steps 2, 3, and 4: System is loading plus water seal, steady state, open
C4
300cm3
20oC
100 kPa
Comp.
P
Water
seal
Loading
Air = ?
0.015
H2O(sat)
Total
p*H2O at 20oC = 2.34 kPa
3
pV 100.0 kPa 300 cm3 ⎛ 1 m ⎞ (kg mol)(K)
n c4 =
=
⎜
⎟
RT
293.15 K ⎝ 100 cm ⎠ 8.314(kPa)m3 )
= 1.231 × 10-5 kg mol = 1.231 × 10-2 g mol C4
Steps 6, 7, 8, and 9:
At P, the mole fraction of water is 2.34/120 = x H O = 0.0195
2
The moles of P come from a C4 balance (C4 is a tie element)
P = 0.821 g mol
8–20
kPa
air
C4
1.231 × 10-2 = 0.015P
mol fr.
2.34
1.00
120
Solutions Chapter 8
Then the air in is
0.821 (1-0.0195 – 0.015) = 0.793 g mol
V=
nRT 0.793×10-3 kg mol 293.15 K 8.314 (kPa)(m3 )
=
p
100 kPa
(kg mol)(K)
V= 1.93 ×10-2 m3 at 100.0 kPa and 20ºC per min
8.4.17
Basis: 1 hr
The question really is: does the Hg condense at 150ºF, or is it still a vapor (and thus not
collected).
The moles of Hg are: 40,000 (0.023 × 10-2)
lb Hg 1 lb mol Hg
= 0.046 lb mol
200 lb Hg
The moles of gas are (ignoring the Hg):
40,000 lb gas 1 lb mol gas
= 1250 lb mol
32 lb gas
y Hg =
0.046
= 3.68 10−5
0.046 + 1250
×
The partial pressure of the Hg in the gas is yp total
(3.68 × 10-5)(14.7) = 5.4 × 10-4 psia
The Hg will remain a vapor and not condense.
8–21
Solutions Chapter 8
8.4.18
Basis: 20,000 ft3 moist air at entering conditions in 1 day
20,000 ft 3 273 800 1 lb mol
= 57.18 lb mol total
280 760 359 ft 3 at SC
(0.532 lb mol H2O)
System: Refrigerator is system
Steps 6 and 7:
Unknowns: W2 and P; balance: H2O and air
Steps 8 and 9:
⎫
⎛ 0.99.12 ⎞
⎛ 0.13 ⎞
H 2 O: 57.18 ⎜
⎟ = W2 (1) + P ⎜
⎟
⎪ lb mol
⎝ 106.63 ⎠
⎝ 106.63 ⎠
⎪
⎬ P = 56.72
⎛ 105.64 ⎞
⎛ 106.5 ⎞
⎪
Air: 57.18 ⎜
⎟ = W2 (0) + P ⎜
⎟ not accurate ⎪ W2 = 0.4613
⎝ 106.63 ⎠
⎝ 106.63 ⎠
⎭
Total: 57.18
= W2
+P
Basis: 30 days
30 (0.4613) (18) = 249 lb H 2O/30 days
8–22
Solutions Chapter 8
8.4.19
Steps 1-4:
air
F(mol)
H2O
25oC
100 kPa
remove 50%
H2O
(saturated)
air
H2O
P
100 kPa
W (mol)
100% H2O
dew point =
16oC
mol fr.
100-p*P/100
p*P/100
1.00
at exit p* = ?
At 16ºC
0.57 in Hg
0.28 psia
*
From
steam
tables
p*
=
1.92
kPa
p H2O ≅ 13.6 mm Hg =1.81 kPa
pair = 100 – 1.92 = 98.08
Thus pair = 100 – 1.81 = 98.18 kPa
Step 5: Basis is 100 Mol F
Step 6: Unknowns: W, P
Step 7: Balances: H2O, air
Steps 8 & 9: Balances around process in mol
Air
⎛ 100 − p*H2O ⎞
⎛ 98.08 ⎞
100 ⎜
=
W(0)
+
P
⎜
⎟ in P
⎟
100
⎝ 100 ⎠
⎝
⎠
H2O
⎛ p*H2O in P ⎞
⎛ 1.92 ⎞
100 ⎜
⎟
⎟ = W(1) + P ⎜
⎝ 100 ⎠
⎝ 100 ⎠
Total
100
Also, p* = f(T)
=W+P
vapor pressure relation
•
The exit gas has one-half the entering H2O, or ½ (1.92) = 0.96 mol thus W = ½
(1.92) = 0.96 mol H2O.
•
The exit air has 98.08 mol dry air.
•
At the exit p H O = p*H O = y H O (100) =
2
2
2
0.96
100 ≅ 0.97 kPa
0.96 + 98.08
8–23
Solutions Chapter 8
•
From the steam tables, this corresponds to about
280 K
(7oC)
, or use Antoine Eq.
Note: Taking ½ of (1.92) = 0.96 and dividing by 100 instead of (pH O +pair ) is wrong. The
answer is close to correct because of the very small amount of water.
2
8.4.20
Basis: 1000m3 sat. air at 99 kPa and 30ºC
p*H2O at 30ºC = 0.6155 psia = 31.8 mm Hg = 4.24 kPa
1000 m3
or
1 kg mol
99 kPa 273 K
= 39.3 kg mol
3
22.415 m a tSC 101.3 kPa 303 K
n=
99 kPa 1000 m3 (kg mol)(K)
=39.3 kg mol
303 K 8.314 (kPa)(m3 )
Initial:
4.24 ⎞
Initial mol H2O = 39.3 ⎛⎜
⎟ = 1.68 kg mol H 2 O
⎝ 99 ⎠
99 − 4.24 ⎞
Initial mol air = 39.3 ⎛⎜
⎟ = 39.3 − 1.68 = 37.62 kg mol air
⎝
99
⎠
Final:
At 14ºC and 133 kPa, the air is still saturated, and nair = 37.62 kg mol
p*14oC H2O = 0.2302 psia = 11.9 mm Hg = 1.59 kPa = p H 2O @14oC
p H2 O
pair
=
n H2 O
n air
so
n H O ⎫ There is the same as
1.59
= 2 ⎬
(133 − 1.59) 37.62 ⎭ an H2O material balance
n H2O = 0.46 kg mol
1.68 – 0.46 = 1.22 kg mol or 22 kg
8–24
Solutions Chapter 8
8.4.21
Steps 2, 3, and 4:
C2H6
CO2
H2O
O2
N2
Combustion
chamber
AIR
100 kPa
(20% excess)
Assume combustion is complete and the gases are ideal. The process is open, steady
state with reaction. The first objective is to make material balances.
C2 H6 +3.5O2 → 2CO2 + 3H2 O
Step 5:
Basis: 100 mol C2H6
Step 4:
The entering O2 is
The excess O2 is (0.20) (3.50) =
Total O2 is
350 mol
70 mol
420 mol
0.79 ⎞
The N2 is 420 ⎛⎜
⎟ =
1580 mol
⎝ 0.21 ⎠
Steps 6 and 7:
Unknowns: 4 (The components of the flue gas)
Balances: C, H, O, N, total of 4
Degrees of freedom = 0
Steps 8 and 9 (balances in moles, in = out):
C:
2N:
H:
O:
2(100) = n CO
1580 = n N
6 (100) = 2n H O
420(2) = 2n CO +n H O +2n O
Total
n CO 2 = 200 mol
2
n N2 = 1580 mol
2
n H2O = 300 mol
2
2
2
n O2 = 700 mol
2150 mol
2
The dewpoint is the temperature at which the flue gas is saturated
p*H2O = y H2O pTotal =
300
(100 kPa) = 14.0 kPa
2150
This pressure of H2O corresponds (from the steam tables) to 325.7 K (52.6o C)
8–25
Solutions Chapter 8
8.4.22
mol %
4.5 CO2
26.0 CO
13.0 H2O
0.5 CH4
56.0 N2
100.0
F
P
Pt = 98 kPa
nCO
2
nH O
2
nN 2
nO 2
Air 10% excess
O2 22.55 mol
N2 84.83
}
Step 4:
Calculate the xs air and add total air to the diagram
Step 5:
Basis: 100 mol synthesis gas
Step 6:
Unknowns: n O ,NCO ,n H O ,n N ,P
(5 total)
Step7:
Balances
(5 independent equations)
Steps 8, 9:
Use element balances
2
2
2
2
C, O, H, N, ∑ n i = P
In
out
C:
4.5
+ 26.0 + 0.5
=
n CO 2
n CO 2 = 31.0
H:
0.5 (4)
+ 13 (2)
=
n H2O (2)
n H2O =
O2:
4.5
+
N2:
56.0
+ 84.83
y H2O =
26.0
+ 22.55 =
2
=
n CO2 +
n N2
n H2O
2
28
2
n O2 = 2.05
n N2 =
140.8
187.9
28 / 2
14
=
187.9 187.9
⎛ 14 ⎞
p*H2 O = p H2 O = 98 ⎜
⎟ =7.3 kPa (or 1.06 psia)
⎝ 187.9 ⎠
This is equivalent to a temperature of 104oF from the steam tables (40ºC)
8–26
Solutions Chapter 8
8.4.23
Basis: 100 mol gases leaving absorber
CH4 + 2O2 →CO2 + 2 H2 O
0.8335mol N2 0.21mol O 2 1mol CO2
= 0.1108 mol CO2
0.79 mol N2 2 mol O2
1 - 0.8335 - 0.1108 = 0.0557 mol H2O; 0.00557 (20) = 1.114 psia

Dew point ≈ 105 F at 1.114 psia
b) At constant temperature of 130 ˚F, the gas must be compressed until the partial
pressure increases and becomes equal to the vapor pressure in order to reach the point of
condensation.
From Steam tables, the vapor pressure of water at 130 ˚F is 2.223 psia.
The total pressure at which the partial pressure of water (mol fraction = 0.0557) will
be 2.223 psia is
pH 2 O = p t y H 2 O
2.223
= 40.1psia
0.0557
8.4.24
Basis: 1 lb mol of benzene free air
Barometric Pressure
=
mm Hg
742
p* benzene at 40ºC
=
181
Partial pressure of air
=
561
Partial pressure of benzene
=
181
At 25 psig total pressure or
p* benzene at 10ºC
Partial pressure of air
⇒ 0.323 lb mol benzene
2052
=
45.4
2006.6
Final partial pressure of benzene in air
=
45.4 ⇒ 0.023 lb mol benzene
8–27
Solutions Chapter 8
0.323 – 0.023
= 0.3 lb mol C6H6 recovered
a.
(0.3)(100)
= 92.9% recovery
(0.323)
b.
2 psig is 103 + 742 = 845 mm Hg
(0.023)(845)
= 18.9 mm Hg partial pressure of benzene in the recycled air.
(1.023)
8.4.25
Steps 2, 3, and 4:
The dewpoint of 140ºF gives p*H2O = 149.3 mm Hg
Step 5:
Basis: 100 lb wet solids in (BDS is bone dry solids)
Steps 8 and 9:
BDS balance: 60 lb BDS = 0.90 (wet solids out)
wet solids out = 66.7 lb
From a total balance 100 – 66.7 = 33.3 lb H2O evaporated = 1.85 lb mol H2O evaporated
For the air,
Assume the air exits at 1.0 atm
Composition in:
H2O is
10
(100)
800
8–28
=
1.25%
Solutions Chapter 8
790
(100)
800
149.3
H2O is
(100)
800
=
Air
=
Air is
Composition out:
98.75%
=
18.7%
81.3%
Overall balance (mol) on the gas phase
Ain + 1.85 = Aout
Air balance (mol) on the gas phase
0.9875 Ain = 0.813 Aout
Ain = 8.619 lb moles wet air in
8.619 lb mol 359 ft 3 660o R 760
= 3940 ft 3 of wet air @ 800 mm Hg and 200ºF
lb mol 492o R 800
8.4.26
Basis: 1 hr
Calculate the amounts of the various components in the gas phase:
Water
H 2O in = 0 (by specification)
H 2O out (saturated):
p* at 300 K = 3.51 kPa = p H 2O
y H2O =
3.51 kPa
= 0.0319
110 kPa
CO2
CO2 in:
y CO2 = 0.00055
CO2 out: y CO2 = 0.125
O2
O2 in: yO2 = 0.21
8–29
Solutions Chapter 8
O2 out: yO2 = 0.0804
Assume the other gas in and out is N2.
The entering gas is
600 m3 120 kPa
(kg mol)(K)
= 28.87 kg mol
300 K 8.314(kPa)(m3 )
The exit gas can be obtained from a N2 balance
28.87(1 − 0 − 0.00055 − 0.21) = n out (1 − 0.0319 − 0.125 − 0.0804)
n out = 29.89 kg mol
In one hour in the steady state the moles of CO2 produced were
29.89(0.125) − 28.87(0.00055) = 3.72 kg mol
The mole of O2 consumed were
28.87(0.21) − 30.14(0.0804) = 3.64 kg mol
RQ =
3.72
= 1.02
3.64
8.4.27
Step 5:
Basis: 100 lb mol flue gas
Steps 2, 3, and 4:
2.5 lb
97.5 lbF
W
H2O
Coal (dry)
F
%
C
80
H
6
O
8
Ash 6
100
P (flue gas)
Air
H2O
CO2
CO
N2
O2
mols fr.
0.140
0.004
0.800
0.056
1.000
N2 0.79
O2 0.21
1.00
H2O, dew point 50oF; p*H2O = 0.3624 in Hg
8–30
Solutions Chapter 8
Steps 6 and 7:
Unknowns: F, A, W
Equations: Material balances: C, H, O (the ash is ignored)
Steps 8 and 9 (balances are in moles; in = out)
To get F:
C: (0.80F)(1/12) = (0.140 + 0.004) 100
To get A:
N2: 80 = A(0.79)
F = 216 lb or 18.0 lb mol
A = 101.27 lb mol
H: Hydrogen from the dry coal plus the water in the coal is (MW H = 1.008):
⎡ 216(0.06) 2.5 216 1 2 ⎤ 1
+
⎢
⎥ = 1
97.5
18.016 ⎦ 2
⎣ 1.008
6.74 lb mol H2O
In A: p*H2O = 0.3624 = yAH2O ptotal = yAH2O (29.90)
yAH2O = 0.0121
The water in A is (0.0121) (101.27) =
1.23 lb mol
Total H2O in P: W =
7.97 lb mol
The dew point is
7.97
(29.92) = 2.21 in. Hg equivalent to 105oF
107.97
8.5.1
(a) F; (b) F;(c) F; (d) T; (e) T; (f) T
8.5.2
For multiple components, the vapor pressure varies with composition as well as with
temperature, and to be precise the composition should be stated.
8–31
Solutions Chapter 8
8.5.3
Henry’s Law is p = Hx, a straight line with no intercept. A plot of the data indicates
Henry’s law fits the data well.
8.5.4
Use Henry’s law p = Hx. The MW of CHC13 is 119.4. The total pressure is 1 atm.
x=
p 0.024(1)
=
= 1.41
× 10−4 mol fraction
H
170
The concentration in mg/L is
Water:
1000
= 55.6 g mol/L
18
The number of moles of CHC13 is
(55.6)(1.41× 10−4 )
= 7.84 ×10−3 g mol/L
−4
1 − 1.41×10
Concentration =
119.4 g 7.84×10-3g mol 1000 mg
= 940 mg/L
g mol
1L
1g
8–32
Solutions Chapter 8
8.5.5
p = Hx where p is in kPa and x is mol fraction
The mole fraction of O2 in the gas phase is y O
p O2
y O2 =
pTotal
=
2
Hx O2
pTotal
The moles of O2 and H2O are:
6 mg
1g
103 L 1 g mol O 2
= 0.1875 g mol/m3 in water
L 1000 mg 1 m3 32 g O 2
O2 :
H 2 O:
x O2 =
3
10×10
18
3
= 55.6 × 103g mol/m 3
0.1875
= 3.37 10−6
55.6 × 103 + 0.1875
pTotal = 1 atm
y O2 =
×
4.02×106 kPa
1
3.37×10-6 mol fr.
= 0.134
mol fr.
101.3 kPa
Basis: 1 L gas
nO =
2
pO
2
RT
V=
(101.3 kPa)(0.134)
1 atm
101.3 kPa
1L
(g mol)(K)
18 g
0.08206(L)(atm) 290 K 1 g mol
8–33
1000 mg
1g
=
0.101 mg
per L
Solutions Chapter 8
8.5.6
Steps 2, 3, and 4
pTotal = 1 atm
and at 60ºF the vapor pressures are:
p*Toluene = 16 mm Hg
p*Benzene = 60 mm Hg
Assume the mixture to be ideal so that Raoult’s Law applies
Step 5:
Basis: 1 mol liquid
Steps 6-9:
yi =
pi
p Total
=
p*i x i
p Total
For toluene:
yTol =
For benzene:
(16)(0.60)
= 0.0126
760
y Bz =
(60)(0.40)
= 0.0316
760
The balance is air.
0.0126
0.0316
0.0442
Toluene
Benzene
Total
The vapor is flamable.
8.5.7
Use Raoult’s Law.
p Total = p*P (1 − x B ) +p*B x B
From Perry the vapor pressures are:
p*Propane = 16.8 atm
p*Butane = 4.8 atm
pTotal = (100/14.7) atm = 6.80 atm
6.80 = 16.8(1 − x B ) + 4.8x B
x Butane = 0.83 assuming the liquid phase is essentially all of the mixture.
8–34
Solutions Chapter 8
8.5.8
Mole fraction in the liquid phase is given by solving the following equations for x Bz :
and yTol =p*Tol xTol /pTotal to get xBz = (pTotal − p*Tol )/(p*Bz − p*Tol )
yBz = p*Bz x Bz /pTotal
The corresponding equilibrium mole fraction in vapor phase (y) is,
yi = p*i x i /pTotal
Using these equations for the various temperatures given, the following data for x and y
are obtained for the total pressure of 1 atm
T(ºC)
80
92
100
110.4
xBz
1.000
0.508
0.256
0.000
yBz
1.000
0.720
0.453
0.000
From the above-calculated data, the following T-x diagram can be drawn.
8–35
Solutions Chapter 8
8.5.9
8.5.10
The mole fractions of each component are needed to apply Raoult’s law. Assuming a
basis of 100 g of solution, we can construct the following:
Water
Methanol
g
Molecular
weight
25
75
18
32
Mol
Mol
fraction
1.39
2.34
3.73
0.37
0.63
1.00
Raoult’s law is used to compute the vapor pressure (p*) of pure methanol based on the
partial pressure required to flash:
p = xp*
p* = p/x = 62/0.63 = 98.4 mm Hg.
Use the Antoine equation to get T
ln(p* ) = 18.5875 −
3626.55
-34.29 + T
T = 293 K (20o C)
8–36
Solutions Chapter 8
8.5.11
p
Basis:
*
Nap
@ 180 ˚F = 460 mm Hg
p* H2 O @ 180 ˚F = 388 mm Hg
= 460 + 388 = 848mm Hg
pMax = pH 2 O + pNaptha
(a)
p* Naptha @ 160 ˚F = 318 mm Hg
p* H2 O @ 160 ˚F = 245 mm Hg
Basis:
6000 lb feed
(5000) (0.928) = 4640 lb pure Naptha
lb mol H 2 O 245
=
lb mol Nap 318
W1 n1 M1
=
; W1
W 2 n 2 M2
1000 - 600
= 0.77
⎛ 18 ⎞
= 4640 ⎝
(0.77) = 600 lb H2O Distilled
107 ⎠
= 400 lb H2 O left
(b)
8.5.12
Steps 2, 3, and 4:
LH O = ?
2
100%
H2O
SO2
Air
y1 = 0.03
G = 84.9 m3
293 K
pTotal = 1 atm
y2 = 0.003 SO
2
Air
290K
pTotal = 1 atm
LH O
2
nH O=?
2
nSO2= ?
saturated
Step 5:
Basis: 1 min (85 m3 entering gas)
8–37
Solutions Chapter 8
Steps 6 and 7:
L
n LH2O , n SO
, Air
2
Unknowns:
Equations. Material balances:
H2O, SO2, Air
yi > Hxi
Steps 8 and 9:
The liquid and vapor are assumed to be in equilibrium at the exit, and the water is
saturated with SO2 so that
y1 = Hx1
0.03 = (43) x SO = so that x SO2 = 0.00070 mol fraction
2
Material balance SO2 (moles):
G (0.03 – 0.003) = L (0.00070 – 0)
b.
g mol H 2 O
L
= 38.6
G
g mol air
The respective flow rates are
G=
85 m3 1 g mol gas
= 3540 g mol gas
0.024 m3
L = (3540) (38.4) = 137 ×103 g mol water
In terms of kg
a.
L = (136) (18) = 2450 kg/min
8–38
Solutions Chapter 8
8.5.13
Use the Antoine equation or the physical property package on the CD to get the vapor
pressure of pentane (P) and heptane (H). Assume ideal liquid and vapor.
Basis: Data given in problem statement
ln p*P (mm) = 15.8333 −
2477.07
T-39.94
p*P = 283.6 mm Hg (5.48 psia)
ln p*H (mm) = 15.8737 −
2911.32
T − 56.51
p*H = 20.6 mm Hg (0.40 psia)
Convert mass fractions to mole fractions:
Basis: 100 g liquid
P
H
g
20
80
100
MW
72.15
86.17
g mol
0.28
0.93
1.21
mol. fr.
0.23
0.77
1.00
Equations to use, i = P and H
∑ yi = 1
a.
yi =
p*i x i
pTotal
p*P
p*H
xP +
xH = 1
pTotal
pTotal
pTotal = (283.6)(0.23) + 20.6(0.77) = 80.95 mm Hg
3.19 in Hg
10.8 kPa
1.57 psia
16
(0.23) = 0.90
4.07
0.5
yH =
(0.77) = 0.10
4.07
1.00
yP =
b.
yi =
p*i x i
pTotal
8–39
Solutions Chapter 8
8.5.14
Data:
MW
Benzene (B) 78.1
Toluene (T) 92.1
p* at 60ºC
51.3kPa
18.5 kPa
Assume ideal gas and liquid. Apply Raoult’s Law. Calculate mole fractions in the
liquid.
b.
xB =
1
78.1
= 0.541
1
1
+
78.1 92.1
xT =
1
92.1
= 0.459
1
1
+
78.1 92.1
a.
The equations to use are
x B p tot = x B p*B
yT p tot = x T p*T
yB + yT = 1
yB + x T = 1
(1 − y B )p tot = x T p*T
⎛ x B p*B ⎞
*
⎜1 −
⎟ p tot = x T p T
p
tot ⎠
⎝
tot
p tot = x B p*B + x T p*T = (0.541)(51.3) + (0.459)(18.5) = 36.2 kPa
(b)
What will be the composition of this first bubble?
yB =
x B p* B
(0.541)(51.3)
=
= 0.766
p tot
36.2
y T = 1 − y=B
0.234
8–40
Solutions Chapter 8
8.5.15
Data:
p* at –31.2ºC (kPa)
Propane (P) 160.0
n-butane (B) 27.6
Assume ideal solution and vapor. Use Raoult’s law
(a)
yP ptotal = x P p*P
yP +yB = 1
yBptotal = x Bp*B
yP +yB = 1
x p p*p
yP =
(1 − y P )p total = (1 − x P )p*B
p total
More manipulations give
xP =
p total − p*B 101.3 − 26.7
=
= 0.56
p*P − p*B 160.0 − 26.7
x B = 1 − 0.56
=
0.44
(0.56)(160.0)
= 0.884
101.3
(b)
yP =
(c)
For propane
TºC
-0.5
-16.3
-31.2
-42.1
x
0.0
0.196
0.560
1.0
yB = 1 − 0.884
=
y
0.0
0.577
0.884
1.0
8–41
0.116
Solutions Chapter 8
8.5.16
Basis: 1.0 lb mol of mixture @ 200ºC
Data:
p*C7 @ 200o F = 14.7 psia
p*C8 @ 200o F = 5.5 psia
Mole fraction C7
Partial Press C7
Partial Press C8
0
0
5.5
5.5
0.2
2.94
4.40
7.34
0.4
5.88
3.30
9.18
0.6
0.8
8.82 11.75
2.20 1.10
11.02 12.75
8–42
1.0
14.7
0
14.7
Solutions Chapter 8
C7 = 0.47: pTot = 11.7 psia
For
pC7 = 10.0 psia
pC8 = 1.7 psia
8.5.17
Assume ideal solution.
For the bubblepoint
pTotal = p*1 (T)x1 + p*2 (T)x 2
(1)
pTotal = 200 psia or 10,342 mm Hg
xi (mol fr.)
0.20
n-pentane
⎡
2477.07 ⎤
p*1 = exp ⎢15.8333 −
(−39.94 + T) ⎥⎦
⎣
(2)
0.80
n-hexane
⎡
2697.55 ⎤
p*2 = exp ⎢15.8366 −
(−48.78 + T) ⎥⎦
⎣
(3)
p*i is mm Hg and T is in K
Solve Equation (1) to get from Polymath T = 447 K
8–43
Solutions Chapter 8
8.5.18
Assume ideal solution
For the dewpoint
1
p Total
=
y1 y 2
+
p*1 p*2
(1)
p Total = 100 psia or 5, 171 mm Hg
yi (mol fr.)
0.20
n-pentane
⎡
2477.07 ⎤
p*1 = exp ⎢15.8333 −
(−39.94 + T) ⎥⎦
⎣
(2)
0.80
n-hexane
⎡
2697.55 ⎤
p*2 = exp ⎢15.8366 −
(−48.78 + T) ⎥⎦
⎣
(3)
p*i is mm Hg and T is in K
Solve Equation (1) to get from Polymath T = 413 K
8.5.19
Calculate the bubble point and dew point temperatures at 39.36 in. Hg = 1000 mm Hg.
Use the Antoine equations to get p* based on the assumption the solution is ideal.
ln p* = A −
B
C+T
1 = Benzene
2 = Toluene
K1 =
p*1
pt
ln(K1 ) = 1n(p*1 ) − 1n(pt ) = 1n(p*1 )− 6.9078
Calculate 1nK1 and lnK2. Solve the bubble point equation and the dew point equation by
a computer code (or Newton’s method) starting with T = 365K (slightly above p* for
benzene).
a. Bubble point temperature
∑ yi = 1 =
K
∑i x i
8–44
Solutions Chapter 8
If solution is ideal
Ki xi =
p*i x i
p*
p*
and 0.5 1 + 0.5 2 − 1 = 0
pt
pt
pt
(1)
Use the Antoine equation to get p*i
Benzene p*1 = 15.9008 −
2788.51
−52.36 + T
Toluene p*2 = 16.0137 −
3096.52
−53.67 + T
Solution of Equation (1) using Polymath: Bubble point temperature is 375 K
b. Dew point temperature
∑ xi = 1 =
y
0.5 0.5
∑i =
+
Ki
Ki
K2
(2)
Solution of Equation (2): Dewpoint temperature is 381.5 K
8–45
Solutions Chapter 8
8.5.20
V
V
F
L
p = 80 psia (5.45 atm)
T = 250oF (121.1oC)
L
Assume ideal liquid and vapor exist.
Vapor pressure data at 121ºC from the Antoine equation (refer to problem P19.23 for the
equations)
p* (atm)
9.07
4.03
Pentane (P)
Hexane (H)
p P = p*P x P = 9.07x P
pH = p*H (1 − x P ) = 4.03(1 − x P )
pTotal = 5.45 = pP + pH = 9.07x P + 4.03(1 − x P )
b.
x P = 0.282
yP =
a.
(0.718 hexane)
pP
p* x
(9.07)(0.282)
= P P =
= 0.469 (0.531 hexane)
pTotal
pTotal
5.45
Basis: F = 1 mol
F=L+V
L=1–V
F (xF) = Lx + Vy
1(0.40) = (1-V) (0.282) + V(0.469)
L = 0.37 mol
V = 0.63 mol
8–46
Solutions Chapter 8
8.5.21
Assume pentane is the major contributor to the gas-phase composition.
yi =
pi
p tot
For Raoult's law,
*
pi = p i x i
*
pi =
so
yi p tot (0.018)(100)
=
= 36kPa
xi
0.05
Using the Antoine equation,
ln (36) =15.8333−
2477.07
−39.94 + T
T = 242 K
8.5.22
Assume that the pressure at the bottom of the lake = pCO2
p = pgh varies with height
Avg. depth = 225 m
pCO2 =
1000 kg 9.8 m 225 m
= 2.21 × 103 kPa + 1.01 × 102 kPa
m3
s2
= 23 Atm
pCO2 = HxCO 2 ⇒
x CO2
=
23atm
= 0.0134
1.7×10 3 atm mol fr
If the entire 200,000 tons were saturated at pCO but not supersaturated
200, 000 ton H2 O 1000 kg 1kgmol
= 1.11 × 107 kg mol
ton
18kg
0.0134 =
n CO2
nCO 2 is mol of CO2
1.11×107 + nCO 2
8–47
Solutions Chapter 8
n CO2 = 1.51 × 105 kg mol
(1.5×10 5 )(8.314)(273)
nRT
6
3
= VCO2 =
= 3.39×10 m
p
101.3kPa
Above would be the worst case. The CO2 would be less; p CO2 average might be a better
choice to use in which case VCO 2 would be, say ½ the calculated amount.
8.5.23
a.
Apply Henry’s law to solve the problem.
p = Hx or x =
p
H
From the internet, H = 43,600 where x is the mole fraction in the liquid and p is in
bars.
Enriched gas
(110)(0.397) kPa 1 bar
100 kPa
x o2 =
= 1.00×10-5 mol fr.
43,600 bar
mol fr.
b.
To compare enriched gas with nonenriched gas
x enriched
x nonenriched
percent excess is:
⎛ p ⎞
⎜ ⎟
⎝ H ⎠enriched
=
⎛ p ⎞
⎜ ⎟
⎝ H ⎠ nonenriched
=
(110)(0.397)
= 1.89
(110)(0.21)
1.89 − 1
(100) = 89%
1
8–48
Solutions Chapter 8
8.5.24
VI
y1I
FI
z1I
VII
yII
2
I
II
LI
x1I = z1II
LII
x1II
1 designates component A
2 designates component B
For each of the separation units we can write a mass balance and an equilibrium
relationship, e.g., Raoult’s Law.
Material balance on Comp. 1
V
y1
F
z1
Fz1 = Vy1 + Lx1
(1)
Raoult’s Law
Sep.
Unit
L
x1
y1pTotal = x1p*1
(2)
(1-y1 )pTotal = (1-x1 )p*2
(3)
Solve for y1 in (1):
⎛ Fz ⎞ ⎛ L ⎞
y1 = ⎜ 1 ⎟ − ⎜ ⎟ x1
⎝ V ⎠ ⎝ V ⎠
(4)
Introduce in (2):
⎡⎛ Fz1 ⎞ ⎛ L ⎞ ⎤
*
⎢⎜ V ⎟ − ⎜ V ⎟ x1 ⎥ pTotal = x1p 1
⎝
⎠
⎠
⎣⎝
⎦
Basis: 1 minute
Introduce the given values
p*1 = 10 kPa
p*2 = 100 kPa
FI = 100 mol
LI = 50 mol
8–49
Solutions Chapter 8
V I = 50 mol
VII= 25 mol
LII = 25 mol
z1 = 0.50
The solution is
p*1 =
10kPa
p*1 =
10kPa
p*2 =
100kPa
p*2 =
100kPa
Unit I
FI =
LI =
VI =
z1I =
Unit II
FII =
LII =
VII =
z1II =
100 mol/s
50 mol/s
50 mol/s
0.5
I
x1 =
I
x2 =
50 mol/s
25 mol/s
25 mol/s
0.75975
II
x1 =
II
0.93391
II
II
0.06609
0.75975 = z1
0.24025 = z 2
x2 =
I
1
y =
0.24025
y
=
0.58559
I
0.75975
y2 =
0.41441
y2 =
II
1
II
I
II
p Total = 31.62 kPa
pTotal = 15.95 kPa
8.6.1
Apply K values to solve the problem. Get the mole fraction of the compound in
the liquid phase.
Basis: 100 g water (5.555 g mol)
liquid
g
MW
g mol
g mol
x (mol fr.)
vapor
K
y (mol fr.)
H 2O
Glycerol
5.5
92.08
0.0597
5.555
0.0106
1.2 × 10-7
1.27 × 10-9
MEK
1.1
72.10
0.0153
5.555
0.00275
3.065
0.00843
Phenol
2.1
94.11
0.0223
5.555
0.00400
0.00485
1.94 × 10-5
The mole fractions in the gas phase are in the far right hand column. Glycerol and phenol
have the lowest concentrations. If volatization is proportional to the vapor phase
concentration, only MEK might be a problem.
8–50
Solutions Chapter 9
9.1.1
Basis: 1 lbm
252 cal 1 lb m
4
= 2.5 × 10 cal / kg
a. 45.0 Btu
1 Btu 0.454 kg
lb m
3
1.055 × 10 J 1 lb m
5
45.0
Btu
= 1.048 × 10 J / kg
b.
1 Btu
0.454 kg
lb m
–4
2.930 × 10 (kW)(hr) 1 lb m
– 2 ( kW)(hr )
45.0
Btu
= 2.91× 10
c.
1 Btu
0.454 kg
lb m
kg
2
7.7816 × 10 (ft) (lb f)
4
d. 45.0 Btu
= 3.5 × 10 ( ft )(lb f ) / lb m
1 Btu
lb m
9.1.2
9.484 × 10
a. 4.184 J
J
(g) (°C)
–4
–41.6 J 9.484 × 10
b.
J
kg
–4
c.
0.59 (kg (m)
3
(s ) (K)
= 0.341
Btu
1g
2.2 × 10
Btu
–3
1°C
lb m 1.8°F
=
1.00
(lb m )(° F )
Btu
kg
1g
= –0.0179
1000 g 2.20 × 10 – 3 lb
lb m
m
1N
–2
(kg) (m) (s )
–1
(J) (m ) 9.484 × 10
J
1N
–4
Btu
Btu
(ft )(hr )(° F)
9.1.3
2
cal
252 cal hr
ft
a. 6000 Btu
= 0.4521
2
2
2
(s) cm 2
(hr) (ft ) Btu 3600 s 30.48 cm
( )
cal
(Δ) 1.8°F
b. 2.3 Btu 252 cal lb
= 2.3
( g)(°C )
(lb) (°F) Btu 453.6 g (Δ) ° C
9–1
1 m 3600 s 1 K
3.2808 ft hr 1.8°F
Solutions Chapter 9
c.
J
200 Btu 1, 055 J hr
(Δ )1.8°F
ft
= 3.46
(s)(cm )(° C)
(hr )( ft )(°F ) Btu 3600 30.45 cm ( Δ )°C
3
d.
10.73 (lb f) (ft ) 144 in. 2
(in. 2 ) (lb mol) (°R)
= 8.31
ft
2
3
Btu
lb mol 1.055×10 J (Δ) 1.8°R
Btu
778 (ft) (lb f) 453.6 g mol
(Δ) ° K
J
(g mol )(° K)
9.1.4
Basis: 1011 watts
108 W 8.6057 ×105 cal 1 hr
1 (min)(cm2 ) 1 m2
103 (W)(hr) 60 min 0.10 32.0 cal (100 cm) 2
= 4.5 × 104 m2 larg e
9.1.5
(a) T; (b) T; (c) F; (d) F; (e) T
9.1.6
(a) T; (b) T; (c) T; (d) T; (e) T; (f) F; (g) F
9.1.7
partial pressure
volume
specific gravity
potential energy
relative saturation
specific volume
surface tension
refractive index
intensive
extensive
intensive
extensive
intensive
intensive
intensive
intensive
9–2
Solutions Chapter 9
9.1.8
a.
b.
c.
d.
Intensive
Intensive
Intensive
Extensive
9.1.9
To convert to J/(min) (cm2) (°C), we set up the dimensional equation as follows
2
h=
2
0.026 G 0.6 Btu 1055 J 1 hr ⎛ 1 ft ⎞ ⎛ 1 in. ⎞ 1.8o F
⎜
⎟ ⎜
⎟
D0.4 (hr) (ft 2 ) (o F) 1 Btu 60 min ⎝ 12 in. ⎠ ⎝ 2.54 cm ⎠ 1o C
= 8.86 × 10-4
G 0.6 J
D0.4 (min)(cm2 )(o C)
9.1.10
a.
1 lb fat 1 kg 7,700 k cal day = 7 days
2.2 lb
kg
500 k cal
b.
1 lb fat 1 kg 7,700 k cal 4.184 k J km 0.6214 mi = 22.75 mi
1 k cal 400 k J 1 km
2.2 lb
kg
c. Total energy consumed is the same. The higher power consumption is compensated
for in the shorter travel time.
9–3
Solutions Chapter 9
9.1.11
975 W A m 2 320 days 24 hr 5 0.21 3.6 × 106 J
m2
1 day 25
1( kW)( hr )
a.
20
= 3 × 10 J
A = 2.5 × 101 1m 2
( 2.5×10 km )
5
2
The capital cost is probably too great and the maintenance too high relative to other
power sources. Also, the location would have to be in the far western united so power
would have to be transmitted over long distances.
b.
3×1020 J =
10,000 Btu 2,000 lb T tons 1055 J 0.70
lb
ton
1 Btu
T = 2.0 ×1010 tons
2.0 ×1010
= 0.012
1.7 ×1012
9.1.12
First substitute for T
o
F
ToF = To C (1.8) + 32
Next change units
k=
{a + b ⎡⎣T
}
(1.8) + 32⎤⎦ Btu 1055 J 1 ft
1 in
1 hr 1.80o F
(hr)(ft)(o F)
1 Btu 12 in 2.54 cm 60 min 10 K
o
C
= 1.05 ⎡⎣a + b(To C (1.8) + 32 ⎤⎦
J
(min)(cm)(K)
or k = 1.05 a + 1.8 bT C + 33.2
o
9–4
Solutions Chapter 9
9.1.13
900,000 J (s)(10-3 ) kW
= 105 kW or 100 MW
-3
9×10 s
1J
The answer is yes .
9.1.14
(a)
Some examples are:
How far would you have to run to lose 1 kg of fat?
32,000 kJ 1 km
= 80 km
400 kJ
about the distance of 2 marathons!
(b)
How many days would you have to reduce your diet from 2400 calories to 1400
calories to lose one kg of fat?
7700 k cal 1 day
= 8 days
1 kg fat 1000 k cal
9.1.15
A main meal is equivalent to about 4000 kJ. A ¼ hour is 900 seconds, so that (700 J/s)
(900 s) / 1000 = 630 kJ, an insufficient amount of time.
9.1.16
A closed system because no mass exchange occurs with the surroundings.
9.1.17
Yes. Watts are power, and if the energy is transferred for a very short period to time, the
power can be quite large. For example, 1J (a very small amount of energy) transferred in
10–6s is 1 MW!
9–5
Solutions Chapter 9
9.1.18
The trend is to advertise in a way that makes people select a product. And sometimes
that just creates confusion.
What the manufacturer is doing is basing the advertisement on the weight of the hot dog.
A hot dog weights about 43 grams and has 8 grams of fat in it. They divided 8 by 43 and
get 19 percent fat. They can round it off to 20 and say it’s 80 percent fat free.
But you are concerned with the percentage of fat in the calories. You want 30 percent or
less to come from fat.
Your figures are right.
9.2.1
Basis: 40.0 Newtons × 6.00 m = 240(N)(m)
(a)
(b)
(c)
(d)
240(N)(m)
1J
= 240 J
1(N)(m)
240(N)(m) 0.738(ft)(lb f )
= 177.0(ft)(lb f )
1(N)(m)
240(N)(m)
1 J 0.239 cal
= 57.4 cal
1(N)(m)
1J
240(N)(m)
1 J 9.478 ×10-4 Btu
= 0.226 Btu
1(N)(m)
1J
9.2.2
The work done by the cylinder-piston system is W = -p(V2-V1)
=−
350 kPa 0.15 − 0.02 m3
1 kJ
= 45.5 kJ
(1 kPa)(1 m3 )
9–6
−
Solutions Chapter 9
9.2.3
No work is done because the boundary of the system remains fixed.
9.2.4
Basis: 2 kg mass
PE = mgh =
12 kg 9.80 m 25 m 1(J)(s2 )
= 2940 J
s2
1(kg)(m2 )
9.2.5
Any correct examples will be satisfactory for (b), (c), and (d). But for (a), liquid
to solid, there are probably no examples. If any, they would be pathological cases.
9.2.6
(1) Solid melting.
9.2.7
Basis: 10 lb water (5 lb vaporized)
The work done by the piston is caused by the water vapor expanding at constant
temperature and pressure. Ignore the volume of the liquid water at the initial conditions,
and assume the initial amount of vapor is also negligible. From state to state 2
ΔV̂vapor = 4.897 ft 3 / lb from the steam tables. For the 5 lb
V2 = (5 lb)(4.897 ft 3 /lb) = 24.49 ft 3
W=−
89.65 lbf 144 in 2 (24.49 ft 3 )
= −3.16 × 105 (lbf )(ft)
2
2
in
1 ft
9.2.8
(a) T; (b) F; (c) T; (d) F; (e) F; (f) F; (g) T
9–7
Solutions Chapter 9
9.2.9
The comment is not a correct use of the term heat, which is a transfer of energy.
9.2.10
Conservation is confused with balancing. Heat is energy transferred from one system to
another, and from the viewpoint of either system (for which an energy balance is made),
heat is not conserved. Thus (a) is wrong, heat is part of a balance; (b) same as (a); (c) heat
is not conserved—it is transferred and is part of the energy balance; (d) the answer should
be no with the explanation offered.
9.2.11
(a)
Wrong. Heat is energy, not material.
(b)
Wrong. Heat is energy by definition.
(c)
Wrong. Heat is energy; cold is related to temperature.
(d)
Wrong. See (c).
(e)
Wrong. Heat is a transfer of energy.
(f)
Heat can be measured, but indirectly via temperature, mass, etc.
(g)
Wrong. See (a).
(h)
OK except for “or can be stored,” which is wrong.
(i)
Burning produces a change in the state of the reactants and products, but the
change is not heat. Heat transfer can result from the change in state.
(j)
Heat cannot be stored and hence destroyed. Heat transfer can be terminated.
9.2.12
A better use of words than “absorbing heat” would be “to transfer heat.” Heat is really
not “concentrated.” The internal energy of the medium increases.
9–8
Solutions Chapter 9
9.2.13
(a) F; (b) F; (c) T; (d) F; (e) T; (f) T but debatable; (g) F; (h) F
9.2.14

Heat transfer rate is Q=hAΔT=h(
π DL)ΔT
=
Q
5J
5 cm 1 m 100π m (120 − 20)oC
7854J/s
100 cm
(s)(m 2 )( oC)
9.2.15
Basis: 1 minute
KE = (1/2 ) mv2
v=
500,000 g 1 cm3
4
= 22,143 cm/min
min
1.15g π(5)2cm2
1 500 kg
KE =
2 min
2
2
⎛ 22,143 cm ⎞ 1(J)(s )
⎜
⎟
2
⎝ min ⎠ 1(kg)(m )
2
2
⎛ 1 min ⎞ ⎛ 1 m ⎞
⎜
⎟ ⎜
⎟ = 6810 J
⎝ 60s ⎠ ⎝ 100 cm ⎠
9.2.16
Basis: 1 lbm H2O
KE = 1/2 mv2
2
1 1 lbm ⎛ 10 ft ⎞ (lbf )(s2 )
KE =
= 1.55 (ft) (lbf )
⎜
⎟
2
⎝ s ⎠ 32.2 (lbm )(ft)
9–9
Solutions Chapter 9
9.2.17
Relative to the reference values in the steam tables and the CD:
(a)
U = 3528 kJ/kg (from the steam tables)
(b)
The quality is 0.13% and U = 1213 kJ/kg from the steam tables
(c)
At 100oC and 1000 kPa, U of the liquid is about U ≅ 532 kJ/kg from the steam
table
9.2.18
Basis: 1 kg steam
1
2
250.5°C
4000 kPa
650°C
10,000 kPa
ΔU = ΔH – ΔpV or use U values directly from SI tables or computer code.
Data here taken from the steam tables.
State 1 (assumed saturated) State 2 (superheated)
250.5°C (523.5 K)
650°C (923 K)
482°F
1202°F
4000 kPa abs.
10,000 kPa abs.
p
580 psia
1450 psia
Vˆ
0.498 m3/kg
0.04117 m3/kg
ΔU
2602 kJ/kg
3338 kJ/kg
ΔH (ref. satd liquid at 0°C)
2801 kJ/kg
3742 kJ/kg
T
9–10
Solutions Chapter 9
9.2.19
Internal energy is the energy contained in the material of a system because of its
molecular arrangement. Heat is energy that flows between the system and the
surroundings because of a temperature gradient. Therefore they are not the same class of
energy.
9.2.20
ΔH = ΔU + Δ (pV)
When a liquid vaporizes Δ (pV) adds to ΔU.
9.2.21
∆H and ∆U are zero because ∆H and ∆U are point (state) functions (variables).
9.2.22
None. The initial and final states for U and H are the same.
9.2.23
The energy balance is ΔΕ = ΔU + ΔPE + ΔKE = Q + W
Q
>0
0
0
0
(a)
(b1)
(b2)
(c)
ΔE
>0
>0
>0
>0
W
0
>0
>0
>0
9.2.24
(a)
(b)
(c)
(d)
(e)
T
F
F
T
F
(f)
(g)
(h)
(i)
(j)
F
F
F
T
T
9–11
ΔU
>0
>0
>0
>0
Solutions Chapter 9
9.2.25
(a)
(b)
(c)
(d)
(e)
F
F
F
T
T
(f)
(g)
(h)
(i)
(j)
T
F
T
F
F
(k)
(l)
(m)
(n)
T
F
T
T
9.2.26
(a) 1; (b) 3; (c) 5; (d) 2; (e) 4
The boiling temperature is at 4 reading to the left scale (100ºC)
The freezing temperature is at 1 reading to the left scale (0ºC)
9.2.27
Tc = 540.2 K
pc = 27 atm
⎛ T ⎞
0.0331⎜ b ⎟ − 0.0327 + 0.0297 log (p c )
kJ
⎝ Tc ⎠
ΔH v = Tb
(g mol) (K)
⎛ T ⎞
1.07 − ⎜ b ⎟
⎝ Tc ⎠
ΔH v = 31.679
kJ
g mol
The percent error is negligible in view of the precision of the data.
9.2.28
Below log is log10 . The critical temperature = 408.1 K
C p = A + Blog Tr
Basis: 1 g mol
B=
Cp1 − Cp2
log Tr1 − log Tr2
;
A = Cp1
Blog Tr1
9–12
−
Solutions Chapter 9
Cp1 = 97.3
Tr1 = 300 / 408.1 = 0.735
log Tr1 = 0.134
C p2 = 149.0
Tr2 = 500 / 408.1 = 1.225
log Tr2 = 0.088
A = 128.5
B = 232.6
at 1000 K: Tr1000 = 1000 / 408.1 = 2.45
−
log Tr1000 = 0.389
Cp = 128.5 + 232.6(0.389) = 219.0 J /(g mol)(K)
(100)
227.6 − 219.0
= 3.8%
227.6
9.2.29
ΔH = 2∫
250 + 273
50 + 273
(27.32 + 0.6226 ×10−2 T − 0.0950 ×10−5 T 2 )dT
ΔH = 11,980 J (The CD gives 11,784 J)
9.2.30
Integrate the heat capacity equation, use Table D3 in the Appendix, or use CD
From the CD ΔH = 2580 J/g mol
9.2.31
2
⎛ πD 2 ⎞
⎟ 5.5 = π (0.25) ( 5.5) = 0.27 cm 3 . The density of Al is
The volume of the wire is ⎜
⎝ 4 ⎠
4
3
19.35 g/cm , so 19.35(0.27) = 5.22g or 0.194 g mol Al. From Perry, Cp = 20.0 +
0.0135T (T in K, Cp in J/(g mol)(°C) and ∆Hfusion = 10,670 J/g mol at 660°C.
⎡660 +273
⎤
ΔH = 0.194 ⎢ ∫ (20.0 + 0.0135T)dT + 10,670 ⎥ = 5560 J
⎣ 25+ 273
⎦
(
)
5560 2.773 × 10 –7 = 1.54 × 10 –3 kWh
9–13
Solutions Chapter 9
9.2.32
Basis: 5 ft3 vessel
Vapor = 4 ft3, Liquid = 1 ft3
Steam tables: Sat'd steam at 1000 psia
3
3
ft
Vv = 0.44596
lb
ft
Vl = 0.02159
lb
lb
ml =
mv =
1 ft
3
lb
0.02159 ft
4 ft
3
3
lb
0.44596 ft
3
mass fr.
=
46.32
0.838
8.97
0.162
55.29
1.000
=
mt =
Quality = 0.162
9.2.33
We will use the steam tables for this problem.
Basis: 1 lb of H2O at 60°F
From steam tables (ref. temp. = 32°F):
Ĥ = 28.07 Btu/lb
at 60°F
Ĥ = 1604.5 Btu/lb
at 1150°F and 240 psig (254.7 psia)
Δ Ĥ = (1604.5 – 28.07) = 1576.4 Btu/lb
ΔH = 1576(8.345) = 13.150 Btu/gal
Note: The enthalpy value that has been used for liquid was taken from the steam tables
for the saturated liquid under its own vapor pressure. Since the enthalpy of liquid water
changes negligibly with pressure, no loss of accuracy is encountered for engineering
purposes if the initial pressure on the water is not stated.
9–14
Solutions Chapter 9
9.2.34
Basis: 3 kg H2O
The initial conditions are obtained from the SI steam tables for saturated liquid. Enthalpy
is a function of the temperature and pressure, but the effect of pressure on liquid water
under these conditions is negligible. Therefore the enthalpy of water at 300K and 101.3
kPa can be approximated by the enthalpy of saturated water at the same temperature but
at the vapor pressure of 3.536 kPa.
The final conditions are presumably a state in which water is all vapor. A check of the
steam tables shows this assumption to be true.
At
T(K)
p(kPa)
300
800
3.536
1500
ˆ
ΔH(kJ/kg)
111.7
3384.3
The enthalpy change is
ΔH =
3 kg (3384.3-111.7) kJ
= 9817.8 kJ
kg
9.2.35
The amount of energy that the water gives up in cooling to 0ºC is not enough to melt all
the ice, since each gram of water will give up 209 J in cooling, while heating 1 g of ice to
0º and melting it requires that 2 × 40 + 335 = 415 J be expended (the heat capacity of ice
is 2 J/(g) (ºC)).
9.2.36
Yes. 40ºF and saturated liquid is an arbitrary reference state for enthalpy that can be
assigned a value of zero. Only enthalpy changes can be computed using the chart; the
enthalpy itself cannot be determined.
9–15
Solutions Chapter 9
9.2.37
Basis: 1.2 ft3 gas at 7.3 atm
p of 1 atm = 15,450 lbf/ft2
pV1.3 = const.
V1 = (1.2)1.3 = 1.27
1.3
p1V1 = (15450)(1.27) = 19,620
1.3
19,620
W = − ∫=pdV − ∫= − 1.3 dV
V
19,620 ∫
dV
⎡ 19,620 ⎤ V2
= − ⎢
1.3
0.3 ⎥
V
⎣ −0.3V ⎦ V1
1.3
1.3
2
V
(V )(p1 ) (1.27)(7.3)
= 1
=
= 9.27
p2
(1.0)
V2 = 5.54 ft 3
V1 = (1.2)0.3 = 1.06
0.3
V2
0.3
W=
= (5.54)0.3 = 1.67
19,620 ⎡ 1
1 ⎤
(19,620)(0.344)
−
=−
= −22,500(ft)(lbf )
⎢
⎥
0.3 ⎣1.67 1.06 ⎦
0.3
9.2.38
The density of air at 27ºC and 1 atm is ρ =
ρ=
101.3 kPa
Power =
(p)(MW)
(RT)
29 kg air
(kg mol)(K)
= 1.18 kg / m3
1 kg mol air 8.314(kPa)(m3 ) 300 K
1
1
1
 2 = (ρAv) v 2 = ρAv 3
mv
2
2
2
9–16
Solutions Chapter 9
Assume air flow through the windmill is equivalent to the average flow in a pipe of
diameter 15m.
v=
20 mi 1.61×103m 1 hr
= 8.94 m/s
hr
1 mi
3600 s
2
⎛ 15 ⎞
π ⎜ m ⎟
3
2
1
1.18
kg
⎛ ⎞
⎝ 2 ⎠ ⎛ 8.94 m ⎞ 1(s )J (s)(W) 1 kW = 74.46 kW
Power = ⎜ ⎟
⎜
⎟
3
2
⎝ s ⎠ 1(kg)(m ) 1 J 1000W
⎝ 2 ⎠ m
Electrical energy = (74.46)(0.30) = 22.3 kW
9.2.39
a.
Basis: 1 lbm of vehicle
25, 000 mi 1 hr 5280 ft
= 36, 600 ft/s
hr
3600 s 1 mi
9 2
1 1.34 × 10 ft 1 lb m 1
7 (ft) (lb f)
K.E. =
= 2.08 × 10
lb m
2
2
gc
s
7
2.08 × 10 (ft) (lb f)
1 Btu
=
K.E. =
778 (ft) (lb f) 2.68 × 10 4 Btu / lb
lb m
m
b.
Heat capacity of, say, 1 Btu/(lb) (°F) gives
1 Btu 20°F
= 20 Btu/lb m .
(lb) (°F)
Essentially all the energy must be transferred to the surroundings
9–17
Solutions Chapter 9
9.2.40
Assume 7 meters is a
constant level difference
hence h = 7m
Filled reservoir
7m
For one-half of the complete cycle (6 hours):
m = ρAh =
Δ(PE) =
3
2
10 kg 23 km 7 m
m
3
3
10 m
1 km
2
= 1.61 × 10
11
kg
1.61×1011 kg 9.80 m 7 m
1J
= 1.10×1013 J
2
(1 kg)(m 2 )
s
s2
For a complete cycle:
2 1.10×1013J 0.85
1 min
= 4.23×108 J/s
370 min 60s
8
or 5.06 × 10 W
9–18
Solutions Chapter 9
9.2.41
The heat capacity of a monoatomic ideal gas is
volume is:
5
R , and the heat capacity at constant
2
5
3
R −R = R
2
2
You can show that C p = C Vˆ + R
ˆ ⎞ ⎡ ∂U
ˆ + ∂ (pV)
ˆ ⎤ ⎛ ∂U ⎞
⎛ ∂H
⎛ ∂V ⎞
Cp = ⎜
⎟ = ⎢
⎥ = ⎜
⎟ + p ⎜
⎟
∂T
⎝ ∂T ⎠p
⎝ ∂T ⎠p ⎣
⎦ p ⎝ ∂T ⎠p
For the ideal gas Û is a function of T only (not of p or V̂ ) so that
⎛ ∂U ⎞ ⎛ ∂U ⎞
⎜
⎟ = ⎜
⎟ = Cv
⎝ ∂T ⎠p ⎝ ∂T ⎠Vˆ
⎛ ∂V̂ ⎞
⎛ R ⎞
p ⎜
⎟ = p ⎜ ⎟ = R
⎝ p ⎠
⎝ ∂T ⎠ p
ΔU = ∫ Cv̂dT = (3/2 R)(50) = 624 J
9.2.42
Use the same reference state, so that
ΔH = ΔU + Δ (pV)
For one mole of an ideal gas: pV = RT
ΔU = ΔH – ΔRT
= 6.05 ×105
J
8.314 J 103g mol [(100 + 273) − (273)]K
–
kg mol ( g mol )( K ) 1 kg mol
= 6.05 × 105 – 8.314 × 105
= −2.26 ×105 J/kg mol
9–19
Solutions Chapter 9
9.2.43
Basis: 100 mol gas
Comp.
CO
H2
CO2
mol
68
30
2
100
ΔĤ (J/g mol) relative to 0°C and 1 atm
CO
H2
CO2
25°C =
1400°C =
298 K
1500 K
1673 K
1750 K
728
39,576
45,723
48,459
718
36,994
42,724
45,275
912
62,676
72,896
77,445
Basis: 1000 m3 at 101 kPa and 1673 K
1000 m3 273 K 1 kg mol
= 7.28 kg mol
1673 K 22.4 m3
Calculation of ΔH:
CO
H2
CO2
kg mol
ΔĤ (kJ/kg mol)
7.28 (.68)
7.28 (.30)
7.28 (.02)
-44,995
-42,010
-71,984
9–20
ΔH(kJ)
-222,743
-91,750
-10,481
Solutions Chapter 9
9.2.44
K
423
°C
150.0
°F
302
353.3
80.1
176
278.7
5.5
42
253.0
-20.0
-4
vapor
↓
→ satd
satd vapor
liquid
↓
Satd liquid
→ satd
solid
↓
solid
Basis: 1 g mol
From Table D.1 in the Appendix.
Benzene properties:
Mol. wt.
78.11
Boiling point
353.26 K
Melting point
278.69 K
ΔHvaporization
30.76 kJ/g mol
ΔH fusion
984 kJ/g mol
K
423
353.26
278.69
253
°C
150
80.11
5.54
-20
Heat Capacity Equation Coefficients:
Benzene(g)
Benzene(1)
a
74.06
62.55
b
32.95 × 10-2
23.9 × 10-2
c
-25.20 × 10-5
-
d
77.57 × 10-9
-
The Cp equation is in °C hence the limits on integration should be in °C. ΔH1 =
∫
353
423
(74.06 + 32.95 ×10−2 T − 25.20 ×10−5 T 2 + 77.57 × 10−9 T3 ) dT = -11,790 J/g mol
9–21
T
°C
K
Solutions Chapter 9
Condensation
ΔH2 = −3.076 ×104 J/g mol
Liquid
ΔH3 = ∫
278.69
353.26
(62.55 + 23.4 ×10−2 T)dT = 10,180J/g
−
mol
Fusion
ΔH4 = −9.84 ×103 J/g mol
Solid
From Perry, 5th edition:
Cp (cal)/(g)(oC)
0oC
-50oC
0.375
0.299
12.3
97.7
If a linear relationship exists between Cp and T,
-11,790
Cp,av = 119 J/(g mol) (o C)
ΔHs = ∫
−20.0
5.5
(119)dT = 3034
− J/g mol
-30,760
-10,180
5
Overall, ΔH = ∑ ΔHi = 65,
− 604 J/g mol
i =1
-9,840
-3,034
-65,604
−65,604 g mol 1000 g
= −8.40×105 J/kg
g mol 78.11 kg
9–22
Cp (J/(g mol)(oC))
Solutions Chapter 9
9.2.45
Basis: liquid water at 32°F and 1 atm
ΔH = 3 ΔH 300°F,1
a.
b.
atm
– ΔH 32°F,1
atm
= 3 (1192 – 0) = 3576 Btu
Basis: liquid water at 40°F and 60 psia
Δ H 32 ° F,60 psia = 0
c.
Δ H = 3 (11192 – 0) =
3576 Btu
Basis: liquid water 40°F and 60 psia
ΔH = ΔH 300°F,60 psia – ΔH 40°F,60 psia = (1181.4 – 8.05)
= 1173.4 Btu
d.
Basis: water-steam mixture of 60% quality
ΔH, Btu/lb
State
1
60% steam - 40% water at 300°F; sat'd steam 1179.7
2
80% steam - 20% water at 300°F; water
ΔH = ΔH 2 – ΔH 1
ΔH 2 = (0.80) (1179.7) + (0.20) (269.6) = 998 Btu/lb
ΔH 1 = (0.60) (1179.7) + (0.40) (269.6) = 816 Btu/lb
Enthalpy change needed = ΔH
ΔH = ΔH 2 – ΔH 1 = 182 Btu / lb
e.
Basis: steam at 500°F and 120 psia
ΔH = ΔH 2 – ΔH 1 = 312.5 – 1276.7 = –964.2 Btu / lb
f.
ΔH = ΔH 2 – ΔH 1 = 487.8 – 1276.7 = 788.9 Btu / lb
9–23
269.6
Solutions Chapter 9
Data from text
ΔHA = 1205.0
ΔHB use tables in pocket
210o F
5 psia
200°F
1148.3
250°F
1171.1
7 psia
1146.7
1170.2
78.17
84.24
38.88
41.96
a. Enthalpy decrease
V̂A =1.0451 ft3 /lb
VB
(h)
40 psia and 267.24°F
sat steam/sat liquid
70 psia/302°F
liquid
b. volume goes up
70 psia/304°F
superheated steam
depending on heat supply
(i)
2.5 ft3 tank of water @ 160 psia and 363.5°F. @ 160 psia 6363.5 total volume is
occupied by sat steam
2.5
Vˆ steam = 2.834 ft 3 / lb ⇒ m steam =
= 0.88 lb
2.834
(
 water = 1-m steam
Vˆ Liq/ water = 0.0182 ft 3 / lb ⇒ m
V water =
)
2.5
3
× 0.0182 = 0.016 ft
2.834
9–24
Solutions Chapter 9
Vsteam = 2.5 – 0.016 = 2.49 ft 3
Of 5 lb H2 O assume x lb is sat steam (5 – x )lb is sat. water
x( 2.834) + ( 5 – x)(0.0182)
= 2.5
⇒ x(2.834 – 0.0182)
= 2.5 – 5(0.0182)
⇒
x
=
2.5 − .0910
= 0.86 lb
2.816
Yes, it is possible
(k)
Enthalpy of 10 lb steam @ 100 psia = 9000 Btu
Enthalpy/lb steam @ 100 psia = 900 Btu
ˆ
ΔH
steam @ 100 psia = 1187.3 Btu
ˆ
ΔH
water @ 100 psia = 298.43 Btu
Let x be fraction of steam
(1–x) is fraction of water
(1187.3)x + (1–x) 298.43 = 900
x(1187.3–298.43) = 900 – 298.43
x=
or 68%
601.57
= 0.68
888.87
9.2.46
a.
State 1
State 2
T = 400 K
T = 900 K
p = 100 kPa
p = 100 kPa
From the steam tables in SI units (interpolation required):
(1) ΔH 1 = 2729.8 kJ/kg
(2) ΔH 2 = 3763.6 kJ/kg
9–25
Solutions Chapter 9
ΔH = 3763.6 – 2729.8 = 1033.8 kJ/kg
1033.8 kJ 18 kg 2 kg mol
4
= 3.7217 × 10 kJ/2kg mol
kg
kg mol
b.
Using the table for the enthalpies of combustion gases:
(1) ΔH = 4284 J/g mol
(2) ΔH = 22, 760 J/g mol
ΔH = 22760 – 4284 = 1.848 ∞ 104 J/g mol = 1.848 ∞ 104 J/kg mol
1.1848 kJ 2 kg mol
4
= 3.695 × 10 kJ/2 kg mol
kg mol
c.
Using the heat capacity equation for steam:
627°C + 273
ΔH =
900
(33.46 + 0.6880 × 10
C p dT =
127° + 273
–2
T + 0.7604 × 10
–5
T
2
400
- 3.593 ∞10–9 T3) dT
= 33.46 (900 – 400) +
+
0.7604 × 10
3
0.6880 × 10
2
–5
(900 3 – 400 3 ) –
–2
2
2
( 900 – 400 )
3.593 × 10
4
–9
4
4
(900 – 400 )
= 20,085 J/g mol = 20,085 kJ/kg mol
4
20,085 × 10 kJ 2 kg mol
=
kg mol
4.017 ×104 kJ/2 kg mol
The steam tables are the most accurate value.
9–26
Solutions Chapter 9
9.2.47
a-1:
V̂ ≅ 0.23 ft 3/lb
a-2:
gas
b-1:
42°F
b-2:
mixture of liquid and vapor
9.2.48
Basis: 1 g mol CO2
ΔH = m∫
100o C
50o C
(a + bT + cT
2
+ dT3 )dT
b
c
d
⎡
⎤
ΔH = m ⎢a(100 − 50) + (1002 − 502 ) + (1003 − 503 ) + (100
− 4 50 4 ) ⎥
2
3
4
⎣
⎦
⎡
4.233 ×10−2
2.887 ×10−5
2
2
(100 − 50 ) −
(1003 − 503 )+
⎢36.11(100 − 50) +
2
3
ΔH = 1 g mol ⎢
−9
⎢ 7.464 ×10
(1004 − 504 )
⎢⎣
4
ΔH = 1956 J
⎤
⎥
⎥
⎥
⎥⎦
From the CO2 chart ΔH ≅ 198-178 = 20 Btu/lb or 2040 J/g mol. From the enthalpy tables
(interpolating) ΔH = 1957 J/g mol. The CD gives 2038 J/g mol.
9–27
Solutions Chapter 9
9.2.49
ˆ − ΔH
ˆ ) = 10(86 − 288) = −
ΔH = 10(ΔH
2020 Btu . The CD gives –1962 J.
2
1
9.2.50
Assume the temperature is 80ºF, and the propane is saturated liquid and vapor. The
corresponding pressure is the saturation pressure, namely about 2.3 atm. Another
temperature would correspond to another pressure. After 80% of the propane is used, the
conditions are still saturated liquid-vapor mixture at 80ºF and 2.3 atm.
9.2.51
Basis: 10 lb mol ideal gas
(a)
Use pV = nRT
ˆ = Δ(RT)
Δ(pV)
3
359 (ft) 1 atm 500°R
3
(1) 10 lb mol
= 36.5( ft ) @ 100 atm, 40°F
lb mol 100 atm 492°R
900°R
(2) 100 atm
= 180 atm
500°R
(3)
ΔH = ( Hˆ 2 – Hˆ 1) 10 = 10 [300 + 8.0 (440) – (300 + 8.0 (40))] = 32, 000 Btu
(b)
The equation for H is based on the reference conditions where Ĥ = 0.
Ĥ = 0 when 300 = -8T or Tref = -37.5 ºF = -38.5ºC. U should have the same
reference temperature.
Δ Uˆ = Δ Hˆ – Δ(p Vˆ ) = Δ Hˆ – ΔRT for an ideal gas
Change units for Ĥ:
First term:
300 Btu 1 lb mol 1055 J
697 J
=
lb mol 454 g mol Btu
g mol
9–28
To F =1.8To C + 32
Solutions Chapter 9
Second term:
o
8.0 Btu 1055 J lb mol (1.8To C + 32) F
J
= (33.5To C +595)
o
(lb mol)( F) Btu 454 g mol
g mol
Ĥ ref T = 0
ΔH = H T − H ref T = H T
ΔRT = (RT)T − (RT)refT
ΔU = 697 + 595 + 33.5 To C − R(TABS − TABS
(T is absolute)
)
REF T
TABS =To C + 273
TASS1REF.T = − 38.5 + 273
ΔU = 1292 + 33.5 To C − 1.987 (To C + 38.5) = 1215 + 31.5 To C
9.2.52
Steady state flow process, no reaction
700 kPa
Dry steam, 100 kPa and 125°C
ˆ = 2725.5 kJ / kg
ΔH
wet steam
ˆ =?
ΔH
The energy balances reduces to ∆H=0 for this process. At 700 kPa saturated (438.1K)
ˆ = 696.7 kJ / kg
ΔH
L
ˆ = 2763.1 kJ / kg
ΔH
v
Let x = vapor fraction
(1 – x)(696.7) + x( 2763.1) = 2725.4
x = 0.98
9.3.1
All of them.
9–29
Solutions Chapter 9
9.3.2
Closed unsteady state system
Q + W = ΔU
60 + W = 220
W = 160 Btu (work done on the system)
9.3.3
(a)
pump
pump
system is the pump. Open, steady state.
Q
Very little
W
Yes
ΔU
No
ΔH
Yes(flow)
ΔPE
No
ΔKE
No
(b)
system is the pump and motor. Open, steady state
A little
Yes
No
Yes
No
No
(c)
System is the ice. Closed, unsteady state.
Yes
No
Yes
9–30
Yes
No
No
Solutions Chapter 9
(d)
∞
system is the mixer and solution. The process is closed,
unsteady state.
Possibly Yes
Yes
Yes
Yes
No
No
9.3.4
a. System:
can and liquid
Q = 0, W ≠ 0
b. System:
motor
Q = 0, W ≠ 0 But motor may get hot.
c. System:
pipe and water
Q = 0, W ≠ 0
9.3.5
a.
boundary
Steam out
Energy out
Water in
Energy in
Q
9–31
Solutions Chapter 9
b.
Q=0
Steam in
Energy in
Turbine
Steam out
Energy out
Work
c.
Work
Battery
Q=0
W ≠0
9.3.6
No. The energy balance is Q + W = ΔU, and for Q = 0 and since ΔU = 0 for an
isothermal process W = 0 .
9.3.7
Closed unsteady state system
Q + W = ΔU
In kJ: (30-5) + 0.5 = Uf – 10
Uf = 35.5 kJ
9–32
Solutions Chapter 9
9.3.8
Closed unsteady state system
Q + W = ΔU
Q=0
W = ΔU = CvΔT
W=
(100 W)(5 hr) 1 kW 3.6 × 103 kJ
= 1.8 ×103 kJ
1000 W 1 kW
mols of air = n =
pV
(100 kPa)(100 m3 ) (kg mol)(K)
=
= 3.97 kg mol
RT
303 K
(8.314)(kPa)(m3 )
1.8 ×103 = 30(3.97)(T − 30)
T = 45o C
9.3.9
Closed unsteady state process
Q + W = ΔU + Δ(PE)
Ignore change in center of mass of the air
Q = 0 assumed
ΔU = 0 assumed (to get maximum elevation)
W = 100 J
100 = Δ(PE) = mgΔh = (0.990 kg) (9.80 m/s2) Δh (m)
Δh = 10.3 m
9–33
Solutions Chapter 9
9.3.10
Closed, unsteady state process
Initial conditions
Final conditions
V = 1ft saturated dry satd. steam (the basis)
V = 1ft (know) + V2
Look up
At p = 1 psia T = 200ºF. The water
is vapor
V̂ = 33.610 ft3/lb
Ĥ = 1145.72 Btu/lb
Û = 1073.96 Btu/lb
From the steam tables:
Ĥ = 1150.15 Btu/lb
Û = 1077.48 Btu/lb
V̂ = 392.713 ft3/lb
p = 11.548 psia
Calculate the lb of steam
m=
1 lb
1 ft 3
= 0.02975 lb
33.610 ft 3
Q + W = ΔU
W = 0 (fixed boundary)
a.
Q = ΔU
b&c.
d.
ΔU = U2 – U1 = (1077.48) (0.02975) – (1073.96) (0.02975) = 0.105 Btu = Q
The volume of the second tank is Vfinal = (392.713) (0.02975) = 11.68
V2 = 11.68 − 1 10.68
= ft 3
9.3.11
No. For each path Q + W = ΔU, and for the cycle 1 to 2 and back ΔU = 0. Addition
gives
(QA + WA ) + (QB + WB ) = ΔU1−2 + ΔU 2−1 =0
(QA + QB ) = − (WA + WB )
Note: The equations in the problem used a different sign for W. A similar analysis
applies to the second equation in the problem.
9–34
Solutions Chapter 9
9.3.12
A closed steady-state system.
The cooling at the maximum efficiency is
0.695 kW 0.948 Btu 3600s
= 2372 Btu/hr
(1 kW)(s) 1 hr
not 5500 Btu/hr.
The main question with the advertisement is: where does the energy go that is removed
from the room if the unit does not require outside venting?
9.3.13
A closed steady-state system
 +W
 = ΔU
 =0
Q
 =−W

Q is negative (heat loss)
Q
3
-3
 = − 2050 × 10 J 1 hr 10 kW = − 0.57 kW
Q
hr
3600 s 1(J)(s)
 = 0.57 kW
W
9.3.14
A closed steady-state system
 +W
 = ΔU
 =0
Q
 =−W

Q
 = 0.25 hp 0.7068 ⎛ Btu ⎞ 60 s = 10.6 Btu/min
W
1 hp ⎜⎝ s ⎟⎠ 1 min
 = − 10.6 Btu/min
Q
9–35
Solutions Chapter 9
9.3.15
Open system, unsteady-state. The system is the volume containing 1 lb of H2O at 27ºC
(liquid). At the end of the process the system contains nothing. The water leaving is at
100ºC because the atmospheric pressure is 760 mm Hg.
Q + W = ΔH
W=0
ˆ (1) − H
ˆ (0)
Q = ΔH = H
out
in
The reference temperature for enthalpy is 0ºC but the water is at
ˆ =H
ˆ
ˆ
27ºC, H
−H
out
100 C
27 C
o
o
Q = (1 kg) (2675.6 – 111.7) kJ/kg = 2558 kJ
9.3.16
Steps 1, 2, 3, and 4:
Figure P22.27 shows the known quantities. No reaction occurs. The process is clearly a
flow process (open system). Assume that the entering velocity of the air is zero.
Step 5:
Basis: 100 kg of air = 1 hr
Steps 6 and 7:
Simplify the energy balance (only one component exists):
ˆ + KE
ˆ + PE)m]
ˆ
ΔE = Q + W − Δ[(H
(1) The process is in the steady state, hence ΔE = 0.
9–36
Solutions Chapter 9
(2) m1 = m2 = m.
ˆ
(3) Δ(PE)(m)
= 0.
(4) Q = 0 by assumption (Q would be small even if the system were not insulated).
(5) v1 = 0 (value is not known but would be small).
The result is
 )m] = ΔH + ΔKE
W = Δ[(Ĥ + KE
We have one equation and one unknown, W. ΔKE and ΔH can be calculated, hence the
problem has a unique solution.
Steps 7, 8, and 9
ΔH =
(509 − 489)kJ 100 kg
= 2000 kJ
kg
ΔKE =
1
2
m(v22 − v12 )
3
1 kJ
⎛ 1 ⎞ 100 kg (60 m )
= ⎜ ⎟
= 180 kJ
2
1000(kg)(m 2 )
s
⎝ 2 ⎠
(s) 2
W = (2000 + 180) = 2180 kJ
(Note: The positive sign indicates work is done on the air.)
To convert to power (work/time),
kW=
2180 kJ 1 kW 1 hr
= 0.61 kW
1 hr 1 kJ 3600 s
s
9–37
Solutions Chapter 9
9.3.17
(a)
Steady state process assumed.
 + PE
 )m ⎤ + Q + W
ΔE = − Δ ⎡(Ĥ + KE
⎣
⎦
 ; ΔPE
 =0
1. Ignore PE
 is small so ΔKE
 =0
2. ΔKE
3. No reaction
4. W = 0
5. (E2 − E1 ) = 0, steady state
ΔH = Q
 + PE
 )m] + Q + W
ΔE = − [(Ĥ + KE
(b)
1.
2.
3.
4.
5.
 ; ΔPE
 =0
Ignore PE
No reaction
(E2 − E1 ) = 0; steady state
Q = 0 – assumed
W=0
 =0
6. ΔKE
ΔH = 0
9.3.18
System: gas
(
)
 + PE
 m⎤+ Q + W
E t − E t = − Δ ⎡ Ĥ + KE
⎥⎦
2
1
⎣⎢
a.
no reaction
 + PE
 )m ⎤ = 0, no flow
-Δ ⎡(Ĥ + KE
⎣
⎦
ΔKE and ΔPE of gas = 0
Ut 2 − Ut1 = Q + W
9–38
Solutions Chapter 9
b.
Here W = 0
Ut 2 − Ut1 = Q
c.
Here Q = 0,
Ut 2 − Ut1 = W
d.
Ut 2 − Ut1 = Q + W
e.
Here, the system is the gas
Q = 0, W = 0
Ut 2 − Ut1 = 0
9.3.19
(a) System is the tank
(b) Closed, unsteady
state process
(c)
2 ft3
8 lb
H2O
100oF
W
The 8 lb H 2O occupy !
#
0.129 ft 3 hence the rest #
##
of space is vapor.
" (d1)
We will ignore the
#
initial vapor (see note #
#
#$
at end of solution)
Basis: 8 lb H2O
" Initial data for H O are (satd liquid)
2
$
o
(d) #
V̂ = 0.01613 ft 3 /lb
T = 100 F
$
p = 0.9487 psia
ΔĤ =67.97 Btu/lb if neglect vapor
%
Basis: 1 hr
#The energy balance is
%
(e) $
ΔE = [ΔU + ΔPE + ΔKE]inside = Q + W − ΔH − ΔPE − ΔKE
%
no change in PE or KE inside
no mass transfer
&
ΔU = W = ΔH − Δ(pV) = ΔH − VΔp
(f) W =
0.25 hp 0.7068 Btu 3600s
= + 636 Btu
1(hp) (s) 1 hr
9–39
Solutions Chapter 9
This proves to be negligible quantity
↓
(g1) 636 = ΔĤ final (8) −
#
%
%
(h) $
%
%
&
(g2)
2 ft 3 (pfinal − 0.9487) lbf 144 in 2
1 Btu
2
2
in 1 ft 748(ft)(lbf )
Since pfinal ,V̂final , and ΔĤ final are in the tables, you
have to get a ΔĤ final , and V̂final vapor given an assumed pfinal
that satisfy the equation. ΔĤ final will be composed of saturated
liquid and vapor. ΔĤ L (8 − m) + ΔĤ v (m) = ΔĤ final . m = lb vapor. (g3)
ˆ (8 − m) V
2 ft 3 = V
+ˆ V (m).
L
ˆ ,V
ˆ , ΔH
ˆ , ΔH
ˆ , and p final are all related in the steam tables.
V
L
v
L
v
_________________
* V̂ of saturated vapor = 350.8 ft3/lb so that 1.87 ft3 represents 0.0053 lb, a negligible
amount relative to 2 lb.
9.3.20
(a)
The system is the mixing point of the 1000°F “air” and water at 70°F
(b)
Open system
(c) and (d)
H2O at 70oF
H = 38.05 Btu/lb
Air at 1000oF
H = 6984 Btu/lb mol
Products (100 lb)
System
H2O @ 400oF, H = 1201.2 Btu/lb
air @ 400oF, H = 2576 Btu/lb mol
The enthalpies of water come from the steam tables. The properties of air come from the
tables in Appendix D7 (with interpolation). The reference temperature for the enthalpy is
32°F.
9–40
Solutions Chapter 9
ΔE = [ΔU + ΔPE + ΔKE]inside = Q + W − ΔH − ΔPE − ΔKE



(e)
•
•
•
•
Thus
0 because steady state
Q is assumed to be 0 because time is short in transit
W = 0 (none specified)
ΔKE = ΔPE = 0 none with mass flow
ΔH = 0
or
ΔHout = ΔHin
(f) The units in the balance will be Btu and lb
( n air ) ΔHˆ air
ˆ (g) = ( n ) ΔH
ˆ + ( n ) ΔH
ˆ
+ ( n H 2O ) ΔH
H 2O
air
air
H 2O
H2O
@ 400o F
@ 400o F
@ 1000o F
(e)
(a) 70o F
(100)(2576)
(6984)
+ (mH2O )(1238.9) = (100)
+ m H2O (38.05)
29
29
9.3.21
 + PE
 )m] + Q + W for all parts of this problem
a. (E 2 − E1 )= − Δ[Ĥ + KE
 ; ΔPE
 =0
1. Ignore PE
 is small so ΔKE
 = 0 (or it can be included)
2. ΔKE
3.
No reaction
4.
W=0
5.
(E2 – E1) = 0; steady state
ΔH = Q
b.
 ; ΔPE
 =0
1. Ignore PE
2.
No reaction
3.
W=0
4.
(E2 – E1) = 0; steady state
5.
Q=0
9–41
Solutions Chapter 9
 )m] = 0
-Δ[Ĥ + KE
c.
 ; ΔPE
 =0
1. Ignore PE
 =0
2. ΔKE
3.
No reaction
4.
(E2 – E1) = 0; steady state
5.
Q = 0 (assume)
ΔH = W
d.
1.
 ΔPE
 =0
Ignore PE;
2.
No reaction
3.
)
(E2 – E1) = 0; steady state (may be not the KE
4.
Q = 0 (assumed)
 )m] = W
Δ[Ĥ + KE
9–42
Solutions Chapter 9
9.3.22
General balance is
 + PE
 )m] + Q + W
ΔE = [ΔU + ΔKE + ΔPE]inside = − Δ[(Ĥ + KE
(1)
(2)
(3)
(4) (5)
(6)
(7) (8)
Assume all reactants are in bomb at start of reaction.
a.
Q
1
2
3
4
5
6
ΔU retain because changes
ΔKE delete - no change
ΔPE delete - no change
7
8
Q retain
W delete (fixed
boundary)
⎫ ΔH
⎪
⎬ ΔKE delete as no mass flow in or out
⎪ ΔPE
⎭
Result: ΔU = Q
b.
Gases
Fuel
W
Q
ΔH = Q - W
1 ⎫⎪
2 ⎬ΔE = 0 because steady state
3 ⎪⎭
4
ΔH
keep
5
delete as negligible
6
delete not a factor
7
Q
keep
8
W
keep (electric work)
Result: ΔH = Q + W
9–43
Solutions Chapter 9
c.
1 ⎫⎪
2 ⎬ΔU = 0
3 ⎪⎭
4
steady state
ΔH is kept
Result: ΔH = 0
5
ΔKE delete negligible
6
ΔPE delete not applicable
7
Q = 0 insulated applicable
8
W=0
no work done
9.3.23
Unsteady state, closed, isothermal process at 25oC
Basis: 1 kg CO2
p1 = 550 kPa
T1 = 25°C
p2 = 3500 kPa
T2 = 25°C
From CO 2 ⎧⎪V̂1 = 0.10 m 3 /kg
⎨
ˆ = 205 kJ/kg
chart ⎪⎩ΔH
V̂2 = 0.0125 m3 /kg
ˆ =170 kJ/kg
ΔH
2
(converted to SI units)
Energy balance
ΔE = Δ[(U + PE + KE)]inside = Q + W − Δ(H + PE + KE)
ΔU = Q + W = ΔH − Δ(pV) so Q = ΔH − Δ(pV) − W
ΔH = (170 − 205)(1) = 35kJ
−
W = 4.106 kJ
⎡ 3500 kPa 0.0125 m 3
Δ(pV) = ⎢
−
kg
⎣⎢
Q = (−35) − (−11.25) − 4.106
=
550 kPa 0.10 m ⎤
⎥
kg ⎦⎥
3
−27.86 kJ / kg CO2
(removed)
9–44
⎡ 3 N
⎤
⎢10 m 2
1J ⎥
⎢
⎥ = −11.25 kJ
⎢ 1 kPa 1(N)(m) ⎥
⎣
⎦
Solutions Chapter 9
9.3.24
Pick the room as the system. The simplified form of the energy balance is (for no mass
flow):
ΔE = ΔU = Q + W = ΔU
where Q = 0 (room is insulated)
W = the electrical energy provided freezer through the system boundary
and is positive. (No volume change occurs). Hence ΔU = CvΔT is positive and ΔT is
positive; i.e. the temperature increases.
9.3.25
This is an unsteady state closed process
Basis: 1 lb water
Data from the CD
Initial Conditions
Final Conditions
T = 327.8ºF
T = 327.8ºF
p = 100 psia
V̂ = 0.018 ft 3 /lb for liquid and 4.435 for vapor
9–45
V̂ = 4.435
Solutions Chapter 9
Ĥ = 298.475 Btu/lb
Û = 298.146 Btu/lb
V = 4.435 ft3
V = 4.435 ft3
At the final conditions the water is saturated vapor
H = 1187.48 Btu/lb
U = 1105.46 Btu/lb
T = 327.75 ºF
p = 100 psia still
ΔH = 1187.48 – 298.475 = 889.0 Btu/lb
ΔU = 1105.46 – 298.146 = 807.3 Btu/lb
The energy balance is
ΔU = Q + W
W = 0 (fixed container)
Q = ΔU = 807.3 Btu/lb
9.3.26
Closed unsteady state system
Use the CD to get data
Basis: 4 kg steam
Initial Conditions
Final Conditions
Given:
T = 500K
p = 700 kPa
Given:
Known:
Look up:
V̂ = 0.320m3/kg
Ĥ = 2903.94 kJ/kg
Û = 2680.51 kJ/kg
Look up (2 phase system) the
quality (trial and error on the CD): x = 0.437
T = 400K
V̂ = 0.320m3/kg
Ĥ = 1486.81 kJ/kg
9–46
Solutions Chapter 9
Calculate V = 0.320(4) = 1.280 m3/kg
ˆ +W
ˆ = ΔU
ˆ
Q
Û = 1408.23 kJ/kg
p = 245.77 kPa
ˆ =0
W
Q = (1408.23 – 2680.51)(4) = (-1272.3)(4) = −5,089 kJ
ˆ = ΔH
ˆ − (ΔpV)
ˆ .
If you do not use the CD, ΔU
9.3.27
A closed, unsteady state system comprised of 2 tanks. Get the steam properties from the
CD.
Basis: 1 lb steam.
State 1
State 2
m = 1 lb
p = 600 psia
V̂ = 0.795 ft3/lb
T = 500oF (960oR)
Û = 1128.19 Btu/lb
Ĥ = 1215.97 Btu/lb
V1 = 0.795 ft3
m = 1 lb*
p = 330.45 psia
V̂ = 1.590 ft3/lb
T = 500oF (960oR)*
Û = 1157.03 Btu/lb
Ĥ = 1253.98 Btu/lb
V2 = 2V1 = 1590 ft3*
* used to get state 2 properties
The closed system is Q + W = ΔU
ΔU = 1157.03 – 1128.19 = 28.84 Btu/lb
W = 0 (fixed boundary)
Q = ΔU = 28.84 Btu/lb
ΔH = 1253.98 – 1215.97 = 38.01 Btu/lb
9–47
Solutions Chapter 9
9.3.28
Basis: 1 lb H2O at 212ºF, 1 atm
Data from the CD
Û
Ĥ
V̂
Initial (liquid)
180.182 Btu/lb
180.226 Btu/lb
0.017
ft3/lb
Final (vapor)
1077.40 Btu/lb
1150.29 Btu/lb
26.773
ft3/lb
p = 1.00 atm
Non-flow process (unsteady-state)
Q + W = ΔU
ΔU = (1 lb)(1077.40 – 180.182) Btu/lb = 897.22 Btu
W = − ∫ pdV = (1 atm)(26.773
– 0.017) ft3/lb (2.72) Btu/(atm)(ft3)
−
= -72.78 Btu
Q = 897.22 – (-72.78) = 970.0 Btu
ΔH = 1150.29 − 180.226
=
970.0 Btu
Flow process (unsteady-state)
ΔU = Q + W –ΔH
Assume the volume of the system is fixed at V = 0.017 ft3 and that W = 0. Also assume
that nothing is left in the system after the water evaporates.
ˆ (0) − U
ˆ
ΔU = U
final
initial (1) = − 180.182 Btu
ˆ
ˆ (0) = 1150.29 Btu
ΔH = H
(1) − H
in
out 1212o F
Q = ΔU + ΔH = − 180.182 + 1150.29 = 970.1 Btu
9–48
Solutions Chapter 9
9.3.29
Treat the system (the cylinder) as an unsteady state flow system.
ΔU = Q + W - ΔH
W=0
Q=0
U t − U t = −(Hout − Hin )
2
1
n f = final moles in cylinder
ˆ =U
ˆ
U
t2
T 1 , 40 atm
ˆ =U
ˆ
U
t1
n i = initial moles in cylinder
298K,1 atm
Ĥ out = 0
ˆ =H
ˆ
H
in
298 K, 50 atm
Let the reference temperature be TR that makes Ĥ R = 0 . Ignore the effect of pressure on
ˆ and H.
ˆ
the values of U
ˆ =H
ˆ + C (T − T )
H
in
R
p
in
R
ˆ =0
H
R
in
ΔU
out
ˆ −n U
ˆ
ˆ
Ut 2 − Ut1 = n f U
t2
i t1 = (n f − n i )Hin − 0
n f [Cv (T − TR ) − RTR ] − n i [Cv (Tin − TR ) − RTR ] = (n f − n i ) ⎣⎡Cp (Tin − TR ) ⎦⎤
= (n f − n i ) [(Cv + R)(Tin − TR )]
Multiply all of the terms out to get (note terms with TR cancel):
CV = C P – R
R 2
= for diatomic gas
CV 5
n f Cv T = n f Cv Tin + n f RTin − n i RTin
Insert pV = nRT to get ni and nf
The equations reduce to
n f Cv T = n f Cv Tin + n f RTin − n i RTin
T ⎡
R p T R ⎤
= ⎢1 + − i
⎥
Tin ⎣ Cv pf Ti Cv ⎦
9–49
Solutions Chapter 9
⎡ 2
T
= ⎢1 +
298 ⎣ 5
⎛ 1 ⎞⎛ T ⎞⎛ 2 ⎞ ⎤
− ⎜ ⎟⎜
⎟⎜ ⎟ ⎥
⎝ 40 ⎠⎝ 298 ⎠⎝ 5 ⎠ ⎦
T = 417 K
9.3.30
The tank is an open unsteady state system. f stands for final, i for initial, in for
in, m for mass in lb.
Data:
at 291°F and 50 psia
1179.39
1099.60
8.653
ΔU = Q + W – ΔH
at 14.7 psia saturated (212°F)
Ĥ Btu/lb
Û Btu/lb
V̂ ft3/lb
W=0
1150.26
1077.37
26.818
Q=0
ΔU = -ΔH
ˆ −m U
ˆ
ˆ
nf U
f
i i = (m f − m i )H in − 0
V = 50 ft 3 so mf =
mi =
50
= 5.78 lb
8.653
50
= 1.86 lb
26.818
ˆ ) − 1.86(1077.37) = (5.78 − 1.86) 1179.39
5.78(U
f
Û f = 1146.56 Btu/lb at 50 psia
Tf = 411o F
(superheated)
9–50
Hout = 0
Solutions Chapter 9
9.3.31
The general energy balance is
ˆ + KE
ˆ + PE)
ˆ m]
ΔE = Q + W − Δ[(H
(1)
(a)
(2) (3)
(4)
(5) (6)
(1)
ΔE = 0
steady state flow process, one overall system (You can also pick
two system, one the H2O and one the benzene).
(2)
Q is the net overall heat transfer to and from the heat exchanger with the
surroundings.
It is essentially 0 if exchanger is well insulated. If you pick 2 systems,
Q is the heat transfer between the benzene and the H2O, and is not 0.
(b)
(3)
W=0
no mechanical work in the process
(4)
ΔH ≠ 0
for all streams
(5)
ΔKE = 0 negligible KE change
(6)
(1)
ΔPE = 0
ΔE = 0
level exchanger assumed.
steady state process
(2)
Q
not zero as isothermal
9–51
Q = ΔH
Solutions Chapter 9
(3)
(4)
W
ΔH ≠ 0
retain
the pressure changes even if T is constant
(5)
ΔKE = 0 pipe diameter remains the same, and the flow rate is the same
(6)
ΔPE = 0
level pump lines
Q + W = ΔH
9.3.32
Unknowns
Relationships (R)
F
P
(1) F = P + V
(2) 0.05F = xPP
ˆ + SΔH = VH + PH
ˆ
(3) FH
F
S
V
P

(4) SΔH = Q

V
S

Q
S
 = UA(T − T )
(5) Q
S
V
TV
XP
Yes, two measurements must be made to balance the unknowns and the relationships. If
x P and TV are measured: solve (R5) for Q . (R4) for S substitute (R2) and (R1) in (R3)
 . Note that measuring
and solve for F. Use (R2) to calculate P and (R1) to calculate V
TV and S is not satisfactory because this combination eliminates either (R4) or (R5) as an
independent relationship. Also note that if you neglect the vapor pressure of the organic,
you can set Tv = 212ºF.
9–52
Solutions Chapter 9
9.3.33
Basis: 1 hour
Input conditions:
500o F ⎫
⎬ h1 = 1264.7 Btu/lb
250 psi ⎭
Exit conditions:
⎫ h SV = 1150.4 Btu/lb
⎬
15% liquid ⎭ h SL =180.07 Btu/lb
14.7 psi
Ĥ2 = 0.15(180.07) + 0.85(1150.4)
= 1004 Btu/lb
The simplified energy balance is:
ˆ =Q+W
mΔH
ΔĤ = 1004 − 1265 = − 261 Btu/lb
ˆ = −261,000 Btu
mΔH
W=−
86.5 hp 1 hr
1 Btu
= −2.20 × 105 Btu
2.93×10-4 (hp)(hr)
Thus, Q = mΔHˆ − W = − 2.61×105 + 2.2×105 = − 4.1×104 Btu
hence, not adiabatic.
9–53
Solutions Chapter 9
9.3.34
Basis: 1 lb air
The energy balance is in the steady state. The system is the value.
 + PE
 )m ⎤ = Q + W
Δ ⎡(Ĥ + KE
⎣
⎦
if ΔKE ≅ 0
W=0
ΔPE = 0
Q ; 0
Hence ΔH = 0
ΔH = ∫ CpdT = 0
(a)
ΔT = 0 or
T2 = 250 K
Since ΔH = 0 (the kinetic energy term was negligible), the steady-state constraint requires
that the mass flow rate be constant:
 = ΔvS = Constant
m
where: ρ = density
v = velocity of the gas
S = cross-sectional area of pipe
Since the diameter of the pipe does not change and density is a function of temperature
and pressure, the gas velocity is also a function temperature and pressure.
vSTP =
100 lb Air 1 lb mol 359 ft 3 1 hr
1
= 7.0 ft/s
1 hr
29 lb Air 1 lb mol 3600 s π(1.5/12 ft)2
Therefore, the velocity downstream of the valve is:
v Outlet =
7.0 ft 250 K 1 atm
= 3.2 ft/s
s 273 K 2 atm
9–54
Solutions Chapter 9
9.3.35
Steady-state process. Basis: 1 second
 +W

 (Ĥ out − Ĥ in ) = Q
m
To get Ĥ in use the steam tables or a Cp equation. Using constant Cp is not the most
accurate method, but for liquids for small temperature changes, it is a good
approximation.
 +W


)(Tout − Tin ) = Q
(m)(C
p
Because the tank is well-mixed, the outlet water temperature is the same as the
temperature inside the tank T.
T = Tout
 (C p ) (T − Tin ) = 100 (20 − T) + 300
m
 =
m
1 kg 1 min
1 kg
=
min 60 s
60 s
Cp = 4180 J/(kg)(K)
4180
(T − 40) = 100 (20 − T) + 300
60
T = 30.0o C
9–55
Solutions Chapter 9
9.3.36
Assumptions
1.
This is a steady-flow process since there is no change with time at any point,
hence ΔE = 0.
2.
Air is an ideal gas since it is at a high temperature and low pressure relative to its
critical point.
3.
The kinetic and potential energy changes are negligible ΔKE ≅ ΔPE ≅ 0.
4.
Constant specific heats at room temperature can be used for air.
The simplified energy balance is
 +W
 = ΔH
 = m(Ĥ − Ĥ )
Q
out
in
 = − 200W
Q
 = 15 kW
W
Assume that with little error
ΔH = CpΔT
with Cp = 1.00 kJ/(kg)(ºC)
Otherwise you will need tables or a data base for the properties of air.
The ideal gas law gives the specific volume of air at the inlet of the duct
V̂1 =
RTm
[(0.287 kPa)(m 3 )/(kg)(K)](290 K)
=
= 0.832 m 3 /kg
pin
100 kPa
9–56
Solutions Chapter 9
The mass flow rate of the air through the duct is determined from
 =
m

V
150 m 3 /min ⎛ 1 min ⎞
1
=
= 3.0 kg/s
0.832 m 3 /kg ⎜⎝ 60 s ⎟⎠
V̂1
Substituting the known quantities, the exit temperature of the air is determined to be
(15 kJ/s) − (0.2 kJ/s) = (3 kg/s)[1.00 kJ/(kg)(oC)](Tout − 17)oC
Tout = 21.9o C
9.3.37
The procedure is correct only for an open, steady state system.
(a)
If the evaporation occurs from an open, unsteady-state system of fixed volume:
ΔU = Q + W –ΔH
W=0
ΔU = Q – ΔH
Data:
Ûliquid = 419.5 kJ/kg
Û vapor = 2506.0 kJ/kg
Ĥliquid = 419.5 kJ/kg
Ĥ vapor = 2675.6 kJ/kg
Assume only 1 kg of water is in the system at t = initial. At t = final, 0 kg of water are in
the system.
U final = 0
Uinitial = (1 kg) (419.5 kJ/kg) = 419.5 kJ
Hout = (1 kg)(2675.6 kJ/kg) = 2675.6 kJ
H in = 0
-419.5 = Q – (2675.6-0)
Q = 2256.1 kJ
(b)
If the evaporation occurs in a steady-state open system, m in is in kg and min − 1 = mout .
The process is isothermal.
9–57
Solutions Chapter 9
ΔU = 0
W=0
OUT
IN
ˆ
ˆ
ˆ
⎤
Q = ΔH = ⎡⎣(m1 − 1)H
liquid + (1)H vapor ⎦ − m1Hin
ˆ =H
ˆ
H
in
liquid
Q = (1)(2675.6 − 419.5) 2256.1
=
kJ
The answer in the solution is ok, but note that the equation ΔU = Q − pΔV refers to a closed
system, not an open one, in which the vessel expands against the atmosphere. This viewpoint is
ok, but an unlikely procedure.
9.3.38
Open, unsteady state system – the adiabatic turbine
 =0
Q
 =0
ΔU
 =0
ΔPE
 = ΔH
 + ΔKE

W
Data from interpolating in the SI steam tables
At 600ºC and 100 kPa
3697.9
0.4011
At 400ºC and 100 kPa
3278.2
3.103
Ĥ(kJ/kg)
3
V̂(m /kg)
The velocities at the inlet and outlet to the turbine are calculated from the volumetric
flow rate = m V̂=
d2
v where d is the pipe diameter. Therefore,
4
⎛
kg ⎞ ⎛
m3 ⎞
(4) ⎜ 2.5 ⎟ ⎜ 0.4011
s ⎠⎝
kg ⎟⎠
⎝
 1V̂1
4m
m
v1 =
=
= 127.7 ,
2
2
s
(3.14159)(0.1 m)
d in
9–58
Solutions Chapter 9
and
v2 =
 2 V̂2
4m
m
= 158.0
2
s
d out
⎡
⎛ 1⎞
⎛ 1 ⎞⎤
ΔKE = ⎜ ⎟ (2.5) ⎢158.02 − 127.7 2 ) ⎜
⎥ kJ/s = 10.82 kJ/s
⎝ 2⎠
⎝ 100 ⎟⎠ ⎦
⎣
 = (2.5)(3278.2 − 3697.9) kJ/s = − 1049.25 kJ/s
ΔH
 = − 1049.25 + 10.82 = 1038.4 kJ/s
W
(1038.4 kW)
9.3.39
This is an unsteady-state closed process. The energy balance reduces to
ΔE = ΔU + ΔPE + ΔKE = Q + W
Q = ΔU = U2 − U
=1
ΔPE = ΔKE = 0
ΔH − Δ(pV)
=
ΔH − (p2V2 − p1V2 )
where 2 (327.8ºF) is the final state and 1 (406ºF) is the initial state
Get the data from the CD in the back of the book. The final state is saturated
ˆ and H
ˆ are in Btu/lb and V
ˆ is ft3/lb.
conditions. U
Ĥ1,L = 381.65
Ĥ1,V = 1202.2
Ĥ 2,L = 298.5
Ĥ 2,V = 1187.5
Û1,L = 380.72
Û1,V = 1116.7
Û 2,L = 298.2
Û 2,V = 1105.5
V̂1,L = 0.019
V̂1,V = 1.743
V̂2,L = 0.0177
V̂2,V = 4.433
From the steam tables, get the initial amount of steam and use as the basis:
⎛ 1 ⎞
100 ⎜
⎟ = 57.4 lb ignoring the liquid volume. (If you do not, you have to get the
⎝ 1.743 ⎠
average volume) and get the final state. Let y = the fraction of liquid at 100 psia
3
3
ˆ
ˆ
V
liquid = 0.0177 ft / lb, Vvapor = 4.433 ft / lb
Basis: 1 lb water
y(0.0177) + (1 − y)(4.433) = 1.74 y = 0.6095 lb (liquid) x = 0.3905 lb(vapor)
9–59
Solutions Chapter 9
Use values of Û from the CD to calculate Q. Otherwise you have to calculate Δ(pV).
Q = [(0.6095)(298.2) + (0.3905)(1187.5)]− [(1116.7)(1)]
Q = −471
=
Btu
or Q −2.71×104 Btu
lb
9.3.40
This is an unsteady state process in a closed system, or a closed system with air and
surroundings with water. From the latter viewpoint the system is the air, and ΔU = Q +
W where Q is the heat transfer to the water from the air.
Q
5 lb H2 O
air
W (positive)
For the system of the air
Q + W = ΔE = ΔU + ΔP + ΔK = ΔHΔ
− (pV)
For the water:
Q = ΔH – Δ( pV ) but Δ ( pV ) is negligible for liquid water so that
T +2.3
Q = ΔH = 5
∫ C pdT
= 5 Cp (T + 2.3 – T ) for constant C p .
T
(or use the steam tables)
C p = 8.0
Btu
(lb mol )(° F)
Btu 2.3°F 1 lb mol H2 O 5 lb H2 O
= 5.11 Btu
(lb mol )
18 lb H2 O
Q is removed from the air, so for the system comprised of the air, Q = –5.11 Btu.
Q = 8.0
9–60
Solutions Chapter 9
For the system of the air
W=
12, 500(ft )(lb f ) 0.0012854 Btu
= 16.07 Btu
1(ft )(lb f )
W is positive when the surroundings do work on the system of the air.
3
Basis: 3 ft air (not essential)
∆U = –5.11 Btu + 16.07 Btu = 10.96 Btu
9.3.41
Steps 1, 2, 3 and 4
Get the needed properties from the CO2 chart or tables (Be sure to use the same reference
value). Let m CO2 be the mass of CO2 in the cylinder at the final state, V̂ is the specific
volume, Ûfinal is the specific internal energy of the CO2 at the end of the filling, and Ĥm
is the specific enthalpy of the CO2 in the pipeline.
At 200 psia, 40°F, Ĥ ≅ 153 Btu / lb and V ≅ 0.57 ft 3 / lb
Step 5
Basis: 3 ft3 of CO2 @ 200 psia (final conditions)
System: The cylinder (open system, unsteady state)
Steps 6 and 7:
Unknowns: V̂CO2 , mCO2 , TCO2
Equations: energy balance, enthalpy chart
ˆ chart to fix the conditions in the cylinder. The energy
We need two points on the p − H
balance reduces to (W = 0, Q assumed to be 0)
ΔU = −ΔH
which gives one point:
9–61
Solutions Chapter 9
ˆ ) − m (U
ˆ
ˆ
ˆ )
mCO2 (U
−(H
final
in itial
initial ) = min (Hin ) mout
out
minitial = 0
mout = 0
m CO2 = m in
ˆ
ˆ
U
final = Hin = 153 Btu / lb
ˆ and
The other point is the pressure p = 200 psia. Unfortunately the CO2 chart is p − H
ˆ , thus requiring a trial and error solution. Assume a T, lookup V̂ , and calculate
not p − U
ˆ =H
ˆ − pV
ˆ (ignore pV
ˆ at the reference sate of –40oF).
U
When Û = 153Btu / lb , then you have determined T. If you used tables of p vs Û in a
handbook, the calculations would be easier.
Ĥ
p
V̂
Û
Assume T = 160oF.
T = 140 F.
o
ˆ − pV
ˆ =
Then H
ˆ − pV
ˆ =
Then H
The calculates are quite approximate but
T = 140 F
o
mCO2
3 ft 3
=
= 4.3 lb
0.70 ft 3 / lb
9–62
182-(200)(0.68)(0.185) = 157
178-(200)(0.70)(0.185) = 153
Solutions Chapter 9
9.3.42
Basis: 1 lb steam
Closed system; hence Q + W = ∆E = ΔU
A
expand
130 psia, 600°F
75 psia,
600°F
const. T
B
cool, const. V
adiabatic
compression
60 psia
Q=0
C
Basis: 1 lb steam
Note: The CD will result in slightly different values for ΔĤ .
A
B
: ΔĤ = 1329.6 – 1326.1 = 3.5 Btu / lb
Use the steam tables or the CD directly for U, or calculate
ˆ = ΔHΔ
ˆ − (pV)=
ˆ
ΔU
= 3.5 –
75 lb f 8.319 ft 3 130 lb f 4.760 ft 3
–
2
lb
2
in
in
lb
= 3.5 – .95 = 2.5 Btu / lb
B
C :
Q̂ = ?
144 in
ft
2
2
1 Btu
778 (ft) (lb f)
Ŵ = ?
Ŵ= – ∫ pdV but pV relation is not known.
W = 0 (at constant volume)
Δ Ĥ = 1232.7 – 1329.6 = –96.9 Btu / lb
2
1 Btu
60 8.319 – 75 8.319 144 in
= –73.8 Btu/lb
Δ Û = – 96.9 –
2 778 (ft) (lb )
f
ft
ˆ = –73.8 Btu / lb
Q̂ = Δ U
9–63
Solutions Chapter 9
C
A :
( )
ˆ = 0 so W
ˆ = ΔU=ΔH–Δ
ˆ
ˆ
ˆ
Q
pV
ΔĤ = 1326.1 – 1232.7 = 93.4 Btu/lb
Δ pV = 130 4.760 – 60 8.319 144
Ŵ = 93.4 – 22.15 = 71.3 Btu/lb
778
= 22.15 Btu/lb
ΔU = 93.4 – 22.15 = 71.3 Btu/lb
9.3.43
To make the process a steady state open process from (1) to (3) assume the system is the
coil plus the compressor. Then
0
0 0
ˆ )m + Q + W]
ΔE = –Δ[(Hˆ + Pˆ + K
or
W = ∆H – Q
Basis: 1 lb water evaporated
ˆ @ (1): (0.98) (1143.3) + (0.02) (161.17) = 1123.6 Btu/lb
Calculate Δ H
From (1) to (2): Balance on the compressor
ˆ – ΔH
ˆ = 1235.2 – 1126.6 = 111.6 Btu/lb
ΔH
2
1
9–64
Qˆ = –6 Btu / lb
Solutions Chapter 9
W = 111.6 –(–6) = 117.6 Btu/lb
From (2) to (3): Balance on the coil
ˆ – ΔH
ˆ = 168 – 1235.2 = – 1067.2 Btu/lb
ΔH
3
2
a.
Q
= 1067.2 = 9.1
W 117.6
ˆ = 0 so Q
ˆ = ΔH
ˆ = -1067.2 Btu/lb
W
(Note: the problem asks for the heat transfer
out as a + value)
⎡ 38.46 ft 3 0.98 (0.016)ft 3 0.02 ⎤ lb steam
V1 = ⎢
+
= 37.69 ft 3/lb
⎥
lb
lb
1067.2
Btu
⎣
⎦
Basis: 1,000,000 Btu/hr
b.
106 Btu 1hr 1 lb steam 37.69 ft 3
= 589 ft 3 / min
hr
60 min 1067.2 Btu lb steam
9–65
Solutions Chapter 9
9.3.44
Basis: 25 kg exit fluid
Q + W = ΔH
W=0
Q = ΔH = ΔH25 –ΔH15 –ΔH10 =ΔH25 – ( 2784.4 )(15) – (82.2 )(10 )
–50 J 25 × 103 g
= –1250 kJ
g exit
⎧ ΔHˆ L = 798.5 kJ / kg
For the exit fluid ⎨
ˆ
⎩ ΔH = 2784.4 kJ / kg
ˆ ~ 10 J )
( note ΔKE
Q=
–4
V
Let x = kg vapor, 25 – x = kg liquid
–1250 =
x kg 2784.4 kJ (25-x) kg 798.5 kJ 15 kg 2784.4 kJ 10 kg 82.2 kJ
–
–
+
kg
kg
kg
kg
– 1250 = 2784.4 x + 19,963 – 798.5 x – 41,766 – 822
x = 10.76
fraction vapor =
10.76
= 0.43
25.0
9–66
Solutions Chapter 9
9.3.45
Closed unsteady state process for each stage.
Q + W = ΔU overall
Calculate the moles of gas:
a.
n=
pV (1 atm)(0.387 ft 3 )(lb mol)(o R)
=
= 1×10-3 lb mol
RT
(530o R)(0.7302)(ft 3 )(atm)
Known values
p (atm)
T (ºF)
n (lb mol)
V
A
1
170 (630ºR)
1 × 10-3
?
B
C
1
70 (530ºR)
1 × 10-3
0.387
10
70 (530ºR)
1 × 10-3
?
(Basis)
D
10
823 (1283ºR)
1 × 10-3
?
For an ideal gas, C v = 5/2 R (has to be looked up or calculated) and U and H are functions
of T only
Also, Δ(pV) = Δ (nRT)
Calculate some of the missing values
b.
VA =
(1×10−3 )(0.7302)(630)
= 0.460 ft 3
1
VC =
(1×10−3 )(0.7302)(530)
= 0.0387 ft 3
10
VD =
(1×10−3 )(0.7302)(1283)
= 0.0937ft 3
10
The work for each stage will be assumed to be ∫ pdV (ideal process)
WAB (constant pressure process)
c.
W = − ∫ pdV = − p AB (VB − VA )
=−
1 atm (0.387 − 0.460)ft 3 2.72 Btu
= 0.199 Btu
(atm)(ft 3 )
9–67
Solutions Chapter 9
WBC (isothermal process)
d.
0.0387
W = − ∫=pdV − ∫=
0.387
0.0387 dV
nRT
dV −nRT
=− ∫
0.387
V
V
= −(1×10
= −3 )(1.987(530)(ln 0.1)
nRT [ln V ]0.387
0.0387
2.42 Btu
WCD (constant pressure process)
e.
W = − ∫ pdV = − pCD (VD − V
= c ) −(10)(0.0937 − 0.0387) = − 0.550 Btu
WDA (ideal adiabatic process)
W = − ∫ pdV = − ∫
nRT
dV However, both T and V vary.
V
Instead use
Q + W = ΔU
Q=O
R = 1.987 Btu/(lb mol) (ºR)
f.
⎛ 5R ⎞
W = ΔU = nC v ΔT = (10−3 ) ⎜
⎟ ( 630 − 1283) = − 3.243 Btu
⎝ 2 ⎠
g.
Woverall = 0.199 + 0.891 – 0.550 – 3.243 = – 1.74 Btu
h.
i.
ΔH = 0 (a cycle)
For the overall process Q + W = ΔU. Because of the cycle, ΔU = 0, and
thus Q = -W = 1.74 Btu
9–68
Solutions Chapter 9
9.3.46
exit stock
T2 = ?
Cooling water out
60oC
W1
C
27.5 kg vapor
100 kg C
9.1% H2O
90.9% Toluene
150oC
Condenser
F
F
liquid
90oC
P
Cooler
C
50,000 kg/day
20oC
Cp = 2.1 kJ/(kg)(oC)
W Cooling water in
20oC
Basis: 1 day
27.5 kg F 50,000 kg C 1 day
= 13,750 kg F
100.0 kg C
day
13,750 kg F 9.1 kg H 2O
= 1,251.25 kg H 2O
100 kg F
13,750 kg F – 1251.25 kg H2O = 12,498.75 kg toluene.
Energy balance at condenser:
The energy balance reduces to ΔH = 0
ΔHC + ΔHF = 0
T2
100
m C ∫ C p dT + m H 2 O ⎡ ∫ C p dT – ΔHvap
20
⎣ 150
111
+ m tol ⎡ ∫ C p dT – ΔH vap
⎣ 150
50, 000 kg
2. 1 kJ
at 111°C
+∫
90
111
liquid 40oC
at 100°C
+∫
C p dT⎤ = 0
⎦
T 2 – 20°C
(kg) (°C)
9–69
90
100
Cp dT ⎤
⎦
Solutions Chapter 9
+ 1, 251.25 kg H 2O
+ 12, 498.75 kg tol
2.1 kJ (100 – 150)°C
(kg) (°C)
1.3 kJ (111 – 150) °C
(kg)(°C)
–
–
2260 kJ
4.2 kJ (90 – 100) °C
+
kg
(kg) (°C)
230 kJ
1.7 kJ (90 – 111)°C
=0
+
kg
(kg) (°C)
105,000 T2 – 2,100,000 – 3,011,758.75 – 3,954,604.5 = 0
(a)
T2 =
9,066,363.25
= 86.3o C
105,000
Enthalpy balance at cooler:
ΔH = 0
out
in
(ΔHΔ
W − H W ) − (ΔHΔ
p − HF ) = 0
mW
+
60
40
40
m W ∫ C p dT – ⎡ m H 2 O ∫
C pdT – m td ∫
C dT⎤ = 0
⎣
20
86.34
86.34 p ⎦
4.2 kJ (60 – 20 )°C 1,251.25 kg H2 O 4.2 kJ (40 – 90)°C
=
(kg)(°C)
(kg)(°C)
12498.75 kg tol 1.7 kJ ( 40 – 90 )°C
( kg)(°C)
m W 168 kJ – 262,762 kJ – 1,062,394 kJ
kg
mW =
1,325,156 kJ
= 7,887 kg H 2O
168 kJ/kg
9–70
Solutions Chapter 9
9.3.47
Basis: 1 lb water/steam
Steady-state, open system for each unit
The data:
ˆ H
ˆ and quality or superheats at the numbered points:
Tabulation of p, T, V,
p (psia)
T (oF)
V̂(ft 3 / lb)
qual./sup.heat
1
14.7
65
0.026
0%
2
250
65
0.016
0%
3
250
401
1.844
100%
4
250
550
2.202
149º
5
40
268
10.14
96.5%
6
0.306
65
0.016
0%
Ĥ(Btu / lb)
33.09
33.79
1201.1
1291
1137
33.05
Calculate the specific volume at point 5:
V̂ = 0.965(10.506) + 0.035(0.01715) = 10.14
(a)
Heat to boiler:
Q = ΔH = H3 − H2 = 1201.1 − 33.8
=
1167.3 Btu / lb
(b)
Heat to superheater:
Q = ΔH = H4 − H3 = 1291 − 1201.1
=
90 Btu / lb
(c)
Heat removed in condenser = H6 − H5 = 33 −1137 = -1104 Btu/lb
(d)
Work delivered by turbine: Q + W = ΔH
W = ΔH = H5 − H4 = 1137 − 1291
=−
Q=0
(e)
154Btu / lb
Work required by the pump between 1 and 2:
Q + W = ΔH
Q=0
hence W = ΔH
Because we do not have values of ΔH for compressed water at 65ºF, we
will use
W = ΔU + Δ(pV)
ΔU = ∫
65o F
65o F
Cv dT = 0
9–71
Solutions Chapter 9
Δ(pV) = p 2 V2 − p1V1 = (p 2 − p1 )V =
Efficiency =
(250 − 14.7) 0.016 144
= 0.70 Btu / lb
778
Net work delivered
154 − 0.70
=
Total heat Supplied 1167.3 + 90
= 0.121
For a water rate of 2000 lb/hr:
hp =
154 Btu 2000 lb steam 1 hr 1.415 hp
= 121
lb
hr
3600s 1 Btu/s
You can improve the efficiency by:
(a) Exhausting the turbine at a lower pressure.
(b) Use higher boiler pressure.
(c) Use higher superheat.
9.3.48
Basis: 100 lb/day liquid C12 feed at 8ºF
Assume
(1) Adiabatic operation of all units
(2) Flow process, steady-state
R
2.5 T/day, 0oF
R' )
P
D )
100 T/day, -30oF
F
100 T/day, 8oF
First find the enthalpy lost or gained by the C12 in the heat exchanger from which we can
get the work done by compressor A on the Freon.
Overall C12 balance through the heat exchanger
F+R=P
100 + 2.5 = 102.5
9–72
Solutions Chapter 9
Energy balance for C12 through the heat exchanger
For a flow process:
Q + W = ΔH
W=0
Q = ΔH
Choose as the reference state liquid C12 at -30ºF
ΔH = H Pout − H Fin − H R in
=
8.1 Btu lb mol 102.5 lb
⎡⎣ −30 − (−30)o F⎤⎦
o
(lb mol)( F) 71 lb 100 lb F
−
8.1 Btu lb mol 100 lb
⎡⎣8 − (−30)o F⎤⎦
o
(lb mol)( F) 71 lb 100 lb F
−
8.1 Btu lb mol 2.5 lb
Btu
⎡0 − (−30)o F⎤⎦ = 0 − 433.5 − 8.55 = −442.1
o
⎣
(lb mol)( F) 71 lb 100 lb F
100 lb F
Consequently
Q = −442.1 Btu /100 lb F
This means heat is lost by the C12 and gained by the Freon
Energy balance for the Freon flowing through compressor A:
Q + W = ΔH
But since the flow of Freon is cyclical
ΔH = 0
Q Freon = −Q
=C12
−(−442.1)
=
442.1 Btu /100 lb F
WA = QFreon = 442.1
Btu
100 lb F
Energy balance for the C12 flowing through the compressor B:
2.5 lb/day @ -30ºF → 2.5 lb/day at 0ºF
Q + W = ΔH Q = 0
9–73
Solutions Chapter 9
ΔH = Enthalpy change across compressor = H R − H R '
Other units
W = 539 Btu/mole
= 1,520,000 Btu/day
ΔH =
+
H R = CpΔT + HΔvap
2.5 lb 8.1
[0 − (−30)]
100 lb F 71
4878 cal 1.8 Btu / lb 1 lb mol 2.5 lb
g mol cal / g mol 71 lb 100 lb F
= 8.55 + 309 = 317.5
Btu
100 lb F
Total work input needed to make the process operational:
hp =
317.5 + 442.1 Btu 2000 lb 100 ton F 778(ft)(lbf )
100 lb F
ton
day
Btu
1 day
(hp)(min)
(24) × (60) min 33,000(ft)(lb f ) 0.30
= 82.9 horsepower (actual power input)
9–74
Solutions Chapter 9
9.3.49
Basis: 1 day; Reference temp = 80ºF
a. Material and Energy Balances:
Overall Material:
In = 1,000,000 lb
Out : residue: 500,000 lb
naptha:
200,000 lb
gasoline:
300,000 lb
Total out
1,000,000 lb
Tower 1:
Material
feed
reflux
total
In
1,000,000
1,500,000
2,500,000
Out
2,000,000
500,000
2,500,000
vapor
residue
total
Energy:
In: feed: ( 1×106 )(0.53)(480 − 90) 106 (100) +
+(106 )(0.45)(500 − 480)
=
6
reflux: (1.5 ×10 )(0.59)(180 − 90)
=
steam:
(by difference)
total:
315,000,000 Btu
79,700,000 Btu
114,500,000 Btu
509,700,000 Btu
Out: residue: (5 ×105 )(0.51)(480 − 90)
vapor:
(2 ×106 )(111)
liquid:
(2 ×106 )(0.59)(250 − 90)
total (Btu):
99,200,000
222,000,000
188,500,000
509,700,000
=
=
=
=
Furnace:
Material
In
1,000,000 lb
Energy
In
6
feed: (10 )(0.53)(200 − 90) = 58,300,000
furnace ht:
257,200,000
total: (Btu)
315,500,000
Condenser I:
Material
In: vapor
H2O: 304,500,000/(1)(120-70)
9–75
Out
1,000,000 lb
Out
315,500,000
=
=
2.0 ×106 lb
6.1×106 lb
Solutions Chapter 9
Out: liquid:
2.0 ×106 lb
Energy
H2 O
In: vapor: 222 ×106 + 188.5
× 106
Out: liquid: (2 ×106 )(0.59)(180 − 90)
Removed by water
Total out:
Preheater I:
Material
In:
liquid = 500 ×106 lb ; Out: vapor
Energy
In: liquid: (5 ×105 )(0.59)(180 − 90)
steam:
Total
Out: vapor:
(5 ×105 )(0.59)(250 − 90)
+(5 ×105 )(111)
+ (5 ×105 )(0.51)(200 − 250)
Tower II
Material
In:
feed = 500,000 lb Out:
reflux = 600,000 lb
Total = 1,100,000 lb
=
=
=
=
6.1×106 lb
410.5 ×106 Btu
106 ×106 Btu
304.5 ×106 Btu
=
410.5 ×106 Btu
=
500 ×106 lb
=
=
26.5 ×106 Btu
78.65 ×106 Btu
=
105.15 ×106 Btu
=
105.15 ×106 Btu
vapor =
residue =
total =
Energy
In:
feed:
reflux: (6 ×105 )(0.63)(120 − 90)
steam: (by difference)
Total: (Btu):
=
=
=
=
900,000
200,000
1,100,000
105,150,000
11,300,000
42,700,000
159,150,000
Out: vapor: (9 ×105 )(0.63)(150 − 90) + 9 105 (118) =×140.05 ×106
residue: (2 ×105 )(0.58)(255 − 90)
= 19.1× 106
Total (Btu):
Condenser II
Material
In:
vapor:
H2O:
Out:
liquid
= 159.15 ×106
= 0.9 ×106 lb
123.0.5 ×106
(1)(110 − 70)
= 3.08 ×106 lb
= 0.9 ×106 lb
9–76
Solutions Chapter 9
H2 O
= 3.08 ×106 lb
Energy
In:
(Btu)
Out: liquid: (9 ×105 )(0.63)(120 − 90)
removed by H2O:
Total (Btu):
=
=
=
=
140.05 ×106
17 ×106
123.05 ×106
140.05 ×106
Heat Exchanger II (Assume Tout = 90ºF)
Material
In
liquid:
300,000 lb
H2O: 5,670,000/(1)(80-70) = 567,000 lb
Energy
In: (300,000)(0.63)(120-90)
Out: liquid:
removed by H2O
liquid:
H2O:
= 5.67 ×106 Btu
= 0.00
= 5.67 ×106 Btu
Heat Exchanger III
Material
In = Out:
Bottoms
200,000 lb
Charge
1,000,000 lb
Energy
In:
(200,000)(0.58)(255-90)
(1 × 106)(0.53)(90-90)
=
=
19.15 ×106
0.0
Total (Btu):
Out: (2 × 105)(0.58)(26.3-90)
(1 × 106)(0.53)(140-90)
Total (Btu):
=
=
=
=
19.15 ×106
−7.35 ×106
26.5 ×106
19.15 × 106
Tout of 26.3ºF does not seem reasonable
Heat Exchanger IV
Material
In = Out
Bottoms
0.5 ×106 lb
Charge
1.0 ×106 lb
9–77
Out
300,000
567,000
Solutions Chapter 9
Energy
In:
(5 × 105)(0.51)(480-90)
(1 × 106)(0.53)(140-90)
Total (Btu):
Out: (5 × 105)(0.51)(257-90)
(1 × 106)(0.53)(200-90)
Total (Btu):
=
=
99.3 ×106
26.5 ×106
=
125.8 ×106
=
=
=
67.5 ×106
58.3 ×106
125.8 × 106
Heat load of furnace = 315.5 × 106 - 58.3 × 106 = 257.2 ×106 Btu
b.
Additional heat if charge to furnace is at 90ºF
Total heat load = 315.5 × 106 Btu
Additional heat = 315.5 × 106 – 257.2 × 106 = 58.3 ×106 Btu
9.3.50
Basis: 1 hour
Open steady-state process
Deaerator Balance
Material Balance
F1 + F2 + 50,000 = 105,000
F2 = 105,000 − 50,000 − F1
Energy Balance
F1 (1164.1) + F2 (48.0) + 50,000(218.9) = 105,000(218.9)
F1 (1164.1) + (55,000 − F1 )(48.0)= 55,000(218.9)
a.
F2 = 8, 422 lb / hr steam
b.
F2 = 46,578 lb / hr make-up feedwater
Boiler Feedwater Pump
lb ⎞ ⎛ 144 in.2 ⎞ ⎛ (hp)(s) ⎞ ⎛ hr ⎞
⎛
lb ⎞ ⎛
ft 3 ⎞ ⎛
 V̂Δp = ⎜ 105,000 ⎟ ⎜ 0.017 ⎟ ⎜ 800 − 30 f2 ⎟ ⎜
m
lb ⎠ ⎝
hr ⎠ ⎝
⎝
in ⎠ ⎝ ft 2 ⎟⎠ ⎜⎝ 550(ft)(lbf ) ⎟⎠ ⎜⎝ 3600s ⎟⎠
= 100 hp
9–78
Solutions Chapter 9
c.
For 55% efficiency: 181.7 hp
d.
⎛
kW ⎞
(181.7 hp) ⎜ 0.7457
⎟ = 135.5 kW
hp
⎝
⎠
e.
(135.5kW)(8760
f.
⎛ 1 ⎞⎛ 1 ⎞⎛ 0.7457 ⎞
Savings = (105, 000)(0.017)(200)(144) ⎜
⎟⎜
⎟⎜
⎟ (8760 )( (0.05) )
⎝ 550 ⎠⎝ 3600 ⎠⎝ 0.55 ⎠
hr
)($0.05 / kWhr) = $59,360 / year
yr
= $15, 420 / year
Steam Drum
(100,000)(1230.7) + (5000)(471.7) − 105,000(218.9) = 102.4 106 Btu×/ hr
g.
Steam Drum Heat Input = 102.4 ×106 Btu / hr
h.
Superheater = (100,000)(1351.8-1230.7) = 12.1×106 Btu / hr
Flash Drum
x = vapor flowrate
(100,000 − x)(218.93) + x(1164.1) = 100,000(330.65)
x=
11,820 lb/hr
=
vapor flowrate
and
88,180 lb/hr
=
liquid flowrate
i.
Steam Lost =11,820 – 8,422 = 3,398 lb / hr
j.
Condensate Lost = 88,180 – 50,000 = 38,180 lb / hr
9–79
Solutions Chapter 9
9.3.51
Basis: one hour operation open, steady-state process.
A material balance initially is necessary to determine amounts and compositions of
streams.
A.
Feed
Component
C2H6
C2H4
B.
Wt.%
65
35
lb
650
350
Product
Have 97% recovery of ethylene
lb C2H4 = (0.97) (350) = 340
Now, since the product is 98% C2H4
340 lb C2 H 4 2 lb C2 H6
= 6.95 lb C2 H6
98 lb C2 H 4
Component
C2H6
Wt.%
2
lb
6.95
9–80
Solutions Chapter 9
C2H4
C.
98
340
Bottoms
lb
lb
C2H4 = 350 – 340 = 10
C2H6 = 650 – 6.95 = 643.05
Component
Wt.%
lb
C2H6
C2H4
98.47
1.53
643.05
10
We are now ready to determine, by a series of energy balances, the quantities
desired.
Energy Balance on System “A” (see Fig.)
⎧Energy out with Product +
⎪
Energy in with Feed + energy in with steam = ⎨Energy out with bottoms +
⎪Energy out with refrigerant
⎩
We will assign an arbitrary enthalpy of zero to the liquid feed as it enters the Pre-heater at
–100ºF. Actually, any reference condition could be chosen since we are only concerned
with enthalpy changes in the balances.
------------------------------------------------------------The feed undergoes at 20ºF rise before it enters the still.
T = -80ºF and
Cp feed = (0.65) (0.65) + (0.35) (0.55) = 0.62 Btu/(lb) (ºF)
Enthalpy in with feed = (lb feed) (CpΔT) = (1000) (0.62) (-80-(-100))
= 12,400 Btu/hr
------------------------------------------------------------Enthalpy in with steam = (lb steam) (ΔHv) at 30 psig,
ΔHv = 945 Btu/lb
Enthalpy in with steam = (S) (945) btu/hr
------------------------------------------------------------From the data given, the temperature of the saturated liquid product is =30ºF since it is
essentially pure ethylene.
Cp = 0.55 Btu/(lb) (ºF)
9–81
Solutions Chapter 9
Enthalpy out with product = (346.95) (0.55) [-30-(-100)] = 13,360 Btu/hr
The bottoms are practically pure ethane so, T = 10ºF,
Cp = 0.65 Btu/(lb) (ºF)
Enthalpy out with bottoms = (653.05) (0.65) [10-(-100)] = 46,700 Btu/hr
------------------------------------------------------------The refrigerant experiences a temperature rise of 25°F as it passes through the condenser.
The heat capacity may be assumed to be 1.0 Btu/(lb) (°F)
Enthalpy out with refrigerant = (lb refrig.) (1.0) (25) m Btu/hr
Now, rewriting the energy balance “A” with a substitution of known terms:
12,400 + 945S = 13,360 + 46,700 + 25m
(1)
It is evident from this first balance that a second balance will be necessary to determine
the unknown quantities.
Energy Balance “B”
Energy in with Overhead =
Energy out with refrigerant +
Energy out with Product +
Energy out with Reflux
Since we are using a reflux ration of 6.1
lb reflux = (6.1) (346.95) = 2,120 lb/hr.
Now, the specific enthalpy of the reflux = the specific enthalpy of the product, since they
are from the same stream, and the differences in enthalpy between the overhead and the
product, or overhead and reflux, is merely the heat of vaporization of ethylene
(neglecting the small amount of C2H6 present), therefore, we can rewrite Energy Balance
“B” as:
(lb Overhead) (ΔHv) = 25 m
(2,120 + 346.95) (135) = 25 m
or m = 13,320 lb/hr.
Now substituting this value into equation (1) we have
12,400 + 945 S = 13,360 + 46,700 + (25) (13,320)
9–82
Solutions Chapter 9
so that S = 403 lb / hr
a.
403 lb steam
hr
= 0.403 lb steam / lb feed
hr
1000 lb feed
b.
13,320 lb refrigerant 1 ft 3 refrig. 7.48 gallons
= 1,993 gallons of refrigerant/hr
hr
50 lb refrig.
ft 3
The bottoms leave the still and enter the Pre-heater at approximately 10ºF, the saturation
temperature of pure C2H6. The final temperature of the bottoms as it leaves the Preheater can be calculated from a simple energy balance around the Pre-heater.
c.
Enthalpy Balance “C”
(lb bottoms) [0.65 Btu/(lb) (°F)] (10 – Tf) = (lb feed)
×[0.62 Btu /(lb)(o F)][−80 − (−100)]
(653.05) (0.54) (10 – Tf) = (1000) (0.62) (20)
Tf = 19.2o F
9–83
Solutions Chapter 10
10.1.1
(b), (d), (f)
10.1.2
(a)
10.1.3
(c)
10.1.4
The ΔH o f is 0 by definition.
10.1.5
ˆ ° of Fe O is calculated as follows.
ΔH
f
2 3
(1)
ˆ ° – Σn ΔH
ˆ°
ΔHRxn = Σni ΔH
fi
i
fi
products
reactants
°
ˆ ° (2 ) – ΔHˆ °
–822.200 kJ = ΔHˆ fFe O – Δ H
fFe
fO
2 3
2
( 32 )
ˆ°
= ΔH
fFe O – 0 – 0
2 3
(2)
ΔHRxn = –284.100 kJ
ˆ°
ˆ°
ˆ°
= ΔH
fFe O – Δ H fFeO (2) – ΔH f O
2 3
2
( 12 )
ˆ ° (2) – 0( 1 )
–284.100 = –822.200 – Δ H
fFeO
2
ˆ
ΔH
fFeO =
°
–538.10
kJ = –269.05 kJ
2
vs – 267 in Appendix F
10–1
Solutions Chapter 10
10.1.6
ΔH°(kJ)
+123.8
+2220.0
C3H8 (g) → C3H6 (g) + H 2 (g)
3CO2 (g) + 4H2O(1) → C3H8 (g) + 5O2 (g)
4 × (−285.8)
4[H 2 (g) + 1/ 2O2 (g) → H 2O(1)]
3 × (−393.4)
3[C(s) + O2 (g) → CO2 (g)]
________________________________________________
+20.1 kJ/g mol
3C(s) + 3H2 (g) → C3H6 (g)
ΔHof of C3H6 = 20.1 kJ/g mol
10.1.7
Basis: 1 g mol C5H2
ΔH (kJ/g mol)
5 CO2 (g) + H 2O(l )
→
add : 5[C(s) + O 2 (g)] →
1
1
C5 H 2 (s)+ 5 O 2 (g)
2110.5
5[CO 2 (g)]
5[-394.1]
2
add: H 2 (g)+ O 2 (g)
→
H 2O(g)
-241.826
add: H 2O(g)
→
H 2 O(l )
-43.911
net: 5C(s)+H 2 (g)
→
C5 H 2 (s)
2
ΔH o f = -143.737 kJ/g mol
10.1.8
(a)
(b)
(c)
NH4 (l ) : -67.20 kJ/g mol from Appendix F
Formaldehyde gas (H2CO): -115.89 kJ/g mol from Appendix F
Acetaldehyde liquid (CH3CHO):
For gas ΔH o f = 166.4
− kJ / g mol from Appendix F
The heat of vaporization at 293.2 K is 25.732 kJ/g mol from the CD. The
value is close enough to ΔHvap (298 K) to use.
ΔHof = -166.4 + 25.732 = -140.7 kJ/g mol
10–2
Solutions Chapter 10
10.2.1
a.
Basis: 1 g mol NH3 ( g)
NH3 ( g)
ˆ ° (kJ / g mol ) :
ΔH
f
°
ΔHrxn =
b.
HCl( g)
+
–46.191
ˆ –
∑ ni ΔH
i
products
→
–92.311
–315.4
ˆ
∑ ni ΔH
i
reac tants
=
(1)( –315.4) – [(1)( –46.191) + (1)( –92.311)]
=
–176.9 kJ / g mol NH3 ( g)
Basis: 1 g mol CH4 (g )
CH4 (g ) + 2 O2 (g ) → CO2 (g)
ˆ (kJ / g mol ) :
ΔH
f
°
ΔHrxn =
c.
NH4 Cl (s)
–74.84
ˆ –
∑ ni ΔH
i
products
0
+ 2 H 2 O ( )
–393.51
–285.840
ˆ
∑ ni ΔH
i
reac tants
=
⎣⎡( 2)( –285.840) + (1)( –393.51)⎦⎤ – ⎣⎡(1)( –74.84) + 0⎤⎦
=
–890.4 kJ / g mol CH4 ( g )
Basis: 1 g mol C6 H 2 (g )
C6 H1 2(g )
ˆ ° (kJ / g mol ) :
ΔH
f
–123.1
→
C6 H 6 ( )
+48.66
+
3H2 (g)
0
ΔH°rxn = [ 0 + (1)(+48.66)] – [(1)(–123.1)] = 171.76 kJ / g mol C 6 H12 (g)
10–3
Solutions Chapter 10
10.2.2
a. Basis: 1 g mol CO2 (g)
CO2 (g) + H2 (g) → CO (g) + H2O (1)
ΔH°rxn = (– 110.52 – 285.84) – (– 393.51) = –2.849 kJ
b.
Basis: 1 g mol CaO (s)
ΔH°rxn = (–986.59 – 924.66) = [(–635.55 – 601.83) - 2 (285.84)] =
−102.19 kJ per mol CaO(s)
Multiply by 2 for the reaction listed −204.38 kJ
c.
Basis: 1 g mol Na2SO4 (s)
ΔH°rxn = (–1090.35 – 110.54) – (–1384.49) = 183.59 kJ
d.
Basis: 1 g mol NaCl (s)
ΔH°rxn = (–92.31 – 1126.33) – (–811.32 – 411.01) = 3.67 kJ
e.
Basis: 1 g mol O2
ΔH°rxn = [2 (–1384.5) + 4 (–92.31)] – [(–411.00) + 2 (–296.90)
+ 2 (–285.84)] = –1561.76
f.
Basis: 1 g mol SO2 (g)
ΔH°rxn = (–811.32) – (–285.84 – 296.90) = –228.58 kJ
g.
Basis: 1 g mol N2
ΔHºrxn = 2 (90.374) = 180.75 kJ
h.
Basis: 1 g mol Na2CO3 (s)
ΔH°rxn = [3 (–1117.13) + (–393.51)] – [–1130.94 + 2 (–373.21)
+ 4 (–296.90)] = –1867.54 kJ
10–4
Solutions Chapter 10
i.
Basis: 1 g mol CS2 (1)
ΔH°rxn = (–1394.69 – 60.25) – (+87.86) = –287.61 kJ
j.
Basis: 1 g mol C2H2 (g)
ΔH°rxn = (+105.02) – (52.28 – 92.31) = –145.05 kJ
k.
Basis: 1 g mol CH3OH (g)
ΔH°rxn = (–115.90 – 241.82) – (–201.25) = –156.48 kJ
l.
Basis: 1 g mol C2H2 (g)
ΔH°rxn = (–140.7) – (226.75 – 285.84) = −81.61 kJ
m.
Basis: 1 g mol C4H10 (g)
ΔH°rxn = (52.28 – 84.67) – (–124.73) = 92.34 kJ
10.2.3
You want to get ΔH for
1
1
2
2
C3H8 (g) + Br2 (g) → C3H 7 Br(g)
+
ΔH
H 2 (g)
(J)
(–b)
C3 H8 ( g) → C3 H6 ( g) + H2 (g )
+126,000
(+a)
HBr( g) + C3 H6 (g) → C3 H7 Br
–84,441
(+c)
H2 (g ) + Br( g) → 2HBr( g)
–36,233
(–d)
Br (g) → Br (l )
–15,355
Total
10–5
−10, 029 J / g mol C3H8
Solutions Chapter 10
10.2.4
Basis: 1 g mol Fe S2
ΔH o rxn = 567.4
− kJ / g mol FeS2
( − 567.4) kJ 1 g mol FeS2 1000 g
kJ
= −4728.3
g mol FeS2 120 g FeS2 1 kg
kg FeS2
Note: The information about conversion does not affect the value of the standard heat of
reaction. It will affect the values in the energy balance.
10.2.5
G+E →
← E+F
ˆ ° is at 25°C, 1 atm
ΔH
Rxn
(
)
ˆ ° = ΣΔ H
ˆ ° – ΣΔH
ˆ ° = 1.040 × 10 9 – 0.990 × 109 J / g mol G
ΔH
Rxn
f
f
prod.
react.
6
= 50 × 10 J / g mol G converted
Since 0.48 reacts, ΔH°Rxn = 0.48( 50) × 10 6 J / g mol G
= 24 × 106 J / g mol G fed (not asked for)
10.2.6
No. The value reported is the heat transfer for constant volume, Qv. The heating value at
constant pressure should be reported for the standard heat of reaction (an enthalpy
change). Assume the reaction is at 25ºC.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l )
Basis: 1 g mol CH4
Q = ΔU = ΔH – Δ(pV)
If the gases can be treated as ideal Δ(pV) = Δ(nRT) = Δn (RT).
Δn = 1 – 2 – 1 = -2
10–6
Solutions Chapter 10
The correction is
ΔnRΤ = -2 (8.314) (298) = -4958 J/g mol
In SI units of 273.15 K and 101.3 kPa, the volume is 22.415 m3 g mol.
The correction is (ignoring the pressure change from 1000 kPa to p2)
−4958
= 221.2J / m3 or 0.22 kJ / m3
22.415
Otherwise you have to calculate p2 V − p1V where p2 = 1000 kPa and p1 = p*H2O + pCO2 .
The value to be reported should be 39.97 – 0.22 = 39.75 kJ / m3
10.2.7
Yes. Calculate the sensible heats using the tables and using a common reference
temperature.
10.2.8
C63H132O15
tristearin
3 C18 H36O2 +
stearic and
3 C3H8O3
glycerol
ˆ o = −[(3)(−964.3) + (3)(−159.16)] − (−3820)(1) = − 449.6 kJ/g mol
ΔH
rxn
10.2.9
a.
If the exit temperature is > 25oC, and the dilutent is at the entering temperature of
the reactants, then less heat has to be removed from the process.
b.
Remains the same as the presumption is in calculating ΔHºrxn that stoichiometric
quantities react to completion.
c.
ˆ o = ΔH
ˆ o with H O(g) = 241.826
At 25ºC, ΔH
−
kJ / g mol H2O.
rxn
f
2
At 500 K, the sensible heats have to be considered, and ½ O2 and H2 together have
a larger ΔH than does H2O so that ΔHrxn at 500 K is less than ΔHrxn at 25ºC.
10–7
Solutions Chapter 10
10.2.10
S (s) + O2 (g)
SO2 (g)
ΔHRxn
600K (327oC)
ΔHReact.
ΔHProd.
298K (25oC)
ΔHoRxn
Basis: 1 g mol S(s) =1 g mol SO2 (g)
ΔHrxn, 600 K = ΔHorxn + ∑ ΔHprod, 600 K − ∑ ΔHreact, 600 K
ΔHo rxn, 298 = ∑ ΔHo f − ∑ ΔHo f
prod
react
= [−296.90) − [0 + 0] = −296.90 kJ / g mol S
= − 296,900 J/g mol S
∑ ΔH
prod,600 K
ΔHSO2 ,600 K = ∫
327
25
(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606 10×−9 T3 )dT
= 13,490 J/g mol
∑ ΔH
(13,530 from Enthalpy Tables)
react,600 K
This assumes s is really a liquid at 600ºC, not a solid as in the equation
ΔHS,600K = ΔH 298−386 + ΔH fusion + H 386−600
Δ
solid
=∫
386
298
liquid
(15.2 + 2.68 ×10−2 T)dT + 10, 000 + 1/ 2∫
600
386
(35.90 + 1.26 10−3×T)dT
½ (S from Perry)
2
= 2144 + 10,000 + 3975
10–8
Solutions Chapter 10
= 16,119 J/g mol
ΔHO2 ,600K = ∫
327
25
(29.10 + 1.158 ×10−2 T − 0.6076 ×10−5 T 2 + 1.311 10
× −9 T3 )dT
= 9,340 J/g mol (8899 from the enthalpy tables)
ΔH rxn,600K = −296,900 + (13, 490 − 25, 429)
= 308,840 J / g mol S
10.2.11
C3H8 (g) → C2 H2 (g) + CH4 (g) + H 2 (g)
Basis: 1 g mol C3H8
ΔH500 = ΔH25 + ∑ ΔHprod − ∑ ΔHreact
ΔH o 25 = 226.75 + (−74.84) + 0 − (−103.85)
= 255.76 kJ / g mol
ΔH = ∫ CpdT = aT + b / 2 T2 + c / 3 T3 + d / 4 T4
ΔHC3H8 = 68.032T + 1.130 ×10−1 T2 − 4.37 ×10−5 T3 + 7.928× 10−9 T4
500
25
= 55.53 kJ / g mol
ΔHC2H2 = 42.43 T + 3.027 ×10−2 T2 − 1.678 ×10−5 T3 + 4.55 ×10−9 T4
500
25
= 25.89 kJ / g mol
ΔHCH4 = 34.31 T + 2.735 ×10−2 T2 + 1.220 ×10−6 T3 + 2.75 ×10−9 T4
500
25
= 23.10 kJ / g mol
ΔHH2 = 28.84 T + 3.825 ×10−5 T2 + 1.096 ×10−6 T3 + 2.175
× 10−10 T4
500
25
= 13.83 kJ / g mol
ΔH500 = (255.76 + 25.89 + 23.10 + 13.83) − 55.53=
10–9
263.05 kJ / g mol
Solutions Chapter 10
10.2.12
All data is from Tables E.1 and F.1 of the Appendix.
a.
o
CH 3OH(g) + 1/ 2O 2 (g) 200
C H 2CO(g) + H 2O(g)
Basis: 1 g mol CH3OH
Comp.
Cp (J /(g mol)(K or oC))
T
ΔĤof (kJ / g mol)
CH3OH(g)
42.39 + 8.301× 10−2 T − 1.87 × 10−5 T 2 − 8.03 × 10−9 T 3
ºC
-201.25
O2(g)
29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC
H2CO(g)
34.28 + 4.268 × 10−2 T − 8.694 ×10−9 T 3
H2O(g)
33.46 + 0.6880 × 10−2 T +0.7604 × 10−5 T 2 − 3.593 × 10−9 T 3 ºC
ºC
0
-115.89
-241.826
25ºC = 298 K
200ºC = 473 K
ΔHo rxn ,25o C = [−115.89 + (−241.826)] − (−201.25) = 156.47 −kJ / g mol
= −156, 470 J / g mol
Products
Add the Cp equations for the two products and integrate (each involves 1 mole).
∑ ΔH prod,200o C = ∫
=
200
25
(67.74 + 4.95 ×10−2 T + 0.7604 ×10−5 T 2 − 1.228 ×10−8 T 3 )dT
12,850 J/g mol
Reactants
ΔHCH3OH = ∫
200
25
=
(42.39 + 8.301×10−2 T − 1.87 ×10−5 T 2 − 8.03 ×10−9 T3 )dT
9000 J/g mol
10–10
Solutions Chapter 10
ΔHO2 =
1 200
(29.10 + 1.158 ×10−2 T − 0.6076 ×10 −5 T 2 + 1.311× 10 −9 T 3 )dT
∫
25
2
= 2,650
ΔHrxn,200o C = −156, 470 + 12,850 − 2,650
146,
= 270 J / g mol
o
SO 2 (g) + 1/ 2 O 2 (g)300
C SO3 (g)
b.
Basis: 1 g mol SO2
DATA:
Comp.
Cp (J /(g mol)(K or oC))
O2(g)
29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC
0
SO2(g)
38.91 + 3.904 × 10−2 T − 3.104 × 10−5 T 2 + 8.606 × 10−9 T 3 ºC
-296.90
SO3
48.50 + 9.188 × 10−2 T − 8.540 ×10 −5 T 2 + 32.40 ×10 −9 T 3 ºC
-395.18
T
ΔĤof (kJ / g mol)
ΔHo rxn ,25o C = −395.18 + 296.90 = −98.28 kJ / g mol = 98, 280 −J / g mol
Products
∑ ΔH prod,300o C = ∫
300
25
(48.50 + 9.188 ×10−2 T − 8.540 ×10−5 T 2 + ×
32.40 10−9 T3 )dT
= 16, 740 J/g mol
Reactants
ΔHSO2 = ∫
300
25
(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606× 10−9 T3 )dT
= 12, 180 J/g mol
ΔH O2 = 8, 470 J / g mol
1
ΔH rxn,300o C = −98, 280 + 16, 740 − 12,180 − (8, 470) = 97,960J
− / g mol SO2
2
10–11
Solutions Chapter 10
10.2.13
Basis: 1 g mol SnO
SnO + 1/ 2O2 → SnO2
ΔHrxn25 = −577.8 − (−283.3 + 0) = 294.5
− kJ / g mol
ΔH90 = ΔH rxn, 25 + ∑ ΔH prod − ∑ ΔH react
C
ΔH = ∫ 90
C dT
25o C p
o
ΔHSnO = 39.33T + 7.575 ×10−3 T=2
363
2.88 kJ
298
ΔHSnO2 = 73.89 T + 5.02 ×10−3 T 2 +
2.16 ×104
T
363
298
= 2.89 kJ
ΔHO2 = 29.1 T + 5.79 ×10−3 T 2 − 2.025 ×10−6 T3 + 3.278 ×10−10=T 4
90
25
1.93 kJ
ΔH90 = −294.5 + 2.89 − (2.88 + 0.5(1.93)) = 295.46
−
kJ / g mol
10.2.14
Basis: 100 g mol CO
Assume pressure is 1 atm, and an open, steady state process.
CO +
1
2
O 2 → CO 2
The energy balance reduces to Q = ΔH. The reference temperature is 25°C.
100 g mol
CO
CO2
400oC
300oC
100 g mol O2
100oC
10–12
Solutions Chapter 10
Data:
Sensible heats
ˆ
from CD : ΔH(kJ
/ g mol)* ΔH(kJ)
mol
o
T( C)
ΔĤ (kJ / g mol)
O2
CO
100
100
100
300
-110.541
2.235
8.294
223.5
-10,225
Out
O2
CO2
50
100
400
400
-393.505
11.715
16.245
585.8
-37,726
Comp.
o
f
In
*You can calculate ΔĤsensible using the tables in the appendix, but the process requires
interpolation.
Q = ΔH = (−37,726 + 585.8) − (−10, 225 + 223.5) = − 27,139 kJ
10.2.15
Basis: 1 g mol yeast
The heat of formation is
3.92 C + 6.5 H + 1.94 O2 → C3.92 H6.5 O1.94
The heat of combustion is for the reaction
C3.92 H6.5 O1.94 + 5.60 O2 → 3.92 CO2 (g) + 3.25 H2 O(l )
ˆ 0 − ∑ n ΔH
ˆ 0 ⎤
−1,518 kJ = − ⎡⎣ ∑ n i ΔH
f ,i
i
f ,i ⎦
product
reactants
ˆ0
⎤
−1,518 = − ⎣⎡(3.92)(−393.51) + (3.25)(−285.840) − 5.60(0) − (1)ΔH
f ,yeast ⎦
ΔĤ 0f,yeast = 963.54
− kJ/g mol
The molecular weight of C3.92 H 6.5O1.94 is 84.63. Per 100 g yeast
ΔĤ0f,yeast = 1139−kJ/100 g yeast
10–13
Solutions Chapter 10
10.2.16
The problem requires more information to solve. To make the solution easier but
approximate, assume the reaction takes place at 25°C and 1 atm. Assume that the energy
balance reduces to Q = ∆H, and that Q represents the desired kJ. Data:
°
ˆ°
ΔH
f , H 2 O = –285.840 ΔĤ f, CO2 = –393.51 , both in kJ/g mol. Ref. temp. = 25°C. Basis: 1
g mol glucose. MW of glucose = 180.
(C6 H12O6 )glucose + 6O2 (g) → 6H 2O(l ) + 6CO2 (g)
ΔH° =ΔH°products –ΔH°reactants
= [6( –285.840) + 6(–393.51)] – [1( –1260) + 6( 0)] = –2816 kJ
0.90 ( –2816 )
= 14.0 kJ/g glucose
180
At 37°C (body temperature), assume only the O2 is used in a reaction
6 g mol O2 22.4L at SC 310K 1 g mol suc.
= 0.85 L/g at 37o C and 1 atm
1 g mol suc. g mol O2 273K 180g suc.
10.2.17
The reaction is
C3H8 (l ) + 5O2 (g) → 3CO2 (g)+ 4H 2O(l )
Need to calculate the heat of formation of C3H8 (l ) and H2O (l ) first
C3H8 (l )
ˆ of ,C3H8 ( l ) = ΔH
ˆ ofC H8(g) − ΔH
ˆ o vap(25o C)
ΔH
3
= −24.820 − 3.820
=−
28.643 k cal / g mol
H2O(l )
ˆ o f ,H2O( l ) = ΔH
ˆ o f ,H 2O(g) − ΔH
ˆ o vap(25o C)
ΔH
10–14
Solutions Chapter 10
= −57.798 − 10.519 = 68.317
− k cal / g mol
Basis: 1 g mol C3H8 (l )
ΔHRxn,C3H8 ( l ) = 4(−68.317) + 3(−94.052) − (−28.643)
-526.781
=
k cal/g mol
10.2.18
a.
Presumably H2, O2, C1, HC1, and H2 are gases.
(1)
6 g mol FeC13(-403.34 kJ/g mol FeC13) + 9 g mol H2O (-241.826 kJ/g mol H2O)
-3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3)-18 g mol HC1(-92.312 kJ/g mol HC1)
= −468.39 kJ
(2)
6 g mol FeC12(-342.67 kJ/g mol FeC12) + 3 g mol C12 (0.0 kJ/g mol C12)
-6 g mol FeC13 (-403.34 kJ/g mol FeC13) = 364.02 kJ
(3)
2 g mol Fe3O4 (-1116.7 kJ/g mol Fe3O4) + 12 g mol HC1 (-92.312 kJ/g mol HC1)
+ 2 g mol H2 (0.0 kJ/g mol H2) – 6 g mol FeC12 (-342.67 kJ/g mol FeC12)
-8 g mol H2O (-241.826 kJ/g mol H2O) = 649.48 kJ
(4)
3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3) – 2 g mol Fe3O4 (-1116.7 kJ/g mol
Fe3O4) – ½ g mol O2 (0.0 kJ/g mol O2) = −233.07 kJ
(5)
6 g mol HC1 (-92.312 kJ/g mol HC1) + 3/2 g mol O2 (0.0 kJ/g mol O2)
-3 g mol H2O (-241.826 kJ/g mol H2O) – 3 C12 (0.0 kJ/g mol C12) = 171.61 kJ
b.
2H2O
2H2 + O2
∑ ΔH
o
n,i
= 483.65 kJ
10–15
Solutions Chapter 10
10.2.19
Basis: 1 g gasoline
ΔH =
40 kJ 0.84
= 33.6 kJ / cm3
g cm3
Methanol:
Assume the reaction is:
ΔĤof (kJ / g mol)
1
CH3OH(l ) + 1
2
O2 (g) → CO2 (g) 2H
+ 2O(g)
0
− 238.64
-393.51
-241.826
ΔHRxn = 638.52
− kJ / g mol
638.52 kJ 1 g mol 0.792 g
= 15.79 kJ / cm3
3
g mol 32.04 g cm
33.6 kJ cm3 methanol
cm3 methanol
=
2.13
cm3 gas 15.79 kJ
cm3 gas
a. (2.13-1) 100 = 113% larg er tan k
Ethanol:
C2 H5OH(l ) + O2 (g) → 2CO2 (g)
+ 3H2O(g)
ΔĤof (kJ / g mol)
− 277.63
0
-393.51 -241.828
ΔHRxn = 1235
− kJ / g mol
1235
1 g mol 0.789 g
= 21.15 kJ / cm3
3
g mol EtOH 46.07 g cm
33.6 kJ cm3 EtOH
= 1.59
cm3 gas 21.15 kJ
b. (1.59) 100 -100 = 59% larg er
10–16
Solutions Chapter 10
10.2.20
Basis: 1 g mol C2H4(g)
C2 H4 (g)
⎛ kJ ⎞
ΔĤ of ⎜
⎟
⎝ g mol ⎠
+
52.283
2H 2 (g)
→
2CH 4 (g)
0
25o C
-74.84
ΔH o rxn = 2(−74.84) − 52.283 = 202.0
− kJ / g mol
The reaction gives Q = ΔU = ΔH – Δ(pV). Assume ideal gases so that Δ(pV) = Δ(nRT) =
Δn(RT)
Q = -202.0 – (-1) (8.31 × 10-3) (298)
Q = −199.5 kJ / g mol
10.2.21
Basis: 1 g mol Cu and 1 g mol H2SO4
Data from Perry
Cu(s) + H2SO4 (l ) → H2 (g) CuSO
+ 4 (s)
ΔH o rxn = (0 − 772.78) − (0 − 810.40)
= 37.61 kJ
For a constant volume process:
Q = ΔU = ΔH – Δ(pV) = ΔH –Δn(RT)
Δn = 1
Q = 37.61 – (1) (8.314 × 10-3) (298) = 35.13 kJ/g mol
Basis: 1 lb mol of each reactant
(37.13)(1000)J 454 g 1 Btu
= 15,978 Btu / lb mol
g mol
1 lb 1055 J
10–17
Solutions Chapter 10
10.2.22
Assume steady state flow process with reaction.
N2(g)
H2(g)
C6H12(g)
C6H6(g)
Q
Ref. temp. = 25°C
C6 H12 (g) → C6 H6 (g) + 3H2 (g)
The standard heat of reactions is
ΔĤoRxn = [1(82,927) + 3(0)] − [1(−123,100)] = 2.06 ×105 J / g mol C6 H12
Material balance:
Basis: 1 g mol C6H12 (g) in
Step 5:
Steps 6 to 9:
Make balances for each species so that the unknowns H2 and C6H6 can be calculated.
Results:
n H2
=
0.70(1)(3)
=
out
=
0.70(1)(1)
= 0.70
n C6H12
out
=
0.30(1)
= 0.30
out
=
0.50
ˆ o f (J/gmol)
ΔH
∫
T
25
Out:
In:
H2(g)
C6H6(g)
C6H12(g)
N2(g)
0
+ 82.927 × 103
- 123.1 × 103
0
C6H12(g)
N2 (g)
-123.1 × 103
0
(a)
2.10
n C6 H 6
n N2
Energy balance
Component
out
n C6H12 in = 1.0
ni
ΔH 300
(kJ)
25
0
0
0
0
2.10
0.70
0.30
0.50
0
0
1.00
0.50
16.865
71.226
-47.552
4.016
44.556
-123.1
0
-123.1
C p dT (or use tables)*; T = 300°C
Q = 1.44 × 105 J / gmol
8031
18,824
35,408
8,031
or
0
0
or
(b)
= 0.50
n N2 in
Q = + 1.68 × 105 J / gmol
*at 25°C exit temp. and entrance temp. these terms are 0.
10–18
Solutions Chapter 10
10.2.23
Basis: 1 gal of each fuel
MW density (g/cm3)
(1) Ethanol : C2 H5OH(l ) + 3O2 (g) → 2CO2 (g)
+ 3H2O(l )
(2) Benzene : C6 H6 (l ) + 7.5O2 (g) → 6CO2 (g) + 3H 2O(l )
(3) Isooctane : C8 H18 (l ) + 12.5O2 (g) → 8CO2 (g) +9H2O(l )
46
78
114
0.789
0.879
0.692
Example calculation
1 gal Bz 0.879 g Bz 3.785 L 1000cm3 1 g mol Bz 6 g mol CO2
Benzene :
cm3Bz
1 gal
1L
78 g Bz 1 g mol Bz
×
44 g CO2 1 lb
= 24.8 lb CO2 / gal Bz
1 g mol CO2 454 g
Other results:
Ethanol: 12.5 lb CO2 / gal EtOH
Isooctane:
17.8 lb CO2 / gal iso
Carry out the standard combustion material balance for stoichiometric air. The ratios are
at SC or the same temperature and pressure.
Example of the calculation for benzene (Reaction 2)
Use data from the CD, heat capacity equations, or tables to get the sensible heats. The
data below are from the CD. Reference T = 25ºC.
IN:
T(ºC)
mol
ΔĤ o f (kJ / g mol)
ΔĤsensible (kJ / g mol) ΔH(kJ)
ΔH(kJ)
C6 H 6 ( l )
O2 (g)
N 2 (g)
25
100
100
1
7.5
28.21
+ 44.66
-
2.235
2.183
CO 2 (g)
H 2O(l )
N 2 (g)
25
25
25
6
3
28.21
-393.505
-285.840
-
-
46.66
16.763
61.553
OUT:
10–19
2361.030
857.520
-
Solutions Chapter 10
ΔHRxn = (−2361.030 − 857.520) − (46.66 + 16.763 + 61.553) = 3093.6
− kJ / g mol Bz
−3093.6 kJ 1 gal Bz 0.879 g Bz 3.785 L 1000 cm3 1 g mol Bz 1 Btu
g mol Bz
cm3 Bz
1 gal
1L
78 g Bz 1.055 kJ
= −1.251× 105 Btu per gal Bz
(exothermic)
Ethanol:
−7.55 ×104 Btu per gal EtOH
Isooctane:
−1.21×105 Btu per gal isooctane
Ethanol
Benzene
Isooctane
lb CO2 /106 Btu
166
198
147
10.2.24
At 25°C and 1 atm H2O is a liquid.
1
H 2 (g) + O2 (g) → H 2 O(l )
2
o
ΔH c (kJ/g mol): − 285.84
0
0
ΔHRxn25oC
ΔHReact
ΔHProd
ΔHRxn0oC
Basis: 1 g mol H2 (g)
ΔHo rxn = −[∑ ΔHo c,Prod − ∑ ΔHo c,React ]
1
= −[(1)(0) − (1)(−285.84) − (= −
)(0)]
2
285.84 kJ / g mol H 2
H2O at 25ºC is a liquid
10–20
Solutions Chapter 10
ΔH Pr od
4.184 J 18 g (−25o C)
= ∫ o Cp dT =
= 1.883
− kJ
25 C
(g)(o C) 1 g mol
0o C
ΔH React = (1) ∫
0o C
25o C
1
0o C
2
25
CpH2 (g) dT + ( ) ∫
1
CpO2 (g) dT = −(718)(1) − (727)( ) = 1.081
− kJ
2
ΔHrxn (0o C) = (−1.883) − (−1.081) + (−285.84) = 286.64 kJ
− / g mol H2
10.2.25
Basis: 1 g mol dry cells
The heat of reaction is
ˆ 0 − ∑ n ΔH
ˆ 0 ⎤
ΔH orxn = − ⎣⎡ ∑ n i ΔH
i,c
i
i,c ⎦
products
reactants
= −[(1)(−1,517) + 2.75(−393.51) + 3.42(
−
285.840)
−6.67(−2,817) − 2.10(0)] = 15, −
213 kJ/g mol cells
For 100 g dry cells
-15,213 kJ
1g mol
100 g dry cells
= -1.800×104 kJ
g mol dry cells 84.58 g dry cells
10.3.1
Steady State, open process. Ref T = 25ºC
Basis: 100 g mol feed
The moles out come from the material balance, and the other data from the CD.
Q = ΔH
ΔH data from the CD.
Comp.
% = g mol
T(ºC)
ΔĤof (kJ / g mol)
IN with gas
10–21
ΔĤsensible (kJ)
ΔH(kJ)
Solutions Chapter 10
CO2
6.4
500
-393.505
20.996
-2384.06
O2
0.2
500
-
15.034
3.04
CO
40.0
500
-110.541
14.690
-3834.04
H2
50.8
500
-
13.834
702.77
N2
2.6
500
-
14.242
37.03
100.0
-5,475.29
In with air
O2
63.28
25
-
0
0
N2
240.65
25
-
0
0
303.93
Out with fg
CO2
46.4
720
-393.505
32.006
-16,773.55
H2O(g)
50.8
720
-241.835
26.133
-10,957.66
N2
240.65
720
-
21.242
5,111.89
O2
18.1
720
-
22.555
408.25
356.0
-22,211.08
Q = ΔH = (−22, 211.08) − (−5, 475.29) = − 16,736 kJ
(Heat leaving)
10–22
Solutions Chapter 10
10.3.2
Steady state, open process. Ref T = 25°C
Basis: 100 g mol P
W
H2O
coke F
40oC
Furnace
P 1100oC
CO2
CO
O2
N2
mol %
16.1
0.8
4.3
78.8
1000
R Refuse 200oC
mass fr.
C
0.90
H
0.06
inert 0.04
1.00
A
Air 40oC
mol fr.
O2 0.21
N2 0.79
1.00
C
inert
mass fr.
0.10
0.90
1.00
Data come from the material balances and the CD.
C p (J / g mol)( o C)
H 20.8
C 14.3
Comp. Percent (g)
g mol
T(ºC) ΔĤof (kJ / g mol)
ΔĤsensible (kJ / gmol) ΔH(kJ)
In F (214.8 g) solid
C
90.0
16.11
40
-
0.214
3.456
H
6.0
12.79
40
-
0.312
3.991
Inert
4.0
-
40
-
ignore
-
100.0
28.90
7.447
In air (99.7 g mol)
O2
20.94
40
-
0.442
9.255
N2
78.76
40
-
0.436
34.339
99.70
43.595
10–23
Solutions Chapter 10
Out in P (g)
CO2
16.1
1100
-393.505
24.744
-5937.05
CO
0.8
1100
-110.541
25.032
-68.41
O2
4.3
1100
-
26.209
112.70
N2
78.8
1100
-
38.891
3064.59
Water out (g)
100.0
6.80 1100
30.174
–2828.17
205.18
Refuse
10 g
-241.826
200
-
1.488 per g
14.88
Q = ΔH = (−2828.17 + 205.18 + 14.88) − (7.447 + 43.595) = − 2659.2 kJ
(heat exiting)
10.3.3
Basis: 1 g mol CaC12 ⋅ 6H 2O(s)
Unsteady state, open process. Reference T = 25ºC.
68o F → 20o C and 86o F → 30o C
ΔU = Q − ΔH H2Oexiting
Assume ΔU ; ΔH inside the system
Cp (J /(g)(o C))
ΔĤof (kJ / g mol)
Data
MW
CaC12 ⋅ 6H 2O(s)
-2607.89
1.34
219
CaC12 ⋅ 2H 2O(s)
-1402.90
0.97
147
Calculate ΔÛ
Comp.
Initial
CaC12 ⋅ 6H 2O(s)
g mol T(ºC)
1
20
MW
219.1
ΔĤof (kJ / g mol)
ΔHsensible (J)
ΔH(kJ)
-2607.89 (1)(1.34)(219.1)(25-20) -2606.422
10–24
Solutions Chapter 10
Final
CaC12 ⋅ 2H 2O(s)
Out
H2O(g)
965.798
1
30
129.0
-1402.90 (1)(0.97)(129.0)(30-25) -1402.274
4
30
18.0
-241.826 (1)(4.18)(18)(30-25)
-
CaC12 ⋅ 6H2O(s) → CaC12 ⋅ 2H2O 4H
+ 2O(g)
Per g mol CaC12 ⋅ 6H2O(s) :
Q = ΔU + ΔH exiting = [(−1402.274) − (−2606.422)] + (− 965.798)
= 238 kJ
or 226 Btu/g mol CaC12 ⋅ 6H 2O(s)
200, 000 Btu 1 g mol CaC12 ⋅ 6H 2O(s) 219 g 1 kg
= 194 kg
226 Btu
1 g mol 1000 g
(427 lb)
10.4.1
(a)
0.005⎞
⎛
+ 4050(0.006)
HHV = 14,544 (.80) + 62,028 ⎝ 0.0003 –
8 ⎠
= 11,640 Btu/lb
LHV = 11,800 – 91.23(0.3) = 11,770 Btu/lb
(b)
HHV = 17,887 + 57.5(30) – 102.2 (0.5) = 19,560 Btu/lb
LHV = 19,560 – 91.23 (12.05) = 18,460 Btu/lb
10.4.2
If the fuel cannot produce H 2O as a product, HHV = LHV, but the concept really refers
to cases in which water is produced so that HHV ≠ LHV.
10–25
Solutions Chapter 10
10.4.3
Both H2 and O2 enter at 0°C, and the products leave at 0°C. The reference temperature is
25°C. Stoichiometric quantities react according to the reaction equation.
H2 (g) + 1/2 O2 (g) → H2O (l )
°
ΔHf (kJ/g mol):
0
0
–285.83
Basis: 1 g mol H2 (g)
ΔHRxn( 0°C) =
ΣΔHPr oducts
ΣProducts
( –285.83) + ∫20°C
5°C
=
–
ΣΔHRe ac tan ts
4.184 J 18g H2 O
dT = –287.71 kJ
( g)(°C ) g mol
Alternately you can use the steam tables or the CD to get the value. The reactants are
gases, hence ΔH can be obtained from tables.
ΣReactants
=
28.84 + 0.00765 × 10 –2 T )dT
(
25°C
(0 + 0) + (1)∫
+ ( 12 )∫
0°C
25°
0°C
(29.10 + 1.158 × 10 –2 T )dT = –1.45 kJ
ΔHRxn( 0°C) = –287.71 – ( –1.45) = –286.26 kJ
10–26
Solutions Chapter 10
10.4.4
Basis: 100 lb mol gas
The reaction for each compound is the stoichiometric amount of O2 going to CO2 gas and
H2 O ( l )
Comp.
CO2
C2 H4
CO
H2
CH4
N2
lb mol
9.2
0.4
20.9
15.6
1.9
52.0
100.0
ΔĤ(Btu/lb mol)
ΔH (Btu)
0
598,000
122,000
123,000
383,000
0
0
239,000
2,544,000
1,918,000
728,000
0
5,429,000
Note: NGI standard state is 60°F
Btu 5,429,000 Btu
1 lb mol 520°R
3 =
100 lb mol 359 ft 3 492°R
ft
3
= 143 Btu / ft at std. conditions of NGI
10.4.5
Basis: 1 m3 gas at 25ºC and 1 atm with 40% relative saturation
Calculate the moles of C9H12 gas. Its pressure at 25ºC is calculated as follows:
( p*H2O = 3.2 kPa)
(3.2) (0.40) = 1.28 kPa, and thus the pressure of the C9H12 (g) is (101.3-1.28) = 100 kPa.
n=
pV 100 kPa 1 m3
1(kg mol)(K)
=
= 0.0404 kg mol
RT
1 atm
298 K 8.314(kPa)(m3 )
Basis: 1 g mol n-propyl benzene (C6H5
⋅
CH2
⋅ C2H5)
The higher heating value is with H2O as a liquid. Also, at 25°C, C9H16 is a liquid; its
normal boiling point is 432 K.
C9H12 (l ) + 12 O2 (g) → 9 CO2 (g) + 6 H2O (l )
10–27
Solutions Chapter 10
ΔH° f
kJ
:
g mol
–38.40
0
–393.51
–285.840
ΔH° rxn = [6 (–285.840 + 9 (–393.51)] – [(1) (–38.40)]
= −5218.2 kJ/g mol C9 H12
HHV= −
−5218.2 kJ 40.4 g mol C9 H12
= 2.11×105 kJ/m3
3
g mol C9 H12
1 m gas
10.4.6
Basis: 100 mol of gas
(a1)
Comp.
mol.wt.
CH4
C2 H6
C3H8
C4H10
Total
16
30
44
58
Mol%=
Vol.%
lb
88
6
4
2
100
1408
180
176
116
1880
–ΔH°c1
kJ/g mol
–ΔH°c2
kJ/g mol
890.35
1559.88
2220.05
2878.52
802.32
1427.84
2044.00
2658.45
mol. wt. mixture = 18.80 lb/lb mol
−ΔHo c1 =H2O(1) to CO2 (g) products;
− ΔHo c2 = H 2O (g) + CO2 (g) products
ΔHº1 rxn = – [ΔHc prod – ΔHºc react] = 1022.90 kJ/g mol
(a1)
ΔH o rxn =
−1022.90 454 g mol Btu
lb mol
Btu
= − 23,600
g mol
lb mol 1.055 kJ 18.80 lb
lb
HHV = 23,600 Btu/lb
(a2)
ΔHºrxn = (–23,600) (18.80) = -439,000 Btu / lb mol
HHV = 439,000 Btu/lb mol
(a3)
ΔHo rxn =
−439,000 Btu
1 lb mol
= − 1160
3
lb mol
379.2 ft at 60o F, 760 mm Hg
10–28
Solutions Chapter 10
HHV
3
= 1160 Btu / ft at 60°F, 760 mm Hg
(b1) ΔH°2 rxn = –925.71 kJ/g mol
ΔH° 2 rxn =
–925.71 454
= –21,200 Btu/lb
1.055 18.80
LHV = 21,200 Btu/lb
(b2) ΔH°2 rxn = (–21,200) (18.80) = -398,000 Btu/lb mol
LHV = 398, 000 Btu / lb mol
(b3) ΔH°2 rxn = (–398,000)/(379.2) = -1051
LHV = 1051 Btu/ft 3 at 60°F and 760 mm Hg
10.4.7
Write the equation for the combustion for 1 gram as:
1 g of Fuel or food + Oxygen → CO2 (g) + H2O(1) + N2 (g)
Both the reactants and the products are at 25ºC and 1 atm. The chemical energy released
by the complete combustion of 1 g of a food or fuel according to the above equation at
standard conditions is the heat of combustion. Assume the ΔHc are additive. Then
ΔHC = (17.1)(28) + (39.5)(10) + (14)(4) = 930 kJ
The High Energy bar has (capital C for calorie)
220 Calories 1000 calories 4.184 J 1 kJ
= 921 kJ
1 Calorie 1 calorie 1000 J
The values are close enough.
10–29
Solutions Chapter 10
10.4.8
When the combustion results in a single component.
10.4.9
Basis: 1 lb mixture
Get the data for the ΔHc in Btu/lb, not Btu/mol.
Component
Mass fr., ωi
Hexachloroethane, C2C16
Tetrachloroethene, C2C14
Chlorobenzene, C6H5C1
Toluene, C7H8
0.0487
0.0503
0.2952
0.6058
ΔHc (Btu / lb)
828
2141
11,876
18,246
ΔH(Btu)
Basis
1200 lb
40
108
3,506
11,053
58.44
60.36
354,24
726.96
Total
1.000
14,707
1200.00
_______________________________________________________________________
The Btu/lb are 14, 707 from the heat of combustion data, a deviation of 3.2% from
15,200. For DuLong’s equation, the component elements are needed.
Basis: 1200 lb
_______________________________________________________________________
C
H
C1
O
MW Total (lb)
_______________________________________________________________________
C2C16
24
0
213
0
237.00
ωi
0.10
0.00
0.90
0.00
lb
5.84
0.00
52.60
0.00
24
0
142
0
ωi
0.14
0.00
0.86
0.00
lb
8.45
0.00
51.91
0.00
72
5
35.5
0
C2C14
C6H5C1
10–30
58.44
166.00
60.36
112.50
Solutions Chapter 10
ωi
0.64
0.04
0.32
0.00
lb
226.71
14.17
113.36
0.00
84
8
ωi
0.91
0.09
0.00
0.00
lb
661.53
65.43
0.00
0.00
Total
902.53
79.60
217.87
0
1200
0.75
0.07
0.18
0
1.00
C7H8
ωi
0
0
354.24
9200
726.91
HHV = 14,455 C + 62,028 H – 7753.5O2 + 4050 S
= (14,455)(0.75) + (62,028)(0.07)
= 15,183 Btu/lb (the higher heating value) a deviation of 0.11%
10.4.10
The reactions are:
C2 H5OH(1) + 3 O2 (g) → 2 CO2 (g)
+ 3 H 2O(1)
(1)
C8H18 (1) + 12.5 O2 (g) → 8 CO2 (g) +9 H2O(1)
(2)
⎡
ˆ o − ∑ n ΔH
ˆ o ⎤⎥
ΔH rxn = − ⎢ ∑ n i ΔH
c,i
i
c,i
reac tan ts
⎣ products
⎦
For (1): −[(2)(0) + (3)(0) − (3)(0) − (1)(−1366.91) = 1366.91
− kJ / g mol C2 H5OH
For (2): −[(8)(0) + (9)(0) − (12.5)(0) − (−1307.53)(4.184)] = 5470.71
− kJ / g mol C8H18
Note: the ΔĤc = 1307.53 k cal / g mol for n-octane liquid is from Perry.
10–31
Solutions Chapter 10
Basis: 1 kg gasahol
Component
mass.fr. = kg
C2H5OH
0.10
C8H18
0.90
Total
1.00
MW
g mol
ΔĤ(kJ)
46.05
2.17
-2,966
114.14
7.89
-43,137
−46,103
If the fuel were all octane
1 kg oc tan e 1 kg mol octane 1000 g −5470.71 kJ
= −47,930 kJ
114.14 kg octane 1 kg g mol octane
−47,930 − (−46,103)
(100) = 3.8%
−47,930
10–32
Solutions Chapter 11
11.1.1
Basis: gas as given in problem statement
(a) 0.030 =
p H2O
pTot − pH2O
=
p H 2O
101.6 − p H2O
p* H2 O at 60 °C
% humidity
so, pH 2 O = 2.96 kPa
= 19.9 kPa
⎛ 2.96 ⎞ ⎡ 101.6 −19.9 ⎤
= 100 ⎝
= 12.2%
19.9 ⎠ ⎢⎣ 101.6 − 2.96 ⎥⎦
(b) Relative humidity
(100)
2.96
19.9
14.9%
=
*
(c) Dewpoint occurs where pH 2 O = 2.96 kPa, or T = 24oC from the steam tables.
11.1.2
*
At 27 °C, pH 2 O = 3.52 kPa. The actual pressure of the water vapor is obtained from
p=
nRT ⎛ 0.636 ⎞ ⎛⎜ 8.314 ⎞⎟ ⎛⎜ 300 K ⎞⎟ ⎛
⎞ = 3.15 kPa
=⎝
⎠
⎝
⎠⎝
⎠ 28.0 ⎠
V
18.01 ⎝
⎛ 3.15 ⎞
100 ⎝
3.52 ⎠
= 89%RH
11–1
Solutions Chapter 11
11.1.3
.5067
---------p
.1217
-----
Initial state
40 80 T °F
*
At 80 °F, pH 2 O = 0.5067 psia from the steam tables
*
At 40 °F, pH 2 O = 0.1217 psia
*
At 58 °F, pH 2 O = 0.2384 psia
(a) 100
0.1217 psia
=
0.5067psia
24%
(b) The mol fraction of H2O vapor is constant on compression unless saturation is
reached. At the
initial conditions
y H 2O =
0.1217
−3
= 8.28 × 10
14.696
⎛ 0.1217⎞
= 0.2437 psia
At 2 atm, pH 2 O = p Tot y H2 O = (14.692)(2) ⎝
14.696 ⎠
0.2437
> 1 hence water condenses and the relative humidity is 100%
0.2384
11–2
Solutions Chapter 11
11.1.4
P
T = 140°F
*
H 2O
= 5.878 in Hg
p = 30 in Hg
H = 0.03 mol H2O/mol BDA ⇒ Basis: 1 mol BDA
mol H 2O
0.03
=
= 0.0291 = y H 2O
mol H 2O + BDA 1 + .03
p H 2O = 30 (.0291) = 0.874 in Hg
pair = 30 (1 – 0.0291) = 29.13 in Hg
(a)
% rel. humidity = 100
p H2O
p*H2O
⎛ 0.874 ⎞
= 100 ⎜
⎟ = 14.9%
⎝ 5.878 ⎠
(b)
Dew point is the temperature at which vapor first condenses on cooling at
constant humidity, hence p* = 0.874 in Hg ~ 75°F.
11–3
Solutions Chapter 11
11.1.5
Steps 1, 2, 3, and 4:
ptot = 275 kPa
Air
30oC
F (mol)
100 kPa
RH = 75.0%
275 kPa
Compressor
Cooler
P(mol)
Air
Saturated
20oC
RH = 100%
0.341 kg H2O
*
pH 2 O
*
= 0.6153 psia
= 4.241 kPa
pH 2 O
pH 2 O = 4.241 (.75) = 3.18 kPa
System: overall
Step 5: Basis: 0.341 kg H2O
Step 6 and 7: Unknowns:
(0.341/18) kg mol
F, P
Balances: air, H 2O
Steps 8 and 9: Balances are in kg mol
Air:
3.18 ⎞
H 2O: F ⎛
⎝ 100 ⎠
V=
⎫
⎪
⎪ Solve anytwo toget
⎬
⎪ F = 0.804 kg mol
0.341 ⎛ 2.34 ⎞ ⎪
=
+P
⎝ 275 ⎠ ⎭
18
0.341
18
⎛ 100 − 3.18 ⎞
⎛ 275− 2.34 ⎞
=P ⎝
F⎝
⎠
275 ⎠
100
Total: F
= P+
0.804 kgmol 8.314 303
3
= 20.3 m at 30°C and 100 kPa
100
11–4
= 0.3388 psia
= 2.34 kPa
Solutions Chapter 11
11.1.6
Basis: Air at T = 66 °F and 21.2 psia; assume V is constant.
(a) Initial
pair ,i V = n air RT 6 6 F
Final
Assume all of the water evaporates, and check later on to see if the assumption is true.
pair , f + pH 2 O = p t so pair , f = pt − p H 2 O
pair , f V = n air RT 1 8 0  F
The material balance is simple: all the initial air = final air.
pair , i
pt − p H 2 O
=
66 + 460 526
21.2
=
=
180+ 460 640 33.0− pH 2 O
pH 2 O = 7.20 psia
Since p* = 7.51 psia, all of the water can evaporate and the air will not be saturated.
(b) Basis: 1 lb H 2O (0.0555 lb mol H 2O )
At the final condition
pH 2 O
pt
=
7.20 n H 2 O 0.0555
=
=
33.0
nt
nt
so n = 0.254 lb mol
nRT ⎛⎜ 0.254 ⎞⎟ ⎛ 10.73 ⎞ ⎛⎜ 640 ⎞⎟
3
= 53.0 ft
=
⎝
⎠
⎝
⎠ 33.0 ⎝
⎠
p
0.17 lb H2 O
1lbH 2O
= 0.254 - 0.0555 = 0.1985 lb mol
=
lb air
0.1985(29)
V=
(c) n air
11–5
Solutions Chapter 11
11.1.7
Steps 1, 2, 3, and 4:
pt = 103.0 kPa
F (kg mol)
P (kg mol)
Air: 22oC
RH = 50%
Air: 72oC
RH = 80%
W = 200 kg mol H2O
18
Data for p*
from the steam tables:
T (°C)
p* (kPa)
22°C (295K)
72°C (345K)
2.622
33.77
These calculations provide the composition of F and P.
In
kPa
Out
kPa
pH 2 O = 0.50 (2.60)
=
1.31
pH 2 O
pair = 103.0 - 1.31
=
101.69 pair
= 102.0 - 27.0
=
103.0
pt
Step 5: Basis:
= 0.80 (33.77) = 27.02
= 76.0
pt
=103.0
W = 200 kg H 2O (1 hour)
Steps 6 and 7: Unknowns are: F, P
Balances are: H2O, air
Steps 8 and 9: This is a steady state process without a reaction
Balances
In
Out
H 2O
F (1.311 0 3) + 2 0 01 8(1.00) = P ( 2 7. 01 0 3)
air:
F (1 0 1. 71 0 3)
F = 32.86 kg mol
Bone dry air:
Step 10:
= P( 7 61 0 3)
P = 43.95 kg mol
F (1.071 0 3) = 32.45 kg mol
Check using total balance
Weight in kg of dry air used = 32.45 (29 = 941 kg
11–6
P ( 7 61 0 3) = 32.43kgmol
Solutions Chapter 11
11.1.8
(a)
Entering
Leaving
80°F = 26.8°C
70°F = 21.0°C
p* at 26.8°C = 26 mm Hg
p* at 21.0°C = 19 mm Hg
Rel. Humidity =
(b)
p
5 mm Hg
100
=
p* 26 mm Hg
= 94.8%
Entering
Leaving
V
p
=
Vtot p tot
Vol fr. =
5 mm Hg
100 = 0.676%
740 mm Hg
H 2O:
Air:100 - 0.0676 = 99.324%
(c)
p 18 mm Hg
100
=
p* 19 mm Hg
= 19.2%
Vol fr. =
H 2O:
Rel. Humidity =
p
p tot
18 mm Hg
100 = 2.43%
740 mm Hg
Air: 100 - 2.43 = 97.57%
Entering
Leaving
Basis: 1 lb mol at 80°F, 740 mm Hg
Vol % = mol %
Basis: 1 lb mol at 70°F, 740 mm Hg
lb
wt %
H 2O: (0.00676)(18)
0.12
0.4
H 2O: (0.0243)(18)
0.44
1.53
Air: (99.324)(29)
28.80
28.92
99.6
100.0
Air: (0.9757)(29)
28.30
28.74
98.47
100.00
(d)
lb
wt %
Entering
Leaving
Basis: 1 lb mol at 80°F, 740 mm Hg
Basis: 1 lb mol at 70°F, 740 mm Hg
Actual: pvap = 5 mm Hg
Actual: pvap = 18 mm Hg
pvapor-free gas = 740 – 5 = 735 mm Hg
pvapor-free air = 740 – 18 = 722 mm Hg
11–7
Solutions Chapter 11
Humidity =
n H 2O
n dry air
=
g mol H 2O
18 g
1 g mol dry air
g mol dry air 1 g mol H 2O
29 g
p H 2O
p dry air
Entering
Humidity =
(e)
Leaving
5 18
g H 2O
= 4.22×10-3
735 29
g dry air
18 18
g H 2O
= 0.0155
722 29
g dry air
Basis: 1000 ft3 of mixture at 740 mm Hg and T
Entering
Leaving
Vol % H2O = 0.676 %
Vol % H2O = 2.43 %
Vol H2O = (0.00676)(1000 ft3) = 6.76 ft3
Vol H2O = (0.0243)(1000 ft3) = 24.3 ft3
@ 26.8 °C and 740 mm Hg
@ 21.0°C and 740 mm HG
3
3
6.76 ft 1 lb mol 492°R 740 mm Hg 18 lb
3 540°R 760 mm Hg lb mol
359 ft
24.3 ft 1 lb mol 740 mm Hg 492°R 18 lb
3 760 mm Hg 530°R lb mol
359 ft
= 0.30 lbH 2 O / 1000 ft 3
(f)
at 80 F and 740 mm Hg
= 1.10 lb H2 O / 1000 ft 3
at 70°F and 740 mm Hg
Entering
Leaving
0.30
0.99324
1.10
0.9757
3
3
= 0.302 lbH 2 O / 1000 ft vapor free air
(g)
= 1.127lb H2 O / 1000 ft vapor freeair
Basis: 800,000 ft3/day at 80°F, 740 mm Hg
Entering
Leaving
Vol dry air = 794, 600 (
Vol % air = 99.324 %
530
)
540
= 779,900 ft3 at 70°F
Vol dry air = (800,000)(0.99324)
11–9
Solutions Chapter 11
= 794,600 ft3
Vol % = 97.57 %
Total vol =
Vol H2O vapor = (800,000)(0.00676)
= 5400 ft3 at 80°F, 740 mm Hg
5400 ft
3
779, 900
3
= 799, 300 ft
0.9757
Vol H2O vap = 799,300 – 779,900
740 492
760 540
= 19,400 ft3 at 70°F, 740 mm Hg
= 4790 ft3 at SC
19, 400
492 740
= 17, 535 at SC
530 760
Water evaporated = 17,535 – 4,790 = 12,745 ft3 at SC
3
12, 745 ft 1 lb mol 18 lb
= 639 lb H 2O/day
3 lb mol
359 ft
Part g could also be worked using part f and the volume of dry air:
Entering
Leaving
794,600 ft3 dry air
779,900 ft3 dry air
0.302 lb H2O/1000 ft3 dry air (see f)
1.127 lb H2O/1000 ft3 dry air (see f)
794, 600 0.302
= 240 lb H 2O
1000
779, 900 1.127
= 879 lb H 2O
1000
Water evaporated = 879 – 240 = 639lb H2 O / day
11.2.1
a.
Dew point = 10ºC
b.
%RH =
c.
H =
p H2O
p H2O
pair
psat
=
100 =
p10o C
p27o C
100 =
1.27 kPa
100 = 38%
3.356 kPa
1.27
18
= 0.79 kg H 2O / kg air
101.3 − 1.27 29
11–10
Solutions Chapter 11
11.2.2
When the air is saturated.
11.2.3
Draw a horizontal line until it intersects with the saturation curve. The
temperature at the intersection is the dew-point temperature.
11.2.4
When the air is saturated (100% relative humidity).
11.2.5
The two temperatures are approximately equal at atmospheric temperatures and
pressure.
11.2.6
If by “air” is meant wet air, H =
0.02
lb H 2O
= 0.204
0.98
lb dry air
From the SI humidity chart
a.
Dew point = 25.2ºC
b.
%RH ≅ 59%
11.2.7
From the SI chart at the intersection of TDB = 30o C and RH = 65%
H = 0.0174 kg H2O / kg dry air
11–11
Solutions Chapter 11
11.2.8
Let A = alcohol and C = CO2. MW of A = 46; MW of C = 44.
At 40ºC, pA* = 134.26 mm Hg or 17.90 kPa
a.
pA = 0.10 (100) = 10 kPa
H =
(46)(10)
= 0.116 g A/g C
(44)(100 − 10)
p
10
(100) =
(100) = 55.9%
p*
17.90
b.
%RS =
c.
Cs = 1.00 + 1.88(H) = 1.00 + 1.88 (0.116) = 1.22 J/(K) (g C)
d.
V̂' = 2.83 ×10−3 TK + 4.56 ×10−3 (H )
= (2.80)(10-3)(313.15) + (4.56)(10-3)(0.116) = 0.877 m3 / kg C
c.
At saturation at 40ºC
H =
(46)(17.90)
= 0.228 g A / g C
(44)(100 − 17.90)
V̂' = (2.80)(10-3) (313.15) + 4.56 × 10-3 (0.228) = 0.878 m 3 / kg C
11.2.9
They are almost parallel to each other.
11–12
Solutions Chapter 11
11.2.10
Condenses water from the air in humid climates.
11.2.11
a.
From Humidity Chart, where tDB = 90ºC (194ºF) and tWB = 46ºC (115ºF) H =
0.049 kg H2O/kg air. Upon cooling to 43ºC (109ºF), no condensation occurs,
therefore H is constant.
0.049 kg H 2O 29 kg air 1 kg mol H 2O
kg mol H 2O
= 0.079
1 kg air
1 kg mol air 18 kg H 2O
kg mol air
b.
c.
⎛ 273 + 43 ⎞
Final pressure = 100 ⎝
= 87.1 kPa
273 + 90 ⎠
At saturation: 0.079 =
p*H2O
*
87.1 − p H2O
; solving p*H2O = 6.38 kPa
At the dew point, the vapor pressure of pure water is equal to 6.38 kPa. Dew
point = 37°C (99°F)
The same answer can be obtained by proceeding to the dew point at constant H on
a humidity chart for the correct pressure.
11–13
Solutions Chapter 11
11.2.12
Basis: 1 lb dry air
Data from the humidity chart.
Initial state:
H = 0.0637 lb H2O/lb dry air
TDB = 180o F and TWB = 120o F
Hsaturated ≅ 120 Btu / lb dry air
δHdeviation ≅ −1.5 Btu / lb dry air
H = 118.5 Btu/lb dry air
Final state
TDB = 115o F,
TWB = ?, H = 0.0657 lb H 2O / lb dry air
Hsaturated ≅ 101 Btu / lb dry air
ΔH = 101 – 118.5 = −17.5 Btu / lb dry air
11.2.13
From the SI psychometric chart at 29ºC and 40% relative humidity read
TWB = 19.3ºC
Assuming the liquid water is supplied at a temperature not much different than the
exit temperature of the air stream, the evaporative cooling process follows a line of
constant wet-bulb temperature, which is the lowest temperature that can be obtained on
an evaporative cooler. That is,
Tmin = TWB = 19.3o C
11–14
Solutions Chapter 11
11.2.14
Basis: 1 lb bone dry air (BDA)
Basis: 100 m3
Data from psychometric chart (BDA = bone dry air).
a.
@ 1, Dew point = 23°C
b.
@ 1, humidity = 0.018 kg H 2O / kg dry air
@1, Relative humidity =
c.
d.
ΔH = Q + W
H 0.018
=
(100 = ) 54.5%
H 2 0.033
W=0
Q = ΔH 2 – ΔH 1 = (141.5 – 3.6)
kJ
kJ
kJ
– (79.0 – 0.3)
= 59.2
kg BDA
kg BDA
kg BDA
V 100 m 3 entering air kg BDA
m= ˆ =
= 112.36 kg BDA
V
0.89 m3
Q=
e.
59.2 kJ 112.36 kg BDA
= 6, 651.7 kJ
kg BDA
Adiabatic cooling by evaporation yields a saturated humidity of 0.041 kg
H2O/kg BDA:
(0.041 − 0.018) kg H 2O 112.36 kg BDA 2.58 kg H2 O
=
3
100 m
kg BDA
110 m3 air
f.
Texit = 37°C
11–15
Solutions Chapter 11
11.3.1
a.
The humidity of the entering air at 225ºF DB and 110ºF WB is obtained from the
humidity chart.
Humidity = 0.031
lb H 2O
lb dry air
Assume the exit air to be saturated at 125ºF.
Humidity = 0.0955
b.
lb H 2 O
lb dry air
Basis: 1 hr
10 tons 1 day 2000 lb
= 835 lb/hr
day 24 hr ton
Water in = (0.1) (835) = 83.5 lb/hr
Water out =
(0.9) (835) lb dry grain
1 lb H 2O
= 7.59 lb/hr
99 lb dry grain
lb H2O removed/hr = water in – water out = 83.5 – 7.59 = 75.9 lb H 2 O / hr
c. Product output = (0.9) (835)
d.
lb H 2 O
lb dry grain
+ 7.59
hr
hr
24 hr = 18, 200 lb / day
1 day
7.59 lb H 2O/hr
lb BDA
= 1175 lb BDA/hr
=
(0.0955 – 0.0310) lb H 2O/lb BDA
hr
Q = ΔH = Enthalpy out – Enthalpy in
= ΔHair + ΔHdry grain + ΔHwater
= 1175
+
lb BDA
Btu
Btu
136.5
– (92.25 – 2.25)
hr
lb BDA
lb BDA
(0.9) (835) lb dry grain 0.18 Btu (110 – 70) °F
hr
(lb) (°F)
11–16
Solutions Chapter 11
+
–
7.59 lb H 2O 1.0 Btu (110 – 32) °F
hr
(lb) (°F)
83.5 lb H 2O 1.0 Btu 2 (70 – 32) °F
hr
(lb) (°F)
4
= 5.46 × 104 + 0.54 × 104 + 0.059 × 104 – 0.32 × 104 = 5.74 × 10 Btu / hr
11.3.2
Assume in this problem
1.
2.
3.
4.
Steady operating conditions
Dry air and water vapor are ideal gases
ΔKE = ΔPE = W = 0
The mixing is adiabatic (Q = 0)
Data from the humidity chart:
Stream 1:
H1 = 110.3 kJ/kg dry air
H = 0.0272 kg H2O/kg dry air
Stream 2:
H2 = 50.9 kJ/kg dry air
H 2 = 0.0130 kg H2O/kg dry air
The specific humidity and the enthalpy of the mixture can be determined from mass and
energy balances for the adiabatic mixing of the two streams:
Basis: 1 kg dry air
Total Mass balance: 8 + 6 = 14 kg total
Water mass balance:
8(0.0272) + 6(0.0130) = 14 (Hmixture)
b.
H = 0.0211 kg H2O/kg dry air
11–17
Solutions Chapter 11
Energy balance (ΔH = 0)
8(110.3) + 6(50.9) = 14 (Hmixture)
H = 84.8 kJ/kg dry air
These two properties fix the state of the mixture. Other properties of the mixture are
determined from the psychometric chart.
a.
T = 30.7o C
c.
RH = 75.1%
11.3.3
Data from Humidity Chart
a.
Hair in = 0.01813 lb H 2 O / lb air
b.
Hair out = 0.031 lb H2 O / lb air
H2O picked up = 0.031 – 0.01813
= 0.0129 lb H2O/lb air
Energy Balance:
ΔHair out – ΔHair in = ΔHwater in – ΔHwater out
Basis: 1 lb dry air
Ref. temp = 85ºF
ΔHH 2 O out
ΔHair out


 
90–85
) + 0.45 Btu 0.031 lb (90–85)
0.24 Btu 1 lb (
( lb)(°F)
( lb)(°F)
11–18
Solutions Chapter 11
ΔHH2O out
ΔHair + H2O in
ΔHH2O in – ΔHH2O out

+1040 Btu 0.0129 lb
lb



–0

1 Btu m lb (102–89 )
=
( lb)(°F)
m = 1.136 lb H2O/lb air or 0.915 lb air / lb H2 O
c. % H 2 O vaporized = 0.0129 100 = 1.14%
1.136
11.3.4
Steps 1, 2, 3, and 4
The process will be assumed to be a steady state, open, continuous one with 5L
flowing in (in the material balance the air is invariant). All of the data for the stream
flows and compositions have been placed on the figure. The lungs will be the system.
P L air, 37oC
5 L air, 25oC
p
H2O
pair
ttot
Lungs
0.95 kPa
96.05 kPa
97.0 kPa
p
H2O
pair
ttot
= p*
H2O
6.27 kPa
90.73 kPa
97.0 kPa
p*25o C = 3.17 kPa
p H2O = 3.17(0.3) = 0.95 kPa
Step 5:
Basis: 5 L air → 1 min
Steps 6, 7, 8, and 9:
Material balances
In: n air =
96.05 kPa
1 atm
5L
1(g mol)(K)
= 0.194 g mol
101.3 kPa 298 K 0.08206(L)(atm)
11–19
Solutions Chapter 11
⎛ 0.95 ⎞
−3
n H2O ⎜
⎟ (0.194) = 1.92 ×10 g mol
95.05
⎝
⎠
Out:
n air = 0.194 g mol
⎛ 6.27 ⎞
n H2O = 0.194 ⎜
⎟ = 0.0134 g mol
⎝ 90.73 ⎠
Energy balance
The energy balance reduces to Q = ΔH. The reference temperature will be 25ºC.
Assume the increase in water vapor comes from water vaporized at 37ºC
ˆ = 2414.3 kJ / kg).
(ΔH
vap
ΔHin: Both the air and water enter at 25ºC so ΔH = 0 for both the input streams.
ΔHout:
ΔHair = 0.194∫
310
298
(27.2 + 0.0041 TK )dT = 66.2 J
310
ΔH H2O = 1.92 ×10−3 ∫+ (34.4 0.0063 TK )dT
298
+ (0.0134 – 1.92 × 10-3) (2414.3)(18) = 499 J
Q = ΔH = 499 + 66.2 = 565 J/min or 33.9 kJ / hr
11–20
Solutions Chapter 11
11.3.5
Basis: 180 kg/hr of product
Material Balance
Dry product =
a.
92 kg dry P 180 kg P
= 166 kg dry P/hr
100 kg P
1 hr
Water removed from solid:
In =
1.25 kg H 2O 166 kg/hr dry P
– (0.08) (180) kg/hr
1.00 kg dry P
= 207.5 – 14.4 = 193 kg H 2 O / hr
Basis: 1 BDA
Water picked up in dryer
out
in
0.0571 kg H 2O
kg BDA
out, 53°
–
0.0083 kg H 2O
kg BDA
in, 21°
= 0.0488
kg H 2O
kg BDA
193 kg H 2 O
kg BDA
=
= 3955 kg BDA / hr
hr
0.0488 kg H 2O / kg BDA
b.
Energy balance
Air:
ΔH (kJ/kg BDA)
air in (21ºC, 52% RH ):
58.2
air out (53ºC, 60% RH):
219.7
ΔH =
161.5 kJ/kg BDA
Basis: 1 hr
ΔH = (161.6) (3955) = 6.387 × 105 kJ
Solid (ref. 21ºC)
11–21
Solutions Chapter 11
solid out: ΔH =
43
21
3
0.18 cal 4.184 J 10 (43 – 21) °C
C pdT = =
= 16.57 kJ/kg P
(g) (°C) cal 10 3
solid in: ΔH = 0 (because of reference temperature)
Basis: 1 hr
ΔH = (16.57) (180) = 2983 kJ
If the dryer and reheater are insulated, then for the system
Q – W = ΔH and
W=0
5
Qreheater = 2983 + 6.387 × 105 = 6.42 × 10 kJ / hr
11.3.6
Initial air: TDB = 38ºC, TWB = 27ºC,
H1 = 0.0175 kg/kg dry air
Air from scrubber: T = 24ºC, RH = 100%, RH2 = 0.0188 kg/kg dry air
Heated to 93ºC:
H3 = 0.0188
From drier: TDB = 49ºC,
H4 = 0.0377
(1000 kg/hr) (0.05) = 50.0 kg H2O to be evaporated
50.0/(0.0377 – 0.0188) = 2650 kg dry air/hr
2650 (0.0377) = 100 kg H2O/hr
⎛ 2650 100 ⎞ 22.4 273 + 49
V4 ⎜
+
= 2560 m3 at 49ºC and 1 atm
⎟
18 ⎠
273
⎝ 29
Heat supplied:
Q = [(2650) (1.00) + 100 (0.200] (93 – 24) = 1.84 × 105 kJ/hr
Water at 93ºC has p* = 79.4 kPa
11–22
Solutions Chapter 11
H3 =
(18) ( 79.4) = 2.25 kg H O/kg dry air
2
( 29) (101.3–79.4 )
[(0.0188/2.25)] (100) = 0.83%
RH air from heater
Answers:
a.
1.
2.
3.
4.
H = 0.0175
H = 0.0188
H = 0.0188
H = 0.0377
b. 1.
2.
3.
4.
42%
100%
0.83%
47%
c. 2650 kg dry air/hr
d. 2560 m3/hr
e. 1.84 × 105 kJ/hr
11.3.7
Step 5:
Basis: 1 hr
Assume:
1.
2.
3.
4.
ΔPE = ΔKE = W = 0,
No reaction occurs,
open, steady state process,
ideal gas behavior
Data:
Entrance air (in)
Ĥ (kJ/kg dry air)
H (kg H2O/kg dry air)
V̂(m3 / kg dry air)
50.5
0.00587
0.88
Assume the properties of the wet penicillin are the same as those of water. ΔHvap at 34ºC
= 2420.25 kJ/kg water. Let P = kg dry penicillin per hour, and F = kg dry air/hr.
Steps 3 and 4
11–23
Solutions Chapter 11
Moist air
TDB= 34oC
TWB = ?
Penicillin
Penicillin
Dryer
34oC
34oC
P
in
ω = 0.80
ωPout = 0.50
Moist air
3300 m3/hr
TDB=34oC
TWB = 17oC
air in =
4500 m3 1 kg dry air
= 5114 kg dry air
0.88 m3
Steps 6 and 7:
The exit conditions for the air are not known, but the H and H are related on the
humidity chart hence only one is unknown. P (dry penicillin is unknown).
The balances are water, dry air, and dry penicillin.
Steps 8 and 9
TWB ( o C)
Assume:
20
21
22
⎛ kg H 2O ⎞
H ⎜
⎟
⎝ kg dry air ⎠
0.009
0.010
0.0115
Ĥ(kJ / kg dry air)
57.0
60.6
64.0
Water balance:
5114 (0.009 – 0.00587) = water evaporated = 16.0 kg
5114 (0.010 – 0.00587) = 21.1 kg
5114 (0.0115 – 0.00587) = 28.8 kg
Energy balance:
5114 (57.0 – 50.5) / (2420.25) = 13.73 kg
5114 (60.5 – 50.5) / (2420.25) = 21.1 kg
5114 (64.0 – 50.5) / (2420.25) = 28.5 kg
The solution is very sensitive to the values read from the psychometric chart. Assume the
final TWB ; 21 or 22o C .
a.
Water evaporated = 28.5 kg
11–24
Solutions Chapter 11
b.
c.
ΔH (21ºC) = 51,140 kJ/kg dry air ⇒ 51,140 kJ / hr
Steps 6 and 7
Unknowns
P, exit TWB
Equations
air, water
Steps 8 and 9
Water balance on air gives water evaporated (5114) (0.010-0.00587) = 21.1 kg H2O
Energy balance: (60.5 – 50.5) (5114) + 21.1 (2420.25) = 0
0.80 → 0.60
Use 21ºC
⎛ 0.80 0.50 ⎞
21.1 kg H2O evap = P ⎜
−
3P
⎟ =
⎝ 0.20 0.50 ⎠
P=
21.1
= 7 kg / hr
3
11.3.8
R
80oF
120oF
Hin=0.0031
lb H2O/lb DA
Hout=0.0292
lb H2O/lb DA
H = 0.0075
lb H2O/lb DA
Basis = 2000 lb dry waste (DW)
Comp.
in %
out
waste in =
H2 O
63.4
22.7
2000 0.773 100
= 4, 220 lb
36.6
DW
36.6
77.3
DW out = 2,000 lb
→
11–25
Solutions Chapter 11
100.0
100.0
H2O evap.
= 2,220 lb
Basis: 1 lb dry air
Air in, 80ºF, 29.92 in. Hg, and a wet bulb temperature = 54ºF:
Hin = 0.0031 lb H2O/lb dry air Ĥ'in = 22.62 − 0.18 =22.54 Btu / lb dry air
Air out, 120ºF, 29.92 Hg, and a wet bulb temperature = 94ºF:
Hout = 0.0292 lb H2O/lb dry air
Ĥ ' = 61.77-0.4 = 61.37 Btu/lb dry air
Overall
b.
Hout
=
0.0292
Hin
=
0.0031
H2O evaporated =
0.0261
air used per ton dust =
lb H2O/lb dry air
lb H2O/lb dry air
(1 + 0.0031)
2220
0.0261
= 85, 400 lb moist inlet air
per ton DW
Air out =
1.00
2220
0.0261
0.0292
(29)
= 2,930 lb mol dry air out
2220
0.0261
(18)
= 138 mol H2O out
3068 total mol out
c.
3068 359 580
= 1,300, 000 ft3 wet air exiting/ton DW
492
Air Recirculated
R = lb air added at mixing point before kiln
Water balance:
(0.0031) (1) + (0.0292) (W) = (0.0075) (1+W)
0.0292 W
R=
a.
=
0.0075
-0.0075 W
-0.0031
0.0217 W
0.0044
0.0044
= 0.203 lb
0.0217
Recirculation =
0.203 100
= 16.8% of gas out is recirculated
1.203
11–26
Solutions Chapter 11
11.3.9
Adiabatic operation removes 425,000 Btu/hr from the process.
Basis: 1 hr
The energy balance is ΔH = 0 overall.
a.
For the process:
TWB = 100o F⎫⎪
⎬ RH = 1.3%
TDB = 234o F ⎪⎭
b.
c.
Temperature leaving the cooler is 61ºF (dew point constant H)
d.
ΔH = (72 − 27)Btu
=
/ lb BDA 45
425, 000 Btu 1 lb BDA
= 9450 BDA / hr
hr
45 Btu
11–27