INSTRUCTOR’S MANUAL
TO ACCOMPANY
ELEMENTARY PRINCIPLES OF
CHEMICAL PROCESSES
Third Edition
Richard M. Felder
Ronald W. Rousseau
with assistance from
Matthew Burke, Swapnil Chhabra, Jun Gao,
Gary Huvard, Concepción Jimenez-Gonzalez, Linda Holm,
Norman Kaplan, Brian Keyes, Amit Khandelwal,
Stephanie Manfredi, Janette Mendez-Santiago, Amy Michel,
Dong Niu, Amitabh Sehgal, James Semler,
Kai Wang, Esther Wilcox, Jack Winnick,
Tao Wu, Jian Zhou
3
INSTRUCTOR’S MANUAL
to accompany
ELEMENTARY PRINCIPLES
OF CHEMICAL PROCESSES
THIRD EDITION
RICHARD M. FELDER
North Carolina State University
RONALD W. ROUSSEAU
Georgia Institute of Technology
JOHN WILEY & SONS
New York Chichester Brisbane
Toronto
Singapore
4
CONTENTS
Notes to the Instructor
iv
Section/Problem Concordance
vi
Sample Assignment Schedule I
ix
Sample Assignment Schedule II
x
Sample Responses to a Creativity Exercise
xi
Transparency Masters
xvi
Compressibility charts
Cox vapor pressure chart
Psychrometric chart – SI units
Psychrometric chart – American engineering units
Enthalpy-concentration chart: H2SO4-H2O
Enthalpy-concentration chart: NH3-H2O
xvii
xxi
xxii
xxiii
xxiv
xxv
Problem Solutions
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12 (Case Study 1)
Chapter 13 (Case Study 2)
Chapter 14 (Case Study 3)
2-1
3-1
4-1
5-1
6-1
7-1
8-1
9-1
10-1
11-1
12-1
13-1
14-1
iii
NOTES TO THE INSTRUCTOR
Problem Assignments
To aid in the structuring of the course, we have provided a section/problem concordance on pp.
vii–ix and two sample assignment schedules on pp. x–xi.
We believe there is far too much material in the textbook to attempt to cover in one semester or
two quarters. The sample assignment schedules therefore cover only Chapters 1-9, and within those
chapters some topics are omitted (e.g. liquid-liquid equilibrium and adsorption on solid surfaces in
Chapter 6 and mechanical energy balances in Chapter 7). The missing sections can substitute for
Chapters 2 and 3 in classes where the content of those chapters has been well covered in chemistry
and/or physics courses), and Chapters 10 (computer flowsheeting) and 11 (transient balances) may
be assigned for extra credit or covered in honors sections or subsequent courses in the curriculum.
We will discuss the case studies in Chapters 12-14 separately.
In the sample assignment schedules, we have designated a number of “bonus problems” which
may be ignored, assigned for extra credit as add-ons to the regular assignments (those will be long
assignments), or assigned for extra credit in lieu of some of the problems in the regular assignments.
The bonus problems may be assigned as individual exercises or students may work on them in pairs.
We have had good experience with the latter approach.
Creativity Exercises
The creativity exercises in the text are designed to stimulate divergent thinking and to induce the
students to think about course material from different perspectives. We have used such exercises
both as extra-credit assignments to individuals or pairs of students or as the foci of in-class
brainstorming sessions. In all cases, we have found that they invariably lead to innovative, clever,
and often amusing ideas; they give students who are by nature creative an opportunity to demonstrate
their talent and they help other students develop creative problem-solving skills; and the students
usually enjoy doing them.
There are no “right answers” to such exercises, and so we have not included solutions in this
manual. However, to provide an idea of the kind of things that students come up with, we have
included on pp. xi–xv a collection of student responses to a creativity exercise given by one of the
authors in a junior course on fluid dynamics.
Transparency Masters
Several of our colleagues have suggested that we include in the text enlarged versions of some of
the figures, such as the psychometric charts, which are difficult to read in reduced format. We have
chosen not to do so, since whether they are inserts or fold-outs such charts tend to be ripped off (one
way or another) or otherwise lost. Instead, we have included in this manual, beginning on p. xvi,
large versions of some of the most commonly used figures. These masters can be used to make
transparencies for lectures; they can also be copied and distributed to the students for use in solving
problem.
Case Studies
The case studies comprise Chapters 12 through 14 of the text. In them, we seek to (1) illustrate
the development of complex chemical processes from basic principles, and to provide a broad
process context for the text material; (2) raise questions that require students to think about topics
strictly beyond the scope of the first course, and to seek out sources of information other than the
text; (3) accustom the students to team project work. We do not organize the activities of case study
iv
teams, nor do we assign team leaders, although we suggest to the students that they do so. This is a
risk, and sometimes it is necessary to step in and get a laggard group started. However, letting the
teams shape their own working relationships and structure their own activities usually is an
enlightening experience to the students.
Problem Solutions
The detailed solutions to 634 of the 635 chapter-end problems constitute the principal content of
this manual. (The solution to the last problem of Chapter 10 is left as an exercise for the professor,
or for anyone else who wants to do it.)
With few exceptions, the conversion factors and physical property data needed to solve the endof-chapter problems are contained in the text. It may be presumed that conversion factors for which
sources are not explicitly cited come from the front cover table; densities, latent heats, and critical
constants come from Table B.1; heat capacity formulas come from Table B.2; enthalpies of
combustion gases come from Tables B.8 and B.9; vapor pressures come from Table B.4 or (for
water) Table B.3; and enthalpies, internal energies, and specific volumes of water at different
temperatures and pressures come from Tables B.5-B.7.
As the reader of the text may have discovered, we believe strongly in the systematic use of the
flow charts in the solution of material and energy balance problems. When a student comes to us to
ask for help with a problem, we first ask to see the labeled flow chart: no flow chart, no help. Other
instructors we know demand fully labeled flow charts and solution outlines at the beginning of every
problem solution, before any calculations are performed. In any case, we find that the students who
can be persuaded to adopt this approach generally complete their assignments in reasonable periods
of time and do well in the course; most of those who continue to resist it find themselves taking
hours to do the homework problems, and do poorly in the course.
Posting Problem Solutions: An Impassioned Plea
It is common practice for instructors to photocopy solutions from the manual and to post them
after the assignments are handed in, or, even worse, to distribute the solutions to the students. What
happens then, of course, is that the solutions get into circulation and reincarnate with increasing
frequency as student solutions. After one or two course offerings, the homework problems
consequently lose much of their instructional value and become more exercises in stenography than
engineering problem solving.
In the stoichiometry course particularly, the concepts are relatively elementary: the main point is
to teach the students to set up and solve problems. A great deal of classroom lecturing on concepts
should therefore not be necessary, and a good deal of the class time can be spent in outlining
problem solutions. The burden should be placed on the students to make sure they know how to do
the problems: to ask about them in class, to make notes on solutions outlined on the board, and to fill
in omitted calculations. Besides being pedagogically superior to solution-posting, this approach
should cut down on the ease with which students can simply copy letter-perfect solutions instead of
doing the work themselves.
Errors
A great deal of time and effort has been expended to make the solutions in this manual as free of
errors as possible. Nevertheless, errors undoubtedly still exist. We will be grateful to any of our
colleagues who send us corrections, no matter how major or minor they may be; we will provide an
errata list on the text Web site (http://www2.ncsu.edu/unity/lockers/users/f/felder/public/EPCP.html)
and make the corrections in subsequent printings of the text.
v
SECTION/PROBLEM CONCORDANCE
Key: i-Routine drill
j-Application
k-Longer or more challenging
*-Computer solution required
CHAPTER 2
Section
Problems
2.1-2.3
2.4
2.5
2.6
2.7
1-5 , 6-7
i
j
k
j
8-10 , 11-12 , 13 , 14-15
i
j
j
16-17 , 18-19 , 20 *
j
j
21-28 , 29 *
i
j
i
j
k
j
k
30-31 , 32-37 , 38 , 39-41 , 42-44 *
CHAPTER 3
Section
Problems
3.1-3-2
3.3
3.4
3.5
1-2 , 3-8 , 9 , 10 , 11 , 12-13
i
j
k
j
14-16 , 17-25 , 26 *, 27-31
i
j
k
32 , 33-46 , 47
i
j
k
k
48 , 49-52 , 53 , 54 *
i
j
k
j
k
j
CHAPTER 4
Section
Problems
4.1-4.3a
4.3b-e
4.4
4.5
4.6a, b
4.6c
4.7a-e
4.7f
4.7g
4.8
1 , 2-5
j
k
k
j
k
k
6-20 , 21 , 22 *, 22-25 , 26 , 27 *
j
k
28-30 , 31
j
k
32-34 , 35-38
i
k
j
39-40 , 41 *, 42-45
k
k
46 , 47-48 *
j
k
49-53 , 54-55 *
j
k
56-57 , 58-59
k
k
60-62 , 63 *
i
j
k
j
k
j
k
k
64-65 , 66 , 67 *, 68-73 , 74-76 , 77-78 , 79 , 80 *
i
j
CHAPTER 5
Section
Problems
5.1
5.2a,b
5.2c
5.3a-c
5.4a,b
5.4c
1 , 2-3 , 4
i
j
5-6 , 7-15
j
k
j
k
j
k
16-21 , 22-23 , 24-30 , 31-34 , 35-46 , 47-54
j
k
j
k
k
55-56 , 57 *, 58-60 , 61-62 , 63 *
i
j
k
j
64-65 , 66-69 , 70 , 71-73
j
k
j
74-77 , 78 ,79-83
i
j
k
vi
CHAPTER 6
Section
6.1
6.2, 6.3
6.4a
6.4b
6.4c
6.4d
6.5a,b
6.5c
6.6
6.7
Problems
j
k
j
1-4 , 5 *, 6-8
j
k
k
j
k
j
k
6.9-29 , 30 , 31 *, 32-34 , 35-36 , 39-41 , 42
j
43-44
i
j
k
45-46 , 47-50 , 51
j
k
j
k
j
k
52-57 , 58 *, 59 , 60 , 61-63 , 64 *
j
k
j
k
65-67 , 68-69 *, 70 , 71-73
i
j
k
74 , 75-80 , 81-83
i
j
k
84-85 , 86 , 87
j
k
j
88-91 , 92 , 93-97
j
k
98-99 , 100-101
CHAPTER 7
Section
Problems
7.1, 7.2
7.3
7.4
7.5
7.6
7.7
1-2 , 3 , 4-6 , 7-8
i
j
9 , 10-11
i
j
i
j
12-13 , 14-17 , 18 , 19-23
j
k
24-28 , 29
j
k
j
k
j
k
k
30-38 , 39 , 40-41 , 42-44 , 45-48 , 49-50 , 51 *
j
k
52-56 , 57-58
i
j
i
j
CHAPTER 8
Section
Problems
8.1-8.3b
8.3c
8.3d
8.3e
8.4a
8.4b
8.4c
8.4d
8.4e
8.5
1-4 , 5-16
i
17-18
j
k
j
k
19-25 , 26 , 27-31 , 32
j
k
33 , 34 *
i
j
35 , 36-41
i
j
42 , 43-44
j
k
j
k
k
k
45-53 , 54 , 55-56 , 57-65 , 66 *, 67-68
i
j
k
j
69-71 , 72-73 , 74 , 75
i
j
k
76-77 , 78-79 , 80
i
j
k
j
k
k
j
81-82 , 83-87 , 88-90 , 91 , 92 , 93-94 *, 95-99
i
j
CHAPTER 9
Section
Problems
9.1
9.2
9.3, 9.4
9.5a
9.5b
9.5c
9.6a
9.6b
1-2 , 3-4
i
5-6
I
j
7-9 , 10
j
k
j
k
j
k
11-17 , 18 *, 19-21 , 22-23 , 24 *, 25-30
k
k
j
k
31-34 , 35 *, 36 , 37-38
j
k
39-44 , 45-47
j
k
j
k
48-50 , 51 , 52-56 , 57-61
j
k
k
k
62-63 , 64-67 , 68 *, 69-70
i
j
vii
CHAPTER 10
Section
Problems
j
10.11-4 , 5
10.2, 10.3
k
k
j
k
6 , 7 *, 8-14 *
CHAPTER 11
Section
Problems
j
k
j
j
j
k
j
11.1, 11.2 1-2 , 3 , 4 , 5 *, 6-9 , 10 , 11-14 , 15-19
j
k
11.320-26 , 27-30
k
viii
SAMPLE ASSIGNMENT SCHEDULE I
Assignment
Read (Ch., Sect.)
Problems Due
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
—
1; 2-2.6
2:3, 6, 9, 12
2.7
2:14, 17, 19, 22
3-3.3
2:30, 33; 3:3, 6
3.4-3.5
3:13, 16, 22, 29
4-4.3a
3:32, 37, 40, 49
4.3b-4.3e
3:47; 4:2, 4, 7
4.4
4:10, 11, 14
4.5
4:18, 25, 28
—
4:26, 29
TEST THROUGH SECTION 4.3
4.6a-b
4:32, 39
4.6c, 4.7a-d
4:37, 43
4.7e-g
4:49, 53
4.8
4:56, 65
5-5.2b
4:69, 73; 5:3
5.2c-5.3
5:7, 12, 15, 19
5.4
5:25, 26, 59
6-6.2
5:35, 40, 66; 6:1
6.3
6:2, 9, 12, 15
6.4a,b
6:23, 33
6.4c
6:37, 47
6.5
6:53, 66, 74
7-7.3
6:36, 70
TEST THROUGH SECTION 6.4
7.4-7.5
7.1, 6, 9, 10
7.6
7:12, 18, 25
8-8.3b
7:24, 28, 30
8.3c-8.3e
7:33, 35; 8:2
—
7:45; 8:5, 6, 8
8.4a-c
8:15, 18, 25
8.4d-e
8:29, 36, 44
—
8:46, 49
8.5
8:51, 55, 73
—
8:61
—
8:79, 81, 86
TEST THROUGH SECTION 8.4d
9-9.4
8: 95, 98; 9:1
9.5a
9:7, 12
9.5b-c
9:14
9.6a-b
9:17
9.6c-d
9:32
—
9:48
—
9:54
—
9:63
ix
Bonus Problems
(*Computer problem)
2.13
2:20*, 36-43
2:38-43; 3.9,11
2:40*; 3.26*
3:53*, 54*
4:22*
4:27*
4:28*, 31, 35-36
4:38, 41*
4:39, 43*
4:47*, 48*
4:54*, 55*
4:58,59
4:60-62, 63*, 67*
4:79-80*; 5:4*, 22-23
5: 31-34, 47-54
5:57*, 61-62, 63*; 6:5*
6:30, 31*, 35, 42
6:51, 58*, 60, 64*
6:68-69, 71-73
6:81, 83, 86-87
6:88-101
7:3, 17
7:29, 39, 42-44
7:49-50, 51*
7:52-58
8:12
8:26, 32, 34*
8:54, 57-59
8:62-68
8:74, 80
8:88-90
8:93-94*
9:9
9:18*
9:22-34
9:35*, 37-38, 45-47
9:51, 57-61, 64-65
9:67-70
SAMPLE ASSIGMENT SCHEDULE II
Assignment
Read (Ch., Sect.)
Problems Due
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
—
1; 2-2.6
2:4, 7, 8, 11
2.7
2:15, 16, 18, 23
3-3.3
2:31, 34; 3:4, 7
3.4-3.5
3:12, 17, 23, 28
4-4.3a
3:32, 39, 43, 50
4.3b-4.3e
3:51; 4:3, 4, 6
4.4
4:9, 12, 15
4.5
4:21, 23, 28
—
4:26, 30
TEST THROUGH SECTION 4.3
4.6a-b
4:33, 40
4.6c, 4.7a-d
4:34, 45
4.7e-g
4:50, 51
4.8
4:57, 64
5-5.2b
4:70, 71; 5:2
5.2c-5.3
5:8, 11, 13, 17
5.4
5:25, 29, 58
6-6.2
5:38, 42, 67; 6:1
6.3
6:6, 8, 10, 17
6.4a,b
6:27, 35
6.4c
6:39, 46
6.5
6:52, 65, 75
7-7.3
6:41, 70
TEST THROUGH SECTION 6.4
7.4-7.5
7:2, 7, 9, 11
7.6
7:13, 18, 22
8-8.3b
7:24, 28, 30
8.3c-8.3e
7:32, 37; 8.1
—
7:47; 8:5, 6, 9
8.4a-c
8:14, 17, 24
8.4d-e
8:30, 37, 43
—
8:45, 50
8.5
8:53, 56, 69
—
8:61
—
8:78, 83, 85
TEST THROUGH SECTION 8.4d
9-9.4
8:96, 97; 9:2
9.5a
9:7, 11
9.5b-c
9:15
9.6a-b
9:16
9.6c-d
9:33
—
9:50
—
9:55
—
9:66
x
Bonus Problems
(*Computer problem)
2.13
2:20*, 36-43
2:38-43; 3.9,11
2:40*; 3.26*
3:53*, 54*
4.22*
4:26, 27*
4: 28*, 31, 35-36
4:38, 41*
4:39, 43*
4:47*, 48*
4:54*, 55*
4:58, 59
4:60-62, 63*, 67*
4:79-80*; 5:4*, 22-23
5: 31-34, 47-54
5:57*, 61-61, 63*; 6.5*
6:30, 31*, 35-36, 42
6:51, 58*, 60, 64*
6:68-69, 71-73
6:81, 83, 86-87
6:88-101
7:3, 17
7:29, 39, 42-44
7:49-50, 51*
7:52-58
8:12
8:26, 32, 34*
8:54, 57-59
8:62-68
8:74, 80
8:88-90
8:93-94*
9:9
9:18*
9:22-34
9:35*, 37-38, 45-47
9:51, 57-61, 64-65
9:67-70
SAMPLE RESPONSES TO A CREATIVITY EXERCISE
The exercise that follows was given to a junior class in fluid dynamics. The students were given
a week, and were told to do the exercise either individually or in pairs. The grading system used is
explained in the statement of the exercise.
Thirty-one individuals and nine pairs submitted responses, for a total of 40 responses from 49
students. Some students found their way to Perry’s Handbook and took ideas from there, which was
perfectly acceptable; many were more inventive, and submitted a wide variety of clever, ingenious,
and humorous responses. The average number of suggested flow measurement techniques was 26;
the high was 53, and the low was 5. A summary of the collected responses with duplicates
eliminated follows the exercise statement.
Exercise
You are faced with the task of measuring the volumetric flow rate of a liquid in a large pipeline.
The liquid is in turbulent flow, and a flat velocity profile may be assumed (so that you need only
measure the fluid velocity to determine the volumetric flow rate). The line is not equipped with a
built-in flowmeter; however, there are taps to permit the injection or suspension of devices or
substances and the withdrawal of fluid samples. The pipeline is glass and the liquid is clear.
Assume that any device you want to insert in the pipe can be made leakproof if necessary, and that
any technique you propose can be calibrated against known flow rates of the fluid.
Come up with as many ways as you can think of to perform the measurement that might have a
chance of working. (Example: insert a small salmon in the pipe, suspend a lure irresistible to
salmon upstream of the insertion point, and time how long it takes the fish to traverse a measured
section of the pipe.) You will get 1 point for every 5 techniques you think of (no fractional points
awarded), up to a maximum of 10 points. Note, however: The techniques must be substantially
different from one another to count. Giving me a pitot tube with 10 different manometer fluids, for
example, will get you nowhere.
Responses
1. Pitot tube.
2. Hot-wire or hot-film anemometer
3. Pass effluent through a venturi meter or orifice meter or nozzle meter or rotameter or ... (no
credit for simply naming the meter if it can’t be easily inserted in the pipeline).
4. Pass effluent into a weir, measure height in notch.
5. Inject dye, measure time for it to traverse a known length.
6. Insert a solid object (such as a balloon, a bucket, a cork, a marble, a bar of Ivory soap, or the 311
book), measure time for it to traverse a known length (or travel alongside it on a bicycle or
moped or pickup truck and note your speed, or attach it to a string on a spool and measure the
rate of rotation of the spool).
7. Insert a series of solid objects, measure rate at which they pass a point (or frequency of collisions
with the pipe wall, or rate of collection on a filter).
xi
8. Inject dye at fixed rate, shine light on the pipe, measure light absorbance downstream (or angle
of refraction or turbidity, or put a sunbather under the pipe and measure his rate of tanning).
9. Measure the energy being consumed by the pump being used to move the fluid.
10. Put a magnet in the flow, measure the magnetic force required to hold it in place (or measure its
velocity along the wall, or have two external magnetic switches triggered by its passage and time
the interval between events, or measure the rate of motion of a compass needle as the magnet
passes).
11. Collect effluent (or a sidestream), measure amount (volume, mass) collected in a known time
interval (or the rate at which the level in the container rises, or the time required to fill a known
volume or to saturate a sponge, or to water a plant or wash a pulp sample, or to saturate a plot of
ground in Ethiopia where they really need it).
12. Direct effluent into a container of salt, see how much dissolves.
13. Direct effluent against a raft in a pond, measure its velocity.
14. Discharge effluent horizontally, letting it fall through a known height, and measure its horizontal
displacement.
15. Discharge effluent horizontally, and measure the force it exerts on a plate.
16. Discharge effluent (or a sidestream) upward, measure height of fountain (or suspend a ping pong
ball at the top, and measure its height)
17. Discharge effluent downward from a flexible hose, and measure height of nozzle above the
ground.
18. Let fluid fill a balloon (or a water bed), measure time required for explosion, or volume increase
in a known time interval.
19. Insert a U-tube at each of two points in the line, use as a manometer (either straight pipe between
points or insert an obstacle to flow, like an orifice or a solid object).
20. Insert paddle wheel (or a water wheel at the outlet), measure rotational speed.
21. Insert propeller, measure rotational speed by counting or automatically.
22. Insert turbine-generator, measure work output (or intensity of light attached to generator).
23. Suspend solid object on a string, measure angle made by string with vertical (or horizontal
displacement of object, or rotation of a lever arm, or angle at which your hand is bent back).
24. Drop in an object denser than the fluid, measure horizontal distance traversed before hitting the
bottom of the channel.
25. Inject from below an object lighter than the fluid (e.g. a bubble), measure horizontal distance
traversed before hitting the top of the channel.
xii
26. Inject from below an object heavier than the fluid, measure its horizontal displacement (or
follow its trajectory using stop-motion photography).
27. Put a flexible fiber (or membrane or easily deformed plate) in the path of the flow, measure its
distension in flow direction, or thickness at which flow is sufficient to break it.
28. Pluck a guitar string in the flow, time its period of vibration.
29. Attach tape to the wall, time its unraveling.
30. Feed effluent into a centrifugal pump or a lobbed-impeller flowmeter, measure rotational speed.
31. Measure height of fluid in a vertical standpipe coming from the top of the pipe.
32. Determine time required for effluent to sink a ship (or to flood out the football coach’s house,
hopefully with some of his players in it).
33. Determine time required for effluent to float a duck out of a well of known depth (or to float an
object of known weight and displacement).
34. Insert a solid object (e.g. a snowball, or a tootsie pop, or Alka Seltzer, or the Wicked Witch of
the West), determine time required to dissolve it (or wear it away, or wash the paint off it).
35. Insert an absorbing object, measure its rate of fluid uptake.
36. Insert solid objects of different sizes, find the one such that the drag force is just sufficient to
initiate motion (or measure rate at which a given object is dragged along the pipe).
37. Measure vibration intensity or amplitude of tube (or noise level, or sound of a bell clapper),
either naturally occurring or after the pipe is struck.
38. Insert an iron bar, measure rate of corrosion.
39. Propel a bullet (or an arrow, or a torpedo, or Nolan Ryan’s fastball) into the pipe outlet, measure
distance it travels before stopping.
40. Determine velocity of immersed submarine moving against (or with) the flow (or Mark Spitz, or
a trout approaching a lure, or a seal approaching food, or a squirrel approaching an acorn, or a
hungry dog approaching a dead rabbit, or a snake approaching a mouse, or a sailor approaching a
mermaid, or a horny male frog approaching erotic pictures of female dancing frogs, or a 311
student after this test approaching free beer).
41. Measure vibrational speed of a trout’s tail swimming against the flow and remaining stationary
(or the rate of flapping of a piece of cloth or the rate of wobble of a nutating disk or the
magnitude of noise generated by “chatterbox” lure).
42. Insert balloon (or piston-fitted cylinder with piston facing upstream, or closed tube with flexible
diaphragm covering opening), measure final gas pressure (or rate of pressure increase or rate of
motion of piston or intensity of whistle if piston drives gas through it).
xiii
43. Insert solid object, measure force required to hold it still (or extension or compression of a spring
or elastic band, or put the object against razor blade and see how long it takes for the blade to
split it).
44. Insert a solid object, determine distance required for downstream wake to disappear.
45. Put in plug, measure force required to hold it in (or distance it travels when it is ejected).
46. Measure the shear force on the pipe wall (with or without a bend in the line), or the extension of
the pipe length due to shear.
47. Measure the rate of heat generation or temperature rise due to friction in the pipeline.
48. Measure rate at which air is drawn through a Buchner funnel (or pitch or intensity at which it is
drawn through a whistle) by the suction created by the flowing fluid.
49. Add heat, measure temperature rise (or rate of evaporation).
50. Insert a hot object, measure its rate of cooling (or a cold object, and measure its rate of heating).
51. Cool, measure temperature drop (or rate of freezing).
52. Pass effluent through a heat exchanger, measure rate of heat transfer.
53. Add an acid or base at a known rate, measure pH downstream or determine amount needed to
change litmus paper color (or add salt and measure change in electrical conductivity, or add a
radioisotope and measure change in activity, or add a phosphorescent substance and measure
luminescence, or add any chemical and measure its concentration by any means).
54. Add sugar at a known rate, measure rate of formation of rock candy downstream.
55. Add a second liquid of different density, measure resulting density change (or add an immiscible
liquid, measure its rate of passage).
56. Add a reactant, determine amount needed to react completely with the fluid (or with another
reactant on a permeable membrane in the flow channel, or inject chlorine ions and measure the
rate of electroplating on a silver electrode).
57. Get a technician to drink the effluent, measure his weight gain after a fixed time (or the time
required for his mouth to fill up, or the time required to drown a rat).
58. Add alcohol (or poison, or salt, or Kool-Aid mix) to the fluid at a known rate, have a technician
drink it (or do it yourself), and determine the time required to feel the effects.
59. Immerse pipe outlet in water, find the depth at which the hydrostatic head is just sufficient to
stop the flow.
60. Insert air tube facing upstream, determine pressure needed to initiate bubbling.
61. Place pipe in wind tunnel, find wind velocity just adequate to stop the flow.
xiv
62. If pipe is only partially filled, put in sailboat, measure wind force needed to hold it stationary.
63. Put another pipe against outlet, find flow in second pipe that just neutralizes unknown flow.
64. Send sound wave through, time passage over known distance (or use Doppler meter, or time
passage of an electrical impulse or a light wave).
65. Put bacteria in line, determine reproduction rate.
66. Put algae in pipe, measure change in COD.
67. Put a spawning fish in the line, measure how far the eggs travel in a given time interval.
68. Measure how long it takes for the effluent to put out a fire of a given size.
69. Pass the fluid spirally into a funnel, measure how long it takes for a drop of dye to disappear.
70. If the fluid is combustible, burn it in a combustion engine and measure the rate of power output.
71. Determine how long you can hold your breath, then jump in and see how far you travel before
you have to breathe (or see if an animal can make it out of the pipe before drowning).
72. Add effluent to bubble bath, measure extent of generation of bubbles.
73. Insert a fish with a monitor in its heart, time how long it takes him to die. (Must kill a lot of fish
to calibrate-don’t tell “Save the Whales.”)
74. Count rate of passage of molecules.
75. Insert a monkey who can insert pegs in holes at a known rate, and count the number of pegs
inserted over a known distance.
76. Install a Japanese flowmeter equipped with lots of flashing lights.
77. Correlate the velocity with the rate at which the student pulls out his hair during the experiment.
78. Hire someone to do it.
79. Break into the pipeline company’s office and steal the flow rate records.
80. Look it up in the Enquirer flow rate tables.
81. Fill a balloon, throw it at your boss, and correlate his anger with the flow rate. (?)
82. Ask your local psychic.
xv
TRANSPARENCY MASTERS
The following pages contain oversized renderings of illustrations taken from the text.
The illustration numbers are listed below with their text page numbers.
You are granted permission to have these illustrations reproduced as transparencies for
your own use in conjunction with the textbook, Felder and Rousseau: ELEMENTARY
PRINCIPLES OF CHEMICAL PROCESSES, 3rd Edition, John Wiley & Sons. Resale is
expressly prohibited.
Transparencies may easily be prepared using either thermal copy (ThermoFax) or
electrostatic copy (Xerox) machines. See the operating manual for your particular copy
machine. Transparency film can be purchased from your usual copy paper supplier.
Other illustrations in the book may also serve as transparency masters. For best results,
it may be necessary to enlarge the illustration to fill the sheet of copy film. Many
electrostatic copiers have this capability.
Figure
Description
Compressibility charts
Cox vapor pressure chart
Psychrometric chart – SI units
Psychrometric chart – Am. Engr. units
Enthalpy conc. chart – H2SO4/H2O
Enthalpy conc. chart – NH3/H2O
xvi
Figure
Number
Text
Page
5.4-1
5.4-2
5.4-3
5.4-4
6.1-4
8.4-1
8.4-2
8.5-1
8.5-2
208
209
210
211
247
382
383
396
400
CHAPTER TWO
2.1 (a)
(b)
(c)
2.2 (a)
3 wk
7d
24 h
3600 s 1000 ms
1 wk 1 d
s
1 h 1
38.1 ft / s 0.0006214 mi 3600 s
3.2808
ft
1d
554 m 4
1h
d ˜ kg 24 h 60 min
1
760 mi
h
1 h
u 10 9 ms
.
18144
25.98 mi / h Ÿ 26.0 mi / h
1 kg 10 8 cm 4
1000 g 1 m 4
m
1 h
0.0006214 mi 3600 s
340 m / s
1 m3
35.3145 ft 3
(b)
921 kg 2.20462 lb m
m3
1
kg
(c)
5.37 u 10 3 kJ 1 min 1000 J
min 60 s
1 kJ
3.85 u 10 4 cm 4 / min˜ g
57.5 lb m / ft 3
1.34 u 10 -3 hp
119.93 hp Ÿ 120 hp
J/s
1
2.3 Assume that a golf ball occupies the space equivalent to a 2 in u 2 in u 2 in cube. For a
classroom with dimensions 40 ft u 40 ft u 40 ft :
40 u 40 u 40 ft 3 (12) 3 in 3 1 ball
n balls
6.48 u 10 6 | 7 million balls
3
3
3
2 in
ft
The estimate could vary by an order of magnitude or more, depending on the assumptions made.
2.4 4.3 light yr 365 d 24 h
1 yr 1 d
3600 s 1.86 u 105 mi
3.2808 ft
1 step
1 h
1
s 0.0006214 mi 2 ft
2.5 Distance from the earth to the moon = 238857 miles
238857 mi
1
m
1 report
0.0006214 mi
0.001 m
4 u 1011 reports
2.6
19 km 1000 m 0.0006214 mi
1000 L
1 L
1 km
1
m 264.17 gal
Calculate the total cost to travel x miles.
Total Cost
Total Cost
American
European
$14,500 $21,700 Equate the two costs Ÿ x
44.7 mi / gal
$1.25 1 gal x (mi)
14,500 0.04464 x
gal 28 mi
x (mi)
$1.25
1 gal
gal 44.7 mi
4.3 u 10 5 miles
2-1
21,700 0.02796 x
7 u 1016 steps
2.7
10 6 cm 3
5320 imp. gal 14 h 365 d
plane ˜ h 1 d 1 yr
ton kerosene
u 10 5
1188
.
plane ˜ yr
0.965 g
220.83 imp. gal
4.02 u 10 9 ton crude oil 1 ton kerosene
1 cm
1 kg
3
1 ton
1000 g
1000 kg
plane ˜ yr
u 10 ton kerosene
.
yr
7 ton crude oil 1188
4834 planes Ÿ 5000 planes
5
2.8 (a)
(b)
(c)
2.9
2.10
2.11
32.1714 ft / s 2
25.0 lb m
25 N
1
9.8066 m / s 2
10 ton
1 lb m
5 u 10 -4 ton
50 u 15 u 2 m 3
500 lb m
(a)
mdisplaced fluid
1
lb f
32.1714 lb m ˜ ft / s 2
1 kg ˜ m / s 2
1N
2.55 kg Ÿ 2.6 kg
980.66 cm / s 2
1000 g
2.20462 lb m
35.3145 ft 3
1 m3
25.0 lb f
85.3 lb m
1 ft 3
1 dyne
1 g ˜ cm / s 2
1 lb f
32.174 ft
2
32.174 lb m / ft ˜ s 2
1 s
FG IJ FG 1 IJ | 25 m
H K H 10K
1 m3
1
| 5 u 10 2
2
11.5 kg
1 kg
2.20462 lb m
mcylinder Ÿ U f V f
9 u 10 9 dynes
U cVc Ÿ U f hSr 2
4.5 u 10 6 lb f
3
U c HSr 2
Uc
Ufh
(30 cm 14.1 cm)(100
. g / cm 3 )
Uc
0.53 g / cm 3
H
30 cm
U c H (30 cm)(0.53 g / cm 3 )
(b) U f
171
. g / cm 3
h
(30 cm - 20.7 cm)
2.12
Vs
SR 2 H
; Vf
3
R
SR 2 H Sr 2 h
;
H
3
3
Ÿ Vf
U fVf
U sVs Ÿ U f
ŸUf
Us
H
H
h3
H2
2
2
2
3
H3
H 3 h3
h
3
r
2
H
2
Us
s
2
Us
Uf
h
R
h
H
r
Ÿr
h
FG IJ SR FG H h IJ
H K 3H HK
h I
SR F
SR H
H
U
J
G
3 H
3
H K
SR 2 H Sh Rh
3
3 H
H
Us
R
1
F hI
1 G J
H HK
2-2
3
Uf
2.13
Say h m
depth of liquid
y
y= 1
dA
y= 1– h
x
Ÿ
2
A(m )
z
h
1m
x = 1– y
2
1 y 2
dA dy ˜
2 1 y 2 dx
dx
1 y 2
d i
ŸAm
dA 2
E
d i
A m2
b g
W N
z z
1
1
2
1 h
1 y 2 dy
1 h
Table of integrals, or trigonometric substitution
S
2
1 h 1 1 h sin 1 1 h
2
1 h
4 m u A( m 2 ) 0.879 g 10 6 cm 2 1 kg 9.81 N
3.45 u 10 4 A
cm 3
kg
1 m3
10 3 g
,
y 1 y 2 sin 1 y
b g
1
E Substitute for A
LS
W b N g 3.45 u 10 M b1 hg
N2
4
b g
b g
g g0
b g
1 1 h
2
b gOPQ
sin 1 1 h
32.174 lb m ˜ ft / s 2 Ÿ 1 slug = 32.174 lb m
1
lb f
1 poundal = 1 lb m ˜ ft / s 2
32.174
(a) (i) On the earth:
175 lb m
1 slug
M
5.44 slugs
32.174 lb m
2.14 1 lb f
1 slug ˜ ft / s 2
W
175 lb m
(ii) On the moon
175 lb m
M
W
175 lb m
32.174 ft
s2
1 poundal
1 lb m ˜ ft / s 2
1 slug
32.174 lb m
32.174 ft
6
s2
5.63 u 10 3 poundals
5.44 slugs
1 poundal
1 lb m ˜ ft / s 2
2-3
938 poundals
2.14 (cont’d)
ma Ÿ a
F/m
0.135 m / s
2
(b) F
2.15 (a) F
25.0 slugs
ma Ÿ 1 fern = (1 bung)(32.174 ft / s 2 )
Ÿ
FG 1IJ
H 6K
1 slug
32.174 lb m
1m
3.2808 ft
5.3623 bung ˜ ft / s 2
1 fern
5.3623 bung ˜ ft / s 2
(b) On the moon: W
On the earth: W
2.16 (a) | (3)(9)
3 bung 32.174 ft
1 fern
2
6 s 5.3623 bung ˜ ft / s 2
3 fern
(3)(32.174) / 5.3623 = 18 fern
27
(2.7)(8.632)
(b)
23
(c) | 2 125 127
(d)
2.365 125.2 127.5
2.17 R |
1 lb m ˜ ft / s 2
1 poundal
355 poundals
40 u 10 4 5 u 8 u 10 4
| 1 u 10 4
|
45
5u 9
4
(3.600 u 10 ) / 45 9 u 10 5
|
| 50 u 10 3 1 u 10 3 | 49 u 10 3 | 5 u 10 4
4.753 u 10 4 9 u 10 2
5 u 10 4
(7 u 10 1 )(3 u 10 5 )(6)(5 u 10 4 )
| 42 u 10 2 | 4 u 10 3
6
(3)(5 u 10 )
Rexact
3812.5 Ÿ 3800 Ÿ 38
. u 10 3
2.18 (a)
A: R
. 72.4
731
0.7 o C
X
. 72.6 72.8 73.0
72.4 731
5
s
(72.4 72.8) 2 (731
. 72.8) 2 (72.6 72.8) 2 (72.8 72.8) 2 (73.0 72.8) 2
51
72.8 o C
0.3o C
. 97.3 58
. oC
B: R 1031
X
. 98.7 1031
. 100.4
97.3 1014
100.2 o C
5
s
(97.3 100.2) 2 (1014
. 100.2) 2 (98.7 100.2) 2 (1031
. 100.2) 2 (100.4 100.2) 2
51
2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-4
2.19 (a)
12
12
¦
¦ ( X 735. )
Xi
i 1
C min=
.
735
12
X 2 s 735
. 2(12
. )
711
.
C max=
X 2 s 735
. 2(12
. )
75.9
X
s
2
i 1
12
.
12 1
(b) Joanne is more likely to be the statistician, because she wants to make the control limits
stricter.
(c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor
temperature (failure of reactor control system), problems with the color measurement
system, operator carelessness
2.20 (a), (b)
1
2
(a) Run
134 131
X
Mean(X) 131.9
Stdev(X) 2.2
127.5
Min
136.4
Max
(b) Run
1
2
3
4
5
6
7
8
9
10
11
12
13
14
X
128
131
133
130
133
129
133
135
137
133
136
138
135
139
Min
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
127.5
3
129
4
5 6 7 8 9 10 11 12 13 14 15
133 135 131 134 130 131 136 129 130 133 130 133
Mean
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
131.9
Max
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
136.4
140
138
136
134
132
130
128
126
0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance
limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q'
2.36 u 10 4 kg ˜ m 2
h
2.10462 lb 3.2808 2 ft 2
kg
m2
(2 u 10 4 )(2)(9)
| 12 u 10 ( 4 3) | 12
. u 10 6 lb ˜ ft 2 / s
36 u 10 3 2
. u 10 lb ˜ ft / s 0.00000148 lb ˜ ft 2 / s
= 148
(b) Q' approximate |
Q' exact
1 h
3600 s
2-5
CpP
2.22 N Pr
k
0.583 J / g ˜ o C 1936 lb m
0.286 W / m ˜ o C
ft ˜ h
1 h 3.2808 ft
1000 g
3600 s
m 2.20462 lb m
(6 u 10 1 )(2 u 10 3 )(3 u 10 3 ) 3 u 10 3
. u 10 3 . The calculator solution is 163
. u 10 3
|
| 15
2
(3 u 10 1 )(4 u 10 3 )(2)
N Pr |
2.23
Re
DuU
P
Re |
(5 u 10 1 )(2)(8 u 10 1 )(10 6 ) 5 u 101 ( 3)
|
| 2 u 10 4 Ÿ the flow is turbulent
3
4
3
(3)(4 u 10)(10 )(4 u 10 )
0.48 ft
2.067 in
1 m
1 m
3
s 3.2808 ft 0.43 u 10 kg / m ˜ s 39.37 in
0.805 g
cm 3
1 kg 10 6 cm 3
1000 g 1 m 3
2.24
(a)
kgd p y
F P IJ FG d uU IJ
2.00 0.600G
D
H UD K H P K
L 100
. u 10 N ˜ s / m
2.00 0.600M
. kg / m )(100
. u 10 m
N (100
1/ 2
1/ 3
p
5
5
3
44.426 Ÿ
k g (0.00500 m) (0100
. )
. u 10 5 m 2 / s
100
OP LM (0.00500 m)(10.0 m / s)(100
. kg / m ) O
PQ
/ s) Q N
(100
. u 10 N ˜ s / m )
1/ 3
2
2
3
5
44.426 Ÿ k g
1/ 2
2
0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the
equation may differ significantly from the conditions in the reactor (out of the range of
empirical data), all of the other variables are subject to measurement or estimation error.
(c)
dp (m)
0.005
0.010
0.005
0.005
0.005
y
0.1
0.1
0.1
0.1
0.1
D (m2/s)
1.00E-05
1.00E-05
2.00E-05
1.00E-05
1.00E-05
P (N-s/m2) U (kg/m3) u (m/s)
1.00E-05
1
10
1.00E-05
1
10
1.00E-05
1
10
2.00E-05
1
10
1.00E-05
1
20
kg
0.889
0.620
1.427
0.796
1.240
2.25 (a) 200 crystals / min ˜ mm; 10 crystals / min ˜ mm 2
(b) r
200 crystals 0.050 in 25.4 mm
10 crystals
min ˜ mm
in
min ˜ mm 2
238 crystals / min Ÿ
b g
(c) D mm
b g
D c in
Ÿ 60r c
(25.4) 2 mm 2
in 2
238 crystals 1 min
4.0 crystals / s
60 s
min
FG
H
25.4 mm
crystals
25.4 D c ; r
min
1 in
b
0.050 2 in 2
g b
200 25.4 D c 10 25.4 D c
g
2
2-6
IJ
K
rc
crystals
60 s
s
1 min
b g
Ÿ r c 84.7 D c 108 D c
2
60r c
2.26 (a) 70.5 lb m / ft 3 ; 8.27 u 10 -7 in 2 / lb f
(b) U
L8.27 u 10
/ ft ) exp M
N
(70.5 lb m
35.3145 ft 3
70.57 lb m
ft
FG lb IJ
H ft K
F lb IJ
PG
H in K
(c) U
3
m
Uc
m
3
f
2
g
cm 3
P'
Ÿ 62.43U c
7
in 2
lb f
9 u 10 6 N
m2
m3
1000 g
3
N
m2
1
6
3
10 cm
3
2.20462 lb m
1 lb m
28,317 cm 3
453593
.
g
1 ft 3
d
OP
Q
. g / cm 3
113
62.43U c
12
m2
39.37 2 in 2
0.2248 lb f
1N
14.696 lb f / in 2
u 10 6 N / m 2
.
101325
145
. u 10 4 P'
id
i
d
P' 9.00 u 10 6 N / m 2 Ÿ U ' 113
. exp[(120
. u 10 10 )(9.00 u 10 6 )] 113
. g / cm 3
cm
d i d i 28,317
1728 in
Ÿ 16.39V ' expb3600t c g Ÿ V '
2.27 (a) V cm 3
i
70.5 exp 8.27 u 10 7 145
. u 10 4 P ' Ÿ U c 113
. exp 120
. u 10 10 P '
V ' in 3
3
3
b g 3600t cbhrg
0.06102 expb3600t cg
16.39V ' ; t s
(b) The t in the exponent has a coefficient of s-1.
2.28 (a) 3.00 mol / L, 2.00 min -1
0 Ÿ C 3.00 exp[(-2.00)(0)] = 3.00 mol / L
t = 1 Ÿ C 3.00 exp[(-2.00)(1)] = 0.406 mol / L
0.406 3.00
Cint
For t=0.6 min:
(0.6 0) 3.00 14
. mol / L
1 0
Cexact 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L
(b) t
For C=0.10 mol/L:
1 0
(010
. 3.00) 0 112
. min
0.406 3
1 0.10
C
1
= 1.70 min
= - ln
ln
=2 3.00
2.00 3.00
t int
t exact
(c)
3.5
C exact vs. t
3
C (mol/L)
2.5
2
(t=0.6, C =1.4)
1.5
1
(t=1.12, C=0.10)
0.5
0
0
1
2
t (min)
2-7
p*
2.29 (a)
(b)
60 20
(185 166.2) 20 42 mm Hg
199.8 166.2
c
MAIN PROGRAM FOR PROBLEM 2.29
IMPLICIT REAL*4(A–H, 0–Z)
DIMENSION TD(6), PD(6)
DO 1 I = 1, 6
READ (5, *) TD(I), PD(I)
1
CONTINUE
WRITE (5, 902)
902
FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X,
*
‘
(C)
(MM HG)’/)
DO 2 I = 0, 115, 5
T = 100 + I
CALL VAP (T, P, TD, PD)
WRITE (6, 903) T, P
903
FORMAT (10X, F5.1, 10X, F5.1)
2
CONTINUE
END
SUBROUTINE VAP (T, P, TD, PD)
DIMENSION TD(6), PD(6)
I=1
1
IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2
I=I+1
IF (I.EQ.6) STOP
GO TO 1
2
P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I))
RETURN
END
DATA
OUTPUT
98.5
1.0
TEMPERATURE
VAPOR PRESSURE
131.8
5.0
(C)
(MM HG)
100.0
1.2
215.5
100.0
105.0
1.8
215.0
98.7
2.30 (b) ln y
b
ln a
ln a bx Ÿ y ae bx
(ln y 2 ln y1 ) / ( x 2 x1 )
ln y bx
(ln 2 ln 1) / (1 2)
ln 2 0.63(1) Ÿ a
4.00 Ÿ y
0.693
4.00e 0.693 x
(c) ln y ln a b ln x Ÿ y ax b
b (ln y 2 ln y1 ) / (ln x 2 ln x1 ) (ln 2 ln 1) / (ln 1 ln 2)
ln a ln y b ln x ln 2 ( 1) ln(1) Ÿ a 2 Ÿ y 2 / x
1
(d) ln( xy ) ln a b( y / x ) Ÿ xy ae by / x Ÿ y (a / x )e by / x [can' t get y f ( x )]
b [ln( xy ) 2 ln( xy ) 1 ] / [( y / x ) 2 ( y / x ) 1 ] (ln 807.0 ln 40.2) / ( 2.0 10
. )
ln a
ln( xy ) b( y / x )
ln 807.0 3 ln( 2.0) Ÿ a
2-8
2 Ÿ xy
2e 3 y / x Ÿ y
3
(2 / x )e 3 y / x
2.30 (cont’d)
(e) ln( y 2 / x )
ln a b ln( x 2) Ÿ y 2 / x
a ( x 2) b Ÿ y [ax ( x 2) b ]1/ 2
b [ln( y 2 / x ) 2 ln( y 2 / x ) 1 ] / [ln( x 2) 2 ln( x 2) 1 ]
(ln 807.0 ln 40.2) / (ln 2.0 ln 10
. ) 4.33
ln( y 2 / x ) b( x 2)
ln a
Ÿ y2 / x
ln 807.0 4.33 ln(2.0) Ÿ a
40.2( x 2) 4.33 Ÿ y
6.34 x 1/ 2 ( x 2) 2.165
2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope
(c)
1
ln( y 3)
(d)
1
( y 1) 2
1 a
1
x Ÿ Plot
vs.
b b
ln( y 3)
a ( x 3) 3 Ÿ Plot
40.2
m, Intcpt
n
x [rect. axes], slope =
a
1
, intercept =
b
b
1
vs. ( x 3) 3 [rect. axes], slope = a , intercept = 0
( y 1) 2
OR
2 ln( y 1) ln a 3 ln( x 3)
Plot ln( y 1) vs. ln( x 3) [rect.] or (y + 1) vs. (x - 3) [log]
3
ln a
Ÿ slope = , intercept = 2
2
(e) ln y
a x b
Plot ln y vs.
(f) log10 ( xy )
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
a( x 2 y 2 ) b
Plot log10 ( xy ) vs. ( x 2 y 2 ) [rect.] Ÿ slope = a, intercept = b
(g)
1
y
OR
ax 1
y
b
x
Ÿ
x
y
ax ax 2 b Ÿ Plot
b
1
Ÿ
x
xy
a
x
vs. x 2 [rect.], slope = a , intercept = b
y
b
1
1
Ÿ Plot
vs. 2 [rect.] , slope = b, intercept = a
2
xy
x
x
2-9
2.32 (a) A plot of y vs. R is a line through ( R 5 , y
0.011 ) and ( R 80 , y
0169
.
).
0.18
0.16
0.14
0.12
y
0.1
0.08
0.06
0.04
0.02
0
0
20
40
60
80
100
R
y
aRb
a
b
U|
V|
W
0.011
.
0169
2.11 u 10 3
80 5
Ÿy
0.011 2.11 u 10 3 5 4.50 u 10 4
ib g
d
2.11 u 10 3 R 4.50 u 10 4
d2.11 u 10 ib43g 4.50 u 10 0.092 kg H O kg
b1200 kg hgb0.092 kg H O kgg 110 kg H O h
(b) R
3
43 Ÿ y
4
2
2
2.33 (a) ln T
b
ln a
(b) T
T
T
T
2
ln a b ln I Ÿ T aI b
(ln T2 ln T1 ) / (ln I 2 ln I 1 )
ln T b ln I
(ln 120 ln 210) / (ln 40 ln 25)
ln 210 ( 119
. ) ln( 25) Ÿ a
9677.6 Ÿ T
119
.
9677.6I 1.19
b9677.6 / T g
. L/s
85 C Ÿ I b9677.6 / 85g
535
175 C Ÿ I b9677.6 / 175g
29.1 L / s
290 C Ÿ I b9677.6 / 290g
19.0 L / s
9677.6I 1.19 Ÿ I
1.19
1.19
o
o
1.19
o
1.19
(c) The estimate for T=175°C is probably closest to the real value, because the value of
temperature is in the range of the data originally taken to fit the line. The value of T=290°C
is probably the least likely to be correct, because it is farthest away from the date range.
2-10
ln ((CA-CAe)/(CA0-CAe))
2.34 (a) Yes, because when ln[(C A C Ae ) / (C A0 C Ae )] is plotted vs. t in rectangular coordinates,
the plot is a straight line.
0
50
100
150
200
0
-0.5
-1
-1.5
-2
t (m in)
Slope = -0.0093 Ÿ k = 9.3 u 10-3 min 1
(b) ln[(C A C Ae ) / (C A0 C Ae )] kt Ÿ C A
0.0495)e ( 9.3u10
(01823
.
CA
3
)(120)
(C A 0 C Ae )e kt C Ae
0.0495 = 9.300 u 10-2 g / L
9.300 u 10-2 g 30.5 gal 23.317 L
8.8 g
L
7.4805 gal
C = m / V Ÿ m = CV
2.35 (a) ft 3 and h -2 , respectively
(b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353
. u 10 2 ) ; or
V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353
. u 10 2
3
3
7 2
(c) V ( m ) 100
. u 10 exp(15
. u 10 t )
CŸ P
2.36 PV k
C / V k Ÿ ln P
ln C k lnV
8.5
lnP
8
7.5
7
6.5
6
2.5
3
lnP = -1.573(lnV ) + 12.736
k
slope
3.5
lnV
( 1573
(dimensionless)
. ) 1573
.
Intercept = ln C 12.736 Ÿ C
G GL
G0 G
G G
1
Ÿ 0
m
G GL
KLC
e12 .736
3.40 u 105 mm Hg ˜ cm4.719
K L C m Ÿ ln
G0 G
G GL
ln K L m ln C
ln ( G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
3
ln(G 0-G)/(G-G L )
2.37 (a)
4
2
1
0
-1
3 .5
4
4 .5
5
ln C
2-11
5 .5
2.37 (cont’d)
m slope 2.483 (dimensionless)
10.045 Ÿ K L
Intercept = ln K L
4.340 u 10 5 ppm-2.483
G 180
. u 103
u 10 3
.
4.340 u 105 (475) 2.483 Ÿ G 1806
3.00 u 10 3 G
C=475 ppm is well beyond the range of the data.
475 Ÿ
(b) C
2.38 (a) For runs 2, 3 and 4:
Z aV b p c Ÿ ln Z ln a b lnV c ln p
b 0.68
ln(35
. ) ln a b ln(102
. ) c ln( 9.1)
Ÿ c 146
.
ln(2.58) ln a b ln(102
. ) c ln(112
. )
ln(3.72) ln a b ln(175
. ) c ln(112
. )
a = 86.7 volts ˜ kPa 1.46 / (L / s) 0.678
. Slope=b, Intercept= ln a c ln p
(b) When P is constant (runs 1 to 4), plot ln Z vs. lnV
2
lnZ
1.5
1
0.5
0
-1
-0.5
0
lnZ = 0.5199lnV + 1.0035
0.5
1
1.5
b
lnV
slope
0.52
Intercept = lna c ln P 10035
.
When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a c lnV
2
lnZ
1.5
1
0.5
0
1.5
1.7
lnZ = -0.9972lnP + 3.4551
1.9
2.1
2.3
c
lnP
slope
0.997 Ÿ 10
.
Intercept = lna b lnV
3.4551
Z
Plot Z vs V b P c . Slope=a (no intercept)
7
6
5
4
3
2
1
0.05
Z = 31.096VbPc
0.1
0.15
0.2
Vb P c
a
slope
311
. volt ˜ kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-12
2.39 (a)
sxy
sxx
a
b
y
(b) a
¦x y
1
n
¦x
1
n
sx
n
1
n
[(0.4)(0.3) (2.1)(19
. ) (31
. )(3.2)] / 3 4.677
i i
i 1
n
(0.32 19
. 2 3.2 2 ) / 3 4.647
2
i
i 1
n
¦
(0.3 19
. 3.2) / 3 18
. ; sy
xi
i 1
sxy sx s y
b g
sxx sx
4.677 (18
. )(1.867)
4.647 (18
. )2
2
sxx s y sxy sx
b g
sxx sx
1
n
n
¦y
i
0.936
(4.647)(1867
. ) (4.677)(18
. )
2
4.647 (18
. )
2
(0.4 2.1 31
. ) / 3 1867
.
i 1
0182
.
0.936 x 0182
.
sxy
4.677
4.647
sxx
x
Ÿ y 10065
.
.
10065
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1
0
0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a
slope = 0.477 L / g ˜ h;
a
Intercept = 0.082 L / g
3
2.5
2
1.5
1
0.5
0
2
1.5
C
1/C
(b) b
1
0.5
0
0
1
2
3
4
5
6
1
t
1/C = 0.4771t + 0.0823
C
2
C-fitted
3
4
5
t
(c) C 1 / (a bt ) Ÿ 1 / [0.082 0.477(0)] 12.2 g / L
t
(1 / C a ) / b
(1 / 0.01 0.082) / 0.477
209.5 h
(d) t=0 and C=0.01 are out of the range of the experimental data.
(e) The concentration of the hazardous substance could be enough to cause damage to the
biotic resources in the river; the treatment requires an extremely large period of time; some
of the hazardous substances might remain in the tank instead of being converted; the
decomposition products might not be harmless.
2-13
2.41 (a) and (c)
y
10
1
0.1
1
10
100
x
ax b Ÿ ln y
(b) y
ln a b ln x; Slope = b, Intercept = ln a
ln y = 0.1684ln x + 1.1258
2
ln y
1.5
1
0.5
0
-1
0
1
2
ln x
3
4
b slope
5
0168
.
Ÿa
Intercept = ln a 11258
.
3.08
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0
(b)
400
600
ln(1-Cp/Cao)
300
600
800
-2
-4
500
600
-4
t
k
0
400
-2
0.0062 s-1
400
200
Lab 1
0
ln(1-Cp/Cao)
ln(1-Cp/Cao)
200
100
0
-6
ln(1-Cp/Cao) = -0.0111t
t
k
0
0
800
ln(1-Cp/Cao)
0
200
0
-1
-2
-3
-4
ln(1-Cp/Cao) = -0.0062t
200
400
Lab 2
0.0111 s-1
600
800
0
-2
-4
-6
ln(1-Cp/Cao)= -0.0064t
-6
ln(1-Cp/Cao) = -0.0063t
t
k
t
Lab 3
0.0063 s-1
k
Lab 4
0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with
the average of the calculated k’s. k 0.0063 s-1
(d) Errors in measurement of concentration, poor temperature control, errors in time
measurements, delays in taking the samples, impure reactants, impurities acting as
catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty
reactor.
2-14
2.43 yi
axi Ÿ I (a )
n
n
¦
¦by
di2
i
i 1
Ÿa
n
¦
i 1
2.44
i 1
axi
g
2
dI
Ÿ
da
n
0
¦ 2b y
i
i 1
g
axi xi Ÿ
n
¦y x
i i
i 1
a
n
¦x
2
i
i 1
n
yi xi /
¦x
2
i
i 1
DIMENSION X(100), Y(100)
READ (5, 1) N
C
N = NUMBER OF DATA POINTS
1FORMAT (I10)
READ (5, 2) (X(J), Y(J), J = 1, N
2FORMAT (8F 10.2)
SX = 0.0
SY = 0.0
SXX = 0.0
SXY = 0.0
DO 100J = 1, N
SX = SX + X(J)
SY = SY + Y(J)
SXX = SXX + X(J) ** 2
100SXY = SXY + X(J) * Y(J)
AN = N
SX = SX/AN
SY = SY/AN
SXX = SXX/AN
SXY = SXY/AN
CALCULATE SLOPE AND INTERCEPT
A = (SXY - SX * SY)/(SXX - SX ** 2)
B = SY - A * SX
WRITE (6, 3)
3FORMAT (1H1, 20X 'PROBLEM 2-39'/)
WRITE (6, 4) A, B
4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/)
C
CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF
RESIDUALS
SSQ = 0.0
DO 200J = 1, N
YC = A * X(J) + B
RES = Y(J) - YC
WRITE (6, 5) X(J), Y(J), YC, RES
5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X
* 'RESIDUALb =', F6.3)
200SSQ = SSQ + RES ** 2
WRITE (6, 6) SSQ
6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3)
STOP
END
$DATA
5
1.0 2.35
1.5
5.53
2.0
8.92
2.5
12.15
3.0 15.38
SOLUTION: a 6.536, b 4.206
2-15
0
2.45 (a) E(cal/mol), D0 (cm2/s)
(b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0.
(c) Intercept = ln D0 = -3.0151 Ÿ D0 = 0.05 cm2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = E / R = -3666 K Ÿ E = (3666 K)(1.987 cal / mol ˜ K) = 7284 cal / mol
-10.0
ln D
-11.0
-12.0
-13.0
-14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet
T
347
374.2
396.2
420.7
447.7
471.2
D
1.34E-06
2.50E-06
4.55E-06
8.52E-06
1.41E-05
2.00E-05
1/T
2.88E-03
2.67E-03
2.52E-03
2.38E-03
2.23E-03
2.12E-03
Sx
Sy
Syx
Sxx
-E/R
ln D0
lnD (1/T)*(lnD)
-13.5
-0.03897
-12.9
-0.03447
-12.3
-0.03105
-11.7
-0.02775
-11.2
-0.02495
-10.8
-0.02296
2.47E-03
-12.1
-3.00E-02
6.16E-06
-3666
-3.0151
D0
7284
E
0.05
2-16
(1/T)**2
8.31E-06
7.14E-06
6.37E-06
5.65E-06
4.99E-06
4.50E-06
CHAPTER THREE
3.1
16 u 6 u 2 m3 1000 kg
| 2 u 10 5 2 103 | 2 u 105 kg
3
m
b
(a) m
8 oz
(b) m
2s
gb gb gd i
106 cm3
1 qt
1g
4 u 106
| 1 u 102 g / s
|
3
3
32 oz 1056.68 qt cm
3 u 10 10
b
gd i
(c) Weight of a boxer | 220 lb m
12 u 220 lb m 1 stone
| 220 stones
Wmax t
14 lb m
dictionary
(d)
SD 2 L
4
V
|
314
4.52 ft 2
.
4
d
800 miles 5880 ft 7.4805 gal 1 barrel
1 mile
1 ft 3
31.5 gal
i d
i
3 u 4 u 5 u 8 u 102 u 5 u 103 u 7
4 u 3 u 10
(e) (i)V |
| 1 u 107 barrels
6 ft u 1 ft u 0.5 ft 28,317 cm3
| 3 u 3 u 104 | 1 u 105 cm3
1 ft 3
(ii)V |
150 lb m
1 ft 3
62.4 lb m
28,317 cm3 150 u 3 u 104
|
| 1 u 105 cm3
3
60
1 ft
(f) SG | 105
.
3.2
(a) (i)
(ii)
(b) U
3.3
(a)
995 kg
1 lb m
0.028317 m3
m3
0.45359 kg
1 ft 3
995 kg / m3
62.43 lb m / ft 3
1000 kg / m3
U H2 O u SG
62.43 lb m / ft 3 u 5.7
50 L
62.12 lb m / ft 3
62.12 lb m / ft 3
360 lb m / ft 3
0.70 u 103 kg 1 m3
35 kg
m3 103 L
(b)
m3 1000 L 1 min
1150 kg
27 L s
min
0.7 u 1000 kg 1 m3
60 s
(c)
10 gal
1 ft 3
0.70 u 62.43 lb m
# 29 lb m / min
2 min 7.481 gal
1 ft 3
3-1
3.3 (cont’d)
(d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline
d
i
d
i
1dcm kerosenei Ÿ 0.82dg kerosenei
d0.70V 0.82idg blendi 0.78 Ÿ V
SG
V 1dcm blend i
Vg cm3gasoline Ÿ 0.70Vg g gasoline
3
0.82 0.78
0.78 0.70
g
3
g
g
Vgasoline
Vkerosene
Volumetric ratio
3.4
In France:
In U.S.:
3.5
50.0 kg
L
0.50 cm3
1 cm3
5 Fr
3
0.5 0 cm
0.50 cm3 gasoline / cm3 kerosene
$1
$68.42
0.7 u 1.0 kg 1L 5.22 Fr
L
$1.20
50.0 kg
1 gal
$22.64
0.70 u 1.0 kg 3.7854 L 1 gal
VB ( ft 3 / h ), m B ( lb m / h )
V ( ft 3 / h), SG
VH ( ft 3 / h ), m H ( lb m / h )
0.850
700 lb m / h
700 lb m
ft 3
1319
. ft 3 / h
h
0.850 u 62.43 lb m
3
VB ft
0.879 u 62.43 lb m
kg / h
m B
54.88V
B
h
ft 3
m H VH 0.659 u 62.43 4114
. VH kg / h
(a) V
d i
bg
d hb
VB VH
b
g
b
g
g
. ft 3 / h
1319
54.88V 4114
. V
m B m H
700 lb m
B
H
3
Ÿ VB 114
. ft / h Ÿ m B 628 lb m / h benzene
VH
. ft 3 / h Ÿ m H
174
. lb m / h hexane
716
(b) – No buildup of mass in unit.
– U B and U H at inlet stream conditions are equal to their tabulated values (which are
o
strictly valid at 20 C and 1 atm.)
– Volumes of benzene and hexane are additive.
– Densitometer gives correct reading.
3-2
3.6
. kg H 2SO 4
1955
(a) V
L
1 kg solution
u 1000
.
.
kg
0.35kg H 2SO 4 12563
445 L
(b)
195.5 kg H 2 SO 4
L
u 1.00 kg
18255
.
195.5 kg H 2 SO 4
0.65 kg H 2 O
L
0.35 kg H 2 SO 4 1.000 kg
470 445
u 100% 5.6%
% error
445
Videal
3.7
b g E Weight of block bdowng
Mass of oil displaced + Mass of water displaced = Mass of block
U b0.542gV U
b1 0.542gV U V
Buoyant force up
oil
c
H 2O
From Table B.1: U c
2.26 g / cm3 , U w
100
. g / cm3 Ÿ U oil
moil U oil u V 3.325 g / cm3 u 35.3 cm3
moil + flask 117.4 g 124.8 g 242 g
3.8
470 L
b g
117.4 g
b
Buoyant force up = Weight of block down
Wblock Ÿ ( UVg ) disp. Liq
Ÿ Wdisplaced liquid
b g
b g
. Ag
Expt. 1: U w 15
UB 2A g Ÿ UB
U w 1.00 g/cm3
bg
UB
Let U w
B
0.75
b g
2U B
15
. g / cm3 Ÿ SG
15
.
soln
density of water. Note: U A ! U w (object sinks)
Ab hsi
d
Ab h p1 hb1
U wVd 1 g
i
d
Subst. (1) for Vd 1 , solve for h p1 hb1
WA WB
pw gAb
h p1 hb1
bg
subst. 2
bi g
A p h p1 Vw
bg
subst. 3 for h p 1 in
b2 g, solve for h
b1
hb1
i
(2)
Volume of pond water: Vw
WA WB
Ÿ h p1
pw g
b
d
Ap h p1 Vd 1 ŸVw
W WB
Vw
A
Ap
pw g
Ap h p1 Ab h p1 hb1
Vw WA WB
Ap
pw gAp
g LM 1
MN A
p
3-3
1
Ab
OP
PQ
(1)
WA WB
weight of displaced water
Before object is jettisoned
for b p 1 hb 1
b g
0.75 g / cm3 Ÿ SG
Archimedes Ÿ
hU 1
hb 1
.
15
2
Volume displaced: Vd 1
hs 1
WA + WB
Uw u
U B 2 A g Ÿ U soln
3.9
g
( UVg ) block
b g
Expt. 2: U soln A g
3.325 g / cm3
(3)
(4)
i
3.9 (cont’d)
hs 2
WB
WA
Let V A
hU2
h b2
Ab h p 2 hb 2
(6)
E
, solve for dh
WB
pw gAb
p2
hb 2
i
(7)
Ap h p 2 Vd 2 V A
Volume of pond water: Vw
b 5g , b 6 g & b 7 g
Vw
Ap h p 2 WB
W
A
pw g p A g
Vw
WB
WA
Ap pw gAp p A gAp
Ÿ hp 2
h p 21
bg
subst. 8
Ÿ
bg
for h p 2 in 7 , solve for hb 2
i
Archimedes Ÿ U WVd 2 g WB
Subst. for Vd 2
solve for
(5)
d
Volume displaced by boat:Vd 2
After object is jettisoned
h p 2 hb 2
WA
U Ag
volume of jettisoned object =
(8)
Vw
WB
WA
WB
Ap pw gAp p A gAp pw gAb
hb 2
(9)
(a) Change in pond level
h p 2 h p1
LM 1
A gNp
b8gb 3g WA
p
A
1
pW
OP
Q
WA b pW p A g b5gF
p A pW gAp
GH
0 !0 VA
VA
0
pW p A pW Ap
IJ F
K GH
I
JK
Ÿ the pond level falls
(b)
Change in boat level
h p 2 h p1
LM 1
A g MN p A
p
A
Ÿ the boat rises
3.10 (a) U bulk
2.93 kg CaCO 3
L CaCO 3
O
L
I PP ! 0
F
I
p FA
1
M
1
1
P
G JK JK P
p A PQ GH A JK M GH p H A
MN
PQ
! 0
b9 gb4 g WA
p
1
pW Ap
O b5gF VA I M
W
0.70 L CaCO 3
L total
b
p
!0
A
p
W
b
2.05 kg / L
2.05 kg 50 L 9.807 m / s2
1N
100
. u 103 N
L
1 kg ˜ m / s2
Neglected the weight of the bag itself and of the air in the filled bag.
(b) Wbag
U bulkVg
(c) The limestone would fall short of filling three bags, because
– the powder would pack tighter than the original particles.
– you could never recover 100% of what you fed to the mill.
3-4
3.11 (a) Wb
122.5 kg 9.807 m / s2
mb g
Ub
(b)
m f mnf
mb
mf
Ÿ mf
xf
mb
d
V f Vnf
mf
Fx
GH U
f
Vb Ÿ
I
JK
1 x f
U nf
f
1 / U b 1 / U nf
(c) x f
(2)
mb x f
mb 1 x f
Ÿ mb
Uf
mf
mnf
mb x f
mb (1 x f )
mb
mnf
U nf
F
GH
mb
Ub
I
JK
1
1
Ÿ xf
U b U nf
mb
Ub
Fx
GH U
f
1 x f
f
U nf
I (V
JK
lungs
Vother )
mb
b
lungs
nf
b
nf
nf
nf
other
b
lungs
b
1 / U f 1 / U nf
F1 1I
GH U U JK
F 1 1 I 1 1 V V
GH U U JK U U
m
F 1 1 I F V V I F 1 1 I F 12. 01. I
J G
GH U U JK GH m JK GH 103
J
11
.
. K H 122.5 K
Ÿx
F1 1I
FG 1 1 IJ
GH U U JK
H 0.9 11. K
Ÿ xf
f
1 / U b 1 / U nf
Vb
Vlungs Vother
U nf
(3)
mb
1
1
Ÿ xf
Ub
U f U nf
(d) V f Vnf Vlungs Vother
mnf
i
1 / 103
. 1 / 1.1
0.31
1 / 0.9 1 / 1.1
1 / U f 1 / U nf
119 L
(1)
(1),(2) Ÿ mnf
b2 g,b3g
Uf
1202 N
1 kg ˜ m / s2
(1202 N - 44.0 N)
Wb WI
Uwg
0.996 kg / L u 9.807 m / s2
1N
mb 122.5 kg
1.03 kg / L
Vb
119 L
Vb
mf
1N
1 kg ˜ m / s2
other
b
f
f
nf
3-5
0.25
Conc. (g Ile/100 g H2O)
3.12 (a)
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0.987
y = 545.5x - 539.03
R2 = 0.9992
0.989
0.991
0.993
0.995
0.997
Density (g/cm3)
5455
. U 539.03
From the plot above, r
(b) For U = 0.9940 g / cm3 ,
m Ile
150 L
h
r
3.197 g Ile / 100g H 2 O
0.994 g 1000 cm3
3.197 g Ile
1 kg
3
cm
L
103.197 g sol 1000 g
4.6 kg Ile / h
(c) The density of H2O increases as T decreases, therefore the density was higher than it
should have been to use the calibration formula. The valve of r and hence the Ile mass
flow rate calculated in part (b) would be too high.
3.13 (a)
Mass Flow Rate (kg/min)
1.20
1.00
y = 0.0743x + 0.1523
R2 = 0.9989
0.80
0.60
0.40
0.20
0.00
0.0
2.0
4.0
6.0
8.0
10.0
12.0
Rotameter Reading
b g
From the plot, R = 5.3 Ÿ m 0.0743 5.3 01523
0.546 l g / min
.
3-6
3.13 (cont’d)
(b)
Rotamete Collection Collected
r Reading Time
Volume
(min)
(cm3)
2
1
297
2
1
301
4
1
454
4
1
448
6
0.5
300
6
0.5
298
8
0.5
371
8
0.5
377
10
0.5
440
10
0.5
453
Mass Flow
Rate
(kg/min)
0.297
0.301
0.454
0.448
0.600
0.596
0.742
0.754
0.880
0.906
b
Difference
Duplicate
(Di)
Mean Di
0.004
0.006
0.004
0.0104
0.012
0.026
g
1
0.004 0.006 0.004 0.012 0.026 0.0104 kg / min
5
. Di ) kg / min 0.610 r 0.018 kg / min
95% confidence limits: (0.610 r 174
Di
There is roughly a 95% probability that the true flow rate is between 0.532 kg / min
and 0.628 kg / min .
3.14 (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
15.0 kmol C 6 H 6
15.0 kmol C 6 H 6
15,000 mol C 6 H 6
78.114 kg C 6 H 6
117
. u 103 kg C 6 H 6
kmol C 6 H 6
1000 mol
15
. u 104 mol C 6 H 6
kmol
lb - mole
453.6 mol
15,000 mol C 6 H 6
6 mol C
1 mol C 6 H 6
15,000 mol C 6 H 6
6 mol H
1 mol C 6 H 6
90,000 mol C 12.011 g C
mol C
90,000 mol H 1.008 g H
mol H
15,000 mol C 6 H 6
33.07 lb - mole C 6 H 6
90,000 mol C
90,000 mol H
1.08 u 106 g C
9.07 u 104 g H
6.022 u 1023
mol
9.03 u 1027 molecules of C6 H 6
3-7
3.15 (a) m
(b) n
175 m3
h
1000 L
m3
2526 kg
1000 mol 1 min
min
0.866 kg
1h
L 60 min
92.13 kg
60 s
2526 kg / min
457 mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm
3.16 (a)
200.0 kg mix 0.150 kg CH 3OH
kg mix
(b) m mix
3.17
M
m N 2
100.0 lb - mole MA
h
0.25 mol N 2
3000 kg
h
kmol CH 3OH 1000 mol
32.04 kg CH 3OH
74.08 lb m MA
m / V
28.02 g N 2
0.75 mol H 2
mol N 2
kmol 0.25 kmol N 2
8.52 kg
kmol feed
936 mol CH 3OH
8715 lb m / h
2.02 g H 2
8.52 g mol
mol H 2
28.02 kg N 2
2470 kg N 2 h
kmol N 2
215 g 65 g 150 g
500 g / 455 mL 110
. g mL
(c) 150 g CaCO 3 / 500 g suspension
3.19
1 lb m mix
1 lb - mole MA 0.850 lb m MA
3.18 M suspension 565 g 65 g 500 g , M CaCO 3
(a) V 455 mL min , m 500 g min
(b) U
1 kmol
0.300 g CaCO 3 g suspension
Assume 100 mol mix.
mC2 H 5OH
10.0 mol C 2 H 5OH
46.07 g C 2 H 5OH
75.0 mol C 4 H 8 O 2
mol C 2 H 5OH
88.1 g C 4 H 8 O 2
461 g C 2 H 5OH
6608 g C 4 H 8 O 2
mol C 4 H 8 O 2
15.0 mol CH 3COOH 60.05 g CH 3COOH
mCH 3COOH
901 g CH 3COOH
mol CH 3COOH
461 g
xC 2 H 5OH
0.0578 g C2 H 5OH / g mix
461 g + 6608 g + 901 g
6608 g
xC 4 H 8 O 2
0.8291 g C 4 H 8 O 2 / g mix
461 g + 6608 g + 901 g
901 g
xCH 3COOH
0113
.
g CH 3COOH / g mix
461 g + 6608 g + 901 g
461 g + 6608 g + 901 g
MW
79.7 g / mol
100 mol
25 kmol EA 100 kmol mix 79.7 kg mix
m
2660 kg mix
75 kmol EA 1 kmol mix
mC4 H 8O 2
3-8
3.20 (a)
Unit
Crystallizer
Filter
Dryer
(b) m gypsu m
Function
Form solid gypsum particles from a solution
Separate particles from solution
Remove water from filter cake
0.35 kg C aSO 4 ˜ 2 H 2 O
L slurry
1 L slurry
0.35 kg CaSO4 ˜ 2H2O
Vgypsum
CaSO 4 in gypsum: m
CaSO 4 in soln.: m
% recovery =
L CaSO4 ˜ 2H2O
0151
. L CaSO4 ˜ 2H2O
2.32 kg CaSO4 ˜ 2H2O
0.35 kg gypsum 136.15 kg CaSO 4
0.277 kg CaSO 4
172.18 kg gypsum
. g L sol
b1 0151
0.35 kg gypsum
(c) m
0 .35 kg C aSO 4 ˜ 2 H 2 O
1.05 kg 0.209 kg CaSO 4
L
100.209 kg sol
0.209 g CaSO 4
0.05 kg sol
0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 u 10 -5 g
u 100%
0.277 g + 0.00186 g
FB:
45.8 L
0.90 kg
min
L
55.2 L
0.75 kg
min
L
3.84 u 10 -5 kg CaSO 4
99.3%
3.21
CSA:
0.00186 kg CaSO 4
kmol
0.5496
75 kg
kmol
0.4600
90 kg
U|
|V
||
W
kmol
0.5496
min
Ÿ
kmol
0.4600
min
1.2
mol CSA
mol FB
She was wrong.
The mixer would come to a grinding halt and the motor would overheat.
3.22 (a)
150 mol EtOH
6910 g EtO H
V
SG
(b) V c
46.07 g EtOH
6910 g EtOH
mol EtOH
0.600 g H 2 O
10365 g H 2 O
0.400 g EtOH
6910 g EtOH
L
789 g EtOH
10365 g H 2 O
L
1000 g H 2 O
(6910 +10365) g
L
0.903
19.1 L
1000 g
( 6910 10365) g mix
% error
L
935.18 g
(19.123 18.472 ) L
u 100%
18.472 L
18.472 L Ÿ 18.5 L
3.5%
3-9
19.123 L Ÿ 19.1 L
3.23
0.09 mol CH 4
M
16.04 g
0.91 mol Air
29.0 g Air
mol
kmol 0.090 kmol CH 4
700 kg
mol
27.83 g mol
2.264 kmol CH 4 h
h
27.83 kg
1.00 kmol mix
2.264 kmol CH 4
0.91 kmol air
22.89 kmol air h
h
0.09 kmol CH 4
5% CH 4 Ÿ
2.264 kmol CH 4
0.95 kmol air
h
0.05 kmol CH 4
b
43.01 kmol air h
g
Dilution air required: 43.01 - 22.89 kmol air 1000 mol 20200 mol air h
h
1 kmol
20.20 kmol Air 29 kg Air
Product gas: 700 kg 1286 kg h
h
h
kmol Air
43.01 kmol Air 0.21 kmol O2 32.00 kg O2
h
kg O2
0.225
h
1.00 kmol Air 1 kmol O 2 1286 kg total
kg
3.24
m m
¦ Mi V i
i
¦ xi U i
A:
B:
mi
M
, U=
Vi
V
mi
, Ui
M
xi
xi
¦U
1
U
¦
1
M
mi2
¦V zU
i
V
mi Vi
1
1
Vi
Correct.
¦
M
M mi
M
U
0.60
0.25
0.15
1.091 Ÿ U 0.917 g / cm 3
0.791 1.049 1.595
¦
i
xi
Ui
3.25 (a) Basis: 100 mol N 2 Ÿ 20 mol CH 4
N total
x CO
x CH 4
(b) M
Not helpful.
100 20 64 32
R|20 u 80
25
ŸS
|T 20 u 40
25
64 mol CO 2
32 mol CO
216 mol
32
64
0.15 m ol C O / m ol , x CO 2
0.30 m ol C O 2 / m ol
216
216
100
20
0.46 mol N 2 / mol
0.09 mol CH 4 / mol , x N 2
216
216
¦y M
i
i
015
. u 28 0.30 u 44 0.09 u 16 0.46 u 28
3-10
32 g / mol
3.26 (a)
Samples Species
MW
k
1
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
Peak
Area
3.6
2.8
2.4
1.7
Mole
Mass
Fraction Fraction
0.156
0.062
0.233
0.173
0.324
0.353
0.287
0.412
2
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
7.8
2.4
5.6
0.4
0.249
0.146
0.556
0.050
0.111
0.123
0.685
0.081
1.170
0.689
2.615
0.233
18.767
20.712
115.304
13.554
3
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
3.4
4.5
2.6
0.8
0.146
0.371
0.349
0.134
0.064
0.304
0.419
0.212
0.510
1.292
1.214
0.466
8.180
38.835
53.534
27.107
4
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
4.8
2.5
1.3
0.2
0.333
0.332
0.281
0.054
0.173
0.324
0.401
0.102
0.720
0.718
0.607
0.117
11.549
21.575
26.767
6.777
5
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
6.4
7.9
4.8
2.3
0.141
0.333
0.329
0.197
0.059
0.262
0.380
0.299
0.960
2.267
2.242
1.341
15.398
68.178
98.832
77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST
INTEGER N, ND, ID, J
READ (5, *) N
CN-NUMBER OF SPECIES
READ (5, *) (MW(J), K(J), J = 1, N)
READ (5, *) ND
DO 20 ID 1, ND
READ (5, *)(A(J), J = 1, N)
MOLT 0. 0
MASST 0. 0
DO 10 J = 1, N
MOL(J) =
MASS(J) = MOL(J) * MW(J)
MOLT = MOLT + MOL(J)
MASST = MASST + MASS(J)
10
CONTINUE
DO 15 J = 1, N
MOL(J) = MOL(J)/MOLT
MASS(J) = MASS(J)/MASST
15
CONTINUE
WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N)
20 CONTINUE
1 FORMAT (' SAMPLE: `, I3, /,
' SPECIES MOLE FR. MASS FR.', /,
10(3X, I3, 2(5X, F5.3), /), /)
3-11
moles
mass
0.540
0.804
1.121
0.991
8.662
24.164
49.416
57.603
END
$DATA
4
16 . 04
0 . 150
30 . 07 0. 287
44 . 09 0. 467
58 . 12
0 . 583
5
3. 6 2. 8 2 . 4 1. 7
7. 8 2. 4 5. 6 0. 4
3. 4 4 . 5 2. 6 0 . 8
4 . 8 2. 5 1. 3
0. 2
6. 4 7. 9 4 . 8 2 . 3
[OUTPUT]
SAMPLE:
1
SPECIES
MOLE FR MASS FR
1
0.156
0.062
2
0.233
0.173
3
0.324
0.353
0.287
0.412
4
SAMPLE: 2
(ETC.)
3.27 (a)
(8.7 u 10 6 u 0.40) kg C
44 kg CO 2
12 kg C
(11
. u 10 6 u 0.26) kg C 28 kg CO
12 kg C
( 3.8 u 10 5 u 0.10) kg C
m
M
16 kg CH 4
12 kg C
1.28 u 10 7 kg CO 2 Ÿ 2.9 u 105 kmol CO 2
6.67 u 10 5 kg CO Ÿ 2.38 u 10 4 kmol CO
5.07 u 10 4 kg CH 4 Ÿ 3.17 u 10 3 kmol CH 4
(1.28 u 10 7 6.67 u 10 5 5.07 u 10 4 ) kg 1 metric ton
1000 kg
¦y M
i
i
0.915 u 44 0.075 u 28 0.01 u 16
13,500
metric tons
yr
42.5 g / mol
3.28 (a) Basis: 1 liter of solution
1000 mL
1.03 g 5 g H 2 SO 4
mL
100 g
mol H 2 SO 4
98.08 g H 2 SO 4
3.28 (cont’d)
3-12
0.525 mol / L Ÿ 0.525 molar solution
(b) t
V
V
55 gal
(c) u
V
A
t
L
u
55 gal
3.7854 L
min
60 s
gal
87 L
min
3.7854 L 10 3 mL 1.03 g
gal
1L
mL
144 s
0.0500 g H 2 SO 4
g
87 L
m 3 1 min
min 1000 L
60 s (S u 0.06 2 / 4 ) m 2
45 m
88 s
0.513 m / s
1 lbm
453.59 g
23.6 lb m H 2 SO 4
0.513 m / s
3.29 (a)
n1 (mol/min)
0.180 mol C6H14/mol
0.820 mol N2/mol
n2 (mol/min)
0.050 mol C6H14/mol
0.950 mol N2/mol
1.50 L C6H14(l)/min
n 3 (mol C6H14(l)/min)
n3
150
. L 0.659 kg 1000 mol
1147
. mol / min
min
L 86.17 kg
UV
W
|RS
|T
Hexane balance: 0.180n1 0050
. n2 1147
. (mol C6 H14 / min) solve n1 838
. mol / min
Ÿ
n2 = 72.3 mol / min
Nitrogen balance: 0.820n1 0950
. n2 (mol N2 / min)
(b) Hexane recovery
30 mL
3.30
n3
.
1147
u 100%
u 100% 76%
n1
. 838
.
0180
b g
1 L 0.030 mol 172 g
0155
. g Nauseum
103 mL
lL
1 mol
3-13
3.31 (a) kt is dimensionless Ÿ k (min -1 )
(b) A semilog plot of C A vs. t is a straight line Ÿ ln CA
ln(CA)
1
0
-1
-2
-3
-4
-5
y = -0.4137x + 0.2512
R2 = 0.9996
0.0
k
ln CAO kt
5.0
t (min)
10.0
0.414 min 1
Ÿ CAO 1286
lb - moles ft 3
ln CAO 02512
.
.
FG 1b - molesIJ Cc mol 28.317 liter
H ft K liter 1 ft
t cbsg 1 min
t bming
t c 60
60 s
(c) C A
A
3
CA
0.06243C Ac
t
3.32 (a)
(b)
(c)
3
2600 mm Hg
1000 mol
b
g
drop primes
Ÿ
b
g
C A mol / L
b
2.14 exp 0.00693t
g
5.30 mol / L
14.696 psi
50.3 psi
760 mm Hg
275 ft H 2 O 101.325 kPa
33.9 ft H 2 O
822.0 kPa
.
3.00 atm 101325
12 m2
u 105 N m2
1 atm
1002 cm2
30.4 N cm2
280 cm Hg 10 mm 101325
.
u 106 dynes cm2 1002 cm2
(d)
1 cm
760 mm Hg
12 m2
(e) 1 atm 0.06243C Ac
C A 0 exp kt
1334
.
exp 0.419t c 60
200 s Ÿ C A
2.26462 lb - moles
20 cm Hg 10 mm
1 cm
1 atm
760 mm Hg
3-14
0.737 atm
3.733 u 1010
dynes
m2
3.32 (cont’d)
(f)
(g)
b
b25.0 14.696gpsi
760 mm Hg
14.696 psi
(h) 325 mm Hg 760 mm Hg
(i)
g
25.0 psig 760 mm Hg gauge
14.696 psig
35.0 psi 760 mm Hg
14.696 psi
3.33 (a) Pg
Ugh
Ÿ h (m)
b
g
1293 mm Hg gauge
b g
2053 mm Hg abs
b
435 mm Hg gauge
g
13.546 g Hg cm3
1 cm
1540 cm CCl 4
10 mm 1595
g CCl 4 cm3
.
0.92 u 1000 kg 9.81 m / s2
m3
0111
. Pg (kPa)
h (m)
1N
1 kPa
2
3
1 kg ˜ m / s 10 N / m2
h
Pg
Pg
moil
68 kPa Ÿ h
UV
(b) Pg Patm
0111
. u 68 7.55 m
FG 0.92 u 1000 kg IJ u FG 7.55 u S u 16
H
4
m K H
3
2
m3
IJ
K
. u 10 6 kg
14
Ptop Ugh
b
g b g
68 101 115 0.92 u 1000 u 9.81 / 103 h Ÿ h
5.98 m
3.34 (a) Weight of block = Sum of weights of displaced liquids
U 1h1 U 2 h2
(h1 h2 ) AU b g h1 AU 1 g h2 AU 2 g Ÿ U b
h1 h2
(b)
Ptop Patm U1gh0 , Pbottom Patm U1g(h0 h1) U2 gh2 , Wb Ub (h1 h2 ) A
Ÿ Fdown (Patm U1gh0 ) A Ub (h1 h2 ) A , Fup [ Patm U1g(h0 h1) U2 gh2 ]A
Fdown Fup Ÿ Ub (h1 h2 ) A U1gh1 A U2 gh2 A Ÿ Wblock Wliquid displaced
3-15
'P
3.35
bP
atm
g
Ugh Pinside
1 atm 1 atm F
m
3.36
P
154 N 65 cm2
2
cm
. g1000 kg
b105
m3
100
. u 104 N u
9.8066 m 150 m 12 m2
1N
s2
1002 cm2 1 kg ˜ m / s2
lb I
.
FG 022481
H 1 N JK
f
2250 lb f
14
. u 62.43 lb m
1 ft 3
2.3 u 106 gal
2.69 u 107 lb m
UV
3
ft
7.481 gal
P0 Ugh
. u 62.43 lb m 32.174 ft 30 ft
1 lb f
12 ft 2
lb f 14
14.7 2 in
ft 3
s2
32.174 lb m ˜ ft / s2 12 2 in 2
32.9 psi
— Structural flaw in the tank.
— Tank strength inadequate for that much force.
— Molasses corroded tank wall
S u 24 2 u 3 in 3 1 ft 3 8.0 u 62.43 lb m
392 lb m
3.37 (a) mhead
4
12 3 in 3
ft 3
392 lb m
32.174 ft / s 2
1 lb f
W mhead g
392 lb f
32.174 lb m ˜ ft / s2
30 14.7 lb f S u 202 in 2
Fnet Finside Fout W
392 lb f 4415 lb f
in 2
4
The head would blow off.
2
Fnet 4415 lb f 32.174 lb m ˜ ft / s
Initial acceleration: a
362 ft / s2
mhead 392 lb m
1 lb f
b
g
(b) Vent the reactor through a valve to the outside or a hood before removing the head.
3.38 (a)
a
2m
1m
b
Pa Ugh Patm , Pb Patm
If the inside pressure on the door equaled Pa , the force on
the door would be F Adoor ( Pa Pb ) UghAdoor
Since the pressure at every point on the door is greater than
Pa , Since the pressure at every point on the door is greater
than Pa , F >UghAdoor
3-16
3.38 (cont’d)
(b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to
fill.
V 5 u 25
. u 2 ft 3
Vtub
. ft 3 / min Ÿ V 5 u 25
. 125
. ft 3 / min
25
|
t
10 min
(i)
For a full room, h
10 m
. m
1N
10 m 2 m2
1000 kg 981
Ÿ F ! 2.0 u 105 N
3
2
2
m
s 1 kg ˜ m / s
The door will break before the room fills
ŸF!
(ii)
d i
dP i
3.39 (a) Pg
g
If the door holds, it will take
3
35.3145 ft 3
Vroom 5 u 15 u 10 m
t fill
V
12.5 ft 3 / min
1 m3
He will not have enough time.
b
g
1h
60 min
31 h
25 m H 2 O
tap
b
junction
101.3 kPa
245 kPa
10.33 m H 2 O
25 5 m H 2 O
101.3 kPa
294 kPa
10.33 m H 2 O
g
(b) Air in the line. (lowers average density of the water.)
(c) The line could be clogged, or there could be a leak between the junction and the tap.
3.40
Pabs
Pgauge
Patm
800 mm Hg
25 mm Hg
800 25 775 mm Hg
b
3.41 (a) P1 U A g h1 h2
Ÿ P1 P2
bU
B
g
P2 U B gh1 U C gh2
g
b
g
U A gh1 U C U A gh2
LMb10. 0.792g g 981 cm 30.0 cm b137
. 0.792g g
cm
s
cm
N
I 123.0 kPa
F 1 dyne I F
101325
.
kPa
uG
J
G
H 1 g ˜ cm / s K H 1.01325 u 10 dynes / cm JK
(b) P1 121 kPa +
3
2
2
3
6
2
3-17
981 cm 24.0 cm
s2
OP
Q
3.42 (a) U T g (500 x )
U W gR Ÿ R
UT
(500 x )
UW
b
g
0.866
u 500 10 424 cm
1.000
0.866
u 500 400 87 cm
If h 400 cm , R =
1.000
If h 10 cm , R =
b
g
UT
(500 x )
U Hg
For Hg, R c
b
g
0.866
. cm
u 500 10 312
13.6
0.866
. cm
If h 400 cm , R =
u 500 400 637
13.6
If h 10 cm , R =
b
g
Use mercury, because the water manometer would have to be too tall.
(b) If the manometer were simply filled with toluene, the level in the glass tube would be at
the level in the tank.
Advantages of using mercury: smaller manometer; less evaporation.
(c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen,
minimizing the risk of combustion.
3.43
Patm
b
P
g
7.23 g
P
I
U gJ b26 cmg
dU U igb26 cmg FGH 7.23
K
m
U f g 7.23 m Ÿ U f
Pa Pb
atm
atm
f
w
w
F756 mmHg 1 m 1000 kg 9.81m/s
GH 7.23m 100 cm m
2
3
Ÿ Pa Pb
3.44 (a) 'h 900 hl
(b) 'h
Ib g
JK
1N
760 mmHg
1m
26 cm
2
5
2
1 kg˜ m/s 1.01325u10 N m 100 cm
81
. mm Hg
75
. psi 760 mm Hg
388 mm Hg Ÿ hl = 900 388=512 mm
14.696 psi
388 25 u 2
338 mm Ÿ Pg =
338 mm Hg
14.696 psi
760 mm Hg
3-18
6.54 psig
3.45 (a) h = L sin T
(b) h
b8.7 cmg sinb15qg
3.46 (a) P
Patm Poil PHg
2.3 cm H 2 O
23 mm H 2 O
920 kg 9.81 m / s2
765 365 m3
393 mm Hg
0.10 m
1N
760 mm Hg
2
1 kg ˜ m / s 1.01325 u 105 N / m2
(b) — Nonreactive with the vapor in the apparatus.
— Lighter than and immiscible with mercury.
— Low rate of evaporation (low volatility).
c
3.47 (a) Let U f
h
manometer fluid density 110
. g cm 3 , U ac
c0.791 g cm h
acetone density
3
Differential manometer formula: 'P
dU
f
i
U ac gh
. 0791
. gg 981 cm h (mm) 1 cm
1 dyne
g b110
cm
s
10 mm 1 g˜ cm/ s
0.02274 hb mmg
V b mL sg
62
87
107
123
138
151
hb mmg
5
10
15
20
25
30
'Pb mm Hgg 0.114 0.227 0.341 0.455 0.568 0.682
b
'P mm Hg
(b) lnV
3
2
2
b g
n ln 'P ln K
6
ln(V)
5.5
y = 0.4979x + 5.2068
5
4.5
4
-2.5
-2
-1.5
-1
-0.5
0
ln( P)
b g
.
ln 'P 52068
.
From the plot above, ln V 04979
3-19
760 mm Hg
1.01325u106 dyne/ cm2
.
. , ln K
Ÿ n = 04979
| 05
5.2068 Ÿ K
183
ml s
bmm Hgg
0.5
3.47 (cont’d)
(c) h
23 Ÿ 'P
b0.02274gb23g
132 mL 0.791 g
s
mL
104 g s
0.523 mm Hg Ÿ V
104 g
s
1 mol
58.08 g
b
g
183 0.523
0.5
132 mL s
. mol s
180
. 544q R / 18
. 303 K 273 30q C
3.48 (a) T 85q F 4597
. 474q R 460 14q F
(b) T 10q C 273 263 K u 18
(c) 'T
(d)
. qF
. qK
85q C 1.8q R
85q C 18
85q C 10
153q R
153q F;
85q K;
. qC
. qC
10
1q C
10
150q R 1q F
150q R 1.0q C
150q R 1.0$ K
150q F;
83.3q K;
1q R
1.8q R
1.8q R
3.49 (a) T
0.0940 u 1000$ FB 4.00 98.0$ C Ÿ T = 98.0 u 1.8 + 32 = 208$ F
(b) 'T ($ C)
'T ($ F)
0.0940'T ($ FB)
15$ C Ÿ 100$ L ; T2
T ($ C) aT ($ L) b
b43 15g$ C
b1000 - 100g$ L
Ÿ T ($ C)
0.94$ C Ÿ 'T (K)
0.94 K
0.94$ C 1.8$ F
. $R
1.69$ F Ÿ 'T ($ R) 169
$
1.0 C
(c) T1
a
83.3q C
43$ C Ÿ1000$ L
F $ CI ;
GH $ L JK
0.0311
b
15 0.0311 u 100
0.0311T ($ L) 119
. and T ($ L)
119
. $C
32.15T ($ C) 382.6
(d) Tbp
. $ FB Ÿ 3232 $ L
88.6$ C Ÿ 184.6 K Ÿ 332.3$ R Ÿ -127.4$ F Ÿ 9851
(e) 'T
50.0$ L Ÿ 1.56$ C Ÿ 16.6$ FB Ÿ 156
. K Ÿ 2.8$ F Ÿ 2.8$ R
3-20
3.50
bT g
(a) V b mVg
bT g
100q C
b H 2O
455q C
m AgCl
b g
aT q C b
a 0.05524 mV q C
5.27 100a b
Ÿ
b 0.2539 mV
24.88 455a b
V mV 0.05524T q C 0.2539
b g
b g
b g
b g
T qC
1810
. V mV 4.596
. mV o136
. mV Ÿ1856
. q C o2508
. qC Ÿ
(b) 100
ln K n ln R
3.51 (a) ln T
b
g
b
g
b2508. 1856. gq C
20 s
326
. qC / s
KR n
T
ln 250.0 110.0
n
dT
dt
1184
.
ln 40.0 20.0
.
ln K ln 1100
. 1184
. (ln 200
. ) 1154
. Ÿ K 3169
. ŸT 3169
. R1184
FG 320 IJ
H 3169
. K
(b) R
1/1.184
49.3
(c) Extrapolation error, thermocouple reading wrong.
0.08206nT
3.52 (a) PV
.
g 14696
b g Pcbpsig14696
.
bg d i
, V L V c ft 3 u
P atm
b g b
n mol
g
453.59 mol
n c lb - moles u
, T($ K)
lb moles
b Pc 14.696g u V c u 28.317
Ÿ
14.696
b
g
b
g
Ÿ P c 14.696 u V c
0.08206 u nc u
28317
. ft 3
L
T c ($ F) 32
.
27315
1.8
LM
N
OP
Q
453.59 (T c 32)
.
u
27315
1
1.8
0.08206 u 14.696 u 453.59
u n c u T c 459.7
.
28.317 u 18
b
b
g
Ÿ Pc 14.696 V c 1073
. nc T c 4597
.
3-21
g
3.52 (cont’d)
b500 14.696g u 35.
10.73 u b85 459.7g
(b) ntot
c
0308
. lb - mole 0.30 lb - mole CO 28 lbm CO
26
. lb m CO
lb - mole
lb - mole CO
mCO
(c) T c
b3000 14.696g u 35. 459.7
10.73 u 0.308
b g
3.53 (a) T q C
b
2733$ F
g
a u r ohms b
UV Ÿ
. a bW
33028
. a b
0 23624
100
0.308 lb - mole
a 10634
.
b 25122
.
b g
ŸT qC
FG kmol IJ n c (kmol) 1 min n c
H s K min 60 s 60
Pcbmm Hgg
1 atm
Pc
Pbatmg
760 mm Hg 760
F m I m 1 min V c
VG J V c
H s K min 60 s 60
b g
. r ohms 25122
.
10634
(b) n
.
b g T cbq Cg 27316
, TK
3
3
b g d
b g
i
.
Pc mm Hg V c m3 min
0016034
.
Pc
V c
n c 12186
Ÿ n c
.
T c q C 27316
60
.
760 T c 27316
60
(c) T
10.634r 25122
.
r1
26159
.
Ÿ T1
26.95q C
Ÿ r2
26157
.
Ÿ T2
26.93q C
r3
44.789 Ÿ T3
2251
. qC
P (mm Hg)
h Patm
h (29.76 in Hg)
FG 760 mm Hg IJ
H 29.92 in Hg K
h1 232 mm Ÿ P1 987.9 mm Hg
Ÿ h2 156 mm Ÿ P2
h3
9119
. mm Hg
74 mm Ÿ P3 829.9 mm Hg
3-22
h 755.9
3.53 (cont’d)
b0.016034gb987.9gb947 60g 0.8331 kmol CH
.
26.95 27316
b0.016034gb9119. gb195g 9.501 kmol air min
(d) n1
n2
min
.
26.93 27316
n1 n2 10.33 kmol min
n3
b
. g
g b10.33gb2251. 27316
b0.016034gb829.9g
n3 T2 27316
.
0.016034 P3
(e) V3
(f)
4
0.8331 kmol CH 4 16.04 kg CH 4
min
kmol
0.21 u 9.501 kmol O2 32.0 kg O2
xCH4
387 m3 min
kg CH 4
min
0.79 u 9.501 kmol N 2 29.0 kg N 2
13.36
min
kmol O2
min
13.36 kg CH 4 min
0.0465 kg CH 4 kg
(13.36 274) kg / min
REAL, MW, T, SLOPE, INTCPT, KO, E
REAL TIME (100), CA (100), TK (100), X (100), Y(100)
INTEGER IT, N, NT, J
READ 5, MW, NT
DO 10 IT=1, NT
READ 5, TC, N
TK(IT) = TC + 273.15
READ 5, (TIME (J), CA (J), J 1 , N)
DO 1 J=1, N
CA J CA J / MW
3.54
b g
b g
b g
bg bg
XbJg TIMEbJg
YbJg 1./CAbJg
1
CONTINUE
CALL LS (X, Y, N, SLOPE, INTCPT)
b g
K IT
SLOPE
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J 1 , N)
WRITE (6, 3) K (IT)
10 CONTINUE
DO 4 J=1, NT
3-23
kmol N 2
274
kg air
min
bg
YbJg
XJ
bg
LOGcKbJgh
1./TK J
3.54 (cont’d)
4
CONTINUE
CALL LS (X, Y, NT, SLOPE, INTCPT)
KO EXP INTCPT
b
2
3
5
10
g
E 8.314 SLOPE
WRITE (6, 5) KO, E
FORMAT (' TEMPERATURE (K): ', F6.2, /
* ' TIME CA', /,
* ' (MIN) (MOLES)', /
* 100 (IX, F5.2, 3X, F7.4, /))
FORMAT (' K (L/MOL – MIN): ', F5.3, //)
FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4)
END
SUBROUTINE LS (X, Y, N, SLOPE, INTCPT)
REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN
INTEGER N, J
SX=0
SY=0
SXX=0
SXY=0
DO 10 J=1,N
SX = SX + X(J)
SY = SY + Y(J)
SXX = SXX + X(J)**2
SXY = SXY + X(J)*Y(J)
CONTINUE
AN = N
SX = SX/AN
SY = SY/AN
SXX = SXX/AN
SXY = SXY/AN
SLOPE = (SXY – SX*SY)/(SXX – SX**2)
INTCPT = SY – SLOPE*SX
RETURN
END
$ DATA
65.0
94.0
10.0
20.0
4
6
8.1
4.3
[OUTPUT]
TEMPERATURE (K): 367.15
TIME CA
(MIN) (MOLS/L)
10.00 0.1246
3-24
30.0
40.0
50.0
3.54 (cont’d)
3.0
2.2
1.8
20.00 0.0662
30.00 0.0462
40.00 0.0338
60.0
1.5
50.00 0.0277
60.00 0.0231
b
g
K L / MOL ˜ MIN : 0.707
110.
10.0
20.0
30.0
40.0
50.0
60.0
6
3.5
1.8
1.2
0.92
0.73
0.61
127.
6
ETC
bat 94qCg
TEMPERATURE (K): 383.15
b
g
K L / MOL ˜ MIN : 1.758
b
g
E bJ / MOLg: 0.6690E 05
K0 L / MOL MIN : 0.2329E 10
3-25
CHAPTER FOUR
4.1
a.
Continuous, Transient
b.
Input – Output = Accumulation
No reactions Ÿ Generation = 0, Consumption = 0
6.00
c.
t
4.2
kg
kg
3.00
s
s
dn
dn
Ÿ
dt
dt
. m3 1000 kg 1 s
100
1 m3 3.00 kg
3.00
333 s
a.
Continuous, Steady State
b.
k
c.
Input – Output – Consumption = 0
Steady state Ÿ Accumulation = 0
A is a reactant Ÿ Generation = 0
0 Ÿ CA
C A0
k
kg
s
f Ÿ CA
0
FG IJ FG IJ V FG m IJ C FG mol IJ kVC FG mol IJ Ÿ C
HsK
H K H K H sK Hm K
m3
mol
V
C A0
s
m3
4.3
3
A
b
v kg / h
m
a.
100 kg / h
0.550 kg B / kg
0.450 kg T / kg
b
A
CA0
kV
1
V
g
0.850 kg B / kg
0.150 kg T / kg
l kg / h
m
A
3
g
Input – Output = 0
Steady state Ÿ Accumulation = 0
No reaction Ÿ Generation = 0, Consumption = 0
0.106 kg B / kg
0.894 kg T / kg
(1) Total Mass Balance: 100.0 kg / h
v m
l
m
(2) Benzene Balance: 0.550 u 100.0 kg B / h
v
Solve (1) & (2) simultaneously Ÿ m
v 0106
l
0.850m
. m
59.7 kg h, m l
40.3 kg h
b.
The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by
masses (kg). The balance equations are also identical (initial input = final output).
c.
Possible explanations Ÿ a chemical reaction is taking place, the process is not at steady state,
the feed composition is incorrect, the flow rates are not what they are supposed to be, other
species are in the feed stream, measurement errors.
4-1
4.4
b.
c.
b
n (mol)
0.500 mol N 2 mol
0.500 mol CH 4 mol
100.0 g / s
b
g
bg C H g g
bg C H g g
xB
d.
3
8
4
10
b
b
b
b
g
U|
R|n blb - mole DA sg
S| 0.21 lb - moleO lb - mole DA V|
T 0.79 lb - mole N lb - mole DAW
g
g
26.45 x E lb - mole C2 H 6 / h
n1 lb - mole H 2 O s
0.014 n kg N 2
100 x E g C2 H 6 1 lb m lb - mole C2 H 6 3600 s
s
h
453593
. g 30 lb m C2 H 6
n E
x E g C2 H 6 g
xP
g
0.500 n mol N 2 28 g N 2 1 kg
mol N 2 1000 g
nO 2
b
0.21n 2 lb - mole O 2 / s
g
g
2
2
x H 2O
2
xO 2
e.
b g
n mol
n N 2O 4
0.400 mol NO mol
b
y NO2 mol NO 2 mol
b
g
0.600 y NO2 mol N 2O 4 mol
4.5
a.
FG
H
lb - mole H 2 O
n1
lb - mole
n1 n 2
FG
H
0.21n 2 lb - mole O 2
lb - mole
n1 n 2
IJ
K
IJ
K
b
n 0.600 y NO 2 mol N 2 O 4
g
Basis: 1000 lbm C3H8 / h fresh feed
(Could also take 1 h operation as basis flow chart would be as below except
that all / h would be deleted.)
1000 lb m C3H8 / h
g
b
b
n 6 lb m / h
n 7 lb m / h
g
0.02 lb m C3H 8 / lb m
0.98 lb m C3H 6 / lb m
g
0.97 lb m C3H 8 / lb m
0.03 lb m C3H 6 / lb m
Still
Compressor
b
n blb
n blb
n blb
g
C H / hg
CH / h g
H / hg
b
b
n1 lb m C3H 8 / h
n1 lb m C3H 8 / h
Reactor
2
m
3
m
4
m
3
4
2
Note: the compressor and the off gas from
the absorber are not mentioned explicitly
in the process description, but their presence
should be inferred.
b
b
n3 lb m CH 4 / h
n 4 lb m H 2 / h
g
g
b
n5 lb m / h
g
Stripper
Absorber
b
b
n blb
n1 lb m C3H 8 / h
g
g
n 2 lb m C3H 6 / h
5
4-2
g
g
n 2 lb m C3H 6 / h
6
m
oil / h
g
4.5 (cont’d)
b. Overall objective: To produce C3H6 from C3H8.
Preheater function: Raise temperature of the reactants to raise the reaction rate.
Reactor function: Convert C3H8 to C3H6.
Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other
components.
Stripping tower function: Recover the C3H8 and C3H6 from the solvent.
Distillation column function: Separate the C3H5 from the C3H8.
4.6
a.
3 independent balances (one for each species)
b.
1 , m 3 , m 5 , x2 , y2 , y4 , z4 )
7 unknowns ( m
– 3 balances
– 2 mole fraction summations
2 unknowns must be specified
c.
y2
1 x2
FG kg A IJ
H hK
b gb g FGH kghA IJK
F kg I
F kg I
5300 G J m 1200 m G J
Overall Balance: m
H hK
H hK
F kg BIJ 1200 y 0.60m FG kg BIJ
5300 x G
B Balance: 0.03m
H hK
H hK
A Balance: 5300 x2
3 1200 0.70
m
1
3
1
z4
4.7
a.
2
4
5
1 0.70 y4
3 independent balances (one for each species)
b.
Water Balance:
400 g 0.885 g H 2O
g
min
Acetic Acid Balance:
bg
b g
R g 0.995 g H O
m
2
R
Ÿm
min
g
F g CH OOH IJ
. gG
b400gb0115
H min K
3
E
Ÿm
356 g min
R 0.096m
E
0.005m
FG g CH OOH IJ
H min K
3
461g min
FG g IJ m m FG g IJ Ÿ m 417 g min
H min K
H min K
F g IJ b0.096gb461g FG g IJ Ÿ 44 g min = 44 g min
. gb400g b0.005gb356g G
b0115
H min K
H min K
C 400
Overall Balance: m
c.
5
R
4-3
E
C
4.7 (cont’d)
d.
CH3COOH
H 2O
some CH3COOH
CH3COOH
H 2O
C4 H9OH
C4 H 9OH
Extractor
Distillation
Column
CH 3COOH
C4 H9OH
4.8
a.
X-large: 25 broken eggs/min
35 unbroken eggs/min
120 eggs/min
0.30 broken egg/egg
0.70 unbroken egg/egg
b.
120
n1 n2
b
g
25 35 n1 n2 eggs min Ÿ n1 n2
b0.30gb120g
c.
Large: n 1 broken eggs/min
n 2 unbroken eggs/min
25 n1
50
U| n
V| Ÿ n
W
1
2
11
39
50 large eggs min
b11 50g 0.22
22% of the large eggs (right hand) and b25 70g Ÿ 36% of the extra-large eggs (left hand)
n1 large eggs broken/50 large eggs
d.
are broken. Since it does not require much strength to break an egg, the left hand is probably
poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.
4.9
a.
b
m1 lb m strawberries
g
b
m3 lb m W evaporated
015
. lb m S / lb m
0.85 lb m W / lb m
c
m2 lb m S sugar
b.
1.00 lb m jam
0.667 lb m S / lb m
0.333 lb m W / lb m
h
3 unknowns ( m1 , m2 , m3 )
– 2 balances
– 1 feed ratio
0 DF
4-4
g
4.10
a.
300 gal
b g
m1 lb m
0.750 lb m C 2 H 5OH / lb m
0.250 lb m H 2 O / lb m
b g
m3 lb m
0.600 lb m C 2 H 5OH / lb m
0.400 lb m H 2 O / lb m
b g
m blb g
4 unknowns ( m1 , m2 ,V40 , m3 )
– 2 balances
– 2 specific gravities
0 DF
V40 gal
2
m
0.400 lb m C 2 H 5 OH / lb m
0.600 lb m H 2 O / lb m
b.
300 gal
m1
1ft 3
0.877 u 62.4 lb m
7.4805 gal
ft 3
Overall balance: m1 m2 m3
C2H5OH balance: 0.750m1 0.400m2
Solve (1) & (2) simultaneously Ÿ m2
1646 lb m
V40
4.11
a.
b
n1 mol / s
(1)
(2)
0.600m3
1646 lb m, , m3
ft 3 7.4805 gal
0.952 u 62.4lb m
1ft 3
3841 lb m
207 gal
3 unknowns ( n1 , n2 , n3 )
– 2 balances
1 DF
g
0.0403 mol C3H 8 / mol
0.9597 mol air / mol
b
2195 lb m
n 2 mol air / s
b
n3 mol / s
g
0.0205 mol C3H 8 / mol
0.9795 mol air / mol
g
0.21mol O 2 / mol
0.79 mol N 2 / mol
b.
Propane feed rate: 0.0403n1
Propane balance: 0.0403n1
Overall balance: 3722 n2
c.
b
g
150 Ÿ n1 3722 mol / s
0.0205n3 Ÿ n3 7317 mol / s
7317 Ÿ n2 3600 mol / s
b
b
g
g
! . The dilution rate should be greater than the value calculated to ensure that ignition is not
possible even if the fuel feed rate increases slightly.
4-5
4.12
a.
b
kg / h
m
g
0.960 kg CH3OH / kg
0.040 kg H 2O / kg
1000 kg / h
0.500 kg CH 3OH / kg
,x )
2 unknowns ( m
– 2 balances
0 DF
0.500 kg H 2O / kg
673 kg / h
b
g
1 x b kg H O / kg g
x kg CH3OH / kg
2
b.
Overall balance: 1000
673 Ÿ m
327 kg / h
m
b g
Methanol balance: 0.500 1000
b g b g
0.960 327 x 673 Ÿ x
Molar flow rates of methanol and water:
673 kg 0.276 kg CH3OH 1000 g mol CH3OH
h
kg
kg 32.0 g CH3OH
0.276 kg CH3OH / kg
5.80 u 103 mol CH3OH / h
673 kg 0.724 kg H 2 O 1000 g mol H 2 O
2.71 u 104 mol H 2 O / h
h
kg
kg 18 g H 2O
Mole fraction of Methanol:
5.80 u 103
. mol CH3OH / mol
0176
5.80 u 103 2.71 u 104
c.
4.13
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the
system is not at steady state.
a.
Product
Feed
Reactor
2253 kg
Reactor effluent
1239 kg
Purifier
2253 kg
R = 583
W aste
b g
R = 388
m w kg
R = 140
Analyzer Calibration Data
1
xp
x p = 0.000145R
1.364546
0.1
0.01
100
R
4-6
1000
4.13 (cont’d)
b.
Effluent: x p
1.3645
1.3645
Product: x p
1.3645
Waste: x p
Efficiency
c.
b g 0.494 kg P / kg
0.000145b583g
0.861 kg P / kg
. kg P / kg
0.000145b140g
0123
0.861b1239g
u 100% 95.8%
0.494b2253g
0.000145 388
Mass balance on purifier: 2253 1239 mw Ÿ mw
P balance on purifier:
Input: 0.494 kg P / kg 2253 kg 1113 kg P
1014 kg
b
gb
g
Output: b0.861 kg P / kggb1239 kgg b0123
.
kg P / kggb1014 kgg
1192 kg P
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation
beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady
state; additional reaction occurs in purifier; normal data scatter.
4.14
a.
b
g
n1 lb- mole/ h
.
00100
lb- mole H2O/ lb- mole
.
09900
lb- mole DA/ lb- mole
b
v dft / hi
b
g
n3 lb- mole/ h
0100
. lb- mole H2O / lb- mole
0900
. lb- mole DA/ lb- mole
g
h
n2 lb- mole HO/
2
3
2
4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF
aR b
96.9 40.0
1.626
50 15
Assume linear relationship: v
v2 v1
R2 R1
Slope: a
va aR1
Intercept: b
v2
b g
b g
40.0 1.626 15
c
1.626 95 15.61 170 ft / h
3
h
b
3
n 2
170 ft 62 .4 lb m lb - mol
h
ft 3
18.0 lb m
589 lb - moles H 2 O / h
DA balance: 0.9900n1 0.900n3
Overall balance: n1 n2 n3
Solve (1) & (2) simultaneously Ÿ n1
b.
15.61
g
5890 lb - moles / h, n 3
(1)
(2)
6480 lb - moles / h
Bad calibration data, not at steady state, leaks, 7% value is wrong, v R relationship is not
linear, extrapolation of analyzer correlation leads to error.
4-7
4.15
a.
b
kg / s
m
100 kg / s
0.600 kg E / kg
g
0.900 kg E / kg
.
kg H 2 O / kg
0100
0.050 kg S / kg
0.350 kg H 2 O / kg
b
kg / s
m
g
, xE , xS )
3 unknowns ( m
– 3 balances
0 DF
b
g
x b kgS / kg g
1 x x b kg H O / kg g
x E kg E / kg
S
E
b.
2
g
0100
. b kgS / kgg
2m Ÿ m 50.0 kg / s
Overall balance: 100
b g x b50g Ÿ x
S balance: 0.050 100
b g
E balance: 0.600 100
kg E in bottom stream
kg E in feed
c.
b
S
S
S
0.300 kg E / kg
b g b g
0.300b50g
kg E in bottom stream
0.25
0.600b100g
kg Ein feed
0.900 50 x E 50 Ÿ x E
b g lnbag b lnbRg
lnb x / x g lnb0.400 / 0100
. g
1491
.
b
lnb R / R g
lnb38 / 15g
lnbag lnb x g b lnb R g lnb0100
. g 1491
. lnb15g
x aRb Ÿ ln x
2
1
2
1
1
1
6.340 Ÿ a 1764
u 103
.
u 103 R1.491
.
x 1764
R
d.
FG x IJ FG 0.900 IJ
H a K H 1764
u 10 K
.
1
b
3
1
1.491
655
.
Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass
fractions – measure against known standard. Impurities in the stream – analyze a sample.
Mixture is not all liquid – check sample. Calibration data are temperature dependent – check
calibration at various temperatures. System is not at steady state – take more measurements.
Scatter in data – take more measurements.
4-8
4.16
a.
4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution
.
L of solution
mol H 2SO 4 1213
kgsolution
b
0.323 kg H 2SO 4 / kgsolution
bg
b.
5 unknowns ( v1 , v2 , v3 , m2 , m3 )
– 2 balances
– 3 specific gravities
0 DF
v1 L
bg
m b kg g
100 kg
0.200 kg H 2SO 4 / kg
0.800 kg H 2 O / kg
.
SG 1139
g
v3 L
3
0.323 kg H 2SO 4 / kg
0.677 kg H 2 O / kg
bg
m b kg g
v2 L
SG
1.213
2
0.600 kg H 2SO 4 / kg
0.400 kg H 2 O / kg
SG 1.498
Overall mass balance: 100 m2
b g
UV Ÿ m
m
0.677m W
m3
Water balance: 0.800 100 0.400m2
v1
v2
v1
v2
c.
4.17
100 kg
L
1139
. kg
44.4 kg
87.80
29.64
L
1498
.
kg
2.96
3
2
44.4 kg
3
144 kg
87.80 L20%solution
29.64 L 60% solution
L 20%solution
L 60% solution
1250 kg P 44.4 kg 60%solution
L
144 kg P
1498
h
kgsolution
.
257 L / h
b g
m1 kg @$18 / kg
0.25 kg P / kg
0.75 kg H2O / kg
100
. kg
017
. kg P/ kg
0.83 kg H2O / kg
b g
m2 kg @$10 / kg
012
. kg P / kg
0.88 kg H 2O / kg
Overall balance: m1 m2
100
.
. m2
Pigment balance: 0.25m1 012
b g
Solve (1) and (2) simultaneously Ÿ m 0.385 kg 25% paint, m
Cost of blend: 0.385b$18.00g 0.615b$10.00g $13.08 per kg
. b$13.08g $14.39 per kg
Selling price: 110
017
. 100
.
1
4-9
2
(1)
(2)
0.615 kg12% paint
4.18
b
a.
gb
m1 kg H 2O 85% of entering water
g
100 kg
0.800 kgS / kg
0.200 kg H 2 O / kg
b g
m b kg H Og
m2 kgS
3
2
b
gb g 17.0 kg H O
0.800b100g 80.0 kgS
0.850 0.200 100
85% drying: m1
Sugar balance: m2
2
Overall balance: 100 17 80 m3 Ÿ m3
3 kg H 2 O
0.0361 kg H 2O / kg
xw
3 80 kg
b
g
17 kg H 2O
80 3 kg
m1
m2 m3
b.
3 kg H 2O
b
g
0.205 kg H 2O / kg wet sugar
1000 tons wet sugar
3 tons H 2 O
100 tons wet sugar
day
30 tons H 2 O / day
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365 days
day
ton WS
ton
lb m
year
c.
b
b
g
1
xw1 x w 2 ... xw10 0.0504 kg H 2 O / kg
10
1
2
2
SD
0.00181 kg H 2 O / kg
xw1 xw ... xw10 x w
9
Endpoints 0.0504 r 3 0.00181
xw
g
b
b
Lower limit
4.19
$8.8 u 10 7 per year
g
g
0.0450, Upper limit
0.0558
d.
The evaporator is probably not working according to design specifications since
x w 0.0361 0.0450 .
a.
v1 m 3
c h
m b kg H O g
1
2
1.00
SG
d i
m b kg suspension g
v3 m 3
3
d i
v2 m
3
SG
1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 )
– 1 mass balance
– 1 volume balance
– 3 specific gravities
0 DF
400 kg galena
SG
7.44
Total mass balance: m1 400
m3
(1)
4-10
4.19 (cont’d)
Assume volume additivity:
b g
m1 kg
400 kg m 3
m3
1000 kg
7440 kg
Solve (1) and (2) simultaneously Ÿ m1
668 kg
v1
4.20
b g
m3 kg
m3
(2)
1480 kg
668 kg H 2 O, m3 1068 kg suspension
3
m
1000 kg
0.668 m 3 water fed to tank
b.
Specific gravity of coal < 1.48 < Specific gravity of slate
c.
The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
a.
b
n1 mol / h
g
b
n2 mol / h
b
0.040 mol H 2 O / mol
0.960 mol DA / mol
g
x mol H 2O / mol
b
g
1 x mol DA / mol
b
n3 mol H 2 O adsorbed / h
g
g
97% of H 2O in feed
. 3.40g kg
b354
Adsorption rate: n3
b
g
0.97 0.04n1 Ÿ n1
97% adsorbed: 156
.
Total mole balance: n1
b g
.
1556
mol H 2 O / h
401
. mol / h
n2 n3 Ÿ n 2
.
Water balance: 0.040 401
4.21
5h
mol H 2O
0.0180 kg H 2 O
. 1556
.
401
b
g
38.54 mol / h
b
x 38.54 Ÿ x 12
1566
.
. u 10 5 mol H 2O / mol
g
b.
The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction
will reach that of the inlet stream, i.e. 4%.
a.
300 lb m / h
0.55 lb m H 2SO 4 / lb m
b
0.45 lb m H 2 O / lb m
b
B lb m / h
m
C lb m / h
m
g
g
0.75 lb m H 2SO 4 / lb m
0.25 lb m H 2 O / lb m
0.90 lb m H 2SO 4 / lb m
010
. lb m H 2 O / lb m
B
Overall balance: 300 m
m C
B
H2SO4 balance: 0.55 300 0.90m
b g
(1)
0.75m C
B
Solve (1) and (2) simultaneously Ÿ m
4-11
C
400 lb m / h, m
(2)
700 lb m / h
4.21 (cont’d)
b.
500 150
A 7.78 RA 44.4
RA 25 Ÿ m
70 25
800 200
B 200
B 15.0 RB 100
m
RB 20 Ÿ m
60 20
ln 100 ln 20
Ÿ x 6.841e0.2682 Rx
ln x ln 20
.
Rx 4 Ÿ ln x 0.2682 Rx 1923
10 4
300 44.4
400 100
44.3, mB 400 Ÿ RB
333
.,
mA 300 Ÿ RA
7.78
15.0
1
55
ln
7.78
x 55% Ÿ Rx
0.268
6.841
b
b
A 150
m
g
g
b
g
IJ
K
FG
H
c.
A m B
Overall balance: m
m C
A 0.90m B
H2SO4 balance: 0.01xm
d
0.75 0.01 6.841e 0.2682 Rx
Ÿ 15.0 RB 100
Ÿ RB
d2.59 0.236e
Check: RA
4.22
a.
44.3, Rx
b
n A kmol / h
0.2682 Rx
iR
b
b
g
0.75 m A m B Ÿ m B
i b7.78 R
A
b0.75 0.01xgm
g
44.4
135
. e0.2682 Rx 813
.
A
7.78 Ÿ RB
e2.59 0.236e
b g
0.2682 7.78
. e
j44.3 135
b g 813
.
0.2682 7.78
g
100 kg / h
b
n P kmol / h
g
0.20 kmol H 2 / kmol
0.80 kmol N 2 / kmol
g
0.50 kmol H 2 / kmol
0.50 kmol N 2 / kmol
MW
b
g
b
A
.
015
.
015
. kmol H 2 / kmol
010
0.90 kmol N 2 / kmol
n B kmol / h
0.75m C
g
0.20 2.016 0.80 28.012
22.813 kg / kmol
100 kg kmol
4.38 kmol / h
h 22.813 kg
Overall balance: n A n B 4.38
H2 balance: 010
. n A 0.50n B 0.20 4.38
Ÿ n P
(1)
b g
Solve (1) and (2) simultaneously Ÿ n A
4-12
(2)
3.29 kmol / h, n B
. kmol / h
110
333
.
4.22 (cont’d)
b.
n P
m P
22.813
Overall balance: n A n B
H2 balance: x A n A x B n B
Ÿ
c.
Trial
1
2
3
4
5
6
7
8
9
10
11
12
n A
b
b
m P
22.813
x P m P
22.813
P xB x P
m
22.813 x B x A
XA
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
g
g
b
b
P xP x A
m
22.813 x B x A
n B
XB
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
0.50
XP
0.10
0.20
0.30
0.40
0.50
0.60
0.10
0.20
0.30
0.40
0.50
0.60
mP
100
100
100
100
100
100
250
250
250
250
250
250
g
g
nA
4.38
3.29
2.19
1.10
0.00
-1.10
10.96
8.22
5.48
2.74
0.00
-2.74
nB
0.00
1.10
2.19
3.29
4.38
5.48
0.00
2.74
5.48
8.22
10.96
13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot
blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture.
d.
4.23
Results are the same as in part c.
Venous blood
195.0 ml / min
175
. mg urea / ml
Arterial blood
200.0 ml / min
190
. mg urea / ml
Dialysate
b
b
Dialyzing fluid
1500 ml / min
a.
g
Water removal rate: 200.0 195.0 5.0 ml / min
b
g
b
g
Urea removal rate: 190
. 200.0 175
. 195.0
b.
g
v ml / min
c mg urea / ml
v 1500 5.0 1505ml / min
38.8 mg urea / min
0.258 mg urea / ml
c
1505 ml / min
4-13
38.8 mg urea / min
4.23 (cont’d)
c.
1 min
10 3 ml 5.0 L
2.7 11
. mg removed
206 min (3.4 h)
ml
38.8 mg removed
1L
b
4.24
a.
g
b
n1 kmol / min
g
b
20.0 kg CO 2 / min
b
n 2 kmol / min
n3 kmol / min
g
g
0.023 kmol CO 2 / kmol
0.015 kmol CO 2 / kmol
20.0 kg CO 2
kmol
0.455 kmol CO 2 / min
44.0 kg CO 2
min
Overall balance: 0.455 n 2 n3
CO2 balance: 0.455 0.015n 2 0.023n3
Solve (1) and (2) simultaneously Ÿ n 2 55.6 kmol / min, n3
n1
b.
4.25
u
150 m
18 s
8.33 m / s
A
1 2
SD
4
. kmol
561
1 min s
m3
Ÿ D 108
. m
min 0123
. kmol 60 s 8.33 m
Spectrophotometer calibration: C
Dye concentration: A
0.60 cm 3
Dye injected
b
g bg
Ÿ 3.0 P g V L
4.26
a.
1000 L B / min
3
1
1
2
1
A 0.9
C 3
. gb018
. g
b3333
g
3.333 A
3.0 P g
5.0 L
b
g
y b kmol SO / kmolg
1 y b kmol A / kmolg
b kg / min g
m
x b kg SO / kgg
1 x b kg B / kg g
n3 kmol / min
2
3
3
4
2
4
4
4-14
561
. kmol / min
0.600 Pg / L
1L
5.0 mg 103 P g
1L
103 cm 3
1 mg
0.600 P g / L Ÿ V
b
g
V d m / min i
n b kmol / min g
y b kmol SO / kmolg
1 y b kmol A / kmolg
2 kg B / min
m
1
018
. ŸC
b
! C Pg / L
kA
(1)
(2)
4.26 (cont’d)
2 , m 4 , x4 , y1 , y3 )
8 unknowns ( n1 , n3 , v1 , m
– 3 material balances
– 2 analyzer readings
– 1 meter reading
– 1 gas density formula
– 1 specific gravity
0 DF
b.
Orifice meter calibration:
A log plot of V vs. h is a line through the points h1 100, V1 142 and h2
d
ln V
b ln h ln a Ÿ V ah
ln V2 V1
ln 290 142
2
d
1
1
2.58 Ÿ a
1
e 2.58
13.2 Ÿ V
13.2h 0.515
Analyzer calibration:
ln y bR ln a Ÿ y aebR
b
ln a
b
ln y 2 y1
h1
R2 R1
ln y1 bR1
E
0.0600
5.00 u 10 4 e 0.0600 R
b g 207.3 m h
b12.2g b150 14.7g 14.7 batmg 0.460 mol / L = 0.460 kmol / m
b75 460g 18. bKg
E
210 mm Ÿ V1 13.2 210
U feed gas
n1
U|
90 20
||
lnb0.00166g 0.0600b20g 7.60V Ÿ y
||
|W
0.00166g
.
g lnb01107
a 5.00 u 10 4
c.
207.3 m 3 0.460 kmol
min
m3
0.515
3
. kmol min
9534
b
g
expb0.0600 u 116
. g
. u 104 exp 0.0600 u 82.4
R1 82.4 Ÿ y1 500
0.0702 kmol SO2 kmol
. Ÿ y3 500
. u 104
R3 116
0.00100 kmol SO2 kmol
2
m
400, V2
b
d h b
g 0.515
lnbh h g lnb400 100g
ln a ln V b ln h lnb142g 0.515 ln 100
b
i
. kg
1000 L B 130
1300 kg / min
min
LB
4-15
3
i
290 .
4.26 (cont’d)
A balance: 1 0.0702 95.34
b
SO2
gb g b1 0.00100gn Ÿ n 88.7 kmol min
x
. g( 64.0 kg / kmol) b0.00100gb88.7g(64) m
balance: b0.0702gb9534
3
3
4 4
(1)
4 (1 x4 )
B balance: 1300 = m
(2)
4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg
Solve (1) and (2) simultaneously Ÿ m
4 x4
SO2 removed = m
4.27
422 kg SO 2 / min
d.
Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a
higher rate of transfer of SO2 from the gas to the liquid phase.
a.
V2 m 3 / min
b
g
y b kmolSO / kmol g
1 y b kmol A / kmolg
d
i
b kg B / min g
m
n 3 kmol / min
3
d
i
n b kmol / min g
y b kmolSO / kmolg
1 y b kmol A / kmol g
V1 m 3 / min
R3
b
g
x b kgSO kg g
1 x b kg B / kg g
4 kg / min
m
1
1
2
3
2
2
4
1
2
4
P1 , T1 , R1 , h1
b.
2 , n3 , y3 , R3 , m
4 , x4 )
14 unknowns ( n1 ,V1 , y1 , P1 , T1 , R1 , h1 ,V2 , m
– 3 material balances
– 3 analyzer and orifice meter readings
– 1 gas density formula (relates n1 and V1 )
2 and V2 )
– 1 specific gravity (relates m
6 DF
b
g b1 y gn
A balance: 1 y1 n1
SO2 balance: y1n1
2
B balance: m
3
y3n3 b1 x gm
4
Calibration formulas:
(1)
3
4
x4 m
64 kgSO 2 / kmol
(2)
(3)
4
y1
5.00 u 104 e0.060 R1
(4)
4 0.060 R3
y3 5.00 u 10 e
V1 13.2h10.515
Gas density formula: n1
b
g
(6)
12.2 P1 14.7 / 14.7
bT 460g / 18.
1
Liquid specific gravity: SG 130
. Ÿ V2
4-16
(5)
V1
(7)
b g
(8)
2 kg
m
m3
h
1300 kg
4.27 (cont’d)
c.
T1
75 °F
y1
0.07 kmol SO2/kmol
P1
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
82.4
x4 (kg SO2/kg)
Trial
1
2
3
4
5
6
7
8
9
10
y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h)
0.10
0.10
0.10
0.10
0.10
0.20
0.20
0.20
0.20
0.20
0.050
0.025
0.010
0.005
0.001
0.050
0.025
0.010
0.005
0.001
0.89
1.95
2.56
2.76
2.92
0.39
0.87
1.14
1.23
1.30
93.25
90.86
89.48
89.03
88.68
93.25
90.86
89.48
89.03
88.68
m4 (kg/h)
1283.45
2813.72
3694.78
3982.57
4210.72
641.73
1406.86
1847.39
1991.28
2105.36
m2 (kg/h)
1155.11
2532.35
3325.31
3584.31
3789.65
513.38
1125.49
1477.91
1593.03
1684.29
3
V 2 ( m /h )
V2 vs. y3
3 .5 0
3 .0 0
2 .5 0
2 .0 0
1 .5 0
1 .0 0
0 .5 0
0 .0 0
0 .0 0 0
0 .0 2 0
0 .0 4 0
0 .0 6 0
y 3 ( k m o l S O 2 / k m o l)
x4 = 0 .1 0
x4 = 0 .2 0
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed
rate ( V2 ).
For a given SO2 removal rate (y3), a higher solvent feed rate ( V2 ) tends to a more dilute
SO2 solution at the outlet (lower x4).
d.
4.28
Answers are the same as in part c.
Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3
3
Overall mass balance Ÿ m
1
Mass balance - Unit 1 Ÿ m
A balance - Unit 1 Ÿ x1
2
Mass balance - mixing point Ÿ m
A balance - mixing point Ÿ x2
C balance - mixing point Ÿ y2
4-17
4.29
a.
100 mol / h
0.300 mol B / mol
0.250 mol T / mol
0.450 mol X / mol
b
n 2 mol / h
b
g
b
g
Column 1 x b mol T / molg
1 x x b mol X / mol g
n b mol / h g
n 4 mol / h
0.940 mol B / mol
0.060 mol T / mol
x B 2 mol B / mol
T2
B2
Column 2
T2
3
0.020 mol T / mol
b
n5 mol / h
b
g
x b mol T / molg
1 x x b mol X / mol g
T5
B5
Column 1
4 unknowns ( n2 , n3 , x B 2 , xT 2 )
–3 balances
– 1 recovery of X in bot. (96%)
0 DF
b
Total mole balance: 100
b g
T balance: 0.250b100g
B balance: 0.300 100
Total mole balance: n2
b.
0.98n3
(1)
n 2 n3
(2)
x B 2 n 2
(3)
xT 2 n 2 0.020n3
(4)
0.940n 4
(5)
n4 n5
(6)
B balance: x B 2 n 2
0.940n 4 x B5 n5
(7)
T balance: xT 2 n 2
0.060n 4 xT 5 n5
(8)
(1) Ÿ n 3 44.1 mol / h
(3) Ÿ x B 2 0.536 mol B / mol
(5) Ÿ n 4 30.95 mol / h
(7) Ÿ x B5 0.036 mol B / mol
b
(2) Ÿ n 2 55.9 mol / h
(4) Ÿ x T 2 0.431 mol T / mol
(6) Ÿ n5 24.96 mol / h
(8) Ÿ x T 5 0.892 mol T / mol
g
0.940 30.95
u 100%
0.300 100
b g
0.892b24.96g
u 100
Overall toluene recovery:
0.250b100g
Overall benzene recovery:
T5
Column 2:
4 unknowns ( n3 , n4 , n5 , y x )
– 3 balances
– 1 recovery of B in top (97%)
0 DF
gb g
Column 2
97% B recovery: 0.97 x B 2 n 2
g
x B5 mol B / mol
0.980 mol X / mol
Column 1
96% X recovery: 0.96 0.450 100
4-18
g
97%
89%
4.30
a.
100 kg / h
0.035 kg S / kg
0.965 kg H 2 O / kg
b
w kg H 2O / h
m
b.
b
b
3 kg / h
m
g
x3 kg S / kg
1
b
1 x3 kg H 2O / kg
b
w kg H 2O / h
0100
. m
g
b g
x b kg S / kgg
1 x b kg H O / kgg
4
g
4
4
g
b
m 10 ( kg / h)
0.050 kg S/kg
0.950 kg H2O/kg
m w ( kg H 2 O / h)
b g
0.050m 10
Salt balance: 0.035 100
Overall balance: 100
H2O yield: Yw
m w m 10
b
b
g
w kg H 2O recovered
m
96.5 kg H 2 O in fresh feed
g
First 4 evaporators
b
b
g
m
4 kg / h
x 4 kg S/ kg
1 x4 kg H2 O / kg
100 kg/ h
0.035 kg S/ kg
0.965 kg H2 O / kg
b g
Yw
0.31
x4
0.0398
b
g
.
4 0100
m w m 4
Salt balance: 0.035 100
c.
g
b
g
. m
4 u 0100
b kg H O / hg
w
Overall balance: 100
2
w kg H 2O / h
0100
. m
Overall process
100 kg/h
0.035 kg S/kg
0.965 kg H2O/kg
b
4 kg / h
m
g
x4 m 4
4-19
2
g
10 kg / h
m
10
g
0.050 kg S / kg
0.950 kg H 2O / kg
b
w kg H 2O / h
0100
. m
g
4.31
b g
a.
2n1 mol
Condenser
0.97 mol B / mol
0.03 mol T / mol
b g
b g
100 mol
0.50 mol B / mol
Still
0.50 mol T / mol
n1 mol
n1 mol (89.2% of Bin feed )
0.97 mol B / mol
0.03 mol T / mol
0.97 mol B / mol
0.03 mol T / mol
b gb
y b mol B / molg
1 y b mol T / mol g
n 4 mol 45% of feed to reboiler
g
B
B
b g
z b mol B / molg
1 z b mol T / molg
n 2 mol
B
b g
x b mol B / molg
1 x b mol T / molg
n3 mol
Reboiler
B
B
B
Overall process:
Condenser:
3 unknowns ( n1 , n3 , x B )
Still: 5 unknowns ( n1 , n2 , n4 , y B , z B )
– 2 balances
– 2 balances
– 1 relationship (89.2% recovery)
3 DF
0 DF
1 unknown ( n1 )
– 0 balances
1 DF
Reboiler:
6 unknowns ( n2 , n3 , n4 , x B , y B , z B )
– 2 balances
– 2 relationships (2.25 ratio & 45% vapor)
3 DF
Begin with overall process.
b.
Overall process
89.2% recovery: 0.892 0.50 100
b gb g
Overall balance: 100
b g
B balance: 0.50 100
0.97 n1
n1 n3
0.97 n1 x B n3
Reboiler
e j
/ b1 x g
yB / 1 yB
Composition relationship:
Percent vaporized: n 4
Mole balance: n2
xB
0.45n 2
(1)
n3 n 4
(2)
(Solve (1) and (2) simultaneously.)
B balance: z B n2
2.25
B
x B n3 y B n 4
4-20
4.31 (cont’d)
c. B fraction in bottoms: x B
Moles of overhead: n1
Recovery of toluene:
4.32
0100
. mol B / mol
Moles of bottoms: n3
46.0 mol
. gb54.02g
b1 x gn u 100% b1 010
u 100%
0.50b100g
0.50b100g
B
3
a.
b
m3 kg H 2O
Mixing point
b g
Evaporator
m1 kg
. kg S / kg
012
0.88 kg H2O / kg
0.88 kg H2O / kg
97%
g
Bypass
100 kg
012
. kg S / kg
54.0 mol
b g
b g
m4 kg
m5 kg
0.58 kg S / kg
0.42 kg H 2O / kg
0.42 kg S / kg
0.58 kg H2O / kg
b g
m2 kg
012
. kg S / kg
0.88 kg H2O / kg
Overall process:
Evaporator:
2 unknowns ( m3 , m5 )
– 2 balances
0 DF
3 unknowns ( m1 , m3 , m4 )
– 2 balances
1 DF
b g
. 100
Overall S balance: 012
Overall mass balance: 100
Bypass:
2 unknowns ( m1 , m2 )
– 1 independent balance
1 DF
Mixing point: 3 unknowns ( m2 , m4 , m5 )
– 2 balances
1 DF
0.42m5
m3 m5
Mixing point mass balance: m4 m2
m5
. m2
Mixing point S balance: 0.58m4 012
(1)
0.42m5
(2)
Solve (1) and (2) simultaneously
Bypass mass balance: 100 m1 m2
b.
m1
90.05 kg, m2
Bypass fraction:
c.
m2
100
9.95 kg, m3
714
. kg, m4
18.65 kg, m5
28.6 kg product
0.095
Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a
stream consisting of 90% solids could be hard to transport.
4-21
4.33
a.
b
4 kg Cr / h
m
b
1 kg / h
m
g
b
2 kg / h
m
g
g
b g
x b kg Cr / kgg
1 x b kg W / kg g
5 kg / h
m
0.0515 kg Cr / kg
0.0515 kg Cr / kg
0.9485 kg W / kg
0.9485 kg W / kg
Treatment
Unit
b
3 kg / h
m
5
5
b g
x b kg Cr / kgg
1 x b kg W / kgg
6 kg / h
m
6
6
g
0.0515 kg Cr / kg
0.9485 kg W / kg
b.
1
m
b
2
6000 kg / h Ÿ m
4500 kg / h maximum allowed value
3 6000 4500 1500 kg / h
Bypass point mass balance: m
4
95% Cr removal: m
b
gb
g
g
0.95 0.0515 4500 220.2 kg Cr / h
5 4500 220.2 4279.8 kg / h
Mass balance on treatment unit: m
0.0515 4500 220.2
Cr balance on treatment unit: x5
0.002707 kg Cr / kg
4779.8
6 1500 4279.8 5779.8 kg / h
Mixing point mass balance: m
b
Mixing point Cr balance: x6
c.
b g
g
b
g
0.0515 1500 0.0002707 4279.8
5779.8
m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h)
1000
1000
0
48.9
951
2000
2000
0
97.9
1902
3000
3000
0
147
2853
4000
4000
0
196
3804
5000
4500
500
220
4280
6000
4500
1500
220
4280
7000
4500
2500
220
4280
8000
4500
3500
220
4280
9000
4500
4500
220
4280
10000
4500
5500
220
4280
4-22
0.0154 kg Cr / kg
x5
m 6 (kg/h)
0.00271
951
0.00271
1902
0.00271
2853
0.00271
3804
0.00271
4780
0.00271
5780
0.00271
6780
0.00271
7780
0.00271
8780
0.00271
9780
x6
0.00271
0.00271
0.00271
0.00271
0.00781
0.0154
0.0207
0.0247
0.0277
0.0301
4.33 (cont’d)
x 6 (kg Cr/kg)
m 1 vs. x 6
0.03500
0.03000
0.02500
0.02000
0.01500
0.01000
0.00500
0.00000
0
2000
4000
6000
8000 10000 12000
m 1 (kg /h )
d.
4.34
Cost of additional capacity – installation and maintenance, revenue from additional
recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon,
regulatory limits on Cr emissions.
a.
b
175 kg H2O / s 45% of water fed to evaporator
b
1 kg / s
m
g
b
b
4 kg K 2SO 4 / s
m
0196
.
kg K 2SO 4 / kg
0.804 kg H 2O / kg
5 kg H 2O / s
m
g
g
b
b
6 kg K 2SO 4 / s
m
Evaporator
7 kg H 2 O / s
m
g
g
g
Crystallizer
Filter
Filter cake
b
2 kg K 2SO 4 / s
10m
g
b
g
RS0.400 kg K SO / kg UV
T0.600 kg H O / kg W
2 kgsoln / s
m
2
Filtrate
b
3 kg / s
m
4
2
g
0.400 kg K 2SO 4 / kg
0.600 kg H 2 O / kg
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s
1, m
2)
Overall process: 2 unknowns ( m
- 2 balances
0 DF
Evaporator:
4, m
5, m
6, m
7 )
4 unknowns ( m
– 2 balances
– 1 percent evaporation
1 DF
Mixing point:
Crystallizer:
1, m
2
Strategy: Overall balances Ÿ m
5
% evaporation Ÿ m
3, m
4
Balances around mixing point Ÿ m
6, m
7
Balances around evaporator Ÿ m
4-23
1, m
3, m
4, m
5 )
4 unknowns ( m
- 2 balances
2 DF
2, m
3, m
6, m
7 )
4 unknowns ( m
– 2 balances
2 DF
U| verify that each
|V chosen subsystem involves
|| no more than two
W unknown variables
4.34 (cont’d)
Overall mass balance: m 1 175 10m 2 m 2
Overall K balance:
. m 1 10m 2 0.400m 2
0196
U|
V|
W
Production rate of crystals 10m 2
45% evaporation: 175 kg evaporated min
0.450m 5
W balance around mixing point: 0.804m 1 0.600m 3
Mass balance around mixing point: m 1 m 3
K balance around evaporator: m 6
m 4
W balance around evaporator: m 5
175 m 7
m 4 m 5
Mole fraction of K in stream entering evaporator =
b.
1
Fresh feed rate: m
c.
b
g
3 kg recycle s
m
1 kg fresh feed s
m
b
m 4
m 4 m 5
221 kg / s
bg
2
Production rate of crystals 10m
Recycle ratio:
m 5
416
. kg K s s
g
352.3
kg recycle
160
.
220.8
kg fresh feed
Scale to 75% of capacity.
Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s
46.3% K, 537%
.
W
d.
Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and
the cooling cost for the crystallizer.
4-24
4.35
a.
Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge
CO2 to the atmosphere.
Absorber function: Separates CO2 from CH4.
Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b.
The top streams are liquids while the bottom streams are gases. The liquids are heavier than
the gases so the liquids fall through the columns and the gases rise.
c.
b
n1 mol / h
g
b
g
n b mol CO / h g
n 5 mol N 2 / h
0.010 mol CO 2 / mol
0.990 mol CH 4 / mol
100 mol / h
0.300 mol CO 2 / mol
b
n 2 mol / h
Absorber
6
g
Stripper
0.005 mol CO 2 / mol
0.700 mol CH 4 / mol
2
b
n 5 mol N 2 / h
g
0.995 mol CH 3OH / mol
b
g
n b mol CH OH / h g
n 3 mol CO 2 / h
4
Overall:
Stripper:
3
3 unknowns ( n1 , n5 , n 6 )
– 2 balances
1 DF
Absorber:
4 unknowns ( n2 , n3 , n4 , n5 )
– 2 balances
– 1 percent removal (90%)
1 DF
b0.700gb100g bmol CH / hg
Overall mole balance: 100b mol / h g n n
Overall CH4 balance:
4
1
Percent CO2 stripped: 0.90n3
Stripper CO2 balance: n3
n1
n6
70.71 mol / h , n2
0.990n1
6
n 6
n 6 0.005n2
Stripper CH3OH balance: n4
d.
4 unknowns ( n1 , n 2 , n3 , n 4 )
– 3 balances
1 DF
0.995n 2
6510
. mol / h , n3
32.55 mol CO 2 / h, n 4
647.7 mol CH 3OH / h ,
29.29 mol CO 2 / h
Fractional CO2 absorption: f CO 2
30.0 0.010n1
30.0
4-25
0.976 mol CO 2 absorbed / mol fed
4.35 (cont’d)
Total molar flow rate of liquid feed to stripper and mole fraction of CO2:
n3
n3 n 4 680 mol / h , x3
0.0478 mol CO 2 / mol
n3 n 4
e.
Scale up to 1000 kg/h (=106 g/h) of product gas:
MW1
b
g
b
0.01 44 g CO 2 / mol 0.99 16 g CH 4 / mol
g
16.28 g / mol
u 10 mol / h
.
bn g d10. u 10 g / hib16.28 g / molg 6142
u 10 mol / h) / (70.71 mol / h)
.
bn g b100 mol / hg (6142
6
4
1 new
8.69 u 104 mol / h
4
feed new
4.36
f.
Ta Ts The higher temperature in the stripper will help drive off the gas.
Pa ! Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g.
The methanol must have a high solubility for CO2, a low solubility for CH4, and a low
volatility at the stripper temperature.
a.
Basis: 100 kg beans fed
e
m kg C H
5
6 14
e
m kg C H
1
6 14
j
300 kg C 6 H14
Ex
j
Condenser
b g
b
g
y b kg oil / kg g
1 x y b kg C H
m2 kg
2
2
2
6
b g
b
g
1 y b kg C H
m4 kg
F
x 2 kg S / kg
14 / kg
13.0 kg oil
87.0 kg S
Ev
y 4 kg oil / kg
6
4
g
14
/ kg
g
g
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg
b
g
0.25 y3 kg C 6 H14 / kg
Overall:
b
m6 kg oil
4 unknowns ( m1 , m3 , m6 , y3 )
– 3 balances
1 DF
Extractor:
g
3 unknowns ( m2 , x2 , y2 )
– 3 balances
0 DF
Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 )
2 unknowns ( m1 , m5 )
– 1 balance
– 2 balances
1 DF
2 DF
Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 )
– 3 balances
– 1 oil/hexane ratio
3 DF
Mixing Pt:
Start with extractor (0 degrees of freedom)
Extractor mass balance: 300 87.0 13.0 kg
4-26
m2
4.36
(cont’d)
Extractor S balance: 87.0 kg S
x2 m2
Extractor oil balance: 13.0 kg oil
y2 m2
Filter S balance: 87.0 kg S 0.75m3
b g
m3 m4 Oil / hexane ratio in filter cake:
Filter mass balance: m2 kg
y3
y2
1 x2 y 2
0.25 y3
Filter oil balance: 13.0 kg oil
b
y3m3 y4 m4
g
Evaporator hexane balance: 1 y4 m4
Mixing pt. Hexane balance: m1 m5
Evaporator oil balance: y4 m4
b.
m5
300 kg C6 H14
m6
b
g
118
. kg oil
0118
.
kg oil / kg beans fed
100 kg beans fed
m1
28 kg C6 H14
0.28 kg C6 H14 / kg beans fed
Fresh hexanefeed
100 100 kg beans fed
m5 272 kg C 6 H14 recycled
Recycle ratio
9.71 kg C6 H14 recycled / kg C6 H14 fed
m1
28 kg C 6 H14 fed
Yield
m6
100
b
b
c.
g
g
Lower heating cost for the evaporator and lower cooling cost for the condenser.
4.37
b
g
m lb m dirt
1
98 lb m dry shirts
3 lb m Whizzo
100 lb
m
2 lb m dirt
98 lb m dry shirts
b
m lb m Whizzo
2
g
b g
Tub
Filter
b g
m lb m
4
013
. lb m dirt / lb m
0.87 lb m Whizzo / lb m
m lb m
3
0.03 lb m dirt / lb m
0.97 lb m Whizzo / lb m
b g
b
b
m lb m
4
1 x lb m dirt / lb m
4
x lb m Whizzo / lb m
4
g
b g
m lb m
5
0.92 lb m dirt / lb m
0.08 lb m Whizzo / lb m
g
Strategy
95% dirt removal Ÿ m1 ( 5% of the dirt entering)
Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling
the chart) Ÿ m2 , m5 (solves Part (a))
4-27
4.37
(cont’d)
b
g
around the filter b m , m , x g , but the tub only involves 2 b m , m g and 2 balances are
Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances
4
3
6
4
allowed for each subsystem. Balances around tub Ÿ m3 , m4
Balances around mixing point Ÿ m6 , x (solves Part (b))
a.
. lb dirt
b0.05gb2.0g 010
Overall dirt balance: 2.0 010
. b0.92gm Ÿ m 2.065 lb dirt
Overall Whizzo balance: m
3 b0.08gb2.065g blb Whizzog
95% dirt removal: m1
m
5
5
2
b.
m
m
Tub dirt balance: 2 0.03m3 010
. 013
. m4
Tub Whizzo balance: 0.97m3 3 0.87m4
Solve (1) & (2) simultaneously Ÿ m3 20.4 lb m , m4 19.3 lb m
Mixing pt. mass balance: 317
. m6 20.4 lb m Ÿ m6 17.3 lb m
Mixing pt. Whizzo balance:
317
. x4 19.3
0.97 20.4 Ÿ x 0.833 lb m Whizzo / lb m Ÿ 833%
. Whizzo, 6.7% dirt
b g b gb g
4.38
317
. lb m Whizzo
a.
2720 kg S
mixer 3
Discarded
C 3L kg L
C 3S kg S
3300 kg S
Filter 3
C 2L kg L
C 2S kg S
F 3L kg L
F 3S kg S
620 kg L
mixer 1
Filter 1
F 1L
F 1S
mixer 2
C 1L kg L
C 1S kg S
kg L
kg S
Filter 2
F 2L kg L
F 2S kg S
To holding tank
b g
g U|
V|
W
F1L 6.2 kg L
0.01 620 F1L Ÿ
mixer filter 1:
620 6.2 C1L Ÿ C1L 6138
. kg L
balance:
F2 L
. F3 L
F2 L 6.2 kg L
mixer filter 2: 0.01 6138
. F3L F2 L C3 L Ÿ C2 L 613.7 kg L
balance: 6138
F3 L 61
. kg L
mixer filter 3: 0.01C2 L F3L
balance:
613.7 = 6.1+ C3L Ÿ C3 L 607.6 kg L
b
4-28
(1)
(2)
4.38 (cont’d)
Solvent
m f 1:
balance:
m f 2:
balance:
m f 3:
balance:
b
g
C1S Ÿ
495 F1S Ÿ
C2 S
. 3300
015
3300
. 495 F3S
015
b
495 F3S
b
g
C2 S F2 S
015
. 2720 C2 S
g
C3S
F3S C3S
2720 + C2S
U|
|V
||
W
C1S 495 kg S
F1S 2805 kg S
C2 S 482.6 kg S
F2 S 2734.6 kg S
Ÿ
C3S 480.4 kg S
F3S 2722.2 kg S
Holding Tank Contents
6.2 6.2 12.4 kg leaf
2805 2734.6
b.
5540 kg solvent
b g
5540 kgS
b g
QR kg
0165
. kg E / kg
Q0 kg
. kg E / kg
Extraction 013
0.15kg F / kg
Unit
0.835 kg W / kg
b g
Q b kg Fg
Steam
Stripper
0.855kg W / kg
QD kg D
0.774 kg W / kg
QB kg
b g
b g
Q b kg Dg
Q b kg Fg
QE kg E
F
0.200 kg E / kg
0.026 kg F / kg
0.013 kg E / kg
0.987 kg W / kg
D
F
b
Q3 kg steam
Mass of D in Product:
1 kg D
620 kg leaf
1000 kg leaf
0.62 kg D
QD
b g
Water balance around extraction unit: 0.835 5540 0.855QR Ÿ QR
Ethanol balance around extraction unit:
0165
.
5540 013
. 5410 QE Ÿ QE 211 kg ethanol in extract
b
c.
g
b
g
b
F balance around stripper
0.015 5410 0.026Q0 Ÿ Q0
b g
g
b
b g
b
4.39
a.
g
b
g
b
C 2 H 2 2 H 2 o C2 H 6
2 mol H 2 react / mol C2 H 2 react
0.5 kmol C2 H 6 formed / mol H 2 react
4-29
g
6085 kg mass of stripper bottom product
W balance around stripper
0.855 5410 QS 0.774 3121 0.987 6085 Ÿ QS
b
5410 kg
3121 kg mass of stripper overhead product
E balance around stripper
013
. 5410 0.200 3121 0.013QB Ÿ QB
b g
g
g
g
3796 kg steam fed to stripper
4.39 (cont’d)
b.
nH 2
15
. 2.0 Ÿ H 2 islimiting reactant
nC2 H 2
15
. mol H 2 fed Ÿ 10
. mol C2 H 2 fed Ÿ 0.75 mol C2 H 2 required (theoretical)
10
. mol fed 0.75 mol required
u 100% 333%
% excess C2 H 2
.
0.75 mol required
c.
4 u 106 tonnes C2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C2 H 6 2 kmol H 2 2.00 kg H 2
300 days 24 h 3600 s tonne 30.0 kg C2 H 6 1 kmol C2 H 6 1 kmol H 2
yr
20.6 kg H 2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH 3 5 O 2 o 4 NO 6 H 2 O
5 lb - mole O 2 react
. lb - mole O 2 react / lb - mole NO formed
125
4 lb - mole NO formed
b.
dn i
O2
100 kmol NH3 5 kmol O 2
h
4 kmol NH3
theoretical
d i
40% excess O 2 Ÿ nO 2
c.
fed
b
125 kmol O 2
140
. 125 kmol O 2
g
175 kmol O 2
b50.0 kg NH gb1 kmol NH / 17 kg NH g 2.94 kmol NH
.
kmol O
b100.0 kg O gb1 kmol O / 32 kg O g 3125
F n I 3125
F n I 5 125
. G
GH n JK 2..94 106
H n JK 4 .
3
3
2
2
2
O2
NH 3
3
3
2
O2
NH 3
fed
stoich
Ÿ O 2 is the limiting reactant
Required NH3:
3125
. kmol O 2 4 kmol NH3
5 kmol O 2
2.50 kmol NH3
2.94 2.50
u 100% 17.6% excess NH3
2.50
nO 2 vO 2 [ Ÿ 0 3125
. 5 [ Ÿ [
Extent of reaction: nO 2
% excess NH3
d i
Mass of NO:
4.41
a.
b g
0
3125
.
kmol O 2 4 kmol NO 30.0 kg NO
5 kmol O 2 1 kmol NO
0.625 kmol
625 mol
75.0 kg NO
By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed.
Automation provides for faster and more accurate response to fluctuations in the feed stream,
reducing the risk of release of H2S and SO2. It also may reduce labor costs.
4-30
4.41 (cont’d)
b.
3.00 u 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2
h
kmol
2 kmol H 2 S
n c
127.5 kmol SO 2 / h
c.
C a lib r a t io n C u r v e
1 .2 0
X ( m o l H2S /m o l)
1 .0 0
0 .8 0
0 .6 0
0 .4 0
0 .2 0
0 .0 0
0 .0
2 0 .0
4 0 .0
6 0 .0
8 0 .0
1 0 0 .0
R a (m V )
0.0199 Ra 0.0605
X
b
d.
n c kmol SO 2 / h
b
n f kmol / h
b
g
x kmol H 2S / kmol
g
Blender
g
Flowmeter calibration:
n f aR f
100 kmol / h , R f
n f
Control valve calibration:
n c
nc
UVn
15 mV W
25.0 kmol / h, R c
60.0 kmol / h , Rc
FG
H
f
UV
W
20
Rf
3
10.0 mV
n c
25.0 mV
IJ b
K
7
5
Rc 3
3
1
7
5 1 20
n f x Ÿ Rc R f 0.0119 Ra 0.0605
2
3
3 2 3
5
10
R f 0.0119 Ra 0.0605 Ÿ Rc
7
7
Stoichiometric feed: n c
b
n f
3.00 u 10 2 kmol / h Ÿ R f
g
3
n f
20
4-31
45 mV
g
4.41 (cont’d)
b gb
b g
e.
gb g
5
10
53.9 mV
45 0.0119 76.5 0.0605 7
7
5
7
Ÿ n c
127.4 kmol / h
53.9 3
3
Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond
range of calibration data, system had not reached steady state yet.
Rc
4.42
165 mol / s
b
x mol C 2 H 4 / mol
b
b
n mol / s
g
1 x mol HBr / mol
g
0.310 mol C2 H 4 / mol
0173
.
mol HBr / mol
0.517 mol C 2 H 5Br / mol
g
C 2 H 4 HBr o C 2 H 5 Br
C balance:
b
g
b
165 mol x mol C2 H 4 2 mol C
s
mol
mol C2 H 4
gb g b
gb g
n 0.310 2 n 0.517 2
. gb1g
b gb g nb0173
Br balance: 165 1 x 1
(2)
Solve (1) and (2) simultaneously Ÿ n 108.77 mol / s, x
b g
Ÿ 1 x
(1)
0.545 mol C 2 H 4 / mol
0.455 mol HBr / mol
Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1),
HBr is the limiting reactant .
bn g b165 mol / sgb0.455 mol HBr / molg 75.08 mol HBr
. gb108.8g
75.08 b0173
u 100%
Fractional conversion of HBr
HBr fed
dn i 75.08 mol C H
dn i b165 mol / sgb0.545 mol C H
C2H 4
stoich
2
75.08
0.749 mol HBr react / mol fed
4
C 2 H 4 fed
2
4
/ mol
g
89.93 mol C2 H 4
89.93 75.08
19.8%
75.08
Extent of reaction: n C2 H5Br n C2 H5Br vC2 H5Br[ Ÿ 108.8 0.517
% excess of C2 H 4
d
i
0
4-32
b
gb
g
bg
0 1[ Ÿ[
56.2 mol / s
4.43
a.
1
O 2 o Cl 2 H 2 O
2
2HCl Basis: 100 mol HCl fed to reactor
b
g
n b mol O g
n b mol N g
n b mol Cl g
n b mol H Og
100 mol HCl
b
n1 mol air
n2 mol HCl
g
0.21 mol O 2 / mol
100 mol HCl 0.5 mol O 2
2
b
g
85 mol HCl react
1 mol Cl 2
2
15 mol HCl
42.5 mol Cl 2
2 mol HCl
N 2 balance:
2
135
. u 25 Ÿ n1 160.7 mol air fed
85% conversion Ÿ 85 mol HCl react Ÿ n2
b85gb1 2g
2
25 mol O 2
2 mol HCl
35% excess air: 0.21n1 mol O 2 fed
n6
4
6
35% excess
n5
2
5
0.79 mol N 2 / mol
bO gstoic
3
42.5 mol H 2 O
b160.7gb0.79g
n4 Ÿ n4
127 mol N 2
O balance:
b160.7gb0.21g mol O
2
2 mol O
1 mol O 2
2 n3 42.5 mol H 2 O
1 mol O
Ÿ n3
1 mol H 2 O
12.5 mol O 2
Total moles:
5
¦nj
239.5 mol Ÿ
j 2
b.
0.063
mol O 2
mol N 2
mol HCl
, 0.052
, 0.530
,
mol
mol
mol
0177
.
mol Cl 2
mol H 2 O
, 0177
.
mol
mol
15 mol HCl
239.5 mol
As before, n1 160.7 mol air fed , n2
15 mol HCl
1
2HCl O 2 o Cl 2 H 2 O
2
ni
E
bn g
i 0
vi [
HCl: 15 100 2[ Ÿ [
42.5 mol
4-33
4.43 (cont’d)
N 2 : n4
b g
0.79b160.7g
Cl 2 : n5
[
42.5 mol Cl 2
H 2 O: n6
[
42.5 mol H 2 O
c.
4.44
1
[ 12.5 mol O 2
2
127 mol N 2
0.21 160.7 O 2 : n3
These molar quantities are the same as in part (a), so the mole fractions would also be the
same.
Use of pure O2 would eliminate the need for an extra process to remove the N2 from the
product gas, but O2 costs much more than air. The cheaper process will be the process of
choice.
b g
Fe O 3H SO o Fe bSO g 3H O
bTiOgSO 2H O o H TiO bsg H SO
H TiO bsg o TiO bsg H O
FeTiO3 2H 2SO 4 o TiO SO 4 FeSO 4 2H 2 O
2
3
2
4
2
4
2
2
2
3
2
4 3
2
3
2
4
2
Basis: 1000 kg TiO2 produced
1000 kg TiO 2
kmol TiO 2
1 kmol FeTiO 3
79.90 kg TiO 2
1 kmol TiO 2
12.52 kmol FeTiO 3 dec.
12.52 kmol FeTiO 3 decomposes
1 kmol FeTiO 3 feed
14.06 kmol FeTiO 3 fed
0.89 kmol FeTiO 3 dec.
14.06 kmol FeTiO 3
b
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO 3
kmol Ti
673.5 kg Ti / M kg ore
g
0.243 Ÿ M
6735
. kg Ti fed
2772 kg ore fed
b
g
Ore is made up entirely of 14.06 kmol FeTiO 3 + n kmol Fe2 O3 (Assumption!)
n
2772 kg ore 638.1 kg Fe 2 O 3
14.06 kmol FeTiO 3 151.74 kg FeTiO 3
kmol FeTiO3
kmol Fe2O 3
. kg Fe2 O 3
6381
4.00 kmol Fe2O 3
159.69 kg Fe2 O 3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4
4012
. kmol H2SO4
1 kmol FeTiO3
1 kmol Fe2O3
b
50% excess: 15
. 4012
. kmol H 2SO 4
Mass of 80% solution:
5902.4 kg H 2 SO 4 / M
g
6018
. kmol H 2SO 4 fed
60.18 kmol H 2SO 4
a
bkg solng
98.08 kg H 2SO 4
1 kmol H 2SO 4
0.80 Ÿ M
4-34
a
5902.4 kg H 2SO 4
7380 kg 80% H 2 SO 4 feed
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
dR
i
0.30 g m 3 and
10, C1
1
ln C bR ln a œ C
b
g
ln 2.67 0.30
d
C g m3
i C c(ftlb
E
48, C2
2
m)
3
b g
b g
ln 2.67 0.0575 48
453.6 g 35.31 ft 3
1 m3
1 lb m
d
d2867 ft sib60 s ming
178
. Ÿa
e 1.78
.
0169
i
u 10 5 e 0.0575 R
1055
.
3
138 ft 3 lb m coal
1250 lb m min
R
d
37 Ÿ C c lb m SO 2 ft 3
i
u 105 eb
1055
.
138 ft 3
8.86 u 105 lb m SO 2
ft 3
1 lb m coal
c.
IK
16,020C c
16,020C ' 0169
. e 0.0575 R Ÿ C c lb m SO 2 ft 3
b.
2.67 g m 3
ae br
0.0575 , ln a
48 10
Ÿ C 0169
. e 0.0575 R
b
FH R
gb g
0.0575 37
0.012 0.018
8.86 u 10
5
lb m SO 2 ft 3
lb m SO 2
compliance achieved
lb m coal
S O 2 o SO 2
1250 lb m coal 0.05 lb m S 64.06 lb m SO 2
min
1 lb m coal
32.06 lb m S
124.9 lb m SO 2 generated min
2867 ft 3 60 s 886
. u 105 lb m SO2
. lbm SO2 min in scrubbed gas
152
s
ft3
1 min
air
1250 lbm coal/min
62.5 lb m S/min
% removal
d.
furnace
ash
b124.9 15.2g lb
scrubbing fluid
stack gas
124.9 lbm SO2 /min
scrubber scrubbed gas
15.2 lb m SO2 /min
liquid effluent
(124.9 – 15.2) lb m SO2 (absorbed)/min
SO 2 scrubbed min
u 100%
124.9 lb m SO 2 fed to scrubber min
m
88%
The regulation was avoided by diluting the stack gas with fresh air before it exited from the
stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal
burned is independent of the flow rate of air in the stack.
4-35
4.46
a.
A B ===== C + D
nA nA [
0
nB
nB [
yA
nC
nC [
yB
nD
nD [
yC
0
0
0
nI
nI
yD
0
¦ ni
Total nT
At equilibrium:
en
en
en
en
bn
bn
yC y D
y A yB
C0
A0
gb
[ gbn
B0
j
[j n
[j n
[j n
[ nT
T
C0 D0
T
g
[ g
[ c n D0 [ c
c
b
c
A0
B0
T
4.87 ( nT ’s cancel)
c
gh b
. [ 2c nC0 nD0 487
. nA0 nB0 [ c nC0nD0 487
. nA0nB0
387
g
0
[a[ 2c b[ c c 0]
?[ c
b.
a 387
.
1
2
b r b 4ac where b nC0 nD0 487
. nA0 nB0
2a
c nC0nD0 487
. nA0nB0
e
b[
b
387
.
F
b gH
1
1174
. r
2 387
.
[e
e2
1174
c
.
. g
b1174
nA0 80, nC0
nC
nA
nB
nC
nD
4.87
2
4.87
b gb gIK Ÿ [
4 387
4.87
.
e1
0.496
2.54 is also a solution but leads to a negative conversion
Fractional conversion: X A
c.
g
nA0 1 nB0 1 nC0 nB0 nI 0 0
Basis: 1 mol A feed
Constants: a
b
j
b
XB
g
n A0 n A
n A0
[ e1
n A0
g
0.496
nJ 0 0
nC0 0
! [ c 70 mol
70 nC0 [ c
n A0 [ c n A0 70 mol
n B0 [ c 80 70 10 mol
nC 0 [ c 70 mol
n D0 [ c 70 mol
yC y D
y A yB
nD0
b gb g
b
gb g
70 70
nC n D
Ÿ
n AnB
n A0 70 10
4-36
4.87 Ÿ n A0
170.6 mol methanol fed
4.46 (cont’d)
Product gas n A
nB
nC
U|
|V
||
W
yA
170.6 70 100.6 mol
yB
10 mol
Ÿ
yC
70 mol
yD
70 mol
nD
0.040 mol CH3COOH mol
0.279 mol CH3COOCH 3 mol
0.279 mol H 2 O mol
250.6 mol
ntotal
4.47
0.401 mol CH3OH mol
d.
Cost of reactants, selling price for product, market for product, rate of reaction, need for
heating or cooling, and many other items.
a.
o CO 2 H 2
CO H 2 O m
(A)
(B)
(C)
(D)
b
g
n b mol H Og
n b mol CO g
n b mol H g
n b mol Ig
. mol
100
0.20 mol CO / mol
. mol CO 2 / mol
010
0.40 mol H 2 O / mol
0.30 mol I / mol
n A mol CO
2
B
2
C
2
D
I
6 unknowns ( n A , nB , nC , nD , n I , [ )
Degree of freedom analysis:
bg
– 4 expressions for ni [
– 1 balance on I
– 1 equilibrium relationship
0 DF
b.
Since two moles are prodcued for every two moles that react,
ntotal out ntotal in 100
. mol
b g b g
b g
n A 0.20 [
nB 0.40 [
nC 010
. [
nD [
n I 0.30
ntot
(1)
(2)
(3)
(4)
(5)
100
. mol
At equilibrium:
yD
c.
nD
[
yC y D
y A yB
b
. [ gb[ g
b010
b0.20 [ gb0.40 [ g
/ molg
nC nD
n A nB
0110
.
mol H 2
The reaction has not reached equilibrium yet.
4-37
0.0247 exp
FG 4020 IJ Ÿ [
H 1123 K
mol
0110
.
4.47
(cont’d)
d.
T (K)
1223
1123
1023
923
823
723
623
673
698
688
x (CO)
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
x (H2O)
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
1123
1123
1123
1123
0.2
0.4
0.3
0.5
0.4
0.2
0.3
0.4
x (CO2)
Keq
Keq (Goal Seek) Extent of Reaction
0.6610
0.6610
0.2242
0.8858
0.8856
0.2424
1.2569
1.2569
0.2643
1.9240
1.9242
0.2905
3.2662
3.2661
0.3219
6.4187
6.4188
0.3585
15.6692
15.6692
0.3992
9.7017
9.7011
0.3785
7.8331
7.8331
0.3684
8.5171
8.5177
0.3724
0
0
0
0
0
0
0
0
0
0
0.1
0.1
0
0
0.8858
0.8858
0.8858
0.8858
0.8863
0.8857
0.8856
0.8867
0.1101
0.1100
0.1454
0.2156
y (H2)
0.224
0.242
0.264
0.291
0.322
0.358
0.399
0.378
0.368
0.372
0.110
0.110
0.145
0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of
carbon monoxide and water also maximizes the extent of reaction.
4.48
a.
A 2B o C
b
ln K e1 / K e 2
1 T1 1 T2
E
4
11458
1 373 1 573
ln K e1 11458 T1
ln A0
b.
b g
g lnd10.5 / 2.316 u 10 i
ln A0 E T K
ln K e
ln 10.5 11458 373 28.37 Ÿ A0
4.79 u 1013
b gh
c
Ke
4.79 u 10 13 exp 11458 T K atm 2 Ÿ Ke (450K) 0.0548 atm1
nA
nB
nC
n A0 [
nB 0 2[
nC 0 [
nT
nT 0
U|
|V
|
2[ |W
bn
bn
bn
yA
yB
Ÿ
yC
bn
gb
gb
gb
g
[ nT 0 2[
nT 0 2[
B 0 2[
nT 0 2[
C0 [
n A0 nB 0 nC 0
T0
A0
g
g
g
At equilibrium,
yC 1
y A y B2 P2
c.
bn
bn
C0
A0
gb
[ gb n
g
2[ g
[ e nT 0 2[ e
e
B0
2
2
e
bg
1
P2
bg
Ke T (substitute for K T from Part a.)
e
Basis: 1 mol A (CO)
1 nB 0
n A0
b
[ e 2 2[ e
1 nC 0
g
2
b1 [ gb1 2[ g
e
e
2
1
4 atm 2
0 Ÿ nT 0
2, P
b g
K e 423
2 atm , T
423K
0.278 atm -2 Ÿ [ 2e [ e 01317
.
4-38
0
4.47 (cont’d)
(For this particular set of initial conditions, we get a quadratic equation. In general, the
equation will be cubic.)
[e
, 0.844
0156
.
Reject the second solution, since it leads to a negative nB .
. g c2 2b0156
. gh Ÿ y
0.500
b1 0156
y
. gh c2 2b0156
. gh Ÿ y
0.408
c1 2b0156
y
. g c2 2b0156
. gh Ÿ y
0.092
b0 0156
n n
[
Fractional Conversion of CO b Ag
n
n
yA
A
B
B
C
C
A0
A
A0
d.
0156
.
mol A reacted / mol A feed
A0
Use the equations from part b.
i)
ii)
iii)
iv)
Fractional conversion decreases with increasing fraction of CO.
Fractional conversion decreases with increasing fraction of CH3OH.
Fractional conversion decreases with increasing temperature.
Fractional conversion increases with increasing pressure.
e.
*
1
2
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI,
FN, FDN, NT, CON, YA, YB, YC
INTEGER NIT, INMAX
TAU = 0.0001
INMAX = 10
A = 4.79E–13
E = 11458.
READ (5, *) YA0, YB0, YC0, T, P
KE = A * EXP(E/T)
P2KE = P*P*KE
C0 = YC0 – P2KE * YA0 * YB0 * YB0
C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0)
C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0))
C3 = 4. * (1. + P2KE)
EK = 0.0
(Assume an initial value [ e 0. 0 )
NIT = 0
FN = C0 + EK * (C1 + EK * (C2 + EK * C3))
FDN = C1 + EK * (2. * C2 + EK * 3. * C3)
EKPI = EK - FN/FDN
NIT = NIT + 1
IF (NIT.EQ.INMAX) GOTO 4
IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2
EK = EKPI
GOTO 1
NT = 1. – 2. * EKPI
YA = (YA0 – EKPI)/NT
YB = (YB0 – 2. + EKPI)/NT
YC = (YC0 + EKPI)/NT
4-39
4.48 (cont’d)
CON = EKPI/YA0
WRITE (6, 3) YA, YB, YC, CON
STOP
WRITE (6, 5) INMAX, EKPI
FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X))
FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/,
*
'CURRENT VALUE = ', F6.3)
END
4
3
$DATA
0.5
RESULTS
YA
0.500
0.5
YB
0.408
0.0
423.
YC
0.092
2.
CON
0.156
Note: This will only find one root — there are two others that can only be found by
choosing different initial values of [ a
4.49
a.
CH 4 O 2 o HCHO H 2O
(1)
CH 4 2O 2 o CO 2 2H 2O
(2)
100 mol / s
b
g
n b mol O / sg
n b mol HCHO / sg
n b mol H O / sg
n b mol CO g
n1 mol CH 4 / s
0.50 mol CH 4 / mol
0.50 mol O 2 / mol
2
2
3
4
5
2
2
7 unknowns ( n1 , n 2 , n3 , n 4 , n5 , [ 1 , [ 2 )
– 5 equations for n [ , [
i
e
1
2
j
2 DF
b.
n1
n2
1
n4
[ 1
[ 2[
n5
[ 2
n3
c.
50 [ 1 [ 2
50 [ 2[
1
(1)
(2)
2
(3)
(4)
2
(5)
Fractional conversion:
Fractional yield:
n3
50
b50 n g
1
50
0.855 Ÿ n3
0.900 Ÿ n1
5.00 mol CH / s
4
42.75 mol HCHO / s
4-40
4.49 (cont’d)
U|
||
V|
||
W
y CH
0.0500 mol CH 4 / mol
Equation 3 Ÿ [ 1 42.75
4
yO
0.0275 mol O 2 / mol
Equation 1 Ÿ [ 2 2.25
2
Equation 2 Ÿ n 2 2.75 Ÿ y HCHO 0.4275 mol HCHO / mol
y H O 0.4725 mol H 2 O / mol
Equation 4 Ÿ n 4 47.25
2
y CO
0.0225 mol CO 2 / mol
Equation 5 Ÿ n5 2.25
2
Selectivity:
42.75 mol HCHO / s
2.25 mol CO 2 / s
19.0 mol HCHO / mol CO 2
4-41
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane
make the second reaction unlikely.
b.
C2H6 + Cl2 o C2H5Cl + HCl, C2H5Cl + Cl2 o C2H4Cl2 + HCl
Basis: 100 mol C2H5Cl produced
c.
n1 (mol C2H6)
100 mol C2H5OH
n2 (mol C2H6)
n3 (mol C2H6)
n4 (mol HCl)
n5 (mol C2H5Cl2)
5 unknowns
–3 atomic balances
2 D.F.
Selectivity: 100 mol C 2 H 5 Cl 14n5 (mol C 2 H 4 Cl 2 ) Ÿ n5
U| Ÿ n 714.3 mol C H in
g n
V
2n 2b100g 2n 2b7.143g|W n 114.3 mol C H out
6b714.3g 5b100g 6b114.3g n 4b7.143g Ÿ n 607.1 mol HCl
2n 100 607.1 2b7.143g Ÿ n 114.3 mol Cl
b
. n1
13% conversion: 1 015
C balance:
H balance:
Cl balance:
7.143 mol C 2 H 4 Cl 2
3
1
3
1
2
6
3
2
6
4
2
4
2
Feed Ratio: 114.3 mol Cl 2 / 714.3 mol C2 H 6
2
016
. mol Cl 2 / mol C2 H 6
Maximum possible amount of C2H5Cl:
114.3 mol Cl 2 1 mol C 2 H 5 Cl
n max
114.3 mol C 2 H 5 Cl
1 mol Cl 2
Fractional yield of C2H5Cl:
4.51
nC2 H5Cl
n max
100 mol
114.3 mol
0.875
d.
Some of the C2H4Cl2 is further chlorinated in an undesired side reaction:
C2H5Cl2 + Cl2 o C2H4Cl3 + HCl
a.
C2H4 + H2O o C2H5OH, 2 C2H5OH o (C2H5)2O + H2O
Basis: 100 mol effluent gas
100 mol
n (mol C H )
2 4
1
n [mol H O (v)]
2
2
n 3 (mol I)
0.433 mol C 2 H 4 / mol
3 unknowns
0.025 mol C 2 H 5 OH / mol
-2 independent atomic balances
0.0014 mol (C H ) O / mol
2 5 2
0.093 mol I / mol
-1 I balance
0 D. F.
0.4476 mol H O (v) / mol
2
(1) C balance: 2n1
b
100 2 0.433 2 0.025 4 0.0014
(2) H balance: 4n1 2n2
(3) O balance: n2
b
b
g
100 4 0.433 6 0.025 10 0.0014 2 0.4476
g
g
100 0.025 0.0014 0.4476
Note; Eq. (1) 2 Eq. (3) 2
(4) I balance: n3 = 9.3
Eq. (2) Ÿ2 independent atomic balances
4-42
4.51 (cont'd)
b.
(1) Ÿ n1 46.08 mol C 2 H 6
(3) Ÿ n2 47.4 mol H 2 O Ÿ Reactor feed contains 44.8% C 2 H 6 , 46.1% H 2 O, 9.1% I
(4) Ÿ n3 9.3 mol I
U|
V|
W
46.08 43.3
u 100% 6.0%
46.08
If all C2H4 were converted and the second reaction did not occur, nC2 H5OH
% conversion of C2H4:
d
d
Ÿ Fractional Yield of C2H5OH: nC2 H5OH / nC2 H5OH
Selectivity of C2H5OH to (C2H5)2O:
2.5 mol C 2 H 5 OH
0.14 mol (C 2 H 5 ) 2 O
c.
4.52
i
max
b2.5 / 46.08g
i
max
46.08 mol
0.054
17.9 mol C 2 H 5OH / mol (C 2 H 5 ) 2 O
Keep conversion low to prevent C2H5OH from being in reactor long enough to form
significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.
bg
bg
bg
bg
CaF2 s H 2 SO 4 l o CaSO 4 s 2HF g
1 metric ton acid
1000 kg acid
0.60 kg HF
1 metric ton acid
1 kg acid
600 kg HF
Basis: 100 kg Ore dissolved (not fed)
100 kg Ore d issolved
0.96 kg CaF 2 /kg
0.04 kg SiO 2/ kg
nA (kg 93% H2 SO4 )
0.93 H2 SO4 kg/ kg
0.07 H2 O kg/ kg
n1
n2
n3
n4
(kg CaSO4)
(kg HF)
(kg H 4SiF6 )
(kg H 2SO 4)
n5 (kg H2 O)
Atomic balance - Si:
b g
0.04 100 kg SiO 2
28.1 kg Si
n3 (kg H 4 SiF6 )
60.1 kg SiO 2
28.1 kg Si
Ÿ n3
146.1 kg H 4 SiF6
Atomic balance - F:
b g
0.96 100 kg CaF2
38.0 kg F
n2 (kg HF)
78.1 kg CaF2
114.0 kg F
9.72 kg H 4 SiF6
Ÿ n2
146.1 kg H 4 SiF6
600 kg HF 100 kg ore diss.
41.2 kg HF
1 kg ore feed
0.95 kg ore diss.
4-43
19.0 kg F
20.0 kg HF
412
. kg HF
1533 kg ore
9.72 kg H 4 SiF6
4.53
a.
C 6 H 6 Cl 2 o C 6 H 5 Cl HCl
C 6 H 5 Cl Cl 2 o C 6 H 4 Cl 2 HCl
C 6 H 4 Cl 2 Cl 2 o C 6 H 3 Cl 3 HCl
Convert output wt% to mol%: Basis 100 g output
species
C6H 6
C 6 H 5 Cl
C 6 H 4 Cl 2
C 6 H 3 Cl 3
g
65.0
32.0
2.5
0.5
Mol. Wt.
78.11
112.56
147.01
181.46
mol
0.832
0.284
0.017
0.003
mol %
73.2
25.0
1.5
0.3
total 1.136
Basis: 100 mol output
n 4 (mol HCl(g ))
n 3 (mol I)
n1 (mol C6 H6 )
65.0 mo l C6 H6
32.0 mo l C6 H 5 Cl
2.5 mo l C 6 H 4 Cl 2
0.5 mo l C6 H 3 Cl 3
n2 (mol Cl 2)
n3 (mol I)
b.
4 unknowns
-3 atomic balances
-1 wt% Cl 2 in feed
0 D.F.
b
100 mol C H
g
H balance: 6b100g 6b65.0g 5b32.0g 4b2.5g 3b0.5g n Ÿ n
38.5 mol HCl
Cl balance: 2n 38.5 32.0 2b2.5g 3b0.5g Ÿ n 38.5 mol Cl
Theoretical C H
38.5 mol Cl b1 mol C H 1 mol Cl g 38.5 mol C H
Excess C H : b100 38.5g 38.5 u 100% 160% excess C H
Fractional Conversion: b100 65.0g 100 0.350 mol C H react / mol fed
6 65.0 32.0 2.5 0.5 Ÿ n1
C balance: 6n1
6
4
4
2
6
6
6
2
6
2
6
2
6
2
6
6
6
6
6
6
6
Yield: (32.0 mol C 6 H 5 Cl) (38.5 mol C 6 H 5 Cl maximum) = 0.831
U|
V|
W
38.5 mol Cl 2 70.91 g Cl 2 1 g gas
2091 g gas
g gas
Ÿ 0.357
mole Cl 2 0.98 g Cl 2
g liquid
Liquid feed: 100 mol C 6 H 6 78.11 g C 6 H 6 / mol C 6 H 6 7811 g liquid
Gas feed:
b
gb
g
c.
Low conversion Ÿ low residence time in reactor Ÿ lower chance of 2nd and 3rd reactions
occurring. Large excess of C 6 H 6 Ÿ Cl 2 much more likely to encounter C 6 H 6
than substituted C 6 H 6 Ÿ higher selectivity.
d.
Dissolve in water to produce hydrochloric acid.
e.
Reagent grade costs much more. Use only if impurities in technical grade mixture affect the
reaction rate or desired product yield.
4-44
4.54
a.
2CO 2 œ 2CO O 2
2A œ 2B C
O 2 N 2 œ 2NO
C D œ 2E
bn
bn
bn
bn
bn
n A0 2[ e1
yA
n B 0 2[ e 2
yB
nC 0 [ e1 [ e 2 Ÿ y C
nD0 [ e2
yD
n E 0 2[ e 2
yE
nA
nB
nC
nD
nE
bn
ntotal = nT 0 [ e1
2[ e1
2[ e1
A0
B0
g bn
g bn
T0
T0
[ e1
[ e1
gb
g
g
D0
[ e1 [ e 2 nT 0 [ e1
1[ e 2 nT 0 [ e1
E0
2[ e 2
C0
gb
g bn
T0
[ e1
g
g
g
n A0 n B 0 nC 0 n D 0 n E 0
T0
g
Equilibrium at 3000K and 1 atm
bn
g bn [ [ g 01071
.
bn 2[ g bn [ g
bn 2[ g
bn [ [ gbn [ g 0.01493
y B2 y C
y 2A
B0
2[ e1
2
A0
e1
C0
2
e1
T0
e2
e1
2
y E2
yC y D
B0
A0
e1
e2
e2
D0
e2
E
f1
b
01071
.
n A0 2[ e1
f2
b
g bn
2
T0
g b
gbn
g bn [ [ g
g bn 2[ g 0
[ e1 n B 0 2[ e1
0.01493 nC 0 [ e1 [ e 2
D0
[ e2
2
C0
e1
e2
2
E0
e2
U|
V| b g
W b g
0 Defines functions
f 1 [ 1 , [ 2 and
f2 [1, [ 2
b.
Given all nio’s, solve above equations for [e1 and [e2 Ÿ nA, nB, nC, nD, nE Ÿ yA, yB, yC, yD, yE
c.
nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 Ÿ [e1 =0.0593, [e2 = 0.0208
Ÿ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a11 d 1 a12 d 2 f 1
a12 f 2 a 22 f 1
d1
a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a11 d 1 a12 d 2 f 1
a12 f 2 a 22 f 1
d1
a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a 21 d 1 a 22 d 2 f 2
a 21 f 1 a11 f 2
d2
a11 a 22 a12 a 21
b[ g
e 2 new
[ e1 d 2
a 21 d 1 a 22 d 2 f 2
a 21 f 1 a11 f 2
d2
a11 a 22 a12 a 21
b[ g
e 2 new
(Solution given following program listing.)
4-45
[ e1 d 2
4.54 (cont’d)
.
1
30
2
3
100
4
120
IMPLICIT REAL * 4(N)
WRITE (6, 1)
FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///)
READ (5, *) NA0, NB0, NC0, ND0, NE0
IF (NA0.LT.0.0)STOP
WRITE (6, 2) NA0, NB0, NC0, ND0, NE0
FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/)
NTO = NA0 + NB0 + NC0 + ND0 + NE0
NMAX = 10
X1 = 0.1
X2 = 0.1
DO 100 J = 1, NMAX
NA = NA0 – X1 – X1
NB = NB0 + X1 + X1
NC = NC0 + X1 – X2
ND = ND0 – X2
NE = NE0 + X2 + X2
NAS = NA ** 2
NBS = NB ** 2
NES = NE ** 2
NT = NT0 + X1
F1 = 0.1071 * NAS * NT – NBS * NC
F2 = 0.01493 * NC * ND – NES
A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS
A12 = NBS
A21 = 0.01493 * ND
A22 = –0.01493 * (NC + ND) – 4.0 * NE
DEN = A11 * A22 – A12 * A21
D1 = (A12 * F2 – A22 * F1)/DEN
D2 = (A21 * F1 – A11 * F2)/DEN
X1C = X1 + D1
X2C = X2 + D2
WRITE (6, 3) J, X1, X2, X1C, X2C
FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5)
IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120
X1 = X1C
X2 = X2C
CONTINUE
WRITE (6, 4) NMAX
FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/)
STOP
YA = NA/NT
YB = NB/NT
YC = NC/NT
YD = ND/NT
YE = NE/NT
WRITE (6, 5) YA, YB, YC, YD, YE
5 FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///)
GOTO 30
END
$DATA
0.3333 0.00 0.3333 0.3333 0.0
0.50
0.0 0.0
0.50
0.0
0.20
0.20 0.20
0.20
0.20
4-46
4.54 (cont’d)
SOLUTION TO PROBLEM 4.54
NA0, NB0, NC0, ND0, NE0 = 0.33 0.00
0.33
ITER = 1 X1A, X2A = 0.10000 0.10000
ITER = 2 X1A, X2A = 0.06418 0.05181
ITER = 3 X1A, X2A = 0.05969 0.02486
ITER = 4 X1A, X2A = 0.05437 0.02213
ITER = 5 X1A, X2A = 0.05931 0.02086
ITER = 6 X1A, X2A = 0.05930 0.02083
YA, YB, YC, YD, YE =
0.33
0.00
X1C, X2C = 0.06418
X1C, X2C = 0.05969
X1C, X2C = 0.05937
X1C, X2C = 0.05931
X1C, X2C = 0.05930
X1C, X2C = 0.05930
2.0270E 01
2.9501E 01
NA0, NB0, NC0, ND0, NE0 = 0.20 0.20
ITER = 1 X1A, X2A = 0.10000
p
ITER = 7 X1A, X2A = –0.02244
0.20
1.1197 E 01
3.9319 E 02
0.20
0.10000
3.5100E 01
0.20
X1C, X2C = 0.00012
–0.08339 X1C, X2C = –0.02244
YA, YB, YC, YD, YE =
2.5051E 01 15868
.
E 01
2.8989 E 01 3.3991E 02
4.55
0.05181
0.02986
0.02213
0.02086
0.02083
0.02083
0.00037
–0.08339
2.6693E 01
(B)
a.
mB0 (kg A / h)
1 kg B/ kg A fed to reactor
( A)
m A 0 (kg A / h)
x RA (kg R / kg A)
( P)
m A 0 (kg A / h)
x RA (kg R / kg A)
m3 (kg A / h)
R o 5
x R 3 (Kg R / kg)
99% conv.
m P (kg P / h)
0.0075 kg R / kg P
f mA0 (kg A / h)
x RA (kg R / kg A)
Reactor
Splitting point
4 unknowns (mA0, mB0, f, xRA)
-1 independent balance (mass)
3 D.F.
4 unknowns (xRA, mB0, m3, xRA)
-1 balance (mass)
-1 conversion
2 D.F
Mixing point
Total process
5 unknowns (f, m3, mP, xRA, xR3)
7 unknowns (mA0, xRA, f, mB0, m3, xR3, mP)
-2 balances (mass, R)
-5 relations
3 D.F.
2 D.F.
4-47
4.55 (cont’d)
b. Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3
(2)
99% conversion of R: xR3 m3 = 0.01 xRA mB0
(3)
Mass balance on mixing point: m3 + f mA0 = mP
R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP
(4)
(5)
Given xRA and mP, solve simultaneously for mA0, mB0, f, m3, xR3
c.
mA0 = 2778 kg A/h
mB0 = 2072 kg B/h
fA = 0.255 kg bypass/kg fresh feed
mP
4850
4850
4850
4850
4850
4850
4850
4850
4850
xRA
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
mA0
3327
3022
2870
2778
2717
2674
2641
2616
2596
mB0
1523
1828
1980
2072
2133
2176
2209
2234
2254
f
0.54
0.40
0.31
0.25
0.21
0.19
0.16
0.15
0.13
mP
2450
2450
2450
2450
2450
2450
2450
2450
2450
xRA
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
mA0
1663
1511
1435
1389
1359
1337
1321
1308
1298
mB0
762
914
990
1036
1066
1088
1104
1117
1127
f
0.54
0.40
0.31
0.25
0.22
0.19
0.16
0.15
0.13
f v s . x RA
f (kg bypass/kg fresh feed)
d.
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
0.02
0.04
0.06
x R A (k g R / k g A )
4-48
0.08
0.10
0.12
4.56
a.
900 kg HCHO 1 kmol HCHO
30.03 kg HCHO
h
30.0 kmol HCHO / h
n (kmol CH OH / h)
1
3
30.0 kmol HCHO / h
n 2 (kmol H 2 / h)
n 3 (kmol CH 3 OH / h)
% conversion:
30.0
n1
0.60 Ÿ n1
50.0 kmol CH 3 OH / h
b.
n (kmol CH OH / h)
1
3
30.0 kmol HCHO / h
30.0 kmol HCHO / h
n 2 (kmol H 2 / h)
n 2 (kmol H 2 / h)
n 3 (kmol CH 3 OH / h)
n (kmol CH OH / h)
3
3
Overall C balance: n1 (1) = 30.0 (1) Ÿ n1 = 30.0 kmol CH3OH/h (fresh feed)
Single pass conversion:
30.0
n1 n3
0.60 Ÿ n3
20.0 kmol CH 3OH / h
n1 + n3 = 50.0 kmol CH3OH fed to reactor/h
4.57
c.
Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and
(2) lower the quantities of unreacted methanol and so will decrease the cost of the
separation. The plot would resemble a concave upward parabola with a minimum
around xsp = 60%.
a.
Convert effluent composition to molar basis. Basis: 100 g effluent:
10.6 g H 2 1 mol H 2
2.02 g H 2
64.0 g CO 1 mol CO
28.01 g CO
5.25 mol H 2
2.28 mol CO
25.4 g CH 3 OH 1 mol CH 3 OH
32.04 g CH 3 OH
0.793 mol CH 3 OH
4-49
H : 0.631 mol H / mol
2
2
Ÿ
CO: 0.274 mol CO / mol
CH OH: 0.0953 mol CH OH / mol
3
3
4.57 (cont’d)
n4 (mol / min)
0.004 mol CH 3OH(v)/ mol
x (mol CO/ mol)
(0.896 - x) (mol H 2 / mol)
Cond.
Reactor
350 mol/ min
n1 (mol CO/ min)
n 2 (mol H 2 / min)
0.631 mol CH 3OH(v)/ mol
n 3 (mol CH 3OH(l) / min)
0.274 mol CO/ mol
CO H 2 o CH 3OH 0.0953 mol H / mol
2
Condenser
Overall process
3 unknowns (n3, n4, x)
2 unknowns (n1, n2)
-3 balances
0 degrees of freedom
-2 independent atomic balances
0 degrees of freedom
Balances around condenser
CO: 350 0.274
H : 350 0.631 n
2
4
CH OH: 350 0.0953
3
U|
V|
W
n
32.1 mol CH 3 OH(l) / min
n x
3
4
Ÿ n
318.7 mol recycle / min
( 0.996 x )
4
x .301 molCO / mol
n 0.004 n
3
4
Overall balances
U|V
|W
n
C: n = n
1
3 Ÿ 1
n
H: 2n = 4n
2
2
2
32.08 mol / min CO in feed
64.16 mol / min H 2 in feed
Single pass conversion of CO:
32.08 318.72 0.3009 350 0.274
u 100% 25.07%
32.08 318.72 0.3009
–
32.08 0
u 100% 100%
32.08
Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.)
–
Impurities in feed. (Re-analyze feed.)
–
Leak in methanol outlet pipe before flowmeter. (Check for it.)
Overall conversion of CO:
b.
4-50
4.58
a.
Basis: 100 kmol reactor feed/hr
n3 (kmol CH4 /h)
100 kmol /h
Reactor
n1 (kmol CH4 /h) 80 kmol CH4 /h
n2 (kmol Cl2 /h) 20 kmol Cl2 /h
n3 (kmol CH4 /h)
n4 (kmol HCl /h)
5n5 (kmol CH3Cl /h)
n5 (kmol CH2Cl 2 /h)
Cond.
Solvent
Absorb
n3 (kmol CH4 /h)
n4 (kmol HCl/h)
n4 (kmol HCl/h)
5n5 (kmol CH3Cl /h)
Still
5n5 (kmol CH3Cl /h)
n5 (kmol CH2Cl 2 /h)
n5 (kmol CH2Cl 2 /h)
Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F.
Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F.
Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F.
Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F.
Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F.
Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F.
Atomic balances around reactor:
½
1) C balance : 80 n 3 5n 5 n 5
°
2) H balance : 320 4n 3 n 4 15n 5 2n 5 ¾ Ÿ Solve for n 3 , n 4 , n 5
°
3) Cl balance : 40 n 4 5n 5 2n 5
¿
CH4 balance around mixing point: n1 = (80 – n3)
Cl2 balance: n2 = 20
b.
Solve for n1
For a basis of 100 kmol/h into reactor
n1 = 17.1 kmol CH4/h
n4 = 20.0 kmol HCl/h
n2 = 20.0 kmol Cl2/h
5n5 = 14.5 kmol CH3Cl/h
n3 = 62.9 kmol CH4/h
c.
(1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h
Scale factor =
19.81 kmol CH 3 Cl/h
14.5 kmol CH 3Cl/h
1.366
n tot 50.6 kmol/h
n
(17.1)(1.366) 23.3 kmol CH 4 /h ½
Ÿ
Fresh feed: 1
¾
n 2 (20.0)(1.366) 27.3 kmol Cl 2 /h ¿ 46.0 mol% CH 4 , 54.0 mole% Cl 2
Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h
4-51
4.59
a.
Basis: 100 mol fed to reactor/h Ÿ 25 mol O2/h, 75 mol C2H4/h
n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)
Seperator
reactor
nC2H4 ( mol C 2H 4 /h)
nO2 (mol O 2 /h)
75 mol C 2H 4 //h
25 mol O2 /h
n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h)
n5 (mol H 2O /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h)
n5 (mol H 2O /h)
Reactor
5 unknowns (n1 - n5)
-3 atomic balances
-1 - % yield
-1 - % conversion
0 D.F.
Strategy: 1. Solve balances around reactor to find n1- n5
2. Solve balances around mixing point to find nO2, nC2H4
(1) % Conversion Ÿ n1 = .800 * 75
(2) % yield: (.200)(75) mol C 2 H 4 u
90 mol C 2 H 4 O
100 mol C 2 H 4
n 3 (production rate of C 2 H 4 O)
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4
(4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5
(5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5
(6) O2 balance (mix pt): nO2 = 25 – n2
(7) C2H4 balance (mix pt): nC2H4 = 75 – n1
Overall conversion of C2H4: 100%
b.
c.
n1 = 60.0 mol C2H4/h
n5 = 3.00 mol H2O/h
n2 = 13.75 mol O2 /h
nO2 = 11.25 mol O2/h
n3 = 13.5 mol C2H4O/h
n4 = 3.00 mol CO2/h
nC2H4 = 15.0 mol C2H4/h
Scale factor =
100% conversion of C2H4
2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O
h
h
44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h
nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h
4-52
3.363
lb mol / h
mol / h
4.60
a.
Basis: 100 mol feed/h
100 mol/h
n1 (mol /h)
32 mol CO/h
64 mol H 2 / h
4 mol N 2 / h
.13 mol N 2 /mol
reactor
n3 (mol CH 3 OH / h)
cond.
500 mol / h
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)
n3 (mol / h)
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)
Purge
Mixing point balances:
total: (100) + 500 = n1 Ÿ n1 = 600 mol/h
N2: 4 + x1 * 500 = .13 * 600 Ÿ x1 = 0.148 mol N2/mol
Overall system balances:
N2: 4 = .148 * n3 Ÿ n3 = 27 mol/h
Atomic C: 32 = n2 + x2*27 Ÿ n2 = 24.3 mol CH3OH/h
Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27 Ÿ x2 = 0.284 mol CO/mol
Overall CO conversion: 100*[32-0.284(27)]/32 = 76%
Single pass CO conversion: 24.3/ (32+.284*500) = 14%
b.
Recycle: To recover unconsumed CO and H2 and get a better overall conversion.
Purge: to prevent buildup of N2.
4.61
a.
2N2 + 3H2 -> NH3
(1-yp) (1-fsp) n1 (mol N2)
(1-yp) (1-fsp) 3n1 (mol H2)
(1-yp) n2 (mol I)
1 mol
(1-XI0)/4 (mol N2 / mol)
3/4 (1-XI0) (mol H2 / mol)
XI0 (mol I / mol)
nr (mol)
n1 (mol N2)
3n1 (mol H2)
n2 (mol I)
Reactor
4-53
(1-fsp) n1 (mol N2)
(1-fsp) 3n1 (mol H2)
n2 (mol I)
nr (mol)
(1-fsp) n1 (mol N2)
(1-fsp) 3n1 (mol H2)
n2 (mol I)
2 fsp n1 (mol NH3)
yp (1-fsp) n1 (mol N2)
yp (1-fsp) 3n1 (mol H2)
yp n2 (mol I)
Condenser
np (mol)
2 fsp n1 (mol NH3)
4.61 (cont’d)
At mixing point:
N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1
I: XI0 + (1-yp) n2 = n2
Total moles fed to reactor: nr = 4n1 + n2
Moles of NH3 produced: np = 2fspn1
Overall N2 conversion:
b.
(1 X I0 ) / 4 y p (1 f sp )n 1
(1 X I0 ) / 4
u 100%
XI0 = 0.01 fsp = 0.20 yp = 0.10
n1 = 0.884 mol N2
nr = 3.636 mol fed
n2 = 0.1 mol I
np = 0.3536 mol NH3 produced
N2 conversion = 71.4%
c.
Recycle: recover and reuse unconsumed reactants.
Purge: avoid accumulation of I in the system.
d.
Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp
results in decreasing nr, increasing np, and increasing fov.
Increasing yp results in decreasing nr, decreasing np, and decreasing fov.
Optimal values would result in a low value of nr and fsp, and a high value of np, this would
give the highest profit.
XI0
0.01
0.05
0.10
0.01
0.01
0.01
0.10
0.10
0.10
fsp
0.20
0.20
0.20
0.30
0.40
0.50
0.20
0.20
0.20
yp
0.10
0.10
0.10
0.10
0.10
0.10
0.20
0.30
0.40
nr
3.636
3.893
4.214
2.776
2.252
1.900
3.000
2.379
1.981
4-54
np
0.354
0.339
0.321
0.401
0.430
0.450
0.250
0.205
0.173
fov
71.4%
71.4%
71.4%
81.1%
87.0%
90.9%
55.6%
45.5%
38.5%
4.62
a.
i - C 4 H 10 C 4 H 8
Basis: 1-hour operation
C 8 H 18
n 2 (n-C 4 H10 )
n 3 (i-C 4 H 10)
n 1 (C 8 H18)
m 4 (91% H 2 SO4 )
D
P
F
decanter
E
Units of n : kmol
Units of m: kg
n 1 (C 8 H18)
n 2 (n-C 4 H10 )
n 3 (i-C 4 H 10)
still
n 5 (n-C 4 H10)
n 6 (i-C 4 H 10)
n 7 (C 8 H18)
m 8 (91% H 2 SO 4 )
reactor
C
B
n 1 (C 8 H18)
n 2 (n-C 4 H10 )
m 4 (kg 91% H 2 SO4 )
40000 kg
A
n 0 kmol
0.25 i-C4 H10
0.50 n-C4 H10
0.25 C4 H 8
n 3 (i-C 4 H 10)
Calculate moles of feed
M
. g b0.25gb5610
. g
b0.75gb5812
0.25 M L C4 H10 0.50 M n C4 H10 0.25 M C4 H 8
57.6 kg kmol
n0
b40000 kggb1 kmol 57.6 kgg
Overall n - C 4 H 10 balance: n2
694 kmol
b0.50gb694g
347 kmol n - C 4 H 10 in product
C 8 H 18 balance:
n1
b0.25gb694g kmol C H
4
8
react 1 mol C 8 H 18
1 mol C 4 H 8
1735
. kmol C 8 H 8 in product
b
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 Ÿ n mol i - C 4 H 10
b
Note: n mol C 4 H 8
g
g b5gb0.25gb694g
A
moles C 4 H 8 at
A=173.5
173.5 at (A), (B) and (C) and in feed
b gb g
i - C 4 H 10 balance around first mixing point Ÿ 0.25 694 n3
Ÿ n3
867.5
694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H10 mol C4 H 8
b
Ÿ n mol i - C4 H10
g b200gb1735. g
C
4-55
34,700 kmol i - C4 H10
867.5 kmol
i -C 4 H 10 at
bA g and b Bg
4.62 (cont’d)
i - C 4 H 10 balance around second mixing point Ÿ 867.5 n6
Ÿ n6
34,700
33,800 kmol C 4 H 10 in recycle E
Recycle E: Since Streams (D) and (E) have the same composition,
b
g
bmoles n - C H g
bmoles C H g n
bmoles C H g n
n5 moles n - C 4 H 10
n2
n7
n1
4
b
g
bmoles i - C H g
n6 moles i - C 4 H 10
E
n3
10 D
8
18 E
6
8
18 D
3
4
Ÿ n7
E
Ÿ n5
16,900 kmol n - C 4 H 10
10 D
8460 kmol C 4 H 18
Hydrocarbons entering reactor:
kg I
b347 16900gbkmol n - C H g FGH 58.12 kmol
JK
kg I
F . kg IJ 1735. kmol C H FG 5610
b867.5 33800gb kmol i - C H g G 5812
.
H kmol K
H kmol JK
kg I
F
8460 kmol C H G 114.22
J 4.00 u 10 kg .
H
kmol K
H SO solution entering reactor 4.00 u 10 kg HC 2 kg H SO baq g
band leaving reactor g
1 kg HC
8.00 u 10 kg H SO baq g
m b H SO in recycleg
n b n - C H in recycleg
8.00 u 10 b H SO leaving reactor g n n b n - C H leaving reactor g
Ÿ m 7.84 u 10 kg H SO baq g in recycle E
4
10
4
10
4
8
6
8
2
18
6
4
2
4
6
2
2
8
4
4
5
4
10
5
4
10
6
2
2
4
6
2
8
m4
4
H 2 SO 4 entering reactor H 2 SO 4 in E
b g
16
. u 10 5 kg H 2 SO 4 aq recycled from decanter
b
g 1480 kmol H SO in recycle
d
ib g
d16. u 10 ib0.09gkg H O b1 kmol 18.02 kgg 799 kmol H O from decanter
Ÿ 16
. u 10 5 0.91 kg H 2 SO 4 1 kmol 98.08 kg
2
4
5
2
2
Summary: (Change amounts to flow rates)
Product: 173.5 kmol C 8 H 18 h , 347 kmol n - C 4 H 10 h
Recycle from still: 694 kmol i - C 4 H 10 h
Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h
Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C 4 H 10 h , 8460 kmol C 8 H 18 h,
7.84 u 10 6 kg h 91% H 2 SO 4 Ÿ 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h
4-56
4.63
a.
A balance on ith tank (input = output + consumption)
Ai kC Ai C Bi mol liter ˜ min V L
v L min C A , i 1 mol L vC
b
g
b
g
b
E y v, note V / v
gbg
W
C Ai kW C Ai C Bi
C A , i 1
B balance. By analogy, C B , i 1
C Bi kW C Ai C Bi
Subtract equations Ÿ C Bi C Ai
C B , i 1 C A, i 1
A
from balances on
bi 1g tank
C B , i 2 C A, i 2 C B 0 C A 0
st
b.
C Bi C Ai
C A , i 1
2
C Ai
C B 0 C A 0 Ÿ C Bi
b
C Ai C B 0 C A0 . Substitute in A balance from part (a).
g
0
2
C Ai kW C Ai C Ai C B 0 C A0 . Collect terms in C Ai
.
, C 1Ai , C Ai
b
g
kW C AL 1 kW C B 0 C A0 C A, i 1
2
Ÿ D C AL
E C AL J
0 where D
0
kW , E
b
g
1 kW C B 0 C A0 , J
C A , i 1
E E 2 4DJ
(Only + rather than r: since DJ is negative and the
2D
negative solution would yield a negative concentration.)
Solution: C Ai
c.
k=
v=
V=
CA0 =
CB0 =
alpha =
beta =
36.2
5000
2000
0.0567
0.1000
14.48
1.6270
N
1
2
3
4
5
6
7
8
9
10
11
12
13
14
gamma
-5.670E-02
-2.791E-02
-1.512E-02
-8.631E-03
-5.076E-03
-3.038E-03
-1.837E-03
-1.118E-03
-6.830E-04
-4.182E-04
-2.565E-04
-1.574E-04
-9.667E-05
-5.939E-05
CA(N)
2.791E-02
1.512E-02
8.631E-03
5.076E-03
3.038E-03
1.837E-03
1.118E-03
6.830E-04
4.182E-04
2.565E-04
1.574E-04
9.667E-05
5.939E-05
3.649E-05
xA(N)
0.5077
0.7333
0.8478
0.9105
0.9464
0.9676
0.9803
0.9880
0.9926
0.9955
0.9972
0.9983
0.9990
0.9994
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6),
(xmin = 0.99, N = 9), (xmin = 0.999, N = 13).
As xmin o 1, the required number of tanks and hence the process cost becomes infinite.
d.
(i) k increases Ÿ N decreases (faster reaction Ÿ fewer tanks)
(ii) v increases Ÿ N increases (faster throughput Ÿ less time spent in reactor
Ÿ lower conversion per reactor)
(iii) V increases Ÿ N decreases (larger reactor Ÿ more time spent in reactor
Ÿ higher conversion per reactor)
4-57
4.64
a.
Basis: 1000 g gas
Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3H8
800
44.09
18.145
77.2%
87.5%
C4H10
150
58.12
2.581
11.0%
12.5%
H2O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol
Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas
b.
C3H8 + 5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O
Theoretical O2:
C3H8:
C 4 H 10 :
5 kmol O 2
100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8
h
100 kg gas 44.09 kg C 3 H 8 1 kmol C 3 H 8
9.07 kmol O 2 / h
6.5 kmol O 2
100 kg gas 15 kg C 4 H 10 1 kmol C 4 H 10
h
100 kg gas 58.12 kg C 4 H 10 1 kmol C 4 H 10
1.68 kmol O 2 / h
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h
Air feed rate:
10.75 kmol O 2 1 kmol Air 1.3 kmol air fed
h
.21 kmol O 2 1 kmol air required
66.5 kmol air / h
The answer does not change for incomplete combustion
4.65
5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H 14
L C 6 H 14
86 kg C 6 H 14
38.3 mol C 6 H 14
4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C 7 H 16
L C 7 H 16
100 kg C 7 H 16
27.36 mol C 7 H 16
C6H14 +19/2 O2 o 6 CO2 + 7 H2O
C6H14 +13/2 O2 o 6 CO + 7 H2O
C7H16 + 11 O2 o 7 CO2 + 8 H2O
C7H16 + 15/2 O2 o 7 CO + 8 H2O
Theoretical oxygen:
38.3 mol C 6 H14
9.5 mol O 2
27.36 mol C 7 H 16 11 mol O 2
mol C 6 H 14
mol C 7 H16
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed
Percent excess air:
840 665
u 100%
665
26.3% excess air
4-58
665 mol O 2 required
4.66
CO 1
O 2 o CO 2
2
H2 1
O2 o H2O
2
130 kmol/h
0.500 kmol N2/kmol
x (kmol CO/mol)
(0.500–x) (kmol H2/kmol)
20% excess air
Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can
calculate the air feed rate without determining x CO . We include its calculation for illustrative
purposes.
b
A plot of x vs. R on log paper is a straight line through the points R1
bR
99.7, x 2
2
@
x
g
0.05 and
10
. .
b ln R ln a
ln x
R
g
10.0, x1
a Rb
38.3 Ÿ x
0.288
.
g 1303
ln a lnb10
. g 1303
.
lnb99.7g 6.00
a expb 6.00g 2.49 u 10 3
b
b
g b
ln 10
. 0.05 ln 99.7 10.0
Ÿx
.
2.49 u 10 3 R 1303
moles CO
mol
Theoretical O : 175 kmol 0.288 kmol CO 0.5 kmol O
2
2
h
kmol
kmol CO
Air fed:
4.67
a.
Theoretical O 2 :
o CO 2 2H 2 O
o 3CO 2 4H 2 O
17% excess air
na (kmol air/h)
0.21 O2
0.79 N2
o 4CO 2 5H 2 O
b g
b g
0.0060 100 kmol C 3 H 8
h
2
100 kmol/h
0.944 CH4
0.0340 C2H6
0.0060 C3H8
0.0050 C4H10
o 2CO 2 3H 2 O
0.944 100 kmol CH 4
h
kmol O
0.5 kmol O
2
2 43.75
kmol H
h
2
1.2 kmol air fed
kmol air
250
1 kmol air required
h
0.212 kmol H
2
kmol
43.75 kmol O required
1 kmol air
2
h
0.21 kmol O
CH 4 2O 2
7
C2 H 6 O2
2
C 3 H 8 5O 2
13
C 4 H 10 O 2
2
175 kmol
h
b g
2 kmol O 2 0.0340 100 kmol C 2 H 6 3.5 kmol O 2
1 kmol CH 4
h
1 kmol C 2 H 6
b g
0.0050 100 kmol C 4 H 10
5 kmol O 2
1 kmol C 3 H 3
h
207.0 kmol O 2 h
4-59
6.5 kmol O 2
1 kmol C 4 H 10
4.67 (cont’d)
Air feed rate: n f
207.0 kmol O 2
1 kmol air
h
0.21 kmol O 2
b
gb
gb
1.17 kmol air fed
kmol air req.
g
b.
na
n f 2 x1 35
. x 2 5x 3 6.5x 4 1 Pxs 100 1 0.21
c.
n f
aR f , ( n f
n a
bRa , (n a
xi
kAi Ÿ
75.0 kmol / h, R f
550 kmol / h, Ra
¦x
k
i
¦A
25) Ÿ n a
1 Ÿ k
i
i
60) Ÿ n f
i
1153 kmol air h
125
. Rf
22.0 Ra
1
¦A
i
i
Ai
Ÿ xi
¦A
, i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10
i
i
4.68
Run
1
2
3
Pxs
15%
15%
15%
Rf
62
83
108
A1
248.7
305.3
294.2
A2
19.74
14.57
16.61
A3
6.35
2.56
4.78
A4
1.48
0.70
2.11
Run
1
2
3
nf
77.5
103.8
135.0
x1
0.900
0.945
0.926
x2
0.0715
0.0451
0.0523
x3
0.0230
0.0079
0.0150
x4
0.0054
0.0022
0.0066
na
934
1194
1592
d.
Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the
flowmeter calibration formulas might not be linear, or the stack gas analysis could be
incorrect.
a.
C4H10 + 13/2 O2 o 4 CO2 + 5 H2O
Basis:
100 mol C4H10
nCO2 (mol CO2)
nH2O (mol H2O)
nC4H10 (mol C4H10)
nO2 (mol O2)
nN2 (mol N2)
Pxs (% excess air)
nair (mol air)
0.21 O2
0.79 N2
D.F. analysis
6 unknowns (n, n1, n2, n3, n4, n5)
-3 atomic balances (C, H, O)
-1 N2 balance
-1 % excess air
-1 % conversion
0 D.F.
4-60
Ra
42.4
54.3
72.4
4.68 (cont’d)
b.
i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2
n air
(650 mol O 2 )(1 mol air / 0.21 mol O 2 )
100% conversion Ÿ n C4H10
n N2
nCO2
n H2O
0 , nO 2
3095 mol air
0
b0.79gb3095 molg 2445 mol
b100 mol C H reactgb4 mol CO mol C H g
b100 mol C H reactgb5 mol H O mol C H g
4
4
10
2
10
2
4
4
10
10
U| 73.1% N
400 mol CO V12.0% CO
500 mol H O |W14.9% H O
2
2
2
2
2
ii) 100% conversion Ÿ nC4H10 = 0
20% excess Ÿ nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2)
Exit gas:
400 mol CO2
10.1% CO2
500 mol H2O
12.6% H2O
130 mol O2
3.3% O2
2934 mol N2
74.0% N 2
iii) 90% conversion Ÿ nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed)
20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2)
Exit gas:
10 mol C4H10
0.3% C4H10
9.1% CO2
360 mol CO2
450 mol H2O (v)
4.69
a.
11.4% H2O
195 mol O2
4.9% O2
2934 mol N2
74.3% N 2
C3H8 + 5 O2 o 3 CO2 + 4 H2O
H2 +1/2 O2 o H2O
C3H8 + 7/2 O2 o 3 CO + 4 H2O
Basis: 100 mol feed gas
100 mol
0.75 mol C3H8
0.25 mol H2
n1 (mol C3H8)
n2 (mol H2)
n3 (mol CO2)
n4 (mol CO)
n5 (mol H2O)
n6 (mol O2)
n7 (mol N2)
n0 (mol air)
0.21 mol O2/mol
0.79 mol N2/mol
Theoretical oxygen:
75 mol C 3 H 8
5 mol O 2
25 mol H 2 0.50 mol O 2
mol C 3 H 8
mol H 2
4.69 (cont’d)
4-61
387.5 mol O 2
Air feed rate: n0
387.5 mol O 2 1 kmol air 125
. kmol air fed
h
0.21 kmol O 2 1 kmol air req' d.
90% propane conversion Ÿ n1
2306.5 mol air
. (75 mol C 3 H 8 ) = 7.5 mol C 3 H 8
0100
(67.5 mol C 3 H 8 reacts)
85% hydrogen conversion Ÿ n2
. (25 mol C 3 H 8 ) = 3.75 mol H 2
0150
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated
mol C 3 H 8 react
95% CO 2 selectivity Ÿ n3
192.4 mol CO 2
0.05(67.5 mol C 3 H 8 react) 3 mol CO generated
mol C 3 H 8 react
5% CO selectivity Ÿ n3
FG
H
H balance: (75 mol C 3 H 8 ) 8
. mol CO
101
IJ
K
mol H
( 25 mol H 2 )(2)
mol C 3 H 8
(7.5 mol C 3 H 8 )(8) (3.75 mol H 2 )(2) n5 ( mol H 2 O)(2) Ÿ n5
2912
. mol H 2 O
mol O
) (192.4 mol CO 2 )(2)
mol O 2
(101
. mol CO)(1) ( 2912
. mol H 2 O)(1) + 2n6 ( mol O 2 ) Ÿ n6
1413
. mol O 2
O balance: (0.21 u 2306.5 mol O 2 )(2
N 2 balance: n7
0.79(2306.5) mol N 2
1822 mol N 2
Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol
= 2468 mol
CO concentration in exit gas =
b.
101
. mol CO
u 10 6
2468 mol
4090 ppm
If more air is fed to the furnace,
(i)
more gas must be compressed (pumped), leading to a higher cost (possibly a larger
pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the
product gas temperature decreases and less steam is produced.
4-62
4.70
a.
C5H12 + 8 O2 o 5 CO2 + 6 H2O
Basis: 100 moles dry product gas
n1 (mol C5H12)
100 mol dry product gas (DPG)
0.0027 mol C5H12/mol DPG
0.053 mol O2/mol DPG
0.091 mol CO2/mol DPG
0.853 mol N2/mol DPG
n3 (mol H2O)
Excess air
n2 (mol O2)
3.76n2 (mol N2)
3 unknowns (n1, n2, n3)
-3 atomic balances (O, C, H)
-1 N2 balance
-1 D.F. Ÿ Problem is overspecified
b.
N2 balance: 3.76 n2 = 0.8533 (100) Ÿ n2 = 22.69 mol O2
C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) Ÿ n1 = 2.09 mol C5H12
H balance: 12 n1 = 12(0.0027)(100) + 2n3 Ÿ n3 = 10.92 mol H2O
O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 Ÿ 45.38 mol O = 39.72 mol O
Since the 4th balance does not close, the given data cannot be correct.
c.
n1 (mol C5H12)
Excess air
n2 (mol O2)
3.76n2 (mol N2)
100 mol dry product gas (DPG)
0.00304 mol C5H12/mol DPG
0.059 mol O2/mol DPG
0.102 mol CO2/mol DPG
0.836 mol N2/mol DPG
n3 (mol H2O)
N2 balance: 3.76 n2 = 0.836 (100) Ÿ n2 = 22.2 mol O2
C balance: 5 n1 = 100 (5*0.00304 + 0.102) Ÿ n1 = 2.34 mol C5H12
H balance: 12 n1 = 12(0.00304)(100) + 2n3 Ÿ n3 = 12.2 mol H2O
O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 Ÿ 44.4 mol O = 44.4 mol O ¸
Fractional conversion of C5H12:
2.344 100 u 0.00304
2.344
0.870 mol react/mol fed
Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 28.75 mol O2
% excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required
u 100% 18.6% excess air
18.75 mol O 2 required
4-63
4.71
a.
12 L CH 3 OH 1000 ml 0.792 g mol
h
L
ml 32.04 g
296.6 mol CH 3 OH / h
CH3OH + 3/2 O2 o CO2 +2 H2O, CH3OH + O2 o CO +2 H2O
n 2 ( mol dry gas / h)
0.0045 mol CH3OH(v)/mol DG
0.0903 mol CO2/mol DG
0.0181 mol CO/mol DG
x (mol O2/mol DG)
(1–x) (mol N2/mol DG)
n 3 ( mol H 2 O(v) / h)
296.6 mol CH3OH(l)/h
n1 (mol O 2 / h)
3.76n1 (mol N 2 / h)
4 unknowns (n1 , n 2 , n 3 , x ) – 4 balances (C, H, O, N2) = 0 D.F.
b.
Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h
C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) Ÿ n 2 = 2627 mol/h
H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 Ÿ n 3 = 569.6 mol H2O / h
O balance : 296.6 2n1 2627[0.0045 2(0.0903) 0.0181 2(0.8871- x)] 569.6
N2 balance: 3.76 n 1 = x ( 2627)
Solving simultaneously Ÿ n1
Fractional conversion:
% excess air:
296.6 2627(0.0045)
296.6
0.960 mol CH 3 OH react/mol fed
574.3 444.9
u 100% 29.1%
444.9
569.6 mol H 2 O
(2627 569.6) mol
Mole fraction of water:
4.72
574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG
0.178 mol H 2 O/mol
c.
Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger
alarm if concentrations are too high
a.
G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH 4 and xC2 H 6 are
the mole fractions of methane and ethane in the fuel, then
b g b
n b molg x b mol CH
gb
g
molgb1 mol C 1 mol CH g
ns mol xC2 H 6 mol C2 H 2 mol 2 mol C 1 mol C 2 H 6
E
s
CH 4
b
bmol CH
4
4
xC2 H 6 mol C2 H 6 mol fuel
xCH 4
4
mol fuel
g
g
20
85
01176
.
mole C 2 H 6 mole CH 4 in fuel gas
4-64
4.72 (cont’d)
b1.134 g H Ogb1 mol 18.02 gg
mole H 2 O
0.50 mol product gas
mole product gas
Basis: 100 mol product gas. Since we have the most information about the product stream
composition, we choose this basis now, and would subsequently scale to the given
fuel and air flow rates if it were necessary (which it is not).
2
Condensation measurement:
0126
.
CH 4 2O 2 o CO 2 2H 2 O
7
C 2 H 6 O 2 o 2CO 2 3H 2 O
2
100 mol dry gas / h
n1 (mol CH4 )
0.1176 n1 (mol C2H6)
n2 (mol CO2)
0.126 mol H2O / mol
0..874 mol dry gas / mol
0.119 mol CO2 / mol D.G.
x (mol N2 / mol)
(0.881-x) (mol O2 / mol D.G.)
n3 (mol O2 / h)
376 n3 (mol N2 / h)
Strategy: H balance Ÿ n ;
1
C balance Ÿ n 2 ;
b gb
H balance: 4n1 6 01176
.
n1
UV
W
N 2 balance
Ÿ n3 , x
O balance
. gb2g Ÿ n
g b100gb0126
1
5.356 mol CH 4 in fuel
Ÿ 0.1176(5.356) = 0.630 mol C2H6 in fuel
b gb
g
C balance: 5.356 2 0.630 n2
. gŸn
b100gb0.874gb0119
2
3.784 mol CO 2 in fuel
Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO 2
Ÿ 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
b100gb0.874gx
b0.881 x g
. g b100gb0.874gb2g 0119
.
b2gb3.784g 2n b100gb0126
N 2 balance: 3.76n3
O balance:
3
Solve simultaneously: n3 18.86 mols O 2 fed , x 0.813
5.356 mol CH 4 2 mol O 2 0.630 mol C 2 H 6 3.5 mol O 2
Theoretical O 2 :
1 mol CH 4
1 mol CH 4
12.92 mol O 2 required
Desired O2 fed:
(5.356 0.630 3.784) mol fuel 7 mol air 0.21 mol O 2
= 14.36 mol O2
1 mol fuel mol air
Desired % excess air:
b.
Actual % excess air:
14.36 12.92
u 100% 11%
12.92
18.86 12.92
u 100% 46%
12.92
Actual molar feed ratio of air to fuel:
(18.86 / 0.21) mol air
9.77 mol feed
4-65
9 :1
4.73
a.
C3H8 +5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O
Basis 100: mol product gas
n1 (mol C3H8)
n2 (mol C4H10)
100 mol
0.474 mol H2O/mol
x (mol CO2/mol)
(0.526–x) (mol O2/mol)
n3 (mol O2)
Dry product gas contains 69.4% CO2 Ÿ
x
0.526 x
69.4
Ÿx
30.6
0.365 mol CO 2 /mol
3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F.
O balance: 2 n3 = 152.6 Ÿ n3 = 76.3 mol O2
n1 7.1 mol C 3 H 8
C balance : 3 n1 4 n 2 36.5 ½
Ÿ 65.1% C 3 H 8 , 34.9% C 4 H10
¾Ÿ
H balance : 8 n1 10 n 2 94.8¿ n 2 3.8 mol C 4 H10
b.
nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C
nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H
Ÿ 27.8%C, 72.2% H
From a:
7.10 mol C 3 H 8
3.80 mol C 4 H10 4 mol C
3 mol C
mol C 4 H10
mol C 3 H 8
7.10 mol C 3 H 8 11 mol (C H) 3.80 mol C 4 H10 14 mol (C H)
mol C 4 H10
mol C 3 H 8
4.74
Basis: 100 kg fuel oil
Moles of C in fuel:
100 kg 0.85 kg C 1 kmol C
kg
12.01 kg C
Moles of H in fuel:
100 kg 0.12 kg H 1 kmol H
12.0 kmol H
kg
1 kg H
Moles of S in fuel:
100 kg 0.017 kg S 1 kmol S
kg
32.064 kg S
1.3 kg non-combustible materials (NC)
4.74 (cont’d)
4-66
7.08 kmol C
0.053 kmol S
u 100% 27.8% C
100 kg fuel oil
7.08 kmol C
12.0 kmol H
0.053 kmol S
1.3 kg NC (s)
20% excess air
n1 (kmol O2)
3.76 n1 (kmol N2)
n2 (kmol N2)
n3 (kmol O2)
n4 (kmol CO2)
(8/92) n4 (kmol CO)
n5 (kmol SO2)
n6 (kmol H2O)
C + O2 o CO2
C + 1/2 O2 o CO
2H + 1/2 O2 o H2O
S + O2 o SO2
Theoretical O2:
7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2
1 kmol C
2 kmol H
1 kmol S
10.133 kmol O 2
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed
O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 Ÿ n3 = 2.3102 kmol O2
C balance: 7.08 = n4+8n4/92 Ÿ n4 = 6.514 mol CO2
Ÿ 8 (6.514)/92 = 0.566 mol CO
S balance: n5 = 0.53 kmol SO2
H balance: 12 = 2n6 Ÿ n6 = 6.00 kmol H2O
N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2
Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol
= 61.16 kmol
Ÿ 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 2
4.75
a. Basis: 5000 kg coal/h; 50 kmol air min
3000 kmol air h
5000 kg coal / h
0.75 kg C / kg
0.17 kg H / kg
0.02 kg S / kg
0.06 kg ash / kg
C + 02 --> CO2
2H + 1/2 O2 -->H2O
S + O2 --> SO2
C + 1/2 O2 --> CO
3000 kmol air / h
0.21 kmol O2 / kmol
0.79 kmol N2 / kmol
n1 (kmol O2 / h)
n2 (kmol N2 / h)
n3 (kmol CO2 / h)
0.1 n3 (kmol CO / h)
n4 (kmol SO2 / h)
n5 (kmol H2O / h)
mo kg slag / h
Theoretical O 2 :
C:
0.75 5000 kg C
b g
1 kmol C
1 kmol O 2
h
12.01 kg C
1 kmol C
4.75 (cont’d)
4-67
312.2 kmol O 2 h
H:
S:
b g
0.17 5000 kg H 1 kmol H 1 kmol H 2 O
101
. kg H
h
1 kmol O 2
2 kmol H
2 kmol H 2 O
210.4 kmol O 2 h
b g
0.02 5000 kg S 1 kmol S 1 kmol O 2
= 3.1 kmol O2/h
h
32.06 kg S 1 kmol S
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h
b
g
O 2 fed = 0.21 3000
Excess air:
630 kmol O 2 h
630 525.7
u 100% 19.8% excess air
525.7
b. Balances:
0.94 0.75 5000 kg C react 1 kmol C
C:
12.01 kg C
h
b gb gb g
Ÿ n 3
H:
S:
N2 :
O:
. n 3
266.8 kmol CO 2 h , 01
. gb5000g kg H
b017
(from part a)
26.7 kmol CO h
1 kmol H 1 kmol H 2 O
101
. kg H
h
n 3 0.1n 3
b
2 kmol H
3.1 kmol O 2 for SO 2
h
g
n 5 Ÿ n5
1 kmol SO 2
1 kmol O 2
420.8 kmol H 2 O h
n 4 Ÿ n 4
31
. kmol SO 2 h
b0.79gb3000g kmol N h n Ÿ n 2370 kmol N h
b0.21g(3000)b2g 2n 2b266.8g 1b26.68g 2b31. g b1gb420.8g
2
2
2
2
1
Ÿ n1
136.4 kmol O 2 / h
Stack gas total
3223 kmol h
Mole fractions:
c.
x CO
26.7 3224 8.3 u 10 3 mol CO mol
xSO2
31
. 3224
9.6 u 10 4 mol SO 2 mol
1
SO 2 O 2 o SO 3
2
SO 3 H 2 O o H 2SO 4
3.1 kmol SO 2 1 kmol SO 3 1 kmol H 2SO 4
h
1 kmol SO 2
1 kmol SO 3
4-68
98.08 kg H 2SO 4
kmol H 2SO 4
304 kg H 2SO 4 h
4.76 a.
Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile
matter
100 g c.a. r. 1.147 g a.d.c.
1.207 g c.a. r.
95.03 g a.d.c
95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
. ggH O
b1.234 1204
2
1.234 g a.d.c.
Total H 2 O
95.03 g a.d.c
4.97 g 2.31 g
2.31 g H 2 O lost in second drying step
7.28 g moisture
0.811g g b v. m. H Og
.
b1347
2.31 g H O
2
2
1.347 g a.d.c.
0.111 g ash
95.03 g a.d.c
1.175 g a.d.c.
8.98 g ash
b100 7.28 35.50 8.98gg
Fixed carbon
3550
. g volatile matter
48.24 g fixed carbon
7.28 g moisture
48.24 g fixed carbon
7.3% moisture
48.2% fixed carbon
35.50 g volatile matter Ÿ
35.5% volatile matter
8.98 g ash
9.0% ash
100 g coal as received
b. Assume volatile matter is all carbon and hydrogen.
C CO 2 o CO 2 :
2H 1 mol O 2
1 mol C
1 mol C 10 3 g
1 mol air
12.01 g C 1 kg 0.21 mol O 2
0.5 mol O 2
1
O2 o H 2O :
2
2 mol H
Air required:
396.5 mol air kg C
1 mol H 10 3 g
1 mol air
1.01 g H 1 kg 0.21 mol O 2
1179 mol air kg H
1000 kg coal 0.482 kg C 396.5 mol air
kg coal
1000 kg 0.355 kg v. m.
kg C
6 kg C
396.5 mol air
kg
7 kg v. m.
kg C
1000 kg 0.355 kg v. m. 1 kg H 1179 mol air
kg
7 kg v. m.
4-69
kg H
3.72 u 105 mol air
4.77
a.
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (mol C)
n2 (mol H)
n3 (mol S)
100 mol dry gas
C + 02 --> CO2
C + 1/2 O2 --> CO
2H + 1/2 O2 -->H2O
S + O2 --> SO2
0.720 mol CO2 / mol
0.0257 mol CO / mol
0.000592 mol SO2 / mol
0.254 mol O2 / mol
n4 (mol O2)
(20% excess)
n5 (mol H2O (v))
½
°
O balance : 2 n 4 100 [ 2(0.720) 0.0257 2 (0.000592) 2 (0.254)] n 5 ¾
°
20 % excess O 2 : (1.20) (74.57 0.0592 0.25 n 2 ] n 4
¿
H balance : n 2
2 n5
Ÿ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O
Total moles in feed: 258.4 mol (C+H+S) Ÿ 28.9% C, 71.1% H, 0.023% S
4.78
Basis: 100 g oil
Stack
SO 2 , N 2 , O 2, CO 2, H 2O
(612.5 ppm SO 2)
x n 3 mol SO 2
(N2 , O2 , CO2 , H 2 O)
0.10 (1 – x ) n 5 mol SO 2
(N2 , O2 , CO2 , H 2 O)
100 g oil
0.87 g C/g
0.10 g H/g
0.03 g S/g
n 1 mol O2
3.76 n 1 mol N2
(25% excess)
furnace
Alkaline solution
(1 – x ) n 5 mol SO 2
(N2 , O2 , CO2 , H 2 O)
n 2 mol N 2
n 3 mol O 2
n 4 mol CO2
n 5 mol SO 2
n 6 mol H 2 O
CO 2 :
H 2 O:
b g
0.87 100 g C
scrubber
0.90 (1 – x ) n 5 mol SO 2
1 mol C 1 mol CO 2
Ÿ n4
12.01 g C 1 mol C
b g
0.10 100 g H 1 mol H 1 mol H 2 O
Ÿ n6
2 mol H
. gH
101
4-70
FG 7.244 mol O IJ
H consumed K
F 2.475 mol O IJ
4.95 mol H OG
H consumed K
7.244 mol CO 2
2
2
2
4.78 (cont’d)
SO 2 :
b g
0.03 100 g S
1 mol S 1 mol SO 2
Ÿ n5
32.06 g S 1 mol S
b
0.0936 mol SO 2
g
FG 0.0956 mol O IJ
H consumed K
2
125
. 7.244 2.475 0.0936 Ÿ 12.27 mol O 2
25% excess O 2 : n1
b
g
12.27 mol O 2 fed 7.244 2.475 0.0936 mol O 2 consumed
O 2 balance: n3
2.46 mol O 2
N 2 balance: n 2
b
b
3.76 12.27 mol
g
4614
. mol N 2
g
SO 2 in stack SO 2 balance around mixing point :
F
H
I
K
b gb
g
x 0.0936 010
. 1 x 0.0936
n5
b
0.00936 0.0842 x mol SO 2
g
Total dry gas in stack (Assume no CO2 , O2 , or N 2 is absorbed in the scrubber)
b
7.244 2.46 4614
. 0.00936 0.0842 x
bCO g bO g
2
bN g
2
2
b
bSO g
g
b
g
5585
. 0.0842 x mol dry gas
2
g
612.5 ppm SO 2 dry basis in stack gas
0.00936 0.0842 x
. 0.0842 x
5585
612.5
Ÿx
. u 10 6
10
0.295 Ÿ 30% bypassed
Basis: 100 mol stack gas
4.79
n 1 (mol C)
n 2 (mol H)
n 3 (mol S)
n 4 (mol O2 )
3.76 n 4 (mol O2 )
a.
C balance: n1
H balance: n2
C + O2 o CO2
1
2H + O2 o H 2 O
2
S + O 2 o SO 2
100 mol
0.7566 N 2
0.1024 CO2
0.0827 H 2 O
0.0575 O 2
0.000825 SO 2
.
b100gb01024
g 10.24 mol C
b100gb0.0827gb2g 16.54 mol H
Ÿ
10.24 mol C
16.54 mol H
0.62
mol C
mol H
The C/H mole ratio of CH 4 is 0.25, and that of C2 H 6 is 0.333; no mixture of the two could
have a C/H ratio of 0.62, so the fuel could not be the natural gas.
b.
b100gb0.000825g 0.0825 mol S
b10.24 mol Cgb12.0 g 1 molg 122.88 g CU| 122.88 7.35 g C g H
b16.54 mol Hgb1.01 g 1 molg 16.71 g H V Ÿ 216.65.71
b0.0825 mol Sgb32.07 g 1 molg 2.65 g S |W 142.24 u 100% 1.9% S
S balance: n 3
4-71
Ÿ No. 4 fuel oil
4.80
a.
Basis: 1 mol CpHqOr
1 mol CpHqOr
no (mol S)
Xs (kg s/ kg fuel)
C + 02 --> CO2
2H + 1/2 O2 -->H2O
S + O2 --> SO2
P (% excess air)
n1 (mol O2)
3.76 n1 (mol N2)
n2 (mol CO2)
n3 (mol SO2)
n4 (mol O2)
3.76 n1 (mol N2)
n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C)
q (mol H) (1 g / mol) =
q (g H)
Ÿ (12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O)
S in feed:
no=
(12 p q 16r) g fuel
Theoretical O2:
X s (g S)
1 mol S
(1 - X s ) (g fuel) 32.07 g S
X s (12 p q 16 r)
(mol S) (1)
32.07(1 - X s )
p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2
1 mol C
2 mol H
2 mol O
(p 1/4 q 1/2 r) mol O 2 required
% excess Ÿ n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n2 = p
(3)
H balance: n5 = q/2
(4)
S balance: n3 = n0
(5)
O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 Ÿ n4 = ½ (r+2n1-2n2-2n3-n5)
(6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air
(1) Ÿ n0 = 0.00616 mol S
(5) Ÿ n3 = 0.00616 mol SO2
(2) Ÿ n1 = 1.16 mol O2 fed
(6) Ÿ n4 = 0.170 mol O2
(3) Ÿ n2 = 0.71 mol CO2
(4) Ÿ n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2
Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas
Dry basis composition
yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2
yO2 = (0.170 / 5.246) * 100% = 3.2% O2
yN2 = (4.36 / 5.246) * 100% = 83.1% N2
ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO2
4-72
CHAPTER FIVE
5.1
Assume volume additivity
m
m0 Ÿ m
=
mt
mass of tank
at time t
empty tank
a.
A
0.400
0.600
ŸU
0.703 kg L 0.730 kg L
1
U
Av. density (Eq. 5.1-1):
A
mass of
A
A
UO
UD
b250 150gkg
b10 3g min
14.28 kg min
0.719 kg L
bm
mass flow rate of liquid
1L
/ min)
/ min) = m(kg
14.28 kg
Ÿ V(L
Ÿ V
19.9 L min
min
0.719 kg
U ( kg / L)
b. m0
5.2
bg
150 14.28 3
m(t) - mt
107 kg
b
g
void volume of bed: 100 cm 3 2335
. 184 cm3
porosity: 50.5 cm3 void 184 cm3 total
bulk density: 600 g 184 cm3
b
50.5 cm 3
0.274 cm3 void cm3 total
3.26 g cm3
g
absolute density: 600 g 184 50.5 cm3
4.49 g cm3
5.3
C 6 H 6 (l )
B (kg / min)
m
VB = 20.0 L / min
(kg / min)
m
(L / min)
V
C 7 H 8 (l )
T (kg / min)
m
(L / min)
V
T
2
. m) 2 015
. m
= 'V SD 'h S (55
V
0.0594 m3 / min
't
4
60 min
4 't
Assume additive volumes
-V
59.4 20.0 L / min = 39.4 L / min
V
V
T
B
b
m
xB
g
0.879 kg 20.0 L 0.866 kg 39.4 L
L
min
L
min
(0.879 kg / L)(20.0 L / min)
0.34 kg B / kg
(517
. kg / min)
B UT ˜ V
T
UB ˜ V
B
m
m
5-1
. kg / min
517
g
a.
b.
1
U sl
U|
V|
W
b
F IF I
e jge jhbmgGH 11 N JK GH 11 Pa JK
U sl
kg
m
s2
m3
kg˜m
s2
N
m2
U sl gh
g
1 xc
xc
Ÿ check units!
Uc
Ul
1
kg crystals / kg slurry kg liquid / kg slurry
kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid
L slurry L crystals L liquid
L slurry
kg slurry kg slurry kg slurry kg slurry
'P
2775
1415 kg / m3
c. i.) U sl
9.8066 0.200
gh
ii.)
1
U sl
xc
xc
Uc
b gb g
b1- x g Ÿ x FG 1 1 IJ FG 1
U
HU U K HU
c
l
F 1
GG 1415 kg / m
H
c
3
l
sl
1
. 1000 kg / m3
12
d
I
iJJK
1
Ul
F
I
GG 2.3d10001kg / m i 12. d10001kg / m iJJ
H
K
iii.) Vsl
msl
U sl
iv.) mc
x c msl
IJ
K
0.316 kg crystals / kg slurry
3
175 kg
1000 L
3
1415 kg / m
m3
1238
. L
b0.316 kg crystals / kg slurrygb175 kg slurryg
55.3 kg crystals
55.3 kg CuSO 4 ˜ 5H 2 O 1 kmol
1 kmol CuSO 4
159.6 kg
249 kg 1 kmol CuSO 4 ˜ 5H 2 O 1 kmol
v.) mCuSO 4
vi.) ml
c
3
b1 x gm b0.684 kg liquid / kg slurrygb175 kg slurryg
c
ml
Ul
vii.)
Vl
h(m)
Ul(kg/m^3)
Uc(kg/m^3)
'P(Pa)
xc
Usl(kg/m^3)
0.2
1200
2300
2353.58
0
1200.00
sl
120 kg
1000 L
m3
b1.2gd1000 kg / m i
3
35.4 kg CuSO 4
120 kg liquid solution
100 L
d.
2411.24
0.05
1229.40
2471.80
0.1
1260.27
2602.52
0.2
1326.92
2747.84
0.3
1401.02
2772.61
0.316
1413.64
2910.35
0.4
1483.87
3093.28
0.5
1577.14
Effect of Slurry Density on Pressure Measurement
0.6
Solids Fraction
5.4
P0 U sl gh1
P0 U sl gh 2 Ÿ 'P = P1 P2
h = h1 h 2
P1
P2
0.5
0.4
0.3
'P = 2775, U = 0.316
0.2
0.1
0
2300.00
2500.00
2700.00
2900.00
Pressure Difference (Pa)
5-2
3100.00
5.4 (cont’d)
e.
b
g d
b
i
b1- x gbkg liquid g, V dm
c
3
l
liquid
g
x c kg crystals
Basis: 1 kg slurry Ÿ x c kg crystals , Vc m3 crystals
d
U c kg / m
3
i
g
i b1-Ux dgbkgkg/ mliquid
i
c
3
l
1 kg
U sl
5.7
a.
1 atm
1 mol
29.0 g
3
0.0064 m air
nRT
P
V=
b3.06L - 2.8Lg u 100%
Assume Patm
1013
.
bar
PV
Ps Vs
n
. gbar
b10 1013
nRT
Ÿn
n s RTs
20.0 m3
V˜
273K
298.2K
b
3.06 L
9.3%
2.8L
nRT Ÿ n
PV
0.0064 m3 mol
4.5 kg m 3
mol 103 g
100
. mol 0.08206 L ˜ atm 373.2 K
mol ˜ K 10 atm
a.
5.8
1 kg
b. % error =
b.
g
3
= 0.08206 m ˜ atm 313.2 K 1 kmol
RT Ÿ V
kmol ˜ K 4.0 atm 103 mol
PV
5.6
l
Assume Patm
U
b
bV V gdm i
c
5.5
1
1 xc
xc
Uc
Ul
3
kmol ˜ K 28.02 kg N 2
20.0 m3
kmol
25 + 273.2 K 0.08314 m3 ˜ bar
g
249 kg N 2
Ts P n s
˜ ˜
T Ps Vs
. gbar
b10 1013
1.013 bar
1 kmol
28.02 kg N 2
kmol
22.415 m STP
3
a.
R=
Ps Vs
n sTs
1 atm 22.415 m3
1 kmol 273 K
b.
R=
Ps Vs
n sTs
1 atm 760 torr 359.05 ft 3
1 lb - mole 1 atm
492 $ R
8.21 u 10 2
5-3
b g
atm ˜ m3
kmol ˜ K
555
torr ˜ ft 3
lb - mole ˜$ R
249 kg N 2
5.9
P = 1 atm +
10 cm H 2 O
1m
1 atm
2
10 cm 10.333 m H 2 O
101
. atm
3
= 2.0 m
T = 25$ C = 298.2 K , V
5 min
m = n mol / min ˜ MW g / mol
b
g
=
m
a.
=
m
b.
b
0.40 m3 min = 400 L min
g
L
28.02
1.01 atm 400 min
PV
˜ MW =
L˜atm
0.08206 mol˜K 298.2 K
RT
400
L
min
28.02
273 K
1 mol
298.2 K 22.4 L STP
b g
g
mol
F mI V dm si
uG J
H s K Ad m i
3
5.10 Assume ideal gas behavior:
458 g min
458 g min
nRT
P
u
Ÿ 2
2
u1
SD 4
2
T2 P1 D12
nR
˜ ˜ ˜ 2
nR
T1 P2 D 2
. 1013
. gbar b7.50 cmg
b180
134 m sec
sec
333.2 K b153
. 1013
. gbar b5.00 cmg
. 100
. g atm 5 L
PV b100
Assume ideal gas behavior: n
0.406 mol
u2
5.11
g
mol
u1
T2 P1D12
T1P2 D 22
2
60.0 m 300.2 K
2
L˜atm
0.08206 mol
RT
˜K
32.0 g mol Ÿ Oxygen
MW 13.0 g 0.406 mol
mass of tank, n g
5.12 Assume ideal gas behavior: Say m t
unknown: MW
g
t
0.009391 mol
b g
std cm3 STP min
V
' V liters 273K 763 mm Hg 103 cm3
1L
' t min 296.2K 760 mm Hg
b g
I
std cm3 STP min
V
5.0
9.0
12.0
139
268
370
U|
|V straight line plot
||I 0.031EV 0.93
W
std
b.
std
V
b g
0.010 mol N 2 22.4 liters STP 103 cm3
min
1 mole
1L
d
mol of gas in tank
gU|V Ÿ n 0.009391 mol
g |W m 37.0256 g
b37.062 37.0256gg 3.9 g mol Ÿ Helium
b
b
m t n g 28.02 g mol
m t n g 44.1 g mol
N 2 : 37.289 g
CO 2 : 37.440 g
5.13 a.
300 K
i
I = 0.031 224 cm3 / min 0.93 7.9
5-4
224 cm3 / min
925.3
'V
't
/ kmol)
g nbkmolgM(kg
Vb Lg
b
FU I
si ˜ G J
HU K
5.14 Assume ideal gas behavior U kg L
d
3
i V dcm
V2 cm s
n P
V RT
PM
! RT
12
3
1
1
V1 P1 M1T2 P2 M 2 T1
12
2
LM
N
cm3 758 mm Hg 28.02 g mol 323.2K
s 1800 mm Hg 2.02 g mol 298.2K
a.
VH 2
b.
M
0.25M CH 4 0.75M C 3H 8
Vg
cm3
350
s
350
OP
Q
12
881 cm3 s
b0.25gb16.05g b0.75gb44.11g
LM b758gb28.02gb323.2g OP
N b1800gb37.10gb298.2g Q
37.10 g mol
12
205 cm3 s
5.15 a.
Reactor
'h
soap
b.
d
2
S 0.012 m
4
i
2
n CO 2
2
PV
= S R 'h
ŸV
't
RT
n CO 2
755 mm Hg
1 atm
11
. u 10-3 m3 / min 1 kmol
3
m ˜atm 760 mm Hg
300 K
1000 mol
0.08206 kmol
˜K
. m 60 s
12
. u 10 3 m3 / min
11
7.4 s min
0.044 mol / min
5.16
air
m
10.0 kg / h
n air (kmol / h)
n (kmol / h)
y CO (kmol CO 2 / kmol)
2
CO = 20.0 m3 / h
V
2
n CO (kmol / h)
2
150 o C, 1.5 bar
Assume ideal gas behavior
10.0 kg 1 kmol
n air
0.345 kmol air / h
h
29.0 kg air
n CO 2
PV
RT
y CO 2 u 100%
. bar
15
100 kPa 20.0 m3 / h
3
1 bar
423.2 K
8.314 mkmol˜kPa
˜K
b
0.853 kmol CO 2 / h
0.853 kmol CO 2 / h
u 100%
0.853 kmol CO 2 / h + 0.345 kmol air / h
g
5-5
.
712%
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior
3
o
1 (kg / min)
m
0.70 kg H 2 O / kg
0.30 kg S / kg
311 m / min, 83 C, 1 atm
n 3 (kmol / min)
0.12 kmol H 2 O / kmol
0.88 kmol dry air / kmol
n 2 (kmol air / min)
(m 3 / min)
V
2
167 o C, - 40 cm H 2 O gauge
4 (kg S / min)
m
a.
n 3
kmol ˜ K
10.38 kmol min
0.08206 m3 ˜ atm
1 atm 311 m3
365.2K
min
10.38 kmol 0.12 kmol H 2 O 18.02 kg
kmol
kmol
min
H 2 O balance: 0.70 m1
1
Ÿm
32.2 kg min milk
b g
b
S olids balance: 0.30 32.2 kg min
4 Ÿm
4
m
b
9.6 kg S min
g
0.88 10.38 kmol min Ÿ n 2
Dry air balance: n 2
2
V
g
9.13 kmol 0.08206 m3 ˜ atm
kmol ˜ K
min
9.13 kmol min air
440K
1033 cm H 2 O
1033 40 cm H 2 O
1 atm
b
g
3
343 m air min
u air (m / min) =
air (m3 / s)
V
A (m2 )
343 m3 1 min
min 60 s
S
4 ˜ (6
m) 2
0.20 m / s
b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor
by the air instead of falling to the conveyor belt.
5.18 SG CO 2
5.19 a.
x CO 2
U CO 2
U air
PM CO2
RT
PM air
RT
0.75 x air
M CO 2
M air
152
.
1 0.75 0.25
Since air is 21% O 2 , x O 2
b.
44 kg / kmol
29 kg / kmol
mCO 2 = n ˜ x CO 2 ˜ M CO 2
(0.25)(0.21)
0.0525 5.25 mole% O 2
b
g
2 u 1.5 u 3 m3 0.75 kmol CO 2 44.01 kg CO 2
1 atm
12 kg
m 3 ˜atm
298
2
K
kmol
kmol
CO
.
0.08206 kmol
2
˜K
More needs to escape from the cylinder since the room is not sealed.
5-6
5.19 (cont’d)
c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a
person entered the room and closed the door, over a period of time the person could die
of asphyxiation. Measures that would reduce hazards are:
1. Change the lock so the door can always be opened from the inside without a key.
2. Provide ventilation that keeps air flowing through the room.
3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount.
4. Install safety valves on the cylinder in case of leaks.
15.7 kg
5.20 n CO 2
1 kmol
0.357 kmol CO 2
44.01 kg
b
Assume ideal gas behavior, negligible temperature change T 19q C 292.2 K
a.
P1V
P2 V
b
n1 RT
n1
Ÿ
n1 0.357 RT
n1 0.357
g
Ÿ n1
b. Vtank
102kPa
3.27 u 103 kPa
0.0115 kmol air in tank
n1RT
P1
0.0115 kmol 292.2 K 8.314 m3 ˜ kPa 103 L
m3
102kPa
kmol ˜ K
15700 g CO 2 + 11.5 mol air ˜ (29.0 g air / mol)
274 L
Uf
c.
P1
P2
g
274 L
58.5 g / L
CO 2 sublimates Ÿ large volume change due to phase change Ÿ rapid pressure rise.
Sublimation causes temperature drop; afterwards, T gradually rises back to room
temperature, increase in T at constant V Ÿ slow pressure rise.
5.21 At point of entry, P1
b10 ft H Ogb29.9 in. Hg 33.9 ft H Og 28.3 in. Hg
2
2
37.1 in. Hg .
At surface, P2
28.3 in. Hg, V2 bubble volume at entry
1
x solid x solution
0.20
0.80
Mean Slurry Density:
3
U sl U solid U solution (1.2)(1.00 g / cm ) (100
. g / cm 3 )
0.967
cm 3
Ÿ U sl
g
1.03 g 2.20 lb 5 u 10 4 ton 10 6 cm 3
1 lb
264.17 gal
cm 3 1000 g
a.
gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg
300 ton
hr
4.3 u 10 3 ton 1000 gal
492 o R 37.1 in Hg
b.
P2 V2
P1V1
nRT
V
Ÿ 2
nRT
V1
% change =
4
3S
P1
Ÿ
P2
4
3S
e j
e j
b2.2 - 2.0g mm u 100
2.0 mm
D2 3
2
D1
2
37.1
Ÿ D 32
28.3
3
10%
5-7
1.31D13
4.3 u 10 3 ton / gal
2440 ft 3 / hr
!D
D1 2 mm
2
2.2 mm
5.22 Let B = benzene
n1 , n 2 , n 3 moles in the container when the sample is collected, after
the helium is added, and after the gas is fed to the GC.
n inj moles of gas injected
n B , n air , n He moles of benzene and air in the container and moles of helium added
n BGC , m BGC moles, g of benzene in the GC
y B mole fraction of benzene in room air
a.
P1V1
n1
P2 V2
n2
P3 V3
n3
n1RT1 (1 { condition when sample was taken): P1 = 99 kPa, T1
99 kPa 2 L
mol ˜ K
kPa
101.3 atm 306 K .08206 L ˜ atm
0.078 mol = n air n B
n 2 RT2 (2 { condition when charged with He): P2 = 500 kPa, T2
mol ˜ K
500 kPa 2 L
101.3 kPa
306
K
.08206
L ˜ atm
atm
mol ˜ K
400 kPa 2 L
kPa
101.3 atm 296 K .08206 L ˜ atm
n BGC u
y B (ppm) =
306K
0.393 mol = n air + n B n He
n 3 RT3 (3 { final condition in lab): P3 = 400 kPa, T3
n inj = n 2 n 3
nB
306K
296K
0.325 mol = (n air n B n He ) n inj
0.068 mol
n2
n inj
0.393 mol m BGC (g B) 1 mol
0.068 mol
78.0 g
nB
u 106
n1
0.0741 ˜ m BGC
u 106
0.078
9 am: y B
(0.950 u 106 )(0.656 u 10 6 )
1 pm: y B
(0.950 u 106 )(0.788 u 10 6 )
5 pm: y B
(0.950 u 106 )(0.910 u 10 6 )
0.0741 ˜ m BGC
0.950 u 106 ˜ m BGC
U|
|
0.749 ppm VThe avg. is below the PEL
|
0.864 ppm|
W
0.623 ppm
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container
so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix
sufficiently and to reach thermal equilibrium.
c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may
be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity
at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is
less dense than the benzene and air; therefore, the sample injected in the GC may be Herich depending on where the sample was taken from the cylinder. (iv) The benzene may
not be uniformly distributed in the laboratory. In some areas the benzene concentration
could be well above the PEL.
5-8
b g
4
S 10 m
3
Moles of gas in balloon
5.23 Volume of balloon
b
n kmol
a.
b.
4189 m3
g
3
4189 m3
492q R 3 atm
1 kmol
535q R 1 atm 22.4 m3 STP
b g
515.9 kmol
He in balloon:
m
b515.9 kmolg ˜ b4.003 kg kmolg
mg
2065 kg 9.807 m
1N
2
s 1 kg ˜ m / s2
dP
dP
iV
iV
gas in balloon
air displaced
Fbuoyant
n gas RT
n air RT
Fbuoyant
2065 kg He
20,250 N
Pair
˜ n gas
Pgas
Ÿ n air
172.0 kmol 29.0 kg 9.807 m 1 N
1 kmol
s2 1 kg2˜m
s
Wair displaced
Since balloon is stationary,
Wtotal
1 atm
˜ 515.9 kmol 172.0 kmol
3 atm
¦F
1
Fcable
Fcable
Fbuoyant Wtotal
c. When cable is released, Fnet
48920 N dAi = 27200 N
Ÿa
b
48,920 N
0
b2065 150gkg
9.807 m 1 N
s2 1 kgs2˜m
M tot a
27200 N
1 kg ˜ m / s2
2065 + 150 kg
N
g
12.3 m s2
d. When mass of displaced air equals mass of balloon helium the balloon stops rising.
Need to know how density of air varies with altitude.
e. The balloon expands, displacing more air Ÿ buoyant force increases Ÿ balloon rises
until decrease in air density at higher altitudes compensates for added volume.
5.24 Assume ideal gas behavior, Patm
a.
b.
1 atm
PN VN 5.7 atm 400 m3 / h
240 ft 3 h
Pc
9.5 atm
Mass flow rate before diversion:
PN VN
240 m3
h
Pc Vc Ÿ Vc
273 K 5.7 atm
1 kmol
303 K 1 atm 22.4 m3 STP
b g
5-9
44.09 kg
kmol
2426
kg C 3 H 6
h
27,20
5.24 (cont’d)
Monthly revenue:
b2426 kg hgb24 h daygb30 days monthgb$0.60 kgg
c.
$1,048,000 month
Mass flow rate at Noxious plant after diversion:
240 m3
hr
273 K 2.8 atm 1 kmol
306 K 1 atm 22.4 m3
Propane diverted
5.25 a. PHe
PCH 4
PN 2
b2426 1180g kg h
44.09 kg
1180 kg hr
kmol
1246 kg h
y He ˜ P = 0.35 ˜ (2.00 atm) = 0.70 atm
y CH 4 ˜ P = 0.20 ˜ (2.00 atm) = 0.40 atm
y N 2 ˜ P = 0.45 ˜ (2.00 atm) = 0.90 atm
b. Assume 1.00 mole gas
4.004 g
0.35 mol He
mol
U|
||
16.05 g I
F
0.20 mol CH G
H mol JK 3.21 g CH V|17.22 g Ÿ mass fraction CH
F 28.02 g IJ 12.61 g N ||
0.45 mol N G
H mol K
W
FG
H
IJ
K
140
. g He
4
2
c.
MW
d.
U gas
4
3.21 g
17.22 g
0186
.
2
g of gas
17.2 g / mol
mol
P MW
m n MW
V
4
d i d i b2.00 atmgb17.2 kg / kmolg
V
RT
e0.08206 jb363.2 Kg
m 3 ˜atm
kmol˜K
. kg / m3
115
5.26 a. It is safer to release a mixture that is too lean to ignite.
If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the
C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard.
b.
fuel-air mixture
n 1 ( mol / s)
y C 3H 8 0.0403 mol C 3 H 8 / mol
n C 3H 8 150 mol C3 H 8 / s
n 3 ( mol / s)
0.0205 mol C 3 H 8 / mol
diluting air
n 2 ( mol / s)
n 1
150 mol C3 H 8
mol
s
0.0403 mol C 3 H 8
Propane balance: 150 = 0.0205 ˜ n 3 Ÿ n 3
5-10
3722 mol / s
7317 mol / s
5.26 (cont’d)
Total mole balance: n 1 n 2
c.
b g
n 3 Ÿ n 2
7317 3722
n 2
. n 2
13
2
V
4674 mol / s 8.314 m3 ˜ Pa 398.2 K
mol ˜ K 131,000 Pa
3595 mol air / s
4674 mol / s
min
1
V
3722 mol 8.314 m3 ˜ Pa 298.2 K
s
mol ˜ K 110000 Pa
y2
150 mol / s
n 1 n 2
b
U|
|V V
|| V
W
118 m3 / s
2
.
141
1
83.9 m3 / s
m3 diluting air
m3 fuel gas
150 mol / s
u 100% 18%
.
3722 mol / s + 4674 mol / s
g
d. The incoming propane mixture could be higher than 4.03%.
If n 2 n 2 min , fluctuations in the air flow rate would lead to temporary explosive
b g
conditions.
5.27
b
gb
Basis: 12 breaths min 500 mL air inhaled breath
g
600 mL inhaled min
o
24 C, 1 atm
6000 mL / min
lungs
n in (mol / min)
0.206 O 2
0.774 N 2
0.020 H 2 O
a.
n in
6000 mL
blood
1L
273K
37 o C, 1 atm
n out (mol / min)
0.151 O 2
0.037 CO 2
0.750 N 2
0.062 H 2 O
1 mol
b g
3
min
10 mL 297K 22.4 L STP
b
gb
g
N 2 balance: 0.774 0.246
O 2 transferred to blood:
0.750n out Ÿ n out
0.246 mol min
0.254 mol exhaled min
. g b mol O
b0.246gb0.206g b0.254gb0151
2
g
min 32.0 g mol
0.394 g O 2 min
CO 2 transferred from blood:
b0.254gb0.037g bmol CO
2
g
min 44.01 g mol
0.414 g CO 2 min
H 2 O transferred from blood:
b0.254gb0.062g b0.246gb0.020g bmol H O ming 18.02 g mol
2
g H 2 O min
0195
.
5-11
5.27 (cont’d)
PVin
PVout
FG n IJ FG T IJ FG 0.254 mol min IJ FG 310K IJ 1078
H n K H T K H 0.246 mol min K H 297K K . mL exhaled ml inhaled
.
g H O lost ming b0.394 g O gained ming 0.215 g min
b0.414 g CO lost ming b0195
Ÿ
b.
n in RTin
n out RTout
Vout
Vin
out
out
in
in
2
2
2
STACK
5.28
Ta (K)
Ts (K)
M s (g/mol) M a (g/mol)
Ps (Pa)
Pc (Pa)
LL(m)
(M)
PM
RT
Ideal gas: U
bUgLg
a.
D
b.
Ms
Pa
755 mm Hg
b g
UgL
PM
Pa M a
gL a s gL
RTs
RTa
stack
. gb44.1g b0.02gb32.0g b0.80gb28.0g
b018
29.0 g mol , Ta
Ma
D
combust.
755 mm Hg
310
. g mol , Ts
OP
Q
655K ,
294 K , L 53 m
1 atm
53.0 m 9.807 m
760 mm Hg
u
LM
N
Pa gL M a M a
R
Ta
Ts
s
2
kmol - K
0.08206 m3 atm
LM 29.0 kg kmol 310. kg kmol OP u FG 1 N IJ
655K
N 294K
Q H 1 kg ˜ m / s K
2
323 N 1033 cm H 2 O
m2 1.013 u 105 N m2
3.3 cm H 2 O
b.
b g
P MW
!
U CCl O
b g
m CCl O
2
c.
MWCCl2O 98.91 g / mol
98.91
3.41
U air
29.0
RT
Phosgene, which is 3.41 times more dense than air, will displace air near the ground.
2
S D in L S
2
Vtube
0.635 cm - 2 0.0559 cm 15.0 cm 3.22 cm3
4
4
5.29 a. U
n CCl 2 O(l)
2
b
Vtube ˜ U CCl O
2
3.22 cm3
gb
g
1L
1 atm
98.91 g / mol
3
L˜atm
296.2 K
10 cm 0.08206 mol˜K
3
. u 1000
.
3.22 cm3 137
g mol
3
cm 98.91 g
5-12
0.0446 mol CCl 2 O
0.0131 g
5.29 (cont’d)
1 atm 2200 ft 3 28.317 L
mol ˜ K
2563 mol air
3
.08206 L ˜ atm
296.2K
ft
PV
RT
n air
n CCl 2 O
0.0446
17.4 u 10 6
2563
n air
17.4 ppm
The level of phosgene in the room exceeded the safe level by a factor of more than 100.
Even if the phosgene were below the safe level, there would be an unsafe level near the
floor since phosgene is denser than air, and the concentration would be much higher in
the vicinity of the leak.
d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or
guidance from an experienced safety officer. He also should have been working under a
hood and should have worn a gas mask.
5.30 CH 4 2O 2 o CO 2 2 H 2 O
7
C 2 H 6 O 2 o 2CO 2 3H 2 O
2
C 3 H 8 5O 2 o 3CO 2 4 H 2 O
1450 m3 / h @ 15o C, 150 kPa
n 1 (kmol / h)
0.86 CH 4 , 0.08 C 2 H 6 , 0.06 C 3 H 8
n 2 (kmol air / h)
8% excess, 0.21 O 2 , 0.79 N 2
n 1
1450 m3
h
273.2K
288.2K
b1013. 150gkPa
101.3 kPa
1 kmol
22.4 m3 STP
b g
152 kmol h
FG
H
IJ OP
K PQ
Theoretical O 2 :
LM F
MN GH
IJ
K
IJ
K
FG
H
152 kmol
2 kmol O 2
3.5 kmol O 2
5 kmol O 2
0.06
0.08
0.86
h
kmol CH 4
kmol C 2 H 6
kmol C 3 H 8
air
Air flow: V
b
g
. 349.6 kmol O 2
108
h
1 kmol Air
0.21 kmol O 2
5-13
b g
22.4 m3 STP
kmol
349.6 kmol h O 2
b g
4.0 u 104 m3 STP h
5.31 Calibration formulas
bT
dP
dV
dV
25.0; R T
g
0; R p
F
0; R p
A
0; R A
27g Ÿ Tbq Cg 0.77R 14.2
g b
0i , d P 20.0, R 6i Ÿ P
bkPag 3.33R
d m hi 200R
0i , d V
2.0 u 10 , R 10i Ÿ V
d m hi 4000R
0h , d V
1.0 u 10 , R
25i Ÿ V
14 , T 35.0, R T
g
T
r
gauge
p
3
F
3
F
F
5
F
3
A
A
A
A
(m 3 / h), T, P
V
F
g
CH 4 2O 2 o CO 2 2 H 2 O
7
C 2 H 6 O 2 o 2CO 2 3H 2 O
2
C 3 H 8 5O 2 o 3CO 2 4 H 2 O
13
C 4 H 10 O 2 o 4CO 2 5H 2 O
2
n F (kmol / h)
x A (mol CH 4 / mol)
x B (mol C 2 H 6 / mol)
x C (mol C 3 H 8 / mol)
x D (mol n - C 4 H 10 / mol)
x E (mol i - C 4 H 10 / mol)
( m 3 / h) (STP)
V
A
n A (kmol / h)
0.21 mol O 2 / mol
0.79 mol N 2 / mol
n F
d
F m3 h
V
i
273.2K
dP 101.3ikPa
1 kmol
g
bT 273.2gK 101.3 kPa
d P 101.3i kmol
0.12031V
FG IJ
T
+
273
b
g H h K
F
b g
22.4 m3 STP
g
Theoretical O 2 :
dn i
o2
Th
b
c
Air feed: n A
dn i
o2
Th
kmol O 2 req.
1 kmol air
h
0.21 kmol O 2
FG P IJ dn i
H 100K
bkmol air hg d22.4 m bSTPg kmoli
4.762 1 A
V
n a
gh
n F 2x A 3.5x B 5x C 6.5 x D x E kmol O 2 req. h
b1 P
g
100 kmol feed
1 kmol req.
x
x
o2
Th
3
b g
22.4n A m 3 STP h
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA
Vf (m3/h) Va (m3/h) Ra
25.0 7.25 0.81 0.08 0.05 0.04 0.02
1450 22506.2 5.63
23.1 32.0 7.5
15 72.2 183.47 1004.74
64.3 5.8 0.58 0.31 0.06 0.05 0.00
1160 29697.8 7.42
7.5 20.0 19.3
23 78.9 226.4 1325.8
52.6 2.45 0.00 0.00 0.65 0.25 0.10
490 22022.3 5.51
46.5 50.0 15.8
33 28.1 155.2 983.1
10.0
1200 30439.2 7.6
21 30.4
3
6 0.02 0.4 0.35 0.1 0.13
15 53.0 248.1 1358.9
13.3
1400 29283.4 7.3
23 31.9
4
7 0.45 0.12 0.23 0.16 0.04
15 63.3 238.7 1307.3
16.7
1800 32721.2 8.2
25 33.5
5
9 0.5 0.3 0.1 0.04 0.06
15 83.4 266.7 1460.8
20.0 10 0.5 0.3 0.1 0.04 0.06
2000 37196.7 9.3
27 35.0
6
15 94.8 303.2 1660.6
5-14
5.32 NO 21 O 2 œ NO 2
1 mol
0.20 mol NO / mol
0.80 mol air / mol
0.21 O 2
0.79 N 2
P0 380 kPa
R|
S|
T
a.
n1 (mol NO)
n 2 (mol O 2 )
n 3 (mol N 2 )
n 4 (mol NO 2 )
Pf (kPa)
U|
V|
W
Basis: 1.0 mol feed
90% NO conversion: n1
O 2 balance: n 2
0.80(0.21) 018
. mol NO 0.5 mol O 2
mol NO
N 2 balance: n 3
0.80(0.79)
0.632 mol N 2
n4
y NO
yO2
018
. mol NO 1 mol NO 2
018
. mol NO 2 Ÿ n f
1 mol NO
0.020 mol NO
mol NO
0.022
0.91 mol
mol
0.086
mol O 2
yN2
mol
Pf V n f RT
Ÿ Pf
=
P0 V n 0 RT
b.
0.020 mol NO Ÿ NO reacted = 0.18 mol
010
. (0.20)
n f = n0
Pf
P0
P0
(1 mol)
0.695
nf
n0
mol N 2
y NO 2
mol
380 kPa
360 kPa
380 kPa
FG 0.91 mol IJ
H 1 mol K
0.0780 mol O 2
n1 n 2 n 3 n 4
.
0198
0.91 mol
mol NO 2
mol
346 kPa
0.95 mol
n i0 X i[
ni
E
n1 (mol NO)
0.20 [
n 2 ( mol O 2 ) (0.21)(0.80) 0.5[
n 3 (mol N 2 ) ( 0.79)(0.80)
n 4 ( mol NO 2 ) [
1 0.5[
nf
0.95 Ÿ [
010
.
Ÿ n1 010
. mol NO, n 2 0118
.
mol O 2 , n 3 0.632 mol N 2 ,
. mol NO 2 Ÿ y NO 0105
. , y O 2 0124
. , y N 2 0.665, y NO 2
n 4 010
NO conversion =
P (atm) =
Kp
b0.20 - n g u 100%
1
3.55 atm
(y NO 2 P)
( y NO P )( y O 2 P)
50%
0.20
360 kPa
101.3 kPa
atm
0105
.
(y NO 2 )
0.5
0.5
( y NO )( y O 2 ) P
0.5
5-15
b
0.105
(0105
. ) 0.124
g b3.55g
0.5
1
0.5
. atm 2
151
5.33
Liquid composition:
49.2 kg M 1 kmol
112.6 kg
0.437 kmol M
29.6 kg D 1 kmol
147.0 kg
0.201 kmol D
Ÿ 0.221 kmol D / kmol
21.2 kg B 1 kmol
78.12 kg
0.271 kmol B
0.298 kmol B / kmol
100 kg liquid Ÿ
0.481 kmol M / kmol
0.909 kmol
a.
Basis: 1 kmol C 6 H 6 fed
V1 (m 3 ) @ 40 o C, 120 kPa
n1 (kmol)
0.920 HCl
0.080 Cl 2
1 kmol C 6 H 6 (78.12 kg)
n 0 (kmol Cl 2 )
n 2 (kmol)
0.298 C 6 H 6
0.481 C 6 H 5Cl
0.221 C 6 H 4 Cl 2
C 6 H 6 Cl 2 o C 6 H 5Cl + HCl
1 kmol C 6 H 6
C balance:
Ÿ n2
6 kmol C
1 kmol C 6 H 6
n 2 0.298 u 6 0.481 u 6 0.221 u 6
100
. kmol
1 kmol C 6 H 6
H balance:
C 6 H 5Cl Cl 2 o C 6 H 5Cl 2 + HCl
6 kmol H
1 kmol C 6 H 6
b
g
n1 0.920 (1)
n 2 0.298 u 6 0.481 u 5 0.221 u 4 Ÿ n1 100
. kmol
Ÿ
b.
100
. kmol 1013
. kPa 0.08206 m3 ˜ atm 313.2 K
kmol ˜ K
120 kPa
1 atm
n1RT
P
V1
217
. m3
7812
. kg B
V1
mB
( m3 / s)
V
gas
d2 =
0.278 m3 / kg B
u(m / s) ˜ A(m 2 )
u˜
4˜V
Sd 2
gas
Ÿ d2 =
S ˜u
4
B0 ( kg B) 0.278 m3
4m
s 1 min 104 cm2
kg B S (10) m 60 s
min
m2
Ÿ d(cm)
b g
B0
2.43 ˜ m
217
. m3
B0 (cm2 )
5.90m
1
2
c. Decreased use of chlorinated products, especially solvents.
5-16
5.34
Vb ( m3 / min) @900$ C, 604 mtorr
60% DCS conversion
n 1 (mol DCS / min)
n 2 (mol N 2 O / min)
n b (mol / min)
n 3 (mol N 2 / min)
n 4 (mol HCl(g) / min)
3.74 SCMM
U|
V|
W
n a ( mol / min)
0.220 DCS
0.780 N 2 O
SiH 2 Cl 2(g) 2 N 2 O (g) o SiO 2(s) 2 N 2(g) 2 HCl (g)
a.
n a
3.74 m3 (STP)
103 mol
167 mol / min
min
22.4 m3 (STP)
mol DCS I
gFGH 0.220 mol
JK b167 mol / ming 14.7 mol DCS / min
mol DCS
DCS reacted: b0.60gb0.220gb167g
22.04 mol DCS reacted / min
min
b
60% conversion: n 1 = 1- 0.60
N 2 O balance: n 2
b g molminN O
0.780 167
N 2 balance: n 3
HCl balance: n 4
n B
b.
n B RT
P
p DCS
x DCS ˜ P =
p N 2O
22.04 mol DCS 2 mol N 2 O
mol DCS
min
22.04 mol DCS 2 mol N 2
mol DCS
min
22.04 mol DCS 2 mol HCl
min
mol DCS
n 1 n 2 n 3 n 4
B
ŸV
2
. mol N 2 O / min
8618
44.08 mol N 2 / min
44.08 mol HCl / min
189 mol / min
189 mol 62.36 L ˜ torr 0.001 m3 1173 K
min
mol ˜ K
L
0.604 torr
2.29 u 104 m3 / min
n 1
14.7 mol DCS / min
˜ 640 mtorr = 49.8 mtorr
P=
n B
189 mol / min
n
86.2 mol N 2 O / min
˜ 640 mtorr = 292 mtorr
x N 2O ˜ P = 2 P =
n B
189 mol / min
r = 3.16 u 10-8 ˜ p DCS ˜ p N 2 O
= r ˜ t ˜ MW
c. h(A)
U SiO2
0.65
b
gb g
3.16 u 10-8 49.8 292
0.65
6.3 u 10 5
mol SiO 2
m2 ˜ s
6.3 u 10 5 mol SiO 2 60 s 120 min 60.09 g / mol 1010 A
2
6
3
m ˜ s min
2.25 u 10 g / m 1 m
(Table B.1)
= 1.2 u 10 A
5
The films will be thicker closer to the entrance where the lower conversion yields higher
pDCS and p N 2 O values, which in turn yields a higher deposition rate.
5-17
5.35
Basis: 100 kmol dry product gas
n1 (kmol C x H y )
m1 (kg C x H y )
kmol dry gasU
R|100
|V
0.105 CO
S|0.053 O
|W
T0.842 N
(m 3 )
V
2
n 2 (kmol air)
0.21 O 2
0.79 N 2
2
2
2
o
n 3 (kmol H 2 O)
30 C, 98 kPa
a.
N 2 balance: 0.79n 2
b
O balance: 2 0.21n 2
C balance:
g
0.842(100) Ÿ n 2 = 106.6 kmol air
b
g b
g
2 0.053 n 3 Ÿ n 3
100 2 0105
.
d
n1 kmol C x H y
i x bkmol Cg
dkmol C H i
x
b
1317
. kmol H 2 O
g
b1g
Ÿ n1x = 10.5
100 0105
.
y
b 2g
n 3 13.17
! n1 y
H balance: n1y = 2n 3
26.34
b g b g yx 2610.34.5 2.51 mol H / mol C
fed: 0.21b106.6 kmol air g 22.4 kmol
in excess = 5.3 kmol Ÿ Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol
Divide 2 by 1 Ÿ
O2
O2
2
% excess =
b.
V2
m1 =
5.3 kmol O 2
u 100%
17.1 kmol O 2
31% excess air
106.6 kmol N 2 22.4 m3 (STP) 1013
. kPa 303 K
kmol
98 kPa 273 K
b
g
b
g
n1x kmol C 12.0 kg n1 y kmol H 101
. kg
kmol
kmol
2740 m3 air
V2
=
m1 152.6 kg fuel
18.0
2740 m3
n1x=10.5
! m1
n1y=26.34
152.6 kg
m3 air
kg fuel
5.36 3N 2 H 4 o 6xH 2 (1 2 x)N 2 (4 4 x)NH 3
a. 0 d x d 1
b.
nN2H4
50 L 0.82 kg 1 kmol
L
32.06 kg
128
. kmol
LM 6x kmol H b1 2xg kmol N b4 4xg kmol NH OP
3 kmol N H
N 3 kmol N H 3 kmol N H
Q
.
128
x + 2.13 kmol
.
b6x 1 2x 4 4xg 1707
3
n product
2
2
. kmol N 2 H 4
128
2
4
5-18
2
4
3
2
4
5.36 (cont’d)
nproduct
2.13
2.30
2.47
2.64
2.81
2.98
3.15
3.32
3.50
3.67
3.84
Vp (L)
15447.92
16685.93
17923.94
19161.95
20399.96
21637.97
22875.98
24113.99
25352.00
26590.01
27828.02
Volume of Product Gas
30000.00
25000.00
20000.00
V (L)
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
15000.00
10000.00
5000.00
0.00
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
c.
Hydrazine is a good propellant because as it decomposes generates a large number of
moles and hence a large volume of gas.
5.37
A (g A / h)
m
c
h
C A (g A / m 3 )
m3 / h
V
air
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in
cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste
containing A poured into sink, A used as cleaning solvent.
A
b. m
c.
d.
FG kg A IJ
H hK
yA
in
mol A
mol air
yA 50 u106
d h
V
air
min
FG kg A IJ
H hK
C e j˜ V
M e j˜ n
A
m
out
gA
A m3
gA
A mol
F I FG
GH JK H
3
m C kg A
V
air
A
h
m3
air
CA
A
m
PV
; nair
RT
k˜Vair
!
IJ
K
yA
A RT
m
k ˜ Vair M A P
A 90
m
. g/ h
A RT
m
kyA MA P
d
0.5 50 u 10
m ˜Pa
8.314 mol
293 K
˜K
3
101.3 u 10 Pa 104.14 g / mol
3
9.0 g / h
6
i
83 m3 / h
Concentration of styrene could be higher in some areas due to incomplete mixing (high
concentrations of A near source); 9.0 g/h may be an underestimate; some individuals
might be sensitive to concentrations PEL.
e. Increase in the room temperature could increase the volatility of A and hence the rate of
air : At higher T, need
evaporation from the tank. T in the numerator of expression for V
a greater air volume throughput for y to be PEL.
5-19
C3H 6 H 2 œ C3H 8
5.38 Basis: 2 mol feed gas
235$ C, P2
a. At completion, n p
P2 V
P1V
35.1 atm
n2
P2 T1
n1
P1 T2
1.29
b
g
2 np
1 mol 508K 32.0 atm
27.3 atm
2 mol 298K
n 2 T2
P1
n1 T1
35.1 atm 298K 2 mol
1.29 mol
32.0 atm 508K
2 np Ÿ np
Ÿ 1- 0.71
n p 2(1 n p )
2 1 1 mol
1 mol , n 2
n 2 RT2
Ÿ P2
n1RT1
b. P2
U|
V|
W
n p (mol C 3 H 8 )
(1 - n p )(mol C 3 H 6 ) n 2
(1 - n p )(mol H 2 )
1 mol C 3 H 6
1 mol H 2
25$ C, 32 atm
0.71 mol C 3 H 8 produced
0.29 mol C 3 H 6 unreacted Ÿ 71% conversion of propylene
c.
n2
1.009
1.028
1.083
1.101
1.156
1.174
1.211
1.229
1.248
1.266
1.285
1.358
1.431
1.468
C3H8 prod.
0.99075
0.9724
0.91735
0.899
0.84395
0.8256
0.7889
0.77055
0.7522
0.73385
0.7155
0.6421
0.5687
0.532
%conv.
99.075
97.24
91.735
89.9
84.395
82.56
78.89
77.055
75.22
73.385
71.55
64.21
56.87
53.2
Pressure vs Fraction Conversion
120
100
80
% conversion
P2 (atm)
27.5
28.0
29.5
30.0
31.5
32.0
33.0
33.5
34.0
34.5
35.0
37.0
39.0
40.0
60
%conv.
40
20
0
25.0
27.0
29.0
31.0
33.0
Pressure (atm)
5-20
35.0
37.0
39.0
41.0
Convert fuel composition to molar basis
5.39
Basis: 100 g Ÿ
UV Ÿ 97.2 mol % CH
017
. mol C H W 2.8 mol % C H
b
g
5 g C H b1 mol 30.07 gg
95 g CH 4 1 mol 16.04 g
2
5.92 mol CH 4
6
500 m3 / h
2
2
6
4
6
n 2 (kmol CO 2 / h)
n 3 (kmol H 2 O / h)
n 1 (mol / h)
0.972 CH 4
n 4 (kmol O 2 / h)
n 5 (kmol N 2 / h)
0.028 C 2 H 6
40$ C, 1.1 bar
(SCMH)
V
air
25% excess air
n 1
1
P1V
RT1
kmol ˜ K
0.08314 m3 ˜ bar
. bar 500 m3
11
313K
h
7
C 2 H 6 O 2 o 2CO 2 3H 2 O
2
CH 4 2O 2 o CO 2 2 H 2 O
Theoretical O 2 =
. kmol h
211
LM
N
21.1 kmol 0.972 kmol CH 4
kmol
h
0.028 kmol C 2 H 6
kmol
b
125
. 431
. kmol O 2
Air Feed:
h
g
3.5 kmol O 2
1 kmol C 2 H 6
1 kmol Air
0.21 kmol O 2
5.40 Basis: 1 m3 gas fed @ 205q C, 1.1 bars Ac
1 m 3 @205$ C, 1.1 bar
2 kmol O 2
1 kmol CH 4
OP
Q
.
431
b g
22.4 m3 STP
1 kmol
kmol O 2
h
5700 SCMH
acetone
n 3 (kmol), 10$ C, 40 bar
condenser
n1 (kmol)
y1 (kmol Ac / kmol)
(1 - y 1 )(kmol air / kmol)
p AC 0.100 bar
y 3 (kmol Ac / kmol)
(1 - y 3 )(kmol air / kmol)
p AC 0.379 bar
n 2 (kmol Ac(l))
a.
n1
y1
100
. m3
273K 110
1 kmol
. bars
478K 10132
bars 22.4 m3 STP
.
b g
0100
.
bar
1.1 bars
0.0909 kmol Ac kmol , y 3
b
gb
g
Air balance: 0.0277 0.910
Mole balance: 0.0277
Acetone condensed
0.0277 kmol
0.379 bar
40.0 bars
(1 9.47 u 10 3 )n 3 Ÿ n 3
0.0254 n 2 Ÿ n 2
0.0254 kmol
0.0023 kmol Ac condensed
0.0023 kmol Ac 58.08 kg Ac
1 kmol Ac
5-21
9.47 u 10 3 kmol Ac kmol
0.133 kg acetone condensed
5.40 (cont’d)
b g
0.0254 kmol 22.4 m3 STP
Product gas volume
b.
.
283K 10132
bars
0.0149 m3
273K 40.0 bars
20.0 m3 effluent 0.0277 kmol feed 0.0909 kmol Ac 58.08 kg Ac
196 kg Ac h
h
kmol feed
kmol Ac
0.0149 m3 effluent
5.41 Basis: 1.00 u 10 6 gal. wastewater day. Neglect evaporation of water.
1.00 u 10 6 gal / day
Effluent gas: 68$ F, 21.3 psia(assume)
n1 (lb - moles H2 O / day)
0.03n1 (lb - moles NH3 / day)
n2 (lb - moles air / day)
n3 (lb - moles NH 3 / day)
300 u 10 6 ft 3 air / day
Effluent liquid
$
n1 (lb - moles H2 O / day)
n4 (lb - moles NH3 / day)
20 C, 21.3 psia
n2 (lb - moles air / day)
a.
Density of wastewater: Assume U
LMn
N
1
lb - moles H 2 O
day
100
. u 106
Ÿ n1
n2
18.02 lb m
1 lb - mole
62.4 lb m ft 3
0.03n1 lb m NH 3
day
OP u 1 ft
1 lb - moleQ 62.4 lb
3
17.03 lb m
m
7.4805 gal
1 ft 3
gal
day
450000 lb - moles H 2 O fed day , 0.03n1
13500 lb - moles NH 3 fed day
. psi 1 lb - mole
300 u 106 ft 3 492 $ R 213
$
day
527.7 R 14.7 psi 359 ft 3 STP
b g
93% stripping: n 3
. u 106 lb - moles air day
113
0.93 u 13500 lb - moles NH 3 fed day 12555 lb - moles NH 3 day
Volumetric flow rate of effluent gas
PVout
PVin
n out RT
Ÿ Vout
n in RT
Vin
n out
n in
300 u 106 ft 3
d1.13 u 10
6
i
12555 lb - moles day
1.13 u 10 lb - moles day
6
day
303 u 106 ft 3 day
Partial pressure of NH 3
y NH 3 P
12555 lb - moles NH 3 day
d1.129 u 10
0.234 psi
5-22
6
i
12555 lb - moles day
u 213
. psi
5.42 Basis: 2 liters fed / min
Cl ads.=
2.0 L soln 1130 g 012
. g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2
L
g soln
40.0 g mol NaOH 2 mol NaOH
60 min
2 L / min @ 23$ C, 510 mm H 2 O
n 1 (mol / min)
y (mol Cl 2 / mol)
(1 - y)(mol air / mol)
0.013
mol
min
n 2 (mol air / mol)
0.013 mol Cl 2 / min
Assume Patm
n 1
2L
b g b10.33 0.510g m H O
10.33 m H 2 O Ÿ Pabs
273K 10.84 m H 2 O
1 mol
b g
0.013 Ÿ y
0150
.
mol Cl 2
,? specification is wrong
mol
(L / min) @ 65o C, 1 atm
V
3
125 L / min @ 25o C, 105 kPa
n 1 ( mol / min)
y1 (mol H 2 O / mol)
1- y1 ( mol dry gas / mol)
0.235 mol C 2 H 6 / mol DG
0.765 mol C 2 H 4 / mol DG
R|b g
S|
T
10.84 m H 2 O
0.0864 mol min
min 296K 10.33 m H 2 O 22.4 L STP
Cl balance: 0.0864y
5.43
2
in
n C 2 H 6 (mol C 2 H 6 / min)
n C 2 H 4 (mol C 2 H 4 / min)
n air (mol air / min)
n H 2 O (mol H 2 O / min)
U|
V|
W
355 L / min air @ 75o C, 115 kPa
n 2 (mol / min)
y 2 ( mol H 2 O / mol)
(1- y 2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y
b
ln a
b
ln y1 y 2
R 2 R1
g lnd0.2 10 i
ln y1 bR 1
dy
bR ln a
ae bR
i
4
90 5
0.08942
bg
ln 10 4 0.08942 5 Ÿ a
6.395 u 105 Ÿ y
6.395 u 105 e 0.08942R
b. n 1
125 L 273K 105 kPa
1 mol
min 298K 101 kPa 22.4 L STP
5.315 mol min wet gas
n 2
355 L 273K 115 kPa
1 mol
min 348K 101 kPa 22.4 L STP
14.156 mol min wet air
R1
86.0 o y1
b g
b g
0.140 , R 2
12.8 o y 2
5-23
2.00 u 104 mol H 2 O mol
5.43 (cont’d)
C H I
b5.315 mol mingFGH b1 0.140g molmolDG IJK FGH 0.235 mol
J
mol DG K
2
C 2 H 6 balance: n C2 H 6
6
1.07 mol C 2 H 6 min
b5.315gb0.860gb0.765g 3.50 mol C H min
Dry air balance: n
b14.156gd1 2.00 u 10 i 14.15 mol DA min
Water balance: n
b5.315gb0.140g b14.156gd1.00 u 10 i 0.746 mol H O min
n
b1.07 3.50 14.15g mol min 18.72 mol min ,
n
b18.72 0.746g 19.47 mol min
19.47 mol min 22.4 L bSTPg 338K 540 liters min
V
C 2 H 4 balance: n C2 H 4
2
4
4
air
4
H 2O
2
dry product gas
total
3
mol
Dry basis composition:
c.
273K
FG 1.07 IJ u 100%
H 18.72 K
5.7% C 2 H 6 , 18.7% C 2 H 4 , 75% dry air
0.746 mol H 2 O
u 1 atm 0.03832 atm
19.47 mol
p H 2O
y H 2Ol ˜ P
y H 2O
0.03832 Ÿ R
FG
H
1
0.03832
ln
0.08942
6.395 u 10 5
IJ
K
71.5
5.44 CaCO 3 o CaO CO 2
n CO 2
1350 m3 273K
1 kmol
h
1273K 22.4 m3 STP
b g
12.92 kmol CO 2
12.92 kmol CO 2 h
1 kmol CaCO 3 100.09 kg CaCO 3
h
1 kmol CO 2
1 kmol CaCO 3
1362 kg limestone
0.17 kg clay
h
0.83 kg limestone
279 kg clay h
Weight % Fe 2 O 3
b g
kg Fe 2 O 3 kg clay
279
0.07
u 100% 18%
Fe 2 O 3
.
1362 279 12.92 44.1
kg limestone
kg clay
b g
kg CO 2 evolved
5-24
1 kg limestone
0.95 kg CaCO 3
1362 kg limestone h
5.45
Basis:
R|864.7 g C b1 mol 12.01 gg 72.0 mol C
|116.5 g H b1 mol 1.01 gg 115.3 mol H
1 kg Oil Ÿ S
||13.5 g S b1 mol 32.06 gg 0.4211 mol S
T5.3 g I
C O 2 o CO 2
1
C + O 2 o CO
2
S O 2 o SO 2
1
2H O 2 o H 2 O
2
5.3 g I
n 1 (mol CO 2 )
n 2 (mol CO)
n 3 (mol H 2 O)
n 4 (mol SO 2 )
n 5 (mol O 2 )
n 6 (mol N 2 )
72.0 mol C
115.3 mol H
0.4211 mol S
5.3 g I
n a (mol), 0.21 O 2 , 0.79 N 2
15% excess air
175$ C, 180 mm Hg (gauge)
a.
Theoretical O 2 :
(0.95)72.0 mol C 1 mol O
2 (0.05)72.0 mol C 0.5 mol O 2
1 mol C
1 mol C
0.4211 mol S 1 mol O
115.3 mol H 0.25 mol O
2
2
1 mol S
1 mol H
Air Fed:
e
1.15 99.45 mol O
2
j
1 mol Air
0.21 mol O
b g
544 mol Air 22.4 liter STP
1 kg oil
mol
544 mol Air
n
2
99.45 mol O
2
a
1 m3
448K 760 mm Hg
16.2 m3 air kg oil
3
10 liter 273K 940 mm Hg
b. S balance: n 4 0.4211 mol SO 2
H balance: 115.3 = 2n 3 Ÿ n 3 57.6 mol H 2 O
C balance:
b g
0.95 72.0 = n1 Ÿ n1
0.05(72.0)
n2
68.4 mol CO 2
3.6 mol CO
429.8 mol N
b g
O balance: 0.21b544g2 = 57.6 + 3.6 + 2(68.4) + 2b0.4211g + 2n Ÿ n
Total moles bexcluding inertsg wet: 559.8 mols dry: 502.2 mols
N 2 balance: 0.79 544 = n 6 Ÿ n 6
2
5
dry basis:
3.6 mol CO
502.2 mol
wet basis:
3.6 mol CO
u 10 6
559.8 mol
7.2 u 10 3
mol CO 0.4211 mol SO 2
,
mol
502.2 mol
6400 ppm CO ,
5-25
5
14.8 mol O 2
8.4 u 10 4
0.4211 mol SO 2
u 10 6
559.8 mol
mol SO 2
mol
750 ppm SO 2
bg
5.46 Basis: 50.4 liters C 5 H 12 l min
50.4 L C 5 H 12 ( l ) / min
b
n 1 kmol C 5 H 12 / min
g
heater
(L / min)
15% excess air, V
air
n 2 kmol air
0.21 O 2
0.79 N 2
336 K, 208.6 kPa (gauge)
a
f
n b kmol air / ming
n 1 kmol C5 H 12 / min
Combustion
chamber
2
n 3 (kmol C 5 H 12 / min)
n 4 (kmol O 2 / min)
n 5 (kmol N 2 / min)
n 6 (kmol CO 2 / min)
n 7 (kmol H 2 O / min)
(L / min)
V
gas
Condenser
(L / min), 275 K, 1 atm
V
liq
= 3.175 kg C 5O12 / min
m
n 3 (kmol C 5O12 / min)
n 7 (kmol H 2 O(l ) / min)
C5 H 12 802 o 5CO 2 6H 2 O
a.
n 1
n 3
50.4 L 0.630 kg
min
.
kg 1 kmol
3175
min 72.15 kg
frac. convert =
n 2
air
V
L
1 kmol
72.15 kg
0.440 kmol min
0.044 kmol / min
0.440 - 0.044 kmol
u 100 90% C5 H12 converted
0.440
b
0.440 kmol C 2 H 12 1.15 8 kmol O 2
min
kmol C 5 H 12
b g
19.28 kmol 22.4 L STP
min
mol
g
1 mol air
0.21 mol O 2
19.28 kmol air min
336K 101 kPa 1000 mol
273K 309.6 kPa kmol
n 4
0.440 kmol C 2 H12 0.15(8 kmol O 2 )
min
kmol C 2 H12
n 5
19.28 mol air 0.79 mol N 2
min
mol air
n 6
0.90(0.440 kmol C5 H12 ) 5 kmol CO 2
kmol C5 H 12
min
gas
V
n 4 (kmol O 2 / min)
n 5 (kmol N 2 / min)
n 6 (kmol CO 2 / min)
173000 L min
0.528 kmol O 2
15.23 kmol N 2 / min
. kmol CO 2 / min
198
0.528 + 15.23 + 1.98 kmol 22.4 L(STP) 548 K 1000 mol
min
mol
273 K kmol
5-26
7.98 u 105 L / min
5.46 (cont’d)
0.9(0.440 kmol C5 H12 ) 6 kmol H 2 O
kmol C5 H 12
min
n 7
2.38 kmol H 2 O(l ) / min
Condensate:
0.044 kmol 72.15 kg
V
C5 H 12
V
H 2O
min
kmol
2.38 kmol 18.02 kg
min
L
kmol
5.04 L min
0.630 kg
L
42.89 L min
1 kg
Assume volume additivity (liquids are immiscible)
liq 5.04 42.89 47.9 L min
V
b.
bg
C5 H 12 l
C 5 H 12 ( l )
bg
H 2O l
bg
H2O l
5.47
n air (kmol / min), 25$ C, 1 atm
0.21 O 2
0.79 N 2
n 0 (kmol / min)
n 1 (kmol H 2S / min) Furnace
H 2 S + 23 O 2 o SO 2 H 2 O
0.20 kmol H 2 S / mol
0.80 kmol CO 2 / mol
Reactor
n 2 (kmol H 2 S / min)
2H 2 S + SO 2 o 3S(g) 2 H 2 O
10.0 m 3 / min @ 380$ C, 205 kPa
n exit (kmol / min)
n 3 (kmol N 2 / min)
n 4 (kmol H 2 O / min)
n 5 (kmol CO 2 / min)
n 6 (kmol S / min)
n exit
n 1
PV
RT
b0.20gn
205 kPa 10.0 m3 / min
m3 ˜kPa
653 K
8.314 kmol
˜K
0
/ 3 = 0.0667n 0 ; n 2
0.377 kmol / min
2 n 1 = 0.133n 0
5-27
5.47 (cont’d)
0.0667 n 0 (kmol H 2 S fed) 15
. kmol O 2 1 kmol air
1 kmol H 2 S 0.21 kmol O 2
(min)
Air feed to furnace: n air
0.4764 n 0 kmol air / min
0.4764 n 0 (kmol air) 0.79 kmol N 2
(min)
min
Overall N 2 balance: n 3
Overall S balance: n 6
0.200n 0 (kmol H 2S) 1 kmol S
(min)
1 kmol H 2S
Overall CO 2 balance: n 5
Overall H balance:
0.3764 n 0 ( kmol N 2 / min)
0.200n 0 (kmol S / min)
0.800n 0 (kmol CO 2 / min)
0.200n 0 (kmol H 2 S) 2 kmol H
(min)
1 kmol H 2 S
n 4 kmol H 2 O 2 kmol H
min
1 kmol H 2 O
Ÿ n 4 = 0.200n 0 (kmol H 2 O / min)
n exit
b
g
n 0 0.376 0.200 + 0.200 + 0.800 = 0.377 kmol / min Ÿ n 0 = 0.24 kmol / min
n air = 0.4764(0.24 kmol air / min)
0114
.
kmol air / min
5.48 Basis: 100 kg ore fed Ÿ 82.0 kg FeS 2 (s), 18.0 kg I.
b
gb
n FeS2 fed = 82.0 kg FeS2 1 kmol / 120.0 kg
100 kg ore
06833
.
kmol FeS2
18 kg I
g
0.6833 kmol FeS2
Vout m3 (STP)
40% excess air
n 1 (kmol)
0.21 O 2
0.79 N 2
V1 m 3 (STP)
n 2 (kmol SO 2 )
n 3 (kmol SO 3 )
n 4 (kmol O 2 )
n5 (kmol N 2 )
m6 (kg FeS2 )
m7 (kg Fe 2 O 3 )
18 kg I
2 FeS 2(s) 112 O 2(g) o Fe 2 O 3(s) 4SO 2(g)
2 FeS 2(s) 152 O 2(g) o Fe 2 O 3(s) 4SO 3(g)
a.
n1
0.6833 kmol FeS2 7.5 kmol O 2 1 kmol air req'd 140
. kmol air fed
2 kmol FeS2 0.21 kmol O 2 kmol air req'd
V1
b17.08 kmolgb22.4 SCM / kmolg
n2
(0.85)(0.40)0.6833 kmol FeS2 4 kmol SO 2
2 kmol FeS2
382 SCM / 100 kg ore
5-28
0.4646 kmol SO 2
17.08 kmol air
5.48 (cont’d)
n3
(0.85)(0.60)0.6833 kmol FeS2 4 kmol SO 2
2 kmol FeS2
n4
b0.21 u 17.08gkmol O
n5
Vout
2
fed 0.6970 kmol SO 3
.4646 kmol SO 2 55
. kmol O 2
4 kmol SO 2
.702 kmol SO 3 7.5 kmol O 2
4 kmol SO 3
1633
.
kmol O 2
b0.79 u 17.08g kmol N 13.49 kmol N
= b0.4646 + 0.6970 + 1.633 + 13.49 g kmol
2
2
22.4 SCM (STP) / kmol
365 SCM / 100 kg ore fed
ySO 2
0.4646 kmol SO 2
u 100%
16.285 kmol
2.9%; y SO3
4.3%; y O 2
10.0%; y N 2
82.8%
b.
e j
Product gas, T o C
Converter
0.4646 kmol SO 2
SO 3
0.697 kmol
21 OO2 2œ
SO
SO 3
. 2 kmol
1633
13.49 kmol N 2
n SO 2 (kmol)
n SO 3 (kmol)
n O 2 (kmol)
n N 2 (kmol)
Let [ (kmol) = extent of reaction
n SO 2 0.4646 [
n SO 3 0.697 [
21 [
n O 2 1633
.
n N 2 13.49
n = 16.28 - 21 [
K p (T) =
U|
|V Ÿ y
|| y
W
SO 2
1
2
P = 1 atm, T = 600o C, K p
Ÿ n SO 2
1
2
-1
9.53 atm 2 Ÿ [
01847
.
kmol Ÿ fSO2
P = 1 atm, T = 400o C, K p
Ÿ n SO 2
c
h
Ÿ
[h
(0.4646 [ )c1633
.
(0.697 [ ) 16.28 21 [
P ˜ y SO 3
P ˜ y SO 2 ( P ˜ y O 2 )
O2
0.4646 [
0.697 [
, y SO 3
1
16.28 - 2 [
16.28 - 21 [
1
2[
.
1633
13.49
, y N2
1
16.28 - 2 [
16.28 - 21 [
1
2
˜P
- 21
1
K p (T) ˜ P 2
0.2799 kmol
.
b0.4646 01847
g kmol SO
2
0.4646 kmol SO 2 fed
-1
397 atm 2 Ÿ [
0.00059 kmol Ÿ fSO 2
1
2
reacted
0.602
0.4587 kmol
0.987
The gases are initially heated in order to get the reaction going at a reasonable rate. Once
the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium
conversion of SO2.
5-29
5.48 (cont’d)
c.
SO 3 leaving converter: (0.6970 + 0.4687) kmol = 1.156 kmol
1.156 kmol SO 3 1 kmol H 2 SO 4 98 kg H 2 SO 4
min
kmol
1 kmol SO 3
Ÿ
Sulfur in ore:
0.683 kmol FeS 2 2 kmol S 32.1 kg S
kmol FeS2 kmol
113.3 kg H 2SO 4
43.8 kg S
100% conv.of S:
438
. kg S
kg H 2SO 4
kg S
2.59
0.683 kmol FeS 2 2 kmol S 1 kmol H 2SO 4 98 kg
kmol FeS2
1 kmol S
kmol
133.9 kg H 2SO 4
43.8 kg S
Ÿ
113.3 kg H 2 SO 4
3.06
133.9 kg H 2SO 4
kg H 2SO 4
kg S
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of
FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter.
5.49 N 2 O 4 œ 2 NO 2
a.
n0
b.
n1
p NO 2
dP
gauge
i
b2.00 atmgb2.00 Lg
b473Kgb0.08206 L ˜ atm mol - Kg
1.00 V
RT0
mol NO 2 , n 2
y NO 2 P
mol N 2 O 4
FG n IJ P , p
Hn n K
FG n IJ P Ÿ K
n
P
n bn n g
Hn n K
bn n gRT Ÿ n n PV / RT b1g
2
1
2
1
1
0.103 mol NO 2
N 2O4
2
Ideal gas equation of state Ÿ PV
p
1
1
2
2
2
1
1
2
2
Stoichiometric equation Ÿ each mole of N 2 O 4 present at equilibrium represents a loss
of two moles of NO 2 from that initially present Ÿ n1 2n 2
Solve (1) and (2) Ÿ n1
b3g ,
2(PV / RT) 0.103
n2
0.103
b2 g
b4 g
0.103 (PV / RT)
Substitute (3) and (4) in the expression for K p , and replace P with Pgauge 1
b2n 0.103g dP
n b0.103 n g
2
t
gauge
t
T(K)
350
335
315
300
Pgauge(atm)
0.272
0.111
-0.097
-0.224
i
1 where n t
dP
gauge
nt
Kp(atm)
0.088568 5.46915
0.080821 2.131425
0.069861 0.525954
0.063037 0.164006
(1/T)
ln(Kp)
0.002857 1.699123
0.002985 0.756791
0.003175 -0.64254
0.003333 -1.80785
i
1 V
RT
t
Ÿ nt
i
T
V 2L
2
1
0
-1
-2
0.0028
y = -7367x + 22.747
R2 = 1
0.003
0.0032
1/T
5-30
d
24.37 Pg 1
Variation of Kp with Temperature
ln Kp
Kp
0.0034
5.49 (cont’d)
c.
A semilog plot of K p vs.
1
is a straight line. Fitting the line to the exponential law
T
yields
ln K p
7367
22.747 Ÿ K p
T
7.567 u 109 exp
FG 7367 IJ Ÿ a 7.567 u 10
H T K b = 7367K
9
atm
10.00 atm
5.50
n 1 (kmol A / h)
n 2 (kmol H 2 / h)
5.00 kmol S / h
n 4 (kmol A / h)
n 5 (kmol H 2 / h)
3n 3 (kmol A / h)
n 3 (kmol H 2 / h)
5.00 kmol S / h
n 4 (kmol A / h)
n 5 (kmol H 2 / h)
Vrcy (SCMH)
Extent of reaction equations: n i
n i0 Q i[
A + H2 l S
A:
n 4
H 2 : n 5
S:
3n 3 [
n [
3
5.00 = [
! n 4
n 5
n tot
U|
V| Ÿ p
10.0 W
3n 3 5.00
n 3 5.00
4 n 3
A
pH2
pS
Kp
pS
pA p H2
b
g
5.00 4n 3 10.0
10.0 3n 3 5.00 n 3 5.00
b
gb
g
yA P =
n 4
P
n tot
y H2 P =
yS P =
Ÿ n 3
.
0100
n 5
P
n tot
3n 3 - 5.00
10.0
4 n 3 10.0
n 3 - 5.00
10.0
4 n 3 10.0
5.00
10.0
4 n 3 10.0
. kmol H 2 / h
1108
n 4 3(1108
. ) - 5.00 28.24 kmol A / h
n 5 1108
. 5.00 6.08 kmol H 2 / h
rcy
V
b28.24 + 6.08g kmol / h d22.4 m (STP) / kmoli
3
5-31
769 SCMH
5.51
n 4 (kmol CO / h)
n 5 (kmol H 2 / h)
Reactor
Separator
100 kmol CO / h
n 1 (kmol CO / h)
n 2 (kmol H 2 / h)
H xs (% H 2 excess)
a.
n 6 (kmol M / h)
n 4 (kmol CO / h)
n 5 (kmol H 2 / h)
n 6 (kmol M / h)
T, P
n 3 (kmol H 2 / h)
T (K), P (kPa)
Balances on reactor Ÿ four equations in n 3 , n 4 , n 5 , and n 6
1 kmol CO
n 6 Ÿ n 4 100 n 6
CO balance n 4 100 kmol CO / h 1 kmol M
100 kmol CO 1 kmol CO
H 2 balance n 3 105
.
210 kmol H 2 / h
h
1 kmol M
2 kmol CO
n 6 Ÿ n 5 210 2 n 6
n 5 210 kmol H 2 / h 1 kmol M
n T n 4 n 5 n 6 100 n 6 210 2 n 6 n 6 310 2 n 6
b g
b1g
b g
b g
b
b
K p T = 500K
g
u 10 4
1390
.
b 2g
g b
g
b3g
F 21.225 + 9143.6 7.492lnb500Kg
I
G
exp
GH +4.076 u 10500bK500Kg -1.161 u 10 b500Kg JJK
-3
2
-8
9.11 u 10 7 kPa -2
yMP
Kp
d
y CO P y H 2 P
i
2
d i
y CO y H 2
b
(1) ( 3)
yM
Ÿ Kp P2
2
!
n 6
310 2 n 6
g
b100 n g b210 2n g
b310 2n g b310 2n g
n b310 2 n g
b100 n gb210 2n g
6
6
Kp P2
b
9.11 u 10 7 kPa -2 5000 kPa
g
22.775
6
6
2
6
Solving for n 6 Ÿ n 6
100 n 6
n 5
210 2 n 6
n 1
n 2
rec
V
bn
4
n 5
75.7 kmol M / h
n 4
24.3 kmol CO / h
58.6 kmol H 2 / h
1 kmol CO
n 6 75.7 kmol CO / h
1 kmol M
2 kmol H 2
n 6 151 kmol H 2 / h
1 kmol M
3
m (STP)
g 22.4 kmol
6
2
2
1860 SCMH
5-32
2
6
6
2
5.51 (cont’d)
b.
P(kPa)
1000
5000
10000
5000
5000
5000
5000
5000
5000
`
T(K) Hxs(%)
500
5
500
5
500
5
400
5
500
5
600
5
500
0
500
5
500
10
Kp(T)E8
9.1E+01
9.1E+01
9.1E+01
3.1E+04
9.1E+01
1.6E+00
9.1E+01
9.1E+01
9.1E+01
ntot
n6(kmol
M/h) (kmol/h) KpcE8
25.55 258.90 9.1E-01
9.00 292.00 2.3E-01
86.72 136.56 9.1E+01
98.93 112.15 7.8E+03
75.68 158.64 2.3E+01
14.58 280.84 4.1E-01
73.35 153.30 2.3E+01
75.68 158.64 2.3E+01
77.77 164.45 2.3E+01
KpP^2
0.91
22.78
91.11
7849.77
22.78
0.41
22.78
22.78
22.78
KpP^2- n1(kmol
KpcP^2
CO/h)
1.3E-05
25.55
2.3E+01
9.00
4.9E-03
86.72
3.2E-08
98.93
3.4E-03
75.68
-2.9E-04
14.58
9.8E-03
73.35
3.4E-03
75.68
-3.1E-03
77.77
n3(kmol n4(kmol n5(kmol
H2/h)
H2/h)
CO/h)
210
74.45 158.90
210
91.00 192.00
210
13.28
36.56
210
1.07
12.15
210
24.32
58.64
210
85.42 180.84
200
26.65
53.30
210
24.32
58.64
220
22.23
64.45
n2(kmol
H2/h)
51.10
18.00
173.44
197.85
151.36
29.16
146.70
151.36
155.55
Vrec
(SCMH)
5227
6339
1116
296
1858
5964
1791
1858
1942
c. Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the
pressure raises costs because more compression is needed.
d. If the temperature is too low, a low reaction rate may keep the reaction from reaching
equilibrium in a reasonable time period.
e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of
methanol, not steady-state measurement errors.
5.52
CO 2 œ CO + 21 O 2
1.0 mol CO 2
1.0 mol O 2
1.0 mol N 2
T = 3000 K, P = 5.0 atm
dp
CO p O 2
d
1
A œ B C
2
1
1
C D E
2
2
1/ 2
i
0.3272 atm
p CO 2
1
O 2 12 N 2 œ NO
2
p NO
K2
0.1222
1/ 2
p N 2 p O2
K1
1/ 2
i
A CO 2 , B CO , C O 2 , D N 2 , E NO
n A0
n C0
n D0
1 , n B0
n E0
5-33
0
[ 1 - extent of rxn 1
[ 2 - extent of rxn 2
5.52 (cont’d)
1 [1
[1
1
1
1 [1 [ 2
2
2
1
1 [2
2
[2
1
6 [1
3 [1
2
2
nA
nB
nC
nD
nE
n tot
U|
|| y n n 2 b1 [ g b 6 [ g
| yy b22[ [b 6 [[ gg b 6 [ g
V|
|| yy 2b 2[ [b 6 g b[6 g [ g
||
W
A
B
A
1
tot
1
1
C
2
y B y1C2 b1 12 1g
p
yA
1
CO 2
1
b
2[ 1 2 [ 1 [ 2
12
12
12
1
K2
d
p NO
pO2 p N 2
i
12
12
1
yE
12 12
yC y D
b
Ÿ 01222
2 [1 [ 2
.
1
p11 2 1 2
g b2 [ g
12
12
2
1
b2 [
b
2[ 2
1
[2
g b2 [ g
12
yC
5.53 a.
12
01222
.
2
2[ 2
(2)
g b6 [ g
2 1 [1
yB
(1)
2
Solve (1) and (2) simultaneously with E-Z Solve Ÿ [ 1
yA
0.3272
1
12
1
yiP
1
2
g b5g
K
p
2b1 [ gb6 [ g
Ÿ 0.3272b1 [ gb6 [ g
2.236[ b2 [ [ g
p CO p1O22
pi
1
2
D
E
1
1
0.2574 mol CO 2 mol y D
0.0650 mol CO mol
yE
0.3355 mol O 2 mol
1
0.20167, [ 2
012081
.
,
0.3030 mol N 2 mol
0.0390 mol NO mol
n 4 (kmol / h)
0.04 O 2
0.96 N 2
PX = C8 H 10 , TPA = C8 H 6 O, S = Solvent
(m 3 / h) @105o C, 5.5 atm
V
3
n 3O (kmol O 2 / h)
n 3N (kmol N 2 / h)
n 3W (kmol H 2 O(v) / h)
(m 3 / h) at 25o C, 6.0 atm
V
2
n 2 (kmol / h)
0.21 O 2
0.79 N 2
condenser
n 3W (kmol H 2 O(v) / h)
(m 3 / h)
V
3W
reactor
n 1 (kmol PX / h)
( n 1 n 3p ) kmol PX / h
s (kg S / h)
m
3 kg S / kg PX
5-34
n 3p ( kmolPX / h)
100 kmol TPA / h
s (kg S / h)
m
separator
100 mol TPA / s
n 3p (kmol PX / h)
s (kg S / h)
m
5.53 (cont’d)
b. Overall C balance:
n 1
c.
FG kmol PX IJ 8 kmol C
H h K kmol PX
O 2 consumed =
100 kmol TPA 8 kmol C
Ÿ n 1
h
kmol TPA
100 kmol TPA 1.5 kmol O 2
h
1 kmol TPA
150
Overall N 2 balance: 0.79n 2
Overall H 2 O balance: n 3W
2
V
3
V
3W
V
3W
g
n 4 RT
U|
V|
W
n 2
n 4
Ÿ
100 kmol TPA 2 kmol H 2 O
h
1 kmol TPA
848 kmol 0.08206 m3 ˜ atm 298 K
h
kmol ˜ K 6.0 atm
n 2 RT
P
bn
150 kmol O 2 / h
kmol O 2
+ 0.04n 4
h
0.96n 4
Overall O 2 balance: 0.21n 2
100 kmol PX / h
200 kmol H 2 O / h
3450 m3 air / h
b200 + 698g kmol 0.08206 m ˜ atm 378 K
3
P
kmol ˜ K 5.5 atm
h
200 kmol H 2 O (l) 18.0 kg 1 m3
h
kmol 1000 kg
d
b100 + 11.1g kmol PX
h
5065 m3 / h
3.60 m3 H 2 O(l) / h leave condenser
i
n 1 100
d. 90% single pass conversion Ÿ n 3p = 0.10 n 1 n 3p ====> n 3p
recycle
m
848 kmol air / h
698 kmol / h
106 kg 4 kg recycle
kg PX
1 kmol PX
111
. kmol PX / h
4.71 u 104 kg recycle / h
e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the
air.
The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX.
f. The stream can be allowed to settle and spearate into water and PX layers, which may then
be separated.
5.54
n 1 (kmol CO / h), n 3 (kmol H 2 / h), 0.10n 2 (kmol H 2 / h)
Separator
n 6 (kmol CO / h)
n 7 (kmol H 2 / h)
n 8 (kmol CO 2 / h)
2 kmol N 2 / h
0.90 n 2
2 kmol N 2 / h
n 1 , n 2 , n 3
2 kmol N 2 / h
0.300 kmol CO / kmol
0.630 kmol H 2 / kmol
0.020 kmol N 2 / kmol
0.050 kmol CO 2 / kmol
Reactor
n 1 (kmol CO / h)
n 2 (kmol H 2 / h)
n 3 (kmol CO 2 / h)
n 4 (kmol M / h)
n 5 (kmol H 2 O / h)
2
5-35 kmol N 2 / h
Separator
n 4 (kmol M / h)
n 5 (kmol H 2 O / h)
5.54 (cont’d)
CO + 2H 2 œ CH 3OH(M)
CO 2 3H 2 œ CH 3OH + H 2 O
a.
Let [ 1 ( kmol / h)
extent of rxn 1, [ 2 ( kmol / h)
extent of rxn 2
n 1 = 30 - [ 1
n 2 = 63 - 2[ 1 3[ 2
CO:
H2:
U|
M:
n = [ [
||
P˜y P˜y
H O: n = [
V| Ÿ dK i P ˜ y P ˜Py ˜ y , dK i bP ˜ y gd P ˜ y i
d i
d
id i
N : n = 2
||
n
100 - 2[ 2[ W
n
2[ 2[ g
n
84.65
(1)
dK i ˜ P = n F n I bb[30[[ gbgb100
63 2[ 3[ g
G J
n H n K
FG n IJ FG n IJ
.
(2)
dK i ˜ P = FH nn IKFH nn IK [ bb[5 [[ gbgb63100 2[2[ 3[2[g g 1259
GH n JK GH n JK
CO 2 : n 3 = 5 - [ 2
2
2
4
1
5
2
2
M
p 1
CO
N2
tot
CO 2
H2
3
H2
4
2
1
tot
p 1
2
1
2
2
2
1
2
tot
tot
2
H 2O
2
1
2
M
p 2
2
1
4
5
tot
tot
3
2
tot
tot
1
2
2
p 2
2
2
1
1
2
2
3
2
1
2
Solve (1) and (2) for [ 1 , [ 2 Ÿ [ 1 = 25.27 kmol / h [ 2 = 0.0157 kmol / h
Ÿ
n 1
30.0 25.27
4.73 kmol CO / h
9.98% CO
n 2
63.0 2(25.27) 3(0.0157) 12.4 kmol H 2 / h
26.2% H 2
n 3
5.0 0.0157 4.98 kmol CO 2 / h
10.5% CO 2
n 4
25.27 0.0157
n 5
0.0157 0.0157 kmol H 2 O / h
n total
49.4 kmol / h
C balance: n 4 25.3 kmol / h
O balance: n 6 2 n 8 n 4 n 5
H balance: 2n 7
b. (n 4 ) process
53.4% M
0.03% H 2 O
UV Ÿ n 25.4 kmol CO / h
25.44 mol / sW n = 0.02 kmol CO / h
2(0.9 n 2 ) 4 n 4 2 n 5
6
8
123.7 Ÿ n 7
237 kmol M / h
Ÿ Scale Factor =
Ÿ
25.3 kmol M / h
237 kmol M / h
25.3 kmol / h
5-36
2
618
. mol H 2 / s
5.54 (cont’d)
237 kmol / h I F 22.4 m (STP) I
gFGH 25.3
J kmol JK 18,700 SCMH
kmol / h K GH
F 237 kmol / h IJ 444 kmol / h
Reactor effluent flow rate: b 49.4 kmol / hgG
H 25.3 kmol / sK
F kmol IJ FG 22.4 m (STP) IJ 9946 SCMH
Ÿ V G 444
H h K H kmol K
3
b
. 0.02 2.0
Process feed: 25.4 618
3
std
. kPa
9950 m 3 (STP) 473.2 K 1013
h
273.2 K 4925 kPa
Ÿ Vactual
c.
=V
V
n
354 m 3 / h 1000 L 1 kmol
444 kmol / h m 3 1000 mol
354 m 3 / h
0.8 L / mol
(5.2-36)
< 20 L / mol ====> ideal gas approximation is poor
V
from n using the ideal gas equation of state is likely
Most obviously, the calculation of V
to lead to error. In addition, the reaction equilibrium expressions are probably strictly
valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
RTc
PV
B
Bo ZB1
1 Ÿ B =
Pc
RT
V
From Table B.1 for ethane: Tc 305.4 K, Pc 48.2 atm
From Table 5.3 -1 Z = 0.098
0.422
0.422
0.333
Bo 0.083 1.6 0.083 1.6
Tr
308.2 K
305.4 K
.
.
0172
0172
4.2 0139
0.0270
.
.
B1 0139
4 .2
Tr
308.2 K
305.4K
RTc
0.08206 L ˜ atm 305.4 K
0.333 0.098 0.0270
B(T) =
B o ZB1
mol ˜ K 48.2 atm
Pc
01745
L / mol
.
b
b
g
e
j
e
j
g
FG
H
b
IJ
K
g
2
mol ˜ K 2 PV
- B = 10.0 atm
V
V V + 0.1745 = 0
308.2K 0.08206 L ˜ atm
RT
=
ŸV
Videal
b.
b
gb
1 r 1 - 4 0.395 mol / L 01745
.
L / mol
RT / P
b
g
2 0.395 mol / L
g
2.343 L / mol, 0.188 L / mol
0.08206 u 308.2 / 10.0 2.53, so the second solution is
likely to be a mathematical artifact.
10.0 atm 2.343 L / mol
PV
z=
L˜atm
RT 0.08206 mol
308.2K
˜K
0.926
5-37
c.
5.56
=
m
V
1000 L
mol 30.0 g 1 kg
MW =
h 2.343 L
mol 1000 g
V
PV
RT
1
b
RTc
B
Ÿ B=
Bo ZB1
Pc
V
12.8 kg / h
g
b
g 513.2 K, P 78.50 atm
T bC H g 369.9 K, P 42.0 atm
From Table 5.3 -1 Z bCH OH g = 0.559, Z bC H g = 0.152
From Table B.1 Tc CH 3OH
c
3
c
8
c
3
Bo (CH 3OH)
0.083 Bo ( C 3 H 8 )
0.422
1.6
Tr
.
0139
B1 (CH 3OH)
0139
.
B1 (C 3 H 8 )
3
0.422
0.083 1.6
Tr
B(CH 3OH) =
0.083 0.083 .
0172
4 .2
Tr
0172
.
4 .2
Tr
RTc
Bo ZB1
Pc
e
e
e
513.2K
0.422
j
0.333
j
1.6
369.9K
.
0172
373.2K
513.2K
0172
.
373.2K
0.619
1.6
373.2K
373.2K
.
0139
0139
.
b
e
8
0.422
369.9K
j
j
0.516
4 .2
0.0270
4 .2
g
0.08206 L ˜ atm 513.2K
0.619 0.559 0.516
mol ˜ K 78.5 atm
RTc
Bo ZB1
B(C 3 H 8 ) =
Pc
b
c
b
g
h
b
c
¦¦y y B
i
i
B ij
B mix
b
L
mol
g
0.08206 L ˜ atm 369.9 K
.
0.0270
0.333 0152
mol ˜ K 42.0 atm
B mix
0.4868
j
ij
ŸB ij
j
d
0.5 B ii B jj
g
h
0.2436
L
mol
i
g
0.5 0.4868 0.2436 L / mol = -0.3652 L / mol
b0.30gb0.30gb0.4868g 2b0.30gb0.70gb0.3652g b0.70gb0.70gb0.2436g
0.3166 L / mol
IJ
K
FG
H
2
mol ˜ K 2 PV
- B = 10.0 atm
V
V V + 0.3166 = 0
mix
373.2K 0.08206 L ˜ atm
RT
=
Solve for V:V
V
ideal
RT
P
b
gb
1 r 1- 4 0.326 mol / L 0.3166 L / mol
b
g
2 0.326 mol / L
0.08206 L ˜ atm 373.2 K
mol ˜ K 10.0 atm
g
2.70 L / mol, 0.359 L / mol
3.06 L / mol Ÿ V
virial
5-38
2.70 L / mol
5.57 a.
15.0 kmol CH 3OH / h 1000 mol 1 m3
135 m3 / h
0.30 kmol CH 3OH / kmol 1 kmol 1000 L
2.70 L / mol
= Vn
V
van der Waals equation: P =
d
d
RT
a2
2
-b
V
V
i
i
2 V
- b Ÿ PV
3 PV
2 b = RTV
2 aV
+ ab
Multiply both sides by V
b
g
+ -Pb - RT V
aV
- ab = 0
PV
3
c3
2
P = 50.0 atm
c2
b-Pb - RTg b50.0 atmgb0.0366 L / molg c0.08206
c1
a = 133
. atm ˜ L2 / mol 2
c0
ab = - 133
. atm ˜ L2 / mol 2 0.0366 L / mol
ib
d
0.0487
hb223 Kg
201
. L ˜ atm / mol
g
atm ˜ L
mol 3
3
RT
P
b. V
ideal
L˜atm
mol˜K
0.08206 L ˜ atm 223 K
mol ˜ K 50.0 atm
0.366 L / mol
c.
T(K)
P(atm)
223
223
223
223
223
1.0
10.0
50.0
100.0
200.0
c3
c2
1.0
10.0
50.0
100.0
200.0
c1
-18.336
-18.6654
-20.1294
-21.9594
-25.6194
c0
1.33
1.33
1.33
1.33
1.33
-0.0487
-0.0487
-0.0487
-0.0487
-0.0487
V(ideal)
V
f(V)
% error
(L/mol) (L/mol)
18.2994 18.2633 0.0000
0.2
1.8299
1.7939 0.0000
2.0
0.3660
0.3313 0.0008
10.5
0.1830
0.1532 -0.0007
19.4
0.0915
0.0835 0.0002
9.6
d. 1 eq. in 1 unknown - use Newton-Raphson.
b1g Ÿ gdV i
b
Eq. (A.2-13) Ÿ a
Eq. (A.2-14) Ÿ ad
(k +1)
Then V
g
b g
3 + -20.1294 V
2 133
50.0V
=0
. V-.0487
wg
wV
2 40.259 V
+ 1.33
150V
solve
g Ÿ d
g
a
(k) d Guess V
(1)
V
1
2
3
4
V
ideal
(k)
V
0.3660
0.33714
0.33137
0.33114
5-39
0.3660 L / mol .
(k +1)
V
0.33714
0.33137
0.33114
0.33114
converged
b
5.58 C 3 H 8 : TC
369.9 K
5.0 m3
75 kg
Specific Volume
d
42.0 atm 4.26 u 106 Pa
PC
44.09 kg 1 kmol
1 kmol 103 mol
i
Z
0152
.
2.93 u 10 3 m3 mol
Calculate constants
0.42747
a
d8.314 m ˜ Pa mol ˜ Ki b369.9 Kg
2
3
4.26 u 10 Pa
6
0.08664
b
d8.314 m ˜ Pa mol ˜ Ki b369.9 Kg
3
b
g
b
e
D
1 0.717 1 298.2 369.9
j
g
2
0.949 m6 ˜ Pa mol 2
6.25 u 10 5 m3 mol
4.26 u 106 Pa
015613
0152
0152
m 0.48508 155171
.
.
.
.
2
0.717
2
115
.
SRK Equation:
d8.314 m ˜ Pa mol ˜ Kib298.2 Kg d2.93 u 10 6.25 u 10 i m mol 2.93 u 10
d
. 0.949 m6 ˜ Pa mol 2
115
3
P
3
ŸP
5
3
3
d
i
i
m3 mol 2.93 u 10 3 6.25 u 10 5 m3 mol
7.40 u 106 Pa Ÿ 7.30 atm
Ideal:
RT
V
P
TC
3
3
8.46 u 106 Pa Ÿ 8.35 atm
2.93 u 10 m mol
3
(8.35 7.30) atm
u 100% 14.4%
7.30 atm
Percent Error:
5.59 CO 2 :
d8.314 m ˜ Pa mol ˜ Kib298.2 Kg
304.2 K
PC
72.9 atm Z
0.225
151.2 K PC 48.0 atm Z 0.004
35.0 L / 50.0 mol 0.70 L mol
P 510
. atm , V
Ar:
TC
Calculate constants (use R
L2 ˜ atm
L
, m 0.826 , b 0.0297
,D
2
mol
mol
L2 ˜ atm
L
.
, m 0.479 , b 0.0224
,D
a 137
2
mol
mol
CO 2 : a
Ar:
bg
f T
0.08206 L ˜ atm mol ˜ K )
3.65
e
RT
a
1 m 1 T TC
V
b
Vb V
d
i
j
2
P=0
Use E-Z Solve. Initial value (ideal gas):
L
L ˜ atm
Tideal 510
. atm 0.70
0.08206
mol
mol ˜ K
b
gFGH
e
1 0.479e1 IJ FG
K H
5-40
IJ
K
j
. j
T 1512
1 0.826 1 T 304.2
435.0 K
2
2
b g
E - Z Solve Ÿ Tmax
5.60 O 2 : TC
CO 2
154.4 K ; PC
455.4 K ,
bT g
431.2 K
max Ar
49.7 atm ; Z
b
g
0.021 ; T 208.2 K 65q C ; P 8.3 atm ;
0.08206 L ˜ atm mol ˜ K
250 kg h ; R
m
0.0221 L mol ; m 0.517 ; D
. L2 ˜ atm mol 2 ; b
SRK constants: a 138
0.840
d i dVRT bi V dVaD bi P = 0=====> V = 2.01 L / mol
E-Z Solve
f V
SRK equation:
ŸV
250 kg
kmol
103 mol 2.01 L
15,700 L h
h
32.00 kg 1 kmol
mol
W
5.61
e
¦F
PCO2 ˜ A - W = 0 where W = mg = 5500 kg 9.81
y
m
s2
j
53900 N
PCO 2 ˜ A
a.
PCO 2
53900 N
W
A piston
S
4
. mg
b015
2
b. SRK equation of state: P =
For CO 2 : Tc
304.2, Pc
1 atm
1.013 u 105 N / m2
Da
RT
-b V
V
+b
V
d i d
. atm
301
i
72.9 atm , Z = 0.225
a = 3.654 m ˜ atm / kmol , b = 0.02967 m3 / kmol, m 0.8263, D (25o C) 1016
.
6
2
. ge3.654
e0.08206 jb298.2 Kg b1016
j
. atm =
301
eV - 0.02967 j V dV + 0.02967i
m 3 ˜atm
kmol˜K
m6 ˜atm
kmol 2
m6
kmol 2
m3
kmol
E-Z Solve
= 0.675 m 3 / kmol
=====> V
b
g 0.030 m
. mg b15
. mg
Vbafter expansiong 0.030 m b015
3
V before expansion
3
mCO 2
2
S
4
44.01 kg
0.0565 m3
V
MW =
3
0.675 m / kmol kmol
V
mCO 2 (initially) =
0.0565 m3
3.68 kg
0.030 m3 44.01 kg
1 atm
PV
MW =
m 3 ˜atm 298.2 K
kmol
RT
0.08206 kmol
˜K
mCO 2 (added) = 3.68 - 0.0540 kg = 3.63 kg
5-41
0.0540 kg
5.61 (cont’d)
c.
W = 53,900 N
V
h
add 3.63 kg CO 2
n o (kmol)
Vo (m3 )
1 atm, 25o C
! n (kmol)
P (atm), 25o C
ho
ho
d(m)
d(m)
Given T, Vo , h, find d
Vo
RT
Initial: n o
Final: V = Vo =V
V
n
Vo bP 1g
o
Sd 2 h
3.63 (kg)
, n = no 44 (kg / kmol)
4
Vo
0.0825
RT
Sd 2 h
4
Vo
0.0825
RT
Da
53,900
W
RT
Ÿ 2
P=
V
+b
A piston V - b V
Sd / 4
d
Da
RT
V
+b
V-b V
i
d i
in b1g Ÿ one equation in one unknown.
Substitute expression for V
b1g
Solve for d .
5.62 a. Using ideal gas assumption:
Pg
nRT
Patm
V
35.3 lb m O 2 1 lb - mole 10.73 ft 3 ˜ psia 509.7 o R
14.7 psia = 2400 psig
32.0 lb m lb - mole ˜ o R 2.5 ft 3
b. SRK Equation of state: P =
Da
RT
-b V
V
+b
V
d i d
3
= 2.5 ft 32.0 lb m / lb - mole
V
35.3 lb m
For O 2 : Tc
277.9 o R, Pc
ft ˜ psi
, b
lb - mole 2
2.27
ft 3
lb - mole
730.4 psi, Z = 0.021
6
.
a = 52038
i
0.3537
ft 3
, m = 0.518, D 50o F
lb - mole
d
i
e10.73
jd509.7 Ri b0.667ge52038.
b2400 + 14.7g psi = V - 0.3537
dV
+ 0.3537i
V
d
i
ft 3 ˜psi
o
lb-mole˜ R
ft 6 ˜psi
lb-mole 2
o
ft 3
lb-mole
= 2.139 ft 3 / lb - mole
E - Z Solve Ÿ V
32.0 lb m
V
2.5 ft 3
mO 2
MW =
3
2.139 ft / lb - mole lb - mole
V
5-42
37.4 lb m
0.667
j
ft 6
lb-mole 2
5.62 (cont’d)
Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can
safely hold.
c. 1.
2.
3.
4.
Pressure gauge is faulty
The room temperature is higher than 50qF
Crack or weakness in the tank
Tank was not completely evacuated before charging and O2 reacted with something in
the tank
5. Faulty scale used to measure O2
6. The tank was mislabeled and did not contain pure oxygen.
5.63 a.
SRK Equation of State: P =
Da
RT
+b
V-b V V
d i d
i
d id
i = PV
dV
- bid V
+ bi RTV
dV
+ bi Dad V
- bi
f dV
i PV
RTV
dDa - b P - bRTiV
- Dab = 0
f dV
i
V
-b V
+b :
Ÿ multiply both sides of the equation by V
3
2
0
2
b.
Problem 5.63-SRK Equation Spreadsheet
Species
Tc(K)
Pc(atm)
Z
a
b
m
CO2
304.2
R=0.08206 m^3 atm/kmol K
72.9
0.225
3.653924 m^6 atm/kmol^2
0.029668 m^3/kmol
0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8
T(K)
200
250
300
300
300
P(atm)
6.8
12.3
6.8
21.5
50.0
alpha
1.3370
1.1604
1.0115
1.0115
1.0115
V(ideal)
2.4135
1.6679
3.6203
1.1450
0.4924
V(SRK)
2.1125
1.4727
3.4972
1.0149
0.3392
f(V)
0.0003
0.0001
0.0001
0.0000
0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part
b.
d. REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP
INTEGER I
CHARACTER A20 GAS
DATA R 10.08206/
READ (5, *) GAS
WRITE (6, *) GAS
10 READ (5, *) TC, PC, W
READ (5, *) T, P
IF (T.LT.Q.) STOP
5-43
5.63 (cont’d)
R 0.42747 *R*R/PC*TC*TC
B 0.08664 *R*TC/PC
W 015613
M 0.48508 W 155171
.
.
b
ALP
d1. M c1 bT / TCg
hi
0.5
g
2.
VP R T / P
DO 20 I 7, 15
V = VP
F = R * T/(V – B) – ALP * A/V/(V + B) – P
FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2.
VP = V – F/FP
IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30
20 CONTINUE
WRITE (6, 2)
2 FORMAT ('DID NOT CONVERGE')
STOP
30 WRITE (6, 3) T, P, VP
3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL')
GOTO 10
END
$ DATA
CARBON
304.2
200.0
250.0
300.0
–1
72.9
6.8
12.3
21.5
0.
DIOXIDE
0.225
RESULTS
CARBON DIOXIDE
200.0 K
6.8 ATM
250.0 K
12.3 ATM
300.0 K
6.8 ATM
300.0 K
21.5 ATM
300.0 K
50.0 ATM
5.64 a.
b.
N 2 : TC
PC
Tr
126.2 K
Ÿ
33.5 atm Pr
He: TC
PC
Tr
5.26 K
Ÿ
2.26 atm Pr
2.11 LITER/MOL
1.47 LITER/MOL
3.50 LITER/MOL
1.01 LITER/MOL
0.34 LITER/MOL
U|
10 atm
40 MPa
V Ÿ z 12.
. |
1178
33.5 atm 1.013 MPa
W
. |U
b200 273.2g b5.26 8g 552
V| Ÿ z 16.
350 b2.26 8g 34.11
W
b40 273.2g 126.2
2.48
Fig. 5.4-4
Fig. 5.4-4
n
Newton’s correction
5-44
5.65 a.
d
U kg / m3
i
m (kg)
V (m3 )
(MW)P
RT
30 kg kmol
9.0 MPa
10 atm
3
m
atm
˜
465 K
0.08206 kmol˜K 1.013 MPa
b.
UV
W
Tr
Pr
.
465 310 15
9.0 4.5 2.0
U
(MW)P
zRT
5.66 Moles of CO 2 :
TC
PC
T
Fig. 5.4-3
Ÿ z
2.27 lb - moles
44.01 lb m CO 2
P PC
V
r
VP
C
RTC
5.67 O : T
2
C
PC
. kg m3
831
100 lb m CO 2 1 lb - mole CO 2
UV
W
PV
znR
0.84
69.8 kg m3
0.84
304.2 K
Ÿ Pr
72.9 atm
, Vr
1507
.
Fig. 5.4-3: Pr
b1600 14.7gpsi
72.9 atm
1 atm
.
1507
14.7 psi
10.0 ft 3
72.9 atm lb - mole˜q R
1k
3
2.27 lb - moles 304.2 k 0.7302 ft ˜ atm 1.8 q R
0.80 Ÿ z 0.85
1614.7 psi
10.0 ft 3
lb - mole˜q R
1 atm
3
0.85
2.27 lb - moles 0.7302 ft ˜ atm 14.7 psi
Tr1
154.4 K
49.7 atm
Pr1
Tr2
Pr2
|UVz
1 49.7 0.02 |W
358 154.4 2.23 U|
Vz
1000 49.7 20.12 |W
298 154.4 1.93
b
1.61 Fig. 5.4 - 4
2
320 q F
g
z 2 T2 P1
z1 T1 P2
V1
V2
127 m3 1.61 358 K
1 atm
0.246 m3 h
h
1.00 298 K 1000 atm
n1 n 2
779q R
0.80
1.00 (Fig. 5.4 - 2)
1
V2
5.68 O 2 : TC
PC
69.8 kg m3
154.4 K
49.7 atm
FG
H
V P1 P2
RT z1 z 2
IJ
K
Tr
b27 273.2g 154.4
Pr1
175 49.7
3.52 Ÿ z1
0.95
Pr2
1.1 49.7
0.02 Ÿ z 2
1.00
1.94
FG
H
(Fig. 5.3-2)
mol ˜ K
175 atm 11
. atm
300.2 K 0.08206 L ˜ atm 0.95
1.00
10.0 L
5-45
IJ
K
74.3 mol O 2
5.69 a.
= V 50.0 mL 44.01 g 4401
V
. mL / mol
n
mol
5.00 g
RT 82.06 mL ˜ atm
1000 K
P=
186 atm
mol ˜ K 440.1 mL / mol
V
b. For CO 2 : Tc 304.2 K, Pc 72.9 atm
T 1000 K
Tr
3.2873
Tc 304.2 K
VP
. mL 72.9 atm
mol ˜ K
4401
c
Vr ideal
RTc
mol 304.2 K 82.06 mL ˜ atm
Figure 5.4 - 4: Vr ideal
P=
c.
zRT
V
128
. and Tr
.
128
3.29 Ÿ z = 1.02
. 82.06 mL ˜ atm
mol 1000 K
102
mol ˜ K 440.1 mL
190 atm
a = 3.654 u 10 6 mL2 ˜ atm / mol 2 , b = 29.67 mL / mol, m 0.8263, D (1000 K )
P=
.
ge
j
c82.06 hb1000 Kg b01077
b440.1- 29.67g
440.1b440.1 + 29.67 g
2
3.654 u 106 mLmol˜atm
2
mL˜atm
mol˜K
mL2
mol 2
mL
mol
01077
.
198 atm
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the
presence of O2.
b. Enough N2 needs to be added to make x O 2
10 u 10 6 . Since the O2 is so dilute at this
condition, the properties of the gas will be that of N2.
. atm, Tr 2.36
Tc 126.2 K, Pc 335
n initial
n1
PV
RT
1 atm
5000 L
L˜atm
0.08206 mol˜K 298.2 K
FG 0.21 mol O IJ
H mol air K
nO2
204.3 mol air
n O2
10 u 10 6 Ÿ n 2
n2
2
204.3 mol
42.9 mol O 2
4.29 u 10 6 mol
5000 L
116
. u 10 -3 L / mol
6
4.29 u 10 mol
. u 10 3 L
. atm
mol ˜ K 335
ideal VPc 116
V
r
RTc
mol 0.08206 L ˜ atm 126.2 K
Ÿ not found on compressibility charts
=
V
Ideal gas: P =
RT
V
0.08206 L ˜ atm
298.2 K
mol ˜ K 116
. u 10 3 L / mol
38
. u 10 3
2.1 u 10 4 atm
The pressure required will be higher than 2.1 u 10 4 atm if z t 1, which from
Fig. 5.3 - 3 is very likely.
n added
ib
d
4.29 u 106 204.3 # 4.29 u 106 mol N 2 0.028 kg N 2 / mol
5-46
g
120
. u 105 kg N 2
5.70 (cont’d)
c.
143
. kmol N 2
143
. kmol N 2
n initial 0.204 kmol
yO
0.21 kmol O 2 / kmol
2
143
. kmol N 2
y1
Fig 5.4-2
N 2 at 700 kPa gauge = 7.91 atm abs. Ÿ Pr
n2
P2 V
zRT
y1
y init n init
1.634
y2
y 1 n init
1.634
y init
0.236, Tr
7.91 atm
5000 L
L˜atm
298.2 K
0.99 0.08206 mol
˜K
b0.21g0.204
.
1634
y init
FG n IJ
K
H 1634
.
2.36 =======> z = 0.99
.
1633
kmol
0.026
2
init
0.0033
FG y IJ
FG n IJ Ÿ n = H y K 4.8 Ÿ Need at least 5 stages
K
H 1634
.
F n IJ
lnG
H 1.634 K
. kmol N gb28.0 kg / kmolg 200 kg N
5b143
ln
n
yn
143
. kmol N 2
y2
n
init
init
init
Total N 2
2
2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is
that it takes longer.
5.71
= MW
a. m
b.
Tc
Pc
SPV
PV
= MW
Ÿ Cost ($ / h) = mS
RT
RT
UV
W
369.9 K = 665.8o R Ÿ Tr 0.85
Ÿ Pr 016
42.0 atm
.
= 60.4
m
PV
zT
ideal
m
z
F 44.09 lb / lb - mol I SPV
GH 0.7302
JK T
m
ft ˜atm
lb-mol˜o R
3
60.4
SPV
T
Fig. 5.4-2
Ÿ z = 0.91
ideal
110
. m
Ÿ Delivering 10% more than they are charging for (undercharging their customer)
5-47
5.72 a.
For N 2 : Tc
After heater: Tr
n =
b. tank =
4.65 ft 3 / min
34,900 gal
133.0 K, Pc
34.5 atm
300 K
2.26
133.0 K
2514.7 psia 1 atm
34.5 atm 14.7 psia
Pr1
U|
|V
5.0|
|W
Fig. 5.4-3
Ÿ z = 1.02
mol ˜ K
2514.7 psia 150 L 1 atm
1022 mol
1.02
300 K 14.7 psia 0.08206 L ˜ atm
After 60h: Tr1
Pr1
b.
0.418 lb - mole / min
b g
Initially: Tr1
n leak
1.02
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks
min
h
day week
0.81 62.4 lb m / ft 3
For CO: Tc
n2
U|
|V Ÿ z
12
. |
|W
. 0.418 lb - mole 10.73 ft 3 ˜ psia 609.7 o R
102
min
lb - mole ˜o R 600 psia
4668 ft 3
n1
335
. atm
609.7 o R
2.68
227.16o R
600 psia 1 atm
Pr
. atm 14.7 psia
335
150 SCFM
359 SCF / lb - mole
= zRTn
V
P
5.73 a.
227.16o R, Pc
126.20 K
300 K
2.26
133.0 K
2258.7 psia 1 atm
34.5 atm 14.7 psia
U|
|V
4.5|
|W
Fig. 5.4-3
Ÿ z = 1.02
2259.7 psia 150 L 1 atm
mol ˜ K
1.02
300 K 14.7 psia 0.08206 L ˜ atm
n n2
173
= 1
. mol / h
60 h
PV
RT
918 mol
200 u 106 mol CO
1 atm
30.7 m3 1000 L
L˜atm
mol air
0.08206 mol
m3
˜K 300 K
n2
y 2 n air
y2
t min
n2
n leak
0.25 mol
1.73 mol / h
0.25 mol
014
. h
Ÿ t min would be greater because the room is not perfectly sealed
c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high
concentration area; (iii) there may be residual CO left from another tank; (iv) the tank
temperature could be higher than the room temperature, and the estimate of gas escaping
could be low.
5-48
5.74 CH 4 : Tc
190.7 K , Pc
45.8 atm
C 2 H 6 : Tc
305.4 K , Pc
48.2 atm
C 2 H 4 : Tc
2831
. K , Pc
50.5 atm
b0.20gb190.7g b0.30gb305.4g b0.50gb2831. g 2713. K
Pseudocritical pressure: P c b0.20gb458
. g b0.30gb48.2g b0.50gb50.5g 48.9 atm
U|
b90 273.2gK 134
Reduced temperature:
T
.
|V Ÿ z 0.71
. K
2713
200
bars
1
atm
Reduced pressure:
P
4.04 |
|W
48.9 atm 1.01325 bars
Mean molecular weight of mixture:
b0.30gM
b0.50gM
M b0.20gM
b0.20gb16.04g b0.30gb30.07g b0.50gb28.05g
Pseudocritical temperature: Tcc
c
r
Figure 5.4-3
r
CH 4
C2H 6
C2 H 4
26.25 kg kmol
V
znRT
P
0.71 10 kg
1 kmol 0.08314 m 3 ˜ bar
26.25 kg
kmol ˜ K
UV
W
5.75 N 2 : Tc 126.2 K, PC 33.5 atm Tcc
N 2 O: Tc 309.5 K, PC 71.7 atm Pcc
M
n
a.
b g
b g
200 bars
b
0.041 m 3 (41 L)
g
0.10 309.5 0.90 126.2 144.5 K
0.10 71.7 0.90 33.5 37.3 atm
b g b g 29.62
5.0 kgb1 kmol 29.62 kgg 0.169 kmol
T b24 273.2 g 144.5 2.06
b g
0.10 44.02 0.90 28.02
169 mol
r
V
r
mol ˜ K
30 L
Pr
V
r
T
30 L
37.3 atm
mol ˜ K
169 mol 144.5 K 0.08206 L ˜ atm
U|
VŸz
0.56|
W
0.97 169 mol 297.2 K 0.08206 L ˜ atm
P
b.
b90 + 273g K
U|V Ÿ z
0.56 b from a.g |W
273 37.3 7.32
b
g
518 K Ÿ 245q C
5-49
g
133 atm Ÿ 132 atm gauge
1.14 Fig. 5.4 - 3
mol ˜ K
273 atm 30 L
1.14 169 mol 0.08206 L ˜ atm
b
0.97 Fig. 5.4 - 3
b g
b g
UV
W
Turbine inlet:
Turbine exit: Tr
Pr
in
Pin V
out
Pout V
Tr
b150 273.2g 96.2
Pr
2000 psi 1 atm
29.0 atm 14.7 psi
373.2 96.2 3.88
g
g
b
96.2 K
29.0 atm
U|
V o z | 1.01
4.69|
W
4.4
Fig. 5.4-1
Ÿ z=1.0
1 29.0 0.03
z in nRTin
Ÿ Vin
z out n RTout
b
0.60 133.0 0.40 33 8
0.60 34.5 0.40 12.8 8
5.76 CO: Tc 133.0 K, Pc 34.5 atm Tcc
H 2 : Tc 33 K, Pc 12.8 atm
Pcc
Vout u
Pout z in Tin
Pin z out Tout
15,000
ft 3 14.7 psia 1.01 423.2K
min 2000 psia 1.00 373.2
126 ft 3 / min
If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00
Ÿ 1% error
UV
W
5.77 CO: Tc 133.0 K, Pc 34.5 atm Tcc
CO 2 : Tc 304.2 K, Pc 72.9 atm Pcc
Initial: Tr
Pr
Final: Pr
303.2 138.1 2.2
2014.7 524.8 3.8
b g
b g
UV o z
W
1889.7 524.8 3.6 Ÿ z1
b
g
0.97 133.0 0.03 304.2 138.1 K
0.97 34.5 0.03 72.9 35.7 atm 524.8 psi
Fig. 5.4-3
1
b g
0.97
0.97
Total moles leaked:
n1 n 2
FG P P IJ V b2000 1875gpsi
0.97
H z z K RT
1
2
1
2
mol ˜ K
303 K 14.7 psi 0.08206 L ˜ atm
30.0 L
1 atm
10.6 mol leaked
b g
Moles CO leaked: 0.97 10.6
Total moles in room:
Mole% CO in room =
10.3 mol CO
24.2 m3 103 L 273 K
1 mol
3
303 K 22.4 L STP
1m
b g
10.3 mol CO
u 100% 10%
CO
.
973.4 mol
5-50
973.4 mol
CO 2H 2 o CH 3OH
5.78 Basis: 54.5 kmol CH 3OH h
n 1 (kmol CO / h)
2n 1 (kmol H 2 / h)
644 K
34.5 MPa
Catalyst
Bed
CO, H 2
Condenser
54.5 kmol CH 3OH (l ) / h
a.
54.5 kmol CH 3OH 1 kmol CO react
n 1
h
b g
2n 1
2 218
1 kmol CH 3OH
1 kmol CO fed
0.25 kmol CO react
b
g
436 kmol H 2 h Ÿ 218 436
CO: Tc
133.0 K
Pc
34.5 atm
H 2 : Tc
33 K
Pc
12.8 atm
218 kmol h CO
654 kmol h (total feed)
Newton’s corrections
Tcc
Pcc
b g b g
1
b3 34.5g 23 b12.8 8g
2
1
133.0 33 8
3
3
Tr
644 71.7 8.98
Pr
34.5 MPa
10 atm
24.5 atm 1.013 MPa
feed
V
717
. K
25.4 atm
U|
V  o z
13.45|
W
Fig. 5.4-4
1
1.18
1.18 654 kmol
644 K
0.08206 m3 ˜ atm 1.013 MPa
120 m3 h
h
34.5 MPa
kmol ˜ K
10 atm
Vcat
120 m3 h
1 m3 cat
25,000 m3 / h
0.0048 m3 catalyst (4.8 L)
b.
CO, H 2
n 4 kmol CO / h
2n 4 kmol H 2 / h
Overall C balance Ÿ n 4
Fresh feed:
54.5 kmol CH 3OH (l ) / h
54.5 mol CO h
54.5 kmol CO h
109.0 kmol H 2 h
163.5 kmol feed gas h
feed
V
1.18 163.5 kmol
644 K
0.08206 m3 ˜ atm 1.013 MPa
h
34.5 MPa
kmol ˜ K
10 atm
5-51
29.9 m3 h
5.79
H 2 : Tc
Pc
(33.3 8) K = 41.3 K
1 - butene: Tc
419.6 K
Pc
39.7 atm
(12.8 8) atm = 20.8 atm
. (413
. K) + 0.85(419.6 K) = 362.8 K
Tc ' 015
. (20.8 atm) + 0.85(39.7 atm) = 36.9 atm
Pc ' 015
= znRT
V
P
0.86 35 kmol 0.08206 m3 ˜ atm 323 K 1 h
h
kmol ˜ K 10 atm 60 min
CH 3 :
Tc
190.7 K Pc
458
. atm
C 2 H 4 : Tc
2831
. K Pc
50.5 atm
C 2 H 6 : Tc
305.4 K Pc
48.2 atm
UV
W
Fig. 5.4-2
Ÿ z = 0.86
133
. m3 / min
d
i FG 100 cmIJ
S b150 m / ming H m K
F I FG IJ d i
GH JK H K
4 133
. m 3 / min
3
2
m = u m A m 2 = u u Sd Ÿ d = 4V
V
Su
4
min
min
5.80
Tr ' 0.89
Pr ' 0.27
U|
V|
. W
35
T=90 o C
. (190.7 K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====>
Tc ' 015
10.6 cm
.
Tr ' 132
P=175 bar
. (458
. atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> Pr '
Pc ' 015
5.4-3
Fig.

o z = 0.67
F I FG IJ d i FG
GH JK H K
H
3
m = u m A m 2 = 10 m
V
s
s
s
n =
5.81
PV
zRT
N2:
IJ FG 60 s IJ S b0.02 mg
K H min K 4
2
.
0188
kmol ˜ K 0188
.
m3 / min
175 bar 1 atm
0.67 1.013 bar .08206 m3 ˜ atm
363 K
Tc
acetonitrile: Tc
126.2 K = 227.16o R Pc
335
. atm
548 K = 986.4 o R
47.7 atm
Pc
Tank 1 (acetonitrile): T1 = 550o F, P1 = 4500 psia Ÿ Tr1
Ÿ n 1 =
P1 V1
z 1 RT1
306 atm 0.200 ft 3
0.80 1009.7 o R
P2 V2
z 2 RT2
10.0 atm 2.00 ft 3
1.00 1009.7 o R
5-52
. kmol / min
163
Fig. 5.4-3
Pr1
.
102
6.4 Ÿ z1 = 0.80
lb - mole ˜ o R
= 0.104 lb - mole
0.7302 ft 3 ˜ atm
Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm Ÿ Tr2
Ÿ n 2 =
m3
min
Fig. 5.4-3
4.4 , Pr2
6.4 Ÿ z 2 = 1.00
lb - mole ˜ o R
= 0.027 lb - mole
.7302 ft 3 ˜ atm
5.81 (cont’d)
. I
.027 I
FG 0104
J 227.16 R = 830 R
J 986.4 R + FGH 00131
H 0131
. K
. K
. I
F 0.027 IJ 33.5 atm = 44.8 atm
FG 0104
47.7 atm + G
J
H 0131
K
H 0131
.
. K
o
Final: Tc '
Pc '
dV i
r
P=
=
ideal
o
o
T=550

Fo
o
Tr ' 122
.
'
Fig. 5.4-2
VP
2.2 ft 3
44.8 atm lb - mole ˜ o R
c
Ÿ z = 0.85
=
=
1.24
RTc ' 0.131 lb - mole 830 o R 0.7302 ft 3 ˜ atm
0.85 0131
.
lb - mole .7302 ft 3 ˜ atm 1009.7 o R
lb - mole ˜ o R 2.2 ft 3
znRT
V
37.3 atm
5.82
3.48 g Ca H b O c , 26.8o C, 499.9 kPa
n c (mol C), n H (mol H), n O (mol O)
1 L @483.4 o C, 1950 kPa
n p (mol)
0.387 mol CO 2 / mol
0.258 mol O 2 / mol
0.355 mol H 2 O / mol
n O 2 (mol O 2 )
26.8o C, 499.9 kPa
a.
d
Volume of sample: 3.42 g 1 cm3 159
. g
i
2.15 cm3
O 2 in Charge:
d
1.000 L 2.15 cm 3 10 3 L km 3
L ˜ atm
0.08206
mol ˜ K
n O2
i
499.9 kPa
1 atm
300 K
101.3 kPa
0.200 mol O 2
Product
1.000 L
1950 kPa
1 atm
L ˜ atm
756.6 K 101.3 kPa
0.08206
mol ˜ K
Balances:
0.310 mol product
np
b
g
O: 2 0.200 n O
b
b
g
C: n C
0.387 0.310
H: n H
2 0.355 0.310
b
gb
g b
g
0.310 2 0.387 2 0.258 0.355 Ÿ n O
0.110 mol O in sample
0.120 mol C in sample
g
0.220 mol H in sample
Assume c 1 Ÿ a 0.120 0.110 1.1 b 0.220 0.110 2
Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20),
etc.
b.
b g
b g
MW 12.01a 1.01b 16.0c 12.01 1.1c 1.01 2c 16.0c
300 MW 350 Ÿ c 10 Ÿ C11 H 20 O 10
5-53
31.23c
bg
C5 H 10 5.83 Basis: 10 mL C5 H 10 l charged to reactor
15
O 2 o 5CO 2 5H 2 O
2
bg
10 mL C5 H 10 l
n1 (mol C5 H 10 )
n 2 (mol air)
0.21 O 2
0.79 N 2
27 o C, 11.2 L, Po (bar)
a.
n1
b
bg
10.0 mL C5 H 10 l
Stoichiometric air: n 2
Po
n 3 (mol CO 2 )
n 4 mol H 2 O(v)
n 5 (mol N 2 )
75.3 bar (gauge), Tad
nRT
V
0.745 g 1 mol
mL
70.13 g
d Ci
o
0.1062 mol C5 H 10
0.1062 mol C5 H 10
7.5 mol O 2
1 mol air
1 mol C 2 H 10
0.21 mol O 2
3.79 mol 0.08314 L ˜ bar 300K
11.2 L
g
mol ˜ K
3.79 mol air
8.44 bars
(We neglect the C5 H 10 that may be present in the gas phase due to evaporation)
Initial gauge pressure 8.44 bar 1 bar
b.
0.1062 mol C 5 H 10
n3
5 mol CO 2
7.44 bar
0.531 mol CO 2
1 mol C 5 H 10
1 mol H 2 O
0.531 mol H 2 O
1 mol CO 2
2.99 mol N 2
0.531 mol CO 2
n4
b g
n5
0.79 3.79
U|
||
V| Ÿ 4.052 mol product gas
||
W
CO 2 : y 3 = 0.531 / 4.052 = 0.131 mol CO 2 / mol, Tc = 304.2 K Pc = 72.9 atm
H 2 O: y 4 = 0.531 / 4.052 = 0.131 mol H 2 O / mol,
N2:
y 5 = 2.99 / 4.052 = 0.738 mol N 2 / mol,
Tc = 647.4 K Pc = 218.3 atm
Tc = 126.2 K Pc = 33.5 atm
. (304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K
Tc ' 0131
. (72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm Ÿ Pr ' 121
.
Pc ' 0131
ideal
V
r
T=
PV
znR
'
VP
c
RTc '
11.2 L 62.9 atm
mol ˜ K
4.052 mol 217.8 K .08206 L ˜ atm
b75.3 1gbars
1.04
112
mol ˜ K
. L
4.052 mol 0.08314 L ˜ bar
5-54
9.7 Ÿ z | 1.04 (Fig. 5.4 - 3)
2439 K - 273 = 2166o C
CHAPTER SIX
6.1
a.
AB: Heat liquid - -V | constant
BC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium
curve as long as some liquid is present. T 100 o C.
CD: Heat vapor - -T increases, V increases .
b. Point B: From Perry’s Handbook, Table 3.8
U H 2 O(l), 80q C 0.9718 g ml , U H 2 O(l), 100q C
bg
10 mL 0.9718 g / mL
H 2 O l , 100q C: V B
0.9584 g / mL
0.9584 g mL
10.14 mL
Point C: H2O (v, 100qC)
n
10 mL 0.9718 g
0.5393 mol
18.02 g
nRTC 0.5393 mol 0.08206 L ˜ atm 373 K
16.5 L
1 atm
mol ˜ K
PC
mL
nRTC Ÿ VC
PCVC
6.2
1 mol
a. Pfinal
243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the
vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the
species at the system temperature.
b. Assuming ideal gas behavior for the vapor,
m(vapor)
m(liquid)
6.3
a.
mol ˜ K
(3.000 - 0.010) L
(30 + 273.2) K
10 mL
0.08206 L ˜ atm
1.489 g
mL
m(vapor) + m(liquid) = 19.5 g
x vapor =
4.59
19.48
b. ln p
B
119.39 g
760 mm Hg
mol
0.235 g vapor / g total
1238.71
45 217
2.370 Ÿ p *
'H v 1
'H v
BŸ
R T
R
ln p2* / p1*
7.09808 ln( p1* ) 1 atm
14.89 g
m total
log 10 p
243 mm Hg
'H v / R
T1
b
d
1
T2
1
T1
10 2.370
i
b
6-1
g
ln 760 / 118.3
b
1
77 .0 273.2 K
K
g b29.54151
273.2gK
ln 118.3 234.5 mm Hg
g
18.49
b
1
29 .5 273.2 K
g
4151K
4.59 g
6.3 (cont’d)
ln p (45o C)
b
4151
18.49 Ÿ p
45 273.2
g
. 234.5
2310
u 100%
234.5
c.
327.7 234.5
u 100%
234.5
b
327.7 mm Hg
39.7% error
g
1
(rect. scale) on semilog paper
T 273.2
Ÿ straight line: slope 7076K , intercept 2167
.
Plot p log scale vs
b
g
ln p mm Hg
' Hv
R
b
7076K Ÿ ' H v
7076 K
8.314 J
g
exp
1 kJ
mol ˜ K 10 3 J
LM 7076 2167
O
. P
N T ( C) 2732.
Q
o
58.8 kJ mol
ln p* = A/T(K) + B
p*(mm Hg)
5
20
40
100
400
760
1/T(K)
0.002834
0.002639
0.002543
0.002410
0.002214
0.002125
ln(p*) p*(fitted)
1.609
5.03
2.996
20.01
3.689
39.26
4.605
101.05
5.991
403.81
6.633
755.13
7
6
5
4
3
2
1
0
1/T
6-2
0.003
0.0028
0.0026
0.0024
y = -7075.9x + 21.666
0.002
o
T( C)
79.7
105.8
120.0
141.8
178.5
197.3
ln(p*)
6.5
7076
2167
. Ÿ p mm Hg
T ( C) 2732
.
o
0.0022
6.4
15%
.
error
FG 118.3 760IJ b45 29.5g 118.3
H 29.5 77 K
p
. mm Hg
2310
o
T( C) p*(fitted)
50
0.80
80
5.12
110
24.55
198
760.00
230 2000.00 Least confidence
(Extrapolated)
6.6
a.
T(°C)
1/T(K)
42.7
58.9
68.3
77.9
88.6
98.3
105.8
3.17u10-3
3.01u10-3
2.93u10-3
2.85u10-3
2.76u10-3
2.69u10-3
2.64u10-3
p*(mm Hg)
=758.9 + hright -hleft
34.9
78.9
122.9
184.9
282.9
404.9
524.9
b. Plot is linear, ln p
'H v
B Ÿ ln p
RT
760 mmHg Ÿ Tb
At the normal boiling point, p
' H v
5143.8 K
19.855
T
8.314 J 5143.8 K
1 kJ
mol ˜ K
10 3 J
116q C
42.8 kJ mol
c. Yes — linearity of the ln p vs 1 / T plot over the full range implies validity.
6.7
a.
b
g
a T 273.2 b Ÿ y
ln p
ax b
y
b
ln p ; x 1 T 273.2
g
Perry' s Handbook, Table 3 - 8:
400 mm Hg Ÿ x1
39.5q C , p1
T1
760 mm Hg Ÿ x 2
56.5q C , p2
T2
50q C Ÿ x 3.0941 u 10
x x1
y1 y 2 y1
x 2 x1
T
FG
H
y
IJ b
K
u 10 3 , y1
31980
.
3
3.0331 u 10 , y 2
5.99146
6.63332
3
g
b
6.39588 Ÿ p 50q C
g
e 6.39588
599 mm Hg
b. 50q C 122q F
12 psi 760 mm Hg
Cox Chart Ÿ p
c.
6.8
7.02447 log p
b
.
11610
50 224
2.7872 Ÿ p
g
Estimate p 35q C : Assume ln p
a
b
ln p2
1
T2
p1
1
T1
b ln p1 a
T1
g
625 mm Hg
14.6 psi
b
g
ln 200 50
1
45 273.2
b g
6577.1
25 273.2
ln 50 1
25 273.2
10 2.7872
613 mm Hg
a
b , interpolate given data.
T K
b g
U| ln p b35q Cg 6577.1 25.97 4.630
35 273.2
|V Ÿ
25.97 |
p b35q Cg e
102.5 mm Hg
|W
6577.1
4 .630
6-3
6.8 (cont’d)
Moles in gas phase: n
150 mL
8.0 u 10
6.9
a.
m 2 S
2Ÿ F
b
4
222
273 K
102.5 mm Hg
1L
1 mol
3
35 + 273.2 K 760 mm Hg 10 mL 22.4 L STP
g
b g
mol
2.
Two intensive variable values (e.g., T & P) must be
specified to determine the state of the system.
1209.6
2.5107 Ÿ p MEK 10 2.5107 324 mm Hg
b. log p MEK 6.97421 55 216.
Since vapor & liquid are in equilibrium p MEK p MEK 324 mm Hg
Ÿ y MEK
p MEK / P
324 1200 0.27 ! 0115
.
The vessel does not constitute an explosion
hazard.
6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with
a flash point of 15qC should always be prevented from contacting air at room temperature. The
other one should be kept from any heating sources when contacted with air.
b. At the LFL, y M 0.06 Ÿ p M p *M 0.06 u 760 mm Hg = 45.60 mm Hg
1473.11
Ÿ T 6.85q C
Antoine Ÿ log 10 45.60 = 7.87863 T + 230
c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point.
6.11 a. At the dew point,
p ( H 2 O) = p( H 2 O) = 500 u 0.1 = 50 mm Hg Ÿ T = 38.1q C from Table B.3.
b. VH2O
30.0 L
0.100 mol H2 O 18.02 g 1 cm3
(50 + 273) K 760 mm Hg 22.4 L (STP)
mol
mol g
273 K
500 mm Hg
1 mol
c. (iv) (the gauge pressure)
6-4
134
. cm3
6.12 a.
T1
58.3q C , p1
T2
110q C , p2
a
b
T K
b g
ln p
b
ln p2
a
b
60 mm Hg
b
g
755 mm Hg b577 222gmm Hg 400 mm Hg
755 mm Hg 747 52 mm Hg
1
T2
p1
ln p1 g
b
a
T1
g
ln 400 60
1
T1
1
110 273.2
b g
ln 60 46614
.
58.31273.2
4661.4
58.3 273.2
18.156
T=130oC=403.2 K
46614
.
18156
.
T
ln p 130q C 6.595 Ÿ p 130q C e 6.595 7314
. mm Hg
ln p
b
b.
g
b
g
Basis: 100 mol feed gas CB denotes chlorobenzene.
n1 mol @ 58.3qC, 1atm
y1 (mol CB(v)/mol) (sat’d)
(1-y1) (mol air/mol)
100 mol @ 130qC, 1atm
y0 (mol CB(v)/mol) (sat’d)
(1-y0) (mol air/mol)
n2 mol CB (l)
b
g
731 mm Hg
760 mm Hg
pCB 130q C Ÿ y o
Saturation condition at inlet: y o P
0.962 mol CB mol
b
60 mm Hg
0.0789 mol CB mol
g
760 mm Hg
Air balance: 100b1 y g n b1 y g Ÿ n b100gb1 0.962g b1 0.0789g 4.126 mol
Total mole balance: 100 n n Ÿ n 100 4.126 9587
. mol CBbl g
pCB 58.3q C Ÿ y1
Saturation condition at outlet: y1 P
o
1
1
1
% condensation:
1
2
2
95.87 mol CB condensed
u 100%
0.962 u 100 mol CB feed
b
g
99.7%
c. Assumptions: (1) Raoult’s law holds at initial and final conditions;
(2) CB is the only condensable species (no water condenses);
(3) Clausius-Clapeyron estimate is accurate at 130qC.
6.13 T
78q F = 25.56q C , Pbar
y H 2O P
b
0.87 p 25.56q C
d i
Dew Point: p Tdp
yp
29.9 in Hg = 759.5 mm Hg , hr
g
Table B.3
y H 2O
b
g
0.0293 759.5
87%
b
0.87 25544
.
mm Hg
759.5 mm Hg
22.22 mm Hg
6-5
Table B.3
g
0.0293 mol H 2 O mol air
Tdp
23.9q C
6.13 (cont’d)
hm
ha
hp
0.0293
1 0.0293
0.0302 mol H 2 O mol dry air
0.0293 mol H 2 O 18.02 g H 2 O
mol dry air
b
p 2556
. qC
mol H 2 O
hm
g
mol dry air
b
29.0 g dry air
P p 2556
. qC
g
0.0182 g H 2 O g dry air
0.0293
u 100 84%
24.544 759.5 24.544
u 100%
6.14 Basis I : 1 mol humid air @ 70q F (21.1q C), 1 atm, hr
hr
50% Ÿ y H 2 O P
Table B.3
Mass of air:
y H 2O
0.50 p H 2 O
b21.1q Cg
0.50 u 18.765 mm Hg
760.0 mm Hg
0.012 mol H 2 O 18.02 g
1 mol
Volume of air:
mol H 2 O
mol
0.012
0.988 mol dry air 29.0 g
1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP
1 mol
28.87 g
24.13 L
Density of air
50%
273.2K
80% Ÿ y H 2 O P
Table B.3
Mass of air:
y H 2O
0.80 p H 2 O
Density of air
80%
b21.1q Cg
0.80 u 18.765 mm Hg
760.0 mm Hg
0.020 mol H 2 O 18.02 g
1 mol
Volume of air:
0.020
mol H 2 O
mol
0.980 mol dry air 29.0 g
1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP
1 mol
28.78 g
24.13 L
273.2K
80% Ÿ y H 2 O P
Table B.3
y H 2O
24.13 L
g L
1193
.
Basis III: 1 mol humid air @ 90q F (32.2q C), 1 atm, hr
hr
24.13 L
g L
1196
.
Basis II: 1 mol humid air @ 70q F (21.1q C), 1 atm, hr
hr
28.87 g
0.80 p H 2 O
80%
b32.2q Cg
0.80 u 36.068 mm Hg
760.0 mm Hg
6-6
0.038
mol H 2 O
mol
28.78 g
6.14 (cont’d)
Mass of air:
0.038 mol H 2 O 18.02 g
Volume of air:
Density
1 mol
0.962 mol dry air 29.0 g
1 mol
b g b273.2 32.2gK
1 mol
28.58 g
25.04 L
22.4 L STP
1 mol
273.2K
28.58 g
25.04 L
g L
1141
.
Increase in T Ÿ increase in V Ÿ decrease in density
Increase in hr Ÿ more water (MW = 18), less dry air (MW = 29)
Ÿ decrease in m Ÿ decrease in density
Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force
on the ball must also be lower. Therefore, the statement is wrong.
6.15 a. hr
50% Ÿ y H 2 O P
Table B.3
0.50 p H 2 O
0.50 u 525.76 mm Hg
760.0 mm Hg
y H 2O
Dew Point: y H 2 O p
d i
p Tdp
Degrees of Superheat
b. Basis:
b90q Cg
b g
0.346 mol H 2 O / mol
262.9 mm Hg
0.346 760
1 m 3 feed gas 10 3 L 273K
m
3
mol
b g
n1 mol @ 25qC, 1atm
y1 (mol H2O (v)/mol) (sat’d)
(1-y1) (mol air/mol)
n2 mol H2O (l)
b g 23.756
P
760
Dry air balance: 0.346b33.6g n b1 0.0313g Ÿ n
Saturation Condition: y1
p H* 2 O 25q C
1
g
p * ( 90q C)
y H 2O
0.0313 mol H 2 O mol
1
Total mol balance: 33.6 = 12.0 + n2 Ÿ n2
b
72.7q C
33.6 mol
363K 22.4 L STP
0.346 H2O mol /mol
0.654 mol air/mol
p 90q C Ÿ P
Tdp
90 72.7 17.3q C of superheat
33.6 mol @ 90qC, 1atm
c. y H 2 O P
Table B.3
12.0 mol
216
. mol H 2 O condense / m 3
525.76 mmHg
0.346
6-7
1520 mm Hg = 2.00 atm
6.16 T
90q F = 32.2q C , p
29.7 in Hg = 754.4 mm Hg , hr
95%
Basis: 10 gal water condensed/min
n condensed
1 ft 3
62.43 lb m
7.4805 gal
ft 3
10 gal H 2 O
y2 (lb-mol H2O (v)/lb-mol) (sat’d)
(1-y2) (lb-mol DA/lb-mol)
40oF (4.4oC), 754 mm Hg
y1 (lb-mol H2O (v)/lb-mol) (sat’d)
(1-y1) (lb-mol DA/lb-mol)
hr=95%, 90oF (32.2oC),
29.7 in Hg (754 mm Hg)
Raoult' s law: y 2 P
4.631 lb-moles H2O (l)/min
b
0.95 p 32.2q C
95% hr at inlet: y H 2 O P
y H 2O
4.631 lb - mole / min
n2 (lb - moles / min)
V1 (ft 3 / min)
n1 (lb - moles / min)
Table B.3
1 lb - mol
18.02 lb m
b
g
0.95 36.068 mm Hg
g
0.045 lb - mol H 2 O lb - mol
754.4 mm Hg
b
g
Table B.3
p * 4.4q C
y2
6.292
754.4
UV RS
W T
n1
Mole balance: n1 n 2 4.631
Ÿ
n 2
Water balance: 0.045n1 0.00834n 2 4.631
0.00834 lb - mol H 2 O lb - mol
125.3 lb - moles / min
120.6 lb - moles / min
125.3 lb - moles 359 ft 3 (STP) (460 + 90) o R 760 mm Hg
Volume in: V =
min
lb - moles
492 o R
754 mm Hg
5.07 u 10 4 ft 3 / min
6.17 a. Assume no water condenses and that the vapor at 15qC can be treated as an ideal gas.
760 mm Hg
p final
(15 273) K
(200 + 273) K
462.7 mm Hg Ÿ ( p H 2 O ) final
0.20 u 462.7
p * (15q C) = 12.79 mm Hg < p H 2 O . Impossible Ÿ condensation occurs.
( pair ) final
P
( pair ) initial
p H 2O pair
b. Basis:
Tfinal
Tinitial
(0.80 u 760) mm Hg u
370.2 12.79
1 L 273 K
383 mm Hg
mol
473 K 22.4 L (STP)
0.0258 mol
6-8
288 K
473 K
370.2 mm Hg
92.6 mm Hg
6.17 (cont’d)
n1 mol @ 15qC,
383.1 mm Hg
y1 (mol H2O (v)/mol) (sat’d)
(1-y1) (mol dry air/mol)
0.0258 mols @ 200qC,
760 mm Hg
0.20 H2O mol /mol
0.80 mol air/mol
n2 mol H2O (l)
b
p H* 2 O 15q C
Saturation Condition: y1
12.79 mm Hg
. mm Hg
3831
P
b
g n b1 0.03339g Ÿ n
Dry air balance: 0.800 0.0258
c.
g
1
Mass of water condensed =
0.02135 mol
1
Total mole balance: 0.0258 = 0.02135 + n2 Ÿ n2
0.00445 mol
0.03339 mol H 2 O mol
0.00445 mol
18.02 g
mol
0.0802 g
6.18 Basis: 1 mol feed
3
n2 (mol), 15.6°C, 3 atm
y 2 (mol H 2 O (v)/mol)(sat'd)
(1 – y 2) (mol DA/mol)
V1 (m )
1 mol, 90°C, 1 atm
0.10 mol H 2O (v)/mol
0.90 mol dry air/mol
heat
100°C, 3 atm
n2 (mol)
3
V2 (m )
n3 (mol) H 2 O( l ), 15.6°C, 3 atm
b
p H* 2 O 15.6q C
Saturation: y 2
g
Table B.3
P
atm
13.29 mm Hg
0.00583
3 atm
760 mm Hg
y2
b g n b1 0.00583g Ÿ n
H O mol balance: 0.10b1g 0.00583b0.9053g n
Dry air balance: 0.90 1
2
2
Fraction H 2 O condensed:
hr
y 2 P u 100%
p 100q C
b
g
0.9053 mol
2
3
Ÿ n3
0.0947 mol condensed
.
0100
mol fed
b
g u 100%
0.00583 3 atm
1 atm
0.0947 mol
0.947 mol condense mol fed
.
175%
n1 mol @ 15qC,
383.1 mm Hg
y1 (mol H2O (v)/mol) (sat’d)
(1-y1) (mol dry air/mol)
0.0258 mols @ 200qC,
760 mm Hg
0.20 H2O mol /mol
0.80 mol air/mol
n2 mol H2O (l)
6-9
6.18 (cont’d)
b g
V2
0.9053 mol 22.4 L STP
mol
V1
1 mol 22.4 L STP
mol
V2
V1
b g
363K 1 m 3
2.98 u 10 2 m 3 feed air @ 90q C
3
273K 10 L
9.24 u 10 3 m 3 outlet air
2.98 u 10 2 m 3 feed air
6.19 Liquid H 2 O initially present:
0.0208
1 0.0208
0.310 m 3 outlet air m 3 feed air
25 L 1.00 kg
b
p H* 2 O 25q C
Saturation at outlet: y H 2 O
Ÿ
373K 1 atm 1 m 3
9.24 u 10 3 m 3 outlet air @ 100q C
273K 3 atm 10 3 L
g
1 kmol
18.02 kg
L
bg
1.387 kmol H 2 O l
23.76 mm Hg
15
. u 760 mm Hg
P
0.0208 mol H 2 O mol air
0.0212 mol H 2 O mol dry air
b g
15 L STP
1 mol
0.670 mol dry air min
min
22.4 L STP
0.670 mol dry air 0.0212 mol H 2 O
0.0142 mol H 2 O min
Evaporation Rate:
min
mol dry air
Flow rate of dry air:
Complete Evaporation:
b g
min
1.387 kmol 10 3 mol
1h
1628 h
0.0142 mol 60 min
kmol
b67.8 daysg
7.069 u 10 3 ft 3 7.481 gal
S
2
5.288 u 104 gal / day
6.20 a. Daily rate of octane use = ˜ 30 ˜ (18 8)
3
4
day
ft
( SG ) C8 H18
b.
'p
1 ft 3
0.703 u 62.43 lb m
5.288 u 10 4 gal
day
7.481 gal
ft 3
0.703 Ÿ
3.10 u 105 lb m C 8 H 18 / day
0.703 u 62.43 lb m
ft
32.174 ft
3
c. Table B.4: pC* 8 H18 (90 o F)
s
2
1 lb f
lb ˜ ft
32.174 m2
s
20.74 mm Hg
(18 - 8) ft
14.696 psi
29.921 in Hg
14.696 lb f / in 2
0.40 lb f / in 2
6.21 in Hg
poctane y octane P
760 mm Hg
Octane lost to environment = octane vapor contained in the vapor space displaced by liquid
during refilling.
Volume:
5.288 u 10 4 gal
1 ft 3
7.481 gal
7069 ft 3
6-10
6.20 (cont’d)
(16.0 + 14.7) psi
7069 ft 3
36.77 lb - moles
10.73 ft 3 ˜ psi / (lb - mole ˜ o R) (90 + 460) o R
pC H
0.40 psi
Mole fraction of C 8 H 18 : y = 8 18
0.0130 lb - mole C 8 H 18 / lb - mole
P
(16.0 + 14.7) psi
pV
RT
Total moles: n
Octane lost
0.0130(36.77) lb - mole 0.479 lb - mole ( 55 lb m
25 kg)
d. A mixture of octane and air could ignite.
*
*
(85o F) = ptol
(29.44 o C) = 35.63 mmHg = ptol
6.21 a. Antoine equation Ÿ ptol
ptol
P
Mole fraction of toluene in gas: y
yPV
RT
0.0469 lb - mole tol
Toluene displaced
35.63 mmHg
760 mmHg
0.0469 lb - mole toluene / lb - mole
yntotal
1 atm
lb - mole
0.7302
1 ft 3
900 gal
ft ˜ atm
3
lb - mole ˜ R
o
(85 460) o R 7.481 gal
92.13 lb m tol
lb - mole
1.31 lb m toluene displaced
b.
Basis: 1mol
0.0469 mol C7H8(v)/mol
0.9531 mol G/mol
nV (mol)
y (mol C7H8(v)/mol)
(1-y) (mol G/mol)
T(oF), 5 atm
Assume G is
noncondensable
nL [mol C7H8 (l)]
90% of C7H8 in feed
90% condensation Ÿ n L
0.90(0.0469)(1) mol C 7 H 8
Mole balance: 1 nV 0.0422 Ÿ nV
Toluene balance: 0.0469(1)
Raoult’s law:
ptol
0.0422 mol C 7 H 8 (l )
0.9578 mol
y (0.9578) 0.0422 Ÿ y
0.004907 mol C 7 H 8 / mol
*
(0.004907)(5 u 760) 18.65 mmHg = ptol
(T )
yP
Antoine equation:
T
B C( A log 10 p * )
A log 10 p
*
1343.943 219.377(6.95334 log 10 18.65)
6.95334 log 10 18.65
6-11
17.12 o C = 62.8 o F
6.22 a. Molar flow rate: n =
VP
RT
100 m 3
h
kmol ˜ K
2 atm
3
82.06 u 10 m ˜ atm (100 + 273) K
-3
6.53 kmol / h
b. Antoine Equation:
1171.530
100 + 224.366
1845 mm Hg
log 10 p *Hex (100q C) = 6.87776 Ÿ p*
p Hex
y Hex ˜ P
0.150(2.00) atm 760 mm Hg
*
Ÿ not saturated
228 mm Hg < p Hex
atm
228 mm Hg Ÿ log 10 228 = 6.87776 -
p *Hex (T )
3.26601
1171.530
T + 224.366
2.35793 Ÿ T
34.8q C
nV (kmol/h)
y (kmol C6H14 (v)/kmol), sat’d
(1-y) (kmol N2/kmol)
T (oC), 2 atm
c.
6.53 kmol/h
0.15 C6H14 (v)
0.85 N2
nL (kmol C6H14 (l)/h)
80% of C6H14 in feed
n L 0.80(0.15)(6.53 kmol / h) = 0.7836 kmol C 6 H 14 (l ) / h
80% condensation:
Mole balance: 6.53 nV 0.7836 Ÿ nV 5.746 kmol / h
Hexane balance:
015
. (6.53) y (5.746) 0.7836 Ÿ y 0.03409 kmol C 6 H 14 / kmol
(0.03409)(2 u 760 mmHg) = 51.82 mmHg = p *Hex (T )
1171530
.
Ÿ T 2.52 o C
6.87776 Antoine equation: log 10 5182
.
T 224.366
Raoult’s law:
p Hex
yP
6.23 Let H=n-hexane
a.
n0 ( kmol / min)
y0 (kmol H(v)/kmol
(1-y0) (kmol N2/kmol)
80oC, 1 atm, 50% rel. sat’n
50% relative saturation at inlet: y o P
Table B.4
yo
Condenser
n1 ( kmol / min)
0.05 kmol H(v)/kmol, sat’d
0.95 kmol N2/kmol
T (oC), 1 atm
1.50 kmol H(l)/min
0.500 p H* (80 o C)
(0.500)(1068 mmHg)
= 0.703 kmolH / kmol
760 mmHg
Saturation at outlet: 0.05 P
p H* (T1 ) Ÿ p H* (T1 )
6-12
0.05(760 mmHg) = 38 mmHg
6.23 (cont’d)
Antoine equation: log 10 38 6.87776 .
1171530
Ÿ T1
T1 224.366
UV RS
W T
Mole balance: n 0 n1 150
.
n 0
Ÿ
N 2 balance: (1 0.703)n 0 0.95n1
n1
N2 volume: VN 2
b.
3.24 o C
2.18 kmol / min
0.682 kmol / min
(0.95)0.682 kmol 22.4 m 3 (STP)
14.5 SCMM
min
kmol
Assume no condensation occurs during the compression
2.18 kmol/min
0.703 H(v)
0.297 N2
80oC, 1 atm
Compressor
V0 ( m3 / min)
2.18 kmol/min
0.703 H(v)
0.297 N2
T0 (oC), 10 atm, 50% R.S.
V1 (m 3 / min)
0.682 kmol/min
0.05 H(v), sat’d
0.95 N2
T1 (oC), 10 atm
Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet:
0.500 p *H (T0 )
u 10 4 mmHg
0.703(7600 mmHg) Ÿ p H* (T0 ) 1068
.
Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p H* (T1 )
Volume ratio:
V1
V0
n1 RT1 / P
n0 RT0 / P
n1 (T1 273.2)
n0 (T0 273.2)
Antoine
Antoine
0.682 kmol / min 1321 K
u
2.18 kmol / min
460 K
T0
187 o C
T1
48.2q C
m 3 out
0.22 3
m in
c. The cost of cooling to 3.24 o C (installed cost of condenser + utilities and other operating
costs) vs. the cost of compressing to 10 atm and cooling at 10 atm.
6.24 a.
Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes.
(SG)nonane
n max
Assume T
n gas
15 L C 9 H 20 (l ) 0.718 u 1.00 kg
L C 9 H 20
kmol
128.25 kg
25o C, P = 1 atm
1
kmol
2 u 10 4 L 273 K
3
298 K 22.4 u 10 L(STP)
6-13
0.818 kmol
0.084 kmol C 9 H 20
6.24 (cont’d)
y max
n max
n gas
0.084 kmol C 9 H 20
0.818 kmol
010
. kmol C 9 H 20 / kmol (10 mole%)
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to
2.9%).
The answer is therefore yes .
The nonane will not spread uniformly—it will be high near the sump as long as liquid is present
(and low far from the sump). There will always be a region where the mixture is explosive at
some time during the evaporation.
b. ln p *
A
B
T
T1
. o C = 299 K, p1*
258
5.00 mmHg
T2
66.0 o C = 339 K, p2*
40.0 mmHg
5269
ln( 40.0 / 5.00)
5269
Ÿ A 5269, B = ln(5.00) +
19.23 Ÿ p * exp(19.23 )
1
1
T ( K)
299
339 299
At lower explosion limit, y 0.008 kmol C 9 H 20 / kmol Ÿ p * (T ) yP (0.008)(760 mm Hg)
A
= 6.08 mm Hg
Formula for p
T
*
302 K = 29 o C
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather
than air is to make sure an explosive mixture of nonane and oxygen is never present in the
tank. Before anyone goes into the tank, a sample of the contents should be drawn and
analyzed for nonane.
6.25 Basis: 24 hours of breathing
n0 (mol H 2 O)
23°C, 1 atm
n1 (mol) @ hr = 10%
0.79 mol N 2/mol
y 1 (mol H 2 O/mol)
+ O2 , CO2
Lungs
O2
Air inhaled: n1
37°C, 1 atm
n2 (mol), saturated
0.75 mol N 2/mol
y 2 (mol H 2 O/mol)
+ O2 , CO2
CO2
12 breaths 500 ml 1 liter
min
273K
1 mol
23 273 K 22.4 liter STP
10 ml
b
b23q Cg
. 2107
. mm Hg
010
3
breath
g
b g
60 min 24 hr
1 hr
1 day
356 mol inhaled day
Inhaled air - -10% r. h.: y1
Inhaled air - -50% r. h.: y1
. p
010
H 2O
b
P
0.50 p
H 2O
g
760 mm Hg
b23q Cg
b
. mm Hg
0.50 2107
P
760 mm Hg
6-14
g
2.77 u 10 3
mol H 2 O
mol
. u 10 2
139
mol H 2 O
mol
6.25 (cont’d)
n2 y 2 n1 y1 Ÿ (n0 ) 10% rh (n0 ) 50% rh
H 2 O balance: n0
(n1 y1 ) 50% (n1 y1 ) 10%
FG 356 mol IJ L(0.0139 0.00277) mol H O OFG 18.0 g IJ
mol PQH 1 mol K
H day K MN
2
71 g / day
Although the problem does not call for it, we could also calculate that n2 = 375 mol exhaled/day,
y2 = 0.0619, and the rate of weight loss by breathing at 23oC and 50% relative humidity is
n0 (18) = (n2y2 - n1y1)18 = 329 g/day.
6.26 a. To increase profits and reduce pollution.
b.
Assume condensation occurs. A=acetone
n 1 mol @ To C, 1 at m
1 mo l @ 90o C, 1 atm
y1 mol A(v)/ mol (sat’d)
(1-y1 ) mol N2 /mo l
0.20 mol A(v)/ mol
0.80 mol N2 /mo l
n 2 mol A(l)
For cooling water at 20oC
d
log 10 p *A 20 o C
i
Saturation: y1 ˜ P
7.02447 d
d
.
11610
20 224
i
p *A 20 o C Ÿ y1
2.26627 Ÿ p *A 20 o C
184.6
760
i
184.6 mmHg
0.243 ! 0.2 , so no saturation occurs.
For refrigerant at –35oC
d
log 10 p *A 35o C
i
7.02447 11610
.
35 224
d
0.88161 Ÿ p *A 35o C
i
7.61 mmHg
7.61
0.0100
d i
760
N mole balance: 1b0.8g n b1 0.01g Ÿ n 0.808 mol
Saturation: y1 ˜ P
2
p *A 35o C Ÿ y1
1
1
Total mole balance: 1 0.808 n2 Ÿ n2
Percentage acetone recovery:
c.
d.
0.192 mol
.
0192
u 100%
1
96%
Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant
The condenser temperature could never be as low as the initial cooling fluid temperature
because heat is transferred between the condenser and the surrounding environment. It
will lower the percentage acetone recovery.
6-15
6.27
Basis:
12500 L
1 mol
273 K 103000 Pa
h
22.4 L(STP) 293 K 101325 Pa
528.5 mol / h
528.5 (mo l/h) @ 20o C, 103 KPa
y0 [mol H2O(v)/mol]
y1–
H2 O(v)/
mol (sat’d)
1 mol
y0 (mol
DA/mol)
(1-y1 ) mol DA/mo l
n o (mol/h) @ 35o C, 103 KPa
y0 [mol H2O(v)/mol]
y1–
H2 O(v)/
mol
o mol
y0 (mol
DA/mol)
)
mol
DA/mo
l
(1-y
o
hr=90%
h r =90%
[molH2O(l)/h
H2O(l)/h]
n 2n2mol
Inlet: y o
d
hr ˜ p H* 2 O 35o C
P
d
p H* 2 O 20 o C
i
i
0.90 u 42.175 mmHg 101325 Pa
103000 Pa
760 mmHg
0.4913 mol H 2 O / mol
17.535 mmHg 101325 Pa
0.02270 mol H 2 O / mol
P
103000 Pa 760 mmHg
Dry air balance: 1 0.04913 no 1 0.02270 528.5 Ÿ no 543.2 mol / h
Outlet: y1
b
g b
gb
g
543.2 mol 22.4 L(STP) 308 K 101325 Pa
13500 L / h
h
mol
273 K 103000 Pa
Total balance: 543.2 528.5 n2 Ÿ n2 14.7 mol / h
Inlet air:
14.7 mol 18.02 g H 2 O 1 kg
h
1 mol H 2 O 1000 g
Condensation rate:
6.28
Basis:
0.265 kg / h
10000 ft 3 1 lb - mol 492 o R 29.8 in Hg
min 359 ft 3 (STP) 550 o R 29.92 in Hg
24.82 lb - mol / min
n1 lb-mole/min
40oF, 29.8 in.Hg
y1 [lb-mole H2O(v)/lb-mole]
1- y1 (lb-mole DA/mol)
24.82 lb-mole/min
90oF, 29.8 in.Hg
y0 [lb-mole H2O(v)/mol
1- y0 (lb-mole DA/mol)
hr = 88%
n1 lb-mole/min
65oF, 29.8 in.Hg
y1 [lb-mole H2O(v)/lb-mole]
1- y1 (lb-mole DA/lb-mole)
n2 [lb-mole H2O(l)/min]
Inlet: y o
Outlet: y1
d
i
b
hr ˜ p H* 2O 90 o F
0.88 36.07 mmHg
P
29.8 in Hg
d
i
p H* 2 O 40 o F
P
b
1 in Hg
25.4 mmHg
6.274 mmHg 1 in Hg
29.8 in Hg 25.4 mmHg
Dry air balance: 24.82 1 0.0419
Total balance: 24.82
g
0.00829 lb - mol H 2 O / lb - mol
g n b1 0.00829g Ÿ n
1
23.98 n2 Ÿ n2
1
23.98 lb - mol / min
0.84 lb - mole / min
6-16
0.0419 lb - mol H 2 O / lb - mol
6.28 (cont’d)
0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal
min
lb mol 62.4 lb m 1 ft 3
Condensation rate:
Air delivered @ 65oF:
181
. gal / min
23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg
min
1 lb mol 492 o R 29.8 in Hg
9223 ft 3 / min
6.29 Basis: 100 mol product gas
no mol, 32oC, 1 atm
yo mol H2O(v)/mol
(1-yo) mol DA/mol
hr=70%
100 mol, T1, 1 atm
100 mol, 25oC,1 atm
y1 mol H2O(v)/mol, (sat’d)
(1-y1) mol DA/mol
y1 mol H2O(v)/mol,
(1-y1) mol DA/mol
hr=55%
(mol HH22O(l))
nn22lb-mol
O(l)/min
Outlet: y1
d
hr ˜ p H* 2 O 25o C
i
b
g
0.55 23.756
P
b g 13.07 p bT g Ÿ T 15.3 C
d32 Ci 0.70b35.663g 0.0328 mol H O / mol
hr ˜ p H* 2O
b
g
Total balance: 1016
. n2
Ratio:
b
2
b
g
100 1 0.0172 Ÿ no
100.0 Ÿ n2
16
. mol 18.02 g 1 kg
1 mol 1000 g
g
1016
. mol
. mol (i.e. removed)
16
0.0288 kg H 2 O
100 1 0.0172 mol 29.0 g 1 kg
1 mol 1000 g
0.0288
2.85
1
760
Dry air balance: no 1 0.0328
kg dry air:
1
o
P
kg H 2 O removed :
o
*
H 2O
Saturation at T1 : 0.0172 760
Inlet: y o
0.0172 mol H 2 O / mol
760
2.85 kg dry air
0.0101 kg H 2 O removed / kg dry air
6-17
6.30 a.
Room air T
y1 P
0.40 P
22q C , P 1 atm , hr
H 2O
g19.827 mm Hg
b22q Cg Ÿ y b0.40760
mm Hg
b
P
H 2O
b50q Cg Ÿ y
9251 mm Hg
839 mm Hg
2
bH ln a œ y
b
0.01044g
.
g lnb01103
H 2 H1
Ÿ y
Basis:
ae bH , y1
b
0.01044, H1
g b
gb g
expb0.054827 H g
ln 0.01044 0.054827 5
7.937 u 10 3
1 m 3 delivered air
b
n o mol, 35o C, 1 at m
yo mol H2 O(v)/ mol
(1-yo ) mol DA/mo l
H=30
01103
.
mol H 2 O mol
5 , y2
01103
.
, H2
48
0.054827
48 5
ln a ln y1 bH1
b.
50q C , P 839 mm Hg , saturated:
ln y
ln y 2 y1
0.01044 mol H 2 O mol
1
Second sample T
y2 P
40% :
b
g
4.8362 Ÿ a exp 4.8362
273K
1 k mol
22 273 K 22.4m 3 STP
g
b g
10 3 mol
1 kmol
7.937 u 10 3
4131
. mol air delivered
41.31 mol, T, 1 at m
41.4 mol, 22o C,1 at m
0.0104 mo l H2 O(v)/mo l, (sat’d)
0.09896 mo l DA/ mol
0.0104 mo l H2 O(v)/mo l
0.09896 mo l DA/ mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage:
y H 2O P
bg
b g b0.01044gb760 mm Hgg
PH*2 O T Ÿ PH 2 O T
ŸT
7.93 mm Hg Ÿ Table 8.3
7.8q C
bg
Part a
30 Ÿ y 0
Humidity of outside air: H
b
Overall dry air balance: n0 1 V0
Overall water balance: n0 y 0
n2
0.0411 mol H 2 O mol
. gb0.9896g
b4131
b g
b1 0.0411g 42.63 mol
b4131
. gb0.0104g Ÿ n b42.63gb0.0411g b4131
. gb0.0104g
g
. 0.9896 Ÿ n0
4131
2
132
. mol H 2 O condensed
Mass of condensed water
132
. mol H 2 O 18.02 g H 2 O 1 kg
1 mol H 2 O 10 3 g
0.024 kg H 2 O condensed m 3 air delivered
6-18
6.31 a.
Basis: n 0 mol feed gas . S solvent , G
solvent - free gas
n1 (mol) @ Tf (qC), P4 (mm Hg)
y1 [mol S(v)/mol] (sat’d)
(1–y1) (mol G/mol)
n0 (mol) @ T0 (qC), P0 (mm Hg)
y0 (mol S/mol)
(1-y0) (mol G/mol)
Td0 (qC) (dew point)
n2 (mol S (l))
Po
n1 n2 Ÿ n1
n0 y 0 f
n 0 n2
(1)
d i
p Tf
d i
p T f Ÿ y1
Fractional condensation of S = f Ÿ n2
b gb g
p Tdo
p Tdo Ÿ y o
Saturation condition at outlet: y1 Pf
Total mole balance: n 0
b g
b g
Inlet dew point = T0 Ÿ y o Po
(2)
Pf
(1)

o n2
bg
Eq. 3 for n1
Ÿ
n 0
n1
b gP
n fp bT g
n0 fp T0
0
0
do
Po
n1 y1 n2
S balance: n0 y 0
(1) - (4)
b g OPFG p dT iIJ n fp bT g
P
PQGH P JK p
L fp eTdo j OP
p dT i M1 MM Po PP
b1 f g p bT g LM1 fp bT g OP p dT i Ÿ P
N
Q
P
MN P PQ P
p T
b1 f g eP do j
b g LMn
P
MN
n 0 p Tdo
0
n 0 fp Tdo
o
f
o
0
f
do
o
f
Ÿ
do
f
do
f
o
o
f
o
b.
Condensation of ethylbenzene fromnitrogen
Antoine constants for ethylbenzene
A= 6.9572
B= 1424.3
C= 213.21
Run T0
P0 Td0
f
1
2
3
4
50
50
50
50
765
765
765
765
40
40
40
40
0.95
0.95
0.95
0.95
Tf
p* (Td0) p*(Tf)
45
40
35
20
21.492
21.492
21.492
21.492
6-19
27.62
21.49
16.56
7.08
Pf
19137
14892
11472
4904
Crefr Ccomp Ctot
2675
4700
8075
26300
107014 109689
83327 88027
64244 72319
27595 53895
(3)
(4)
6.31 (cont’d)
When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to
c.
increase the fractional condensation. When you decrease Tf, less compression is required to
achieve a specified fractional condensation.
d.
6.32 a.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr).
However, since less compression is required at the lower temperature, Ccomp is lower at the
lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises
the compression cost. The sum of the two costs is a minimum at an intermediate
temperature.
Basis : 120 m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule.
b g P batmg bT g
Cmpd. Tc K
33.2
H2
CO
133.0
304.2
CO 2
CH 4 190.7
Tcc
Pcc
Tr
Pr
V
c corr .
c
12.8
34.5
72.9
458
.
bP g
c corr
413
.
20.8
bApply Newton' s corrections for H g
2
b g b g b g b g 133.4 K
0.40b20.8g 0.35b34.5g 0.20b72.9g 0.05b458
. g 37.3 atm
1273 K 133.4 K 9.54 U Generalized compressibility charts (Fig. 5.3 - 2)
V
Ÿ z 102
35.0 atm 37.3 atm 0.94 W
.
¦yT
¦y P
i ci
i
ci
8.314 N ˜ m
1.02
35 atm
120 m 3
min
0.40 413
. 0.35 133.0 0.20 304.2 0.05 190.7
1273 K
mol ˜ K
mol 1 kmol
3.04 u 10 m 3 10 3 mol
3
1 atm
101325 N m
3
3.04 u 10 3 m 3 mol
39.5 kmol min
n1 (kmol/min), 261 K, 35atm
yNaOH sat’d
yH2
yCH4 (2% of feed)
yCO
1.2(39.5) kmol/min MeOH(l)
39.5 kmol/min, 283K, 35 atm
0.40 mol H2/mol
0.35 mol CO/mol
0.20 mol CO2/mol
0.05 mol CH4/mol
n2 (kmol/min), liquid
yMeOH
yCO2
yCH4 (98% of feed)
6-20
6.32 (cont’d)
p
Saturation at Outlet: y McOH
MeOH
b261K g
b.
n MeOH
g
mm Hg
g
35 atm 760 mm Hg atm
4
mol MeOH mol
n MeOH
n MeOH
n
+ H 2 nCH 4
A
input
b
7 .878631473.11 12 2300
b
P
4.97 u 10
y McOH
10
A
0.02 of input
nCO
A
input
n MeOH 39.5 0.40 0.02 0.05 0.35
E
n MeOH
0.0148 kmol min MeOH in gas
The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy
required to pump it would be wasted.
6.33
n0 (kmol/min wet air) @ 28qC, 760 mmHg
n1 (kmol/min wet air) @ 80qC, 770
y1 (mol H2O/mol)
(1-y1) (mol dry air/mol)
50% rel. sat.
y2 (mol H2O/mol)
(1-y2) (mol dry air/mol)
Tdew point = 40.0oC
1500 kg/min wet pulp
m 1 (kg/min wet pulp)
0.75 /(1 + 0.75) kg H2O/kg
1/1.75 kg dry pulp/kg
0.0015 kg H2O/kg
0.9985 kg dry pulp/kg
Dry pulp balance: 1500 u
1
1 0.75
50% rel. sat’n at inlet: y1 P
m 1 (1 0.0015) Ÿ m 1
0.50 p H* 2 O (20 o C) Ÿ y1
858 kg / min
0.50(28.349 mm Hg) / (760 mm Hg)
= 0.0187 mol H 2 O / mol
o
40 C dew point at outlet: y 2 P
p H* 2 O (40 o C)
Ÿ y2
(55.324 mm Hg) / ( 770 mm Hg)
= 0.0718 mol H 2 O / mol
Mass balance on dry air:
n 0 (1 0.0187) n1 (1 0.0718)
(1)
Mass balance on water:
n 0 ( 0.0187 )(18.0 kg / kmol ) 1500( 0.75 / 1.75)
Solve (1) and (2) Ÿ n 0
622.8 kmol / min, n1
n 1 ( 0.0718)(18) 858( 0.0015) ( 2 )
658.4 kmol / min
Mass of water removed from pulp: [1500(0.75/1.75)–858(.0015)]kg H2O = 642 kg / min
Air feed rate: V0
622.8 kmol 22.4 m 3 (STP) (273 + 28) K
1.538 u 10 4 m 3 / min
kmol
273 K
min
6-21
6.34
Basis: 500 lb m hr dried leather (L)
n1 (lb - moles / h)@130o F, 1 atm
n0 (lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole)
(1- y1 )(lb - moles dry air / lb - mole)
0 (lb m / h)
m
500 lb m / h
0.06 lb m H2 O(l) / lb m
0.61 lb m H2 O(l) / lb m
0.39 lb m L / lb m
0.94 lb m L / lb
b0.94gb500g Ÿ m
Dry leather balance: 0.39m0
0
Humidity of outlet air: y1 P
0.50 p
b gb
H 2 O balance: 0.61 1205 lb m hr
n1
g
H 2O
1205 lb m wet leather hr
b130q Fg Ÿ y
a
0.06 500 lb
E
m
b1 0.0756g(517.5) lb - moles hr
478.4 lb - moles 359 ft bSTPg b140 460gq R
3
6.35 a.
hr
fb
hr 0.0756
mol H 2 O
mol
g
0.0756n1 lb - moles H 2 O 18.02 lb m
hr
1 lb - mole
517.5 lb - moles hr
Dry air balance: n0
Vinlet
0.50(115 mm Hg)
760 mmHg
1
1 lb - mole
492q R
478.4 lb - moles hr
2.09 u 10 5 ft 3 hr
Basis: 1 kg dry solids
n 1 (kmol)N 2, 85°C
n 2 (kmol) 80°C, 1 atm
y 2 (mol Hex/mol)
(1 – y 2) (mols N 2 /mol)
70% rel. sat.
dryer
1.00 kg solids
0.78 kg Hex
condenser
n 3 (kmol) 28°C, 5.0 atm
y 3 (mol Hex/mol) sat'd
(1 – y 3) (mols N 2 /mol)
n 4 (kmol) Hex(l)
0.05 kg Hex
1.00 kg solids
Mol Hex in gas at 80q C
b0.78 0.05gkg
kmol
8.47 u 10 3 kmol Hex
86.17 kg
Antoine eq.
70% rel. sat.: y 2
0.70 p
B
hex
b80q Cg b0.70g10
P
b
6.87776 1171.530 80 224 .366
760
6-22
g
0.984 mol Hex mol
6.35 (cont’d)
n2
8.47 u 10 3 kmol Hex
1 kmol
0.984 kmol Hex
b1 0.984g0.0086
N 2 balance on dryer: n1
0.0086 kmol
u 10 4 kmol
1376
.
Antoine Eq.
p
Saturation at outlet: y 3
B
hex
b28q Cg
10
b
6.877761171.580 28 224 .366
b g
0.0452 mol Hex mol
5 760
P
Overall N 2 balance: 1.376 u 10 -4
b
g
. u 10 4 kmol
144
n3 1 0.0452 Ÿ n3
Mole balance on condenser: 0.0086 144
. u 10 4 n4 Ÿ n4
Fractional hexane recovery:
g
0.0085 kmol cond. 86.17 kg
0.78 kg feed
kmol
0.0085 kmol
0.939 kg cond. kg feed
b. Basis: 1 kg dry solids
0.9n
heater
3
0.9n 3 (kmol) @ 28°C, 5.0 atm
y3
(1 – y3)
y 3 (mol Hex/mol) sat'd
(1 – y 3) (mol N 2/mol)
n 2 (kmol) 80°C, 1 atm
y 2 (mol Hex/mol)
(1 – y 2) (mols N2 /mol)
70% rel. sat.
n 1 (kmol)N 2
85°C
dryer
1.00 kg solids
0.78 kg Hex
condenser
0.1n3
n4 (kmol) Hex(l)
0.05 kg Hex
1.00 kg solids
Mol Hex in gas at 80q C: .8.47 u 10 3 0.9n3 (0.0452)
N2 balance on dryer: n1 0.9n3 (1 0.0452)
Overall N2 balance: n1
n3 (kmol)
y3
(1 – y3)
n2 (0.984)
n2 (1 0.984)
( 2)
0.3n3 (1 0.0452)
R|n
Equations (1) to (3) Ÿ Sn
|Tn
(1)
(3)
5
2
u 10 kmol
1447
.
0.0086 kmol
3
u 10 5 kmol
1515
.
1
Saved fraction of nitrogen =
u 10 5
1.376 u 10 -4 1447
.
u 100%
1.376 u 10 4
90%
Introducing the recycle leads to added costs for pumping (compression) and heating.
6-23
6.36 b.
m 1 (lbm/h)
300 lbm/h wet product
0.2
.
0167
lb m T(l) / lb m
1 0.2
0.833 lb m D / lb m
0.02 / (102
. ) 0.0196 lb m T(l)
0.9804 lb m D / lb m
Dryer
n3 (lb-mole/h) @ 200
n1 (lb-mole/h)
O
F,
y3 (lb-mole T/lb-mole)
y1 (lb-mole T(v)/lb-mole)
(1-y3)( lb-mole N2/lb-mole)
(1–y1) (lb-mole N2/lbmole)
T=toluene
Heater
70% r.s.,150oF, 1.2 atm
D=dry solids
n3 (lb-mole/h)
y3 (lb-mole T(v)/lb-mole)
(1-y3) (lb-mole N2/lb-mole)
Condenser
Eq.@ 90OF,
1atm
n2 ( lb-mole T(l)/h
Strategy: Overall balanceŸ m 1 & n 2 ;
Relative saturationŸy1;, Gas and liquid equilibriumŸy3
Balance over the condenserŸ n1 & n 3
UV RS
W T
m 1 u 0.0196 n 2 u 92.13
Toluene Balance: 300 u 0167
.
m 1 255 lb m / h
Ÿ
m 1 u 0.9804
Dry Solids Balance: 300 u 0833
.
n 2 0.488 lb - mole / h
70% relative saturation of dryer outlet gas:
pC* 7 H8
y1 P
O
O
(150 F = 6556
.
C) = 10
(6.95334-
0.70 pC* 7 H8 (150 O F) Ÿ y1
1343.943
)
65.56 219 .37
0.70 pC*7 H8
P
172.47 mmHg
(0.70)(172.47)
. u 760
12
.
lb - mole T(v) / lb - mole
01324
Saturation at condenser outlet:
pC* 7 H8
y3
O
O
(90 F = 32.22 C) = 10
pC* 7 H8
P
40.90
760
(6.95334-
1343.943
)
32 .22 219 .37
40.90 mmHg
0.0538 mol T(v) / mol
UV RS
W T
0.488 n 3 u 0.0538
Condenser Toluene Balance: n1 u 01324
.
n1
Ÿ
Condenser N 2 Balance: n1 u (1 01324
.
) n 3 u (1 0.0538)
n 3
6-24
5875
lb - mole / h
.
5.387 lb - mole / h
6.36 (cont’d)
Circulation rate of dry nitrogen = 5.875 u (1 - 0.1324) =
5.097 lb - mole lb - mole
h
28.02 lb m
lb m / h
0182
.
Vinlet
6.37
b g
(200 460)q R
492q R
5.387 lb - moles 359 ft 3 STP
hr
1 lb - mole
C 6 H 14 Basis: 100 mol C 6 H 14
2590 ft 3 h
19
O 2 o 6CO 2 7H 2 O
2
n1 (mol) dry gas, 1 atm
0.821 mol N 2 /mol D.G.
0.069 mol CO2 /mol D.G.
0.021 mol CO/mol D.G.
0.086 mol O 2 /mol D.G.
0.00265 mol C 6 H 14/mol D.G.
n2 (mol H 2 O)
100 mol C 6 H 14
n0 (mol) air
0.21 mol O 2 /mol
0.79 mol N 2 /mol
b g n LMM0b.069g 0b.021g 6b0b.00265g gOPP Ÿ n 5666 mol dry gas
Q
N
100 0.00265b5666g mol reacted
u 100% 85.0%
Conversion:
C balance: 6 100
1
CO 2
1
CO
C 6 H 14
100 mol fed
N 2 balance: 0.79n0
Theoretical air:
Excess air:
b g
0.821 5666 Ÿ n0
100 mol C 2 H 14
5888 4524
u 100%
4524
b g
H balance: 14 100
Dew point: y H 2 O
5888 mol air
19 mol O 2
1 mol air
2 mol C 2 H 14
0.21 mol O 2
30.2% excess air
b gb
g
2n2 5666 14 0.00265 Ÿ n2
595
595 + 5666
4524 mol air
d i
p Tdp
760 mm Hg
6-25
595 mol H 2 O
d i
Ÿ p Tdp
Table B.3
72.2 mm Hg Ÿ Tdp
451
. qC
6.38 Basis: 1 mol outlet gas/min
n0 ( mol / min)
y 0 ( mol CH 4 / mol)
(1 y0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa
y1 (mol CO 2 / mol)
y 2 (mol H 2 O / mol)
(1 y1 y 2 ) mol N 2 / mol
n1 (mol O 2 / min)
3.76n1 (mol N 2 / min)
CH 4 2O 2 o CO 2 2H 2 O
pCO 2
80 mmHg Ÿ y1
80 mmHg 101325 Pa
105000 Pa 760 mmHg
b
100% O2 conversion : 2no yo 7 no 1 yo
b
C balance: no yo 2no 1 yo
. n1
N2 balance: 376
C2 H6 2
g
g
7
O 2 o 2CO 2 3H 2 O
2
0.1016 mol CO 2 / mol
(1)
n1
01016
.
(2)
1 y1 y2
b
H balance: 4no yo 6no 1 yo
g
(3)
2 y2
(4)
R|n 0.0770 mol
| y 0.6924 mol CH / mol
Solve equations 1 to 4 Ÿ S
mol O
.
||n 01912
mol H O / mol
.
Ty 01793
Dew point:
01793
.
b105000g Pa 760 mmHg 1412. mmHg Ÿ T
p dT i
o
4
o
*
H2 O
1
2
2
2
dp
dp
101325 Pa
b
6.39 Basis: 100 mol dry stack gas
n P (mol C 3 H 8)
n B (mol C 4H10 )
n out (mol)
0.21 O2
0.79 N2
P = 780 mm Hg
Stack gas: Tdp = 46.5°C
100 mol dry gas
0.000527 mol C 3 H 8/mol
0.000527 mol C 4H 10/mol
0.0148 mol CO/mol
0.0712 mol CO 2/mol
+ O 2, N 2
nw (mol H2O)
C 3 H 8 5O 2 o 3CO 2 4H 2 O
C 4 H 10 6-26
g
58.8 o C Table B.3
13
O 2 o 4CO 2 5H 2 O
2
Dew point
But y w
46.5q C Ÿ y w P
nw
100 n w
p
w
b46.5q Cg Ÿ y
0.0995 Ÿ n w
77.6 mm Hg
780 mm Hg
w
0.0995
mol H 2 O
mol
11.0 mol H 2 O
b100g b0.000527gb3g b0.000527gb4g 0.0148 0.0712
Ÿ 3n 4n
8.969 b1g
. gb2g
H balance: 8n 10n b100g b0.000527gb8g b0.000527gb10g b110
Ÿ 8n 10n
23149
.
b2 g
R| 56% C H
mol C H U
.
Rn 1454
ŸS
Solve b1g & b2g simultaneously: Ÿ S
V
mol C H W
.
|T 44% C H
Tn 1152
bAnswers may vary r 2% due to loss of precisiong
C balance: 3n p 4n B
p
B
p
B
p
B
p
3
8
3
8
B
4
10
4
10
6.40 a.
L1 (lb - mole C 10 H 22 / h)
L 2 (lb - mole / h)
x 2 (lb - mole C 3 H 8 / lb - mole)
1 x 2 (lb - mole C 10 H 22 / lb - mole)
G 1 (lb - mole / h)
G 2 = 1 lb - mole / h
0.07 (lb - mole C 3 H 8 / lb - mole)
0.93 (lb - mole N 2 / lb - mole)
y1 (lb - mole C 3 H 8 / lb - mole)
1 y1 (lb - mole N 2 / lb - mole)
Basis: G 2
1 lb - mole h feed gas
b gb g
N 2 balance: 1 0.93
b
g
98.5% propane absorption Ÿ G 1 y1
b1g & b2g Ÿ G
1
b
g 0.93
b1 0.985gb1gb0.07g Ÿ G y
G 1 1 y1 Ÿ G 1 1 y1
1 1
. u 10 3
105
b1g
b2g
u 10 3 mol C 3 H 8 mol
1128
.
0.93105 lb - mol h , y1
Assume G 2 L 2 streams are in equilibrium
From Cox Chart (Figure 6.1-4), p *C3 H8 (80 o F ) 160 lb / in 2
Raoult' s law: x 2 p
C3H 8
b80q Fg
b gb g
Propane balance: 0.07 1
Decane balance: L1
0.07 p Ÿ x 2
b0.07gb10. atmg
G 1 y1 L 2 x 2 Ÿ L 2
10.89 atm
b
Ÿ
gd h b
d L / G h
1
2 min
0.006428
gd
mol H 2 O
mol
u 10 3
0.07 0.93105 1128
.
gb
g
10.66 lb - mole h
10.7 mol liquid feed / mol gas feed
6-27
i
0.006428
10.726 lb - mole h
1 x 2 L 2
1 0.006428 10.726
b
10.89 atm
6.40 (cont’d)
b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed
rate and fractional absorption], or
n C3H 8
10.726 lb - mole 0.006428 lb - mole C 3 H 3
h
lb - mole
0.06895 lb - mol C 3 H 8 h
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22/h
Ÿ x2
b
0.06895 lb - mole C 3 H 8 h
0.06895 + 12.8 lb - moles h
g
0.00536 lb - mole C 3 H 8 / lb - mole
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and
hence the cost) of the column, but increases the raw material (decane) and pumping costs.
All three costs would have to be determined as a function of the feed ratio.
6.41 a. Basis: 100 mol/s liquid feed stream
Let B n - butane , HC = other hydrocarbons
o
100 mol/s @ 30 C, 1 atm
n4 (mol/s) @ 30qC, 1 atm
xB =12.5 mol B/s
87.5 mol other hydrocarbon/s
y4 (mol B/mol)
(1-y4) (mol N2/mol)
88.125 mol/s
n3 (mol N2/s)
0.625 mol B/s (5% of B fed)
87.5 mol HC/s
p *B (30 o C) # 41 lb / in 2
Raoult' s law: y 4 P
2120 mm Hg (from Figure 6.1-4)
b
95% n-butane stripped: n 4 ˜ 0.3487
Total mole balance: 100 n 3
mol gas fed
Ÿ
mol liquid fed
b. If y 4
0.8 u 0.3487
x B p B* (30 o C) 0125
.
u 2120
760
P
12.5 0.95 Ÿ n 4 34.06 mol / s
x B p *B (30 o C) Ÿ y 4
0.3487
g b gb g
34.08 88125
.
Ÿ n 3
22.20 mol / s
100 mol / s
22.20 mol / s
0.2220 mol gas fed / mol liquid fed
0.2790 , following the same steps as in Part (a),
b
g b gb g
12.5 0.95 Ÿ n 4 42.56 mol / s
95% n-butane is stripped: n 4 ˜ 0.2790
Total mole balance: 100 n 3 42.56 88.125 Ÿ n 3 30.68 mol / s
30.68 mol / s
mol gas fed
Ÿ
0.3068 mol gas fed / mol liquid fed
100 mol / s
mol liquid fed
c. Increasing the nitrogen feed rate can cause an increase of nitrogen consumption, but more
butane can be recovered. Therefore, costs of nitrogen (including compression) and
butane are needed to determine the most cost-effective value of the gas/liquid feed rate.
6-28
6.42 Basis: 100 mol NH 3
Preheated
air
100 mol NH 3
780 kPa sat'd
N2
O2
converter
n3
n4
n5
n6
n 1 (mol) O 2
3.76 n 1 (mol) N 2
n 2 (mol) H 2 O
1 atm, 30°C
h r = 0.5
a. i)
b g
NH 3 feed: P
P Tsat
Antoine:
log 10 6150
b
g
820 kPa
55 wt% HNO 3 (aq )
n 8 (mol HNO 3 )
n 9 (mol H 2O)
n7 (mol H 2O)
6150 mm Hg 8.09 atm
b
g
7.55466 1002.711 Tsat 247.885 Ÿ Tsat
Table B.1 Ÿ
VNH 3
absorber
(mol NO)
(mol N 2)
(mol O 2 )
(mol H 2O)
Pc
Tc
18.4q C 291.6 K
UV
W
. atm Ÿ Pr 8.09 / 1113
.
1113
0.073
Ÿz
. K Ÿ Tr 2916
. / 4055
. 0.72
4055
b
0.92 100 mol
g
8.314 Pa
2916
. K
mol - K 820 u 10 3 Pa
0.92
(Fig. 5.3-1)
0.272 m 3 NH 3
Air feed: NH 3 2O 2 o HNO 3 H 2 O
n1
100 mol NH 3
2 mol O 2
mol NH 3
b
hr ˜ p * 30q C
Water in Air: y H 2 O
p
Ÿ 0.02094
200 mol O 2
g
.
0.500 u 31824
0.02094
760
n2
Ÿ n2 20.36 mol H 2 O
n2 4.76 200
A
( 4 .76 mol air mol O 2 )
Vair
b g
b g
4.76 200 20.36 mol 22.4 L STP
1 mol
303K 1 m 3
24.2 m 3 air
273K 10 3 L
ii) Reactions: 4 NH 3 5O 2 o 4 NO 6H 2 O , 4 NH 3 3O 2 o 2 N 2 6H 2 O
Balances on converter
NO: n3
97 mol NH 3
4 mol NO
4 mol NH 3
6-29
97 mol NO
6.42 (cont’d)
b g
N 2 : n4
3.76 2.00 mol +
O 2 : n5
200 mol Ÿ n total
2 mol N 2
4 mol NH 3
97 mol NH 3
5 mol O 2
4 mol NH 3
3 mol NH 3
3 mol O 2
4 mol NH 3
H 2 O: n6
3 mol NH 3
20.36 mol +
7535
. mol N 2
76.5 mol O 2
6 mol H 2 O
170.4 mol H 2 O
4 mol NH 3
100 mol NH 3
(97 7535
. 76.5 170.4) mol
= 1097 mol converter effluent
8.8% NO, 68.7% N 2 , 7.0% O 2 , 15.5% H 2 O
iii) Reaction: 4 NO 3O 2 2 H 2 O o 4 HNO 3
HNO 3 bal. in absorber: n8
H 2 O in product: n9
97 mol NO react 4 mol HNO 3
4 mol NO
97 mol HNO 3
63.02 g HNO 3
mol
97 mol HNO 3
1 mol H 2 O
45 g H 2 O
55 g HNO 3 18.02 g H 2 O
277.56 mol H 2 O
b
gb g
H balance on absorber: 170.4 2 2n7
b.
b
gb gb
97 277.6 2 mol H
Ÿ n7
155.7 mol H 2 O added
VH 2O
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3
1 m3
1 mol
1 g 10 6 cm 3
V NH 3
Vair
VH 2O
bg
2.81 u 10 3 m 3 H 2 O l
97 mol HNO 3
63.02 g HNO 3 277.6 mol H 2 O 18.02 g HNO 3
mol
mol
11115 g 11115
.
kg
M acid in old basis
Scale factor
g
b1000 metric tonsgb1000 kg metric tong
8.997 u 10 4
11.115 kg
d8.997 u 10 id0.272 m NH i 2.45 u 10 m NH
d8.997 u 10 id24.2 m airi 2.18 u 10 m air
d8.997 u 10 id2.81 u 10 m H Oi 253 m H Oblg
3
4
3
4
3
3
3
4
4
6
3
3
3
3
2
6-30
2
6.43 a.
Basis: 100 mol feed gas
G = NH3 -free gas
100 mol
0.10 mol NH3 /mol
0.90 mol G/mol
Absorber
n 1 (mol H2 O( l))
n 2 (mol)
in equilibrium
y A (mol NH 3 /mol)
at 10°C(50°F)
(mol
H
O/mol)
yW
2
and 1 atm
(mol
G/mol)
yG
(mol)
n3
x A (mol NH 3 /mol)
(1 – x A) (mol H 2O/mol)
Composition of liquid effluent . Basis: 100 g solution
Perry, Table 2.32, p. 2-99: T = 10oC (50oF), U = 0.9534 g/mL Ÿ 0.120 g NH3/g solution
Ÿ
12.0 g NH 3
88.0 g H 2 O
= 0.706 mol NH 3 ,
= 4.89 mol NH 3
(17.0 g / 1 mol)
(18.0 g / 1 mol)
Ÿ 12.6 mole% NH 3 ( aq), 87.4 mole% H 2 O(l)
Composition of gas effluent
p NH 3
T
o
50 F, x A
 o
.
0126
Perry
yA
Ÿ yW
yG
. / 14.7 0.0823 mol NH 3 mol
121
.
/ 14.7 0.0105 mol H 2 O mol
0155
1 y A yW 0.907 mol G mol
b gb g n y Ÿ n b100gb0.90g b0.907g 99.2 mol
184
. mol NH
absorbed b100gb0.10g b99.2gb0.0823g
G balance: 100 0.90
NH 3
p H 2O
p total
g U|
.
0155
psiabTable 2 - 21gV
|W
14.7 psia
b
. psia Table 2 - 23
121
% absorption
2
G
2
in
3
out
1.84 mol absorbed
u 100% 18.4%
100 010
. mol fed
b gb g
b. If the slip stream or densitometer temperature were higher than the temperature in the
contactor, dissolved ammonia would come out of solution and the calculated solution
composition would be in error.
6.44 a.
15% oleum: Basis - 100kg
15 kg SO 3 85 kg H 2 SO 4
1 kmol H 2 SO 4
98.08 kg H 2 SO 4
Ÿ 84.4% SO 3
6-31
1 kmol SO 3
1 kmol H 2 SO 4
80.07 kg SO 3
1 kmol SO 3
84.4 kg
6.44 (cont’d)
b.
Basis 1 kg liquid feed
n o (mol), 40o C, 1.2 at m
n1 (mol), 40o C, 1.2 at m
0.90 mol SO3 /mol
0.10 mol G/ mol
y1 mol SO3 /mo l
(1-y1 ) mol G/ mol
Equilib riu m @ 40o C
i)
ii)
1 kg 98% H2 SO4
m1 (kg) 15% oleu m
0.98 kg SO3
0.02 kg H2 O
0.15 kg SO3 /kg
0.85 kg H2 SO4 /kg
y1
b
pSO3 40q C, 84.4%
H balance:
g
P
0.98 kg H 2 SO 4
115
.
760
151
. u 10 3 mol SO 3 mol
2.02 kg H
2.02 kg H
0.02 kg H 2 O
98.08 kg H 2 SO 4
18.02 kg H 2 O
0.85 m1 H 2 SO 4
2.02 kg H
Ÿ m1 128
. kg
98.08 kg H 2 SO 4
But since the feed solution has a mass of 1 kg,
0.28 kg SO 3 10 3 g 1 mol
SO 3 absorbed 128
. 10
. kg
. mol
350
kg 80.07 g
Ÿ 3.5 mol n0 n1
. u 10 3 n1
G balance: 0.10n0 1 151
b
g
d
i
E
. mol
n0 389
n1 0.39 mol
V
b g
3.89 mol
22.4 L STP
1 kg liquid feed
mol
313K 1 atm 1 m 3
273K 1.2 atm 10 3 L
8.33 u 10 -2 m 3 kg liquid feed
6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen.
b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid
here.
c. Raoult’s law can be used for water, and Henry’s law can be used for CO2.
b g
p b100q Cg
6.46 p B 100q C
10
T
10
Raoult' s Law: y B
c6.90565 1211.033 b100 220.790gh 1350.5 mm Hg
c6.95334 1343.943 b100 219.377gh 556.3 mm Hg
0.40b1350.5g
0.0711 mol Benzene mol
P x p Ÿ y
10b760g
0.60b556.3g
0.0439 mol Toluene mol
y
10b760g
B
B
B
T
y N2
1 0.0711 0.0439 0.885 mol N 2 mol
6-32
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 - 138
b
Ÿ H N 2 80q C
g
12.6 u 10 4 atm mole fraction
b0.003gd12.6 u 10 i 378 atm
. mm Hg
3551
1 atm
H O - Raoult' s law: p b80q Cg
760 mm Hg
Ÿ p
d x id p i b0.997gb0.467g 0.466 atm
Ÿ pN2
4
xN2 H N2
2
H 2O
H 2O
Total pressure: P
H 2O
p N 2 p H 2O
Mole fractions: y H 2 O
y N2
378 0.466 378.5 atm
p H 2O P
1 y H 2O
b
6.48 H 2 O - Raoult' s law: p H 2 O 70q C
Ÿ p H 2O
Methane Henry' s law: p m
Total pressure: P
0 / 466 / 378.5 123
. u 10 3 mol H 2 O mol gas
0.999 mol N 2 mol gas
g
233.7 mm Hg
1 atm
0.3075 atm
760 mm Hg
x H 2O pH 2O
b1 x gb0.3075g
m
xm ˜ Hm
p m p H 2O
x m ˜ 6.66 u 10 4 (1 x m )(0.3075) 10
146
. u 10 4 mol CH 4 / mol
Ÿ xm
6.49 a.
0.467 atm
H 2O
Moles of water: n H 2 O
1000 cm 3
1g
mol
55.49 mol
3
cm 18.02 g
Moles of nitrogen:
nN2
(1 - 0.334) u 14.1 cm 3 (STP)
1 mol
1L
22.4 L (STP) 1000 cm
3
4.192 u 10 4 mol
Moles of oxygen:
n O2
(0.334) ˜ 14.1 cm 3 (STP)
mol
L
22.4 L (STP) 1000 cm
3
2.102 u 10 4 mol
Mole fractions of dissolved gases:
nN2
4.192 u 10 4
x N2
n H 2 O n N 2 nO 2 55.49 4.192 u 10 4 2.102 u 10 4
7.554 u 10 6 mol N 2 / mol
x O2
nO 2
n H 2 O n N 2 nO 2
2.102 u 10 4
55.49 4.192 u 10 4 2.102 u 10 4
3.788 u 10 6 mol O 2 / mol
6-33
6.49 (cont’d)
Henry' s law
p N2
Nitrogen: H N 2
x N2
u 10 5 atm / mole fraction
1046
.
0.21 ˜ 1
3.788 u 10 6
pO2
Oxygen: H O 2
0.79 ˜ 1
7.554 u 10 6
x O2
u 10 4 atm / mole fraction
5544
.
b. Mass of oxygen dissolved in 1 liter of blood:
2.102 u 10 -4 mol 32.0 g
m O2
Mass flow rate of blood: m blood
6.726 u 10 3 g
mol
0.4 g O 2
min
1 L blood
6.72 u 10 -3 g O 2
59 L blood / min
c. Assumptions:
(1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much
greater);
(2) The temperature of blood is 36.9qC.
6.50 a.
bg
Basis: 1 cm 3 H 2 O l
1 g H 2 O 1 mol
H2O
 
o
(SG)
1.0
18.0 g
b g
0.0901 cm 3 STP CO 2
 
o
( SC) CO2
0.0555 mol H 2 O
0.0901
1 mol
22,400 cm 3 STP
4.022 u 10 6 mol CO 2
b g
d4.022 u 10 i mol CO 7.246 u 10 mol CO
d0.0555 + 4.022 u 10 i mol
1 atm
Ÿ H b20q Cg
13800 atm mole fraction
7.246 u 10
6
b.
pCO 2
1 atm Ÿ x CO2
pCO 2
x CO 2 H CO2
2
5
2
6
mol
5
CO 2
b g
For simplicity, assume n total | n H 2 O mol
x CO2
nCO 2
c. V
pCO 2 H
12 oz
b35. atmg b13800 atm mole fractiong
1L
33.8 oz
0.220 g CO 2
0.220 g CO 2
10 3 g H 2 O 1 mol H 2 O
1L
1 mol CO 2
44.0 g CO 2
18.0 g H 2 O
2.536 u 10 4 mol CO 2 mol
2.536 u 10 4 mol CO 2
44.0 g CO 2
1 mol H 2 O
1 mol CO 2
b g b273 37gK
22.4 L STP
1 mol
6-34
273K
0127
.
L 127 cm 3
6.51 a.
– SO2 is hazardous and should not be released directly into the atmosphere, especially if
the analyzer is inside.
– From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in
the liquid, which increases with time. If the water were never replaced, the gas leaving
the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole
fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the
gas phase.
b
g
b
b. Calculate x mol SO 2 mol in terms of r g SO 2 100 g H 2 O
b
b
Basis: 100 g H 2 O 1 mol 18.02 g
r (g SO 2 ) 1 mol 64.07 g
FG
H
g
g
555
. mol H 2 O
0.01561r (mol SO 2 )
IJ
K
mol SO 2
0.01561r
mol
555
. 0.01561r
From this relation and the given data, pSO 2
Ÿ xSO 2
g
0 mmHg œ xSO2
0 mol SO 2 mol
1.4 x 10–3
2.8 x 10–3
4.2 x 10–3
5.6 x 10–3
42
85
129
176
A plot of pSO 2 vs. xSO 2 is a straight line. Fitting the line using the method of least squares
(Appendix A.1) yields
dp
HSO 2 xSO2
SO 2
c.
100 ppm SO 2 Ÿ ySO2
Ÿ pSO2
ySO2 P
,
H SO 2
u 104
.
3136
mm Hg
mole fraction
100 mol SO 2
106 mols gas
d10. u 10 ib760 mm Hgg
Henry's law Ÿ xSO 2
Since xSO 2
i
4
pSO 2
H SO 2
0.0760 mm Hg
0.0760 mm Hg
3.136 u 104 mm Hg mole fraction
2.40 u 106 mol SO 2 mol
is so low, we may assume for simplicity that Vfinal | Vinitial
nfinal | ninitial
bg
140 L 103 g H 2 O l
1L
1 mol
18 g
7.78 u 103 moles
7.78 u 103 mol solution 2.40 u 10 6 mol SO 2
1 mol solution
0.0187 mol SO 2 dissolved
134
. u 10 4 mol SO 2 L
140 L
Ÿ nSO 2
140 L , and
0.0187 mol SO 2 dissolved
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the
solution to react with the absorbed SO2.
6-35
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene):
x S pS Ÿ xS
yT P
xT pT Ÿ xT
10
b
6.95334 1343.943 T 219 .377
b
6.92409 1420.0 T 206
b
6.92409 1420.0 85.9 206
1 xS
b g
P b85q Cg
6.53 PB 85q C
T
g
0.350(150 mm Hg)
0.65(150)
10
b
6.92409 1420.0 T 206
1 0.852
g
g
0.35(150)
b
g
10
10
85.9q C (Determine using E - Z Solve or a spreadsheeet)
ŸT
xT
10
0.65(150)
1 x S xT
xS
0.650(150 mm Hg)
yS P
6.95334 1343.943 T 219 .377
0.852 mol styrene / mol
g
0148
.
mol toluene / mol
b
g
6.95334 1343.943 b85 219 .377 g
10
10
6.905651211.033 85 220.790
Raoult's Law: y B P
x B PB Ÿ
. mm Hg
8817
. mm Hg
3451
b g
b g
b g
b g
0.35 8817
.
0.0406 mol Benzene mol
10 760
0.65 3451
.
0.0295 mol Toluene mol
10 760
1 0.0406 0.0295 0.930 mol N 2 mol
yB
yT
yN 2
6.54 a. From the Cox chart, at 77q F, p*P
140 psig, p*nB
35 psig, p*iB
51 psig
Total pressure P = x p ˜ p *p + x nB ˜ p *nB + x iB ˜ p *iB
0.50(140) 0.30(35) 0.20(51) 150.8 psig
P 200 psig, so the container is technically safe.
b. From the Cox chart, at 140q F, pP*
*
300 psig, pnB
*
90 psig, piB
120 psig
Total pressure P = 0.50(300) 0.30(90) 0.20(120) # 200 psig
The temperature in a room will never reach 140oF unless a fire breaks out, so the container
is adequate.
b g
P b120q Cg
6.55 a. Antoine: Pnp 120q C
ip
10
10
b
6.852211064.630 120 232 .000
b
6.78967 1020.012 120 233.097
g
g
6725 mm Hg
7960 mm Hg
When the first bubble of vapor forms,
xnp
0.500 mol n - C5 H 12 (l) / mol
*
Total pressure: P = xnp ˜ pnp
+ xip ˜ pip*
xnp
0.500 mol i - C5 H 12 (l) / mol
0.50(6725) 0.50(7960)
6-36
7342 mm Hg
6.55 (cont’d)
*
xnp ˜ pnp
ynp
0.500(6725)
7342
P
1 ynp
yip
0.458 mol n - C5 H 12 (v) / mol
1 0.458 0.542 mol i - C5 H 12 (v) / mol
When the last drop of liquid evaporates,
0.500 mol n - C5 H 12 (v) / mol
ynp
ynp P
*
pnp (120o C)
xnp xip
yip
yip P
*
pip (120o C)
0.5 mol i - C5 H 12 (v) / mol
0.500 P 0.500 P
6725
7960
1Ÿ P
7291 mm Hg
0.5 * 7291 mm Hg
0.542 mol n - C5 H 12 (l) / mol
6725 mm Hg
1 xnp 1 0.542 0.458 mol i - C5 H 12 (l) / mol
xnp
xip
b. When the first drop of liquid forms,
0.500 mol n - C5 H 12 (v) / mol
ynp
yip
0.500 mol i - C12 H 12 (v) / mol
P = (1200 + 760) = 1960 mm Hg
0.500 P
0.500 P
*
*
pnp (Tdp ) pip (Tdp )
xnp xip
Ÿ Tdp
10
b
6.852211064 .63 63.1 232 .000
10
Pip
10
xnp
0.5 * 1960 mm Hg
*
(631
. o C)
pnp
b
g
6.78967 1020.012 63.1 233.097
1 xnp
980
10
6.78967 1020.012 /( Tdp 233.097)
. oC
631
Pnp
xip
980
6.852211064 .63/( Tdp 232 )
1756 mm Hg
g
2218 mm Hg
0.558 mol n - C5 H12 / mol
1 0.558 0.442 mol i - C5 H 12 / mol
When the last bubble of vapor condenses,
xnp
0.500 mol n - C5 H 12 (l) / mol
xip
0.500 mol i - C5 H12 (l) / mol
*
Total pressure: P = xnp ˜ pnp
+ xip ˜ pip*
Ÿ 1960 (0.5)10
Ÿ T 62.6q C
ynp
yip
*
(62.6o C)
xnp ˜ pnp
1 ynp
b
6.852211064 .63 T 232 .000
P
1 0.442
g (0.5)106.78967 1020.012 /( T
bp 233.097)
0.5(1731)
0.442 mol n - C5 H 12 (v) / mol
1960
0.558 mol i - C5 H 12 (v) / mol
6-37
1
6.56 B = benzene, T = toluene
nv ( mol / min) at 80o C, 3 atm
10 L(STP)/min
nN 2 ( mol / min)
xB [mol B(l)/mol]
yN2 (mol N2/mol)
yB [mol B(v)/mol]
xT [mol T(l)/mol]
yT [mol T(v)/mol]
10.0 L(STP) / min
= 0.4464 mol N 2 / min
22.4 L(STP) / mol
nN 2
b g
p b80q Cg
Antoine: p B 80q C
T
b
g
6.95334 1343.943 b80 219 .377 g
10
10
6.905651211.033 80 220.079
757.7 mm Hg
. mm Hg
2912
a. Initially, xB = 0.500, xT = 0.500.
Raoult's law:
b
g
b
g
yB
x B pB (80o C)
P
0.500 757.7
3(760)
yT
xT pT (80o C)
P
0.500 2912
.
3(760)
mol B(v) mol
0166
.
0.0639 mol T(v) mol
0.0639) Ÿ n v
.
N 2 balance: 0.4464 mol N 2 / min = nv (1 0166
mol BI
FG 0.5797 mol IJ FG 0166
.
J
H
K
H
mol K
min
FG 0.5797 mol IJ FG 0.0639 mol BIJ
H
mol K
min K H
Ÿ nB0
nT0
0.5797 mol / min
mol B(v)
min
0.0962
0.0370
mol T(v)
min
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (=
1–xB) increases.
c. Since xB decreases, yB (= xBpB*/P) also decreases. Since xT increases, yT (= xTpT*/P) also
increases.
6.57 a. P
d i
d i
xhex phex Tbp xhep phep Tbp
760 mm Hg
0.500 10
, yi
xi
yi P
d i
pi Tdp
Ÿ
¦x
i
i
P
6.87776 1171.53/( Tbp 224 .366)
E-Z Solve or Goal Seek Ÿ Tbp
b.
d i
xi pi Tbp
P
¦
0.500 10
80.5q C Ÿ yhex
yi
i
d i
pi Tdp
1
6-38
, Antoine equation for Pr
6.90240 1268.115/( Tbp 216.9 )
0.713, yhep
0.287
6.57 (cont’d)
760 mmHg
LM
N10
0.30
6.87776 1171.53/( Tdp 224 .366)
E-Z Solve or Goal Seek Ÿ Tdp
6.58
a.
f (T )
P
N
¦
xi pi* (T )
OP
Q
0.30
10
6.90240 1268.115/( Tdp 216.9 )
711
. q C Ÿ xhex
0 Ÿ T , where
0.279, xhep
1
0.721
FG A B IJ
T C K
10H
i
pi* (T )
i
i
i 1
yi (i 1,2,, N )
xi pi* (T )
P
b.
Calculation of Bubble Points
A
B
Benzene
6.90565 1211.033
Ethylbenzene 6.95719 1424.255
Toluene
6.95334 1343.943
C
220.79
213.206
219.377
P(mmHg)= 760
xB
0.226
0.443
0.226
When x B
When x EB
When xT
xEB
0.443
0.226
0.226
Tbp(oC)
108.09
96.47
104.48
xT
0.331
0.331
0.548
b
g
dT i
1 pure benzene , Tbp
b
g
bp C H
6 6
1 pure ethylbenzene , Tbp
b
g
1 pure toluene , Tbp
. oC
801
dT i
dT i
bp C H
7 8
pEB
pT
f(T)
pB
378.0 148.2 233.9 -0.086
543.1 51.6 165.2
0.11
344.0 67.3 348.6
0.07
bp C H
8 10
136.2 o C Ÿ Tbp , EB ! Tbp ,T ! Tbp , B
110.6o C
Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp)
than Mixture 2, and so (Tbp)1 > (Tbp)2 . Mixture 3 contains more toluene (lower bp) and
less ethylbenzene (higher bp) than Mixture 1, and so (Tbp)3 < (Tbp)1. Mixture 3 contains
more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3 >
(Tbp)2
6-39
a. Basis: 150.0 L/s vapor mixture
6.59
n 1 (mol/s) @ T(oC), 1100 mm Hg
n 0 (mol/s) @ 120qC,
0.600 mol B(v)/mol
0.400 mol H(v)/mol
0.500 mol B(v)/mol
0.500 mol H(v)/mol
n 2 (mol/s)
x2 [mol B(l)/mol]
(1- x2) [mol H(l)/mol]
Gibbs phase rule: F = 2 + m - S = 2 + 2 - 2 = 2
Since the composition of the vapor and the pressure are given, the information is enough.
Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law
for butane and hexane
150.0 L 393 K
b. Molar flow rate of feed: n 0
s
mol
273 K 22.4 L (STP)
9.640 mol / s
Raoult' s law for butane: 0.600(1100) = x 2 ˜ 10 6.82485 943.453/( T 239.711)
Raoult's law for hexane: 0.400(1100) = (1- x 2 ) ˜ 106.877761171.530 /( T 224.366)
(1)
Mole balance on butane: 1(0.5) = n 1 ˜ 0.6 n 2 ˜ x 2
(2)
Mole balance on hexane: 1(0.5) = n1 ˜ 0.4 n 2 ˜ (1 x 2 )
c. From Raoult's law, 1 =
1100(0.4)
1100(0.6)
.
1171530
943.453
)
) 10 **(6.87776 10 **(6.82485 T 239.711
T 224.366
Ÿ T = 57.0 q C
x2
1100(0.6)
10
6.82485 943.453/( 57 .0 239 .711)
015
. mol butane / mol
Solving (1) and (2) simultaneously Ÿ n1
d.
6.60
0.778 mol C 4 H 10 / s; n 2
0.222 molC 6 H 14 / s
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure;
(2) Raoult’s law is accurate;
(3) Ideal gas law is valid.
P = n-pentane, H = n-hexane
170.0 kmol/h, T1a (oC), 1 atm
85.0 kmol/h, T1b (oC), 1
n 0 (kmol/h)
0.98 mol P(l)/mol
0.02 mol H(l)/mol
0.45 kmol P(l)/kmol
0.45 kmol H(l)/kmol
n 2 (kmol/h) (l),
x2 (kmol B(l)/kmol)
(1- x2) (kmol H(l)/kmol)
6-40
6.60 (cont’d)
a. Molar flow rate of feed: n 0 (0.45)(0.95) 85(0.98) Ÿ n 0
Total mole balance : 195 85.0 n 2 Ÿ n 2
195 kmol / h
110 kmol / h
Pentane balance: 195( 0.45) 85.0(0.98) 110 ˜ x 2 Ÿ x 2
0.0405 mol P / mol
b. Dew point of column overhead vapor effluent:
Eq. 6.4 - 7, Antoine equation
Ÿ
0.98(760)
10
6.852211064.63/ ( T1a 232.000)
0.02(760)
10
6.877761171.530/( T1a 224.366)
1 Ÿ T1a
37.3o C
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior,
Vvapor
170 kmol 0.08206 m 3 ˜ atm (273.2 + 37.3) K
h
kmol ˜ K
1 atm
4330 m 3 / h
Flow rate of liquid distillate product.
Table B.1 Ÿ U P = 0.621 g / mL, U H = 0.659 g / mL
Vdistillate
0.98(85) kmol P 72.15 kg P
L
h
kmol P 0.621 kg P
+
0.02(85) kmol H 86.17 kg H
L
h
kmol H 0.659 kg H
9.9 u 10 3 L / h
c. Reboiler temperature.
0.04 ˜ 10 6.852211064.63/ ( T2 232.000) 0.96 ˜ 10 6.877761171.530 /( T2 224 .366)
760 Ÿ T2 = 66.6q C
Boilup composition.
y2
x 2 p P* (66.6 o C)
P
Ÿ (1 - y 2 )
0.04 ˜ 10 6.852211064.63/ ( 66.6232 )
760
0.102 mol P(v) / mol
0.898 mol H(v) / mol
d. Minimum pipe diameter
F I
GH JK
m3
V
s
Ÿ Dmin
u max
FG mIJ u SD
H sK 4
4Vvapor
S ˜ u max
2
min
(m2 )
4 4330 m 3 / h 1 h
S 10 m / s 3600 s
0.39 m (39 cm)
Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation,
constant temperature and pressure in the pipe connecting the column and the condenser,
column operates at steady state.
6-41
Condenser
6.61 a.
F (mol)
x 0 (mol butane/mol)
V (mol)
0.96 mol butane/mol
R (mol)
x 1 (mol butane/mol)
T
P
Partial condenser: 40q C is the dew point of a 96% C 4 H 10 4% C 5 H 12 vapor mixture at
P Pmin
Total condenser: 40q C is the bubble point of a 96% C 4 H 10 - 4% C 5 H 12 liquid mixture at
P Pmin
yi P
1
xi
Ÿ Pmin
Dew Point: 1
pi 40q C
yi pi 40q C
¦
¦
b
(Raoult's Law)
g
b g
Antoine Eq. for p bC H g
5
P
¦y P ¦x p
i
75 kmol / h , R V
Total balance: F
b
i
15
. Ÿ R
75 u 15
. kmol / h 112.5 kmol / h
b
b
b
g
Butane balance: 187.5x 0
yi
xi
Raoult' s law:
d
i
*
o
p B (85 C)
187.5 kmol / h
UV P 2596 mm Hg
gb gW x 0.8803 mol butane mol
112.5b0.8803g 0.96b75g Ÿ x
0.9122 mol butane mol reflux
1
0
yA xA
yB xB
pA P
pA
pB P
pB
110.76 mm Hg
FG 6.95719 1424.255 IJ
85 213.206 K
p *EB (85o C) 10 H
F
g
pi
Ÿ D AB
P
FG 6.92409 1420.0 IJ
85 206 K
10 H
0.96 kmol butane kmol
75 112.5 187.5 kmol / h
Total balance as in b. R 112.5 kmol / h
b. p S* 85o C
g
i
Equilibrium: 0.96 P x1 2830.70
Raoult' s law 0.04 P 1 x1 867.22
6.62 a.
867.22 mmHg
2595.63 mm Hg partial condenser
Feed and product stream compositions are identical: y
c.
2830.70 mmHg
b40q Cg
0.96b2830.70g 0.04b867.22g 2752.16 mm Hg b total condenser g
Bubble Point: P
b. V
12
1
0.96 2830.70 0.04 867.22
Ÿ Pmin
g
FG 6.82485 943.453 IJ
40 239 .711 K
10 H
FG 6.85221 1064.63 IJ
40 232 .00 K
10 H
Antoine Eq. for pi C 4 H10
i
b
¦
FG 6.90565 1211.033 IJ
85 220.790 K
10 H
15174
. mm Hg
88167
. mm Hg
6-42
D AB
6.62 (cont’d)
p S*
D S,EB
110.76
15174
.
*
p EB
p B*
0.730 , D B,EB
*
p EB
88167
.
15174
.
5810
.
Styrene ethylbenzene is the more difficult pair to separate by distillation
because D S,EB is closer to 1 than is D B,EB .
c.
yi xi
yj xj
D ij
d. D B , EB
ŸD
yi xi
Ÿ yi
(1 yi ) 1 xi
b
ij
g
x B D B , EB
Ÿ yB
5810
.
1 (D B , EB 1) x B
d
D ij xi
i
1 D ij 1 xi
581
. xB
, P
1 4.81x B
x B p *B (1 x B ) p *EB
bg
bg
. mol B l mol
0.0
0.2
0.4
0.6
0.8
10
. mol B v mol
0.0 0.592 0.795 0.897 0.959 10
mmHg
152 298
444 5900 736 882
xB
yB
P
6.63 a.
y j 1 yi
x j 1 xi
Since benzene is more volatile, the fraction of benzene will increase moving up the
column. For ideal stages, the temperature of each stage corresponds to the bubble point
temperature of the liquid. Since the fraction of benzene (the more volatile species)
increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: n l
y0
150 mol / h, n v
0.55 mol B mol Ÿ 0.45 mol S mol ;
0.65 mol B mol Ÿ 0.35 mol S mol
¦ x p bT g
Bubble point T : P
P1
200 mol / h ; x1
i
(0.400 u 760) mmHg
 o T1
E-Z Solve
Ÿ y1
i
b0.55g10
6.905651211.033/ ( T 220.79 )
b g
0.45 10 6.92409 1420 /( T 206 )
67.6 o C
bg
x1 p B T
P
B balance: y 0 n v x 2 n l
b g
0.55 508
0.920 mol B mol Ÿ 0.080 mol S mol
0.400 u 760
y1 n v x1n l Ÿ x 2 0.910 mol B mol Ÿ 0.090 mol S mol
Stage 2:
Solve

o T2
(0.400 u 760) mmHg 0.910 p B* (T2 ) 0.090 p S* (T2 ) E-Z
b
55.3o C
g
.
0.910 3310
0.991 mol B mol Ÿ 0.009 mol S mol
760 u 0.400
B balance: y1 n v x 3 n l y 2 n v x 2 n l Ÿ x 3 | 1 mol B mol Ÿ | 0 mol S mol
y2
c. In this process, the styrene content is less than 5% in two stages. In general, the
calculation of part b would be repeated until (1–yn) is less than the specified fraction.
6-43
6.64 Basis: 100 mol/s gas feed. H=hexane.
200 mol oil/s
n l (mol/s)
x +i 1 (mol H/mol)
n F (mol/s)
y F (mol H/mol)
1 – y F (mol N 2/mol)
n 2 (mol/s)
x 2 (mol H/mol)
1 – x 2 (mol Oil/mol)
100 mol/s
0.05 mol H/mol
0.95 mol N /mol
2
Stage i
99.5% of H in feed.
b g b g
b gb g
y F nF
U|V Ÿ n
|W y
Mole Balance: 100 200 95.025 n2 Ÿ n2
b g
Hexane Balance: 0.05 100
n L
b
g
1
200 205 Ÿ n L
2
B
n v (mol/s)
y –i 1 (mol H/mol)
n l (mol/s)
x i (mol H/mol)
1 y F nF
N 2 balance: 0.95 100
99.5% absorption: 0.05 100 0.005
a.
n v (mol/s)
y i (mol H/mol)
b
F
95.025 mol s
F
2.63 u 10 4 mol H(v) mol
205 mol s
0.0243 mol H(l) mol
g b g
1
b100 95.025g Ÿ n 97.52 mol s
2
2.63 u 10 4 95.025 x1 204.99 Ÿ x1
202.48 mol s , n G
G
Antoine
y1
b.
b
g
x1 p H 50q C / P
b
g
0.0243 403.73 / 760 0.0129 mol H(v) mol
H balance on 1st Stage: y 0 n v x 2 n l
y1 n v x1 n l Ÿ x 2
0.00643 mol H(l) mol
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i.
d.
Hexane Absorption
P=
y0=
nGf=
A=
760
0.05
95.025
6.87776
T
30
p*(T)
187.1
i
0
1
2
3
x(i)
2.43E-02
3.07E-03
5.57E-04
PR=
x1=
nL1=
B=
y(i)
5.00E-02
5.98E-03
7.56E-04
1.37E-04
1
0.0243
204.98
1172
ye= 2.63E-04
nG= 97.52 nL=
C= 224.366
T
50
p*(T)
405.3281
i
0
1
2
3
4
5
x(i)
2.43E-02
6.42E-03
1.84E-03
6.63E-04
3.60E-04
6-44
y(i)
5.00E-02
1.29E-02
3.43E-03
9.81E-04
3.53E-04
1.92E-04
202.48
T
70
p*(T)
790.5304
i
0
1
2
3
4
5
...
21
x(i)
y(i)
5.00E-02
2.43E-02 2.52E-02
1.23E-02 1.28E-02
6.38E-03 6.63E-03
3.38E-03 3.52E-03
1.89E-03 1.96E-03
...
...
3.80E-04 3.96E-04
6.64 (cont’d)
e. If the column is long enough, the liquid flowing down eventually approaches equilibrium
with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in
equilibrium with the mole fraction in the entering gas is 3.80x10–4 mol H/mol, which is
insufficient to bring the total hexane absorption to the desired level. To reach that level at
70oC, either the liquid feed rate must be increased or the pressure must be raised to a
value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The
solution is Pmin 951 mm Hg.
0.30 at T
6.65 a. Intersection of vapor curve with y B
b. T
100q C Ÿ x B
b
104q C Ÿ 13% B(l), 87%T(l)
g
b
0.24 mol B mol liquid , y B
0.46 mol B mol liquid
g
0.727 mol nV
Ÿ
0.273 mol
nL
mol vapor
mol liquid
n V (mol vapor)
0.46 mol B(v)/mol
n L (mol liquid)
0.24 mol B(l)/mol
Basis: 1 mol
0.30 mol B(v)/mol
Balances
UV
W
nL
Total moles: 1 nV n L
Ÿ
nV
B:
0.30 0.46nV 0.24n L
c. Intersection of liquid curve with x B
0.3 at T
6.66 a.
P
798 mm Hg, y B
0.50 mol B(v) mol
b.
P
690 mm Hg, y B
015
. mol B(l) mol
c.
P
750 mm Hg, y B
0.24 mol B(v) mol , x B
0.375
98q C Ÿ 50% B(v), 50%T(v)
0.43 mol B(l) mol
nV (mol)
0.43 mol B/mol
nL (mol)
0.24 mol B/mol
3 mol B
7 mol T
UV
W
nV
Mole bal.: 10 nV n L
Ÿ
nL
B bal.:
3 0.43nV 0.24n L
316
. mol
n
Ÿ v
6.84 mol
nl
0.46
mol vapor
mol liquid
Answers may vary due to difficulty of reading chart.
d. i)
P 1000 mm Hg Ÿ all liquid . Assume volume additivity of mixture components.
V
3 mol B 78.11 g B 10 3 L
7 mol T 92.13 g T 10 3 L
10
. L
mol B
0.879 g B
mol T
0.866 g T
ii) 750 mmHg. Assume liquid volume negligible
6-45
6.66 (cont’d)
3.16 mol vapor
V
0.08206 L ˜ atm
373 K
760 mm Hg
mol ˜ K 750 mm Hg
1 atm
0.6 L
97.4 L
(Liquid volume is about 0.6 L)
iii) 600 mm Hg
v
10 mol vapor 0.08206 L ˜ atm
373K
760 mm Hg
mol ˜ K 600 mm Hg
1 atm
388 L
6.67 a. M = methanol
n V (mol)
y (mol M (v)mol)
n L (mol)
x (mol M (l)/mol)
n f (mol)
x F (mol M (l)/mol)
UV
W
nV n L
Ÿ x F nV x F n L
MeOH balance: x F n f ynV xn L
Mole balance:
xF
b. Tmin
0.4, x
nf
0.23, y
75o C, f
0.62 Ÿ f
0 , Tmax
0.4 0.23
0.62 0.23
87 o C, f
ynV xn L Ÿ f
0.436
1
6.68 a.
Txy diagram
(P=1 atm)
80
75
T(oC)
70
Vapor
65
liquid
60
55
50
0
0.2
0.4
0.6
Mole fraction of Acetone
b.
xA
0.47; y A
0.66
6-46
0.8
1
nV
nL
xF x
yx
6.68 (cont’d)
c. (i) x A
0.34; y A
0.55
UV
W
(ii) Mole bal.: 1 nV n L
Ÿ nV
B bal.:
0.50 0.55nV 0.34n L
(iii) U A( l )
Ÿ Ul
MA
0.791 g / cm 3 , U E(l)
0.789 g / cm 3
0.790 g / cm 3
46.07 g / mol
b0.34gb58.08g b1 0.34gb46.07g
Ÿ Ml
0.238 mol liquid
Ÿ 76.2 mole% vapor
b0.34gb0.791g b1 0.34gb0.789g
58.08 g / mol, M E
0.762 mol vapor, n L
5015
. g / mol
Basis: 1 mol liquid Ÿ (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor
(1 mol)(5015
. g / mol)
Liquid volume: Vl
63.48 cm 3
3
(0.790 g / cm )
Vapor volume:
3.2 mol 22400 cm 3 (STP) (65 + 273)K
88,747 cm 3
mol
273K
88,747
u 100% 99.9 volume% vapor
Volume percent of vapor
88747 63.48
Vv =
d. For a basis of 1 mol fed, guess T, calculate nV as above; if nV z 0.20, pick new T.
T
65 qC
64.5 qC
e.
xA
0.34
0.36
Raoult' s law: yi P = xi pi* Ÿ P
760 0.5 u 10
7.02447 1161/( Tbp 224 )
yA
0.55
0.56
fV
0.333
0.200
x A p *A x E p E*
0.5 u 10
8.04494 1554.3/ ( Tbp 222.65)
Ÿ Tbp
66.25o C
0.5 u 10 7.02447 1161/( 66.25 224)
0.696 mol acetone / mol
760
'Tbp
.
66.25 618
u 100% 7.20% error in Tbp
The actual Tbp 618
. oCŸ
Tbp (real)
.
618
y
xp *A
P
yA
0.674 Ÿ
'y A
y A (real)
0.696 0.674
u 100%
0.696
error in y A
.
316%
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane
and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s
law to be valid for acetone mole fractions that are not very close to 1.
6-47
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oC
The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when
xC = 0 and at 61.0oC when xC = 1. (See solution to part c.)
b.
Txy Diagram for an Ideal Binary Solution
A
B
C
6.90328 1163.03
227.4
Chloroform
6.90565 1211.033
220.79
Benzene
760
P(mmHg)=
x
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
T
80.10
78.92
77.77
76.66
75.58
74.53
73.51
72.52
71.56
70.62
69.71
68.82
67.95
67.11
66.28
65.48
64.69
63.93
63.18
62.45
61.73
y
0
0.084
0.163
0.236
0.305
0.370
0.431
0.488
0.542
0.593
0.641
0.686
0.729
0.770
0.808
0.844
0.879
0.911
0.942
0.972
1
p1
0
63.90
123.65
179.63
232.10
281.34
327.61
371.15
412.18
450.78
487.27
521.68
554.15
585.00
614.02
641.70
667.76
692.72
716.27
738.72
760
p2
760
696.13
636.28
580.34
527.86
478.59
432.30
388.79
347.85
309.20
272.79
238.38
205.83
175.10
145.94
118.36
92.17
67.35
43.75
21.33
0
p1+p2
760
760.03
759.93
759.97
759.96
759.93
759.91
759.94
760.03
759.99
760.07
760.06
759.98
760.10
759.96
760.06
759.93
760.07
760.03
760.05
760
T xy diagram
(P =1 atm )
85
75
V apor
o
T( C)
80
70
Liquid
65
60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
M ole fraction of chloroform
6-48
0.8
0.9
1
6.69 (cont’d)
d.
Txy diagram
(P=1 atm)
85
T(oC)
80
yc
xc
75
70
x
65
y
60
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
Mole fraction of choloroform
'T
Tactual
Raoult’s law: Tbp = 71o C, y = 0.58 Ÿ
'y
y actual
71 75.3
u 100% 5.7% error in Tbp
75.3
0.58 0.60
u 100% 3.33% error in y
0.60
Benzene and chloroform are not structurally similar compounds (as are, for example,
pentane and hexane or benzene and toluene). There is consequently no reason to expect
Raoult’s law to be valid for chloroform mole fractions that are not very close to 1.
6.70 P | 1 atm 760 mm Hg
760 0.40 u 10
d i b
g d i
x m pm* Tbp 1 x m p *P Tbp
7 .878631411/( Tbp 230 )
0.60 u 10
7 .74416 1437 .686/( Tbp 198.463)
Solve
E-Z

o T
79.9 o C
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and
surface tension effects on the boiling point are negligible.
The liquid temperature will rise until it reaches 79.9 qC, where boiling will commence. The
escaping vapor will be richer in methanol and thus the liquid composition will become richer
in ethanol. The increasing fraction of the less volatile component in the residual liquid will
cause the boiling temperature to rise.
6-49
6.71 Basis: 1000 kg/h product
nH4 (mol H 2 /h)
scrubber
E = C2 H5 OH ( M = 46.05)
A = CH 3 CHO ( M = 44.05)
n3 (mol/h)
y A3 (mol A/mol), sat'd
y E3 (mol E/mol), sat'd
y H3 (mol H 2 /h)
vapor, –40°C
P = 760 mm Hg
Fresh feed
n0 (mol E/h)
nA1 (mol A/h)
nE1 (mol E/h)
280°C
reactor
nA2 (mol A/h)
nE2 (mol E/h)
nH2 (mol H 2/h)
condenser
nC (mol/h)
0.550 A
0.450 E
liquid, –40°C
Scrubbed
Hydrocarbons
nA4 (mol A/h)
nE4 (mol E/h)
still
Product
1000 kg/h
np (mol/h)
0.97 A
0.03 E
nr (mol/h)
0.05 A
0.95 E
Strategy
a.
d i
x
Calculate molar flow rate of product n p from mass flow rate and composition
x
Calculate y A3 and y E3 from Raoult’s law: y H3
the still involve fewest unknowns (n c and n r )
x
Total mole balance about still
Ÿ n c , n r
A balance about still
x
A, E and H 2 balances about scrubber Ÿ n A4 , n E4 , and n H4 in terms of n 3 .
Overall atomic balances on C, H, and O now involve only 2 unknowns ( n 0 , n 3 )
x
Overall C balance
Ÿ n 0 , n 3
Overall H balance
x
x
x
x
A balance about fresh feed-recycle mixing point Ÿ n A1
E balance about fresh feed-recycle mixing point Ÿ n E1
A, E, H 2 balances about condenser n A2 , n E2 , n H2
All desired quantities may now be calculated from known molar flow rates.
1 y A3 y E3 . Balances about
UV
W
UV
W
Molar flow rate of product
M
0.97 M A 0.03 M E
b0.97gb44.05g b0.03gb46.05g
n p
1000 kg 1 kmol
h
44.11 kg
22.67 kmol h
b g
p b 40q Cg
0.550 p b 40 q Cg
Table B.4 (Antoine) Ÿ
pA* 40q C
*
E
Raoult’s law Ÿ y A3
*
A
P
44.11 g mol
38.9 mm Hg
0.343 mm Hg
0.550(38.9)
760
6-50
0.02814 kmol A / kmol
6.71 (cont’d)
y E3
b
0.450 p E* 40 q C
g
0.450( 0.343)
2.03 u 10 4 kmol E kmol
760
0.9716 kmol H 2 kmol
P
1 y A3 y E3
y H3
UV
W
n r
Mole balance about still: n c n p n r Ÿ n c 22.67 n r
Ÿ
A balance about still: 0.550n c 0.97(22.67) 0.05n r
n c
29.5 kmol / h recycle
52.1 kmol / h
A balance about scrubber: n A4
n 3 y A3
0.02815n 3
(1)
E balance about scrubber: n E4
n 3 y E3
2.03 u 10 4 n 3
(2)
0.9716n 3
(3)
H 2 balance about scrubber: n H4
n 3 y H3
Overall C balance:
n 0 (mol E) 2 mol C
h
bn gb2g bn gb2g d0.97n ib2g d0.03n ib2g
A4
1 mol E
Ÿ n 0
E4
p
p
n A 4 n E 4 22.67
(4)
Overall H balance:
6n 0
b gb g b gb g
2n H4 4n A4 6n E4 n p 0.97 4 0.03 6
(5)
Solve (1)–(5) simultaneously (E-Z Solve):
n 0
23.3 kmol E / h (fresh feed), n H 4
22.6 kmol H 2 / h (in off - gas)
n 3 = 23.3 kmol / h, n A 4 = 0.66 kmol A / h, n E 4 = 0.0047 kmol E / h
A balance about feed mixing point: n A1
0.05n r
E balance about feed mixing point: n E1
n 0 0.95n r
E balance about condenser: n E2
1.47 kmol A h
n 3 y E3 0.450n c
51.33 kmol E h
23.47 kmol E h
Ideal gas equation of state:
. 5133
. g kmol 22.4 m bSTP g b273 + 280gK
b147
3
Vreactor feed
b. Overall conversion
h
n 0 0.03n p
Single-pass conversion
1 kmol
u 100%
n 0
n E1 n E2
u 100%
n E1
b gb
23.33 0.03 22.67
n E4
0.0047 kmol E h
6-51
2396 m 3 h
g u 100%
23.33
513
. 235
.
u 100%
51.3
Feed rate of A to scrubber: n A4 = 0.66 kmol A / h
Feed rate of E to scrubber:
273K
54%
97%
6.72 a. G = dry natural gas, W = water
n 3 (lb - mole G / d)
n 4 (lb - mole W / d)
10 lb m W / 10 6 SCF gas
90 o F, 500 psia
Absorber
n 7 (lb - mole W / d)
FG lb - mole TEG IJ
H d K
F lb - mole W IJ
n G
H d K
4.0 u 10 6 SCF / d
4 u 80 = 320 lb m W / d
n1 (lb - mole G / d)
n 2 [lb - mole W(v) / d]
n 5
Distillation
Column
6
FG lb- mole TEGIJ
H d K
F lb - mole WIJ
n G
H d K
n5
8
Overall system D. F. analysis:
5 unknowns (n1 , n 2 , n 3 , n 4 , n 7 )
2 feed specifications (total flow rate, flow rate of water)
1 water content of dried gas
2 balances (W, G)
0 D. F.
320 lb m W 1 lb - mole
Water feed rate: n 2
d
18.0 lb m
17.78 lb - moles W / d
Dry gas feed rate:
4.0 u 106 SCF 1 lb - mole
lb - moles W
. u 104 lb - moles G / d
17.78
1112
n1
d
d
359 SCF
n 3 Ÿ n 3
Overall G balance: n1
.
1112
u 10 4 lb - moles G / d
Flow rate of water in dried gas:
n 4
(n 3 n 4 ) lb - moles
d
3 1.112 u10
n
o n 4
4
359 SCF gas 10 lb m W 1 lb - mole W
lb - mole 10 6 SCF
18.0 lb m
2.218 lb - mole W(l) / d
Overall W balance:
n 7
(17.78 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-52
280
lb m W
u
d
F 1 ft I
GH 62.4 lb JK
3
m
4.5
ft 3 W
d
6.72 (cont’d)
b. Mole fraction of water in dried gas =
n 4
n 3 n 4
yw
2.218 lb - moles W / d
(2.218 + 1.112 u 10 4 ) lb - moles / d
. u 10 4
199
lb - moles W(v)
lb - mole
Henry’s law: ywP = Hwxw Ÿ
( x w ) max
(199
. u 10 4 )(500 psia)(1 atm / 14.7 psia)
0.398 atm / mole fraction
0.0170
lb - mole dissolved W
lb - mole solution
c. Solvent/solute mole ratio
n5
n2 n4
37 lb m TEG 1 lb - mole TEG 18.0 lb m W
lb m W
150.2 lb m TEG
1 lb m W
4.434
lb - mole TEG
lb - mole W absorbed
Ÿ n5 4.434(17.78 2.22) 69.0 lb - moles TEG / d
xw = 0.80(0.0170) = 0.0136
n6
lb - mole W
5 69.0
. lb - mole W/ d
=
n
o n6 0951
lb - mole
n5 n6
Solvent stream entering absorber
m
0.951 lb - moles W 18.0 lb m
69.0 lb - moles TEG
d
lb - mole
d
150.2 lb m
lb - mole
= 1.04 u 104 lb m / d
W balance on absorber
. 095
. 222
. ) lb - moles W/ d = 16.51 lb - moles W/ d
n8 (1778
1651
. lb - moles W / d
Ÿ xw
019
. lb - mole W / lb - mole
(16.51 + 69.9) lb - moles / d
c. The distillation column recovers the solvent for subsequent re-use in the absorber.
6.73 Basis: Given feed rates
G2
G1
G3
100 mol/h
200 mol air/h
n1 (mol/h)
0.96 H2
0.999 H 2
0.04 H2 S, sat'd
0.001 H 2S
1.8 atm
absorber
stripper
40°C
L2
0°C
L1
(mol/h)
(mol/h)
n3
n4
0.002 H 2S
x 3 (mol H 2 S/mol)
0.998 solvent
(1 – x 3) (mol solvent/mol)
0°C
heater
6-53
G4
200 mol air/h
n2 mol H 2S/mol
0.40°C, 1 at m
n3 (mol/h)
x 3 (mol H 2S/mol)
(1 – x 3) (mol solvent/mol)
40°C
6.73 (cont’d)
Equilibrium condition: At G1, p H 2S
p H 2S
Ÿ x3
b0.04gb18. atmg
0.072 atm
27 atm mol fraction
H H 2S
0.072 atm
2.67 u 10 3 mole H 2 S mole
Strategy: Overall H 2 and H 2 S balances Ÿ n1 , n 2
n 2 air flow rate Ÿ volumetric flow rate at G4
H 2 S and solvent balances around absorber Ÿ n 3 , n 4
0.998n 4 solvent flow rate
b100gb0.96g 0.999n Ÿ n
Overall H S balance: b100gb0.04g 0.001n n
Overall H 2 balance:
1
2
961
. mol h
1
1
n1 96.1
Ÿ n 2
2
3.90 mol H 2 S h
Volumetric flow rate at stripper outlet
VG4
b200 + 3.90gmol
h
b g b273 40gK
22.4 liters STP
1 mol
273 K
5240 L hr
H 2 S and solvent balances around absorber:
b100gb0.04g 0.002n 0.001n
0.998n
n d1 2.67 u 10 i
4
1
n 3 x 3 Ÿ n 4
1335
. n 3 1952
3
4
3
Solvent flow rate
0.998n 4
5820 mol solvent h
6.74 Basis: 100 g H 2 O
Sat'd solution @ 30°C
100 g H 2 O
11.1 g NaHCO3
Sat'd solution @ 60°C
100 g H 2 O
16.4 g NaHCO 3
ms (g NaHCO3 ( s))
bg
. ms Ÿ ms
NaHCO 3 balance Ÿ 16.4 111
% crystallization
5.3 g NaHCO 3 s
5.3 g crystallized
u 100%
16.4 g fed
32.3%
6.75 Basis: 1275 kg/h feed solution
m1 (kg H2 O(v )/h)
875 kg/h
x 0 (kg KOH/kg)
(1 – x0) (kg H 2O/kg)
Sat'd solution 10°C
m2 (kg H2 O(1)/h)
1.03 m2 (kg KOH/h)
m3 (kg KOH-2H 2O( s)/h)
60% of KOH in feed
6-54
|UV Ÿ n
W|
3
| n 4
5830 mol h
6.75 (cont’d)
Analysis of feed: 2KOH H 2 SO 4 o K 2 SO 4 2H 2 O
x0
bg
22.4 mL H 2 SO 4 l
1L
0.85 mol H 2 SO 4
3
5 g feed soln
10 mL
L
0.427 g KOH g feed
b
gb g
60% recovery: 875 0.427 0.60
h
b
737.2 kg KOH ˜ 2H 2 O h 288.3 kg H 2 O h
56.11 kg KOH
b g
KOH balance: 0.503 1275
56.11 g KOH
1 mol KOH
224.2 g KOH h
448.9 kg KOH 92.15 kg KOH ˜ 2H 2 O
m3
2 mol KOH
1 mol H 2 SO 4
448.9 0.97m2 Ÿ m2
b
g
198.4 kg h
g
Total mass balance: 875 737.2 197
. 198.4 m1 Ÿ m1
147.0 kg H 2 O h evaporated
6.76 a.
45
R 0 30
g A dissolved
CA 0 0.200 0.300
mL solution
Plot CA vs. R Ÿ CA = R / 150
CA
b. Mass of solution:
500 mol 1.10 g
550 g (160 g A, 390 g S)
ml
The initial solution is saturated at 10.2 qC.
160 g A
. g A 100 g S @ 10.2q C
0.410 g A g S 410
Solubility @ 10.2 qC
390 g S
17.5 150 g A 1 mL soln
At 0qC, R 17.5 Ÿ CA
0.106 g A g soln
. g soln
110
mL soln
Thus 1 g of solution saturated at 0qC contains 0.106 g A & 0.894 g S.
0106
.
gA
Solubility @ 0qC
g A g S 118
0118
.
. g A 100 g S @ 0q C
0.894 g S
390 g S 11.8 g A
Mass of solid A: 160 g A 114 g A s
100 g S
bg
c.
g A remaining in soln
0.5 u 390 g S 11.8 g A
160 114 g A 100 g S
b
g A initial
g
bg
23.0 g A s
6.77 a. Table 6.5-1 shows that at 50oF (10.0oF), the salt that crystallizes is MgSO 4 ˜ 7 H 2 O , which
contains 48.8 wt% MgSO4.
b. Basis: 1000 kg crystals/h.
m 0 (g/h) sat’d solution @ 130oF
m 1 (g/h) sat’d solution @ 50oF
0.23 g MgSO4/g
0.77 g H2O/g
0.35 g MgSO4/g
0.65 g H2O/g
1000 kg MgSO4∙7H2O(s)/h
6-55
6.77 (cont’d)
m 0
Mass balance: m 0 m 1 1000 kg / h
MgSO 4 balance: 0.35m0 0.23m 1 0.488(1000) kg MgSO 4 / h Ÿ m
1
The crystals would yield 0.488 u 1000 kg / h = 488
2150 kg feed / h
1150 kg soln / h
kg anhydrous MgSO 4
h
6.78 Basis: 1 lbm feed solution.
Figure 6.5-1 Ÿ a saturated KNO3 solution at 25oC contains 40 g KNO3/100 g H2O
Ÿ x KNO 3
40 g KNO 3
(40 + 100) g solution
0.286 g KNO 3 / g = 0.286 lb m KNO 3 / lb m x
1 lbm solution @ 80oC
0.50 lbm KNO3/lbm
0.50 lbm H2O/lbm
m1 m2
KNO3 balance: 0.50 lb m KNO3
m1(lbm) sat’d solution @ 25oC
0.286 lbm KNO3/lbm soln
0.714 lbm H2O/lbm soln
m2 [lbm KNO3(s)]
Mass balance: 1 lb m
0.286m1 m2
Ÿ
m1 = 0.700 lb m solution / lb m feed
m2 0.300 lb m crystals / lb m feed
0.300 lb m crystals / lb m feed
= 0.429 lb m crystals / lb m solution
0.700 lb m solution / lb m feed
Solid / liquid mass ratio =
6.79 a. Basis: 1000 kg NaCl(s)/h.
Figure 6.5-1 Ÿ a saturated NaCl solution at 80oC contains 39 g NaCl/100 g H2O
Ÿ x NaCl
39 g NaCl
(39 + 100) g solution
0.281 g NaCl / g = 0.281 kg NaCl / kg
m 2 [kg H 2 O(v) / h]
m 0 (kg/h) solution
m 1 (kg/h) sat’d solution @ 80oC
0.281 kg NaCl/kg soln
0.719 kg H2O/kg soln
1000 kg NaCl(s)/h
0.100 kg NaCl/kg
0.900 kg H2O/kg
0 m
1 m
2
Mass balance: m
1 m
2
NaCl balance: 0.100 kg NaCl 0281
. m
Solid / liquid mass ratio =
Ÿ
1 = 0.700 lb m solution / lb m feed
m
2 0.300 lb m crystals / lbm feed
m
0.300 lb m crystals / lbm feed
= 0.429 lb m crystals / lb m solution
0.700 lbm solution / lbm feed
The minimum feed rate would be that for which all of the water in the feed evaporates to
produce solid NaCl at the specified rate. In this case
6-56
6.79 (cont’d)
0 ) min
. (m
0100
1000 kg NaCl / h Ÿ (m 0 ) min
Evaporation rate: m 2
10,000 kg / min
9000 kg H 2 O / h
Exit solution flow rate: m 1
0
m 2 [kg H 2 O(v) / h]
b.
m 0 (kg/h) solution
m 1 (kg/h) sat’d solution @ 80oC
0.281 kg NaCl/kg soln
0.719 kg H2O/kg soln
1000 kg NaCl(s)/h
0.100 kg NaCl/kg
0.900 kg H2O/kg
40% solids content in slurry Ÿ 1000
0
NaCl balance: 0.100m
kg NaCl
= 0.400( m 1 ) max Ÿ ( m 1 ) max
h
0
. (2500) Ÿ m
0281
0 2500 m
2 Ÿ m
2
Mass balance: m
2500
kg
h
7025 kg / h
4525 kg H2 O evaporate / h
6.80 Basis: 1000 kg K 2 Cr2 O 7 (s) h . Let K = K 2 Cr2 O 7 , A = dry air, S = solution, W = water.
Composition of saturated solution:
0.20 kg K
0.20 kg K
kg K kg soln
01667
.
Ÿ
kg W
1+ 0.20 kg soln
b
g
n2 (mol / h)
y2 (mol W(v) / mol)
m e [kg W(v) / h)
(1 y2 )(mol A/ mol)
90o C, 1 atm, Tdp 392
. oC
1 (kg / h)
m
f (kg / h)
m
f m
r (kg / h)
m
CRYSTALLIZERCENTRIFUGE
0.210 kg K/ kg
0.90 kg K(s) / kg
DRYER
1000 kg K(s) / h
0.10 kg soln / kg
0.1667 kg K / kg
0.790 kg W(l) / kg
0.8333 kg W/ kg
na (mol A / h)
m r (kg recycle / h)
0.1667 kg K / kg
0.8333 kg W / kg
Dryer outlet gas: y 2 P
b
g
*
39.2q C Ÿ y 2
pW
f
Overall K balance: 0.210m
53.01 mm Hg
760 mm Hg
1000 kg K h Ÿ m f
6-57
0.0698 mol W mol
4760 kg h feed solution
6.80 (cont’d)
b
gb
1 01667
.
010
. m 1
K balance on dryer: 0.90m
g
1000 kg h Ÿ m 1 1090 kg h
Mass balance around crystallizer-centrifuge
f m r
m
1 m r Ÿ me
m e m
95% solution recycled Ÿ m r
4760 1090 3670 kg h water evaporated
b0.10 u 1090g kg
h not recycled
95 kg recycled
5 kg not recycled
2070 kg h recycled
Water balance on dryer
. gb1090g kg W h
b0.8333gb010
18.01 u 10
3
kg mol
0.0698n 2 Ÿ n 2
7.225 u 104 mol h
Dry air balance on dryer
na
b1 0.0698g7.225 u 10
4
b g
mol 22.4 L STP
h
1 mol
6-58
b g
151
. u 10 6 L STP h
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm
n 2w (kmol H 2 O )(sat' d)
100 kg Feed
0.07 kg Na 2 CO 3 / kg
Reactor
Reactor
0.93 kg H 2 O / kg
e
n 2c (kmol CO 2 )
n 2a (kmol Air)
70 o C, 3 atm(absolute)
n1 (kmol)
Filtrate
0.70 kmol CO 2 / kmol
m3 ( kg NaHCO 3 (s))
0.30 kmol Air / kmol
Filter
R|m (kg solution) U|
S|0.024 kg NaHCO / kgV|
T0.976 kg H O / kg W
4
3
2
m5 (kg)
0.024 kg NaHCO 3 / kg
0.976 kg H 2 O / kg
Filter cake
m6 (kg)
0.86 kg NaHCO 3 (s) / kg
R|0.14 kg solution U|
S|0.024 kg NaHCO / kgV|
T0.976 kg H O / kg W
3
2
Degree of freedom analysis:
Reactor
6 unknowns (n1, n2, y2w, y2c, m3, m4)
–4 atomic species balances (Na, C, O, H)
–1 air balance
–1 (Raoult's law for water)
0 DF
Filter
2 unknowns
–2 balances
0 DF
Na balance on reactor
100 kg 0.07 kg Na 2 CO3
(m3 0.024m4 ) kg NaHCO3
46 kg Na
kg
106 kg Na 2 CO3
Ÿ 3.038 0.2738(m3 0.024m4 )
(1)
Air balance: 0.300 n1
n2 a
23 kg Na
84 kg NaHCO3
( 2)
C balance on reactor :
n1 (kmol) 0.700 kmol CO 2
kmol
12 kg C
100 kg 0.07 kg Na 2 CO 3
12 kg C
1 kmol CO 2
kg
106 kg Na 2 CO 3
12
(n2c )(12) (m3 0.024m2 )( ) Ÿ 8.40n1 0.7924 12n2c 01429
.
(m3 0.024m4 )
(3)
84
H balance :
2
1
2
) (n2 w )(2) ( m3 0.024m4 )( ) 0.976m4 ( )
18
84
18
( 4)
Ÿ 10.33 2n2 w 0.01190(m3 0.024m4 ) 0.1084m4
(100)(0.93)(
6-59
6.81(cont'd)
O balance (not counting O in the air):
16
48
n1 (0.700)(932) 100 (0.07)(
) 100 (0.93)( )
18
106
16
48
(n2 w )(16) n2 c ( 32) (m3 0.024m4 )( ) 0.976m4 ( )
18
84
Ÿ 22.4n1 85.84 16n2 w 32n2 c 0.5714(m3 0.024m4 ) 0.8676m4
(5)
Raoult's Law :
yw P
p w* (70 o C) Ÿ
Ÿ n2 w
01025
.
(n2 w
n2 w
n2 w n 2 c n 2 a
n2 c n2 a )
233.7 mm Hg
(3 * 760) mm Hg
(6)
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to
converge).
n1
n2w
n2a 0.2426 kmol air,
n2c 0.500 kmol CO 2 ,
0.8086 kmol,
m3 8.874 kg NaHCO 3 (s),
m4 92.50 kg solution
0.0848 kmol H 2 O(v),
NaHCO3 balance on filter:
m3 0.024m4
0.024m5 m6 [0.86 (014
. )(0.024)]
m3 8.874
11.09
0.024m 5 0.8634m 6
(7)
m4 92 .50
Mass Balance on filter: 8.874 92.50 1014
.
Solve (7) & (8) Ÿ
Scale factor
m5
m6
91.09 kg filtrate
10.31 kg filter cake
500 kg / h
8.867 kg
m5 m6
(8)
Ÿ (0.86)(10.31) 8.867 kg NaHCO 3 (s)
56.39 h 1
(a) Gas stream leaving reactor
U|
V|
W
R|
|S
||
T
46.7kmol / h
n 2w (0.0848)(56.39) 4.78 kmol H 2 O(v) / h
0.102 kmol H 2 O(v) / kmol
n 2c (0.500)(56.39) 28.2 kmol O 2 / h
Ÿ
0.604 kmol CO 2 / kmol
n 2a (0.2426)(56.39) 13.7 kmol air / h
0.293 kmol Air / kmol
V2
n 2 RT
P
(b) Gas feed rate: V1
(46.7 kmol / h)(0.08206
3 atm
m 3 atm
)(343 K)
kmol ˜ K
438 m 3 / h
56.39 u 0.8086 kmol 22.4 m 3 (STP)
1h
17.0 SCMM
h
kmol
60 min
6-60
6.81(cont'd)
(c) Liquid feed: (100)(56.39) 5640 kg / h
To calculate V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution.
(d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual
solution would contain less than 2.4% NaHCO3.
(e)
Henry's law
Benefit: Higher pressure Ÿ greater pCO2
higher concentration of CO 2 in solution
Ÿ higher rate of reaction Ÿ smaller reactor needed to get the same conversion Ÿ lower cost
Penalty: Higher pressure Ÿ greater cost of compressing the gas (purchase cost of compressor,
power consumption)
6.82
600 lb m / h
Dissolution
Dissolution
0.90 MgSO4 ˜ 7H 2 O Dissolution
Tank
Tank
Tank
0.10 I
1 (lb m H 2 O / h)
m
Filter I
R|m (lb soln / h) U
S|0.32 kg MgSO / kg
T0.68 kg H O / kg W
2
4
4
6000 lb m I / h
6 (lb m / h)
m
o
R|300 lb soln / h|U
S|0.32 MgSO V|
T0.68 H O W
m
m
2
6000 lb m I / h
2
3 (lb m so ln/ h)
m
0.32 MgSO 4
110 F
0.23 lb m MgSO 4 / lb m
0.68 H 2 O
0.77 lb m H 2 O / lb m
4 (lb m MgSO 4 ˜ 7 H 2 O / h
m
Filter II
R|m (lb
|S0.23 lb
||0.77 lb
T
5
m
soln)
MgSO 4 / lb m
m
H 2 O / lb m
m
U|
|V
||
W
Crystallizer
4 (lb m MgSO 4 ˜ 7 H 2 O)
m
R|0.05m
S|0.23 lb
T0.77 lb
(lb m soln)
4
m
MgSO 4 / lb m
m
H 2 O / lb m
U|
V|
W
a. Heating the solution dissolves all MgSO4; filtering removes I, and cooling recrystallizes
MgSO4 enabling subsequent recovery.
(b) Strategy: Do D.F analysis.
6-61
6.82(cont'd)
UV
W
Overall mass balance
Ÿ m 1 , m 4
Overall MgSO 4 balance
( MW) MgSO4
UV
W
Diss. tank overall mass balance
Ÿ m 2 , m 6
Diss. tank MgSO 4 balance
(24.31 32.06 64.00) 120.37, ( MW) MgSO4 ˜7H2O
(12037
. 7 * 18.01) 246.44
Overall MgSO4 balance:
60,000 lb m 0.90 lb m MgSO 4 ˜ 7H 2 O
120.37 lb m MgSO 4
h
lb m
246.44 lb m MgSO 4 ˜ 7H 2 O
(300 lb m / h)(0.32 lb m MgSO 4 / lb m ) m 4 (120.37 / 246.44) 0.05m 4 (0.23)
Ÿ m 4
5.257 x10 4 lb m crystals / h
m 4 5.257 x104 lb m / h
Overall mass balance: 60,000 m 1
. m 4
6300 105
m 1
1494 lb m H 2 O / h
c.
Diss. tank overall mass balance:
Diss. tank MgSO 4 balance:
Ÿ
60,000 m 1 m 6 m 2 6000
54,000(120.37 / 246.44) 0.23m 6
m 2
. x10 5 lb m / h
1512
m 6
9.575x10 4 lb m / h recycle
Recycle/fresh feed ratio
9.575x10 4 lb m / h
1494 lb m / h
0.32m 2
UV
W
64 lb m recycle / lb m fresh feed
6.83 a.
n 1 (kmol CO 2 / h)
Cryst
Filter
1000 kg H 2SO 4 / h (10 wt%)
1000 kg HNO 3 / h
w (kg H 2 O / h)
m
Filter cake
5 (kg / h)
m
2 (kg CaSO4 / h)
m
3 (kg Ca(NO3 )2 / h)
m
0.96 kg CaSO 4 (s) / kg
4 (kg H2O/ h)
m
0.04 kg soln / kg
0 (kg CaCO 3 / h)
m
0 (kg solution / h)
2m
8 (kg soln / h)
m
0 (kg solution / h)
2m
0 (kg CaCO 3 / h)
m
R| X (kg CaSO / kg) U|
500 X (kg H O / kg)
Solution composition: S
|T(1 501X )(kg Ca(NO ) / kg)V|W
a
4
a
a
6-62
2
3 2
6.83 (cont’d)
b. Acid is corrosive to pipes and other equipment in waste water treatment plant.
c. Acid feed:
1000 kg H 2 SO 4 / h
(2000 m w ) kg / h
0.10 Ÿ m w
8000 kg H 2 O / h
Overall S balance:
1000 kg H 2 SO 4
h
32 kg S
5 (kg / h) (0.96 0.04 X a ) (kg CaSO 4 )
m
98 kg H 2 SO4
m8 (kg / h) X a (kg CaSO 4 )
kg
32 kg S
kg
136 kg CaSO 4
32 kg S
136 kg CaSO 4
5 (0.96 0.04 X a ) 0.2353m
8 Xa
Ÿ 3265
. 0.2353m
(1)
Overall N balance:
1000 kg HNO 3
14 kg N
h
63 kg HNO 3
0.04m 5 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 )
kg
8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 )
m
kg
28 kg N
164 kg Ca(NO 3 ) 2
28 kg N
164 kg Ca(NO 3 ) 2
Ÿ 222.2 0.00683m 5 (1 501X a ) 0171
. m 8 (1 501X a )
(2)
Overall Ca balance:
5 (kg / h) (0.96 0.04X a ) (kg CaSO 4 )
m
40 kg Ca
kg
136 kg CaSO 4
5 (kg / h)
(1 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m
40 kg Ca
0 (kg / h)
m
40 kg Ca
100 kg CaCO 3
kg
164 kg Ca(NO 3 ) 2
8 (kg / h) X a (kg CaSO 4 )
m
40 kg Ca
kg
136 kg CaSO 4
8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 )
m
40 kg Ca
kg
164 kg Ca(NO 3 ) 2
Ÿ 0.40m 0
0.294m 5 (0.96 0.04 X a ) 0.00976m 5 (1 501 X a )
0.294m 8 X a 0.244m 8 (1 501 X a )
(3)
Overall C balance :
0 (kg / h)
m
Ÿ 0.01m 0
12 kg C
100 kg CaCO 3
n1
n 1 (kmol CO 2 / h)
(4)
6-63
1 kmol C
1 kmol CO 2
12 kg C
1 kmol C
6.83 (cont’d)
Overall H balance :
1000 (kg H 2SO4 )
2 kg H
1000 kg HNO3
h
98 kg H 2 SO4
5 X a 5556
8 Xa
Ÿ 92517
2.22m
.
. m
(5)
1 kg H
h
w (kg / h)
m
2 kg H
63 kg HNO3
18 kg H 2 O
5 (kg / h) 500 X a (kg H 2 O) 2 kg H
8 (kg / h) 500 X a (kg H 2 O) 2 kg H
m
0.04m
kg
18 kg H 2 O
kg
18 kg H 2 O
Solve eqns. (1)-(5) simultaneously, using E-Z Solve.
m 0
1812.5 kg CaCO 3 (s) / h,
m 5
n1
18.1 kmol CO 2 / h(v),
Xa
Recycle stream
2 * m 0
1428.1 kg / h,
m 8
9584.9 kg soln / h,
0.00173 kg CaSO 4 / kg
3625 kg soln / h
.
CaSO
R| 0.00173(kg CaSO / kg) U| R||0173%
S| 500 * 0.00173(kg H O / kg) V| Ÿ S|86.5% H O
T(1 501* 0.00173)(kg Ca(NO ) / kg)W |T13.3% Ca(NO )
4
4
2
2
3 2
d.
3 2
U|
|V
||
W
From Table B.1, for CO2:
Tc
Pc 72.9 atm
304.2 K ,
T (40 273.2) K
Ÿ Tr
. ,
103
Tc
304.2
Pr
30 atm
72.9 atm
0.411
From generalized compressibility chart (Fig. 5.4-2):
z
0.86 Ÿ V
0.86 0.08206 L ˜ atm 313.2 K
L
0.737
mol ˜ K
30 atm
mol CO 2
zRT
P
Volumetric flow rate of CO2:
V
e.
n1 * V
18.1 kmol CO 2
0.737 L
1000 mol
h
mol CO 2
1 kmol
1.33x10 4 L / h
Solution saturated with Ca(NO3)2:
Ÿ
1 501X a (kg Ca(NO 3 ) 2 / kg)
500Xa (kg H 2 O / kg)
1.526 Ÿ X a
0.00079 kg CaSO 4 / kg
Let m 1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without
crystallization.
6-64
6.83 (cont’d)
Overall S balance:
5 (kg / h) (0.96 (0.04)(0.00079)) (kg CaSO4 )
m
1000 kg H 2SO4
32 kg S
h
98 kg H2SO4
kg
8 (kg / h) 0.00079 (kg CaSO4 )
m
kg
32 kg S
136 kg CaSO4
32 kg S
136 kg CaSO4
5 0.000186m
8
Ÿ 326.5 0.226m
(1')
Overall N balance:
1 (kg HNO3 )
m
14 kg N
h
63kg HNO3
5 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 ) 2 )
0.04m
kg
28 kg N
164 kg Ca(NO3 ) 2
8 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 )2 )
m
kg
28 kg N
164 kg Ca(NO3 ) 2
1 000413
5 0103
8
Ÿ 0222
m
. m
.
. m
(2')
Overall H balance:
1 kg HNO 3
m
1000 (kg H 2 SO 4 )
2 kg H
1 kg H
h
98 kg H 2 SO 4
h
63 kg HNO 3
5 (kg / h) 500(0.00079) (kg H 2 O)
8000 (kg / h)
2 kg H
0.04m
2 kg H
18 kg H 2 O
kg
18 kg H 2 O
8 (kg / h) 500(0.00079) (kg H 2 O)
m
2 kg H
kg
18 kg H 2 O
1 0.00175m
5 0.0439m 8
Ÿ 909.30 0.0159m
Solve eqns (1')-(3') simultaneously using E-Z solve:
m 1
1155
. x10 4 kg / h;
5
m
x10 3 kg / h;
1424
.
Maximum ratio of nitric acid to sulfuric acid in the feed
. x10 4 kg / h
1155
1000 kg / h
. kg HNO 3 / kg H 2 SO 4
115
6-65
8
m
2.484 x10 4 kg / h
(3' )
6.84
Moles of diphenyl (DP):
Moles of benzene (B):
0.363
619
. 0.363
Ÿ x DP
bg
bg
b
'Tbp
'Tbp
o
55
. 3.6 19
. qC
o
30,765
801
82.0 q C
. 185
.
RTbp2
'H mp
Ÿ Tbp
2
2
0.0q C, 'Tfp
114.0 mm Hg
m
v
6.85 Tfp
g
g b0.0554g 3.6 K = 3.6 C Ÿ T
9837
8.314b273.2 801
.g
. K = 1.85 C
b0.0554g 185
RTb02
x DP
'H
Ÿ Tb
0.945 120.67 mm Hg
8.314 273.2 55
.
2
RTm0
x DP
'H
m
'Tbp
U|
|V
619
. mol |
|W
0.0544 mol DP mol
b
(1 x DP ) p B* T
p B* T
'Tm
56.0 g
0.363 mol
154.2 g mol
550.0 ml 0.879 g 1 mol
ml
78.11 g
4.6q C
xu
b
g b0.0445g
8.314 273.2 100.0
40,656
2
. qC
13
. 101.3 q C
100.0 13
30
. q C Ÿ xu
'Tbp 'H mp
RTbp2
b
g
8.314b373.2g
3.0 40,656
2
mol urea mol
0105
.
Initial mass of urea:
1000 g soln
1 mol soln
0.0445 mol U 60.06 g U
60.06 0.0445 18.02 0.9555 g soln
mol soln
mol U
b
g
b
g
134 g urea
Moles of H 2 O:
b1000 134g g
Final moles of urea:
Added Urea:
1 mol
18.02 g
m
m 48.0
5.63 mol 60.06 g
mol
48.0 mol H 2 O
Ÿ m 5.63 mol urea
0105
.
134 g
204 g urea
6-66
b0.5150 gg b1101. g molg
b0.5150 gg b1101. g molg b100.0 gg b94.10 g molg
6.86 x aI
'TmI
RTm20
xs Ÿ
'TmII
'H m
'Tm
Ÿ
Ÿ
x sII
b1 0.00523g mol solvent
b g
b
0.49q C
0.41q C
'H vI
B , ln p s* Tbs
RTb 0
b g
6380 J mol
b g
Ÿ ln Ps* Tb 0 ln P0* Tbs
*
s
FG
H
*
0
bs
6.88
m 1 (g styrene)
90 g ethylbenezene
100 g EG
90 g ethylbenzene
30 g styrene
Styrene balance: m1 m2
m 2 (g styrene)
100 g EG
30 g styrene
m2
100 m2
6.38 kJ / mol
'H v 1
'H v Tbs Tb 0
1
#
R Tb 0 Tbs
R
Tb20
b0
Equilibrium relation:
mol solute
mol solution
'H vII
B
RTbs
IJ
K
p bT g b1 x g p bT g Ÿ lnb1 x g | x
b g
0.00523
2
Assume 'H vI # 'H vII ; T0 Ts # T02
b. Raoult’s Law:
0.00438
'TmI
g b0.00523g
0.49
'TmII
x sI
8.314 273.2 5.00
RTm20
xs
'Tm
ln p s* Tb 0
x sII
94.10 g solvent 0.4460 g solute
8350
. g solute mol
1 mol solvent 95.60 g solvent
0.00523 mol solute
'H m
6.87 a.
x sI
0.00438 mol solute mol
.
019
FG m IJ
H 90 m K
1
1
solve simultaneously
m1
25.6 g styrene in ethylbenzene phase
m2
4.4 g styrene in ethylene glycol phase
6-67
'H v 'Tb
RTb20
Ÿ 'Tb
RTb20
x
'H v
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h
0.05 kg A / kg
0.95 kg C / kg
95.0 kg C / h
2 kg A / h
m
3 kg A / h
m
1 kg P / h
m
1 kg P / h
m
a. 90% extraction: m 3
(0.09)(0.05)(100 kg / h) = 4.5 kg A / h
Balance on oleic acid: (0.05)(100)
Equilibrium condition:
.
015
m 2 4.5 kg A / h Ÿ m 2
0.5 / (n1 0.5)
Ÿ n1
4.5 / (4.5 95)
0.5 kg A / h
73.2 kg P / h
b. Operating pressure must be above the vapor pressure of propane at T=85oC=185oF
*
500 psi 34 atm
Figure 6.1-4 Ÿ p propane
c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental
hazards.
6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass
fraction of acetic acid in water.
Basis: 100 kg feed.
A=Acetic acid, W=H2O, H=Hexane, B=Benzene
100 (kg)
m1 (kg)
0.30 kg A / kg
0.10 kg A / kg
0.70 kg W / kg
0.90 kg W / kg
m2 (kg A)
m H (kg H)
m H (kg H) or m B (kg B)
or m B (kg B)
Balance on W:
100 * 0.70 m1 * 0.90 Ÿ m1
Balance on A:
100 * 0.30 m2 77.8 * 0.10 Ÿ m2
77.8 kg
22.2 kg
Equilibrium for H:
KH
m2 / (m2 m H )
xA
22.2 / (22.2 m H )
0.10
0.017 Ÿ m H
22.2 / (22.2 mB )
0.10
0.098 Ÿ m B
1.30 x10 4 kg H
Equilibrium for B:
KB
m2 / (m2 m B )
xA
2.20 x10 3 kg B
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and
environmental considerations.
6-68
6.91
a. Basis: 100 g feed Ÿ 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water
40 g A
60 g W
e 1 (g A)
60 g W
25°C
100 g H
100 g H
r 1 (g A)
25°C
75 g H
b
xA in H phase / xA in W phase
0.343 x mass fraction
Balance on A stage 1:
Equilibrium condition stage 1:
Balance on A stage 2:
Equilibrium condition stage 2:
75 g H
r 2 (g A)
g
U| e
Ÿ
0.343V| r
W
U| r
Ÿ
0.343V| e
W
40 e1 r1
r1 100 r1
e1
b
g
b60 e g
1
1
1
27.8 e2 r2
r2 75 r2
e2
b g
b60 e g
2
2
20.6 g A remaining
u 100%
40 g A fed
% acetone not extracted
e 2 (g A)
60 g W
2
27.8 g acetone
12.2 g acetone
7.2 g acetone
20.6 g acetone
515%
.
b.
40 g A
60 g W
e1 g A
60 g W
r1 g A
175 g H
175 g H
Balance on A stage 1:
Equilibrium condition stage 1:
% acetone not extracted
c.
U| r
b
g 0.343V Ÿ e
|W
b60 e g
40.0 e1 r1
r1 175 r1
e1
1
555%
.
19.4 g A
60 g W
20.6 g A
m (g H)
m (g H)
20.6 / (m 20.6)
19.4 / (60 19.4)
17.8 g acetone
22.2g acetone
1
22.2 g A remaining
u 100%
40 g A fed
40 g A
60 g W
Equilibrium condition:
1
0.343 Ÿ m
225 g hexane
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane
over process lifetime) – (cost of an equilibrium stage x number of stages). The most costeffective process is the one for which F is the highest.
6-69
6.92
a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
Broth
Mixing tank
100 kg
0.015 P
0.985 Ac
m1 (kg BA)
Extraction Unit I
Acid
D.F. analysis:
Extraction Unit I
3 unknown (m1, m2p, m3p)
–1 balance (P)
–1 distribution coefficient
–1 (90% transfer)
0 DF
m3P (kg P)
98.5 (kg Ac)
pH=2.1
m4 (kg Alk)
Extraction
II
m6P (kg P)
m1 (kg BA)
m5P (kg P)
m4 (kg Alk)
pH=5.8
Extraction Unit II (consider m1, m3p)
3 unknowns
–1 balance (P)
–1 distribution coefficient
–1 (90% transfer)
0 DF
b. In Unit I, 90% transfer Ÿ m3 P 0.90(15
. ) 135
. kg P
P balance:
15
. m2 P 1.35 Ÿ m2 P 015
. kg P
. / (135
. m1 )
135
Ÿ m1 34.16 kg BA
pH=2.1 Ÿ K 25.0
. / (015
. 98.5)
015
In Unit II, 90% transfer: m5 P 0.90(m3 P ) 1.215 kg P
m3 P 1215
P balance:
.
m6 P Ÿ m6 P 0.135 kg P
m6 P / (m6 P 34.16)
pH=5.8 Ÿ K 0.10
Ÿ m4 29.65 kg Alk
1.215 / (1.215 m4 )
m1 34.16 kg BA
0.3416 kg butyl acetate / kg acidified broth
100 100 kg broth
m4 29.65 kg Alk
0.2965 kg alkaline solution / kg acidified broth
100 100 kg broth
Mass fraction of P in the product solution:
m5 P
1215
.
P
xP
m4 m5 P (29.65 + 1.215) kg
0.394 kg P / kg
c. (i). The first transfer (low pH) separates most of the P from the other broth constituents,
which are not soluble in butyl acetate. The second transfer (high pH) moves the
penicillin back into an aqueous phase without the broth impurities.
(ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the
aqueous phase.
(iii).The penicillin always moves from the raffinate solvent to the extract solvent.
6-70
6.93
W = water, A = acetone, M = methyl isobutyl ketone
xW
xA
xM
0.20
0.33
0.47
U|
V| Ÿ
W
Figure 6.6-1
Phase 1: x W
0.07, x A
0.35, x M
0.58
Phase 2: x W
0.71, x A
0.25, x M
0.04
Basis: 1.2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2.
H 2 O Balance:
Acetone balance:
R|
S|
T
m1 0.95 kg in MIBK - rich phase
. * 0.20 0.07m1 0.71m2
12
Ÿ
m2 0.24 kg in water - rich phase
. * 0.33 0.35m1 0.25m2
12
6.94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK
Overall system composition:
U|
b
g
V
3500 g b20 wt% A, 80 wt% M g Ÿ 700 g A, 2800 g M |W
2200 g A U
|
Ÿ 3500 g WV Ÿ 25.9% A, 41.2% W, 32.9% M
2800 g M |W
5000 g 30 wt% A, 70 wt% W Ÿ 1500 g A, 3500 g W
Phase 1: 31% A, 63% M, 6% W
Phase 2: 21% A, 3% M, 76% W
Fig. 6.6-1
Let m1=total mass in phase 1, m2=total mass in phase 2.
H 2 O Balance:
Acetone balance:
R|
S|
T
m1 4200 g in MIBK - rich phase
3500 0.06m1 0.76m2
Ÿ
m2 4270 g in water - rich phase
2200 0.31m1 0.21m2
6.95 A=acetone, W = H2O, M=MIBK
32 lb m / h
41.0 lb m / h
x AF (lb m A / lb m )
x A,1 , x W,1 , 0.70
x WF (lb m W / lb m )
2 lb m / h
m
x A ,2 , x W ,2 , x M ,2
1 (lb m M / h)
m
Figure 6.6-1Ÿ Phase 1: x M
Phase 2: x w ,2
Overall mass balance:
MIBK balance:
0.700 Ÿ x w ,1
0.05; x A,1
0.25 ;
0.81; x A ,2
0.81; x M ,2
0.03
UV
W
m 1
32.0 lb m / h m 1 410
. lb m h m 2
Ÿ
m 1 410
m 2
. * 0.7 m 2 * 0.03
6-71
281
. lb m MIBK / h
19.1 lb m h
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK
System 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W
x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol W
UV
W
maq,1
maq ,1 morg ,1 100
Ÿ
Acetone balance: maq ,1 * 0.275 morg ,1 * 0.375 33.33
morg ,1
Mass balance:
. kg
417
58.3 kg
System 2: x a,org = 0.100 mol A, x m,org = 0.870 mol M, x w,org = 0.030 mol W
x a,aq = 0.055 mol A, x m,aq = 0.020 mol M, x w,aq
= 0.925 mol W
UV
W
m aq,2
maq ,2 morg ,2 100
Mass balance:
Ÿ
9
.
Acetone balance: maq ,2 * 0.055 morg ,2 * 0100
morg,2
b. K a ,1
x a ,org ,1
x a ,aq ,1
0.375
136
. ;
0.275
K a ,2
x a ,org ,2
x a ,aq ,2
22.2 kg
77.8 kg
0100
.
182
.
0.055
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK
into water.
c.
E
x a ,org / x w ,org
aw,1
0.375 / 0.075
12.3; E
aw,2
0.275 / 0.675
x a ,aq / x w ,aq
If water and MIBK were immiscible, x w ,org
d.
0Ÿ E
aw
0100
.
/ 0.040
0.055 / 0.920
418
.
of
Organic phase= extract phase; aqueous phase= raffinate phase
E a ,w
( x a / x w ) org
( x a ) org / ( x a ) aq
( x a / x w ) aq
( x w ) org / ( x w ) aq
Ka
Kw
When it is critically important for the raffinate to be as pure (acetone-free) as possible.
6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK
e 1
(kg / h)
x1A (kg A / kg)
x 2A (kg A / kg)
x1W (kg W / kg)
x 2W (kg W / kg)
x1M (kg M / kg)
x 2M (kg M / kg)
200 kg / h
0.30 kg A / kg
e 2
(kg / h)
r1
Stage I
0.70 kg M / kg
(kg / h)
y1A (kg A / kg)
Stage II
Stage
IIS
y1M (kg M / kg)
6-72
(kg / h)
y 2A (kg A / kg)
y 2W (kg W / kg)
y1W (kg W / kg)
300 kg W / h
r2
300 kg W / h
y 2M (kg M / kg)
6.97(cont'd)
Overall composition of feed to Stage 1:
b200gb0.30g
200 60
U| 500 kg h
140 kg M h V Ÿ
12% A, 28% M, 60% W
300 kg W h |W
60 kg A h
Figure 6.6-1 Ÿ
Extract: x1A
Raffinate: y1A
Mass balance
Acetone balance:
0.095, x1W
0.880, x1M
0.15, y1W
0.025
0.035, y1M
R|
S|
T
e1
500 e1 r1
Ÿ
r1
60 0.095e1 015
. r1
0.815
273 kg / h
227 kg / h
Overall composition of feed to Stage 2:
U| 527 kg h
. g 34 kg A h
b227gb015
b227gb0.815g 185 kg M h V Ÿ 6.5% A, 35.1% MIBK, 58.4% W
b227gb0.035g 300 308 kg W h|W
Figure 6.6-1 Ÿ
Extract: x 2 A
Raffinate: y 2 A
Mass balance:
Acetone balance:
0.04, x 2 W
0.94, x 2 M
0.085, y 2 W
0.02
0.025, y 2 M
R|
S|
T
e2
527 e2 r2
Ÿ
r2
34 0.04e2 0.085r2
0.89
240 kg / h
287 kg / h
Acetone removed:
[60 (0.085)(287)] kg A removed / h
60 kg A / h in feed
0.59 kg acetone removed / kg fed
Combined extract:
Overall flow rate = e1 e2
273 240 513 kg / h
( x1 A e1 x 2 A e2 ) kg A
0.095 * 273 0.04 * 240
513
0.069 kg A / kg
Water:
( x1w e1 x 2 w e2 ) kg W
e1 e2
0.88 * 273 0.94 * 240
513
0.908 kg W / kg
MIBK:
( x1 M e1 x 2 M e2 ) kg M
(e1 e2 ) kg
Acetone:
0.025 * 273 0.02 * 240
513
6-73
0.023 kg M / kg
6.98. a.
1.50 L / min
25o C, 1atm, rh = 25%
n0 (mol / min)
M (g gel)
Ma (g H2O)
y0 (mol H2 O / mol)
(1- y0 ) (mol dry air / mol)
n 0
PV
RT
(1 atm)(1.50 L / min)
(0.08206 L ˜ atm / mol ˜ K)(298 K)
r.h.=25%Ÿ
pH2O
0.25
pH* 2O (25o C)
Silica gel saturation condition:
Water feed rate:
Ÿ m H
X*
12.5
p H 2O
p H* 2 O
0.25 p H* 2 O (25o C )
y0
0.06134 mol / min
p
12.5 * 0.25 3125
.
0.25(23.756 mm Hg)
760 mm Hg
0.06134 mol 0.00781 mol H 2 O 18.01 g H 2 O
2O
min
mol
mol H 2 O
g H 2 O ads
100 g silica gel
0.00781
mol H 2 O
mol
0.00863 g H 2 O / min
Adsorption in 2 hours (0.00863 g H 2 O / min)(120min) 1.035 g H 2 O
Saturation condition:
1.035 g H 2 O
M (g silica gel)
3.125 g H 2 O
ŸM
100 g silica gel
33.1 g silica gel
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and
T are constant.
b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a
significant fraction of the water in the entering air and relatively little oxygen and nitrogen.
The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel
reaches its capacity. If air were still fed to the column past this point, no further
dehumidification would take place. To keep this situation from occurring, the gel is
replaced at or (preferably) before the time when it becomes saturated.
6.99 a.
Let c = CCl4
Relative saturation
0.30 Ÿ
pc
*
pc (34 o C)
Ÿ pc
0.30 * (169 mm Hg) 50.7 mm Hg
b. Initial moles of gas in tank:
n0
P0V0
RT0
1 atm
50.0 L
.
mol
1985
0.08206 L ˜ atm / mol ˜ K 307 K
Initial moles of CCl4 in tank:
nc0
y c 0 n0
pc 0
n0
P0
50.7 mm Hg
u 1.985 mol
760 mm Hg
6-74
0.1324 mol CCl 4
6.99 (cont’d)
50% CCl4 adsorbed Ÿ nc
0.500nc 0
n0 nads
Total moles in tank: n tot
0662 mol CCl 4 (= nads)
(1.985 0.0662) mol = 1.919 mol
Pressure in tank. Assume T = T0 and V = V0.
n tot RT0
V0
P
nc
n tot
yC
. )(0.08206)(307)
FG (1919
I F 760 mm Hg IJ
atmJ G
H
K H atm K
50.0
0.0662 mol CCl 4
.
1919
mol
Ÿ pc
nc
nc 1853
.
Ÿ n tot
mol CCl 4
mol
0.0345(760 mm Hg) = 26.2 mm Hg
n0 nc 0
c. Moles of air in tank: na
yc
0.0345
0.001
nc nair
(1.985 0.1324) mol air = 1.853 mol air
mol CCl 4
Ÿ nc
mol
u 10 3 mol CCl 4
.
1854
mol
.
1854
LM n RT OP = 1854
u 10 mol
.
50.0 L
N V Q
3
pc
yc P
735 mm Hg
0.001
0
tot
0
0.08206 L ˜ atm 307 K 760 mm
mol ˜ K
1 atm
0.710 mm Hg
X*
FG g CCl IJ
H g carbon K
4
0.0762 pc
Ÿ X*
1 0.096 pc
Mass of CCl4 adsorbed
mads
(nc 0 nc )( MW ) c
0.0762( 0.710)
1 0.096(0.710)
0.001854) mol CCl 4
(01324
.
20.3 mol CCl 4 adsorbed
20.3 g CCl 4 ads
Mass of carbon required: mc
g CCl 4 ads
0.0506
g carbon
a.
X*
E
Ÿ ln X *
K F p NO
2
ln(PNO2)
6.100
0.0506
0
E
ln K F E ln p NO
2
1
2
ln(X*)
6-75
153.85 g
1 mol CCl 4
400 g carbon
y = 1.406x - 1.965
2
1.5
1
0.5
0
-0.5
-1
-1.5
g CCl 4 adsorbed
g carbon
3
6.100 (cont’d)
Ÿ X*
.
ln p NO2 1965
.
1406
ln X *
.406
e 1.965 p 1NO
2
.406
p 1NO
.
0140
2
0.140 (kg NO 2 / 100 kg gel)(mm Hg) 1.406 ; E
KF
1406
.
S * (0.05m) 2 (1m) 10 3 L 0.75 kg gel
5.89 kg gel
1m 3
L
Maximum NO2 adsorbed :
b. Mass of silica gel : mg
p NO 2 in feed
mads
0.010(760 mm Hg)
0.140(7.60) 1.406 kg NO 2
100 kg gel
7.60 mm Hg
5.89 kg gel
0.143 kg NO 2
Average molecular weight of feed :
MW
0.01( MW ) NO2 0.99( MW ) air
(0.01)(46.01) (0.99)(29.0)
29.17
kg
kmol
Mass feed rate of NO2:
m
8.00 kg
1 kmol
0.01 kmol NO 2
46.01 kg NO 2
h
29.17 kg
kmol
kmol NO 2
Breakthrough time:
tb
0.143 kg NO 2
0.126 kg NO 2 / h
1.13 h
0.126
kg NO 2
h
68 min
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for
another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish
at 2.26 h. Since the first column would still be in the regeneration stage, a third column
would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first
column would be available for another run. The first few cycles are shown below on a
Gantt chart.
Run
Regenerate
Column 1
0
1.13
2.63
3.39
4.52
6.02
Column 2
1.13
2.26
3.76
4.52
5.65
Column 3
2.26
6-76
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add mA (kg A)
mS (kg S)
mS (kg S)
mI (kg I)
m I0 (kg I)
R0 (color units / kg S)
R (color units / kg S)
Come to equilibrium
V (L)
V (L)
mA (kg A)
mIA (kg I adsorbed)
Assume
x
no sucrose is adsorbed
solution volume (V) is not affected by addition of the carbon
m
a. R(color units/kg S) kCi (kg I / L) = k I
(1)
V
x
Ÿ 'R = k (Ci 0 Ci )
mIA mI 0 mI
k
( mI 0 m I )
V
'R
kmIA
V
'R
x100%
R0
km IA / V
m
x100 100 IA
kmI 0 / V
mI 0
m IA
Equilibrium adsorption ratio: X i*
mA
Normalized percentage color removal:
% removal of color
X
% removal ( 3) 100 m IA / m I 0
=
m A / mS
m A / mS
m
Ÿ X = 100X *i S Ÿ X i*
mI 0
Freundlich isotherm X i*
Ÿ X=
100mS K F
E
RE
9.500
9.000
mI 0
X
100mS
E;
8.500
8.000
0.000 1.000 2.000 3.000
6-77
(3)
(4)
(5)
R
KF ( )E
k
K F' R E
y = 0.4504x + 8.0718
ln R
(2)
m IA mS
m A mI 0
mI 0
X
100mS
(1),( 5)
K F Ci E
mI 0 k
100
A plot of ln X vs. ln R should be linear: slope
ln v
6.101
intercept = lnK 'F
6.101 (cont’d)
ln X
0.4504 ln p NO2 8.0718 Ÿ X
Ÿ K F'
3203, E
X
3203R 0.4504
0.4504
b. 100 kg 48% sucrose solution Ÿ m S
95% reduction in color
e 8.0718 R 0.4504
480 kg
Ÿ R = 0.025(20.0) = 0.50 color units / kg sucrose
K F' R E
3203(0.50) 0.4504 2344
% color reduction
97.5
Ÿ 2344 =
Ÿ mA
m A / mS
m A / 480
6-78
20.0 kg carbon
CHAPTER SEVEN
7.1
. u 10 4 kJ 0.30 kJ work
1h
1 kW
0.80 L 35
2.33 kW Ÿ 2.3 kW
h
L
1 kJ heat
3600 s 1 k J s
2.33 kW 10 3 W 1.341 u 10 3 hp
312
. hp Ÿ 3.1 hp
1W
1 kW
7.2
All kinetic energy dissipated by friction
(a) E k
mu 2
2
5500 lbm 552 miles 2
2
h2
52802 ft 2
12 mile 2
12 h 2
36002 s 2
1 lbf
9.486 u 10 4 B
32.174 lbm ˜ ft / s2 0.7376 ft ˜ lb f
715 Btu
(b)
3 u 10 8 brakings 715 Btu 1 day
1h
1W
1 MW
4
2
day
braking 24 h 3600 s 9.486 u 10 Btu / s 10 6 W
7.3
(a) Emissions:
1000 sacks
Paper Ÿ
Plastic Ÿ
2000 sacks
1000 sacks
(0.0045 0.0146) oz
(724 905) Btu
sack
Plastic Ÿ
2000 sacks
1 lb m
sack 16 oz
sack
Energy:
Paper Ÿ
(0.0510 0.0516) oz
(185 464) Btu
sack
2617 MW
6.41 lb m
1 lb m
16 oz
2.39 lb m
1.63 u 10 6 Btu
1.30 u 10 6 Btu
(b) For paper (double for plastic)
Materials
for 400 sacks
Raw Materials
Acquisition and
Production
Sack
Production and
Use
7-1
1000 sacks
400 sacks
Disposal
7.3 (cont’d)
Emissions:
Paper Ÿ
400 sacks
Plastic Ÿ
800 sacks
0.0510 oz
1 lb m 1000 sacks
sack 16 oz
0.0516 oz
1 lb m
4.5 lb m
sack 16 oz
Ÿ 30% reduction
0.0045 oz
1 lb m 2000 sacks
sack 16 oz
0.0146 oz
1 lb m
2.05 lb m
sack 16 oz
Ÿ 14% reduction
Energy:
Paper Ÿ
400 sacks
Plastic Ÿ
(c) .
724 Btu
sack
800 sacks
185 Btu
sack
3 u 10 8 persons
905 Btu
1000 sacks
119
. u 10 6 Btu; 27% reduction
sack
464 Btu
2000 sacks
sack
1 sack
1 day
1h
person - day
24 h
3600 s
1.08 u 10 6 Btu; 17% reduction
649 Btu
1J
1 MW
-4
1 sack 9.486 u 10 Btu 10 6 J / s
2,375 MW
. (2,375 MW) = 404 MW
Savings for recycling: 017
(d) Cost, toxicity, biodegradability, depletion of nonrenewable resources.
7.4
(a) Mass flow rate: m
Stream velocity: u
Kinetic energy: E k
3.00 gal
1 ft 3
(0.792)(62.43) lb m
min
7.4805 gal
1 ft 3
1728 in 3
3.00 gal
b g
7.4805 gal 3 0.5
min
mu 2
2
1
0.330 lb m
s
d7.70 u 10
3
. g ft
b1225
2
2
s
2
2
in
2
3
f
(b) Heat losses in electrical circuits, friction in pump bearings.
7-2
1 ft
1 min
12 in
60 s
ft s
.
1225
ft ˜ lb f
1
1
lb f
7.70 u 103
2
s
2 32.174 lb m ˜ ft / s
. u 10 hp I
iFGH 01341
.7376 ft ˜ lb / sJK
ft ˜ lb f / s
1 min
0.330 lb m s
60 s
140
. u 105 hp
7.5
(a) Mass flow rate:
b
g
42.0 m S 0.07 m
m
s
2
mu
2
E k
2
10 3 L 273 K 130 kPa
1 mol
3
573 K 101.3 kPa 22.4 L STP
1m
b g
4
42.0 2 m2
1N
1J
2
2
s
1 kg ˜ m / s N ˜ m
127.9 g 1 kg
2
s 1000 g
(b)
b g
127.9 g 1 mol 273 K 101.3 kPa 22.4 L STP
s
573 K
29 g
130 kPa
29 g
mol
113 J s
1 m3
4
10 L S (0.07)2 m2
3
1 mol
127.9 g s
49.32 m s
127.9 g 1 kg
49.32 2 m2
1N
1J
1558
. J/s
2
2
s 1000 g
s
1 kg ˜ m / s2 N ˜ m
'E k = E k (400 $ C) - E k (300 $ C) = (155.8 - 113) J / s = 42.8 J / s Ÿ 43 J / s
2
mu
2
E k
(c) Some of the heat added goes to raise T (and hence U) of the air
7.6
(a)
'E p
mg'z
7.4805 gal
mu 2
'E p Ÿ
2
(b) E k
.
62.43 lb m 32174
ft 10 ft
1 ft 3
1 gal
1 ft
3
b g
mg 'z Ÿ u
s2
b g
2 g 'z
12
1 lbf
32.174 lbm ˜ ft / s2
LM2FG 32.174 ft IJ b10 ftgOP
NH sK Q
2
834
. ft ˜ lb f
12
25.4
ft
s
(c) False
7.7 (a)
'E k Ÿ positive When the pressure decreases, the volumetric flow rate increases, and
hence the velocity increases.
'E Ÿ negative The gas exits at a level below the entrance level.
p
b g
2
5 m S 1.5 cm 2
(b) m
1 m3
4
s
10 cm
273 K
2
10 bars
1 kmol
16.0 kg CH 4
3
1 kmol
.
303 K 101325
bars 22.4 m
bSTPg
0.0225 kg s
d h
P d AV h
Pout AVout
in
in
V
nRT
Ÿ out
nRT
Vin
Ÿ uout
7.8
'z
'E p mg
uin
Pin
Pout
uout (m / s) ˜ A(m2)
Pin
Ÿ
Pout
uin (m / s) ˜ A(m2)
bar
b g 109 bar
5ms
Pin
Pout
5555
ms
.
105 m3 103 L 1 kg H2 O 981
1J
2.778 u 107 kW˜ h
1N
. m 75 m
h 1 m3
s2
1 kg ˜ m/ s2 1 N ˜ m
1J
1L
204
. u 104 kW˜ h h
The maximum energy to be gained equals the potential energy lost by the water, or
2.04 u 10 4 kW ˜ h 24 h 7 days
h
1 day 1 week
7-3
3.43 u 10 6 kW ˜ h week (more than sufficient)
7.9
(b) Q W
'U 'E k 'E p
b
g
0 b no height changeg
'E k
0 system is stationary
'E p
Q W
'U , Q 0, W ! 0
(c) Q W
'U 'E k 'E p
b
g
b
g
Q 0 adiabatic , W 0 no moving parts or generated currents
'E k 0 system is stationary
'E p 0 no height change
'U
(d). Q W
b
b
g
g
0
'U 'E k 'E p
b
g
W 0 no moving parts or generated currents
'E k 0 system is stationary
'E p 0 no height change
'U , Q 0
Q
b
b
g
g
Even though the system is isothermal, the occurrence of a chemical
reaction assures that 'U z 0 in a non-adiabatic reactor. If the
temperature went up in the adiabatic reactor, heat must be transferred
from the system to keep T constant, hence Q 0 .
7.10 4.00 L, 30 °C, 5.00 bar Ÿ V (L), T (°C), 8.00 bar
(a). Closed system:
'U 'E k 'E p
RS'E
|T'E
'U
(b)
Constant T Ÿ 'U 0 Ÿ Q W
(c) Adiabatic Ÿ Q
bg
S 3
0 Ÿ 'U
Q W
b
b
0 initial / final states stationary
0 by assumption
k
p
g
g
Q W
7.65 L ˜ bar
W
8.314 J
0.08314 L ˜ bar
765 J
transferred from
gas to
surroundings
7.65 L ˜ bar > 0, Tfinal ! 30q C
1 m2
2.83 u 10 3 m 2
4
2
10 cm
(a) Downward force on piston:
7.11 A
Fd
cm 2
Patm A mpiston+weight g
1 atm 1.01325 u 105 N / m2 2.83 u 10 3 m2
atm
7-4
24.50 kg 9.81 m
s
2
1N
1 kg ˜ m / s2
527 N
7.11 (cont’d)
Upward force on piston: Fu
APgas
d2.83 u 10
3
i d
m 2 Pg N m 2
i
Equilibrium condition:
Fu
Fd Ÿ 2.83 u 10 3 m2 ˜ P0
V0
303 K
1.01325 u 105 Pa 0.08206 L ˜ atm
nRT 1.40 g N 2 1 mol N 2
0.677 L
P0
28.02 g 1.86 u 105 Pa
1 atm
mol ˜ K
527 Ÿ P0
(b) For any step, 'U 'E k 'E p
Step 1: Q | 0 Ÿ 'U
Step 2: 'U
that V
186
. u 10 5 N m 2
Q W Ÿ 'U
'E k 0
'E p 0
186
. u 105 Pa
Q W
W
Q W As the gas temperature changes, the pressure remains constant, so
nRT Pg must vary. This implies that the piston moves, so that W is not zero.
Tfinal Ÿ 'U
Overall: Tinitial
0Ÿ Q W
0
In step 1, the gas expands Ÿ W ! 0 Ÿ 'U 0 Ÿ T decreases
(c) Downward force Fd
u 10 id2.83 u 10 i b4.50gb9.81gb1g
. gd101325
.
b100
3
5
331 N (units
as in Part (a))
Final gas pressure Pf
Since T0
Tf
F
A
30q C , Pf V f
Distance traversed by piston
ŸW
331 N
2.83 u 10 3 m 2
P0V0 Ÿ V f
'V
A
.
mg
b331 Ngb0142
Fd
116
. u 10 5 N m 2
V0
P0
Pf
b1.08 0.677g L
. u 10
b0.677 Lg 186
116
. u 10
5
5
Pa
Pa
1 m3
10 3 L
2.83 u 10 3 m2
108
. L
m
.
0142
47 N ˜ m 47 J
Since work is done by the gas on its surroundings, W
47 J Ÿ Q
Q W 0
47 J
(heat transferred to gas)
7.12 V
H
32.00 g 4.684 cm3
mol
U PV
103 L
0.1499 L mol
106 cm3
41.64 atm 0.1499 L
8.314
J / (mol ˜ K)
1706 J mol mol
0.08206 L ˜ atm / (mol ˜ K)
g
7-5
2338 J mol
d
i
7.13 (a) Ref state U
(b) 'U
0 Ÿ liquid Bromine @ 300 K, 0.310 bar
U final U initial
0.000 28.24
d i
' H
'U ' PV
'U P'V (Pressure Constant)
0.310 bar
' H 2824
. kJ mol 79.94g L
.
b00516
8.314 J
1 kJ
307
. kJ mol
mol 0.08341 L ˜ bar 103 J
. kJ Ÿ 154 kJ
b5.00 molgb30.7 kJ / molg 15358
U independent of P Ÿ U b300 K, 0.205 bar g U b300 K, 0.310 bar g 28.24 kJ mol
U d340 K, P i U b340 K, 1.33 bar g 29.62 kJ mol
n' H
'H
(c)
28.24 kJ mol
f
'U U final U initial
E
'U
29.62 28.24 1380
kJ mol
.
= P' V'
Ÿ V'
= PV
/ P'
V changes with pressure. At constant temperature Ÿ PV
(T = 300K, P = 0.205 bar) = (0.310 bar)(79.94 L / mol)
V'
0.205 bar
5.00 L
1 mol
n
0.0414 mol
120.88 L
'U n'U 0.0414 mol 138
. kJ / mol 0.0571 kJ
b
gb
'U 'E k 'E p
0
0
Q W ŸQ
120.88 L / mol
g
0.0571 kJ
0
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is
being neglected; internal energy is not completely independent of pressure.
7.14 (a) By definition H U PV ; ideal gas PV
RT Ÿ H
U RT
b g U bT g Ÿ H bT , Pg U bT g RT H bT g independent of P
U T , P
(b) ' H
'H
'U R'T
n' H
3500
cal 1.987 cal 50 K
mol ˜ K
mol
b2.5 molgb3599 cal / molg
7.15 'U 'E k 'E p
Q Ws
b
b
3599 cal mol
8998 cal Ÿ 9.0 u 10 3 cal
g
' E k 0 no change in m and u
' E p 0 no elevation change
Ws P'V since energy is transferred from the system to the surroundings
'U
Q W Ÿ 'U
g
b
Q P'V Ÿ Q
'U P'V
7-6
g
' (U PV )
'H
b
b
7.16. (a) ' E k
'Ep
g
0 u1 u 2 0
0 no elevation change
g
'P 0 (the pressure is constant since restraining force is constant, and area is constrant)
Ws P'V the only work done is expansion work
H 34980 355
. T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J
125 u 103 Pa
1 m3
785 cm3
PV
n=
0.0295 mol
RT 8.314 m3 ˜ Pa / mol ˜ K 400 K 106 cm3
-H
) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol)
Q = 'H = n(H
b
.
g
2
1
2
83.8 J = 0.0295 35.5T2 - 35.5(400) Ÿ T2
480 K
0.0295 mol 8.314 m3 ˜ Pa 106 cm3 480 K
941 cm3
mol ˜ K
125 u 105 Pa
1 m3
125 u 105 N (941 - 785)cm3 1 m3
P'V
19.5 J
m2
106 cm3
'U P'V Ÿ 'U Q 'PV 838
. J 19.5 J 64.3 J
nRT
P
i) V
ii ) W
iii ) Q
(b) 'Ep
0
7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could
occur, the temperature would drop during these periods.)
(b) 'U ' E p ' E R Q 't W 't
'E p
Q
0, ' E k
0.90 u 1.4 W
0, W
1 J s
1W
0, U ( t
0)
0
1.26 J s
U (J ) 126
. t
Moles in tank: n
U
U
n
PV RT
. t (J)
126
0.0859 mol
1 atm
b
2.10 L
25 273 K
g
1
mol ˜ K
0.08206 L ˜ atm
0.0859 mol
14.67t
Thermocouple calibration: T
aE b
b g
T qC
T 0 , E 0.249
T 100 , E 5.27
b g
. E mV 4.51
181
U 14.67t
0 440 880 1320
T 181
. E 4.51 25 45 65 85
(c) To keep the temperature uniform throughout the chamber.
(d) Power losses in electrical lines, heat absorbed by chamber walls.
(e) In a closed container, the pressure will increase with increasing temperature. However, at
the low pressures of the experiment, the gas is probably close to ideal Ÿ U f T only.
bg
Ideality could be tested by repeating experiment at several initial pressures Ÿ same
results.
7-7
7.18 (b) 'H 'E k 'E p
Q W s (The system is the liquid stream.)
c
c
h
' E k 0 no change in m and u
' E p 0 no elevation change
Ws 0 no moving parts or generated currents
c
' H
h
h
Q , Q ! 0
(c) 'H 'E k 'E p
Q W s (The system is the water)
c
h
' H 0 T and P ~ constant
' E k 0 no change in m and u
Q 0 no ' T between system and surroundings
c
'E p
c
b
g
Q W s (The system is the oil)
c
' E k 0 no velocity change
h
Q W s Q 0 (friction loss); W s 0 (pump work).
(e) 'H 'E k 'E p
Q W s (The system is the reaction mixture)
c
h
' E k ' E p 0 given
'Ws 0 no moving parts or generated current
c
' H
h
W s , W s ! 0 for water system
(d) 'H 'E k 'E p
'H ' E p
h
h
Q , Q pos. or neg. depends on reaction
7.19 (a) molar flow:
1 mol
125
. m3 273 K 122 kPa
423 K 101.3 kPa 22.4 L STP
min
b g
' H ' E k ' E p
103 L
43.4 mol min
1 m3
Q W s
c
h
' E k ' E p 0 given
Ws 0 no moving parts
c
Q
' H
n' H
h
43.37 mol 1 min
min
60s
3640 J
kW
mol 10 3 J / s
2.63 kW
(b) More information would be needed. The change in kinetic energy would depend on the
cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet
and outlet pipes would be needed to answer this question.
7-8
b g
7.20 (a) H 104
. T q C 25
H in kJ kg
P=110 kPa
H out 1.04 34.0 25 9.36 kJ kg
H in 1.04 30.0 25 5.20 kJ kg
' H 9.36 5.20 4.16 kJ kg
' H ' E k ' E p
n (mol/s)
N2
34 oC
Q W s
c
Q =1.25 kW
h
h
' E k ' E p 0 assumed
Ws 0 no moving parts
c
' H n' H
Q
Ÿ n
Q
' H
1.25 kW
kg
1 kJ / s 103 g 1 mol
10.7 mol s
4.16 kJ
kW
1 kg 28.02 g
b g
10.7 mol 22.4 L STP
Ÿ V =
s
mol
303 K 1013
. kPa
273 K 110 kPa
2455
. L / s Ÿ 246 L s
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is
assumed to depend linearly on temperature and to be independent of pressure, errors in
measured temperature and in wattmeter reading.
7.21 (a) H
H 2 H 1
T2 T1
H1 aT1
aT b a
b
H
0 Ÿ Tref
130.2
5.2
b g
Table B.1 Ÿ S . G.
b
U kJ kg
g
b
Ÿ U kJ kg
b gb g
b
g
b g
5.2T q C 130.2
25q C
bg
C 6 H 14 l
H PV
U|
V|
W
129.8 258
.
5.2
Ÿ H kJ kg
50 30
258
. 5.2 30 130.2
0.659 Ÿ V
1 m3
152
. u 10 3 m 3 kg
659 kg
b5.2T 130.2gkJ / kg
1J
1 kJ
1 atm 1.0132 u 105 N / m2 1.52 u 10 3 m3
1
atm
1
kg 1 N ˜ m 103 J
g
5.2T 130.4
(b) Energy balance: Q
'Ek , 'E p , W 0
'U
20 kg [(5.2 u 20 - 130.4) - (5.2 u 80 - 130.4)] kJ
1
kg
Average rate of heat removal
6240 kJ 1 min
5 min
60 s
7-9
20.8 kW
6240 kJ
7.22
m (kg/s)
260°C, 7 bars
H = 2974 kJ/kg
u0 = 0
m (kg/s)
200°C, 4 bars
H = 2860 kJ/kg
u (m/s)
'H 'E k 'E p
'E p
Q
Q Ws
Ws
0
2
mu
H out H in
m
'E k 'H Ÿ
2
d
d
u 2 2 H in H out
i
b
i
g
(2) 2974 2860 kJ 103 N ˜ m 1 kg ˜ m / s2
kg
1 kJ
1N
7.23 (a) 5 L/min
2.28 u 105
5 L/min
100 mm Hg (gauge)
0 mm Hg (gauge)
Qout
Qin
Since there is only one inlet stream and one outlet stream, and m in
Eq. (7.4-12) may be written
m
'z Q W s
m 'U m ' PV ' u 2 mg
2
d i
'U
a
'u
Pout Pin
mV
2
Heat input: Q in
Efficiency:
V' P
Q in
f
V'P
0 assume for incompressible fluid
'z
0
W s
0 all energy other than flow work included in heat terms
Q
(b) Flow work: V'P
m out { m ,
d i
0 given
m 'PV
V'P
m2
Ÿ u 477 m / s
s2
Q in Q out
Q in Q out
5L
min
b100 0gmm Hg
5 ml O 2
min
20.2 kJ
1 ml O 2
66.7 J min
1.01 u 105 J min
1 atm
8.314 J
66.7 J min
760 mm Hg 0.08206 liter ˜ atm
10 3 J
101
. u 105 J min
1 kJ
u 100 0.066%
7-10
Q W s ; 'E k , 'E p , W s
7.24 (a) 'H 'E k 'E p
b
b
g
0 Ÿ 'H
Q
H 400q C, 1 atm 3278 kJ kg (Table B.6)
H 100q C, sat' d Ÿ 1 atm 2676 kJ kg (Table B.5)
g
100 kg H 2 O(v) / s
100 kg H 2 O(v) / s
o
400 o C, 1 atm
100 C, saturated
Q (kW)
100 kg
Q
s
b3278 2676.0gkJ
10 3 J
kg 1 kJ
(b) 'U 'E k 'E p
b
U 100q C, 1 atm
ŸQ
g
Q W ; 'E k , 'E p , W 0 Ÿ 'U Q
2507 kJ kg U 400q C, 1 atm 2968 kJ kg (Tables B.5 & B.7)
b
g
b
m'U
'U
6.02 u 10 7 J s
g
d
100 kg 2968 2507 kJ kg 10 3 J kJ
i
4.61 u 10 7 J
The difference is the net energy needed to move the fluid through the system (flow work).
c b g h 83.9 kJ kg (Table B.5)
H bsteam, 20 bars, sat' d g 2797.2 kJ kg (Table B.6)
7.25 H H 2 O l , 20q C
m [kg H 2 O(l) / h]
m [kg H 2 O(v) / h]
o
20 C
20 bar (sat' d)
Q = 0.65(813 kW)
(a) 'H 'E k 'E p
'H
m
(b) V
Q
' H
Q W s ; 'E k , 'E p , W s
528 kW
0 Ÿ 'H
Q
m 'H
528 kW
b
kg
1 kJ / s 3600 s
2797.2 83.9 kJ 1 kW
1 h
g
b701 kg hg d0.0995 m kgi
A
3
701 kg h
69.7 m 3 h sat' d steam @ 20 bar
Table B.6
(c) V
nRT
P
701 kg / h
18.02 g / mol
103 g / kg 485.4 K 0.08314 L ˜ bar
mol ˜ K
20 bar
1 m3
103 L
78.5 m3 / h
The calculation in (b) is more accurate because the steam tables account for the effect of
pressure on specific enthalpy (nonideal gas behavior).
(d) Most energy released goes to raise the temperature of the combustion products, some is
transferred to the boiler tubes and walls, and some is lost to the surroundings.
7-11
c bg
7.26 H H 2 O l , 24q C, 10 bar
h
100.6 kJ kg (Table B.5 for saturated liquid at 24oC; assume H
independent of P).
b
g
2776.2 kJ kg (Table B.6) Ÿ ' H
H 10 bar, sat' d steam
[kg H2 O(v) / h]
m
2776.2 100.6 2675.6 kJ kg
[kg H2 O(v) / h]
m
15,000 m3 / h @10 bar (sat'd)
o
24 C, 10 bar
Q (kW)
15000 m 3
m
h
kg
01943
m3
.
7.72 u 10 4 kg h
A
b Table 8.6 g
d
Energy balance 'E p , W s
'E k
'E k
E kfinal E kinitial
f
mu
2
i
0 : 'H 'E k
E kinitial |0
7.72 u 10 4 kg
'E k
E kfinal
d15,000 m hi
3
2
2
015
. 2 S 4 m2
h
2
Q
1
1
h3
1J
1 kg ˜ m 2 / s 2
2 3600 3 s 3
A
A S D2 4
5.96 u 10 5 J / s
1h
7.72 u 10 4 kg 2675.6 kJ
5.96 u 10 5 J 1 kJ
h
kg 3600 s
s 10 3 J
Q m 'H 'E k
57973 kJ s 5.80 u 10 4 kW
7.27 (a)
228 g/min
25oC
228 g/min
T(oC)
Q ( kW)
Energy balance: Q
'E x , 'E p , Ws 0
=0
b g
'H Ÿ Q W
b g 0.263Q bW g
T bq C g
25
H bJ gg 0.263QbW g 0
(b) H bbT 25g
Ÿ H bJ gg 3.34 T bq Cg 25
228 g 1 min ( H out H in ) J
min
60 s
g
Ÿ H out J g
26.4 27.8 29.0 32.4
4.47 9.28 13.4
Fit to data by least squares (App. A.1)
b
7-12
¦ H
i
i
24.8
bT 25g ¦ bT 25g
i
i
i
2
3.34
7.27 (cont’d)
(c) Q
' H
b
g
350 kg 10 3 g 1 min 3.34 40 20 J
min
kg
60 s
g
kW ˜ s
390 kW heat input to liquid
10 3 J
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads.
7.28
m w [ kg H 2 O(v) / min]
3 bar, sat' d
m w [ kg H 2 O(l) / min]
27 o C
Q ( kW)
m e [ kg C 2 H 6 / min]
16 o C, 2.5 bar
(a) C H mass flow: m
2 6
e
H ei
m e [ kg C 2 H 6 / min]
93 o C, 2.5 bar
795 m 3 10 3 L 2.50 bar
min
m 3 289 K
2.487 u 103 kg min
1
1 kg
mol 1000 g
0.08314 L - bar
941 kJ kg , H ef
1073 kJ kg
Energy Balance on C 2 H 6 : 'E p , W s 0, 'E k # 0 Ÿ Q
LMb
N
g OPQ
kJ
kg
Q 2.487 u 103
1073 941
min
kg
b
bliquid, 27q Cg
(b) H s1 3.00 bar, sat' d vapor
H s2
K - mol 30.01 g
g
'H
2.487 u 103 kJ 1 min
5.47 u 103 kW
min 60 s
2724.7 kJ kg (Table B.6)
1131
. kJ kg (Table B.5)
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
d
condensing steam equals the heat transferred to the ethane 5.47 u 10 3 kW
Energy balance on H 2 O: Q
Ÿ m
Ÿ Vs
Q
H s2 H s1
'H
5.47 u 10 3 kJ
s
b2.09 kg / sg d0.606 m kgi
A
3
d
m H s2 H s1
i
kg
. 2724.7 kJ
1131
b
i
g
2.09 kg s steam
1.27 m 3 s
Table B.6
Too low. Extra flow would make up for the heat losses to surroundings.
(c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it
would require heat flow from the ethane to the steam over some portion of the exchanger.
(Observe the two outlet temperatures)
7-13
7.29
250 kg H2 O(v )/min
40 bar, 500°C
H 1 (kJ/kg)
W
b
b
s
Heat
exchanger
250 kg/min
5 bar, T 2 (°C), H 2 (kJ/kg)
Turbine
=1500 kW
250 kg/min
5 bar, 500°C
H3 (kJ/kg)
Q(kW)
g
H 2 O v , 40 bar, 500q C : H 1 3445 kJ kg (Table B.7)
H 2 O v , 5 bar, 500q C : H 3 3484 kJ kg (Table B.7)
g
(a) Energy balance on turbine: 'E p
'H
d
W s Ÿ m H 2 H 1
i
3445 kJ 1500 kJ
s
kg
H
0, Q
0, 'E k # 0
W s Ÿ H 2
H 1 W s m
min 60 s
3085 kJ kg
250 kg 1 min
3085 kJ kg and P 5 bars Ÿ T = 310q C (Table B.7)
(b) Energy balance on heat exchanger: 'E p
d
H 3 H 2
Q 'H m
i
250 kg
d
Q W s Ÿ m s H 3 H 1
Q
'H 'W s
0, 'E k # 0
b3484 3085gkJ
1 min
1 kW
kg
60 s
1 kJ / s
min
(c) Overall energy balance: 'E p
'H
0, W s
1663 kW
0, 'E k # 0
i
Q W s
250 kg
b3484 3445gkJ
1 min
min
kg
60 s
1500 kJ 1 kW
s
1 kJ / s
1 kJ / s
1 kW
1663 kW —
b
g
H Obv , 5 bar, 310q Cg: V
(d) H 2 O v , 40 bar, 500q C : V1
2
2
0.0864 m 3 kg (Table B.7)
0.5318 m 3 kg (Table B.7)
u1
250 kg 1 min 0.0864 m 3
min
60 s
kg
u2
250 kg min 0.5318 m 3
min
60 s
kg
'E k
m 2
u2 u12
2
1
0.5 S 4 m 2
1
0.5 S 4 m 2
2
250 kg 1 1 min
min
2
0.26 kW << 1500 kW
. ms
183
2
. ms
113
. g
b113. g b183
60 s
2
s
7-14
2
2
m2
1 kW ˜ s
1N
1 kg ˜ m / s
2
10 3 N ˜ m
7.30 (a) 'E p , 'E k , W s
b
0 Ÿ Q
'H Ÿ hA Ts To
h 8 Ÿ To
Ts =34.2
(b) Clothed:
g
b
300 kJ h Ÿ 18
. h Ts To
g
300
kJ
h
13.4q C
Nude, immersed: h 64 Ÿ To
Ts =34.2
316
. q C (Assuming Ts remains 34.2qC)
(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal
insulator —i.e., in the absence of wind, h is low.) For a given To , the skin temperature
must drop to satisfy the energy balance equation: when Ts drops, you feel cold.
7.31 Basis: 1 kg of 30°C stream
1 kg H2O(l)@30oC
3 kg H2O(l)@Tf(oC)
2 kg H2O(l)@90oC
b
g b
g
2
1
30 o C 90 $ C 70 $ C
3
3
(b) Internal Energy of feeds: U 30q C, liq.
U 90q C, liq.
(a) T f
b
b
U|V
376.9 kJ kg|W
g
g
125.7 kJ kg
(Table B.5 - neglecting effect of P on H )
Energy Balance: Q - W = 'U + 'E p + 'E k
b
g
Q =W = 'E p = 'E k 0
b
Ÿ 3U f (1 kg) 125.7 kJ / kg (2 kg) 376.9 kJ / kg
7.32
Ÿ U f
293.2 kJ kg Ÿ T f
Diff.
70.05 70.00
u 100%
70.05
P 5 bars
(a) Table B.6
T
52.5 m3 H2O(v)/h
.
m(kg/h)
5 bar, T(oC)
.
Q (kW)
1518
. q C , H L
. gb140 kg hg
b015
Energy balance: Q 'H
0
0.07% (Any answer of this magnitude is acceptable).
. kJ kg , H V
6401
bar, sat'd) = 0.375 m 3 / kg Ÿ m
V(5
(b) H 2 O evaporated
0
70.05q C (Table B.5)
.
m(kg/h)
. kg H2 O( v)/kg
0.85
0.15 kg H2 O( l)/kg
5 bar, saturated, T(oC)
g
'U
21 kg
h
52.5 m3
h
1
2747.5 kJ kg
kg
0.375 m 3
140 kg h
21 kg h
b2747.5 640.1gkJ
1h
1 kW
12 kW
kg 3600 s 1 kJ s
7-15
7.33 (a) P 5 bar
Table B.6
(b) Inlet: T=350°C, P=40 bar
Outlet: T=75°C, P=5 bar
uin
Vin
Ain
uout
Vout
Aout
1518
. o C . At 75°C the discharge is all liquid
Tsaturation
Table B.7
H in = 3095 kJ / kg , Vin = 0.0665 m 3 / kg
H out = 314.3 kJ / kg , Vout = 1.03 u 10 -3 m 3 / kg
Table B.7
200 kg 1 min 0.0665 m 3 / kg
S (0.075) 2 / 4 m 2
min 60 s
200 kg 1 min
min 60 s
0.00103 m 3 / kg
S (0.05) 2 / 4 m 2
5018
. m/s
175
. m/s
m
m ( H 2 H 1 ) (u22 u12 )
2
200 kg 1 min (314 - 3095) kJ 200 kg 1 min (1.752 - 50.18 2 ) m 2
W s Q min 60 s
kg
2 min 60 s
s2
13,460 kW
Energy balance: Q W s | 'H 'E k
7.34 (a) Assume all heat from stream transferred to oil
1.00 u 10 4 kJ 1 min
Q
167 kJ s
min 60 s
100 kg oil/min
135°C
m (kg H2O(v)/s)
25 bars, sat'd
Energy balance on H 2 O: Q
'E p , 'E k , W s
H l , 25 bar, sat' d
m
Q
H out H in
d
m H out H in
i
0
962.0 kJ kg , H v , 25 bar, sat' d
167 kJ
s
Time between discharges:
(b) Unit Cost of Steam:
'H
100 kg oil/min
185°C
m (kg H2O(l)/s)
25 bars, sat'd
b
kg
962.0 2800.9 kJ
g
2800.9 kJ kg (Table B.6)
0.091 kg s
1s
1 kg
1200 g
13 s discharge
discharge 0.091 kg 10 3 g
$1
10 Btu
b2800.9 83.9g kJ
6
kg
0.9486 Btu
$2.6 u 10 3 / kg
kJ
Yearly cost:
1000 traps 0.091 kg stream 0.10 kg last 2.6 u 10 3$ 3600 s 24 h 360 day
trap ˜ s
kg stream
kg lost
h
day
year
$7.4 u 105 / year
7-16
7.35 Basis: Given feed rate
200 kg / h H 2 O (v )
10 bar, sat'd
H 2776.2 kJ / kg
n3 (kg / h H 2 O ( v ))
10 bar, 250 o C
H 2443 kJ / kg
n2 (kg / h H 2 O (v ))
10 bar, 300 o C
H 3052 kJ / kg
Q (kJ / h)
H from Table B.6 (saturated steam) or Table B.7 (superheated steam)
Mass balance: 200 n 2
Energy balance: Q
n 3
'H
'E K , 'E p , W 0
(a) n 3
300 kg h
(b) Q 0
(1), (2)
(1)
n 2
(1)
b g
b
g b
g
n 3 2943 200 2776.2 n 2 3052 Q in kJ h
n 2
( 2)
100 kg h
406 kg h , n 3
7.36 (a) Tsaturation @ 1.0 bar = 99.6 °C Ÿ T f
H 2 O (1.0 bar, sat' d) Ÿ H l
Q
(2)
3.23 u 10 4 kJ h
606 kg h
99.6 $ C
417.5 kJ / kg, H v
2675.4 kJ / kg
$
H 2 O (60 bar, 250 C) 1085.8 kJ / kg
Mass balance: mv ml 100 kg
Energy balance: 'H 0
(1)
'E K , Q , 'E p , W 0
Ÿ mv H v ml H l m1 H 1
(1,2)
ml
70.4 kg, mv
(b) Energy balance: 'H
'E p , Q , W 0
mv H v ml H l (100 kg)(1085.8 kJ / kg) = 0
29.6 kg Ÿ y v
'E k Ÿ H 2 H 1
29.6 kg vapor
100 kg
'E k Ÿ H 2
0.296
(2)
kg vapor
kg
H 1 'E k
The system undergoes expansion, and assuming the same pipe diameter 'E k ! 0 Ÿ 'H 0
Since some of the water will be evaporated, the temperature will still be the saturation
temperature at the given final pressure, so T is unchanged.
Less water will evaporate because some of the energy that went to evaporate the water will
instead be converted to kinetic energy.
(c) P f
39.8 bar (pressure at which the water is still liquid, but has the same enthalpy as the
feed)
7-17
7.36 (cont’d)
(d) Since enthalpy does not change, then when Pf t 39.8 bar the temperature cannot increase,
because a higher temperature would increase the enthalpy. Also, when Pf t 39.8 bar , the product
is only liquid Ÿ no evaporation occurs.
0.4
Tf (C)
y
0.3
0.2
0.1
0
0
20
40
60
300
250
200
150
100
50
0
1
80
5
10
15
20
25
30
36 39.8 60
Pf (bar)
Pf (bar)
7.37 10 m3, n moles of steam(v), 275°C, 15 bar Ÿ 10 m3, n moles of water (v+l), 1.2 bar
10.0 m3 H2O (v)
10.0 m3
min (kg)
275oC, 1.5 bar
mv [kg H2O (v)]
ml [kg H2O (l)]
Q
Table B.6
(a) P=1.2 bar, saturated,
(b) Total mass of water: min =
Mass Balance: mv ml
10 m 3
104.8 $ C
T2
1 kg
0.1818 m 3
55 kg
55.0
Volume additivity: Vv Vl
Ÿ mv
1.2 bar, saturated
7.0 kg, ml
10.0 m 3
mv (1428
.
m 3 / kg) ml (0.001048 m 3 / kg)
48.0 kg condensed
(c) Table B.7 Ÿ U in = 2739.2 kJ / kg; Vin = 0.1818 m 3 / kg
3
Table B.6 Ÿ U l = 439.2 kJ / kg; Vl = 0.001048 m / kg
U = 2512.1 kJ / kg;
V = 1.428 m 3 / kg
R|
S|
T
v
v
Energy balance: Q = 'U = mv U v ml U l minU in
'E p , 'E k , W 0
[(7.0)(2512.1 kJ / kg) + (48.0)(439.2) - 55 kg (2739.2)] kJ
= 1.12 u 10 5 kJ
7.38 (a) Assume both liquid and vapor are present in the valve effluent.
1 kg H 2 O ( v ) / s
15 bar, Tsat 150 o C
m l [ kg H 2 O ( l ) / s]
m v [ kg H 2 O ( v ) / s]
1.0 bar, saturated
7-18
7.38 (cont’d)
(b) Table B.6 Ÿ Tsat'n (15 bar) = 198.3o C Ÿ Tin 348.3o C
Table B.7 Ÿ H in H (348.3$ C, 15 bar) | 3149 kJ / kg
Table B.6 Ÿ H l (1.0 bar, sat' d) = 417.5 kJ / kg; H v (1.0 bar, sat' d) = 2675.4 kJ / kg
Energy balance: 'H 0 Ÿ m H m H m H
0
l
'E p , 'E k ,Q , Ws 0
Ÿ m in H in
m l H l m v H v
l
v
m v m l
in
v
in
3149 kJ / kg m l (417.5) (1 m l )(2675.4)
There is no value of m l between 0 and 1 that would satisfy this equation. (For any value
in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase
assumption is therefore incorrect; the effluent must be pure vapor.
m in m out 1
(c) Energy balance Ÿ m out H out m in H in
3149 kJ / kg = H (1 bar, Tout )
Table B.7
Tout | 337 $ C
(This answer is only approximate, since 'E k is not zero in this process).
7.39 Basis: 40 lb m min circulation
(a) Expansion valve
R = Refrigerant 12
40 lbm R(l)/min
40 lb m / min
x v lb m R ( v ) / lb m
93.3 psig, 86°F
H = 27.8 Btu/lb m
(1 x v ) lb m R( l ) / lb m
H v 77.8 Btu / lb m , H l
Energy balance: 'E p , W s , Q 0, neglect 'E k Ÿ 'H
bg
40 X v lb m R v
min
b
g
E
¦ n H ¦ n H
i
i
out
bg
77.8 Btu 40 1 X v lb m R l
lb m
min
Xv
9.6 Btu / lb m
9.6 Btu 40 lb m
min
lb m
b
i
0
i
in
27.8 Btu
lb m
0
g
0.267 26.7% evaporates
(b) Evaporator coil
40 lb m R( v )/min
11.8 psig, 5°F
H = 77.8 Btu/lbm
40 lbm /min
0.267 R(v )
0.733 R( l )
11.8 psig, 5°F
H v = 77.8 Btu/lbm , H l = 9.6 Btu/lbm
Energy balance: 'E p , W s
0, neglect 'E k Ÿ Q
b gb
g
bg
40 lb m 77.8 Btu 40 0.267 lb m R v
Q
min
lb m
min
2000 Btu min
7-19
'H
b gb
g
bg
77.8 Btu
40 0.733 lb m R l
lb m
min
9.6 Btu
lb m
7.39 (cont’d)
(c) We may analyze the overall process in several ways, each of which leads to the same
result. Let us first note that the net rate of heat input to the system is
Q
Q evaporator Q condenser
2000 2500
500 Btu min
and the compressor work Wc represents the total work done on the system. The system is
b g
closed (no mass flow in or out). Consider a time interval 't min . Since the system is at
steady state, the changes 'U , 'E k and 'E p over this time interval all equal zero. The
total heat input is Q 't , the work input is W 't , and (Eq. 8.3-4) yields
c
Q 't Wc 't
0 Ÿ W c
500 Btu 1 min
u 10 3 hp
.
1341
. hp
118
min
60 s 9.486 u 10 4 Btu s
Q
7.40 Basis: Given feed rates
n1 (mol / h)
nC3H 8 (mol C 3 H 8 / h)
nC4 H10 (mol C 4 H 10 / h)
227 o C
0.2 C 3 H 8
0.8 C 4 H 10
0 o C, 1.1 atm
n2 (mol / h)
0.40 C 3 H 8
0.60 C 4 H 10
25o C, 1.1 atm
Q (kJ / h)
Molar flow rates of feed streams:
300 L 1.1 atm
1 mol
n1
hr 1 atm 22.4 L STP
b g
14.7 mol h
200 L 273 K 1.1 atm
1 mol
9.00 mol h
hr 298 K 1 atm 22.4 L STP
14.7 mol 0.20 mol C 3 H 8 9.00 mol 0.40 mol C 3 H 8
Propane balance Ÿ n C 3H 8
h
mol
h
mol
6.54 mol C 3 H 8 h
Total mole balance: n C4 H10 (14.7 9.00 6.54) mol C 4 H 20 h 17.16 mol C 4 H 20 h
b g
n 2
Energy balance: 'E p , W s
Q 'H
¦ N H ¦ N H
i
i
i
out
in
b0.40 u 9.00g mol C H
( H i
0, neglect 'E k Ÿ Q
3
h
8
i
'H
6.54 mol C3 H8 20.685 kJ 17.16 mol C4 H10
h
mol
h
b
g
27.442 kJ
mol
1.772 kJ 0.60 u 9.00 mol C4 H10 2.394 kJ
587 kJ h
mol
h
mol
0 for components of 1st feed stream)
7-20
510 m 3 273 K 10 3 L
1 mol
3
min 291 K m
22.4 L STP
(a)
1 kmol
10 3 mol
b g
7.41 Basis:
.
n 0 (kmol/min)
38°C, h r = 97%
x 0 (mol H 2 O/mol)
(1 – x 0) (mol dry air/mol)
. kmol min
214
21.4 kmol/min
18°C, sat'd
x 1 (mol H 2 O/mol)
(1 – x 1) (mol dry air)
.
n 2 (kmol H 2O(l )/mol)
18°C
b
hr PH 2 O 38q C
Inlet condition: x o
P
b
PH 2 O 18q C
Outlet condition: x1
P
Dry air balance: 1 0.0634 n o
b
g
b
0.97 49.692 mm Hg
760 mm Hg
g
g
0.0634 mol H 2 O mol
15.477 mm Hg
0.0204 mol H 2 O mol
760 mm Hg
1 0.0204 21.4 Ÿ n o 22.4 kmol min
g b
b
g
g
b
g
Water balance: 0.0634 22.4 n 2 0.0204 214
. Ÿ n 2 0.98 kmol min
0.98 kmol 18.02 kg
18 kg / min H 2 O condenses
min
kmol
b
(b). Enthaphies: H air 38q C
g
b
g
0.0291 38 25
0.3783 kJ mol
b g 0.0291b18 25g 0.204 kJ mol
bv, 38q Cg 2570.8 kJkg 101 kgg 18.02molg 46.33 kJ molU|
||
2534.5 kJ 1 kg 18.02 g
45.67 kJ mol VTable B.5
bv, 18q Cg
kg 10 g
mol
||
1 kg 18.02 g
75.5 kJ
. kJ mol
bl, 18q Cg kg 10 g mol 136
|W
H air 18q C
H H 2 O
H H 2 O
H H 2 O
3
3
3
Energy balance:
'E , W 0, 'E # 0
p
s
b1 0.0204gd214. u 10 ib0.204g
b0.0204gd214
. u 10 ib45.67g d0.98 u 10 ib136
. g b1 0.0634gd22.4 u 10 ib0.3783g
b0.0634gd22.4 u 10 ib46.33g 5.67 u 10 kJ min
Q
k
'H
¦ n H ¦ n H
i
out
i
i
3
i
Ÿ Q
in
3
3
Ÿ
3
3
4
5.67 u 10 4 kJ 60 min 0.9486 Btu 1 ton cooling
min
h
kJ
12000 Btu
7-21
270 tons of cooling
7.42 Basis: 100 mol feed
n2 (mol), 63.0°C
0.98 A(v )
0.02 B(v )
A - Acetone
B - Acetic Acid
Qc (cal)
0.5 n2 (mol)
0.98 A(l )
0.02 B(l )
100 mol, 65.0°C
0.65 A(l )
0.35 B(l )
56.8°C
n5 (mol), 98.7°C
0.544 A(v )
0.456 B(v )
0.5 n2 (mol)
0.98 A(l )
0.02 B(l )
n5 (mol), 98.7°C
0.155 A(l )
0.845 B(l )
Qr (cal)
(a) Overall balances:
UV
W
Total moles: 100 0.5n2 n5
n2
A: 0.65 100 0.98 0.5n2 0155
. n5 n5
b g
b
g
120 mol
40 mol
b g
b g
0.155b40g 6.2 mol A
0.845b40g 338
. mol B
Product flow rates: Overhead 0.5 120 0.98 58.8 mol A
0.5 120 0.02 12
. mol B
Bottoms
'H
Overall energy balance: Q
'E , W 0 , ' E # 0
p
2
¦ n H ¦ n H
i
out
x
i
i
i
in
interpolate in table
bg
B
bg
b g b g b g
(b) Flow through condenser: 2b58.8g 117.6 mols A
2b12
. g 2.4 mols B
interpolate in table
B
b g
Ÿ Q 58.8 0 1.2 0 6.2 1385 33.8 1312 65 271 53 257
Energy balance on condenser: Qc
'E , W 0 , ' E # 0
p
Qc
b
3
k
g
b
117.6 0 7322 2.4 0 6807
2.63 u 10 4 cal
'H
g
8.77 u 10 5 cal heat removed from condenser
Assume negligible heat transfer between system & surroundings other than Qc & Qr
Qr
7.43
Q Qc
d
2.63 u 10 4 8.77 u 10 5
i
1.96 kg, P1= 10.0 bar, T1
9.03 u 105 cal heat added to reboiler
2.96 kg, P3= 7.0 bar, T3=250oC
1.00 kg, P2= 7.0 bar, T2
Q= 0
7-22
7.43 (cont’d)
(a) T2 T ( P 7.0 bar, sat' d steam) = 165.0 o C
H 3 ( H 2 O(v ), P = 7.0 bar, T = 250 o C) 2954 kJ kg (Table B.7)
H 2 ( H 2 O(v ), P = 7.0 bar, sat' d) 2760 kJ kg (Table B.6)
Energy balance
'E , Q, W , 'E # 0
p
s
k
0 2.96H 3 196
. H 1 10
. H 2 Ÿ 196
. H 1
'H
2.96 kg(2954 kJ / kg) - 1.0 kg(2760 kJ / kg)
Ÿ H 1 (10.0 bar, T1 ) 3053 kJ / kg Ÿ T1 # 300 C
$
(b) The estimate is too low. If heat is being lost the entering steam temperature would have to
be higher for the exiting steam to be at the given temperature.
7.44
T1 T ( P 3.0 bar, sat' d.) = 133.5$ C
Vl ( P 3.0 bar, sat' d.) = 0.001074 m 3 / kg
V ( P 3.0 bar, sat' d.) = 0.606 m 3 / kg
(a)
Vapor
P=3 bar
v
Liquid
0.001074 m 3 1000 L 165 kg
Vl
177.2 L
kg
m3
V space 200.0 L - 177.2 L = 22.8 L
22.8 L
mv
(b) P
1 m
1 kg
1000 L 0.606 m 3
mtotal
0.0376 kg
165.0 0.0376 165.04 kg
T1 T ( P 20.0 bar, sat' d.) = 212.4 $ C
Vl ( P 20.0 bar, sat' d.) = 0.001177 m 3 / kg; Vv ( P
V
m V m V Ÿ m V ( m
m )V
total
l l
Ÿ
v v
200.0 L
1m
l l
total
l
20.0 bar, sat' d.) = 0.0995 m 3 / kg
v
3
ml kg(0.001177 m 3 / kg) + (165.04 - ml ) kg(0.0995 m 3 / kg)
1000 L
164.98 kg; mv
Ÿ ml
V=200.0 L
Pmax=20 bar
3
20.0 bar;
Pmax
m=165.0 kg
0.06 kg
3
0.001177 m 1000 L 164.98 kg
194.2 L;
kg
m3
(0.06 - 0.04) kg 1000 g
mevaporated
20 g
kg
Vl
(c) Energy balance Q = 'U
'E , W , ' E # 0
p
U l ( P
U ( P
l
s
U(P
V space
20.0 bar, sat' d) U ( P
200.0 L - 194.2 L = 5.8 L
3.0 bar, sat' d)
k
20.0 bar, sat' d.) = 906.2 kJ / kg; U v ( P 20.0 bar, sat' d.) = 2598.2 kJ / kg
3.0 bar, sat' d.) = 561.1 kJ / kg; U v ( P 3.0 bar, sat' d.) = 2543 kJ / kg
Q 0.06 kg(2598.2 kJ / kg) + 164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg)
165.0 kg (561.1 kJ / kg) = 5.70 u 10 4 kJ
Heat lost to the surroundings, energy needed to heat the walls of the tank
7-23
7.44 (cont’d)
(d) (i) The specific volume of liquid increases with the temperature, hence the same mass of
liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower
density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of
the changes mentioned above.
(e) – Using an automatic control system that interrupts the heating at a set value of pressure
– A safety valve for pressure overload.
– Never leaving a tank under pressure unattended during operations that involve
temperature and pressure changes.
7.45 Basis: 1 kg wet steam
(a) 1 kg H2 O 20 bars
0.97 kg H2 O(v)
0.03 kg H2 O(l)
H1 (kJ/kg)
1 kg H 2O,(v) 1 atm
H2 (kJ/kg)
Q=0
b
b
Q
U|VbTable B.7g
908.6 kJ kg |W
H
'H 0 Ÿ H
0.97b2797.2g 0.03b908.6g
g
g
Enthalpies: H v , 20 bars, sat' d
H l , 20 bars, sat' d
Energy balance on condenser:
'E , 'E , Q , W =0
p
1 kg H2 O
Tamb, 1 atm
K
2797.2 kJ kg
2
1
3
Ÿ H 2 = 2740 kJ / kg
Table B.7
T | 132 o C
(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches
its saturation temperature at 1 atm, it begins to condense, so that T 100q C . The white
plume is a mist formed by liquid droplets.
bg
1 quart
1 m3
1000 kg
32 oz 1057 quarts
m3
(For simplicity, we assume the beverage is water)
7.46 Basis:
bg
8 oz H 2 O l
0.2365 kg H2O (l)
18°C
m (kg H2O (s))
32°F (0°C)
0.2365 kg H 2 O l
(m + 0.2365) (kg H2O (l))
4°C
Assume P 1 atm
Internal energies (from Table B.5):
U H 2 O(l), 18q C 755
. kJ / kg; U H 2 O(l), 4q C
b
g
b
g 168. kJ / kg; U bH O(s), 0q Cg = -348 kJ / kg
Energy balance bclosed systemg: Ÿ 'U ¦ n U ¦ n U 0
2
i
'E p , ' E k , Q , W
0
out
i
i
i
in
Ÿ (m 0.2365) kg(16.8 kJ / kg) = 0.2365 kg(75.5 kJ / kg) + m kg (-348 kJ / kg)
Ÿ m 0.038 kg = 38 g ice
7-24
0 o C, H
7.47 (a) When T
0, Ÿ Tref
0o C
(b) Energy Balance-Closed System: 'U
'E , ' E , Q , W
k
p
0
0
25 g Fe, 175°C
25 g Fe
1000 g H2O
Tf (°C)
1000 g H2O(l)
20°C
d i
b
d i b g
4.13dT 175ical 4.184 J
g
U Fe T f U H 2 O T f U Fe 175q C U H 2 O 20q C, 1 atm
'U Fe
25.0 g
Table B.5 Ÿ 'U H 2 O
cal
1.0 L 10 3 g
eU dT i 83.9j J
1 L
g
H 2O
f
d i
Ÿ 432T f 1000U H 2 O T f 160
. u 10 5
Ÿ
Tf q C
30
f Tf
2.1 u 10
d i
40
4
2.5 u 10
7-25
0
432 T f 175 J
f
g
0 or 'U Fe 'U H 2 O
d i
f Tf
35
4
e
d i
j
1000 U H 2 O T f 83.9 J
0
34
1670 2612
Interpolate
Tf
34.6q C
7.48
II
I
H 2 O(v )
760 mm Hg
100°C
H 2 O(v )
(760 + 50.1) mm Hg
Tf
Ÿ 1.08 bar sat'd
Ÿ Tf = 101.8°C (Table 8.5)
H 2 O( l ), Tf
Ÿ
H 2 O( l ), 100 °C
Tf
T0
Energy balance - closed system:
'E p , 'E K , W , Q
v -vapor
0 mvII U vII mlII U lII mbIIU bII mvI U vI mlIU lI mbI U bI
'U
Vl
Vv
U l
U
0
b
I 101
. bar, 100q C
1044
.
1673
419.0
2506.5
bL kgg
bL kgg
bL kgg
bL kgg
v
Initial vapor volume: VvI
1046
.
1576
426.6
2508.6
50 kg
20.0 L 5.0 L b
b
0.36b101.8g
5.0 L 1.044 L kg
Final energy of bar: U bII
g
g
1L
bg
14.4 L H 2 O v
8.92 kg
bg
8.61 u 10 3 kg H 2 O v
bg
4.79 kg H 2 O l
36.6 kJ kg
Assume negligible change in volume & liquid Ÿ VvII
b
Final vapor mass: mvII
b -block
. bar, 101.8q Cg
g II b108
Initial vapor mass: mvI = 14.4 L 1673 L kg
Initial liquid mass: mlI
l -liquid
14.4 L 1576 L kg
g
14.4 L
bg
9.14 u 10 3 kg H 2 O v
Initial energy of the bar:
d
b
H 2 O evaporated
mvII mvI
g
b
g b g
b
g
b
gi
1
914
. u 10 3 2508.6 4.79 426.6 5.0 36.6 8.61 u 10 3 2506.5 4.79 419.0
5.0 kg
44.1 kJ kg
44.1 kJ / kg
(a) Oven Temperature: To
122.5q C
0.36 kJ / kg ˜ o C
U bI
9.14 u 10 3 kg - 8.61 u 10 3 kg = 5.30 u 10 4 kg = 0.53 g
(b) U bI 44.1 8.3 5.0 458
. kJ kg
To 458
. 0.36 127.2q C
(c) Meshuggeneh forgot to turn the oven on ( To 100q C )
7-26
weight of piston
atmospheric pressure
area of piston
7.49 (a) Pressure in cylinder
P
30.0 kg
400.0 cm2
Ÿ Tsat
b100 cmg
1 bmg
9.807 N
2
2
kg
2
10
. bar
105 N m2
1 atm 1.013 bar
108
. bar
atm
1018
. qC
Heat required to bring the water and block to the boiling point
Q
'U
d b
g
b
gi
d b g
b426.6 83.9gkJ 3.0 kg
7.0 kg
[0.94(1018
. 20)]kJ
kg
kg
2630 kJ < 3310 kJ Ÿ Sufficient heat for vaporization
(b) T f
Tsat
b
mw U wl 108
. bar, sat' d U wl l, 20q C m Al U Al Tsat U Al 20q C
gi
2630 kJ
V 1046
.
L kg , U l 426.6 kJ kg
1018
. q C . Table B.5 Ÿ l
Vv 1576 L kg , U v 2508.6 kJ kg
7.0 kg H 2 O(l )
H 426.6 kJ / kg
= 1.046 L / kg
V
mv (kg H 2 O(v ))
1576 L/kg, 2508.6 kJ/kg
T { 101.8°C
P { 1.08 bars
1.046 L/kg, 426.6 kJ/kg
ml (kg H 2 O(l ))
Q (kJ)
W (kJ)
(Since the Al block stays at the same temperature in this stage of the process, we can
ignore it -i.e., U in U out )
Water balance: 7.0 ml mv (1)
Work done by the piston: W
LM w P
NA
atm
u
OPb A' zg
Q
F' z
w piston Patm A ' z
b1.08 bar g 1576m
P' V Ÿ W
8.314 J / mol ˜ K
1 kJ
0.08314 liter - bar / mol ˜ K 10 3 J
Energy balance: 'U
v
b
gb g
1.046ml 1.046 7.0 L
b170.2m
v
g
0.113ml 0.7908 kJ
Q W
'U
Q
W
Ÿ 2508.6mv 426.6m L 426.6 7 ( 3310 2630) (170.2mv 0113
. m L 0.7908)
Ÿ 2679mv 426.7m L 3667 0 (2)
Solving (1) and (2) simultaneously yields mv 0.302 kg , ml 6.698 kg
Liquid volume
Vapor volume
bg
.
L kgg
b6.698 kggb1046
b0.302 kggb1576 L kgg
Piston displacement: 'z
(c) Tupper
7.01 L liquid
476 L vapor
b gb g
'V
A
7.01 476 7.0 1046
.
L 10 3 cm3
b
g b
1
1190 cm
1 L 400 cm2
Ÿ All 3310 kJ go into the block before a measurable amount is transferred to the
water. Then 'U AL
g
Q Ÿ 3.0 kg 0.94 Tu 20 kJ kg
o
neglected. In fact, the bar would melt at 660 C.
7-27
3310 Ÿ Tu
1194q C if melting is
7.50
100
. L H 2 O(v ), 25o C
m v1 (kg)
not all the liquid
UV Assume
Eq. at
W Tis vaporized.
, P . m kg H O vaporized.
m v2 [kg H 2 O(v)]
= m v1 me
f
o
f
2
e
m L2 [kg H 2 O(l)]
= m L1 me
4.00 L H 2 O(l ), 25 C
m L1 (kg)
Q=2915 kJ
Initial conditions: Table B.5 Ÿ U L1
L kg P 0.0317 bar
104.8 kJ kg , VL1 1003
.
25q C, sat' d Ÿ U v1 2409.9 kJ kg , V L1 43,400 L kg
T
b1.00 lg b43400 l kgg
mv1
2.304 u 10 5 kg , m LI
b4.00 lg b1.003 l kgg
3.988 kg
Energy balance:
d
i d i b
g d i d
ib
Q Ÿ 2.304 u 10 5 me U v T f 3.988 me U L T f 2.304 u 10 5 2409.9
'U
b
g
3.988 (104.8)
d
2915 kJ
g d i
i dEi b
3333 d2.304 u 10 iU 3.988U
Ÿ 2.304 u 10 5 me U v T f 3.988 me U v T f
3333
5
Ÿ me
F
GG
H
v
L
I
J d i b
A JK A
5.00 d2.304 u 10 iV 3.988V
g d i
Vtan k Ÿ 2.304 u 10 5 me VL T f 3.988 me VL T f
V L Vv
(1)
U v U L
kg
5.00 L
liters kg
5
Ÿ me
b1g b2g Ÿ f dT i
f
v
b2g
L
Vv VL
3333 2.304 u 10 5 U v T f 3.988U L T f
U U
d
i d i
d
v
L
d i
i
5.00 2.304 u 10 5 Vv 3.988VL
V V
v
d i Find T
Table 8.5
Procedure: Assume T f
Tf
.
2014
198.3
195.0
196.4
U v
.
25938
2592.4
2590.8
.
25915
bg
or Eq b 2 g
Eq 1
me
U L
856.7
842.9
828.5
834.6
Ÿ U v , U L , Vv , VL Ÿ f T f
Vv
123.7
.
1317
140.7
136.9
0
L
f
VL
f
.
. u 10 2
1159
512
193
.
. u 10 2
1154
.
. u 10 2
1149
134
4.03 u 10 4 Ÿ T f # 196.4q C, Pf
.
1151
2.6 u 10 3 kg Ÿ 2.6 g evaporated
7-28
d i
such that f T f
0
14.4 bars
g
7.51.
Basis: 1 mol feed
B = benzene
T = toluene
nV (mol vapor)
y B(mol B(v)/mol)
(1 – y B ) (mol T(v)/mol)
1 mol @ 130°C
z B (mol B(l)/mol)
(1 – z B )(mol T(l)/mol)
in equilibrium
at T(°C), P(mm Hg)
nL (mol liquid)
x B(mol B(l)/mol)
(1 – x B ) (mol T(l)/mol)
(a) 7 variables: (nV , y B , n L , x B , Q, T , P )
–2 equilibrium equations
–2 material balances
–1 energy balance
2 degrees of freedom. If T and P are fixed, we can calculate nV , y B , n L , x B , and Q.
(b) Mass balance: nV n L 1 Ÿ nV 1 n2
Benzene balance: z B nV y B n L x B
bg d
C H bv g: dT
C H bl g: dT
C H bv g: dT
C6 H 6 l : T
i d
0, H
6
6
80, H
7
8
0, H
7
8
89, H
10.85 Ÿ H BL
id
4161
. , T 120, H
i d
111, H
0 , T
i d
i
i
18.58 Ÿ H TL
T
01674
.
i
g
g b g
(3)
01045
.
T 33.25
52.05 Ÿ H TV
0, neglect 'E k
b
01356
.
T
45.79 Ÿ H BV
111, H
49.18 , T
Energy balance: 'E p , Ws
Q 'H
i
80, H
0 , T
(1)
(2)
b
g
(5)
T 37.57
01304
.
bg
b g
nV y B H BV nV 1 y B H TV n L x B H BL n L 1 x B H TL 1 z B H BL TF
1 1 z H T
b gb
Raoult' s Law:
B
TL
yB P
x B p *B
(90 C) 10
1021 mmHg
[ 6.95334 1343.943/( 90 219 .377)]
406.7 mmHg
Adding equations (8) and (9) Ÿ
P x B p*B + (1 x B ) pT* Ÿ x B
nL
P pT*
pB*
P pT*
pT*
p*B
x B pB*
pT*
652 406.7
0.399 mol B(l) / mol
1021- 406.7
0.399(1021 mmHg)
0.625 mol B(v) / mol
P
652 mmHg
zB xB
0.5 0.399
0.446 mol vapor
y B x B 0.625 0.399
1 nV 1 0.446 0.554 mol liquid
yB
Solving (1) and (2) Ÿ nV
(7)
(8)
p *B (90 o C) 10[ 6.905651211.033/( 90 220.79 )]
o
(6)
F
(1 - y B ) P (1 x B ) pT*
Antoine Equation. For T= 90°C and P=652 mmHg:
pT*
(4)
7-29
(9)
7.51 (cont’d)
Substituting (3), (4), (5), and (6) in (7) Ÿ
.
(90) 33.25] 0.446(1 0.625)[01304
.
(90) 37.57]
Q 0.446(0.625)[01045
0.554(0.399)[01356
.
(90)] 0.554(1 0.399)[01674
.
(90)] 0.5[01356
.
(130)]
0.5[01674
.
(130)] Ÿ Q 814
. kJ / mol
(c). If P<Pmin, all the output is vapor. If P>Pmax, all the output is liquid.
(d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714
mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure,
there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium
vapor: enthalpy out < enthalpy in.
zB
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
T
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
P
652
714
582
590
600
610
620
630
640
650
660
670
680
690
700
710
pB
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
1021
pT
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
406.7
xB
0.399
0.500
0.285
0.298
0.315
0.331
0.347
0.364
0.380
0.396
0.412
0.429
0.445
0.461
0.477
0.494
(e). Pmax = 714 mmHg, Pmin = 582 mmHg
nV vs. P
1
0.8
nV
0.6
0.4
0.2
0
582
632
682
732
P (mm Hg)
nV
0.5 @ P # 640 mmHg
7-30
yB
0.625
0.715
0.500
0.516
0.535
0.554
0.572
0.589
0.606
0.622
0.638
0.653
0.668
0.682
0.696
0.710
nV
0.446
-0.001
0.998
0.925
0.840
0.758
0.680
0.605
0.532
0.460
0.389
0.318
0.247
0.176
0.103
0.029
nL
0.554
1.001
0.002
0.075
0.160
0.242
0.320
0.395
0.468
0.540
0.611
0.682
0.753
0.824
0.897
0.971
Q
8.14
-6.09
26.20
23.8
21.0
18.3
15.8
13.3
10.9
8.60
6.31
4.04
1.78
-0.50
-2.80
-5.14
'P 'u 2
g'z
U
2
7.52 (a). Bernoulli equation:
0
'P
U
d0.977 u 10
g'z
(9.8066 m / s 2 ) 6 m 58.8 m 2 / s 2
5
i
15
. u 10 5 Pa 1 N / m 2
m3
1.12 u 10 3 kg
Pa
bg
46.7
m2
s2
'u 2
46.7 58.8 m 2 / s 2 Ÿ u22 u12 2 12.1 m 2 / s 2
2
2
5.00 m 2 / s 2 (2)(12.1) m 2 / s 2 0.800 m 2 / s 2 Ÿ u2 0.894 m / s
b
Bernoulli Ÿ
g
d
i
b g
d
(b). Since the fluid is incompressible, V m 3 s
Ÿ d1
d
u2
u1
d2
b6 cmg
0.894 m s
5.00 m s
d i b g
i
7.53 (a). V m 3 s
' P 'u 2
U
2
S d 12 u1 4 S d 22 u2 4
2.54 cm
d i b g
A2 m 2 u2 m s Ÿ u2
A1 m 2 u1 m s
(b). Bernoulli equation ('z
i
u1
A1
A2
A1 4 A2
u2
4u1
0)
0Ÿ 'P
P2 P1
d
U u22 u12
i
2
Multiply both sides by 1
2
2
16u1
Substitute u 2
2
Multiply top and bottom of right - hand side by A1
2
note V
P1 P2
(c) P1 P2
2
V
d
i
U Hg U H 2 O gh
b g
2 S 7.5
2 2
15
Ÿ V
Ÿ V 2
m4
9.8066 m
108 cm4
s2
cm4 1
2
15UV 2
2 A12
15U H 2 OV 2
2 A12
2
A1 u1
0.044 m 3 s 44 L s
7-31
F
GH
I
JK
2 A12 gh U Hg
1
15
U H 2O
38 cm
1m
102 cm
b13.6 1g
1955
.
u 10 3
m6
s2
7.54 (a). Point 1 - surface of fluid . P1
Point 2 - discharge pipe outlet . P2
'U
U
31
.
. gbar
b1013
g'z
9.8066 m
s2
7 m , u1
31
. bar , z1
1 atm , z 2
b
1.013 bar
g
10 5 N
1
m3
m 2 ˜ bar 0.792 u 10 3 kg
7 m
Bernoulli equation Ÿ
b g
b g
0ms
0 m , u2
?
2635
. m 2 s2
68.6 m 2 s 2
'u 2
2
'u
'P
g'z
U
u2 0
2
2
b2635. 68.6g m
2
s2
332.1 m 2 s 2
2
u22
2(332.1 m 2 s 2 )
V
S (100
. 2 ) cm 2
4
664.2 m 2 s 2 Ÿ u2
258
. m/s
60 s
2580 cm 1 L
122 L / min
1
s 10 3 cm 3 1 min
(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation,
becomes increasingly significant as the valve is closed.
7.55 Point 1 - surface of lake . P1
Point 2 - pipe outlet . P2
u2
V
A
95 gal
min
FG L
H
Z
2z
sin 30q
IJ
K
W
Shaft work: s
m
0
bg
1
2
S 0.5 u 1049
in 2
.
b
g
bP P g
0.041b2 z g ft ˜ lb lb
0
0 , u1
z ft
1 atm , z 2
1 ft 3
7.4805 gal
Pressure drop: ' P U
Friction loss: F
1 atm , z1
1
144 in 2
1 ft 2
1 min
60 s
35.3 ft s
2
f
0.0822 z (ft ˜ lb f lb m )
m
-8 hp 0.7376 ft ˜ lb f / s
1.341 u 10
3
1 min 7.4805 gal
95 gal
hp
1 ft
3
1 ft 3
60 s
62.4 lb m
1 min
333 ft ˜ lb f lb m
Kinetic energy: ' u 2 2
b35.3g
2
0 2 ft 2
2
Potential energy: g'z
b
g
Eq. 7.7 - 2 Ÿ
32.174 ft
s2
s
2
bg
z ft
'P 'u 2
g'z F
U
2
1
lb f
32.174 lb m ˜ ft / s 2
1 lb f
32.174 lb m ˜ ft / s 2
19.4 ft ˜ lb f lb m
b
z ft ˜ lb f lb m
W s
Ÿ 19.4 z 0.082 z 333 Ÿ z
m
7-32
g
290 ft
7.56 Point 1 - surface of reservoir . P1 1 atm (assume), u1 0 , z1 60 m
Point 2 - discharge pipe outlet . P2 1 atm (assume), u2 ? , z 2 0
'P U
0
'u 2
2
u22
2
dV Ah
2
V 2 (m 6 / s 2 )
2
9.8066 m
s2
W s
m
b g
bN ˜ m kgg
S 35
(2)
3.376V 2
g'z
1
65 m
1N
1 kg ˜ m / s 2
2 2
cm 4
10 8 cm 4
1 N
1 m4
1 kg ˜ m / s 2
637 N ˜ m kg
s
0.80 u 10 6 W 1 N ˜ m / s
V m3
W
1 m3
800 V N ˜ m kg
1000 kg
b
d i
Mechanical energy balance: neglect F b Eq. 7.7 - 2g
W s
Ÿ 3.376V 2 637
m
'P 'u 2
g'z
U
2
800 T E ŸV
V
g
127
. m 3 60 s
76.2 m 3 min
s
1 min
Include friction (add F ! 0 to left side of equation) Ÿ V increases.
7.57 (a). Point 1: Surface at fluid in storage tank, P1
0 , z1
1 atm , u1
Point 2 (just within pipe): Entrance to washing machine. P2
u2
600 L
min S 4.0 cm
b
'u 2
2
'P
U
0;
g'z
9.807 m
s2
u22
2
3
10 cm
4 1 L
g
b7.96 m sg
2
2
c0 Hbmgh
Bernoulli Equation:
2
3
1 atm , z 2
0;
'u 2
2
317
.
1 J
1 kg ˜ m 2 / s 2
1 J
1 kg ˜ m 2 / s 2
'P 'u 2
g'z
U
2
J
; g'z
kg
Mechanical energy balance: W s
. J / kg
317
9.807 H (J / kg)
0Ÿ H
3.23 m
Ÿ W s
600 L 0.96 kg 1 min
L
60 s
min
b
g
9.807 3.23 0
m
0
1 min
1m
7.96 m s
60 s 100 cm
(b). Point 1: Fluid in washing machine. P1 1 atm , u1 | 0 , z1 0
Point 2: Entrance to storage tank (within pipe). P2 1 atm , u2
'P
U
b g
H m
LM 'P 'u
NU 2
2
317
.
J
; F
kg
g'z F
b31.7 + 31.7 + 72g J
OP
Q
7.96 m s , z 2
72
3.23 m
J
kg
1 kW
1.30 kW
kg 10 3 J s
(work applied to the system)
Rated Power 130
. kW 0.75 1.7 kW
7-33
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity.
xi 0.95 0.05
l
1
Density of 95% solution:
Ÿ U 124
0.804
. kg liter
U
U
kg
126
100
.
.
b Eq. 6.1-1g
i
¦
Density of 35% solution:
Mass of 95% solution:
0.35 0.65
.
1.26 100
1
U
0.9278
l
Ÿ U 1.08 kg liter
kg
1000 liters 1.24 kg
1240 kg
liter
G = glycerol
W = water
1240 kg (1000 L)
0.95 G
0.05 W
m2 (kg)
0.60 G
0.40 W
23 m
m1 (kg)
0.35 G
0.65 W
5 cm I.D.
UV Ÿ m
b gb g b gb g b0.60gbm gW m
Mass balance: 1240 m1 m2
Glycerol balance: 0.95 1240 0.35 m1
2
1740 kg
Volume of 35% solution added
Ÿ Final solution volume
1
b1000 1610g L
2
1740 kg 35% solution
2980 kg 60% solution
1L
1610 L
1.08 kg
2610 L
Point 1. Surface of fluid in 35% solution storage tank. P1 1 atm , u1
Point 2. Exit from discharge pipe. P2 1 atm , z 2 23 m
u2
1610 L
13 min
'P U
g'z
1
1 m 3 1 min
2
3
10 L
60 s S 2.5 cm 2
b g
'u 2
2
'u22
2
9.8066 m
s2
23 m
0,
. g
b1051
b
Mechanical energy balance Eq. 7.7 - 2
W s
m
LM 'P 'u
NU 2
2
g'z F
OP
Q
m2 / s2
(2)
1N
1 kg ˜ m / s 2
1740 kg 1 min
13 min
60 s
Mass flow rate: m
2
10 4 cm 2
1 m2
0 , z1
0
.
1051
ms
1 N
1 kg ˜ m / s 2
225.6 N ˜ m kg , F
0.552 N ˜ m kg
50 J kg 50 N ˜ m kg
2.23 kg s
g
2.23 kg
b0.552 + 225.6 + 50gN ˜ m
s
kg
0.62 kW Ÿ 0.62 kW delivered to fluid by pump.
7-34
1J
1 kW
1 N ˜ m 10 3 J s
CHAPTER EIGHT
8.1
a.
U (T ) 25.96T 0.02134T 2 J / mol
U (0 o C) 0 J / mol U (100 o C) 2809 J / mol
Tref
o C) = 0)
0 o C (since U(0
b.
We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to
U (100 o C) .
c.
Q W
'U 'E k 'E p
'E k
Q
d.
'U
0 , 'E p
z
dU
dT
V
[25.96 0.04268T ] J / (mol˜ o C)
z
T2
'U
0
(3.0 mol)[(2809 0) J / mol] 8428 J Ÿ 8400 J
F wU I
GH wT JK
Cv
0, W
F
GG 25.96T 0.04268 T2 OPP
Q
H
100
Cv (T )dT
2
(25.96 0.04268T )dT
0
T1
(3.0 mol) ˜ [25.96(100 0) 0.02134(100 2 0)] (J / mol)
a.
27.0 0.0291T [J / (mol˜q C)]
z
z
100
'H
c.
z
C v dT
25
H is a state property
a.
Cv [ kJ / (mol˜ o C)]
PV
RT
Q1
n'U 1
z
T2
2
OP
PQ
100
2784 J mol
25
100
C p dT d.
n
0.0891
100
25
8.3
100
25
35.3T
C p dT
25
100
'U
8428 J Ÿ 8400 J
b35.3 0.0291T g[J / (mol˜q C)] b8.314 [J / (mol ˜ K)]gb1 K 1q Cg
C p R Ÿ Cv
Cv
Ÿ Cv
b.
0
I
JJ J / mol
K
(3.0 mol) ˜ 'U (J / mol)
'U
8.2
100
RdT
'H R'T
b
gb
g
2784 8.314 100 25
2160 J mol
25
0.0252 1547
.
u 10 5 T 3.012 u 10 9 T 2
(2.00 atm)(3.00 L)
(0.08206[atm ˜ L / (mol ˜ K)](298 K)
z
0.245 mol
1000
(0.245 mol) ˜
z
z
0.0252 dT ( kJ / mol)
6.02 kJ
25
1000
Q2
n'U 2
(0.245) ˜
u 10 5 T ] dT
[0.0252 1547
.
7.91 kJ
25
1000
Q3
n'U 3
(0.245) ˜
u 10 5 T 3.012 u 10 9 T 2 ] dT
[0.0252 1547
.
25
6.02 - 7.67
u 100%
7.67
7.91- 7.67
% error in Q2 =
u 100%
7.67
% error in Q1 =
.
215%
.
313%
8-1
7.67 kJ
8.3 (cont’d)
b.
Cv R
Cp
u 10 5 T 3.012 u 10 9 T 2 ) 0.008314
(0.0252 1547
.
C p [ kJ / (mol˜ o C)]
u 10 5 T 3.012 u 10 9 T 2
0.0335 1547
.
z
T2
'H
Q
n C P dT
T1
z
1000
(0.245 mol) ˜
[0.0335 1547
.
u 10 5 T 3.012 u 10 9 T 2 ] dT [kJ / (mol˜ o C)]
9.65 u 10 3 J
25
Piston moves upward (gas expands).
c.
8.4
a.
b.
The difference is the work done on the piston by the gas in the constant pressure process.
b g b313 Kg
b g 0.1360 [kJ / (mol ˜ K)]
b g b40q Cg 0.07406 32.95 u 10 b40g 25.20 u 10 b40g 77.57 u 10 b40g
dC i
dC i
p C H l
6 6
0.06255 23.4 u 10 5 313
5
8
12
2
3
p C H v
6 6
0.08684 [kJ / (mol˜ o C)]
c.
dC i b g b313 Kg
d.
'H C6 H 6 bv g
e.
8.5
b g
'H Cb sg
b g
0.01118 1095
.
u 10 5 313 4.891 u 10 2 313
p C s
2
0.009615 [ kJ / (mol ˜ K)]
32.95 u 10 5 2 2520
. u 10 8 3 77.57 u 10 12 4
0.07406T T T T
3
2
4
0.01118T 1.095 u 10 5 2
T 4.891 u 10 2 T 1
2
OP
PQ
OP
PQ
300
3171
. kJ mol
40
573
3.459 kJ / mol
313
H 2 O (v, 100 o C, 1 atm) o H 2 O (v, 350 o C, 100 bar)
a. H 2926 kJ kg 2676 kJ kg 250 kJ kg
z
350
b.
H
0.03346 0.6886 u 10 5 T 0.7604 u 10 8 T 2 3.593 u 10 12 T 3 dT
100
8.845 kJ mol Ÿ 491.4 kJ kg
Difference results from assumption in (b) that H is independent of P. The numerical difference
is 'H for H 2 O v, 350q C, 1 atm o H 2 O v, 350q C, 100 bar
b
8.6
b.
g
b
z
g
80
dC i
p n C H (l)
6 14
0.2163 kJ / (mol˜ o C) Ÿ 'H
[0.2163] dT
1190
. kJ / mol
25
The specific enthalpy of liquid n-hexane at 80oC relative to liquid n-hexane at 25oC is 11.90 kJ/mol
c.
dC i
p n C H (v) [ kJ
6 14
z
/ (mol˜ o C)]
013744
.
40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 12 T 3
0
'H
[0.13744 40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 12 T 3 ] dT = 110.7 kJ / mol
500
The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110.7 kJ/mol. The
specific enthalpy of hexane vapor at 0oC relative to hexane vapor at 500oC is –110.7 kJ/mol.
8-2
8.7
b g 181. T cbq Fg 32 0.5556T cbq Fg 17.78
C bcal mol˜q Cg 6.890 0.001436 0.5556T cbq Fg 17.78 6.864 0.0007978T cbq Fg
cal
453.6 mol 1 Btu
1q C
C c b Btu lb - mole˜q Fg
C
b1.00gC
mol˜q C 1 lb - mole 252 cal 1.8q F
E
T qC
p
p
p
b
g
C p Btu lb - mole˜q F
8.8
dC i
p CH CH OH(l)
3
2
Q
'H
bT g
p
drop primes
b g
6.864 0.0007978T q F
0.1031 .
0.1031g
b01588
T
FG
H
100
0.1031 0.000557T [kJ / (mol˜ o C)]
55.0 L 789 g 1 mol
0.000557 2
.
T
T
01031
s
1 L 46.07 g
2
OP
Q
78.5
20
kJ mol
= 941.9 u 7.636 kJ / s = 7193 kW
8.9
kJ mol
a.
Q
'H
b5,000 mol sg ˜
z
200
0.03360 1367
.
u 10 5 T 1.607 u 10 8 T 2 6.473 u 10 12 T 3 dT
100
17,650 kW
b.
Q
'U
'H 'PV
b
'H nR'T
gb
gb
17,650 kJ 5.0 kmol ˜ 8.314 [kJ / (kmol ˜ K) ] ˜ 100 K
13,490 kJ
c.
8.10 a.
b.
The difference is the flow work done on the gas in the continuous system.
Qadditional heat needed to raise temperature of vessel wall + heat that escapes from wall to
surroundings.
C p is a constant, i.e. C p is independent of T.
Q
Cp
Q
m'T
(16.73 - 6.14) kJ
mC p 'T Ÿ C p
Q
m' T
(2.00 L)(3.10 K) 659 g
Table B.2 Ÿ C p
8.11
H
U PV
1L
0.216 kJ / (mol˜ o C)
86.17 g 10 3 J
1 mol
FG wH IJ FG wU IJ
H wT K H wT K
F wU I dU F wU I
depends only on T, G
H wT JK dT GH wT JK
PV RT
! H
U RT
aw wT f
= 0.223 kJ / (mol ˜ K)
0.216 kJ / (mol ˜ K)
P
!
p
But since U
1 kJ
p
8-3
R Ÿ Cp
p
FG wU IJ
H wT K
{ Cv Ÿ C p
V
R
p
Cv R
g
8.12 a.
dC i
75.4 kJ / (kmol˜ o C) =75.4 kJ/(kmol.oC) V = 1230 L ,
p H O(l)
2
n
VU
M
1230 L 1 kg 1 kmol
1 L 18 kg
zd
T2
n˜
Q
b.
Q
t
i
Cp
H 2 O(l)
68.3 kmol
dT
68.3 kmol 75.4 kJ (40 29) o C 1 h
8h
3600 s
kmol˜ o C
T`
t
Q total
Q to the surroundings Q to water , Q to the surroundings
z
1.967 kW
.
1967
kW
40
Q total
c.
n ˜ C P ( H2 O) dT
Qto water
t
Q to water
t
7.212 kW Ÿ E total
7.212 kW u 3 h = 21.64 kW ˜ h
7.212 kW u 1 h u $0.10 / (kW ˜ h) = $0.72
Cost keeping temperature constant for 1 hour
$2.16 $0.72
d.
$2.88
If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher
cost.
8.13 a.
'H N
2 (25
b.
'H H
2 (800
c.
'H CO
d.
'H O
8.14 a.
5.245 kW
21.64 kW ˜ h u $0.10 / (kW ˜ h) = $2.16
Cost heating up from 29 o C to 40 o C
Cost total
68.3 kmol 75.4 kJ / (kmol˜ o C) 11 o C
3h
3600 s / h
29
o
C) o N 2 (700 o C)
H N
F) o H 2 (77 o F)
H H
o
2 (300
2 (970
o
o
H O
m
300 kg / min n
Q
n ˜ 'H
n ˜
z
T2
T1
2 (77
o
o
2 (0
o
F)
C)
F)
H CO
C) o CO 2 (1250 o C)
F) o O 2 (0o F)
2 (700
H N
H H
2 (1250
H O
o
2 (25
2 (800
C)
o
o
C)
F)
b20.59 0g 20.59 kJ mol
b0 5021g 5021 Btu / lb - mol
. g 5148
. kJ mol
b63.06 1158
b539 6774g 7313 Btu / lb - mol
H CO
2 (970
o
F)
2 (300
o
C)
300 kg 1 min 1000 g 1 mol
min 60 s 1 kg 28.01 g
178.5 mol / s
C p dT
z
50
(178.5 mol / s) ˜
b
[0.02895 0.411 u 10 5 T 0.3548 u 10 8 T 2 2.22 u 10 12 T 3 ] dT [kJ / mol]
450
g
(178.5 mol / s) 12.076 [kJ / mol] = 2,156 kW
b.
Q
n ˜ 'H
n ˜ H (50o C) H ( 450o C)
8.15 a.
n
250 mol / h
i)
Q
250 mol (2676 3697) kJ 1 kg
1 h 18.02 g
h
1 kg
1000 g 3600 s 1 mol
n'H
Q n'H n ˜
z
T2
T1
ii)
=
(178.5 mol / s)(0.73 -12.815[kJ / mol]) = 2,157 kW
1.278 kW
C p dT
250 mol 1 h
h
3600 s
z
100
600
0.6880 u 10 5 T 07604
u 108 T 2 3593
[003346
.
.
. u 1012 T 3 ] 1.274 kW
8-4
8.15 (cont’d)
b
g
250 mol
˜ 2.54 20.91 [kJ / mol] 1276
.
kW
3600 s
Method (i) is most accurate since it is not based on ideal gas assumption.
The work done by the water vapor.
Q
iii)
b.
c.
8.16 Assume ideal gas behavior, so that pressure changes do not affect 'H .
n
200 ft 3 492 o R 1.2 atm 1 lb - mol
h
537 o R 1 atm 359 ft 3 (STP)
Q
n'H
(0.6125
50 kg 1.14 kJ
8.17 a.
b
b50 10gq C
dC i
p Na O
2 3
dC i
dC i
p C H O(l)
6 14
105.99 g
2085 2280
u 100%
2280
b
g b
d i 2b0.026g 0.0075 3b0.017g
b50 10gq C 2085 kJ
3 Cp
1 mol
mol˜q C
% error
p C
Na
O
0.1105 kJ mol˜q C
8.6% error
g b
g
6 0.012 14 0.018 1 0.025
p CH COCH (l)
3
3
1833 Btu / h
2280 kJ
d i dC i
| 2 Cp
50,000 g 0.1105 kJ
8.18
g
lb - mole
) ˜ (2993 0) [Btu / lb - mole]
h
kg˜q C
b.
0.6125 lb - mole / h
0.349 kJ / (mol˜ o C) (Kopp’s Rule)
0.1230 18.6 u 10 5 T kJ (mol˜q C)
Assume 'H mix # 0
p CH 3 COCH 3
C pm
e
p C 6 H 14 O
j
b
g
0.30 0.1230 +18.6 u 10 5 T kJ 1 mol
0.70 0.349T kJ 1 mol
mol˜q C
58.08 g
mol˜q C
102.17 g
[0.003026 9.607 u 10 7 T] kJ (g˜q C)
z
'H
20
[0.003026 9.607 u 10 7 T] dT
0.07643 kJ g
45
8.19 Assume ideal gas behavior, 'H mix # 0
g
g b g 26.68 mol
'H
dC i dT 10.08 kJ / mol, 'H
dC i dT 14.49 kJ / mol
2
L1
OF 1000 g IJ FG 1 mol IJ 433 kJ kg
H M b14.49 kJ / mol g b10.08 kJ / molgPG
3
3
N
QH 1 kg K H 26.68 g K
Mw
O2
b
1
2
16.04 32.00
3
3
z
350
25
p O
2
CH 4
8-5
z
350
25
p CH
4
8.20
1000 m 3 1 min 273 K
1 kmol
0.6704 kmol s 670.4 mol / s
min
60 s 303 K 22.4 m 3 STP
Energy balance on air:
Table B.8 for 'H
670.4 mol 0.73 kJ 1 kW
489.4 kW
Q 'H n 'H
Q
s
mol
1 kJ s
n
489.4 kW heating 1 kW solar energy
Solar energy required
0.3 kW heating
1 m2
1627 kW 1000 W
Area required
8.21
b g
1 kW
900 W
1631 kW
1813 m 2
C 3 H 8 5O 2 o 3CO 2 4H 2 O
n fuel
1.35 u 10 5 SCFH 1 lb - mol
h
359 ft 3
n air
.
376 lb mol 5 lb - mol O 2
1 lb - mol air 115
h
1b - mol C 3 H 8 0.211b - mol O 2
376
lb - mol
h
1.03 u 10 4
lb mol
h
z
T2
Q = 'H = n ˜ C p dT
T1
FG
H
= 1.03 u 10 4
IJ
K
lb mol
˜
h
z
302
[0.02894 0.4147 u 10 5 T 0.3191 u 10 8 T 2 1965
.
u 10 12 T 3 ] dT
0
1.03 u 10 lb - mol 8.954 kJ 453.593 mol 9.486 u 10 -1 Btu
= 3.97 u 10 7 Btu / h
h
mol
lb - mol
kJ
4
=
8.22 a.
Basis: 100 mol feed (95 mol CH4 and 5 mol C2H6)
7
CH 4 2O 2 o CO 2 2H 2 O
C 2 H 6 O 2 o 2CO 2 3H 2 O
2
nO2
125
. ˜
LM 95 mol CH
MN
4
2 mol O 2
5 mol C 2 H 6 3.5 mol O 2
1 mol CH 4
1 mol C 2 H 6
OP
PQ
259.4 mol O 2
Product Gas:
CO 2 : 95(1) + 5(2) = 105 mol CO 2
H 2 O: 95(2) + 5(3) = 205 mol H 2 O
O 2 : 259.4 - 95(2) - 5(3.5) = 51.9 mol O 2
N 2 : 3.76(259.4) = 975 mol N 2
Energy balance (enthalpies from Table B.8)
H
18.845 42.94
H
'H
CO 2
(CO , 450o C)
(CO , 900 o C)
2
'H
H 2O
H
(H
'H
O2
H
(O
'H
N2
H
(N
Q = 'H
Q
2 O,
2,
2,
24.09 kJ / mol
2
450 o C)
H
(H
450o C)
H
(O
450o C)
H
(N
2,
2 O,
900 o C)
1512
. 33.32
18.20 kJ / mol
900o C)
13.375 28.89
15.51 kJ / mol
900 o C)
12.695 27.19
14.49 kJ / mol
2,
105(-24.09) 205(-18.20) 51.9(-15.51) 975(-14.49)
21,200 kJ / 100 mol feed
8-6
8.22 (cont’d)
(40 o C)
b. From Table B.5: H
liq
167.5 kJ / kg; H
vap (50 bars)
2794.2 kJ / kg;
= n(2794.2 -167.5) = 21200 Ÿ n = 8.07 kg / 100 mol feed
Q = n ˜ 'H
c.
From part (b), 8.07 kg steam is produced per 100 mol feed
1250 kg steam 0.1 kmol feed 1 h
4.30 u 10 3 kmol / s
n feed
h
8.07 kg steam 3600 s
723 K
4.30 mol feed 1336.9 mol product gas 8.314 Pa ˜ m 3
s
100 mol feed
mol ˜ K
1.01325 u 10 5 Pa
Vproduct gas
3.41 m 3 / s
d. Steam produced from the waste heat boiler is used for heating, power generation, or
process application. Without the waste heat boiler, the steam required will have to be
produced with additional cost to the plant.
Assume 'H mix # 0 Ÿ 'H
8.23
d i
Kopp’s rule: C p
15.0 L 879 g 1 mol
˜
L 78.11 g
'H C6 H6
8.24
10(12) 12(18) 2(25)
C10 H12 O2
e
386 J mol˜ o C
20.0 L 1021 g 1 kJ 2.35 J (71 25) o C
L 10 3 J g˜ o C
'H C10 H12O2
'H
'H C10 H12O2 'H C6 H6
2207 1166
LM
N
z
348
e
2.35 J g˜ o C
j
2207 kJ
[0.06255 23.4 u 10 5 T] dT
298
OP
Q
1166 kJ
3373 kJ
100 mol C3 H8 @ 40 o C, 250 kPa
a.
j
100 mol C3 H8 @ 240 o C, 250 kPa
VP 1 (m3 )
VP 2 (m3 )
mw kg H2 O(v) @ 300 o C, 5.0 bar
mw kg H2 O(l, sat‘d) @ 5.0 bar
Vw2 (m3 )
Vw1 (m3 )
b. References: H2O (l, 0.01 oC), C3H8 (gas, 40 oC)
C 3 H 8 : H in
0 kJ / mol; H out
z
240
C pC H dT 19.36 kJ mol (Cp from Table B.2)
40
H 2 O: H in
c.
'H C3 H8
Q
'H
3 8
3065 kJ / kg (Table B.7); H out
19.36 kJ / mol, 'H w
100'H C 3H 8 mw 'H w
b
(2747.5 3065) kJ / kg
0 Ÿ mw
From Table B.7: Vsteam 5.0 bar, 300q C
b
VC3 H8 40q C, 250 kPa
6.09 kg steam
g
2747.5 kJ / kg (Table B.6)
g
318 kJ / kg
6.09 kg
0.522 m 3 kg
0.008314 m 3 ˜ kPa (mol ˜ K) 313 K
250 kPa
3
0.522 m steam
1 mol C 3 H 8
0.0104 m 3 mol C 3 H 8
3.06 m 3 steam m 3 C 3 H 8
d.
100 mol C 3 H 8
1 kg steam
0.0104 m 3 C 3 H 8
Q mw 'H w 6.09 kg u (-318 kJ / kg) = -1940 kJ
e.
A lower outlet temperature for propane and a higher outlet temperature for steam.
8-7
8.25
a.
5500 L(ST P)/ min CH3 OH (v) 65o C
n 2 mol/min CH3 OH (v) 260o C
n 2 (mol/ min)
mw kg/ min H2 O(l, sat‘d) @ 90o C
mw kg/ min H2 O(v, s at‘d) @ 300o C
Vw 2 (m3 /min )
b.
8.26 a.
Q
Vw 1 (m3 /min )
kg I F
kJ I F 1 min I F 1 kW I
FG113
H . min JK GH 2373.9 kg JK GH 50 sec JK GH 1 kJ / sJK
44.7 kW
n 2 mol/s (30o C)
0.020 mol H2 O(v)/ mo l
y 2 mol
(molCO/s
CO/mol)
(molCO
CO2 /s
(0.980-y 2 ) mol
2/mol)
100 mol/s (30o C)
0.100 mol H2 O(v)/ mo l
0.100 mol CO/ mol
0.800 mol CO2 /mol
m3 kg humid air/s (50o C)
o
m4 kg humid air/s (30
(48oC)
C)
H 2O(v) only
(0.002 /1.002 ) kg H2 O(v)/kg humid air
(1.000 /1.002 ) kg dry air/ kg humid air
y 4 kg H2 O(v)/kg humid air
(1-y 4 ) kg dry air/kg humid air
Basis: 100 mol gas mixture/s
5 unknowns: n2, m3, m4, y2, y4
– 4 independent material balances, H2O(v), CO, CO2 , dry air
– 1 energy balance equation
0 degrees of freedom (all unknowns may be determined)
b.
(1) CO balance:
(100)(0.100) = n 2 y 2
(2) CO 2 balance: (100)(0.800) = n 2 (1 y 2 )
|UV Ÿ n
W|
2
9184
. mol / s, x 2
01089
mol CO / mol
.
1000
.
m4 (1 y 4 )
1002
.
0.002
(100)(0.100)(18)
(0.020)(18)
m 3
m 4 y 4
9184
(4) H 2 O balance:
.
1002
1000
1000
.
References: CO, CO2, H2O(v), air at 25oC ( H values from Table B.8 )
substance
n in ( mol / s)
n out ( mol / s)
H in (kJ / mol)
(3) Dry air balance: m3
H2O(v)
CO
CO2
10
10
80
0.169
0.146
0.193
91.84(0.020)
10
80
H out (kJ / mol)
0.169
0.146
0.193
H2O(v)
dry air
m3(0.002/1.002)(1000/18)
m3(1.000/1.002) (1000/29)
0.847
0.727
m4y4(1000/18)
m4(1-y4) (1000/29)
0.779
0.672
(5) Energy balance:
.
IJ FG 1000 IJ (0.727)
FG 0.002 IJ FG 1000 IJ (0.847) m FG 1000
K H 29 K
H 1002
H 1.002 K H 18 K
.
F 1000IJ m (1 y )(0.672)FG 1000 IJ
= 91.84( 0.020)( 0.169) m y (0.779)G
H 29 K
H 18 K
10(0.169) m3
3
4
4
4
8-8
4
8.26 (cont’d)
Solve Eqs. (3)–(5) simultaneously Ÿ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2O/kg
2.55 kg humid air / s
100 mol gas / s
Mole fraction of water :
8.27 a.
29 kg DA 1 kmol H 2 O
.0963
(1-.0564) kg dry air kmol DA 18 kg H 2 O
0.0963 kmol H 2 O
(1 0.0963) kmol humid air
p H 2O
Relative humidity:
kg humid air
mol gas
00564
kg H 2 O
.
Ÿ
c.
0.0255
p H* 2 O
.
00878
e48 Cj
kmol DA
kmol H 2 O
kmol humid air
(0.0878)(760 mm Hg )
u 100%
83.71 mm Hg
o
kmol H 2 O
79.7%
The membrane must be permeable to water, impermeable to CO, CO2, O2, and N2, and both durable
and leakproof at temperatures up to 50oC.
b
p * 57q C
g
129.82 mm Hg
mol H 2 O mol
0171
.
P
760 mm Hg
p
3
28.5 m STP
1 mol
1270 mol h Ÿ 217.2 mol H 2 O h
h
0.0224 m 3 STP
y H 2O
b g
1270 217.2
b g
1053
mol dry gas
h
given
b3.91 kg H O hg
2
R| 89.5 mol CO h
|110.5 mol CO h
!S
|| 5.3 mol O h
T847.6 mol N h
2
percentages
2
2
1270 mol/h, 620°C
425°C
m (kg H2 O( l )/h), 20°C
References for enthalpy calculations:
e
j
CO, CO 2 , O 2 , N 2 at 25qC (Table B.8); H 2 O l, 0.01o C (steam tables)
substance
n in
CO
CO 2
89.5
110.6
5.3
847.6
O2
N2
bg
H Obl g
3.91
m
H2O v
H in
18.22
27.60
19.10
18.03
3749
83.9
n out
89.5
110.6
5.3
847.6
H out
12.03
17.60
12.54
11.92
3.91 m 3330
---
2
'H
¦ n H ¦ n H
i
out
i
i
i
0 Ÿ 8504 3246m
0Ÿm
U| n in mol h
V| H in kJ mol
W
UV n in kg h
W H in kJ kg
2.62 kg h
in
b. When cold water contacts hot gas, heat is transferred from the hot gas to the cold water lowering
the temperature of the gas (the object of the process) and raising the temperature of the water.
8-9
. gb5.294 mm Hg g
b015
8.28 2qC, 15% rel. humidity Ÿ p H 2 O
dy i
H 2O
inhaled
n inhaled
b0.7941g b760g
0.7941 mm Hg
1.045 u 10 3 mol H 2 O mol inhaled air
5500 ml 273 K 1 liter
1 mol
3
min
275 K 10 ml 22.4 liters STP
b
p * 37q C
Saturation at 37 qC Ÿ y H 2 O
g
760 mm Hg
b g
47.067
760
0.2438 mol air inhaled min
0.0619 mol H 2 O mol exhaled dry gas
0.2438 mol/min 2oC
n2 kmol/min 37oC
1.045 x 10-3 H2O
0.999 dry gas
0.0619 H2O
0.9381 dry gas
n1 mol H2O(l)/min 22o C
Mass of dry gas inhaled (and exhaled)
b0.999gb0.2438g
Dry gas balance:
b0.2438gb0.999gmol dry gas
29.0 g
min
mol
0.9381 n 2 Ÿ n 2
7.063 g min
0.2596 mols exhaled min
u 10 j n
.
b0.2438ge1045
b0.2596gb0.0619g Ÿ n 0.0158 mol H O min
References for enthalpy calculations: H Obl g at triple point, dry gas at 2 qC
3
H 2 O balance:
1
2
1
2
substance
m in
Dry gas
H2O v
7.063
0.00459
0.285
bg
H Obl g
2
Q
¦ m H ¦ m H
'H
H out
36.75
2569
—
m out
7.063
0.290
—
m in g min
H in J g
m H 2 O
n H 2 O 18.02
H H 2 O from Table 8.4
H dry gas
i
i
out
8.29 a.
H in
0
2505
92.2
i
i
in
bg
75 liters C 2 H 5 OH l
966.8 J 60 min 24 hr
1.39 u 10 6 J day
min
1 hr
1 day
789 g 1 mol
liter 46.07 g
bg
1284 mol C 2 H 3 OH l
e
j
3054 mol H Obl g
(C p ) CH 3OH
.
01031
0.557 u 10 3 T kJ / (mol˜ o C) (fitting the two values in Table B.2)
55 L H 2 O l
1000 g 1 mol
liter 18.01 g
bg
2
(C p ) H 2O
b
0.0754 kJ mol˜q C
g
1284 mol C2H5 OH(l) (70.0oC)
1284 mol C2H5 OH (l) (To C)
3054 mol H2O(l) (20.0oC)
3054 mol H2O(l) (To C)
ze
T
b
0 1284
'U # 'H liquids
Q
0 adiabatic
g
j
0.557 u 10 T dT 3054
.
01031
gUV Ÿ E Integrate, solve quadratic equation
W T = 44.3 C
Q
b
3
70
o
8-10
zb
T
25
g
0.0754T dT
b
105
. T 2
g
8.29 (cont’d)
b. 1.
2.
3.
4.
5.
6.
7.
Heat of mixing could affect the final temperature.
Heat loss to the outside (not adiabatic)
Heat absorbed by the flask wall & thermometer
Evaporation of the liquids will affect the final temperature.
Heat capacity of ethanol may not be linear; heat capacity of water may not be constant
Mistakes in measured volumes & initial temperatures of feed liquids
Thermometer is wrong
8.30 a.
1515 L/s air
500o C, 835 tor,
Tdp=30o C
1515 L/s air , 1 atm
110 g/s H2O(v)
110 g/s H2O, T=25oC
Let n1 (mol / s) be the molar flow rate of dry air in the air stream, and n 2 (mol / s) be the molar
flow rate of H2O in the air stream.
1515 L 835 mm Hg
mol ˜ K
26.2 mol / s
n1 + n 2
s
773 K
62.36 L ˜ mm Hg
n 2
mmHg
.
p * (30 o C) 31824
0.0381 mol H 2 O / mol air
=y=
n1 + n 2
835 mmHg
Ptotal
. mol H 2 O / s
Ÿ n1 25.2 mol dry air / s; n 2 10
o
References: H2O (l, 25 C), Air (v, 25oC)
substances
n in (mol / s)
n out (mol / s)
H in (kJ / mol)
dry air
25.2
H2O(v)
14.37
zd
zd
1.0
100
25
500
100
H2O(l)
'H
Cp
6.1
H2 O ( l )
25.2
7.1
dT H vap
H2 O ( v )
25
z
--
z
b.
c.
Q
b gb
p H O( l )
2
vap
100
p H O( l )
2
25
Integrate, solve : T
100
25
vap
z
z
z
g b gFGH dC i
100
25
p H O ( l ) dT
2
Cp
i
air
H2 O ( l )
H2 O ( v )
dT
dT H vap
dT
--
T
100
i
i
Cp
Cp
p H O ( v ) dT
2
IJ
K
500
100
139 o C
25.2 14.37 100
.
100
100
b25.2gFGH dC i dT IJK b7.1gFGH dC i dT H dC i
F dC i dT H dC i dTIJ
b25.2gb14.37g b1.00gG
K
H
p air
zd
zd
25
T
0
T
zd
T
25
dT
n out ˜ H out n in ˜ H in
0
z
i
i
Cp
H out (kJ / mol)
H vap p H O(v )
2
zd
500
100
Cp
i
H2 O ( v )
0
dT
IJ
K
423 kW
When cold water contacts hot air, heat is transferred from the air to the cold water mist lowering
the temperature of the gas and raising the temperature of the cooling water.
8-11
8.31
250 kg NH 3 10 3 g
Basis:
h
1 mol
1 kg
1h
8.48 mol NH 3 /s
25°C
n 1 (mol air/s)
T °C
n 2 (mol/s)
0.100 NH 3
0.900 air
600°C
Q = –7 kW
NH 3 balance: 8.48
0100
. n2 Ÿ n2
b0.900gb84.8g
Air balance: n1
8.48 mol NH 3 s
17.03 g 3600 s
84.8 mol s
76.3 mol air s
bg
References for enthalphy calculations: NH 3 g , air at 25qC
NH 3
H in
0.0
z
C p from
dC i dT Ÿ H 25.62 kJ mol
Air: C bJ mol ˜q Cg 28.94 0.4147 u 10 T bq Cg
H
z C dT LMMN28.94bT 25g 0.004147FGH T2 252 IJK OPPQ molJ u 101 kJJ
e2.0735 u 10 T 0.02894T 0.7248jbkJ molg
H out
600
25
p NH
3
out
Table B.2
2
p
2
T
in
25
p
3
6
z
H out
100
25
C p dT
Energy balance: Q
7 kJ s
2
17.39 kJ mol
'H
E¦
out
ni H i ¦ ni H i
in
b8.48 mols NH sgb25.62 kJ molg b76.3 mols air sgb17.39 kJ molg
b8.48gb0.0g b76.3ge2.0735 u 10 T 0.02894T 0.7248j
3
6
E
1582
.
u 10 4 T 2 2.208T 1606
8.32 a.
2
0ŸT
2
693q C (–14,650°C)
Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane
Stack gas (ToC)
Stack gas (900oC)
100 mol/s
0.95 mol M/mol
0.05 mol E/mol
Furnace
n3
n4
n5
n6
mol
mol
mol
mol
CO2/s
H2O/s
O2/s
N2/s
Heat
Exchanger
n3
n4
n5
n6
mol
mol
mol
mol
CO2/s
H2O/s
O2/s
N2/s
air (245oC)
20 % excess air (20oC)
n1 mol O2/s
n2 mol N2/s
n1 mol O2/s
n2 mol N2/s
CH 4 2O 2 o CO 2 2H 2 O
b g
C 2 H 6 7 / 2 O 2 o 2CO 2 3H 2 O
8-12
8.32 (cont’d)
LM 95 mol M 2 mol O
N s 1 mol M
2
nair
.
12
nair
1185 mol air / s
4.76 mol air 5 mol E 3.5 mol O 2 4.76 mol air
1 mol E
mol O 2
s
mol O 2
n1
0.21 u 1185
0.79 u 1185
n3
95 mol M 1 mol CO 2 5 mol E 2 mol CO 2
1 mol E
1 mol M
s
s
105 mol CO 2 / s
n4
95 mol M 2 mol H 2 O 5 mol E 3 mol H 2 O
1 mol E
1 mol M
s
s
205 mol H 2 O / s
n5
249 n6
n2
249 mol O 2 / s, n2
936 mol N 2 / s
. mol O 2
95 mol M 2 mol O 2 5 mol E 35
1 mol M
1 mol E
s
s
936 mol N 2 / s
Energy balance on air:
Table B.8: H air (20o C)
415
. mol H 2 O / s
0144
.
kJ / mol, H air (245o C)
e
Q = n air H air (245o C) - H air (20o C)
j
OP
Q
b
6.509 kJ / mol
1180 6.509 0144
.
g
7851 kW
Energy balance on stack gas:
Q
6
FH
¦ ni
'H
7851 n3
z
i 3
T
900
dC i
z
T
900
p CO
2
dC i dT IK
p i
dT n4
z
T
900
dC i
p H O ( v ) dT
2
z
n5
T
900
dC i
p O
2
dT n6
z
T
900
dC i
p N
2
dT
Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z SolveŸ T
b.
mol
350 m 3 (STP)
1000 L 1 h
h
22.4 L(STP) m 3 3600 s
792 o C
4.34 mol / s
4.34 mol / s
0.0434
100 mol / s
0.0434 7851 341 kW
Scale factor =
b g
Q c
8.33 a.
Q
b.
z
'H
600
0
'H
C p dT
n'H
b
g b
g
100
. 4 351
. 38.4 42.0 2 36.7 40.2 43.9
335
3
150 mol 23100 J 1 kW
3465 kW
s
mol 1000 J / s
23100 J mol
T, y
The method of least squares (Equations A1-4 and A1-5) yields (for X
z
b g
00334
.
. u 105 T q C kJ (mol ˜q C) Ÿ Q 150
1732
Cp
600
0
Cp )
0.0334 1732
. u 105 T dT
3474 kW
The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is
forced in (b).
8.34 a.
ln C p
b
ln a
bT 1 2 ln a Ÿ C p
ln C p 2 C p1
T2 T1
e
j
a exp bT 1 2 ,
T1
0.0473
ln C p1 b T1
14475
Ÿa
.
e
1.4475
8-13
7.1 , C p1
U|
|V Ÿ C
0.235|
|W
0.329 ,
p
17.3 , C p2
T2
e
0.235 exp 0.0473T 1 2
j
0.533
8.34 (cont’d)
b.
z
150
e
0.235 exp 0.0473T
1800
1
20
30
40
2
200
12
jdT
b0.235gb2g RSexpe0.473T
0.0473 T
jLMNT
12
1
.0473
OPUV
QW
150
1730 cal g
1800
DIMENSIONS CP(101), NPTS(2)
WRITE (6, 1)
FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/)
NPTS(1) 51
NPTS(2) 101
DO 200K 1, 2
N NPTS (K)
NM1 N – 1
NM2 N – 2
DT (150.0 – 1800.0)/FLOAT (NM1)
T 1800.0
DO 20 J 1, N
CP (J) 0.235*EXP(0.0473*SQRT(T))
T T DT
SUMI 0.0
DO 30 J 2, NM1, 2
SUMI SUMI CP(J)
SUM2 0.0
DO 40 J 3, NM2, 2
SUM2 SUM2 CP (J)
DH DT*(CP(1) 4.0 SUM1 2.0 SUM2 CP(N))/3.0
WRITE (6, 2) N, DH
FORMAT (1H0, 5XI3, 'bPOINT INTEGRATIONbbbDELTA(H)b ', E11.4,'bCAL/G')
CONTINUE
STOP
END
Solution: N
N
11 Ÿ 'H
101 Ÿ 'H
1731 cal g
1731 cal g
Simpson's rule with N
8.35 a.
12
U|
M .W . 62.07 g / mol V Ÿ Q
|
56.9 kJ / mol W
'H
11 thus provides an excellent approximation
m 175 kg / min
'H
175 kg 1000 g 1 mol 56.9 kJ 1 min
min
kg 62.07 g mol 60 s
2670 kW
v
b. The product stream will be a mixture of vapor and liquid.
c. The product stream will be a supercooled liquid. The stream goes from state A to state B as shown
in the following phase diagram.
P
B
A
T
8-14
(T )
68.74 o C, 'H
v b
Table B.1 Ÿ Tb
8.36 a.
28.85 kJ / mol
is not a function of pressure
Assume: n - hexane vapor is an ideal gas, i.e. 'H
bC H g
B 'H
bC H g
6
6
bC H g
A 'H
b g
 o bC H g
'H
Total
 
o
14 l, 20o C
1
z
z
68.74
20
200
'H 2
6
0.2163 dT
68.74
b g
b.
'H
c.
U 200 o C , 2 atm
10.54 24.66 28.85 64.05 kJ / mol
64.05 kJ / mol
e
j
H PV
Assume ideal gas behavior Ÿ PV
U 64.05 3.93 6012
. kJ / mol
b g
'H v tb
100.00q C
e
j
B 'H
H O el, 100 Cj
H 2 O l, 50o C
e
50 o C
'H
v
j
o
3.93 kJ / mol
e
j
A 'H
H O e v, 100 Cj
 
o
2
RT
40.656 kJ mol
H 2 O v, 50 o C
1
z
z
14 v, 68.74 o C
013744
40.85 u 10 5 T 23.92 u 10 8 T 2 57.66 u 10 9 T 3 dT
.
24.66 kJ / mol
'H 1 'H 2 'H v Tb
Total
Tb
2
10.54 kJ / mol
'H 2
'H
8.37
14 v, 200 o C
T
'H
v
b
14 l, 68.74 o C
'H 1
6
2
e
100 o C
'H
v
j
 o
o
2
100
'H 1
C pH 2 Obl g dT
3.77 kJ mol
25
25
'H 2
C pH 2 Obv g dT
. kJ mol
169
100
B
Table B.1
b
'H v 50q C
g
Steam table:
.
3.77 40.656 169
b2547.3 104.8gkJ
kg
42.7 kJ mol
18.01 g 1 kg
1 mol 10 g
44.0 kJ mol
The assumption of ideal gas behavior might account for the difference between the two values.
8.38
1.75 m3 879 kg kmol min
2.0 min m 3 78.11 kg 60 s
e
j
B 'H
C H e v, 80.1 Cj
C 6 H 6 v, 580 o C

o
164.1 mol / s , Tb
1
o
6
6
e
j
A 'H
el, 80.1 Cj
C 6 H 6 l, 25o C
2

o C 6 H 6
o
8-15
b g
801
. q C , 'H v Tb
30.765 kJ mol
8.38 (cont’d)
z
z
80.1
'H 1
77.23 kJ mol
C pC6 H 6 bv g dT
580
298
'H 2
7.699 kJ mol
C pC 6 H 6 bl gdT
353.1
'H
Q
54.164 kJ / mol
e j
b164.1 mol / sgb54.164 kJ / molg 8888 kW
'H 1 'H v 801
. o C 'H 2
n'H
'H
B
Antoine
8.39
UV
W
35q C
Ÿ yCCl 4
15% relative saturation
( 'H v ) CCl 4
Table B.1
30.0
0.15
kJ
Ÿ Q
mol
b
PV 25q C
g
015
.
1 atm
'H
176.0 mm Hg
760 mm Hg
0.0347 mol CCl 4 mol
10 mol 0.0347 mol CCl 4
min
mol
30.0 kJ
mol CCl 4
10.4 kJ min
Time to Saturation
6 kg carbon 0.40 g CCl 4
g carbon
8.40 a.
b
g
1 mol CCl 4
1 mol gas
1 min
153.84 g CCl 4
0.0347 mol CCl 4
10 mol gas
b
g
CO 2 g, 20q C o CO 2 s, 78.4q C : 'H
z
78.4
20
45.0 min
b gdT 'H sub b78.4q Cg
dC i
p CO g
2
In the absence of better heat capacity data; we use the formula given in Table B.2 (which is
strictly applicable only above 0qC ).
78.4
kJ
'H |
.03611 4.233 u 10 5 T 2.887 u 10 8 T 2 7.464 u 10 12 T 3 dT
20
mol
z
FG IJ
H K
6030
Q
cal
4.184 u 10 3 kJ
mol
1 cal
n'H
'H
28.66 kJ mol
300 kg CO 2
10 3 g
1 mol
28.66 kJ removed
h
1 kg
44.01 g
mol CO 2
(or 6.23 u 107 cal hr or 72.4 kW )
b. According to Figure 6.1-1b, Tfusion=-56oC
Q
n'H
'H
where, 'H
Q
n
z
56
20
LMz dC i
N
p CO (v) dT
2
56
20
e j z dC i
dT 'H e56 Cj z
dC i dT OQP
dC i
p CO (v)
2
'H v 56 o C o
v
78.4
56
p CO (l) dT
2
78.4
56
8-16
p CO (l)
2
. u 10 5 k J h
195
a bT
Cp
8.41 a.
U|
. 0.01765T b Kg
V| Ÿ C bJ mol ˜ Kg 4512
. |W
a 53.94 b0.01765gb500g 4512
NaCl b s, 300 Kg o NaClb s, 1073 Kg o NaClbl , 1073 Kg
O J 30.21 kJ
L b4512
. 0.01765T gdT P
C dT 'H b1073 Kg M
'H
Q mol mol
N
53.94 50.41
500 300
b
0.01765
p
z
ps
300
7.44 u 10
b.
'U
Q
z
1073
n
z
m
4
10 3 J
300
1 kJ
J mol
b
1073
300
1073
Cv dT 'U m 1073 K
g
Cv |C p
'U m # 'Hm
Q | 'H
n'H
200 kg 10 3 g
1 kg
t
c.
8.42
2.55 u 10 8 J
s
1 mol
74450 J
58.44 g
mol
1 kJ
0.85 u 3000 kJ 10 3 J
'H v
136.2q C
35.98 kJ mol , Tb
Chen's rule:
100 s
409.4 K , Pc
b0.088gb409.4 Kg
Trouton's rule: 'H v | 0.088Tb
2.55 u 10 8 J
37.0 atm , Tc
b
619.7 K (from Table B.1)
36.0 kJ mol 0.1% error
g
FG T IJ 0.0327 0.0297 log P OP
HT K
PQ 35.7 kJ mol (–0.7% error)
'H |
FT I
. G J
107
HT K
F 619.7 373.2 IJ 38.2 kJ mol
Watson’s correlation : 'H b100q Cg | 35.98G
H 619.7 409.4 K
LM
MN
b
Tb 0.0331
v
10
c
c
b
c
0.38
v
8.43
b g b g
Trouton's Rule Ÿ 'H b200q Cg 0.088b200 + 273.2g = 41.6 kJ mol
C H Nbl , 25q Cg o C H Nbl , 200q Cg o C H Nbv , 200q Cg
C 7 H 2 N : Kopp's Rule Ÿ C p | 7 0.012 12 0.018 0.033 0.333 k J (mol ˜q C)
v
7
12
z
200
'H
25
7
b
12
7
g
12
kJ
kJ
41.6
C p dT 'H r 200q C | 0.333(200 25)
mol
mol
8-17
100 kJ mol
8.44 a.
b g
Watson Correction:
b.
1211.033
220.790
6.90565 log 100
Antoine equation: Tb q C
b g
F 562.6 299.3IJ
'H b261
. q Cg 30.765G
H 562.6 3531. K
b
Clausius-Clapeyron: ln p
c.
0.008314
R|
S|
T
g
b
118
. q C ; Tb 150 mm Hg
'H v
C Ÿ 'H v
RT
b
' H v (80.1°C)
C 6 H 6 ( l , 80.1°C)
zd
zd
i
C p dT
26.1
26.1
'H 2
i
'H v 261
. qC
g
U|
V|
W
b
g
35.2q C
ln p 2 p1
g
1 T2 1 T1
34.3 kJ mol
' H2
C 6 H 6 (v , 80.1°C)
7.50 kJ mol
C p dT
80.1
b
l
33.6 kJ mol
C 6 H 6 (v , 26.1°C)
' H 1
80.1
g
R
ln 150 50
kJ
mol ˜ K 1 308.4 K 1 285.0 K
C 6 H 6 ( l , 26.1°C)
'H 1
0.38
v
Antoine equation: Tb 50 mm Hg
'H v
261
. qC
v
4.90 kJ mol
7.50 30.765 4.90
33.4 kJ mol
8.45 a. Tout = 49.3oC. The only temperature at which a pure species can exist as both vapor and liquid at 1
atm is the normal boiling point, which from Table B.1 is 49.3oC for cyclopentane.
b. Let n f , n v , and n l denote the molar flow rates of the feed, vapor product, and liquid product
streams, respectively.
Ideal gas equation of state
n f
1550 L 273 K
s
1 mol
44.66 mol C 5 H 10 (v) / s
423 K 22.4 L(STP)
55% condensation: n l
0.550(44.66 mol / s) = 24.56 mol C 5 H 10 (l) / s
Cyclopentane balance Ÿ n v
(44.66 24.56) mol C 5 H 10 / s = 20.10 mol C 5 H 10 (v) / s
Reference: C5H10(l) at 49.3oC
n in
(mol/s)
H in
(kJ/mol)
n out
(mol/s)
H out
(kJ/mol)
C5H10 (l)
—
—
24.56
0
C5H10 (v)
44.66
H f
20.10
H v
Substance
Hi
'H v z
Ti
49.3o C
8-18
C p dT
8.45 (cont’d)
Substituting for 'H v from Table B.1 and for C p from Table B.2
Ÿ H
38.36 kJ / mol, H
27.30 kJ / mol
f
v
Energy balance: Q
8.46 a.
¦n
out H out
¦n
. u 10 3 kJ / s = 116
. u 10 3 kW
116
in H in
Basis: 100 mol humid air fed
n 2 (mol), 20o C, 1 atm
y 2 (mol H2 O/ mol), sat’d
1-y 2 (mol d ry air/ mo l)
100 mol
y 1 (mol H2 O/ mol)
1-y 1 (mol d ry air/ mo l)
50o C, 1 at m, 2o superheat
n 3 (mol H2 O(l))
There are four unknowns (n2, n3, y1 and y2) and four equations (two independent material
balances, 2oC superheat and saturation at outlet). All the unknowns can be calculated.
b.
b
p 48q C
2q C superheat Ÿ y1
g
p
saturation at outlet Ÿ y2
b
p 20q C
g
p
b gb g n b1 y g
H O balance: b100gb y g bn gb y g n
dry air balance: 100 1 y1
2
1
2
2
2
b
c.
2
3
b
g
g
References: Air 25q C , H 2 O l, 20q C
100
H 2
20
=
100
20
50
H 2
20
25
air
Cp
dT
H 2 O(l)
50
25
n in mol
n2 ˜ y 2
H 4
H in kJ mol
n3
0
b
1
g
3
0.02894 0.4147 u 10 5 T 0.3191 u 10 8 T 2 1.965 u 10 12 T 3 dT
e
j
dT 'H v 100 o C z
50
100
dC i
p H O(v) dT
2
0.0754 dT 40.656 Cp
100
20
g
H out
H
n out
.
u 10 12 T 3 dT
0.03346 0.688 u 10 5 T 0.7604 u 10 8 T 2 3593
100
H 3
b
zd i z
z d i
z
z
zd i
z d i
25
H 2
100 ˜ 1 y1
2
Cp
100 ˜ y1
Air
bg
H Ob l g
50
n2 ˜ 1 y 2
n in
H2O v
H 1
H in
H
Substance
air
Cp
dT
H 2 O(l)
e
j
dT 'H v 100 o C z
20
100
dC i
p H O(v) dT
2
8-19
8.46 (cont’d)
c.
'H
Q
¦ ni H i ¦ ni H i
out
in
Ÿ
d.
100 mol 8.314 Pa ˜ m3
323 K
mol ˜ K
1.01325 u 105 Pa
Vair
¦ n H ¦ n H
i
Q
Vair
out
i
i
i
in
100 mol 8.314 Pa ˜ m 3
323 K
mol ˜ K
u 10 5 Pa
101325
.
b
p 48q C
2q C superheat Ÿ y1
g
p
83.71 mm Hg
760 mm Hg
b
p 20q C
saturation at outlet Ÿ y2
g
p
0.110 mol H 2 O mol
17.535 mm Hg
760 mm Hg
b gb
g n b1 0.023g Ÿ n
H O balance: b100gb0110
. g b9110
. gb0.023g n Ÿ n
.
dry air balance: 100 1 0110
2
. mol
9110
2
3
2
0.023 mol H 2 O mol
8.90 mol H 2 O 0.018 kg
3
1 mol
0.160 kg H 2 O condensed
Q
'H
¦ ni H i ¦ ni H i
out
Vair
100 mol 8.314 Pa ˜ m 3
323 K
mol ˜ K
101325
u 10 5 Pa
.
Ÿ
Ÿ
e.
f.
480.5 kJ
in
0.160 kg H 2 O condensed
3
2.65 m air fed
480.5 kJ
3
2.65 m air fed
2.65 m 3
0.0604 kg H 2 O condensed / m 3 air fed
181 kJ / m 3 air fed
Solve equations with Maple.
Q
8.47 Basis:
181 kJ 250 m 3 air fed 1 h 1 kW
h
3600 s 1 kJ / s
m 3 air fed
226 m 3
min
273 K
10 3 mol
b g
309 K 22.415 m 3 STP
12.6 kW
8908 mol humid air min . DA = Dry air
Q ( kJ / min)
8908 mol / min
y 0 [ mol H 2 O(v) / mol]
(1- y 0 )(mol DA / mol)
n1 ( mol / min)
y1 [ mol H 2 O(v) / mol]
(1- y1 )(mol DA / mol)
36 o C, 1 atm, 98% rel. hum.
10 o C, 1 atm, saturated
n 2 [ mol H 2 O(l) / min], 10 o C
8-20
8.47 (cont’d)
a. Degree of freedom analysis
5 unknowns – (1 relative humidity + 2 material balances + 1 saturation condition at outlet
+ 1 energy balance) = 0 degrees of freedom.
b36Bq Cg Ÿ y
Table B.3
0.98 p w*
b. Inlet air: y 0 P
p (10 o C) / P
Outlet air: y1
b
0
0.98(44.563 mm Hg)
760 mm Hg
b9.209 mm Hgg b760 mm Hgg
g
Air balance: 1 0.0575 (8908 mol / min)
b1 0.0121gn
1
Ÿ n1
References:
IJ
K
H Obl, triple point g, air b77q Fg
Substance
n in
FG
H
H 2 O balance: 0.0575 8908
Air
0.0575 mol H 2 O(v) mol
0.0121 mol H 2 O(v) mol
8499 mol / min
mol
mol
= 0.0121(8499
) n 2 Ÿ n 2
min
min
409 mol H 2 O(l) min
2
H in
n out
H out
8396 0.3198 8396 0.4352 n in mol min
bg
H Ob l g
H2O v
H in kJ / mol
512
462
103
453
409
0.741
2
Air: H from Table B.8
H 2 O: H ( kJ / kg) from Table B.5 u (0.018 kg / mol)
Energy balance:
Q 'H
¦
ni H i out
¦
ni H i
. u 105 kJ 60 min 9.486 u 104 Btu
196
min
in
1h
1 ton
12000 Btu h
0.001 kJ
930 tons
8.48
Basis:
746.7 m 3 outlet gas / h 3 atm
1 kmol
100 kmol/h at 0°C, 3 atm
yout (kmol C 6 H 14( v)/kmol), saturated
(1 – yout) (kmol N2 /kmol)
n 2 kmols/h n C6 H14 ( v), 0°C
n 1 (kmol/h) at 75°C, 3 atm
yin (kmol C 6 H 14( v)/kmol), 90% sat'd
(1 – yin) (kmol N 2 /kmol)
Antoine: log p v
y out
y in
100.0 kmol / h
b g
1 atm 22.4 m 3 STP
b g
p v 0q C
6.87776 45.24
3 760
1171530
.
224.336 T
b g
b
p v 0q C
45.24 mm Hg, p v 75q C
b g 0.0198 kmol C H kmol ,
0.90 p b75q Cg b0.90gb920.44g
kmol C H
0.363
3b760g
kmol
P
P
6
14
v
6
8-21
14
g
920.44 mm Hg
8.48 (cont’d)
b
g
b
N 2 balance: n1 1 0.363
g
100 1 0.0198 Ÿ n1
b
153.9 kmol h
. kmol C H bl g h
gb g b100gb0.0198g n Ÿ n 5389
Percent Condensation: b5389
. kmol h condenseg b0.363 u 153.9 gb kmol h in feed g u 100%
C 6 H 14 balance: 153.9 0.363
2
References: N2(25oC), n-C6H14(l, 0oC)
Substance
n
H
n
in
N2
98000
bg
bl g
n - C 6 H 14 r
n - C 6 H 14
N 2 : H
b
in
out
55800 44.75
2000
33.33
53800
0.0
z
68.7
g
C p T 25 , n C 6 H 14 (v): H
6
'H
n in mol h
i
i
z
T
b g
C p" dT 'H v 68.7 C pv dT
68.7
( 2.64 u 10 6 kJ h)(1 h / 3600 s) Ÿ 733 kW
¦ n H ¦ n H
out
96.5%
H in kJ mol
0
Energy balance: Q
14
H out
98000 0.726
146
.
2
i
i
in
8.49 Let A denote acetone.
Q ( kW)
W s
25.2 kW
n1 (mol / s) @ 18 o C, 5 atm
y1 [mol A(v) / mol], sat' d
(1 y1 )( mol air / mol)
142 L / s @ 150 o C, 1.3 atm
n 0 ( mol / s)
y 0 [mol A(v) / mol], sat' d
(1 y 0 )( mol air / mol)
n 2 [ mol A(l) / s]@18 o C, 5 atm
a.
Degree of freedom analysis:
b.
Ideal gas equation of state
P0V0
RT0
(1) n 0
6 unknowns ( n 0 , n1 , n 2 , y 0 , y1 , Q )
–2 material balances
–1 equation of state for feed gas
–1 sampling result for feed gas
–1 saturation condition at outlet
–1 energy balance
0 degrees of freedom
Raoult’s law
(2) y1
p *A ( 18 o C)
5 atm
(Antoine equation for p *A )
Feed stream analysis
(3)
y0
FG mol A IJ
H mol K
[(4.973 4.017) g A][1 mol A / 58.05 g]
[(3.00 L) P0 / RT0 ] mol feed gas
8-22
8.49 (cont’d)
Air balance
Acetone balance
n 0 (1 y 0 )
(1 y1 )
(4) n1
n 0 y 0 n1 y1
(5) n 2
o
o
Reference states: A(l, –18 C), air(25 C)
n in
( mol / s)
H in
( kJ / mol)
n in
( mol / s)
H in
( kJ / mol)
A(l)
n 2
0
A(v)
n 0 y 0
H A0
n1 y1
H A1
air
n 0 (1 y 0 )
H a 0
n1 (1 y1 )
H a1
Substance
(6) H A(v) (T )
z
56o C
18 C
o
Table B.2
(7)
z
(C p ) A(l) dT + ( 'H v ) A +
T
56 o C
(C p ) A(v) dT
Tab le B.1
Ta ble B.2
H air (T ) from Table B.8
(8) Q
W s ¦ n
out H out
¦ n
(W s
in H in
25.2 kJ / s)
c.
(1) Ÿ n 0
5.32 mol feed gas / s
(3) Ÿ y 0
0.147 mol A(v) / mol feed gas
(4) Ÿ n1
4.57 mol outlet gas / s
(5) Ÿ n 2
0.75 mol A(l) / s
(6) Ÿ H A 0
56.07 kJ / mol, H A1
33.34 kJ / mol
(7) Ÿ H a 0
3.666 kJ / mol, H a1
1245
.
kJ / mol
(8) Ÿ Q
8.50 a.
6.58 u 10 3 mol A(v) / mol outlet gas
(2) Ÿ y1
90.2 kW
b g
3 m S u 35
2
cm 2
s
1 m2
273 K
10 4 cm 2
b273 40gK
850 mmHg
1 kg ˜ mol
b g
760 mmHg 22.4 m 3 STP
10 3 mol
1 kg ˜ mol
50.3 mol s
H = n-hexane
assume P=850 mmHg
50.3 mol/s, 850 mmHg
x0 mo l H/mo l
(1-x0 ) mo l a ir/ mol
40o C, Tdp =20o C
n 2 mol H(v)/ mol, sat’d @ ToC
n 3 mol air/mol
n 1 (mols H( l )/s)
(90% of H in feed)
8-23
8.50 (cont’d)
Degree-of-freedom analysis
5 unknowns (n1, n2, n3, x0 and T)
– 2 independent material balances
– 1 saturation condition
– 1 60% recovery equation
– 1 energy balance
0 degrees of freedom
All unknowns can be calculated.
b.
Antoine equation, Table B.4
dT i
b
*
25q C
pH
25 q C Ÿ x 0
dp feed
P
151 mm Hg
850 mm Hg
0178
.
mol H mol
. g mols H feed
b50.3gb0178
0.600
Ÿ n1
60% recovery
g
bg
5.37 mols H l s
s
. g 358
. mols Hbv g s
b0.400gb50.3gb0178
. g 413
. mols air s
b50.3gb1 0178
n2
Air balance: n3
Mole fraction of hexane in outlet gas:
n2
n2 n3
b
358
.
358
. 41.3
g
bg
pH* T
bg
Ÿ pH* T
850 mm Hg
67.8 mm Hg Ÿ T
Antoine equation: p H*
b
67.8 mm Hg
7.8 q C
g
Reference states: C 6 H 14 l, 7.8q C , air (25qC)
bg
Obl g
C 6 H14 O v
C 6 H 14
8.95
H in
37.5
3.58
H out
32.7
—
—
5.37
0
41.3
0.435
41.3
–0.499
n in
Substance
Air
z
n out
68 . 74
bg
C 6 H14 O v : H
b
g
z
T
C pl dT ' H v 68.74 q C 7 .8
n in mol/s
H in kJ/mol
C pv dT ,
68 .74
C p from Table B.2
' H v from Table B.1
Air: H from Table B.8
Energy balance: Q
'H
¦ n H ¦ n H
i
i
i
out
c.
u˜ A
u'˜ A' ; A
S ˜ D2
; D'
4
257 kJ s 1 kW cooling
i
1 kJ s
in
|UV
W|
1
D Ÿ u' 4 ˜ u
2
8-24
12.0 m / s
257 kW
8.51
n v ( mol / min) @ 65o C, P0 (atm)
y[ mol P(v) / mol], sat' d
(1- y )(mol H(v) / mol)
100 mol / s @80 o C, 5.0 atm
0.500 mol P(l) / mol
0.500 mol H(l) / mol
Q ( kJ / s)
n l ( mol / min) @ 65o C, P0 (atm)
0.41 mol P(l) / mol
0.59 mol H(l) / mol
a. Degree of freedom analysis
5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance
= 0 degrees of freedom
Antoine equation (Table B.4) Ÿ p *P (65 o C) = 1851 mm Hg, p *H (65 o C) = 675 mm Hg
Raoult' s law for pentane and hexane
0.410 p *P (65 o C) = yP0
0.590 p *H (65 o C) = (1 y ) P0
0.656 mol P(v) / mol
y
Ÿ
1157 mm Hg (1.52 atm)
P0
Total mole balance: 100 mol = n v n l
Ÿ
Pentane balance: 50 mole P = 0.656n v + 0.410n l
nv RT
P0
Ideal gas equation of state: Vv
Fractional vaporization: f
36.6 mol
n v
36.6 mol vapor / s
n l
63.4 mol liquid / s
0.08206 L ˜ atm
mol ˜ K
s
36.6 mol vapor / s
100 mol / s
0.366
mol vaporized
mol fed
References: P(l), H(l) at 65 o C
H in
Substance n in
P(v)
P(l)
50
H(v)
H(l)
50
H out
n out
24.0 24.33
2.806 26.0
0
Liquid: H (T) =
H in kJ / mol
12.6 29.05
3.245 37.4
Vapor: H (T ) =
n in mol s
z
z
Tb
65o C
0
C pl dT 'H v (Tb ) z
T
Tb
C pv dT
T
65o C
C pl dT
Tb and 'H v from Table B.1, C p from Table B.2
Energy balance:
Q
¦ n
out H out
¦ n
in H in
8-25
1040 kW
(65 + 273)K
. atm
152
667 L / s
8.52 a.
B=benzene; T=toluene
n 2 mol/s 95o C
1320 mo l/s 25o C
0.735 mol B/ mo l
0.265 mol T/ mol
0.500 mol B/ mo l
0.500 mol T/ mol
n 3 mol/s 95o C
0.425 mol B/ mo l
0.575 mol T/ mol
Q
UV RS
W T
n
Total mole balance: 1320 n 2 n 3
Ÿ 2
n3
Benzene balance: 1320(0.500) = n 2 (0.735) n 3 (0.425)
319 mol / s
1001 mol / s
References: B(l, 25oC), T(l, 25oC)
n in (mol / s) H in ( kJ / mol) n out (mol / s) H out ( kJ / mol)
Substance
B(l)
B(v)
T(l)
T(v)
Q
660
-660
--
¦ n H ¦ n H
i
i
i
out
b.
0
-0
-i
425
234
576
85
5.361
36.39
10.43
44.19
2.06 u 10 4 kW
in
e
Antoine equation (Table B.4) Ÿ p *B 95 o C
Raoult' s law
Benzene:
Toluene:
j
e
1176 torr , p T* 95 o C
j
476.9 torr
|UV Ÿ P z P'
1035 torr |W
b0.425gb1176g b0.735g P Ÿ P
b0.575gb476.9g b0.265g P' Ÿ P'
680 torr
Ÿ Analyses are inconsistent.
Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the
samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is
invalid at the system conditions (not likely).
b5gb7.5g b12gb9.6g 17
bg
C H Obl g — C
b5gb12g b12gb18g 25
. gb113 273g 42.1 kJ mol
Trouton’s rule — Eq. (8.4-3): 'H
b0109
Eq. (8.4-5) Ÿ 'H
b0.050gb52 273g 16.25 k J mol
8.53 Kopp’s rule (Table B.10):
C 5 H 12 O s — C p
5
12
p
v
m
Basis:
235 m 3
273 K
1 kmol
b g
389 K 22.4 m 3 STP
h
10 3 mol
1h
1 kmol
3600 s
2.05 mol s
Neglect enthalpy change for the vapor transition from 116qC to 113qC.
b
g
b
g
Obs, 25q Cg
b
C 5 H 12 O v , 113q C o C 5 H 12 O l , 113q C o C 5 H 12 O v , 52q C
b
g
o C 5 H 12 O s, 52q C o C 5 H 12
8-26
g
170 J mol
301 J mol
8.53 (cont’d)
'H
b
g
b
g
kJ
J
1 kJ
16.2
b301gb61g b170gb27g
u
mol
mol 10 J
'H v C pl 52 113 'H m C ps 25 52
42.1
kJ
mol
813
. kJ mol
3
2.05 mol 813
. kJ
n'H
'H
Required heat transfer: Q
s
1 kW
mol
1 kJ s
167 kW
8.54
Basis: 100 kg wet film Ÿ
a.
95 kg dry film
5 kg acetone
95 kg DF
5 kg C3 H 6 O( l )
Tf 1 = 35°C
n 1 mol air
Ta1 , 1.01 atm
4.5 kg acetone exit in gas phase
95 kg DF
0.5 kg C3 H 6 O( l )
Tf 2
n 1 mol air
4.5 kg C3 H 6 O( v) (40% sat'd)
Ta2 = 49°C, 1.0 atm
Antoine equation (Table B.4) Ÿ p C* 3H 6O
4.5 kg C 3 H 6 O
Ÿy=
0.5 kg acetone remain in film
90% A evaporation
1 kmol
10 3 mol
58.08 kg
kmol
b
bg
77.5 mol C 3 H 6 O v in exit gas
0.40 59118
. mm Hg
77.5
77.5 n1
59118
. mm Hg
b g
g Ÿn
171.6 mol 22.4 L STP
1
760 mm Hg
b
g
405
.
mol
b
g b
95 kg DF
kg DF
g
References: Air 25q C , C 3 H 6 O l , 35q C , DF 35q C
b.
H out
Substance
n in
H in
n out
DF
95
0
95
86.1
0
8.6
—
—
77.5
32.3
dC i
171.6
0.70
bg
Obv g
C 6 H 14 O l
C 6 H 14
171.6
Air
z
Ta1
25
z
86
H A(v)
dC i dT 'H
p l
35
p air dT
zd
49
v
Cp
i dT ,
v
d
0.129 dT
i
35i
1.33 T f 2 35
b
H DF
f2
'H
¦ n H ¦ n H
i
i
i
out
Ÿ
z
25
Ta1
i
dC i
120q C Ÿ
p air dT
z
Ta1
25
n in mol
H in kJ/mol
g
C p T 35
d
dC i
z
i
. (T f 2 35) 2623.4 1716
.
126.4 T f 2 35 111
in
Ta1
n in kg
H in kJ/kg
86
Energy balance
c.
b g
L STP
d
i
.
127.5 T f 2 35 26234
p air dT
.
1716
d
i
2.78 kJ mol Ÿ T f 2 35 q C
8-27
16.8q C
Ta 1
25
dC i
p air dT
0
8.54 (cont’d)
T& E
34q C Ÿ Ta1
d. T f 2
e.
8.55
T&E
36q C Ÿ Ta1
506q C , T f 2
552q C
In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed
phase drops. In this problem, the heat transferred from the air goes to (1) vaporize 90% of the
acetone in the feed; (2) raise the temperature of the remaining wet film above what it would be
if the process were adiabatic. If the feed air temperature is above about 530 qC, enough heat is
transferred to keep the film above its inlet temperature of 35 qC; otherwise, the film temperature
drops.
b
g
200 psia | 100q F (Cox chart – Fig. 6.1-4)
Tset p
a. Basis:
3.00 u 10 3 SCF 1 lb - mole
h
359 SCF
8.357 lb ˜ mole h C 3 H 8
8.357 lb-mole C3H8(v)/h
200 psia, 100oF
8.357 lb-mole C3H8(l)/h
200 psia, 100oF
Q
- mole H 2 O(l) / h
m(lb
70oF
- mole H 2 O(l) / h
m(lb
85oF
The outlet water temperature is 85oF. It must be less than the outlet propane temperature;
otherwise, heat would be transferred from the water to the propane near the outlet, causing
vaporization rather than condensation of the propane.
b. Energy balance on propane:
Table B.1
Q
'H
n'H v
B
8.357 lb moles 18.77 kJ 0.9486 Btu 453.593 mol
h
mol
kJ
1 lb ˜ mole
6.75 u 10 4
Energy balance on cooling water: Assume no heat loss to surroundings.
Q
'H
p 'T Ÿ m
mC
8.07 u 10 5 Btu lb m ˜q F
h
1.0 Btu 15 q F
53,800
lb m cooling water
h
8.56
m 2 [ kg H 2 O(v) / h]@30 o C, 1 atm
1000 kg/h, 30oC
0.200 kg solids/kg
0.800 kg H2O(l)/kg
m 3 ( kg / h) @ 30 o C
0.350 kg solids / kg
0.650 kg H 2 O(l) / kg
m 1 [ kg H 2 O(l) / h], 1.6 bar, sat' d
m 1 [ kg H 2 O(v) / h], 1.6 bar, sat' d
a.
Solids balance: 200
0.35m3
H 2 O balance: 800
m2 0.65 5714
.
b
g
Ÿ m3
571.4 kg h slurry
Ÿ m2
428.6 kg h H 2 O v
8-28
bg
Btu
h
8.56 (cont’d)
References: Solids (30qC), H 2 O (l, triple point)
Substance
n in
n out
H out
H in
0
200
0
200
Solids
H2O l
125.7
371.4
125.7
800
2529.2
428.6
—
—
H Ov
2
bg
bg
E.B. Q
'H
2791.7
m1
H 2 O , lb˜bars
¦ n H ¦ n H
i
i
out
i
b
b
g
H H 2 O from steam tables
857.6
m1
0 Ÿ 1.030 u 10 6 1833m1
i
g
n kg h
H kJ kg
0 Ÿ m1
562 kg steam h
in
b.
b562.0 428.6g
c.
The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
133 kg h additional steam
8.57 Basis: 15,000 kg feed/h.
A = acetone, B = acetic acid, C = acetic anhydride
Q c (kJ/h)
2 n 1 (kg A(v )/h)
329 K
condenser
n 1 (kg A(l )/h)
303 K
n 1 (kg A(l )/h)
303 K
15000 kg/h
0.46 A
0.27 B
0.27 C
348 K, 1 atm
still
1% of A in feed
n 2 (kg A(l )/h)
n 3 (kg B( l )/h)
Q r (kJ/h) n 4 (kg C( l )/h)
398 K
reboiler
a.
b0.01gb0.46gb15,000 kg hg 69 kg A h
Acetic acid balance: n
b0.27gb15,000g 4050 kg B h
Acetic anhydride balance: n
b0.27gb15,000g 4050 kg h
6831 kg h `
Acetone balance: b0.46gb15,000g n 69 Ÿ n
n 2
3
4
1
1
Distillate product: 6831 kg acetone h
Bottoms product:
b.
b69 4050 4050g kg h
8169 kg h
0.8% acetone
49.6% acetic acid
49.6% acetic anhydride
Energy balance on condenser
8-29
8.57 (cont’d)
b
g
b
g
b
C 3 H 6 O v , 329 K o C 3 H 6 O l , 329 K o C 3 H 6 O l , 303 K
b
'H
z
303
g C dT
b2 u 6831gkg
'H v 329 K pl
b gb g
520.6 2.3 26
g
580.4 kJ kg
329
Q c
c.
'H
n'H
h
580.4 kJ
kg
7.93 u 10 6 kJ h
Overall process energy balance
Reference states: A(l), B(l), C(l) at 348 K (All H m
Substance
n out
n in
H
H
out
in
b
g — 0 6831 –103.5
115.0
69
0
—
A bl, 398 Kg
4050 109.0
0
—
B bl, 398 Kg
4050 113
0
—
C bl, 398 Kg
J
C | b4 u 12g b6 u 18g b3 u 25g
Acetic anhydride (l):
mol˜q C
n in kg/h
H in kJ/kg
A l, 303 K
1 mol 10 3 g 1 kJ
102.1 g 1 kg 10 3 J
p
bg
b
H T
Q
0)
2.3 kJ kg˜q C
g
C p T 348 (all substances)
'H Ÿ Q c Q r
¦ n H ¦ n H
i
i
i
out
in
A
i
Ÿ Q r
Q c ¦ n H e7.93 u 10
i
i
6
out
. u 10 6 kJ h
813
0
(We have neglected heat losses from the still.)
d.
H 2 O (saturated at | 11 bars): 'H v
Q r
8.58
n H 2 O 'H v Ÿ n H 2 O
1999 kJ kg (Table 8.6)
813
. u 10 kJ h
1999 kJ kg
6
4070 kg steam h
Basis: 5000 kg seawater/h
a.
S = Salt
n 3 (kg H 2 O(l )/h @ 4 bars)
2738 kJ/kg
5000 kg/h @ 300 K
0.035 S
0.965 H 2O( l)
113.1 kJ/kg
n 5 (kg H 2 O(l )/h @ 4 bars)
605 kJ/kg
b. S balance on 1st effect:
n 4 kg H 2 O(v )/h @ 0.2 bars
2610 kJ/kg
n 2 (kg H 2 O(v )/h @ 0.6 bars)
2654 kJ/kg
n 1 (kg/h @ 0.6 bars)
0.055 S
0.945 H 2 O(l )
360 kJ/kg
b0.035gb5000g
Mass balance on 1st effect: 5000
0.055n1 Ÿ n1
3182 n 2 Ÿ n 2
8-30
n 3 (kg/h @ 0.2 bars)
x (kg S/kg)
(1 – x) (kg H2 O(l )/hr)
252 kJ/kg
n 2 (kg H 2 O(l )/h @ 0.6 bars)
360 kJ/kg
3182 kg h
1818 kg h
j
2.00 u 105 kJ h
8.58 (cont’d)
Energy balance on 1st effect:
'H
b gb
n 5
n1 3182
n2 1818
c.
g b gb g b gb
2534 kg H Obv g h
g b gb g
.
0 Ÿ n 2 2654 n1 360 n 5 605 2738 5000 1131
0
2
Mass balance on 2nd effect: 3182
n 3 n 4 (1)
b'H 0g
bn gb2610g bn gb252g bn gb360 2654g bn gb360g
E n 3182, n 1818
Energy balance on 2nd effect:
4
3
2
1
1
5.316 u 10 6
0
2
252n 3 2610n 4
(2)
Solve (1) and (2) simultaneously:
n 3
1267 kg h brine solution
n 4
1915 kg h H 2 O v
bg
n 2 n 4
Production rate of fresh water
b
gb g
Overall S balance: 0.035 5000
d.
e.
b1818 1915g
1267 x Ÿ x
3733 kg h fresh water
kg salt kg
0138
.
The entering steam must be at a higher temperature (and hence a higher saturation pressure) than
that of the liquid to be vaporized for the required heat transfer to take place.
n 5 (kg H 2 O(v )/h)
2738 kJ/kg
5000 kg/h
0.035 S
0.965 H 2 O(l )
113.1 kJ/kg
3733 kg/h H2 O(v ) @ 0.2 bar
2610 kJ/kg
n 1 (kg brine/h @ 0.2 bar
252 kJ/kg
Q3
Mass balance: 5000
n 5 (kg H 2 O(l )/h)
605 kJ/kg
3733 n1 Ÿ n1
1267 kg h
d'H 0i
b3733gb2610g b1267gb252g n b605 2738g b5000gb1131. g
Ÿ n
4452 kg H Obv g h
Energy balance:
5
5
0
2
Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or
the construction and maintenance of the second effect?
8-31
8.59 a.
583 kg h
0.30
5000 583 4417 kg fresh water h
Fresh water produced: n L 7 n L1
b.
Final result given in Part (d).
c.
Salt balance on i th effect:
n Li x Li
bn g b x g
L i 1
b0.035gb5000g
x L1 n L1 Ÿ n L1
Salt balance: x L 7 n L 7
L i 1
bn g b x g
L i 1
Ÿ x Li
L i 1
(1)
nT Li
Energy balance on i th effect:
b g e H j
0 Ÿ n vi H vi n v
'H
Ÿ
b g
n v
b g e H j
n H bn g e H j
e H j e H j
L 1
n vi H vi
L 1
v
Li
v
Li
i 1
b g effect:
bn g bn g Ÿ bn g
Mass balance on i 1
n Li
L 1
v i 1
L i 1
n Li H Li n L
L i 1
L
L
L 1
L
L 1
b g e H j
n v
L 1
v
L 1
0
(2)
i 1
L 1
th
L i 1
b g
n Li n v
(3)
i 1
d.
Fresh steam
Effect 1
Effect 2
Effect 3
Effect 4
Effect 5
Effect 6
Effect (7)
P
(bar)
2.0
0.9
0.7
0.5
0.3
0.2
0.1
1.0
T
(K)
393.4
369.9
363.2
354.5
342.3
333.3
319.0
300.0
nL
(kg/h)
--584
1518
2407
3216
3950
4562
5000
8-32
xL
--0.2997
0.1153
0.0727
0.0544
0.0443
0.0384
0.0350
nV
(kg/h)
981
934
889
809
734
612
438
---
HL
(kJ/kg)
504.7
405.2
376.8
340.6
289.3
251.5
191.8
113.0
HV
(kJ/kg)
2706.3
2670.9
2660.1
2646.0
2625.4
2609.9
2584.8
---
8.60 a.
b g | dC i
dC i d C i
p v
20 cal (mol˜q C) ; C v
p l
p v
v
b
R | 10 2
g molcal˜q C
8 cal (mol˜q C)
b.
n0 (mol N2)
n0 (mol N2)
o
3.00 L@ 93 C, 1 atm
n2 [mol A(v)]
85oC, P(atm)
n1 (mol A(l)
n3 [mol A(l)]
o
85oC, P(atm)
0.70 mL, 93 C
3.00 L
n0
b
273 K
1 mol
273 93 K 22.4 L STP
70.0 mL
n1
g
b g
bg
0.90 g 1 mol
mL 42 g
Energy balance Ÿ 'U
0Ÿ
15
. mol A l
¦ n U ¦ n U
i
i
i
out
c.
0.100 mol N 2
i
0
in
b g b gb
g
References: N 2 g , A l 85q C, 1 atm
Substance n in U in n out U out
010
.
39.8 010
.
0
n in mol
N2
15
.
160 n3
0
Al
U in cal mol
Av
n 2 20050
bg
bg
b
g
b
g C b93 85g
Abv , 85q Cg: U
20b90 85g 20,000 10b85 90g
. gb39.8g b15
. gb160g 0 Ÿ n
'U 0 Ÿ n b20050g b010
A l, 93q C and N 2 g , 93q C : U
v
20050 cal mol
A( v )
v1
0.012 mol A 42 g A
Ÿ
d.
mol A
v1
0.012 mol A evaporate
0.51 g evaporate
Ideal gas equation of state
bn
P
0
g
n 2 RT
V
0.112 mol
3.00 liters
b273 85gK
0.08206 L ˜ atm
1.097 atm
mol ˜ K
Raoult’s law
b
p A 85q C
g
yAP
n2
P
n 0 n2
0.012 mol 1.097 atm
0.112 mol
8-33
0.117 atm
b
89.3 mmHg
g
8.61 (a) i)
b4.4553 3.2551gkg
FG mIJ
HV K
Expt 1 Ÿ
2.000 L
liquid
0.600
b g
kg
Ÿ SG
L
b3.2571 3.2551gkg 0.0020 kg
2.000 L 273 K b763 500gmm Hg
ii) Expt 2 Ÿ Mass of gas
Moles of gas
363 K
Molecular weight
2.000 liters 10 3 cm 3
1 liter
iii) Expt. 1 Ÿ n
bliquid g
0.600
2.0 g
1 mol
22.4 liters STP
760 mm Hg
b2.0 gg b0.0232 molg
liquid
b g
0.0232 mol
86 g mol
0.600 g 1 mol
14 mol
cm 3
86 g
Energy balance: The data show that Cv is independent of temperature
Q
'U nCv 'T
b g
Ÿ Cv
b g
Ÿ Cv
Q
n'T
liquid
liquid
800 J
14 mols 2.4 K
b
gb g
800 J
b14 molsgb2.4 Kg
24 J mol ˜ K@284.2 K
24 J mol ˜ K@331.2 K
{ 24 J mol ˜ K
bg
Expt. 2 Ÿ n 0.0232 mol from ii
b vapor g
Cv
a bT Ÿ Q
0.0232
z
T2
T1
(a bT )dT
LM
N
LM
N
LM
N
b
0.0232 a (T2 T1 ) (T22 T12 )
2
b
130
. J = 0.0232 a(366.9 - 363.0) + (366.9 2 363.02 )
2
b
130
. J = 0.0232 a(492.7 - 490.0) + (492.7 2 490.02 )
2
b g
Ÿ Cv
vapor
(J / mol ˜ K)
bg
OP U|
Q |V Ÿ a
OP| b
Q|W
4.069
0.05052
4.069 0.05052T K
iv) Liquid: C p | Cv { 24 J mol ˜ K
Vapor: Assuming ideal gas behavior, C p
Cv R Cv 8.314 J mol ˜ K
b
Ÿ C p J mol ˜ K
v) Expt. 3 Ÿ T
T
T
T
315K ,
334 K ,
354 K ,
379 K ,
p
p
p
p
b763 564gmm Hg
401 mm Hg
761 mm Hg
1521 mm Hg
8-34
g
bg
4.245 0.05052T K
199 mm Hg
OP
Q
8.61 (cont’d)
Plot p (log scale) vs. 1 T (linear scale); straight line fit yields
3770
17.28 or p 3196
u 107 exp 3770 T
ln p
.
T K
b
bg
1
A T
Part v b
760 mm Hg Ÿ
vi) p
vii)
'H v
R
A 3770b Kg
Part v
Ÿ 'H v
b g
17.28 ln 760
3770
2.824 u 10 3 K 1 Ÿ Tb
b3770 Kgb8.314 J mol ˜ Kg Ÿ 'H
1 mol
s
510 K 22.4l STP
Let A denote the drug
(b) Basis:
g
3.5 L feed 273 K
b g
v
31,300 J mol
0.0836 mol s feed gas
.
0.0836 mol/s @ 510 K
0.20 A
0.80 N 2
n 1 [mol A(v)/s]
.
n 2 [mol N 2 /s]
T(K), saturated with A
.
Q(kW)
n 3 (mols A( l )/s), 90% of A in feed
T(K)
b0.800gb0.0836 mol sg 0.0669 mol N s
90% condensation: n b0.900gb0.200 u 0.0836g 0.01505 mol Abl g s
n b0100
. gb0.200 u 0.0836g 167
. u 10 mol Abv g s
N 2 balance: n2
2
3
3
1
Partial pressure of A in outlet gas:
pA
b
n1
P
n1 n2
g
1
T
T
. u 10 3 mol
167
(760 mm Hg) 18.5 mm Hg
0.0686 mol
E Part (a) - (v)
17.28 lnb18.5g
3770
. u 103 K 1
381
262 K
bg
Reference states: N 2 , A l at 262 K
nin
H in
nout
substance
0.0669 7286
0.0669
N2
Av
0.0167 37575 167
. u 10 3
Al
0.01505
bg
bg
H out
n in mol s
0
31686 H in J mol
0
8-35
354 K
bg
pA T
8.61 (cont’d)
b
g
N 2 510 K : H N 2 (510K) - H N 2 (262 K) = H N 2 (237 o C) - H N 2 ( 11o C)
Table B.8
B
[6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
b
A(v, 262K): H
g
b
g
C pl Tb 262 'H v 359 K z
262
Tb
C pv dT
Part (a) results for Tb , C pl , C pv , 'H v
H
L
T O
24b354 262g 31300 M4.245 0.05052 P
31686 J mol
2 Q
N
H C bT 262g 'H b354 K g z C dT 37575 J mol
2 262
354
A(v, 510K):
510
pl
b
v
Energy balance: Q
'H
Tb
¦ ni H i ¦ ni H i
out
in
pv
1060 J s 1 kW cooling
1.06 kW
103 kJ s
8.62 a. Basis: 50 kg wet steaks/min
D.M. = dry meat
m1 (kg H 2 O(v )/min) (96% of H 2 O in feed)
60°C
50 kg/min @ –26°C
0.72 H 2O( s)
0.28 D.M.
Q (kW)
m2 (kg D.M./min)
m3 (kg H 2 O(l )/min)
50°C
96% vaporization:
m 1 0.96 0.72 u 50 kg min
b
g 34.56 kg H O bvg min
m
0.04b0.72 u 50 kg ming 144
. kg H O bl g min
Dry meat balance:
m
b0.28gb50g 14.0 kg D. M. min
Reference states: Dry meat at 26q C , H Obl, 0q Cg
2
3
2
2
2
substance
dry meat
H 2 O s, 26q C
H 2 O l , 50q C
H 2 O v , 60q C
b
b
b
g
g
H in
m out H out
0
14.0 105 m in kg min
H in kJ kg
36.0 390
144
209
.
34.56 2599
m in
14.0
g
kJ 76q C
105 kJ kg
b g 1.38
kg ˜ Cq
H Obs, 26q Cg: H Obl , 0q Cg o H Obs, 0q Cg o H Obs, 26q Cg
b
Dry meat: H 50q C
2
g
C p 50 26
2
2
8-36
2
8.62 (cont’d)
z
b g
'H
6.01 kJ
26
'H m 0q C C p dT
1 mol 10 3 g 2.17 kJ
18.02 g 1 kg kg˜q C
mol
A
0
26q C
390 kJ kg
Table B.1
b
g
b
g
b
b50 0gq C
H 2 O l, 50q C : H 2 O l , 0q C o H 2 O l , 50q C
z
0.0754 kJ
mol q C
50
'H
C p dT
b
1 mol 1000 g
18.02 g 1 kg
A
Table B.2
0
g
b
g
g
b
g
209 kJ kg
b
g
b
H 2 O v , 60q C : H 2 O l , 0q C o H 2 O l , 100q C o H 2 O v , 100q C o H 2 O v , 60q C
'H
0.0754 kJ
b100 0gq C
40.656
mol˜q C
A
Ad
Table B.2
kJ
mol
Table B.1 'H v
46.830 kJ 1 mol 1000 g
mol
18.02 g 1 kg
zd
60
100
i
Cp
i
H 2 O(v)
A
g
dT
Table B.2
2599 kJ kg
Energy balance:
Q
'H
¦
mi H i out
¦
mi H i
in
1.06 u 10 5 kJ 1 min 1 kW
1760 kW
min
60 s 1 kJ s
8.63 Basis: 20,000 kg/h ice crystallized. S = solids in juice. W = water
.
.
m
1 (kg/h) juice
0.12 solids(S)
0.88 H 2O( l )(W)
20°C
Qf
preconcentrate
.m
.
m5 (kg/h) product
Slurry(10% ice), –7°C
2 (kg/h)
freezer
filter
x 2 (kg S/kg)
0.45 kg S/kg
20,000 kg W( s )/h
.
m4 kg residue/h
0.55 kg W/kg
(1 – x 2) (kg W/kg)
20,000 kg W( s )/h
0.45 kg S/kg
.
m4 (kg/h), 0.45 S, 0.55 W
kg W( l )/kg
.m (kg/h),0.55
0°C
3
separator
20,000 kg W( s )/h
0.45 kg S/kg
0.45 kg W( l )/kg
(a) 10% ice in slurry Ÿ
20000
m 4
10
Ÿ m 4
90
180000 kg h concentrate leaving freezer
UV
W
m 1
Overall S balance: 012
. m 1 0.45m 5
Ÿ
m 5
Overall mass balance: m 1 m 5 20000
27273 kg h feed
7273 kg h concentrate product
Mass balance on filter: 20000 m 4 m 5 20000 m 6
Ÿ
m 4 180000
m 5 7273
m 6
Mass balance on mixing point:
27273 172730 m 2 Ÿ m 2
2.000 u 105 kg h preconcentrate
8-37
172730 kg h recycle
8.63 (Cont’d)
S balance on mixing point:
012
.
27273 0.45 172730
b gb
g b gb
g
2.000 u 105 X 2 Ÿ X 2 ˜ 100%
40.5% S
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:
fresh feed and recycle streams; output; slurry leaving freezer)
bg
Refs: S, H 2 O l at 7q C
m in
H in
m out
H out
substance
m kg h
12% soln 27273 108
H kJ kg
0
45% soln 172730 28 180000
20000 337
H 2O s
b
b
bg
g
g
b g 4.00 T b7g kJ kg
'H b T q Cg | 'H b0q Cg
Solutions: H T
Ice: H
m
m
6.0095 kJ mol Ÿ 337 kJ kg
D Table B.1
E.B. Q c
1452
u 107 kJ
.
1h
1 kW
4030 kW
h
3600 s 1 kJ s
¦ m i H i ¦ m i H i
'H
out
in
d
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. (C )
p hf
24.5 kmol/h @ 10oC, P (bar)
0.35 kmol B(l)/h
2.62 kJ / kg˜ o C
i
24.5 kmol/h @ 180oC
0.35 kmol B(l)/h
Q ( kW)
m (kg HF / h), T( o C)
m (kg HF / h), 215 o C
From the Cox chart (Figure 6.1-4)
d
p B* 10 o C
b.
i
d
22 psi, pI* 10 o C
p min
p B pI
d
i
i
32 psi
x B p B* x I pI*
28.5 psi
d
i
FG 1.01325 bar IJ
H 14.696 psi K
d
Hv
H1

o B v, 10 o C '

o B v, 180 o C
B l, 10 o C '
d
i
d
i
d
Hv
H2

o I v, 10 o C '

o I v, 180 o C
I l, 10 o C '
196
. bar
i
i
Assume temperature remains constant during vaporization.
Assume mixture vaporizes at 10oC i.e. won’t vaporize at respective boiling points
as a pure component.
8-38
8.64 (cont’d)
References: B(l, 10oC), I(l, 10oC)
substance
nin mol / h
H in kJ / mol
B (l)
8575
0
B (v)
--I (l)
15925
0
I (v)
---
b
z
z
g
d H i d'H i dC i
d H i d'H i dC i
out
v
B
B
b
180
n out mol / h
g
b
H out kJ / mol
-8575
-15925
g
-42.21
-41.01
42.21 kJ / mol
p B
10
b
g
180
out
v
I
i
i
out
'H
i
4101
. kJ / mol
b
g
b
g
8575 42.21 15825 4101
.
i
in
u 10 6 kJ / h
1015
.
d
Q 1015
u 10 6 kJ / h = m hf 2.62 kJ / kg˜ o C
.
m hf
d.
p I
10
¦ n H ¦ n H
'H
c.
I
i b215 45g C
o
2280 kg / h
b2540 kg / hg 2.62 kJ / dkg˜ Ci b215 45g C
o
u 10 6 kJ / h
1131
.
o
u 10 6 1015
u 10 6
Heat transfer rate 1131
.
.
116
. u 10 5 kJ / h
e. The heat loss leads to a pumping cost for the additional heating fluid and a greater
heating cost to raise the additional fluid back to 215oC.
f. Adding the insulation reduces the costs given in part (e). The insulation is
probably preferable since it is a one-time cost and the other costs continue as long
as the process runs. The final decision would depend on how long it would take
for the savings to make up for the cost of buying and installing the insulation.
8.65 (a) Basis: 100 g of mixture, SGBenzene=0.879: SGToluene=0.866
ntotal
Vtotal
dx i
f
50 g
50 g
mol
(0.640 0.542) mol 1183
.
78.11 g / mol 92.13 g / mol
50 g
50 g
114.6 cm3
3
0.879 g / cm
0.866 g / cm3
C6H 6
0.640 mol C 6 H 6
1.183 mol
Actual feed:
T
0.541 mol C 6 H 6 mol
.
mol mixture
32.5 m3 106 cm3 1183
1h
. mol / s
9319
3
3
h
1 m
114.6 cm mixture 3600 s
90q C Ÿ pC 6 H 6
Raoult' s law: ptot
1021 mm Hg , pC 7 H 8
407 mm Hg (from Table 6.1-1)
x C6 H 6 pC 6 H 6 x C 7 H 8 pC7 H 8
739.2 mmHg
1 atm
760 mmHg
8-39
b0.541gb1021g b0.459gb407g
0.973 atm Ÿ P0 ! 0.973 atm
8.65 (cont’d)
75q C Ÿ pC 6 H 6
(b) T
Raoult's law Ÿ ptank
648 mm Hg , pC 7 H 8
244 mm Hg (from Table 6.1-1)
b0.439gb648g b0.561gb244g
xC6 H 6 pC6 H 6 xC7 H 8 pC7 H 8
b284 137gmm Hg = 421 mmHg Ÿ P
284 mm Hg
0.675 mol C H bv g mol
421 mm Hg
0.554 atm
tank
yC 6 H 6
6
6
n v (mol/s), 75°C
0.675 C 6H 6 (v )
0.554 atm 0.325 C 7H 8 (v )
n L (mol/s), 75°C
0.439 C6 H6 (l )
0.541 C7H8 (l )
93.19 mol/s
0.541 C 6H 6( l )
0.459 C 7 H 8 (l )
90°C, P0 atm
UV
W
nv
. = nv n L
Mole balance: 9319
Ÿ
nL
.
0.675nv 0.439n L
C 6 H 6 balance: 0.541 9319
b
gb
g
bg
40.27 mol vapor s
52.92 mol liquid s
bg
(c) Reference states: C 6 H 6 l , C 6 H 6 l at 75q C
Substance
C6 H 6 v
n out
H out
.
27.18 310
50.41 2.16 23.23
0
13.09 35.3
42.78 2.64 29.69
0
n in
H in
n in mol s
bg
H in kJ mol
C H bl g
C H bv g
C H bl g
C H bl , 90q Cg: H b0144
. gb90 75g 2.16 kJ mol
C H bl , 90q Cg: H b0176
. gb90 75g 2.64 kJ mol
C H bv , 75q Cg: H b0144
. gb801
. 75g 30.77 z 0.074 0.330 u 10
A
6
6
7
8
7
8
6
6
7
8
6
6
75
b
'H v 80 .1qC
g
80.1
3
T dT
310
. kJ mol
b
g
C 7 H 8 v , 75q C : H
. gb110.6 75g 33.47 z
b0176
75
110.6
0.0942 0.380 u 10 3 T dT
35.3 kJ mol
Energy balance: Q
'H
¦ n H ¦ n H
i
out
i
i
in
i
1082 kJ 1 kW
1082 kW
s 1 kJ s
(d) The feed composition changed; the chromatographic analysis is wrong; the heating rate
changed; the system is not at steady state; Raoult’s law and/or the Antoine equation are
only approximations; the vapor and liquid streams are not in equilibrium.
(e) Heat is required to vaporize a liquid and heat is lost from any vessel for which T>Tambient.
If insufficient heat is provided to the vessel, the temperature drops. To run the experiment
isothermally, a greater heating rate is required.
8-40
8.66 a. Basis: 1 mol feed/s
n V mo l vapor/s @ T, P
1 mo l/s @ TFo C
y mol A/mol
(1-y) mol B/mo l
xF mol A/mol
(1-xF) mol B/mo l
n L mol vapor/s @ T, P
vapor and liquid streams
in equilibrium
x mol A/mol
(1-x) mol B/mo l
bg b g bg
bT g Ÿ y x ˜ p bT g
Raoult's law Ÿ x ˜ p A T 1 x ˜ p B T
pA
y˜P
x ˜ pA
b x gb1g
F
Energy balance: 'H
bg
bT g p bT g
P pB T
pA
(1)
B
A
(2)
P
Mole balance: 1 n L nV Ÿ nV
A balance:
PŸ x
1 n L
(4)
for nv from (4)

o n L
y ˜ nV x ˜ n L Substitute
¦ n H ¦ n H
i
out
i
i
y xF
yx
(3)
0
i
(5)
in
b.
ref(deg.C) = 25
Compound
n-pentane
n-hexane
A
B
C
6.85221 1064.63 232.000
6.87776 1171.53 224.366
xF
Tf(deg.C)
P(mm Hg)
HAF(kJ/mol)
HBF(kJ/mol)
0.5
110
760
16.6
18.4
0.5
110
1000
16.6
18.4
0.5
150
1000
24.4
27.0
T(deg.C)
pA*(mm Hg)
pB*(mm Hg)
x
y
nL(mol/s)
nV(mol/s)
HAL(kJ/mol)
HBL(kJ/mol)
HAV(kJ/mol)
HBV(kJ/mol)
DH(kJ/s)
51.8
1262
432
0.395
0.656
0.598
0.402
5.2
5.8
31.4
42.4
0.00
60.0
1609
573
0.412
0.663
0.648
0.352
6.8
7.6
32.5
43.7
0.00
62.3
1715
617
0.349
0.598
0.393
0.607
7.3
8.0
32.8
44.1
0.00
al
0.195
0.216
8-41
av
0.115
0.137
bv
3.41E-04
4.09E-04
Tbp
36.07
68.74
DHv
25.77
28.85
8.66 (cont’d)
c.
C*
C*
1
C*
C*
C*
C*
2
20
25
3
30
PROGRAM FOR PROBLEM 8.66
IMPLICIT REAL (N)
READ (5, 1) A1, B1, C1, A2, B2, C2
ANTOINE EQUATION COEFFICIENTS FOR A AND B
FORMAT (8F10.4)
READ (5, 1) TRA, TRB
ABRITARY REFERENCE TEMPERATURES (DEG.C.) FOR A AND B
READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2
READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2
CP(LIQ, KS/MBL-DEG.C.), NORMAL BOILING POINT (DEG.C), HEAT
OF
VAPORIZATION
(KJ/MOL), COEFFICIENTS OF CP(VAP., KJ/MOL-DEG.C) CV1 CV2*T(DEG.C)
READ (5, 1) XF, TF, P
MOLE FRACTION OF A IN FEED, FEED TEMP.(DEG.C), EVAPORATOR
PRESSURE (MMHG)
WRITE (6, 2) TF, XF, P
FORMAT (1H0, 'FEEDbATb', F6.1, 'bDEG.CbCONTAINSb', F6.3,'
bMOLESbA/MOLEbT
*OTAL'//1X'EVAPORATORbPRESSUREb ', E11.4, 'bMMbHG'/)
ITER 0
DT 0.5
HAF CAL*(TF – TRA)
HBF CBL*(TF – TRB)
F1 XF*HAF (1.0 – XF)*HBF
F2 CAL*(TBPA – TRA) DHVA – CAV1*TBPA – 0.5*CAV2*TBPA**2
F3 CBL*(TBPB – TRB) DHVB – CBV1*TBPB – 0.5*CBV2*TBPB**2
T TF
INTER ITER 1
IF(ITER – 200) 30, 30, 25
WRITE (6, 3)
FORMAT (1H0, 'NO CONVERGENCE')
STOP
PAV 10.0** (A1 – B1/(T C1))
PAV 10.0** (A2 – B2/(T C2))
XL (P – PBV)/(PAV – PBV)
XV XL*PAV/P
NL (XV – XF)/(XV – XL)
NV 1.0 – NL
IF (XL.LE.00.OR.XL.GE.1.0.OR.NL.LE.0.0.OR.NL.GE.1.0) GO TO 45
HAL CAL*(T – TRA)
HBL CBL*(T – TRB)
HAV F2 CAV1*T 0.5*CAV2*T**2
HBV F3 CBV1*T 0.5*CBV2*T**2
8-42
8.66(cont’d)
DELH NL *(XL*HAL (1.0 – XL)*HBL) NV*(XV*HAV (1.0 –
XV)*HBV) – F1
WRITE (6, 4) T, NL, NV, DELH
4
FORMAT (1Hb, 5X' Tb ', F6.1, 3X' NLb ', F7.4, 3X' NVb ', F7.4,
3X'DELHb ',* E11.4)
WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV
5
FORMAT (1Hb, 5X' PAV, PBVb ', 2F8.1, 3X' XL, HAL, HBLb ', F7.4,
2E13.4,3X' XV, HAV, HBVb ', F7.4, 2E13.4/)
IF (DELH) 50, 50, 40
40
DHOLD DELH
TOLD T
45
T T – DT
GO TO 20
50
T (T*DHOLD – TOLD*DELH)/(DHOLD – DELH)
PAV 10.0**(A1 – B1/(T C1))
PBV 10.0**(A2 – B2/(T C2))
XL (P – PBV)/(PAV – PBV)
XV XL * PAV/P
NL (XV – XF)/(XV – XL)
NV 1.0 – NL
WRITE (6, 6) T, NL, XL, NV, XV
6
FORMAT (1H0, 'PROCEDUREbCONVERGED'//3X'EVAPORATORb
TEMPERATUREb ', F6.
*1//3X' LIQUIDbPRODUCTb--', F6.3, 'bMOLEbCONTAININGb', F6.3,
'bMOLEbA/
*MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6.3,
MOLEbCONTAININGb,' F6.3,
*'bMOLEbA/MOLEb TOTAL')
STOP
END
$DATA (Fields of 10 Columns)
.
6.85221 1064.63 232.0 6.87776 117153
224.366
25.0
25.0
.
.
0195
36.07 25.77 0115
0.000341
.
0.216
68.74 28.85 0137
0.000409
0.500
110.0 760.0
Solution:
Tevaportor 52.2q C
d
i
d
i
nL
0.552 mol, x C5 H12
nv
0.448 mol, x C5 H12
liquid
vapor
0.383 mol C 5 H 12 mol liquid
0.644 mol C 5 H 12 mol liquid
8-43
8.67 Basis:
2500 kmol product 1 kmol condensate
10,000 kmol h fed to condenser
h
.25 kmol product
.
m1 , (kg/h) at T1
1090 kmol/h C 3 H 8 ( l )
7520 kmol/h i -C 4H10 ( l )
P (mm Hg) 1390 kmol/h -C H ( )
n 4 10 l
Tout
1090 kmol/h C 3 H 8 (v )
7520 kmol/h i -C 4H10 (v )
1390 kmol/h n -C 4H10 (v )
saturated vapor at Tf, P
.
m1 (kg/h) at
2 T2
0 o C , T1
(a) Refrigerant: Tout
Antoine constants
C3H 8
i C 4 H 10
n C 4 H 10
A
7.58163
6.74808
6.83029
Calculate P for Tout
P
6 o C .
T2
B
1133.65
882.8
945.0
C
283.26
240.0
240.0
Tbubble pt.
¦ xi pi* b0q Cg
b
g
b
g
b
g
0109
.
3797 mm Hg 0.752 1174 mm Hg 0139
.
774 mm Hg
i
Ÿ P 1404 mm Hg
d i
Tdp Ÿ f Tf
Dew pt. Tf
1 P¦
i
d i
yi
*
d i
Pi Tf
0 trial & error to find Tf
Tf
f Tf
5q C 0.00162 Ÿ linear interpretation Tf
4q C 0.0333
bg
4.95q C
bg
Refs: C 3 H 8 l , C 4 H 10 l at 0 qC, Refrigerant @ –6qC
b g
Assume: 'H v Tb , Table B.1
substance
C3H 8
i C 4 H 10
n C 4 H 10
Refrigerant
E.B.:
'H
nin
1090 19110
7520 21740
1390 22760
m 1
¦ ni H i ¦ ni H i
out
H in
0
p
nout
H out
1090
7520
1390
0
0
0
n (kmol/h)
H (kJ/kmol)
m 1
151
m (kg/h)
H (kJ/kmol)
8-44
2
z
v
4 .95
p
0
0 Ÿ 151m 1 2.16 u 106
in
U| H bvaporg 'H b0q Cg V| C dTbTable B.2g
W
0 Ÿ m 1
UV H
W
'H v
143
. u 106 kg h refrigerant
8.67 (cont’d)
¦ xi pi* b40q Cg
P
34qC , T1
40q C , T2
(b) Cooling water: Tout
b
g
25qC
b g
b g
0109
.
11,877 0.752 3983 0139
.
2853
4653 mm Hg
i
i
*
i
i
Refs:
TE
d i 1 P¦ p dyT i 0 Ÿ T 45.7q C
C H bl g , C H bl g @ 40qC, H Obl g @ 25qC.
f Tf
3
8
'H
4
f
f
10
2
0 Ÿ 37.7m 1 2.17 u 10
0 Ÿ m 1
8
5.74 u 10 6 kg H 2 O / h
(c) Cost of refrigerant pumping and recompression, cost of cooling water pumping, cost of
maintaining system at the higher pressure of part (b).
8.68 Basis: 100 mol leaving conversion reactor
H 2 O(v )
3.1 bars, sat'd
n 3 (mol O 2 )
3.76 n 3 (mol N 2 )
H 2 O( l )
45°C
conversion 100 mol, 600°C, 1 atm
145°C
100°C
reactor
0.199 mol HCHO/mol
n 4 (mol H 2 O( v))
0.0834 mol CH 3OH/mol
0.303 mol N 2/mol
n 1 (mol CH3 OH(l )) n 2 (mol CH 3 OH(l ))
0.0083 mol O 2/mol
m w1 (kg H 2 O(l )) m w2 (kg H 2 O(l ))
0.050 mol H 2/mol
n 8 (mol CH 3 OH(l ))
0.356 mol H 2O( v)/mol 3.1 bars, sat'd
30°C
Q (kJ)
CH 3 OH( l ), 1 atm, sat'd
n 6a (mol HCHO)
2.5n 8 (mol CH 3 OH)
distillation
absorption
n 6b (mol CH 3 OH( l ))
(l )
n 6c (mol H 2 O( l ))
sat'd, 1 atm
Product solution
88°C, 1 atm
n 7 (mol)
m w3 (kg H 2 O(l ))
0.37 g HCHO/g (x 1 mol/min)
Absorber off-gas
n 5a (mol N 2 )
0.01 g CH 3OH/g (x 2 mol/min)
30°C
n 5b (mol O 2 )
0.82 g H 3 O/g (x 3 mol/min)
n 5 c (mol H 2 )
n 5d (mol H 2 O(v )), sat'd
n 5 e (mol HCHO(v )), 200 ppm
27°C, 1 atm
a. Strategy
C balance on conversion reactor Ÿ n2 , N 2 balance on conversion reactor Ÿ n3
H balance on conversion reactor Ÿ n4 , (O balance on conversion reactor to check
consistency)
N 2 balance on absorber Ÿ n5a , O 2 balance on absorber Ÿ n5b
H 2 balance on absorber Ÿ n5e
UV
W
H 2 O saturation of absorber off - gas
Ÿ n5d , n5b
200 ppm HCHO in absorber off - gas
8-45
8.68 (cont’d)
HCHO balance on absorber Ÿ n6a , CH 3 OH balance on absorber Ÿ n6b
Wt. fractions of product solution Ÿ x1 , x 2 , x 3
HCHO balance on distillation column Ÿ n7
CH 3 OH balance on distillation column Ÿ n8
CH 3 OH balance on recycle mixing point Ÿ n1
Energy balance on waste heat boiler Ÿ mw1 , E.B. on cooler Ÿ mw2
Energy balance on reboiler Ÿ Q
C balance on conversion reactor:
n2 19.9 mol HCHO 8.34 mol CH 3 OH
28.24 mol CH 3 OH
N 2 balance on conversion reactor:
3.76n3
30.3 Ÿ n3
8.06 mol O 2 , 3.76 u 8.06 30.3 mol N 2 feed
H balance on conversion reactor:
bg
bg
bg
bg bg
bg
n4 2 28.24 4 19.9 2 8.34 4 5 2 35.6 2 Ÿ n4
20.7 mol H 2 O fed
O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error)
N 2 balance on absorber: 30.3 n5a Ÿ n5a
30.3 mol N 2
O 2 balance on absorber: 0.83 n5b Ÿ n5b
0.83 mol O 2
H 2 balance on absorber: 5.00 n5c Ÿ n5c
5.00 mol H 2
H 2 O saturation of off - gas:
yw
b
p w* 27q C
Ÿ n5d
P
g LM 26.739 mm Hg
N 760 mm Hg
n5d
30.3 0.83 5.00 n5d n5e
g U|
|
200 ppm HCHO in off gas:
V|
n
200
Ÿ
2|
|W
. n n
3613
10
OP
Q
b
. n5d n5e 1
0.03518 3613
solve
Ÿ
n5d
n5e
.
mol H 2 O
1318
7.49 u 10 3 mol HCHO
5e
6
5d
Moles of absorber off-gas
5e
n5a n5b n5c n5e
HCHO balance on absorber: 19.9
37.46 mol off - gas
n6a 7.49 u 10 3 Ÿ n6a 19.89 mol HCHO
CH 3 OH balance on absorber: 8.34
n 6b Ÿ n 6b
Product solution
8.34 mol CH 3 OH
U|
|V
||
W
%MW
x1
.
Basis - 100 g Ÿ 37.0 g HCHO Ÿ 1232
mol HCHO
1.0 g CH 3 OH Ÿ 0.031 mol CH 3 OH Ÿ x 2
x3
62.0 g H 2 O Ÿ 3.441 mol H 2 O
8-46
0.262 mol HCHO mol
0.006 mol CH 3 OH mol
0.732 mol H 2 O mol
8.68 (cont’d)
HCHO balance on distillation column (include the condenser + reflux stream within the
system for this and the next balance):
0.262n7 Ÿ n7
19.89
75.9 mol product
CH 3 OH balance on distillation column:
b g
0.006 75.9 n8 Ÿ n8
8.34
7.88 mol CH 3 OH
CH 3 OH balance on recycle mixing point:
n1 n8
n2 Ÿ n1
28.24 7.83 20.36 mol CH 3 OH fresh feed
Summary of requested material balance results:
bg
n1
20.4 mol CH 3 OH l fresh feed
n2
75.9 mol product solution
n3
7.88 mol CH 3 OH l recycle
n4
37.5 mol absorber off - gas
bg
Waste heat boiler:
b
g
b
g
bg
Refs: HCHO v, 145q C , CH 3 OH v, 145q C ; N 2 , O 2 , H 2 , H 2 O v at 25qC for product
b
g
gas, H 2 O l, triple point for boiler water
substance
nin
H in
nout
H out
HCHO
CH 3 OH
19.9
8.34
30.3
0.83
5.0
35.6
22.55
32.02
17.39
18.41
16.81
20.91
19.9
8.34
30.3
0.83
5.0
35.6
0
n (mol)
0
3.51
3.60 H (kJ/mol)
3.47
4.09
mw1
566.2
mw1 2726.3 m (kg)
2 H (kJ/kg)
N2
O2
H2
H2O
H2O
(boiler)
E.B. 'H
¦ n H ¦ n H
i
out
i
i
i
0 Ÿ 1814 2160mw1
in
8-47
UV H C dT
W
U|
|V H C bT g T 25
||
W
UV H from steam tables
W
z
T
p
145
p
0 Ÿ mw1
0.84 kg 3.1 bar steam
8.68 (cont’d)
b
g
Gas cooler: Same refs. as above for product gas, H 2 O l, 30q C for cooling water
substance
nin
H in
nout
H out
HCHO
CH 3 OH
19.9
8.34
30.3
0.83
5.0
35.6
0
0
3.51
3.60
3.47
4.09
19.9
8.34
30.3
0.83
5.0
35.6
–1.78
–2.38
2.19
2.24
2.16
2.54
mw2
0
mw2
62.76
N2
O2
H2
H2O
H2O
(coolant)
¦ n H ¦ n H
E.B. 'H
i
i
i
out
i
b.
Q
m (kg)
H (kJ/kg)
0 Ÿ 1581
. 62.6mw 2
H
4.184
0 Ÿ mw 2
b
g
kJ
T 30 q C
kg˜q C
2.52 kg cooling water
in
b35. gb7.88g
n'H b1 atmg b27.58 molgb35.27 kJ molg
Condenser: CH 3 OH condensed
E.B.:
n (mol)
H (kJ/mol)
n8 2.5n8
27.58 mol CH 3 OH condensed
v
973 kJ (transferred from condenser)
3.6 u 10 4 tonne / y
10 6 g
1 yr
1d
1 metric ton 350 d 24 h
b0.37gd4.286 u 10 i
Ÿ b0.01gd4.286 u 10 i
b0.62gd4.286 u 10 i
4.286 u 10 6 g h product soln
6
u 10 6 g HCHO h Ÿ 5.281 u 10 4 mol HCHO h
.
1586
6
4.286 u 10 6 g CH 3 OH h Ÿ 1338 mol CH 3 OH h
6
u 105 mol H 2 O h
.
2.657 u 10 6 g H 2 O h Ÿ 1475
Ÿ 2.016 u 10 5 mol h Ÿ Scale factor =
2.016 u 10 5 mol h
75.9 mol
U|
V|
|W
2657 h 1
8.69 (a) For 24°C and 50% relative humidity, from Figure 8.4-1,
Absolute humidity = 0.0093 kg water / kg DA, Humid volume | 0.856 m 3 / kg DA
Specific enthalpy = (48 - 0.2) kJ / kg DA = 47.8 kJ / kg DA , Dew point = 13o C, Twb
17 o C
(b) 24 o C (Tdb )
(c) 13o C (Dew point)
(d) Water evaporates, causing your skin temperature to drop. Tskin | 13o C (Twb ). At 98%
R.H. the rate of evaporation would be lower, Tskin would be closer to Tambient , and you
would not feel as cold.
8-48
8.70
141 ft 3 . DA = dry air.
V room
mDA =
ha
140 ft 3
lb - mol˜ o R 29 lb m DA 1 atm
lb - mol 550 o R
0.7302 ft 3 ˜ atm
0.205 lb m H 2 O
101
. lb m DA
0.0203 lb m H 2 O / lb m DA
90 o F, ha
From the psychrometric chart, Tdb
hr
Tdb
35q C
Tab
27q C
8.72 a. Tdb
Ÿ hr
b. Mass of dry air: mda
c.
80.5o F
14.3 ft 3 / lb DA
V
m
H
44.0 011
. # 43.9 Btu / lb m
33%, ha
0.0148 kg H 2 O kg dry air
77.3o F
55% He wins
40q C, Tdew point
Mass of water:
0.0903
Twb
67%
Tdew point
8.71
101
. lb m DA
Fig. 8.4-1
20q C
Ÿ
hr
Twb
255
. qC
1 m 3 1 kg dry air
2.2 u 10 3 kg dry air
10 3 L
0.92 m 3
n from Fig. 8.4-1
2.00 L
2.2 u 10 3 kg dry air 0.0148 kg H 2 O 10 3 g
1 kg dry air
1 kg
0.033 g H 2 O
77.4 kJ kg dry air
b
g b
g
H b20q C, saturated g | 57.5 kJ kg dry air (both values from Fig. 8.4-1)
2.2 u 10 kg dry air b57.5 77.4g kJ 10 J
'H
44 J
H 40q C, 33% relative humidity | 78.0 0.65 kJ kg dry air
3
3
40o 20
kg dry air
1 kJ
d. Energy balance: closed system
2.2 u 10 3 kg dry air 10 3 g 1 mol 0.033 g H 2 O 1 mol
18 g
1 kg 29 g
Q 'U n'U n 'H R'T 'H nR'T
n
d
= 44 J i
0.078 mol 8.314 J
mol ˜ K
b20 40gq C
8-49
1K
1q C
0.078 mol
31 J(23 J transferred from the air)
8.73 (a)
400 kg 2.44 kg water
97.56 kg air
10 kg H 2 O min
400 kg dry air min
min
(b) ha
Fig. 8.4-1
(c) Tdb
(d)
10.0 kg water evaporates / min
0.025 kg H 2 O kg dry air , Tdb
b116 11. g
H
115 kJ kg dry air , Twb
400 kg dry air
min
33q C, hr
0.0077 kg H 2 O kg dry air , H
10q C , saturated Ÿ ha
b0.0250 0.0077g kg H O
2
50q C
32%, Tdew point
28.5q C
29.5 kJ kg dry air
6.92 kg H 2 O min condense
kg dry air
bg
References: Dry air at 0q C, H 2 O l at 0q C
substance
m out
m in
H out
H in
Air
bg
H 2O l
b
400
115
400
29.5
—
—
4.1
42
g
b
g
m air in kg dry air/min, m H 2 O in kg/min
H air in kJ/kg dry air, H H 2 O in kJ/kg
H 2 O l , 0q C o H 2 O l , 20q C :
75.4
H
Q
'H
J
1 mol
mol˜q C 18 g
¦ m i H i ¦ m i H i
out
(e)
in
b10 0gq C
1 kJ 103 g
42 kJ kg
103 J 1 kg
34027.8 kJ 1 min 1 kW
567 kW
min 60 s 1 kJ / s
T>50°C, because the heat required to evaporate the water would be transferred from the
air, causing its temperature to drop. To calculate (Tair)in, you would need to know the
flow rate, heat capacity and temperature change of the solids.
8.74 a. Outside air: Tdb 87q F , hr 80% Ÿ ha 0.0226 lb m H 2 O lb m D.A. ,
H 455
. 0.01 455
. Btu lb m D.A.
Room air: Tdb 75q F , hr 40% Ÿ ha 0.0075 lb m H 2 O lb m D.A. ,
H 26.2 0.02 26.2 Btu lb m D.A.
Delivered air: Tdb 55q F , ha 0.0075 lb m H 2 O lb m D.A.
Ÿ H 214
. 0.02 214
. Btu lb m D.A. , V 13.07 ft 3 lb m D.A.
1,000 ft 3 1 lb m D.A.
76.5 lb m D.A. min
Dry air delivered:
min
13.07 ft 2
H 2 O condensed:
76.5 lb m D.A.
min
b0.0226 0.0075g lb
H 2O
lb m D.A.
8-50
m
12
. lb m H 2 O min condensed
8.74 (cont’d)
The outside air is first cooled to a temperature at which the required amount of water is
condensed, and the cold air is then reheated to 75qF. Since ha remains constant in the
second step, the condition of the air following the cooling step must lie at the intersection
of the ha 0.0075 line and the saturation curve Ÿ T 49q F
b
g
References: Same as Fig. 8.4-2 [including H 2 O l, 32q F ]
substance
m out H
m in H
out
in
Air
b
g
H 2 O l, 49q F
76.5 45.5 76.5 21.4 m air in lb m D.A./min
—
—
1.2 17.0 H in Btu/ lb D.A.
air
m
m
in lb /min, H
H 2O
Q
'H
b76.5g 214. 455. + 1.2(17.0) (Btu)
min
m
H 2O
in Btu/ lb m
60 min 1 ton cooling
12,000 Btu h
1h
9.1 tons cooling
b. Examination of the flow chart shows that the overall energy balance must be the same as
the system without recirculation, except for all extensive variables being scaled down by a
factor of 1/7. (Air must be delivered to the room at the same rate, and 6/7 of it is
recycled.) Consequently
Net cooling: Q
9.1 tons / 7 1.3 tons
6
u 100 86%
7
Once the system reaches steady state, most of the air passing through the conditioner is
cooler than the outside air, and (more importantly) much less water must be condensed
(only the water in the fresh feed).
Percent Saved:
c. Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup
of carbon dioxide in the laboratory.
8.75 Basis: 1 kg wet chips. DA = dry air, DC = dry chips
Outlet air: Tdb=38oC, Twb=29oC
m2a (kg DA)
m2w [kg H2O(v)]
Inlet air: 11.6 m3(STP), Tdb=100oC
m1a (kg DA)
1 kg wet chips, 19oC
0.40 kg H2O(l)/kg
0.60 kg DC/kg
m3c (kg dry chips)
m3w [kg H2O(l)]
T (oC)
(a) Dry air: m1a =
Outlet air:
Tdb 38q C, Twb
b
b g
11.6 m 3 STP DA
1 kmol
22.4 m 3 STP
b g
g
8.4-1

o H 2
29q C Fig.
8-51
29.0 kg
15.02 kg DA = m 2a
1 kmol
95.3 kJ kg D.A. , ha2
0.0223 kg H 2 O kg D.A.
8.75 (cont’d)
Water in outlet air: m2 w
b
ha2 m2 a
0.0223 15.02
g
0.335 kg H 2 O
(b) H 2 O balance: 0.400 kg = 0.335 kg + m3w Ÿ m3w
0.065 kg H 2 O
Moisture content of exiting chips:
0.065 kg water
u 100%
(1 0.335) kg chips
9.8% 15% ? meets design specification
bg
(c) References: Dry air, H 2 O l , dry chips @ 0qC.
substance
min
H in
Air
H 2O l
dry chips
15.02
0.400
0.600
100.2
79.5
39.9
bg
H out
mout
15.02 95.3 mair in kg DA, H air in kJ/kg DA
0.065 4.184T m in kg DC, H in in kJ/kg DC
0.6 2.10T
Energy Balance:
'H
Tdb 45q C
hr 10%
b. Twb 210
. qC
hr 60%
¦m
8.76 a.
H 2 O added:
8.77
¦m
Tas
Twb
out H out
in H in
0 Ÿ 129.3 1532
. T
210
. qC
ha
0ŸT
84.4q C
0.0059 kg H 2 O kg DA
Fig. 8.4-1
Tdb
26.8q C
Fig. 8.4-1
15 kg air
1 kg D.A.
min
1.0059 kg air
Inlet air: Tdb 50q C
Tdew pt. 4q C
Fig. 8.4-1
V
ha
ha
0.0142 kg H 2 O kg DA
b0.0142 0.0059g kg H O
2
1 kg D.A.
0.92 m 3 kg D.A. , Twb
012
. kg H 2 O min
22q C
0.0050 kg H 2 O kg D.A.
11.3 m 3 1 kg D.A.
12.3 kg D.A. min
min
0.92 m3
Outlet air: Twb
Tas
22q C
saturated
Evaporation:
12.3 kg D.A.
min
ŸT
22q C ha
0.0165 kg H 2 O kg D.A.
b0.0165 0.0050g kg H O
2
kg D.A.
8-52
014
. kg H 2 O min
8.78 a. Tdb 45q C
Tdew point 4q C
Twb
UV
W
bh g
0.0050 kg H 2 O kg D.A.
a in
20.4q C, V
Twb
Fig. 8.4-1
b g
20.4q C, saturated Ÿ ha
Tas
0.908 m 3 kg D.A.
0.0151 kg H 2 O kg D.A.
out
b. Basis: 1 kg entering sugar (S) solution
m1 (kg D.A.)
0.0050 kg H2O/kg DA
m1 (kg D.A.)
0.0151 kg H2O(v)/kg
m2 (kg)
0.20 kg S/kg
0.80 kg H2O/kg
1 kg
0.05 kg S/kg
0.95 kg H2O/kg
b gb g b0.20gm Ÿ m 0.25 kg
Water balance: bm gb0.0050g b1gb0.95g bm gb0.0151g b0.25gb0.80g
Sugar balance: 0.05 1
2
2
1
1
m1
Ÿ
B
A
1 lb m D.A.
ha1 (lb m H 2O)
T d = 20°F
h r = 70%
Inlet air (A):
C
D
Spray 1 lb m D.A.
1 lb m D.A.
ha2(lb m H 2O) chamber h a3(lb m H 2O)
T d = 75°F
H2 O
Coil
bank
Tdb 20q F
hr 70%
Outlet air (D):
0.908 m 3
67 m 3
1 kg D.A.
74 kg dry air
V
8.79
74 kg dry air
UV
W
Tdb 70q F
hr 35%
1 lb m D.A.
h a3(lb m H 2O)
T d = 70°F
h r = 35%
ha1 | 0.0017 lb m H 2 O lb m D.A.
V | 12.2 ft 3 lb m D.A.
Fig. 8.4-2
UV
W
Coil
bank
Fig. 8.4-2
a. Inlet of spray chamber (B):
ha 3
0.0054 lb m H 2 O lb m D.A.
UV
W
ha 0.0017 lb m H 2 O lb m D.A.
Ÿ Twb
Tdb 75q F
49.5q F
The state of the air at (C) must lie on the same adiabatic saturation curve as does the state
at (B), or Twb 49.5q F . Thus,
Outlet of spray chamber (C):
At point C, Tdb
b.
bh
a3
UV
W
ha 0.0054 lb m H 2 O lb m D.A.
Ÿ hr
Twb 49.5q F
52%
58.5q F
g
ha1 lb m H 2 O evaporate
lb m DA
lb m DA
V A ft 3 inlet air
d
8-53
i
b0.0054 0.0017g
12.2
3.0 u 10 4
lb m H 2 O
ft 3 air
8.79 (cont’d)
c. QBA
QDC
(20 - 6.4) Btu / lb m dry air
H B H A #
= 1.1 Btu / ft 3
12.2 ft 3 / lb m dry air
(23 - 20) Btu / lb m dry air
'H H D H C #
= 0.25 Btu / ft 3
3
12.2 ft / lb m dry air
'H
d.
70%
52%
35%
C
D
A
B
58.5
20
70
75
8.80 Basis: 1 kg D.A.
a.
1 kg D.A.
ha1(kg H 2 O/kg D.A.)
Tdb = 40°C, Tab = 18°C
1 kg D.A.
ha2(kg H 2 O/kg D.A.)
20°C,
m w kg H2 O
Tdb
40q C
Ÿ ha1 0.0039 kg H 2 O kg D.A.
Twb 18q C
Tdb 20q C
Ÿ ha 2
Outlet air:
Twb 18q C adiabatic humidification
Inlet air:
b
g
0.0122 kg H 2 O kg D.A.
b gb g b1gbh g Ÿ m b0.0122 0.0039gkg H O kg D.A.
Overall H 2 O balance: mw 1 ha1
a2
n
0.0083 kg H 2 O kg D.A.
b.
ma (lb m H2 O/h)
T=15o C, sat’d
1250 kg/h
T=37o C, h r=50%
mc (lb m H2O/h)
liquid, 12°C
Qc (Btu/h)
8-54
2
8.80 (cont’d)
Inlet air:
Tdb 37q C
hr 50%
Moles dry air: m a
Outlet air: Tdb
RSh 0.0198 kg H O kg DA
TH b88.5 - 0.5g kJ kg DA 88.0 kJ kg DA
Fig. 8.4-1
Ÿ
2
a1
1
1250 kg
1 kg DA
h
1.0198 kg
RSh
TH
Fig. 8.4-1
15q C, sat' d
0.0106 kg H 2 O kg DA
a
Ÿ
Overall water balance Ÿ m c
1226 kg DA h
2
42.1 kJ kg DA
1226 kg DA
b0.0198 0.0106g kg H O
h
kg DA
2
113
. kg H 2 O h withdrawn
bg
Reference states for enthalpy calculations: H 2 O l , dry air at 0oC. (Cp)H2O(l) = 1
b
z
g
H 2 O l , 12q C : H
kJ
kg ˜ o C
12
C p dT
50.3 kJ / kg
0
Overall system energy balance:
Q c
'H
¦ m H ¦ m H
i
i
out
i
i
in
LM113. kg H O 50.3 kJ 1226 kg DA b42.1 88g kJ OPFG 1 h IJ FG 1 kW IJ
h
kg DA QH 3600 s K H 1 kJ / s K
N h kg H O
2
2
155
. kW
'H
8.81
8.82 a.
b.
400 mol NH 3
78.2 kJ
31,280 kJ
mol NH 3
b
g
b
g
b
g
HCl g , 25q C , H 2 O l , 25q C o HCl 25q C, r 5 .
B.11
'H 'H s 25q C, r 5 Table

o 'H 64.05 kJ mol HCl
b
g
HClbaq, r = fg o HClbr 5g, H Obl g
'H 'H b25q C, n 5g 'H b25q C, n fg
. g kJ mol HCl 1109
. kJ / mol HCl
b64.05 7514
2
s
s
8-55
8.83 Basis: 100 mol solution Ÿ 20 mol NaOH, 80 mol H2O
80 mol H 2 O
20 mol NaOH
Ÿr
4.00 mol H 2 O mol NaOH
bg
Refs: NaOH(s), H 2 O l @25q C
bg
bg
b
H2O l
NaOH r
g
4.00
b
H NaOH, r
g
out
H out
n in mol
H in kJ mol
34.43 m n in mol NaOH
nout
20.0
34.43 kJ mol NaOH (Table B.11)
4.00
¦ ni H i ¦ ni H i
'H
H in
0.0
0.0
nin
20.0
80.0
substance
NaOH s
688.6 kJ 9.486 u 104 Btu
10 3 kJ
(20)( 34.43)
in
653.2 Btu
103 g
20.0 40.00 80.0 18.01 g 2.20462 lb m
b
Q
g
b
653.2 Btu
132.3 Btu lb m product solution
g
8.84 Basis: 1 liter solution
1 L 8 g - eq
n H 2SO4
L
L
1000 mol H 2 O
18.02 kg H 2 O
2
n H 2O
Ÿr
46.49 mol H 2 O
4 mol H 2 SO 4
n H 2SO 4
d
FG 0.09808 kg IJ
H 1 mol K
0.392 kg H 2 SO 4
1.230 kg solution
0.392gkg H O
.
b1230
n H 2O
4 mol H 2 SO 4 u
2 g - eq
1 L 1.230 kg
mtotal
1 mol
i
11.6
46.5 mol H 2 O
mol H 2 O
mol H 2 SO 4
d
i
d
H 2 SO 4 aq, r = f,25o C o H 2 SO 4 aq, r = 11.6, 25o C + H 2 O l , 25o C
'H 1
'H s (r 116
. ) 'H s (r
f)
LMn
N
H b H SO , r 116
. , 60q Cg
n
RS4 mol H SO
1
4 mol H SO T
Table B.11
H 2SO 4 'H1
2
4
2
2
4
H 2SO 4
4
m
( 67.6 9619
. )
z
60
25
OP
Q
28.6
kJ
mol H 2 SO 4
C p dT kJ
( mol H 2 SO 4 )
28.6 kJ
1.230 kg
mol H 2 SO 4
60.9 kJ mol H 2 SO 4
8-56
i
3.00 kJ
kg˜q C
b60 25gq CUV
W
d
i
0.30 2.00 nH 2 O Ÿ nH 2 O
8.85 2 mol H 2SO 4
4.67 mol H 2 O Ÿ r
4.67
2
2.33
mol H 2 O
mol H 2SO 4
a. For this closed constant pressure system,
b. msolution
g
2 mol H 2SO 4
2.33
mol H 2SO 4
mol
b
g
b280.6 150gg
88.6 kJ +
Basis:
44.28 kJ
2 mol H 2SO 4
98.08 g H 2SO 4
0 Ÿ nH 2SO 4 'H s 25q C, r
'H
8.86 a.
b
nH 2SO 4 'H s 25q C, r
'H
Q
2.33 m
e j
1 L product solution 1.12 103 g
z
88.6 kJ
4.67 mol H 2 O 18.0 g H 2 O
mol
T
25
C p dT
3.3 J
g˜q C
280.2 g
0
bT 25gq C
1 kJ
1000 J
0ŸT
1120 g solution
L
1 L 8 mol HCl 36.47 g HCl
L
mol HCl
292 g HCl
46.0 mol H2O(l, 25°C)
8.0 mo l HCl(g , 20°C, 790 mm Hg)
1120 g 292 g
828 g H 2 O
828 g H 2 O
mol
46.0 mol H 2 O
18.0 g
n
1 L HCl (aq)
46.0 mol H 2 O
8.0 mol HCl
5.75 mol H 2 O mol HCl
Assume all HCl is absorbed
Volume of gas:
b g
8 mol 293 K 760 mm Hg 22.4 L STP
273 K 790 mm Hg
mol
b g
185 liter STP gas feed L HCl solution
b. Ref: 25qC
H in
nin
H 2O l
46.0
0.0
HCl g
8.0
0.15
8.0
59.07
b
HCl n
bg
bg
g
5.75
nout
H out
substance
n in mol
H in kJ mol
8-57
87q C
8.87 (cont’d)
b
g
H HCl, n
b
g
'H s 25q C, n
5.75
5.75 64.87 kJ mol e
H HCl, 20o C
c.
Q
'H
Q
0
0.15
z
j
20
25
z
40
1
nHCl
mC p dT
25
1120 g 0.66 cal
b40 25gq C
4.184 J
kJ
cal
103 J
g˜q C
8 mols
0.02913 0.1341 u 10 5 T 0.9715 u 10 8 T 2 4.335 u 10 12 T 3 dT
= -0.15 kJ / mol
471 kJ L product
'H
H
e j b
g
8 H 8 015
.
64.87 b
g
o
1120 g 0.66 cal T 25 C 4.184 J 1 kJ
cal 1000 J
8 mol g˜o C
192 o C
T
8.87 Basis: Given solution feed rate
.
.
n a (mol air/min)
200°C, 1.1 bars
n a (mol air/min)
.
n 1 (mol H 2O( v)/min)
saturated @ 50°C, 1 atm
150 mol/min solution
0.001 NaOH
0.999 H 2O
25°C
b
.
n 2 (mol/min) @ 50°C
0.05 NaOH
0.95 H 2O
gb g 0.05n Ÿ n 3.0 mol min
H O balance: b0.999gb150g n 0.95b3.0g Ÿ n 147 mol H O min
n
P p b50q Cg
Raoult’s law: y P
92.51 mm Hg Ÿ
n n
NaOH balance: 0.001 150
2
2
2
1
1
2
Table B.4
1
H 2O
H 2O
1
n1 147
P 760
a
b g
1061 mol 22.4 L STP
Vinlet air
min
1 mol
bg
.
473 K 1013
bars
273 K
1061
mol air
min
37,900 L min
1.1 bars
bg
'H b25q Cg
n a
References for enthalpy calculations: H 2 O l , NaOH s , air @ 25q C
999 mol H 2 O Table B.11
Ÿ
1 mol NaOH
0.1% solution @ 25qC: r
5% solution @ 50qC: r
Solution mass: m
b
H 50q C
g
b
95 mol H 2 O
5 mol NaOH
19 mol H 2 O
mol NaOH
b
Ÿ 'H s 25q C
1 mol NaOH 40.0 g 19 mol H 2 O 18.0 g
1 mol
1 mol
g
'H s 25q C m
42.81
42.47 kJ mol NaOH
s
z
g
382
42.81
kJ
mol NaOH
g solution
mol NaOH
50
25
C p dT
382 g
4.184 J
kJ
mol NaOH mol NaOH 1 g˜q C
8-58
b50 25gq C
1 kJ
103 J
2.85 kJ
8.87 (cont’d)
Air @ 200qC: Table B.8 Ÿ H
515
. kJ mol
Air (dry) @ 50qC: Table B.8 Ÿ H
b
0.73 kJ mol
b2592 104.8g kJ
g
H 2 O v , 50q C : Table B.5 Ÿ H
1 kg
kg
10 g 1 mol
substance
NaOH aq
H 2O v
nin
H in
nout
015
.
42.47
015
.
147
H out
2.85 n in mol min
44.81 H in kJ mol
Dry air
1061
.
515
1061
0.73
b g
bg
Energy balance: Q
¦ ni H i ¦ ni H i
'H
b neglect 'E g
out
n
18.0 g
3
44.81 kJ mol
1900 kJ min transferred to unit
in
8.88 a. Basis: 1 L 4.00 molar H2SO4 solution (S.G.
1.231)
4.00 mol H 2 SO 4
1231 392.3 838.7 g H 2 O
1 L 1231 g
1231 g Ÿ
Ÿ
392.3 g H 2 SO 4
L
46.57 mol H 2 O
B.11
. mol H 2 O / mol H 2 SO 4 Table
Ÿ r 1164

o 'H s
b
g
b
Ref: H 2 O l , 25q C , H 2 SO 4 25q C
g
nin
H in
46.57 0.0754 T 25
4.00
0
substance
H 2O l
67.6 kJ / mol H 2 SO 4
nout
H out
bg
b g
H SO bl g
H SO b25q C, n 1164
. g
4.00 67.6
Q 'H 0 4.00b 67.6g 46.57b0.0754gbT 25g Ÿ T 52q C
2
2
4
4
n in mol
H in kJ mol
(The water would not be liquid at this temperature Ÿ impossible alternative!)
b
g
b
b. Ref: H 2 O l , 25q C , H 2 SO 4 25q C
substance
H 2O l
H2O s
H 2 SO 4 (l )
H 2 SO 4 25q C, n 1164
.
bg
bg
b
b
'H m H 2 O, 0q C
g
g
nin
H in
nl
0.0754 0 25
n s 6.01 0.0754 0 25
4.00
0
b
g
b
g
g
nout
H out
n in mols
H in kJ mol
4.00 67.61
6.01 kJ mol
A
Table B.1
UV Ÿ n
'H 0 4.00b 67.61g n b 1885
. g b46.57 n gb 7.895gW n
. g H Ob"g 547.3 g H Ob sg@0q C
Ÿ 2914
n" n s
46.57
l
2
l
2
8-59
l
s
. mol liquid H 2 O
1618
30.39 mol ice
8.89 P2 O 5 3H 2 O o 2H 3 PO 4
mol H 3 PO 4
a.
b
g u 100% ,
n 14196
.
wt% P2 O 5
mt
B
2n
wt% H 3 PO 4
g H 3 PO 4 mol
B
b98.00g u 100%
mc
A
g total
where n
total mass .
mol P2 O 5 and mt
wt% H 3 PO 4
b
g wt% P O
2 98.00
.
14196
2
.
1381
wt% P2 O 5
5
b. Basis: 1 lb m feed solution 28 wt% P2 O 5 Ÿ 38.67 wt% H 3 PO 4
m1 (lbm H2 O(v )), T , 3.7 psia
1 lb msolution, 125°F
0.3867 lb mH 3PO 4
0.6133 lb mH 2O
m2 (lbm solution), T
0.5800 lb mH 3PO 4/lb
0.4200 lb mH 2O/lb m
H 3 PO 4 balance: 0.3867
m
0.5800m2 Ÿ m2 0.667 lb m solution
Total balance: 1 m1 m2 Ÿ m1
bg
0.3333 lb m H 2 O r
bg
Evaporation ratio: 0.3333 lb m H 2 O v lb m feed solution
c. Condensate:
b
P 37
. psia 0.255 bar
Table B.6
Ÿ Tsat
m
V
g
. o C =149 o F, Vliq
654
ft 3 / m3
.
0.00102 m3 353145
kg
100 tons feed 2000 lb m 1 lb m H 2 O
day
46.3 lb m
min
1 ton
0.0163 ft 3
lb m
2.205 lb m / kg
1 day
(24 u 60) min
3 lb m
46.3 lb m / min
7.4805 gal
5.65 gal condensate / min
ft 3
Heat of condensation process:
46.3lbm H2O(v)/min
46.3lbm H2O(l)/min
(149+37)°F, 3.7 psia
149°F, 3.7 psia
.
Q (Btu/min)
8-60
0.0163
ft 3
lb m H 2 O(l)
8.89 (cont’d)
R|
||H
Table B.6 Ÿ S
||H
|T
Q m 'H
F
GG
H
H2 O ( l ) (149
(46.3
o
b
g
F = 65.4 o C) = (274 kJ / kg) 0.4303
LM
N
lb m
Btu
) (118 1141)
lb m
min
OP
Q
I
JJ
kJ
kg K
Btu
o
o
H2 O ( v ) (186 F = 85.6 C) = (2652 kJ / kg) 0.4303
lb m
1141 Btu / lb m
118 Btu / lb m
47,360 Btu / min
Ÿ 4.74 u 10 4 Btu min available at 149 o F
bg
bg
d. Refs: H 3 PO 4 l , H 2 O l @77q F
min
H in
mout
H out
m in lb m
.
100
13.95
0.667 34.13 H in Btu lb m
0.3333 1099
substance
H 3 PO 4 28%
H 3 PO 4 42%
H 2O v
b g
b g
bg
b
H H 3 PO 4 , 28%
b
0.705 Btu
lb m ˜q F
5040 Btu
lb - mole H 3 PO 4
b125 77gq F
0.705 Btu
lb m ˜q F
H H 3 PO 4 , 42%
g
g
0.3867 lb m H 3 PO 4
1.00 lb m soln
13.95 Btu lb m soln
5040 Btu
lb - mole H 3 PO 4
b186.7 77gq F
1 lb - mole H 3 PO 3
98.00 lb m H 3 PO 4
1 lb - mole H 3 PO 4
98.00 lb m H 3 PO 4
0.5800 lb m H 3 PO 4
1.00 lb m sol.
34.13 Btu lb m soln
b g H b3.7psia, 186q Fg H bl, 77q Fg b2652 104.7g kJ kg Ÿ 1096 Btu lb
H H 2 O
At 27.6 psia (=1.90 bar), Table B.6 Ÿ 'H v
'H
¦ ni H i ¦ ni H i
out
Ÿ
Ÿ
in
2206 kJ / kg = 949 Btu / lb m
375 Btu = msteam 'H v Ÿ msteam
0.395 lb m steam 100 u 2000 lb m H 3 PO 4
lb m 28% H 3 PO 4
day
375 Btu
949 Btu / lb m
0.395 lb m steam
1 day
3292 lb m steam / h
24 h
lb m steam
3292 lb m steam
.
118
(46.3 u 60) lb m H 2 O evaporated / h
lb m H 2 O evaporated
8-61
m
8.90 Basis: 200 kg/h feed solution.
A = NaC 2 H 3 O 2
.
n 1 (kmol H2 O(v )/h)
50°C, 16.9% of H 2O
in feed
200 kg/h @ 60°C
.
n 0 (kmol/h)
0.20 A
0.80 H 2O
Product slurry @ 50°C
.
n 2 (kmol A-3H 2 O(v )/h)
.
n 3 (kmol solution/h)
0.154 A
0.896 H 2 O
Q (kJ/hr)
0.200 M A 0.800 M H 2 O
a. Average molecular weight of feed solution: M
b0.200gb82.0g b0.800gb18.0g
Molar flow rate of feed: n0
200 kg
1 kmol
h
30.8 kg
6.49 kmol h
. gb0.80gb6.49 kmol hg
b0169
b0.20gb6.49 kmol hg E n bkmol Ah ˜ 3 H Og
b. 16.9% evaporation Ÿ n1
A balance:
2
2
Ÿ n2 0.154n3
b gb
H 2 O balance: 0.80 6.49 kmol h
g
0.846n3 Ÿ 3n2 0.846n3
bg
0.877 kmol H 2 O v h
1 mole A
. n3
0154
1 mole A ˜ 3 H 2 O
1.30
0.877 (1)
b
n2 kmol A ˜ 3 H 2 O
h
g
3 moles H 2 O
1 mole A ˜ 3 H 2 O
4.315
bg b g
113
. kmol A ˜ 3H 2 O s h
1095
kmol solution h
.
Mass flow rate of crystals
1.13 kmol A ˜ 3H 2 O 136 kg A ˜ 3H 2 O
h
1 kmol
b
bg
154 kg NaC 2 H 3 O 2 ˜ 3H 2 O s
h
gb g
bg
200 kg feed 154 kg crystals 0.877 18.0 kg H 2 O v
h
h
h
c.
b2g
bg
Solve 1 and 2 simultaneously Ÿ n2
n3
Mass flow rate of product solution
30.8 kg k
bg
30 kg solution h
bg
References for enthalpy calculations: NaC 2 H 3 O 2 s , H 2 O l @25q C
b
Feed solution: nH
nH
g
n A 'H s 25q C m
b0.20g6.49 kmol A
h
z
60
25
C p dT (form solution at 25q C , heat to 60q C )
171
. u 104 kJ 200 kg 3.5 kJ
hr
kg˜q C
kmol A
8-62
b60 25gq C
2300 kJ h
8.90 (cont’d)
Product solution: nH
b
g
n A 'H s 25q C m
b0.154g1.095 kmol A
z
50
25
. u 10 4 kJ 30 kg 3.5 kJ
171
h
kg˜q C
kmol A
h
259 kJ h
Crystals: nH
n A 'H hydration m
z
25
C p dT (hydrate at 25q C , heat to 50q C )
bg
3.66 u 10 4 kJ 154 kg 1.2 kJ
h
kg˜q C
kmol
h
36700 kJ h
b
g
z
LM
N
n 'H v 25
b neglect 'E g
'H
C p dT (vaporize at 25q C , heat to 50q C )
b gb
i
g
4.39 u 10 4 32.4 50 25 kJ
¦ n H ¦ n H
i
out
R
b50 25gq C
OP
Q
50
0.877 kmol H 2 O
h
Energy balance: Q
b50 25gq C
50
1.13 kmol A ˜ 3H 2 O s
H 2 O v , 50q C : n'H
C p dT
i
39200 kJ h
b259 36700 39200g b2300g
i
kJ h
in
60 kJ h (Transfer heat from unit)
8.91 50 mL H 2SO4
bg
84.2 mL H 2 O l
U|
|V Ÿ r
84.2 g H Oblg Ÿ 4.678 mol H Oblg|
|W
g
.
1834
917
. g H2SO4 Ÿ 0.935 mol H 2SO4
mL
100
. g
mL
2
500
. mol H 2 O mol H 2SO4
2
Ref: H 2 O , H 2SO 4 @ 25 qC
H ( H 2 O(l ), 15o C) [0.0754 kJ / (mol ˜ o C)](15 25) o C = 0.754 kJ / mol
b
H H 2 SO 4 , r
g
substance
bg
H 2O l
H 2SO 4
b
H 2SO 4 r
(91.7 + 84.2) g
2.43 J
kJ
mol 0.935 mol H 2 SO 4 g˜q C
( 69.46 0.457T )( kJ / mol H 2 SO 4 )
58.03
5.00
nin
H in
g
Energy Balance: 'H
1 kJ
10 3 J
H out
nout
4.678 –0.754 —
—
0.0
0.935
0.935
—
4.00 —
bT 25gq C
n in mol
H in kJ/mol
—
—
b69.46 0.457T g n bmol H SO g
3
b
g
b
g
0 0.935 69.46 0.457T 4.678 0.754 Ÿ T
Conditions: Adiabatic, negligible heat absorbed by the solution container.
8-63
4
144 q C
8.92 a.
mA (g A) @ TA0 (oC)
nA (mol A)
nS (mol solution) @ Tmax (oC)
mB (g B) @ TB0 (oC)
nB (mol B)
Refs: A(l), B(l) @ 25 qC
substance nin H in nout
nA H
—
A
H out
A
—
n in mol
H in J / mol
B
nB H B
—
—
S
—
nA
H S (J mol A)
—
mA (g A)
, nB
M A (g A / mol A)
Moles of feed materials: n A (mol A) =
mB
MB
Enthalpies of feeds and product
H A
m A C pA ( T A 0 25 o C), H B
r (mol B mol A) = n B n A
H S
FG J IJ
H mol A K
nA
m B C pB ( TB 0 25 o C)
mB / M B
mA / M A
LMn
1
M
( mol A) M
MM
N
A ( mol
FG J IJ
H mol A K
F J I u (T
)( g soln) u C G
H g soln ˜ C JK
A) u 'H m ( r )
(m A m B
ps
max
o
25)( o
OP
PP
C) P
PQ
1
n A 'H m ( r ) ( m A m B ) C ps ( Tmax 25)
nA
Ÿ H S
Energy balance
'H
n A H S n A H A n B H B
Ÿ
0
bg
b
g
b
g
mA 'Hm r (m A mB )C ps (Tmax 25) m A C pA TA0 25 mB C pB TB0 25
MA
b
g
b
g
m A C pA TA0 25 mB C pB TB0 25 Ÿ Tmax
25 0
bg
mA 'Hm r
MA
(m A mB )C ps
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container,
negligible dependence of heat capacities on temperature between 25oC and TA0 for A, 25oC
and TB0 for B, and 25oC and Tmax for the solution.
b.
mA
100.0 g
MA
mB
225.0 g
MB
C ps
3.35 J (g˜q C)
40.00 TA 0
18.01 TB 0
b
25q C
C pA
40q C C pB
g
'H m n 5.00
g UV Ÿ r
4.18 J (g˜q C) |W
b
? irrelevant
37,740 J mol A Ÿ Tmax
8-64
5.00
125q C
mol H 2 O
mol NaOH
8.93 Refs: Sulfuric acid and water @ 25 qC
b.
substance
nin
H in
H2SO4
H2O
H 2 SO 4 aq
1
r
—
M A C pA T0 25
M w C pw T0 25
—
b g
H out
nout
b
b
g
g
—
—
1
n in mol
—
H in J/mol
—
'H m r M A rM w C ps Ts 25
bg b
g b
g
(J/mol H2SO4)
'H
bg b
g b g
b g
'H br g b98 18r gC bT 25g (98C 18rC )bT
1
25 (98C 18rC )bT 25g 'H br g
( 98 18r )C
m
Ÿ Ts
b
g
0 'H m r M A rM w C ps Ts 25 M A C pa T0 25 rM w C pw T0 25
s
ps
pa
pa
pw
pw
0
0
g
25
m
ps
c.
H2O(l)
H2SO4
r
0.5
1
1.5
2
3
4
5
10
25
50
100
Cp
(J/mol-K)
75.4
185.6
Cp
(J/g-K)
4.2
1.9
Cps
1.58
1.85
1.89
1.94
2.1
2.27
2.43
3.03
3.56
3.84
4
'H m (r )
-15,730
-28,070
-36,900
-41,920
-48,990
-54,060
-58,030
-67,030
-72,300
-73,340
-73,970
Ts
137.9
174.0
200.2
205.7
197.8
184.0
170.5
121.3
78.0
59.6
50.0
250
Ts
200
150
100
50
0
0.1
1
10
100
r
d. Some heat would be lost to the surroundings, leading to a lower final temperature.
8-65
8.94 a.
Ideal gas equation of state n A0
P0V g / RT0
b
gd
Vl ( L) u SG B u 1 kg / L 10 3 g / kg
Total moles of B: n B 0 ( mol B)
i
(2)
M B (g / mol B)
n Av n Al
Total moles of A: n Ao
Henry’s Law: r
(1)
FG mol A(l) IJ
H mol B K
(3)
ks pA Ÿ
bc
n Al
n B0
0
c1T
g n V RT
Av
(4)
g
Solve (3) and (4) for nAl and nAv.
b
nB 0 RT
c0 c1T
Vg
g
LM1 n RT bc c T gOP
MN V
PQ
n
LM1 n RT bc c T gOP
MN V
PQ
n Al
(5)
B0
0
1
g
Ao
n Av
(6)
B0
0
1
g
Ideal gas equation of state
n Av RT
Vg
P
n A0 RT
Vg nB 0 RT c0 c1T
( 6)
b
g
(7)
b g bg
Refs: A g , B l @ 298 K
substance
U in
g
298g
U eq
b
g
n Ao
M A CvA T0 298
n Av
M A CvA T 298
nB0
M B CvB
—
—
Solution
—
n Al
U 1 (kJ/mol A)
0
—
b
g b
n in mol
U in kJ/mol
g
1
'U s n Al M A n B 0 M B Cvs T 298
n Al
E.B.: 'U
0
¦ n U ¦ n U
i
out
0
b
bT
neq
bg
Bb l g
Ag
U 1
nin
cn
Av CvA
ŸT
i
i
i
in
b
g hb
g
b
d'U i bn C n C gbT 298g
n C bn M n M gC
gb
g
n Al M A nB M B Cvs T 298 n Al 'U s n Ao CvA n B CvB T0 298
298 n Al
s
Av
Ao
vA
Al
vA
B
A
8-66
0
vB
B
B
vs
8.94 (cont’d)
b.
Vt
20.0
MA
47.0
CvA
0.831
MB
26.0
CvB
3.85
SGB
1.76
Vl
3.0
3.0
3.0
3.0
3.0
3.0
3.0
3.0
T0
300
300
300
300
330
330
330
330
P0
1.0
5.0
10.0
20.0
1.0
5.0
10.0
20.0
Vg
17.0
17.0
17.0
17.0
17.0
17.0
17.0
17.0
nB0
203.1
203.1
203.1
203.1
203.1
203.1
203.1
203.1
nA0
0.691
3.453
6.906
13.811
0.628
3.139
6.278
12.555
c0
c1
0.00154 -1.60E-06
T
301.4
307.0
313.9
327.6
331.3
336.4
342.8
355.3
nA(v)
0.526
2.624
5.234
10.414
0.473
2.359
4.709
9.381
Dus
-174000
Cvs
3.80
nA(l)
0.164
0.828
1.671
3.397
0.155
0.779
1.569
3.174
P
0.8
3.9
7.9
16.5
0.8
3.8
7.8
16.1
c.
C*
REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS
REAL NA0, T, DEN, P, NAL, NAV, NUM, TN
INTEGER K
R = 0.08206
1
READ (5, *) NB
IF (NB.LT.0) STOP
READ (1, *) T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS
WRITE (6, 900)
NA0 = P0 * VG/R/T0
T = 1.1 * T0
K=1
10
DEN = VG/R/T/NB + C + D * T
P = NA0/NB/DEN
NAL = (C + D * T) * NA0/DEN
NAV = VG/R/T/NB * NA0/DEN
NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298)
DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS
TN = 298 + NUM/DEN
WRITE (6, 901) T, P, NAV, NAL, TN
IF (ABS(T – TN).LT.0.01) GOTO 20
K=K+1
T = TN
IF (K.LT.15) GOTO 10
WRITE (6, 902)
STOP
20
WRITE (6, 903)
GOTO 1
900
FORMAT ('T(assumed)
P
Nav Nal T(calc.)'/
*
'
(K)
(atm) (mols) (mols)
(K)')
901
FORMAT (F9.2, 2X, F6.3, 2X, F7.3, 2X, F7.3, 2X, F7.3, 2)
902
FORMAT (' *** DID NOT CONVERGE ***')
903
FORMAT ('CONVERGENCE'/)
END
$ DATA
300
291
10.0
15.0
1.54E–3
–2.6E–6
–74
35.0
18.0
0.0291
0.0754
4.2E–03
8-67
Tcalc
301.4
307.0
313.9
327.6
331.3
336.4
342.8
355.3
8.94 (cont’d)
300
291
35.0
–1
50.0
18.0
Program Output
T (assumed)
(K)
321.10
296.54
296.57
15.0
0.0291
1.54E–3
0.0754
–2.6E–6
4.2E–03
–74
P
(atm)
8.019
7.415
7.416
Nav
(mols)
4.579
4.571
4.571
Nal
(mols)
1.703
1.711
1.711
T(calc.)
(K)
296.542
296.568
296.568
P
(atm)
40.093
39.676
39.680
Nav
(mols)
22.895
22.885
22.885
Nal
(mols)
8.573
8.523
8.523
T(calc.)
(K)
316.912
316.942
316.942
Convergence
T (assumed)
(K)
320.10
316.91
316.94
8.95
Q=0
350 mL 85% H2SO4
ma(g), 60 oF, U=1.78
30% H2SO4
ms(g), T(oF)
H2O, Vw(mL),
mw(g), 60 oF
a.
Vw
350 mL feed
178
g
.
1 mL feed
0.85(70 / 30) 015
. g H 2 O added 1 mL water
g feed
1 g water
1140 mL H 2 O
b.
Fig. 8.5 -1 Ÿ H a | 103 Btu / lb m ;
Water: H water | 65 Btu / lb m
Mass Balance: mp=mf+mw=(350 mL)(1.78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g
Energy Balance:
'H
Ÿ H product
c.
T ( H
mf H f mw H w
mp
(623)( 103) (1140)( 65)
5.7 Btu / lb m
1765
0 mp H product ma H a mw H w Ÿ H s
5.7 Btu / lb m ,30%) | 160o F
d. When acid is added slowly to water, the rate of temperature change is slow: few isotherms
are crossed on Fig. 8.5-1 when xacid increases by, say, 0.10. On the other hand, a change
from xacid=1 to xacid=0.9 can lead to a temperature increase of 200°F or more.
8-68
8.96 a. 2.30 lb m 15.0 wt% H 2 SO 4
10 Btu / lb
@ 77 o F Ÿ H
1
m
U|
||
V| o m ( lb
||
W
adiabatic mixing
3
m2 (lb m ) 80.0 wt% H 2 SO 4
120 Btu / lb
@ 60o F Ÿ H
2
60.0 wt% H 2 SO4 @ To F, H
3
|UV Ÿ |RSm
mass balance: 2.30b0.150g m b0.800g m (0.600) W| |m
T
Total mass balance:
H 2 SO 4
m
m)
2.30 + m2
m3
2
2
b. Adiabatic mixing Ÿ Q = 'H
3
3
517
. lb m (80%)
7.47 lb m (60%)
0
. gb 120g
b7.47gH b2.30gb10g b517
0 Ÿ H 3
3
. Btu / lb m
861
E Figure 8.5 - 1
T = 140 o F
c.
d
i
H 60 wt%, 77 o F
130 Btu / lb m
d
i
Q m3 H 60 wt%, 77 o F H 3
b7.475gb130 861. g
328 Btu
d. Add the concentrated solution to the dilute solution . The rate of temperature rise is
much lower (isotherms are crossed at a lower rate) when moving from left to right on
Figure 8.5-1.
Fig. 8.5-2
8.97 a.
b.
x NH 3
0.30
y NH 3
0.96 lb m NH 3 lb m vapor , T
Basis: 1 lb m system mass
Ÿ 0.90 lb m liquid
Ÿ 010
. lb m vapor
Mass fractions: zNH 3
b0.27 0.096glb
m
NH 3
1 lb m
1 0.37
Enthalpy: H
0.90 lb m liquid
25 Btu
1 lb m
1 lb m liquid
8-69
x NH 3 0.30
80q F
0.27 lb m NH 3
0.63 lb m H 2 O
x NH 3 0.96
0.096 lb m NH 3
0.004 lb m H 2 O
0.37 lb m NH 3 lb m
0.63 lb m H 2 O lb m
0.10 lb m vapor
670 Btu
1 lb m
1 lb m vapor
44 Btu lb m
8.98
T
140q F
Fig. 8.5-2
Vapor: 80% NH 3 , 20% H 2 O
Liquid: 14% NH 3 , 86% H 2 O
C
A
B
Basis: 250 g system mass
Ÿ mv ( g vapor), mL ( g liquid)
Mass Balance: mv mL
Liquid: mNH 3
.60
.80 x NH3
250
NH3 Balance: 0.80m g 014
. mL
Vapor: mNH 3
.14
(0.60)( 250) Ÿ mv
175 g, mL
75g
b0.80gb175 gg 140 g NH , 35 g H O
. gb75 gg 10.5 g NH , 64.5 g H O Liquid
b014
3
2
3
2
8.99 Basis: 200 lb m feed h
m v (lb m h)
xv(lbm NH3(g)/lbm)
H v ( Btu lb m )
200 lbm/h
0.70 lbm NH3(aq)/lbm
0.30 lbm H2O(l)/lbm
m l (lb m h)
H f
xl[lbm NH3(aq)/lbm]
50 Btu lb m
in equilibrium
at 80oF
H l ( Btu lb m )
Q ( Btu h)
Figure 8.5-2 Ÿ Mass fraction of NH 3 in vapor: xv
0.96 lb m NH 3 lb m
Mass fraction of NH 3 in liquid: xl
0.30 lb m NH 3 lb m
Specific enthalpies: H v
650 Btu lb m , H l
UV
W
m v
Mass balance:
200 m v m l
Ÿ
m l
Ammonia balance: 0.70 200 0.96m v 0.30m l
b gb g
30 Btu lb m
120 lb m h vapor
80 lb m h liquid
Energy balance: Neglect 'E k .
Q
'H
¦ m H
i
out
i
m f H f
120 lb m
h
86,000
650 Btu 80 lb m
lb m
h
Btu
h
8-70
30 Btu 200 lb m
lb m
h
50 Btu
lb m
CHAPTER NINE
4 NH 3 (g) 5O 2 (g) o 4NO(g) + 6H 2 O(g)
'H o 904.7 kJ / mol
9.1
r
a.
When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25qC and 1 atm react to form 4 g-moles of
NO(g) and 6 g-moles of water vapor at 25qC and 1 atm, the change in enthalpy is -904.7 kJ.
b.
Exothermic at 25qC. The reactor must be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the molecular bonds of the
reactants is less than the energy released when the product bonds are formed.
c.
5
O 2 (g) o 2NO(g) + 3H 2 O(g)
2
Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half.
904.7
'H ro 452.4 kJ / mol
2
3
5
NO(g) + H 2 O(g) o NH 3 (g) O 2 (g)
2
4
Reversing the reaction reverses the sign of the heat of reaction. Also reducing the stoichiometric
coefficients to one-fourth reduces the heat of reaction to one-fourth.
( 904.7)
'H ro 226.2 kJ / mol
4
d.
e.
2 NH 3 (g) NH 3
m
n NH 3
Q
340 g / s
340 g
=
'H
1 mol
20.0 mol / s
17.03 g
o
n NH 3 'H
20.0 mol NH 3 904.7 kJ
r
s
Q NH 3
s
1 mol NH 3
. u 10 4 kJ / s
18
The reactor pressure is low enough to have a negligible effect on enthalpy.
f.
9.2
Yes. Pure water can only exist as vapor at 1 atm above 100qC, but in a mixture of gases, it can
exist as vapor at lower temperatures.
C 9 H 20 (l) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l)
'H o 6124 kJ / mol
r
a.
When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25qC and 1 atm react to form 9 g-moles of
CO2(g) and 10 g-moles of water vapor at 25qC and 1 atm, the change in enthalpy is -6124 kJ.
b.
Exothermic at 25qC. The reactor must be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the molecular bonds of the
reactants is less than the energy released when the product bonds are formed.
c.
Q
=
'H
0
n C9 H 20 'H
r
25.0 mol C 9 H 20
Q C9 H 20
s
6124 kJ
1 kW
1 mol C 9 H 20 1 kJ / s
9-1
153
. u 105 kW
9.2 (cont'd)
Heat Output = 1.53u105 kW.
The reactor pressure is low enough to have a negligible effect on enthalpy.
d.
C 9 H 20 (g) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l)
'H o 6171 kJ / mol
(1)
C 9 H 20 (l) 14O 2 (g) o 9CO 2 (g) +10H 2 O(l)
'H o 6124 kJ / mol
(2)
r
r
(2) (1) Ÿ C 9 H 20 (l) o C 9 H 20 (g)
6124 kJ / mol ( 6171 kJ / mol) = 47 kJ / mol
'H o (C H ,25$ C)
v
e.
9
20
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6qC, but in a mixture of gases, it
can exist as a vapor at lower temperatures.
9.4 (cont'd)
9.3
a.
Exothermic. The reactor will have to be cooled to keep the temperature constant. The temperature
would increase under adiabatic conditions. The energy required to break the reactant bonds is less
than the energy released when the product bonds are formed.
b.
b g 192 O bgg o 6CO bgg 7H Obgg b1g 'H ?
19
1.791 u 10 Btu lb - mole
'H
C H blg O bg g o 6CO bgg 7 H Oblg b2g 'H
2
e 'H j
13,550 Btu lb - mole
C H bgg o C H blg b3g 'H
H Oblg o H Obg g b4g 'H
e'H j 18,934 Btu lb - mole
b1g b2g b3g 7 u b4g Ÿ 'H 'H 'H 7'H 1.672 u 10 Btu lb - mole
C 6 H 14 g 6
14
6
14
2
2
6
2
o
r
2
2
2
2
3
14
v
4
2
v
o
r
2
C 2 H 14
H 2O
Hess's law
6
1
c.
M O2 =32.0
Ÿ
m 120 lb m / s
Q
'H
bg
nO 2 'H ro
vO 2
n
2
3
4
3.75 lb - mole / s.
.
u 10 6 Btu
3.75 lb - mole 1672
s
bg
a.
b
6.27 u 106 Btu / s from reactor
1 lb - mole O 2
bg
bg
bg
CaC 2 s 5H 2 O l o CaO s 2CO 2 g 5H 2 g , 'H ro
9.4
6
g
69.36 kJ kmol
Endothermic. The reactor will have to be heated to keep the temperature constant. The temperature
would decrease under adiabatic conditions. The energy required to break the reactant bonds is more
than the energy released when the product bonds are formed.
b.
'U ro
'H ro RT
OP
LM
MM ¦ Q ¦ Q PP
PQ
MN
i
gaseous
products
i
69.36
gaseous
reactants
52.0 kJ mol
9-2
kJ 8.314 J 1 kJ 298 K
mol mol ˜ K 10 3 J
b7 0g
9.4 (cont’d)
'U ro is the change in internal energy when 1 g - mole of CaC2 (s) and 5 g - moles of H2 O(l) at 25$ C and
1 atm react to form 1 g - mole of CaO(s), 2 g - moles of CO2 (g) and 5 g - moles of H2 (g) at 25$ C
and 1 atm.
c.
'U
Q
nCaC 2 'U ro
150 g CaC 2
1 mol
52.0 kJ
121.7 kJ
64.10 g 1 mol CaC 2
vCaC 2
Heat must be transferred to the reactor.
Hess's law
9.5
a.
Given reaction
Ÿ
(1) – (2)
'H ro
'H ro1 'H ro2
b1226 18,935g Btu lb - mole
17,709 Btu lb - mole
Hess's law
b.
Given reaction
Ÿ
(1) – (2)
'H ro
'H ro1 'H ro2
b121,740 104,040g Btu lb - mole
17,700 Btu lb - mole
9.6
bg
(1) 2 u 2
Hess's law
Ÿ
b326.2 kJ molg 2b2858. kJ molg
'H ro
a.
Reaction (3)
b.
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose
methanol with only reaction (3) occurring.
F B I
2G 90.37 kJ molJ
GH
JK
245.4 kJ mol
Table B.1
9.7
bg
bg
bg
a.
N 2 g O 2 g o 2NO g ,
b.
n C 5 H 12 g bg
6
o
f
14
e j
NO(g)
180.74 kJ mol
11
O 2 g o 5CO g 6H 2 O l
2
6 'H fo
'H fo
CO(g)
o
f
CO 2
n C 5 H 12 g
H 2O l
2
o
r
d.
2
'H fo
bg
bg
bg
'H
5e 'H j
e j bg e j b g
. g b 146.4g kJ mol 2121.2 kJ mol
b5gb110.52g b6gb28584
19
C H blg O bgg o 6CO bgg 7H Obgg
2
'H
6e 'H j
7e 'H j
b g e 'H j b g
. g b 173.0g kJ mol 3855 kJ mol
b6gb393.5g 7b24183
o
r
c.
'H ro
2
o
f
2
H 2O g
o
f
C 6 H 14 l
Na 2SO 4 (l) 4CO(g) o Na 2S(l) 4CO 2 (g)
'H ro
e'H j
4e 'H j
e j e'H j
( 373.2 6.7) b4gb 3935
. g b 1384.5 24.3g 4( 110.52
o
f
Na 2S( l )
4 'H fo
CO 2 ( g )
o
f
9-3
Na 2SO 4 ( l )
o
f
CO(g )
kJ mol
138.2 kJ mol
9.8
a.
e'H j
e'H j
'H ro1
o
f
'H ro2
Ÿ e 'H j
385.76 52.28 333.48 kJ mol
e j
276.2 92.31 333.48 35.03 kJ mol
b g e 'H j b g e 'H j
b1g b2g Ÿ 385.76 35.03 420.79 kJ mol
C 2 H 2 Cl 4 ( l )
o
f
Given reaction
Q
'H
o
f
C 2 HCl 3 l
b.
c.
'H fo
o
f
C2 H 4 (g)
o
f
HCl g
300 mol C 2 HCl 3
420.79 kJ
h
mol
C 2 H 2 Cl 4 ( l )
C 2 H 2 Cl 4 (l )
b
126
. u 105 kJ h
35 kW
g
Heat is evolved.
9.9
b.
c.
5
'H co 1299.6 kJ mol
O 2 (g) o 2CO 2 (g) + H 2 O(l)
2
The enthalpy change when 1 g-mole of C2H2(g) and 2.5 g-moles of O2(g) at 25qC and 1 atm react
to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25qC and 1 atm is -1299.6 kJ.
C 2 H 2 ( g) e j
'H co
2 'H fo
B
CO 2 ( g )
e j
'H fo
b g e 'H f j C H b g g
o
H 2O l
2
2
Table B.1
b
g b
d'H i
(i) 'H ro
o
f
B
Table B.1
C2 H 6 ( g )
kJ
g mol
o
c
B
Table B.1
C 2 H 2 ( g) d
'H fo
i
C2H 2 (g)
1299.6
kJ
mol
bg
C2 H2 g
kJ
b84.67g b226.75g mol
d'H i
(ii) 'H ro
d.
g b
2 393.5 285.84 226.75
d
2 'H co
i
3114
.
H2 (g)
d
'H co
kJ
mol
i
bg
C2H 6 g
kJ
. ) b 1559.9g
b1299.6g 2(28584
mol
5
O 2 (g) o 2CO 2 (g) + H 2 O(l)
2
1
O 2 (g) o H 2 O(l)
2
7
C 2 H 6 (g) O 2 (g) o 2CO 2 (g) + 3H 2 O(l)
2
H 2 ( g) 3114
.
kJ
mol
(1) 'H co1
1299.6 kJ mol
(2) 'H co2
. kJ mol
28584
(3) 'H co3
1559.9 kJ mol
The acetylene dehydrogenation reaction is (1) + 2 u (2) (3)
Hess's law
Ÿ
'H ro
'H co1 2 u 'H co2 'H co3
b1299.6 2(285.84) (1559.9)g kJ mol
9-4
311.4 kJ / mol
9.10
a.
bg
C 8 H 18 l bg
bg
25
O 2 (g) o 8CO 2 g 9H 2 O g
2
'H ro
4850 kJ / mol
When 1 g-mole of C8H18(l) and 12.5 g-moles of O2(g) at 25qC and 1 atm react to form 8 g-moles
of CO2(g) and 9 g-moles of H2O(g), the change in enthalpy equals -4850 kJ.
b.
Energy balance on reaction system (not including heated water):
0ŸQ
'E k , 'E p , W
Q
Ÿ
'U co
'H co
114.2 g
OP
LM
MM ¦ Q ¦ Q PP
PQ
MN
i
5079 kJ mol 1 mol C 8 H 18
gaseous
reactants
8.314 J
1 kJ
b8 9 12.5g
298 K
mol ˜ K 10 J
3
5090 kJ mol
% difference =
'H co
'U co (kJ)
i
gaseous
products
c.
89.4 kJ
5079 kJ mol
'U co RT
Ÿ 'H co
g
1 mol
2.01 g C 8 H 18 consumed 1 mol C8 H 18
'U Ÿ 89.4 kJ
b
75.4 u 10 3 kJ 21.34q C
18.0 u 10 3 kg
mol.$ C
1.00 kg
mH 2O (Cp ) H 2 O(l) 'T
g
75.4 u 10 3 kJ / mol.$ C
(Cp ) H 2 O(l) from Table B.2
Q
b
n mol C 8 H 18 consumed 'U co kJ mol
'U
e j
Ÿ e 'H j
8 'H fo
( 5090) ( 4850)
u 100 = 4.7 %
5090
b g 9e'H ff j H Obg g e 'H f jC H blg
o
CO 2 g
o
f
bg
C8 H 18 l
b
o
2
8
g b
g
18
8 393.5 9 24183
. 5090 kJ / mol
234.5 kJ / mol
There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only
product.
9-5
9.11
a.
bg
bg
n C 4 H10 g o i C 4 H10 g
Basis: 1 mol feed gas
0.930 mol n-C4H10
(nn- C4H10)out
0.050 mol i-C4H10
( ni-C4H10)out
0.020 mol HCl
0.020 mol HCl
149qC
Q(kJ/mol)
(n n-CH 4 H10 ) out
0.930(1 0.400)
(n i-CH 4 H10 ) out
0.050 0.930 u 0.400
0.560 mol
(n n-C4H10 ) out (n n-C4H10 ) in
[
Q n-C
e'H j
'H ro
c.
References: n C 4 H10
o
f
i C 4 H 10
0.560 0.930
1
(mol)
Table B.1
4
b kJ molg
1
i C 4 H 10
z
B
Table B.2
149
25
Cp
O kJ
dT P
PQ mol
Q = 'H ['H r o
i
For 325 mol/h fed, Q =
b
'H r 149q C
g
9.8 kJ mol
b kJ molg
(mol)
0.600
H 1
0.400
H 2
H 2
14.29 kJ mol
i
kJ mol
H out
n out
¦ n H ¦ n H
out
g
10
H in
n C 4 H 10
LM
MN
b
134.5 124.7
o
r
n C 4 H 10
H 1
H 1
0.370 mol
e j
Ÿ 'H
bgg, i C H bgg at 25q C
'H fo
n in
substance
0.420 mol
4 H10
b.
d.
149qC
i
i
LM
MN
z
b gb
B
Table B.2
149
25
C p dT
OP kJ
PQ mol
g b gb
14.14 kJ mol
g
9.8 1 14.142 1 14.287 kJ
9.95 kJ
in
9.95 kJ 325 mol feed
1h
1 kW
1 mol feed
h
3600 s 1 kJ / s
9.95 kJ
(0.930 0.560) mol n - C 4 H 10 react
9-6
26.9 kJ / mol
0.90 kW
9.12
a.
1 m3 at 298K, 3.00 torr
Products at 1375K, 3.00 torr
n0 (mol)
0.111 mol SiH4/mol
0.8889 mol O2/mol
n1 (mol O2)
n2 (mol SiO2)
n3 (mol H2)
SiH 4 (g) + O 2 (g) o SiO 2 (s) + 2H 2 (g)
1 m3
Ideal Gas Equation of state : no
273 K 3.00 torr
1 mol
298 K 760 torr 22.4 u 10 -3 m 3
.
01017
m
nio Q i [
ni
SiH 4 : 0 = 0.1111(0.1017 mol) [ Ÿ [
O 2 : n1
SiO 2 : n2
.
0.8889(01017
mol) [
[
0.0113 mol
0.0790 mol O 2
0.0113 mol SiO 2
H 2 : n3 = 2[ = 0.0226 mol H 2
b.
'H ro
( 'H fo )SiO 2 (s) ( 'H fo )SiH 4 ( g)
= [851 ( 61.9)] kJ mol
789.1 kJ / mol
References : SiH 4 (g), O 2 (g), SiO 2 (g), H 2 (g) at 298 K
nin
nout
Substance
H
H
in
SiH 4
O2
out
(mol h) (kJ mol) (mol h) (kJ mol)
0
0.013
0.904
0
0.0790
H
1
SiO 2
H2
H 2
H
0.0113
0.0226
3
B
Table B.8
O 2 (g,1375K): H 1
H O 2 (1375K) = 46.14 kJ / mol
z
1375
SiO 2 (s,1375K): H 2
79.18 kJ / mol
(C p )SiO 2 (s) dT
298
B
Table B.8
H 2 (g,1375K): H 3
c.
Q
'H
H H 2 (1375K) = 41.08 kJ / mol
[ 'H ro ¦ ni H i ¦ ni H i
out
= 3.45 kJ 27.5 m
Q
m3
h
3.45 kJ / m 3 feed
in
3
1h
1 kW
3600 s 1 kJ / s
9-7
0.0264 kW (transferred from reactor)
9.13
a.
bg bg
bg
bg
Fe 2 O 3 s 3C s o 2 Fe s + 3CO g , 'H r (77 $ F)
Basis:
2.111 u 105 Btu lb - mole
2000 lb m Fe 1 lb - mole
3581
. lb - moles Fe produced
55.85 lb m
53.72 lb - moles CO produced
17.9 lb - moles Fe 2 O 3 fed
53.72 lb - moles C fed
17.9 lb-moles Fe2O3 (s)
77q F
35.81 lb-moles Fe (l)
2800q F
53.72 lb-moles C
77q F
53.72 lb-moles CO(g)
570q F
Q (Btu/ton Fe)
b.
bg bg bg
bg
References: Fe 2 O 3 s , C s , Fe s , CO g at 77q F
Substance
b
g
Fe 2 O 3 s,77q F
b g
Febl,2800q Fg
CObg,570q Fg
H in
nin
(lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole)
17.91
0
C s,77q F
Fe(l,2800$ F): H 1
CO(g,570$ F): H 2
H out
nout
53.72
0
.
3581
H
53.72
H 2
z
2794
77
dC i
b gdT 'Hm b2794q Fg p Fe s
H CO (570$ F)
z
1
dC i
2800
2794
b gdT
p Fe l
28400 Btu lb - mole
3486 Btu lb - mole
A
FH interpolating
I
from Table B.9 K
c.
Q
'H
nFe 'H ro
Q Fe
¦ n H ¦ n H
i
out
i
i
i
in
. ge2.111 u 10 j
b3581
. gb28400g b53.72gb3486g 0
b3581
5
2
d.
4.98 u 106 Btu / ton Fe produced
Effect of any pressure changes on enthalpy are neglected.
Specific heat of Fe(s) is assumed to vary linearly with temperature from 77qF to 570qF.
Specific heat of Fe(l) is assumed to remain constant with temperature.
Reaction is complete.
No vaporization occurs.
9-8
9.14
a.
bg
C 7 H 16 g o C 6 H 5CH 3 (g) 4 H 2 (g)
Basis: 1 mol C7H16
1 mol C7H16
1 mol C6H5CH3
400qC
4 mol H2
400q C
Q (kJ/mol)
bg b g
References: C s , H 2 g at 25q C
b.
H in
substance nin
H out
nout
bmolg bkJ molg bmolg bkJ molg
C 7 H 16
b
C 7 H 16
1
H 1
C7 H8
1
H2
4
g
g,400q C : H 1
( 'H f$ ) C7 H 16 ( g) LM
MN
z
H2
H
3
B
Table B.2
400
25
C p dT
OP
PQ
( 187.8 + 104.1) kJ / mol = 83.7 kJ / mol
b
g
C 6 H 5 CH 3 g,400q C : H 2
( 'H f$ ) C 6 H 5CH 3 ( g)
L
M
MN
z
B
Table B.2
400
25
C p dT
OP
PQ
(+50 + 60.2) kJ / mol = 110.2 kJ / mol
B
Table B.8
b
g
H 2 g,400q C : H 3
c.
H H 2 (400$ C)
10.89 kJ mol
1 mol C 7 H 16 reacted Ÿ [ = 1 mol
Q
'H
['H ro ¦ n H ¦ n H
i
out
i
i
i
in
ob1gb237.8g b1gb60.2g b4gb10.86g b1gb104.1g tkJ
d.
'H r ( 400 $ C) =
237.3 kJ
1 mol C 7 H16 react
237.3 kJ / mol
9-9
237.3 kJ (transferred to reactor)
9.15
a.
bCH g Obgg o CH bgg H bgg CObgg
3 2
4
2
Moles charged: (Assume ideal gas)
2.00 liters 273 K 350 mm Hg
b g
1 mol
b g
0.01286 mol CH 3 2 O
873 K 760 mm Hg 22.4 liters STP
Let x
b g
fraction CH 3 2 O decomposed (Clearly x<1 since Pf 3 P0 )
0.01286(1 – x ) mol (CH3 )2O
0.01286 x mol CH4
0.01286 x mol H2
0.01286 x mol CO
0.01286 mol
(CH 3)2 O
600°C, 350 mm Hg
Total moles in tank at t
b.
Pf V
n f RT
P0V
n0 RT
Ÿ
nf
Pf
n0
P0
Ÿ
b g
b
0.01286 1 x 3x
2h
b
0.01286 1 2 x
g
875
Ÿx
350
0.01286
600°C
875 mm Hg
g
0.01286 1 2 x mol
0.75 Ÿ 75% decomposed
af b g b g
References:C s , H 2 g , O 2 g at 25$ F
H in
nin
substance
H out
nout
(mol) ( kJ / mol)
( mol)
(kJ / mol)
0.01286
H 1
H 1
0.25 u 0.01286
H
0.75 u 0.01286
b g bg
CH bg g
H bg g
CObgg
CH 3 2 O g
4
2
2
0.75 u 0.01286
0.75 u 0.01286
bCH g O(g,600 C): H
( 'H fo ) bCH 3 g
$
3 2
1
O
2
118 kJ mol
CH 4 (g,600 C): H 2
$
( 'H fo ) CH 4
L
M
MN
B
z
LM
MN
z
B
given
873
298
C p dT
B
Table B.2
600
25
C p dT
H 3
H
4
OP J 1 kJ
PQ mol u 10 J
3
OP 1
PQ 10
3
( 180.16 62.40) kJ / mol
74.85 29.46 45.39 kJ mol
Table B.8
H 2 (g,600$ C): H 3
H H 2 (600$ C)
16.81 kJ mol
Table B.1
Table B.8
Table B.8
CO(g,600$ C): H 4
( 'H fo ) CO
B
H CO (600 $ C)
110.52 17.57 kJ mol
B
92.95 kJ mol
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are :
c.
'H
[
'H
¦n
out H out
¦n
( n CH 4 ) out ( n CH 4 ) in
Q CH 4
b
g
in H in
1.5507 ( 1.5143) kJ
0.75 u 0.01286
mol
1
b
[ 'H r 600q C Ÿ 'H r 600q C
g
0.009645 mol
0.0364 kJ
0.009645
9-10
0.0364 kJ
3.77 kJ / mol
9.15 (cont’d)
b
'U r 600q C
g
b
g
¦Q
'H r 600q C RT [
i
gaseous
products
3.77 kJ mol 9.16
b
[ 'U r 600q C
d.
Q
a.
SO 2 (g) Basis :
g
8.314 J
¦Q
i
]
gaseous
reactants
1 kJ
873 K
b1 1 1 1g
18.3 kJ mol
mol ˜ K 10 J
3
(0.009645 mol)( 18.3 kJ / mol)
0.176 kJ (transferred from reactor)
1
O 2 (g) o SO 3 (g)
2
l00 kg SO 3
10 3 mol SO 3
min
80.07 kg SO 3
1249 mol SO 3 min
n 0 mol SO2 /min
450°C
100% excess
ni1 mol O2 /min
3.76 n 1 mol O2 /min
450°C
1249 mol SO 3/min
s
n 0 mol SO2 /min
n 3 mol O 2/min
3.76 n 1 mol O2
550°C / i
mw (kg H 2O( l) /h)
25°C
mw (kg H 2O(l) /h)
40°C
Assume low enough pressure for H to be independent of P.
SO 3 balance :
bGeneration
output
n0 (mol SO 2 fed) 0.65 mol SO 2 react 1 mol SO 3 produced
g
Ÿ n0
1922 mol SO 2
0.5 mol O 2 reqd
min
1 mol SO 2
b g
N 2 balance : 3.76 1922
b
7227 mol / min in & out
b1 +1g mol O
673 mol SO
g
b2gb1922g b2gb1922g b3gb1249g b2gb673g 2n
.
Extent of reaction : [
fed
2
min out
3
Ÿ n3
.
(n SO 2 ) out (n SO 2 ) in
( 'H fo )SO 3 ( g) ( 'H fo )SO 2 ( g)
1922 mol O 2 min fed
1298 mol / min out
Q SO 2
B
673 1922
1
1249 mol / min
Table B.1
'H ro
mol SO 3
min
g
b
.
2
1 mol O 2 reqd
1922 1 0.65 mol s
65% conversion : n2
b.
1249
1 mol SO 2 react
1922 mol SO 2 / min fed
100% excess air: n1
O balance:
1 mol SO 2 fed
min
395.18 ( 296.9)
9-11
99.28 kJ / mol
9.16 (cont'd)
bg bg bg
bg
References : SO 2 g , O 2 g , N 2 g , SO 3 g at 25$ C
H in
nin
Substance
nout
H out
( mol / min) ( kJ / mol) ( mol / min) ( kJ / mol)
H 1
H 4
1922
673
H 2
H 5
1922
1298
H
H
7227
7227
SO 2
O2
N2
3
SO 3
z
SO 2 (g,450 C) : H 1
$
6
B
1249
H 7
Table B.2
450
19.62 kJ / mol
C p dT
25
B
Table B.8
$
H 2
H O 2 (450$ C)
$
H 3
H N 2 (450 C)
O 2 (g,450 C)
13.36 kJ / mol
B
Table B.8
N 2 (g,450 C)
$
Out :
z
SO 2 (g,450 C) : H 4
$
B
12.69 kJ / mol
Table B.2
550
24.79 kJ / mol
C p dT
25
B
Table B.8
$
H 5
H O 2 (550 $ C)
$
H 6
H N 2 (550$ C)
O 2 (g,550 C)
16.71 kJ / mol
B
Table B.8
N 2 (g,550 C)
z
SO 3 (g,550$ C) : H 7
Q
' H
[ 'H ro B
1581
. kJ / mol
Table B.2
550
35.34 kJ / mol
C p dT
25
¦ n H ¦ n H
i
i
i
out
i
in
b1249gb98.28g b673gb24.796g b179.8gb16.711g b7227gb15.808g b1249gb35.336g b1922gb19.623g
1922b13.362 g b7227gb12.691g
. u 104 kJ / min
8111
c.
Assume system is adiabatic, so that Q lost from reactor
Q
'H
LM
MM A e
N
Table B.5
Ÿ 8.111 u 10 4
d.
j
e
m w H w l, 40$ C H w l, 25$ C
kJ
min
m w
A
Table B.5
OP
jP
PQ
Q gained by cooling water
FG kg IJ 167.5 104.8 kJ Ÿ m
H min K
kg
w
1290 kg min cooling water
If elemental species were taken as references, the heats of formation of each molecular species would
have to be taken into account in the enthalpy calculations and the heat of reaction term would not have
been included in the calculation of 'H .
9-12
bg
CO(g) H 2 O v o H 2 (g) CO 2 (g) ,
9.17
B
Table B.1
'H ro
a.
e j
'H fo
CO 2 ( g)
3
e j
'H fo
CO(g)
e j
'H fo
bg
H 2O v
4115
.
b g
3
kJ
mol
b g
Basis : 2.5 m STP product gas h 1000 mol 22.4 m STP
n 0 (mol CO/h)
25°C
n 2 (mol H 2 O(v)/h)
150°C
Q r (kW)
H balance on reactor : 2n2
Q c (kW)
b0.40gb1116. mol hg 44.64 mol CO h
111.6 b2gb0.40g b2 gb0.20g mol h Ÿ n
2
44.64 mol CO 1 mol H 2 O
Steam theoretically required
h
1 mol CO
b66.96 44.64g mol h u 100%
% excess steam
44.64 mol h
CO 2 balance on condenser : n3
H 2 balance on condenser: n4
b
gŸ
bg
66.96 mol H 2 O v h
44.64 mol H 2 O
50% excess steam
b0.40gb1116. mol hg 44.64 mol CO
b0.40gb111.6 mol hg 44.64 mol H h
2
h
2
Saturation of condenser outlet gas:
pw 15q C
n 3 (mol CO2 /h)
n 4 (mol H 2 /h)
n 5 (mol H 2 O(v)/h),sat'd
15°C, 1 atm
n 6 (mol H 2 O( l )/h)
15°C, 1 atm
111.6 mol/h
condenser
0.40 mol H 2/mol
0.40 mol CO 2/mol
0.20 mol H 2O( v)/h
500°C
reactor
C balance on reactor : n1
1116
. mol h
b
g
n5 mol H 2 O h
44.64 + 44.64 + n5 mol h
12.788 mm Hg
Ÿ n5
760 mm Hg
b
gb g
H O balance on condenser: b111.6gb0.20gmol H O h 153
. n
y H 2O
p
2
2
Ÿ n6
b.
6
20.8 mol H 2 O h condensed = 0.374 kg / h
Energy balance on condenser
$
References : H 2 ( g), CO 2 (g) at 25 C, H 2 O at reference point of steam tables
Substance
CO 2 (g)
H 2 (g)
bg
bg
H 2O v
H 2O l
n in
n out
H in
H out
mol / h kJ / mol mol / h kJ / mol
44.64
H 1
H 4
44.64
44.64
44.64
H 2
H 5
22.32
153
H 3
H 6
.
20.80
H 7
Enthalpies for CO2 and H2 from Table B.8
CO 2 (g,500 $ C) : H 1
H 2 ( g,500 $ C) : H 2
H CO 2 (500 $ C)
2134
. kJ / mol
H H 2 (500 $ C) 1383
. kJ / mol
9-13
bg
153
. mol H 2 O v h
9.17 (cont’d)
H 2 O(v,500 $ C) : H 3
CO 2 (g,15$ C) : H 4
H CO2 (15$ C)
H 2 ( g,15$ C) : H 5
H H 2 (15$ C)
H 2 O(v,15$ C) : H 6
2529
H 2 O(l,15$ C) : H 7
62.9
Q
i
i
out
FG
H
FG
H
IJ
K
IJ
K
45.52 kJ mol
. kJ mol
113
b49.22 29718. g kJ
i
1h
1 kW
0.812 kW
3600 s 1 kJ s
h
in
bheat transferred from condenser g
c.
62.86 kJ mol
0.432 kJ / mol
18.0 kg
kJ
u
kg 10 3 mol
i
IJ
K
0.552 kJ / mol
18.0 kg
kJ
u
kg
10 3 mol
¦ n H ¦ n H
'H
FG
H
18 kg
kJ
u
kg
10 3 mol
3488
Energy balance on reactor :
References : H 2 (g), C(s), O 2 (g) at 25q C
Substance
CO(g)
H 2 O ( v)
bg
bg
H2 g
H in
nin
H out
nout
( mol / h) ( kJ / mol) ( mol / h) ( kJ / mol)
H 1
44.64
66.96
H2
22.32
H 3
44.64
H
4
CO 2 g
CO(g,25$ C) : H 1
( 'H f$ ) CO
H 5
44.64
Table B.1
110.52 kJ / mol
H 2 O(v,150 $ C) : H 2 = ( 'H f$ ) H 2 O(v) H H 2 O (150 $ C)
Tables B.1, B.8
H 2 O(v,500 $ C) : H 3 = ( 'H f$ ) H 2 O(v) H H 2 O (500 $ C)
Tables B.1, B.8
H 2 (g,500 $ C) : H 4
H H 2 (500 $ C)
CO 2 (g,500 $ C) : H 5
Q
'H
i
out
i
i
in
i
224.82 kJ mol
Table B.8
1383
. kJ / mol
( 'H $f ) CO 2 H CO2 (500 $ C)
¦ n H ¦ n H
237.56 kJ mol
Tables B.1, B.8
372.16 kJ / mol
2101383
1h
1 kW
. ( 20839.96) kJ
h
3600 s 1 kJ s
0.0483 kW
bheat transferred from reactor g
d.
Benefits
Preheating CO Ÿ more heat transferred from reactor (possibly generate additional steam for plant)
Cooling CO Ÿ lower cooling cost in condenser.
9-14
9.1 8
b.
References : FeO(s), CO(g), Fe(s), CO 2 ( g) at 25o C
nin
nout
H in
H out
( mol) ( kJ / mol) ( mol) ( kJ / mol)
1.00
n1
H 1
0
n0
H0
n2
H 2
n
H
Substance
FeO
CO
Fe
CO 2
Q [ 'H ro ¦n
out H out
¦n
3
3
n4
H 4
in H in
Ÿ Q [ 'H ro n1 H 1 n2 H 2 n3 H 3 n4 H 4 n0 H 0
Fractional Conversion : X
CO consumed :
Ÿ n2
(100
. n1 )
Ÿ n1
.
100
1 X
(1 n1 ) mol FeO consumed
1 mol CO
1 mol FeO consumed
n0 (1 n1 )
(1 n1 ) mol CO
n0 X
Fe produced : n3 =
(1 n1 ) mol FeO consumed
1 mol Fe
1 mol FeO consumed
1 mol CO 2
(1 n1 ) mol FeO consumed
CO 2 produced : n4 =
1 mol FeO consumed
Extent of reaction : [
z
(nCO ) out (nCO ) in
n2 n0
Q CO
1
(1 n1 ) mol Fe = X
(1 n1 ) mol CO 2
X
T
H i
C pi dT
for i
0,1,2,3,4
25
H 0
0.02761 (T0 298) 2.51 u 10 6 (T0 2 298 2 )
Ÿ H
( 8.451 0.02761 T 2.51 u 10 6 T 2 ) kJ / mol
0
u 10 6 T 2 3188
u 10 2 / T ) kJ / mol
( 17.0814 0.0528 T 31215
.
.
Ÿ H 1
(0.02761 ( T 298) 2.51 u 10 6 (T 2 298 2 )
Ÿ H 2
H 3
8.451 0.02761 T 2.51 u 10 6 T 2 ) kJ / mol
u 10 5 (T 2 298 2 )
.
0.01728 (T 298) 1335
Ÿ H 3
H 4
0
.
.
0.0528 (T 298) 31215
u 10 6 (T 2 298 2 ) 3188
u 10 2 (1 / T 1 / 298)
H 1
H 2
0
u 10 5 T 2 ) kJ / mol
( 6.335 0.01728 T 1335
.
0.04326(T 298) 0.573 u 10 5 (T 2 298 2 ) 818
. u 10 2 (1 / T 1 / 298)
Ÿ H
0.04326 T 0.573 u 10 5 T 2 818
( 16145
.
. u 10 2 / T ) kJ / mol
4
9-15
X
9.18 (cont'd)
c.
n0
2.0 mol CO, T0
Ÿ n1
1 0.7
Summary: H 0
H 3
550 K, and X
2 0.7 13
. , n3
0.3, n2
kJ / mol, H 1
1520
.
7.207 kJ / mol, H 4
'H ro
Q
350 K, T
0.700 mol FeO reacted / mol FeO fed
0.7, n4
kJ / mol, H 2
1348
.
0.7, [
0.7
7.494 kJ / mol,
.
kJ / mol
1087
1.648 kJ / mol
(0.7)( 1.648) (0.3)(1.348) (1.3)(7.494) (0.7)(7.207) (0.7)(1087
. ) (2)(1520
. )
ŸQ
1186
kJ / mol
.
d.
no
X T
1 298
1 400
1 500
1 600
1 700
1 800
1 900
1 1000
Xi
no To
1 298
1 400
1 500
1 600
1 700
1 800
1 900
1 1000
X
1
1
1
1
1
1
1
1
T
700
700
700
700
700
700
700
700
Xi
no
1
1
1
1
1
1
1
1
To
400
400
400
400
400
400
400
400
n1
0
0
0
0
0
0
0
0
n2 n3 n4
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
H0
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
n2 n3 n4
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
H0
0
0
0
0
0
0
0
0
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
n2 n3 n4
1
0
0
0.9 0.1
0
0.8 0.2
0
0.7 0.3
0
0.6 0.4
0
0.5 0.5
1
0.4 0.6
1
0.3 0.7
1
0.2 0.8
1
0.1 0.9
1
0
1
1
1
1
1
1
1
1
1
1
n1
1
1
1
1
1
1
1
1
To
400
400
400
400
400
400
400
400
400
400
400
X
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
T
500
500
500
500
500
500
500
500
500
500
500
Xi
1
1
1
1
1
1
1
1
1
1
1
no
0.5
0.5
0.5
1
1
To
400
400
400
400
400
X
0.5
0.5
0.5
0.5
0.5
T
400
400
400
400
400
Xi
0.5
0.5
0.5
0.5
0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
n1
n1 n2
0.5
0.5
0.5
0.5
0.5
n3 n4
0.5
1
0.5
0.5
0.5
0.5
9-16
H1
H2
0
5.335
10.737
16.254
21.864
27.555
33.321
39.159
0
2.995
5.982
9.019
12.11
15.24
18.43
21.67
H3
0
2.713
5.643
8.839
12.303
16.033
20.031
24.295
H4
0
4.121
8.553
13.237
18.113
23.152
28.339
33.663
Q
-19.48
-12.64
-5.279
2.601
10.941
19.71
28.895
38.483
0
2.995
5.982
9.019
12.11
15.24
18.43
21.67
H1
21.864
21.864
21.864
21.864
21.864
21.864
21.864
21.864
H2
12.11
12.11
12.11
12.11
12.11
12.11
12.11
12.11
H3
12.303
12.303
12.303
12.303
12.303
12.303
12.303
12.303
H4
18.113
18.113
18.113
18.113
18.113
18.113
18.113
18.113
Q
13.936
10.941
7.954
4.917
1.83
-1.308
-4.495
-7.733
H0
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
H1
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
H2
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
H3
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
H4
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
Q
-3.188
-3.399
-3.61
-3.821
-4.032
-4.244
-4.455
-4.666
-4.877
-5.088
-5.299
H0
2.995
2.995
2.995
2.995
2.995
H1
5.335
5.335
5.335
5.335
5.335
H2
2.995
2.995
2.995
2.995
2.995
H3
2.713
2.713
2.713
2.713
2.713
H4
4.121
4.121
4.121
4.121
4.121
Q
-3.653
-3.653
-3.653
-3.653
-3.653
9.18 (cont'd)
400
400
400
400
400
400
0.5
0.5
0.5
0.5
0.5
0.5
400
400
400
400
400
400
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
2.995
2.995
2.995
2.995
2.995
2.995
50
40
30
20
10
0
-10
-20
-30
5.335
5.335
5.335
5.335
5.335
5.335
2.995
2.995
2.995
2.995
2.995
2.995
2.713
2.713
2.713
2.713
2.713
2.713
10
5
0
-5
-10
500
1000
0
1500
500
0
-1
-3
Q
Q
-2
-4
-5
-6
0
0.2
1000
1500
T o (K )
T (K )
0.4
0.6
0.8
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
1
0
0.5
1
X
a.
-3.653
-3.653
-3.653
-3.653
-3.653
-3.653
15
0
9.19
4.121
4.121
4.121
4.121
4.121
4.121
20
Q
Q
1
1
1
2
2
2
1.5
2
2.5
no
Fermentor capacity : 550,000 gal
Solution volume : (0.9 u 550,000)
495,000 gal
R|0.071 lb C H OH / lb solution
S|
T0.86 lb H O / lb solution
m
2
5
m
Final reaction mixture : 0.069 lb m (yeast, other species) / lb m solution
2
Mass of tank contents :
m
1 ft 3
65.52 lb m
7.4805 gal
1 ft 3
495,000 gal
Mass of ethanol produced :
4.336 u 106 lb m solution 0.071 lb m C 2 H 5OH
Ÿ
3.078 u 10 lb m C 2 H 5OH
Ÿ
307827 lb m C 2 H 5OH
5
4335593 lb m
lb m solution
1 lb - mole C 2 H 5OH
46.1 lb m C 2 H 5OH
6677 lb - mole C 2 H 5OH
1 ft 3 C 2 H 5OH
7.4805 gal
49.67 lb m C 2 H 5OH
1 ft 3
Makeup water required : 495,000 gal 3.078 u 105 lb m C 2 H 5
46,360 gal C 2 H 5OH
46,360 gal C 2 H 5OH
25 gal mash
2.6 gal C 2 H 5OH
9-17
4.9 u 104 gal
9.19 (cont'd)
46,360 gal C2 H5OH
1 bu
1 acre 1 batch 24 h 330 days
acres
175
. u 105
Acres reqd. :
1 batch
2.6 gal C2 H5OH 101 bu 8 h 1 day 1 year
year
b.
C12 H 22 O11 (s) 12O 2 (g) o 12CO 2 (g) 11H 2 O(l)
'H co
'H co 12 'H fo (CO2 ) 11'H fo ( H 2 O) 'H fo (C12 H 22 O11 )
. kJ / mol
56491
Ÿ 'H fo (C12 H 22 O11 ) 221714
. kJ / mol
C12 H 22 O11 (s) H 2 O(l) o 4C2 H 5OH(l) 4CO2 (g)
'H ro
4 'H fo (C 2 H 5 OH) 4 'H fo (CO 2 ) 'H fo (C12 H 22 O11 ) 'H fo ( H 2 O) = 184.5kJ / mol
1815
. kJ 453.6 mol 0.9486 Btu
Ÿ 'H ro
7.811 u 104 Btu / lb - mole
1 mol
1 lb - mole
1 kJ
c.
Moles of maltose :
4.336 u 106 lb m solution 0.071 lb C 2 H 5OH 1 lb - mole C2 H 5 OH 1 lb - mole C12 H 22 O11
46.1 lb C2 H 5OH
4 lb - mole C2 H 5 OH
1 lb m solution
1669 lb - moles C12 H 22 O11 Ÿ [ nC10H22O11
1669 lb - moles
Q = ['H r mC p (95$ F - 85$ F)
Btu
Btu
) (4.336 u 106 lb m )(0.95 $ )(10$ F)
lb - mole
lb - F
7
= 8.9 u 10 Btu ( heat transferred from reactor)
= (1669 lb - moles)(7.811 u 104
9.20
d.
Brazil has a shortage of natural reserves of petroleum, unlike Venezuela.
a.
4NH 3 5O 2 o 4NO 6H 2 O,
3
O 2 o N 2 3H 2 O
2
References: N 2 g , H 2 g , O 2 (g), at 25q C
2NH 3 bg bg
NH 3
Air
NO
H in
nin
Substance
nout
H out
(mol min) (kJ mol) (mol min) (kJ mol)
100
H 1
900
H2
90
H
3
H 2O
150
N2
716
H 4
H
O2
69
H 6
H i
'H foi b
z
T
25
C pi dT
g
NH 3 g, 25q C : H 1
b
5
g
Air g, 150q C : H 2
Table B.1
( 'H fo ) NH 3
B
46.19 kJ mol
Table B.8
B
3.67 kJ mol
9-18
9.20 (cont'd)
b
g
NO g, 700q C : H 3
b
90.37 25
Table B.1,
Table B.8
g
B
H 2 O g, 700q C : H 4
b
g
Q
g
'H
111.97 kJ mol
216.91 kJ mol
B
20.59 kJ mol
Table B.8
O 2 g, 700q C : H 6
b.
B
C p dT
Table B.8
N 2 g, 700q C : H 5
b
z
Table B.1,Table B.2
700
B
21.86 kJ mol
¦ n H ¦ n H
i
i
out
i
i
4890 kJ min u (1 min / 60s) 815
. kW
in
(heat transferred from the reactor)
9.21
c.
If molecular species had been chosen as references for enthalpy calculations, the extents of each
reaction would have to be calculated and Equation 9.5-1b used to determine 'H . The value of Q
would remain unchanged.
a.
Basis: 1 mol feed
I=Inert
1 mol at 310qC
0.537 C2H4 (v)
0.367 H20 (v)
0.096 N2(g)
Products at 310qC
n1 (mol C2H4 (v))
n2 (mol H2O(v))
0.096 mol N2 (g)
n3 (mol C2H5OH (v))
n4 (mol (C2H5)2O) (v))
C 2 H 4 ( v) H 2 O(v) œ C 2 H 5OH(v)
b
g
2C 2 H 5OH(v) œ C 2 H 5 2 O(v) H 2 O(v)
5% ethylene conversion:
b0.537gb0.05g
Ÿ n1
0.02685 mol C 2 H 4 consumed
b0.95gb0.537g
0.510 mol C 2 H 4
90% ethanol yield:
n3
0.02685 mol C 2 H 4 consumed 0.9 mol C 2 H 5OH
C balance :
0.02417 mol C 2 H 5OH
1 mol C 2 H 4
b2gb0.537g b2gb0.510g b2gb0.02417g 4n
Ÿ n4
4
O balance : 0.367
.
n2 0.02417 1415
u 10 3 Ÿ n2
af b g b g
bg
0.3414 mol H 2 O
References: C s , H 2 g , O 2 g at 25$ C, I g at 310 $ C
9-19
b
g
1.415 u 10 3 mol C 2 H 5 2 O
9.21 (cont'd)
substance
H in
nin
0.3414
0
0.096
C 2 H 5 OH
0.02417
1.415 u 10 3
H 4
Ÿ
b52.28 16.41g
2
4
o
f C2 H 4
1
g
b
0.510
2
5 2
H 2 O g, 310q C : H 2
z
300
25
C p dT
g
( 'H fo ) C 2 H 5OH(g) bC H g Obg, 310q Cg: H e'H j
o
f
4
5 2
(C 2 H 5 )O(l)
z
310
25
'H
¦ n H ¦ n H
i
i
out
C p dT
b
i
i
0
H
3
Ÿ
Table B.1
Table B.8
Ÿ
Table B.1
Table B.2
g
'H v 25q C 204.2 kJ mol
Energy balance: Q
2
Table B.1 for 'H fo
Table B.2 for Cp
( 'H fo ) H 2 O(v) H H 2 O(v) (310$ C)
C 2 H 5OH g, 310q C : H 3
2
(kJ / mol)
H 1
H
0.096
bC H g O
C H bg, 310q Cg : H
( 'H )
b.
(mol)
I
H2O
b
H out
(mol) (kJ / mol)
0.537
H 1
H
0.367
C2H4
2
nout
z
68.69 kJ mol
. 9.93g
b24183
b235.31 24.16g
310
25
C p dT
231.90 kJ mol
21115
. kJ mol
b272.8 26.05 42.52g
1.3 kJ Ÿ 1.3 kJ transferred from reactor mol feed
in
To suppress the undesired side reaction. Separation of unconsumed reactants from products and
recycle of ethylene.
9.22
C 6 H 5CH 3 O 2 o C 6 H 5CHO H 2 O
C 6 H 5CH 3 9O 2 o 7CO 2 4H 2 O
Basis: 100 lb-mole of C 6 H 5CH 3 fed to reactor.
100 lb-moles C 6 H5 CH 3
n0 (lb-moles O 2 )
3.76n 0 (lb-moles N 2 )
350°F, 1 atm
V0 (ft3 )
reactor
Q(Btu)
jacket
mw(lbm H2 O( l )), 80°F
Vp (ft3 ) at 379°F, 1 atm
n1 (lb-moles C6 H5 CH3 )
n2 (lb-moles O 2 )
3.76n0 (lb-moles N 2 )
n3 (lb-moles C 6 H5 CHO)
n4 (lb-moles CO2 )
n5 (lb-moles H2 O)
mw(lbm H2 O( l )), 105°F
Strategy:
All material and energy balances will be performed for the assumed basis of 100 lb-mole
C 6 H 5CH 3 . The calculated quantities will then be scaled to the known flow rate of water in
b
g
the product gas 29.3 lb m 4 h .
9-20
9.22 (cont'd)
Plan of attack: % excess air Ÿ n0
Ideal gas equation of state Ÿ V0
13% C6 H5CHO formation Ÿ n3 Ideal gas equation of state Ÿ V p
0.5% CO2 formation Ÿ n4
E.B. on reactor Ÿ Q
C balance Ÿ n1
E.B. on jacket Ÿ mw
H balance Ÿ n5
Scale V0 , Vp , Q, mw by n5 actual / n5 basis
O balance Ÿ n2
b g
100% excess air:
n0
100 lb - moles C 6 H 5CH 3
b
N 2 feed & output
g
b1 1gmole O
1 mol O 2 reqd
1 mole C 6 H 5CH 3
b g
3.76 200 lb - moles N 2
fed
1 mol O 2 reqd
2
b g
200 lb - moles O 2
752 lb - moles N 2
100 lb - moles C6 H5CH 3 0.13 mol C6 H5CH 3 react 1 mole C6 H5CH 3
1 mole C6 H5CH 3 fed 1 mole C6 H5CH 3
13% o C6 H5CHO Ÿ n3
= 13 lb - moles C7 H 6 O
b100gb0.005glb - moles C H CH
0.5% o CO2 Ÿ n4
6
5
3
react
7 moles CO2
35
. lb - moles CO2
1 mole C6 H5CH 3
mol C mole C7H8
C balance:
H balance:
O balance:
a100faB7f lb - moles C 7n a13fa7f a3.5fa1f Ÿ n 86.5 lb - moles C H CH
b100gb8glb - moles H b86.5gb8g b13gb6g 2n Ÿ n 15.0 lb - moles H Obvg
b200gb2glb - moles O 2n b13gb1g b35. gb2g b15gb1g Ÿ n 182.5 lb - moles O
1
5
b100 + 200 + 752glb - moles
FG 86.5 182.5
H
C7 H 8
Vp
O2
C 7 H 8O
CO 2
H 2O
b g b350 460g R
$
359 ft 3 STP
1 lb - moles
$
492 R
IJ
K
N2
bg b g b g b g
e
6.218 u 105 ft 3
359 ft 3
b379 460g R
1 lb - mole
492 $ R
13 3.5 15 752 lb - moles
Energy balance on reactor (excluding cooling jacket)
$
j
References : C s , H 2 g , O 2 g , N 2 g at 25$ C 77 $ F
nin
H in
nout
H out
C 6 H 5CH 3
100
86.5
O2
200
182.5
N2
752
H 1
H 2
H
H 4
H 5
H
C 6 H 5CHO
13
CO 2
3.5
H 7
H
H 2O
15
H 9
substance
blb - molesg bBtu lb - moleg blb - molesg bBtu lb - moleg
3
9-21
3
2
2
Ideal gas law – outlet:
5
2
5
2
Ideal gas law inlet:
V0
6
1
752
6
8
6.443 u 105 ft 3
9.22 (cont'd)
Enthalpies:
LM B
OP
430.28 Btu lb - mole
Btu
31
C H CH (g,T): H bT g 'H b kJ molg u
bT 77g FP
MM
1 kJ mol
1b - mole˜q F
PQ
N
Table B.1
6
5
C 6 H 5CH 3 (g,350$ F): H 1
C H CH (g,379 $ F): H
6
5
$
o
f
3
2.998 u 104 Btu lb - mole
3.088 u 104 Btu lb - mole
4
3
bg
b
C 6 H 5 CHO(g,T): H T
Ÿ H 7
g
$
17200 31 T 77 F Btu lb - mole
7.83 u 103 Btu lb - mole
B
Table B.9
e
$
e
$
e
$
e
$
j
H O 2 (350 F)
j
H N 2 (350 F)
j
H O 2 (379 $ F)
j
H N 2 (379 $ F)
O 2 g,350 F : H 2
$
B
1.972 u 10 3 Btu / lb mole
Table B.9
N 2 g,350 F : H 3
$
B
u 10 3 Btu / lb mole
1911
.
Table B.9
O 2 g,379 F : H 5
2.186 u 103 Btu / lb mole
B
Table B.9
N 2 g,379 F : H 6
2.116 u 10 3 Btu / lb mole
B
Table B.1 and B.9
e
j
H Obg,379q Fg: H
CO 2 g,379 F : H 8
( 'H f$ ) CO2 ( g)
H CO2 (379 F)
9
( 'H f$ ) H 2 O ( g)
H H 2 O (379 F)
$
2
$
1664
u 105 Btu / lb mole
.
B
Table B.1 and B.9
$
1016
u 105 Btu / lb mole
.
Energy Balance :
Q
'H
¦ n H ¦ n H
i
i
i
out
i
2.376 u 10 6 Btu
in
Energy balance on cooling jacket:
Q
'H
mw
Q
z
105
80
b g
mw lb m u 1.0
bn g
bn g
5 actual
V0
Vp
b.
Q
w
m
b
29.3 lb m H 2 O
g
bg
9.504 u 10 4 lb m H 2 O l
1b - mole H 2 O
1
18.016 lb m H 2 O 15.0 lb - moles H 2 O
d6.218 u 10 ft id0.02711 h i
d6.443 u 10 ft id0.02711 h i
d2.376 u 10
1.0 Btu (lb m ˜ $ F)
Btu
$
u 105 80 F Ÿ mw
lb m ˜q F
4h
5 basis
a.
b g dT
p H O l
2
2.376 u 10 4 Btu , C p
2.376 u 10 6 Btu
Scale factor:
dC i
0.02711 h 1
b g
h b product g
5
3
1
169
. u 10 4 ft 3 h feed
5
3
1
175
. u 10 4 ft 3
id
i
6.44 u 10 4 Btu / h
id
i
2577 lb m
1 ft 3
7.4805 gal 1 h
. gal H 2 O min
515
h
62.4 lb m
1 ft 3 60 min
6
Btu 0.02711 h 1
. u 10
d9504
Btu 0.02711 h 1
4
9-22
9.23
a.
CaCO 3 ( s) o CaO(s) +CO 2 (g)
CaO(s)
900qC
CaCO3(s)
25qC
CO2(g)
900qC
Q (kJ)
1000 kg
Basis : 1000 kg CaCO 3
10.0 kmol CaO(s) produced
1 mol
10.0 kmol CaCO 3 Ÿ 10.0 kmol CO 2 (g) produced
0.100 kg
10.0 kmol CaCO 3 (s) fed
References: Ca(s), C(s), O2(g) at 25qC
nin
nout
H in
H out
Substance (mol) (kJ / mol) (mol) (kJ / mol)
CaCO 3
10.0
H 1
CaO
H2
10.0
CO 2
10.0
H 3
Table B.1
CaCO 3 (s, 25 C) : H 1
o
B
( 'H fo ) CaCO 3 ( s)
z
1206.9 kJ / mol
Table B.1,
Table B.2
1173
CaO(s, 900 o C) : H 2
( 'H fo ) CaO ( s) B
C p dT
( 635.6 48.54) kJ / mol
587.06 kJ / mol
298
Table B.1,
Table B.8
CO 2 (g, 900 o C) : H 3
Energy balance: Q
( 'H fo ) CO2 ( g) H CO2 (900 o C)
'H
F n H n H I
GH ¦ ¦ JK
i
out
b.
i
i
i
B
( 393.5 42.94) kJ / mol
2.7 u 10 6 kJ
in
Basis : 1000 kg CaCO3 fed Ÿ 10.0 kmol CaCO3
CaCO 3 ( s) o CaO(s) + CO 2 (g)
2CO + O 2 o 2CO 2
10 kmol CaCO3
25 oC
200 kmol at 900oC
0.75 N2
0.020 O2
0.090 CO
0.14 CO2
10 kmol CaCO 3 react Ÿ n1
Product gas at 900oC
n2 (kmol CO2 )
n3 (kmol N2 )
n4 (kmol CO)
n1 [kmol CaO(s)]
10.0 kmol CaO
9-23
350.56 kJ / mol
9.23 (cont'd)
n2
(0.14)(200) 10.0 kmol CaCO 3 react 1 kmol CO 2
n3
(0.75)(200)
150 kmol N 2
1 kmol O 2
4 kmol O 2 react 2 kmol CO 2
1 kmol O 2
C balance: (10.0)(1) + (200)(0.09)(1) + (200)(0.14)(1) = 46(1) + n4 (1) Ÿ n4
46 kmol CO 2
10.0 kmol CO
References : Ca(s), C(s), O 2 (g), N 2 (g) at 25$ C
H in
nin
H out
nout
Substance (mol) (kJ / mol) (mol) (kJ / mol)
CaCO 3
10.0
H 1
CaO
10
587.06
CO 2
28
46
CO
18
350.56
H
350.56
H
O2
4.0
N2
150
10
1
H 2
H
1
H
150
3
3
Table B.1,
Table B.8
CO(g, 900 o C) : H 1
( 'H fo ) CO( g) H CO (900 o C)
O 2 (g, 900 C) : H 2
H O 2 (900 o C)
B
( 110.52 27.49) kJ / mol
83.03 kJ / mol
Table B.8
o
B
28.89 kJ / mol
Table B.8
N 2 (g, 900o C) : H 3
Q
'H
H N 2 (900 o C)
F n H n H I
GH ¦ ¦ JK
i
i
i
out
i
9.24
a.
27.19 kJ / mol
0.44 u 106 kJ
in
% reduction in heat requirement
c.
B
2.7 u 106 0.44 u 10 6
2.7 u 10 6
u 100
838%
.
The hot combustion gases raise the temperature of the limestone, so that less heat from the outside
is needed to do so. Additional thermal energy is provided by the combustion of CO.
A+Bo C
(1)
2C o D + B
(2)
Basis: 1 mol of feed gas
1.0 mol
x AO (mol A / mol)
n A (mol A)
x BO (mol B / mol)
n B (mol B)
x IO (mol I / mol)
nC (mol C)
n D (mol D)
n I (mol I)
T ( $ C)
Fractional conversion:
fA
mol A consumed
mol A feed
9-24
x AO n A
Ÿ nA
x AO
x AO (1 f A )
9.24 (cont'd)
C generated: n0
Ÿ nC
x A0 (mol A fed)
f A (mol A consumed) YC (mol C generated)
mol A fed
mol A consumed
x AO f A YC
D generated: nD = 0.5 u mol C consumed = (1 2) u (mol A consumed mol C out)
Ÿ nD
(1 2)( x AO f A nC )
Balance on B: mol B out = mol B in mol B consumed in (1) + mol B generated in (2)
= mol B in mol A consumed in (1) + mol D generated in (2)
Ÿ n B x BO x AO f A n D
Balance on I: mol I out = mol I in Ÿ n I
x IO
b.
Species Formula
A
C2H4(v)
B
H2O(v)
C2H5OH(v)
C
D
C4H10)O(v
I
N2(g)
c.
9.25
a.
Tf
310
Tp
310
Species
A
B
C
D
I
n(in)
(mol)
0.537
0.367
0
0
0.096
Q(kJ) =
-1.31
DHf
52.28
-241.83
-235.31
-246.75
0
a
0.04075
0.03346
0.06134
0.08945
0.02900
b
1.15E-04
6.88E-06
1.57E-04
4.03E-04
2.20E-05
xB0
0.367
xI0
0.096
xA0
0.537
H(in)
(kJ/mol)
68.7
-231.9
-211.2
-204.2
9.4
n(out)
(mol)
0.510
0.341
0.024
0.001
0.096
c
-6.89E-08
7.60E-09
-8.75E-08
-2.24E-07
5.72E-09
fA
0.05
d
1.77E-11
-3.59E-12
1.98E-11
0
-2.87E-12
YC
0.90
H(out)
(kJ/mol)
68.7
-231.9
-211.2
-204.2
9.4
For T f = 125o C, Q = 7.90 kJ . Raising Tp, lowering fA, and raising YC all increase Q.
CH 4 ( g) O 2 ( g) o HCHO(g) + H 2 O(g)
n3 (mol HCHO)
n4 (mol H2O)
10 L, 200 kPa
n0 (mol feed gas) at 25qC
0.851 mol CH4/mol
0.15 mol O2 /mol
n5 (mol CH4)
T (qC), P(kPa), 10L
Q (kJ)
9-25
9.25 (cont'd)
Basis : n0
200 kPa 1000 Pa 10 L 10 3 m 3
1 kPa
1 mol K
8.314 m 3 Pa
1L
298 K
0.8072 mol feed gas mixture
0.8072 mol feed gas mixture Ÿ (0.85)(0.8072) = 0.6861 mol CH 4 ,
Ÿ (0.15)(0.8072) = 0.1211 mol O 2
1 mol CH 4
0.1211 mol O 2 fed
1 mol O 2 fed
CH 4 consumed :
Ÿ n5
(0.6861 01211
.
) mol CH 4
0.5650 mol CH 4
1 mol HCHO
HCHO produced : n3
mol CH 4
.
01211
01211
.
mol CH 4 consumed
1 mol CH 4 consumed
0.1211 mol CH 4 consumed
1 mol H 2 O
H 2 O produced : n4
1 mol CH 4 consumed
Extent of reaction : [
( nO 2 ) out ( nO 2 ) in
0 0.1211
Q O2
1
0.1211 mol HCHO
.
01211
mol H 2 O
0.1211 mol
References : CH 4 (g), O 2 (g), HCHO(g), H 2 O(g), at 25o C
Substance
CH 4
z
T
U i
U in
nin
nout
mol kJ mol mol kJ mol
0.6861
0
0.5650
U 1
O2
01211
.
0
HCHO
01211
.
H 2O
01211
.
z
U out
U 2
U
3
T
(Cv ) i dT
25
(C p R ) i dT
1,2,3
i
25
Using (C p ) i from Table B.2 and R = 8.314 u 10 3 kJ / mol ˜ K:
U 1
U
(0.02599 T 2.7345 u 10 5 T 2 0.1220 u 10 8 T 3 2.75 u 10 12 T 4 0.6670) kJ / mol
U 3
(0.02515 T 0.3440 u 10 5 T 2 0.2535 u 10 8 T 3 0.8983 u 10 12 T 4 0.6309) kJ / mol
(0.02597 T 2.1340 u 10 5 T 2 2.1735 u 10 12 T 4 0.6623) kJ / mol
2
Q
100 J 85 s
s
1 kJ
8.5 kJ
1000 J
Table B.1
'H ro
('H fo ) HCHO ('H fo ) H2O ('H fo ) CH4
B
. ) (24183
. ) (74.85)g kJ / mol
b(11590
28288
. kJ / mol
'U ro
'H ro RT (
¦
Qi gaseous
products
282.88 kJ / mol ¦Q
i
)
gaseous
reactants
8.314 J
298 K (1 + 1 1 1)
1 kJ
10 3 J
mol K
9-26
282.88 kJ / mol
9.25 (cont'd)
Energy Balance :
Q ['U ro ¦ (n
i ) out (U i ) out
¦ (n
i ) in (U i ) in
. kJ / mol) +0.5650 U 1 01211
.
.
(0.1211)(28288
U 2 01211
U 3
Substitute for U 1 through U 3 and Q
u 10 5 T 2 0.09963 u 10 8 T 3 1926
u 10 12 T 4 4329
0 0.02088 T 1845
.
.
. kJ / mol
Solve for T using E - Z Solve Ÿ T 1091o C 1364 K
1L
0.8072 mol 8.314 m3 ˜ Pa 1364 K
mol ˜ K
10 L 10 3 m3
Ÿ P nRT / V
915 u 103 Pa 915 kPa
Add heat to raise the reactants to a temperature at which the reaction rate is significant.
b.
c.
9.26
Side reaction : CH 4 2O2 o CO2 2H 2 O. T would have been higher (more negative heat of
reaction for combustion of methane), volume and total moles would be the same, therefore
P nRT / V would be greater.
bg
bg
1
O 2 (g) o C 2 H 4 O g
2
C 2 H 4 3O 2 o 2CO 2 2H 2 O
a.
C2 H 4 g Basis: 2 mol C 2 H 4 fed to reactor
n 6 (mol CO2 )
n 7 (mol H 2 O(l ))
25°C
Qr (kJ)
heat
n1 (mol C 2H 4)
n2 (mol O 2 )
25°C
reactor
2 mol C2 H4
1 mol O2
450°C
n 3 (mol C 2H 4)
n 4 (mol O 2 )
separation
process
n 3 (mol C 2H 4)
n 4 (mol O 2 )
n 5 (mol C 2H 4O)
n 6 (mol CO2 )
n 7 (mol H 2 O)
450°C
25% conversion Ÿ 0.500 mol C 2 H 4 consumed Ÿ n 3
70% yield Ÿ n5
150
. mol C 2 H 4
0.500 mol C 2 H 4 consumed 0.700 mol C 2 H 4 O
C balance on reactor:
Water formed: n 7
O balance on reactor:
1 mol C 2 H 4
. g b2gb0.350g n
b2gb2g b2gb150
0.300 mol CO 2
b2gb1g
1 mol H 2 O
1 mol CO 2
n 5 (mol C 2H 4O(g))
25°C
6
Ÿ n6
0.350 mol C 2 H 4 O
0.300 mol CO 2
0.300 mol H 2 O
0.375 mol O
b gb g
0.300 b2gb0.350g Ÿ n 0.500 mol C H
n
b2gb0.300g b0.300g b0.350g Ÿ n 0.625 mol O
2n 4 0.350 2 0.300 0.300 Ÿ n4
Overall C balance: 2n1
n 6 2n5
Overall O balance: 2n 2
2n 6 n 7
1
5
2
2
2
9-27
4
2
9.26 (cont'd)
Feed stream: 44.4% C 2 H 4 , 55.6% O 2
Reactor inlet: 66.7% C 2 H 4 , 33.3% O 2
Recycle stream: 80.0% C 2 H 4 , 20.0% O 2
Reactor outlet: 53.1% C 2 H 4 , 13.3% O 2 , 12.4% C 2 H 4 O, 10.6% CO 2 , 10.6% H 2 O
0.350 mol C 2 H 4 O 44.05 g
Mass of ethylene oxide
b.
1 kg
bg
0.0154 kg
10 3 g
1 mol
bg bg
References for enthalpy calculations : C s , H 2 g , O 2 g at 25q C
bg
H i T
'H ofi z
z
T
25
'H 0f C p dT for C 2 H 4
T 273
298
C p dT for C 2 H 4 O
'H ofi H i ( table B.8)
bg
for O 2 , CO 2 , H 2 O g
bg
'H of for H 2 O l
Overall Process
H in
nin
Substance
Reactor
H out
nout
(mol) (kJ / mol) (mol) (kJ / mol)
52.28
0.500
C2 H4
H out
(mol) (kJ / mol) (mol) (kJ / mol)
79.26
150
.
79.26
2
C2 H 4
O2
1
0350
.
5100
.
C2 H 4 O
0.350
19.99
0300
.
3935
.
CO 2
0.300
374.66
0300
.
28584
.
H2O g
0.300
226.72
0625
.
C2 H4 O
CO2
H2 O l
0
bg
¦ n H ¦ n H
'H
Energy balance on process: Q
i
i
i
out
1337
.
0.375
1337
.
248 kJ
i
in
¦ n H ¦ n H
'H
Energy balance on reactor: Q
i
out
c.
nout
O2
bg
H in
nin
substance
i
i
236 kJ
i
in
Scale to 1500 kg C 2 H 4 O day :
C 2 H 4 O production for initial basis
Ÿ Scale factor
(0.350 mol)(
1500 kg day
0.01542 kg
44.05 kg
10 3 mol
0.01542 kg C 2 H 4 O
)
9.73 u 10 4 day 1
U|
V| M b0.500gb28.05 g C H molg b0.625gb32.0 g O molg
W = 34.025 u 10 kg
kgje9.73 u 10 day j 3310 kg day (44.4% C H , 55.6% O )
day j 1 day 1 hr
1 kW
279 kW
In initial basis, fresh feed contains
0.500 mol C 2 H 4
0.625 mol O 2
2
Qprocess
e34.025 u 10
b248 kJ ge9.73 u 10
Qreactor
b236 kJge9.73 u 10
Fresh feed rate
4
2
3
3
4
1
2
4
1
24 hr 3600 s 1 kJ s
4
day 1
j
1 day
1 hr
1 kW
24 hr 3600 s 1 kJ s
9-28
265 kW
4
2
9.27
a.
Basis:
1200 lb m C 9 H 12
1 lb - mole
h
120 lb m
Overall process :
n 1 (lb-moles/h)
0.75 C 3H 6
0.25 C 4H 10
10.0 lb - moles cumene produced h
n 3 (lb-moles C3 H 6 /h)
n 4 (lb-moles C4 H10 /h)
10.0 lb-moles C9 H12 /h
n 2 (lb-moles C6 H 6 /h)
bg
bg
bg
b
Benzene balance: n 2
binput
consumption
g
output consumption
h
1 mole C 9 H12 produced
10.0 lb - moles C 6 H 6
78.1 lb m C 6 H 6
h
1 lb - mole
1 mole C 3 H 6
h
1 mole C 9 H12
g
b
Mass flow rate of C 3 H 6 / C 4 H 10 feed
UV Ÿ n 16.67 lb - moles h
2.50 lb - moles C H h
n
gW
b0.75gb16.67glb - moles C H 42.08 lb C H
1
3
b0.25gb16.67glb - moles C H
4
10
m
58.12 lb m C 4 H 10
1 lb - mole
h
1 mole fresh feed
10.0 lb-moles C9H12/h
2.50 lb-moles C3H6/h
4.17 lb-moles C4H10/h
30.0 lb-moles C6H6/h
400oF
40 lb - moles C 6 H 6 h
46.7 lb-moles/h
21.4% C9H12
5.4% C3H6
8.9% C4H10
64.3% C6H6
6.67 lb - moles h
UV Ÿ 37.5%
C H
hW
62.5% C H
2.50 lb - moles C 3 H 6 h
4.17 lb - moles C 4 H 10
3
6
4
10
Heat exchanger :
Reactor effluent at 400°F
10.0 lb-moles C9H12 /h
2.50 lb-moles C3H6 /h
4.17 lb-moles C4H10 /h
30.0 lb-moles C6H6 /h
200°F
40.0 lb-moles C6H6 /h
77°F
T (°F)
Energy balance: 'H
(Assume adiabatic)
0Ÿ
9-29
ni H i , out H i , in
¦ e
6
768 lb m h
a3 1fmoles fed to reactor
Overhead from T1 Ÿ
3
1 lb - mole
10.0 lb - moles fresh feed
40.0 lb-moles C6H6/h
6
6
h
h
16.67 lb-moles/h @ 77oF
0.75 C3H6
0.25 C4H10
b.
3
3
Reactor :
Benzene feed rate
781 lb m C 6 H 6 h
10.0 lb - moles C 9 H12
Ÿ 0.75n1 n3 10
20% C 3 H 6 unreacted Ÿ n3 0.20 0.75n1
39520 Btu lb - mole
10.0 lb - moles C 9 H12 produced 1 mole C 6 H 6 consumed
Propylene balance: 0.75n1 n3 binput
g
'H r 77q F
C 3 H 6 l C 6 H 6 l o C 9 H 12 l ,
j ¦ n C bT
i
pi
out
Tin
g
i
0
9.27 (cont'd)
0Ÿ
Energy balance: 'H
¦ n e H
i
i , out
H i , in
j ¦ n C bT
i
pi
out
Tin
g
0
i
(Assume adiabatic)
LM10 lb - moles C H
h
N
9
C 3H 6
120 lb m
12
OPe200 F 400 Fj b2.B50gb42.08gb0.57ge200 F 400
˜ FQ
0.40 Btu
1 lb - mole
$
$
$
$
1b m
C 4 H10
B
b gb
gb ge
j b A gb
gb ge
j
4.17 5812
.
. 0.45 200 $ F 400$ F
0.55 200$ F 400$ F 30.0 7811
C6H6 in
effluent
b A gb
gb ge
j
40.0 7811
. 0.45 T 77 $ F
0Ÿ T
323q F
C6H6 fed
to reactor
(Refer to flow chart of Part b: T 323q F )
References : C 3 H 6 l , C 4 H 10 l , C 6 H 6 l , C 9 H 12 l at 77q F
H i Btu lb - mole C pi Btu lb m ˜q F M i lb m lb - mole T 77 q F
b
bg
g
bg
b
bg
H in
n in
Substance
bg
g b
gb
gb g
H out
n out
(lb - mole / h) (Btu / lb - mole) (lb - mole / h) (Btu / lb - mole)
0
2.50
7750
12.0
C3H 6
C 4 H10
4.17
0
4.17
10330
C6H 6
40.0
C 9 H12
8650
30.0
11350
10.0
15530
Energy balance on reactor :
n C 9 H12 'H ro
Q 'H
ni H i vC 9 H12
out
¦
¦ n H
i
i
in
b10.0gb39520g b2.50gb7750g b4.17gb10330g b30.0gb11350g b10.0gb15530g
b1g
b40.0gb8650g 183000 Btu h b heat removalg
9.28
Basis :
a.
100 kg C 8 H 8 10 3 g
h
1 kg
1 mol
104.15 g
960 mol h styrene produced
C 8 H 10 (g) o C 8 H 8 (g) H 2 (g)
Overall system
n 2 (mol H2 /h)
Fresh feed
n 1 (mol C8H10/h)
960 mol C8H8 /h
Fresh feed rate: n1
bC H
8
10
balance
g
H 2 balance : n 2
960 mol C 8 H 8 1 mol C 8 H10
h
960 mol C 8 H10
h
1 mol C 8 H 8
1 mol H 2
1 mol C 8 H10
9-30
960 mol C 8 H10 h fresh feed
960 mol H 2 h
9.28 (cont'd)
Reactor :
.
n5 (mol C8H10 /h)
v
n 4 (mol H2O(v)/s)
960 (mol C8H8 /s)
960 (mol H2 /s)
560°C
n 3 (mol C8H10 /h)
n 4 (mol H2O( v )/h)
600°C
Qc (kJ/h)
35% 1-pass conversion Ÿ
Ÿ n 3
b
0.35n3 mol C 8 H 10 react
2740 mol C 8 H10
h
h fed to reactor
a2740 960f
Ÿ Recycle rate
g
1 mol C 8 H 8
1 mol C 8 H 10
960 mol C 8 H 8 h
1780 mol C 8 H10 h recycled
Reactor feed mixing point
2740 mol C8H10(v)/h
500oC
2740 mol C8H10(v)/h
n4 [mol H2O(v)/h]
600oC
n4 [mol H2O(v)/h]
700oC
2740 'H C8H10 n 4 'H H 2 O
Energy balance: 'H
b Neglect Q, 'E g
k
LM
OP J 1 kJ
MM b118 0.30T gdT PP mol ˜ C u 10 J
N
Q
z
'H C8 H10
'H H 2 O
b g
0 kJ h
600
$
500
28.3 kJ mol
3
Cp
Table B.8
Ÿ
3.9 kJ mol
P 1 bar
a2740fa28.3f n a3.9f
0 Ÿ n 4
4
1.99 u 10 4 mol H 2 O / h
bg
Ethylbenzene preheater A :
b.
bg
960 mol fresh feed 1780 mol recycled 2740 mol EB l
at 25q C
h
h
h
2740 mol EB v
Ÿ
at 500q C
h
136
500
'H
C pi dT 'H v 136q C C pv dT 20.2 36.0 77.7 kJ mol 133.9 kJ mol
bg
z
a
25
Q A
'H
f
z
a
136
2740 mol C 8 H10
133.9 kJ
h
mol C 8 H10
f
b
3.67 u 10 5 kJ h preheater
bg
19400 mol h H Obl, 25q Cg o 19400 mol h H Ob v, 700q C, 1 atmg
Table B.5 Ÿ H bl, 25q Cg 104.8 kJ kg ;
Table B.7 Ÿ H b v, 700q C, 1 atm | 1 bar g 3928 kJ kg
Steam generator F :
2
2
9-31
g
9.28 (cont'd)
Q F
19400 mol H 2 O 18.0 g
'H
1.34 u 10
a3928 104.8fkJ
3
h
6
1 kg
1 mol 10 g
b
kJ h steam generator
kg
g
bg
Reactor C :
bg
bg bg
bg
References: C8 H 8 v , C8 H 10 v , H 2 g , H 2 O v at 600q C
e
H i 560 $ C
j
zd
560
600
i
C pv i dT for C 8 H 10 , C 8 H 8
| H (T) for H 2 , H 2 O (interpolating from Table B.8)
n in
n out
Substance
H
H
out
in
C 8 H10
(mol h ) (kJ mol) (mol h ) (kJ mol)
0
1780
11.68
2740
0
19900
C8H8
19900
960
1.56
10.86
H2
960
119
.
H2O
Energy balance :
Q c
'H
960 mol C 8 H 8 produced
124.5 kJ
h
1 mol C 8 H 8
a
5.61 u 10 4 kJ h reactor
c.
¦ n H ¦ n H
i
out
i
i
in
H2 1
O2 o H 2O
2
CH 3 OH
O2 , N 2
H2
product gas
145°C
separation
units
a.
n f (mol/h) at 145°C, 1 atm
0.42 mol CH 3 OH/mol
0.58 mol air/mol
0.21 mol O2 /mol air
0.79 mol N2 /mol air
n s mol H2 O(v )/h
saturated at 145°C
b.
i
This is a poorly designed process as shown. The reactor effluents are cooled to 25$ C , and
then all but the hydrogen are reheated after separation. Probably less cooling is needed, and
in any case provisions for heat exchange should be included in the design.
CH 3OH o HCHO H 2 ,
9.29
f
reactor
reactor
product gas, 600°C
n 1 (mol CH3 OH/h)
waste
n 2 (mol O 2 /h)
heat
n 3 (mol N 2 /h)
boiler
n 4 (mol HCHO/h)
0.37 kg HCHO/h
n 5 (mol H 2 /h)
0.63 kg H 2O/h
n 6 (mol H 2 O/h)
mb (kg H2 O(v)/h)
mb (kg H2 O(v)/h)
30°C
sat'd at 3.1 bars
In the absence of data to the contrary, we assume that the separation of methanol from
formaldehyde is complete.
Methanol vaporizer:
bg
e j
The product stream, which contains 42 mole % CH 3OH v , is saturated at Tm $ C and 1 atm.
9-32
9.29 (cont'd)
b g b gb
pm Tm Ÿ 0.42 760 mmHg
ym P
 o p m
Antoine equation
c.
g
b g
319.2 mmHg = pm Tm
319.2 mmHg Ÿ Tm
44.1$ C
Moles HCHO formed :
36 u 106 kg solution 0.37 kg HCHO
350 days
1 kg solution
1 kmol
1 day
30.03 kg HCHO
24 h
but if all the HCHO is recovered, then this equals n4 , or n4
52.80
kmol HCHO
h
52.80 kmol HCHO h
70% conversion :
52.80 kmol HCHO
1 kmol CH 3OH react
h
Ÿ n f
1 kmol CH 3OH fed
1 kmol feed gas
1 kmol HCHO formed 0.70 kmol CH 3OH react 0.42 kmol CH 3OH
n f
179.59 kmol h
Methanol unreacted:
n1
b0.42gb179.59gkmol CH OH fed b1 0.70g kmol CH OH fed
3
3
h
1 kmol CH 3OH fed
b179.6 kmol hgb0.58gb0.79g
N 2 balance: n3
22.63
kmol CH 3OH
h
82.29 kmol N 2 h
Four reactor stream variables remain unknown — n s , n2 , n5 , and n6 — and four relations are
available — H and O balances, the given H 2 content of the product gas (5%), and the energy
balance. The solution is tedious but straightforward.
b
gb gb g
H balance: 179.6 0.42 4 2ns
Ÿ n s
b
b22.63gb4g b52.8gb2g 2n 2n
5
n 5 n 6 52.80
gb gb g b
(1)
g
b gb g
O balance: 179.6 0.42 1 179.6 (0.58) 0.21 2 n s
Ÿ n s
(2)
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25q C
H
'H fo z
Table B.2
T
25
(22.63)(1) 2n 2 (52.80)(1) n 6
2n 2 n 6 43.75
n 5
22.63 n 2 82.29 52.89 n 5 n 6
H 2 content:
6
B
C p dT
or Table B.8 for O 2 , N 2 and H 2
9-33
0.05 Ÿ 19n 5 n 2 n 6
157.72
(3)
9.29 (cont'd)
substance
n in
kmol / h
H in
kJ / kmol
n out
kmol / h
H out
kJ / kmol
CH 3 OH
75.43
195220
22.63
163200
O2
N2
H 2O
HCHO
.
2188
82.29
3620
3510
n2
82.29
ns
237740
n6
52.80
18410
17390
220920
88800
n5
16810
H2
Energy Balance :
'H
¦n H ¦n H
i
i
out
i
0 Ÿ 18410n2 16810n5 220920n6 237704ns
i
7.406 u 106
(4)
in
We now have four equations in four unknowns. Solve using E-Z Solve.
bg
n s
58.8 kmol H 2 O v
h
18.02 kg
1 kmol
n2
2.26 kmol O 2 h , n5
1060 kg steam fed h
. kmol H 2 h , n6
1358
98.00 kmol H 2 O h
Summarizing, the product gas component flow rates are 22.63 kmol CH3OH/h, 2.26 kmol O2/h,
82.29 kmol N2/h, 52.80 kmol HCHO/h, 13.58 kmol H2/h, and 98.02 kmol H2O/h
Ÿ
d.
272 kmol h product gas
8% CH 3 OH, 0.8% O 2 , 30% N 2 , 19% HCHO, 5% H 2 , 37% H 2 O
Energy balance on waste heat boiler. Since we have already calculated specific enthalpies of all
components of the product gas at the boiler inlet (at 600qC), and for all but two of them at the
boiler outlet (at 145qC), we will use the same reference states for the boiler calculation
bg b g b g b g
H Oblg at triple point for boiler water
Reference States: C s , H 2 g , O 2 g , N 2 g at 25q C for reactor gas
2
Substance
CH 3 OH
n in
kmol / h
22.63
H in
kJ / kmol
163200
n out
mol
22.63
H out
kJ / mol
195220
2.26
82.29
98.02
18410
17390
220920
88800
2.26
82.29
98.02
3620
3510
237730
111350
O2
N2
H 2O
52.80
13.58
HCHO
H2
H 2O
mb
( kg / h)
16800
52.80
13.58
125.7
(kJ / kg)
mb
(kg / h)
Energy Balance :
'H
¦ n H ¦ n H
i
out
b
i
i
in
i
0
g
Ÿ mb 2726.1 125.7 4.92 u 10 6
Ÿ mb
0
1892 kg steam h
9-34
3550
2726.1
( kJ / kg )
9.30
a.
C 2 H 4 HCl o C 2 H 5Cl
bg
Basis:
1600 kg C 2 H 5Cl l
103 g
1 mol
h
1 kg
64.52 g
C
n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4( g)/h)
condenser
n 5 (mol C 2H 6( g)/h)
n 6 (mol C 2H 5Cl( g)/h)
50°C
A
n 1 (mol HCl(g)/h)
0°C
n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4( g)/h)
n 5 (mol C 2H 6( g)/h)
0°C
n 6 (mol C 2H 5Cl( g)/h)
reactor
B
24800 mol h C 2 H 5Cl
n 2 (mol/h) at 0°C
0.93 C 2H 4
0.07 C 2H 6
D
( n 6 – 24,800) (mol C 2H 5 Cl( l)/h)
0°C
24,800 mol C2 H5 Cl(l )/h
Product composition data:
n3
0.015n1
n4
0.015 0.93n2
n5
0.07n2
b
g
b1g
b2 g
b3g
0.01395n2
Overall Cl balance :
b
n1 mol HCl h
g
1 mol Cl
1 mol HCl
bn gb1g b24800gb1g
b4g
3
Solve (4) simultaneously with (1) Ÿ n1
25180 mol h
2518
. kmol HCl fed / h
bg
378 mol HCl g h
n3
Overall C balance :
b gb g b gb g
n2 0.93 2 n2 0.07 2
b gb
g
2n4 2n5 2 24800
LM
N
From Eqs. (2) and (3) Ÿ 2n2 0.93 0.07 0.0139 0.07
b.
c.
n2
27070 mol fed h
n3
378 mol HCl h
n4
0.01395 27070 = 378 mol C 2 H 4 h
n5
0.07 27070 = 1895 mol C 2 H 6 h
b
b
g
OP b2gb24800g
Q
27.07 kmol h of Feed B
g
bg
bg
U| 2.65 kmol / h of Product C
V| 14.3% HCl, 14.3% C H , 71.4% C H
W
2
bg
bg
References : C 2 H 4 g , C 2 H 6 g , C 2 H 5 Cl g , HCl g at 0 $ C
e
j
e
j
C 2 H 4 g, 50$ C : H
C 2 H 4 g, 50$ C : H
z
z
50
C p dT
Table B.2
Ÿ
2.181 kJ mol
0
50
C p dT
Table B.2
Ÿ
2.512 kJ mol
0
9-35
4
2
6
9-30 (cont’d)
e
z
j
HCl g, 50$ C : H
Table B.2
50
Ÿ 1.456 kJ mol
C p dT
0
e j ' H e0 Cj 24.7 kJ mol
C H Cleg, 50 Cj: H
C dT 2.709 kJ mol
C 2 H 5Cl l, 0$ C : H
$
v
z
$
2
5
50
pv
0
HCl
nin
mol
25180
H in
kJ / mol
0
nout
mol
378
H out
kJ / mol
1456
.
C2 H 4
25175
0
378
2.181
C2 H 6
1895
0
1895
2.512
C 2 H 5Cl
n6 24800
24.7
n6
2.709
substance
Energy balance:
'H
Ÿ
0Ÿ
e j
nA 'H r 0$ C
QA
¦ n H ¦ n H
i
i
i
out
b25180 378gmol HCl react
64.5 kJ
h
1 mol HCl
b
gb
2.627n 6 n 6 248000 24.7
g
b gb
g b gb
g b gb
.
378 1456
378 2.181 1895 2.512
0 Ÿ n6
g
282475 mol C 2 H 5 Cl h in reactor effluent
282475 mol condensed 24800 mol product
h
h
kmol recycled
257.6
h
C 2 H 5Cl recycled
d.
0
i
in
257600
mol
h
C p is a linear function of temperature.
'H v is independent of temperature.
100% condensation of ethylbenzene in the heat exchanger is assumed.
Heat of mixing and influence of pressure on enthalpy is neglected.
Reactor is adiabatic.
No C2H4 or C2H6 is absorbed in the ethyl chloride product.
9.31
a.
4NH3(g) + 5O2(g) Æ 4NO(g) +6H2O(g)
'H ro
904.7 kJ / mol
Basis : 10 mol/s Feed gas
4 mol / s NH 3
6 mol / s O 2
n3 (mol O 2 )
n4 (mol NO)
Tin = 200o C
n5 (mol H 2 O)
Tout
O 2 consumed :
5 mol O 2
4 mol NH 3 fed
4 mol NH 3
s
NO produced : n 4
5 mol / s Ÿ n 3
4 mol NO produced 4 mol NH 3 fed
4 mol NH 3
s
9-36
(6 1) mol O 2 / s
= 4 mol NO / s
1 mol O 2 /
9.31 (cont'd)
6 mol H 2 O produced 4 mol NH 3 fed
H 2 O produced : n5
4 mol NH 3
Extent of reaction : [ =
s
(n NH 3 ) out (n NH 3 ) in
04
Q NH 3
4
= 6 mol H 2 O / s
1 mol / s
b.
Well-insulated reactor, so no heat loss
No absorption of heat by container wall
Neglect kinetic and potential energy changes;
No shaft work
No side reactions.
c.
References : NH 3 ( g), O 2 (g), NO(g), H 2 O(g) at 25o C, 1atm
Substance
NH 3 (g)
O 2 ( g)
NO(g)
nin
nout
H in
H out
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
4.00
H 1
6.00
100
.
H2
H 3
4.00
H
4
H 2 O(g)
z
H 5
6.00
Table B.2
200
H 1
(C p ) NH 3 dT
B
Table B.8
6.74 kJ / mol,
H 2
H O 2 (200 o C)
B
5.31 kJ / mol
25
Using (C p ) i from Table B.2 :
H 3
(0.0291 Tout 0.5790 u 10 5 Tout 2 0.2025 u 10 8 Tout 3 0.3278 u 10 12 Tout 4 0.7311) kJ / mol
H 4
(0.0295 Tout 0.4094 u 10 5 Tout 2 0.0975 u 10 8 Tout 3 0.0913 u 10 12 Tout 4 0.7400) kJ / mol
H 5
(0.03346 Tout 0.3440 u 10 5 Tout 2 0.2535 u 10 8 Tout 3 0.8983 u 10 12 Tout 4 0.8387) kJ / mol
Energy Balance: 'H
'H
[ 'H ro 5
¦
0
(ni ) out ( H i ) out i 3
Ÿ 'H
E
'H
2
¦ (n
i ) in ( Hi ) in
i 1
[ 'H ro (1.00) H 3 (4.00) H 4 (6.00) H 5 (4.00) H 1 (6.00) H 2
o
Substitute for [ , 'H r , and H 1 through H 6
(0.3479 Tout 4.28 u 10 5 Tout 2 2.114 u 10 8 Tout 3 4.697 u 10 12 Tout 4 ) 972.24 kJ / mol
E - Z Solve Ÿ Tout
2261 o C
d.
z
T
If only the first term from Table B.2 is used, H i
(C pi )dT
C pi (T 25)
25
H 1
H
4
0.03515(200 25)
0.0295(Tout
615
. kJ / mol, H 2
25), H 5
5.31 kJ / mol, H 3
0.03346(Tout 25)
9-37
0.0291(Tout 25),
9-31 (cont’d)
E. B. 'H
[ 'H ro (1.00) H 3 (4.00) H 4 (6.00) H 5 (4.00) H 1 (6.00) H 2
E
o
Substitute for [ ( = 1 mol / s), 'H r (
0 = 0.3479 Tout 969.86 Ÿ Tout
e.
9.32
0
904 .7 kJ / mol) and H 1 through H 6
2788 o C Ÿ % error =
2788 o C 2261o C
2261 o C
u 100
23%
If the higher temperature were used as the basis, the reactor design would be safer (but more
expensive).
Basis : 100 lb m coke fed
Ÿ 84 lb m C Ÿ 7.00 lb - moles C fed Ÿ 7.00 lb - moles CO 2 fed
7.00 lb-moles CO2
400°F
7.00 lb-moles(84 lbm)C/hr
16 lb mash/hr
77°F
585,900 Btu
a.
bg
bg
bg
e77 Fj e'H j b g 2e'H j b g
393.50 b2gb 282.99g kJ
n 1 (lb-moles CO)
n 2 (lb-moles CO2 )
1830°F
n 3 lb-moles C( s )/hr
16 lb mash/hr
1830°F
C s CO 2 g o 2CO g ,
'H ro
$
o
c
25q C
o
c
CO 2 g
CO g
0.9486 Btu 453.6 mols
mol
Let x
1 lb - mole
74,210 Btu lb - mole
fractional conversion of C and CO 2 :
E
b
7.00 x lb - moles C reacted
n1
1 kJ
b g
7.00b1 x g lb - moles Cbsg
g
2 lb - moles CO formed
1 lb - mole C reacted
14.0 x lb - moles CO
7.00 1 x lb - moles CO 2
n2
n3
bg
bg bg
$
References for enthalpy calculations: C s , CO 2 g , CO g , ash at 77 F
b
g
CO bg,1830q Fg: H
CObg,1830q Fg: H
CO 2 g,400q F : H
2
b
g
Solid 1830q F : H
H CO 2 (400$ F)
Table B.9
Ÿ
H CO 2 (1830$ F)
H CO (1830$ F)
0.24 Btu
lb m ˜$ F
3130 Btu lb - mole
Table B.9
Ÿ
20,880 Btu lb - mole
Table B.9
Ÿ 13,280 Btu lb - mole
b1830 77gq F
420 Btu lb m
Mass of solids (emerging)
b g
7.00 1 x lb - moles C
12.0 lb m
16 lb m
1 lb - mole
9-38
b100 84 xg lb
m
9.32 (cont'd)
nin
nout
H in
H out
(lb moles) (Btu lb - mole) (lb moles) (Btu lb - mole)
7.00
3130
7.00 1 x
20,890
substance
b g
CO 2
CO
(lb m )
(Btu lb m )
14.0 x
(lb m )
13,280
(Btu lb m )
solid
100
0
100 84 x
420
Extent of reaction: n CO
( n CO ) o Q CO[ Ÿ 14.0 x
2[ Ÿ [ ( lb - moles) = 7.0 x
Energy balance:
Q
[ 'H ro 'H
¦ n H ¦ n H
i
i
i
out
585,900 Btu
a fa f
a14.0 x fa13,280 f a100 84 x fa420 f a7.00 fa3130 f
E
7.0 x (lb - moles)
74,210 Btu
lb - mole
7.00 1 x 20,880
0.801 Ÿ 80.1% conversion
x
b.
i
in
Advantages of CO. Gases are easier to store and transport than solids, and the product of the
combustion is CO2, which is a much lower environmental hazard than are the products of
coke combustion.
Disadvantages of CO. It is highly toxic and dangerous if it leaks or is not completely burned,
and it has a lower heating value than coke. Also, it costs something to produce it from coke.
9.33
Basis :
17.1 m 3 10 3 L 273 K 5.00 atm
1 mol
a f
h
1 m 3 298 K 1.00 atm 22.4 L STP
CO g 2 H 2 g o CH 3OH g ,
bg
'H ro
bg
e'H j
o
f
bg
b g e 'H j
CH 3OH g
o
f
CO(g)
3497 mol h feed
90.68 kJ mol
3497 mol/h
n 1 (mol CH 3 OH /h)
0.333 mol CO/mol
n 2 (mol CO/h)
0.667 mol H 2/mol
n 3 (mol H 2 /h)
25°C, 5 atm
127°C, 5 atm
Q = –17.05 kW
Let f fractional conversion of CO (which also equals the fractional conversion of H 2 , since
CO and H 2 are fed in stoichiometric proportion).
f mol react
1166 f mol CO react
mol feed
1166 f mol CO react 1 mol CH 3 OH
CH 3OH produced : n1
1166 f mol CH 3 OH h
1 mol CO
CO remaining : n 2 1166 1 f mol CO h
CO reacted :
3497 0. 333 mol CO feed
a f
9-39
9.33 (cont'd)
b3497gb0.667g mol H fed 1166 f mol CO react
2332b1 f g mol H h
Reference states : CO(g), H bgg , CH OHbgg at 25qC
H 2 remaining : n3
2
2 mol H 2 react
1 mol CO react
2
3
2
H in
n in
Substance
H out
n out
bmol hg bkJ molg bmol hg bkJ molg
1166
0
1166a1 f f
H
2332
0
2332a1 f f
H
CO
1
H2
2
CH 3 OH
H 3
1166 f
B
Table B.8
e
$
e
$
j
H CO (127 $ C)
j
H H 2 (127 $ C)
CO g,127 C : H 1
2.99 kJ mol
B
Table B.8
H 2 g,127 C : H 2
z
C p dT
25
Energy balance : Q
'H
B
Table B.2
122
CH 3OH(g,127 C): H 3
$
2.943 kJ mol
[ 'H ro 5.009 kJ / mol
¦ n H ¦ n H
i
i
out
Ÿ
17.05 kJ 3600 s
s
1h
b
Ÿ 1102
u 10 5 f
.
n2
n3
n tot
9.34
a.
(1166 f )( 90.68)
7.173 u 10 4 Ÿ f
i
b
kJ
1166 1 f
h
g b2.993g 1166 f b5.009g bkJ hg
2332 1 f
n1
i
in
b
g b2.99g
g
0.651 mol CO or H 2 converted mol fed
b g 759.1 mol h
1166b1 0.651g 406.9 mol h
2332b1 0.651g 813.9 mol h
E
1166 0.651
1980
mol
Ÿ Vout
h
bg bg
b g
1980 mol 22.4 L STP
h
bg
1 mol
bg
CH 4 g 4S g o CS2 g 2 H 2 S g , 'H r 700q C
400 K 1.00 atm
1 m3
273 K 5.00 atm 103 L
13.0 m 3 h
274 kJ mol
Basis : 1 mol of feed
1 mol at 700°C
0.20 mol CH 4/mol
0.80 mol S/mol
Reactor
Q = –41 kJ
Product gas at 800°C
n1 (mol CS2)
n2 (mol H 2S)
n3 (mol CH 4)
n4 (mol S (v))
Let f fractional conversion of CH 4 (which also equals fractional conversion of S, since the
species are fed in stoichiometric proportion)
9-40
9.34 (cont'd)
0.20 f , Extent of reaction = [ (mol) = 0.20 f
Moles CH 4 reacted
b
g
n3
0.20 1 f mol CH 4
n4
0.80 mol S fed n1
n2
b
0.20 f mol CH 4 react
g
4 mol S react
1 mol CH 4 react
1 mol CS 2
0.20 f mol CH 4 react
1 mol CH 4
2 mol H 2 S
0.20 f mol CH 4 react
1 mol CH 4
b
g
0.80 1 f mol S
0.20 f mol CS 2
0.40 f mol H 2 S
bg
References: CH 4 (g), S g , CS2 (g), H 2 S(g) at 700qC (temperature at which 'H r is known)
substance nin
H in
bmolg bkJ molg bmolg bkJ molg
0.20
0
0.20b1 f g
H
0.80
0
0.80b1 f g
H
CH 4
1
S
2
CS 2
0.20 f
H 2S
0.40 f
H out
H out
nout
C pi 800 700 Ÿ
b
H 3
H
4
g
CH 4 g, 800q C : H 1 7.14 kJ / mol
S g, 800q C : H
3.64 kJ / mol
b
g
b
b
g
g
2
CS 2 g, 800q C : H 3
H S g, 800q C : H
2
4
3.18 kJ / mol
4.48 kJ / mol
Energy balance on reactor:
Q
'H
[ 'H r ¦ n H ¦ n H
i
out
i
i
i
in
41
kJ
s
b0.20 f gb274.0g 0.20b1 f gb7.140g 0.80b1 f gb3.640g 0.20 f b3180
. g 0.40 f b4.480g
b1g
Ÿ f
0.800
b.
0.04 mol CH4
0.16 mol S(l )
0.16 mol CS2
0.32 mol H2 S
200qC
0.20 mol CH4
0.80 mol S(l )
150°C
0.20 mol CH4
0.80 mol S( g)
T (°C)
Q (kJ)
preheater
0.04 mol CH4
0.16 mol S(g )
0.16 mol CS2
0.32 mol H2S
800°C
9-41
0.20 mol CH4
0.80 mol S(l )
700°C
9.34 (cont'd)
System: Heat exchanger-preheater combination. Assume the heat exchanger is adiabatic, so
that the only heat transferred to the system from its surroundings is Q for the preheater.
bg
References : CH 4 (g), S l , CS2 (g), H 2S(g) at 200qC
Substance
n in
H in
n out
H out
0.20
H 1
H
0.20
H 7
0.04
0
0.16
0.16
0
H8
0
0.32
0
bmolg bkJ molg bmolg bkJ molg
bCH g
bCH g
Sblg
Sbgg
4 150q ,700q
4 800q,200q
0.04
2
H 3
H
0.80
0.16
CS 2
0.16
H 2S
0.32
0.80
4
H 5
H
6
a
f
dC i a f aT 200f for Salf
dC i a f FGH 444.6 200IJK 'H bT g dC i b g aT 444.6f for Sbgg
H i
C pi T 200 for all substances but S
p Sl
p Sl
b
b
v
83.7 k J mol
Tb
g
g
CH 4 g, 150q C : H 1
CH g, 800q C : H
b
b
4
g
g
4
2
b
103.83 kJ / mol
b g ¦ n H ¦ n H
i
g
g
b
g
6
19.08 kJ / mol
26.88 kJ / mol
CH 4 g, 700q C : H 7 35.7 kJ / mol
S g, 700q C : H 8 100.19 kJ / mol
1.47 kJ / mol
out
b
b
CS 2 g, 800q C : H 5
H S g, 800q C : H
42.84 kJ / mol
Energy balance: Q kJ
c.
p Sg
3.57 kJ / mol
2
S l, 150q C : H 3
S g, 800q C : H
b
i
i
i
ŸQ
g
59.2 kJ Ÿ 59.2 kJ mol feed
in
The energy economy might be improved by insulating the reactor better. The reactor effluent will
emerge at a higher temperature and transfer more heat to the fresh feed in the first preheater,
lowering (and possibly eliminating) the heat requirement in the second preheater.
9-42
9.35
Basis : 1 mol C 2 H 6 fed to reactor
1 mol C H
2
6
1273 K, P atm
a.
n (mols) @ T (K), P atm
n C 2H6 (mol C 2H 6)
n C 2H4 (mol C 2H 4)
n H 2 (mol H 2)
x C2H 4 x H 2
C2 H 6 œ C2 H 4 H 2 , K p
b
7.28 u 10 6 exp[ 17,000 / T ( K )]
P
x C2 H 6
Fractional conversion = f mols C 2 H 6 react mol fed
U| x
b1 f gbmol C H g||
f b mol C H g
V| Ÿ x
f b mol H g
|| x
1 f b molsg
W
[ (mol)
C2 H 6
nH2
n
2
2
6
4
C2H 4
2
f
x C2H 4 x H 2
Kp
x C2 H 4
e1 f jK
b.
f2P
1 f 1 f
b
F K I
GH P K JK
12
gb
g
f
2
1 f
2
P
b2 g
p
2
p
2
2
f PŸ f
2
H2
b1 f g P
b1 f g
b1 f g
PŸ Kp
g
1 f mol C 2 H 6
1 f
mol
f mol C 2 H 4
1 f
mol
f mol H 2
1 f mol
f
n C2 H 6
n C2 H 4
(1)
p
bg
bg
bg
References : C 2 H 6 g , C 2 H 4 g , H 2 g at 1273 K
Energy balance:
e H j
e H j
i
i
b
g ¦ n H ¦ n H
0 Ÿ [ 'H r 1273 K 'H
i
b
0 inlet temperature
in
out
z
i
i
out
i
in
reference temperature
g
T
1273
C pi dT
energy balance
b
g b
f 'H r 1273 K kJ 1 f
z
g dC i
T
1273
p
dT f
z
T
1273
C2H 6
dC i
p C H dT
2 4
rearrange, reverse limits and change signs of integrals
1 f
f
b
g
'H r 1273K z
1273
z
dC i
dC i
p C H dT
2 4
T
z
1273
T
dC i
1273
T
p C H
2 6
dT
bg
I T
1 f
f
bg
I T Ÿ 1 f
bg
fI T Ÿ f
b g b4g
1
1 I T
9-43
p H dT
2
b3g
f
z
T
1273
dC i
p H dT
2
0
9.35 (cont'd)
145600 bg
IT
zb
g
1273
z
T
bg
c.
1273
3
T
1273
T
ŸI T
z
e26.90 4.167 u 10 T jdT
. 0.1392T gdT
b1135
9.419 0.1147T dT 3052 36.2T 0.05943T 2
127240 113
. T 0.0696T 2
F K I
F K I
GH 1 K JK 1 I1bT g Ÿ GH 1 K JK 1 I1bT g \ bT g 0
I bT g given by expression of Part b. K bT g given by Eq. (1)
12
12
p
p
p
p
p
d.
P
(atm)
0.01
0.05
0.1
0.5
1
5
10
T
(K)
794
847.4
872.3
932.8
960.3
1026
1055
f
0.518
0.47
0.446
0.388
0.36
0.292
0.261
Kp
(atm)
0.0037
0.0141
0.025
0.0886
0.1492
0.4646
0.7283
Phi
Psi
0.93152 -0.0001115
1.12964 -0.0002618
1.24028 0.00097743
1.57826
3.41E-05
1.77566
4.69E-05
2.42913 -2.57E-05
2.83692 -7.54E-05
Plot of T vs ln P
Plot of f vs. ln P
1100
0.6
0.5
1000
f
T(K)
0.4
900
0.3
0.2
800
0.1
0
700
-3
-2
-1
0
1
-3
2
-2
ln P(atm)
e.
-1
0
ln P(atm)
C **PROGRAM FOR PROBLEM 9-35
WRITE (5, 1)
1
FORMAT ('1', 20X, 'SOLUTION TO PROBLEM 9-35'//)
T 1200.0
TLAST 0.0
PSIL 0.0
9-44
1
2
9.35 (cont'd)
C **DECREMENT BY 50 DEG. AND LOOK FOR A SIGN IN PSI
DO 10I 1, 20
CALL PSICAL (T, PHI, PSI)
IF ((PSIL*PSI).LT.0.0) GO TO 40
TLAST T
PSIL PSI
T T – 50.
10
CONTINUE
40
IF (T.GE.0.0) GO TO 45
WRITE (3, 2)
2
FORMAT (1X, 'T LESS THAN ZERO -- ERROR')
STOP
C **APPLY REGULA-FALSI
45
DO 50 I 1, 20
IF (I.NE.1) T2L T2
T2 (T*PSIL-TLAST*PSI)/(PSIL-PSI)
IF (ABS(T2-T2L).LT.0.01) GO TO 99
CALL PSICAL (T2, PHIT, PSIT)
IF (PSIT.EQ.0) GO TO 99
IF ((PBIT*PBIL).GT.0.0) PSIL PSIT
IF ((PSIT*PSIL).GT.0.0) TLAST T2
IF ((PSIT*PSI).GT.0.0) PSI PSIT
IF ((PSIT*PSI).GT.0.0) T T2
50
CONTINUE
IF (I.EQ.20) WRITE (3, 3)
3
FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS')
93
STOP
END
SUBROUTINE PSICAL (T, PHI, PSI)
REAL KF
PHI (3052 36.2*T 36.2*T 0.05943*T**2)/(127240. – 11.35*T
* – 0.0636*T**2)
KP 7.28E6*EXP(-17000./T)
FBI SQRT((KP/(1. KP)) – 1./12. PHI)
WRITE (3, 1) T, PSI
1
FORMAT (6X, 'T ', F6.2, 4X, 'PSI ', E11,4)
RETURN
END
OUTPUT: SOLUTION TO PROBLEM 9-35
T 1200.00 PSI 0.8226E 00
T 1150.00 PSI 0.7048E 00
T 1100.00 PSI
T 1050.00 PSI
0.5551E 00
0.3696E 00
T
950.00
PSI
01619
.
E 00
0.3950E 01
T
959.80
PSI
0.1824 E 02
T
960.25
PSI
0.7671E 04
T
960.27
PSI
0.3278E 05
T 1000.00 PSI
Solution: T
960.3 K, f
0.360 mol C 2 H 6 reacted mol fed
9-45
2CH 4 o C 2 H 2 3H 2
9.36
C 2 H 2 o 2C(s) + H 2
bg
Basis: 10 mol CH 4 g fed s
n1 (mol CH 4 / s)
10.0 mol CH 4 (g) / s
o
n 2 (mol C 2 H 2 / s)
n 3 (mol C(s) / s)
1500 C
1500 o C
975 kW
a.
b
60% conversion Ÿ n1
g
10 1 0.600
4.00 mol CH 4 s
b g 4b1g 2n n Ÿ 2n n 6
10b4 g 4b4g 2n 2n Ÿ 2n 2n
C balance: 10 1
H balance:
2
4
2
2
3
(1)
4
2
24
3
(2)
bg
References for enthalpy calculations : C(s), H 2 g at 25qC
e
Hi
'H fo
Substance
bg
bg
H bg g
Cbsg
CH 4 g
C2 H 2 g
2
Energy Balance: Q
j
i
b
g
C pi 1500 25 , i
CH 4 , C 2 H 2 , C, H 2
H in
n in
( mol s) ( kJ mol) ( mol s) (kJ mol)
10
41.68
.
4
4168
303.45
n 2
45.72
n 3
'H Ÿ 975 kJ / s
n 4
n 2
Solve (1) - (3) simultaneously Ÿ n 3
n 4
b.
32.45
¦ n H ¦ n H
i
out
Yield of acetylene
H out
n out
i
i
i
(3)
in
2.50 mol C 2 H 2 / s
9.50 mol H 2 / s
100
. mol C / s
2.50 mol C 2 H 2 s
6.00 mol CH 4 consumed s
0.417 mol C 2 H 2 mol CH 4 consumed
If no side reaction,
n1
10.0(1 0.600)
n 3
0 Ÿ n 2
4.00 mol CH 4 / s
3.00 mol C 2 H 2 / s, n 4
9.00 mol H 2 / s
Yield of acetylene
3.00 mol C 2 H 2 s
6.00 mol CH 4 consumed s
Reactor Efficiency
0.417
0.500
0.834
9-46
0.500 mol C 2 H 2 mol CH 4 consumed
9.37
bg
bg
bg bg
CObgg H Ob vg o CO bgg H bgg
C 3 H 8 g 3H 2 O v o 3CO g 7H 2 g
2
2
2
Basis : 1 mol C 3 H 8 fed
Heating gas
4.94 m3 at 1400°C, 1 atm
n g (mol)
n g (mol), 900°C
4.94 m 3 10 3 L
ng
1m
273 K
3
Product gas, 800°C
n 1 (mol C 3H 8) = 0
n 2 (mol H 2O)
n 3 (mol CO)
n 4 (mol CO2 )
n 5 (mol H 2)
a
1 mol C 3H 8(g )
6 mol H 2 O( g )
125°C
1 mol
1673 K 22.4 L
35.99 mol heating gas
Let [ 1 and [ 2 be the extents of the two reactions.
n1 0
n1
1 [1 Ÿ [1
n2
6 3[ 1 [ 2 Ÿ n 2
1 mol
[1 1
[1 1
3[ 1 [ 2 Ÿ n 3
n3
n4
[2
n5
7[ 1 [ 2 Ÿ n5
[1 1
3 [ 2
7[2
3[2
bg bg
References : C(s), H 2 g , O 2 g at 25qC, heating gas at 900qC
z
T
H i
'H fio C pi dT
for C 3 H 8
25
Table B.8 for CO 2 , H 2 , H 2 O, CO
z
T
C p dT
b
g
C p T 900 for heating gas
900
n in
H in
n out
H out
C3H8
mol
1
kJ / mol
95.39
mol
0
kJ / mol
H 2O
6
238.43 3 [ 2
212.78
CO
3[2
86.39
CO 2
356.15
H2
[2
7[2
35.99
Substance
heating gas 35.99
200.00
22.85
0
Energy Balance :
¦ n H ¦ n H
i
i
i
i
out
in
n4
1 mol CO 2 , n5
0 Ÿ [2
2.00 mol Ÿ n 2
1 mol H 2 O, n 3
1 mol CO,
9 mol H 2 Ÿ 7.7 mol % H 2 O, 7.7% CO, 15.4% CO 2 , 69.2% H 2
9-47
9.38
a.
Any C consumed in reaction (2) is lost to reaction (1). Without the energy released by reaction
(2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic
reactor and hence the reaction rate would drop.
b.
Basis : 1.00 kg coal fed (+0.500 kg H20)
0.500 kg H20
Ÿ
1.0 kg coal
0.105 kg H2O/kg coal
0.226 kg ash/kg coal
0.669 kg combustible / kg coal
R|
|S0.812 kg C / kg combustible
||0.134 kg O / kg combustible
T0.054 kg H / kg combustible
nf1 (mol C)
nf2 (mol O)
nf3 (mol H)
nf4 (mol H2O)
0.226 kg ash
U|
|V
||
W
n f 1 = [ (1.00)(0.669)(0.812) kg C][1 mol C / 12.01 u 10 3 kg] = 45.23 mol C
n f 2 = (1.00)(0.669)(0.134) / 16.0 u 10 3
5.6 mol O
n f 3 = (1.00)(0.669)(0.054) / 1.01 u 10 3
35.77 mol H
n f 4 = [ (0.500 + 0.105) kg][1 mol H 2 O / 18.016 u 10 3 kg] = 33.58 mol H 2 O
n0 (mol O2) 25qC
Product gas at 2500qC
n1 (mol CO2)
n2 (mol CO)
n3 (mol H2)
n4 (mol H2O)
1 kg coal + H2O, 25qC
45.23 mol C
5.60 mol O
35.77 mol H
33.58 mol H2O
0.226 mol kg ash
0.226 kg slag
2500qC
Reactive oxygen (O) available
(2n 0 5.60) mol O
Oxygen consumed by H ( 2H + O o H 2 O) :
35.77 mol H 1 mol O
2 mol H
Ÿ Reactive O remaining = (2n0 5.60) 17.88
CO 2 formed ( C + 2O o CO 2 ) : n1
C balance : 45.23 = n1 n 2
O balance : 2n 0 5.60 3358
.
(2n0 12.28) mol O
(2n 0 12.28) mol O 1 mol CO 2
2 mol O
n1 n0 6.24
Ÿ
17.88 mol O
(n 0 6.24) mol CO 2
(51.5 n 0 ) mol CO
n2
2n1 n 2 n 4
n1 n0 6.24
Ÿ
n4
(n 0 0.16) mol H 2 O
n2 51.5 n0
H balance : 35.77 + 2(33.58) = 2n 3 2n 4
n4 n0 0.16
Ÿ
9-48
n3
(51.3 n 0 ) mol H 2
9.38 (cont'd)
c.
1 kg coal contains 45.23 mol C and 35.77 mol H
Ÿ 1 kg coal + nO 2 o 45.23 CO 2 (35.77 / 2) mol H 2 O (l)
'H r
21,400 kJ = 45.23( 'H fo ) CO 2 (35.77 / 2)( 'H fo ) H 2 O(l) ( 'H fo ) coal
Ÿ ( 'H fo ) coal
1510 kJ / kg
Re ferences : Coal, ash, CO, CO 2 , H 2 , H 2 O (l) at 25 o C
H in
nin
Substance
(mol)
( kJ / mol)
n 0 6.24
(mol)
CO 2
H2
H 2O
.
3358
0
Coal
Ash(slag)
1 kg
(in coal)
0
0
CO
H out
n out
. n0
515
( kJ / mol)
H 1
H
51.3 n 0
.
n 0 016
H 3
H
0.266 kg
H 5 ( kJ / kg)
2
4
H i
C pi (2500 25), i
H 1
0.0508(2475)
125.7 kJ / mol CO 2
H 2
0.0332(2475)
82.17 kJ / mol CO
H 3
0.0300(2475)
74.25 kJ / mol H 2
H 4
(C p ) H 2 O(l) (100 25) 'H v (100 o C) + (C p ) H 2 O(v) (2500 100)
1,3
(0.0754)(75) 40.656 kJ / mol + (0.0395)(2400) = 141.1 kJ / mol H 2 O
H 5
( 'H m ) ash 1.4(2475)
710 1.4(2475)
4175 kJ / kg ash
Reaction 1 : 1 kg coal + n1 O 2 o 45.23 mol CO 2 +17.88 mol H 2 O(l)
o
'H r1
21,400 kJ / 45.23 mol CO 2
Reaction 2 : 1 kg coal + n 2 O 2 o 45.23 mol CO + 17.88 mol H 2 O(l)
o
'H r2
45.23( 'H fo ) CO 17.88( 'H fo ) H 2 O(l) ( 'H fo ) Coal
8600 kJ / 45.23 mol CO
Energy Balance
'H
(n 0 6.24) mol CO 2 u (
¦n
out
H out ¦n
8600 kJ
21,400 kJ
) (51.5 n 0 ) mol CO u (
)
45.23 mol CO 2
45.23 mol CO
in H in
0
Ÿ 4731
. (n 0 6.24) 190.1(51.5 n 0 ) (n 0 6.24)125.7 (51.5 n 0 )82.17 (51.3 n0 )74.25
+ (n 0 016
. )1411
. (0.266)(4175)
Ÿ n0
0
9.0 mol O 2
9-49
9.39
Mass of H 2 SO 4
3 m3 103 L 1 mol H 2 SO 4
1 m3
L
Mass of solution
3 m3 103 L 103 mL
1 m3
L
FG mol H O IJ
H mol H SO K
e'H j b
161
. u 10 5 mol H 2 O
3000 mol H 2 SO 4
2
2
f
4
2.941 u 105 g H 2 SO 4
H 2SO 4 aq., r 53.6
e'H j
o
f
g
bg
H 2SO 4 l
A
53.6 mol H 2 O mol H 2 SO 4
e j
'H s
b
H 2SO 4 aq ., r 53.6
A
Table B.1
H
FG 98.02 gIJ
H 1 mol K
1.064 g
. u 10 6 g solution
3192
1 mL
1 mol
u 10 6 2.941 u 10 5 )g H 2 O(
(3192
.
) 1.61 u 10 5 mol H 2 O
18.02 g
Ÿ Moles of H 2 O
n
3000 mol H 2SO 4
kJ
b811.32 73.39g mol
g
884.7 kJ mol
Table B.11
(3000 mol H 2 SO 4 )(-884.7 kJ / mol H 2 SO 4 ) = -2.65 u 10 6 kJ
9.40
e'H j
HCl (aq): 'H fo
o
f
bg
HCl g
e
'H so
j
Tables B.1, B11
.
92.31 7514
f
167.45 kJ mol
Tables B1, B.11
e
NaOH (aq): 'H fo
j
'H fo
bg
NaOH s
e
'H so
j
B
426.6 42.89
f
B
Table B.1
NaCl (aq):
e
'H fo
b g
'H fo
b g
j
bg
NaCl s
e
'H so
b g
j
g b
HClbgg NaOHbsg o NaClbsg H Oblg
Given
. 4.87
4110
f
bg
HCl aq NaOH aq o NaCl aq H 2 O l
'H ro
406.1 28584
. 167.45 469.49
b
B
g
469.49 kJ mol
406.1 kJ mol
55.0 kJ mol
2
¦
'H ro
v i 'H fo products
¦
v i 'H fo
reactants
b
g b
g
411.0 285.84 92.31 426.6
kJ mol
177.9 kJ mol
The difference between the two calculated values equals
'H
'H
.
'H
{e j
s
9.41
a.
e j
s
NaCl
b g
HCl
e j
s
NaOH
b g
}
b g
bg
H 2 SO 4 aq 2NaOH aq o Na 2 SO 4 aq 2H 2 O l
Basis: 1 mol H 2 SO 4 soln Ÿ
Ÿ
Ÿ
0.10 mol H 2 SO 4 u 98.08 g mol
g
0.90 mol H 2 O u 18.02 g mol
16.22 g H 2 O
26.03 g soln 1 cm 3
0.10 mol H 2 SO 4
127
. g
b
b
g
9.808 g H 2 SO 4
U|
V|
W
20.49 cm 3
2 mol NaOH
1 mol H 2 SO 4
1 liter caustic soln 10 3 cm 3
3 mol NaOH
9-50
1L
b g
66.67 cm 3 NaOH aq
9.41 (cont'd)
66.67 cm 3 NaOH(aq)
Volume ratio
3.25 cm 3 caustic solution / cm 3 acid solution
3
20.49 cm H 2 SO 4 (aq)
b.
b g
9 mol H 2 O / 1 mol H 2 SO 4
H 2 SO 4 aq : r
kJ
877 kJ mol H SO
b811.32 65.23g mol
. g cm j 75.34 g , and
NaOH(aq) : The solution fed contains e66.67 cm je113
(0.2 mol NaOH)b40.00 g molg 8.00 g NaOH
Ÿ b75.34 8.00g g H O Ÿ b67.39 g H Ogb1 mol 18.02 gg 3.74 mol H O
e'H j
o
f
soln
e'H j
o
f
b g e 'H f j H SO baq., r 9 g
o
H 2 SO 4 l
2
2
3
2
Ÿr
soln
e'H j
Na SO baq g:
e'H j e'H j
2
4
o
f
o
f
o
f
o
f
soln
3
2
2
3.74 mol H 2 O 0.20 mol NaOH
e'H j
18.7 mol H 2 O / mol NaOH
b g e 'H s j NaOH bsgbaq., r
o
NaOH s
bg
Na 2SO 4 s
e
'H fo
j
b g
Na 2SO 4 aq
18.7
g
kJ
b426.6 42.8g mol
kJ
. g
b1384.5 117
mol
(n H 2SO4 ) fed Q H 2SO4 [ Ÿ 0
Extent of reaction: (n H 2SO ) final
4
469.4 kJ mol NaOH
1385.7 kJ mol Na 2 SO 4
0.10mol (1)[ Ÿ [
010
. mol
Energy Balance:
'H
Q
['H ro
[ ( 'H fo ) Na 2SO4 ( aq) 2( 'H fo ) H 2 O ( l) ( 'H fo ) H 2SO 4 ( aq) 2( 'H fo ) NaOH ( aq)
. ) ( 876.55) (2)( 469.4)
= (0.10 mol) 1385.7 + 2( 28584
9.42
kJ
mol
14.2 kJ
Table B.1,
given
a.
NaCl(aq): 'H fo
e'H j
o
f
bg
NaCl s
e
'H so
j
B
f
b411.0 4.87gkJ / mol
406.1 kJ mol
NaOH(aq):
Table B.1
B
b426.6 42.89gkJ / mol
e'H j b g e'H j
1
1
NaClbaq g H Oblg o H bgg Cl bgg NaOHbaq g
2
2
'H
469.5 b 406.1g b 28584
. g kJ mol 222.44 kJ mol
'H fo
2
o
f
o
s
NaOH s
2
f
469.5 kJ mol
2
o
r
b.
8500 ktonne Cl 2
yr
10 3 tonne 10 3 kg 10 3 g
1 ktonne
1 tonne
10 3 J 2.778 u 10 7 kW ˜ h
1 kJ
1J
4
4
1 kg
1 mol Cl 2
222.44 kJ
70.91 g Cl 2
0.5 mol Cl 2
1 MW ˜ h
10 3 kW ˜ h
9-51
148
. u 10 7 MW ˜ h / yr
9.43
a.
bg
bg
b
CaCl 2 s 10H 2 O l o CaCl 2 aq , r
bg
bg
g
b
g
˜ 6H Obsg
CaCl 2 ˜ 6H 2 O s 4 H 2 O l o CaCl 2 aq , r
2
2
o
r1
64.85 kJ mol
'H ro2
32.41 kJ mol
2
2
o
r2
o
r1
o
f
o
f
b
9.44
10
'H ro1
(3)
b1g b2g Ÿ CaCl bsg 6H Oblg o CaCl
Ÿ 'H
'H 'H b Hess' s law g 97.26 kJ mol
From (1), 'H
e'H j b g e'H j b g
Ÿ e 'H j
b
g b64.85 794.96g kJ mol 859.81 kJ mol
o
r3
b.
b1g
b2 g
10
Basis: 1 mol NH 4
o
f
CaCl 2 aq , r 10
CaCl 2 aq , r 10
g SO produced
4
2
2 mol NH3 (g)
75qC
1mol H2SO4 (aq)
25qC
bg
CaCl 2 s
1 mol (NH4)2SO4 (aq)
25qC
b g b
2NH 3 g H 2 SO 4 aq o NH 4
g SO baqg
4
2
a.
References : Elements at 25qC
b
z
g
b
'H fo C p dT
25
g
e'H j b
g SO baq, 25q Cg: H e'H jb
H 2 SO 4 aq , 25q C : H
bNH
I
FG 4619
. J kJ / mol
K
H . 183
75
NH 3 g, 75q C : H
4 2
o
f
H 2SO 4 aq
o
f
4
907.51 kJ mol H 2 SO 4 (Ta.ble B.1)
g
NH 4
44.36 kJ mol (Table B.1, B.2)
b
11731
. kJ mol NH 4
g SO b aq g
2
g SO
4
2
4
(Table B.1)
Energy balance:
Q
'H
¦ n H ¦ n H b1gb11731. g b2gb44.36g b1gb907.51g kJ
i
i
out
177 kJ Ÿ 177
b.
i
i
in
kJ withdrawn
mol NH 4 2 SO 4 produced
b
g
1 mole % (NH 4 ) 2 SO 4 solution Ÿ
1 mol (NH 4 ) 2 SO 4
132 g
mol
99 mol H 2 O 18 g
132 g (NH 4 ) 2 SO 4
1782 g H 2 O
mol
1914 g solution
The heat transferred from the reactor in part (a) now goes to heat the product solution from
25$ C to Tfinal Ÿ 177 kJ =
c.
1.914g
1 kg
=
4.184 kJ (T 25) $ C
3
10 g
$
kg C
Ÿ Tfinal
47.1$ C
In a real reactor, the final solution temperature will be less than the value calculated in part b, due
to heat loss to the surroundings. The final temperature will therefore be less than 47.1oC.
9-52
9.45
a.
b g
b g
b g
b g
H 2 SO 4 aq 2 NaOH aq o Na 2 SO 4 aq 2 H 2 O aq
1 mol H 2SO 4
49 mol H 2O
25°C
Basis : 1 mol H 2 SO 4 fed
1 mol Na 2SO 4
89 mol H 2O
40°C
2 mol NaOH
38 mol H 2O
25°C
bg b g bg b g
H SO eaq , r 49, 25 Cj:
L
O
'H b r 49gPb kJ molg 49e 'H j
nH b1 mol H SO gMe 'H j
b
g
N
Q
b1g 811.3 73.3 884.6 kJ 49e'H j b g
NaOHeaq , r 19, 25 Cj:
L
O
'H b r 19gPb kJ molg 38e 'H j
nH b2 mol NaOH gMe 'H j
bg
N
Q
b2g 426.6 42.8 938.8 kJ 38e'H j b g
Na SO eaq , r 89, 40 Cj:
Reference states : Na s , H 2 g , S s , O 2 g at 25qC
$
2
4
2
o
f
4
o
s
H 2SO 4 l
o
f
o
f
bg
H 2O l
H 2O l
$
o
f
o
s
NaOH s
o
f
o
f
bg
H 2O l
H 2O l
$
2
4
1 kmol Na 2 SO 4
142.0 kg
1 kmol
nH
b1 mol Na SO
2
4
gLMNe'H j
o
f
b
89 kmol H 2 O 18.02 kg
0.142 kg,
Na 2SO 4
kg Ÿ 1746
kg
1604
.
.
1 kmol
e
'H so
j
Na 2SO 4
g
OP 89e'H j
Q
o
f
b g mC p b40 25g
H 2O l
'H fo 1384.5 kJ mol Table B.1
'H so 1.2 kJ mol
4 .814 kJ (kg˜$ C)
m=1.746 kg, C p | C p
H 2O l
FH IK
nH
bg
e
1276 kJ 89 'H fo
'H
Energy balance: Q
j
bg
H 2O l
¦
ni H i out
Mass of acid fed
1 mol H 2 SO 4
98.08 g H 2 SO 4
1 mol
Ÿ
b.
Q
M acid
¦
285.84 kJ mol
e
ni H i
547.4 2 'H fo
j
in
49 mol H 2 O 18.02 g H 2 O
1 mol
bg
H 2O l
24.3 kJ
981 g = 0.981 kg
24.3 kJ
Ÿ 24.8 kJ / kg acid transferred from reactor contents
0.981 kg acid
If the reactor is adiabatic, the heat transferred from the reactor of Part(a) instead goes to heat
the product solution from 40qC to T f
Ÿ 24.3 u 10 J
3
kg 4.184 kJ
1746
.
$
kg ˜ C
9-53
dT
f
i
$
40 C
Ÿ Tf
43 $ C
9.46
a.
b g
b g
b g
bg
H 2 SO 4 aq 2NaOH aq o Na 2 SO 4 aq 2H 2 O l
H 2SO4 solution: :
75 ml of 4M H 2SO 4 solution Ÿ
4 mol H 2 SO 4
1L
1 L acid soln
3
75 mL
0.30 mol H 2SO 4
10 mL
b75 mLgb1.23 g mLg 92.25 g, (0.3 mol H SO )b98.08 g molg 29.42 g H SO
Ÿ b92.25 29.42g g H O Ÿ b62.83 g H Ogb1 mol 18.02 gg 3.49 mol H O
2
2
Ÿr
o
f
soln
e'H j
o
f
2
2
e j
'H fo
bg
H 2SO 4 l
4
2
3.49 mol H 2 O 0.30 mol H 2 SO 4
e'H j
4
11.63 mol H 2 O / mol H 2 SO 4
Table B.1,
Table B.11
B
b
kJ
b811.32 67.42g mol
g
H 2SO 4 aq ., r 11.63
878.74 kJ mol H 2 SO 4
NaOH solution required:
0.30 m ol H 2 SO 4 2 m ol NaOH
1 L NaOH (aq)
10 3 m L
1 m ol H 2 SO 4
12 m ol NaOH
1L
. g mLg
b50.00 mLgb137
68.5 g
12 mol NaOH
50 mL
1L
1 L NaOH(aq) 10 3 mL
b
g
40 g/ mol NaOH
Ÿ
0.60 mol NaOH
b
gb
Ÿ 68.5 24.00 g H 2 O Ÿ 44.5 g H 2 O 1 mol 18.02 g
Ÿr
o
f
soln
e'H j
o
f
g
24.00 g NaOH
2.47 mol H 2 O
4.12 mol H 2 O
mol NaOH
2.47 mol H 2 O 0.6 mol NaOH
e'H j
b g
50.00 m L NaOH aq
b g e'Hs j NaOHbsgbaq., r
o
NaOH s
4.12
kJ
. g
b426.6 3510
mol
g
46170
. kJ mol NaOH
b g
e'H j e'H j
Na 2SO 4 aq :
o
f
mtotal
o
f
soln
b g e'H f j Na SO baq g
o
Na 2SO 4 s
2
4
kJ
. g
b1384.5 117
mol
1385.7 kJ mol Na 2SO 4
total mass of reactants or products = (92.25g H 2SO4 soln + 68.5g NaOH) = 160.75g = 0.161 kg
Extent of reaction: (nH 2SO 4 ) final
(nH 2SO 4 ) fed Q H 2SO 4 [ Ÿ 0
0.30 mol (1)[ Ÿ [
0.30 mol
Standard heat of reaction
'H ro
e'H j
o
f
b g 2e'H f j H Oblg e'H f j H SO baq g 2e 'H f j NaOHbaq g
o
Na 2SO 4 aq
o
2
o
2
4
Energy Balance : Q 'H ['H ro mtotal C p (T 25) $ C
F
GH
(0.30 mol)(1552
. kJ / mol) (0161
. kg) 4.184
b.
Volumes are additive.
Heat transferred to and through the container wall is negligible.
9-54
I (T 25) C = 0 Ÿ T
J
kg C K
kJ
$
$
94 $ C
9.47
Basis : 50,000 mol flue gas/h
50,000 mol/h
0.00300 SO2
0.997 N 2
50°C
n4 (mol SO2 /h)
n5 (mol N 2 /h)
35°C
n1 (mol solution/h)
0.100 (NH 4 ) 2 SO 3
0.900 H 2O( l )
25°C
n2 (mol NH4 HSO3 /h)
1.5n2 (mol (NH 4)2 SO3 /h)
n3 (mol H 2 O(l )/h)
35°C
a
fb
g 15.0 mol SO h
n a0.997fb50,000 mol hg 49,850 mol N h
N balance:
. gbn g n b15
. gb2gn Ÿ n 20n U| n 5400 mol h
NH balance: b2gb0100
V Ÿ 270 mol NH HSO
. n b0.00300gb50,000g 150
. n 15
. n |W n
0100
S balance:
270 mol NH HSO produced
1 mol H O consumed
H O balance: n b0900
. gb5400g h
2 mol NH HSO produced
4725 mol H Oblg h
90% SO2 removal: n 4
0.100 0.00300 50,000 mol h
2
2
5
2
+
4
1
2
2
1
1
2
4
1
2
4
2
2
3
3
h
2
3
2
4
3
2
Heat of reaction:
e 'H j
e j
b g e 'H j b g b g e 'H j
2b 760g b890g b 296.90g b 285.84g kJ mol 47.3 kJ mol
References : N bgg, SO bgg, b NH g SO baq g, NH HSO baq g, H Oblg at 25qC
. kJ mol ( C from Table B.2)
SO eg, 50 Cj: H z dC i dT 101
'H ro
2 'H fo
o
f
NH 4 HSO 4 aq
2
2
NH 4
4 2
o
f
SO 3 aq
2
3
4
o
f
SO 2 ( g )
3
H 2 O(l)
2
50
$
2
e
zd
j
SO 2 g, 35$ C : H
e j
N eg, 35 Cj:
$
2
Cp
25
N 2 g, 50$ C : H
H
p
p SO
2
25
35
i
SO 2
0.40 kJ mol
dT
0.73 kJ mol (Table B.8)
0.292 kJ mol
Entering solution: H
0
Effluent solution at 35qC
b g
m g h
nH
270 mol NH 4 HSO 3
99 g
h
mol
b
g
1.5 u 270 mol NH 4 2 SO 3 116 g 4725 mol H 2 O 18 g
g
159,000
h
1 mol
h
h
mol
mC p 'T
159,000 g 4 J
h
g˜q C
Extent of reaction:
(n NH 4 HSO 3 ) out
a35 25fq C
1 kJ
10 3 J
6360 kJ / h
(n NH 4 HSO3 )in Q NH 4 HSO 3 [ Ÿ 270 mol / h = 0 + 2[ Ÿ [ = 135 mol / h
9-55
9.47 (cont'd)
Energy balance: Q
Q
9.48
a.
135 mol
47.3 kJ
h
mol
effluent solution
6360
bg
[ 'H ro 'H
¦ n H ¦ n H
i
i
out
i
in
N 2 out
a fa f a
fa
SO 2 out
15 0.40 49,850 0.292
a
fa f a
fa
i
f
fa f
50,000 0.003 1.01 49,850 0.73
bg
22,000 kJ
1h
h
1 kW
3600 s 1 kJ s
6.11 kW
bg
CH 4 g + 2O 2 g o CO 2 (g) 2H 2 O v
at 25qC
Table B.1 HHV
B
HHV
'H co
890.36 kJ / mol, LHV
B
e j
2 'H v
b
g
890.36 2 44.01 kJ mol
H 2O
802.34 kJ mol CH 4
bg
7
C 2 H 4 g + O 2 (g) o 2CO 2 (g) 3H 2 O(v)
2
HHV
a
f
a
f
1559.9 3 44.01 kJ mol 1427.87 kJ mol C 2 H 6
1559.9 kJ / mol, LHV
bg
C 3 H 8 g + 5O 2 (g) o 3CO 2 (g) 4H 2 O(v)
HHV
b HHV g
2220.0 4 44.01 kJ mol
2220.0 kJ / mol, LHV
natural gas
2043.96 kJ mol C 3 H 8
b0.875gb890.36 kJ molg b0.070gb1559.9 kJ molg b0.020gb2200.00 kJ molg
933 kJ mol
b LHV g
natural gas
b0.875gb802.34 kJ molg b0.070gb1427.87 kJ molg b0.020gb2043.96 kJ molg
= 843 kJ mol
b.
g I
g I
gFGH mol
JK
JK b0.070 mol C H gFGH 30.07 mol
g I
g I
1 kg
F
F
b0.020 mol C H gG 44.09
H mol JK b0.035 mol N gGH 28.02 mol JK ] u 10 g 0.01800 kg
b
1 mol natural gas Ÿ [ 0.875 mol CH 4 16.04
3
Ÿ
c.
9.49
2
8
6
2
843 kJ
1 mol
mol
0.01800 kg
3
46800 kJ kg
The enthalpy change when 1 kg of the natural gas at 25oC is burned completely with oxygen at
25oC and the products CO2(g) and H2O(v) are brought back to 25oC.
B
Table B.1
bg
C s + O 2 (g) o CO 2 (g), 'H co
e j
'H fo
CO 2 ( g)
B
393.5 kJ 1 mol 10 3 g
mol
12.01 g 1 kg
Table B.1
bg
S s + O 2 (g) o SO 2 (g),
'H co
MSO2 32 .064
e j
'H fo
296.90 kJ mol
SO 2
B
Ÿ
9261 kJ / kg S
B
Table B.1
bg
bg
1
H 2 g + O 2 (g) o H 2 O l , 'H co
2
32,764 kJ kg C
e j
'H fo
M H 2 1. 008
bg
H 2O l
9-56
285.84 kJ mol H 2
B
Ÿ 141,790 kJ kg H
9.49 (cont'd)
a.
x0 (kg O) 2 kg H
total H – H in H 2 O ; latter is kg coal 16 kg O
H available for combustion
Eq. (9.6-3) Ÿ HHV
A
FG
H
32,764 C 141,790 H in water
IJ
K
O
9261S
8
This formula does not take into account the heats of formation of the chemical constituents of
coal.
b.
C
0. 758 , H
1 kg coal Ÿ
0. 082 , S
b
0. 016 Ÿ HHV
Dulong
31,646 kJ kg coal
0.016 kg S 64.07 kg SO 2 formed
Diluting the stack gas lowers the mole fraction of SO2, but does not reduce SO2 emission rates. The
dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it.
bg
'H co
CH 4 + 2O 2 o CO 2 2H 2 O l , HHV
9.50
g
0. 0320 kg SO 2 kg coal
32.06 kg S burned
0.0320 kg SO 2 kg coal
1.01 u 10 6 kg SO 2 kJ
31,646 kJ kg coal
I
c.
0. 051, O
b
g
890.36 kJ mol Table B.1
bg
7
C 2 H 6 + O 2 o 2CO 2 3H 2 O l , HHV 1559. 9 kJ mol
2
1
CO + O 2 o CO 2 , HHV 282. 99 kJ mol
2
2.000 L
273.2K
2323 mm Hg
1 mol
Initial moles charged:
25 + 273.2 K 760 mm Hg 22.4 L STP
(Assume ideal gas)
a
f
a f
0.25 mol
Average mol. wt.: (4.929 g) (0.25 mol) = 19.72 g / mol
Let x1
MW
g
b
gh
19.72 Ÿ x b16.04 g mol CH g x b30.07g b1 x x gb28.01g 19.72 b1g
963.7 kJ mol Ÿ x b890.36g x b1559.9g b1 x x gb282.99g 963.7 b2g
mol CH 4 mol gas , x2
1
HHV
c b
mol C 2 H 6 mol gas Ÿ 1 x1 x2 mol CO mol gas
4
2
1
2
1
2
1
2
Solving (1) & (2) simultaneously yields
x1
9.51
a.
0.725 mol CH 4 mol, x 2
0.188 mol C 2 H 6 mol, 1 x1 x 2
Basis : 1mol/s fuel gas
CH 4 (g) 2O 2 (g) o CO 2 (g) 2H 2 O(v), 'H co
C 2 H 6 (g) 7
O 2 (g) o 2CO 2 (g) 3H 2 O(v), 'H co
2
0.087 mol CO mol
890.36 kJ / mol
1559.9 kJ / mol
Excess O2, 25qC
n2 , mol CO 2
n3 , mol H 2 O
n4 , mol O 2
25qC
1 mol/s fuel gas, 25qC
85% CH4
15% C2H6
9-57
9.51 (cont’d)
1 mol / s fuel gas Ÿ 0.85 mol CH 4 / s , 0.15 mol C 2 H 6 / s
Theoretical oxygen
2 mol O 2
0.85 mol CH 4
1 mol CH 4
s
3.5 mol O 2
0.15 mol C 2 H 6
1 mol C 2 H 6
s
2.225 mol O 2 / s
Assume 10% excess O 2 Ÿ O 2 fed = 1.1 u 2.225 = 2.448 mol O 2 / s
. gb2g Ÿ n
b0.85gb1g b015
H balance : 2n
. gb6g Ÿ n
b0.85gb4g b015
10% excess O Ÿ n
b01. gb2.225g mol O
C balance : n 2
115
. mol CO 2 / s
2
3
2.15 mol H 2 O / s
3
2
4
Extents of reaction: [ 1
2
/s
0.223 mol O 2 / s
0.85 mol / s, [ 2
n CH 4
bg
n C 2 H 6
bg bg bg
015
. mol / s
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
(We will use the values of
'H co
bg
given in Table B.1, which are based on H 2 O l as a
combustion product, and so must choose the liquid as a reference state for water)
Substance
e
'H v 25o C
nout
H out
CH 4
mol
0.85
kJ mol
0
mol
kJ mol
C2 H 6
015
.
0
O2
2.225
0
0.223
0
CO 2
115
.
0
H 2O v
2.15
H1
bg
H 1
H in
nin
j
Energy Balance :
Q nCH 4 'H co
e j
44.01 kJ / mol
CH 4
e j
nC 2 H 6 'H co
C2 H 6
¦ n H ¦ n H
i
out
i
i
i
in
. mol / s C H gb 1559.9 kJ molg
b0.85 mol / s CH gb890.36 kJ molg b015
b2.15 mol / s H Ogb44.01 kJ / molg 896 kW
4
2
6
2
Ÿ Q
b.
896 kW (transferred from reactor)
Constant Volume Process. The flowchart and stoichiometry and material balance calculations are
the same as in part (a), except that amounts replace flow rates (mol instead of mol/s, etc.)
1 mol fuel gas Ÿ 0.85 mol CH 4 , 0.15 mol C 2 H 6
Theoretical oxygen 2.225 mol O 2
Assume 10% excess O 2 Ÿ O 2 fed = 1.1 u 2.225 = 2.448 mol O 2
. gb2g Ÿ n
b0.85gb1g b015
H balance : 2n b0.85gb4g b015
. gb6g Ÿ n
10% excess O Ÿ n
b0.1gb2.225g mol O
C balance : n2
3
2
115
. mol CO 2
2
4
3
2.15 mol H 2 O
2
0.223 mol O 2
9-58
9.51 (cont'd)
bg
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
For a constant volume process the heat released or absorbed is determined by the internal
energy of reaction.
nin
nout
U in
U out
Substance
mol kJ mol mol kJ mol
CH 4
0.85
0
C2 H 6
015
.
0
O2
2.225
0
0.223
0
CO 2
115
.
H 2O v
2.15
0
U 1
bg
e
U 1
'U v 25 o C
j
e
j
'H v 25 o C RT
Eq. (9.1-5) Ÿ 'U co
'H co RT (
¦Q
i
e
j
o
c
i
1 kJ
mol K 1000 J
4153
.
kJ
mol
)
298 K (1 + 2 1 2)
1 kJ
890.36
3
10 J
8.314 J 298 K (3 + 2 35
. 1)
1 kJ
b1559.9 kJ molg mol K
C2H 6
298 K
gaseous
reactants
.314 J
b890.36 kJ molg 8mol
K
CH 4
e'U j
¦Q
gaseous
products
Ÿ 'U co
8.314 J
44.01 kJ / mol 3
10 J
kJ
mol
156114
.
kJ
mol
Energy balance:
Q
j n e'U j ¦ n U ¦ n U
. mol / s C H gb 156114
. kJ mol g
b0.85 mol / s CH gb890.36 kJ molg b015
. kJ / mol g 902 kJ
b2.15 mol / s H Ogb4153
e
n CH 4 'U co
'U
CH 4
C2 H 6
o
c
C2 H 6
i
i
i
out
4
i
in
2
6
2
Ÿ Q
c.
9.52
a.
902 kJ (transferred from reactor)
Since the O2 (and N2 if air were used) are at 25qC at both the inlet and outlet of this process, their
specific enthalpies or internal energies are zero and their amounts therefore have no effect on the
calculated values of 'H and 'U .
n fuel ( 'H co )
W s Q l (Rate of heat release due to combustion = shaft work + rate of heat loss)
V (gal)
28.317 L
h
7.4805 gal
100 hp
0.700 kg 10 3 g 49 kJ
L
1 kg
g
1J/s
1 kJ
3600 s
1.341 u 10 3 hp
10 3 J
h
15 u 10 6 kJ
Ÿ V
298
h
2.5 gal / h
b.
The work delivered would be less since more of the energy released by combustion would go into
heating the exhaust gas.
c.
Heat loss increases as Ta decreases.
Lubricating oil becomes thicker, so more energy goes to overcoming friction.
9-59
9.53
a.
Energy balance: 'U
a
f
b
n lb m fuel burned
0Ÿ
b
g
'U co (Btu)
gb
lb m
ga
b
Ÿ 0.00215 'U co 4.62 lb m 0.900 Btu lb m ˜q F 87.06q F 77.00q F
Ÿ 'U co
b.
g
0
764.0
kJ
mol
mCv Tout 77q F
f
0
19500 Btu lb m
The reaction for which we determined 'U co is
1 lb m oil + aO 2 ( g) o bCO 2 ( g) + cH 2 O(v)
(1)
The higher heating value is 'H r for the reaction
1 lb m oil + aO 2 ( g) o bCO 2 ( g) + cH 2 O(l)
o
Eq. (9.1-5) on p. 441 Ÿ 'H c1
(2)
o
RT (b c a)
'U c1
o
Eq. (9.6-1) on p. 462 Ÿ 'H c2
o
'H c1
c'H v (H 2 O, 77$ F)
( HHV )
( LHV )
To calculate the higher heating value, we therefore need
9.54
a.
a
lb - moles of O 2 that react with 1 lb m fuel oil
b
lb - moles of CO 2 formed when 1 lb m fuel oil is burned
c
lb - moles of H 2 O formed when 1 lb m fuel oil is burned
bg
bg
3
2
e'H j
'H ro
CH 3 OH v + O 2 (g) o CO 2 (g) 2H 2 O l
o
c
bg
CH 3OH v
Basis : 1 mol CH 3OH fed and burned
Q2(kJ)
1 mol CH 3OH( l )
25°C, 1.1 atm
vaporizer
1 mol CH 3OH( v)
100°C
1 atm
Q 1 (kJ)
reactor
n 0 (mol O2 )
3.76n 0 (mol N2 )
100°C
Overall C balance:
1 mol CH 3OH
% excess air
b
b20.83gb0.809g Ÿ n 4.482 mol O
b1 mol CH OHgb15. mol O mol CH OHg
0
3
2
(4.482 15
. ) mol O 2
u 100%
15
. mol O 2
3
nw
uP
nw n p
a
15
. mol O 2
200% excess air
gb
(An atomic O balance Ÿ 9.96 mol O
20.83 mol dry gas
2
H balance: 1 mol CH 3OH 4 mol H 1 mol CH 3OH
pw
gb g
n p 0.048 1 Ÿ n p
1 mol CH 3OH
N 2 balance: 3.76n0
Theoretical O 2 :
b
1 mol C
Effluent at 300°C, 1 atm
n p (mol dry gas)
0.048 mol CO 2/mol D.G.
0.143 mol O 2/mol D.G.
0.809 mol N 2/mol D.G.
n w (mol H 2O)
g
bg
nw 2 Ÿ nw
2 mol H 2 O
9.96 mol O , so that the results are consistent.)
2 mol H 2 O
u 760 mm Hg
2 20.83 mol
f
9-60
66.58 mm Hg
d i
pw Tdp
Table B.3
Ÿ
Tdp
44.1q C
9.54 (cont'd)
b.
Energy balance on vaporizer:
'H
Q1
LM
1 mol M
MN
OP kJ
C dT
C
PP mol 40.33 kJ
A
A
A
Q
References : CH OHa vf, N (g), O (g), CO (g), H Oal f at 25q C
n'H
Substance
CH 3 OH
N2
af
dT 'H v pl
25
Table B.2
z
100
3
2
H in
(mol)
1.00
(kJ / mol)
3.603
pv
64.7
Table B.2
Table B.1
n in
16.85
2
2
2
H out
n out
(mol)
(kJ / mol)
2.187
16.85
8.118
4.482
2.235
2.98
8.470
CO 2
1.00
11578
.
H 2O
2.00
5358
.
O2
H T
z
64.7
H i for N 2 , O 2 , CO 2 (Table B.8)
'H v 25$ C H i for H 2 O v (Eq. 9.6 - 2a on p. 462, Table B.8)
z
d
af
i
af
T
25
C p dT for CH 3 OH v (Table B.2)
bg
(Note: H 2 O l was chosen as the reference state since the given value of 'H co presumes liquid
water as the product.)
Extent of reaction: (nCH 3OH ) out
(nCH 3OH ) in Q CH 3OH [ Ÿ 0 1 mol [ Ÿ [
Energy balance on reactor: Q2
['H co ¦ n H ¦ n H
i
out
i
b1gb764.0g b16.85gb8.118gb4.482gb2.235g kJ
i
i
in
bTable B.1g
534 kJ Ÿ 534 kJ transferred from reactor
9.55
a.
3
CH 4 2O 2 o CO 2 2H 2 O
CH 4 O 2 o CO 2H 2 O
2
Basis : 1000 mol CH 4 h fed
Q(kJ/h)
1000 mol CH4 /h
25°C
Stack gas, 400°C
n1 (mol CH4 /h)
n2 (mol O2 /h)
3.76n0 (mol N2 /h)
n3 (mol CO/h)
10 n3 (mol CO/h)
n4 (mol H2 O/h)
n0 (mol O2 /h)
3.76n0 (mol N2 /h)
100°C
a f
90% combustion Ÿ n1 0.10 1000 100 mol CH 4 h
Theoretical O2 required = 2000 mol/h
10% excess O2 Ÿ O 2 fed = 1.1(2000 mol / h) = 2200 mol / h
9-61
1 mol
9.55 (cont’d)
C balance:
1000 mol CH 4 h 1 mol C mol CH 4
b
gb
Ÿ 10n 3
g b100gb1g n b1g 10n b1g Ÿ n
3
3
3
. mol CO h
818
818 mol CO 2 h
a fa f a100fa4f 2n Ÿ n 1800 mol H O h
O balance: a2200 fa2 f 2 n a81.8fa1f a818fa2 f a1800fa1f Ÿ n
441 mol O
References :Casf, H bgg, O bgg, N bgg at 25 C
H balance: 1000 4
4
4
2
2
2
h
2
$
2
Bo
Table B.1
'H f z
2
n in
H in
n out
H out
CH 4
1000
74.85
100
57.62
O2
2200
2.24
441
.
1172
N2
8272
2.19
8272
.
1115
CO
.
818
99.27
CO 2
818
377.2
H 2O
1800
228.63
Substance
H
2
bmol hg bkJ molg bmol hg bkJ molg
B
Table B.2
T
C p dT for CH 4
25
B
Table B.8
'H fo H i (T) for others
Energy balance: Q
'H
¦ n H ¦ n H
i
i
out
b.
i
7.344 u 10 5 kJ h
a
204 kW
f
in
(iii)
A (increases) Ÿ Q A
%XS A Ÿ Q B (more energy required to heat additional O and N to 400 C, therefore
less energy transferred.)
S
A Ÿ Q A (reaction to form CO2 has a greater heat of combustion and so releases
(iv)
more thermal energy)
Tstack Ÿ Q (more energy required to heat combustion products)
(i)
(ii)
9.56
i
Tair
o
2
2
CO 2 CO
A
B
CH 4 2O 2 o CO 2 2H 2 O, C 2 H 6 7
O 2 o 2 CO 2 3H 2 O
2
Basis : 100 mol stack gas. Assume ideal gas behavior.
n1 (mol CH4 )
n2 (mol C2 H6 )
Vf (m3 at 25°C, 1 atm)
100 mol at 800°C, 1 atm
0.0532 mol CO 2 /mol
0.0160 mol CO/mol
0.0732 mol O2 /mol
0.1224 mol H 2 O/mol
0.7352 mol N2 /mol
n3 (mol O2 )
3.76n3 (mol N2 )
200°C, 1 atm
9-62
9.56 (cont’d)
a.
N 2 balance: 3.76n3
b100gb0.7352gmol N Ÿ n 19.55 mol O fed
C balance: n b1g n b2 g b100gb0.0532gb1g b100gb0.0160gb1gU| n 3.72 mol CH
V| Ÿ n 1.60 mol C H
H balance: n b4g n b6g b100gb0.1224gb2g
W
. gmol fuel gas 22.4 LbSTPg 298.2 K 1 m
b3.72 160
V
0130
m
.
1
2
3
2
1
4
1
2
2
2
2
6
3
3
f
3.72 mol CH 4
Theoretical O 2
1 mol
273.2 K 103 L
2 mol O 2
1 mol CH 4
1.60 mol C 2 H 6
3.5 mol O 2
1 mol C 2 H 6
UV Ÿ 69.9 mole% CH
1.60 mol C H W 30.1 mole% C H
b19.55 13.04gmol O in excess u 100% 50% excess air
3.72 mol CH 4
Fuel composition:
2
4
2
6
6
2
% Excess air:
13.04 mol O 2 required
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25q C
b.
n in
H in
n out
H out
CH 4
mol
3.72
kJ / mol
74.85
mol
kJ / mol
C2 H 6
1.60
84.67
O2
19.55
5.31
7.32
25.35
N2
.
7352
.
513
.
7352
.
2386
CO
1.60
86.39
CO 2
5.32
356.1
H 2O
Substance
Bo
Table B.1
H
'H f z
12.24 212.78
Table B.2, for
CH 4 , C2 H6
B
T
C p dT
25
Table B.8
B
= 'H fo H i (T) for O 2 , N 2 , CO, CO 2 , H 2 O v
bg
Energy balance:
Q
'H
¦ n H ¦ n H
i
out
i
i
in
2764 kJ
i
0.130 m 3 fuel
9-63
2.13 u 104 kJ m 3 fuel
13.04 mol O 2
b0.730gb50000glb
9.57
Basis : 50000 lb m coal fed h Ÿ
mC
1b - mole C
h
3039 1b - mole C h
12.01 lb m
.
2327 lb - moles H h (does not include H in water)
b0.047gb50000g 101
b0.037gb50000g 32.07 57.7 lb - moles S h
b0.068gb50000g 18.02 189 lb - moles H O h
. gb50000g 5900 lb ash h
b0118
2
m
a.
50,000 lb m coal/h
3039 lb-moles C/h
2327 lb-moles H/h
57.7 lb-moles S/h
189 lb-moles H 2O/h
5900 lb m ash/h
77°F, 1 atm (assume)
Stack gas at 600°F, 1 atm (assume)
n 2 (lb-moles CO2 /h)
n 3 (lb-moles H2 O/h)
n 4 (lb-moles SO 2 /h)
n 5 (lb-moles O2 /h)
n 6 (lb-moles N2 /h)
m 7 (lb m fly ash/h)
n 1 (lb-moles air/h)
0.210 O 2
0.790 N 2
77°F, 1 atm (assume)
m 8 (lbm slag/h) at 600°F
0.287 lb mC/lb m
0.016 lb m S/lb m
0.697 lb mash/lb m
Feed rate of air :
b
O 2 required to oxidize carbon C + O 2 o CO 2
g
3039 lb - moles C 1 lb - mole O 2
h
1 lb - mole C
3039 lb - moles O 2 h
Air fed: n1
1.5 u 3039 lb - moles O 2 fed
1 mole air
h
0.210 mole O 2
21710 lb - moles air h
b
8
30% ash in coal emerges in slag Ÿ 0.697m
Ÿ m 7
b g
0.700 5900
b
g
b
gb
g
n 2 0.287 2540 12.01
M CO 2 44.01
2978 lb - moles CO 2 h
b
g b gb g
H balance: 2327 lb - moles H h 189 2
Ÿ n 3
1.31 u 10 5 lb m CO 2 h
2n 3
M H 2 O 18.02
1352.5 lb - moles H 2 O h
2.44 u 10 4 lb m H 2 O h
b0.790g21710 lb - moles h 17150 lb - moles N
57.7blb - moles S hg b1gn 0.016 b2540g 32.06
N 2 balance: n 6
S balance:
Ÿ n 4
M N 2 28.02
2
h
4.81 u 105 lb m N 2 h
4
M SO 2 64.2
56.4 lb - moles SO 2 h
b gb g b gb
3620 lb m SO 2 h
gb g b2978gb2g b1352.5gb1g b56.4gb2g 2b n g
O balance: 189 1 0.21 21710 2
bcoal g
Ÿ n 5
2540 lb m slag / h
4130 lb m fly ash h
C balance: 3039 lb - moles C h
Ÿ n 2
g
0.30 5900 lb m h Ÿ m 8
b air g
bCO g
2
943 lb - moles O 2 h Ÿ 30200 lb m O 2 h
9-64
eH O j
2
bSO g
2
5
O2
9.57 (cont'd)
Summary of component mass flow rates
Stack gas at 600q F, 1 atm
2978 lb - moles CO 2 h Ÿ
131000 lb m CO 2 h
1352.5 lb - moles H 2 O h Ÿ 24400 lb m H 2 O h
56.4 lb - moles SO 2 h Ÿ
3620 lb m SO 2 h
943 lb - moles O 2 h Ÿ
30200 lb m O 2 h
17150 lb - moles N 2 h Ÿ
48100 lb m N 2 h
674,350 lbm stack gas/h
4130 lb m fly ash h
b
gb g
Check: 50000 21710 29
b679600g
Ÿ
in
b
b.
g
œ 676900
œ 674350 2540
out
(0.4% roundoff error)
out
22480 lb - moles h at 600q F , 1 atm (excluding fly ash)
Total molar flow rate
ŸV
in
a f
1060q R
22480 lb - moles 359 ft 3 STP
h
1.74 u 10 7 ft 3 h
492q R
1 lb - mole
References: Coal components, air at 77qF Ÿ ¦ ni H i
0
in
674350 lb m
h
Stack gas: nH
Slag: nH
2540 lb m
0.22 Btu
h
lb m ˜q F
Energy balance: Q
'H
b600 77gq F
7.063 Btu 1 lb - mole
lb - mole˜q F 28.02 lb m
b600 77gq F
b
8.90 u 10 7 Btu h
2.92 u 105 Btu h
g ¦ n H ¦ n H
n coal burned 'H co 77q F i
i
i
out
5 u 104 lb m
18
. u 10 4 Btu
h
lb m
i
in
e
j
8.90 u 107 2.92 u 105 Btu h
8.11 u 108 Btu h
b0.35ge8.11 u 10 jBtu
8
Power generated
Q
e8.11 u 10
Ÿ
Q
HHV
8
Btu h
j b5000 lb
1W
3600 s 9.486 u 10
h
c.
1 hr
m
1. 62 u 10 4 Btu lb m
1.80 u 104 Btu lb m
coal h
g
4
1 MW
Btu s 10 6 W
. MW
831
162
. u 104 Btu lb m coal
0. 901
Some of the heat of combustion goes to vaporize water and heat the stack gas.
d.
Q HHV would be closer to 1. Use heat exchange between the entering air and the stack gas.
9-65
9.58
b.
Basis : 1 mol fuel gas/s
n 0 (mol O 2 s)
3.76n 0 (mol N 2 s)
o
Stack gas, Ts ( C)
o
n O2 (mol O 2 / s)
3.76n 0 (mol N 2 / s)
Ta ( C)
o
n CO (mol CO / s)
rn CO (mol CO 2 / s)
1 mol / s @ 25 C
x m (mol CH 4 / mol)
n H 2 O (mol H 2 O / s)
n Ar (mol Ar / s)
x a (mol Ar / mol)
(1 x m x a ) (mol C 2 H 6 / mol)
CH 4 2O 2 o CO 2 2 H 2 O
C2 H 6 7
2
O 2 o 2CO 2 3H 2 O
Pxs
) 2 x m 3.5(1 x m x a )
100
x m 2(1 x m x a )
C balance: x m 2(1 x m x a ) (1 r )n CO Ÿ n CO
(1 r )
(1 Percent excess air: n 0
H balance: 4 x m 6(1 x m x a )
O balance: 2n 0
2n H 2 O Ÿ n H 2 O
2 x m 3(1 x m x a )
2n O 2 n CO 2 r n CO n H 2 O Ÿ n O 2
n 0 n CO (1 2r ) / 2 n H 2 O / 2
References : C(s), H2(g), O2(g), N2(g) at 25qC
nin
H in
nout
Substance
(1 xm x A )
0
0
A
O2
xA
no
0
H
N2
CO
3.76no
H 2
xA
nO2
CO 2
H 2O
CH 4
xm
C2 H 6
1
z
H out
H
3
H 4
H
3.76no
nCO
5
H 6
H
r nCO
nH 2 O
7
H 8
Ta or Ts Table B.2
H i
c.
( 'H f ) i B
C p ,i dT
25
Given : x m
Ÿ no
0.85, x a
0.05, Px s
5%, r
0.0955, n H 2 O
2.153, n CO
150 o C, Ts
10.0, Ta
2.00, nO 2
700 o C
.
01500
H 1 (kJ / mol) = 8.091, H 2 = 29.588, H 3 = 0.702, H 4 = 3.279,
H = 166.72, H = 8.567, H = 345.35, H = 433.82
5
6
Energy balance: Q
¦ n
7
out H out
¦ n
8
in H in
9-66
655 kW
9.58 (cont'd)
Xa
Pxs
r
Ta
Ts
Q
0
150
150
150
150
150
150
700
700
700
700
700
700
10
10
10
10
10
150
150
150
150
150
700
700
700
700
700
-
-200 0
996
90
813
- 22
631
-90
-869
- 99
0.1
0.2
0.3
0.4
0.5
5
5
5
5
5
10
10
10
10
10
150
150
150
150
150
700
700
700
700
700
-893
8
- 1
860
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
5
5
5
5
1
2
3
4
5
10
20
50
100
150
150
150
150
150
150
150
150
150
700
700
700
700
700
700
700
700
700
-722
-796
-834
-856
-871
-905
-924
-936
-941
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
10
10
10
10
10
25
100
150
200
250
700
700
700
700
700
-852
-883
-905
-926
-948
0.1
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
5
10
10
10
10
10
10
150 500
150 600
150 700
150 800
150 900
150 1000
-1014
-960
-905
-848
-790
-731
9-67
0.2
0.4
0.6
0.8
1
1.2
80
100
120
0.4
0.5
0.6
80
100
120
200
250
300
800
1000
1200
-400
Q
-600
-800
-1000
-1200
Xm
0
- 0
200
Q 400
600
-
20
40
-800
1000
60
Pxs
-850
-860 0
0.1
0.2
0.3
-870
Q -880
-890
-900
-910
x
0
0
20
40
60
-200
-400
Q
10
10
10
10
10
10
-600
-800
-1000
r
-800
-850
Q
0.0
5
0.0
5
1
0.0
5
1
0.0
5
1
0.0
5
1
0.0
5
1
0.1
5
0.1 10
0.1 20
0.1 50
0.1 100
0
50
100
150
-900
-950
-1000
Ta
0
0
200
400
600
-500
Q
d.
-1000
-1500
Ts
9.59
a.
Basis:
H
207.4 liters 273.2 K 1.1 atm
s
1 mol
b g
10.0 mols s fuel gas to furnace
278.2 K 1.0 atm 22.4 liters STP
C 6 H 14 ; M
CH 4
mw (kg H 2O( l)/s)
25°C
Qc (kW)
n0 (mol/s)
y 0 (mol C 6H 14/mol)
(1 – y 0) (mol CH 4/mol)
60°C, 1.2 atm
Tdp = 55°C
10.0 mol/s at 5°C, 1.1 atm
y2 (mol C 6H 14/mol)
(1 – y2) (mol CH 4/mol)
sat'd with C 6H 14
condenser
nb (mol C 6H 14( l)/s)
mw (kg H 2O( v)/s)
10 bars, sat'd
Stack gas at 400°C, 1 atm
n3 (mol O 2/s)
n4 (mol N 2/s)
n5 (mol CO 2/s)
n6 (mol H 2O( v)/s)
reactor
na (mol air/s) @ 200°C
0.21 mol O /mol
2
0.79 mol N /mol
2
100% excess
Antoine Eq.
Tdp
p DH
55q C Ÿ y 0 P
Ÿ y0
B
b55q Cg
483.3 mm Hg
1.2 u 760 mm Hg
483.3 mm Hg
0.530 mol C 6 H 14 mol Ÿ 0.470 mol CH 4 mol
Saturation at condenser outlet:
p H 5q C
58.89 mm Hg
y2
P
1.1 u 760 mm Hg
b g
0.070 mol C 6 H 14 mol
b
Methane balance on condenser: n0 1 y0
Hexane balance on condenser: n0 y0
g
b
n b 10.0 y2
bg
9.78 mol C 6 H 14 l
Volume of condensate
g
10.0 1 y2
Ÿ
y2 0.070
Ÿ
n b
n0 19.78
y0 0.530
y2 0.070
n0
19.78 mol s
9.78 mol C 6 H 14 s condensed
cm 3
86.17 g
s
y0 0.530
0.93% mol CH 4 mol
1L
3
0.659 g 10 cm
mol
A
A
Table B.1
3600 s
3
1h
Table B.1
4600 L C 6 H 14 (l ) h
b.
e
j
e
References : CH 4 g, 5$ C , C 6 H 14 l, 5$ C
Substance
CH 4
bg
bl g
C 6 H 14 v
bg
H out
nout
(mol / s) (kJ / mol) (mol / s) (kJ / mol)
9.30
1985
9.30
0
.
10.48
41212
.
0.70
32.940
9.78
0
C 6 H 14
CH 4 g : H
H in
nin
j
z
B
Table B.2
T
5
bg
C 6 H 14 v : H
C p dT
Condenser energy balance: Q c
'H
¦ n H ¦ n H
i
out
9-68
i
i
in
i
z
B
Table B.1
Tb
5
B
Table B.1
C pR dT 'H v 427 kW
z
T
Tb
C pv dT
9.59 (cont'd)
CH 4 2O 2 o CO 2 2H 2 O , C 6 H 14 Theoretical O 2 :
2
O 2 o 6CO 2 7H 2 O
9.30 mol CH 4
2 mol O 2
s
1 mol CH 4
b g 2 u bO g
balance: 0.79b240.95g n Ÿ n
100% excess Ÿ O 2
N2
19
2 theor.
fed
4
0.70 mol C 6 H 14
9.5 mol O 2
s
1 mol C 6 H 14
Ÿ 0.21na
2 u 25.3 Ÿ na
25.3 mol O 2 s
240.95 mol air s
190.35 mol N 2 s
4
C balance:
9.30 mol CH 4
1 mol C
s
1 mol CH 4
Ÿ n5
135
. mol CO 2 s
0.70 mol C 6 H 14
b
6 mol C
n5 mol CO 2
g
1 mol C
1 mol CO 2
1 mol C 6 H 14
H balance:
b9.30 mol CH sgb4 mol H mol CH g b0.70gb14g n b2g Ÿ n 235. mol H O
1
Since combustion is complete, bO g
bO g
bO g Ÿ n 25.3 mol O s
2
References : Cbsg, H bgg, O bgg, N bgg at 25q C for reactor side, H Oblg at triple point for
4
4
6
2 remaining
2
2
6
2 excess
2
2 fed
2
3
2
2
steam side (reference state for steam tables)
H in
n in
Substance
H out
n out
mol / s
9.30
kJ / mol mol / s
75.553 kJ / mol
C 6 H 14 v
0.70
170.07 O2
50.6
5.31
25.3
.
1172
N2
190.35
.
513
190.35
.
1115
CO 2
.
135
377.15
.
235
228.60
CH 4
bg
bg
H Ob boiler water g
H 2O v
2
bg
m w (kg / s) 104.8
Table B.1 and B.2
B
=
H T
z
m w ( kg / s) 2776.2
T
'H fo C p dT
for CH 4 , C 6 H 14
25
Table B.1 and B.8
B
=
bg
bg
'H fo H i T for O 2 , N 2 , CO 2 , H 2 O v
Energy balance on reactor (assume adiabatic):
'H
¦ n H ¦ n H
i
out
i
i
i
b
g
0 Ÿ 8468 m w 2776.2 104.8
in
9-69
0 Ÿ m w
3.2 kg steam s
9.60
a.
CH 4 2O2 o CO2 2H 2O
Basis: 450 kmol CH 4 fed h
n a ( kmol air / h)@25 o C
o
0.21 kmol O 2 / kmol
0.79 kmol N 2 / kmol
Stack gas@300 C
n 1 (kmol CO 2 / h)
n 2 (kmol H 2 O / h)
Q ( kJ / h)
o
450 kmol CH 4 / h @ 25 C
n 3 (kmol O 2 / h)
n 4 (kmol N 2 / h)
m w [kg H 2 O(l) / h]
m w [kg H 2 O(v) / h]
o
17 bar, 250 o C
25 C
450 kmol CH 4
h
Air fed: n a
2 kmol O 2 req' d 1.2 kmol O 2 fed
1 kmol air
1 kmol CH 4
1 kmol O 2 req' d 0.21 kmol O 2
5143 kmol air h
450 kmol / h CH 4 react Ÿ n1
450 kmol CO 2 h , n 2
u 10
.
b0.79ge5143
N 2 balance: n 4
6
mol h
j
900 kmol H 2 O h
4060 kmol N 2 h
Molecular O 2 balance:
b0.21gb5143g mol Oh
y
450 kmol CO h U
|
900 kmol H O h |
|V Ÿ y
y
4060 kmol N h |
| y
180 kmol O h |
W
n 3
2
fed
450 kmol CH 4 react 2 mol O 2
h
1 mol CH 4
CO 2
0.0805
2
H 2O
.
0161
2
N2
0.726
2
O2
0.0322
2
180 kmol O 2 h
5590 kmol / h
Mean heat capacity of stack gas
Cp
¦y C
i
pi
. gb0.0343g b0.726gb0.0297g b0.0322gb0.0312g
b0.0805gb0.0423g b0161
0.0315 kJ mol ˜ $ C
Energy balance on furnace (combustion side only)
bg
bg bg
bg
bg
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25$ C
n in
H in
n out H out
Substance (kmol / h) (kJ / kmol) (kJ / h)
CH 4
450
0
Air
5143
0
Stack gas
H p
Extent of reaction:
[
n CH 4
450 kmol / h
9-70
9.60 (cont’d)
H p
n 2 ( 'H v ) H
o
2 O(25 C)
n stack gas (C p ) stack gas (Tstack gas 25o C)
3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44.01 kJ h
1 kmol
mol
h
1 kmol
7
= 5.63 u 10 kJ / h
Q
[ ( 'H co ) CH 4 'H
¦ n H ¦ n H
i
i
out
=
i
0.0315 kJ (300- 25) o C
mol ˜ o C
i
in
FG 450 kmol IJ FG1000 mol IJ FG 890.36 kJ IJ 5.63 u 10
H h K H kmol K H
mol K
7
kJ
h
3.44 u 10 8
kJ
h
Energy balance on steam boiler
LM FG IJ OP LMb2914 105g kJ OP
kg Q
N H KQ N
Table B.7 Table B.6
Q
kg
kJ
m w 'H w Ÿ + 3.44 u 10 8
m w
h
h
Ÿ m w 123
. u 10 5 kg steam / h
b.
n a (mol air/h) at
45 kmol CH
25°C
4
/h
mw (kg H 2O/h)
Liquid, 25°C
T a (°C)
Stack gas
n 1 (mol CO 2/h)
n 1 (mol CO 2/h)
air
n 2 (mol H 2O/h)
n 2 (mol H O/h)
2
furnace
n 3 (mol O 2/h)
n 3 (mol O 2/h)
preheater
n 4 (mol N 2/h)
n 4 (mol N 2/h)
300°C
150°C
mw (kg H 2O/h)
vapor, 17 bars
n a (mol air/h) at 25°C
250°C
0.21 O 2
0.79 N 2
E.B. on overall process: The material balances and the energy balance are identical to those of part
(a), except that the stack gas exits at 150oC instead of 300oC.
b g b g b g b g bg
b
g
H Oalf at triple point (steam table reference) (steam tube side)
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25 $ C furnace side
2
Substance
n in
(kmol / h)
H in
(kJ / kmol)
n out H out
(kJ / h)
CH 4
Air
Stack gas
450
5143
0
0
H p
H 2O
H p
n 2 ( 'H v ) H
o
2 O(25 C)
m w ( kg / h) 105 kJ / kg
m w ( kg / h) 2914 kJ / kg
n stack gas (C p ) stack gas (Tstack gas 25o C)
3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44.01 kJ h
1 kmol
mol
h
1 kmol
7
= 2.99 u 10 kJ / h
9-71
0.0315 kJ (150- 25) o C
mol ˜ o C
9.60 (cont’d)
[ ( 'H co ) CH 4 'H
¦ n H ¦ n H
i
i
i
out
0
i
in
FG 450 kmol IJ FG1000 mol IJ FG 890.36 kJ IJ 2.99 u 10 kJ
H h K H kmol K H
mol K
h
L F kg I OL
kJ O
Mm G J P Mb2914 105g P 0 Ÿ m = 1.32 u 10
H
K
h
kg
Q
QN
N
0
Energy balance on preheater: 'H d 'H i
d 'H i
5590 kmol 10 mol 0.0315 kJ b150 300g C
nC 'T
b 'H g
Ÿ
7
w
w
stack gas
stack gas
c.
b 'H g
stack gas
b 'H g
H air
5133
kJ / mol
.
h
$
$
mol ˜ C
1 kmol
Table B.8
2.64 u 10 7
kJ
h
2.64 u 10 7 kJ / h 1 kmol = 5.133 kJ
5143 kmol / h 10 3 mol
mol
n a H air (Ta ) Ÿ H air (Ta )
air
kg steam / h
air
3
p
5
199 o C
Ta
The energy balance on the furnace includes the term ¦n
in H in .
If the air is preheated and the
stack gas temperature remains the same, this term and hence Q become more negative, meaning
that more heat is transferred to the boiler water and more steam is produced. The stack gas is a
logical heating medium since it is available at a high temperature and costs nothing.
9.61
Basis: 40000 kg coal h Ÿ
b0.76 u 40000gkg C
10 3 g 1 mol C
h
Assume coal enters at 25q C
1 kg
. j
b0.05 u 4000gkg H h e10 101
b0.08 u 4000gkg O h e10 16.0j
. u 40000g 4400 kg ash h
b011
12.01 g
3
198
. u 10 6 mol H h
3
2.00 u 10 5 mol O h
4400 kg ash/h, 450°C
Q to steam
40,000 kg coal/h
2.531 u10 6 mol C/h
1.98 u10 6 mol H/h
2.00 u10 5 mol O/h
4400 kg ash/h
25°C
Flue gas at 260°C
n3 (mol dry gas/h)
0.078 mol CO 2 /mol D.G.
0.012 mol CO/mol D.G.
0.114 mol O 2/mol D.G.
0.796 mol N 2/mol D.G.
n4 (mol H2 O/h)
furnace
Preheated air at Ta (°C)
a.
2.531 u 10 6 mol C h
air at 30°C, 1 atm, hr = 30%
n1 (mol O2 /h)
3.76n1 (mol N2 /h)
n2 (mol H2 O/h)
preheater
Cooled flue gas at 150°C
n3 (mol dry gas/h)
0.078 CO 2
0.012 CO
0.114 O 2
0.796 N 2
n4 (mol H2 O/h)
Overall system balances
C balance: 2.531 u 10 6
N 2 balance: 3.76n1
0.078n 3 0.012n 3 Ÿ n 3
b0.796ge2.812 u 10 j Ÿ n
7
1
224 u 10 mol N 2 h
7
9-72
2.812 u 10 7 mol h dry flue gas
b ge
5.95 u 10 6 mol O 2 h 3.76 5.95 u 10 6
j
9.61 (cont'd)
30% relative humidity (inlet air):
B
Table B.3
b
g
0.30 p H 2 O 30q C Ÿ
y H 2O P
Ÿ n 2
n 2
b760 mm Hgg
5.95 u 10 2.24 u 10 n 2
6
7
b
mm Hg
0.300 31824
.
3.61 u 10 5 mol H 2 O h
Volumetric flow rate of inlet air:
V
e5.95 u 10
6
b g
j
224 u 10 7 3.61 u 10 5 mol 22.4 liters STP
h
Air/fuel ratio:
10 liters
161
. SCM air kg coal
e
j
H balance: 198
. u 10 6 mol H h 2 3.61 u 10 5 mol H h
H in coal
2n 4 Ÿ n 4
1.351 u 10 6 mol H 2 O h
H in water vapor
1357
u 10 6 mol H 2 O h
.
H 2 O content of stack gas
b.
6.43 u 10 5 SCMH
3
1 mol
6.43 u 10 5 m 3 air h
40000 kg coal h
1 m3
u 10
.
e1357
6
j
2.812 u 10 7 mol h
u 100%
4.6% H 2 O
Energy balance on stack gas in preheater
bg
References : CO 2 , CO, O 2 , N 2 , H 2 O v at 25 $ C
n in
mol h
2.193 u 10 6
0.337 u 10 6
3.706 u 10 6
22.38 u 10 6
1.357 u 10 6
Substance
CO 2
CO
O2
N2
H 2O
H in
kJ mol
4.942
3669
3758
3655
4266
H i (T ) from Table B.8 for inlet
Q
¦ n H ¦ n H
i
i
out
i
i
n out
mol h
2.193 u 10 6
0.337 u 10 6
3.206 u 10 6
72.38 u 10 6
1.351 u 10 6
H i (T ) =
b
z
H out
kJ mol
9.738
6.961
7.193
6.918
8135
Table B.2
B
Cp
dT for outlet
g
101
. u 10 8 kJ h Heat transferred from stack gas
in
Air preheating
1.01 u10 8 kJ/hr
2.83 u10 7 mol dry air/h
3.61 u10 5 mol H 2 O/h
30°C
2.83 u10 7 mol dry air/h
3.61 u10 5 mol H 2 O/h
T a (°C)
(We assume preheater is adiabatic, so that Qstack gas
Energy balance on air:
Q
'H Ÿ 1.01 u 10
8
kJ hr
¦
z
Ta
z
Qair )
Ta
ni (C p ) i dT
30
30
9-73
Table B.2
n dry air
z
T
Table B.2
a
B
B
(C p ) dry air dT n H 2 O (C p ) H 2 O dT
30
g
9.61 (cont'd)
Ÿ 101
. u 10 8 8.31 u 10 5 (Ta 30) 59.92(Ta2 30 2 ) 0.031(Ta3 30 3 ) 1.42 u 10 5 (Ta4 30 4 )
Ÿ Ta 150 $ C
c.
4400 kg ash/h at 450°C
40,000 kg coal/h
25°C
Flue gas at 260°C
2193u 106 mol CO 2 /h
0.337 u10 6 mo l CO/h
3.206 u 106 mo l O2 /h
22.38 u 106 mo l N2 /h
1.351 u 106 mo l H2 O( v)/h
5.95 u10 6 mo l O2 /h
2.24 u10 7 mo l N2 /h
3.61 u10 5 mo l H2 O( v)/h
150°C(= 149.8°C)
m (kg H 2 O(l )/h)
50°C
m (kg H 2 O(v )/h)
30 bars, sat'd
bg
References for energy balance on furnace: CO 2 , CO, O 2 , N 2 , H 2 O l , coal at 25q C
bg
(Must choose H 2 O l since we are given the higher heating value of the coal.)
H in
0
3.758
3.655
48.28
substance
n in
40000
Coal
Ash
5.95 u 10 6
O2
N2
2.24 u 10 6
CO 2
CO
3.61 u 10 5
H 2O
n out
4400
3.206 u 10 6
2.24 u 10 7
2.193 u 10 6
0.337 u 10 6
1.351 u 10 6
H out
n kg h
412.25 H kJ kg
7.193
6.918
n mol h
9.738 H kJ mol
6.961
52.14
b
b
b
b
g
g
g
g
(Furnace only — exclude boiler water)
Heat transferred from furnace
Q
n coal 'H io ¦ n H ¦ n H
i
i
out
i
i
in
FG 4 u 10 kg IJ FG 2.5 u 10 kJ IJ FG 2.74 u 10
H
h KH
kg K G
H
4
8.76u 10
4
8
3
1.22 u 10 8
A
H of preheated air
I kJ
JJ kg
K
kJ h
8
8
Heat transferred to boiler water: 0.60(8.76x10 kJ/h) = 5.25x10 kJ/h
b g m FGH kgh IJK H cH Oblg, 30b, sat' dh H eH Oblg, 50 Cj
O kJ L
kJ h m M2802.3 209.3P
MN A A PQ kg Ÿ m 2.02 u 10 kg steam h
Energy balance on boiler: Q kJ h
Ÿ 5.25 u 10 8
$
2
2
5
Table B.6
Table B.5
9-74
9.62
1
CO O 2 o CO 2 , 'H co
Basis : 1 mol CO burned.
1 mol CO2
(n 0 – 0.5) mol O2
3.76n 0 mol N2
1400°C
1 mol CO
n 0 mol O2
3.76n 0 mol N2
25°C
a.
Oxygen in product gas: n1
282.99 kJ mol
2
b
g
n0 mol O 2 fed 1 mol CO react 0.5 mol O 2
1 mol CO
References: CO, CO 2 , O 2 , N 2 at 25$ C
n in
n out
H in
Substance mol
mol
kJ mol
b g b
CO
O2
N2
CO 2
1
n0
3.76n 0
g b g b
H out
kJ mol
n 0 0.5
3.76n 0
1
H 1
H 2
H 3
0
0
0
B
n0 0.5
g
Table B.8
b
g
N bg,1400q Cg: H
CO bg,1400q Cg: H
O 2 g,1400q C : H 1
2
H O 2 (1400$ C)
H N 2 (1400$ C)
2
2
47.07 kJ mol
B
Table B.8
44.51 kJ mol
B
Table B.8
H CO 2 (1400$ C)
3
7189
. kJ mol
E.B.:
nCO 'H co 'H
¦ n H ¦ n H
i
i
i
out
Ÿ n0
b
g
b1 mol COgb0.5 mol O
2
mol CO
g
a.
0
0.500 mol O 2
1.094 mol fed 0.500 mol reqd.
u 100% 119% excess oxygen
0.500 mol
Increase %XS air Ÿ Tad would decrease, since the heat liberated by combustion would go into
heating a larger quantity of gas (i.e., the additional N 2 and unconsumed O 2 ).
Excess oxygen:
9.63
g
1.094 mol O 2
Theoretical O 2
b.
b
282.99 47.07 n0 0.5 44.51 3.76 n0 71.89
i
in
Basis : 100 mol natural gas Ÿ 82 mol CH 4 , 18 mol C 2 H 6
CH (g) 2O (g) o CO (g) 2H O(v), 'H o 890.36 kJ / mol
4
2
2
2
c
7
C 2 H 6 (g) O 2 (g) o 2CO 2 (g) 3H 2 O(v), 'H co
2
82 mol CH4
18 mol C2H6
298 K
1559.9 kJ / mol
Stack gas at T(qC)
n2 (mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
n5 (mol N2)
n0 (mol air) at 423 K
0.21 O2 (20% XS)
0.79 N2
9-75
9.63 (cont’d)
2 mol O 2
Theoretical oxygen
82 mol CH 4
1 mol CH 4
1.2 u 227 mol O 2
Air fed : n1
1 mol air
0.21 mol O 2
b82.00gb1g b18.00gb2g Ÿ n
b82.00gb4g b18.00gb6g Ÿ n
C balance : n2
3
b0.79gb1297.14g
Extents of reaction: [ 1
nCH 4
bg
227 mol O 2
1 mol C 2 H 6
1297.14 mol air
218.00 mol H 2 O
b0.2gb227g mol O
20% excess air, complete combustion Ÿ n4
N 2 balance : n5
18 mol C 2 H 6
118.00 mol CO 2
2
H balance : 2n3
3.5 mol O 2
45.40 mol O 2
2
1024.63 mol N 2
82 mol, [ 1
nC2 H 6
18 mol
bg bg bg
bg
Reference states: CH 4 g , C2 H 6 g , N 2 g , O 2 g , H 2 O l at 298 K
(We will use the values of 'H co given in Table B.1, which are based on H 2 O l as a combustion
bg
product, and so must choose the liquid as a reference state for water.)
bg
b
H i T
g
C pi T 298 K for all species but water
b
g
b
g
'H v,H 2 O 298 K C p ,H 2 O T 298 K for water
Substance
n in
mol
H in
kJ mol
CH 4
C2 H 6
O2
N2
CO 2
H 2O v
82.00
18.00
272.40
1024.63
0
4.14
3.91
bg
Energy balance : 'H
e
[ 1 'H co
b
j
CH 4
H out
kJ mol
n out
mol
45.40
0.0331 T 298
1024.63
0.0313 T 298
118.00
0.0500 T 298
218.00 44.013 0.0385 T 298
b
b
b
b
g
g
g
g
0
e
[ 2 'H co
j
C2 H 6
gb
¦ n H ¦ n H
i
i
out
i
i
0
in
g b
gb g b
gb
g b
g
gb
Ÿ 82.00 mol CH 4 890.36 kJ mol 18.00 mol C 2 H 6 1559.90 kJ mol
45.40 0.0331 1024.63 0.0313 118.00 0.0500 218.00 0.0385 T 298
b.
b gb g b
gb
b218.00gb44.01g (272.40)( 4.14) (1024.63)( 3.91)
Solving for T using E - Z Solve Ÿ T
gb
g
0
2317 K
Increase % excess air Ÿ Tout decreases. (Heat of combustion has more gas to heat)
% methane increases Ÿ Tout might decrease. (lower heat of combustion, but heat released goes
into heating fewer moles of gas.)
9-76
bg
bg
af
bg
C 3 H 6 O g 4O 2 g o 3CO 2 g 3H 2 O l , 'H io
9.64
Basis :
b g
1410 m 3 STP feed gas
103 mol
3
1 min
b g
22.4 m STP
min
1821.4 kJ mol
1049 mol s feed gas
60 s
Stochiometric proportion:
b
1 mol C 3 H 6 O Ÿ 4 mol O 2 Ÿ 4 u 3.76 15.04 mol N 2 Ÿ 1 4 15.04
yC 3 H 6O
1 mol C 3 H 6 O
20.04 mol
0. 0499
mol C 3 H 6 O
, yO 2
mol
Preheat
Feed gas
1049 mol/s
0.0499 C 3 H 6 O
0.1996 O22
0.1496
0.7505 N 2
T f (°C), 150 mm Hg
Rel. satn = 12.2%
4
20.04
Ÿp
b.
20.04 mol
0.1996 mol O 2 mol
Reaction
Cooling
Feed gas
562°C
Q1 (kW)
12.2% Ÿ y C 3H 6O P
b0.0499gb1500 mm Hgg
Q2 (kW)
d i
0122
.
p C3H 6O T f
Table B.4
613.52 mm Hg
0.122
Product gas
350°C
Product gas
n 1 (mol CO 2 /s)
n 2 (mol H 2O/s)
n 3 (mol N 2/s)
T a (°C)
a.
Relative saturation
g
50.0 o C
Tf
b
gb
g 52.34 mol C H O s
b1049gb0.1996g 209.4 mol O s
b1049gb0.7505g 787.3 mol N s
n b52.34gb3g 157.0 mol CO s U 14.25 mole% CO
Ÿ Product contains n
b52.34gb3g 157.0 mol H O s|V Ÿ 14.25% H O
|W 71.5% N
787.3 mol N s
n
References : C H Obgg, O , N , H Oblg, CO at 25 C
Feed contains 1049 mol s 0.0499 C 3 H 6 O mol
3
6
2
2
1
2
2
2
3
2
2
2
2
$
3
6
2
2
2
2
H out
( kJ / mol)
Ta
n in
H in
n out
( mols) (kJ / mol) ( mols)
(562 $ C)
Substance
C3H 6O
O2
N2
CO 2
H 2O
52.34
209.4
787.3
67.66
17.72
16.65
787.3
157.0
157.0
0.031 Ta 25
0.050 Ta 25
44.013 0.038 Ta 25
H H 2 O
'H v 25$ C H H 2 O ( T)
b
b
d
b
g
g
g
i
Energy balance on reactor:
'H
b
nC 3H 6O 'H co g FGH
¦ n H ¦ n H
i
out
Ÿ 5234 mol s 18211
.
i
i
IJ
K
b g
0 kJ s
i
in
b
g
b
g
kJ
38.22 Ta 25 157.0 44.013 2.036 u 10 4
mol
9-77
0 Ÿ Ta
2871q C
9.64 (cont'd)
c.
Preheating step: References: C 3 H 6 g , O 2 , N 2 at 25q C
bg
n in
H in
n out
H out
( mol / s) (kJ / mol) (mol / s) ( kJ / mol)
(50 $ C)
(562 $ C)
52.34
315
52.34
67.66
.
209.4
0.826
209.4
17.72
787.3
0.775
787.3
16.65
Substance
C3H 6O
O2
N2
E.B. Ÿ Q 1
¦ n H ¦ n H
i
i
i
out
1.94 u 10 4 kW
i
in
af
Cooling step. References: CO 2 (g), H 2 O v , N 2 (g) at 25$ C
n
H in
n out
H out
Substance in
( mol) (kJ / mol) (mol) (kJ / mol)
e2871 Cj
e350 Cj
$
CO 2
H 2O
N2
E.B. Ÿ Q2
157.0
157.0
787.3
142.3
108.15
88.23
¦ n H ¦ n H
i
i
out
i
i
$
157.0
157.0
787.3
16.25
12.35
10.08
9.64 u 10 4 kW
in
Exchange heat between the reactor feed and product gases.
9.65
a.
Basis : 1 mol C5H12 (l)
C5 H 12 (l) 8O 2 (g) o 5CO 2 (g) 6H 2 O(v),
'H co
1 mol C5H12 (l)
n2(mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
Tad(oC)
n0 (mol O2) , 75qC
30% excess
Theoretical oxygen
30% excess Ÿ n0
C balance: n2
H balance: 2n3
3509.5 kJ / mol
1 mol C 5 H 12
8 mol O 2
1 mol C 5 H 12
8 mol O 2
1.3 u 8 10.4 mol O 2
b1gb5g Ÿ n 5 mol CO
b1gb12g Ÿ n 6 mol H O
2
2
3
2
30% excess O2, complete combustion Ÿ n4
bg b g
bg
b0.3gb8g mol O
2
Reference states: C5 H 12 l , O 2 g , H 2 O l , CO 2 (g) at 25o C
2.4 mol O 2
bg
(We will use the values of 'H c0 given in Table B.1, which are based on H 2 O l as a
combustion product, and so must choose the liquid as a reference state for water)
9-78
9.65 (cont'd)
substance
C 5 H 12
O2
CO 2
H in
nin
nout
H out
mol
1.00
kJ mol mol kJ mol
0
10.40
2.40
H 1
H 2
5.00
H
3
H 2O
z
6.00
H 4
T
H i
(C p ) i dT
i
25
e
j
2,3
z
T
'H v 25 o C (C p ) H 2 O(v) dT for H 2 O(v)
25
Table B.8
H 1
B
H O 2 (75 o C)
148
. kJ / mol
=
Substituting ( C p ) i from Table B.2 :
kJ
mol
kJ
0.9158)
mol
H 2
(0.0291 Tad 0.579 u 10 5 Tad 2 0.2025 u 10 8 Tad 3 0.3278 u 10 12 Tad 4 0.7311)
H 3
(0.03611 Tad 2.1165 u 10 5 Tad 2 0.9623 u 10 8 Tad 3 1866
.
u 10 12 Tad 4
H 4
.
. )
44.01 (0.03346 Tad 0.3440 u 10 5 Tad 2 0.2535 u 10 8 Tad 3 08983
u 10 12 Tad 4 0838
kJ
mol
. (0.03346 Tad 0.3440 u 10 5 Tad 2 0.2535 u 10 8 Tad 3 0.8983 u 10 12 Tad 4 )
4317
kJ
mol
Ÿ H 4
Energy balance : 'H
e j
nC5H12 'H co
0
C5 H 12 ( l)
¦ n H ¦ n H
i
i
i
out
i
0
in
(1 mol C5 H 12 )( 3509.5 kJ / mol) (2.40) H 2 (5.00) H 3 (6.00) H 4 (10.40)( H 1 )
0
Substitute for H 1 through H 4
'H
(0.4512 Tad 14.036 u 10 5 Tad 2 3.777 u 10 8 Tad 3 4.727 u 10 12 Tad 4 ) 3272.20 kJ / mol = 0
Ÿ f (Tad )
Check :
3272.20 0.4512 Tad 14.036 u 10 5 Tad 2 3.777 u 10 8 Tad 3 4.727 u 10 12 Tad 4
3272.20
4.727 u 10 12
6.922 u 1014
Solving for Tad using E - Z Solve Ÿ Tad
b.
Terms
1
2
3
4414 o C
Tad
% Error
7252
64.3%
3481 –21.1%
3938 –10.8%
9-79
0
9.65 (cont’d)
c.
d.
T
7252
5634
4680
4426
4414
f(T)
6.05E+03
1.73E+03
3.10E+02
1.41E+01
3.11E-02
f'(T) Tnew
3.74 5634
1.82 4680
1.22 4426
1.11 4414
1.11 4414
The polynomial formulas are only applicable for T d 1500qC
9.66
5.5 L/s at 25qC, 1.1 atm
n 1(mol CH4/s)
Adiabatic
25% excess air
Reactor
n 2 (mol O2/s)
3.76 n 2 (mol N2/s) n 4 (mol CO2/s)
150qC, 1.1 atm
n 3 (mol O2/s)
3.76 n 2 (mol N2/s)
n 5 (mol H2O/s)
T(qC), 1.05 atm
2CH 4 2O 2 o CO 2 2 H 2 O
550
. L 273 K 1.1 atm
Fuel feed rate :
s
2 u 0.247
Theoretical O 2
mol
298 K 1.0 atm 22.4 L(STP)
0.247 mol CH 4 / s
0.494 mol O 2 / s
25% excess air Ÿ n 2 125
. (0.494 ) 0.6175 mol O 2 / s ,
Ÿ 3.76 u 0.6175 2.32 mol N 2 / s
Complete combustion Ÿ [
n1 = 0.247 mol / s, n 4
n 3
0.247 mol CO 2 / s, n 5
0.6175 mol O 2 fed / s 0.494 mol consumed / s
0.124 mol O 2 / s
Re ferences: CH 4 , O 2 , N 2 , CO 2 , H 2 O at 25 o C
n in
n out
H in
H out
Substance
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
CH 4
O2
N2
CO 2
H 2O
0.247
0.6175
2.32
H 1
H (O 2, 150 o C)
Table B.8
H 2
H ( N 2, 150 o C)
Table B.8
( 'H co ) CH 4
z
0
H 1
H
2
3.78 kJ / mol
3.66 kJ / mol
890.36 kJ / mol
T
H i
C pi dT , i
35
25
9-80
.
0124
2.32
0.247
0.497
H 3
H 4
H 5
H
6
0.494 mol H 2 O / s
9.66 (cont'd)
z
T
H b
( 'H v ) H
$
2 O(25 C)
(C p ) H 2 O(v) dT
25
a.
Energy Balance
'H [ ( 'H o )
c
CH 4
¦ n
out H out
Table B.2 for C pi ( T ), ( 'H v ) H O
2
¦ n
0
in H in
44.01 kJ / mol
0.247( 890.36) 0.494(44.01) 0.0963(T 25) 174
. u 10 5 (T 2 25 2 ) 0.305 u 10 8 (T 3 25 3 )
1.61 u 10 12 (T 4 25 4 ) 0.6175(3.78) 2.32(3.66) 0
Ÿ 211.4 0.0963Tad 1.74 u 10 5 Tad 2 0.305 u 10 8 Tad 3 161
. u 10 12 Tad 4
ŸT
b.
1672 C
In product gas,
T
1672 o C, P
. u 760
105
0.155 mol H 2 O / mol
Table B.3
pH* 2 O (Tdp ) Ÿ pH* 2 O
Raoult's law: y H 2 O P
a.
798 mmHg
0.494 mol / s
(0124
.
2.32 0.247 0.494 ) mol / s
y H 2O
9.67
0
o
(0.155)(798)
124 mmHg
Ÿ
Tdp
56$ C
CH 4 (l) 2O 2 (g) o CO 2 (g) + 2H 2 O(v)
Basis : 1 mol CH 4
1 mol CH 4
Theoretical oxygen
30 % excess air Ÿ 130
. (2.00)
2 mol O 2
2.00 mol O 2
1 mol CH 4
2.60 mol O 2 , Ÿ 3.76 u 2.60
9.78 mol N 2
n2 (mol CO2)
n3 (mol H2O)
n4 (mol O2 )
1 mol CH4
2.60 mol O2
9.78 mol N2
25q C, 4.00 atm
Complete combustion Ÿ n2
1.00 mol CO 2 , n3
2.00 mol O 2 consumed Ÿ n 4
2.00 mol H 2 O
(2.60 2.00) mol O 2
Internal energy of reaction: Eq. (9.1-5) Ÿ
'U co
'H co
0.60 mol O 2
I
F
J
G
RT G ¦ Q ¦ Q J
JK
GH
e
Ÿ 'U co
z
j
CH 4
T
U
25
FG 890.36 kJ IJ 8.314 J
H
mol K mol K
Ideal Gas
(Cv )dT
Ÿ
z
298 K (1 + 2 1 2)
T
(C p Rg )dT
25
If C p is independent of T Ÿ U
(C p Rg )(T 25o C)
9-81
i
i
gaseous
products
gaseous
reactants
1 kJ
10 3 J
890.36
kJ
mol
9.67 (cont’d)
b.
bg bg bg
bg
Reference states: CH 4 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
bg
(We will use the values of 'H c0 given in Table B.1, which are based on H 2 O l as a
combustion product, and so must choose the liquid as a reference state for water.)
U in
nin
Substance
nout
U out
mol kJ mol mol
100
0
.
O2
2.60
0
N2
9.78
0
kJ mol
0.60
U 1
9.78
U
CO 2
1.00
H 2O v
2.00
CH 4
bg
2
U 3
U
4
Part a
U i
B
(C p Rg )(T 25) for all species except H 2 O(v)
'U v 25 o C (C p Rg )(T 25) 'H v 25 o C Rg Tref (C p Rg )(T 25) for H 2 O(v)
e
j
e
j
8.314 u 10 3 kJ / mol yields
Substituting given values of (C p ) i and Rg
U 1
(0.033 8.314 u 10 3 )(T 25) kJ / mol
(0.02469T 0.6172) kJ / mol
U 2
(0.032 8.314 u 10 3 )( T 25) kJ / mol
(0.02369T 0.5922) kJ / mol
U 3
(0.052 8.314 u 10 3 )(T 25) kJ / mol
(0.04369T 10922
.
) kJ / mol
LM44.01 kJ FG 8.314 u 10
N mol H
U 4
Ÿ U 4
OP
Q
IJ
K
kJ
kJ
(298 K) + (0.040 8.314 u 10 3 )( T 25)
mol ˜ K
mol
kJ
kJ
kJ
.
(0.03167T 40.74)
4153
+ (0.052 8.314 u 10 3 )(T 25)
mol
mol
mol
3
Energy Balance
Q
j ¦ n U ¦ n U
(100
. )b 890.36 kJ / molg (0.60)U
e
n CH 4 'U co
ŸQ
CH 4
i
i
out
i
0
i
in
1
(9.87)U 2 (100
. )U 3 (2.00)U 4
0
Substituting U 1 through U 4
0.3557 T 816.19
0Ÿ T
Ideal Gas Equation of State Ÿ
c.
Pf
Pi
2295 o C
Tf
Ÿ Pf
Ti
FG (2295 273) K IJ u 4.00 atm
H (25 273) K K
– Heat loss to and through reactor wall
– Tank would expand at high temperatures and pressures
9-82
34.5 atm
9.68
b.
1 mol natural gas
yCH 4 (mol CH 4 / mol)
nCO 2 (mol CO 2 )
yC2 H 6 (mol C 2 H 6 / mol)
nH 2 O (mol H 2 O)
yC3H 8 (mol C 3 H 8 / mol)
nN 2 (mol N 2 )
nO 2 mol O 2 )
Humid air
na (mol air)
ywo (mol H20(v)/mol)
(1-ywo) (mol dry air/mol)
0.21 mol O2/mol DA
0.79 mol N2/mol DA
Basis : 1 g-mole natural gas
CH 4 (g) 2O 2 (g) o CO 2 (g) H 2 O(v)
7
O 2 (g) o 2CO 2 (g) 3H 2 O(v)
2
C 3 H 8 (g) 5O 2 (g) o 3CO 2 (g) 4 H 2 O(v)
C 2 H 6 (g) Theoretical oxygen :
2 mol O2
1 mol CH 4
yCH4 (mol CH 4 )
3.5 mol O2
1 mol C2 H 6
yC2H6 (mol C2 H 6 )
5 mol O2
1 mol C3 H8
yC3H8 (mol C 3 H 8 )
( 2 yCH 4 35
. yC 2 H 6 + 5 y C 3 H 8 )
Excess oxygen: 0.21n a (1 y wo )
FG
H
Ÿ na = 1 IJ
K
FG1 P IJ ( 2 y
H 100 K
xs
CH 4
3.5 y C 2 H 6 + 5 y C 3H 8 ) mol O 2
Pxs
1
( 2 y CH 4 3.5 y C 2 H 6 + 5 y C 3H 8 )
mol air
100
0.21(1 y w0 )
Feed components
(n O 2 ) in
0.21n a (1 y wo ), (n N 2 ) in
0.79n a (1 y wo ), (n H 2O ) in
na y wo
N 2 in product gas: n N 2 = (n N 2 ) in mol N 2
CO2 in product gas :
nCO2
1 mol CO2 nCH4 (mol CH4 ) 2 mol CO2 nC2H6 (mol C2 H6 ) 3 mol CO2 nC3H8 (mol C3 H8 )
1 mol CH4
1 mol C2 H 6
1 mol C3 H8
(nCH4 2nC2H6 3nC3H8 ) mol CO2
H 2 O in product gas :
nH2O
1 mol H 2 O nCH4 (mol CH 4 ) 3 mol H 2 O nC2H6 (mol C2 H 6 ) 4 mol O2 nC3H8 (mol C3 H8 )
1 mol CH 4
1 mol C2 H 6
1 mol C3 H8
[2nCH4 3nC2H6 4nC3H8 + na (1- ywo )] mol H 2 O
O 2 in product gas : n O 2
Pxs
( 2n CH 4 35
. n C 2 H 6 + 5 nC3 H 8 ) mol O 2
100
9-83
9.68 (cont’d)
c.
References : C(s), H 2 (g) at 25o C
z
T
H
CH 4 ( T)
( ' H of ) CH 4 (C p ) CH 4 dT
25
Using ( ' H fo )CH 4 from Table B.1 and (C p )CH 4 from Table B.2
. kJ / mol +
H CH (T) 7485
4
F
GG
H
z
I
JJ
K
T
u 108 T 2 1100
.
. u 1012 T 3 ) dT kJ / mol
(0.03431+5.469 u 105 T 03661
25
Ÿ H CH (T ) [7572
. + 3.431 u 10 2 T + 2.734 u 10 5 T 2 0122
. u 10 8 T 3 2.75 u 10 12 T 4 ] kJ / mol
4
7
¦
6
(ni ) out ( Hi ) out ¦(n )
i in ( Hi ) in
i 4
H i
i 1
3
ai bi T ci T 2 d i T ei T 4
6
¦
H out
n out
mol kJ / mol mol kJ / mol
n1
H 1
n2
H 2
n3
H3
n4
H 4
n7
H 7
n5
H 5
n8
H 8
n6
n9
H 9
n10
H 10
CH 4
C2 H 6
C3H 8
O2
N2
CO 2
H 2O
'H
H in
n in
Substance
3
¦
(ni ) in ( Hi ) in
i 1
(ni ) in H i (Tf ) i 1
6
¦(n )
i in Hi (Ta )
i 4
7
Ÿ 'H
¦
3
(ni ) out (ai bi T ci T 2 d i T 3 ei T 4 ) out i 4
7
Ÿ 'H
¦ (n )
i out ai
i 1
¦ (n )
i out bi T ¦ (n )
¦ (n )
2
i 1
i 4
3
i out ci T
)
i 1
¦(n )
i out d i T
i 1
6
i in Hi (Tf
¦ (n )
i in Hi (Ta )
i 4
D 0 D 1T D 2 T D 3T D 4 T 4
2
7
where D 0
¦
3
(ni ) out ai i 1
7
D1
3
¦
(ni ) in H i (Tf ) i 1
¦ (n )
¦ (n )
i out bi
i 4
D2
¦(n )
i out ci
i 1
7
¦ (n )
i out d i
D4
¦ (n )
i out ei
i 1
i 1
.
9-84
i in Hi (Ta )
7
i 1
7
D3
6
6
¦ (n )
i in Hi (Ta )
i 4
7
i 1
7
7
7
¦
(ni ) in H i (Tf ) 3
¦ (n )
i out ei T
i 1
4
9.68 (cont’d)
d.
Run 1
yCH4
0.75
yC2H6
0.21
yC3H8
0.04
Tf
40
Ta
150
Pxs
25
ywo 0.0306
nO2i
3.04
nN2
11.44
nH2Oi
0.46
HCH4
-74.3
HC2H6
-83.9
HC3H8
-102.7
HO2i
3.6
HN2i
3.8
HH2Oi
-237.6
nCO2
1.29
nH2O
2.75
nO2
0.61
nN2
11.44
Tad 1743.1
alph0
-1052
alph1 0.4892
alph2 0.0001
alph3 -3.00E-08
alph4 3.00E-12
Delta H 3.00E-07
Species
CH4
C2H6
C3H8
O2
N2
H20
CO2
a
-75.72
-85.95
-105.6
-0.731
-0.728
-242.7
-394.4
Run 2
0.86
0.1
0.04
40
150
25
0.0306
2.84
10.67
0.43
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.18
2.61
0.57
10.67
1737.7
-978.9
0.4567
1.00E-04
-3.00E-08
3.00E-12
9.00E-06
Run 3
0.75
0.21
0.04
150
150
25
0.0306
3.04
11.44
0.46
-70
-77
-93
3.6
3.8
-237.6
1.29
2.75
0.61
11.44
1750.7
-1057
0.4892
0.0001
-3.00E-08
3.00E-12
-4.00E-07
Run 4
0.75
0.21
0.04
40
250
25
0.0306
3.04
11.44
0.46
-74.3
-83.9
-102.7
6.6
6.9
-234.1
1.29
2.75
0.61
11.44
1812.1
-1099
0.4892
0.0001
-3.00E-08
3.00E-12
-1.00E-04
Run 5
0.75
0.21
0.04
40
150
100
0.0306
4.87
18.31
0.73
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.29
3.02
2.44
18.31
1237.5
-1093
0.7512
0.0001
-4.00E-08
4.00E-12
-1.00E-05
b
x 10^2
3.431
4.937
6.803
2.9
2.91
3.346
3.611
c
x 10^5
2.734
6.96
11.3
0.11
0.579
0.344
2.117
d
x 10^8
0.122
-1.939
-4.37
0.191
-0.203
0.254
-0.962
e
x 10^12
-2.75
1.82
7.928
-0.718
0.328
-0.898
1.866
9-85
Run 6
0.75
0.21
0.04
40
150
25
0.1
3.04
11.44
1.61
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.29
3.9
0.61
11.44
1633.6
-1058
0.5278
0.0001
-2.00E-08
2.00E-12
6.00E-04
9.69
n 14 (mol CH4 /h)
25°C
Preheaters
Absorber off-gas
n 8 (mol H 2 /h)
n 12 (mol N 2 /h)
0.988 n 9 (mol CO/h)
0.950 n 6 (mol CH4 /h)
0.006 n 7 (mol C 2H 2/h)
n 13 (mol C(s )/h)
Converter
converter product quench
Tad (°C)
Feed gas, 650°C
n 14 (mol CH4 /h)
0.96n 15 (mol O2 /h)
0.04n 15 (mol N2 /h)
0.917 n 1 (mol DMF/h)
Lean solvent
Converter
product
38°C
filter
absorber
n 6 (mol CH4 /h)
n 7 (mol C 2H 2/h)
n 8 (mol H 2 /h)
n 9 (mol CO/h)
n 10 (mol CO2 /h)
n 11 (mol H 2 O/h)
n 12 (mol N 2 /h)
n 13 (mol C(s )/h)
n 15 (mol/h)
0.96 mol O2 /mol
0.04 mol N2 /mol
25°C
Basis:
5000 kg/h Product gas
n 1 (mol/h)
0.991 mol C2 H2 ( g)/mol
0.00059 mol H2 O/mol
0.00841 mol CO2 /mol
n 6 (mol CH4 /h)
n 7 (mol C 2H 2/h)
n 8 (mol H 2 /h)
n 9 (mol CO/h)
n 10 (mol CO2 /h)
n 11 (mol H 2 O/h)
n 12 (mol N 2 /h)
Rich solvent
n 1 (mol/h)
0.0155 mol C2 H2 /mol
0.0063 mol CO2 /mol
0.00055 mol CO/mol
0.00055 mol CH4 /mol
0.0596 mol H2 O/mol
0.917 mol DMF/mol
Average M.W. of product gas:
b
g
b
g
b
g
0.991 26.04 0.00059 18.016 0.00841 44.01
M
Molar flow rate of product gas: n0
5000 kg 103 g
day
1 kg
1 mol
1 day
26.19 g
24 h
7955 mol h
Stripper balances: C 2 H 2 Ÿ n1 , CO Ÿ n2 , CH 4 Ÿ n3 , H 2 O Ÿ n4 , CO 2 Ÿ n5
Absorber balances: CH 4 Ÿ n6 , C 2 H 2 Ÿ n7 , CO Ÿ n9 , CO 2 Ÿ n10 , H 2 O Ÿ n11
13 , n14 ,
converter H balance Ÿ n8
Converter O balance Ÿ n15 , converter N 2 balance Ÿ n12
Stripper balances:
C 2 H 2 : 0.0155n1
b
g
0.991 7955 mol h Ÿ n1
b
5.086 u 105 mol h
ge
j n Ÿ n 79.7 mol CO h
CH : b0.00055ge5.086 u 10 j n Ÿ n 79.7 mol CH h
H O: b0.0596ge5.086 u 10 j n b0.00059gb7955g Ÿ n = 30308 mol H O h
CO : b0.0068ge5.086 u 10 j n b0.00841gb7955g Ÿ n = 3392 mol CO h
CO: 0.00055 5.086 u 105
2
2
5
4
3
4
3
5
2
4
4
2
5
2
5
Absorber balances
CH 4 : n6
b
5
ge
0.950n6 0.00055 5.086 u 105
9-86
j
Stripper off-gas
n 2 (mol CO/h)
n 3 (mol CH4 /h)
n 4 (mol H 2O(v )/h)
n 5 (mol CO2 /h)
2619
. g mol
Material balances -- plan of attack (refer to flow chart):
RS5.67% soot formationUV Ÿ n
Tconverter C balance W
stripper
n6 Ÿ 5595 mol CH 4 h
2
9.69 (cont'd)
b0.0155ge5.086 u 10 j 0.006n Ÿ n 7931 mol C H h
0.988n b0.00055ge5.086 u 10 j Ÿ n = 23311 mol CO h
CO: n
CO : n
b0.0068ge5.086 u 10 j 3458 mol CO h
H O: n
b0.0596ge5.086 u 10 j 30313 mol H O h
n
b0.0567gn (mol CH ) 1 mol C Ÿ n 0.0567n b1g
Soot formation:
h
1 mol CH
5
C 2 H 2 : n7
7
7
2
2
5
9
9
9
5
2
10
2
5
2
11
2
13
14
4
13
14
4
Converter C balance:
b5595 mol CH hgb1 mol C mol CH g b7931gb2g b23311gb1g b3458gb1g n
Ÿn
n 48226 b2g
2899 mol C bsg h , n
51120 mol CH h
Solve (1) & (2) simultaneously Ÿ n
51120 mol CH
4 mol H
Converter H balance:
b5595gb4g b7931gb2g 2n b30313gb2g
h
1 mol Ch
n14
4
14
13
4
13
13
14
CH 4
4
4
C2 H2
H 2O
H2
8
4
Ÿ n8
b
gb g
Converter O balance: 0.96n15 2
52816 mol H 2 h
23311 mol CO
1 mol O
h
1 mol CO
b
CO 2
gb g b
H 2O
gb g
3458 2 30313 1
Ÿ n15 31531 mol h
b gb
g
Converter N 2 balance: 0.04 31531 n12 Ÿ n12
a.
Feed stream flow rates
VCH 4
VO 2
b.
1261 mol N 2 h
51120 mol CH 4
h
b
31531 mol O 2 N 2
h
Gas feed to absorber
5595 mol
7931 mol
23311 mol
3458 mol
30313 mol
52816 mol
1261 mol
b g
0.0244 m 3 STP
1 mol
g
1145 SCMH CH 4
b g
0.0244 m 3 STP
1 mol
b
706 SCMH O 2 N 2
g
U|
||
|V
||
||
W
CH 4 h
C2 H 2 h
CO h
CO 2 h
4.5 mole% CH 4 , 6.4 % C 2 H 2 , 18.7% CO ,
H 2 O h Ÿ 125 kmol h , 2.8% CO 2 , 24.3% H 2 O , 42.4% H 2 , 1.0% N 2
H2 h
N2 h
1.2469 u 10 5 mol h
Absorber off-gas
52816 mol H 2 h
1261 mol N 2 h
23031 mol CO h
64.1 mole% H 2 , 1.5% N 2 , 27.9% CO,
5315 mol CH 4 h Ÿ 82.5 kmol h, 6.4% CH , 0.06% C H
4
2 2
41.6 mol C 2 H 2 h
8.2471 u 10 4
U|
||
V|
|
mol h|W
9-87
9.69 (cont'd)
Stripper off-gas
279.7 mol CO h
279.7 mol CH 4 h
30308 mol H 2 O h Ÿ 34.3 kmol h, 0.82% CO, 0.82% CH 4 , 88.5% H 2 O, 9.9% CO 2
3392 mol CO h
3.4259 u 10 4
U|
|V
||
mol h W
c.
DMF recirculation rate
d.
Overall product yield
mol I F 1 kmol I
FG
JG
J 466 kmol DMF h
H
h K H 10 mol K
b0.991gb7955g mol C H in product gas 0154
mol C H
.
0.917 5.086 u 10 5
3
2
2
2
51120 mol CH 4 in feed h
2
mol CH 4
The theoretical maximum yield would be obtained if only the reaction 2CH 4 o C 2 H 2 3H 2
occurred, the reaction went to completion, and all the C 2 H 2 formed were recovered in the
product gas. This yield is (1 mol C2H2/2 mol CH4) = 0.500 mol C2H2/2 mol CH4.
The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308.
e.
Methane preheater
Q CH 4
'H
n14
z
Table B.2
650
B
d i
Cp
25
dT
CH 4
51120 mol 32824 J
1h
1 kJ
h
mol
3601 s 10 3 J
466 kW
Oxygen preheater
Table B.8
Q O 2
'H
Table B.8
B
B
0.96n15 H (O 2 ,650 $ C) 0.04n15 H ( N 2 ,650 $ C)
kJ OF 1 h I
FG 31531 mol IJ LMb0.96 u 20135
.
0.04 u 18.99g
PG J
K
H
h N
mol ˜ C QH 3600 s K
References : Cbsg, H bgg, O bgg, N bgg at 25q C
Substance
n
H b650q Cg n
H bT g
$
f.
2
in
CH 4
51120
O2
30270
2
in
2
out
42.026
20.125
18.988
5595
out
74.85 z
out
Ta
25
z
z
z
z
z
z
z
Ta
C p dT
N2
1261
C2 H 2
H2
52816
CO
23311 110.52 C p dT
CO 2
3458
H 2O
. C p dT
30313 24183
bg
2899
Cs
1261
35
176 kW
7931
C p dT
226.75
Ta
25
9-88
C p dT
C p dT
393.5 C p dT
C p dT
b g
H b kJ molg
n mol h
9.69 (cont’d)
z
T
H i
'H i0 C pi dT
kJ mol
¦ n H
i
¦ n H
i
.
1575
u 10 6 kJ h
i
in
i
out
25
kJ mol ˜qC
z
LM5595dC i 1261dC i 7931dC i
N
OP 1 kJ dT
52816dC i 23311dC i 3458dC i
3013dC i
b g Q 10 J
1 kJ
dC i b g u 10 J dT
9.888 u 10 6 kJ h Tout
p CH
4
25
z
p H
2
p CO
p N
2
p CO
2
p C H
3 2
p H O v
2
3
Tad 273
p C s
298
3
We will apply the heat capacity formulas of Table B.2, recognizing that we will probably
push at least some of them above their upper temperature limits
¦ n H
i
z
Tad 273
298
i
Tad
4
j
T 2 1.0162 u 10 7 T 3 dT
25
out
¦ n H
z
e3902 1.2185 5.9885 u 10
F 32.411 0.031744T 14179
u 10 I
.
JKdT
GH
T
9.888 u 10 6 kJ h i
i
6
2
1.000 u 10 7 3943Ta 0.6251Ta2 1.996 u 10 4 Ta3 2.5405 u 10 8 Ta4 out
Energy balance: 'H
¦ n H ¦ n H
i
out
b g
Ÿ f Tc
i
i
i
0
in
. u 10 4 Tc3 2.5405 u 10 8 Tc4 8.485 u 106 3943Tc 0.6251Tc2 1996
E-Z Solve
Tc
.
u 10 6
1418
Ta 273
2032 o C.
9-89
1418
. u 106
Tc 273
0
9.70 a.
n1[ kg W(v) / d]
W = H2O
100o C
o
24,000 kg sludge / d, 22 C
0.35 solids, 0.65 W(l)
DRYER
Q 2
F
n2 (kg conc. sludge / d), 100o C
0.75 solids, 0.25 W(l)
INCINERATOR
Waste gas
n3 [kg W(v) / d]
4B, sat'd
C
n3 [kg W(l) / d]
n3 [kg W(l) / d]
4B, sat'd
Q 3 ( kJ / d)
BOILER
20o C
D
0.10
kmol C2 H 6
110oC
kmol
Q 4 (kJ / d)
Q1
n4 ( kg oil / d)
0.87 C
0.10 H
0.0084 S
0.0216 ash
n6 (kg gas / d)
kmol CH 4
0.90
kmol
Stack gas
CO2 , H 2O(v)
125o C
SO2
O2 , N 2
ash
n7 (kg air / d)
25o C
Q 0 ( kJ / d)
E n5 (kg air / d)
25o C
9-90
9.70 (cont'd)
Solids balance on dryer:
0.35 u 24,000 kg / d 0.75n2 Ÿ n2
11200 kg / d Ÿ
n1 11200 Ÿ n1
Mass Balance on dryer: 24,000
F 11.2 tonnes / d (conc. sludge)
12,800 kg / d
Energy balance on sludge side of dryer:
References : H 2 O(l,22 $ C), Solids(22 $ C)
nin
nout
H in
H out
(mol d ) (kJ mol) (mol d) (kJ mol)
8400
0
8400
H 1
15600
0
2800
H 2
12800
H 3
Substance
Solids
H 2 O(l)
H 2O(v)
H 1
H
2.5(100 22)
(419.1 92.2)
326.9 kJ / kg
H 1
( H
(2676 92.2)
2584 kJ / kg
2
water
195.0 kJ / kg
from Table B.5)
¦ m H ¦ m H Ÿ Q
Q 2
i
i
out
i
in
7
3.56 u 10
0.55
Q steam
3.56 u 10 7 kJ day
2
i
6.47 u 10 7 kJ / d Ÿ Q 3
2.91 u 10 7 kJ / d
Energy balance on steam side of dryer:
6.47 u 10 7
FG IJ
H K
'H v for
H 2 O(sat'd, )
B
kJ
kg
= n3
u 2133
d
d
FG kJ IJ F 1 tonne I Ÿ n
H kg K GH 10 kg JK
3
3
30.3 tonnes / d (boiler feedwater)
Energy balance on steam side of boiler:
(30300
Q1
kg
kJ
)(2737.6 83.9)
d
kg
8.04 u 10 7 kJ / d
62% efficiency Ÿ Fuel heating value needed =
Ÿ n4
130
. u 108 kJ / d
3.75 u 104 kJ / kg
8.04 u 10 7
0.62
13
. u 108 kJ / d
3458 kg / d Ÿ D = 3.5 tonnes / day (fuel oil)
Air feed to boiler furnace: C + O 2 o CO 2 , 4H + O 2 o 2H 2 O,
(nO 2 ) theo
3458
S + O 2 o SO 2
LM
N
kJ
kgC 1 kmol C 1 kmol O 2
1 1
1 1
(0.87
)(
)(
) + (0.10)( )( ) (0.0084)( )( )
kg
kg
1 4
12 kg
1 kmol C
32 1
338 kmol O 2 / d
Air fed (25% excess) = 1.25(4.76
Ÿ
kmol air
kmol O2
kmol air
)(338
) 2011
kmol O2
d
d
2011 kmol 29 kg 1 tonne
Ÿ E = 58.3 tonnes/ d (air to boiler)
d
kmol 103 kg
9-91
OP
Q
9.70 (cont’d)
2011 kmol 29.25 kJ (125 25) $ C
Energy balance on boiler air preheater: Q o
d
kmol $ C
Ÿ Q o
588
. u 10 6 kJ / d
Supplementary fuel for incinerator:
n6
11.2 tonne sludge 195 SCM
d
MWgas
M gas
tonne
1 kmol
22.4 SCM
0.90 MWCH 4 0.10 MWC2 H 6
97.5 kmol d
(0.90)(16) (0.10)(30)
17.4 kg kmol
= 1.7 tonne / d (natural gas)
(97.5)(17.4) Ÿ G
CH 4 2O 2 o CO 2 + 2H 2 O, C 2 H 6 7
O 2 o 2CO 2 + 3H 2 O
2
Air feed to incinerator:
11200 75 kg sol 19000 kJ 3.2 kg air 1 kmol
(O 2 ) th for sludge:
d
kg
kg sol
104 kJ
29 kg
(O 2 ) th for gas: 97.5
n7
2(1761 210)
Ÿ 1.88 u 104
kmol O 2
kmol air
u 4.76
d
kmol O 2
OP
Q
210 kmol O 2 d
. u 104 kmol air / d
188
kmol
kg
= 545 tonnes / d ( incinerator air)
u 29
ŸH
d
kmol
Air preheater : Q 4
b.
LM
N
kmol CH 4 2 kmol O 2
kmol
0.90
u
(010
. )(3.5)
d
kmol
kmol CH 4
1761 kmol O 2 d
1.88 u 104 kmol
29.2 kJ
$
(110 25) $ C
Ÿ Q 4
4.7 u 10 7 kJ / d
d
kmol C
Cost of fuel oil, natural gas, fuel oil and air preheating, pumping and compression, piping, utilities,
operating personnel, instrumentation and control, environmental monitoring. Lowering environmental
hazard might justify lack of profit.
c.
Put hot product gases from boiler and/or incinerator through heat exchangers to preheat both air streams.
Make use of steam from dryer.
d.
Sulfur dioxide, possibly NO2, fly ash in boiler stack gas, volatile toxic and odorous compounds in gas
effluents from dryer and incinerator.
9-92
CHAPTER TEN
10.1 b. Assume no combustion
n 1 (mol gas),T1 (°C)
x 1 (mol CH4 /mol)
x 2 (mol C2 H6 /mol)
1 – x 1 – x 2 (mol C3 H8 /mol)
n 3 (mol), 200°C
y 1 (mol CH4 /mol)
y 2 (mol C2 H6 /mol)
y 3 (mol C3 H8 /mol)
1 – y 1 – y 2 – y 3 (mol air/mol)
n 2 (mol air), T2 (°C)
Q (kJ)
b
b
g
11 variables
n1 , n2 , n3 , x1 , x 2 , y1 , y 2 , y 3 , T1 , T2 , Q
5 relations
4 material balances and 1 energy balance
6 degrees of freedom
g
ln , n , x , x , T , T q
A feasible set of design variables:
1
2
1
2
1
2
Calculate n3 from total mole balance, y1 , y 2 , and y 3 from component balances,
Q from energy balance.
An infeasible set:
ln , n , n , x , x , T q
1
2
3
1
2
1
Specifying n1 and n2 determines n3 (from a total mole balance)
c.
n 2 (mol gas), T 2 , P
y 2 (mol C 6 H 14/mol)
1 – y 2 (mol N 2 /mol)
n 1 (mol gas), T 1 , P
y 1 (mol C 6 H 14/mol)
1 – y 1 (mol N 2 /mol)
n 3 (mol C 6 H 14( l )/mol), T 2 , P
Q (kJ)
b
d
g
n1 , n2 , n3 , y1 , y 2 , T1 , T2 , Q, P
9 variables
4 relations
2 material, 1 energy, and 1 equilibrium: y 2 P
5 degrees of freedom
A feasible set:
b gi
PC*6 H14 T2
ln, y , T , P, n q
1
1
3
Calculate n2 from total balance, y 2 from C 6 H 14 balance, T2 from Raoult’s law:
[ y2 P
b g
PC6 H 4 T2 ], Q from energy balance
An infeasible set:
ln , y , n , P, T q
2
2
3
2
Once y 2 and P are specified, T2 is determined from Raoult’s law
10- 1
b
g
10.2 10 variables n1 , n2 , n3 , n4 , x1 , x 2 , x 3 , x 4 , T , P
2 material balances
2 equilibrium relations: [ x 3 P x 4 PB* T , 1 x 3 P
6 degrees of freedom
bgb
g b1 x g P bT g]
4
*
C
ln , n , n , x , x , Tq
a. A straightforward set:
1
3
4
1
4
Calculate n2 from total material balance, P from sum of Raoult's laws:
P
bg b
g bg
x 4 p B T 1 x 4 Pc T
x 3 from Raoult's law, x 2 from B balance
b. An iterative set:
ln , n , n , x , x , x q
1
2
3
1
2
3
Calculate n4 from total mole balance, x 4 from B balance.
Guess P, calculate T from Raoult's law for B, P from Raoult’s law for C, iterate until
pressure checks.
c. An impossible set:
ln , n , n , n , T , Pq
1
2
3
4
Once n1 , n2 , and n3 are specified, a total mole balance determines n4 .
bg
bg
bg
bg
10.3 2BaSO 4 s 4C s o 2BaS s 4CO 2 g
a.
100 kg ore, T 0 (K)
xb (kg BaSO4 /kg)
n 1 (kg C)
n 2 (kg BaS)
n 3 (kg CO2 )
n 4 (kg other solids)
T f (K)
n 0 (kg coal), T 0 (K)
xc (kg C/kg)
Pex (% excess coal)
Q (kJ)
d
i
11 variables n0 , n1 , n2 , n3 , n4 , x b , x c , T0 , T f , Q, Pex
5 material balances C, BaS, CO 2 , BaSO 4 , other solids
1 energy balance
1 reaction
1 relation defining Pex in terms of n0 , x b , and x c
5 degrees of freedom
n
b
b. Design set: x b , x c , T0 , T f , Pex
g
s
Calculate n0 from x b , x c , and Pex ; n1 through n4 from material balances,
Q from energy balance
10-2
10.3 (cont’d)
l
q
c. Design set: x B , x c , T0 , n2 , Q
Specifying x B determines n2 Ÿ impossible design set.
l
q
d. Design set: x B , x c , T0 , Pex , Q
Calculate n2 from x B , n3 from x B
n0 from x B , x c and Pex
n1 from C material balance, n4 from total material balance
T f from energy balance (trial-and-error probably required)
10.4 2C 2 H 5 OH O 2 o 2CH 3 CHO 2H 2 O
2CH 3 COH O 2 o 2CH 3 CHOOH
n f (mol solution), T 0
x ef (mol EtOH/mol)
1 – x ef (mol H 2 O/mol)
n e (mol EtOH), T
n ah (mol CH 3 CHO)
n ea (mol CH 3 COOH)
n w (mol H 2O)
n ax (mol O 2)
n n (mol N 2)
n w (mol air), Pxs , T 0
0.79 n air (mol N 2 )
0.21 n air (mol O 2 )
Q (kJ)
(Pxs = % excess air)
a.
d
13 variables n f , naw , ne , neh , nea , n w , nex , n0 , x ef , T0 , T , Q, Pxs
6 material balances
1 energy balance
1 relation between Pxs , n f , x ef , and nair
2 reactions
i
7 degrees of freedom
n
b. Design set: n f , x ef , Pxs , ne , nah , T0 , T
s
Calculate nair from n f , x ef and Pxs ; nn from N 2 balance;
naa and nw from n f , x ef , ne , nah and material balances;
nex from O atomic balance; Q from energy balance
n
c. Design set: n f , x ef , T0 , nair , Q, ne , n w
s
Calculate Pxs from n f , x ef and nair ; n’s from material balances; T from energy
balance (generally nonlinear in T)
l
q
d. Design set: nair , nn , . Once nair is specified, an N 2 balance fixes nn
10- 3
10.5
a.
n 1 (mol CO)
n 2 (mol H2 )
reactor
n 3 (mol C3 H6 )
n 4 (mol C3 H6 )
n 5 (mol CO)
n 6 (mol H2 )
n 7 (mol C7 H8 O)
n 8 (mol C4 H7 OH)
n 9 (kg catalyst)
Flash
tank
n10 (kg catalyst)
n 16(mol C3 H6 )
n 11(mol C3 H6 )
n 12(mol CO)
n 17(mol CO)
Separation n 18(mol H2 )
n 13(mol H2 )
n 14(mol C7 H8 O)
n 15(mol C4 H7 OH)
n 19(mol C7 H8 O)
n 20(mol C4 H7 OH)
n 21(mol H2 )
Hydrogenator
n 22(mol H2 )
n 20(mol C4 H7 OH)
b
g
b
g
Reactor:
10 variables n1 n16
6 material balances
2 reactions
6 degrees of freedom
Flash Tank:
12 variables n4 n15
6 material balances
6 degrees of freedom
Separation:
b
10 variables n11 n20
5 material balances
g
5 degrees of freedom
Hydrogenator:
b
5 variables n19 n23
3 material balances
1 reaction
g
3 degrees of freedom
Process:
20 Local degrees of freedom
14 ties
6 overall degrees of freedom
The last answer is what one gets by observing that 14 variables were counted two times
each in summing the local degrees of freedom. However, one relation also was counted
twice: the catalyst material balances on the reactor and flash tank are each n9 n10 . We
must therefore add one degree of freedom to compensate for having subtracted the same
relation twice, to finally obtain 7 overall degrees of freedom (A student who gets this one
has done very well indeed!)
b. The catalyst circulation rate is not included in any equations other than the catalyst balance
(n9 = n10). It may therefore not be determined unless either n9 or n10 is specified.
10-4
b
10.6 n C 4 H 10 o i C 4 H 10 n B i B
n 1 (mol n-B)
mixer
n 2 (mol n-B)
n 3 (mol i-B)
g
reactor
n 4 (mol n-B)
n 5 (mol i-B)
n 6 (mol)
x 6 (mol n-B/mol)
(1 – x 6)(mol i-B/mol)
still
n r (mol)
x r (mol n-B/mol)
(1 – x r)(mol i-B/mol)
b
a. Mixer:
5 variables n1 , n2 , n3 , nr , x r
2 material balances
g
3 degrees of freedom
b
g
Reactor:
4 variables n2 , n3 , n3 , n5
2 material balances
1 reaction
3 degrees of freedom
Still:
6 variables n4 , n5 , n6 , x 6 , nr , x r
2 material balances
b
g
4 degrees of freedom
Process:
b. n1
10 Local degrees of freedom
6 ties
4 overall degrees of freedom
100 mol n C 4 H 10 , x 6
b gb g
b
.
4 0.885 4 mol C Ÿ n6
n6 0115
100 mol n - B fed
Mixer n-B balance: 100 0.85nT
n2
b1g
35% S.P. conversion: n4
Still n – B balance:
n4
0.65n2 Ÿ n4
b1g
65 0.5525nr
b2 g
b2g
. gb100g 0.85n Ÿ n 179.83 mol
b0115
mol recycle
.
b179.83 mol recycleg b100 mol fresh feedg 179
mol fresh feed
n6 x 6 nr x r Ÿ 65 0.5525nr
Recycle ratio
0.85 mol n C 4 H 10 mol
100 mol overhead
gb g b gb g
. gmol n - B unreacted
100 mol n - B fed b100gb0115
u 100% 88.5%
Overall C balance: 100 4
Overall conversion
0.115 mol n C 4 H 10 mol , x r
r
10.6 (cont’d)
10- 5
r
c.
nr
n2
n3
100 0.85n r
n r 1 0.85
n4
0.65n 2
n5
n2 n 3 n 4
b
k 1 k 2 k 3
100.0 132.3 1515
.
185.0 212.5 228.8
g
15.0 19.85 22.73
120.25 138.1 148.7
UV
W
n 4 n5 n 6 n r
n6
Ÿ
n 4 0.115n 6 0.85n r
nr
Error:
0.595
0.595 1
q
nrb
94.21 102.8
80.76 88.55
132.3
1515
. 163.0
179.83 163.0
u 100 9.3% error
179.83
. 132.3
1515
132.3 100.0
d. w
79.75
67.69
0.595
1470
.
b
g c b
ghb g
g
1470
132.3 1 1470
1515
.
.
.
Error:
179.8 179.8
u 100 01%
.
error
179.8
3
179.8
e. Successive substitution, Iteration 32: nr = 179.8319 Æ nr = 179.8319
Wegstein, Iteration 3: nr = 179.8319 Æ nr = 179.8319
S1
10.7
SF
Split
S2
a.
1
2
3
4
5
6
7
8
A
X1 =
nA
nB
nC
nD
T(deg.C)
B
C
0.6
Molar flow rates (mol/h)
SF
S1
85.5
51.3
52.5
31.5
12.0
7.2
0.0
0.0
315
315
Formula in C4: = $B$1*B4
Formula in D4: = B4-C4
10.7 (cont’d)
10-6
D
S2
34.2
21.0
4.8
0.0
315
b.
C **CHAPTER 10 -- PROBLEM 7
DIMENSION SF(8), S1(8), S2(8)
FLOW 150.
N 3
SF(1) 0.35*FLOW
SF(2) 0.57*FLOW
SF(3) 0.08*FLOW
SF(8) 315.
X1 0.60
CALL SPLIT (SF, S1, S2, X1, N)
WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B)
WRITE (6, 900)' STREAM 2', S2(1), S2(2), S2(3), S2(B)
900
FORMAT (A10, F8.2,' mols/h n-octane', /,
* 10X, F8.2,' mols/h iso-octane', /,
* 10X, F8.2,' mols/h inerts', /,
*
10X, F8.2,' K')
END
C
C
C
100
SUBROUTINE SPLIT
SUBROUTINE SPLIT (SF, S1, S2, X1, N)
DIMENSION SF(8), S1(8), S2(8)
D0 100 J 1, N
S1(J) X1*SF(J)
S2(J) SF(J) – S1(J)
S1(8) SF (8)
S2(8) SF (8)
RETURN
END
Program Output: Stream 1 3150
. mols h n-octane
51.30 mols h iso-octane
7.20 mols h inerts
315.00 K
Stream 2 21.00 mols h n-octane
34.20 mols h iso-octane
4.80 mols h inerts
315.00 K
10- 7
10.8
a. Let Bz = benzene, Tl = toluene
Antoine equations: p *Bz 10 6.905651211.033/( T 220.790) (= 1350.491)
pTl*
Raoult' s law: x Bz
10 6.953341343.943/( T 219.377) (= 556.3212)
( P pTl* ) / ( p *Bz - pTl* ) (= 0.307) , y Bz
Total mole balance: 100 = nv nl
Benzene balance:
40 = y Bz nv x Bz nl
*
/ P ( 0.518)
x Bz p Bz
UV
W
Fractional benzene vaporization: f B
40 100x Bz
(= 44.13), nl
y Bz x Bz
nv y Bz / 40 (= 0.571)
Fractional toluene vaporization: f T
nv (1 y Bz ) / 60 (= 0.354)
Ÿ nv
100 nv (= 55.87)
The specific enthalpies are calculated by integrating heat capacities and (for vapors)
adding the heat of vaporization.
Q ¦ nout H out ¦ nin H in (= 1097.9)
b. Once the spreadsheet has been prepared, the goalseek tool can be used to determine the
bubble-point temperature (find the temperature for which nv=0) and the dew-point
temperature (find the temperature for which nl =0). The solutions are
Tbp 96.9 o C, Tdp 103.2 o C
C ** CHAPTER 10 PROBLEM B
DIMENSION SF(3), SL(3), SV(3)
DATA A1, B1, C1/6.90565, 1211.033, 220.790/
DATA A2, B2, C2/6.95334, 1343.943, 219.377/
DATA CP1, CP2, HV1, HV2/ 0.160, 0.190, 30.765, 33.47/
COMMON A1, B1, C1, A2, B2, C2, CP1, CP2, NV1, NV2
FLOW 1.0
SF(1) 0.30*FLOW
SF(2) 0.70*FLOW
T 363.0
P 512.0
CALL FLASH2 (SF, SL, SV, T, P, Q)
WRITE (6, 900) 'Liquid Stream', SL(1), SL(2), SL(3)
WRITE (6, 900) 'Vapor Stream', SV(1), SV(2), SV(3)
900
FORMAT (A15, F7.4,' mol/s Benzene',/,
* 15X, F7.4, mol/s Toluene',/,
* 15X, F7.2, 'K')
WRITE (6, 901) Q
10.8 (cont’d)
c.
10-8
901
FORMAT ('Heat Required', F7.2,' kW')
END
C
C
C
C
C
C
SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q)
REAL NF, NL, NV
DIMESION SF(3), SL(3), SV(3)
COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2
Vapor Pressure
PV1 10.**(A1 – B1/(T – 273.15 C1))
PV2 10.**(A2 – B2/(T – 273.15 C2))
Product fractions
XL1 (P – PV2)/(PV1 – PVS)
XV1 XL1*PM/P
Feed Variables
NF SF(1) SF(2)
XF1 SF(1)/NF
Product flows
NL NF*(XF1 – XV1)/(XL1 – XV1)
NV NF – NL
SL(1) XL1*NL
SL(2) NL – SL(1)
SY(1) XY1*NY
SY(2) NV – SY(1)
SL(3) T
SV(3) T
Energy Balance
Q CP1*SF(1)*SF(1) CP2*SF(2)
Q Q*(T – SF(3)) (NV1*XV1 HV2*(1 – XV1))*NV
RETURN
END
b1g
XF b I g NF XLb I g NL XV b I g
Energy Balance: Q bT TF g ¦ CPb I g c XLb I g
NL NV
10.9 a. Mass Balance: NF
N
NV
I
1,2 n 1
bg
NL XV I
I 1
NV
¦ HV b I g
N
b g b3g
XV 1
I 1
b g
where: XL N
1
N 1
¦ XLb I g
b g
XV N
1
I 1
I 1
bg
XV I
P
¦ XV b I g
I 1
¦ XLb I g
N
Raoult’s law: P
N 1
b g b4 g
XLb I g PV b I g
PV I
10.9 (cont’d)
10- 9
I
1,2, N 1
b5g
NV
h
b 2g
b g 10 d Ab I g Bb I g cCb I g T hi I 1,2, N 1
3 3b N 1g N 4 variables b NF , NL, NV , XF ( I ), XL( I ), XV ( I ), PV ( I ), TF , T , P, Qg
where: PV I
N mass balance
1 energy balances
N equilibrium relations
N Antoine equations
N 3 degrees of freedom
m
b gr
Design Set TF , T , P , NF , XF I
Eliminate NL form (2) using (1)
Eliminate XV(I) form (2) using (5)
Solve (2) for XL(I)
XL I
XF I NF NF NV PV I P 1
bg
bg
c bg
d
hi b6g
Sum (6) ove I to Eliminate XL(I)
b g
f NV
1 NF
¦ XF b I g
N
d NF NV c PV b I g P 1hi
0
I 1
Use Newton's Method to solve (7) for NV
Calulate NL from (1)
XL(I) from (2)
XV(I) from (5)
Q from (3)
C ** CHAPTER 10 - - PROBLEM 9
DIMENSION SF(8), SL(8), SV(8)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, NV
DATA A/6.85221, 6.87776, 6.402040, 0., 0., 0., 0./
DATA B/1064.63, 1171.530, 1268.115, 0., 0., 0., 0./
DATA C/232.00, 224.366, 216.900, 0., 0., 0., 0./
DATA CP/0.188, 0.216, 0.213, 0., 0., 0., 0./
DATA NV/25.77, 28.85, 31.69, 0., 0., 0., 0./
FLOW 1.0
N*3
SF(1) 0.348*FLOW
SF(2) 0.300*FLOW
SF(3) 0.352*FLOW
SF(4) 363
SL(4) 338
SV(4) 338
P*611
CALL FLASHN (SF, SL, SV, N, P, Q)
WRITE (6, 900)' Liquid Stream', (SL(I), I 1, N 1)
WRITE (6, 900)' Vapor Stream', (SV(I), I 1, N 1)
10.9 (cont’d)
b.
10-10
b 7g
900
901
C
C
100
200
C
300
C
500
400
900
C
FORMAT (A15, F7.4,' mols/s n-pentane', /,
* 15X, F7.4,' mols/s n-hexane', /,
* 15X, F7.4,' mols/s n-hephane', /,
*
15X, F7.2,' K')
WRITE (6, 901) Q
FORMAT ('Heat Required', F7.2, 'kW')
END
SUBROUTINE FLASHIN (SF, SL, SV, N, P, Q)
REAL NF, NL, NV, NVP
DIMENSION SF(8), SL(8), SV(8)
DIMENSION XF(7), XL(7), XV(7), PV(7)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, HV
TOL 1,5 – 6
Feed Variables
NF 0.
DO 100 I 1, N
NF NF SF(I)
DO 200 I 1, N
XF(I) SF(I)/NF
TF SF (N 1)
T SL (N 1)
TC T – 273.15
Vapor Pressures
DO 300 I 1, N
PV(I) 10.**(A(I) – B(I)/(TC C(I)))
Find NV -- Initial Guess NF/2
NVP NF/2
DO 400 ITER 1, 10
NV NVP
F –1.
FP 0.
DO 500 I 1, N
PPM1 PV(I)/P – 1.
F F NF*XF(I)/(NF NV*PPM1)
FP = FP – PPM1*XF(I)/(NF NV*PPM1)**2.
NVP NV – F/FP
IF (ABS((NVP – NV)/NVP).LT.TOL) GOTO 600
CONTINUE
WRITE (6, 900)
FORMAT ('FLASHN did not converge on NV')
STOP
Other Variables
10.9 (cont’d)
10- 11
600
700
800
NL NF – NVP
DO 700 I 1, N
XL(I) XF(I)*NF/(NF NV**(PV(I)/P – 1))
SL(I) XL(I)*NL
XV(I) XL(I)*PV(I)/P
SV(I) SF(I) – SL(I)
Q1 0.
Q2 0.
DO 800 I 1, N
Q1 Q1 CP(I)*SF(I)
Q2 Q2 HV(I)*XV(I)
Q Q1*(T – TF) Q2*NVP
RETURN
END
Program Output: Liquid Stream 0.0563 mols
0.1000 mols
0.2011 mols
338.00 K
Vapor Stream 0.2944 mols
0.2000 mols
0.1509 mols
338.00 K
Heat Required 13.01 kW
s n-pentane
s n-hexane
s n-heptane
s n-pentane
s n-hexane
s n-heptane
10.10
a.
Q (kW)
n v (mol / s)
x v ( mol A(v) / mol)
1 x v ( mol B(g) / mol)
T (K), P(mm Hg)
n F (mol / s)
x F (mol A(v) / mol)
1 x F (mol B(g) / mol)
TF (K), P (mm Hg)
n l (mol A(l) / s)
10.10 (cont’d)
10 variables (n F , x F , TF , P , n v , x v , T , n l , p *,A , Q )
–2 material balances
10-12
–1
–1
–1
5
Antoine equation
Raoult’s law
energy balance
degrees of freedom
b.
References: A(l), B(g) at 25oC
Substance
nin
H in
n out
A(l)
—
—
n l
A(v)
n F x F
n v x v
B(g)
n F (1 x F )
H 1
H
2
n v (1 x v )
H out
H
3
H 4
H
5
Given n F and x F (or n AF and n BF ), TF , P , y c (fractional condensation),
Fractional condensation Ÿ n l y c n F x F
Mole balance Ÿ n v
A balance Ÿ x v
n F n l
(n F x F n l ) / n v
Raoult' s law Ÿ p *A
xv P
B
C
A log 10 p *A
'H v C pv (TF 25), H 2 C pg (TF 25), H 3
Antoine' s equation Ÿ T
Enthalpies: H 1
H 4
Energy balance: Q
'H v C pv (T 25), H 5
C pl (T 25),
C pg (T 25)
¦ n out H out ¦ n in H in
c.
nAF
0.704
nV
0.3664
Cpv
0.050
nBF
0.296
xV
0.1921
Cpg
0.030
nF
1.00
A
7.87863
H1
37.02
xF
0.704
B
1473.11
H2
1.05
TF
333
C
230
H3
0.2183
P
760
pA*
146.0
H4
35.41
yc
0.90
T
300.8
H5
0.0839
nL
0.6336
Cpl
0.078
Q
–23.7
Greater fractional methanol condensation (yc) Ÿ lower temperature (T). (yc = 0.10 Ÿ
T = 328oC.)
10.10 (cont’d)
10- 13
e.
C **CHAPTER 10 -- PROBLEM 10
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
DATA A, B, C / 7.87863, 1473.11, 230.0/
DATA CPL, HV, CPV, CPG,/ 0.078, 35.27, 0.050, 0.029/
FLOW = 1.0
SF(1) = 0.704*FLOW
SF(2) = FLOW – SF(1)
YC = 0.90
P = 1.
SF(3) = 333.
CALL CNDNS (SF, SV, SL, P, YC, Q)
WRITE (6, 900) SV(3)
WRITE (6, 401) 'Vapor Stream', SV(1), SV(2)
WRITE (6, 401) 'Liquid Stream', SL(1)
WRITE (6, 902)Q
900
FORMAT ('Condenser Temperature', F7.2,' K')
901
FORMAT (A15, F7.3,' 'mols/s Methyl Alcohol', /,
* 15X, F7.3, 'mols/s air')
902
FORMAT ('Heat Removal Rate', F7.2,' kW')
END
C
SUBROUTINE CNDNS (SF, SV, SL, P, YC, Q)
REAL NF, NL, NV
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
C
Inlet Stream Variables
NF = SF(1) + SF(2)
TF = SF(3)
XF = SF(1)/NF
C
Solve Equations
NL = YC * XF * NF
NV = NF - NL
XV = (XF*NF - NL)/NV
PV = P * XV * 760.
T = B/(A - LOG(N)/LOG (10.)) - C
T = T + 273.15
Q = ((CPV * XV + CPG * (1 - XY)) * NV + CPL * NL) * (T - TF) - NL * HV
C
Output Variables
SL(1) = NL
S2(2) = T
SV(1) = XV*NV
SV(2) = NV - SV(1)
SV(3) = T
RETURN
END
10-14
10.11
K 1 A1 K 2 A2 K 3 A3 K m Am
0
a. Extent of reaction equations:
b g X ] NU b IX g
SPb I g SF b I g NU b I g [ I
[
[ SF IX
1,2, N
Energy Balance: Reference states are molecular species at 298K.
b
g
¦ HF b I g
SF N 1
TF
'H r
b
g
SP N 1
TP
bg
N
NU I
I 1
b
g
Q [ 'H r TP 298
¦ SPb I g
N
bg
CP I (TF 298)
I 1
¦ SF b I g
N
bg
CP I
I 1
b. C 3 H 8 5O 2 o 3CO 2 4H 2 O
Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O
270 m 3
h
1 atm
mol ˜ K
1000 liter
h
3.348 mol C 3 H 8 s [=SF(1)]
3
273K 0.08206 liter ˜ atm
m
3600 s
b
3.348 mol C3 H 8 1.2 5 mol O2
sec
mol C3 H 8
g
20.09 mol O2 s [= SF(2)] Ÿ 7554
. mol N 2 s [= SF(3)]
X C3 H8
0.90 Ÿ n C3 H8
010
. (3.348)
[
b g
b g
[ SF IX
Nu
nin (SF)
X
Xi
nout (SP)
Cp
Tin
Hin
Tout
Hout
HF
DHr
Q
X ] NU IX
0.3348 mol C 3 H 8 / s in product gas [= SP(1)]
= –(3.348 mol/s)(0.90)/(–1) = 3.013 mol/s
1-C3H8
-1
3.348
2-O2
-5
20.09
3-N2
0
75.54
4-CO2 5-H2O(v)
3
4
0.3348
0.1431
5.024
0.033
75.54
0.0308
9.0396
0.0495
12.0528
0.0375
17.9
4.1
3.9
6.2
4.7
107.6
-103.8
24.8
0
23.2
0
37.2
-393.5
28.2
-241.83
0.90
3.01
423
1050
-2044
-4006
For the given conditions, Q
4006 kJ / s . As Tstack increases, more heat goes into the
stack gas so less is transferred out of the reactor: that is, Q becomes less negative.
10- 15
10.11 (cont’d)
C **CHAPTER 10 PROBLEM 11
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
DATA NU/–1., –5, 0., 3., 4., 0., 0./
DATA CP/0.1431, 0.0330, 0.0308, 0.0495, 0.0375, 0., 0./
DATA HF/–103.8, 0., 0., –393.5, –241.83, 0., 0./
COMMON CP, HF
SF(1) = 3.348
SF(2) = 20.09
SF(3) = 75.54
SF(4) = 0.
SF(5) = 0.
SF(6) = 423.
SP(6) = 1050.
IX = 1
X = 0.90
N=5
CALL REACTS (SF, SP, NU, N, X, IX, Q)
WRITE (6, 900) (SP(I), I = 1, N + 1), Q
900
FORMAT ('Product Stream', F7.3, ' mols/s propane', /,
* 15X, F7.3,' mols/s oxygen', /,
* 15X, F7.3,' mols/s nitrogen', /,
*
* 15X, F7.3,' mols/s carbon dioxide', /,
* 15X, F7.3,' mols/s water', /,
* 15X, F7.2,'K', /,
Heat required', F8.2, 'kW')
END
C
SUBROUTINE REACTS (SF, SP, NU, N, X, IX, Q)
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
COMMON CP, HF
C
Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
C
Solve Material Balances
DO 100 I = 1, N
100
SP(I) = SF(I) + EXT = NU(I)
C
Heat of Reaction
HR = 0
DO 200 I = 1, N
200
HR = HR + NF(I)*NU(I)
C
Product Enthalpy (ref * inlet)
HP = 0.
DO 300 I = 1, N
10-16
10.11 (cont’d)
300
HP = HP + SP(I)*CP(I)
HP = HP + (SP(N + 1) – SF (N + 1))
Q = EXT * HR + HP
RETURN
END
10.12 a. Extent of reaction equations:
[ SF IX X NU IX
b g
b g
SPb I g SF b I g NU b I g [
1, N
I
Energy Balance: Reference states are molecular species at feed stream temperature.
Q
'H
N
bg bg
N
b g z CPb I gdT
0 Ÿ 0 [ ¦ NU I HF I ¦ SP I
['H r ¦ nout H out
i 1
I 1
T
Tfeed
CP(I) = ACP(I) + BCP(I)*T + CCP(I)*T2 + DCP(I)*T3
bg
f T
[
¦ NU b I g * HF b I g AP
N
(T Tfeed ) I 1
CP
DP
3
4
(T 3 Tfeed
)
(T 4 Tfeed
)
3
4
¦ SPb I g
0
bg
N
where: AP
BP
2
(T 2 Tfeed
)
2
ACP I , and similarly for BP, CP, & DP
I 1
Use goalseek to solve f (T )
0 for T [= SP(N+1)]
b. 2CO O 2 o 2CO 2
Temporary basis: 2 mol CO fed
b
2 mol CO 1.25 1 mol O2
2 mol CO
g
125
. mol O2 Ÿ 4.70 mol N 2
Ÿ Total moles fed = (2.00 + 1.25 + 4.70) mol = 7.95 mol
Scale to given basis:
(23.0
10 3 mol
1h
kmol
)
)(
)(
3600 s 1 kmol
h
7.95 mol
mol CO fed s
.
SF (1) 1607
0.8036 Ÿ SF (2) 1.004 mol O 2 fed s
SF (3) 3.777 mol N 2 fed s
10- 17
10.12 (cont’d)
Solution to Problem 10.12
Nu
nin (SF)
X
Xi
nout (SP)
ACP
BCP
CCP
DCP
AP
BP
CP
DP
Tfeed
DHF
DHr
T
f(T)
1-CO
-2
1.607
2-O2
-1
1.004
3-N2
0
3.777
4-CO2
2
0
0.45
0.36
0.88385
0.02895
4.11E-06
3.55E-09
-2.22E-12
0.642425
3.777 0.72315
0.0291
0.029 0.03611
1.16E-05 2.20E-06 4.23E-05
-6.08E-09 5.72E-09 -2.89E-08
1.31E-12 -2.87E-12 7.46E-12
0.1799
5.00E-05
-2.90E-11
-6.57E-12
650
-110.52
0
0
-393.5
-566
1560
-4.7E-08
The adiabatic reaction temperature is 1560 o C .
As X increases, T increases. (The reaction is exothermic, so more reaction means
more heat released.)
d.
C ** CHAPTER 10 -- PROBLEM 12
DIMENSION SF(8), SP(B), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
DATA NU / –2., –1., 0., 2., 0., 0., 0./
DATA ACP/ 28.95E-3, 29.10E-3, 29.00E-3, 36.11E-3, 0., 0., 0./
DATA BCP/ 0.4110E-5, 1.158E-5, 0.2199E-5, 4.233E-6, 0., 0., 0./
DATA CCP/ 0.3548E-B, –0.6076E-8, 0.5723E-8, –2.887E-8, 0., 0., 0./
DATA DCP/ –2.220 E-12, 1.311E-12, –2.871E-12, 7.464E-12, 0., 0., 0./
DATA HF / –110.52, 0., 0., –393.5, 0., 0., 0./
SF(1) = 1.607
SF(2) = 1.004
SF(3) = 3.777
SF(4) = 0.
SF(5) = 650.
IX = 1
X = 0.45
N=4
CALL REACTAD (SF, SP, NU, N, X, IX)
WRITE (6, 900) (SP(I), I = 1, N + 1)
10-18
10.12 (cont’d)
900
C
C
C
100
C
200
C
300
C
400
900
FORMAT ('Product Stream', F7.3, ' mols/s carbon monoxide', /,
* 15X, F7.3, 'mols/s oxygen', /.
* 15X, F7.3, 'mols/s nitrogen', /.
*
* 15X, F7.3, 'mols/s carbon dioxide', /,
15X, F7.2, 'C')
END
SUBROUTINE REACTAD (SF, SP, NU, N, X, IX)
DIMENSION SF(8), SP(8), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
TOL = 1.E-6
Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
Solve Material Balances
DO 100 I = 1, N
SP(I) = SF(I) + EXT*NU(I)
Heat of Reaction
HR = 0
DO 200 I = 1, N
HR = HR + HF(I) * NU(I)
HR = HR * EXT
Product Heat Capacity
AP = 0.
BP = 0.
CP = 0.
DP = 0.
DO 300 I = 1, N
AP = AP + SP(I)*ACP(I)
BP = BP + BP(I)*BCP(I)
CP = CP + SP(I)*CCP(I)
DP = DP + SP(I)*DCP(I)
Find T
TIN = SF (N + 1)
TP = TIN
D0 400 ITER = 1, 10
T = TP
F = HR
FP = 0.
F = F +T*(AP + T*(BP/2. + T*(CP/3. + T*DP/4.)))
* –TIN*(AP + TIN*(BP/2. + TIN*(CP/3. + TIN*DP/4.)))
FP = FP + AP + T *(BP + T*(CP + T*DP))
TP = T – F/FP
IF(ABS((TP – T)/T).LT.TOL) GOTO 500
CONTINUE
WRITE (6, 900)
FORMAT ('REACTED did not converge')
STOP
10- 19
10.12 (cont’d)
500
SP(N + 1) = T
RETURN
END
Program Output:
0.884 mol/s carbon monoxide
0.642 mol/s oxygen
3.777 mol/s nitrogen
0.723 mol/s carbon dioxide
T = 1560.43 C
10-20
10.13
37.5 mol C2H4O
a.
Separator
50 mol C2H4
50 mol O2
208.3333 mol C2H4
50 mol O2
Reactor
166.6667
18.75
37.5
8.333333
8.333333
mol C2H4
mol O2
mol C2H4O
mol CO2
mol H2O
8.333333
18.75
8.333333
8.333333
mol C2H4
mol O2
mol CO2
mol H2O
Xsp = 0.2
Ysp = 0.9
158.3333 mol C2H4 (Ra)
158.3333 mol C2H4 (Rc)
Rc-Ra =
0
Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is recalculated recycle rate.
Use goalseek to find the value of Ra that drives (Rc-Ra) to zero.
b.
Xsp
0.2
0.2
0.3
0.3
Ysp
0.72
1
0.75333
1
Yo
0.6
0.833
0.674
0.896
no
158.33
158.33
99.25
99.25
The second reaction consumes six times more oxygen per mole of ethylene consumed. The lower the single pass ethylene oxide
yield, the more oxygen is consumed in the second reaction. At a certain yield for a specified ethylene conversion, all the oxygen in
the feed is consumed. A yield lower than this value would be physically impossible.
10-21
21
10.14
C ** CHAPTER 10 -- PROBLEM 14
DIMENSION XA(3), XC(3)
N=2
EPS = 0.001
KMAX = 20
IPR = 1
XA(1) = 2.0
XA(2) = 2.0
CALL CONVG (XA, XC, N, KMAX, EPS, IPR)
END
C
SUBROUTINE FUNCGEN(N, XA, XC)
DIMENSION XA(3), XC(3)
XC(1) = 0.5*(3. – XA(2) + (XA(1) + XA(2))**0.5
XC(2) = 4. – 5./(XA(1) + XA(2))
RETURN
END
C
SUBROUTINE CONVG (XA, XC, N, KMAX, EPS, IPR)
DIMENSION XA(3), XC(3), XAH(3), XCM(3)
K=1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
DO 100 I = 1, N
XAM(I) = XA(I)
XA(I) = XC(I)
100
XCM(I) = XC(I)
110
K=K+1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
D0 200 I = 1, N
IF (ABS ((XA(I) - XC(I))/XC(I)).GE.EPS) GOTO 300
200
CONTINUE
C
Convergence
RETURN
300
IF(K.EQ.KMAX) GOTO 500
DO 400 I = 1, N
W = (XC(I) – XCM(I))/(XA(I) – XAM(I))
Q = W/(W – 1.)
IF (Q.GT.0.5) Q = 0.5
IF (Q.LT.–5) Q = –5.
XCM(I) = XC(I)
XAM(I) = XA(I)
400
XA(I) = Q = XAM(I) + (1. – Q)*XCM(I)
GOTO 110
500
WRITE (6, 900)
900
FORMAT (' CONVG did not converge')
STOP
END
10.14 (cont’d)
10-22
Elementary Principles of Chemical Processes
23
C
SUBROUTINE IPRNT (K, XA, XC, N)
DIMENSION XA(3), XC(3)
IF (K.EQ.1) WRITE (6, 400)
IF (K.NE.1) WRITE (6, *)
DO 100 I = 1, N
100
WRITE (6, 901) K, I, XA(I), XC(I)
RETURN
900
FORMAT (' K Var Assumed Calculated')
901
FORMAT (I4, I4, 2E15.6)
END
Program Output: K Var
Assumed
Calculated
1
1 0.200000E + 01 0.150000E + 01
1
2
0.200000E + 01 0.275000E + 01
2
2
1
2
0.150000E + 01 0.115578E + 01
0.275000E + 01 0.282353E + 01
3
3
8
1
2
0.395135E + 00 0.482384E + 00
0.283152E + 01 0.245041E + 01
1
0.113575E + 01 0.113289E + 01
8
2
0.269023E + 01 0.269315E + 01
4
9
1
2
0.113199E + 01 0.113180E + 01
0.269186E + 01 0.269241E + 01
10- 23
CHAPTER ELEVEN
11.1 a.
The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:
Mp
xp
M
1 M p / M
Therefore, the leakage rate of hydrogen peroxide is m
b.
Balance on mass: Accumulation = input – output
E
dM
dt
t
0 m
1
m
0, M
M 0 (mass in tank when leakage begins)
Balance on H 2 O 2 : Accumulation = input – output – consumption
E
dM p
dt
t
11.2 a.
0 x p0 m
1
m
0, M p
FG M IJ kM
H MK
p
p
M p0
Balance on H3PO4: Accumulation = input
Density of H3PO4: U 1834
.
g / ml.
Molecular weight of H3PO4: M 98.00 g / mol .
dn p
Accumulation =
(kmol / min)
dt
20.0 L 1000 ml 1.834 g
mol
1 kmol
Input =
min
L
ml
98.00 g 1000 mol
0.3743 kmol / min
E
dn p
0.3743
dt
0, n p0
t
z
z
np
b.
t
dn p
7.5
xp
c.
150 u 0.05 7.5 kmol
.
015
0.3743 dt Ÿ n p
7.5 0.3743t (kmol H 3PO 4 in tank )
0
np
np
n
n0 n p n p 0
7.5 0.3743t
Ÿt
150 0.3743t
7.5 0.3743t
150 0.3743t
kmol H 3PO 4
kmol
. min
471
11-1
11.3 a.
w
m
b
a bt t
g b
w
0, m
750 & t
g
b
w kg h
1000 Ÿ m
w
5, m
g
bg
750 50t h
Balance on methanol: Accumulation = Input – Output
M kg CH 3OH in tank
dM
dt
b
f m
w
m
E
b
dM
450 50t kg h
dt
t 0, M 750 kg
z zb
M
t
dM
b.
g
1200 kg h 750 50t kg h
750
g
g
450 50t dt
0
E
M 750
450t 25t 2
E
M
750 450t 25t 2
Check the solution in two ways:
(1) t 0, M 750 kg Ÿ satisfies the initial condition;
(2)
c.
dM
dt
M
t
d.
dM
dt
450 50t
0Ÿt
0
Ÿ reproduces the mass balance.
450 50 9 h Ÿ M
450 r
b450g 4b25gb750g Ÿ t = –1.54 h, 19.54 h
2b25g
2
1 m3
t
2775 kg (maximum)
750 450t 25t 2
3.40 m 3 103 liter 0.792 kg
M
750 450(9 ) 25(9 )2
2693
450 r
1 liter
2693 kg (capacity of tank)
750 450t 25t 2
b450g 4b25gb750 2693g Ÿ t
2b 25g
2
719
. h,10.81 h
Expressions for M(t) are:
R|750 + 450t - 25t b0 d t d 719
. and 10.81 d t d 19.54g (tank is filling or draining)
M(t) = S2693
(tank is overflowing)
( 719
. d t d 10.81)
||T0
(tank is empty, draining
(19.54 d t d 20.54)
2
as fast as methanol is fed to it)
11-2
11.3 (cont’d)
3000
2500
M(kg)
2000
1500
1000
500
0
0
5
10
15
20
t(h)
11.4 a.
492q R
10.0 ft 3
Air initially in tank: N 0
1 lb - mole
b g
532q R 359 ft 3 STP
0.0258 lb - mole
Air in tank after 15 s:
Pf V
N f RT
P0V
N 0 RT
Ÿ Nf
0.0258 lb - mole 114.7 psia
P0
14.7 psia
15 s
b
g
0.0117 lb - moles s ; t
z z
N
Integrate balance:
t
dN
0.0258
0.0117 lb - mole air s
n dt Ÿ N
0, N
0.0258 lb - mole
b
0.0258 0.0117t lb - mole air
0
Check the solution in two ways:
(1) t = 0, N = 0.0258 lb - mole Ÿ satisfies the initial condition
( 2)
d.
0.2013 lb - mole
Balance on air in tank: Accumulation = input
dN
dt
c.
Pf
b0.2013 0.0258g lb - mole air
Rate of addition: n
b.
N0
dN
dt
0.0117 lb - mole air / s Ÿ reproduces the mass balance
t 120 s Ÿ N
O 2 in tank
b
gb g
0.0258 0.0117 120
b g
0.21 143
.
143
. lb - moles air
0.30 lb - mole O 2
11-3
g
11.5 a.
Since the temperature and pressure of the gas are constant, a volume balance on the gas
is equivalent to a mole balance (conversion factors cancel).
Accumulation = Input – Output
dV 540 m 3
1h
Q w m 3 min
dt
h
60 min
e
t
0, V
b
3.00 u 103 m 3 t
z zb
V
t
b.
0 corresponds to 8:00 AM
g
e j
9.00 Q w dt Ÿ V m 3
dV
3.00 u103
0
Let Q w i
tabulated value of Q w at t
10
Q w dt #
Q w1 Q w 25 4
3
i
0
2488 m 3
V
g
z
t
3.00 u 103 9.00t Q w dt t in minutes
0
LM
MN
z
240
j
b g
10 i 1
i
O
PQ
¦ Q w i 2 ¦ Q w i P
24
24
2, 4 , b g
3.00 u 103 9.00 240 2488
i 3, 5, 1, 2, , 25
b
g b
2672 m 3
c.
Measure the height of the float roof (proportional to volume).
The feed rate decreased, or the withdrawal rate increased between data points,
or the storage tank has a leak, or Simpson’s rule introduced an error.
d.
REAL VW(25), T, V, V0, H
INTEGER I
DATA V0, H/3.0E3, 10./
READ (5, *) (VW(I), I = 1, 25)
V= V0
T=0.
WRITE (6, 1)
WRITE (6, 2) T, V
DO 10 I = 2, 25
T = H * (I – 1)
V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I))
WRITE (6, 2) T, V
10 CONTINUE
1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')
2 FORMAT (F8.2, 7X, F6.0)
END
$DATA
11.4 11.9
12.1
11.8
11.5
11.3
Results:
TIME (MIN) VOLUME (CUBIC METERS)
0.00
3000.
10.00
2974.
20.00
2944.
230.00
2683.
240.00
2674.
Vtrapezoid
2674 m 3 ; VSimpson
2672 m 3 ;
2674 2672
u 100%
2672
Simpson’s rule is more accurate.
11-4
g
10
114
. 9.8 4 124.6 2 113.4
3
0.07%
11.6 a.
b.
b
Q out L min
g
bg
kV L
0.200V Q out
Ÿ Q out
V 300
Q out 60
20.0 L min Ÿ Vs
100 L
Balance on water: Accumulation = input – output (L/min).
(Balance volume directly since density is constant)
dV
20.0 0.200V
dt
t 0, V 300
c.
dV
dt
0
200 0.200Vs Ÿ Vs
100 L
V
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
20.0 0.200(300) 40.0. As t increases, V decreases. Ÿ dV / dt 20.0 0.200V
becomes less negative, approaches zero as t o f . The curve is therefore concave up.
t
z
V
d.
dV
20
.
0
0.200V
300
Ÿ
t
dt
0
FG
H
1
20.0 0.200V
ln
40.0
0.200
Ÿ 0.5 0.005V
V
z
b g
. 100
101
b
IJ
K
t
g
b
exp 0.200t Ÿ V
b
100.0 200.0 exp 0.200t
g
101 L 1% from steady state Ÿ
b
g
101 100 200 exp 0.200t Ÿ t
b
ln 1 200
g
0.200
11-5
26.5 min
g
11.7 a.
A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t
D 2385 kg week ) and ( t 6 weeks, D 755 kg week ).
bt ln a œ D
ln D
ln D2 D1
t 2 t1
b
g
E
b g b
8.007 Ÿ a
e8.007
3000
Inventory balance: Accumulation = –output
b
z z
I
t
dI
18 ,000
11.8 a.
gb g
ln 2385 0.230 1
dI
3000e 0.230t kg week
dt
t 0, I 18,000 kg
c.
0.230
6 1
3000e 0.230t
D
b.
b
ln 755 2385
ln D1 bt1
ln a
ae bt
t
g
3000e 0.230t dt Ÿ I 18,000
0
fŸ I
3000 0.230t
e
0.230
t
0
ŸI
4957 13,043e 0.230t
4957 kg
Total moles in room: N
Molar throughput rate: n
SO 2 balance ( t
1100 m 3
273 K
103 mol
b g
295 K 22.4 m 3 STP
700 m 3
273 K
103 mol
b g
295 K 22.4 m 3 STP
min
45,440 mol
28,920 mol min
0 is the instant after the SO 2 is released into the room):
b gb
N mol x mol SO 2 mol
g
mol SO 2 in room
Accumulation = –output.
b g
d
Nx
dt
nx
dx
45, 440 dt
Ÿ
t
b.
0.6364 x
N
n 28,920
0, x
15
. mol SO 2
45,440 mol
330
. u 10 5 mol SO 2 mol
The plot of x vs. t begins at (t=0, x=3.30u10-5). When t=0, the slope (dx/dt) is
. u 10 5 210
. u 10 5 . As t increases, x decreases.Ÿ
0.6364 u 330
dx dt 0.6364 x becomes less negative, approaches zero as t o f . The curve
is therefore concave up.
11-6
1 week,
x
11.8 (cont’d)
0
t
c.
Separate variables and integrate the balance equation:
z
x
3.30 u10
5
dx
x
z
t
0.6364dt Ÿ ln
0
x
. u 10 5
330
0.6364t Ÿ x
. u 10 5 e 0.6364t
330
Check the solution in two ways:
(1) t = 0, x = 3.30 u 10-5 mol SO 2 / mol Ÿ satisfies the initial condition;
(2)
d.
e.
dx
dt
0.6364 u 330
. u 10 5 e 0.6364 t
45,440 moles x mol SO 2
CSO2
1100 m
3
mol
i)
t
2 min Ÿ CSO 2
ii)
x
10 6 Ÿ t
382
. u 10 7
e
0.6364 x Ÿ reproduces the mass balance.
1 m3
3
10 L
mol SO 2
liter
ln 10 6 3.30 u 10 5
0.6364
u 10 6 e 0.6364t mol SO 2 / L
4131
. u 10 2 x 13632
.
j
55
. min
The room air composition may not be uniform, so the actual concentration of the SO2
in parts of the room may still be higher than the safe level. Also, “safe” is on the average;
someone would be particularly sensitive to SO2.
11-7
11.9 a.
Balance on CO: Accumulation=-output
N ( mol ) x ( mol CO / mol) = total moles of CO in the laboratory
PQ p
kmol
Molar flow rate of entering and leaving gas: n (
)
h
RT
PQ p
kmol
kmol CO
Rate at which CO leaves: n (
=
)x
x
h
kmol
RT
CO balance: Accumulation = -output
PQ p
dx
P
d ( Nx )
Q p x
xŸ
dt
NRT
dt
RT
FG
H
FG
H
E PV
dx
dt
0, x
t
z
c.
dx
x
0.01
Q p
V
kmol CO
kmol
0.01
Q p
V
z
tr
dt Ÿ t r
0
e
tr
d.
x
b g
V
ln 100 x
Q p
350 m 3
350
ln 100 u 35 u 10 6
700
V
IJ
K
NRT
x
b.
IJ
K
j
2.83 hrs
The room air composition may not be uniform, so the actual concentration of CO
in parts of the room may still be higher than the safe level. Also, “safe” is on the
average; someone could be particularly sensitive to CO.
Precautionary steps:
Purge the laboratory longer than the calculated purge time. Use a CO detector
to measure the real concentration of CO in the laboratory and make sure it is
lower than the safe level everywhere in the laboratory.
11.10 a.
Total mass balance: Accumulation = input – output
b
dM
dt
b.
m
kg min
m
g
0 Ÿ? M is a constant
200 kg
Sodium nitrate balance: Accumulation = - output
x = mass fraction of NaNO 3
b g
E
d xM
dt
dx
dt
m
x
M
t 0, x
b
kg min
xm
g
m
x
200
90 200 0.45
11-8
11.10 (cont’d)
c.
0.45
m
50 kg / min
m 100 kg / min
m
x
200 kg / min
0
t(min)
dx
m
x 0 , x decreases when t increases
dt
200
dx
becomes less negative until x reaches 0;
dt
Each curve is concave up and approaches x = 0 as t o f;
dx
becomes more negative Ÿ x decreases faster.
dt
increases Ÿ
m
z
x
d.
dx
x
0.45
z
t
0
m
x
dt Ÿ ln
0.45
M
FG
H
m
tŸx
200
0.45 exp mt
200
IJ
K
Check the solution:
(1) t = 0, x = 0.45 Ÿ satisfies the initial condition;
m
mt
m
dx
0.45 u
x Ÿ satisfies the mass balance.
exp( ) (2)
dt
200
200
200
0.45
0.4
m
0.35
50 kg / m in
m
0.3
100 kg / m in
m
x
0.25
200 kg / m in
0.2
0.15
0.1
0.05
0
0
5
10
15
20
t(min)
e.
100 kg min Ÿ t
m
90% Ÿ x f
0.045 Ÿ t
99% Ÿ x f
0.0045 Ÿ t
99.9% Ÿ x f
d
i
2 ln x f 0.45
4.6 min
9.2 min
0.00045 Ÿ t
138
. min
11-9
25
11.11 a.
e je
Mass of tracer in tank: V m 3 C kg m 3
j
Tracer balance: Accumulation = –output. If perfectly mixed, C out
b g
b
d VC
Q C kg min
dt
b.
z
t
C
dC
C
m0 V
c.
g
z
FG
H
Q
C
dt Ÿ ln
0V
m0 V
t
b
Q
V
g
ln 0.050 0.223
2 1
E
e30 m
V
Q
C
V
m0
0, C
V
FG IJ
H K
Qt
ŸC
V
m0
Q t
exp V
V
e
3
min
j e
0.223 u 10 3 & t
1, C
j
0.050 u 10 3 .
2, C
min 1
1495
.
j e1.495 min j
1
. m3
201
In tent at any time, P=14.7 psia, V=40.0 ft3, T=68qF=528qR
ŸN
b.
IJ
K
PV
RT
m(liquid)
10.73
40.0 ft 3
14.7 psia
ft 3 ˜ psia
lb - mole ˜ R
o
528 o R
Molar throughout rate:
60 ft 3 492q R 16.0 psia
1 lb - mole
n in n out n
min 528q R 14.7 psia 359 ft 3 STP
Moles of O2 in tank= N (lb - mole) u
FG lb - mole O IJ
H lb - mole K
b g
0.1038 lb - mole
0.1695 lb - mole min
2
Balance on O2: Accumulation = input – output
b g
d Nx
dt
c.
C
Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption
of perfect mixing) through t
11.12 a.
dC
dt
V is constant
C tank
z
0.35n xn Ÿ 01038
.
x
dx
0.21 0.35 x
Ÿ
0.35 x
0.14
x
0.27 Ÿ t
z
t
dx
dt
163
. dt Ÿ ln
0
e 1.63t Ÿ x
LM FG
N H
b
b
dx
163
. 0.35 x
01695
0.35 x Ÿ dt
.
t 0, x 0.21
b
b0.35 xg
g
0.35 0.21
g
1.63t
0.35 014
. e 1.63t
1
0.35 0.27
ln
163
.
0.35 0.21
IJ OP
KQ
0.343 min (or 20.6 s)
11-10
g
11.13 a.
b gb
Mass of isotope at any time V liters C mg isotope liter
Balance on isotope: Accumulation = –consumption
b g kCFGH Lmg˜ sIJKV bLg
dC
kC
dt
t 0, C C 0
Cancel V
d
VC
dt
g
Separate variables and integrate
z z
C
C0
dC
C
0
0
C
0.01C0
12
k
ln 2
2.6 hr
2.6 hr Ÿ k
t1 2
0
kdt Ÿ ln
0.5C0 Ÿ t 1 2
C
b.
FG C IJ kt Ÿ t lnbC C g
k
HC K
lnb0.5g
ln 2
Ÿt
t
t=-ln(C/C0)/k
k
0.267 hr 1
t
b g
ln 0.01
0.267
17.2 hr
11.14 A o products
a.
Mole balance on A: Accumulation = –consumption
b g
d C AV
t
0, C A
Ÿ
z
CA
CA0
b.
bV constant; cancelsg
kC AV
dt
C A0
dC A
CA
z
t
kdt Ÿ ln
0
FG C IJ
HC K
b g
kt Ÿ C A
A
C A0 exp kt
A0
Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies
b
assumption of first-order) through t
kt ln C A 0
ln C A
b
ln 0.0185 0.0262
k
120.0 213
.
g
213
. , CA
g b
0.0262 & t
3.53 u 10 3 min 1 Ÿ k
120.0, C A
35
. u 10 3 min 1
11.15 2 A o 2 B C
a.
Mole balance on A: Accumulation = –consumption
b g
d C AV
kC 2AV
dt
t
Ÿ
0, C A
z
bV constant; cancelsg
C A0
CA
dC A
CA0
C A2
z
t
1
1
kdt Ÿ 0
C A C A0
kt Ÿ C A
LM 1
NC
A0
11-11
O
kt P
Q
1
g
0.0185 .
11.15 (cont’d)
b.
CA
0.5C A 0 Ÿ nA
0.5n A 0
b0.5n
b0.5n
nB
nC
A0
A0
total moles
c.
1
1
0.5C A 0 C A 0
gb
g
mol A react.gb1 mol C 2 mol A react.g
mol A react. 2 mol B 2 mol A react.
125
. n A 0 Ÿ P1 2
125
.
n A 0 RT
V
n A0
V
P0
Ÿ t1 2
RT
0.5n A0
0.25n A0
125
. P0
Plot t 1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2nd order
d
decomposition) through t 1 2
RT
k
Slope:
1 0.135 & t 1 2
b1015 Kgb0.08206 L ˜ atm mol ˜ Kg
1 0.683
0.582 L mol ˜ s
143.2 s ˜ atm
F
GH
FG IJ
H K
i
209, 1 P0
143.2 s ˜ atm
I
JK
t 1 2 P0
RT
E
Ÿ ln
exp
k 0 P0
RT
RT
t1 2
i d
1060, 1 P0
1060 209
1 0135
1 0.683
.
Ÿk
d.
1
; but C A0
kC A 0
kt 1 2 Ÿ t 1 2
ln
1 E 1
k0 R T
Plot t 1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.
bg
t 1 2 s , P0
1 atm, R
bg
0.08206 L ˜ atm / (mol ˜ K), T K
d
Data fall on straight line through t 1 2 P0 RT
dt
1 2 P0
e.
0.6383, 1 T
b
g
1
k0
T
C A0
b
g
ln 0.6383 980 K Ÿ k
i
E
29,940 K
1 1050 1 900
i
1 900 &
1 1050
R=8.314 J/ (mol ∙K)
ln 0.6383 74.0
E
R
ln
RT
74.0, 1 T
29,940
1050
FG
H
k 0 exp b
28.96 Ÿ k 0
E
RT
IJ
K
2.49 u 10 5 J mol
3.79 u 1012 L (mol ˜ s)
0.204 L (mol ˜ s)
g
. atm
0.70 120
b0.08206 L ˜ atm mol ˜ Kgb980 Kg
.
1045
u 10 2 mol L
90% conversion
CA
0.10C A0 Ÿ t
LM
N
1 1
1
k C A C A0
4222 s
OP
Q
LM
N
1
1
1
3
0.204 1.045 u 10
1045
.
u 10 2
70.4 min
11-12
OP
Q
RT
kP0
11.16 A o B
a.
Mole balance on A: Accumulation = –consumption(V constant)
dC A
dt
t
z
k 1C A
1 k2CA
0, C A
C A0
1 k2C A
dC A
CA0
k1C A
b.
CA
b
z
b
C
k
1
ln A 2 C A C A 0
k 1 C A0 k 1
t
dt Ÿ
0
b
g
Plot t C A C A0 vs. ln C A / C A 0
b
y
t
C A0 C A
g bC
A0
g
b
g
k2
C
1
C A0 C A ln A
k1
k 1 C A0
t Ÿ t
g
C A on rectangular paper:
x
k
1 ln C A C A 0
2
k 1 C A0 C A
k1
g ;
b
g
1
slope
FG
H
intercept
y1
IJ FG
K H
x1
y2
IJ
K
x2
Data fall on straight line through 116.28, 0.2111 & 130.01, 0.2496
130.01 116.28
0.2496 0.2111
1
k1
b
2.80 u 10 3 L (mol ˜ s)
g
0115
.
L mol
g
b
k2
k1
356.62 Ÿ k 1
130.01 356.62 0.2496
4100
. Ÿ k2
11.17 CO Cl 2 Ÿ COCl 2
a.
3.00 L
273 K
.
mol gas
b g 012035
0.60b0.12035 molg 3.00 L 0.02407 mol L CO U
|Vinitial concentrations
molg 3.00 L 0.01605 mol L Cl |
0.40b012035
.
W
0.02407 C bt g U
| Since 1 mol COCl formed requires 1 mol of each reactant
0.01605 C bt g V|
W
303.8 K 22.4 L STP
bC g
dC i
C bt g
C bt g
CO i
Cl 2
CO
Cl 2
b.
i
2
p
2
p
Mole balance on Phosgene: Accumulation = generation
d i
d VC p
V=3.00 L
8.75CCO CCl 2
d1 58.6C
dt
c.
1 mol
Cl 2
34.3C p
i
t
z
0
24.3C i
.
d1941
dC
d0.02407 C id0.01605 C i
g
p
p
d1.941 24.3C i
0.75 0.01605
2
0.01204
dt
p
t
b
p
11-13
p
id
2.92 0.02407 C p 0.01605 C p
2
Cl 2 limiting; 75% conversion Ÿ C p
1
2.92
d
dC p
0, C p
0
0.01204 mol L
2
i
11.17 (cont’d)
d.
11.18 a.
REAL F(51), SUM1, SUM2, SIMP
INTEGER I, J, NPD(3), N, NM1, NM2
DATA NPD/5, 21, 51/
FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C)
DO 10 I = 1, 3
N = NPD(I)
NM1 = N – 1
NM2 = N – 2
DO 20 J = 1, N
C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1)
F(J) = FN(C)
20
CONTINUE
SUM1 = 0.
DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
30
CONTINUE
SUM2 = 0.
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + F(J)
40
CONTINUE
SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2)
T = SIMP/2.92
WRITE (6, 1) N, T
10 CONTINUE
1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES')
END
RESULTS
5 POINTS — 91.0 MINUTES
21 POINTS — 90.4 MINUTES
51 POINTS — 90.4 MINUTES
t 90.4 minutes
e j e
Moles of CO 2 in liquid phase at any time V cm 3 C A mols cm 3
j
Balance on CO 2 in liquid phase: Accumulation = input
b g
e
d
VC A
dt
kS C *A C A
FjG molsIJ Ÿ dCdt kSV eC
H s K t 0, C 0
A
*
A
CA
j
yV
A
y A P is constant, C *A
Separate variables and integrate. Since p A
z
dC A
CA
0
C *A
Ÿ ln
CA
C *A C A
C *A
z
t
0
e
kS
dt Ÿ ln C *A C A
V
C
kS expb g
t Ÿ 1 *A
V
CA
j
CA
CA 0
kS
t
V
e kSt V Ÿ C A
1 C A C *A
11-14
p A H is also a constant.
e
C *A 1 e kSt V
j
11.18 (cont’d)
b.
t
LM
MN
C
V
ln 1 *A
kS
CA
V
OP
PQ
5000 cm 3 , k
5L
78.5 cm 2 , C A
0.020 cm s , S
0.62 u 10 3 mol / cm 3
a0.30fa20 atmf d9230 atm ˜ cm moli 0.65 u 10 mol cm
e5000 cm j lnF1 0.62 u 10 I 9800 s Ÿ 2.7 hr
b0.02 cm sge78.5 cm j GH 0.65 u 10 JK
C *A
3
t
3
3
yAP H
3
3
3
2
(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals
the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than
this value, leading to a significantly lower t than that to be calculated)
11.19 A o B
a.
Total Mass Balance: Accumulation input
dM d ( UV )
Uv
dt
dt
E
dV
dt
t
v
0, V
0
A Balance: Accumulation input – consumption
dN A
C A 0 v ( kC A )V C =N /V
A
A
dt
dN A
dt
t
b.
z z
z
V
c.
dN A
dt
Steady State:
dV
0
NA
0
Ÿ
t
ŸV
vdt
0
dN A
C A 0 v kN A
C A0 v
k
vt
z
t
dt
0
IJ t Ÿ C v kN
C v
K
C v
1 expb kt g t o f Ÿ N
k
FG
H
C v kN A
1
ln A 0
k
C A 0 v
Ÿ NA
CA
0Ÿ NA
A0
e kt
A
A0
A0
NA
V
A
C A 0 [1 exp( kt )]
kt
11-15
C A0 v
k
C Ao v kN A
0, N A
0
11.19 (cont’d)
When the feed rate of A equals the rate at which A reacts, NA reaches a steady value.
NA would never reach the steady value in a real reactor. The reasons are:
Ÿ t o f , V o f.
(1) In our calculation, V = vt
But in a real reactor, the volume is limited by the reactor volume;
(2) The steady value can only be reached at t o f. In a real reactor, the reaction time is finite.
d.
C A0 [1 exp( kt )]
t of
kt
lim C A
C A0
t of kt
lim
t of
lim
0
From part c, t o f, N A o a finite number, V o f Ÿ C A
11.20 a.
dT
dt
M
MC v
Q W
(3.00 L)(1.00 kg / L) = 3.00 kg
Cv
Cp
W
dT
dt
NA
o0
V
(0.0754 kJ / mol ˜ o C)(1 mol / 0.018 kg) = 4.184 kJ / kg ˜ o C
0
0.0797Q (kJ / s)
t = 0, T = 18 o C
z z
100o C
b.
dT
o
18 C
c.
11.21 a.
240 s
0.0797Q dt Ÿ Q
0
100 18
240 u 0.0797
4.287
kJ
s
4.29 kW
Stove output is much greater.
Only a small fraction of energy goes to heat the water.
Some energy heats the kettle.
Some energy is lost to the surroundings (air).
Energy balance: MC v
dT
dt
Q W
M
20.0 kg
C v | C p ( 0.0754 kJ / mol ˜ C)(1 mol / 0.0180 kg) = 4.184 kJ / (kg ˜ C)
0.97 (2.50) 2.425 kJ s
Q
W 0
o
o
a f
dT
dt
b g
0.0290 q C s , t
0, T
25q C
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
z z
T
b.
t
dT
o
25 C
c.
T
0.0290dt Ÿ T
bg
25q C 0.0290t s
0
100q C Ÿ t
b100 25g 0.0290
2585 s Ÿ 43.1 min
No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal
boiling point).
11-16
11.22 a. Energy balance on the bar
MCv
dTb
dt
b
Q W
UA Tb Tw
g
B
Table B.1
e60 cm je7.7 g cm j
3
M
Cv
0.46 kJ (kg ˜q C), Tw
0.050 J (min ˜ cm ˜q C)
A
2 2 3 2 10 3 10 cm
25q C
2
a fa f a fa f a fa f
b
gb
2
0.02635 Tb 25 q C min
0, Tb
t
dTb
dt
462 g
U
dTb
dt
b.
3
g
95q C
d
i
0.02635 Tbf 25 Ÿ Tbf
0
2
112 cm
25q C
95
85
75
Tb( oC)
65
55
45
35
25
15
5
0
t
z
z
Tb
c.
t
dTb
T 25
95 b
0.02635dt
0
FG T 25IJ 0.02635t
H 95 25K
Ÿ T bt g 25 70 expb 0.02635t g
Ÿ ln
b
b
Check the solution in three ways:
(1) t = 0, Tb
25 70 95o C Ÿ satisfies the initial condition;
dTb
70 u 0.02635e 0.02635t 0.02635(Tb 25) Ÿ reproduces the mass balance;
dt
(3) t o f, Tb 25o C Ÿ confirms the steady state condition.
(2)
Tb
30q C Ÿ t
100 min
11-17
11.23
12.0 kg/min
25oC
12.0 kg/min
T (oC)
Q (kJ/min) = UA (Tsteam-T)
a.
dT
dt
Energy Balance: MCv
b
g
b
p 25 T UA Tsteam T
mC
g
M 760 kg
m 12.0 kg min
dT / dt
Cv | C p
UA
0, T
25o C
2.30 kJ (min ˜q C)
. kJ (min ˜q C)
115
a
f
Tsteam sat' d; 7.5bars
b.
. 0.0224T ( o C min), t
150
Steady State:
dT
dt
167.8q C
0 150
. 0.0224Ts Ÿ Ts
67q C
T( oC)
67
25
0
t
z
Tf
c.
dT
150
. 0.0224T
25
U changed. Let x
dT
dt
z
FG
H
IJ
K
1
150
. 0.0224T
ln
ŸT
0.0224
0.94
150
. 0.94 exp( 0.0224t )
0.0224
49.8q C
(UA) new . The differential equation becomes:
0.3947 0.096 x (0.01579 5.721x )T
55
25
dt Ÿ t
0
40 min. Ÿ T
t
d.
z
t
0.3947 0.096 x (0.01579 5.721 u 104 x )T
Ÿ
Ÿx
'U
U initial
1
0.01579 5.721 u 104
z
40
dT
dt
0
L 0.3947 0.096x e0.01579 5.721 u 10 xj u 55 OP
ln M
x M 0.3947 0.096 x e0.01579 5.721 u 10 x j u 25 P
PQ
NM
4
4
14.27 kJ / (min˜o C)
'(UA)
(UA)initial
14.27 115
.
u 100%
115
.
241%
.
11-18
40
11.24 a.
Energy balance: MCv
dT
Q W
dt
W 0, Cv
M
. J g ˜q C
177
350 g, Q 40.2W 40.2 J s
b gU|V Ÿ T
|W T
dT
0.0649 q C s
dt
t 0, T 20q C
b.
bg
20 0.0649t s
40q C Ÿ t
. min
308 s Ÿ 51
The benzene temperature will continue to rise until it reaches Tb 801
. q C ; thereafter the heat
input will serve to vaporize benzene isothermally.
. 20
801
Time to reach Tb neglect evaporation : t
926 s
0.0649
Time remaining: 40 minutes 60 s min 926 s 1474 s
Evaporation: 'H v
30.765 kJ mol 1 mol 78.11 g 1000 J kJ 393 J g
b
g
b
b
gb
g
gb
g
g s
.
b40.2 J sg / b393 J gg 0102
.
g sgb1474 sg 200 g
Benzene remaining 350 g b0102
Evaporation rate
c.
11.25 a.
1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask.
2. Put an open flask on the burner. Benzene vaporizesŸ toxicity, fire hazard.
Use a covered container or work under a hood.
3. Left the burner unattended.
4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles.
5. Rubbed his eyes with his hand. Wash with water.
6. Picked up flask with bare hands. Use lab gloves.
7. Put hot flask on partner’s homework. Fire hazard.
Moles of air in room: n
60 m 3 273 K
1 kg - mole
b g
2.58 kg - moles
283 K 22.4 m 3 STP
dT
Q W
Energy balance on room air: nCv
dt
Q m s 'H v H 2 O, 3bars, sat' d 30.0 T T0
W 0
dT
s 'H v 30.0 T T0
m
nCv
dt
N 2.58 kg - moles
b
g
b
b
dT
dt
t
g
g
Cv 20.8 kJ (kg - mole˜q C)
'H v 2163 kJ kg from Table B.6
T0
b
g
0q C
b
40.3m s 0.559T q C hr
0, T
g
10q C
(Note: a real process of this type would involve air escaping from the room and a constant pressure
being maintained. We simplify the analysis by assuming n is constant.)
11-19
11.25 (cont’d)
0 Ÿ 40.3m s 0.559T
b. At steady-state, dT dt
24q C Ÿ m s
T
c.
0.333 kg hr
Separate variables and integrate the balance equation:
z
Tf
10
z
dT
40.3m s 0.559T
t
0
z
dt
m s
0.333
E
t
Integral energy balance
'U
Q
bt
b.
LM
MN
b60 20gq C
4.00 u 104 kJ
20 min
1 kW
60 s
dT
dt
Q
. kW
333
1 kJ s
dT
dt
250 kg
M
Cv 4 .00 kJ kg˜q C
z z
T
Integrate:
t
dT
20o C
t
z
t
0.001 Q dT Ÿ T
20 o C Qdt
0
0
b
g
b
g
2 34 37 41 47 54 62 70 80 90 100
b g
jb
e
Ÿ T 600 s 20o C + 0.001 oC / kJ 34830 kJ
c.
Past 600 s, Q
100 b
g
10 kW
t 600 s
60 s
LM
M 20 0.001M Qdt
T 20 0.001 Qdt
MM
N
0.001 F t
600 I
Ÿ t bs g
Ÿ T 54.8 G
6 H6
2 JK
z
0
2
85q C Ÿ t
2
54.8q C
OP
t P
dt
6 P
PP
Q
600
t
0
600
34830
34830 kJ
t 6
z z
t
T
g
b
g
12000 T 24.8
850 s 14 min, 10 s Ÿ explosion at 10:14 10 s
11-20
bg
0.001Q t
0, T
Evaluate the integral by Simpson's Rule (Appendix A.3)
600 s
30
33 4 33 35 39 44 50 58 66 75 85 95
Qdt
3
0
z
4.8 hr
20 min
4.00 u 104 kJ 1 min
Differential energy balance: MCv
b g OP
b g PQ
13.4 0.559 23
1
ln
0.559 13.4 0.559 10
kg˜q C
Required power input: Q
t
g
0 to t
250 kg 4.00 kJ
MCv 'T
dT
23
10 13.4 0.559T
T f 23q C
11.26 a.
0.559T
40.3
0 Ÿ m s
20q C
11.27 a. Total Mass Balance:
Accumulation=Input– Output
E
dM tot
dt
d( UV)
dt
i m
o Ÿ
m
U=constant
8.00U 4.00U
dV
4.00 L / s
dt
t 0, V0 400 L
KCl Balance:
Accumulation=Input-OutputŸ
dM KCl
dt
i , KCl m
o, KCl Ÿ
m
dV
dC
ŸV
C
dt
dt
b.
8 4C
d( CV )
dt
dV dt
4
100
. u 8.00 4.00C
dC 8 8C
dt
V
t 0, C 0 0 g / L
(i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant).
V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow
and V stays constant at 2000.
V
2000
400
0
t
(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02.
As t increases, C increases and V increases (or stays constant)Ÿ dC/dt=(8-8C)/V becomes
less positive, approaches zero as to f. The curve is therefore concave down.
C
1
0
t
c.
dV
dt
4Ÿ
z
V
dV
400
z
t
4 dt Ÿ V
0
400 4t
11-21
11.27 (cont’d)
8 8C
1 C
dC
V 400 4 t
V
dt 50 0.5t
C dC
t
dt
C
t
Ÿ ln(1 C ) 0 2 ln(50 0.5t ) 0
0 1 C
0 50 + 0.5t
50 0.5t
Ÿ ln(1- C)-1 2 ln
ln(1 0.01t )2
50
1
1
(1 0.01t )2 Ÿ C 1 Ÿ
1- C
(1 0.01t )2
dC
dt
z
z
When the tank overflows, V
400 4t
C = 1-
2000 Ÿ t
1
b1 + 0.01 u 400g
2
400 s
0.96 g / L
11.28 a. Salt Balance on the 1st tank:
Accumulation=-Output
E
d(CS1V1 )
dt
CS1v Ÿ
dCS1
dt
CS1
v
V1
0.08CS1
CS1 ( 0 ) 1500 500
3g/L
Salt Balance on the 2nd tank:
Accumulation=Input-Output
E
d(CS2V2 )
dt
CS 1v CS 2 v Ÿ
dCS2
dt
CS 2 ( 0 )
( CS 1 CS 2 )
v
V2
0.08( CS1 CS 2 )
v
V3
0.04( CS 2 CS 3 )
0 g/L
Salt Balance on the 3rd tank:
Accumulation=Input-Output
E
d(CS3V3 )
dt
CS 2 v CS 3v Ÿ
dCS3
dt
CS 3 ( 0 )
( CS 2 CS 3 )
0 g/L
b.
CS1, CS2, CS3
3
CS1
CS2
CS3
0
t
11-22
11.28 (cont’d)
The plot of CS1 vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is 0.08 u 3 0.24 .
As t increases, CS1 decreases Ÿ dCS1/dt=-0.08CS1 becomes less negative, approaches zero as
to f. The curve is therefore concave up.
The plot of CS2 vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0.08( 3 0) 0.24 .
As t increases, CS2 increases, CS1 decreases (CS2 < CS1)Ÿ dCS2/dt =0.08(CS1-CS2) becomes less
positive until dCS2/dt changes to negative (CS2 > CS1). Then CS2 decreases with increasing t as well
as CS1. Finally dCS2/dt approaches zero as tof. Therefore, CS2 increases until it reaches a
maximum value, then it decreases.
The plot of CS3 vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0.04( 0 0) 0 .
As t increases, CS2 increases (CS3 < CS2)Ÿ dCS3/dt =0.04(CS2-CS3) becomes positive Ÿ CS2
increases with increasing t until dCS3/dt changes to negative (CS3 > CS1). Finally dCS3/dt
approaches zero as tof. Therefore, CS3 increases until it reaches a maximum value then it
decreases.
c.
3
CS1, CS2, CS3 (g/L)
2.5
2
CS1
1.5
CS2
1
CS3
0.5
0
0
20
40
60
80
100
120
140
160
11.29 a. (i) Rate of generation of B in the 1st reaction: rB1
2r1
0.2C A
t (s)
(ii) Rate of consumption of B in the 2nd reaction: rB 2
b. Mole Balance on A:
Accumulation=-Consumption
E
d ( C AV )
dt
01
. C AV Ÿ
dC A
01
. CA
dt
t 0, C A0 100
. mol / L
Mole Balance on B:
Accumulation= Generation-Consumption
E
d ( CBV )
dt
0.2C AV 0.2CB2V Ÿ
dCB
0.2C A 0.2CB2
dt
t 0, CB 0 0 mol / L
11-23
r2
0.2C B2
11.29 (cont’d)
c.
2
CA, CB, CC
CC
1
CB
CA
0
t
. u 1 01
. .
The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is 01
As t increases, CA decreases Ÿ dCA/dt=-0.1CA becomes less negative, approaches zero as
tof. CAo0 as tof. The curve is therefore concave up.
The plot of CB vs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 0.2(1 0)
As t increases, CB increases, CA decreases
( CB2
< CA)Ÿ dCB/dt
=0.2(CA- CB2
0.2 .
) becomes less positive
( CB2
until dCB/dt changes to negative
> CA). Then CB decreases with increasing t as well as CA.
Finally dCB/dt approaches zero as tof. Therefore, CB increases first until it reaches a maximum
value, then it decreases. CBo0 as tof.
The plot of CC vs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0.2( 0)
increases, CB increases Ÿ dCc/dt
=0.2 CB2
0 . As t
becomes positive also increases with increasing t
Ÿ CC increases faster until CB decreases with increasing t Ÿ dCc/dt =0.2 CB2 becomes less positive,
approaches zero as tof so CC increases more slowly. Finally CCo2 as tof. The curve is therefore
S-shaped.
CA, CB, CC (mol/L)
d.
2.2
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
CC
CB
CA
0
10
20
30
t (s)
11-24
40
50
11.30 a. When x =1, y =1 .
y
ax x 1, y 1
a
Ÿ 1
Ÿ a 1 b
1 b
xb
b.Raoult’s Law:
pC5 H12
p *C5 H12 ( 46 o C ) 10
Antoine Equation:
o
0.7 u 1053
760
xp *C5 H12 ( 46 C )
Ÿy
xp *C5 H12 ( 46 o C ) Ÿ y
yP
P
( 6.85221
1064.63
)
46 232.00
xp *C5 H12 ( 46o C )
P
1053 mm Hg
0.970
R| y ax x=0.70, y=0.970 0.970 0.70a (1) R|a
ŸS
0.70 b
S| x b
TFrom part (a), a = 1+ b(2) T|b
.
1078
0.078
c. Mole Balance on Residual Liquid:
Accumulation=-Output
E
dN L
nV
dt
t 0, N L 100 mol
Balance on Pentane:
Accumulation=-Output
E
d( N L x)
dt
nV y Ÿ x
ax
dx
dN L
NL
nV
xb
dt
dt
dN L / dt nV
E
dx
dt
t
FG
H
nV
ax
x
NL x b
IJ
K
0, x = 0.70
d.Energy Balance: Consumption=Input
E
nV 'H vap
Q
t
From part (c),
'H vap 27.0 kJ /mol
0, N L
dN L
nV
dt
nV
Q 27.0
NL
Qt
100 27.0
100 mol
NL
nV
b
Q
27.0 kJ / mol
100 nV t 100 Substitute this expression into the equation for dx/dt from part (c):
11-25
g
Qt
27.0
11.30 (cont’d)
dx
dt
FG
H
nV
ax
x
NL x b
IJ
K
FG
H
Q 27.0
ax
x
xb
Qt
100 27.0
IJ
K
x(0) = 0.70
e.
1
0.9
0.8
y (Q=1.5 kJ/s)
0.7
x, y
0.6
x (Q=1.5 kJ/s)
0.5
0.4
x (Q=3 kJ/s)
0.3
y (Q=3 kJ/s)
0.2
0.1
0
0
200
400
600
800
1000 1200 1400 1600 1800
t(s)
f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a
run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The
higher the heating rate, the faster x and y decrease.
11-26
CHAPTER THIRTEEN
Problem
13.1
Methanol
Production
Rate
430,000 metric tons/year
51,114 kg/h
1,597 kmol/h
Process
stoichiometry:
CH4 + H20 ---> CH30H + H2
So that the required feed rates (with given assumptions) are
CH4 Feed Rate =
Steam Feed Rate =
Problem
1,597 kmol/h
1,597 kmol/h
cubic
meters/min
13.2
1
P =
r.h. = (pH20/p*H20
p*H20 =
0.0424
pH20 =
0.02968
yH20
0.029292
Basis:
1 mole of
component moles
N2
0.79
0.21
02
0.030176
Water
Total
1.03
component
N2
02
Total
Basis:
atm
@ 30C) x 100% =
bar
bar
70%
dry air (79 mole% N2, 21 mole% 02)
Mw
mass
mole frac
28
22.12
32
6.72
0.0293
18
0.54
29.38
1.0000
moles
Mw
0.79
0.21
1.00
mass
22.12
6.72 .
28.84 Difference in avg molecular weight
is due to presence of water; the
difference is slight.
1 km01 of CH4 burned
flow rate of air/km01
Problem
596 standard
28,75lkg/h
28
32
nat gas burned
13.3
Composition of effluent gas from burners
mole frac
mass frac
Component
km01
kg
3.2
0.0103
02
0.100
0.0088
N2
7.900
0.6990
221.2
0.7139
44.0
0.1420
co2
1.000
0.0885
0.1337
H20
2.302
0.2037
41.4
1.0000
309.8
1.0000
total
11.302
p*H20 @ 150C =
4.74 bar
< pH20
Therefore, there is no condensation in cooling the exhaust gases
to 15OC, which means the effluent gas and stack gas have the same
composition.
Volumetric
flow rate
effluent gas
stack gas
density of air =
density of stack gas =
specific gravity =
1,143 m3/kmol CH4 burned
392 m3/kmol CH4 burned
1.1471 kg/m3
0.7899 kg/m3
1 0.6886 relative to ambient air
13-1
I
Problem
13.4
Reformer
Temperature
Reformer
Temperature
Reformer
Pressure
Equilibrium
Temperature
855
1128
15.8
1128
C
K
atm
K
1.6 Mpa
CH4 + Hz0 ---> CO + 3Hz
Production
rates
CH4
H20
CO
CO2
H2
feed rate CH4 x (1 - fractional conversion)
feed rate H20 - fractional conversion x feed rate CH4
fractional conversion x feed rate CH4
feed rate CO2
feed rate + 3 x feed rate CH4 x fractional conversion
(a) methane:steam of 3:l
Stoichiometric
Feed
I
(kmol/h)(
1600
4800
0
0
0
6400
Table:
CH4
H20
co
co2
H2
Total
(kmol/h)
170
3,370
1,430
0
4,291
9,261
574.3668
KP~
Ratio1 574.366846
fractional conversion of CH4* =
=
Product
MolFrac
(kg/h)
0.0183
2,716
0.3639
60,655
0.1544
40,047
0.0000
0
0.4633
8,582
1.0000
112,000
MassFrac
0.0242
0.5416
0.3576
0.0000
0.0766
1.0000
lO"(-(11,769/T(K))+l3.1927)
( (Yco x YH2A3) / (YCH4 x YH20)
6.8939
-3.57E-07 Converge* = ((Kpl/Ratiol)-1)lOO
* the Goal Seek tool is used to adjust the fractional
conversion until Converge is close to zero
Methane Conversion =
) p2
1
Product Flow Rate =-I
(b)
methane:steam of 1:l
Stoichiometric
Table:
CH4
H20
co
co2
H2
Total
fractional
Feed
I
(kmol/h)l
1600
1600
0
0
0
3200
574.3668
Kpl
Ratio1 1653.61852
conversion of CH4 =
(kmol/h)
471
471
1,129
0
3,387
5,458
=
lO"(-(11,769/T(K))+l3.1927)
=
( kc0 x YH2A3)
/ (YCH4
x YH20)
Jp
2
0.7055
-65.266061Converge = ((Kpl/Ratiol)-1)lOO
Methane Conversion --I-[
Product Flow Rate ='
Product
MolFrac
(kg/h)
MassFrac
0.0863
7,538
0.1386
0.0863
8,480
0.1559
0.5810
0.2068
31,608
0.0000
0
0.0000
0.6205
6,773
0.1245
54.400
1.0000
1.0000
5,458 kmol/h
54,400 kg/h
13-2
Problem 13.4 (cont'd)
methane:steam
of 2:l
Stoichiometric
Table:
CH4
H20
co
co2
H2
Total
Feed
I
(kmol/h)l,
1600
3200
0
0
0
4800
Product
(kmol/h) MolFrac
MassFrac
(kg/h)
0.0476
248
0.0330
3,960
1,848
0.2462
33,256
0.3997
0.1802
37,869
0.4552
1,352
0
0.0000
0
.o.oooo
4,057
0.5406
8,115
0.0975
7,505
1.0000
83,200
1.0000
574.3668
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
874.51201
=
( (ycO x YH2^3) / (YCH4 x YH20) )p2
fractional conversion of CH4 = 0.84529344
-34.321446 Converge = ((Kpl/Ratiol)-I)100
Methane
Conversion
=1
Product Flow Rate =
methane:steam
of
Stoichiometric
7,505 kmol/h
83,200 kg/h
4:l
Table:
CH4
H20
co
co2
H2
Total
Feed
I
(kmol/h)l
1600
6400
0
0
0
8000
Product
MassFrac
(kmol/h) MolFrac
(kg/h)
0.0147
130
0.0118
2,074
4,930
0.4506
88,733
0.6302
1,470
0.1344
41,171
0.2924
0.0000
0
0.0000
0
0.4032
8,822
0.0627
4,411
10,941
1.0000
140,800
1.0000
=
574.3668
lO"(-(11,769/T(K))+l3.1927)
Kpl
=
Ratio1 411.494126
( (Yco x YH2^3) / (YCH4 x YH~o) jp2
fractional conversion of CH4 = 0.91899524
39.5808125 Converge = ((Kpl/Ratiol)-I)100
Methane
Conversion = 1
Product Flow Rate =I[
Summary
Moles
Steam:Mole
CH4
1
2
3
4
Methane
H2
Conversion Produced
6,773
70.6%
84.5%
8,115
89.4%
8,582
91.9%
8,822
13-3
H2:CO
3.0000
3.0000
3.0000
3.0000
Problem
13.5
Reformer Temperature
Reformer Temperature
Reformer Pressure
Equilibrium
Temperature
Production
1
855
1128
15.79
1128
C
K
atm
K
1.6 Mpa
CH, + Hz0 ---> CO + 3H,
CO f Hz0 ---> COz f H,
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of CO2
co fractional conversion x feed rate CH4
- production rate of CO2
co2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
methane:steam of 3:l
Stoichiometric
Feed
I
(kmol/h)[
CH4
1600
H20
4800
co
0
co2
0
H2
0
Total
6400
(kmol/h)
176
3,156
1,205
219
4,492
9,249
Product
MolFrac
(kg/h)
0.0190
2,811
0.3413
56,814
0.1303
33,740
0.0237
9,650
0.4857
8,985
1.0000
112,000
MassFrac
0.0251
0.5073
0.3013
0.0862
0.0802
1.0000
574.4
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
574.4
=
( (Yco x YHZh3) / (YCHQ x YH20) )P2
0.2590
=
10"(1,197.8/T(K)-1.6485)
KP~
0.2590
=
Ratio2
(Y co2 x Y t I 2 ) / (Yco x YH20)
fractional conversion of CH4*
0.89021022
Converge 1" -4.3538E-05
= ((Kpl/Ratiol)-1)*100
moles of CO2 formed**
219.322711
= ((Kp2/Ratio2)-l)*lOO
Converge 2"" -0.00028801
* Use Goal Seek tool to adjust CH4 conversion so that Converge 1 goes to zero.
** Use Goal Seek tool to adjust CO2 formation so that Converge 2 goes to zero.
CH4 Conv =
CH4 to CO =
-1
Product Flow Rate =mI
CO with water-gas shift reaction:CO
without water-gas shift reaction =
184.381
13-4
Problem 13.5 (cont'd)
methane:steam of 1:l
Stoichiometric
Feed
I
(Iunol/h)l
CH4
1600
H20
1600
co
0
co2
0
H2
0
Total
3200
*(kmol/h)
482
445
1,082
37
3,392
5,437
Product
MolFrac
(kg/h)
0.0886
7,706
0.0818
8,007
0.1990
30,286
0.0068
1,617
0.6239
6,784
1.0000
54,400
MassFrac
0.1416
0.1472
0.5567
0.0297
0.1247
1.0000
=
KP~
574.4
lO"(-(11,769/T(K))+l3.1927)
=
Ratio1
1662.1
( (Yco x Y,2^3 11 (YCH4 x YH20) ) p2
=
KP~
0.2590
10A(1,197.8/T(K)-1.6485)
=
Ratio2
0.2590
(Y co2 x Y,,) / (Yco x YIi20)
fractional conversion of CH4*
0.69900166
Converge 1" -65.4436238
= ((Kpl/Ratiol)-1)*100
moles of CO2 formed**
36.7477842
Converge 2** -5.79043-05
= ((Kp2/Ratio2)-I)*100
CH4 Conv =
CH4 to CO =
Product Flow Rate =I1
methane:steam of 2:l
Stoichiometric
Feed
I
(kmol/h)l
CH4
1600
3200
H20
co
0
0
co2
H2
0
Total
4800
(kmol/h)
257
1,726
1,213
130
4,160
7,487
=
574.4
Kpl
=
876.6
Ratio1
=
0.2590
KP~
=
Ratio2
0.2590
fractional conversion of CH4*
Converge l* -34.4757385
moles of CO2 formed**
Converge 2** -1.52263-05
5:;4: g;Fh
Product
MolFrac
(kg/h)
0.0343
4,108
0.2306
31,074
0.1620
33,961
0.0174
5,737
0.5557
8,320
1.0000
83,200
lO^(-(11,769/T(K))+13.1927)
( (yco x ~~2~3) 1 (Y c H 4 x YHZO) ) P2
10"(1,197.8/T(K)-1.6485)
(Y c o2 x YH2) / (Yco x YH20)
0.83954105
= ((Kpl/Ratiol)-I)*100
130.381595
= ((Kp2/Ratio2)-l)*lOO
CH4 Conv =
CH4 to CO =
Product Flow Rate -Im[
13-5
MassFrac
0.0494
0.3735
0.4082
0.0690
0.1000
1.0000
Problem 13.5 (cont'd)
methane:steam of 4:l
Stoichiometric
Feed
I
(kmol/h)
CH4
H20
co
co2
H2
Total
1600
6400
0
0
0
8000
1
,(kmol/h)
133
4,634
1,169
299
4,700
10,934
Product
MolFrac
(kg/h)
0.0122
2,126
0.4238
83,418
0.1069
32,722
0.0273
13,134
0.4298
9,400
140,800
1.0000
MassFrac
0.0151
0.5925
0.2324
0.0933
0.0668
1.0000
574.4
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
410.9
=
( (Yco x ~~2 ~x 1 ! (YCH4 x YH20) ) p2
0.2590
=
10"(1,197.8/T(K)-1.6485)
Kp2
Ratio2
0.2590
=
(Y co2 x YH2) / (Yco x YH20)
fractional conversion of CH4*
0.91695885
Converge 1* 39.77214561
= ((Kpl/Ratiol)-l)*lOO
moles of CO2 formed**
298.507525
Converge 2**
-l.O4E-05
= ((Kp2/Ratio2)-l)*lOO
CH4 Conv =
CH4 to CO =
Product Flow Rate =mI
Moles
Methane
H2
H2:CO
Steam:Mole CH4 Conversion Production Production
CO:H2
1
69.9%
3,392
3.1359 0.3188883
84.0%
4,160
3.4300 0.2915462
2
91.7%
4,492
3.7280 0.2682379
3
4,700
4.0217 0.2486487
4
89.0%
13-6
Problem
13.6
Reformer
Reformer
Temperature
Temperature
Pressure
855 C
1128 K
,15.79 atm
1.6 MPa
Production
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of CO2
CO fractional conversion x feed rate CH4
- production rate of CO2
CO2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
Stoichiometric
Table:
COMPONENT
FEED
PRODUCT
MolFrac
CH4
1600
176
0.0190
4800
H20
3,156
0.3413
co
0
1,205
0.1303
co2
0
219
0.0237
H2
0
4,492
0.4857
Total
6400
9,249
1.0000
574.4
=
Kpl
574.4
=
Ratio1
0.2590
=
Kp2
Ratio2
0.2590
=
fractional conversion of CH4*
Converge 1* -4.35353-05
moles of CO2 formed**
Converge 2** -0.00028798
Reformer
Reformer
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
T
T
lO"(-(11,769/T(K))+l3.1927)
( (Yco x Y,,^3) / (YCHI x YHZO) )P2
10"(1,197.8/T(K)-1.6485)
(Y CO2 x YH2) / (YCO x YH20)
0.8902102
((Kpl/Ratiol)-l)*lOO
219.=32271
=
((Kp2/Ratio2)-l)*lOO
750 c
1023 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
576
3,510
759
266
3,338
8,448
MOLFRAC
0.0681
0.4155
0.0898
0.0314
0.3951
Kpl
48.79
Ratio1
48.79
Kp2
0.3329
Ratio2
0.3329
fractional conversion of CH4*
0.64015
Converge 1" - 6 . 4 5 1 9 3 - 0 5 =
moles of CO2 formed**
265.59039
Converge 2** 0.000608333
=
13-7
nCO/nCH4
nH2/nCH4
nCO/nC02
0.474156
2.086444
2.856465
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-1)*100
Problem 13.6 (cont'd)
Reformer
Reformer
Stoichiometric
Temperature
Temperature
Pressure
800 C
1073 K
.15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
357
3,312
999
244
3,975
8,887
MOLFRAC
0.0401
0.3727
0.1124
0.0275
0.4472
Kpl
167.6
Ratio1
167.6
Kp2
0.2936
Ratio2
0.2936
fractional conversion of CH4*
0.7771058
Converge l* 0.000238464
((Kpl/Ratiol)-l)*lOO
moles of CO2 formed**
244: 4398
Converge 2** -2.48073-05
=
((Kp2/Ratio2)-l)*lOO
Reformer
Reformer
Stoichiometric
CH4
H20
co
co2
H2
Total
Temperature
Temperature
Pressure
900 c
1173 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
88
3,086
1,311
201
4,738
9,424
MOLFRAC
0.0093
0.3275
0.1391
0.0214
0.5027
1.0000
Kpl
1443.6
Ratio1
1443.6
KP~
0.2359
Ratio2
0.2359
fractional conversion of CH4*
0.9451022
Converge l* -4.1853-05
=
moles of CO2 formed**
201.39252
Converge 2"" -0.00020159
=
13-8
((Kpl/Ratiol)-I)*100
((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer
Reformer
Stoichiometric
Temperature
Temperature
Pressure
950 c
' 1223 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
39
3,054
1,377
185
4,869
9,523
MOLFRAC
0.0040
0.3207
0.1445
0.0194
0.5113
1.0000
Kpl
3712.3
Ratio1
3712.3
KP~
0.2142
Ratio2
0.2142
fractional conversion of CH4*
0.9759079
Converge l* 0.000662212
=
moles of CO2 formed**
184.93694
Converge 2** -2.0226E-05
=
1.60 MPa
T (Cl
nCO/nCH4
nH2/nCH4
nCO/nCO2
750
0.474
2.086
2.856
800
0.624
2.484
4.087
13-9
855
0.753
2.808
5.494
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
900
0.819
2.961
6.509
950
0.860
3.043
7.443
Problem 13.6 (cont'd)
Reformer
Reformer
Temperature
Temperature
Pressure
855 C
1128 K
*11.84 atm
1.2 MPa
Production
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of Co2
CO fractional conversion x feed rate CH4
- production rate of CO2
CO2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
Stoichiometric
Table:
COMPONENT
FEED
PRODUCT
MolFrac
CH4
1600
116
0.0124
H20
4800
3,098
0.3307
co
0
1,266
0.1352
co2
0
218
0.0232
H2
0
4,670
0.4985
6400
9,368
Total
1.0000
574.4
=
Kpl
Ratio1
574.4
=
0.2590
=
Kp2
Ratio2
0.2590
=
fractional conversion of CH4*
Converge 1* 5.08455E-05
moles of CO2 formed**
Converge 2** -4.7046E-05
Reformer T
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
lO"(-(11,769/T(K))+13.1927)
( (Yco x YH2&3) / (YCH4 x YH20) )P2
10"(1,197.8/T(K)-1.6485)
(Y CO2 x yi-12) / (YCO x YH20)
0.9275941
((Kpl/Ratiol)-1)*100
217.=65358
=
((Kp2/Ratio2)-1)*100
1023 K
11.84 atm
1.2 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
472
3,405
861
267
3,651
8,656
MOLFRAC
0.0545
0.3934
0.0994
0.0309
0.4218
KP~
48.79
Ratio1
48.79
0.3329
KP~
0.3329
Ratio2
fractional conversion of CH4*
0.7049568
Converge 1* -0.00022618
moles of CO2 formed**
267 .=23789
Converge 2** 0.000779089
=
13-10
nCO/nCH4 0.537933
nH2/nCH4 2.281894
3.2207
nCO/nCO2
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Temperature
Reformer Temperature
Pressure
Stoichiometric
800 c
1073 K
*11.84 atm
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
265
3,221
1,092
243
4,249
9,071
MOLFRW
0.0292
0.3551
0.1204
0.0268
0.4685
KP~
167.6
Ratio1
167.6
KP~
0.2936
Ratio2
0.2936
fractional conversion of CH4*
0.8346471
Converge 1* -0.00043805
moles of CO2 formed**
243.14362
Converge 2** -2.3364E-05
=
Reformer Temperature
Reformer Temperature
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
1.2 MPa
900 c
1173 K
11.84 atm
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
1.2 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
54
3,054
1,346
200
4,839
9,492
MOLFRAC
0.0057
0.3217
0.1418
0.0211
0.5098
1.0000
KP~
1443.6
Ratio1
1443.6
Kp2
0.2359
Ratio2
0.2359
fractional conversion of CH4*
0.966348
Converge l* -0.00083839
=
moles of CO2 formed**
200.31077
Converge 2** -2.46013-05
=
13-11
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO