Solution Felder 3ª Ed (Completo)

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3
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
4
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
iii
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 xvii
Cox vapor pressure chart xxi
Psychrometric chart – SI units xxii
Psychrometric chart – American engineering units xxiii
Enthalpy-concentration chart: H2SO4-H2O xxiv
Enthalpy-concentration chart: NH3-H2O xxv
Problem Solutions
Chapter 2 2-1
Chapter 3 3-1
Chapter 4 4-1
Chapter 5 5-1
Chapter 6 6-1
Chapter 7 7-1
Chapter 8 8-1
Chapter 9 9-1
Chapter 10 10-1
Chapter 11 11-1
Chapter 12 (Case Study 1) 12-1
Chapter 13 (Case Study 2) 13-1
Chapter 14 (Case Study 3) 14-1
iv
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
v
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 end-
of-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.
vi
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 1-5
i
, 6-7
 j
2.4 8-10
i
, 11-12
j
 , 13
k
, 14-15
j
2.5 16-17
i
, 18-19
j
, 20
 j*
2.6 21-28
j
, 29
 j*
2.7 30-31
i
, 32-37
j
, 38
k
, 39-41
j
, 42-44
k*
CHAPTER 3
Section Problems
3.1-3-2 1-2
i
, 3-8
j
, 9
k
, 10
j
, 11
k
, 12-13
j
3.3 14-16
i
, 17-25
j
, 26
k*, 27-31
j
3.4 32
i
, 33-46
j
, 47
k
3.5 48
i
, 49-52
j
, 53
k
, 54
k*
CHAPTER 4
Section Problems
4.1-4.3a 1
i
, 2-5
j
4.3b-e 6-20
j
, 21
k
, 22
k*, 22-25
j
, 26
k
, 27
k*
4.4 28-30
j
, 31
k
4.5 32-34
j
, 35-38
k
4.6a, b 39-40
i
, 41
k*, 42-45
j
4.6c 46
k
, 47-48
k*
4.7a-e 49-53
j
, 54-55
k*
4.7f 56-57
j
, 58-59
k
4.7g 60-62
k
, 63
k*
4.8 64-65
i
, 66
j
, 67
k*, 68-73
j
, 74-76
k
, 77-78
j
, 79
k
, 80
k*
CHAPTER 5
Section Problems
5.1 1
i
, 2-3
j
, 4
k
5.2a,b 5-6
i
, 7-15
j
5.2c 16-21
j
, 22-23
k
, 24-30
j
, 31-34
k
, 35-46
j
, 47-54
k
5.3a-c 55-56
j
, 57
k*, 58-60
j
, 61-62
k
, 63
k*
5.4a,b 64-65
i
, 66-69
j
, 70
k
, 71-73
j
5.4c 74-77
j
, 78
k
,79-83
j
vii
CHAPTER 6
Section Problems
6.1 1-4
j
, 5
k*, 6-8
j
6.2, 6.3 6.9-29
j
, 30
k
, 31
k*, 32-34
j
, 35-36
k
, 39-41
j
, 42
k
6.4a 43-44
j
6.4b 45-46
i
, 47-50
j
, 51
k
6.4c 52-57
j
, 58
k*, 59
j
, 60
k
, 61-63
j
, 64
k*
6.4d 65-67
j
, 68-69
k*, 70
j
, 71-73
k
6.5a,b 74
i
, 75-80
j
, 81-83
k
6.5c 84-85
i
, 86
j
, 87
k
6.6 88-91
j
, 92
k
, 93-97
j
6.7 98-99
j
, 100-101
k
CHAPTER 7
Section Problems
7.1, 7.2 1-2
i
, 3
j
, 4-6
i
, 7-8
j
7.3 9
i
, 10-11
j
7.4 12-13
i
, 14-17
j
, 18
i
, 19-23
j
7.5 24-28
j
, 29
k
7.6 30-38
j
, 39
k
, 40-41
j
, 42-44
k
, 45-48
j , 49-50
k
, 51
k*
7.7 52-56
j
, 57-58
k
CHAPTER 8
Section Problems
8.1-8.3b 1-4
i
, 5-16
j
8.3c 17-18
i
8.3d 19-25
j
, 26
k
, 27-31
j
, 32
k
8.3e 33
j
, 34
k*
8.4a 35
i
, 36-41
j
8.4b 42
i
, 43-44
j
8.4c 45-53
j
, 54
k
, 55-56
j
, 57-65
k
, 66
k*, 67-68
k
8.4d 69-71
i
, 72-73
j
, 74
k
, 75
j
8.4e 76-77
i
, 78-79
j
, 80
k
8.5 81-82
i
, 83-87
j
, 88-90
k
, 91
j
, 92
 k
, 93-94
k*, 95-99
j
CHAPTER 9
Section Problems
9.1 1-2
i
, 3-4
j
9.2 5-6
i
9.3, 9.4 7-9
I
, 10
 j
9.5a 11-17
j
, 18
k*, 19-21
j
, 22-23
k
, 24
j*, 25-30
k
9.5b 31-34
k
, 35
k*, 36
j
, 37-38
 k
9.5c 39-44
j
, 45-47
k
9.6a 48-50
j
, 51
k
, 52-56
j
, 57-61
k
9.6b 62-63
j
, 64-67
k
, 68
k*, 69-70
k
viii
CHAPTER 10
Section Problems
10.11-4
j
, 5
k
10.2, 10.3 6
k
, 7
j*, 8-14
k*
CHAPTER 11
Section Problems
11.1, 11.2 1-2
j
, 3
k
, 4
j
, 5
j*, 6-9
j
, 10
k
, 11-14
j
, 15-19
k
11.320-26
j
, 27-30
k
ix
SAMPLE ASSIGNMENT SCHEDULE I
Assignment Read (Ch., Sect.) Problems Due Bonus Problems
(*Computer problem)
 
1 — 
2 1; 2-2.6 2:3, 6, 9, 12 
3 2.7 2:14, 17, 19, 22 2.13
4 3-3.3 2:30, 33; 3:3, 6 2:20*, 36-43
5 3.4-3.5 3:13, 16, 22, 29 2:38-43; 3.9,11
6 4-4.3a 3:32, 37, 40, 49 2:40*; 3.26*
7 4.3b-4.3e 3:47; 4:2, 4, 7 3:53*, 54*
8 4.4 4:10, 11, 14 4:22*
9 4.5 4:18, 25, 28 4:27*
10 — 4:26, 29 4:28*, 31, 35-36
11 TEST THROUGH SECTION 4.3 
12 4.6a-b 4:32, 39 4:38, 41*
13 4.6c, 4.7a-d 4:37, 43 4:39, 43*
14 4.7e-g 4:49, 53 4:47*, 48*
15 4.8 4:56, 65 4:54*, 55*
16 5-5.2b 4:69, 73; 5:3 4:58,59
17 5.2c-5.3 5:7, 12, 15, 19 4:60-62, 63*, 67*
18 5.4 5:25, 26, 59 4:79-80*; 5:4*, 22-23
19 6-6.2 5:35, 40, 66; 6:1 5: 31-34, 47-54
20 6.3 6:2, 9, 12, 15 5:57*, 61-62, 63*; 6:5*
21 6.4a,b 6:23, 33 6:30, 31*, 35, 42
22 6.4c 6:37, 47 6:51, 58*, 60, 64*
23 6.5 6:53, 66, 74 6:68-69, 71-73
24 7-7.3 6:36, 70 6:81, 83, 86-87
25 TEST THROUGH SECTION 6.4 
26 7.4-7.5 7.1, 6, 9, 10 6:88-101
27 7.6 7:12, 18, 25 7:3, 17
28 8-8.3b 7:24, 28, 30 7:29, 39, 42-44
29 8.3c-8.3e 7:33, 35; 8:2 7:49-50, 51*
30 — 7:45; 8:5, 6, 8 7:52-58
31 8.4a-c 8:15, 18, 25 8:12
32 8.4d-e 8:29, 36, 44 8:26, 32, 34*
33 — 8:46, 49 8:54, 57-59
34 8.5 8:51, 55, 73 8:62-68
35 — 8:61 8:74, 80
36 — 8:79, 81, 86 8:88-90
37 TEST THROUGH SECTION 8.4d 
38 9-9.4 8: 95, 98; 9:1 8:93-94*
39 9.5a 9:7, 12 9:9
40 9.5b-c 9:14 9:18*
41 9.6a-b 9:17 9:22-34
42 9.6c-d 9:32 9:35*, 37-38, 45-47
43 — 9:48 9:51, 57-61, 64-65
44 — 9:54 9:67-70
45 — 9:63 
x
SAMPLE ASSIGMENT SCHEDULE II
Assignment Read (Ch., Sect.) Problems Due Bonus Problems
(*Computer problem)
 
1 — 
2 1; 2-2.6 2:4, 7, 8, 11 
3 2.7 2:15, 16, 18, 23 2.13
4 3-3.3 2:31, 34; 3:4, 7 2:20*, 36-43
5 3.4-3.5 3:12, 17, 23, 28 2:38-43; 3.9,11
6 4-4.3a 3:32, 39, 43, 50 2:40*; 3.26*
7 4.3b-4.3e 3:51; 4:3, 4, 6 3:53*, 54*
8 4.4 4:9, 12, 15 4.22*
9 4.5 4:21, 23, 28 4:26, 27*
10 — 4:26, 30 4: 28*, 31, 35-36
11 TEST THROUGH SECTION 4.3 
12 4.6a-b 4:33, 40 4:38, 41*
13 4.6c, 4.7a-d 4:34, 45 4:39, 43*
14 4.7e-g 4:50, 51 4:47*, 48*
15 4.8 4:57, 64 4:54*, 55*
16 5-5.2b 4:70, 71; 5:2 4:58, 59
17 5.2c-5.3 5:8, 11, 13, 17 4:60-62, 63*, 67*
18 5.4 5:25, 29, 58 4:79-80*; 5:4*, 22-23
19 6-6.2 5:38, 42, 67; 6:1 5: 31-34, 47-54
20 6.3 6:6, 8, 10, 17 5:57*, 61-61, 63*; 6.5*
21 6.4a,b 6:27, 35 6:30, 31*, 35-36, 42
22 6.4c 6:39, 46 6:51, 58*, 60, 64*
23 6.5 6:52, 65, 75 6:68-69, 71-73
24 7-7.3 6:41, 70 6:81, 83, 86-87
25 TEST THROUGH SECTION 6.4 
26 7.4-7.5 7:2, 7, 9, 11 6:88-101
27 7.6 7:13, 18, 22 7:3, 17
28 8-8.3b 7:24, 28, 30 7:29, 39, 42-44
29 8.3c-8.3e 7:32, 37; 8.1 7:49-50, 51*
30 — 7:47; 8:5, 6, 9 7:52-58
31 8.4a-c 8:14, 17, 24 8:12
32 8.4d-e 8:30, 37, 43 8:26, 32, 34*
33 — 8:45, 50 8:54, 57-59
34 8.5 8:53, 56, 69 8:62-68
35 — 8:61 8:74, 80
36 — 8:78, 83, 85 8:88-90
37 TEST THROUGH SECTION 8.4d 
38 9-9.4 8:96, 97; 9:2 8:93-94*
39 9.5a 9:7, 11 9:9
40 9.5b-c 9:15 9:18*
41 9.6a-b 9:16 9:22-34
42 9.6c-d 9:33 9:35*, 37-38, 45-47
43 — 9:50 9:51, 57-61, 64-65
44 — 9:55 9:67-70
45 — 9:66 
xi
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).
xii
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.
xiii
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).
xiv
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 requiredfor 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.
xv
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.
xvi
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 Figure Text
Description Number Page
Compressibility charts 5.4-1 208
5.4-2 209
5.4-3 210
5.4-4 211
Cox vapor pressure chart 6.1-4 247
Psychrometric chart – SI units 8.4-1 382
Psychrometric chart – Am. Engr. units 8.4-2 383
Enthalpy conc. chart – H2SO4/H2O 8.5-1 396
Enthalpy conc. chart – NH3/H2O 8.5-2 400
2-1
CHAPTER TWO
2.1 (a)
3 24 3600
1
18144 109
 wk 7 d h s 1000 ms
1 wk 1 d h 1 s
 ms u.
(b)
38 3600
25 98 26 0
.
. .
1 ft / s 0.0006214 mi s
 3.2808 ft 1 h
 mi / h mi / h Ÿ
(c)
554 1 1
1000 g
3
 m 1 d h kg 10 cm
 d kg 24 h 60 min 1 m
85 10 cm g
4 8 4
4
4 4
˜ u ˜. / min
2.2 (a)
760 mi
3600
340
 1 m 1 h
 h 0.0006214 mi s
 m / s 
(b)
921 kg
35.3145 ft
57 5
2.20462 lb 1 m
 m 1 kg
 lb / ftm
3
3 3 m
3 .
(c)
537 10 1000 J 1
1
119 93 120
3.
.
u u Ÿ kJ 1 min .34 10 hp
 min 60 s 1 kJ J / s
 hp hp
-3
2.3 Assume that a golf ball occupies the space equivalent to a 2 2 2 in in in u u cube. For a
classroom with dimensions 40 40 40 ft ft ftu u :
nballs
3 3
6 ft (12) in 1 ball
ft in
6 10 7 million balls u u u |40 40 40
2
48
3
3 3 3
.
The estimate could vary by an order of magnitude or more, depending on the assumptions made.
2.4 4 3 24 3600 s
1 0 0006214
.
.
 light yr 365 d h 1.86 10 mi 3.2808 ft 1 step
 1 yr 1 d h 1 s mi 2 ft
7 10 steps5 16u u
2.5 Distance from the earth to the moon = 238857 miles
238857 mi 1
4 1011
 1 m report
0.0006214 mi 0.001 m
 reports u
2.6
19 0 0006214 1000
26417
44 7
500
25 1
14 500 0 04464
700
25 1
21 700 0 02796
 km 1000 m mi L
1 L 1 km 1 m gal
 mi / gal
Calculate the total cost to travel miles.
Total Cost 
 gal (mi)
 gal 28 mi
Total Cost 
 gal (mi)
 gal 44.7 mi
 
Equate the two costs 4.3 10 miles 
American
European
5
.
.
.
$14,
$1.
, .
$21,
$1.
, .
 
 � �
 � �
Ÿ u
x
x
x
x
x
x
2-2
2.7
5320 14 10
1000
1188 105
 imp. gal h 365 d cm 0.965 g 1 kg 1 ton
 plane h 1 d 1 yr 220.83 imp. gal 1 cm g 1000 kg
 
ton kerosene
plane yr
6 3
3˜
 u ˜.
4 02 10
1188 10
4834 5000
9
5
.
.
u ˜
u
 Ÿ
 ton crude oil 1 ton kerosene plane yr
 yr 7 ton crude oil ton kerosene
 planes planes
2.8 (a)
250
250
.
.
 lb 32.1714 ft / s 1 lb 
32.1714 lb ft / s
 lbm2
f
m
2 f˜ 
(b)
25
2 55 2 6
 N 1 1 kg m / s
9.8066 m / s 1 N 
 kg kg
2
2
˜ Ÿ. .
(c)
10 1000 g 980.66 cm 1
9 109
 ton 1 lb / s dyne
5 10 ton 2.20462 lb 1 g cm / s
 dynesm
2
-4
m
2u ˜ u
2.9
50 15 2 853 32174
1
4 5 106
u u
˜ u
 m 35.3145 ft lb ft 1 lb
 1 m 1 ft s 32.174 lb ft s
 lb
3 3
m f
3 3 2
m
2 f
. .
/
.
2.10
500 lb
5 10
1
2
1
10
252m
3
m
3 1 kg 1 m
2.20462 lb 11.5 kg
 m| u FHG
I
KJ
F
HG
I
KJ |
2.11 (a)
m m V V h r H r
h
H
f f c c f c
c
f
displaced fluid cylinder
3
3 cm cm g / cm
30 cm
 g / cm
 Ÿ Ÿ 
 � 
U U U S U S
U U
2 2
30 141 100
053
( . )( . )
.
(b) U Uf c Hh 
( )( . )
.
30 0 53
171
 cm g / cm
(30 cm - 20.7 cm)
 g / cm
3
3
H
h
Uf
Uc
2.12
V
R H
V
R H r h R
H
r
h
r
R
H
h
V
R H h Rh
H
R
H
h
H
V V
R
H
h
H
R H
H
H
h
H
H
H h h
H
s f
f
f f s s f s
f s s s
 � Ÿ 
Ÿ � FHG
I
KJ �
F
HG
I
KJ
 Ÿ �FHG
I
KJ 
Ÿ 
�
 � � FHG
I
KJ
S S S
S S S
U U U S U S
U U U U
2 2 2
2 2 2 3
2
2 3
2
2
3
2
3
3 3 3
3 3 3
3 3 3
3 3
1
1
; ; 
 
UfUs
R
r
h
H
2-3
2.13 Say h m� � depth of liquid
��������������������������������������
A ( ) m 2 h 
1 m 
Ÿ 
y 
x 
y = 1 
y = 1 – h 
x = 1 – y 2 
d A 
dA dy dx y dx
A m dA y dy
A y y y h h h
y
y
h h
h
 ˜ �
Ÿ �
 � � � � � � � �
� �
�
� �
�
�
�
z
z z
E
1
1
2
2
1
1
2
1
1
2 1
1
1
2 1
2
2
2 1
2 1
1
2
1 1 1 1
d i
d i b g b g b g
 Table of integrals, or trigonometric substitution
m2 sin sin
S
W N
A
A
A
g g
b g 
u
 u
E
4 0 879 1
1 10 345 10
2
3 4
0
m m g 10 cm kg 9.81 N
cm m g kg
 Substitute for 
2 6
3 3
( ) .
.,
W h h hNb g b g b g b g u � � � � � �LNM
O
QP
�3 45 10
2
1 1 1 14
2 1. sin
S
2.14 1 1 32174 1
1
1
32174
 lb slug ft / s lb ft / s slug = 32.174 lb
 poundal = 1 lb ft / s lb
f
2
m
2
m
m
2
f
 ˜ ˜ Ÿ
˜ 
.
.
(a) (i) On the earth:
M
W
 
 ˜ u
175 lb 1
544
175 1
1
m
m
m
2
m
2
3
 slug
32.174 lb
 slugs
 lb 32.174 ft poundal
 s lb ft / s
5.63 10 poundals
.
(ii) On the moon
M
W
 
 ˜ 
175 lb 1
5 44
175 1
1
m
m
m
2
m
2
 slug
32.174 lb
 slugs
 lb 32.174 ft poundal
 6 s lb ft / s
938 poundals
.
2-4
2.14 (cont’d)
( ) /b F ma a F m Ÿ ˜
 
355 pound 1 1als lb ft / s 1 slug m
 25.0 slugs 1 poundal 32.174 lb 3.2808 ft
 0.135 m / s
m
2
m
2
2.15 (a) F ma Ÿ FHG
I
KJ ˜
Ÿ ˜
1
1
6
53623
1
 fern = (1 bung)(32.174 ft / s bung ft / s
 
 fern
5.3623 bung ft / s
2 2
2
) .
(b) On the moon: 
3 bung 32.174 ft 1 fern
 6 s 5.3623 bung ft / s
 fern
On the earth: = 18 fern
2 2W
W
 ˜ 
 
3
3 32174 53623( )( . ) / .
2.16 (a) | 
 
( )( )
( . )( . )
3 9 27
2 7 8 632 23
(b) | u | u uu | u
u u
� �
�
� �
40 10
45
5 8 10
5 9
1 10
3600 10 45 9 10
4 4
4
4 5( . ) /
(c) | � 
� 
2 125 127
2 365 1252 127 5. . .
(d) | u � u | u | u
u � u u
50 10 1 10 49 10 5 10
4 753 10 9 10 5 10
3 3 3 4
4 2 4.
2.17 R
Rexact
| u u uu | u | u
 Ÿ Ÿ u
�( )( )( )( )
( )( )
. .
7 10 3 10 6 5 10
3 5 10
42 10 4 10
3812 5 3800 38 10
1 5 4
6
2 3
3
2.18 (a) 
A: C
C
 C
o
o
o
R
X
s
 � 
 � � � � 
 � � � � � � � � ��
 
731 72 4 0 7
72 4 731 72 6 72 8 730
5
72 8
72 4 72 8 731 72 8 72 6 72 8 72 8 72 8 730 72 8
5 1
0 3
2 2 2 2 2
. . .
. . . . .
.
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
B: C
C
 C
o
o
o
R
X
s
 � 
 � � � � 
 � � � � � � � � ��
 
1031 97 3 58
97 3 1014 987 1031 1004
5
1002
97 3 1002 1014 1002 987 1002 1031 1002 1004 1002
5 1
2 3
2 2 2 2 2
. . .
. . . . .
.
( . . ) ( . . ) ( . . ) ( . . ) ( . . )
.
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-5
2.19 (a)
X
X
s
X
X s
X s
i
i i 
�
� 
 � � 
 � � 
 
¦ ¦
1
12
2
1
12
12
735
735
12 1
12
2 735 2 12 711
2 735 2 12 759
.
( . )
.
. ( . ) .
. ( . ) .
C
C
min=
max=
(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)
 (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'
. u ˜
�2 36 10 4 kg m 2.10462 lb 3.2808 ft 1 h
 h kg m 3600 s
2 2 2
2
(b) Q
Q
'
( )( )( )
. /
' . / /
(
approximate
2
exact
2 2
 lb ft s
= lb ft s 0.00000148 lb ft s
| u u | u | u ˜u ˜ ˜
�
� � �
�
2 10 2 9
3 10
12 10 12 10
148 10
4
3
4 3) 6
6
(a) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
X 134 131 129 133 135 131 134 130 131 136 129 130 133 130 133
Mean(X) 131.9
Stdev(X) 2.2
Min 127.5
Max 136.4
(b) Run X Min Mean Max
1 128 127.5 131.9 136.4
2 131 127.5 131.9 136.4
3 133 127.5 131.9 136.4
4 130 127.5 131.9 136.4
5 133 127.5 131.9 136.4
6 129 127.5 131.9 136.4
7 133 127.5 131.9 136.4
8 135 127.5 131.9 136.4
9 137 127.5 131.9 136.4
10 133 127.5 131.9 136.4
11 136 127.5 131.9 136.4
12 138 127.5 131.9 136.4
13 135 127.5 131.9 136.4
14 139 127.5 131.9 136.4
126
128
130
132
134
136
138
140
0 5 10 15
2-6
2.22 N
C
k
C
C
N
p
o
oPr
Pr
. .
.
( )( )( )
( )( )( )
. . .
 ˜ ˜ ˜
| u u uu u |
u | u u
�
�
P 0583 1936 3 2808
0 286
6 10 2 10 3 10
3 10 4 10 2
3 10
2
15 10 163 10
1 3 3
1 3
3
3 3
 J / g lb 1 h ft 1000 g
 W / m ft h 3600 s m 2.20462 lb
 The calculator solution is 
m
m
2.23
Re
. . .
.
Re
( )( )( )( )
( )( )( )( )
(
 u ˜
| u uu u |
u | u Ÿ
�
� �
�
� �
DuU
P
0 48 2 067 0805
0 43 10
5 10 2 8 10 10
3 4 10 10 4 10
5 10
3
2 10
3
1 1 6
3 4
1 3)
4
 ft 1 m in 1 m g 1 kg 10 cm
 s 3.2808 ft kg / m s 39.37 in cm 1000 g 1 m
 the flow is turbulent 
6 3
3 3
2.24
(a)
k d y
D D
d u
k
k
g p p
g
g
 � FHG
I
KJ
F
HG
I
KJ
 � u ˜u
L
NM
O
QP u ˜
L
NM
O
QP
 Ÿ u Ÿ 
�
� �
�
2 00 0 600
2 00 0 600
100 10
100 100 10
0 00500 10 0 100
100 10
44 426
0 00500 0100
100 10
44 426 0
1 3 1 2
5
5
1 3
5
1 2
5
. .
. .
.
( . )( . / )
( . )( . )( . )
( . )
.
( . ( . )
. /
. /
/ /
/ /
P
U
U
P
 N s / m
 kg / m m s
 m m / s kg / m
 N s / m
 m)
 m s
.888 m s
2
3 2
3
2
2
(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
empiricaldata), all of the other variables are subject to measurement or estimation error.
(c)
dp (m) y D (m
2/s) P (N-s/m2) U (kg/m3) u (m/s) kg
0.005 0.1 1.00E-05 1.00E-05 1 10 0.889
0.010 0.1 1.00E-05 1.00E-05 1 10 0.620
0.005 0.1 2.00E-05 1.00E-05 1 10 1.427
0.005 0.1 1.00E-05 2.00E-05 1 10 0.796
0.005 0.1 1.00E-05 1.00E-05 1 20 1.240
2.25 (a) 200 crystals / min mm; 10 crystals / min mm2˜ ˜
(b) r ˜ � ˜
 Ÿ 
200 10
4 0
 crystals 0.050 in 25.4 mm
 min mm in
 crystals 0.050 in (25.4) mm
 min mm in
 238 crystals / min 
238 crystals 1 min
60 s
 crystals / s
2 2 2 2
2 2
min
.
(c) D
D
Dmm
in mm
1 in
b g b g c c254 254. . ; r r rcrystals
min
crystals 60 s
s 1 min
F
HG
I
KJ c c60
Ÿ c c � c Ÿ c c � c60 200 25 4 10 254 84 7 1082 2r D D r D D. . .b g b g b g
2-7
2.26 (a) 705. / ; lb ft 8.27 10 in / lbm
3 -7 2
fu
(b) U u u u
L
NM
O
QP
 
�
( . / ) exp
. /
.
. .
.
.
705
8 27 10 9 10
101325 10
7057 353145 1000
10 2 20462
113
7 6
6
6
 lb ft
 in N 14.696 lb in
 lb m N / m
 lb ft 1 m g
 ft m cm lb
 g / cm
m
3
2
f
2
f
2 2
m
3 3
3 3 3
m
3
(c) U U Ulb
ft
g lb cm
cm g 1 ft
m
3
m
3
3 3
F
HG
I
KJ c c
1 28 317
453593
62 43
,
.
.
P P P
lb
in
N .2248 lb m
m N 39.37 in
f
2
f
2
2 2 2
F
HG
I
KJ u
�' . '
0 1
1
145 10
2
4
Ÿ c u u Ÿ c u� � �62 43 705 8 27 10 145 10 113 120 107 4 10. . exp . . ' . exp . 'U Ud id i d iP P
P' . ' . exp[( . )( . )] . u Ÿ u u �9 00 10 113 120 10 9 00 10 1136 10 6 N / m g / cm2 3U
2.27 (a) V
V
Vcm
in 28,317 cm
 in
3
3 3
3d i d i 
'
. '
1728
16 39 ; t ts hrb g b g c3600
Ÿ c Ÿ c16 39 3600 0 06102 3600. ' exp ' . expV t V tb g b g
(b) The t in the exponent has a coefficient of s-1.
2.28 (a) 300. mol / L, 2.00 min-1
(b) t C
C
 Ÿ 
Ÿ 
0 3 00
3 00
.
.
 exp[(-2.00)(0)] = 3.00 mol / L
t = 1 exp[(-2.00)(1)] = 0.406 mol / L
For t=0.6 min: C
C
int
. .
( . ) . .
.
 �� � � 
 
0 406 300
1 0
0 6 0 300 14
300
 mol / L
 exp[(-2.00)(0.6)] = 0.9 mol / Lexact
For C=0.10 mol/L: t
t
int
exact
 min
= -
1
2.00
ln
C
3.00
= -
1
2
ln
0.10
3.00
= 1.70 min
 � � � � 
1 0
0 406 3
010 300 0 112
.
( . . ) .
(c)
 
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2
t (min)
C
 (
m
o
l/
L
)
(t=0.6, C=1.4)
(t=1.12, C=0.10)
Cexact vs. t
2-8
2.29 (a) p*
. .
( . ) �� � � 
60 20
199 8 166 2
185 166 2 20 42 mm Hg
(b) 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 ln
(ln ln ) / ( ) (ln ln ) / ( ) .
ln ln ln . ( ) . .
y a bx y ae
b y y x x
a y bx a y e
bx
x
 � Ÿ 
 � � � � �
 � � Ÿ Ÿ �
2 1 2 1
0.693
2 1 1 2 0 693
2 0 63 1 4 00 4 00
(c) ln ln ln
(ln ln ) / (ln ln ) (ln ln ) / (ln ln )
ln ln ln ln ( ) ln( ) /
y a b x y ax
b y y x x
a y b x a y x
b � Ÿ 
 � � � � �
 � � � Ÿ Ÿ 
2 1 2 1 2 1 1 2 1
2 1 1 2 2
(d) ln( ) ln ( / ) ( / ) ( )]
[ln( ) ln( ) ] / [( / ) ( / ) ] (ln . ln . ) / ( . . )
ln ln( ) ( / ) ln . ln( . ) ( / )
/ /
/ /
xy a b y x xy ae y a x e y f x
b xy xy y x y x
a xy b y x a xy e y x e
by x by x
y x y x
 � Ÿ Ÿ 
 � � � � 
 � � Ÿ Ÿ Ÿ 
 [can' t get 
2 1 2 1
3 3
807 0 40 2 2 0 10 3
807 0 3 2 0 2 2 2
2-9
2.30 (cont’d)
(e) ln( / ) ln ln( ) / ( ) [ ( ) ]
[ln( / ) ln( / ) ] / [ln( ) ln( ) ]
(ln . ln . ) / (ln . ln . ) .
ln ln( / ) ( ) ln . . ln( . ) .
/ . ( ) . ( )
/
.33 / .
y x a b x y x a x y ax x
b y x y x x x
a y x b x a
y x x y x x
b b2 2 1 2
2
2
2
1 2 1
2
2 4 1 2 2 165
2 2 2
2 2
807 0 40 2 2 0 10 4 33
2 807 0 4 33 2 0 40 2
40 2 2 6 34 2
 � � Ÿ � Ÿ �
 � � � �
 � � 
 � � � Ÿ 
Ÿ � Ÿ �
2.31 (b) Plot vs. on rectangular axes. Slope Intcpt2 3y x m n �,
(c)
1
3
1 1
3ln( ) ln( )y b
a
b
x
y
x� � Ÿ �Plot vs. [rect. axes], slope = 
1
b
, intercept = 
a
b
(d) 
1
1
3
1
1
3
2
3
2
3
( )
( )
( )
( ) , ,
y
a x
y
x a� � Ÿ � � Plot vs. [rect. axes] slope = intercept = 0
OR
2 1 3 3
1 3
2
ln( ) ln ln( )
ln( ) ln( )
ln
y a x
y x
a
� � � �
� �
Ÿ � �
Plot vs. [rect.] or (y +1) vs. (x - 3) [log] 
 slope =
3
2
, intercept =
(e) ln
ln
y a x b
y x y x
 �
 Plot vs. [rect.] or vs. [semilog ], slope = a, intercept = b
(f) 
 Plot vs. [rect.] slope = a, intercept = b
log ( ) ( )
log ( ) ( )
10
2 2
10
2 2
xy a x y b
xy x y
 � �
� Ÿ
(g) Plot vs. [rect.] slope = , intercept =
 OR 
b
 Plot 
1
 vs. 
1
 [rect.] , slope = intercept = 
1
1 1
2 2
2 2
y
ax
b
x
x
y
ax b
x
y
x a b
y
ax
b
x xy
a
x xy x
b a
 � Ÿ � Ÿ
 � Ÿ � Ÿ
,
,
2-10
2.32 (a) A plot of y vs. R is a line through ( R 5 , y 0 011. ) and ( R 80 , y 0169. ).
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 20 40 60 80 100
R
y
y a R b a
b
y R
 � �� u
 � u u
U
V|
W|
Ÿ u � u
�
� �
� � 
0169 0 011
80 5
2 11 10
0 011 2 11 10 5 4 50 10
2 11 10 4 50 10
3
3 4
3 4
. .
.
. . .
. .
d ib g
(b) R y Ÿ u � u � �43 2 11 10 43 4 50 10 0 0923 4. . .d ib g kg H O kg2
1200 0 092 110 kg kg h kg H O kg H O h 2 2b gb g. 
2.33 (a) ln ln ln
(ln ln ) / (ln ln ) (ln ln ) / (ln ln ) .
ln ln ln ln ( . ) ln( ) . . .
T a b T a
b T T
a T b a T
b � Ÿ 
 � � � � �
 � � � Ÿ Ÿ �
I I
I I
I I
2 1 2 1
119
120 210 40 25 119
210 119 25 9677 6 9677 6
(b) T T
T C
T C
T C
 Ÿ 
 Ÿ 
 Ÿ 
 Ÿ 
�9677 6 9677 6
85 9677 6 85 535
175 9677 6 175 29 1
290 9677 6 290 19 0
119 1 19
119
119
119
. . /
. / .
. / .
. / .
. .
.
.
.
I I
I
I
I
b g
b g
b g
b g
o
o
o
 L / s
 L / s
 L / s
(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.
2-11
2.34 (a) Yes, because when ln[( ) / ( )]C C C CA Ae A Ae� �0 is plotted vs. t in rectangular coordinates,
the plot is a straight line.
-2
-1.5
-1
-0.5
0
0 50 100 150 200
t (min)
ln
 (
(C
A
-C
A
e
)/
(C
A
0
-C
A
e
))
Slope = -0.0093 k = 9.3 10 min-3Ÿ u �1
(b) ln[( ) / ( )] ( )
( . . ) .
. .
( .3
C C C C kt C C C e C
C e
C m V m CV
A Ae A Ae A A Ae
kt
Ae
A
� � � Ÿ � �
 � � u
Ÿ u 
�
� u �
0 09 1001823 0 0495 0 0495
305 23317
3 )(120) -2
-2
 = 9.300 10 g / L
= / =
9.300 10 g gal L
 L 7.4805 gal
8.8 g
2.35 (a) ft and h , respectively3 -2
(b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln( . )353 10 2u � ; or
V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 10 2. u �
(c) V ( ) . exp( . )m t3 2 u u� �100 10 15 103 7
2.36 PV C P C V P C k Vk k Ÿ Ÿ �/ ln ln ln
lnP = -1.573(lnV ) + 12.736
6
6.5
7
7.5
8
8.5
2.5 3 3.5 4
lnV
ln
P
 slope (dimensionless)
 Intercept = ln mm Hg cm4.719
k
C C e
 � � � 
 Ÿ u ˜
( . ) .
. ..736
1573 1573
12 736 3 40 1012 5
2.37 (a)
G G
G G K C
G G
G G
K C
G G
G G
K m CL
L
m
L
L
m
L
L
�
� Ÿ
�
� Ÿ
�
� �0
0 01 ln ln ln
 
ln (G 0-G )/(G -G L)= 2.4835 lnC - 10 .045
-1
0
1
2
3
3.5 4 4.5 5 5.5
ln C
ln
(G
0
-G
)/
(G
-G
L
)
2-12
2.37 (cont’d)
m
K KL L
 
 � Ÿ u �
slope (dimensionless)
Intercept = ln ppm-2.483
2 483
10 045 4 340 10 5
.
. .
(b) C
G
G
G Ÿ � uu � u Ÿ u
�
�
� �475
180 10
3 00 10
4 340 10 475 1806 10
3
3
5 2 3.
.
. ( ) ..483
C=475 ppm is well beyond the range of the data.
2.38 (a) For runs 2, 3 and 4:
Z aV p Z a b V c p
a b c
a b c
a b c
b c Ÿ � �
 � �
 � �
 � �
� ln ln ln � ln
ln( . ) ln ln( . ) ln( . )
ln( . ) ln ln( . ) ln( . )
ln( . ) ln ln( . ) ln( . )
35 102 91
2 58 102 112
3 72 175 112
 
b
c
 
Ÿ �
˜
0 68
146
.
. 
a = 86.7 volts kPa / (L / s)1.46 0.678
(b) When P is constant (runs 1 to 4), plot ln �Z vs. lnV . Slope=b, Intercept= ln lna c p�
lnZ = 0.5199lnV + 1.0035
0
0.5
1
1.5
2
-1 -0.5 0 0.5 1 1.5
lnV
ln
Z
b
a c P
 
� 
slope 
Intercept = ln
052
10035
.
ln .
When �V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln ln �a c V�
lnZ = -0.9972lnP + 3.4551
0
0.5
1
1.5
2
1.5 1.7 1.9 2.1 2.3
lnP
ln
Z
c slope
a b V
 � Ÿ
� 
0 997 10
34551
. .
ln � .
 
Intercept = ln
Plot Z vs �V Pb c . Slope=a (no intercept)
Z = 31.096VbPc
1
2
3
4
5
6
7
0.05 0.1 0.15 0.2
VbPc
Z
a slope ˜311. volt kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-13
2.39 (a)
s
n
x y
s
n
x
s
n
x s
n
y
a
s s s
s s
xy i i
i
n
xx i
i
n
x i
i
n
y i
i
n
xy x y
xx x
 � � 
 � � 
 � � � � 
 �� 
�
 
 
 
¦
¦
¦ ¦
1
0 4 0 3 21 19 31 3 2 3 4 677
1
0 3 19 3 2 3 4 647
1
0 3 19 32 3 18
1
0 4 2 1 31 3 1867
4 677 18 1
1
2
1
2 2 2
1 1
2
[( . )( . ) ( . )( . ) ( . )( . )] / .
( . . . ) / .
( . . . ) / . ; ( . . . ) / .
. ( . )( .
b g
867
4 647 18
0 936
4 647 1867 4 677 18
4 647 18
0182
0 936 0182
2
2 2
)
. ( . )
.
( . )( . ) ( . )( . )
. ( . )
.
. .
� 
 �� 
�
� 
 �
b
s s s s
s s
y x
xx y xy x
xx xb g
(b) a
s
s
y xxy
xx
 Ÿ 4 677
4 647
10065 10065
.
.
. .
y = 1.0065x
y = 0.936x + 0.182
0
1
2
3
4
0 1 2 3 4
x
y
2.40 (a) 1/C vs. t. Slope= b, intercept=a
(b) b a ˜ slope = 0.477 L / g h Intercept = 0.082 L / g;
1/C = 0.4771t + 0.0823
0
0.5
1
1.5
2
2.5
3
0 1 2 3 4 5 6
t
1/
C
0
0.5
1
1.5
2
1 2 3 4 5
t
C
C C-fitted
(c) C a bt
t C a b
 � Ÿ � 
 � � 
1 1 0 082 0 477 0 12 2
1 1 0 01 0 082 0 477 209 5
/ ( ) / [ . . ( )] .
( / ) / ( / . . ) / . .
 g / L
 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.
2-14
2.41 (a) and (c)
 
1
10
0.1 1 10 100
x
y
(b) y ax y a b x ab Ÿ �ln ln ln ; Slope = b, Intercept = ln 
 
ln y = 0.1684ln x + 1.1258
0
0.5
1
1.5
2
-1 0 1 2 3 4 5
ln x
ln
 y
b
a a
 
 Ÿ 
slope 
Intercept = ln 
0168
11258 308
.
. .
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0
(b)
Lab 1
ln(1-Cp/Cao) = -0.0062t
-4
-3
-2
-1
0
0 200 400 600 800
t
ln
(1
-C
p
/C
a
o
)
 
Lab 2
ln(1-Cp/Cao) = -0.0111t
-6
-4
-2
0
0 100 200 300 400 500 600
t
ln
(1
-C
p
/C
a
o
)
k 0 0062. s-1 k 0 0111. s-1
 
Lab 3
ln(1-Cp/Cao) = -0.0063t
-6
-4
-2
0
0 200 400 600 800
t
ln
(1
-C
p
/C
ao
)
Lab 4
ln(1-Cp/Cao)= -0.0064t
-6
-4
-2
0
0 200 400 600 800
t
ln
(1
-C
p
/C
a
o
)
k 0 0063. s-1 k 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.
2-15
2.43 
 
y ax a d y ax
d
da
y ax x y x a x
a y x x
i i i
i
n
i i
i
n
i i
i
n
i i i
i
n
i
i
n
i i
i
n
i
i
n
 Ÿ � Ÿ � Ÿ � 
Ÿ 
 
 
¦ ¦ ¦ ¦ ¦
¦ ¦
I I( )
/
2
1
2
1 1 1
2
1
1
2
1
0 2 0b g b g
2.44 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 b �6 536 4 206. , .
2-16
2.45 (a) E(cal/mol), D0 (cm
2/s)
(b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0.
(c) Intercept = ln = -3.0151 = 0.05 cm / s2D D0 0Ÿ .
Slope = = -3666 K = (3666 K)(1.987 cal / mol K) = 7284 cal / mol� Ÿ ˜E R E/
ln D = -3666(1/T) - 3.0151
-14.0
-13.0
-12.0
-11.0
-10.0
2
.0
E
-0
3
2
.1
E
-0
3
2
.2
E
-0
3
2
.3
E
-0
3
2
.4
E
-0
3
2
.5
E
-0
3
2
.6
E
-0
3
2
.7
E
-0
3
2
.8
E
-0
3
2
.9
E
-0
3
3
.0
E
-0
3
1/T
ln
 D
(d) Spreadsheet
T D 1/T lnD (1/T)*(lnD) (1/T)**2
347 1.34E-06 2.88E-03 -13.5 -0.03897 8.31E-06
374.2 2.50E-06 2.67E-03 -12.9 -0.03447 7.14E-06
396.2 4.55E-06 2.52E-03 -12.3 -0.03105 6.37E-06
420.7 8.52E-06 2.38E-03 -11.7 -0.02775 5.65E-06
447.7 1.41E-05 2.23E-03 -11.2 -0.02495 4.99E-06
471.2 2.00E-05 2.12E-03 -10.8 -0.022964.50E-06
Sx 2.47E-03
Sy -12.1
Syx -3.00E-02
Sxx 6.16E-06
-E/R -3666
ln D0 -3.0151
D0 7284
E 0.05
3-1
CHAPTER THREE
3.1 (a) m u u | u | u16 6 2 1000 2 10 5 2 10 2 103 5 m kg
m
 kg
3
3 b gb gb gd i
(b) �m | uu | u
8 10
2 32
4 10
3 10 10
1 10
6 6
3
2 oz 1 qt cm 1 g
 s oz 1056.68 qt cm
 g / s
3
3 b gd i
(c) Weight of a boxer 220 lbm|
Wmax t
u |12 220 220 stones lb 1 stone
14 lb
m
m
(d) dictionary
V
D L 
| u u u u u u uu u | u
S 2 2
2 3
7
4
314 4 5
4
3 4 5 8 10 5 10 7
4 3 10
1 10
. . ft 800 miles 5880 ft 7.4805 gal 1 barrel
1 mile 1 ft 31.5 gal
 barrels
2
3
d i d i
(e) (i)V | u u | u u | u6 3 3 10 1 104 5 ft 1 ft 0.5 ft 28,317 cm
1 ft
cm
3
3
3
(ii)V | | u u | u150 28 317 150 3 10
60
1 10
4
5 lb 1 ft cm
62.4 lb 1 ft
 cmm
3 3
m
3
3,
(f) SG | 105.
3.2 (a) (i) 
995 1 0 028317
0 45359 1
6212
 kg lb m
m kg ft
 lb / ftm
3
3 3 m
3.
.
. 
(ii) 
995 62 43
1000
6212
 kg / m lb / ft
 kg / m
 lb / ft
3
m
3
3 m
3. . 
(b) U U u u H O SG2 62 43 57 360. . lb / ft lb / ftm 3 m 3
3.3 (a)
50
10
35
3
 L 0.70 10 kg 1 m
 m L
 kg
3 3
3
u 
(b)
1150
1 60
27
 kg m 1000 L 1 min
0.7 1000 kg m s
 L s
3
3min u 
(c)
10 1 0 70 62 43
2 7 481 1
29
 gal ft lb
 min gal ft
 lb / min
3
m
3 m
. .
.
u #
3-2
3.3 (cont’d)
(d) Assuming that 1 cm3 kerosene was mixed with Vg (cm
3) gasoline
V Vg gcm gasoline g gasoline
3d i d iŸ 0 70.
1 082cm kerosene g kerosene3d i d iŸ .
SG
V
V
V
g
g
g 
�
� Ÿ 
�
� 
0 70 0 82
1
0 78
0 82 0 78
0 78 0 70
05
. .
.
. .
. .
.
d id i
d i
g blend
cm blend3
0 cm
3
Volumetric ratio
 cm
 cm
 cm gasoline / cm kerosenegasoline
kerosene
3
3
3 3 V
V
050
1
050
.
.
3.4 In France: 
50 0 5
0 7 10 1 5 22
42
. $1
. . .
$68.
 kg L Fr
 kg L Fru 
In U.S.: 
50 0 1 20
0 70 10 3 7854 1
64
. $1.
. . .
$22.
 kg L gal
 kg L galu 
3.5
(a) � .V u 
700
1319
 lb ft
 h 0.850 62.43 lb
 ft / hm
3
m
3
�
�
. �m VB B u 
3
m
3 B
ft 0.879 62.43 lb
 h ft
V kg / h
d i
b g b g54 88
� � . . . �m V VH H H u d hb g b g0 659 62 43 4114 kg / h
� � . /V VB H� 1319 ft h3
� � . � . �m m V VB H B H� � 54 88 4114 700 lbm
Ÿ � . / � /V mB B Ÿ 114 ft h 628 lb h benzene3 m
� . / � . /V mH H Ÿ 174 716 ft h lb h hexane3 m
(b) – No buildup of mass in unit.
– U B and UH at inlet stream conditions are equal to their tabulated values (which are
strictly valid at 20
o
C and 1 atm.)
– Volumes of benzene and hexane are additive.
– Densitometer gives correct reading.
� ( ) , � ( )V mH Hft / h lb / h3 m
� ( ), � ( )V mB Bft / h lb / h3 m
700 lb / hm
�( ), .V SGft / h3 0850
3-3
3.6 (a) V u 
1955 1
0 35 12563 1000
445
.
. . .
 kg H SO kg solution L
kg H SO kg
 L
2 4
2 4
(b)
Videal
2
2 4 2
2 4
 kg H SO L
 kg
 
 
 kg H SO kg H O L
 kg H SO kg
 L
 u
� 
195 5
18255 100
1955 0 65
0 35 1000
470
4.
. .
. .
. .
% . error � u 470 445
445
100% 5 6%
3.7 Buoyant force up Weight of block downb g b g E
Mass of oil displaced + Mass of water displaced = Mass of block
U U Uoil H O c20542 1 0 542. .b g b gV V V� � 
From Table B.1: g / cm , g / cm g / cm3 3 o
3U U Uc w il Ÿ 2 26 100 3325. . .
m Voil oil
3 3 g / cm cm g u u U 3 325 35 3 117 4. . .
moil + flask g g g � 117 4 124 8 242. .
3.8 Buoyant force up = Weight of block downb g b g
Ÿ Ÿ W W Vg Vgdisplaced liquid block disp. Liq block( ) ( )U U
Expt. 1: U U U Uw B B wA g A g15 2 152.
.b g b g Ÿ u
U Uw B BSG
 Ÿ 1 0 75 0 75.00 . . g/cm 3
3
 g / cm b g
Expt. 2: U U U Usoln soln 3 soln g / cmA g A g SGB Bb g b g b g Ÿ Ÿ 2 2 15 15. .
3.9
W + W hsA B
hb
hU1
Before object is jettisoned
1
1
Let Uw density of water. Note: U UA w! (object sinks)
Volume displaced: V A h A h hd b si b p b1 1 1 �d i (1)
Archimedes Ÿ �Uw d A BV g W W1
weight of displaced water
	
�
Subst. (1) for Vd1 , solve for h hp b1 1�d i
h h
W W
p gAp b
A B
w b
1 1� � (2)
Volume of pond water: V A h V V A h A h hw p p d
i
w p p b p b � Ÿ � �1 1 1 1 1
b g
d i
for 
subst. 2
b h
w p p
A B
w
p
w
p
A B
w pp b
V A h
W W
p g
h
V
A
W W
p gA
1 1
1 1
�
 � � Ÿ � �b g (3)
 2 solve for 
subst. 3 for in
 
b g
b g b g
, h
h
b
w
p
A B
w p bb
p
h
V
A
W W
p g A A
1
1
1
1 1 � � �L
N
MM
O
Q
PP (4)
3-4
3.9 (cont’d)
W
hs
B
hb
hU2
After object is jettisoned
WA
2
2
Let VA volume of jettisoned object = W g
A
AU (5)
Volume displaced by boat:V A h hd b p b2 2 2 �d i (6)
Archimedes Ÿ EUW d BV g W2
Subst. forVd 2 , solve for h hp b2 2�d i
h h
W
p gAp b
B
w b
2 2� (7)
Volume of pond water: V A h V V V A h
W
p g
W
p gw p p d A w p p
B
w
A
A
 � � � �2 2
5 6 7
2
b g b g b g, & 
Ÿ � �
h
p
w
p
B
w p
A
A pp
h
V
A
W
p gA
W
p gA21
2
solve for
(8)
Ÿ � � �
for in 7 solve for 
subst. 8
h h
b
w
p
B
w p
A
A p
B
w bp b
h
V
A
W
p gA
W
p gA
W
p gA2 2
2b g
b g
,
(9)
(a) Change in pond level
h h
W
A g p p
W p p
p p gA
V
p p
V
p Ap p
A
p A W
A W A
A W p
A
W A
A
W p
2 1
8 3 5
0 0
1 1
0� �LNM
O
QP 
� FHG
I
KJ
F
HG
I
KJ �
�
! �
b g b g b gb g
�� �� 
�� ��
Ÿ the pond level falls
(b) Change in boat level
h h
W
A g p A p A p A
V
A
p
p
A
Ap p
A
p A p W p W b
A
p
A
W
p
b
2 1
9 4 5
0 0
1 1 1
1 1 0� � �L
N
MM
O
Q
PP 
F
HG
I
KJ � �
F
HG
I
KJ
F
HG
I
KJ
L
N
MMMM
O
Q
PPPP
!
�
! !
b g b g b g
�� 
 ��� ���
Ÿ the boat rises
3.10 (a) U bulk 3 3
3
2.93 kg CaCO 0.70 L CaCO
 L CaCO L total
 kg / L 2 05.
(b) W Vgbag bulk ˜ uU
2 05 1
100 103
.
.
 kg 50 L 9.807 m / s N
 L 1 kg m / s
 N
2
2
Neglected the weight of the bag itself and of the air in the filled bag.
(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.
3-5
3.11 (a) W m gb b ˜ 
122 5
1202
. kg 9.807 m / s 1 N
1 kg m / s
 N
2
2
V
W W
gb
b I
w
 � ˜u U
(
.
1202
0 996 1
119
 N - 44.0 N) 1 kg m / s
 kg / L 9.807 m / s N
 L
2
2
Ub b
b
m
V
 122 5 103. . kg
119 L
 kg / L
(b) m m mf nf b� (1)
x
m
m
m m xf
f
b
f b f Ÿ (2)
 ( ),( )1 2 1Ÿ �m m xnf b fd i (3)
V V V
m m m
f nf b
f
f
nf
nf
b
b
� Ÿ � U U U
Ÿ � �FHG
I
KJ Ÿ �
F
HG
I
KJ �
2 3 1 1 1 1 1b g b g,
m
x x m
xb
f
f
f
nf
b
b
f
f nf b nfU U U U U U U Ÿ 
�
�x f
b nf
f nf
1 1
1 1
/ /
/ /
U U
U U
(c) x f
b nf
f nf
 �� 
�
� 
1 1
1 1
1 103
0 31
/ /
/ /
/ .
.
U U
U U
1/ 1.1
1/ 0.9 1/ 1.1
(d) V V V V Vf nf lungs other b� � � 
m m
V V
m
m
x x
V V m
f
f
nf
nf
lungs other
b
b
m f mbx f
mnf mb x f
b
f
f
f
nf
lungs otherb
b nf
U U U
U U U U
� � � 
� �FHG
I
KJ � � �
F
HG
I
KJ
 
 �
 
 
( )
( )
1
1 1 1
Ÿ �FHG
I
KJ � �
�
x
V V
mf f nf b nf
lungs other
b
1 1 1 1
U U U U
Ÿ 
�FHG
I
KJ �
�F
HG
I
KJ
�FHG
I
KJ
 
�FHG
I
KJ �
�F
HG
I
KJ
�FHG
I
KJ
 x
V V
m
f
b nf
lungs other
b
f nf
1 1
1 1
1
103
1
11
12 01
122 5
1
0 9
1
11
0 25
U U
U U
. .
. .
.
. .
.
3-6
3.12 (a)
From the plot above, r �5455 539 03. .U
(b) For = g / cm , 3.197 g Ile / 100g H O3 2U 0 9940. r 
� . .mIle 150 0 994 4 6 L g 1000 cm 3.197 g Ile 1 kg
 h cm L 103.197 g sol 1000 g
 kg Ile / h
3
3
(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)
 From the plot, = . l g / minR m5 3 0 0743 5 3 01523 0 546Ÿ � � . . . .b g
y = 0.0743x + 0.1523
R2 = 0.9989
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.0 2.0 4.0 6.0 8.0 10.0 12.0
Rotameter Reading
M
as
s 
F
lo
w
 R
at
e 
(k
g
/m
in
)
y = 545.5x - 539.03
R2 = 0.9992
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0.987 0.989 0.991 0.993 0.995 0.997
Density (g/cm3)
C
o
n
c.
 (
g
 Il
e/
10
0 
g
 H
2O
)
3-7
3.13 (cont’d)
(b)
Rotamete
r Reading
Collection
Time
(min)
Collected
Volume
(cm3)
Mass Flow
Rate
(kg/min)
Difference
Duplicate
(Di)
Mean Di
2 1 297 0.297
2 1 301 0.301 0.004
4 1 454 0.454
4 1 448 0.448 0.006
6 0.5 300 0.600
6 0.5 298 0.596 0.004 0.0104
8 0.5 371 0.742
8 0.5 377 0.754 0.012
10 0.5 440 0.880
10 0.5 453 0.906 0.026
Di � � � � 15 0 004 0 006 0 004 0 012 0 026 0 0104. . . . . .b g kg / min
95% confidence limits: ( . . . .0 610 174 0 610 0 018r rDi ) kg / min kg / min
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)
15 0
117 103
.
.
 kmol C H 78.114 kg C H
 kmol C H
 kg C H6 6 6 6
6 6
6 6 u
(b)
15 0 1000
15 104
.
.
 kmol C H mol 
 kmol 
 mol C H6 6 6 6 u
(c)
15 000
33 07
,
.
 mol C H lb - mole
453.6 mol
 lb - mole C H6 6 6 6 
(d)
15 000 6
1
90 000
,
,
 mol C H mol C
 mol C H
 mol C6 6
6 6
 
(e)
15 000 6
1
90 000
,
,
 mol C H mol H
 mol C H
 mol H6 6
6 6
 
(f)
90 000
108 106
,
.
 mol C 12.011 g C
 mol C
 g C u
(g)
90 000
9 07 104
,
.
 mol H 1.008 g H
 mol H
 g H u
(h)
15 000
9 03 1027
,
.
 mol C H 6.022 10
 mol
 molecules of C H6 6
23
6 6
u u
3-8
3.15 (a) �m 175 2526 m 1000 L 0.866 kg 1 h
 h m L 60 min
 kg / min
3
3
(b) �n 2526 457 kg 1000 mol 1 min
 min 92.13 kg 60 s
 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 0150
936
. . kg mix kg CH OH kmol CH OH 1000 mol
kg mix 32.04 kg CH OH 1 kmol
 mol CH OH3 3
3
3 
(b) �mmix 100.0 lb - mole MA 74.08 lb MA 1 lb mixh 1 lb - mole MA 0.850 lb MA / h
m m
m
m8715 lb
3.17 M � 0 25 28 02 0 75 2 02 8 52. . . . . mol N g N
 mol N
 mol H g H
 mol H
 g mol2 2
2
2 2
2
� . .mN2
3000 0 25 28 02
2470 kg kmol kmol N kg N
h 8.52 kg kmol feed kmol N
kg N h2 2
2
2
3.18 Msuspension g g g � 565 65 500 , MCaCO3 g g g � 215 65 150
(a) �V 455 mL min , �m 500 g min
(b) U � / � .m V 500 110 g / 455 mL g mL
(c) 150 500 0 300 g CaCO g suspension g CaCO g suspension3 3/ . 
3.19 Assume 100 mol mix.
mC H OH
2 5 2 5
2 5
2 52 5
10.0 mol C H OH 46.07 g C H OH
 mol C H OH 
 g C H OH 461
mC H O
4 8 2 4 8 2
4 8 2
4 8 24 8 2
75.0 mol C H O 88.1 g C H O
 mol C H O 
 g C H O 6608
mCH COOH
3 3
3
33
15.0 mol CH COOH 60.05 g CH COOH
 mol CH COOH 
 g CH COOH 901
xC H OH 2 52 5
461 g
 g + 6608 g + 901 g 
 g C H OH / g mix 
461
0 0578.
xC H O 4 8 24 8 2
6608 g
 g + 6608 g + 901 g 
 g C H O / g mix 
461
08291.
xCH COOH 33
901 g
 g + 6608 g + 901 g 
 g CH COOH / g mix 
461
0113.
MW 461 79 7 g + 6608 g + 901 g 
100 mol
 g / mol.
m 25
75
2660
 kmol EA 100 kmol mix 79.7 kg mix
 kmol EA 1 kmol mix
 kg mix
3-9
3.20 (a)
Unit Function
Crystallizer Form solid gypsum particles from a solution
Filter Separate particles from solution
Dryer Remove water from filter cake
(b) mgypsum
4 2
4 2
 L slurry kg CaSO H O
L slurry
 kg CaSO H O ˜ ˜1 0 35 2 0 35 2. .
Vgypsum
4 2 4 2
4 2
4 2
 kg CaSO H O L CaSO H O
2.32 kg CaSO H O
 L CaSO H O ˜ ˜ ˜ ˜
035 2 2
2
0151 2
.
.
CaSO in gypsum: 
 kg ypsum 136.15 kg CaSO
172.18 kg ypsum
 kg CaSO4
4
4m
g
g
 0 35 0 277. .
CaSO in soln.: 
 L sol 1.05 kg kg CaSO
 L 100.209 kg sol
 kg CaSO4
4
4m � 1 0151 0209 000186. . .b g
(c) m u0 35 0 209
0 95
384
. .
.
.
 kg gypsum 0.05 kg sol g CaSO
 kg gypsum 100.209 g sol
10 kg CaSO4 -5 4
% .recovery =
0.277 g + 3.84 10 g
0.277 g + 0.00186 g
-5u u 100% 99 3%
3.21
CSA: 
45.8 L 0.90 kg kmol
 min L 75 kg
 
kmol
min
FB: 
55.2 L 0.75 kg kmol
 min L 90 kg
 
kmol
min
 
mol CSA
mol FB
 
 
U
V
||
W
||
Ÿ 
0 5496
0 4600
0 5496
0 4600
12
.
.
.
.
.
She was wrong.
The mixer would come to a grinding halt and the motor would overheat.
3.22 (a)
150
6910
 mol EtOH 46.07 g EtOH
 mol EtOH
 g EtOH 
6910 g EtO
10365
H 0.600 g H O
0.400 g EtOH
 g H O2 2 
V � Ÿ6910 g EtO 10365 19 123 19 1H L
789 g EtOH
 g H O L
1000 g H O
 L L2
2
. .
SG (6910 +10365) g L
 L 1000 g191
0 903
.
.
(b) c � ŸV ( ) .6910 10365 18 472 g mix L
935.18 g
 L 18.5 L
%
( . . )
.
.error
 L
 L
 � u 19 123 18 472
18 472
100% 3 5%
3-10
3.23 M � 0 09 0 91 27 83. . . mol CH 16.04 g
 mol
 mol Air 29.0 g Air
 mol
 g mol4
700 kg kmol 0.090 kmol CH
 h 27.83 kg 1.00 kmol mix
2.264 kmol CH h4 4 
2 264. kmol CH 0.91 kmol air
h 0.09 kmol CH
22.89 kmol air h4
4
 
5% CH
2.264 kmol CH 0.95 kmol air
h 0.05 kmol CH
43.01 kmol air h4
4
4
Ÿ 
Dilution air required: 43.01 - 22.89 kmol air
 h 1 kmol
 mol air h
b g 1000 mol
20200 
Product gas: 700 20.20 kmol 29 1286
 kg
 h
 Air kg Air
h kmol Air
 kg h� 
43.01 kmol Air 0.21 kmol O 32.00 kg O h
h 1.00 kmol Air 1 kmol O 1286 kg total
0.225
kg O
kg
2 2
2
2 
3.24 x
m
M
m
Vi
i i
i
 , , = M
Vi
U U
A
m
M
m
V M
m
Vi
i i
i
i
i
: x Not helpful. iU U¦ ¦ ¦ z1 2
B
x m
M
V
m M
V
V
M
i
i
i i
i
i: Correct.U U¦ ¦ ¦ 1 1
1 0 60
0 791
0 25
1 049
0 15
1595
1 091 0 917U U U � � Ÿ ¦ xii
.
.
.
.
.
.
. . g / cm 3
3.25 (a) Basis 100 mol N 20 mol CH
 mol CO
 mol CO 
2 4
2
: Ÿ Ÿ
u 
u 
R
S|
T|
20
80
25
64
20
40
25
32
Ntotal � � � 100 20 64 32 216 mol
x xCO CO 2 mol CO / mol , mol CO mol2 
32
216
0 15
64
216
0 30. . /
x xCH 4 N 24 2 mol CH mol , mol N mol 
20
216
0 09
100
216
0 46. / . /
(b) M y Mi i u � u � u � u ¦ 015 28 0 30 44 0 09 16 0 46 28 32. . . . g / mol
3-11
3.26 (a)
Samples Species MW k Peak MoleMass moles mass
Area Fraction Fraction
1 CH4 16.04 0.150 3.6 0.156 0.062 0.540 8.662
C2H6 30.07 0.287 2.8 0.233 0.173 0.804 24.164
C3H8 44.09 0.467 2.4 0.324 0.353 1.121 49.416
C4H10 58.12 0.583 1.7 0.287 0.412 0.991 57.603
2 CH4 16.04 0.150 7.8 0.249 0.111 1.170 18.767
C2H6 30.07 0.287 2.4 0.146 0.123 0.689 20.712
C3H8 44.09 0.467 5.6 0.556 0.685 2.615 115.304
C4H10 58.12 0.583 0.4 0.050 0.081 0.233 13.554
3 CH4 16.04 0.150 3.4 0.146 0.064 0.510 8.180
C2H6 30.07 0.287 4.5 0.371 0.304 1.292 38.835
C3H8 44.09 0.467 2.6 0.349 0.419 1.214 53.534
C4H10 58.12 0.583 0.8 0.134 0.212 0.466 27.107
4 CH4 16.04 0.150 4.8 0.333 0.173 0.720 11.549
C2H6 30.07 0.287 2.5 0.332 0.324 0.718 21.575
C3H8 44.09 0.467 1.3 0.281 0.401 0.607 26.767
C4H10 58.12 0.583 0.2 0.054 0.102 0.117 6.777
5 CH4 16.04 0.150 6.4 0.141 0.059 0.960 15.398
C2H6 30.07 0.287 7.9 0.333 0.262 2.267 68.178
C3H8 44.09 0.467 4.8 0.329 0.380 2.242 98.832
C4H10 58.12 0.583 2.3 0.197 0.299 1.341 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), /), /)
3-12
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
4 0.287 0.412
SAMPLE: 2
(ETC.)
3.27 (a)
(8.
. .
7 10
12
1 28 10 2 9 10
6
7 5u u u Ÿ u0.40) kg C 44 kg CO
 kg C
 kg CO kmol CO2 2 2
( .
. .
11 10
6 67 10 2 38 10
6
5 4u u u Ÿ u0.26) kg C 28 kg CO
12 kg C
 kg CO kmol CO
( .
. .
3 8 10
5 07 10 317 10
5
4 3u u u Ÿ u0.10) kg C 16 kg CH
12 kg C
 kg CH kmol CH4 4 4
m u � u � u ( . . . ) ,1 28 10 6 67 10 5 07 10 13 500
7 5 4 kg 1 metric ton
1000 kg
 
metric tons
yr
M y Mi i u � u � u ¦ 0 915 44 0 075 28 0 01 16 42 5. . . . g / mol
3.28 (a) Basis: 1 liter of solution
1000
0 525 0 525
 mL 1.03 g 5 g H SO mol H SO
 mL 100 g 98.08 g H SO
 mol / L molar solution2 4 2 4
2 4
 Ÿ. .
3.28 (cont’d)
3-13
(b) t
V
V
 � min
55 60
144
 gal 3.7854 L min s
 gal 87 L
 s
55
23 6
 gal 3.7854 L 10 mL 1.03 g 0.0500 g H SO 1 lbm
 gal 1 L mL g 453.59 g
 lb H SO
3
2 4
m 2 4 .
(c) u
V
A
 u 
�
( / )
.
87
4
0 513
 L m 1 min
 min 1000 L 60 s 0.06 m
 m / s
3
2 2S
t
L
u
 45 88 m
0.513 m / s
 s
3.29 (a)
� . .n3
150
1147 L 0.659 kg 1000 mol
 min L 86.17 kg
 mol / min
 Hexane balance: 0 (mol C H / min)
Nitrogen balance: 0.820 (mol N
6 14
2
. � . � .
� . � / min)
180 0050 1147
0950
1 2
1 2
n n
n n
 �
 
UVW Ÿ
 RS|T|
solve mol / min
= 72.3 mol / min
� .
�
n
n
1
2
838
(b) Hexane recovery u u ��
.
. .
n
n
3
1
100%
1147
0180 838
100% 76%b g
3.30
30 mL 1 L 0.030 mol 172 g
10 mL l L 1 mol 
 g Nauseum3 0155.
0.180 mol C6H14/mol
0.820 mol N2/mol
1.50 L C6H14(l)/min
�n3 (mol C6H14(l)/min)
�n2 (mol/min)
0.050 mol C6H14/mol
0.950 mol N2/mol
�n1 (mol/min)
3-14
3.31 (a) kt k is dimensionless (min-1Ÿ )
(b) A semilog plot of vs. t is a straight lineCA Ÿ ln lnC C ktA AO �
k �0 414 1. min
ln . .C CAO AO
3 lb - moles ft Ÿ 02512 1286
(c) C C CA A A
1b - moles
ft
mol liter 2.26462 lb - moles
liter 1 ft mol3 3
F
HG
I
KJ c c
28 317
1000
0 06243
.
.
t
t s
t
C C kt
A A
min
exp
b g b g c c
� � �
1
60
60
0
 min
 s
 
0 06243 1334 0 419 60 214 0 00693. . exp . . exp .c � c Ÿ �C t C tA Ab g b g b g
drop primes
mol / L
t CA Ÿ 200 5 30 s mol / L.
3.32 (a)
2600
50 3
 mm Hg 14.696 psi
760 mm Hg
 psi .
(b)
275 ft H O 101.325 kPa
33.9 ft H O
 kPa2
2
 822 0.
(c)
3.00 atm N m m
1 atm cm
 N cm
2 2
2
2101325 10 1
100
30 4
5 2
2
.
.
u 
(d)
280 cm Hg 10 mm dynes cm cm
1 cm m
dynes
m
2 2
2 2
101325 10 100
760 mm Hg 1
3733 10
6 2
2
10. .
u u
(e) 1
20 1
0 737 atm
 cm Hg 10 mm atm
1 cm 760 mm Hg
 atm� .
y = -0.4137x + 0.2512
R2 = 0.9996
-5
-4
-3
-2
-1
0
1
0.0 5.0 10.0
t (min)
ln
(C
A
)
3-15
3.32 (cont’d)
(f)
25.0 psig 760 mm Hg gauge
14.696 psig
1293 mm Hg gauge
b g b g 
(g)
25.0 psi 760 mm Hg
14.696 psi
2053 mm Hg abs
� 14 696.b g b g
(h) 325 435 mm Hg 760 mm Hg mm Hg gauge� � b g
(i)
35.0 psi 760 mm Hg 1 cm 13.546 g Hg cm
14.696 psi 10 mm g CCl cm
 cm CCl
3
4
3 41595
1540
.
 
3.33 (a) P gh
h
g u ˜U
0 92 1000. kg 9.81 m / s (m) 1 N 1 kPa
 m 1 kg m / s 10 N / m
2
3 2 3 2
Ÿ h Pg (m) (kPa)0111.
P hg Ÿ u 68 0111 68 7 55 kPa m. .
m Voil uFHG
I
KJ u u u
F
HG
I
KJ uU S0 92 1000 7 55
16
4
14 10
2
6. . . 
kg
m
 m kg
3
3
(b) P P P ghg atm top� � U
68 101 115 0 92 1000 9 81 103� � u u. . /b g b g h Ÿ h 5 98. m
3.34 (a) Weight of block = Sum of weights of displaced liquids
( )h h A g h A g h A g
h h
h hb b1 2 1 1 2 2
1 1 2 2
1 2
� � Ÿ ��U U U U
U U
(b)
 , , 
 , 
 
top atm bottom atm b
down atm up atm
down up block liquid displaced
P P gh P P g h h gh W h h A
F P gh A h h A F P g h h gh A
F F h h A gh A gh A W W
b
b
b
 � � � � �
Ÿ � � � � � �
 Ÿ � � Ÿ 
U U U U
U U U U
U U U
1 0 1 0 1 2 2 1 2
1 0 1 2 1 0 1 2 2
1 2 1 1 2 2
( ) ( )
( ) ( ) [ ( ) ]
( )
h
Pg
3-16
3.35 'P P gh P � �atm insideUb g
 �1 atm 1 atm � ˜
105 1000.b g kg 9.8066 m 150 m 1 m 1 N
m s 100 cm 1 kg m / s
2 2
3 2 2 2 2
F u u FHG
I
KJ 
154
100 10
022481
1
22504
 N 65 cm
cm
 N
 lb
 N
 lb
2
2
f
f.
.
3.36 m V u u uU 14 62 43 2 69 107. . . lb 1 ft 2.3 10 gal
 ft 7.481 gal
 lbm
3 6
3 m
P P gh �0 U
 � u ˜14 7
14 62 43 12
.
. .lb
in
 lb 32.174 ft 30 ft 1 lb ft
 ft s 32.174 lb ft / s 12 in
f
2
m f
2
3 2
m
2 2 2
 32 9. psi
— Structural flaw in the tank.
— Tank strength inadequate for that much force.
— Molasses corroded tank wall
3.37 (a) mhead
3 3
m
3 3 3 m
 in 1 ft 8.0 62.43 lb
12 in ft
 lb u u u S 24 3
4
392
2
W m g
s ˜ head
m f
m
2 f
 lb 32.174 ft lb
32.174 lb ft / s
 lb
392 1
392
2/
F F F Wnet inside out
f
2 2
2 f f
 lb 20 in
 in
 lb lb � � � u � 30 14 7
4
392 4415
.b g S
The head would blow off.
Initial acceleration: a
F
m
 ˜ net
head
f m
2
m f
2 lb ft / s
392 lb lb
 ft / s
4415 lb 32 174
1
362
.
(b) Vent the reactor through a valve to the outside or a hood before removing the head.
3.38 (a)
P gh P P Pa atm b atm � U , 
If the inside pressure on the door equaled Pa , the force on
the door would be F A P P ghAdoor a b door � ( ) U
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 ,