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At the end of the semester, students are expected to
II. INSTRUCTOR AND MEETING
TIMES
Instructor: Professor Wei-Yin Chen
Office: Room 140, Anderson Hall
Telephone: 915-5651
E-mail: cmchengs@olemiss.edu
Class: 4:15 pm - 5:45 pm, Tuesdays and Thursdays
Recitation Sessions: 1:30 p.m. - 2:30 p.m. Monday (to be discussed)
Office Hour: 3:00 - 5:30 pm Mondays, Wednesdays and Fridays, or by appointment
There will be two hourly exams, and one final exam. Homework will also
be assigned every week, and the exam problems will be in similar nature of, but
not be verbatim from, the homework assignments.
Homework 200 points
Hourly Exams 200 points
Final Examination 200 points
___________________________________________
Total 600 points
Total grade will be given according to a scale similar to the following:
600 < A < 480
479 < B < 420
419 < C < 360
359 < D < 300
299 < F
Depending on examination difficulty, this scale may be relaxed.
Because the level of difficulties of the materials, attendance is very important for the course. In accordance with Departmental policy, students missing more than 3 class periods are subject to direct grade penalties. Penalties may be assessed without regard to the student's performance.
Students are expected to keep up with the material as it is presented and submit assignments on time; most students find this difficult without regular class attendance. Successful learning involves the following important steps
1. Attend all classes on time and listen carefully
2. Do not fall sleep
3. Do not hesitate to ask questions in or out of the
classroom
4. Do not play crosswords
5. Practice, practice and practice
VII. TENTATIVE COURSE
SCHEDULE
|
Number
of Week |
Week
of |
Topics |
Reading
Assignments |
|
1 |
August
22 |
Introduction
& review of first law of thermodynamics |
Chapter
1 and 2 |
|
2 |
August
29 |
Volumetric
properties of pure fluids |
Chapters
3 |
|
3 |
September
5 |
Volumetric
properties of pure fluids |
Chapter
3 |
|
4 |
September
12 |
Heat
effects (instructor in Pittsburgh) |
Chapter
4 |
|
5 |
September
19 |
Second
law of thermodynamics, Exam I |
Chapter
5 |
|
6 |
September
26 |
Thermodynamic
properties of fluids |
Chapter
6 |
|
7 |
October
3 |
Thermodynamic
properties of fluids |
Chapter
6 |
|
8 |
October
10 |
VLE:
introduction |
Chapter
10 |
|
9 |
October
17 |
Solution
thermodynamics: theory |
Chapter
11 |
|
10 |
October
24 |
Solution
thermodynamics: theory; Exam II |
Chapter
11 |
|
11 |
October
31 |
Solution
thermodynamics: applications |
Chapter
12 |
|
12 |
November
7 |
Solution
thermodynamics: applications (AIChE annual meeting, instructor in Salt Lake
City) |
Chapter
12 |
|
13 |
November
14 |
Solution
thermodynamics: applications and Chemical-reaction equilibria |
Chapters
12 & 13 |
|
14 |
November
21 |
Fall
Break :) |
|
|
15 |
November
28 |
Chemical-reaction
equilibria |
Chapter
13 |
|
16 |
December
5 |
Final
Exam - TBA |
Download MathCad programs: MCPH, ICPH, MCPS, ICPS, MDCPH, IDCPH, MDCPS,
IDCPS, HRB, SRB, and PHIB: (v.2001i format)
http://home.olemiss.edu/~cmchengs/CHE421/programs.mcd
Chapter 2: #1, #11, #27, #30, #33, and the problem below:
"A constant-volume bomb of capacity 0.028 m^3 contains an ideal gas at a
pressure of 10 bar and a temperature of 311 K. Connected to the bomb is a
capillary tube through which the gas may slowly leak out into the
atmosphere. Surrounding the bomb and capillary is a water bath, which
keeps the bomb and the its contains at 311 K. Find the quantity of heat
exchanged between the bath when gas no longer escapes from the bomb. The
gas has a constant Cp of 30 kJ / (kmol K)."
Ans:
#1 a) 1.71 kJ, b) 1.71 kJ, c) 20.02
degC, d) -1.71 kJ, e) 0
#11 0.929 hr
#27 578.82 R
#30 a) 11014 kJ, b) -18.62 kJ
#33 Ws=-172.99 Btu/lbm (Is the kinetic energy change
negligible? yes!), or power=39.52 hp
Extra problem: Q = 25.2 kJ
Chapter 3: #5, #10, #21, #32, #44, #47, #54, Use Lee / Kesler as
#32(f)
Ans:
#5 W=18.302 atm ft^3
#10 a) 2982 kJ, b) 5105 kJ, c) 7635
kJ, d) 13200 kJ, e) 1100 kJ
#21 148.8 C
#32 a) V = 1919 cm^3/mol, Z=0.929,
b) V =
1924 cm^3/mol, Z=0.932,
c) V =
1916.5 cm^3/mol,
d) V =
1918 cm^3/mol, Z=0.928,
e) V =
1900.6cm^3/mol, Z=0.92.
f) V
from Lee/Kesler = ?
#44 m_liq = 127.6 kg, m_vap = 2.3 kg
#47 Use one of the following methods:
a) trial-and-error with Lee/Kesler: 79.73 bar
b) R-K EOS: 85.29 bar
#54 Use of the following methods
a) Figure 3.16: 0.629 gm/cm^3
b) Lee/Kesler: 0.638 gm/cm^3
Chapter 4: #2(a), #9(for n-pentane only), #21(b,c), #22(a), #26, #29,
#31
Ans:
#2(a) 1374.5 K
#9 a) 358.6 J per gm, 5.05%, b) 358.6, -0.4%
#21 b) -905.468 kJ, c) -71.660 kJ
#22(a) -109.795 kJ
#26 -333.509 kJ
#29 heat lost from the furnace = 70.612 kJ, heat
transferred to the heat exchanger = 766.677 kJ
#31 -115,653 J
Chapter 5: #7, #8, #21, #28(b); and #30
Ans:
#7 max power = 5.336*10^6 kW, Q =
8.538*10^6 kW
#8 entropy change of water = 1.305 kJ/(kg K),
entropy change of reservoir = -1.121 kJ/(kg K), total entropy change = 0.184
kJ/(kg K).
#21 2.914 J / mol*K
#28(b) entropy change = 2657.5 J / K
#30 total entropy change = 3.42 J/(mol K); it is
positive, thus, the process is possible.
Chapter 6: #7, #17, #29, #49, #57, #58, #63, #64, #65, #81,
#87(a) Derive
equations #24, #34 and #36 in the Chapter 6 of the notes.
Ans:
#7 DS = -2.661 J / (kg K), DH
= 551.7 J / kg
#17 a) entropy change = 100.34 J/(mol K),
b) 102.14 J/(mol K).
#29 0.299 Btu/(lbm R)
#49 a) Gl = -89.94 Btu/lbm, Gv =
-89.91 Btu/lbm
b) DHlv/T = DSlv = 1.055 Btu/(lbm R)
c) VR = -0.235 ft^3/lbm, HR = -28.5 Btu/lbm, SR = -0.0273
Btu/(lbm R)
d)
DSlv = 1.056 Btu/(lbm R)
Generalized Charts: VR = -0.189 ft^3/lbm, HR = -19.03 Btu/lbm, SR =
-0.0168 Btu/(lbm R)
#57 T = 408.91 K, DS = 24.699 J/(mol K)
#58 DH = -801.9 J/mol, DS = -20.639 J/mol*K
#63 T = 381.41 K, Ws = 5678 J/mol
#64 Wmax = -1224.3 kJ/kg
#65 Frac = 0.4058, DSsurr = 50.234 kW/K*s,
DSsystem = -50.234 kW/K*s, DStotal = ?
#81 Msteam = 3.933 lbm/s, DS = 2.064 BTU/R*s
#87(a) VR = -192 cm^3/mol, HR = -1.429
kJ/mol, SR = -0.002 kJ/(mol K)
Chapter 10: #2(a), #16, #26, #29
Ans:
#2 (a) Please see the graphical results at the end of this section.
#16 (a) 43.846 to 56.745 kPa, (b) P = 51.892 kPa, molar fraction of
vapor = 0.379 (c) no azeotrope, please explain.
#26 and #29 Similar to Examples 10.4 through 10.6 in the
textbook. Please use spreadsheet for your calculations.
Chapter 11: #1, #2, #17, #21, #23(a), #24(b), #25
Ans:
#1 DS = 5.079 J / (mol K)
#2 DS = 38.27 J / K
#17 f = 217.14 bar, G^E / RT = -0.323
#21 f / f^sat = 1.079
#23 (a) f = 2.445 bar
#24 (b) See the graphical results at the end of this
section.
#25 (a) f1 = 10.053 bar, f2 = 17.059 bar, phi1 =
0.957, phi2 = 0.875,
(b) f1
= 9.978 bar, f2 = 17.022 bar, phi1 = 0.95, phi2 = 0.873.
Chapter 12: #3, #13, #15, #18, Prove that the two forms of Margules
equation can be transformed into each other, #26, #36, #37, #47, #50
For #13 through #18, calculations should be conducted for
the 1-propanol (1) / water (2) system
Ans:
#3: a) root-mean-square (RMS) error = 3.805kPa, b) RMS =
3.137kPa, c) RMS = 1.615kPa,
d) RMS = 1.632kPa, e)
1.629kPa, f) 1.566kPa. Please draw the fihures.
#13 See the graphical results at the end of this section.
#15 See the graphical results at the end of this section.
#18 (a) T_bubble = 361.07 K, y1 = 0.419, y2 = 0.581
(b) T_dew = 364.29
K, x1 = 0.048, x2 = 0.952
(c) V = 0.805, x1
= 0.089, y1 = 0.351
(d) T_azeotrope =
360.85 K, x1 = 0.4561
#26 V1bar = 124.76 cm^3 / mol, V2bar = 93.36 cm^3 / bar
#36 x_LiCl = 0.1012
#37 It drops, approximately, from -67 to -79.5 kJ/mol when n
increases from 10 to 1000. Draw the figure.
#47 205 F
#50 By trial-and-error: m = 120 lb, (or 0.37 wt%), enthalpy
of the mixture = -5.5 Btu /lb.
Chapter 13: #3, #4, #11, #16, and #32
Ans:
#3 See the results shown at the end of this section.#4 See the
graphical results at the end of this section.
#11 yHCl = 0.3508, yO2 = 0.0397, yH2O = 0.3048, yCl2 =
0.3048.
#16 (a) fractional conversion is 0.777, (b) 646.8 K.
#32 (a) No, but why? (b) No, but why? (c) ratio = 4.1
(d) No, but why?
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