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
Office Hour: 3:00 - 5:30 pm 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 9 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 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 24 | Introduction & review of first law of thermodynamics | Chapter 1 and 2 |
| 2 | August 31 |
Volumetric properties of pure fluids | Chapters 3 |
| 3 | September 7 |
Volumetric properties of pure fluids | Chapter 3 |
| 4 | September 14 |
Heat effects | Chapter 4 |
| 5 | September 21 |
Second law of thermodynamics, Exam I | Chapter 5 |
| 6 | September 28 | Thermodynamic properties of fluids | Chapter 6 |
| 7 | October 5 |
Thermodynamic properties of fluids | Chapter 6 |
| 8 | October 12 |
VLE: introduction | Chapter 10 |
| 9 | October 19 | Solution thermodynamics: theory ( instructor in Taiwan) | Chapter 11 |
| 10 | October 26 | Solution thermodynamics: theory; Exam II | Chapter 11 |
| 11 | November 2 |
Solution thermodynamics: applications | Chapter 12 |
| 12 | November 9 |
Solution thermodynamics: applications | Chapter 12 |
| 13 | November 16 | Solution thermodynamics: applications and Chemical-reaction
equilibria (AIChE annual meeting, instructor in Philadelphia) |
Chapters 12 & 13 |
| 14 | November 23 |
Fall Break :) | |
| 15 | November 30 |
Chemical-reaction equilibria | Chapter 13 |
| 16 | December 7 |
Final Exam - 4:00 PM, Wednesday, December 10 |
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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