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CHE 220  THERMODYNAMICS  Penn State Study Resources
 Penn State University (Penn State, PSU)
 PROF. MARANNAS

Lecture_24after
School: Penn State
Course: THERMODYNAMICS
CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 24:examples of open system energy balances Flow calorimetry Mixing Adiabatic turbine Sudden expansion 1 Example 1: a flow calorimeter operates at steady state and conditions noted

Sol14
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 14 Due: Will not be collected Reading: Sections 6.76. Note: the nal will cover chapters 16, up to and including section 6.9. 1 QB heat in 1 2 boiler work in work out pump turbine WB WT condenser 3 4 heat out QC Problem

Hw14
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 14 Due: Will not be collected Reading: Sections 6.76. Note: the nal will cover chapters 16, up to and including section 6.9. Problems: 6.27, 6.29. 6.30, 6.31, 6.39 1

Hw13
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 13 Due: April 24 Reading: 6.6 Problems: 6.12, 6.15, 6.17, 6.20 1

Exam4
School: Penn State
Course: THERMODYNAMICS
Problem 1 A rigid pressurized cylinder contains steam at 22 bar, 400 C. The cylinder is placed in a heat bath at 100 C and is allowed to reach thermal equilibrium. a) Calculate the amount of heat. b) Calculate the entropy generation. c) Calculate the volu

Hw12
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 12 Due: Friday April 18 Reading: Chapter 6, 6,16.5 Problems: 6.2, 6.4, 6.5, 6.8 1

Hw7
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 7 Due: Will not be collected Reading: 3.83.10; review of chapter 3. Problems 3.6, 3.17, 3.28, 3.31, 3.35, 3.38 1

Hw6
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 6 Due: February 27 Reading: Chapter 3, sections 3.63.7; read ahead 3.83.9. Problems: 3.18, 3.22, 3.26, 3.29 1

Hw8
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 8 Due: March 20 Happy Spring Break! Reading: Chapter 4: 4.1, 4.2 Problems 4.3, 4.4, 4.5, 4.7. 1

Hw4
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 4 Due: February 13 Reading: 2.32.12, review of chapter 3. Read ahead sections 3.13.3 Problems: 2.15, 2.18, 2.20, 2.25. 1

Hw5
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 5 Due: February 21 Reading: 3.33.6 Problems: 3.3, 3.4, 3.10, 3.14 1

Fig_mollier
School: Penn State
Course: THERMODYNAMICS
4000 Mollier diagram of steam 3900 C 0 T. Matsoukas 2014 75 700 C 3800 650 C 3700 600 C 3600 550 C 1000 bar 800 bar bar 3500 400 bar bar 600 3400 40 b bar 0.1 5b ar 0.2 8b ar 350 C 300 C 0.0 2900 6b 3000 0.0 ar 0.0 3100 400 C bar 0.1 100 60 b ar 150 bar

LK_Tables
School: Penn State
Course: THERMODYNAMICS
Lee Kesler Tables Tr = 0.5 z0 z1 h0 h1 s0 s1 0 1 Pr 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.4 1.6 1.8 2 2.5 3 3.5 4 0.0207 0.0413 0.0619 0.0825 0.1031 0.1236 0.1442 0.1647 0.1851 0.2056 0.2261 0.2465 0.2873 0.3280 0.3687 0.4092 0.5103 0.6110 0.711

Sol2
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 2 Due: January 30 Reading: Chapter 2: 2.12.3 1 Problem 1 Solution a) 25 C, 1 bar: liquid, because the temperature is below the saturation temperature at 1 bar (99.63 C). 10 bar, 80 C: liquid, because the temperature is b

Sol9
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 9 Due: March 27 Reading: 4.34.11, review of chapter 4 1 Problem 1 Solution Outline: For reversible isothermal process, Q D TS The work is then obtained as W D U Q Numerical substitutions (in standard steamtable units) T

Sol13
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 13 Due: April 24 Reading: 6.6 1 Problem 1 ./problemdbase/ch06_12/problem.tex. Solution 1 2 P .bar/ 35: 3: T . C/ 500 200 H .kJ=kg/ 3451:5 2866 S .kJ=kg K/ 7:1592 7:3132 a) m.H2 P P H1 / D W ) mD P P W H2 H1 D 1:28 kg=s

Sol12
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 12 Due: Friday April 18 Reading: Chapter 6, 6,16.5 1 Problem 1 Solution a) W D 45 J=mol K H D CP .T2 Q D H T1 / D W D 9083 J=mol 4583 J=mol b) S D Cp ln T2 T1 R ln P2 D P1 1:01778 J=mol K d) Wideal D H T0 S D 2 8779:55 J

Sol11
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 11 Due: Will not be collected Reading: Chapter 5, sections 5.85.11 1 Problem 1 Solution a) The residual properties can be calculated as follows: H S R R P D V P Z @V dP T @T P @V dP @T P 0 Z D 0 R P with the integrations

Sol3
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 3 Due: (Will not be collected) Reading: Review of sections 2.1, 2.2; 2.32.5 1 Problem 1 Solution a) At 1 bar, 210 C, the specic volume of steam is found by interpolation to be V D 2219:2 cm3 /g D 2:2192 m3 /kg Since the

Sol4
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 4 Due: February 13 Reading: 2.32.12, review of chapter 3. Read ahead sections 3.13.3 Problems: 2.15, 2.18, 2.20, 2.25. Problem 1 Solution a) The second virial coefcient is directly related to the slope of an isotherm on

Sol5
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 5 Due: February 21 Reading: 3.33.6 1 Problem 1 Solution a) For this constantpressure expansion, the amount of work is 1 1 W D P0 .Vs Vl / D P0 s l where P0 D 1 bar is the atmospheric pressure, V is the specic volume, is

Sol6
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 6 Due: February 27 Reading: Chapter 3, sections 3.63.7; read ahead 3.83.9. 1 Problem 1 Solution From the steam tables at 6 bar we nd (in standard steam table units): V U H Liquid 0:0011 669:72 670:5 Vapor 0:3156 2566:8 2

Sol10
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 10 Due: April 3 Reading: review of chapter 4; 5.15.7. 1 Problem 1 Solution The idealgas heat capacity of nitrogen is ig CP .t/ D 8:231 10 12 t 4 C 1:3053 10 8 t 3 C 5:8198 10 7 t2 2:17 10 3 t C 29:423 ig with T in K and

Sol8
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 8 Due: March 20 Happy Spring Break! Reading: Chapter 4: 4.1, 4.2 1 Problem 1 Solution Entropy change of ice: The process is conducted under constant pressure (dQ D dH ), therefore, the entropy change will be calculated a

Sol7
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 7 Due: Will not be collected Reading: 3.83.10; review of chapter 3. 1 Problem 1 Solution a) At the initial state: P1 D 5 bar; T1 D 151:84 C; V1 D 0:3748 m3 =kg; U1 D 2560:7 kJ=kg The mass in the tank is m D V tank =V1 D

Sol1
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 1 Due: January 23 Reading: Chapter 1 1 Problem 1 Solution We will need the following unit conversions: 1 ft D 0:3048 m; 1 lb D 0:454 kg; 1 lbmol D 454 mol (Note on the conversion mol to lbmol: one mol has a mass equal to

Hw9
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 9 Due: March 27 Reading: 4.34.11, review of chapter 4 Problems 4.8, 4.12, 4.18, 4.20 1

Hw3
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 3 Due: (Will not be collected) Reading: Review of sections 2.1, 2.2; 2.32.5 Problems: 2.3, 2.6, 2.8, 2.9, 2.11, 2.12 The rst four problems of this assignment cover review material from sections 2.12.2. The last two prob

Hw2
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 2 Due: January 30 Reading: Chapter 2: 2.12.3 Problems: 2.1, 2.2, 2.4 and 2.7 1

Lecture_17
School: Penn State
Course: THERMODYNAMICS
Lecture 17  1 step model processes Many "real" thermodynamic processes (apart from irreversibility) can be modeled as composed of a succession of "onestep" processes. Important examples of these buildingblock processes are: workfree (no change in

Lecture_16
School: Penn State
Course: THERMODYNAMICS
Lecture 16  Internal energy For a system of a single component, U must be a function of T, P (or T, v or P, v), since there are only two independent state functions Expand the differential of this state function, then, in T and v: dU = (U/T )V dT +

Lecture_6
School: Penn State
Course: THERMODYNAMICS
Lecture 6  Changing the State Recall "equation of state": for pure systems, only two independent properties, so a third may be given in terms of the other two. For example, v = f (T, p). How does v change when we change T and p? Recall that vt is a

Lecture_5
School: Penn State
Course: THERMODYNAMICS
Lecture 5  States on P T and P V diagrams It's easier to visualize slices of the P V T surface. P T , P V , and V T diagrams. Each have their uses. P V diagram A projection of the full P V T diagram onto the P V plane. ("Look down the T axis with o

Lecture_4
School: Penn State
Course: THERMODYNAMICS
Part 2  States of Pure Substances in Terms of Measurables Lecture 4  PvT surfaces for pure components Gibbs phase rule: We seek the number N of independent properties needed to specify a state with P coexisting phases, with C total components prese

Che_220_lecture_2_after
School: Penn State
Course: THERMODYNAMICS
CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 2: The NittyGritty Concepts in thermodynamics Interactions of boundaries Properties and processes Choosing the system is an important step in solving problems in thermodynamics. S

Che_220_lecture_1_after
School: Penn State
Course: THERMODYNAMICS
CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 1: What is thermodynamics? Thermodynamics and its role in chemical engineering Difference between 220 and 320 Work as energy chemical engineering has three main components. Che 33

Lecture__8_after
School: Penn State
Course: THERMODYNAMICS
CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 8: Virial Equations Taylor series expansion Expansion for Z in density Virial coefficients Expansion for Z in pressure 1 The Taylor series expansion is used to estimate the value

Lecture_18
School: Penn State
Course: THERMODYNAMICS
Part VI  Heat effects from property changes Lecture 18  Enthalpy Example: cooling at constant pressure. Suppose we cool isobutylene at 80bar and 752.2K to 500K. a) Compute the work done. W = P V = P (v2  v1 ). To compute initial and final v, nee

Lecture_19
School: Penn State
Course: THERMODYNAMICS
Lecture 19  Isenthalpic expansion JouleThomson process: Model for expansion of a gas through a small hole. Consider a insulated cylinder, with two opposing pistons, held with pressures P1 and P2 < P1 , and with a third, porous piston between the ot

Hw11
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 11 Due: Will not be collected Reading: Chapter 5, sections 5.85.11 Problems 5.2, 5.8, 5.12, 5.15 1

Hw10
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 10 Due: April 3 Reading: review of chapter 4; 5.15.7. Problems 4.13, 4.14, 5.3, 5.4. 1

Exam3
School: Penn State
Course: THERMODYNAMICS
Problem 1 The data below are for saturated liquid X : 190 C P D 1:9 bar T sat D SL D 83:54 J=mol K SV D 147:97 J=mol K L CP V CP D 32:9 J=mol K D 58 J=mol K Use these data to answer the following questions: a) Calculate the entropy of vaporization. b) Cal

Exam2
School: Penn State
Course: THERMODYNAMICS
CH E 220 Exam #2 March 3, 2014 6:307:45 PM Name: Student ID: Open books No personal notes or homework solutions Answers without explanation do not count Organize your work neatly State your assumptions clearly Good luck! Please sign your name under the pl

Hw1
School: Penn State
Course: THERMODYNAMICS
ChE 220  Spring 2014 Homework Set 1 Due: January 23 Reading: Chapter 1 Problems 1.1, 1.2, 1.4 and 1.6 in the book. Problem 1.4 part (a) asks for a plot of the intermolecular potential. This plot should be done on the computer, it should be properly annot

Exam1
School: Penn State
Course: THERMODYNAMICS
CH E 220 Exam #1 February 3 6:307:45 PM Name: Student ID: Open books No personal notes or homework solutions Answers without explanation do not count Organize your work neatly State your assumptions clearly Good luck! Please sign your name under the pledg

Lectures_1213
School: Penn State
Course: THERMODYNAMICS
Part IV  Work Lecture 12  PV work Recall the notion of a thermodynamic process: change of state via some external intervention. Common processes include: Adding or removing heat Expanding or compressing the system Often, the process involves requ

Lecture_29
School: Penn State
Course: THERMODYNAMICS
Lecture 29  Limits to efficiency. Now one might expect that there should be some limitations on the efficiency of a Carnot engine, beyond the constraints of conservation of energy. For example, you might or might not be persuaded that the following

Lecture_28
School: Penn State
Course: THERMODYNAMICS
Part VIII  2nd law Lecture 28  Carnot cycles The main results of classical thermodynamics were developed in the beginning of the 19th century, during the second industrial revolution. The "second" industrial revolution refers to the development of

Lecture_21
School: Penn State
Course: THERMODYNAMICS
Lecture 21  Multistep processes w/o phase change. Haile analyzes an example process, that of compressing and heating CO2 from 25C and 1bar, to 500C and 10bar. He considers four different processes to carry out the required state change. (sketch) You

Lecture_20
School: Penn State
Course: THERMODYNAMICS
Lecture 20  Sensible and latent heat Addition and removal of heat is a very important process in chemical engineering. For example: Chemical reaction rates typically depend exponentially on temperature, so yield depends on heating/cooling; Reactio

Lecture_23_after
School: Penn State
Course: THERMODYNAMICS
CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 23: Systems in Steady Flow The first law for open systems Steady state energy balance Steady state vs. equilibrium examples 1 Chemical plants have mass flow across boundaries. What