• 6 Pages lecture_24after
    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

  • 8 Pages sol14
    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 1-6, 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

  • 1 Page hw14
    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 1-6, up to and including section 6.9. Problems: 6.27, 6.29. 6.30, 6.31, 6.39 1

  • 1 Page hw13
    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

  • 4 Pages exam4
    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

  • 1 Page hw12
    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

  • 1 Page hw7
    Hw7

    School: Penn State

    Course: THERMODYNAMICS

    ChE 220 - Spring 2014 Homework Set 7 Due: Will not be collected Reading: 3.8-3.10; review of chapter 3. Problems 3.6, 3.17, 3.28, 3.31, 3.35, 3.38 1

  • 1 Page hw6
    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

  • 1 Page hw8
    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

  • 1 Page hw4
    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 3.1-3.3 Problems: 2.15, 2.18, 2.20, 2.25. 1

  • 1 Page hw5
    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

  • 1 Page fig_mollier
    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

  • 3 Pages LK_Tables
    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

  • 6 Pages sol2
    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

  • 5 Pages sol9
    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 steam-table units) T

  • 6 Pages sol13
    Sol13

    School: Penn State

    Course: THERMODYNAMICS

    ChE 220 - Spring 2014 Homework Set 13 Due: April 24 Reading: 6.6 1 Problem 1 ./problem-dbase/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

  • 5 Pages sol12
    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

  • 5 Pages sol11
    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

  • 10 Pages sol3
    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

  • 7 Pages sol4
    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 3.1-3.3 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

  • 6 Pages sol5
    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 constant-pressure 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

  • 5 Pages sol6
    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

  • 6 Pages sol10
    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 ideal-gas 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

  • 5 Pages sol8
    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

  • 7 Pages sol7
    Sol7

    School: Penn State

    Course: THERMODYNAMICS

    ChE 220 - Spring 2014 Homework Set 7 Due: Will not be collected Reading: 3.8-3.10; 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

  • 7 Pages sol1
    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

  • 1 Page hw9
    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

  • 1 Page hw3
    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 2.1-2.2. The last two prob

  • 1 Page hw2
    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

  • 2 Pages Lecture_17
    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 "one-step" processes. Important examples of these building-block processes are: workfree (no change in

  • 2 Pages Lecture_16
    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 +

  • 3 Pages Lecture_6
    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

  • 4 Pages Lecture_5
    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

  • 5 Pages Lecture_4
    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

  • 7 Pages che_220_lecture_2_after
    Che_220_lecture_2_after

    School: Penn State

    Course: THERMODYNAMICS

    CHE 220 Introduction to Chemical Engineering Thermodynamics Lecture 2: The Nitty-Gritty Concepts in thermodynamics Interactions of boundaries Properties and processes Choosing the system is an important step in solving problems in thermodynamics. S

  • 6 Pages che_220_lecture_1_after
    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

  • 8 Pages march_28.2008
    March_28.2008

    School: Penn State

    Course: THERMODYNAMICS

  • 7 Pages march_24.2008
    March_24.2008

    School: Penn State

    Course: THERMODYNAMICS

  • 6 Pages march_21__2008
    March_21__2008

    School: Penn State

    Course: THERMODYNAMICS

  • 7 Pages lecture__8_after
    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

  • 2 Pages Lecture_18
    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

  • 2 Pages Lecture_19
    Lecture_19

    School: Penn State

    Course: THERMODYNAMICS

    Lecture 19 - Isenthalpic expansion Joule-Thomson 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

  • 1 Page hw11
    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

  • 1 Page hw10
    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

  • 2 Pages exam3
    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

  • 3 Pages exam2
    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

  • 1 Page hw1
    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

  • 3 Pages exam1
    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

  • 5 Pages Lectures_12-13
    Lectures_12-13

    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

  • 4 Pages Lecture_29
    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

  • 4 Pages Lecture_28
    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

  • 3 Pages Lecture_21
    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

  • 3 Pages Lecture_20
    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

  • 10 Pages lecture_23_after
    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

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