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

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

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

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

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

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

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

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

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 +