Online study resources available anywhere, at any time
High-quality Study Documents, expert Tutors and Flashcards
Everything you need to learn more effectively and succeed
We are not endorsed by this school |
We are sorry, there are no listings for the current search parameters.
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam #1 February 13, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the O
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
School: Penn State
Kristine Schoppe Chemistry 213 18 March 2008 Separation of Fluorene/Fluorenone Mixture by Column Chromatography Introduction Chromatography is the process of separating compounds using their chemical differences along with the differences of two phas
School: Penn State
%Homework1: exercise1 Pc = 24.9; %pressure(bar) P = 0.9*Pc; Pr = P/Pc; Tc = 568.7; R = 8.314e-5; %gas constant (bar. m^3/(mol. k) b1 = 0.118119; b2 = 0.265728; b3 = 0.15479; b4 = 0.030323; c1 = 0.0236744; c2 = 0.0186984; c3 = 0; c4 = 0.042724; d1 = 1.5548
School: Penn State
Problem Set #7 CHE 360 Due: Thursday, 4/16/2015 Consider the 2D Poisson equation on 11 square domain. 2u 2u 50exp(2 x) for 0 x 1 and 0 y 1 x 2 y 2 The boundary conditions are illustrated in the figure below. y u( x,1) 1 ux (1, y) 0 ux (0, y) 0 u( x,0) 0
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
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
School: Penn State
Chemical Engineering Exam #2 Equation Sheet Product Separation and Recycle (p. 135): Overall Conversion: Single-Pass Conversion: Recycle Ratio: Combustion: Percent Excess Air (p. 145): Ch. 5 Density of a mixture of liquids (p. 189): (eqn. 5.1-1) (eqn. 5.1
School: Penn State
Chemical Engineering Exam #2 Equation Sheet Ch. 4 General Balance Equation (p. 85): For continuous steady-state: For batch process: initial input + generation = output + consumption If balanced quantity is total mass, set generation = 0 and consumption =
School: Penn State
Chemical Engineering Exam #1 Equation Sheet Interpolation: Specific Gravity: Density & Flow Rate: Mass Fraction: Mole Fraction: Molality: Normality: Average Molecular Weight: Hydrostatic Pressure: Pressure: General Manometer Equation: Differential Manomet
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
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
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
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
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
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
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam #1 February 13, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the O
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 02, 45 minutes 2014 Feb 07 Ver. C Last name First name Signature, indicating that you performed this exam with full academic integrity. 2. Remote Engine Start. (40 points total) Introduction. Many cars allow remote s , which has t
School: Penn State
Course: MASS BALANCES
1. CURL UP AND DYE [70] Background: The hair care industry relies upon careful metering of dye to offer a diverse selection of hair dye, including Ash brown, Auburn brown, and Oak brown. Should the dye concentration in a hair color kit drift outside a rig
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 03, 45 minutes 2014 Feb 28 Ver. A/B key Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1. MULTIPLE CHOICE [20] FOR VERSION A, 1) A gas cylinder contains (by mole) 8% eth
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2013 Final Quiz, 1 hour 50 minutes 2013 May 1 Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1 SCORE SHEET 1. _ / 40 pts. 2. _ / 15 pts. 3. _ / 40 pts. 4. _ / 5 pts. _ TOTAL / 100
School: Penn State
Course: MASS BALANCES
ChE 210 Fall 2012 Quiz 07, 50 minutes 2012nov17 Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1 The Fischer-Tropsch process is a useful series of catalyzed reactions for converting hydrogen gas
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
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
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
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
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
School: Penn State
Kristine Schoppe Chemistry 213 18 March 2008 Separation of Fluorene/Fluorenone Mixture by Column Chromatography Introduction Chromatography is the process of separating compounds using their chemical differences along with the differences of two phas
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam #1 February 13, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the O
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
School: Penn State
Kristine Schoppe Chemistry 213 18 March 2008 Separation of Fluorene/Fluorenone Mixture by Column Chromatography Introduction Chromatography is the process of separating compounds using their chemical differences along with the differences of two phas
School: Penn State
%Homework1: exercise1 Pc = 24.9; %pressure(bar) P = 0.9*Pc; Pr = P/Pc; Tc = 568.7; R = 8.314e-5; %gas constant (bar. m^3/(mol. k) b1 = 0.118119; b2 = 0.265728; b3 = 0.15479; b4 = 0.030323; c1 = 0.0236744; c2 = 0.0186984; c3 = 0; c4 = 0.042724; d1 = 1.5548
School: Penn State
Problem Set #7 CHE 360 Due: Thursday, 4/16/2015 Consider the 2D Poisson equation on 11 square domain. 2u 2u 50exp(2 x) for 0 x 1 and 0 y 1 x 2 y 2 The boundary conditions are illustrated in the figure below. y u( x,1) 1 ux (1, y) 0 ux (0, y) 0 u( x,0) 0
School: Penn State
dx = 2.5; Tinf = 25; Tb = 50; L = 0:dx:50; x0 = ones(length(L),1); x1 = fsolve(@(x) fin(x),x0); figure(1) plot(L,x1*(Tb-Tinf)+Tinf) x2 = fsolve(@(x) fin2(x),x0); figure(2) plot(L,x2*(Tb-Tinf)+Tinf)
School: Penn State
Homework #5: Part e) C(t+dt) = C(t) + dt * f f is dC/dt which is a nonlinear vector function. We linearize f: C(t+dt) = C(t) + dt *A*C(t) = (I+dt*A)*C(t) where I is a 2x2 identity matrix. C(t + dt) = [ 1 0 1 ] () After N steps we have: C(t+N*dt) = (I+dt
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 1 Assigned: Friday, January 30 Due: Thursday, February 5th on-line (at the beginning of the lecture) Note: Please use the le notation to upload the le on angel, label clearly
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 3 Assigned: Thursday, February 19 Due: Thursday, February 26th on-line (at the beginning of the lecture) Note: Please use the le notation to upload the le on angel, label cle
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 6 Assigned: Friday, March 27st Due: Thursday, April 2nd The temperature prole in a horizontal straight pin n of uniform cross section (Fig. 1) described by the heat Figure 1:
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 2 Assigned: Saturday, February 14 Due: Thursday, February 19 (at the beginning of the lecture) Note: Please include your name in the rst page, label clearly the problems and
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 3 Assigned: Thursday, February 19 Due: Thursday, February 26th on-line (at the beginning of the lecture) Note: Please use the le notation to upload the le on angel, label cle
School: Penn State
clc clear all close all tf = 6; dt = .3; Caf=10; ta=5; k=.12; df = @(C)[(Caf-C(1)/ta - k*C(1);-C(2)/ta + k*C(1)]; C0 = [20 0]'; % Euler explicit t = 0 : dt : tf; Cee = zeros(2,length(t); Cee(:,1) = C0; for i = 1 : length(t)-1 Cee(:,i+1) = Cee(:,i) + dt*d
School: Penn State
> k10 = 0.152; k12 = 0.207; k13 = 0.040; k21 = 0.092; k31 = 0.0048; > A=[-(k12+k13+k10) k21 k31;k12 -k21 0;k13 0 -k31] A= -0.3990 0.0920 0.0048 0.2070 -0.0920 0 0.0400 0 -0.0048 > eig(A) ans = -0.4523 -0.0398 -0.0037
School: Penn State
School: Penn State
HW #4: Part C: %We want to solve the ODE between t=0 and t=20. Time increments are 0.1 tf = 100; dt = .1; %We define a function equal to dC/dt A = [ -2.6 0.2 0.3 0.2; 0.8 -0.2 0 0; 0.7 0 -.3 0; 0.5 0 0 -.2]; I = 3; B = [0.1; 0 ; 0 ; 0]; dC = @(C) A*C+B*I;
School: Penn State
If we run the file HW1_Q1, we get the following output in the command window: > HW1_Q1 ans = The liquid/gas nitrogen volume per is 6.796621e-05 and 6.254211e-04 m^3/mol, respectively part b) In the command window, we call the function written in file HW1_
School: Penn State
ChE 360 - Mathematical Methods in Chemical Engineering, Spring 2015 Homework Set 5 Assigned: Saturday, March 19 Due: Thursday, March 26th on-line (at the beginning of the lecture) Note: Please use the le notation to upload the le on angel, label clearly t
School: Penn State
M = 50; N = 50; Lx = 1; Ly = 1; u0=ones(N,M); u=fsolve(@hotplate,u0); %Discretized Variables: x = linspace(0,Lx,N); y = linspace(0,Ly,M); surf(x,y,u) %This estimates the value of u(.5,.5) u_m=(u(25,25)+u(26,26)/2; % ans = % % 1.7901
School: Penn State
function f=fin2(x) % parameters k = 10; h = 25; epsilon = 0.98; sigma = 5.67e-8; fin_length = .50; dx = .025/fin_length; D = .03; Tinf = 25; Tb = 50; a4 = -epsilon*4*sigma*(Tb-Tinf)^3*fin_length^2/(k*D); a3 = -4*epsilon*4*sigma*(Tb-Tinf)^2*Tinf*fin_length
School: Penn State
function [P] = HW1_Q2Pa(V,T) Pc = 24.9; %pressure(bar) Tc = 568.7; R = 8.314e-5; %gas constant (bar. m^3/(mol. k) b1 = 0.118119; b2 = 0.265728; b3 = 0.15479; b4 = 0.030323; c1 = 0.0236744; c2 = 0.0186984; c3 = 0; c4 = 0.042724; d1 = 1.55488e-5; d2 = 6.236
School: Penn State
%Homework1: exercise1 R = 8.3144621; %P = 2.2e6; %P = 2e5; P = 1.2e6; T = 110; a = 0.1408; b = 3.913e-5; f = @(v) P*v^3 - (P*b + R*T)*v^2 + a*v - a*b; v11 = 1e-2; v12 = 1e-2+1e-4; f1 = f(v11); f2 = f(v12); while abs(f2) > 1e-8 df = (f1 - f2)/(v11-v12); 0e
School: Penn State
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 02, 45 minutes 2014 Feb 07 Ver. C Last name First name Signature, indicating that you performed this exam with full academic integrity. 2. Remote Engine Start. (40 points total) Introduction. Many cars allow remote s , which has t
School: Penn State
Course: MASS BALANCES
1. CURL UP AND DYE [70] Background: The hair care industry relies upon careful metering of dye to offer a diverse selection of hair dye, including Ash brown, Auburn brown, and Oak brown. Should the dye concentration in a hair color kit drift outside a rig
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 03, 45 minutes 2014 Feb 28 Ver. A/B key Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1. MULTIPLE CHOICE [20] FOR VERSION A, 1) A gas cylinder contains (by mole) 8% eth
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2013 Final Quiz, 1 hour 50 minutes 2013 May 1 Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1 SCORE SHEET 1. _ / 40 pts. 2. _ / 15 pts. 3. _ / 40 pts. 4. _ / 5 pts. _ TOTAL / 100
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
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
School: Penn State
Chemical Engineering Exam #2 Equation Sheet Product Separation and Recycle (p. 135): Overall Conversion: Single-Pass Conversion: Recycle Ratio: Combustion: Percent Excess Air (p. 145): Ch. 5 Density of a mixture of liquids (p. 189): (eqn. 5.1-1) (eqn. 5.1
School: Penn State
Chemical Engineering Exam #2 Equation Sheet Ch. 4 General Balance Equation (p. 85): For continuous steady-state: For batch process: initial input + generation = output + consumption If balanced quantity is total mass, set generation = 0 and consumption =
School: Penn State
Chemical Engineering Exam #1 Equation Sheet Interpolation: Specific Gravity: Density & Flow Rate: Mass Fraction: Mole Fraction: Molality: Normality: Average Molecular Weight: Hydrostatic Pressure: Pressure: General Manometer Equation: Differential Manomet
School: Penn State
Course: MASS BALANCES
<?xml version="1.0" encoding="UTF-8"?> <Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>64fcd3f80bbdb5d86abbd5166a90cb8b012b15c7.ppt</Key><RequestId>6 37B9894BA62134D</RequestId><HostId>Qiz4oXLPvjkFD+yTVlBN+01ueyOU+ujI
School: Penn State
Course: MASS BALANCES
Brushing-up Math 251 for ChE 210 1. Separable equations Many first-order differential equations can be reduced to the form g(y) y' = f(x) By algebraic manipulations. Since y' = dy / dx, we find it convenient to write g(y) dy = f(x) dx (2) (1) This
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
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
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
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
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
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
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
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
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
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 +
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
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
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
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
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
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
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
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam #1 February 13, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the O
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 02, 45 minutes 2014 Feb 07 Ver. C Last name First name Signature, indicating that you performed this exam with full academic integrity. 2. Remote Engine Start. (40 points total) Introduction. Many cars allow remote s , which has t
School: Penn State
Course: MASS BALANCES
1. CURL UP AND DYE [70] Background: The hair care industry relies upon careful metering of dye to offer a diverse selection of hair dye, including Ash brown, Auburn brown, and Oak brown. Should the dye concentration in a hair color kit drift outside a rig
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2014 Quiz 03, 45 minutes 2014 Feb 28 Ver. A/B key Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1. MULTIPLE CHOICE [20] FOR VERSION A, 1) A gas cylinder contains (by mole) 8% eth
School: Penn State
Course: MASS BALANCES
ChE 210 Spring 2013 Final Quiz, 1 hour 50 minutes 2013 May 1 Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1 SCORE SHEET 1. _ / 40 pts. 2. _ / 15 pts. 3. _ / 40 pts. 4. _ / 5 pts. _ TOTAL / 100
School: Penn State
Course: MASS BALANCES
ChE 210 Fall 2012 Quiz 07, 50 minutes 2012nov17 Last name _ First name _ Signature, indicating that you performed this exam with full academic integrity. _ 1 The Fischer-Tropsch process is a useful series of catalyzed reactions for converting hydrogen gas
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
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
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
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
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam 3 April 17, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the Offic
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442: Polymer Processing Technology Exam #2 March 20, 2013 R. H. Colby 100 points total Name: _ I acknowledge that cheating will result in a grade of zero for the examination, and may result in failing the course and/or referral to the Offi
School: Penn State
Course: MASS BALANCES
<?xml version="1.0" encoding="UTF-8"?> <Error><Code>NoSuchKey</Code><Message>The specified key does not exist.</Message><Key>c94e630bda18a6d805a576cdf813b26b7c81cf46.doc</Key><RequestId>1 1061F767494826E</RequestId><HostId>e9xhBjcWFKHzHAHyUi9uvEKAjqmS6Ol6
School: Penn State
Course: MASS BALANCES
Name: ChE 210 Spring 2008 Self-Test (will not be graded) The vessel is a vertical cylinder with a cross sectional area A. At the bottom of the vessel is a hole with an area Ao that has plug in it. The initial liquid level has been set to ho. When th
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
School: Penn State
Course: Polymer Processing Technology
MatSE 448 / ChE 442 Spring 2013 HW# 2: Due Wed. Jan. 23 1. Compare and contrast the expected solid liquid transitions of amorphous polysulfone and semicrystalline isotactic polybutene, describing the physical changes that occur in the polymers on heating
School: Penn State
Kristine Schoppe Chemistry 213 18 March 2008 Separation of Fluorene/Fluorenone Mixture by Column Chromatography Introduction Chromatography is the process of separating compounds using their chemical differences along with the differences of two phas