soln03 - 10.34 Fall 2006 Homework #3 - Solutions Problem 1:...

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10.34 – Fall 2006 Homework #3 - Solutions Problem 1: Peak Temperature In this problem, there are two situations that we are concerned with: 100% conversion and equilibrium conversion. The 100% case is relatively easy to solve, since one only needs to solve the enthalpy with all of the CO being converted to products (it is easy because the outlet flow rates are known a priori. The equilibrium case is slightly more complicated because one must also solve the equilibrium condition: −∆ G P n G [ H ][ C O ] K A = exp RT Rxn = K C R T 0 exp R Rxn = [ C 2 ] [ H 2 ] 2 Since are dealing with an equimolar reaction, the dN term goes away. We can also treat the concentrations as molar flow rates in this case, since the volume terms in the concentrations would also cancel. Note that dG is temperature dependent. As was done previously, the molar flow rates (concentrations) of the species can be written in terms of a single extent of reaction variable, so we are only left with T and extent as variables. Part A: 500 600 700 800 900 1000 1100 1200 1300 1400 1500 -50 -40 -30 -20 -10 0 10 20 Thermochemistry for the Water-Gas Shift Reaction Temperature [K] Thermochemistry [kJ/mole] dH(Rxn) T*dS(Rxn) dG(Rxn) Parts B and C: Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Part D: Physically Realizable Reactor Conditions 1600 1400 1200 1000 800 600 1500 1000 80 60 40 20 T In [K] 500 0 CO Conversion [%] T [K] Out 100 1500 1400 1300 1200 1100 1000 900 800 700 600 Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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Problem 2: Finite Differences A. The problem is almost completely formulated in the problem statement. The value of heat supplied by the laser to the water is given by H = 0.1 I 0 (10% of laser is absorbed uniformly across the width w). w Now we can write the system of equations in the matrix form AT = b , where A is the coefficient matrix and T is the vector of temperatures. The rule for making the matrix A and vector b is the following A(1,1) = 1; b(1) = Ta; A(i-1,i) = 1; A(i,i) = -2; A(i+1,i) = 1; b(i) = 0.1 × I 0 ( y ) 2 kw A(n+1,n+1) =1; b(n+1) = Ta; Please look at the commented code Problem3a for the actual code. Once we make the A and b , we can solve for T using the back-slash command T = A b \ . The sample output is given below >> u=problem2a; The maximum temperature is T =362.5 K The y-value at which temperature is maximum =0.005 cm >> The graph of temperature v/s width is Cite as: William Green, Jr., course materials for 10.34 Numerical Methods Applied to Chemical Engineering, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
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300 310 320 330 340 350 360 370 Temperature (K) Temperature as a function of width for uniform laser absorption 0 0.2 0.4 0.6 0.8 1 width (cm) B. This problem is no longer linear. The temperature and intensity of light are related to each other and we have to know one before we can solve for the other. Let us
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This note was uploaded on 11/27/2011 for the course CHEMICAL E 10.302 taught by Professor Clarkcolton during the Fall '04 term at MIT.

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soln03 - 10.34 Fall 2006 Homework #3 - Solutions Problem 1:...

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