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Unformatted text preview: 10.34 – Fall 2006 Homework #6  Solutions Problem 1 – Chemostat ODE This problem is similar to the Quiz 1 case, with some changed parameters and slight structural changes to the equations. The ODEs that should have been solved were: dN cells = k N cells [ Nutrients ] − V 1 N cells dt 1 + c Nutrients 1 + d P flow V ( 1 [ ])( [ ]) rxtr V Nutr − Nutr − k N cells Nutr − × + c ( Cell Mult ) ⎤ dt V Rxtr d Nutr [ ] = 1 { flow ( [ ] [ In ]) ⎣ ⎡ 2 ([ ] 1 10 − 6 ) 2 ⎦ } d P k N d P [ ] = 1 ⎧ ⎪ 3 cells exp ( − [ ]) ⋅ ( Nutrients ] − 0.01 ) 2 − V [ ] P ⎫ ⎪ dt V Rxtr ⎨ ⎪ ⎩ ( 1 + c 1 [ Nutrients ]) [ flow ⎬ ⎪ ⎭ Note that the reactor volume must be introduced into the equations written in terms of concentration. The steady state of the seed reactor was found by integrating the ODE’s to a long time so that the steadystate condition will be reached. The fsolve function should also work to accomplish this. The 230 L reactor part was solved by using two subsequent ODE solves. The first was used to solve the problem from t = 0 Æ t = t nutrient , and the second (with the nutrient solution flowing into the reactor) was used to solve from t nutrient Æ t final . SteadyState Solution to the 150 L CSTR Number of Cells in Reactor = 7.403565e+007 Concentration of Cells in Reactor = 4.935710e+005 Concentration of Nutrients (M) = 0.051391 Concentration of Product (M) = 0.032908 SteadyState Solution to the 230 L CSTR Number of Cells in Reactor = 1.273906e+008 Concentration of Cells in Reactor = 5.538720e+005 Concentration of Nutrients (M) = 0.033451 Concentration of Product (M) = 0.018212 **Part A and C: see subsequent page for plots Part B: Time to within 1% of steady state:  you need to take care to calculate the correct time for the case(s) that overshoot the SS value, i.e. you do not want the time point to get within 1% of the SS Time to 99% of steady state (sec and hrs, respectively) t nut (hr) ode45 ode23s ode45 ode23s 12947 12935 3.60 3.59 2 9897 9881 2.75 2.74 8 31813 31810 8.84 8.84 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]. Part D: (Note that the tolerances used in the solution were an RelTol of 1e10 and an AbsTol of 1e12. The wall time and number of steps will change with the tolerance) Total wall time of Total number of Time per integrator t nut (hr) ode45 ode23s solver (sec) ode45 ode23s integrator steps ode45 ode23s step (msec) 0.3305 12.7283 1938 10576 0.171 1.204 2 0.2904 11.9472 1822 10120 0.159 1.181 8 0.4406 18.2162 2766 15560 0.159 1.171 You can see in this case (and with this tolerance set) that the explicit solver is much faster than the stiff solver because it takes fewer steps and takes less timeperstep. The explicit method takes about an order of magnitude longer per timestep....
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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.
 Fall '04
 ClarkColton

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