Differential Equations Ordinary and Partial Differential Equations Math Forum

 April 18th, 2016, 11:08 AM #1 Newbie   Joined: Apr 2016 From: malmö Posts: 1 Thanks: 0 problems solving diff eq Hi, I have a differential equation: dC/dt=kLa(Csat-C)-OUR I need to solve this equation for steady state and non-steady state situations, where the answers should be: C(t)=Cst (1-e^-kla∙t) and C(t)=C(0)+(Csat-OUR/(kLa)-C(0))(1-e^(-kLa∙t)) I get how to solve the equation during steady state situations where: dC/dt=0 and OUR= kLa(Csat-Cst ) applies. Substitution, variable separation and integration (from C=0 to C, from t=0 and t) gives: ∫dC/(Cst-C) )=kLa∫dt〗 -[ln(Cst-C)]=kla[t] ((Cst-C))/Cst =e^(-kla∙t) C(t)=Cst (1-e^(-kla∙t)) However I do not get how to equate the answer for the non-steady state solution. I get stuck since I cannot separate the variables from each other : ∫dC/(Csat-C) =(kLa-(OUR/(C_sat-C))∫dt How should this be done? Great full for help Best L --------------------------------------------------------------------------- kLa=volumetric mass transfer coefficiet Csat=saturation concentration for dissolved oxygen C=oxygen concentration in liquid phase OUR=oxygen uptake rate by microorganisms,can be considered a constant Cst= steady state oxygen concentration in liquid phase April 23rd, 2016, 08:57 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Yes, finding the "steady state" solution is just a matter of taking dC/dt= 0 and then solving the algebraic equation, kLA(Csat- C)- OUR= 0 for C: kLACsat- kLAC- OUR= 0 so kLAC= kLACsat- OUR and C= (kLACsat- OUR)/kLA= Csat- OUR/kLa. The initial equation, dC/dt=kLa(Csat-C)-OUR, which is the same as dC/dt+ kLaC= kLaCsat- OUR, is a linear first order equation. There is a simple method for finding an integrating factor and then a solution. Another method, that I prefer, is to first solve the associated homogenous equation, dC/dt= -kLaC. That separates as dC/C= -kLa dt which integrates to ln(C)= -kLa t+ P where P is the "constant of integration"and, solving for C, C= e^(-kLa t+ P)= P'e^(-kLa t) where P'= e^P. Since this is linear we just need to add one function that satisfies the entire equation. Since the right side, kLaCsat- OUR, is a constant, we try a constant function for C. If C= A, a constant, then dC/dt= 0 so the equation becomes 0+ kLaA= kLaCsat- OUR and A= (kLaCsat- OUR)/kLa= Csat- OUR/kLa. The general solution to the equation is C= P'e^(-kLa t)+ (kLaCsat- OUR)/kLa= P'e^(-kLa t)+ Csat- OUR/kLa. Last edited by Country Boy; April 23rd, 2016 at 09:00 AM. Tags diff, problems, solving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post daveollie Applied Math 0 December 5th, 2013 11:04 PM Niko Bellic Calculus 2 July 8th, 2013 11:01 AM simons545 Calculus 4 January 3rd, 2013 12:13 PM lu5t Applied Math 0 April 30th, 2012 10:59 AM meti20 Calculus 1 August 28th, 2010 12:34 AM

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