My Math Forum problems solving diff eq

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 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.

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