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April 18th, 2016, 10:08 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(CsatC)OUR I need to solve this equation for steady state and nonsteady state situations, where the answers should be: C(t)=Cst (1e^kla∙t) and C(t)=C(0)+(CsatOUR/(kLa)C(0))(1e^(kLa∙t)) I get how to solve the equation during steady state situations where: dC/dt=0 and OUR= kLa(CsatCst ) applies. Substitution, variable separation and integration (from C=0 to C, from t=0 and t) gives: ∫dC/(CstC) )=kLa∫dt〗 [ln(CstC)]=kla[t] ((CstC))/Cst =e^(kla∙t) C(t)=Cst (1e^(kla∙t)) However I do not get how to equate the answer for the nonsteady state solution. I get stuck since I cannot separate the variables from each other : ∫dC/(CsatC) =(kLa(OUR/(C_satC))∫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, 07:57 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(CsatC)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 08:00 AM. 

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