An ordinary differential equation (ODE) links a quantity to its rate of change over time. Every mass balance on a bioreactor is an ODE. Solving the basic growth ODE Start with dX/dt = μX. Separate variables: dX/X = μ dt. Integrate: ln X = μt + C. Using X = X₀ at t = 0 gives X(t) = X₀e^(μt) . The chemostat In a chemostat fresh medium enters at flow F into volume V; D = F/V is the dilution rate. The biomass balance is dX/dt = (μ − D)X. At steady state dX/dt = 0, so μ = D . If D exceeds μmax the cells wash out. The full batch model dX/dt = μ(S)X dS/dt = −(μ/Yx/s)X − msX dP/dt = (αμ + β)X μ(S) = μmax·S/(Ks + S) These equations are coupled: biomass uses substrate, which changes μ, which changes biomass. With Monod kinetics there is no neat formula, so we solve them numerically. Dissolved oxygen balance dCL/dt = KLa(C* − CL) − OUR KLa is th...
Every number in a fermenter – growth, substrate use, product formation – is a rate (a derivative) or an accumulated total (an integral). Before touching reactor design, you need a small calculus toolkit. Differentiation: the speed of change Differentiation measures an instantaneous rate. In bioprocessing the most important derivative is the specific growth rate , μ = (1/X)(dX/dt): how fast biomass grows per unit of biomass already present. Rule Form Bioprocess use Chain rule d/dx f(g(x)) = f′(g)·g′ ln(X/X₀) = μt gives (1/X)dX/dt = μ Quotient rule d/dx(u/v) = (v·u′ − u·v′)/v² Differentiating the Monod equation Log rule d/dx ln x = 1/x Slope of ln X vs time equals μ Exponential rule d/dx eˣ = eˣ Exponential growth dX/dt = μX Integration: adding it all up Integration accumula...