# Chemostat with Recycle: Biomass Mass Balance Let's break the problem into two parts as requested: --- ## a) **Mass Balance for Biomass around the Chemostat** ### **Definitions and Nomenclature** - \( F \): Fresh feed flow rate (L/h) - \( X_ \): Biomass concentration in feed (assume sterile, so \( X_ = \)) - \( X \): Biomass concentration in reactor (g/L) - \( S \): Substrate concentration in reactor (not directly needed here) - \( F_s \): Product stream flow rate (L/h) - \( x \): Biomass concentration in product stream - \( R \): Recycle flow rate (L/h) - \( x_R \): Biomass concentration in recycle stream - **Recycle ratio:** \( r = \frac{R}{F} \) - **Cell concentration ratio (separator):** \( \alpha = \frac{x_R}{x} \) ### **Step 1: Mass Balance for Biomass** At steady state, **accumulation = **. \[ \text{Input} + \text{Generation} = \text{Output} \] #### **Inputs:** - Feed: \( F X_ \) (since \( X_ = \), this term = ) - Recycle: \( R x_R \) #### **Generation:** - Biomass generated by growth: \( \mu X V \) - \( \mu \): specific growth rate (\( \mathrm{h}^{-1} \)) - \( V \): reactor volume #### **Outputs:** - Product stream: \( F_s x \) #### **Total Output from Reactor:** - Product + Recycle: \( F_s + R = F + R \) (since total outflow = total inflow for steady state) #### **Mass Balance Equation:** \[ \underbrace{F X_}_{\text{Feed}} + \underbrace{R x_R}_{\text{Recycle}} + \underbrace{\mu X V}_{\text{Generation}} = \underbrace{F_s x}_{\text{Product}} + \underbrace{R x_R}_{\text{Recycle output (but this returns to input)}} \] **But since the recycle is not leaving the system, only the product stream leaves, so the output is only \( F_s x \):** \[ F X_ + R x_R + \mu X V = F_s x + R x_R \] **\( R x_R \) cancels on both sides:** \[ F X_ + \mu X V = F_s x \] But with **sterile feed** (\( X_ = \)): \[ \boxed{ \mu X V = F_s x } \] --- ## b) **Show \(\mu\) is proportional to \(D\) (dilution rate) with a recycle** ### **Step 1: Relationships Between Streams** - **Recycle ratio:** \( r = \frac{R}{F} \) - **Cell concentration ratio:** \( \alpha = \frac{x_R}{x} \) - **Total outflow from reactor:** \( F + R = F_s + R \implies F_s = F \) (since recycle returns to reactor) - **Dilution rate:** \( D = \frac{F}{V} \) ### **Step 2: Express Everything in Terms of F, r, α, x, and X** **Separator splits product stream (\(F_s x\)) into:** - Recycle: \( R x_R \) - Net product: \( (F_s - R)x \) (if recycle is removed before measuring \(F_s\)) But usually, separator takes total output, splits off R at concentration x_R, remainder leaves at x. From separator: - \( F_s x = R x_R + (F_s - R)x \) - \( x_R = \alpha x \) - So: \[ F_s x = R x_R + (F_s - R)x = R \alpha x + (F_s - R)x = R \alpha x + F_s x - R x \] \[ F_s x - F_s x = R \alpha x - R x = R (\alpha - 1)x \] So this is just for checking; our main equation is from above. ### **Step 3: Main Mass Balance** Previously we got: \[ \mu X V = F_s x \] But we want everything in terms of reactor quantity \( X \) and dilution rate \( D \). #### **Relate \( x \) to \( X \):** In a well-mixed reactor, output cell concentration (\( x \)) is the same as reactor (\( X \)), so \( x = X \). #### **Relate \( F_s \) to \( F \) and \( r \):** Total outflow from reactor is \( F + R \), but only \( F_s \) exits the process (non-recycled). By definition, \( R = r F \). From separator: - \( F_s = F + R \) - So \( F_s = F + r F = F (1 + r) \) #### **Plug into Mass Balance:** \[ \mu X V = F_s x = F (1 + r) X \] \[ \mu X V = F (1 + r) X \] Divide both sides by \( X V \): \[ \mu = \frac{F (1 + r)}{V} \] Recall: \( D = \frac{F}{V} \): \[ \boxed{ \mu = D (1 + r) } \] --- ## **Final Answers** ### **a) Biomass Mass Balance:** \[ \boxed{ \mu X V = F_s x } \] ### **b) Proportionality of Growth Rate to Dilution Rate:** \[ \boxed{ \mu = D (1 + r) } \] Where: - \( \mu \): specific growth rate - \( D = \frac{F}{V} \): dilution rate - \( r = \frac{R}{F} \): recycle ratio --- **This shows that with recycle, the specific growth rate is proportional to the dilution rate, scaled by the recycle ratio.**