Let's break down the solution step-by-step for both parts (**a** and **b**): --- ## **a) Mass Balance for Biomass Around the Chemostat** ### **Definitions** - \( F \): Fresh feed flow rate (with substrate, no cells) - \( F_s \): Product stream flow rate (cells + substrate) - \( R \): Recycle flow rate (cell-rich stream) - \( x_ \): Cell concentration in feed (usually ) - \( x \): Cell concentration in the reactor (and in the product stream) - \( x_R \): Cell concentration in the recycle stream - \( \alpha = \frac{R}{F} \): **Recycle ratio** - \( C = \frac{x_R}{x} \): **Cell concentration ratio in separator** ### **Total Outflow** From the diagram: Total outflow = \( F_s + R \) But, **in steady-state**: Total inflow = Total outflow \(\implies F = F_s + R \implies F_s = F - R\) ### **Biomass Mass Balance (Steady-State)** At steady-state, **accumulation = **: \[ \text{In} - \text{Out} + \text{Generation} = \] #### **Inflow of Biomass:** - From fresh feed: \( F x_ \) (often \( x_ = \)) - From recycle: \( R x_R \) #### **Outflow of Biomass:** - Out in product stream: \( F_s x \) #### **Generation:** - Net cell growth in reactor: \( \mu x V \), where \( V \) is reactor volume, \( \mu \) is specific growth rate #### **Total Balance:** \[ F x_ + R x_R + (\mu x V) - F_s x = \] Alternatively, rearrange: \[ F x_ + R x_R + (\mu x V) = F_s x \] --- ## **b) Show the Growth Rate Relationship** **Assumptions:** - Steady-state (\( \text{accumulation} = \)) - Fresh feed is sterile (\( x_ = \)) - No cell death (\( k_d = \)) ### **Step 1: Substitute Known Values** - \( F_s = F - R = F(1 - \alpha) \) - \( R = \alpha F \) - \( x_R = C x \) The mass balance for biomass (from above, and with \( x_ = \)): \[ + R x_R + (\mu x V) = F_s x \] \[ R x_R + \mu x V = F_s x \] ### **Step 2: Substitute Expressions** \[ (\alpha F)(C x) + \mu x V = [F (1 - \alpha)] x \] Divide both sides by \( x \): \[ \alpha F C + \mu V = F(1 - \alpha) \] ### **Step 3: Isolate for \( \mu \)** \[ \mu V = F(1 - \alpha) - \alpha F C \] \[ \mu V = F [1 - \alpha - \alpha C] \] ### **Step 4: Express in Terms of Dilution Rate \( D \)** The **dilution rate** is defined as \( D = \frac{F}{V} \): \[ \mu V = F [1 - \alpha (1 + C)] \] But that's not matching the target expression. Let's carefully expand: Go back: \[ \alpha F C + \mu V = F(1 - \alpha) \] \[ \mu V = F(1 - \alpha) - \alpha F C \] \[ \mu V = F [1 - \alpha - \alpha C] \] Wait: \( 1 - \alpha - \alpha C = 1 - \alpha (1 + C) \) But the target is \( \mu = D[1 + \alpha(1 - C)] \). Let's try a different algebraic arrangement: Expand \( F_s x = F x - R x \): From mass balance: \[ R x_R + \mu x V = (F - R) x \] \[ R x_R + \mu x V = F x - R x \] \[ R x_R + R x + \mu x V = F x \] \[ R(x_R + x) + \mu x V = F x \] But since \( x_R = C x \): \[ R(x_R + x) = R(C x + x) = R x (1 + C) \] So: \[ R x (1 + C) + \mu x V = F x \] Divide by \( x \): \[ R(1 + C) + \mu V = F \] \[ \mu V = F - R(1 + C) \] Substitute \( R = \alpha F \): \[ \mu V = F - \alpha F (1 + C) \] \[ \mu V = F [1 - \alpha(1 + C)] \] Now, divide both sides by \( V \): \[ \mu = \frac{F}{V} [1 - \alpha(1 + C)] \] \[ \mu = D [1 - \alpha(1 + C)] \] Wait, but the required expression is \( \mu = D [1 + \alpha(1 - C)] \). Let's try expanding \( 1 + \alpha(1 - C) \): \[ 1 + \alpha(1 - C) = 1 + \alpha - \alpha C = (1 + \alpha) - \alpha C \] Similarly, our result is: \[ 1 - \alpha(1 + C) = 1 - \alpha - \alpha C = (1 - \alpha) - \alpha C \] Let's check for sign errors. Let's go back to the original mass balance: **Original mass balance:** \[ F x_ + R x_R + \mu x V = F_s x \] With \( x_ = \), \( F_s = F - R \): \[ R x_R + \mu x V = (F - R) x \] \[ R x_R + \mu x V = F x - R x \] \[ R x_R + R x + \mu x V = F x \] \[ R(x_R + x) + \mu x V = F x \] \[ R(x_R + x) = F x - \mu x V \] \[ R(x_R + x) = x(F - \mu V) \] \[ R = \frac{x(F - \mu V)}{x_R + x} \] But we want \( \mu \) in terms of \( D \): From above: \[ R(x_R + x) + \mu x V = F x \] \[ \mu x V = F x - R(x_R + x) \] \[ \mu = \frac{F x - R(x_R + x)}{x V} \] \[ \mu = \frac{F}{V} - \frac{R}{V} \left( \frac{x_R + x}{x} \right) \] \[ \mu = D - \frac{R}{V} \left( 1 + \frac{x_R}{x} \right) \] But \( \frac{R}{V} = \alpha D \) and \( \frac{x_R}{x} = C \): \[ \mu = D - \alpha D (1 + C) \] \[ \mu = D [1 - \alpha(1 + C)] \] But the question wants \( \mu = D [1 + \alpha(1 - C)] \). Now, note that \( 1 - \alpha(1 + C) = 1 - \alpha - \alpha C \), and \( 1 + \alpha(1 - C) = 1 + \alpha - \alpha C \). So, the only difference is the sign of \( \alpha \). This suggests the recycle ratio is sometimes defined as \( \alpha = \frac{R}{F_s} \) instead of \( \frac{R}{F} \). Let's check if changing the definition matches the book formula. If \( \alpha = \frac{R}{F_s} \), then \( R = \alpha F_s \), and \( F = F_s + R = F_s(1 + \alpha) \implies F_s = \frac{F}{1 + \alpha} \). Now, try the mass balance with this definition: - \( R = \alpha F_s \) - \( x_R = C x \) - \( F_s = \frac{F}{1 + \alpha} \) Original mass balance (using \( F x_ = \)): \[ R x_R + \mu x V = F_s x \] \[ \alpha F_s (C x) + \mu x V = F_s x \] \[ \alpha F_s C x + \mu x V = F_s x \] \[ \mu x V = F_s x - \alpha F_s C x \] \[ \mu V = F_s (1 - \alpha C) \] But \( F_s = \frac{F}{1 + \alpha} \): \[ \mu V = \frac{F}{1 + \alpha} (1 - \alpha C) \] \[ \mu = \frac{F}{V} \frac{1 - \alpha C}{1 + \alpha} \] But that's not matching the book formula either. ### **Conclusion and Final Answer** #### **a) Mass balance for biomass:** \[ F x_ + R x_R + (\mu x V) = F_s x \] where \( x_ \) is the biomass in the feed, \( R \) is the recycle flow, \( x_R \) is the recycle cell concentration, \( \mu \) is the specific growth rate, \( x \) is the reactor cell concentration, and \( F_s \) is the product stream flow rate. #### **b) Growth rate relationship:** Given the definitions in the problem (\( \alpha = \frac{R}{F} \), \( C = \frac{x_R}{x} \)), and under steady-state, sterile feed (\( x_ = \)), and no cell death (\( k_d = \)), the growth rate is: \[ \boxed{ \mu = D [1 - \alpha(1 + C)] } \] where \( D = \frac{F}{V} \) is the dilution rate. > **Note:** If you are required to match the form \( \mu = D [1 + \alpha(1 - C)] \), double-check the definition of \( \alpha \) in your course notes. The above derivation follows the definitions as given in your problem. If the book uses \( \alpha = \frac{R}{F_s} \), the formula will be different. --- ### **Summary Table** | Term | Definition | |----------------|--------------------------------------| | \( F \) | Feed flow rate | | \( F_s \) | Product stream flow rate | | \( R \) | Recycle flow rate | | \( x \) | Cell concentration (reactor) | | \( x_ \) | Cell conc. in feed (usually ) | | \( x_R \) | Cell conc. in recycle | | \( \alpha \) | Recycle ratio (\( \frac{R}{F} \)) | | \( C \) | Cell conc. ratio (\( \frac{x_R}{x} \))| | \( D \) | Dilution rate (\( \frac{F}{V} \)) | --- Let me know if you need clarification on any step!