Let's solve the problem step by step. --- ## **Given Data** - Initial enzyme concentration, \( C_ = 1 \) g/L - Overall flow rate, \( F = 100 \) L/h - Centrifuge separation factor, \( C_{L1} = 5 \) - Centrifuge flow ratio, \( \frac{L}{F} = .6 \) (where \( L \) is the supernatant flow to membrane) - Membrane filter rejects 95% of enzymes (rejection, \( R = .95 \)) --- ## **(a) Enzyme concentration \( C_L \) after centrifuge** ### **Step 1: Centrifuge Mass Balance** - Feed to centrifuge: - Flow: \( F = 100 \) L/h - Concentration: \( C_ = 1 \) g/L - Supernatant (goes to membrane): - Flow: \( L = .6F = .6 \times 100 = 600 \) L/h - Concentration: \( C_L \) (to be found) #### **Separation Factor Definition** \[ C_{L1} = \frac{C_}{C_L} \implies C_L = \frac{C_}{C_{L1}} = \frac{1}{5} = .2 \text{ g/L} \] --- ## **(b) Permeate flow rate \( P \) and permeate enzyme concentration \( C_2 \)** ### **Step 2: Membrane Flow and Concentration** - Feed to membrane: - Flow: \( L = 600 \) L/h - Concentration: \( C_L = .2 \) g/L - Membrane rejection: \( R = .95 \) #### **Permeate Flow (\( P \)):** Assume no significant volume change (water passes freely), so permeate flow equals feed to membrane: \[ P = L = 600 \text{ L/h} \] #### **Permeate Enzyme Concentration (\( C_2 \)):** \[ C_2 = (1 - R)C_L = (1 - .95) \times .2 = .05 \times .2 = .01 \text{ g/L} \] --- ## **(c) Final flow rate and enzyme concentration after membrane** ### **Step 3: Retentate Concentration** - Retentate flow (\( Q \)): Since 95% of enzyme is retained, almost all water passes to permeate, but let's calculate enzyme mass balance: - Enzyme in to membrane: \[ 600 \, \text{L/h} \times .2 \, \text{g/L} = 120 \, \text{g/h} \] - Enzyme in permeate: \[ 600 \, \text{L/h} \times .01 \, \text{g/L} = 6 \, \text{g/h} \] - Enzyme in retentate: \[ 120 - 6 = 114 \, \text{g/h} \] - If **retentate flow** = \( Q \), \( Q = L - P \) But if nearly all water passes, \( Q \) is negligible (retentate is very concentrated). If not specified, assume all water passes, so retentate flow is very small. --- ## **(d) Overall percentage of enzyme separated from solution** ### **Step 4: Percentage Separated** - **Total enzyme in feed:** \( 100 \times 1 = 100 \) g/h - **Total enzyme in permeate:** \( 600 \times .01 = 6 \) g/h - **Total enzyme removed:** \( 100 - 6 = 994 \) g/h \[ \text{Percentage separated} = \frac{994}{100} \times 100\% = 99.4\% \] --- ## **Summary of Answers** ### **a) Enzyme concentration after centrifuge (\( C_L \)):** \[ \boxed{.2 \text{ g/L}} \] ### **b) Permeate flow rate (\( P \)) and enzyme concentration (\( C_2 \)):** \[ \boxed{P = 600 \text{ L/h}}, \quad \boxed{C_2 = .01 \text{ g/L}} \] ### **c) Final flow rate and enzyme concentration after membrane:** - **Retentate enzyme mass:** \( 114 \) g/h - **Retentate flow:** Very small (not specified, as most water passes) - **Concentration:** Very high (since almost all enzyme is retained in little volume) ### **d) Overall percentage of enzyme separated:** \[ \boxed{99.4\%} \] --- **Let me know if you need step-by-step explanation for the retentate concentration or further details!**

# Step-by-Step Solution ## Given Data - Initial enzyme concentration: \( C_ = 1 \) g/L - Overall flow rate: \( F = 100 \) L/h - Centrifuge separation factor: \( C_{L1} = 5 \) - Centrifuge flow ratio: \( \frac{L}{F} = 0.6 \) (where \( L \) is the supernatant flow to membrane) - Membrane filter rejection: \( R = 0.95 \) ## (a) Enzyme Concentration \( C_L \) After Centrifuge ### Step 1: Centrifuge Mass Balance The feed to the centrifuge has a flow of \( F = 100 \) L/h and a concentration of \( C_ = 1 \) g/L. The supernatant, which goes to the membrane, has a flow of \( L = 0.6F = 0.6 \times 100 = 60 \) L/h and a concentration \( C_L \) to be determined. Using the separation factor definition: \[ C_{L1} = \frac{C_}{C_L} \implies C_L = \frac{C_}{C_{L1}} = \frac{1}{5} = 0.2 \text{ g/L} \] ## (b) Permeate Flow Rate \( P \) and Permeate Enzyme Concentration \( C_2 \) ### Step 2: Membrane Flow and Concentration The feed to the membrane has a flow of \( L = 60 \) L/h and a concentration \( C_L = 0.2 \) g/L. The membrane rejection is \( R = 0.95 \). For the permeate flow rate (\( P \)): \[ P = L = 60 \text{ L/h} \] For the permeate enzyme concentration (\( C_2 \)): \[ C_2 = (1 - R)C_L = (1 - 0.95) \times 0.2 = 0.05 \times 0.2 = 0.01 \text{ g/L} \] ## (c) Final Flow Rate and Enzyme Concentration After Membrane ### Step 3: Retentate Concentration The enzyme mass balance for the feed to the membrane is: \[ 60 \, \text{L/h} \times 0.2 \, \text{g/L} = 12 \, \text{g/h} \] The enzyme in the permeate is: \[ 60 \, \text{L/h} \times 0.01 \, \text{g/L} = 0.6 \, \text{g/h} \] The enzyme retained in the retentate is: \[ 12 - 0.6 = 11.4 \, \text{g/h} \] If the retentate flow (\( Q \)) is assumed negligible because most water passes to the permeate, it can be considered very concentrated. ## (d) Overall Percentage of Enzyme Separated from Solution ### Step 4: Percentage Separated The total enzyme in the feed is: \[ 100 \times 1 = 100 \, \text{g/h} \] The total enzyme in the permeate: \[ 60 \times 0.01 = 0.6 \, \text{g/h} \] The total enzyme removed: \[ 100 - 0.6 = 99.4 \, \text{g/h} \] The overall percentage separated is: \[ \text{Percentage separated} = \frac{99.4}{100} \times 100\% = 99.4\% \] ## Summary of Answers **a)** Enzyme concentration after centrifuge \( C_L \): \[ \boxed{0.2 \text{ g/L}} \] **b)** Permeate flow rate \( P \) and enzyme concentration \( C_2 \): \[ \boxed{P = 60 \text{ L/h}}, \quad \boxed{C_2 = 0.01 \text{ g/L}} \] **c)** Final flow rate and enzyme concentration after membrane: - Retentate enzyme mass: \( 11.4 \) g/h - Retentate flow: Very small (negligible) **d)** Overall percentage of enzyme separated: \[ \boxed{99.4\%} \] Let me know if you need any further clarification or details!