Let's solve each question step by step, explaining the reasoning and giving the final answers. --- ### 1. **Crystal Production Rate Formula** Given: Crystallizer produces hydrated crystals \( R \cdot nH_2O \) from an undersaturated solution. Asked: Derive solute mass balance and formula for crystal production rate \( P \) in terms of feed rate \( F \), feed concentration \( x_F \), mother liquor solubility \( x_{ML} \), and ratio of molecular weights \( R_{MW} \). **Step 1:** **Solute Mass Balance** Let: - \( F \) = feed rate (kg/hr) - \( x_F \) = solute concentration in feed (kg solute/kg feed) - \( x_{ML} \) = solute concentration in mother liquor (kg solute/kg solution) - \( P \) = rate of solid (crystal) production (kg/hr) - \( R_{MW} \) = ratio of MW of hydrated crystal to solute (i.e., \( R_{MW} = \frac{\text{MW (hydrated crystal)}}{\text{MW (solute)}} \)) Solute in feed = \( F x_F \) Solute in mother liquor = \( (F - P \cdot R_{MW}) x_{ML} \) Solute in crystals = \( P \cdot R_{MW} \) (since each kg of crystal contains \( R_{MW} \) kg solute) **Step 2:** **Set up the Mass Balance** Input = Output \( F x_F = (F - P \cdot R_{MW}) x_{ML} + P \) **Step 3:** **Solve for \( P \)** Expand and collect terms: \( F x_F = F x_{ML} - P \cdot R_{MW} x_{ML} + P \) \( F x_F - F x_{ML} = -P R_{MW} x_{ML} + P \) \( F (x_F - x_{ML}) = P [1 - R_{MW} x_{ML}] \) \( P = \frac{F (x_F - x_{ML})}{1 - R_{MW} x_{ML}} \) **Final Answer:** The crystal production rate is \[ P = \frac{F(x_F - x_{ML})}{1 - R_{MW} x_{ML}} \] --- ### 2. **"Lever Rule" for Bypass Operation** Given: Mixed feed is split into bypass and processed stream. Asked: Prove that bypass fraction \( f_{bypass} \) equals the ratio of concentration differences. **Step 1:** Let: - \( F \) = total feed - \( P \) = processed stream - \( B \) = bypass stream - \( x_F \) = feed concentration - \( x_P \) = processed stream concentration - \( x_{out} \) = final mixed concentration Total output: \( F = P + B \) **Step 2:** **Mass Balance** Total solute: \( F x_F = P x_P + B x_F \) Outflow concentration: \[ x_{out} = \frac{P x_P + B x_F}{F} \] **Step 3:** **Bypass Fraction** Bypass fraction, \( f_{bypass} = \frac{B}{F} \) Express \( x_{out} \) in terms of \( f_{bypass} \): \[ x_{out} = (1 - f_{bypass}) x_P + f_{bypass} x_F \] Solving for \( f_{bypass} \): \[ x_{out} = x_P + f_{bypass} (x_F - x_P) \] \[ x_{out} - x_P = f_{bypass} (x_F - x_P) \] \[ f_{bypass} = \frac{x_{out} - x_P}{x_F - x_P} \] **Final Answer:** The bypass fraction is \[ f_{bypass} = \frac{x_{out} - x_P}{x_F - x_P} \] --- ### 3. **Semi-batch Reactor Control Volume and Equations** Given: Gas is sparged into a liquid phase in a semi-batch reactor. Asked: a) Write total mass balance b) Formulate component mass balance c) Solve ODE for time to reach a concentration d) Express the rate for instantaneous selectivity **a) Total Mass Balance** Let \( V \) = liquid volume (constant), \( F_{in} \) = gas flow rate, \( C_A \) = concentration of A in liquid Since only gas is added and the liquid volume is nearly constant, total mass balance is trivial if gas dissolution is negligible. **b) Differential Component Mass Balance** For species A: \[ \frac{d(C_A V)}{dt} = r_A V + F_{in} C_{A,in} \] If A is not in the gas (i.e., only reacts in liquid): \[ \frac{dC_A}{dt} = r_A \] **c) Solve for Time to Reach Concentration** If \( r_A = -k C_A \): \[ \frac{dC_A}{dt} = -k C_A \] Integrate from \( C_{A} \) to \( C_{A} \): \[ \int_{C_{A}}^{C_{A}} \frac{dC_A}{C_A} = -k \int_^t dt \] \[ \ln\left(\frac{C_A}{C_{A}}\right) = -k t \] \[ t = \frac{1}{k} \ln\left(\frac{C_{A}}{C_A}\right) \] **d) Instantaneous Selectivity** If two reactions compete: \[ S = \frac{r_{desired}}{r_{undesired}} \] --- ### 4. **Kirchhoff's Law and Enthalpy of Reaction** Given: \[ \Delta H_T = \Delta H_{ref} + \int_{T_{ref}}^T \Delta C_P (T) dT \] Derive and integrate the law for \( d(\Delta H)/dT = \Delta C_P \). The change in enthalpy with temperature at constant pressure is: \[ \frac{d(\Delta H)}{dT} = \Delta C_P \] Integrate both sides from \( T_{ref} \) to \( T \): \[ \Delta H_T - \Delta H_{ref} = \int_{T_{ref}}^{T} \Delta C_P(T) dT \] Thus, \[ \Delta H_T = \Delta H_{ref} + \int_{T_{ref}}^{T} \Delta C_P(T) dT \] --- ### 5. **Degree of Freedom Analysis (Splitter and Mixer)** Given: Relationship between number of independent variables \( N_v \), number of independent material balances \( N_c \), and number of specified process variables \( N_s \). The general formula is: \[ \text{Degrees of freedom} = N_v - N_c \] For a splitter, composition variables are not independent because the split streams have the same composition as the feed. --- ## **Summary of Final Answers** 1. **Crystal production rate:** \( P = \frac{F(x_F - x_{ML})}{1 - R_{MW} x_{ML}} \) 2. **Bypass fraction ("Lever Rule"):** \( f_{bypass} = \frac{x_{out} - x_P}{x_F - x_P} \) 3. **Semi-batch reactor:** - Total mass balance: \( \frac{dV}{dt} = \) if volume is constant - Component: \( \frac{dC_A}{dt} = r_A \) - Time to reach \( C_A \): \( t = \frac{1}{k} \ln\left(\frac{C_{A}}{C_A}\right) \) - Instantaneous selectivity: \( S = \frac{r_{desired}}{r_{undesired}} \) 4. **Kirchhoff's Law (enthalpy):** \( \Delta H_T = \Delta H_{ref} + \int_{T_{ref}}^T \Delta C_P (T) dT \) 5. **Degree of freedom:** \( \text{Degrees of freedom} = N_v - N_c \), for a splitter, composition variables are not independent.