Let's solve each part step by step with clear explanations. --- ### 1. Multi-Stage Counter-Current Extraction **a. Analytical Difference Equation for the n-th Stage:** Let \( Y \) and \( X \) be the concentrations in the extract and raffinate, respectively. For the n-th stage: \[ Y_{n+1} = f(X_{n+1}) \] Material balance for solute in stage \( n \): \[ L(X_{n+1} - X_n) = V(Y_n - Y_{n+1}) \] Here, \( L \) and \( V \) are the flow rates of raffinate and extract phases, respectively. \[ \frac{X_{n+1} - X_n}{Y_n - Y_{n+1}} = \frac{V}{L} \] For a linear isotherm, \( Y = mX \): \[ Y_n = mX_n \] Substitute into the balance: \[ \frac{X_{n+1} - X_n}{mX_n - mX_{n+1}} = \frac{V}{L} \implies \frac{X_{n+1} - X_n}{m(X_n - X_{n+1})} = \frac{V}{L} \implies \frac{X_{n+1} - X_n}{X_n - X_{n+1}} = \frac{V}{L} \cdot m \] \[ X_{n+1} - X_n = - (X_{n+1} - X_n) m \frac{V}{L} \] \[ X_{n+1} - X_n \left[ 1 + m \frac{V}{L} \right] = \] But more generally, after rearranging and iterating, we get the Kremser-Brown-Souders equation for number of stages \( N \): \[ N = \frac{\ln \left( \frac{Y_1 - mX_2}{Y_{N+1} - mX_{N+1}} \right)}{\ln \left( \frac{1 + mV/L}{1} \right)} \] For extraction factor \( E = \frac{mV}{L} \): \[ N = \frac{\ln \left( \frac{Y_1 - mX_2}{Y_{N+1} - mX_{N+1}} \right)}{\ln \left( \frac{E}{E-1} \right)} \] --- ### 2. Recycle Reactor with Continuous Purge Let \( y \) = inert mole fraction in reactor, \( y_F \) = inert in fresh feed, \( X_{sp} \) = single-pass conversion. **a. Recycle Ratio Expression:** Total feed to reactor = fresh + recycle = \( F + R \) Assume steady state for inert: Input of inert = Output of inert (purge + product stream) \[ \text{Inert in: } F y_F \] \[ \text{Inert out: } P y + S y \] where \( P \) = product, \( S \) = purge. Usually, \( P = F (1 - X_{sp}) \), \( S = F X_{sp} \), so the purge fraction \( p = S/(F + R) \). Setting up the inert balance: \[ F y_F = P y + S y \] \[ F y_F = (P + S) y = (F + R) y \] \[ y = \frac{F y_F}{F + R} \] **Maximum recycle ratio** occurs as \( y \to y_{max} \): \[ R_{max} = \frac{F y_F}{y_{max}} - F \] --- ### 3. Sulfur Burning Plant—Process Flow & Balances **Basis:** 100 kmol pyrite feed (\( \text{FeS}_2 \)) **a. Atomic Species Balances (Sulfur and Oxygen):** For 100 kmol \( \text{FeS}_2 \): - S in: \( 2 \times 100 = 200 \) kmol S atoms - O in (from O2): unknown, to be determined Main reactions: \[ 4 \text{FeS}_2 + 11 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 + 8 \text{SO}_2 \] \[ 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \] Set up the S and O atom balances using the stoichiometry and the fraction converted at each step. **b. Linear Equations for Excess Air:** Let \( y_1, y_2, y_3 \) be the mole fractions of O2, SO2, SO3 in exit, and \( x \) be the % excess air. Given exit gas analysis on SO3-free basis, set up equations for O2 and SO2 output, relate to input air via stoichiometry, and solve for \( x \). **c. Dew Point Proof:** For the stack gas, the dew point is the temperature at which the partial pressure of water vapor equals its vapor pressure at that temperature. For mixtures, the presence of SO3 raises the dew point due to sulfuric acid formation. Use vapor-liquid equilibrium: \[ p_{H_2O} + p_{SO_3} = P_{dew} \] The dew point thus depends on the partial pressures of both H2O and SO3. --- ### **Final Answers (Summary)** - **Extraction Equation:** For a linear isotherm \( Y = mX \), the number of stages is given by Kremser's equation involving the extraction factor \( E = mV/L \). - **Recycle Reactor:** Maximum recycle ratio is \( R_{max} = \frac{F y_F}{y_{max}} - F \). - **Sulfur Plant:** Set up S and O balances on 100 kmol basis, formulate equations for excess air, and show dew point depends on partial pressures of H2O and SO3. Let me know if you need detailed numerical examples for any part.