Let's solve each question step by step: --- ### **1. Completely Mixed Tank with Changing Volume** #### **Given:** - Initial volume, \( V_0 = 0.5\,m^3 \) - Initial salt, \( S_0 = 25\,kg \) - Feed concentration, \( C_{in} = 200\,kg/m^3 \) - Feed rate, \( F_{in} = 0.015\,m^3/min \) - Outlet rate, \( F_{out} = 0.010\,m^3/min \) - Final volume, \( V_{final} = 1.0\,m^3 \) #### **Let:** - \( S(t) \): salt in the tank at time \( t \), in kg - \( V(t) \): volume in the tank at time \( t \) - \( C(t) = \frac{S(t)}{V(t)} \): salt concentration in tank #### **Step 1: Write the Differential Equations** **Volume balance:** \[ \frac{dV}{dt} = F_{in} - F_{out} = 0.015 - 0.010 = 0.005\,m^3/\text{min} \] \[ V(t) = 0.5 + 0.005 t \] When \( V(t) = 1.0 \): \[ 1.0 = 0.5 + 0.005 t \implies t_{full} = 100\,\text{min} \] **Salt balance:** \[ \frac{dS}{dt} = \text{(salt in)} - \text{(salt out)} = (C_{in} F_{in}) - (C(t) F_{out}) \] \[ \frac{dS}{dt} = (200 \times 0.015) - \left(\frac{S(t)}{V(t)} \times 0.010\right) \] \[ \frac{dS}{dt} + \frac{0.010}{V(t)} S(t) = 3.0 \] #### **Step 2: Integrating Factor Method** This is a linear ODE: \[ \frac{dS}{dt} + p(t) S = q(t) \] where \( p(t) = \frac{0.010}{V(t)} \), \( q(t) = 3.0 \). The integrating factor is: \[ IF = \exp\left(\int \frac{0.010}{V(t)} dt\right) \] Since \( V(t) = 0.5 + 0.005 t \): \[ \int \frac{0.010}{0.5 + 0.005 t} dt = 2 \int \frac{1}{0.5 + 0.005 t} \cdot 0.005 \, dt = 2 \ln(0.5 + 0.005 t) \] So, \[ IF = (0.5 + 0.005 t)^2 \] #### **Step 3: Multiply Both Sides by Integrating Factor** \[ (0.5 + 0.005 t)^2 \frac{dS}{dt} + 0.010 (0.5 + 0.005 t) S = 3.0 (0.5 + 0.005 t)^2 \] \[ \frac{d}{dt} \left[ S (0.5 + 0.005 t)^2 \right] = 3.0 (0.5 + 0.005 t)^2 \] Integrate both sides from \( t = 0 \) to \( t \): \[ S (0.5 + 0.005 t)^2 - S_0 (0.5)^2 = 3.0 \int_0^t (0.5 + 0.005 t')^2 dt' \] #### **Step 4: Solve the Integral** Let \( u = 0.5 + 0.005 t' \), \( du = 0.005 dt' \), \( dt' = du/0.005 \): When \( t' = 0 \), \( u = 0.5 \) When \( t' = t \), \( u = 0.5 + 0.005 t \) \[ \int_0^t (0.5 + 0.005 t')^2 dt' = \int_{0.5}^{0.5+0.005 t} u^2 \frac{du}{0.005} = \frac{1}{0.005} \int_{0.5}^{0.5+0.005 t} u^2 du = 200 \left[ \frac{u^3}{3} \right]_{0.5}^{0.5+0.005t} = \frac{200}{3} \left[ (0.5+0.005t)^3 - (0.5)^3 \right] \] #### **Step 5: Combine Everything** \[ S (0.5 + 0.005 t)^2 = 6.25 + 200 \left[ (0.5+0.005t)^3 - (0.5)^3 \right] \] \[ S (0.5 + 0.005 t)^2 = 6.25 + \frac{200}{3} \left[ (0.5+0.005t)^3 - (0.5)^3 \right] \] \[ S(t) = \frac{6.25 + \frac{200}{3} \left[ (0.5+0.005t)^3 - (0.5)^3 \right]}{(0.5+0.005t)^2} \] #### **Step 6: Evaluate at \( V = 1.0 \), \( t = 100\,min \)** \[ V(t_{full}) = 0.5 + 0.005 \times 100 = 1.0\,m^3 \] \[ u = 1.0 \] \[ S(100) = \frac{6.25 + \frac{200}{3} (1^3 - 0.5^3)}{1.0^2} \] \[ 0.5^3 = 0.125 \] \[ 1^3 - 0.125 = 0.875 \] \[ S(100) = 6.25 + \frac{200}{3} \times 0.875 = 6.25 + \frac{200 \times 0.875}{3} \] \[ 200 \times 0.875 = 175 \] \[ \frac{175}{3} \approx 58.33 \] \[ S(100) = 6.25 + 58.33 = 64.58\, kg \] #### **Step 7: Final Concentration** \[ C_{final} = \frac{S(100)}{V(100)} = \frac{64.58}{1.0} = \boxed{64.6\, \text{kg/m}^3} \] --- ### **2. Tank Heating with Heat Loss** #### **Given:** - Mass, \( m = 45\,kg = 45,000\,g \) - Heater, \( Q_{in} = 25,000\,cal/min \) - Specific heat, \( C_p = 0.5\,cal/g^\circ C \) - Heat loss, \( Q_{loss} = 800(T - 25) \,cal/min \) - \( T_{init} = 15^\circ C \), \( T_{amb} = 25^\circ C \) #### **Let:** - \( T(t) \): temperature at time \( t \) (°C) #### **Step 1: Energy Balance** \[ \frac{d}{dt}(m C_p T) = Q_{in} - Q_{loss} \] \[ m C_p \frac{dT}{dt} = 25,000 - 800(T - 25) \] \[ 45,000 \times 0.5 \frac{dT}{dt} = 25,000 - 800(T - 25) \] \[ 22,500 \frac{dT}{dt} = 25,000 - 800T + 20,000 \] \[ 22,500 \frac{dT}{dt} = 45,000 - 800T \] \[ \frac{dT}{dt} + \frac{800}{22,500} T = 2 \] #### **Step 2: Standard Linear ODE** \[ \frac{dT}{dt} + aT = b \] where \( a = \frac{800}{22,500} = 0.03556 \), \( b = 2 \) #### **Step 3: Integrating Factor** \[ IF = e^{a t} \] \[ \frac{dT}{dt} + aT = b \implies \frac{d}{dt}(T e^{a t}) = b e^{a t} \] Integrate: \[ T(t) e^{a t} - T_0 = \int_{0}^{t} b e^{a t'} dt' \] \[ T(t) e^{a t} - T_0 = \frac{b}{a} (e^{a t} - 1) \] \[ T(t) = T_0 e^{-a t} + \frac{b}{a}(1 - e^{-a t}) \] Plug in values: - \( T_0 = 15 \) - \( b = 2 \) - \( a = 0.03556 \) \[ \frac{b}{a} = \frac{2}{0.03556} \approx 56.27 \] \[ T(t) = 15 e^{-0.03556 t} + 56.27 (1 - e^{-0.03556 t}) \] \[ T(t) = 56.27 - 41.27 e^{-0.03556 t} \] #### **Final Answer:** \[ \boxed{T(t) = 56.3 - 41.3 \, e^{-0.0356 t} \quad \text{(in } ^\circ C,\, t \text{ in min)}} \] --- ### **3. CSTR Reaction Conversion and Flow Rate** #### **Given:** - \( V = 0.5\,m^3 \) - \( k = 0.5\, m^3/kmol \cdot hr \) - \( C_{A0} = 4\,kmol/m^3 \) - Desired conversion: \( X = 0.40 \) #### **Step 1: CSTR Mole Balance** For a first-order reaction: \[ \text{For } A \to \text{products} \] \[ F_{in} C_{A0} - F_{out} C_A = V r_A \] Assume steady-state, \( F_{in} = F_{out} = F \). \[ C_A = C_{A0} (1 - X) = 4 \times 0.6 = 2.4\,kmol/m^3 \] \[ \text{First-order:} \quad r_A = -k C_A \] \[ F (C_{A0} - C_A) = V k C_A \] \[ F (C_{A0} - C_A) = V k C_A \] \[ F = \frac{V k C_A}{C_{A0} - C_A} \] \[ C_{A0} - C_A = 4 - 2.4 = 1.6 \] \[ F = \frac{0.5 \times 0.5 \times 2.4}{1.6} \] \[ 0.5 \times 0.5 = 0.25 \] \[ 0.25 \times 2.4 = 0.6 \] \[ F = \frac{0.6}{1.6} = 0.375\, m^3/hr \] #### **Final Answer:** \[ \boxed{F = 0.375\, m^3/\text{hr}} \] --- ### **4. Batch Reactor Intermediate \( B \) as a Function of Time** **Reaction:** \( A \xrightarrow{k_1} B \xrightarrow{k_2} C \), with \( k_1 > k_2 \) **Initial:** \( C_A(0) = C_{A0} \), \( C_B(0) = 0 \), \( C_C(0) = 0 \) #### **Step 1: Write the Differential Equations** \[ \frac{dC_A}{dt} = -k_1 C_A \] \[ \frac{dC_B}{dt} = k_1 C_A - k_2 C_B \] \[ \frac{dC_C}{dt} = k_2 C_B \] #### **Step 2: Solve for \( C_A \)** \[ C_A = C_{A0} e^{-k_1 t} \] #### **Step 3: Plug into \( C_B \) Equation** \[ \frac{dC_B}{dt} + k_2 C_B = k_1 C_{A0} e^{-k_1 t} \] This is a linear ODE;