Let's solve each question step-by-step: --- ### **1. Salt Concentration in a Tank** **Given:** - Initial solution volume in tank, \( V_0 = 0.5 \) m³ - Initial salt in tank, \( S_0 = 25 \) kg - Tank maximum volume, \( V_{max} = 1.0 \) m³ - Inlet salt concentration, \( C_{in} = 200 \) kg/m³ - Inlet volumetric flow rate, \( q_{in} = 0.015 \) m³/min - Outlet volumetric flow rate, \( q_{out} = 0.010 \) m³/min **Let:** - \( V(t) \): volume in the tank at time \( t \) - \( S(t) \): salt in the tank at time \( t \) - \( C(t) = \frac{S(t)}{V(t)} \): salt concentration in the tank #### **Step 1: Write the mass balance equations** \[ \frac{dV}{dt} = q_{in} - q_{out} = 0.015 - 0.010 = 0.005 \text{ m}^3/\text{min} \] So, \[ V(t) = V_0 + 0.005t \] The tank is full when \( V(t) = 1.0 \) m³: \[ 1.0 = 0.5 + 0.005t \implies t = \frac{0.5}{0.005} = 100 \text{ min} \] #### **Step 2: Salt balance** \[ \frac{dS}{dt} = \text{Salt in} - \text{Salt out} = (q_{in}C_{in}) - (q_{out}C(t)) \] \[ \frac{dS}{dt} = 0.015 \times 200 - 0.010 \times \frac{S(t)}{V(t)} \] \[ \frac{dS}{dt} = 3 - 0.010\frac{S(t)}{V(t)} \] #### **Step 3: Substitute \( V(t) \) and solve the ODE** \[ \frac{dS}{dt} + \frac{0.010}{V_0 + 0.005t} S = 3 \] This is a first-order linear ODE. Integrating factor: \[ IF = \exp\left( \int \frac{0.010}{V_0 + 0.005t} dt \right) = \exp\left( 0.010 \int \frac{dt}{0.5 + 0.005t} \right) \] Let \( u = 0.5 + 0.005t \implies du = 0.005dt \implies dt = du/0.005 \) \[ \int \frac{dt}{0.5 + 0.005t} = \int \frac{1}{u} \frac{du}{0.005} = \frac{1}{0.005} \int \frac{du}{u} = 200\ln(u) \] So, \[ IF = \exp(0.010 \times 200 \ln(u)) = \exp(2\ln(u)) = u^2 = (0.5 + 0.005t)^2 \] Multiply both sides by \( IF \): \[ \frac{d}{dt}\left[S (0.5 + 0.005t)^2 \right] = 3 (0.5 + 0.005t)^2 \] Integrate both sides from \( t=0 \) to \( t \): \[ S(t) (0.5 + 0.005t)^2 = S_0 (0.5)^2 + 3 \int_0^t (0.5 + 0.005\tau)^2 d\tau \] Calculate the integral: Let \( u = 0.5 + 0.005\tau \implies du = 0.005d\tau \implies d\tau = du/0.005 \) When \( \tau = 0 \), \( u = 0.5 \) When \( \tau = t \), \( u = 0.5 + 0.005t \) \[ \int_0^t (0.5 + 0.005\tau)^2 d\tau = \int_{u=0.5}^{u=0.5+0.005t} u^2 \frac{du}{0.005} = \frac{1}{0.005} \int_{0.5}^{0.5+0.005t} u^2 du = \frac{1}{0.005} \left[ \frac{u^3}{3} \right]_{0.5}^{0.5+0.005t} \] \[ = \frac{1}{0.005} \left( \frac{(0.5+0.005t)^3}{3} - \frac{(0.5)^3}{3} \right) = 200 \left( \frac{(0.5+0.005t)^3 - (0.5)^3}{3} \right) \] #### **Step 4: Substitute back and solve for \( S(t) \) at \( t = 100 \) min (\( V = 1.0 \) m³)** \[ S(t) (0.5 + 0.005t)^2 = 25 \times (0.5)^2 + 3 \times 200 \times \frac{(0.5+0.005t)^3 - (0.5)^3}{3} \] \[ S(t) (0.5 + 0.005t)^2 = 6.25 + 200[(0.5+0.005t)^3 - 0.125] \] At \( t = 100 \) min, \( 0.5 + 0.005 \times 100 = 1.0 \): \[ S(100) \cdot (1.0)^2 = 6.25 + 200[(1.0)^3 - 0.125] \] \[ S(100) = 6.25 + 200(1 - 0.125) = 6.25 + 200 \times 0.875 = 6.25 + 175 = 181.25 \text{ kg} \] Volume at this time: \( V(100) = 1.0 \) m³ So, **final concentration**: \[ \boxed{C = \frac{181.25}{1.0} = 181.3 \text{ kg/m}^3} \] --- ### **2. Tank Heating with Heat Loss** Given: - Mass, \( m = 45 \) kg \( = 45,000 \) g - Heater input, \( Q_{in} = 25,000 \) cal/min - Specific heat, \( C_p = 0.5 \) cal/g°C - Heat loss coefficient, \( h = 800 \) cal/min·°C - Ambient temperature, \( T_{a} = 25^\circ \)C - Initial temperature, \( T_0 = 15^\circ \)C Let \( T(t) \) be the tank temperature at time \( t \). #### **Step 1: Energy balance** \[ \text{Accumulation} = \text{Input} - \text{Output} \] \[ mC_p \frac{dT}{dt} = Q_{in} - h(T - T_a) \] \[ 45,000 \times 0.5 \frac{dT}{dt} = 25,000 - 800(T - 25) \] \[ 22,500 \frac{dT}{dt} + 800(T - 25) = 25,000 \] \[ \frac{dT}{dt} + \frac{800}{22,500}(T - 25) = \frac{25,000}{22,500} \] \[ \frac{dT}{dt} + 0.03556(T - 25) = 1.111 \] #### **Step 2: Solve the first-order ODE** Let \( x = T - 25 \) (\( dx/dt = dT/dt \)) \[ \frac{dx}{dt} + 0.03556x = 1.111 \] Integrating factor: \[ IF = e^{0.03556 t} \] Multiply both sides: \[ \frac{d}{dt}(x e^{0.03556 t}) = 1.111 e^{0.03556 t} \] Integrate both sides: \[ x e^{0.03556 t} = \int 1.111 e^{0.03556 t} dt + C = 1.111 \frac{e^{0.03556 t}}{0.03556} + C \] \[ x = \frac{1.111}{0.03556} + Ce^{-0.03556 t} \] \[ x = 31.28 + Ce^{-0.03556 t} \] Recall \( x = T - 25 \): \[ T(t) = 25 + 31.28 + Ce^{-0.03556 t} = 56.28 + Ce^{-0.03556 t} \] Initial condition: \( T(0) = 15 \implies 15 = 56.28 + C \) \[ C = 15 - 56.28 = -41.28 \] #### **Final answer:** \[ \boxed{ T(t) = 56.28 - 41.28 e^{-0.03556 t} } \] where \( t \) is in minutes. --- ### **3. CSTR Reaction Conversion and Flow Rate** Given: - CSTR volume, \( V = 0.5 \) m³ - \( k = 0.5 \) m³/kmol·hr - \( C_{A0} = 4 \) kmol/m³, \( C_{R0} = 2 \) kmol/m³ - Desired conversion of A: \( X = 0.4 \) (40%) #### **Step 1: Molar balance for A** For a CSTR: \[ F(C_{A0} - C_A) = kVC_A \] \[ F = \text{feed rate} \ (\text{m}^3/\text{hr}) \] \[ X = \frac{C_{A0} - C_A}{C_{A0}} \implies C_A = C_{A0}(1 - X) \] So, \[ F(C_{A0} - C_A) = kVC_A \] \[ F(C_{A0} - C_{A0}(1-X)) = kV C_{A0}(1-X) \] \[ F(C_{A0}X) = kV C_{A0}(1-X) \] \[ F = \frac{kV(1-X)}{X} \] #### **Step 2: Plug in the numbers** \[ F = \frac{0.5 \times 0.5 \times (1-0.4)}{0.4} = \frac{0.25 \times 0.6}{0.4} = \frac{0.15}{0.4} = 0.375 \ \text{m}^3/\text{hr} \] #### **Final answer:** \[ \boxed{0.375 \ \text{m}^3/\text{hr}} \] is the required feed rate to achieve 40% conversion. --- ### **4. Batch Reactor Intermediate Concentration** **A → B → C** (first-order, \( k_1 > k_2 \)) - Initial: \( [A]_0 = C_{A0} \), \( [B]_0 = [C]_0 = 0 \) - \( k_1 \): rate constant for \( A \to B \) - \( k_2 \): rate constant for \( B \to C \) - Goal: Find \( [B](t) \) #### **Step 1: Write the ODEs** \[ \frac{d[A]}{dt} = -k_1 [A] \] \[ \frac{d[B]}{dt} = k_1 [A] - k_2 [B] \] \[ \frac{d[C]}{dt} = k_2 [B] \] #### **Step 2: Solve for [A](t)** \[ [A](t) = C_{A0} e^{-k_1 t} \] #### **Step 3: Plug into [B]'s equation** \[ \frac{d[B]}{dt} + k_2 [B] = k_1 C_{A0} e^{-k_1 t} \] This is a linear ODE; integrating factor is \( e^{k_2 t} \): \[ e^{k_2 t} \frac{d[B]}{dt} + k_2 e^{k_2 t} [B] = k_1 C_{A0} e^{k_2 t} e^{-k_1 t} \] \[ \frac{d}{dt}\left( [B] e^{k_2 t} \right) = k_1 C