Let's break down and solve the problem step by step. ### **Given Data** - **Waste acid (Stream 1):** - Contains: 55 wt% H₂SO₄, 20 wt% HNO₃, rest water - **Sulfuric acid stream (Stream 2):** - 95 wt% H₂SO₄, rest water - **Nitric acid stream (Stream 3):** - 90 wt% HNO₃, rest water - **Final mixture (Product Stream):** - 26 wt% HNO₃, 60 wt% H₂SO₄, rest water - Total flow rate = 1000 kg/h Let: - \( x_1 \) = Flow rate of waste acid (kg/h) - \( x_2 \) = Flow rate of sulfuric acid stream (kg/h) - \( x_3 \) = Flow rate of nitric acid stream (kg/h) --- ## **a) Overall and Component Material Balances** ### **Overall Mass Balance** \[ x_1 + x_2 + x_3 = 1000 \] --- ### **Component Mass Balances** #### **Sulfuric Acid (H₂SO₄)** \[ \text{Sulfuric acid in:} \quad 0.55x_1 + 0.95x_2 + 0x_3 \] \[ \text{Sulfuric acid out:} \quad 0.60 \times 1000 = 600\, \text{kg} \] So, \[ 0.55x_1 + 0.95x_2 = 600 \] --- #### **Nitric Acid (HNO₃)** \[ \text{Nitric acid in:} \quad 0.20x_1 + 0x_2 + 0.90x_3 \] \[ \text{Nitric acid out:} \quad 0.26 \times 1000 = 260\, \text{kg} \] So, \[ 0.20x_1 + 0.90x_3 = 260 \] --- #### **Water** You can check the water balance as well, but with three unknowns and three equations above, it's not needed. --- ## **b) Determine Individual Flow Rates** We have three equations: 1. \( x_1 + x_2 + x_3 = 1000 \) 2. \( 0.55x_1 + 0.95x_2 = 600 \) 3. \( 0.20x_1 + 0.90x_3 = 260 \) Let's solve step-by-step. --- ### **Step 1: Express \( x_2 \) and \( x_3 \) in terms of \( x_1 \)** From (2): \[ 0.55x_1 + 0.95x_2 = 600 \implies 0.95x_2 = 600 - 0.55x_1 \implies x_2 = \frac{600 - 0.55x_1}{0.95} \] From (3): \[ 0.20x_1 + 0.90x_3 = 260 \implies 0.90x_3 = 260 - 0.20x_1 \implies x_3 = \frac{260 - 0.20x_1}{0.90} \] --- ### **Step 2: Substitute \( x_2 \) and \( x_3 \) into (1):** \[ x_1 + x_2 + x_3 = 1000 \] \[ x_1 + \frac{600 - 0.55x_1}{0.95} + \frac{260 - 0.20x_1}{0.90} = 1000 \] --- ### **Step 3: Simplify** Let’s combine terms: \[ x_1 + \frac{600 - 0.55x_1}{0.95} + \frac{260 - 0.20x_1}{0.90} = 1000 \] Let’s denote: - \( A = x_1 \) - \( B = \frac{600 - 0.55A}{0.95} \) - \( C = \frac{260 - 0.20A}{0.90} \) So, \[ A + B + C = 1000 \] Let’s get a common denominator, but first, express all in terms of \( x_1 \): Expand: \[ x_1 + \frac{600}{0.95} - \frac{0.55x_1}{0.95} + \frac{260}{0.90} - \frac{0.20x_1}{0.90} = 1000 \] Calculate each term: - \( \frac{600}{0.95} = 631.579 \) - \( \frac{0.55x_1}{0.95} = 0.5789x_1 \) - \( \frac{260}{0.90} = 288.889 \) - \( \frac{0.20x_1}{0.90} = 0.2222x_1 \) So, \[ x_1 + 631.579 - 0.5789x_1 + 288.889 - 0.2222x_1 = 1000 \] Combine \( x_1 \) terms: \[ x_1 - 0.5789x_1 - 0.2222x_1 = x_1 - 0.8011x_1 = 0.1989x_1 \] Combine constants: \[ 631.579 + 288.889 = 920.468 \] So, \[ 0.1989x_1 + 920.468 = 1000 \] \[ 0.1989x_1 = 1000 - 920.468 = 79.532 \] \[ x_1 = \frac{79.532}{0.1989} = 399.8 \approx \boxed{400\, \text{kg/h}} \] --- ### **Step 4: Solve for \( x_2 \) and \( x_3 \)** #### **\( x_2 \):** \[ x_2 = \frac{600 - 0.55x_1}{0.95} \] With \( x_1 = 400 \): \[ x_2 = \frac{600 - 0.55 \times 400}{0.95} = \frac{600 - 220}{0.95} = \frac{380}{0.95} = 400\, \text{kg/h} \] --- #### **\( x_3 \):** \[ x_3 = \frac{260 - 0.20x_1}{0.90} \] \[ x_3 = \frac{260 - 0.20 \times 400}{0.90} = \frac{260 - 80}{0.90} = \frac{180}{0.90} = 200\, \text{kg/h} \] --- ## **Final Answers** ### **a) The material balance equations are:** \[ \boxed{ \begin{align*} &x_1 + x_2 + x_3 = 1000 \\ &0.55x_1 + 0.95x_2 = 600 \\ &0.20x_1 + 0.90x_3 = 260 \\ \end{align*} } \] --- ### **b) The individual flow rates are:** \[ \boxed{ \begin{align*} \text{Waste acid (x}_1\text{):} & \quad 400\, \text{kg/h} \\ \text{Sulfuric acid (x}_2\text{):} & \quad 400\, \text{kg/h} \\ \text{Nitric acid (x}_3\text{):} & \quad 200\, \text{kg/h} \\ \end{align*} } \] --- ## **Check:** Sum: \( 400 + 400 + 200 = 1000 \) kg/h ✅ **Sulfuric acid:** \[ 0.55 \times 400 + 0.95 \times 400 = 220 + 380 = 600 \text{ kg} \quad (\text{matches desired output}) \] **Nitric acid:** \[ 0.20 \times 400 + 0.90 \times 200 = 80 + 180 = 260 \text{ kg} \quad (\text{matches desired output}) \] --- ## **Summary Table** | Stream | Flow rate (kg/h) | H₂SO₄ (wt%) | HNO₃ (wt%) | Water (wt%) | |-----------------------|------------------|-------------|------------|-------------| | Waste acid (x₁) | 400 | 55 | 20 | 25 | | Sulfuric acid (x₂) | 400 | 95 | 0 | 5 | | Nitric acid (x₃) | 200 | 0 | 90 | 10 | | **Final mixture** | 1000 | 60 | 26 | 14 | --- **If you need the detailed calculation for water content, let me know!**