Let's break down and solve the problem step by step: --- ## **Problem Restatement** - **Tomato pulp** has 6% solids by weight. - **Evaporator** produces **tomato paste with 28% solids**. - **Evaporator's capacity**: **100 kg/h (feed rate- **Question**: Can we make 28% solid paste by either: - (a) Feeding all pulp to the evaporator? - (b) First producing 33% solid paste and blending it with 6% pulp to get 28% solid paste? --- ## **Step 1: Mass Balance for Option (a)** ### **a. Feeding all pulp to the evaporator** Let: - Feed rate to evaporator = \( F \) (kg/h) = **100 kg/h** - Feed solids = \( S_f = 6\% \) - Product solids = \( S_p = 28\% \) - Product mass = \( P \) (kg/h) **Solids in feed:** \( \text{Solids in feed} = F \times .06 = 100 \times .06 = 60 \) kg solids/h **All solids leave in product:** \( \text{Product solids} = P \times .28 \) So, \( 60 = P \times .28 \implies P = \frac{60}{.28} = 214.29 \) kg/h --- ### **Evaporator Output** - Feed (100 kg/h) → Product (214.29 kg/h of 28% solids paste) --- ## **Step 2: Mass Balance for Option (b)** ### **b. First produce 33% paste, then blend with 6% pulp to get 28% paste** Let: - **Evaporator** makes **33% solids paste** from 6% pulp. - **Blending** some of this paste with additional 6% pulp to get 28% solids paste. Let: - Amount of 33% paste = \( x \) kg - Amount of 6% pulp = \( y \) kg - Final blend = \( x + y \) kg at 28% solids **Solids in blend:** \( .33x + .06y = .28(x+y) \) Expand: \( .33x + .06y = .28x + .28y \) \( .33x + .06y - .28x - .28y = \) \( .05x - .22y = \) \( .05x = .22y \) \( \frac{x}{y} = \frac{.22}{.05} = 4.4 \) So, for every 4.4 kg of 33% paste, blend 1 kg of 6% pulp. --- #### **Evaporator Capacity** Evaporator must process **6% pulp** to make **33% paste**. Let the fresh 6% pulp fed to evaporator = \( F \) (kg/h) Let evaporator product (33% paste) = \( x \) (kg/h) **Solids in feed:** \( .06F \) **Solids in product:** \( .33x \) Set equal: \( .06F = .33x \implies F = 5.5x \) Since the evaporator can process **100 kg/h** of **6% pulp**: - \( F \leq 100 \) - \( x = F/5.5 \leq 100/5.5 = 181.82 \) kg/h (33% paste produced) **Corresponding amount of 6% pulp to blend:** \( y = x/4.4 = 181.82/4.4 = 41.32 \) kg/h **Total final blend:** \( x + y = 181.82 + 41.32 = 223.14 \) kg/h of 28% paste. --- ## **Step 3: Compare Both Options** ### **Option (a):** - 214.29 kg/h of 28% paste ### **Option (b):** - 223.14 kg/h of 28% paste --- ## **Step 4: Feasibility** Both options are feasible, but **Option (b) produces slightly more 28% paste** because blending allows for more efficient use of the evaporator's capacity. --- ## **Final Answers** ### **a. By feeding all pulp to the evaporator:** **Feasible.** You can produce **214.29 kg/h** of 28% paste. ### **b. By first producing 33% paste and blending with 6% pulp:** **Feasible.** You can produce **223.14 kg/h** of 28% paste. --- ## **Summary Table** | Option | Process | 28% Paste Produced (kg/h) | Feasible? | |--------|--------------------------------------|---------------------------|-----------| | (a) | All pulp to evaporator | 214.29 | Yes | | (b) | 33% paste then blend with 6% pulp | 223.14 | Yes | **Option (b) is more productive.**