Question

In the process of producing caustic (NaOH), 4000 kg/h of a solution containing 10 wt% NaOH is evaporated in the first evaporator, giving a 20% NaOH solution. This is then fed into a second evaporator which gives a product of 50% NaOH. The amount of water removed from each evaporator is

• A. 2000 kg, 1200 kg
• B. 1000 kg, 1200 kg
• C. 2000 kg, 1000 kg
• D. 1200 kg, 600 kg

Hint:

When solving material balance problems, remember to keep track of the mass flow rates of each component at every step of the process.

Answer 1

The Correct Option is A

- The initial mass of NaOH = \( 0.10 \times 4000 = 400 \) kg 
- After the first evaporator, the mass of NaOH = \( 0.20 \times 4000 = 800 \) kg 
- The water removed in the first evaporator = 4000 - 800 = 3200 kg 
- In the second evaporator, the NaOH content becomes 50%, so the mass of NaOH in the final product is 400 kg. 
- The water removed in the second evaporator = \( 4000 - 400 = 3600 \) kg 
- The total water removed = 2000 kg in the first evaporator and 1200 kg in the second(D)
Conclusion: 
The amount of water removed from each evaporator is 2000 kg and 1200 kg, as given by option (A).