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Heat Transfer - Engineering

Q1:

A multiple effect evaporator has a capacity to process 4000 kg of solid caustic soda per day, when it is concentrating from 10% to 25% solids. The water evaporated in kg per day is

A 6000

B 24000

C 60000

D 48000

ANS:B - 24000

To find the amount of water evaporated in kg per day in a multiple effect evaporator, we first need to calculate the initial and final masses of caustic soda, as well as the corresponding masses of water. Given:

  • Initial concentration of caustic soda = 10% solids
  • Final concentration of caustic soda = 25% solids
  • Mass of caustic soda processed per day = 4000 kg
  1. Initial mass of caustic soda: Initial mass = Mass of caustic soda / (1 - Initial solids concentration) Initial mass = 4000 kg / (1 - 0.10) = 4000 kg / 0.90 = 4444.44 kg
  2. Final mass of caustic soda: Final mass = Mass of caustic soda / (1 - Final solids concentration) Final mass = 4000 kg / (1 - 0.25) = 4000 kg / 0.75 = 5333.33 kg
  3. Mass of water initially present: Initial mass of water = Initial mass - Mass of caustic soda processed per day Initial mass of water = 4444.44 kg - 4000 kg = 444.44 kg
  4. Mass of water in the final solution: Final mass of water = Final mass - Mass of caustic soda processed per day Final mass of water = 5333.33 kg - 4000 kg = 1333.33 kg
  5. Mass of water evaporated per day: Mass of water evaporated per day = Initial mass of water - Final mass of water Mass of water evaporated per day = 444.44 kg - 1333.33 kg = -888.89 kg
The negative sign indicates that the mass of water has decreased, which is expected as water is being evaporated. Therefore, the correct answer is 888.89 kg.