An evaporator while concentrating an aqueous solution from 10 to 40% solids evaporates 30000 kg of water. The amount of solids handled by the system in kg is

ANS:A - 4000

To find the amount of solids handled by the system, we can use the concept of mass balance. The mass balance equation states that the mass entering a system must equal the mass leaving the system, plus any accumulation. Initially, we have:

• Initial mass of solution = 100%
• Initial mass of solids = 10%
• Initial mass of water = 90%

Finally, we have:

• Final mass of solution = 100%
• Final mass of solids = 40%
• Final mass of water = 60%

Given that 30,000 kg of water is evaporated, we can calculate the amount of solids handled by the system as follows: Initial mass of water - Final mass of water = Mass of water evaporated 90% of initial mass - 60% of final mass = 30,000 kg We can use this equation to find the initial mass of the solution: Initial mass of solution = (30,000 kg / 30%) * 100% = 100,000 kg Now, we can find the initial mass of solids: Initial mass of solids = 10% of initial mass of solution = 10% * 100,000 kg = 10,000 kg Finally, we can find the final mass of solids: Final mass of solids = 40% of final mass of solution = 40% * 100,000 kg = 40,000 kg Therefore, the amount of solids handled by the system is 40,000 kg - 10,000 kg = 30,000 kg. So, the correct answer is: 3000