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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
Correct answer is option 'B'. Can you explain this answer?
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A multiple effect evaporator has a capacity to process 4000 kg of soli...
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If 10% solution contains:
Caustic soda = 4000 kg.
Then;
Mass of water = 40,000 kg.
&
If 25% solution contains:
Caustic soda = 4000 kg.
Then;
Mass of water = 16,000 kg.
So, water evaporated is,
Water evaporated = 40,000 - 16,000.
= 24,000 kg.
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A multiple effect evaporator has a capacity to process 4000 kg of soli...
Calculation of Water Evaporated in Multiple Effect Evaporator
Given data:
Capacity of multiple effect evaporator = 4000 kg/day
Initial concentration of caustic soda = 10%
Final concentration of caustic soda = 25%
Step 1: Calculate the amount of caustic soda processed per day
Amount of caustic soda processed per day = Capacity of multiple effect evaporator = 4000 kg/day
Step 2: Calculate the amount of water in the feed
Amount of water in the feed = (Amount of caustic soda processed per day) / (Initial concentration of caustic soda)
= (4000 kg/day) / (0.1) = 40000 kg/day
Step 3: Calculate the amount of caustic soda in the feed
Amount of caustic soda in the feed = (Amount of caustic soda processed per day) - (Amount of water in the feed)
= (4000 kg/day) - (40000 kg/day) = 36000 kg/day
Step 4: Calculate the amount of caustic soda in the product
Amount of caustic soda in the product = (Amount of caustic soda in the feed) / (Final concentration of caustic soda)
= (36000 kg/day) / (0.25) = 144000 kg/day
Step 5: Calculate the amount of water in the product
Amount of water in the product = (Amount of water in the feed) - (Amount of water evaporated)
= (40000 kg/day) - (Amount of water evaporated)
Step 6: Calculate the amount of water evaporated
Amount of water evaporated = (Amount of water in the feed) - (Amount of water in the product)
= (40000 kg/day) - [(Amount of caustic soda in the product) / (Final concentration of caustic soda)]
= (40000 kg/day) - [(144000 kg/day) / (0.25)]
= 6000 kg/day
Therefore, the amount of water evaporated in the multiple effect evaporator is 24000 kg/day.
Given data:
Capacity of multiple effect evaporator = 4000 kg/day
Initial concentration of caustic soda = 10%
Final concentration of caustic soda = 25%
Step 1: Calculate the amount of caustic soda processed per day
Amount of caustic soda processed per day = Capacity of multiple effect evaporator = 4000 kg/day
Step 2: Calculate the amount of water in the feed
Amount of water in the feed = (Amount of caustic soda processed per day) / (Initial concentration of caustic soda)
= (4000 kg/day) / (0.1) = 40000 kg/day
Step 3: Calculate the amount of caustic soda in the feed
Amount of caustic soda in the feed = (Amount of caustic soda processed per day) - (Amount of water in the feed)
= (4000 kg/day) - (40000 kg/day) = 36000 kg/day
Step 4: Calculate the amount of caustic soda in the product
Amount of caustic soda in the product = (Amount of caustic soda in the feed) / (Final concentration of caustic soda)
= (36000 kg/day) / (0.25) = 144000 kg/day
Step 5: Calculate the amount of water in the product
Amount of water in the product = (Amount of water in the feed) - (Amount of water evaporated)
= (40000 kg/day) - (Amount of water evaporated)
Step 6: Calculate the amount of water evaporated
Amount of water evaporated = (Amount of water in the feed) - (Amount of water in the product)
= (40000 kg/day) - [(Amount of caustic soda in the product) / (Final concentration of caustic soda)]
= (40000 kg/day) - [(144000 kg/day) / (0.25)]
= 6000 kg/day
Therefore, the amount of water evaporated in the multiple effect evaporator is 24000 kg/day.
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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 isa)6000b)24000c)60000d)48000Correct answer is option 'B'. Can you explain this answer?
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