A reversed osmosis plant desalinates 5×106 L/day of feedwater containing 1,500mg/L of salts, yielding 3 x106 L/day of product water with 75 mg/L of salt.
What is reverse osmosis?
What is desalination?
What would the salt concentration be in the brine?
membrane
Feedwater Product water
5×106 L/day 3×106 L/day
1,500 mg/L 75 mg/L
Brine
Reverse osmosis is a process where water is forced through a semipermeable membrane to remove impurities and contaminants such as salts, minerals, and other particles. It's like the water doing an extreme makeover to become pure and clean.
Desalination is the process of removing salt and other minerals from seawater or brackish water to make it suitable for drinking or other purposes. It's like turning salty water into a fancy drinking water cocktail.
As for the salt concentration in the brine, I'd say it's probably high enough to preserve a truckload of potato chips. But in technical terms, it will depend on the efficiency of the reverse osmosis process and the specific characteristics of the feedwater. So, the brine might need a "Caution: High Sodium" warning label.
Reverse osmosis is a process that involves the filtering and purification of water by using a semi-permeable membrane. This membrane allows the passage of water molecules while blocking the contaminants, including salts and other impurities.
Desalination is the process of removing salt and other minerals from seawater or brackish water to make it suitable for human consumption or other uses.
To determine the salt concentration in the brine, we can use the principle of mass balance. Since the feedwater contains 1,500 mg/L of salts and produces 3×106 L/day of product water with 75 mg/L of salt, we can calculate the salt concentration in the brine.
First, we need to find the total amount of salt in the feedwater and the product water:
Total salt in feedwater = 5×106 L/day × 1,500 mg/L = 7.5×109 mg/day
Total salt in product water = 3×106 L/day × 75 mg/L = 2.25×108 mg/day
Next, we can subtract the total salt in the product water from the total salt in the feedwater to find the amount of salt in the brine:
Salt in brine = Total salt in feedwater - Total salt in product water
= 7.5×109 mg/day - 2.25×108 mg/day
= 7.275×109 mg/day
Now, we need to find the volume of the brine. Since the feedwater volume is 5×106 L/day and the product water volume is 3×106 L/day, the remaining volume must be the brine:
Volume of brine = Feedwater volume - Product water volume
= 5×106 L/day - 3×106 L/day
= 2×106 L/day
Finally, we can calculate the salt concentration in the brine by dividing the amount of salt in the brine by the volume of the brine:
Salt concentration in brine = Salt in brine / Volume of brine
= 7.275×109 mg/day / 2×106 L/day
= 3,637.5 mg/L
Therefore, the salt concentration in the brine is 3,637.5 mg/L.
Reverse osmosis is a process used for water treatment that removes impurities and contaminants by utilizing a semi-permeable membrane. In reverse osmosis, pressure is applied to the feedwater, forcing it through the membrane, which allows water molecules to pass through while blocking the larger salts and other impurities.
Desalination, on the other hand, is the process of removing the salt and other minerals from seawater or brackish water, making it suitable for drinking or other purposes. Reverse osmosis is one of the commonly used methods for desalination.
To find the salt concentration in the brine, we can use the principle of mass balance. The total amount of salt in the feedwater should equal the total amount of salt in the product water and the brine.
Given that the feedwater has a salt concentration of 1,500 mg/L and a flow rate of 5×106 L/day, the total amount of salt in the feedwater would be:
Total salt in feedwater = Concentration × Flow rate = 1,500 mg/L × 5×106 L/day
Similarly, the total amount of salt in the product water with a concentration of 75 mg/L and a flow rate of 3×106 L/day would be:
Total salt in product water = Concentration × Flow rate = 75 mg/L × 3×106 L/day
To find the salt concentration in the brine, we subtract the amount of salt in the product water from the total amount of salt in the feedwater, and divide it by the flow rate of the brine:
Salt concentration in brine = (Total salt in feedwater - Total salt in product water) / Brine flow rate
Substituting the values from the given information:
Salt concentration in brine = (1,500 mg/L × 5×106 L/day - 75 mg/L × 3×106 L/day) / (5×106 L/day - 3×106 L/day)
Further simplifying the equation will give us the salt concentration in the brine.