Multi-stage flash distillation is used to desalinate seawater so it is fit for h

Question

Multi-stage flash distillation is used to desalinate seawater so it is fit for human consumption. A flash distillation tank is composed of small chambers (called "stages"), each at a lower pressure than the previous stage. Salt water enters the first chamber, some fresh water flash evaporates, condenses, and is removed from the distillation tank. The remaining salt water moves to the next chamber, some fresh water flash evaporates, condenses, and is removed from the distillation tank. The remaining salt water moves to the next chamber, and the process repeats until very salty water is expelled as waste. Brine that is 35,000 ppm salt is desalinated through a 5-stage flash distillation process. The pressure in the stages is such that 8% by mass of the water entering the stage evaporates and leaves as fresh water. If the desalination plant wants to produce 1400.0 gallons of drinking water per day, how many gallons of brine must the plant take in per day? The density of fresh water is 1.000 g/cm3. The density of brine is 1.025 g/cm3.

Explanation / Answer

Let P kg of brine solution enter the system conc of salt = 35,000 ppm or 35,000 gm of salt in 1,000,000 gm of water mass% of salt = (35,000)/(35,000+1,000,000) = 3.382% mass% of water = 96.62% given 8% of water entering the system leaves in stage 1 water in = 0.9662P water flashed = 0.08(0.9662P) = 0.077296P water to stage 2 = 0.888904P in stage 2 water in = 0.888904P water flashed = 0.08(0.888904P) = 0.07111232P water to stage 3 = 0.81779168P in stage 3 water in = 0.81779168P water flashed = 0.08(0.81779168P) = 0.065423334P water to stage 4 = 0.752368345P in stage 4 water in = 0.752368345P water flashed = 0.08(0.752368345P) = 0.060189467P water to stage 5 = 0.692193982P in stage 5 water in = 0.692193982P water flashed = 0.08(0.692193982P) = 0.055375518P water to stage 5 = 0.636818463P total water leaving the system = sum of water leaving all stages = 0.3294P given daily freshwater production = 1400 gallon 1400 gallon = 5299576.8 cm^3 = 5299576.8 gm = 5299.5768 kg this is equal to 0.3294P so P = 16088.6Kg so mass of waste brine = P - 0.3294P = 0.6706P = 10789 Kg = 23785.7 lb

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