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 2800.0 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. a. What mass of waste brine is expelled in one day?

Answer 1

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 = 2800 gallon

2800 gallon = 1.0599 x 10^7 cm^3 = 1.0599 x 10^7 g = 10599 kg

this is equal to 0.3294P

0.3294P = 10599

so P = 32176.6 Kg

=> P = 8500.158 gallons

8500.158 gallons of brine must the plant take per day.

so mass of waste brine = P - 0.3294P = 0.6706P = 21577.6 Kg