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 9.5% by mass of the water entering the stage evaporates and leaves as fresh water. If the desalination plant wants to produce 3500.0 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/cm^3. The density of brine is 1.025 g/cm^3

Explanation / Answer

Here, consider initially P kg of brine solution enter the system

Now, concentration of salt = 35,000 ppm

this means -
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 that 9.5% of water entering the system leaves

Now consider stage - 1:
water in = 0.9662P
water flashed = 0.095(0.9662P) = 0.091789P
water to stage 2 = 0.874411P

Now consider stage 2:
water in = 0.874411P
water flashed = 0.095(0.874411P) = 0.083069045P
water to stage 3 = 0.791341955P

Now, in stage 3:
water in = 0.791341955P
water flashed = 0.095(0.791341955P) = 0.0751774857P
water to stage 4 = 0.716164469275P

In stage 4:
water in = 0.716164469275P
water flashed = 0.095(0.716164469275P) = 0.068035624581125P
water to stage 5 = 0.648128844693875P

In stage 5:
water in = 0.648128844693875P
water flashed = 0.095(0.648128844693875P) = 0.061572240245918125P
water to stage 5 = 0.586556604447956875P

total water leaving the system = sum of water leaving all stages = 0.333294P (Please check the additions on the calculator)

given daily freshwater production = 3500 gallon

Now, 3500 gallon = 1.5911 x 10^7 cm^3 = 1.5911 x 10^7 gm = 15911 kg

And this is equal to 0.333294P

so P = 47739 Kg

Therefore, mass of waste brine = P - 0.333294P = 31828 Kg

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