ION_EXCHANGE_SOLUTION

# ION_EXCHANGE_SOLUTION - Ion Exchange Solution 1 Column of 1...

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Ion Exchange Solution 1: Column of 1” Diameter , 12” depth Volume = 5L Ca = 1,500 mg/L R-Ca + 2Na + R-Na + Ca ++ Cation Exchange (Reaction) Total Ca ++ = (1500 L mg )*(5L)*( mg meq 20 ) = 20 7500 meq = 375 meq Volume = (1”) 2 ( 4 Π )(12”) 3 3 ) 12 ( in ft ( 3 485 . 7 ft gal )( gal L 785 . 3 )( ) 1000 L ml Resin M L = 154.4 mL Resin capacity C sp = 375 meq/154.4mL = Solution 2: part b) Ca ++ = 70mg/L = 3.5 meq/L Mg ++ = 9.7mg/L = 0.8 meq/L Ca ++ + Mg ++ = 4.3 meq/L (4.3meq/L)(50 mg CaCO 3 /1 meq) = 215 mg/L Since 1 grain = 64.8 mg (215 mg/L)*(1 grain/64.8mg) = 3.32 grains/L of hardness need to be removed Hence: 1*10 6 gal * Gal L ft Gal grains ft Gal L L grains 1 785 . 3 * 1 485 . 7 * 000 , 10 1 * 1 785 . 3 * 1 32 . 3 3 3 Amount of resin needed per MG treated = 35600 L Salt needed to treat 1 MG : 1*10 6 gal * grains lb Gal L L grains 100 5 . 0 * 1 785 . 3 * 1 32 . 3 = 62,831 lb of salt Cost/lbs = \$ 0.20 ( Given, if given as \$ 2/lb it is incorrect, correct is \$0.2/lbs) Cost =( \$ 0.2/lbs)*(62,831 lbs/MG) = \$ 12,566/MG 1

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Solution 3: Meq Ca ++ Wet resin = 150ml Vol effluent = 4.7L C a ++ in effluent = 1220 mg/L Mass of ion adsorbed: L 7 . 4 * mg 40 Ca 2 * 1 1220 + + + + meq L Ca mg = 286.7 meq Ca ++ Resin capacity: =
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ION_EXCHANGE_SOLUTION - Ion Exchange Solution 1 Column of 1...

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