Three brine solutions, B1, B2, and B3 are mixed. B1 is one-half of the total mix
ID: 1494977 • Letter: T
Question
Three brine solutions, B1, B2, and B3 are mixed. B1 is one-half of the total mixture. Brine B1 is 2.5% salt. B2 is 4.5% salt. B3 is 5.5% salt. To this mixture is added 35 lbm of dry salt, while 280 lbm of water is removed leaving 3200 lbm of 5.1% brine. To solve this problem: 1) list the various mass balance equations; 2) show the “A” and “B” arrays that you would enter into Matlab or Excel to solve these simultaneous equations; 3) using Matlab or Excel, determine to initial amounts of B1, B2, and B3. (Hint: solve first for B1 since it is ½ of the total mixture; then you have 2 unknown and need 2 equations)
accompanied by a sketch of the process with inputs, outputs, knowns/unknowns, and compositions identified. Describe the process (batch, flow,etc.).
Explanation / Answer
3200 lbm of 5.1% brine
broken down is
3200(0.051) = 163.2 lb salt
3200(1 - 0.051) = 3036.8 lb water
subtracting out the salt that was added
163.2 - 35 = 128.2 lb salt
adding in the water that was evaporated
3036.8 + 280 = 3316.8 lb water
which means that after the mixing of B1, B2 and B3, there was
3326.8 + 128.2 = 3445 lb brine
half of that was B1
B1 = 1722.5 lb <===ANS
which had
1722.5(0.025) = 43.0625 lb salt
That means B2 and B3 contributed
128.2 - 43.0625 = 85.1375 lb salt
B2 + B3 = 1722.5 lb
B2 = 1722.5 - B3
analyzing the salt
0.045B2 +0.055B3 = 85.1375
0.045(1722.5 - B3) + 0.055B3 = 85.1375
77.5125 - 0.045B3 + 0.055B3 = 85.1375
0.01B3 = 7.625
B3 = 762.5 lb <===ANS
B2 = 1722.5 - 762.5
B2 = 960 lb <===ANS
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