Given: A tank with a volume (?) of 5,000 gal contains salt solution at an initia

Question

Given: A tank with a volume (?) of 5,000 gal contains salt solution at an initial concentration (C0*) of 70,000 mg/L, but the concentration in the tank (C) needs to be reduced to 500 mg/L. Two approaches are being considered as indicated below (Alternatives 1 and 2).
Alt. 1: Flush tank with clean tap water while tank is mixed. Flowrate into the tank (QIn) and flowrate out of tank (QOut) are each 40 gpm (both flows are limited by the drain pipe capacity). If you do not derive equations used (which is not required), identify the source of the ones used with specificity; Define all variables used.
Alt. 2: First decant a portion of the salt solution at a flowrate (QOut) of 40 gpm, then replace the decanted volume with clean tapwater at a flowrate into the tank (QIn) of 100 gpm.
Required: a) Determine which method will accomplish the job in the shortest time.

b) Determine the difference in clean water used/ wastewater generated (gal.) by each method.

c) Of course our time is valuable and the importance of accomplishing the task quickly is obvious. Rather than discussing time, discuss, from the standpoint of sustainability, the importance of:
1) conserving (not wasting) clean water, and
2) minimizing the generation of waste brine (salty water) in the Central San Joaquin Valley.
Be thoughtful; Simplistic answers will not receive much credit. Include in your discussion:
the long-term outlook of groundwater reserves in the Central Valley
the challenges we face with respect to disposal of waste brine solution; what are the legal ways of doing this in the Valley? Note that it would not be legal to put brine in the sewer or percolate it into the ground due to the negative impact both would have on the groundwater.

Explanation / Answer

For some reason I can't post my image, but you can find it here (http://db.tt/arTlEdM4).

For Alternative #1: It will take ~620 minutes (~10 hrs) to equalize the solution, cost about 25,000 gals of clean water, and waste as much.

For Alternative #2: It will take about 175 min (~3 hrs), cost 5,000 gals of water, and waste as much.

The formula I used is call the mass-flow equation and I found a good explanation here (http://faculty.atu.edu/mfinan/2924/cal116.pdf)

If you need a detailed explanation for Part C I can take a crack at it... we have similar water conservation problems in Florida but you might be more familar with the CA terrain and your prof expectations. Please let me know if you need help here too.

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