SolutionsHW#2

SolutionsHW#2 - SOLUTIONS HOMEWORK NO. 2 CE 3400...

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Unformatted text preview: SOLUTIONS HOMEWORK NO. 2 CE 3400 Introduction of Environmental Engineering This homework assignment was worth a total of 10 points. You received one point for every problem you attempted. You received one additional point if the solution was correct and all components of the problem were included (black box diagram, assumptions, equations, etc.) 1. A pickle-packing plant produces and discharges a waste brine solution with a salinity of 13,000 mg/L NaCl at a rate of 100 gal/min. The waste brine enters a stream with a flow rate above the brine discharge point of 1.2 mgd and a salinity of 20 mg/L. Below the discharge point is a prime sport fishing spot, and the fish are intolerant to salt concentrations greater than 200 mg/L. What must the effluent concentration of salt be to keep the level of salinity in the stream at or below 200 mg/L? Problem : Determine what the concentration of waste brine effluent (C E ) must be in order to meet the requirement of no more than 200 mg/L salinity downstream in the area of sport fishery. Assumptions : Steady state conditions, so Q U + Q E = Q D No production or consumption of water or salinity/TSS (so production/consumption terms in materials balance equation = 0) Known values : everything but C E and Q D Conversions required : convert mgd to gpm (Q U ) SOLUTION: Convert 1.2 mgd to gpm: 1 mgd = 694.444 gpm So 1.2 mgd = 833.3 gpm ~ 833 gpm Many of you converted gpm to mgd instead of mgd to gpm. Since there was one value in gpm and one value in mgd, you could go either way in your conversion. Also, some of you convert both units to metric. That is OK but not needed as the units did not affect the solution which was in mg/L. I have shown the solution to this problem using both conversion types. Calculate Q D : Since steady state is assumed, so Q U + Q E = Q D So Q D = 833 gpm + 100 gpm = 933 gpm Set up material balance: = ( Q E C E + Q U C U ) ( Q D C D ) + Rearranging : ( Q E C E + Q U C U ) = Q D C D So : C E = Q D C D Q U C U Q E = [(933 gpm )(200 mg / L )] [(833 gpm )(20 mg / L )] 100 gpm C E = 1699 mg/L, so effluent concentration from pickle factory must be 1700 mg/L SAME PROBLEM AGAIN BUT IN UNITS OF MGD INSTEAD OF GPM. A pickle-packing plant produces and discharges a waste brine solution with a salinity of 13,000 mg/L NaCl at a rate of 100 gal/min. The waste brine enters a stream with a flow rate above the brine discharge point of 1.2 mgd and a salinity of 20 mg/L. Below the discharge point is a prime sport fishing spot, and the fish are intolerant to salt concentrations greater than 200 mg/L. What must the effluent concentration of salt be to keep the level of salinity in the stream at or below 200...
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SolutionsHW#2 - SOLUTIONS HOMEWORK NO. 2 CE 3400...

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