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 picklepacking 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:
0
=
(
Q
E
C
E
+
Q
U
C
U
)
−
(
Q
D
C
D
)
+
0
−
0
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
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SAME PROBLEM AGAIN BUT IN UNITS OF MGD INSTEAD OF GPM.
A picklepacking 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?
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 Spring '10
 Colberg
 Environmental Engineering, Trigraph, Sewage treatment, steady state conditions, lb/day

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