How Do You Calculate Pollutant Concentration in Outflow?

[Myung]

Homework Statement

Hi I need help on my Environmental Engineering subject!

Here is the problem:

An industry, on the banks of a lake, discharges a pollutant into the lake. The river flow is 10 m3/s, the waste flow is 0.1 m3/s, the pollutant’s concentration in the river is 20 mg/l and in the waste flow it is 3000 mg/l.

What is the flow rate out and what is the concentration of the pollutant in the outflow?



HERE is the Graph!
http://img263.imageshack.us/img263/1844/screenshot20110809at511.png

Homework Equations

*The effectiveness of an environmental process in removing a contaminant can be determined using the mass balance technique:

d/dt (M) = d/dt(in) - d/dt(out)

*The mass of contaminant per unit of time may be calculated as :

mass/time = (concentration)(flowrate)

*The mass flow rate is given by the equation:

d/dt M = CinQin - CoutQout

The Attempt at a Solution

My attempt was to get the flow rate out first with simply adding Q(in flow) and Q(wasteflow)

10 m3/s + 0.1 m3/s = Q(out flow)
= 10.1 m3/s

Now I'm confused with the concentration part I don't know how to apply the said equation above to this problem

 

[KataKoniK]

.
Hi there! It's great that you're seeking help for your Environmental Engineering subject. Let's take a look at the problem and see if we can solve it together.

First, let's define some variables:
Qin = River flow (10 m3/s)
Qwaste = Waste flow (0.1 m3/s)
Cin = Concentration of pollutant in river (20 mg/l)
Cwaste = Concentration of pollutant in waste flow (3000 mg/l)
Qout = Flow rate out
Cout = Concentration of pollutant in outflow

Now, let's use the mass balance technique to solve for the flow rate out (Qout). We know that the mass of contaminant per unit of time is equal to the concentration multiplied by the flow rate. So, we can set up the following equation:

d/dt M = d/dt(in) - d/dt(out)

Where M is the mass of contaminant, and d/dt(in) and d/dt(out) represent the change in mass over time for the inflow and outflow, respectively.

Next, we can use the equation for mass flow rate (d/dt M = CinQin - CoutQout) and substitute in the values we know:

d/dt(in) - d/dt(out) = CinQin - CoutQout

We know that d/dt(in) = CinQin (since there is no change in concentration or flow rate in the river), so we can substitute that in:

CinQin - d/dt(out) = CinQin - CoutQout

Now, we can solve for d/dt(out):

d/dt(out) = CoutQout

We also know that the total flow rate out (Qout) is equal to the sum of the river flow and waste flow, so we can substitute that in as well:

d/dt(out) = Cout(Qin + Qwaste)

Substituting in the values we know, we get:

d/dt(out) = Cout(10 m3/s + 0.1 m3/s)

Now, we can solve for Cout by rearranging the equation:

Cout = d/dt(out) / (10.1 m3/s)

We can also use the equation for mass flow rate to solve for Cout:

d/dt M = CinQin - CoutQout
3000 mg/s x