Linear Differential Problem(Pollutants into a lake)

[paraboloid]

Hi,

This is my problem:

Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant,
evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/V .
Assume that water containing a concentration k of pollutant enters the lake at a rate r ,
and that water leaves the lake at the same rate. Suppose that pollutants are also added
directly to the lake at a constant rate P.

If at time t = 0 the concentration of pollutant is c0 , find an expression for the concen-
tration c(t) at any time. What is the limiting concentration as t → ∞?


I have no clue how to solve this notably because pollutants are flowing in and being added. I've been at it for an hour and half with nothing to show so I thought I'd turn to the forums.

Any help would be great,
Thanks

 

[lanedance]

you nee dto write down a differential equation then attempt to solve

so you know:
- initial concentration
- the incoming rate
- the outogoing rate

so make the assumption everything in the lake is perfectly mixed, so will be the same as the outgoing concentration & try to write down the rate of chaneg of contaminant in the lake

it may be easier to do in terms of total volume of contaimnant, though is the same

 

[Susanne217]

Hi,

This is my problem:

Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant,
evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/V .
Assume that water containing a concentration k of pollutant enters the lake at a rate r ,
and that water leaves the lake at the same rate. Suppose that pollutants are also added
directly to the lake at a constant rate P.

If at time t = 0 the concentration of pollutant is c0 , find an expression for the concen-
tration c(t) at any time. What is the limiting concentration as t → ∞?


I have no clue how to solve this notably because pollutants are flowing in and being added. I've been at it for an hour and half with nothing to show so I thought I'd turn to the forums.

Any help would be great,
Thanks

So we view this this \frac{dc}{dt} = \frac{Q}{V} with the intial condition

c(0) = c_0

and isn't this equal to

dc = \frac{Q}{V} dt and integrating on both sides

c(t) = (\frac{Q}{V})\cdot t + k and inserting the intial condition is that then

k = c_0 giving the result c(t) = (\frac{Q}{V})\cdot t + c_0

 

[lanedance]

think about what that solution is telling you, c(t) increases without out bound, which doesn't make a heap of sense...

its probably easier to write in terms of the contaminant in the lake, then solve for concentration later.

So at time t=0, assume we know the pollutant in the lake is
Q(0) = Q_0

the rate of change of the total pollutant can be broken into 2 contributions, one for the incoming pollutant and one for the outgoing:
\frac{dQ(t)}{dt} = (\frac{dQ(t)}{dt})_{in} + (\frac{dQt)}{dt})_{out}

see if you can write down what they are in terms of k, r and c(t) or Q(t)
(\frac{dQ(t)}{dt})_{in}
(\frac{dQt)}{dt})_{out}=?

that will give you the differential equation you need to solve for Q and c will follow

 

[lanedance]

So we view this this \frac{dc}{dt} = \frac{Q}{V} with the intial condition

just to point out where this went wrong you need to start at
c(t) = \frac{Q(t)}{V(t)}

then differentiating w.r.t. t gives
c'(t) = \frac{dc(t)}{dt} = \frac{d}{dt}(\frac{Q(t)}{V(t)})

now is we assume V(t) = V is constant, which we can as the same amount of water is going in as out, it becomes:
c'(t) = \frac{d}{dt}(\frac{Q(t)}{V}) = \frac{dQ(t)}{dt}\frac{1}{V} = \frac{Q'(t)}{V}

then Q(t) isn't constant, but is a function of t, so we can't simply integrate