Consider the figure below, where two ponds are connected and fed by a single str

Question

Consider the figure below, where two ponds are connected and fed by a single stream flowing through them. Pond A holds 500,000 gal of water, while Pond 13 holds 200,000 gal of water. The fresh water (0 lb of toxin per gal) stream flows through these ponds at a rate of 1000 gal/hr. Assume at some time, say t = 0, that 1000 lb of a toxin is spilled into Pond A and that it disperses rapidly enough so that a well-stirred assumption is reasonable (i.e. we will assume that the distribution of toxin in the lake is always uniform). Write and solve a DE to determine the amount x(t) of toxin in pond A at time t. Write and solve a DE to determine the amount y(t) of toxin in pond B at time t. What is the maximum Amount of toxin ever present in Pond B?

Explanation / Answer

Note- Pond A and B has been referred as Pond 1 and 2 in below solution.

The rate of change of the toxin mass in each pond equals mass flow in minus mass flow out. The mass flow of toxin associated with stream equals volumetric flow rate times concentration:

Mi = FCi
where F is the constant flow rate of 1000 gal/hr and Ci(t) the concentration in the pond
Well stirred assumption means the concentration at every instant is the same everywhere in the pond, so the outflowing stream has a concentration of
Ci = Qi/Vi
(Note that volume Vi in the pond is constant)
So the outflow of pond i caries a toxin mass stream of
Mi = (F/Vi)Qi
The first pond gains no toxin from the fresh water input but looses toxin by outflowing water. So:
dQ/dt = - (F/V)Q
The second pond gets the drain from first pond as input, so
dQ/dt = (F/V)Q - (F/V)Q
with
k = (F/V) = 1000 gal/hr / 500,000 gal = 0.002 hr¹
k = (F/V) = 1000 gal/hr / 200,000 gal = 0.005 hr¹
You get the system of differential equations:
dQ/dt = - kQ
dQ/dt = kQ - kQ

Initially pond 1 contain 1000 lb of toxin, i.e.
Q(0) = Q = 1000
while pond 2 is full of pure water
Q(0) = 0

(b)
First equation can be solved by separation of variables [1]
dQ/dt = - kQ
=>
(1/Q) dQ = - k dt
=>
(1/Q) dQ = - kdt
=>
ln(Q) = - kt + C
C is the constant of integration.
When you apply the initial condition for Q you find
ln(Q) = - k0 + C
<=>
C = ln(Q)
=>
ln(Q) = ln(Q) - kt
<=>
Q(t) = e^( ln(Q) - kt) = e^(ln(Q)) e^( -kt) = Q e^( -kt)

When substitute this to the second equation you get an ODE for Q, which can be solved by integrating factor method [2]
dQ/dt = kQ e^( -kt) - kQ
=>
(dQ/dt) + kQ = kQ e^( -kt)
=>
v(t) = k dt = kt
=>
Q(t) = e^(-v(t)) e^(v(t)) kQ e^( -kt) dt
= e^(-kt) e^(kt) kQ e^( -kt) dt
= e^(-kt) kQe^( (k -k)t) dt
= e^(-kt) [ (k/(k -k)))Qe^((k -k)t) + C) ]
As for Q(t) the constant of integration can be found by applying the initial condition
0 = e^(-k0) [ (k/(k -k)))Qe^((k -k)0) + C) ]
=>
0 = (k/(k -k)))Q + C
=>
C = -(k/(k -k)))Q
=>
Q(t) = e^(-kt) [ (k/(k -k)))Qe^((k -k)t) - (k/(k -k)))Q) ]
Q(t) = (k/(k -k)))Q[e^(-kt) e^((k -k)t) - e^(-kt) ]
Q(t) = (k/(k -k)))Q( e^( -kt) - e^(-kt) )
With numerical values:
Q(t) = 1000 e^( -0.002t)
Q(t) = 666.7( e^( -0.002t) - e^(-0.005t) )
where Q and Q in lb and t in hr

(c)
At a maximum the first derivative of Q w.r.t. to time is zero:
dQ/dt = 0
=>
(k/(k -k)))Q( -ke^( -kt) + ke^(-kt) ) = 0
<=>
-ke^( -kt) + ke^(-kt) = 0
<=>
ke^( -kt) = ke^(-kt)
<=>
e^( -kt)/e^(-kt) = k/k
<=>
e^( (k - k)t ) = k/k
=>
t = ln(k/k) / (k - k)
= ln(0.005/0.002) / (0.005 hr¹ - 0.002 hr¹ )
= 305.43 hr

thus Q2(305.43) = 506.71 lb

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