A 25-gallon storage tank contains 15 gallons of water in which 6 pounds of calci

Question

A 25-gallon storage tank contains 15 gallons of water in which 6 pounds of calcium are
dissolved initially. Suppose calcium water containing 2 pounds of calcium per gallon
is pumped into the top of the storage tank at the rate of 4 gallons per minute, while
a well-mixed solution leaves the bottom of the storage tank at a rate of 2 gallons per
minute.
a) Write and solve a di?erential equation for the amount of calcium in the storage
tank as a function of time. Note that you must explain how you found the ODE, and
provide the details of how you solved it to get full credit.
b) Determine when the storage tank will be full.
c) How much calcium is in the storage tank when the storage tank is full?

Explanation / Answer

The capacity of the tank is 25 gal. Let's call the amount of water V and the amount of calcium m.

V(0) = 15
m(0) = 6
dVin/dt = 4 gal/min
dVout/dt = 2 gal/min
dmin/dt = 2*4 = 8 lbs/min

Now, we need to find dmout/dt. To do this, we will multiply the rate of water leaving, dVout/dt, by the density of water, m(t)/V(t). First, lets find an expression for V(t)

dV/dt = dVin/dt - dVout/dt
dV/dt = 4 - 2 = 2
dV = 2 dt
dV = 2 dt
V(t) = 2t + C
V(0) = C = 15
V(t) = 2t + 15
Note that the tank will fill when the capacity reaches 25 gal. So the tank will be full when when t = 5. So now we can find an expression for dmout/dt:
dmout/dt = (m(t)/V(t))*dVout/dt
dmout/dt = [m(t)/(2t+15)]*2 = 2m/(2t+15)

Our total expression for the change in calcium is:
dm/dt = dmin/dt - dmout/dt
dm/dt = 8 - 2m/(2t+15) This is our ODE
m' + 2m/(2t+15) = 8
m' + m/(t + 15/2) = 8

Let's solve for the homogenous solution:
m' + m/(t + 15/2) = 0
dm/dt = -m/(t+15/2)
1/m dm = -1/(t+15/2) dt
1/m dm = -1/(t+15/2) dt
ln(m) = -ln(t+15/2) + C
Realize that the coefficient of the logarithm becomes an exponent of its argument. Then take the exponential of both sides:
m = exp(ln((t+15/2)-1 + C)
m = exp(ln((t+15/2)-1)*exp(C)
exp(C) gets absorbed into C
m = C/(t+15/2)

Now let's find the particular solution:
m' + m/(t + 15/2) = 8
Since we want the left side to equal a constant, lets pick m(t) = A*(t+15/2). It will cancel out the denominator on the m term and its derivative will be constant.
A + A = 8
2A = 8
A = 4
So, the particular solution is 4(t+15/2) = 4t + 30

Our total solution is:
m(t) = 4t + 30 + C/(t + 15/2)

Invoke initial conditions: m(0) = 6
m(0) = 6 = 30 + C/(15/2)
-24 = 2C/15
C = -180
The total solution is then:
m(t) = 4t + 30 - 180/(t + 15/2)
or
m(t) = 4t + 30 - 360/(2t+15)

Now we want to know how much calcium, m(t) will be in the tank when it's full (t = 5)
m(5) = 20 + 30 - 360/(10+15)
m(5) = 50 - 360/25
m(5) = 50 - 14.4
m(5) = 35.6 lbs.

Here are the answers in the order requested:
(a) m(t) = 4t + 30 - 180/(t + 15/2) or
m(t) = 4t + 30 - 360/(2t+15)

(b) 5 minutes

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