Math , i dont know how to solve it?

Question

In an oil refinery, a storage tank contains 2000 gal of gasoline that initially has 100lb of an additive dissolved in it, In preparation for winter weather, gasoline containing 2lb of additive per gallon is pumped into the tank at a rate of 40gal/min, the well mixed solution is pumped out at a rate of 45gal/min. how much of the additive is in the tank 20 minutes after the pumping process begins?

Answer 1

Classical first order Differential Equation:
It was slightly confusing to decide what our variable(s) should be (volume/mass of additive/or concentration)?
Really there's only one variable here, we just have to decide which to take that make it easiest:
initial volume V(0) = 2000gal
initial mass M(0) =100lb
and concentration C(t)=M(t)/V(t)

between time t and t+1 (might as well measure t in minutes):
V(t+1) = V(t) +40-45 = V(t)-5 = 2000 - 5t
M(t+1) = M(t) - 45*C(t) + 40*2
= M(t) - 45*M(t)/V(t) + 40*2
= M(t) - M(t)*(45/(2000-5t)) + 80

Then dM/dt = 80 - M(t)*(45/(2000-5t))

So try M(t) = Ae^kt/(2000-5t),
dM/dt = Ae^kt[ k/(2000-5t) + 1/(-5)]
dM/dt = Ae^kt[ k/(2000-5t) + -1/5 ]
dM/dt = Ae^kt[ (-2000+5t + k)/(2000-5t) ]

Answer 2

The rate of adding additive to the tank is 80 lb/min (2lb/gal at 40 gal/min).
The total volume in the tank at time t minutes is 2000 - 5t gallons, since it starts off with 2000 and 5 gal/min more is pumped out than in.
If x(t) is the amount of additive at time t (with x(0) = 100), then the rate at which it is being pumped out is x(t).45/(2000 - 5t), since the concentration of additive in the tank is x(t) / (2000 - 5t) lb/gal (it's "well-mixed" according to the question, so we can assume a uniform concentration of additive).
So the net rate of increase of the additive is
dx/dt = 80 - 45x / (2000-5t).
We can rewrite this as
dx/dt + (45/(2000-5t)) x = 80
which is a first-order linear ODE. So we compute the integrating factor:
∫45/(2000-5t) dt = 45 ln (2000-5t) / (-5) = -9 ln (2000 - 5t)
so the integrating factor is e^(-9 ln (2000-5t)) = (2000-5t)^(-9).
This gives us
d/dt ((2000-5t)^(-9). x) = 80 (2000-5t)^(-9)
=> (2000-5t)^(-9) x = ∫80 (2000-5t)^(-9)
= 80 (2000-5t)^(-8) / ((-8).(-5)) + c
= 2 (2000-5t)^(-8) + c
So x = 2(2000-5t) + c(2000-5t)^9
= 4000 - 10t + c (2000-5t)^9
x(0) = 100, so 4000 + c(2000)^9 = 100
=> c = -3900/(2000^9).
So x(t) = 4000 - 10t - 3900 (1 - t/400)^9.
Thus the required answer is x(20) = 4000 - 200 - 3900 (1-1/20)^9 = 1342 lb. to the nearest pound.

Answer 3

If there are 2000 gallons and 100 lbs of additive in the tank at the start of pumping then there are 40 gallons/minute of 2 lb per gallon going in for 20 minutes we now have 40 x 20 or 800 gallons of gasoline and 800 x 2 =1600 lbs of additive added to the tank.

Assuming there is no pumping out in that 20 minutes the tank now has 2,800 gallons of fuel and (100 + 1600) = 1700 lbs of additive.

The gasoline now in the tank has 2,800 gallons with 1700 lbs of additive. This is 1700 lbs/ 2800 gallons =0.61 lbs/gallon.

After 20 minutes of pumping out at 45 gallons/minute we have taken out 900 gallons of gasoline and 900(0.61l) = 549 lbs of additive .

We have left in the tank 1900 gallons at 0.61 lbs/gallon of additive

1900 x 0.61 =1159 lbs of additive left