Differential Equation: Mixture s

Getting the Solution

24. A tank contains 80 gallons (gal) of pure water. A brine solution with 2 lb/gal of salt enters at 2 gal/min, and the well – stirred mixture leaves at the same rate. Find (a) the amount of salt in the tank at anytime and (b) the time at which the brine leaving will contain 1 lb/gal salt.

ans: (a) s=160[1-e^(-t/40)] ; (b) t=40 ln 2 min

25. For the tank in Exercise 24, determine the limiting value for the amount of salt in the tank after a long time. How much time must pass before the amount of salt in the tank reaches 80% of this limiting value?

ans: (a) s=160 lb ; (b) t= 64 min

(a)

s(t) = amount of salt (in lbs) at time t (in min)

Salt enters tank at a rate of: 2lb/gal * 2 gal/min = 4 lb/min
Salt exits tank at a rate of: s lb/80 gal * 2 gal/min = (s/40) lb/min

ds/dt = 4 − s/40
ds/dt = (160−s)/40
ds/(160−s) = 1/40 dt

Integrate both sides:
−ln|160−s| = t/40 + C₀
ln|160−s| = C₁ − t/40
160−s = C e^(−t/40)
s = 160 − C e^(−t/40)

Initially, tank contains pure water ----> s(0) = 0
0 = 160 − C e^0
C = 160

s(t) = 160 − 160 e^(−t/40)
s(t) = 160 (1 − e^(−t/40))

(b)

Concentration of water leaving at time t = concentration of water in tank at time t
s(t)/80 = 1
s(t) = 80
160 (1 − e^(−t/40)) = 80
1 − e^(−t/40) = 1/2
e^(−t/40) = 1/2
e^(t/40) = 2
t/40 = ln(2)
t = 40 ln(2)

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25.

(a)

Since concentration of salt entering tank is 2 lb/gal, then tank can never get above 2 lb/gal
This is the limiting value (of the concentration)
Limiting value of salt in tank = 2 lb/gal * 80 gal = 160 lb

Alternatively, just take lim[t→∞] s(t)
lim[t→∞] s(t) (160 (1 − e^(−t/40)))
= 160 (1 − 0)
= 160

(b)

s(t) = 80% of 160
160 (1 − e^(−t/40)) = 0.8 * 160
1 − e^(−t/40) = 0.8
e^(−t/40) 0.2
e^(t/40) = 1/0.2 = 5
t/40 = ln(5)
t = 40 ln(5) ≈ 64.377516497