2. English Brine Initially, 50 lb of salt is dissolved in a tank containing 300

Question

2. English Brine Initially, 50 lb of salt is dissolved in a tank containing 300 gal of water. A salt solution with 2 lb/gal he mixture, after stirring, flows from the tank at the same rate concentration is poured into the tank at 3 gal/min. T the brine is entering the tank. (a) Find the amount of salt in the tank as a function of time. (b) Determine the concentration of salt in the tank at any time. (c) Determine the steady-state amount of salt in the tank. (d) Find the steady-state concentration of salt in the tank. (e) Use a graphing calculator or computer to sketch the graphs of the future amount of salt in the tank and the concentration of salt in the tank.

Explanation / Answer

(a)
(accumulation, lb/min) = (rate in, lb/min) - (rate out, lb/min)
dQ/dt = 2*3 - (Q/300)*3 = 6 - Q/100
dQ/(6-Q/100) = dt
-100*ln(6-Q/100) = t + D
ln(6-Q/100) = -t/100 + C
6-Q/100 = B*exp(-t/100)
600-Q=A*exp(-t/100)
Q = 600-A*exp(-t/100)
initial condition ==> t=0, Q=50 ==> A=550
Q = 600-550*exp(-t/100) (lb)

B)
C = Q/300
C = (600-550*exp(-t/100))/300
C = 2-11/6*exp(-t/100) (lb/gal)

C)
steady-state ==> dQ/dt=0
dQ/dt = 6 - Q/100
0 = 6 - Q/100
Q = 600 lb

D)
steady-state ==> Q=600 (from part C)
C = Q/300
C = 600/300
C = 2 lb/gal......

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