Differential Equations

Question

A tank is filled with 1000 liters of pure water. Brine containing 0.05 kg of salt per liter enters the tank at 10 liters per minute. Another brine solution containing 0.07 kg of salt per liter enters the tank at 9 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 19 liters per minute.

A. Determine the differential equation which describes this system. Let S(t) denote the number of kg of salt in the tank after t minutes. Then dS/dt = ? .B. Solve the differential equation for S(t).

Answer 1

Clearly S(0) = 0 as the tank starts with pure water.

The first brine solution enters at 10 liters per minute , 0.05 kg salt per liter, so puts in 10 x 0.05 = 0.5 kg salt per minute

Second brine solution puts in 0.07 x 9 = 0.63 kg salt per minute (same argument)

After t minutes volume of water in the tank is 1000 + 10t + 9t - 19t = 1000 (constant)

At time t rate of outflow of salt will be (19/1000) S (t) per minute.

so dS/dt = 0.5 + 0.63 - 0.019 S

dS/dt = 1.13 - 0.019 S

dS / 1.13 - 0.019 S = dt

Integrate both sides

(1/-0.019) ln ¦ 1.13 - 0.019s ¦ = t + C

S(0) = 0 so C = ( 1/-0.019) ln (1.13)

So -52.63 ln ¦ 1.13 - 0.019 s¦ = t - 52.63

ln ¦ 1.13 - 0.019s ¦ = ( t / -52.63) + 1

if s > 59.5 then 1.13 - 0.019s < 0 so ¦ 1.13 - 0.019s¦ = 0.019s - 1.13

Then ln (0.019s - 1.13) = (t / -52.63) + 1

0.019s - 1.13 = e^{1 - (t/52.63)}

s = (1/0.019) [ 1.13 + e ^{1 - (t/52.63)}

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