A cistern (water tank) contains 200 litres of pure water. Brine (salty water) that contains 0.02 kg/L of salt is added at a rate of 3 L/min. Brine from a second source with 0.04 kg/L of salt is added at a rate of 1 L/min. Assume that the cistern is instantaneously well-mixed. The cistern is drained at a rate
of 4 L/min. If Q(t) is the amount of salt (in kg) at time t (in min), how much salt is in the cistern after t minutes?
Originally Posted by calculuskid1
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CB
In our problem, the level of water remains the same, as we haveand
flowing in and
flowing out.
If we call the salt level, then we have
To find, we subtract the rate of salt flowing out from the rate flowing in:
The rate flowing in is
Since the fluid is well-mixed, the rate flowing out will be in proportion toas the current salt level is to the entire resevoir. Therefore,
For
Now, all that remains is to solve the differential equation using the initial conditions given.
from the above, we get the equation : ds/dt = 0.1 - s/50
thus, ds/dt + 1/50 s = 0.1 ==> this is linear first degree equation .....(1)
find the integration factor, miu(t) = exp(integration of 1/50 dt)
= exp(1/50 t)
miu(t) x (1) : exp(1/50 t) x (ds/dt + 1/50 s) = 0.1 exp(1/50 t)
d/dt[exp(1/50 t).s] = 0.1 exp(1/50 t)
d[exp(1/50 t).s] = integration of [0.1 exp(1/50 t)] dt
= 0.1 integration of [exp(1/50 t)] dt
exp(1/50 t).s = 0.1/(1/50) exp(1/50 t) + c
thus s(t) = 5 + c exp-(1/50 t) ...(2)
applying initial condition : s(0) = 0 kg in (2) :
0 = 5 + c exp-(1/50 (0))
c= -5 in (2)
thus, the solution : s(t) = 5 - 5 exp-(1/50 t)
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