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?
In our problem, the level of water remains the same, as we have and 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 to as the current salt level is to the entire resevoir. Therefore,
For , we obtain
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)