A 1000 liter tank initially contains 600 1 iters of water with 50 grams of conta

Question

A 1000 liter tank initially contains 600 1 iters of water with 50 grams of contaminate dissolved in it. Contaminated water with a concentration of 10 grams per liter flows into the tank at a rate of 5 liters per hour and a well-mixed solution flows out at a rate of 2 liters per hour. This continues until the amount of contaminate in the tank is 4000 grams or the tank overflows; whichever comes first. At that point in time, the concentration of the incoming contaminated water is decreased to 4 grams per liter (the flow rate remains the same) and the well-mixed solution flows out of the tank at a rate of 5 liters per hour. How much contaminate is in the tank 80 hours after the concentration inflow is changed?

Explanation / Answer

Let V be volume of water in tank at any time t,
Let C be concentration of contaminate in tank at any time t,
Volume of tank = Vmax = 1000 liter
Initial volume = Vi = 600 liter
Initial concentration = Ci = 50 / 600 = 0.08333 gm/lit
Concentration of In flow of water = 10 gm/lit
Rate of inflow of contaminated water = Fin = 5 liter/hour
Rate of out flow of contaminated water = Fout = 2 liter/hour
So now formulating the problem
doing mass balnce for water
so dV/dt = Fin - Fout
dV/dt = 5 - 2 = 3 lit/hr
so, V = 3t + constant
at time t=0 , V = Vi = 600
so, V = 3t + 600
Now doing component balance for contaminate
d(CV)/dt = Fin*Cin - Fout *Cout
C dV/dt + V*dC/dt = (5*10) - (3*C)
C*3 + (3t +600) *dC/dt= 50 -3C
dC/dt = (50 -6C)/(3t +600)
dC/(50 -6C)= dt/(3t +600)
Integrating
at time t=0 , C = Ci = 0.08333 gm/lit
so int (0.08333 to C) dC/(50 -6C) = int (0 to t) dt/(3t +600)
(-1/6) log ((25 -3C)/24.75) = (1/3) log ((t+200)/200)
so, C = {25 -(24.75*((t+200)/200)^-2)}/3
so, time required for volume to completely fill up
1000 = 3t + 600
t = 133.33 hours
At any time t , contaminate in tank = C*V
contaminate in tank = (3t + 600)*{25 -(24.75*((t+200)/200)^-2)}/3
so for contaminate in tank = 4000 gms
(3t + 600)*{25 -(24.75*((t+200)/200)^-2)}/3 =4000
t=94.476 for contaminate in tank = 4000 gms
so time required for contaminate in tank = 4000 gms is less than time required for tank to overflow.
so this continues untill time t= 94.476 hours
so now volume = 3t + 600 = (3*94.476)+600 = 883.428 liters
Concentration of contaminate in tank = 4000/883.428 = 4.5278 gm/lit
Now concentration of incoming water = Cin = 4 gm/lit
Inflow = Fin = 5 lit/hr
Out flow of water = 5 lit/hr
now formulating the problem again
doing mass balnce for water
so dV/dt = Fin - Fout
dV/dt = 5 - 5 = 0 lit/hr
so, V = constant
at time t=0 , V = 883.428
so, V = 883.428
Now doing component balance for contaminate
d(CV)/dt = Fin*Cin - Fout *Cout
C dV/dt + V*dC/dt = (5*4) - (5*C)
C*0 + (883.428) *dC/dt= 20 -5C
dC/dt = (20 -5C)/(883.428)
dC/(20 -5C)= dt/(883.428)
Integrating
at time t=0 , C = 4.5278 gm/lit
int ( 4.5278 to C) dC/(20 -5C) = int (0 to t) dt/(883.428)
(-1/5) log ((4-C) / -0.5278) = t/(883.428)
C = 4-(-0.5278*(10^(-0.00565977t )))
Now after time t= 80 hours
C = 4-(-0.5278*(10^(-0.00565977*80 )))
C = 4.186 gm/lit
Now total contaminate present in tank = C*V = 4.186 * 883.428 = 3698.029608 gm
How much contaminate is in the tank 80 hours after the concentration inflow is changed = 3698.03 gm is the answer

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