Answer to Question #253916 in Differential Equations for DEE

Question #253916

A 2000 L tank initially contains 40 kg of salt dissolved in 1000 L of water. A brine solution containing 0.02 kg/L of salt flows into the tank at a rate of 50 L/min. The solution is kept thoroughly mixed, and the mixture flows out at a rate of 25 L/min. (a) Find the quantity of salt in the tank at any time t > 0) prior to overflow. (b) Find the time of overflow.


1
Expert's answer
2021-11-08T16:05:54-0500

Solution;

(a)

Let S(t) be the amount of salt in the tank at any time t.

Then;

S(0)=0

Let V(t) be the volume in the tank at any time t

Then V(0)=1000

But ,net flow is 50L/min -25L/min.

Hence;

V(t)=1000+25tV(t)=1000+25t

Rate of salt going in;

dSdtin=0.02kgL×50Lmin\frac{dS}{dt}_{in}=\frac{0.02kg}{L}×\frac{50L}{min} =1kgmin1\frac{kg}{min}

Rate of salt going out;

dSdtout=25S1000+25t\frac{dS}{dt}_{out}=\frac{25S}{1000+25t}

Hence the net rate of salt in tank is;

dSdt=125S1000+40\frac{dS}{dt}=1-\frac{25S}{1000+40}

Rearrange as follows;

dSdt+251000+25tS=1\frac{dS}{dt}+\frac{25}{1000+25t}S=1

The integrating factor;

I.F=e251000+25tdtI.F=e^{\frac{25}{1000+25t}dt} =t+40t+40

Hence;

ddx(40+t)S=(40+t)×1\frac{d}{dx}(40+t)S=(40+t)×1

Integrating both sides;

S(40+t)=(40+t)dt=40t+t22+CS(40+t)=\int(40+t)dt=40t+\frac{t^2}{2}+C

Resolve as follows;

S(t)=t2+80t2(40+t)+C40+tS(t)=\frac{t^2+80t}{2(40+t)}+\frac{C}{40+t}

But ;

S(0)=40S(0)=40

Applying this initial conditions;

S(0)=40=0+C40S(0)=40=0+\frac{C}{40}

Hence ;

C=40×40=1600C=40×40=1600

Hence,the quantity of salt in the tank at t>0 prior to overflow is described as;

S(t)=t2+80t2(40+t)+160040+tS(t)=\frac{t^2+80t}{2(40+t)}+\frac{1600}{40+t}

(b)

Time for overflow;

V(t)=1000+25tV(t)=1000+25t

At time of overflow,the tank is full hence;

V(t)=2000LV(t)=2000L

2000=1000+25t2000=1000+25t

25t=100025t=1000

t=40t=40 minutes.


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