Answer:
So the tank contains 1866.67 lbs of salt after 250 minutes.
Step-by-step explanation:
The tank salt concentration at any moment is given as
[tex]SC_{tank}=\frac{Q}{current \,amount\, of\, water\, solution} =\frac{Q}{400 + 2t}[/tex]
Now the rate of change of concentration is given as
[tex]\frac{dQ}{dt} = rate_{in} - rate_{out} =SC_{in} \times FR_{in} - SC_{tank} \times FR_{out}[/tex]
Here
SC_in is given as 2 lb/gal
FR_in is given as 3 gal/min
SC_tank is given in above equation
FR_out is given as 1 gal/min
Putting all in the equation gives
[tex]\frac{dQ}{dt} = SC_{in} \times FR_{in} - SC_{tank} \times FR_{out}\\\frac{dQ}{dt} =2 \times 3- \frac{Q}{400+2t}\times 1\\\frac{dQ}{dt} =6- \frac{Q}{400+2t}\times 1\\\frac{dQ}{dt} +\frac{Q}{400+2t}=6[/tex]
Multiplying by the integrating factor of [tex]\sqrt{400+2t}[/tex] on both sides
[tex](\sqrt{400+2t})\frac{dQ}{dt} +(\sqrt{400+2t})\frac{Q}{400+2t}=6(\sqrt{400+2t}\\(\sqrt{400+2t})\frac{dQ}{dt} +\frac{{Q}}{\sqrt{400+2t}}=6(\sqrt{400+2t})\\\frac{d}{dt}(Q(\sqrt{400+2t}))=6(\sqrt{400+2t})[/tex]
Integrating both sides with respect to t gives
[tex]Q\left(t\right)=\frac{\left(2\left(2t+400\right)^{\frac{3}{2}}+c_1\right)\sqrt{400+2t}}{400+2t}[/tex]
For the initial condition Q(0)=100 lbs so the equation is given as
[tex]100=\frac{(0+c_1)\sqrt{400}}{400+0}\\c_1=\frac{(400 \times 100)}{\sqrt{400}}=2000[/tex]
So the equation is
[tex]Q\left(t\right)=\frac{\left(2\left(2t+400\right)^{\frac{3}{2}}+2000\right)\sqrt{400+2t}}{400+2t}[/tex]
Solving for the t=250
[tex]Q\left(250\right)=\frac{\left(2\left(2(250)+400\right)^{\frac{3}{2}}+2000\right)\sqrt{400+2(250)}}{400+2(250)}\\Q\left(250\right)=1866.67 lbs[/tex]
