Salt flows into the tank at a rate of
(1/2 lb/gal) * (6 gal/min) = 3 lb/min
and flows out at a rate of
(Q(t)/60 lb/gal) * (6 gal/min) = 6Q(t) lb/min
The net rate of change of the amount of salt in the tank at time [tex]t[/tex] is then governed by
[tex]\dfrac{\mathrm dQ}{\mathrm dt}=3-6Q[/tex]
Solve for [tex]Q[/tex]:
[tex]\dfrac{\mathrm dQ}{\mathrm dt}+6Q=3[/tex]
[tex]e^{6t}\dfrac{\mathrm dQ}{\mathrm dt}+6e^{6t}Q=3e^{6t}[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}[e^{6t}Q]=3e^{6t}[/tex]
[tex]e^{6t}Q=\dfrac{e^{6t}}2+C[/tex]
[tex]\implies Q=\dfrac12+Ce^{-6t}[/tex]
The tank starts with 10 lb of salt, so that Q(0) = 10. This gives us
[tex]10=\dfrac12+C\implies C=\dfrac{19}2[/tex]
so that the amount of salt in the tank at time [tex]t[/tex] is given by
[tex]\boxed{Q(t)=\dfrac{1+19e^{-6t}}2}[/tex]
The quantity of salt at time t is, [tex]m_{salt}=60(30-29.833e^{\dfrac{-t}{10}} )[/tex] where t is measured in minutes.
From the conservation of the mass
[tex]m_m-m_{out}=\dfrac{dm}{dt}[/tex]
Where mass flow is the product of salt concentration and water volume flow.
The model of the tank according to the statement is:
[tex]0.5\times 6-6c=V\dfrac{dc}{dt}[/tex]
Where:
[tex]c[/tex] - The salt concentration in the tank, as well at the exit of the tank, was measured.
[tex]\dfrac{dc}{dt}[/tex]- Concentration rate of change in the tank, measured in .
[tex]V[/tex] - Volume of the tank, measured in gallons.
The following first-order linear non-homogeneous differential equation is found:
[tex]V\dfrac{dc}{dt} +6c=3[/tex]
[tex]60\dfrac{dc}{dt} +6c=3[/tex]
[tex]\dfrac{dc}{dt} +\dfrac{1}{10} c=3[/tex]
This equation is solved as follows:
[tex]e^{\frac{t}{10} }(\dfrac{dc}{dt} +\dfrac{1}{10} c)=3e^{\frac{t}{10}[/tex]
[tex]\dfrac{d}{dt} (e^{\frac{t}{10} }c)=3e^\frac{t}{10} dt[/tex]
[tex]e^\frac{t}{10}c=3\int {e^{\frac{t}{10} } \, dt[/tex]
[tex]e^\frac{t}{10} c=30e^\frac{t}{10} +c[/tex]
[tex]c=30+ce^\frac{-t}{10}[/tex]
The initial concentration in the tank is:
[tex]c_0=\dfrac{10}{60} =0.167 \ \frac{lb}{gallon}[/tex]
[tex]0.167=30+c[/tex]
value of the constant will be
[tex]0.167=30+c[/tex]
[tex]c=-29.833[/tex]
The solution of the differential equation will be
[tex]c(t)=30-29.833e^{\frac{-t}{10}}[/tex]
Now, the quantity of salt at time t will be
[tex]m_{salt}=V_{tank}c(t)[/tex]
[tex]m_{saly}60(30-29.833e^{\frac{-t}{10}} )[/tex]
Thus the quantity of salt at time t is, [tex]m_{salt}=60(30-29.833e^{\dfrac{-t}{10}} )[/tex] where t is measured in minutes.
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