Respuesta :
Answer:
a.i [tex]\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[/tex]
ii. time at which vat will be empty
[tex]T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}[/tex] when dQ/dt = 0
b. i Amount of salt at time t
Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]
ii. Concentration at t = T
c(T) = [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ + (rin - rout)T])
Step-by-step explanation:
a.
i. We determine the differential equation for the solution in the vat at time,t.
Let Q be the quantity of salt in the vat at any time,t.
So, dQ/dt = rate of change of quantity of salt in the vat. = rate of change of quantity of salt into the vat - rate of change of quantity of salt out of the vat.
Now, since a concentration of salt, k enter the vat at a rate of rin, rate of change of quantity of salt into the vat = krin.
Let V(0) = V₀ be the volume of water int the vat at time t = 0. Now, the volume of water in the vat increases at a rate of (rin - rout). The increase in volume after a time t is (rin - rout)t. So the volume after a time, t is V(t) = V₀ + (rin - rout)t. The concentration of this liquid is thus Q(t)/V(t) = Q(t)/V₀ + (rin - rout)t. Now, the rate of change of quantity of salt out of the vat is thus [Q(t)/V₀ + (rin - rout)t]rin = Q(t)rout/[V₀ + (rin - rout)t].
So, dQ(t)/dt = krin - Q(t)rout/[V₀ + (rin - rout)t].
[tex]\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[/tex]
ii Time T at which vat will be empty
At time T, when the vat is empty, dQ(T)/dt = 0.
So
[tex]\\\frac{dQ(T)}{dt} + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\0 + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}[/tex]
b. i
i. We solve the differential equation to find the amount of salt at time,t
The integrating factor is ex
[tex]exp(\int\limits {\frac{r_{out} }{V_{0} + (r_{in} - r_{out})t} } \, dt ) = exp(\frac{r_{out}}{(r_{in} - r_{out})} \int\limits {\frac{ (r_{in} - r_{out})}{V_{0} + (r_{in} - r_{out})t} } \, dt )\\= exp(\frac{r_{out}}{(r_{in} - r_{out})} ln [V_{0} + (r_{in} - r_{out})t} ]) \\\= [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})[/tex]
Multiplying both side of the equation by the integrating factor, we have
[tex][V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{dQ(t)}{dt} + [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})[/tex]dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt = krin[V(0) + (rin - rout)texp(rout/(rin - rout))]
Integrating both sides we have
∫(dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt)dt = ∫(krin[V(0) + (rin - rout)texp(rout/(rin - rout))])dt
Let exp(rout/(rin - rout) = A
Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + C
At t = 0, Q(t) = Q(0),
So,
Q(0)[V(0) + A(rin - rout)×0] = AkrinV(0)t + A(rin - rout)×0²/2 + C
C = Q(0)V(0)
Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + Q(0)V(0)
Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]
The above is the amount of salt at time ,t.
ii. The concentration of the last drop of salt at time, t = T
To find the concentration of the last drop of salt at time t = T, we insert T into Q(t) to find its quantity and insert t = T into V(t) = V₀ + (rin - rout)t.
So
Q(T) = [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]
Q(T) = [[exp(rout/(rin - rout)]krin[V(0)T + [exp(rout/(rin - rout)](rin - rout)T²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)T]
the volume at t = T is
V(T) = V₀ + (rin - rout)T.
The concentration at t = T is c(T) = Q(T)/V(T) = [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]/V₀ + (rin - rout)T.
= [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/([V(0) + A(rin - rout)T][V₀ + (rin - rout)T])
= [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ + (rin - rout)T])