Answer:
a) [tex]y(t)=0.65\frac{Kg}{min}(tmin)[/tex]
b) [tex]y(40)=26Kg[/tex]
Step-by-step explanation:
Data
Brine a (Ba)
[tex]V_{Ba}=5\frac{Lt}{min}\\ Concentration(Bca)=0.05\frac{Kg}{Lt}[/tex]
Brine b (Bb)
[tex]V_{Bb}=10\frac{Lt}{min}\\ Concentration(Bcb)=0.04\frac{Kg}{Lt}[/tex]
we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)
, then the amount of salt (y) left in the tank after (t) minutes: [tex]y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}[/tex]
Finally:
a) [tex]y(t)=0.65\frac{Kg}{min}(tmin)[/tex]
b) [tex]y(40)=0.65\frac{Kg}{min}(40min)=26Kg[/tex]
being y(t) the amount of salt (y) per unit of time (t)
