Answer:
x(t) = 581.66 lb
Step-by-step explanation:
From the given information:
Consider the salt quantity in the tank at time t be x(t) lb.
∴ the rate of change of salt content in the tank is [tex]\dfrac{dx}{dt}= Rate \ of \ \ inflow - rate \ of \ \ outflow[/tex]
where:
the rate of inflow = salt conc. × flow rate = 3 lb/gallon × 7 gallons
=21 lb/s
rate of outflow = salt conc. in the tank × flow rate
= x/250 × 9
= 9x/250 lb/s
∴
[tex]\dfrac{dx}{dt} = 21 - \dfrac{9x}{250}[/tex]
[tex]\dfrac{dx}{dt} + \dfrac{9x}{250}= 21[/tex]
This is a 1st order linear differentiation,
The integrating factor if [tex]e^{^{ \int \dfrac{9}{250}\ dt}} = e^{^{ \dfrac{9t}{250} }}[/tex]
∴
[tex]e^{^{ \dfrac{9t }{250}}} \ \ x(t) = \int 21 e^{\dfrac{9t}{250}} \ dt + C[/tex]
[tex]e^{^{ \dfrac{9t }{250}}} \ \ x(t) = 21 \times \dfrac{250}{9}e^{\dfrac{9t}{250}} + C[/tex]
[tex]x(t) = \dfrac{1750}{3}+Ce^{^{\dfrac{-9t}{250}}}[/tex] at t = (0) and x(0) = 100 lb
Hence;
[tex]100 = \dfrac{1750}{3}+C[/tex]
[tex]C = \dfrac{1750}{3} -100[/tex]
[tex]C = - \dfrac{1450}{3}[/tex]
∴[tex]x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9t}{250}}}[/tex]
after time t = 1 minute i.e 60 s
[tex]x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9 \times 60}{250}}}[/tex]
[tex]x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-540}{250}}}[/tex]
x(t) = 581.66 lb
