Question

A 250-gal tank is initially filled with brine (salt-water mixture) containing 100 lb of salt. Brine containing 3 lb of salt per gallon enters the tank at the rate of 7 gal/s, and the well-mixed brine in the tank flows out at the rate of 9 gal/s. Find the amount x(t) of salt in the tank after 1 minute.

Answer 1

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 $\dfrac{dx}{dt}= Rate \ of \ \ inflow - rate \ of \ \ outflow$

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

∴

$\dfrac{dx}{dt} = 21 - \dfrac{9x}{250}$

$\dfrac{dx}{dt} + \dfrac{9x}{250}= 21$

This is a 1st order linear differentiation,

The integrating factor if $e^{^{ \int \dfrac{9}{250}\ dt}} = e^{^{ \dfrac{9t}{250} }}$

∴

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = \int 21 e^{\dfrac{9t}{250}} \ dt + C$

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = 21 \times \dfrac{250}{9}e^{\dfrac{9t}{250}} + C$

$x(t) = \dfrac{1750}{3}+Ce^{^{\dfrac{-9t}{250}}}$     at  t = (0) and x(0) = 100 lb

Hence;

$100 = \dfrac{1750}{3}+C$

$C = \dfrac{1750}{3} -100$

$C = - \dfrac{1450}{3}$

∴$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9t}{250}}}$

after time t = 1 minute i.e 60 s

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9 \times 60}{250}}}$

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-540}{250}}}$

x(t) = 581.66 lb