Differential equations, tank problem: Determine a DE for the amount of salt A(t)
ID: 1720461 • Letter: D
Question
Differential equations, tank problem:
Determine a DE for the amount of salt A(t) in a tank at time t>0
The tank initially holds 100 gallons
50 pounds of salt are dissolved in the initiall 100 pounds
inflow rate = 3 gal/min with 2 lbs/gal of salt thoroughly mixed in it
outflow rate = 3.5 gal/min
DA/dt = Rin - Rout
Rin = 6 lbs/min
dA/dt = 6 - Rout
Rout = 3.5 gal/min * A/(300-(.5)t)
DA/dt = 6 - 2A/(600-t)? is this correct? it is a chegg question/answer: ( https://www.chegg.com/homework-help/differential-equations-with-boundary-value-problems-8th-edition-chapter-1.3-problem-11e-solution-9781111827069 )
how is the original 50 pounds of salt dissolved in the 300 gallon tank/mixed accounted for in this equation?
thank you
Explanation / Answer
First of all originally 50 pounds were dissolved in 100 gallons of water( not 300 gallons)
and this information will be used as a boundary condition while solving the differential equation
let me solve the question for you
I am going to use S instead of A for quantity of salt
Rin = 3 gal/min* 2 lbs / gal
R out = 3.5 gal / min * S/ 100 lbs/gal
dS/dt= Rin - Rout
= 6 - (3.5 S/ 100)= 6- 7S/200
dS/ ( 6 - 7S/200) = dt
integrating both sides we get [ we take 6 - 7s/200 = z => - ds= 200/7 dz]
-200/7* ln(6 - 7S/200) = t + c
now applying boundary condition
when t= 0, s=50
-200/7*ln(6 - 350/200) = 0 + c
=> c = -200/7*ln(6 - 350/200)
therefore the final equation is -200/7*ln(6 - 7S/200) = t -200/7* ln(6 - 350/200)
=> ln(6 - 7S/200) = - 7 t / 200 + ln(6 - 350/200)
=> 6 - 7 S / 200 = e- 7 t / 200 + ln(6 - 350/200)
=> 6 - 7 S / 200 = e- 7 t / 200 * eln(6 - 350/200)
=> 6 - 7 S / 200 = e- 7 t / 200 * (6 - 350/200)
=> 7 S / 200 = 6 - e- 7 t / 200 * (6 - 350/200)
=> S = 200( 6 - e- 7 t / 200 * (6 - 350/200)) / 7
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