Question

a) Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds ofsalt have been dissolved. Pure water is pumped into the tank at a rate of 3gal/min, and whenthe solution is well stirred, it is then pumped out at the same rate. Determine a differentialequation for the amount of salt, , in the tank at time .

Answer 1

Answer:

x  =  50*e∧ -t/100

Step-by-step explanation:

We assume:

1.-That the volume of mixing is always constant 300 gallons

2.-The mixing is instantaneous

Δ(x)t   =  Amount in  - Amount out

Amount  =  rate * concentration*Δt

Amount in  =  3 gallons/ min * 0  =  0

Amount out  = 3 gallons/min *  x/ 300*Δt

Then

Δ(x)t/Δt  =  - 3*x/300    Δt⇒0   lim Δ(x)t/Δt  =  dx/dt

dx/dt  =  - x/100

dx/ x  =  - dt/100

A linear first degree differential equation

∫ dx/x   =  ∫ - dt/100

Ln x  =  - t/100  +  C

initial conditions to determine C

t= 0     x =  50 pounds

Ln (50) = 0/100 * C

C =  ln (50)

Then final solution is:

Ln x  =  - t/100  + Ln(50)   or

e∧ Lnx   =  e ∧ ( -t/100 + Ln(50))

x  =  e∧ ( -t/100) * e∧Ln(50)

x  = e∧ ( -t/100) * 50

x  =  50*e∧ -t/100