A container holds 500L of brine. A brine solution enters the tank at a rate of 2

Question

A container holds 500L of brine. A brine solution enters the tank at a rate of 2 liters per minute and exits at the same rate. The concentration of the brine entering the container is 0.2kg/l and there is 30 kg of salt initially. After 10 minutes, the brine solution entering is shut off and another valve is opened simultaneously and a new brine solution, which is double the concentration of the old solution, enters the tank. Use Laplace transformations to find the amount of salt in the tank, in kilometers at any time t. Then use this solution to find the amount of salt after 20 minutes.

I'm not sure how to work with the heaviside equation at the very end. This is what i've got so far.

a(0)=30kg

da/dt = 0.4kg/min , 0<=t<10

0.8kg/min , t>=10

da/dt +a/250 = 0.4+0.4U(t-10)

Laplace{da/dt + 1/250*a} = Laplace{0.4+0.4U(t-10)}

SA(s) + a(0) + 1/250 A(s) = 0.4/s + (0.4(e^-10s))/s

A(s) [S+1/250] = 0.4/s + (0.4(e^-10s))/s + 30

A(s) = 0.4/ s(s+1/250) + (0.4(e^-10s))/s(s+1/250) + 30/(s+1/250)

Inverse Laplace to all of that =

A(t) = 0.4 + 0.4U(t-10)[250-250e^-(t/250)] + 30e^-t/250

Here's where i'm stuck.

does A(20) = 0.4 + 0.4U(20-10)[250-250e^-(20/250)] + 30e^-20/250

A(20) = 104.977g of salt in the solution?

Explanation / Answer

You got the solution as :

A(t) = 0.4 + 0.4U(t-10)[250-250e^-(t/250)] + 30e^-t/250

its simple because

so for t >10 ; U(t-10) is equal to 1

so for t = 20

A(20) = 0.4 + 0.4*[250-250e^-(20/250)] + 30e^-20/250

A(20) = 35.7818 kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.