Question

A 400 gallon tank initially contains 300 gallons of water containing 0.8 lb of salt. Suppose water containing 0.6 lb of salt per gallon flows into the top of the tank at a rate of 2 gallons per minute. The water in the tank is kept well-mixed, and 7 gallons per minute are removed from the bottom of the tank. a. Write an initial value problem for S(t), the amount of salt in the tank at time t. Do not solve the equation!! b. Write an expression in terms of S for the amount of salt in the tank when the tank is one-quarter full. Show all your work on the paper you scan and submit. You do not need to type anything in the box below.

Answer 1

a. dS(t)/dt + [7S(t)/(300 - 5t)] = 1.2 with S(0) = 0.8

b. S(t) =
`\frac{120e^{(7t)/(100) } }{7}` - 16.34
`e^{-(7t)/(100)`

Explanation:

a. Write an initial value problem for S(t), the amount of salt in the tank at time t. Do not solve the equation!!

The net mass flow rate of salt dS(t)/dt = mass flow in - mass flow out

mass flow in = concentration of flow in × flow rate in = 0.6 lb/gal × 2 gal/min = 1.2 lb/min

Let S(t) be the mass of salt in the tank at time, t. The concentration of salt in the tank at time, t, C(t) = S(t)/volume of tank at time t ,V

V = V₀ + (r - r')t where V₀ = initial volume of water in tank = 300 gal, r = volume flow rate in = 2 gal/min, r'= volume flow rate out = 7 gal/min and t = time

So, V = V₀ + (r - r')t

V = 300 + (2 - 7)t

V = 300 - 5t

So, C(t) = S(t)/V = S(t)/(300 - 5t) lb/gal

Now, mass flow rate out = concentration in tank × volume flow rate out = S(t)/(300 - 5t) lb/gal × 7 gal/min = 7S(t)/(300 - 5t) lb/min

So, dS(t)/dt = mass flow in - mass flow out

dS(t)/dt = 1.2 lb/min - 7S(t)/(300 - 5t) lb

dS(t)/dt = 1.2 - 7S(t)/(300 - 5t)

So, our differential equation is

dS(t)/dt + [7S(t)/(300 - 5t)] = 1.2 with S(0) = 0.8 since we initially have 0.8 lb of salt

b. Write an expression in terms of S for the amount of salt in the tank when the tank is one-quarter full.

When the tank is one-quarter full, V = volume of tank/4 = 400 gallons/4 = 100 gallons

We now solve our differential equation

dS(t)/dt + 7S(t)/100 = 1.2

Using the integrating factor method, I.F =
`e^{\int\limits{(7)/(100) \, dt } =`
`e^{(7t)/(100) }`

`e^{(7t)/(100)`dS(t)/dt +

d[
`e^{(7t)/(100)`S(t)]/dt = 1.2

d[
`e^{(7t)/(100)`S(t)] = [1.2

integrating both sides, we have

∫d[
`e^{(7t)/(100)`S(t)] = ∫[1.2

Let u = 7t/100. So, du/dt = 7/100 and dt = 100du/7

∫[1.2
`e^{(7t)/(100)`]dt = ∫[1.2 × 100
`e^(u)`]du/7 = (120/7)∫

`e^{(7t)/(100)`S(t) =
`\frac{120e^{(7t)/(100) } }{7}` + C

S(t) =
`\frac{120e^{(7t)/(100) } }{7}` ÷
`e^{(7t)/(100)` +
`e^{-(7t)/(100)`

S(t) =
`\frac{120e^{(7t)/(100) } }{7}` + C
`e^{-(7t)/(100)`

From our initial condition, S(0) = 0.8. So,

S(0) =
`\frac{120e^{(7(0))/(100) } }{7}` + C
`e^{-(7(0))/(100)`

S(0) =
`\frac{120e^{(0)/(100) } }{7}` + C
`e^{-(0)/(100)`

S(0) =
`(120e^(0))/(7)` + C
`e^(0)`

0.8 =
`(120)/(7)` + C

C = 0.8 - 120/7

C = 0.8 - 17.14

C = - 16.34

So,

S(t) =
`\frac{120e^{(7t)/(100) } }{7}` - 16.34
`e^{-(7t)/(100)`