A tank contains 100 kg of salt and 1000 L of water. A solution of a concentratio

Question

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration =  (kg/L)

(b) Find the amount of salt in the tank after 1 hours.
amount =  (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration =  (kg/L)

I know (a) .1 and that (c) .05

I have tried many times and really thought I was doing it right. Please show all work so I can figure out where I went wrong.

Thanks

Explanation / Answer

(a) The initial concentration in the tank = 100/1000 = 1/10 = 0.1kg/L

(b) To find amount of salt in tanbk after 1 hour or 60 minutes

Let us define s(t) = amount of salt in tank after t minutes

We will try to find s(t)

Now s'(t) = Rate at which amount of salt is changing in the tank = (rate of salt going in ) - (rate of salt going out)

Amount of salt entering the tank per minute = (0.05kg/L)(10L/min) = 0.5 kg/min

Amount of salt leaving the tank per minute =

Now we have s(t)/1000 L amount of salt in tank and which is kept mixed and the rate at which water leaves the tanekl is 10L/min

So (s(t)/1000L)(10L/min) = s(t)/100 kg/min = Amount of salt leaving the tank per minute

So we get s'(t) = 0.5-1/100(s(t)

This is a differntial equation in s(t)

Using variable seperable form, we get:

ds/(0.5-s/100) = dt => 100ds/(50-s) = dt => -100ds/(s-50) = dt

Now integrating both sides we get,

-100ln|s-50| = t+C =>ln| s-50| = -1/100(t+C) => s=50 + e-(t+C)/100 or s = 50+Ce-t/100

s(0) = 100 (given)

So 100= 50+C => C = 50

Hence we get s(t) = 50+50e-t/100

So now we need to find s(60) = 50+50e-60/100 = 50+50e-0.6 = 50+ 50(1.8221) = 141.105 kgs

(C) As time approaches infinity, e-t/100 approaches 0 and so s(t) approaches 50+0 = 50

Hence concentartion of salt as time approaches infinity = (50/1000) kg/L = 0.05kg/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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