Question

5, A swimming pool initially contains 10,000 gal of water which is 1% chlorine. Pure water flows into the pool at the rate ot 20 gal/min and leaves at the rate of 30 gal/min. What is the concentration of chlorine in the pool after 2 hour? When theoretically will the tank become empty? Ans.

Answer 1

First notice that the volume of water in the pool is initially 10000 gallons, and for each minute afterwards, 20 gallons of water are added and 30 are lost, for a net change of (-10) gallons per minute. So the volume of the pool is V(t) = 10000 -10t.

Let A(t) = the number of gallons of chlorine in the pool at time t minutes after the start. A(0) = (10000)(1%)=100 gallons.

The amount of chlorine in per a minute is = 0.

The amount of chlorine out per a minute is = (the concentration of chlorine at time t)*(the rate of water going out)

= (A(t) / V(t)) * (30) = 30 A(t) / (10000-10t) = 3A(t) / (1000-t).

So the rate of concentration is  .

First notice that the volume of water in the pool is initially 10000 gallons, and for each minute afterwards, 20 gallons of water are added and 30 are lost, for a net change of (-10) gallons per minute. So the volume of the pool is V(t) = 10000 -10t. Let A(t) = the number of gallons of chlorine in the pool at time t minutes after the start. A(0) = (10000)(1%) = 100 gallons. The amount of chlorine in per a minute is = 0 The amount of chlorine out per a minute is = (the concentration of chlorine at time t)*(the rate of water going out) = (A(t)/V(t)) * (30) = 30 A(t)/(10000-10t) = 3A(t)/(1000-t) so the rate of concentration is d A (t)/dt = -3 A(t)/1000 - t Implies d A (t)/A (t) = -3 dt/1000 - t use integration on both sides , we get in (A (t)) = 3 ln (1000 - t) + c Implies A (t) = B (1000)

use integration on both sides, we get  

A(0)= 100 gives B=10^(-7).

The concentration of chlorine after 2 hours(120 minutes) is .

The tank will be empty, when the volume is zero. i.e., 10000-10t=0 gives that tank will be empty after t = 1000 minutes.