00:01
The question says that a tank contains 90 kg of salt and 1 ,000 liter of water.
00:06
Solution of concentration 0 .05 ks of salt per liter.
00:10
It does a tank at the rate of 10 liter per minute.
00:13
Then the solution is mixed and brings from the tank at the same rate.
00:18
We need to find out the initial concentration of solution.
00:21
Then we need to find out the amount of salt after 4 .5 hours.
00:25
And at last we need to find out the concentration of salt in the solution at time approaches.
00:31
Infinity okay so let's start so first of all we'll find out the initial concentration okay so we know that the concentration concentration is given by amount of salt divided by the amount of water initially okay salt was 90 kg and water was thousand liter so it comes out to be 0 .09 kg per liter so that will be my initial concentration okay so now we need to find out the amount of salt after 4 .5 hour so let add time t okay amount of salt v pt okay okay so and dp divided by d t it will denote the rate of change of amount of salt okay the rate of change of amount of salt correct so we need to find out the net rate so it will be rating minus rate out so rate inside inside it is coming with a rate of 10 liter per minute and concentration is 045 okay minus it is going at and its concentration then will be equal to pt divided by the thousand liter times ten correct so what it will be so it will be 0 .45 minus it becomes p of t divided by 100 correct so this is what i am having as of now now what i can do if i have rearrange it what i am going to get dp divided by d t plus p of t and here i am having one divided by hundred is equal to 0 .45 correct so what i can do now you see now this is what this is a linear differential equation linear differential equation and we know how to solve a linear differential equation okay so just we need to solve so what i will do i will try to find out the integrating factor so integrating factor integrating factor okay so it will be e raised to integral of the constant coefficient of t of t is what one divided by hundred d t correct one divided by hundred d t so what it will be will be be t raised to now t integral we know sorry it will not be t it is just one divided by hundred times of d correct so here we are getting 1 divided by 100 d t.
04:16
So it will be 100, 1 divided by 100 times of t, correct.
04:23
Now the solution we know, so hence the solution, hence the solution is given by integrating factor or p of t times the integrating factor.
04:39
Okay equal to integral of 0 .45.
04:49
Okay.
04:51
Times the integrating factor p.
04:54
Raised to t divided by 100 correct and dear t.
05:00
So this is what we need to solve.
05:03
So what it will be, let us see.
05:06
So t multiplied with e raised to t divided to t divided.
05:10
By 100 equal to now you see it will be 0 .45 and e raised to t divided by 100 is nothing but t raised to t divided by 100 divided by 1 divided by 100 okay plus the constant of integration so this will be equal to this becomes 45 t raised to t divided by hundred plus c and here we are having p of t he raised to t divided by hundred correct so this is what i am getting now now what i can do here let us see that so this implies that p of t will be equal to p raised to minus of t divided by hundred times 45, t raised to t divided by and red, plus the constant of integration.
06:14
If i multiply throughout, so what i will get, so this will give me p of t to be equal to 45 plus c, t raised to minus of t divided by and red, correct.
06:31
So this will be the amount of salt we need to find out...