Question

@dan815

Answer 1

In example 2.3.7, find the maximum input flow rate such that the tank will just reach its capacity of 160 liters when the salt concentration reaches 1gram/liter. How many minutes does it take?

Answer 2

Is the tank initially empty ?

Answer 3

Nope, initially, its volume is 150 liters

Answer 4

I don't know how to link them together. We have, the volume at time t is V(t) = 150 +0.5t.
So that to get Volume = 160, just solve for t
160 = 150 + 0.5t , hence t= 20 minutes

Answer 5

why are you fixing the input flow rate ?

Answer 6

So that the fixed time is 20 minutes. But to get the concentration 1g/l, we need find the concentration of the input flow

Answer 7

Looks you're interpreting the question wrong

Answer 8

Because of the input flow rate is 2.5l/min fixed

Answer 9

Concentration of salt in the input flow is fixed here.
Input flowrate is variable

Answer 10

Ok, guide me more, please

Answer 11

say the input flow rate is \(r\)

Answer 12

Do we get the differential equation : \[x' + \left(\dfrac{4}{300+t}\right)x = 3r \]

where \(r\) is the input flow rate ?

Answer 13

The time taken for the tank to be full depends on the input flow rate.
It doesn't depend on the concentration.

Answer 14

Yes,

Answer 15

try to solve above diff eqn

Answer 16

but the input flow rate is fixed, right? it is given by 2.5 l/ min

Answer 17

Solving the ODE is no problem to me. Making it is my problem :(

Answer 18

It isn't fixed in your problem

Answer 19

It is fixed in the attachment though

Answer 20

ok, got you

Answer 21

In your problem input flow rate is not fixed. It is precicely the thing that you need to find :

In example 2.3.7, `find the maximum input flow rate` such that the tank will just reach its capacity of 160 liters when the salt concentration reaches 1gram/liter.

Answer 22

I have already provided the diff eqn for you

Answer 23

nvm, got it

Answer 24

what is the given salt concentration ?

Answer 25

ok good

Answer 26

ok, now I solve it. :)

Answer 27

I'm getting the solution as
\[x(t) = \dfrac{3r}{5}(300+t)+\dfrac{C}{(300+t)^4}\]

Answer 28

plugin the initial condition and solve \(C\)

Answer 29

then?

Answer 30

Still have t and r.

Answer 31

use below equation with the previous solution to simultaneously solve \(r,t\) :
\[160 = 150 + (r-2)t\]

Answer 32

GGGGGGGGGGGGGot it. Thanks a ton. :)

Answer 33

I do appreciate.

Answer 34

np :)

Answer 35

@rsadhvika
If the volume of the tank at time t is V= 150 + (r-2) t
then the ODE becomes
\(\dfrac{dx}{dt}=3r-\dfrac{x}{(150+(r-2)t)}\)
Am I right?

Answer 36

Ahh right !

Answer 37

Oh, but then,when the tank is full, we have 10 = (r-2)t
and the ODE is simple. Am I right?

Answer 38

that comes after solving the diff eqn

Answer 39

\[\dfrac{dx}{dt}=3r-\dfrac{\color{red}{2}x}{(150+(r-2)t)}\]

Answer 40

Why can't we apply it at the same time?
Like this. After the tank full, we know that the desired concentration of salt is 1. And we know x(0) =20, x (time of full) =1. so dx = 1-x

Answer 41

1g/l means the amount of salt at the end is 160g
hence dx = 140

Answer 42

I'm not sure that works
diff eqn models the entire behavior, not just the value at one point in time

Answer 43

If I can't apply the condition at the same time, the ODE becomes extremely hard to solve. :(

Answer 44

\[\dfrac{dx}{dt}=3r-\dfrac{\color{red}{2}x}{(150+(r-2)t)}\]

How is this eqn hard to solve ?

Answer 45

integrating function \(\mu (t) = e^{\int 2/(150 +(r-2)t) dt}\) = not "cute" at all. :)