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I need help with this question please, I couldn't copy the question so I made an URL for it hope it's fine with you!

Regards.

https://ibb.co/LntWz6f

Aug 22, 2019
edited by Melody  Aug 23, 2019

#1
+107018
+1

I don't know where to start but I will watch and learn if someone else wants to answer.

I just found this.  Looks like it should help, I have not watched it yet but it looks relevant.

Aug 23, 2019
#2
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I will try my best but if you find any other hints please share them.

Guest Aug 23, 2019
#3
+107018
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The solution is being added quicker than it is being drained away. So the amount in the container is constantly increasing.

At a given point in time (t minutes after start) the amount in the tank will be  5000+50t-30t = 5000+20t

At the start    5000L  of brine   which has 100Kg sale

Entering     0.3Kg/L      at   50L/min

Leaving        The concentation in the tank at any given moment    at     30L/min

Let A(t)  be the amount of salt in the tank immediately after any given time t (minutes).

I will call this A for short but remember that it is a function of time.

$$\frac{dA}{dT}=\text{rate in - rate out}\\ \frac{dA}{dT}=\text{concentation in* rate of flow - concentation out *rate of flow}\\ \frac{dA}{dT}= \left(\frac{0.3Kg}{1L}*\frac{50L}{1minute}\right)- \left(\frac{A Kg}{(5000+20t)L}*\frac{30L}{1minute}\right)\\ \frac{dA}{dT}= \left(\frac{15Kg}{1minute}\right)- \left(\frac{30A Kg}{(5000+20t)minute}\right)\\ \frac{dA}{dT}= 15- \frac{30A}{5000+20t}\;\;\quad\frac{Kg}{min}$$

I am not sure if it is finished.  I probably should go further ?  what do you think ?

I don't have time right now.

By the way, the video question was simpler than this one but that is the only resource that I used to help me.

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ok I have looked more and I do not know how to go further.  Maybe the question does not require me too.

Does anyone else know how to take if further?

Aug 23, 2019
edited by Melody  Aug 23, 2019
edited by Melody  Aug 23, 2019