agreseza
2021-12-30
Answered

In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?

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Orlando Paz

Answered 2021-12-31
Author has **42** answers

Let $Q=$ amount of salt in the mixture

$100+(3-2)t=100+t=$ volume of mixture at any time t

$\frac{q}{100+t}=$ concentration of salt

$\frac{dq}{dt}=$ rate of gain-rate of loss

$\frac{dq}{dt}=0-2\left(\frac{q}{100+t}\right)$

$\int \frac{dq}{dt}=-2\int \frac{dt}{100+t}+c$

$\mathrm{ln}q=-2\mathrm{ln}(100+t)+c$

$\mathrm{ln}q=-\frac{2}{2}{\mathrm{ln}(100+t)}^{2}+c$

When$t=0:q=50kg$

$\mathrm{ln}50=-{\mathrm{ln}(100+1)}^{2}+c$

$c=13.12236$

When$t=60min,Q=?$

$\mathrm{ln}q=-{\mathrm{ln}(100+60)}^{2}+13.12236$

$\mathrm{ln}q=2.972$

$q=19.53kg$

When

When

veiga34

Answered 2022-01-01
Author has **32** answers

Suppose that the amount of salt in the mixture be Q.

The volume of mixture at time t will be as follows:

Recall the fact that the concentration of salt will be given as follows:

Also, the rate of gain minus rate of loss will be

Here rate of gain is 0 and rate of loss is

Initially the amount of salt in the mixture is 55 kg. That is,

Substitute

By the end one 1 hours, t will be 60 minutes.

Substitute

Thus, the amount of salt in tank after one hour is 21.4913 Kg.

Vasquez

Answered 2022-01-09
Author has **460** answers

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