Answer to Question #278925 in Differential Equations for Dhel

Question

Tank A initially contains 200 liters of brine containing 225 N of salt. Eight liters of fresh water per minute enter A and the mixture, assumed uniform, passes from A to B, initially containing 200 liters of fresh water, at 8 liters per minute. The resulting mixture, also kept uniform, leaves B at the rate of 8 liters per minute. Find the amount of salt in tank B after one hour.

Application of First Order of Differential Equation

Answer 1

Let $A(t) =$ amount, in N of salt in tank A at time $t.$ Then we have

$\dfrac{dA}{dt}=$(rate of salt into tank A) − (rate of salt out of tank A)

$\dfrac{dA}{dt}=0-\dfrac{8A}{200}$

So we get the differential equation

$\dfrac{dA}{A}=-\dfrac{dt}{25}$

Integrate

$A(t)=Ce^{-t/25}$

Given $A(0)=225.$ Then

$A(t)=225e^{-t/25}$

Let $B(t) =$ amount, in N of salt in tank B at time $t.$ Then we have

$\dfrac{dB}{dt}=$(rate of salt into tank B) − (rate of salt out of tank B)

$\dfrac{dB}{dt}=\dfrac{8}{200}(225e^{-t/25})-\dfrac{8}{200}B$ $\dfrac{dB}{dt}+0.04B=e^{-0.04t}$ $\dfrac{dB}{dt}+0.04B=0$ $\dfrac{dB}{B}=-0.04dt$ $B=c_1e^{-0.04t}$ $\dfrac{dB}{dt}=\dfrac{dc_1}{dt}e^{-0.04t}-0.04c_1e^{-0.04t}$ $\dfrac{dc_1}{dt}e^{-0.04t}-0.04c_1e^{-0.04t}+0.04c_1e^{-0.04t}$ $=e^{-0.04t}$ $\dfrac{dc_1}{dt}=1$ $c_1=t+c_2$ $B(t)=(t+c_2)e^{-0.04t}$

Given $B(0)=0.$ Then

$B(t)=te^{-0.04t}$

Hence

$B(60)=60e^{-0.04(60)}\approx5.443 \ N$