Question

A nitric acid solution flows at a constant rate of 6 L/min into a large tank that initially held 200 L of a 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 L/min. If the solution entering the tank is 20% nitric acid, determine the volume of nitric acid in the tank after t min. When will the percentage of nitric acid in the tank reach 10%?

Answer 1

Answer:

0.88 s

Step-by-step explanation:

Let x(t) be the volume of nitric acid  measured in liters after t minute. Therefore the rate of change in the volume of nitric acid (dx / dt) is:

dx / dt = rate of input of nitric acid - rate of output of nitric acid

rate of input of nitric acid = 6 L/min * 20% nitric acid / 100% solution = 1.2 L/min

rate of output of nitric acid = 8 L/min * x L / 200 L = 0.04x L/min

$\frac{dx}{dt}=1.2-0.04x\\\\\frac{dx}{dt}+0.04x=1.2 \\\\The\ integrating\ factor(IF)=e^{\int\limits {0.04} \, dt }=e^{0.04t}\\\\multiply\ through\ by \ IF:\\\\e^{0.04t} \frac{dx}{t}+e^{0.04dt}(0.04x)=e^{0.04t}(1.2)\\\\integrating:xe^{0.04t}=1.2\int\limits e^{0.04t}dt\\\\xe^{0.04t}=30e^{0.04t}+C\\\\x(t)=30+Ce^{-0.04t}\\\\At\ t=0,x=1\\\\x(0)=30+Ce^{-0.04*0}\\\\1=30+C\\\\C=-29\\\\x(t)=30-29e^{-0.04t}\\\\At \ 10\%,x=10%\ of\ 200L=2\\\\2=30-29e^{-0.04t}\\\\29e^{-0.04t}=28$

$29e^{-0.04t}=28\\\\e^{-0.04t}=0.9655\\\\t=0.88\ s$

The percentage of nitric acid in the tank reach 10% at 0.88 s