Consider a flask that contain 3 liters of salt water. Suppose that water containing 25 grams per liters of salt is pumped into the flask at the rate of 2 liters per hour, and the mixture, being steadily stirred, is pumped out of the flask at the same rate. Find a differential equation satisfied by the amount of salt ๐(๐ก) in the flask at time ๐ก.
Let a = 25 grams/liter, V = 3 liters, u = 2 liters/hour
"df=a\\cdot u\\cdot dt-\\frac{f}{V}u\\cdot dt\\\\\n\\frac{df}{dt}=u(a-\\frac{f}{V})\\\\\n\\int_0^f\\frac{d(aV-f)}{aV-f}=-\\int_0^t\\frac{u}{V}dt\\\\\naV-f=(aV-f(0))e^{-\\frac{u}{V}t}\\\\\nf=aV(1-e^{-\\frac{u}{V}t})=75(1-e^{-\\frac{2}{3}t})~(g)"
Comments
Leave a comment