# Ordinary Differential Equation - tank with inflow and outflo

• samg1
In summary, the tank initially contains 60 kg of salt and 2000 L of water. A solution with a concentration of 0.015 kg/L enters the tank at a rate of 6 L/min and mixes with the existing solution. At the same rate, the mixed solution drains from the tank. After 3 hours, the amount of salt in the tank is (65.4/360) - (65.4/360 - 60/2000)e^(-360t/2000) kg. As time approaches infinity, the concentration of salt in the solution in the tank approaches 65.4/360 kg/L.
samg1

## Homework Statement

A tank contains 60 kg of salt and 2000 L of water. A solution of a concentration 0.015 kg of salt per liter enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in kg at t = 3 hours
Find the concentration of salt in the solution in the tank as time approaches infinity.

## The Attempt at a Solution

My attempt:C(t) = C = The concentration of salt in (kg/L) at time t in hours.
dC/dt = (flow_in - flow_out + 60kg/2000L)
flow_in = (0.015kg/L *6L/min*60min/hour) / 2000L // the concentration that flows in
flow_out = C*6L/min*60min/hour)/ 2000L // the concentration that flows out
2000L dC/dt = (0.015kg/L *6L/min*60min/hour) - C*6L/min*60min/hour) + 60kg
2000 dC/dt = 5.4 - 360C + 60
2000 dC/dt = (65.4 - 360C)
2000/ (65.4-360C) dC = dt*Integrate both sides*Left side :

let u = 65-360C
du = -360 dC

du / -360 = dC

2000/-360 ∫1/u du

-----------

(-2000/360) ln(65.4-360C) = t + c // where c is a constant.
ln(65.4 - 360C) = 360t/-2000 + c
e^ln(65.4 - 360C) = e^ 360t/-2000 + c
65.4 - 360C = ce^(360t/-2000)
-360C = ce^(360t/-2000) - 65.4
360C = 65.4 - ce^(360t/-2000)
C = (65.4/360) - ce^(-360t/2000)C(0) = 60/2000 Plug in this to find the constant c
60/2000 = (65.4/360) - c
c = 65.4/360 - 60/2000
C = (65.4/360) - (65.4/360 - 60/2000)e^(-360t/2000)

Got it. We can delete this post!

## 1. What is an ordinary differential equation?

An ordinary differential equation (ODE) is a mathematical equation that describes the relationship between a function and its derivatives. It involves one independent variable and one or more dependent variables, and it expresses how the dependent variable changes as the independent variable changes.

## 2. How does a tank with inflow and outflow relate to an ordinary differential equation?

In an ordinary differential equation, the independent variable represents time, and the dependent variable represents the amount of substance in the tank. The rate of change of the amount of substance in the tank is described by the inflow and outflow rates, which are typically represented by the first derivative of the dependent variable with respect to time.

## 3. What is the importance of understanding ordinary differential equations in a scientific context?

Ordinary differential equations are used to model many natural phenomena, such as population growth, chemical reactions, and fluid flow. They are essential in understanding and predicting how these systems change over time, making them a fundamental tool for scientists in various fields.

## 4. Can you provide an example of an ordinary differential equation for a tank with inflow and outflow?

One example could be the following differential equation: dV/dt = 10 - 2V, where V represents the volume of water in the tank and t represents time. This equation describes the change in volume of water in the tank over time, with a constant inflow rate of 10 units per time and an outflow rate that is proportional to the current volume of water in the tank.

## 5. How do you solve an ordinary differential equation for a tank with inflow and outflow?

The solution to an ordinary differential equation depends on the specific equation and initial conditions. In general, it involves finding a function that satisfies the equation and initial conditions. This can be done analytically, using mathematical techniques such as separation of variables or integrating factors, or numerically, using computer algorithms.

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