Differential Equations Water Evaporation

In summary, the problem discusses a pond with 1,000,000 gal of water and a unknown amount of chemical. The chemical enters at a rate of 300 gal/hour and exits at the same rate, while pure water is also evaporating at a rate of one gallon per hour. The resulting differential equation for the amount of chemical in the pond at any given time is \frac{d amount(t)}{dt} = 300(\frac{1}{100} - \frac{amount(t)}{(1000000 - 1*t)}). This takes into account the change in concentration of the chemical over time. The solution for the first part of the problem involves defining a(t) as the amount of chemical at any
  • #1
TranscendArcu
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Homework Statement



A pond contains 1,000,000 gal of water and an unknown amount of chemical. Water containing .01 gram of this chemical per gallon flows into the pond at a rate of 300 gal/hour. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume the chemical is uniformly distributed throughout the pond. The differential equation for the amount of chemical in the pond at any time is:

[itex]\frac{d amount(t)}{dt} = 300(\frac{1}{100} - \frac{amount(t)}{1000000} )[/itex] (Finding this equation was the first part of the problem).

Suppose that in addition to the flow into and out of the pond described above, that pure water is evaporating from the pond at the rate of one gallon per hour. Write down a differential equation for the amount of chemical in the pond at any given time.

The Attempt at a Solution


I'm tempted just to write the following:

[itex]\frac{d amount(t)}{dt} = 300(\frac{1}{100} - \frac{amount(t)}{(1000000 - 1*t)}) [/itex]

Does this not take into account the change in concentration of the chemical in the water over time? I'm not sure if there's something more sophisticated going on here.
 
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  • #2
Maybe I can get my work checked for the first part of the problem, too. I said:

Define a(t) to be the amount of chemical at any given time, t. The inflow of the chemical into the pond does so with concentration 1 gram/100 gallons. Chemical outflow has concentration a(t) / 106 gallons. Thus, I reasoned that the change in concentration will be given by the difference between inflow and outflow, multiplied by the flow rate, or 300. I wrote,

[itex]\frac{da(t)}{dt} = 300 (\frac{1}{100} - \frac{a(t)}{10^6})[/itex], which gives me my equation.

Does that seem right?
 
  • #3
That looks fine, just remember that t is in hours. These are called "mixture" problems, and the general form of them is usually:
[tex]\frac{dA}{dt} = (rate in) - (rate out)[/tex]
Where A(t) is some function representing amount of stuff in the container.
 

Related to Differential Equations Water Evaporation

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model physical phenomena and solve problems in various fields such as physics, engineering, and economics.

2. How is a differential equation used to model water evaporation?

The differential equation for water evaporation is based on the principle of mass conservation, which states that the rate of change of a system's mass is equal to the net flow of mass into or out of the system. In the case of water evaporation, the rate of change of water mass in a given area is equal to the difference between the rate of evaporation and the rate of precipitation.

3. What factors affect the rate of water evaporation in a differential equation?

The rate of water evaporation is affected by several factors, including temperature, humidity, wind speed, and surface area of the water. These factors can be incorporated into the differential equation as parameters to accurately model the process of water evaporation.

4. Can a differential equation for water evaporation be solved analytically?

In most cases, a differential equation for water evaporation cannot be solved analytically. Instead, numerical methods and computer simulations are used to approximate the solution. This is due to the complex nature of the equation and the many variables involved in the process of water evaporation.

5. How accurate is a differential equation model for water evaporation?

The accuracy of a differential equation model for water evaporation depends on the accuracy of the parameters and initial conditions used. With proper data and calibration, the model can provide a good approximation of the actual process of water evaporation. However, there may be some discrepancies due to the simplified assumptions made in the model.

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