Can anyone explain more about exponential growth of the concentration on nutrients in an estuary when water is flowing into and out of the estuary at the same rate?

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- Nov 4th 2012, 12:34 AMdeb107exponential growth
Can anyone explain more about exponential growth of the concentration on nutrients in an estuary when water is flowing into and out of the estuary at the same rate?

- Nov 4th 2012, 12:38 AMMarkFLRe: exponential growth
Are you given initial conditions and the concentration of nutrients flowing in?

- Nov 4th 2012, 12:41 AMdeb107Re: exponential growth
the flow in rate is 100m^3/day with concentration of 90g/m^3.

Initial concentration in the estuary is 20g/m^3

vol 1000m^3

flow out rate is also 100m^3/day - Nov 4th 2012, 12:47 AMMarkFLRe: exponential growth
First, you want to determine the mass of nutrient initially present. Amount per volume times volume equals amount. What do you find?

- Nov 4th 2012, 12:50 AMdeb107Re: exponential growth
20g/m^3 x 1000m^3 = 20000g/m^3 initial concentration

and I have 90 x 100/day = 9000 flowing in - Nov 4th 2012, 01:08 AMMarkFLRe: exponential growth
You have the correct initial mass of 20,000 g, but you don't want to call this a concentration, it is simply an amount. Also, you are correct that 9000 grams is flowing in. So, let's then use smaller numbers and say the initial mass is 20 kg and 9 kg is flowing in per day.

Let's let $\displaystyle N(t)$ represent the mass of nutrient present in the estuary at time $\displaystyle t$.

Now, we know the time rate of change of $\displaystyle N(t)$ is equal to the rate in minus the rate out. We have already determined the rate in. The rate out will be a function of $\displaystyle N(t)$. We will asume the concentration of nutrient is uniform in the estuary. That is, the concentration of nutrient in any part of the estuary at time $\displaystyle t$ is just $\displaystyle N(t)$ divided by the volume of fluid in the estuary. Since the flow in is equal to the flow out, this volume remains constant at $\displaystyle 1000\text{ m}^3$.

Hence, the output rate of nutrient is:

$\displaystyle \left(100\frac{\text{m}^3}{\text{day}} \right)\left(\frac{N(t)}{1000}\,\frac{\text{kg}}{ \text{m}^3} \right)=\frac{N(t)}{10}\,\frac{\text{kg}}{ \text{day}}$

So, we now have enough information to model $\displaystyle N(t)$ with the IVP:

$\displaystyle \frac{dN}{dt}=9-\frac{N(t)}{10}$ where $\displaystyle N(0)=20$

Can you proceed from here? - Nov 4th 2012, 01:28 AMdeb107Re: exponential growth
would this mean that on day zero, my flow out amount is 9 -20/10 = 7?

we are supposed to write down and solve an appropriate differential equation for N(t) along with the appropriate initial condition. That would somehow be my initial amount plus the flow in rate, minus the flow out rate? My apologies for sounding so thick! - Nov 4th 2012, 01:34 AMMarkFLRe: exponential growth
The initial output rate of nutrient would be 7 kg/day, but that is really only applicable at the very start of the first day, the initial moment.

I have already stated the appropriate differential equation/initial condition. This is called an initial value problem (IVP).

Now, to solve the ordinary differential equation (ODE), do you recognize what type of equation it is, and if so, what you need to do to solve it? - Nov 4th 2012, 01:54 AMMarkFLRe: exponential growth
By the way, are you supposed to find the mass of nutrient at time

*t*, or the concentration? Since the concentration is just mass/volume and the volume is constant it is simple to convert between the two, I was just curious. - Nov 4th 2012, 01:57 AMdeb107Re: exponential growth
we are supposed to find the amount at time t and then the concentration after a long time

- Nov 4th 2012, 01:58 AMMarkFLRe: exponential growth
Okay, good, then we are set to do just that.

- Nov 4th 2012, 01:59 AMdeb107Re: exponential growth
it's an exponential growth equation.

I need the initial amount plus the flow in amount = 20 +9^t minus the flow out amount - Nov 4th 2012, 01:03 AMMarkFLRe: exponential growth
No, the ODE is linear, as we may write it in the form:

$\displaystyle \frac{dN}{dt}+\frac{1}{10}N(t)=9$

Do you recognize this form? Do you recall how to transform the equation so that we have the product of a differentiation on the left? - Nov 4th 2012, 01:25 AMdeb107Re: exponential growth
I'm just reading this from my text book. The equation is already in the required form(dN/dt + P(t)N= Q(t)

- Nov 4th 2012, 01:30 AMMarkFLRe: exponential growth
Yes, so what do we need to do next?