<p>Steady state is when the size of the reservoirs doesn't change with time (because inflow=outflow). The size of each reservoir is the number inside the boxes and the flow rates are the numbers beside the arrows.
If you calculate the total inflow/outflow from each reservoir, you can use the relationship you did enunciate at the start of the question to easily calculate residence times.</p>

<p>However, according to the diagram those reservoirs are not in steady state. So either the question is ill posed, or you are supposed to make some approximations (or a more complex calculation, but I don't think that's the case).</p>

CC BY-SA 3.0
2018-04-11T00:10:37.200Z
Thanks. A TA corrected this and stated: This is due to the fact that each flux rate is rounded to one significant digit, and scientists do not have a complete understanding of the nitrogen cycle. The purpose of this question is to help you gain an understanding of the relative residence times and their orders of magnitude. Therefore, you may use either the inflows or the outflows for each of the calculations that you are asked to perform, and we will accept either answer.
yre
CC BY-SA 3.0
2018-04-11T00:12:42.377Z
So the formula for residence time is = (amount in reservoir) / (outflow or inflow at a steady state).... Then RT_ATMO = (3x10^20) /  (7x10^12) ? Is that right? and then for fixed nitrogen on land. RT_FNLAND = (7x10^15) / (5x10^12)
yre
CC BY-SA 3.0
2018-04-11T00:13:15.050Z
Also, the units are in years from what I am told.
yre
CC BY-SA 3.0
2018-04-11T00:19:55.720Z
No, you have to add all the inflows or outflows, so for the atmosphere it would be $14 \times 10^{12}$ or  $12 \times 10^{12}$ but NOT $7 \times 10^{12}$
Camilo Rada
CC BY-SA 3.0
2018-04-11T01:24:11.890Z
Ok, then it would be RT_ATMO=(3x10^2) / (14x10^12) = 2.14285714e-11YEARS?
yre
CC BY-SA 3.0
2018-04-11T01:26:54.703Z
Nop, check your numbers. The atmospheric reservoir is not $3 \times 10^2$ it is **$3 \times 10^{20}$**
Camilo Rada
CC BY-SA 3.0
2018-04-11T01:31:13.063Z
Opps. Other than that error is the rest correct?
yre
CC BY-SA 3.0
2018-04-11T01:38:15.967Z
yes, it is correct
Camilo Rada
CC BY-SA 3.0
2018-04-11T01:42:09.370Z
Now that you got it, it would be good if you can answer your own question nicely, explaining the concepts, the reasoning and the math. That way the question will be useful for others in the future.
Camilo Rada
CC BY-SA 3.0
2018-04-11T02:07:04.263Z
I fixed you calculation there, you didn't fix the result after fixing the size of the reservoir
Camilo Rada
CC BY-SA 3.0
2018-04-10T22:33:54.323Z
2018-04-10T22:48:43.463Z
2018-04-10T22:48:43.463Z
Camilo Rada
<p>The residence time for an element in a reservoir can be calculated by the reservoir size at steady state divided by the inflow or outflow rate. Given the following diagram, I need to calculate the mean residence times for nitrogen in the following stocks:</p>

<ol>
<li>Atmosphere</li>
<li>Fixed Nitrogen on Land</li>
</ol>

<p><a href="https://i.sstatic.net/aA2E2.png" rel="noreferrer"><img src="https://i.sstatic.net/aA2E2.png" alt="enter image description here"></a></p>

<p>But I am confused by this question, what exactly is its steady state? what exactly is the reservoir size and what it the inflow and outflow?</p>

<p>Thank you</p>

<p>EDIT: Problem-solved with the help of Camilo Rada. The way you calculate residence time is by using the formula:</p>

<p>$\text{Residence time} = \frac{\text{Reservoir size}}{\text{in-flow or outflow rate}}$</p>

<p>In the example I looked at the box in the diagram for atmosphere's content of residence, being $3 \times 10^{20}$ units, that will serve as our total reservoir size, then we can either choose the outgoing or incoming flow (as an approximation as the reservoir is not in steady state), I chose the outgoing, so we add them together: $7 \times 10^{12} + 7 \times 10^{12} = 14 \times 10^{12}$ units per year. Now plug that into the formula:</p>

<p>$RT_{ATMO} = \frac{3 \times 10^{20}}{14 \times 10^{12}} = 21,428,571 \, \text{years} = \, \backsim 21 \, \text{million years}$</p>

<p>It would be the same process for each box.</p>

<atmosphere><geography><homework>
How to calculate residence time for an element in a reservoir?
yre
# How to calculate residence time for an element in a reservoir?

(Asked by: username_0 on 2018-04-10)

The residence time for an element in a reservoir can be calculated by the reservoir size at steady state divided by the inflow or outflow rate. Given the following diagram, I need to calculate the mean residence times for nitrogen in the following stocks:

1. Atmosphere
2. Fixed Nitrogen on Land

[![enter image description here](https://i.sstatic.net/aA2E2.png)](https://i.sstatic.net/aA2E2.png)

But I am confused by this question, what exactly is its steady state? what exactly is the reservoir size and what it the inflow and outflow?

Thank you

EDIT: Problem-solved with the help of username_1. The way you calculate residence time is by using the formula:

$\text{Residence time} = \frac{\text{Reservoir size}}{\text{in-flow or outflow rate}}$

In the example I looked at the box in the diagram for atmosphere's content of residence, being $3 \times 10^{20}$ units, that will serve as our total reservoir size, then we can either choose the outgoing or incoming flow (as an approximation as the reservoir is not in steady state), I chose the outgoing, so we add them together: $7 \times 10^{12} + 7 \times 10^{12} = 14 \times 10^{12}$ units per year. Now plug that into the formula:

$RT\_{ATMO} = \frac{3 \times 10^{20}}{14 \times 10^{12}} = 21,428,571 \, \text{years} = \, \backsim 21 \, \text{million years}$

It would be the same process for each box.

**Tags:** atmosphere, geography, homework


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## 1 answers

### Answer by username_1 on 2018-04-10. Score: 4 | Accepted answer

Steady state is when the size of the reservoirs doesn't change with time (because inflow=outflow). The size of each reservoir is the number inside the boxes and the flow rates are the numbers beside the arrows.
If you calculate the total inflow/outflow from each reservoir, you can use the relationship you did enunciate at the start of the question to easily calculate residence times.

However, according to the diagram those reservoirs are not in steady state. So either the question is ill posed, or you are supposed to make some approximations (or a more complex calculation, but I don't think that's the case).

> **Comments:**
> **username_0**: Thanks. A TA corrected this and stated: This is due to the fact that each flux rate is rounded to one significant digit, and scientists do not have a complete understanding of the nitrogen cycle. The purpose of this question is to help you gain an understanding of the relative residence times and their orders of magnitude. Therefore, you may use either the inflows or the outflows for each of the calculations that you are asked to perform, and we will accept either answer.  
> **username_0**: So the formula for residence time is = (amount in reservoir) / (outflow or inflow at a steady state).... Then RT_ATMO = (3x10^20) /  (7x10^12) ? Is that right? and then for fixed nitrogen on land. RT_FNLAND = (7x10^15) / (5x10^12)  
> **username_0**: Also, the units are in years from what I am told.  
> **username_1**: No, you have to add all the inflows or outflows, so for the atmosphere it would be $14 \times 10^{12}$ or  $12 \times 10^{12}$ but NOT $7 \times 10^{12}$  
> **username_0**: Ok, then it would be RT_ATMO=(3x10^2) / (14x10^12) = 2.14285714e-11YEARS?  
> **username_1**: Nop, check your numbers. The atmospheric reservoir is not $3 \times 10^2$ it is **$3 \times 10^{20}$**  
> **username_0**: Opps. Other than that error is the rest correct?  
> **username_1**: yes, it is correct  
> **username_1**: Now that you got it, it would be good if you can answer your own question nicely, explaining the concepts, the reasoning and the math. That way the question will be useful for others in the future.  
> **username_1**: I fixed you calculation there, you didn't fix the result after fixing the size of the reservoir  

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