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

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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 Camilo Rada. 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.


1 Answer 1


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).

  • $\begingroup$ 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. $\endgroup$
    – yre
    Commented Apr 11, 2018 at 0:10
  • $\begingroup$ 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) $\endgroup$
    – yre
    Commented Apr 11, 2018 at 0:12
  • $\begingroup$ Also, the units are in years from what I am told. $\endgroup$
    – yre
    Commented Apr 11, 2018 at 0:13
  • $\begingroup$ 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 $7 \times 10^{12}$ $\endgroup$ Commented Apr 11, 2018 at 0:19
  • $\begingroup$ Ok, then it would be RT_ATMO=(3x10^2) / (14x10^12) = 2.14285714e-11YEARS? $\endgroup$
    – yre
    Commented Apr 11, 2018 at 1:24

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