# How to calculate residence time for an element in a reservoir?

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

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.

• 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 Apr 11 '18 at 0:19