Interpret declining storage and infeasible budgets

Outputs can exceed this month’s precipitation

A monthly budget does not start with an empty basin. The fictional watershed begins with 100 mm of total stored water, expressed as an equivalent depth over 2 km². If this month’s ET and stream outflow exceed this month’s precipitation, some previously stored water can supply the difference.

Select P = 40 mm and actual ET = 60 mm. Stream outflow remains 40 mm. Outputs total 100 mm, while precipitation supplies only 40 mm. Storage change is −60 mm and ending storage is 40 mm. This does not violate conservation: the missing 60 mm comes from the starting store in the accounting.

Water can remain in surface and subsurface storage and later contribute to flow. A watershed need not stop discharging water whenever rain stops. The browser does not assign the decline to a particular store or flow path. USGS discussion of watershed storage and streamflow.

Separate a negative change from a negative amount

ΔS is ending storage minus starting storage. It may be negative when storage declines. S_end is the calculated total remaining water and cannot physically be negative. These quantities have different interpretations even though both are displayed in millimeters.

At P = 0 and ET = 60 mm, the fixed stream outflow adds another 40 mm of output. The starting 100 mm exactly supplies the 100 mm total, so ΔS = −100 mm and S_end = 0. Draw the beginning and ending stores on paper: a complete decline is different from a negative store.

A zero endpoint does not establish that a real basin could maintain those fluxes throughout the month. Water availability at the right place and time, hydraulic connections and changing ET or streamflow all matter. This model has only monthly totals, so its nonnegative endpoint is a necessary accounting check, not a sufficient physical process simulation.

Preserve an infeasible arithmetic result

Keep precipitation at zero and increase actual ET to 120 mm. Requested outputs now total 160 mm, but the supplied starting water is only 100 mm. Arithmetic gives S_end = −60 mm. The model reports an infeasible budget with a 60 mm deficit under its assumptions. That deficit corresponds to 120,000 m³ over the 2 km² area.

Do not interpret −60 mm as negative physical water. Do not silently replace the result with zero either: the zero would hide 60 mm of unexplained output. The negative arithmetic endpoint is useful because it exposes that the requested fluxes and assumptions cannot all hold together.

List several ways a real ledger might require revision: an output estimate could be too large, initial storage could be underestimated, or a boundary inflow could have been omitted. The calculated deficit alone does not identify which explanation is correct. Retain the original entries while seeking evidence rather than choosing whichever adjustment makes the sum convenient.

Distinguish sensitivity from causation

With outputs held fixed, adding 10 mm of precipitation raises the arithmetic endpoint by 10 mm. With precipitation fixed, adding 10 mm of ET lowers it by 10 mm. Verify both statements with captures around a feasible case. These are sensitivities of the supplied equation; they do not show how a real watershed responds dynamically.

The model has no storage-capacity ceiling, no daily sequencing and no pumping term. A positive remaining amount therefore cannot establish how much water could be withdrawn sustainably. The lesson’s claim is narrower: it checks whether the selected totals are arithmetically compatible with the stated starting amount and boundary.

Compare three kinds of result

Create a table for the declining, zero and negative-endpoint cases above. Include precipitation, total output, ΔS, S_end and a sentence interpreting the result. Keep the numerical result and its physical interpretation in separate columns so a minus sign cannot silently change meaning.

Now ask a partner to inspect only the interpretations and reconstruct the sign of each quantity. If working alone, cover the arithmetic and perform the same check. A correct explanation must preserve both conservation and the model’s limits.

Transfer: diagnose and bound a deficit

For P = 20 mm and actual ET = 100 mm, find the endpoint and deficit. If all other supplied quantities remain fixed, how much additional precipitation would bring the arithmetic endpoint to zero?

Check the storage-change transfer

S_end = 100 + 20 − 100 − 40 = −20 mm. The 20 mm deficit is 40,000 m³. An additional 20 mm of precipitation makes total P = 40 mm and the endpoint zero. This repairs the arithmetic example; it is not evidence that additional rain occurred or that the monthly flow timing is physically achievable.