Issue 080 - Hydrology - Runoff rate
How much water fell on eastern Spain?
Storm Adam brought torrential rain to eastern Spain; reporting attributed 632 mm in 36 hours at Pego, Alicante, to AEMET.
Earth, climate, and environmentAbout 1 minute
Sources checked October 19, 2026. Figures and circumstances may have changed.
The problem
Estimate the average runoff flow in cubic meters per second if rainfall like Pego's 632 mm falls across a plausible local drainage basin and its surface runoff drains over one day.
Choose a basin area and runoff fraction. First estimate total rainfall volume and mass, then divide runoff volume by the drainage time. Treat this as a scenario, since a town's rain gauge does not measure rainfall uniformly across an entire watershed.
Optionally, choose an average descent to the coast and estimate the runoff's gravitational potential energy and average power.
Fermi estimates target scale rather than precision. The Fermi reference collects useful facts and equations.
Grounding facts
Europa Press reported AEMET's 632 liters per square meter in 36 hours at Pego on October 19, 2026. One millimeter of rainfall equals one liter per square meter, so the reported depth is 0.632 meters. Other reports use different gauges and time windows; this snapshot retains the prompt's stated measurement.
Regional drainage is more complex than a town boundary. The local public body MACMA describes the Gallinera, Atzúbia and Pego valleys feeding the Pego-Oliva wetland and its aquifer. A published study gives the nearby Girona basin as about 118 square kilometers; that is a regional scale comparison, not Pego's measured catchment.
We retain 100 square kilometers as an illustrative drainage area, not a delineated basin. USGS explains that infiltration depends on rainfall intensity, soil, prior saturation and land cover. No fixed daily absorption allowance applies to all Spanish soils.
The 40% runoff fraction below is a scenario assumption, not a measured coefficient for Storm Adam. A 20% to 80% sensitivity range shows the uncertainty. The one-day drainage time is also assumed; it is distinct from the 36-hour rainfall window and does not determine the peak flow.
After checking sources
Checked answer and calculation
illustrative basin area
= 100 km2 = 1 x 10^8 m2
rainfall depth
= 632 mm = 0.632 m
rainfall volume
= area x depth
~= 6.3 x 10^7 m3
rainfall mass
~= 6.3 x 10^7 m3 x 1,000 kg/m3
~= 6.3 x 10^10 kg
~= 63 million metric tonnes
runoff volume at 40%
~= 2.5 x 10^7 m3
average flow over one day
~= 2.5 x 10^7 m3 / 86,400 s
~= 300 m3/s
~= 300,000 L/sThat is tens of billions of kilograms of rain and an illustrative runoff flow of hundreds of cubic meters per second. These are conditional estimates, not a measured storm water budget or a measured river discharge.
Keeping the same area and rainfall, 20% to 80% runoff gives about 150 to 600 m3/s. A smaller basin, lower basin-average rainfall, or longer drainage time lowers that result proportionally. Several channels and wetland storage could distribute or delay the water; a storm peak could be much higher than its daily average.
For a local engineering comparison, Spain's official 2006 Gallinera flood-control assessment discusses channel-design flows around 95 m3/s and spillway flows reaching several hundred m3/s. That places our scenario in a serious flood-flow range, but those design values are neither ordinary river flow nor measurements of this event.
optional energy scenario: average descent = 100 m
runoff mass ~= 2.5 x 10^10 kg
potential energy = m g h
~= 2.5 x 10^10 x 10 x 100
~= 2.5 x 10^13 J
average gravitational power over one day
~= 2.5 x 10^13 / 86,400
~= 300 MWThis is potential energy dissipated along the drainage paths, not usable electrical output. The average descent is an additional assumption.
Matt's volume and flow arithmetic are consistent with 0.63 m and 0.5 m, despite the original text labeling them centimeters. Likewise, the final dimensional expression should be m3 divided by seconds, giving m3/s. His 600 m3/s is the upper end of our sensitivity range. Available evidence does not establish that his runoff fraction was wrong; it establishes that it was insufficiently grounded.
Before checking sources
Matt's original estimate
This is Matt's original, unverified estimate, including its unit slips.
I'm not familiar with the region, but I assumed Pego is a large town, or about 10^8 m^2 of surface area.
632 mm is 63.2 cm; I expect most Spanish soils can absorb 10cm of rainfall daily pretty well, and any excess of that could reasonably be expected to become runoff, so over that 36 hours I assumed about 50 cm of the rainfall became runoff.
0.63cm * 10^8 m^2 means a rainfall of 63 million cubic meters, or about 63 million tonnes of rainfall. 0.5 cm * 10^8 m^2 means about 50 million cubic meters of that rainfall may have become runoff.
there's 86400 seconds in a day, if all of that runoff cleared within a day, the average flow rate is 5 * 10^7 m^2 / 8.6 * 10^4 seconds = about 600 m^3/2 or about 600,000 L/s
Reasoning score
Matt's reasoning score: 60 / 100
Useful facts: 20/30. The area was a reasonable regional scale and the water conversions were useful. The absorption allowance was unsupported; the basin and rainfall footprint remain uncertain, so this is a provisional scenario-based judgment.
Reasoning approach: 15/30. Area times depth, then runoff volume divided by time, is correct. Town area, drainage area and a single-gauge rainfall total needed clearer separation, and runoff cannot be inferred from a universal soil allowance.
Math: 5/10. The numerical results follow the intended meter values, but written units switch to centimeters, square meters and m3/2. Those slips obscure an otherwise consistent calculation.
Final estimate: 20/30. About 600 m3/s is within a factor of two of the 300 m3/s scenario and within its sensitivity range. This checks scale, not accuracy against an observed discharge.
Post-check reflection
Matt's reflection
Looks like my territory assumption wasn't terrible, and I shouldn't have assumed the rainfall was uniform over the region, but the biggest error was percentage of the rainfall that might have become runoff. My answer was high as a result, but still within an order of magnitude.
Recommended memory peg
1 mm of rain over 1 km2 equals 1,000 m3 of water, or one million kilograms. For runoff, multiply by a runoff fraction and divide by time; keep basin-average rainfall separate from a point measurement.