SME
Hydrology
Surface runoff
Use this for “A 2 km² basin with C = 0.6 gets 50 mm/h of rain. What is its peak discharge?”
Diagram
Qₚ
=
CiA
3.6
=
0.6
×
50
×
2
3.6
Qₚ = 16.7 m³/s
24 of 40 drops run off to the outlet (C = 0.6); 16 soak in.
The basin is drawn to scale: its outline holds 2 km².
This is an example. Type your own values to start a new problem.
Qₚ (m³/s)
=
C
×
i (mm/h)
×
A (km²)
÷
3.6
Formulas
Qₚ
=
C
×
i
×
A
3.6
16.7
=
0.6
×
50
×
2
3.6
Assumptions
•
A small basin, under about 10 km²: the rational method is for small areas.
•
The rain lasts at least as long as water takes to cross the basin, so all of it sends water at once.
•
C is the share of the rain that runs off: pavement near 0.9, lawns near 0.2, woods near 0.15.
Step-by-step
WE KNOW
C = 0.6, i = 50 mm/h, A = 2 km²
FIND
peak discharge (Qₚ)
Step 1 · Find peak discharge (Qₚ)
Use Qₚ = C × i × A ÷ 3.6
The
share
C
that
runs
off,
times
the
rain
rate
and
the
area;
÷
3.6
turns
mm/h
×
km²
into
m
3
/s.
Qₚ
=
0.6
×
50
×
2
3.6
Qₚ
=
30
×
2
3.6
Qₚ
=
60
3.6
Qₚ = 16.7 m³/s
Check
Put the numbers back in. Both sides should match.
16.7
=
0.6
×
50
×
2
3.6
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Storm runoff by the curve number
Use this for “A basin with curve number 80 gets 75 mm of rain in a storm. How much runs off?”, or the curve number from a storm’s rain and runoff.
Channel flow by Manning’s equation
Use this for “A channel 10 m wide flows 2 m deep on a slope of 0.001 with n = 0.035. What are its speed and discharge?”, or the depth that carries 50 m³/s.
A storm hydrograph
Use this for “Rain is heaviest at hour 2 and the stream peaks at hour 5. What is the lag time, and why is there one?”