SME
Hydraulics & Hydrology · Problem type
Peak flow by the rational method
Use this for “A 20 ha suburban catchment with C = 0.6 gets a 50 mm/h design storm. What peak flow must the outlet carry?”
Diagram
Q
=
CiA
360
=
0.6
×
50
×
20
360
=
1.667
m³/s:
of
all
the
rain,
the
share
C
runs
off
(360
turns
mm/h
on
hectares
into
m
3
/s).
This is an example. Type your own values to start a new problem.
Q (m³/s)
=
×
i (mm/h)
×
A (ha)
360
Formulas
Q
=
C
×
i
×
A
360
1.667
=
0.6
×
50
×
20
360
Assumptions
•
A small catchment, under about 80 ha (0.8 km²) for most design manuals.
•
The storm lasts at least the time of concentration, so the whole area drains to the outlet at once.
•
i in mm/h and A in hectares: dividing by 360 gives m³/s (1 ha = 10,000 m²).
Step-by-step
WE KNOW
C = 0.6, i = 50 mm/h, A = 20 ha
FIND
peak flow (Q)
Step 1 · Find peak flow (Q)
Use Q = C × i × A ÷ 360
The
share
C
of
the
rain
falling
on
the
whole
area
runs
off;
÷
360
turns
mm/h
×
ha
into
m
3
/s.
Q
=
0.6
×
50
×
20
360
Q
=
30
×
20
360
Q
=
600
360
Q
=
1.667
m³/s
Check
Put the numbers back in. Both sides should match.
1.667
=
0.6
×
50
×
20
360
✓
Topic check
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