Hydraulics & Hydrology

Rainfall–runoff

Use this for “A watershed with curve number 80 gets 4 in of rain in a storm. How many inches run off?”, or the curve number from a storm’s rain and runoff.

Diagram

The storm, splitIₐ = 0.5 inF = 1.46 inQ = 2.04 inPP = 4 inRunoff Q against rain P (in)002244CN = 80dashed: all of it runs off
S
=
1000
CN
−
10
=
1000
80
−
10
=
2.5
in;
Iₐ
=
0.2S
=
0.5
in.
Q
=
(P − Iₐ)
2
÷
(P
+
0.8S)
Q
=
(4 − 0.5)
2
÷
(4
+
0.8
×
2.5)
=
2.04
in
The rest soaks in: F = P − Iₐ − Q = 4 − 0.5 − 2.04 = 1.46 in, so Iₐ + F + Q = P.
This is an example. Type your own values to start a new problem.
CN
Curve number
example
S
Potential retention
calculated · in
Iₐ
Initial abstraction
calculated · in
P
Storm rainfall
example · in
Q
Runoff depth
calculated · in
F
Infiltration
worked out (not typed) · in

Formulas

S
=
1,000
CN
−
10
2.5
=
1,000
80
−
10
Iₐ
=
0.2
×
S
0.5 = 0.2 × 2.5
Q
=
(
P
− Iₐ)
2
÷
(
P
+
0.8
×
S
)
2.04
=
(4 − 0.5)
2
÷
(4
+
0.8
×
2.5)
S
=
5
×
(
P
+
2
×
Q
−
√
4 ×
Q
2
+ 5 ×
P
×
Q
)
2.5
=
5
×
(4
+
2
×
2.04
−
√
4 ×
2.04
2
+ 5 × 4 × 2.04
)
F
=
P
−
Iₐ
−
Q
1.46 = 4 − 0.5 − 2.04

Assumptions

•
Depths in inches, as NRCS TR-55 publishes the method; the 1000 and 10 hold only in inches.
•
Average soil moisture before the storm (the curve number of a normal antecedent condition).
•
Iₐ = 0.2S, the ratio of the method: nothing runs off until the first Iₐ of rain is caught.

Step-by-step

WE KNOW
CN = 80, P = 4 in
FIND
potential retention (S), initial abstraction (Iₐ), runoff depth (Q), infiltration (F)
Step 1 · Find potential retention (S)
Use S = 1,000 ÷ CN − 10
The curve number turned into the most rain the ground could still hold, in inches.
S
=
1,000
80
−
10
S
=
12.5
−
10
S
=
2.5
in
Step 2 · Find initial abstraction (Iₐ)
Use Iₐ = 0.2 × S
The
rain
caught
by
leaves
and
hollows
before
any
runs
off
is
a
fifth
of
S
.
Iₐ = 0.2 × 2.5
Iₐ = 0.5 in
Step 3 · Find runoff depth (Q)
Use Q = (P − Iₐ)² ÷ (P + 0.8 × S)
The
rain
past
Iₐ,
squared,
over
the
rain
plus
0.8
S
:
more
of
each
extra
inch
runs
off
as
the
ground
fills.
4 > 0.5: some runs off
Q
=
(4 − 0.5)
2
÷
(4
+
0.8
×
2.5)
Q
=
3.5
2
6
Q
=
12.25
6
Q
=
2.04
in
Step 4 · Find infiltration (F)
Use F = P − Iₐ − Q
What is left of the rain after Iₐ and the runoff soaks into the ground.
F
=
4
−
0.5
−
2.04
F
=
1.46
in
Check
Put the numbers back in. Both sides should match.
2.5
=
1,000
80
−
10
✓
0.5 = 0.2 × 2.5 ✓
2.04
=
(4 − 0.5)
2
÷
(4
+
0.8
×
2.5)
✓
2.5
=
5
×
(4
+
2
×
2.04
−
√
4 ×
2.04
2
+ 5 × 4 × 2.04
)
✓
1.46 = 4 − 0.5 − 2.04 ✓
Topic check
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