Calculate runoff depth, runoff coefficient, total abstractions, and the phi-index for a uniform storm using rainfall and watershed runoff data.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
runoff depth and φ-index (the motivating exam problem)
Given
rainfall depth
75 mm
storm duration
4 h
watershed area
5 km^2
runoff volume
200000 m^3
Assumptions
Rainfall and infiltration are uniform in time and space over the storm (the constant-φ idealization).
Solution steps
Runoff volume → equivalent depth
Spreading the outlet volume over the watershed gives the runoff as a depth, comparable with the rainfall.
Q = V / A
Q = 200000 m^3 / 5 km^2 = 40 mm
= 40 mm
Abstractions and the φ-index
Whatever did not run off was abstracted (infiltration and storage); spreading it evenly over the storm gives the φ-index.
φ = (P − Q)/t
losses = 35 mm over 4 h → φ = 8.75 mm/h
= 8.75 mm/h
Results
Direct runoff depth Q
40mm
Total abstractions P − Q
35mm
φ-index (average infiltration rate)
8.75mm/h
Runoff coefficient Q/P
0.5333
Where this answer was checked
source
Q = V/A; φ = (P − Q)/t — 75 mm over 4 h on 5 km²; runoff volume 200,000 m³
verified by
hand-recomputed
derivation
Q = 200000/5e6 = 0.04 m = 40 mm. Losses = 35 mm over 4 h → φ = 8.75 mm/h. C = 40/75 = 0.5333.
Example 2
US depth-direct: 3 in storm, 1.2 in runoff over 6 h
Given
rainfall depth
3 in
storm duration
6 h
watershed area
2 mi^2
runoff depth
1.2 in
Assumptions
Rainfall and infiltration are uniform in time and space over the storm (the constant-φ idealization).
Solution steps
Abstractions and the φ-index
Whatever did not run off was abstracted (infiltration and storage); spreading it evenly over the storm gives the φ-index.
φ = (P − Q)/t
losses = 1.8 in over 6 h → φ = 0.3 in/h
= 0.3 in/h
Results
Direct runoff depth Q
1.2in
Total abstractions P − Q
1.8in
φ-index (average infiltration rate)
0.3in/h
Runoff coefficient Q/P
0.4
Where this answer was checked
source
φ = (P − Q)/t with the depth given — US units
verified by
hand-recomputed
derivation
losses = 1.8 in over 6 h → φ = 0.3 in/h; C = 0.4.
Example 3
metric ha-scale storm
Given
rainfall depth
50 mm
storm duration
2 h
watershed area
80 ha
runoff volume
16000 m^3
Assumptions
Rainfall and infiltration are uniform in time and space over the storm (the constant-φ idealization).
Solution steps
Runoff volume → equivalent depth
Spreading the outlet volume over the watershed gives the runoff as a depth, comparable with the rainfall.
Q = V / A
Q = 16000 m^3 / 0.8 km^2 = 20 mm
= 20 mm
Abstractions and the φ-index
Whatever did not run off was abstracted (infiltration and storage); spreading it evenly over the storm gives the φ-index.
φ = (P − Q)/t
losses = 30 mm over 2 h → φ = 15 mm/h
= 15 mm/h
Results
Direct runoff depth Q
20mm
Total abstractions P − Q
30mm
φ-index (average infiltration rate)
15mm/h
Runoff coefficient Q/P
0.4
Where this answer was checked
source
Q = V/A on a small catchment — 50 mm over 2 h on 80 ha, 16,000 m³ runoff
verified by
hand-recomputed
derivation
A = 800,000 m²; Q = 16000/8e5 = 0.02 m = 20 mm; φ = 30/2 = 15 mm/h.