CivilSolve

Hydrology

Runoff Depth & φ-Index

Storm water accounting for a uniform storm: converts a measured runoff volume to an equivalent depth over the watershed (or takes the depth directly), gives the runoff coefficient C = Q/P, the total abstractions, and the φ-index — the constant infiltration rate (P − Q)/t that explains the losses. Use for: 'calculate the runoff depth and φ-index', losses/abstraction problems, runoff coefficient from a storm. Not for: varying-intensity hyetographs (the φ-index then needs iteration on which periods contribute).

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Inputs

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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

  1. 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

  2. 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

  1. 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

  1. 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

  2. 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.