NRCS (SCS) Rainfall-Runoff

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

8 formulas

10 exam-style examples

~60 min

All Hydrology and Water Resources lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers NRCS (SCS) Rainfall-Runoff within Hydrology and Water Resources. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what nrcs (scs) rainfall-runoff describes physically and when it applies.
State every one of the 8 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: acre-in/hr ≈ cfs makes the rational formula work in US units.

Lecture

Why this section exists. NRCS (SCS) Rainfall-Runoff is the part of Hydrology and Water Resources that lets you connect a watershed, aquifer or detention facility to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a rainfall-runoff or well-drawdown calculation with one lookup. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. acre-in/hr ≈ cfs makes the rational formula work in US units. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: nrcs (scs) rainfall-runoff.
Capstone Studio instructional photograph

Hydrology and Water Resources — NRCS (SCS) Rainfall-Runoff: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a watershed, aquifer or detention facility. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 8 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

^ h	Quantity produced by "^ h" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = P - 0.2S" — read its definition and unit from the handbook line directly above the equation.
S	Quantity produced by "S= 10" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = precipitation (inches)" — read its definition and unit from the handbook line directly above the equation.
CN	Quantity produced by "CN = curve number" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

P + 0.8S
1, 000
CN -
1, 000
S + 10
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Rational-method peak flow for a mixed catchment

A 14 ha site is 6 ha pavement (C = 0.90) and 8 ha lawn (C = 0.20). The 10-year intensity for t_c = 15 min is 95 mm/h. Find the peak flow.

Given

A₁ = 6 ha at C = 0.90
A₂ = 8 ha at C = 0.20
i = 95 mm/h

Find

Q_p (m³/s)

Start with the thinking

Area-weight C before applying the formula.
In SI, Q = CiA/360 with i in mm/h and A in hectares.

Step-by-step solution

Weighted runoff coefficient
$C = [0.90 (6) + 0.20 (8)] /14 = (5.40 + 1.60) /14$
Evaluate
$C = 0.500$
Rational formula
$Q = C i A /360$
Substitute
$Q = 0.500 (95) (14) /360 = 665/360$
Result
$Q_{p} = 1.85 m^{3} / s$

Answer: Q_p ≈ 1.85 m³/s

Why the other options are there

3.33 m³/s (C = 0.90 applied to the whole site)
0.74 m³/s (C = 0.20 applied to the whole site)

Reference: FE Reference Handbook — Hydrology — Rational method

Example 2

Detention volume from a simple mass balance

A storm produces an inflow of 1.85 m³/s for 25 minutes while the allowable release is 0.60 m³/s. Estimate the required storage.

Given

Q_in = 1.85 m³/s
Q_out = 0.60 m³/s
Duration = 25 min = 1,500 s

Find

Storage volume

Start with the thinking

Storage is the area between the inflow and outflow over the storm.
Convert minutes to seconds before multiplying.

Step-by-step solution

Net rate — ΔQ = 1.85 − 0.60 = 1.25 m³/s
Duration
$t = 25 (60) = 1, 500 s$
Volume — V = ΔQ·t = 1.25(1,500)
Result
$V = 1, 875 m^{3} (ab o u t 1, 900 m^{3} w i t h f r ee b o a r d)$

Answer: V ≈ 1,875 m³

Why the other options are there

2,775 m³ (outflow ignored)
31 m³ (minutes used as seconds)

Reference: FE Reference Handbook — Hydrology — Detention storage

Example 3

Runoff depth from the NRCS curve number method — NRCS (SCS) Rainfall-Runoff

A watershed with curve number CN = 88 receives 3.5 in of rainfall in a design storm. Compute the direct runoff depth.

Given

CN = 88
P = 3.5 in
I_a = 0.2S

Find

Runoff depth Q (in)

Start with the thinking

Potential retention S grows as CN falls — pervious soils store more.
No runoff occurs until rainfall exceeds the initial abstraction.

Step-by-step solution

Retention
$S = 1000/ C N - 10$
Substituting
$S = 1000/88 - 10 = 1.36 in$
Initial abstraction
$I_{a} = 0.2 S = 0.2 (1.36) = 0.27 in$
Runoff
$Q = (P - I_{a})^{2} / (P + 0.8 S)$
Substituting
$Q = (3.5 - 0.27)^{2} / (3.5 + 0.8 \times 1.36)$
Evaluate
$Q = 2.269 in$

Answer: Q ≈ 2.27 in of direct runoff

Why the other options are there

3.23 in (abstraction subtracted only)
2.67 in (I_a omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 4

Design rainfall intensity and depth from an IDF curve — NRCS (SCS) Rainfall-Runoff

A local IDF relation is i = 77/(t + 22)^0.75 with i in in/hr and t in minutes. For a storm duration of 48 minutes, find the design intensity and the total rainfall depth.

Given

a = 77
b = 22
c = 0.75
t = 48 min

Find

i and total depth P

Start with the thinking

The duration used in an IDF curve is normally set equal to the time of concentration.
Depth is intensity multiplied by duration in consistent time units.

Step-by-step solution

Formula
$i = a / (t + b)^{c}$
Denominator
$(t + b)^{c} = (48 + 22)^{0} .75 = 24.200$
Substituting
$i = 77/24.200 = 3.182 in / h r$
Depth
$P = i \times t$
Substituting
$P = 3.182 in / h r \times (48/60) h r = 2.545 in$

Answer: i ≈ 3.18 in/hr; P ≈ 2.55 in

Why the other options are there

152.7 in (minutes not converted to hours)
1.10 in/hr (exponent ignored)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 5

Runoff depth from the NRCS curve number method — NRCS (SCS) Rainfall-Runoff (2)

A watershed with curve number CN = 85 receives 5.3 in of rainfall in a design storm. Compute the direct runoff depth.

Given

CN = 85
P = 5.3 in
I_a = 0.2S

Find

Runoff depth Q (in)

Start with the thinking

Potential retention S grows as CN falls — pervious soils store more.
No runoff occurs until rainfall exceeds the initial abstraction.

Step-by-step solution

Retention
$S = 1000/ C N - 10$
Substituting
$S = 1000/85 - 10 = 1.76 in$
Initial abstraction
$I_{a} = 0.2 S = 0.2 (1.76) = 0.35 in$
Runoff
$Q = (P - I_{a})^{2} / (P + 0.8 S)$
Substituting
$Q = (5.3 - 0.35)^{2} / (5.3 + 0.8 \times 1.76)$
Evaluate
$Q = 3.646 in$

Answer: Q ≈ 3.65 in of direct runoff

Why the other options are there

4.95 in (abstraction subtracted only)
4.19 in (I_a omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 6

Design rainfall intensity and depth from an IDF curve — NRCS (SCS) Rainfall-Runoff (2)

A local IDF relation is i = 101/(t + 17)^0.85 with i in in/hr and t in minutes. For a storm duration of 37 minutes, find the design intensity and the total rainfall depth.

Given

a = 101
b = 17
c = 0.85
t = 37 min

Find

i and total depth P

Start with the thinking

The duration used in an IDF curve is normally set equal to the time of concentration.
Depth is intensity multiplied by duration in consistent time units.

Step-by-step solution

Formula
$i = a / (t + b)^{c}$
Denominator
$(t + b)^{c} = (37 + 17)^{0} .85 = 29.685$
Substituting
$i = 101/29.685 = 3.402 in / h r$
Depth
$P = i \times t$
Substituting
$P = 3.402 in / h r \times (37/60) h r = 2.098 in$

Answer: i ≈ 3.40 in/hr; P ≈ 2.10 in

Why the other options are there

125.9 in (minutes not converted to hours)
1.87 in/hr (exponent ignored)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 7

Runoff depth from the NRCS curve number method — NRCS (SCS) Rainfall-Runoff (3)

A watershed with curve number CN = 81 receives 5.8 in of rainfall in a design storm. Compute the direct runoff depth.

Given

CN = 81
P = 5.8 in
I_a = 0.2S

Find

Runoff depth Q (in)

Start with the thinking

Potential retention S grows as CN falls — pervious soils store more.
No runoff occurs until rainfall exceeds the initial abstraction.

Step-by-step solution

Retention
$S = 1000/ C N - 10$
Substituting
$S = 1000/81 - 10 = 2.35 in$
Initial abstraction
$I_{a} = 0.2 S = 0.2 (2.35) = 0.47 in$
Runoff
$Q = (P - I_{a})^{2} / (P + 0.8 S)$
Substituting
$Q = (5.8 - 0.47)^{2} / (5.8 + 0.8 \times 2.35)$
Evaluate
$Q = 3.702 in$

Answer: Q ≈ 3.70 in of direct runoff

Why the other options are there

5.33 in (abstraction subtracted only)
4.38 in (I_a omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 8

Design rainfall intensity and depth from an IDF curve — NRCS (SCS) Rainfall-Runoff (3)

A local IDF relation is i = 132/(t + 24)^0.70 with i in in/hr and t in minutes. For a storm duration of 46 minutes, find the design intensity and the total rainfall depth.

Given

a = 132
b = 24
c = 0.70
t = 46 min

Find

i and total depth P

Start with the thinking

The duration used in an IDF curve is normally set equal to the time of concentration.
Depth is intensity multiplied by duration in consistent time units.

Step-by-step solution

Formula
$i = a / (t + b)^{c}$
Denominator
$(t + b)^{c} = (46 + 24)^{0} .70 = 19.569$
Substituting
$i = 132/19.569 = 6.745 in / h r$
Depth
$P = i \times t$
Substituting
$P = 6.745 in / h r \times (46/60) h r = 5.171 in$

Answer: i ≈ 6.75 in/hr; P ≈ 5.17 in

Why the other options are there

310.3 in (minutes not converted to hours)
1.89 in/hr (exponent ignored)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 9

Runoff depth from the NRCS curve number method — NRCS (SCS) Rainfall-Runoff (4)

A watershed with curve number CN = 65 receives 5.0 in of rainfall in a design storm. Compute the direct runoff depth.

Given

CN = 65
P = 5.0 in
I_a = 0.2S

Find

Runoff depth Q (in)

Start with the thinking

Potential retention S grows as CN falls — pervious soils store more.
No runoff occurs until rainfall exceeds the initial abstraction.

Step-by-step solution

Retention
$S = 1000/ C N - 10$
Substituting
$S = 1000/65 - 10 = 5.38 in$
Initial abstraction
$I_{a} = 0.2 S = 0.2 (5.38) = 1.08 in$
Runoff
$Q = (P - I_{a})^{2} / (P + 0.8 S)$
Substituting
$Q = (5.0 - 1.08)^{2} / (5.0 + 0.8 \times 5.38)$
Evaluate
$Q = 1.654 in$

Answer: Q ≈ 1.65 in of direct runoff

Why the other options are there

3.92 in (abstraction subtracted only)
2.69 in (I_a omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Example 10

Design rainfall intensity and depth from an IDF curve — NRCS (SCS) Rainfall-Runoff (4)

A local IDF relation is i = 63/(t + 21)^0.80 with i in in/hr and t in minutes. For a storm duration of 60 minutes, find the design intensity and the total rainfall depth.

Given

a = 63
b = 21
c = 0.80
t = 60 min

Find

i and total depth P

Start with the thinking

The duration used in an IDF curve is normally set equal to the time of concentration.
Depth is intensity multiplied by duration in consistent time units.

Step-by-step solution

Formula
$i = a / (t + b)^{c}$
Denominator
$(t + b)^{c} = (60 + 21)^{0} .80 = 33.635$
Substituting
$i = 63/33.635 = 1.873 in / h r$
Depth
$P = i \times t$
Substituting
$P = 1.873 in / h r \times (60/60) h r = 1.873 in$

Answer: i ≈ 1.87 in/hr; P ≈ 1.87 in

Why the other options are there

112.4 in (minutes not converted to hours)
0.78 in/hr (exponent ignored)

Reference: FE Reference Handbook — Hydrology and Water Resources → NRCS (SCS) Rainfall-Runoff

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a watershed, aquifer or detention facility, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

NRCS (SCS) Rainfall-Runoff contains 8 relations; you must be able to find this page in under 15 seconds.
Exam style: a rainfall-runoff or well-drawdown calculation with one lookup.
Unit rule: acre-in/hr ≈ cfs makes the rational formula work in US units.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

acre-in/hr ≈ cfs makes the rational formula work in US units
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.