13.2 Surface water hydrology

Achievable FE Civil

13. Water resources engineering

Surface water hydrology

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This chapter covers the following:

Hydrologic cycle and watersheds
Storm/flood frequency probabilities
Hydrologic mass balance problems
Rainfall computation and analysis
Runoff computation and analysis
Hydrograph and unit hydrograph

Hydrologic cycle and watersheds

The hydrologic cycle describes how water continuously moves through the atmosphere, across the land surface, and through the subsurface. Key processes include precipitation, evaporation, transpiration, infiltration, runoff, and groundwater flow.

A common way to summarize the cycle is with a mass balance, which accounts for where precipitation goes over a chosen time period:

$P = E + T + I + R + Δ S$

Where:

$P$ = Precipitation
$E$ = Evaporation
$T$ = Transpiration
$I$ = Infiltration
$R$ = Runoff
$Δ S$ = Change in storage

A watershed sets the physical boundary for many hydrologic calculations. All surface runoff generated inside a watershed drains to a common outlet, so watershed properties strongly influence the hydrologic response. Important characteristics include area, topography, soil properties, and land use.

Watersheds

A watershed is a geographic area that drains all precipitation to a common outlet.

Important features include:

Area
Shape
Slope
Land use

These features affect how quickly runoff forms, how much runoff is generated, and how large peak flows can become.

Storm/flood frequency probabilities

Storm and flood frequency analysis uses historical records and statistical methods to estimate how likely extreme hydrologic events are, and how large they may be.

Results are often reported using a return period (also called a recurrence interval). The return period is an average time between events of a given magnitude, not a guarantee that events occur on a fixed schedule.

General probability

$p = \frac{n _{i}}{N}$

Where:

$p$ = probability of occurrence of a flood flow of class $i$ (variate $i$ )
$n_{i}$ = number of items in the $i$ th class
$N$ = total number of items in a series

Average recurrence interval (ARI)

$p (x \geq x_{T}) = p = \frac{1}{T}$

Where:

$p$ = probability of a single occurrence in a given storm period
$x$ = magnitude of event
$x_{T}$ = design level event
$T$ = storm/flood return period (years)

Risk or annual exceedance probability (AEP)

This expression gives the probability of exceeding the design level at least once over a multi-year period:

$p (x \geq x_{T} at least once in n years) = p = 1 - (1 - \frac{1}{T})^{n}$

Where:

$p$ = probability of exceeding a flow or intensity in a given period
$n$ = number of years

Reliability (probability of nonexceedance)

This is the complementary idea: the probability that the design level is not exceeded in any year over the period:

$p (x < x_{T} each year for n years) = p = (1 - \frac{1}{T})^{n}$

Where:

$p$ = probability of not exceeding a flow or intensity in a given period

Example: Risk of exceedance

A storm drain is designed for the $T = 100$ -year event. What’s the probability that this design event is exceeded at least once during a $30$ -year design life?

$p = 1 - (1 - \frac{1}{100})^{30} = 1 - (0.99)^{30} \approx 1 - 0.740 = 0.260$

Answer: About a $26%$ chance the $100$ -year event is exceeded at least once over $30$ years.

Watch out: In the ARI, AEP, and reliability formulas above, carry full precision through every intermediate step and round only the final reported answer. Rounding $1/ T$ or $(1 - \frac{1}{T})^{n}$ early can shift you to a distractor answer choice.

Hydrologic mass balance problems

Hydrologic mass balance is based on conservation of mass: over a chosen time period, the difference between inflows and outflows must equal the change in storage within the system.

You can apply this idea to a lake, a soil column, a watershed, or a groundwater basin, under steady-state or transient conditions.

The mass balance equation applied to a control volume:

$Input - Output = Δ Storage$

Mathematically:

$P + Q_{in} - E - Q_{o u t} = Δ S$

Where:

$Q_{in}$ , $Q_{o u t}$ = Inflow and outflow (surface or subsurface)
$Δ S$ = Change in storage over time

Example

Given:

$P = 80 mm$
$E = 20 mm$
$Q_{in} = 100 mm$
$Q_{o u t} = 140 mm$

Then:

$Δ S = P + Q_{in} - E - Q_{o u t} = 80 + 100 - 20 - 140 = 20 mm$

Rainfall computation and analysis

Rainfall computation and analysis focuses on describing precipitation in ways that are useful for hydrologic modeling. Common descriptors include intensity, duration, depth, and temporal distribution.

These data support tools such as intensity-duration-frequency (IDF) relationships, hyetographs, and design storms, which are then used as inputs for runoff estimation and flood prediction.

Rainfall estimation methods

Arithmetic mean method:

$P = \frac{1}{n} i = 1 \sum n P_{i}$

Thiessen polygon method:

$P = i = 1 \sum n w_{i} P_{i}$

Isohyetal method: Use area-weighted average between isohyets.

Rainfall intensity (IDF relationship)

These equations relate design rainfall intensity to storm duration and (in some forms) return period.

Equation 1

$i = \frac{c}{T _{d}^{e} + f}$

Equation 2

$i = \frac{c T ^{m}}{T _{d}^{e} + f}$

Where:

$i$ = design rainfall intensity (in./hr)
$T_{d}$ = duration of event (min)
$m$ = rainfall coefficient
$T$ = return period (years)
$c, e, f$ = return-period coefficients

Runoff computation and analysis

Runoff computation transforms precipitation into surface runoff by accounting for hydrologic losses such as infiltration, interception, evaporation, and surface storage.

Depending on watershed size and available data, you may use analytical methods, empirical methods, or a combination. Runoff analysis is used to estimate peak discharge, runoff volume, and the timing of flow.

Rational method

$Q = C I A$

Where:

$Q$ = peak discharge (ft³/sec)
$C$ = runoff coefficient
$I$ = rainfall intensity from an IDF curve for a duration of $t_{c}$ (in./hr)
$A$ = watershed area (acres)

The conversion factor from acre-in./hr to ft³/sec is approximately 1.0 (1 acre-in./hr = 1.008 ft³/sec), so $Q = C I A$ needs no extra constant in US customary units.

Watch out: The rational method has both a US customary form ( $Q = C I A$ , with $I$ in in./hr, $A$ in acres, and a conversion factor of about 1.0) and an SI form ( $Q = C i A /360$ , with $i$ in mm/hr and $A$ in hectares). Identify which unit system a problem uses before you start, and don’t mix in./hr with mm/hr or acres with hectares in the same calculation.

Time of concentration

$t_{c}$ = time of concentration: time required for runoff to travel from the hydraulically most distant point of the watershed to the point of interest (min). Multiple watersheds may require summation of times to determine an overall time for the system.

$t_{c}$ depends on the length of the flow path, the slope along that path, surface roughness (land cover), and watershed shape - a longer, flatter, or rougher path increases $t_{c}$ .

One common way to estimate $t_{c}$ is the Kirpich formula (US customary units):

$t_{c} = 0.0078 L^{0.77} S^{- 0.385}$

Where:

$t_{c}$ = time of concentration (min)
$L$ = length of the longest flow path (ft)
$S$ = average watershed slope (ft/ft)

Example: Kirpich time of concentration

A watershed’s longest flow path is $L = 1, 000 ft$ , with an average slope of $S = 0.01 ft/ft$ . Find $t_{c}$ .

$t_{c} = 0.0078 (1000)^{0.77} (0.01)^{- 0.385} = 0.0078 (204.3) (5.89) \approx 9.4 min$

Answer: $t_{c} \approx 9.4$ minutes

Weighted composite runoff coefficient

Use this when the drainage area includes multiple land covers (and therefore multiple runoff coefficients):

$C_{w} = \frac{A _{1} C _{1} + A _{2} C _{2} + \dots + A _{n} C _{n}}{A _{1} + A _{2} + \dots + A _{n}}$

Where:

$C_{w}$ = weighted/composite runoff coefficient for whole drainage area

Example

This example uses the SI form of the rational method:

$Q = \frac{C i A}{360}$

where $i$ is in mm/hr, $A$ is in hectares, and the constant $360$ converts the mixed mm·ha/hr units into m³/s.

Given: $C = 0.75$ , $i = 60 mm/hr$ , $A = 2 ha$

$Q = \frac{0.75 \times 60 \times 2}{360} = \frac{90}{360} = 0.25 m^{3} / s$

SCS curve number (CN) method

Used to estimate direct runoff from rainfall.

$Q = \frac{( P - 0.2 S ) ^{2}}{P + 0.8 S}, for P > 0.2 S$

Where:

$S = \frac{1000}{CN} - 10 (inches)$

In SI units, the same retention in millimeters is $25.4$ times that: $S = \frac{25400}{CN} - 254$ (mm).

$Q$ = direct runoff depth (in. or mm, same units as $P$ )
$CN$ = Curve number (dimensionless)
$P$ = Precipitation (in. or mm)
$S$ = Potential maximum retention after runoff begins (in. or mm)

Hydrograph and unit hydrograph

A hydrograph shows how discharge at a specific location changes over time, often in response to a storm.

A unit hydrograph is a standardized way to represent watershed response: it describes the direct runoff hydrograph produced by 1 cm (unit depth) of effective rainfall applied uniformly over the watershed for a specified duration. Both tools are widely used in flood analysis and watershed modeling.

Hydrograph

A hydrograph shows variation of stream discharge with time. Key parts include:

Rising limb
Peak flow
Falling limb
Baseflow

Unit hydrograph (UH)

A unit hydrograph is the direct runoff hydrograph resulting from 1 cm (unit depth) of effective rainfall uniformly distributed over the watershed for a specified duration.

Assumes:

Linearity
Time invariance

Example: Applying the unit hydrograph

A watershed’s 3-hour unit hydrograph (the response to 1 cm of effective rainfall in 3 hours) is:

Time (hr) UH (m³/s)

0 0

3 10

6 25

9 20

12 5

15 0

Step 1: scale for a single 2 cm burst. Because the unit hydrograph is linear, doubling the effective rainfall depth doubles every ordinate. For 2 cm of effective rainfall in one 3-hour period:

Time (hr) Direct runoff (m³/s)

0 0

3 20

6 50

9 40

12 10

15 0

Step 2: superpose a second, lagged burst. Now suppose a second 3-hour period produces 1 cm of effective rainfall, starting 3 hours after the first burst. Its response is the unit hydrograph itself, shifted 3 hours later. Adding the two responses at each time (superposition) gives the combined direct runoff hydrograph:

Time (hr) Burst 1 (2 cm) Burst 2 (1 cm, lagged 3 hr) Total direct runoff (m³/s)

0 0 0 0

3 20 0 20

6 50 10 60

9 40 25 65

12 10 20 30

15 0 5 5

18 0 0 0

Answer: The combined hydrograph peaks at $65 m^{3} / s$ at $t = 9$ hr. This result reflects both the linearity assumption (scaling ordinates for a 2 cm burst) and the superposition/lagging needed to combine responses from successive rainfall periods.

Time (hr)	UH (m³/s)
0	0
3	10
6	25
9	20
12	5
15	0

Time (hr)	Direct runoff (m³/s)
0	0
3	20
6	50
9	40
12	10
15	0

Time (hr)	Burst 1 (2 cm)	Burst 2 (1 cm, lagged 3 hr)	Total direct runoff (m³/s)
0	0	0	0
3	20	0	20
6	50	10	60
9	40	25	65
12	10	20	30
15	0	5	5
18	0	0	0

Hydrologic cycle and watersheds

Hydrologic cycle: continuous movement of water via precipitation, evaporation, transpiration, infiltration, runoff, groundwater flow
Mass balance equation: $P = E + T + I + R + Δ S$
Watershed: area draining to a common outlet; key features - area, shape, slope, land use

Storm/flood frequency probabilities

Return period (recurrence interval): average time between events of a given size
Probability formulas:
- $p = \frac{n _{i}}{N}$ (general probability)
- $p = \frac{1}{T}$ (probability of event in one year)
- $p = 1 - (1 - \frac{1}{T})^{n}$ (probability of exceedance in $n$ years)
- $p = (1 - \frac{1}{T})^{n}$ (probability of nonexceedance in $n$ years)

Hydrologic mass balance problems

Conservation of mass: Input - Output = Change in storage
Mass balance equation: $P + Q_{in} - E - Q_{o u t} = Δ S$
Applies to lakes, watersheds, soil columns, groundwater basins

Rainfall computation and analysis

Rainfall descriptors: intensity, duration, depth, temporal distribution
Estimation methods:
- Arithmetic mean: $P = \frac{1}{n} \sum P_{i}$
- Thiessen polygon: $P = \sum w_{i} P_{i}$
- Isohyetal: area-weighted average between isohyets
IDF relationships: $i = \frac{c T ^{m}}{T _{d}^{e} + f}$ (relates intensity, duration, return period)

Runoff computation and analysis

Runoff = precipitation minus losses (infiltration, evaporation, etc.)
Rational method: $Q = C I A$
- $C$ = runoff coefficient, $I$ = rainfall intensity, $A$ = area
- Weighted $C$ : $C_{w} = \frac{\sum A _{i} C _{i}}{\sum A _{i}}$
Time of concentration ( $t_{c}$ ): time for runoff to reach outlet from furthest point
SCS Curve Number method:
- $Q = \frac{( P - 0.2 S ) ^{2}}{P + 0.8 S}$ , $S = \frac{25400}{CN} - 254$

Hydrograph and unit hydrograph

Hydrograph: graph of stream discharge vs. time (rising limb, peak, falling limb, baseflow)
Unit hydrograph (UH): direct runoff hydrograph from 1 cm effective rainfall, assumes linearity and time invariance
- Used to scale and combine runoff responses for different rainfall events

Next | 13.3 Ground water hydrology

Surface water hydrology

This chapter covers the following:

Hydrologic cycle and watersheds
Storm/flood frequency probabilities
Hydrologic mass balance problems
Rainfall computation and analysis
Runoff computation and analysis
Hydrograph and unit hydrograph

Hydrologic cycle and watersheds

A common way to summarize the cycle is with a mass balance, which accounts for where precipitation goes over a chosen time period:

$P = E + T + I + R + Δ S$

Where:

$P$ = Precipitation
$E$ = Evaporation
$T$ = Transpiration
$I$ = Infiltration
$R$ = Runoff
$Δ S$ = Change in storage

Watersheds

A watershed is a geographic area that drains all precipitation to a common outlet.

Important features include:

Area
Shape
Slope
Land use

These features affect how quickly runoff forms, how much runoff is generated, and how large peak flows can become.

Storm/flood frequency probabilities

Storm and flood frequency analysis uses historical records and statistical methods to estimate how likely extreme hydrologic events are, and how large they may be.

General probability

$p = \frac{n _{i}}{N}$

Where:

$p$ = probability of occurrence of a flood flow of class $i$ (variate $i$ )
$n_{i}$ = number of items in the $i$ th class
$N$ = total number of items in a series

Average recurrence interval (ARI)

$p (x \geq x_{T}) = p = \frac{1}{T}$

Where:

$p$ = probability of a single occurrence in a given storm period
$x$ = magnitude of event
$x_{T}$ = design level event
$T$ = storm/flood return period (years)

Risk or annual exceedance probability (AEP)

This expression gives the probability of exceeding the design level at least once over a multi-year period:

$p (x \geq x_{T} at least once in n years) = p = 1 - (1 - \frac{1}{T})^{n}$

Where:

$p$ = probability of exceeding a flow or intensity in a given period
$n$ = number of years

Reliability (probability of nonexceedance)

This is the complementary idea: the probability that the design level is not exceeded in any year over the period:

$p (x < x_{T} each year for n years) = p = (1 - \frac{1}{T})^{n}$

Where:

$p$ = probability of not exceeding a flow or intensity in a given period

Example: Risk of exceedance

A storm drain is designed for the $T = 100$ -year event. What’s the probability that this design event is exceeded at least once during a $30$ -year design life?

$p = 1 - (1 - \frac{1}{100})^{30} = 1 - (0.99)^{30} \approx 1 - 0.740 = 0.260$

Answer: About a $26%$ chance the $100$ -year event is exceeded at least once over $30$ years.

Hydrologic mass balance problems

Hydrologic mass balance is based on conservation of mass: over a chosen time period, the difference between inflows and outflows must equal the change in storage within the system.

You can apply this idea to a lake, a soil column, a watershed, or a groundwater basin, under steady-state or transient conditions.

The mass balance equation applied to a control volume:

$Input - Output = Δ Storage$

Mathematically:

$P + Q_{in} - E - Q_{o u t} = Δ S$

Where:

$Q_{in}$ , $Q_{o u t}$ = Inflow and outflow (surface or subsurface)
$Δ S$ = Change in storage over time

Example

Given:

$P = 80 mm$
$E = 20 mm$
$Q_{in} = 100 mm$
$Q_{o u t} = 140 mm$

Then:

$Δ S = P + Q_{in} - E - Q_{o u t} = 80 + 100 - 20 - 140 = 20 mm$

Rainfall computation and analysis

These data support tools such as intensity-duration-frequency (IDF) relationships, hyetographs, and design storms, which are then used as inputs for runoff estimation and flood prediction.

Rainfall estimation methods

Arithmetic mean method:

$P = \frac{1}{n} i = 1 \sum n P_{i}$

Thiessen polygon method:

$P = i = 1 \sum n w_{i} P_{i}$

Isohyetal method: Use area-weighted average between isohyets.

Rainfall intensity (IDF relationship)

These equations relate design rainfall intensity to storm duration and (in some forms) return period.

Equation 1

$i = \frac{c}{T _{d}^{e} + f}$

Equation 2

$i = \frac{c T ^{m}}{T _{d}^{e} + f}$

Where:

$i$ = design rainfall intensity (in./hr)
$T_{d}$ = duration of event (min)
$m$ = rainfall coefficient
$T$ = return period (years)
$c, e, f$ = return-period coefficients

Runoff computation and analysis

Runoff computation transforms precipitation into surface runoff by accounting for hydrologic losses such as infiltration, interception, evaporation, and surface storage.

Rational method

$Q = C I A$

Where:

$Q$ = peak discharge (ft³/sec)
$C$ = runoff coefficient
$I$ = rainfall intensity from an IDF curve for a duration of $t_{c}$ (in./hr)
$A$ = watershed area (acres)

The conversion factor from acre-in./hr to ft³/sec is approximately 1.0 (1 acre-in./hr = 1.008 ft³/sec), so $Q = C I A$ needs no extra constant in US customary units.

Time of concentration

$t_{c}$ = time of concentration: time required for runoff to travel from the hydraulically most distant point of the watershed to the point of interest (min). Multiple watersheds may require summation of times to determine an overall time for the system.

$t_{c}$ depends on the length of the flow path, the slope along that path, surface roughness (land cover), and watershed shape - a longer, flatter, or rougher path increases $t_{c}$ .

One common way to estimate $t_{c}$ is the Kirpich formula (US customary units):

$t_{c} = 0.0078 L^{0.77} S^{- 0.385}$

Where:

$t_{c}$ = time of concentration (min)
$L$ = length of the longest flow path (ft)
$S$ = average watershed slope (ft/ft)

Example: Kirpich time of concentration

A watershed’s longest flow path is $L = 1, 000 ft$ , with an average slope of $S = 0.01 ft/ft$ . Find $t_{c}$ .

$t_{c} = 0.0078 (1000)^{0.77} (0.01)^{- 0.385} = 0.0078 (204.3) (5.89) \approx 9.4 min$

Answer: $t_{c} \approx 9.4$ minutes

Weighted composite runoff coefficient

Use this when the drainage area includes multiple land covers (and therefore multiple runoff coefficients):

$C_{w} = \frac{A _{1} C _{1} + A _{2} C _{2} + \dots + A _{n} C _{n}}{A _{1} + A _{2} + \dots + A _{n}}$

Where:

$C_{w}$ = weighted/composite runoff coefficient for whole drainage area

Example

This example uses the SI form of the rational method:

$Q = \frac{C i A}{360}$

where $i$ is in mm/hr, $A$ is in hectares, and the constant $360$ converts the mixed mm·ha/hr units into m³/s.

Given: $C = 0.75$ , $i = 60 mm/hr$ , $A = 2 ha$

$Q = \frac{0.75 \times 60 \times 2}{360} = \frac{90}{360} = 0.25 m^{3} / s$

SCS curve number (CN) method

Used to estimate direct runoff from rainfall.

$Q = \frac{( P - 0.2 S ) ^{2}}{P + 0.8 S}, for P > 0.2 S$

Where:

$S = \frac{1000}{CN} - 10 (inches)$

In SI units, the same retention in millimeters is $25.4$ times that: $S = \frac{25400}{CN} - 254$ (mm).

$Q$ = direct runoff depth (in. or mm, same units as $P$ )
$CN$ = Curve number (dimensionless)
$P$ = Precipitation (in. or mm)
$S$ = Potential maximum retention after runoff begins (in. or mm)

Hydrograph and unit hydrograph

A hydrograph shows how discharge at a specific location changes over time, often in response to a storm.

Hydrograph

A hydrograph shows variation of stream discharge with time. Key parts include:

Rising limb
Peak flow
Falling limb
Baseflow

Unit hydrograph (UH)

A unit hydrograph is the direct runoff hydrograph resulting from 1 cm (unit depth) of effective rainfall uniformly distributed over the watershed for a specified duration.

Assumes:

Linearity
Time invariance

Example: Applying the unit hydrograph

A watershed’s 3-hour unit hydrograph (the response to 1 cm of effective rainfall in 3 hours) is:

Time (hr) UH (m³/s)

0 0

3 10

6 25

9 20

12 5

15 0

Step 1: scale for a single 2 cm burst. Because the unit hydrograph is linear, doubling the effective rainfall depth doubles every ordinate. For 2 cm of effective rainfall in one 3-hour period:

Time (hr) Direct runoff (m³/s)

0 0

3 20

6 50

9 40

12 10

15 0

Step 2: superpose a second, lagged burst. Now suppose a second 3-hour period produces 1 cm of effective rainfall, starting 3 hours after the first burst. Its response is the unit hydrograph itself, shifted 3 hours later. Adding the two responses at each time (superposition) gives the combined direct runoff hydrograph:

Time (hr) Burst 1 (2 cm) Burst 2 (1 cm, lagged 3 hr) Total direct runoff (m³/s)

0 0 0 0

3 20 0 20

6 50 10 60

9 40 25 65

12 10 20 30

15 0 5 5

18 0 0 0

Answer: The combined hydrograph peaks at $65 m^{3} / s$ at $t = 9$ hr. This result reflects both the linearity assumption (scaling ordinates for a 2 cm burst) and the superposition/lagging needed to combine responses from successive rainfall periods.

Time (hr)	UH (m³/s)
0	0
3	10
6	25
9	20
12	5
15	0

Time (hr)	Direct runoff (m³/s)
0	0
3	20
6	50
9	40
12	10
15	0

Time (hr)	Burst 1 (2 cm)	Burst 2 (1 cm, lagged 3 hr)	Total direct runoff (m³/s)
0	0	0	0
3	20	0	20
6	50	10	60
9	40	25	65
12	10	20	30
15	0	5	5
18	0	0	0

Achievable FE Civil

13. Water resources engineering