By: ANKEAT359 : Water Resources Engineering

School of Environmental EngineeringUNIVERSITI MALAYSIA PERLIS

EAT 359/3 : Water Resources Engineering

SURFACE WATER

AIN NIHLA KAMARUDZAMAN

ainnihla@unimap.edu.my

Phone: 019 3848233

At the end of this lesson, the student shall be able to learn:

1. What is the difference between overland flow, interflow and base flow components contributing to stream flow generation.

2. What are hydrograph and hyetographs.

3. Methods to separate base flow from stream hydrograph to find out the Direct Runoff Hydrograph.

4. What is Unit Hydrograph and its assumptions

5. Application of the Unit Hydrograph to find the Direct Runoff Hydrograph

Surface Runoff

• A part of precipitation goes to atmosphere by evaporation and transpiration.

• The remaining part goes to the stream or river of the catchments as:

1. Surface water flow or overland flow2. Interflow or sub surface flow3. Groundwater flow

• The runoff is defined as a part of precipitation which is not evapotranspirated.

• Two type of runoff: surface and sub surface

• Surface runoff is a major component of water cycle.

• Theoretically, surface runoff is the net amount of rainfall after subtracted by evapotranspiration and infiltration.

• In reality, surface runoff is equivalent to river or stream flow (Q in m3/s, lps or ft3/s) of the catchment.

Figure 1 shows how runoff in different forms occurs after precipitation.

Hydrograph Analysis

• Hydrograph – a record and graphical representations of discharge as a function of time at a specific location.

• After a rainfall/storm, discharges, Q are measured at the outlet at different time interval (1 hr or 2 hr or 4 hr).

• Function:

• Reservoir design

• Assessing influence of flood control structures in reducing flood peaks

• Determining the duration and frequency of flooding

• Separating contributions from ground water and surface water

• Concept of Hydrograph:

1. Natural Hydrograph

2. Unit Hydrograph

3. Synthetic Hydrograph

Natural Hydrograph

• An observed Q - t relationship of a catchment due to rainfall event.

• A rainfall event produces a single hydrograph.

• A natural hydrograph has important characteristics;

• Base flow recession

• Rising limb

• Falling limb

• Peak flow

• Inflection points

Peak Flood

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

ch

arg

e (

m3/s

)

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

Rainfall shown in

mm, as a bar graph

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

Discharge in m3/s,

as a line graph

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

The rising of flood

water in the river

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

Peak flow

Peak flow

Maximum discharge in the

river (peak discharge)

• The most important parts of hydrograph.

• Peak flow information –useful for design of hydraulic structures.

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

Peak flow The falling of flood

water in the river

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)

Base flow

mm

4

3

2

Peak flow

Base flow

Normal discharge of

the river

0 12 24 36 48 30 72

Hours from start of rain storm

3

2

1

Dis

charg

e (

m3/s

)mm

4

3

2

Lag timeLag Time

The gap between the peak

rainfall and the peak

discharge

Factors Affecting Runoff Hydrograph

Physiographic Factors Climatic Factors

1. Basin Characteristics:

a) Shape

b) Size

c) Slope

d) Nature of the valley

e) Elevation

f) Drainage density

2. Infiltration Characteristics:

a) Land use and cover

b) Soil type and geological conditions

c) Lakes, swamps and other storage

3. Channel Characteristics:

a) Cross-section

b) Roughness

c) Storage capacity

1. Storm Characteristics:

a) Precipitation

b) Intensity

c) Duration

d) Magnitude

e) Movement of storm

2. Initial loss

3. Evapotranspiration

Assignment

Discuss how the shape of runoff hydrograph fluctuates with the catchment characteristics and rainfall characteristics?

Due date: 03/04/2017

Base Flow Separation

Natural hydrograph consists of two main components; runoff component and base flow component.

The direct runoff is obtained by separating the base flow from the natural hydrograph.

Remember:

The base flow is the initial flow of the river before the rain comes.

It is produced from previous season (rainfall) and also considered to be mostly from the ground water contribution.

Four Techniques for base flow separation:

A. Empirical Formula

B. Straight-line method

C. Fixed base length method

D. Variable slope method Direct runoff = Observed streamflow – Base flow

Direct runoff

Direct runoffonly

A. Empirical Formula

• Where;

A = catchment area (km2)

b = 0.83

N = time intervals (days) from the

peak to the point B

NQP

𝑁 = 𝑏 𝐴0.2

• Point A represents the beginning of the direct runoff.

• Point B represents the end of the direct runoff.

Note: The value of N is only approximate and the position of B should be decided by considering a number of hydrographs for the catchment.

B. Straight-line Method

Draw a horizontal line segment (A‐B) from

start of runoff to intersection with recession

curve (representing the end of the direct

runoff)

• Segment A-B separate the base flow and direct runoff.

C. Fixed Base Line Method

STEP 1

Draw line segment (A – C)

extending base flow

recession to a point

directly below the

hydrograph peak (Point C)

STEP 2

Draw line segment (C – B)

by a straight line

• Segment A-C and C-B separate the base flow and direct runoff

Peak flow

D. Variable slope method

STEP 1

Draw line segment (A‐C)

extending base flow

recession to a point

directly below the

hydrograph peak

STEP 2

Draw line segment (B‐E)

extending base flow

recession backward to a

point directly below the

inflection point

STEP 3

Draw line segment (C‐E)

Peak flow

• The surface runoff hydrograph obtained after the base-flow separation is also known as Direct Runoff Hydrograph (DRH).

Effective Rainfall

The Effective Rainfall Hyetograph (ERH) is obtained by subtracting the initial loss and infiltration losses from the hyetograph of a storm.

It is also known as Hyetograph of Excess Rainfall

Effective rainfall definition;

• Not retained on land surface

• Not infiltrated into soil

Both DRH and ERH represent the same total quantitybut in different units.

ERH is usually in cm/hr plotted against time.

Effective Rainfall = Excess Rainfall = Total Rainfall – (Initial Losses + Infiltration)

Effective Rainfall = Total Rainfall – Abstractions

Excess Rainfall

Total Rainfall

Infiltration IndexABSTRACTION

• Water absorbed by infiltration

• Parameters of infiltration equations can be determined by f-index.

• Abstractions/losses – difference between total rainfall hyetograph and effective rainfall hyetograph.

f -INDEX

• Constant rate of abstraction yielding effective rainfall hyetograph with depth equal to depth of direct runoff

• Accounts for total abstraction.f -index

Total rainfall (P) − f. 𝑡𝑟 = Depth of direct runoff = Effective Rainfall = Excess Rainfall

Where,

𝑃 =

1

𝑁

𝐼𝑖 ∆t ; tr = duration of rainfall

#Example 1

Rainfall of magnitude 3.8 cm and 2.8 cm occurring on two consecutive 4-h durations on a catchment area 27 km2 produced the following hydrograph of flow at the outlet of the catchment. Estimate the effective rainfall and f-index?

Time from start of rainfall (hr)

-6 0 6 12 18 24 30 36 42 48 54 60 66

Observed flow (m3/s)

6 5 13 26 21 16 12 9 7 5 5 4.5 4.5

Solution #Example 1

The hydrograph is plotted to scale. It is seen that the storm hydrograph has a base flow component.

Solution #Example 1…cont.

1. Base flow separation:

A. Using Straight-line method; from observation of the hydrograph data, the streamflow at the start of the rising limb of the hydrograph is 5 m3/s.

Base flow = 5 m3/s

Thus, use Constant base flow of 5 m3/s

Solution #Example 1…cont.

B. Using Empirical Formula; N = 38.5 hr

12 hr

𝑁 = 0.83 𝐴0.2 = 0.83 27 0.2

= 1.6 days ×24 hr

1 day

= 38.5 hr

So the base flow starts at 0th hour and ends at the point (12 + 38.5) hours

50.5 hrs ( say 48 hrs approx.)

Thus, Constant base flow of 5 m3/s

2. Estimate the volume of direct runoff due to rainfall;

A. Using the area under the direct runoff hydrograph

∆t

Q1

Q2

Q3

Q4

Q5

Q6

Q7

A1 A3A2 A6A5A4 A7 A8

Area under the graph = total volume of direct runoff (in m3)

= 𝑄. ∆𝑡

Total volume of direct runoff;

=1

2𝑄1∆𝑡 +

1

2𝑄1 + 𝑄2 ∆𝑡 +

1

2𝑄2 + 𝑄3 ∆𝑡 +

1

2𝑄3 + 𝑄4 ∆𝑡 +

1

2𝑄4 + 𝑄5 ∆𝑡 +

1

2𝑄5 + 𝑄6 ∆𝑡 +

1

2𝑄6 + 𝑄7 ∆𝑡 +

1

2𝑄7∆𝑡

=1

28 +1

28 + 21 +

1

221 + 16 +

1

216 + 11 +

1

211 + 7 +

1

27 + 4 +

1

24 + 2 +

1

22 × 6 hr ×

60 min

1 hr×60 𝑠

1 min

= 1.49 × 106 m3

Calculating the volume of DRH

B. Using table; Estimate the volume of direct runoff due to rainfall;Time from start of

rainfall(hr)

Observed Hydrograph

(m3/s)

Base flow(m3/s)

Direct Runoff Hydrograph

(m3/s)

(1) (2) (3) (4) = (2) – (3)

-6 6 5 1

0 5 5 0

6 13 5 8

12 26 5 21

18 21 5 16

24 16 5 11

30 12 5 7

36 9 5 4

42 7 5 2

48 5 5 0

54 5 5 0

60 4.5 5 0

66 4.5 5 0

Separate the base flow from the observed streamflow hydrograph in order to obtain the Direct Runoff Hydrograph (DRH).Note: Direct Runoff Hydrograph = Observed hydrograph – base flow (Column 4)

Consider for calculation of direct runoff volume

Intervals, ∆t = 6 hr;

The volume of direct runoff (VDRH) due to rainfall;

𝑉𝐷𝑅𝐻 =

𝑛=1

7

𝑄𝑛∆𝑡 = ∆𝑡

𝑛=1

7

𝑄𝑛

= 8+21+16+11+7+4+2m3

s× 6 hr ×

60 min

1 hr×60 s

1 min

= 1.49 × 106m3

3. Determine the effective rainfall;

VDRH in equivalent unit of depth

(Depth of direct runoff)

=1.49×106m3

27 km2×103 m1 km

2=0.0552 m = 5.52cm

= Volume of direct runoff (VDRH) = Effective Rainfall Catchment area (A)

Total volume of DRH

4. Calculate φ-index;

Total rainfall, P = (3.8 + 2.8) cm = 6.6 cm

Duration of rainfall, tr = 2 (4 hr) = 8 hr

f-index = Total rainfall – Effective rainfall

Duration of rainfall, tr

=6.6 − 5.52 cm

8 hr= 0.135 cm/hr

Total rainfall (P) − f. 𝑡𝑟 = Depth of Direct runoff = Effective Rainfall = Excess Rainfall

5. The effective rainfall hyetograph (ERH)

Effective rainfall; For 1st 4 hrs = 3.8 – 0.135 cm/hr (4 hr) = 3.26 cm

For 2nd 4 hrs = 2.8 – 0.135 cm/hr (4 hr) = 2.26 cm

6. Plot the effective rainfall hyetograph (ERH) and Direct runoff hydrograph (DRH)

QuestionThe observed streamflow and rainfall data are tabulated below. The catchment area is 282.6 km2. Using the straight line method for base flow separation.

Time from start of rainfall

(hr)

Observed Hydrograph

(m3/s)

Time(hr)

Observed Rainfall(cm/hr)

0 150 0 – 1 0.251 150 1 – 2 2.752 350 2 – 3 2.753 800 3 – 4 0.254 12005 9006 7507 5508 3509 225

10 15011 150

Question

i. Estimate the volume of direct runoff.

ii. Determine the excess rainfall.

iii. Calculate the abstraction index, f.

Unit Hydrograph (UH)

The theory was 1st proposed by Sherman (1932).

Unit hydrograph is a direct runoff hydrograph (DRH) produced from 1 in (usually taken as 1 cm in SI units) of effective rainfall (net rainfall) occurred uniformly over the entire catchment at uniform rate and for specific duration (D hours).

Simple Linear Model – used to find out the volume of direct runoff due to 1 cm of effective rainfall.

UH - Basic Assumptions

1. The effective rainfall has a constants intensity within the effective duration.

2. The effective rainfall is uniformly distributed over the entire catchment area.

3. The base of time duration of the DRH due to effective rainfall of given duration is constant. Base period of hydrograph with different rainfall intensities remain approximately same.

4. The ordinates of DRH due to effective rainfall with different intensities but same duration are directly proportional.

5. For a given catchment, the hydrograph resulting from a given effective rainfall reflects the unchanging characteristics of the catchment.

Derivation of UH from a simple Flood Hydrograph

Steps required to derive UH are:

STEP 1: From the given flood hydrograph, separate the base flow by any one of the

methods.

STEP 2: Determine the volume of direct runoff hydrograph by the formula:

Volume of DRH = ∑Q∆t

STEP 3: Divide the volume of DRH by the catchment area to get DRH in equivalent unit of depth (in cm), i.e. effective rainfall or excess rainfall.

STEP 4: Divide the ordinates of DRH by the effective rainfall to obtain ordinates of UH.

STEP 5: Plot the ordinates of UH against time to get the UH of the catchment.

#Example 2

Obtain a Unit Hydrograph for a basin with area 315 km2 using the rainfall and discharge data tabulated below.

Time from start of rainfall

(hr)

Observed Discharge

(m3/s)

Time(hr)

ObservedRainfall(cm/hr)

0 100 0 – 1 0.51 100 1 – 2 2.52 300 2 – 3 2.53 700 3 – 4 0.54 10005 8006 6007 4008 3009 200

10 10011 100

Discharge Data Rainfall Data

#Solution Example2..

1. Separate the base flow from the observed discharge hydrograph in order to obtain the Direct Runoff Hydrograph (DRH). Use the straight-line method to separate the base flow.

From observation of the hydrograph data, the discharge at the start of the rising limb of the hydrograph is 100 m3/s. Thus, use constant 100 m3/s as the base flow.

2. Compute the volume of direct runoff. This volume must be equal to the volume of the Effective Rainfall Hyetograph (ERH).

Time from start of rainfall

(hr)

Observed Discharge

(m3/s)

Base flow(m3/s)

Direct Runoff Hydrograph

(m3/s)

(1) (2) (3) (4) = (2) – (3)

0 100 100 0

1 100 100 0

2 300 100 200

3 700 100 600

4 1000 100 900

5 800 100 700

6 600 100 500

7 400 100 300

8 300 100 200

9 200 100 100

10 100 100 0

11 100 100 0

Separate the base flow from the observed streamflow hydrograph in order to obtain the ordinates of Direct Runoff Hydrograph (DRH).Note: Direct Runoff Hydrograph = Observed discharge – base flow

Consider for calculation of direct runoff volume

The volume of direct runoff (VDRH) due to rainfall;

∆t = 1 hr

3. Express VDRH in equivalent units of depth by dividing with a catchment area;

VDRH in equivalent unit of depth

(Depth of direct runoff)

𝑉𝐷𝑅𝐻 =

𝑛=1

8

𝑄𝐷𝑅𝐻𝑛∆𝑡 = ∆𝑡

𝑛=1

8

𝑄𝐷𝑅𝐻𝑛

= 200 + 600 + 900 + 700 + 500 + 300 + 200 + 100m3

s× 1 hr ×

60 min

1 hr×60 s

1 min

= 1.26 × 107m3

= Volume of direct runoff (VDRH) = Effective Rainfall Catchment area (A)

=1.26×107m3

315 km2×103 m1 km

2= 0.04 m = 4.0 cm

4. Derive ordinates of Unit Hydrograph by normalizing the DRH. Normalizing implies dividing the ordinates of the DRH by the VDRH in equivalent units of depth (Effective Rainfall).

Time from start of rainfall

(hr)

Observed Hydrograph

(m3/s)

Direct Runoff Hydrograph, (DRH)

(m3/s)

UnitHydrograph(m3/s/cm)

(1) (2) (3) (4) = (3)/4 cm

0 100 0 0

1 100 0 0

2 300 200 50

3 700 600 150

4 1000 900 225

5 800 700 175

6 600 500 125

7 400 300 75

8 300 200 50

9 200 100 25

10 100 0 0

11 100 0 0

𝑂𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑈𝐻 =Ordinate of DRH

Effective rainfall=𝐷𝑅𝐻

4 𝑐𝑚

5. Determine the duration, D of the Effective Rainfall hyetograph (ERH) associated with the UH obtained in Q4. In order to do this:

a) Determine the volume of losses, VLosses which is equal to the difference between the volume of observed rainfall, VGRH (gross rainfall) and the volume of the direct runoff hydrograph, VDRH in equivalent units of depth;

∆t = 1 hr;

VLosses = VGRH - VDRH = [(0.5 + 2.5 + 2.5 + 0.5) cm/hr x 1 hr] – 4 cm = 2 cm

b) Compute the f-index equal to the ratio of the volume of losses to the rainfall duration, tr. Thus,

Duration of rainfall, tr = 4 hr

f−index=VLossestr=2 cm4 hr= 0.5 cm/hr

c) Determine the Effective Rainfall hyetograph (ERH) by subtracting the infiltration (e.g., f-index) from the gross rainfall (GRH):

As observed in the table, the duration of the effective rainfall hyetograph is 2 hours. Thus, D = 2 hours, and the Unit Hydrograph obtained in Q4 is a 2-hour Unit Hydrograph.

Time(hr)

Observed Rainfall (GRH)

(cm/hr)

Excess Rainfall hyetograph

(ERH) (cm/hr)

(1) (2) (3) = (2) – f-index

0 – 1 0.5 0.0

1 – 2 2.5 2.0

2 – 3 2.5 2.0

3 – 4 0.5 0.0

Question

The following are the ordinates of the hydrograph of flow from a catchment area of 500 km2 due to a 6-hr rainfall. Assuming the base flow to be zero;

Time(hr)

Observed flow(m3/s)

0 06 100

12 25018 20024 15030 10036 7042 5048 3554 2560 1566 572 0

(1) Estimate the volume of direct runoff

(2) Calculate the effective rainfall

(3) Derive the ordinates of 6-hr unit hydrograph.

Soil Conservation Service (SCS) Method for Abstractions

• Soil Conservation Service (SCS) method is an experimentally derived methodfor computing the abstractions from storm rainfall using information about soils, vegetative cover, hydrologic condition and antecedent moisture conditions.

• The method is based on the simple relationship that;

P = Pe + Ia + Fa

• Where,– P is total rainfall – Pe is effective rainfall– Ia is initial losses – Fa is continuing abstraction

Time

Pre

cip

itati

on

pt

aI aF

eP

aae FIPP

Basic equation to calculate the effective rainfall or the depth of direct runoff from a storm by SCS method;

This equation shows the relationship between effective rainfall, Pe and total rainfall, P. Where, S = potential maximum storage.

From study of small experimental catchment, an empirical relation was developed.

So,

SIP

IPP

a

ae

-

-

2

SIa 2.0

SP

SPPe

8.0

2.02

-

Plotting P and Pe from many catchment, the SCS found curves of the type shown as below;

Figure : Solution of the SCS runoff equation (Source : Soil Conservation Service, 1972)

• The curve number, CN (dimensionless) and S (inches) are related by

100)CN0Units;American(

101000

-CN

S

100)CN30Units;SI(

25425400

- CNCN

S

• Curve number, CN:

– Impervious surfaces: CN = 100

– Natural surfaces: CN < 100

Curve Number

• S and CN depend on Antecedent Moisture Conditions (AMC).

• Normal conditions, AMC(II)

• Dry conditions, AMC(I)

• Wet conditions, AMC(III)

)(058.010

)(2.4)(

IICN

IICNICN

-

)(13.010

)(23)(

IICN

IICNIIICN

SIICN

10

1000)(

SCS Curve Numbers depend on Soil Conditions

GroupMinimum Infiltration

Rate (in/hr)Soil Type

A 0.3 – 0.45High infiltration rates. Deep, well drained sands and gravels

B 0.15 – 0.30

Moderate infiltration rates. Moderately deep, moderately well drained soils with moderately coarse textures (silt, silt loam)

C 0.05 – 0.15Slow infiltration rates. Soils with layers, or soils with moderately fine textures (clay loams)

D 0.00 – 0.05Very slow infiltration rates. Clayey soils, high water table, or shallow impervious layer

#Example 3

Calculate the effective rainfall from 5 inches of rainfall on a 1000 acre watershed. The hydrologic soil group is 50% Group B and 50% Group C. Antecedent moisture condition II is assumed. The land use is :

40% residential area that is 30% impervious

12% residential area that is 65% impervious

18% paved roads with curbs and storm sewers

16% open land with 50% fair grass cover and 50% good grass cover

14% parking lots, plaza, school and so on (all impervious)

Given information: SCS Method

• Rainfall: 5 in.

• Area: 1000-ac

• Soils: • Class B: 50%• Class C: 50%

• Antecedent moisture: AMC(II)

• Land use• Residential

• 40% with 30% impervious cover• 12% with 65% impervious cover

• Paved roads: 18% with curbs and storm sewers

• Open land: 16%• 50% fair grass cover• 50% good grass cover

• Parking lots, etc.: 14%

• Pe = ?

• Given P = 5 inches, hence need to find S

• S = (1000/CN) – 10

• Weighted CN = total product /100

SP

SPPe

8.0

2.02

-

Solution:

Hydrologic Soil Group

B C

Land use % CN Product % CN Product

Residential (30% imp cover) 20 72 1440 20 81 1620

Residential (65% imp cover) 6 85 510 6 90 540

Paved roads with curbs and storm sewers

9 98 882 9 98 882

Open land: good cover 4 61 244 4 74 296

Open land: Fair cover 4 69 276 4 79 316

Parking lots, etc 7 98 686 7 98 686

Total 50 4038 50 4340

CN values come from Table 5.5.2

1. Calculate the weighted curve number;

Note:Product = % x CN

Weighted Curve Number, CN = 4038+4340100

= 83.78

Maximum potential storage,

The effective rainfall,

S =1000CN− 10

=100083.78− 10

=1.93 in

SP

SPPe

8.0

2.02

-

in 3.2593.18.05

93.12.052

-

#Example 4

Re-compute the effective rainfall from #Example 3 if the wet conditions of antecedent moisture condition III are applicable.

Solution:

Find a curve number for AMC III equivalent to CN = 83.8 under AMC II using;

)(13.010

)(23)(

IICN

IICNIIICN

)8.83(13.010

8.8323

3.92

Maximum potential storage,

The effective rainfall,

S =1000CN− 10

=100092.3− 10

=0.83 in

SP

SPPe

8.0

2.02

-

in 4.1383.08.05

83.02.052

-

The change in effective rainfall caused by the change in antecedent moisture

conditions is 4.13 – 3.25 = 0.88 in, a 27 percent increases.