Hydro Graph

CE 205 – Engineering Hydrology Lecture Note by Dr. Uditha Ratnayake

Hydrograph Analysis

Hydrology of rivers is required by engineers for estimation of water, design of dams, diversions, and flood control through reservoirs or dykes, etc. Information is gathered through a network of stream gauges. Hydrograph analysis deals with the study of runoff records at a stream gauge. Hydrograph analysis is often combined with rainfall analysis to investigate how a watershed responds to rainfall. In many cases, hydrometric information is not available. This is especially true for small watersheds. In such situations, rainfall information must be combined with rainfall-runoff models Streamflow measurement Flow rate is measured in units of cms (cubic meter per second) or cfs (cubic feet per second). Direct measurement of flow rates requires knowledge of the complete cross sectional velocity profile, which varies with flow rate. While it is tedious to measure flow rate directly, it is straightforward to measure river stage, for example by a gauge. Therefore, flow rates are measured only a few times, enough to establish a rating curve that describes the relationship between flow rate and stage. Regular measurement of stage is then combined with the rating curve to produce time series of streamflow. Components of the hydrograph The hydrograph describes flow as a function of time usually known as a time series of flow. The interest may lie in the hydrograph of a long period of several years or only few selected rainfall events of few hours or days. The latter situation frequently occurs in the development of a rainfall-runoff relationship for a watershed. It is customary to consider two components of the hydrograph:

1. Direct runoff the flow that results directly from the rainfall event. Usually after considering the associated losses from the gross rainfall. The volumes of effective rainfall and the volume of direct runoff should be equal.

2. Base flow flow that is unrelated to the rainfall event. The rainfall-runoff relationship describes the time distribution of direct runoff as a function of excess rainfall (gross rainfall minus losses). Therefore, in developing the rainfall-runoff relationship for a watershed based on observed hyetographs and hydrographs, one must first subtract the baseflow from the hydrograph. Even after long periods without rain, water still flows in many streams and rivers. This flow is the result of seepage from groundwater aquifers into the stream channel. In larger rivers, baseflow can be significant. In periods without rain, the baseflow in a stream will slowly decline as a result of the draw down of the groundwater aquifers. This phenomenon is called baseflow recession. It is often assumed that baseflow declines exponentially. Baseflow separation involves dividing the hydrograph into a direct runoff component and a baseflow component. Unit Hydrograph Rainfall-runoff modeling is the essence of much engineering hydrology. Because flow data are rarely available, design event are usually determined by a combination of rainfall information and rainfall-runoff relationships. When it comes to derive a rainfall-runoff relationship the concept of the runoff caused by a unit rainfall or in other words the unit hydrograph plays a major role. The unit hydrograph of time T is based on a hypothetical case of 1 unit (1 mm) of rain falling uniformly over the whole catchment during a time interval T. The unit hydrograph gives the runoff response of the catchment to that rain. The basic assumptions of the unit hydrograph are:

• The time base of the hydrograph remains the same irrespective of the rain intensity • The unit hydrograph is linear (proportionality and superposition applies) • The unit hydrograph is time invariant

1


Unit Hydrograph Derivation To derive the unit hydrograph from a simple rainstorm divide the direct discharge values of time-discharge curve by the height of effective rainfall to get the unit hydrograph.

For example if the total effective rain volume is 5.4 mm then divide all direct discharge values of time-discharge curve by 5.4 to get the unit hydrograph. A detailed example follows.

Suppose that there are M pulses of excess rainfall and N pulses of direct runoff is observed in a storm. Then N-M+1 values will be needed to define the unit hydrograph. The discrete convolution equation, given below, it allows the computation of direct runoff Qn given excess rainfall Pm and the unit hydrograph Un-m+1.

Q P Un m n

m

n M

= − +=

≤

∑ 11

m

The reverse process, called deconvolution, is needed to derive a unit hydrograph given data on Pm and Qn. Suppose that there are M pulses of excess rainfall and N pulse of direct runoff in the storm considered; then N equation can be written for Qn, with n = 1,2,….,N. The equations will consists N – M + 1 unknown values of the unit hydrograph. Few of the equations will be redundant, because there are more equations (N) than unknowns (N – M + 1). The following table shows the set of equations for discrete time convolution

Q1 = P1U1 Q2 = P2U1 + P1U2 Q3 = P3U1 + P1U2 + P1U3

…….. QM = PMU1 + PM-1U2 + + ……+ P1UM

QM+1 = 0 + PMU2 + ……+ P2UM + P1UM+1 …….

QN – 1 = 0 + 0 + + ……+ 0 + 0 + + ……+ PMUN-M +1 PM-1UN-M+1 QN = 0 + 0 + + ……+ 0 + 0 + + ……+ 0 PMUN-M+1

Example An observed hydrograph is given below with the corresponding excess rainfall. The time interval is 6 hours between readings. Observed hydrograph for this event is shown in the figure. ½ Hours Excess

Rainfall (mm) Observed

Direct Discharge

(m3/s) 1 10 102 30 703 20 2004 4605 17806 38807 31608 11209 620

10 34011 15012 7013 20

0500

10001500200025003000350040004500

0 5 10 15

Time Interval (1/2hr)

Obs

erve

d D

isch

arge

(cm

s)

1st define the number of equations. There are 3 pulses of rainfall so M = 3. There are 13 pulses of observed direct runoff so N = 13. The total number of unit hydrograph ordinates are N – M +1 = 13 – 3 + 1 = 11 ordinates. So we have to solve 11 linear equations as follows.

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Q1 = P1U1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 Q2 = P2U1 + P1U2 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 Q3 = P3U1 + P2U2 + P1U3 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 Q4 = 0 + P3U2 + P2U3 + P1U4 + 0 + 0 + 0 + 0 + 0 + 0 + 0 Q5 = 0 + 0 + P3U3 + P2U4 + P1U5 + 0 + 0 + 0 + 0 + 0 + 0 Q6 = 0 + 0 + 0 + P3U4 + P2U5 + P1U6 + 0 + 0 + 0 + 0 + 0 Q7 = 0 + 0 + 0 + 0 + P3U5 + P2U6 + P1U7 + 0 + 0 + 0 + 0 Q8 = 0 + 0 + 0 + 0 + 0 + P3U6 + P2U7 + P1U8 + 0 + 0 + 0 Q9 = 0 + 0 + 0 + 0 + 0 + 0 + P3U7 + P3U8 + P1U9 + + 0 + 0 Q10 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + P3U8 + P2U9 + P1U10 + 0 Q11 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + P3U9 + P2U10 + P1U11

Explanation of the Table What the unit hydrograph says is that each pulse of runoff (1-13 in this example) is generated by some linear combination of the excess rainfall. For example, the very first pulse of rainfall (0.5 inches) caused the very first pulse of runoff (5 ft3/s). Because it is the only rainfall that occurred during that time interval, it alone is responsible for the runoff that is occurring. Similarly, The second pulse of rainfall is caused by rainfall pulse 1 and 2 (0.5 and 1.2 inches) because they alone occurred during that time interval. So the unit hydrograph is merely a solution to a set of linear equations that determine the contributions of rainfall over time to the direct runoff hydrograph. The unit hydrograph becomes “normalized” during the deconvolution process to represent the flow that would occur from one unit of rainfall occurring during the 1st time interval.

Now plug in the numbers. 10 = 10U1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 70 = 30U1 + 10U2 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 200 = 20U1 + 30U2 + 10U3 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 460 = 0 + 20U2 + 30U3 + 10U4 + 0 + 0 + 0 + 0 + 0 + 0 + 0 1780 = 0 + 0 + 20U3 + 30U4 + 10U5 + 0 + 0 + 0 + 0 + 0 + 0 3880 = 0 + 0 + 0 + 20U4 + 30U5 + 10U6 + 0 + 0 + 0 + 0 + 0 3160 = 0 + 0 + 0 + 0 + 20U5 + 30U6 + 10U7 + 0 + 0 + 0 + 0 1120 = 0 + 0 + 0 + 0 + 0 + 20U6 + 30U7 + 10U8 + 0 + 0 + 0 620 = 0 + 0 + 0 + 0 + 0 + 0 + 20U7 + 30U8 + 10U9 + + 0 + 0 340 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 20U8 + 30U9 + 10U10 + 0 150 = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 20U9 + 30U10 + 10U11

We now have 11 linear equations with 11 unknowns. Each unknown is an ordinate of the unit hydrograph. We have to solve these in a step-wise fashion starting with U1. Let’s work them out. EQ 1. 10 = 10U1 U1 = 10/10 = 1 cms direct runoff / mm excess rainfall

EQ 2. 70 = 30U1 + 10U2 U2 = (70 – 30 U1)/10 = (70 – 30(1))/10 = 4

EQ 3. 200 = 20U1 + 30U2 + 10U3 U3 = (200 –20U1 – 30U2)/10 = (200 – 20(1) – 30(4))/10 = 6

EQ 4. 460 = 20U2 + 30U3 + 10U4 U4 = (460 - 20U2 - 30U3)/10 = (460 – 20(4) – 30(6))/10 = 20

EQ 5. 1780 = 20U3 + 30U4 + 10U5 U5 = (1780 -20U3 - 30U4)/10 = (1780 – 20(6) – 30(20))/10 = 106

EQ 6. 3880 = 20U4 + 30U5 + 10U6 U6 = (3880 - 20U4 - 30U5)/10 = (3880 - 20(20) – 30(106)/10 = 30

EQ 7. 3160 = 20U5 + 30U6 + 10U7 U7 = (3160 - 20U5 - 30U6)/10 = (3160 – 20(106) – 30(30)/10 = 14

EQ 8. 1120 = 20U6 + 30U7 + 10U8 U8 = (1120 - 20U6 - 30U7)/10 = (1120 – 20(30) – 30(14)/10 = 10

EQ 9. 620 = 20U7 + 30U8 + 10U9 U9 = (620 - 20U7 - 30U8)/10 = (620 – 20(14) – 30(10)/10 = 4

EQ 10 340 = 20U8 + 30U9 + 10U10 U10= (340 - 20U8 - 30U9)/10 = (340 – 20(10) – 30(4)/10 = 2

EQ 11 150 = 20U9 + 30U10 + 10U11 U11= (150 - 20U9 - 30U10)/10 = (150 – 20(4) – 30(2) / 10 = 1

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The resulting unit hydrograph is shown below in tabular and graphical form. This hydrograph represents the flow that would result from 1 mm of rainfall occurring during the 1st time interval (i.e. first half an hour).

Time Interval (1/2 hr)

Unit Hydrograph (cms)

1 1 2 4 3 6 4 20 5 106 6 30 7 14 8 10 9 4

10 2 11 1

0

20

40

60

80

100

120

0 2 4 6 8 10 1


Uni

t hyd

rogr

aph

ordi

nate

(cm

s)

2

Using the Unit Hydrograph to derive a direct runoff hydrograph The unit hydrograph can be used to determine the direct runoff hydrograph for any rainfall amount with any time distribution. When it is needed to derive the time-discharge curve due to multiple storms when storm duration is nT (n is integer) use proportionality and principle of superposition to get the total hydrograph. The process is called convolution. Let’s use the following rainfall distribution to calculate the direct runoff hydrograph. Derive the hydrograph for an excess rainfall of 10mm, 5mm in the first and second half hours. Use the above unit hydrograph. The table for calculation looks like this.

Unit Hydrograph Ordinates Direct Runoff Time

Interval Rainfall U1 U2 U3 U4 U5 U6 U7 U8 U9 U10 U11 1 P1 P1U1 0 0 0 0 0 0 0 0 0 0 P1U1 2 P2 P2U1 P1U2 0 0 0 0 0 0 0 0 0 P2U1 + P1U2 3 0 P2U2 P1U3 0 0 0 0 0 0 0 0 P2U2 + P1U3 4 0 0 P2U3 P1U4 0 0 0 0 0 0 0 P2U3 + P1U4 5 0 0 0 P2U4 P1U5 0 0 0 0 0 0 P2U4 + P1U5 6 0 0 0 0 P2U5 P1U6 0 0 0 0 0 P2U5 + P1U6 7 0 0 0 0 0 P2U6 P1U7 0 0 0 0 P2U6 + P1U7 8 0 0 0 0 0 0 P2U7 P1U8 0 0 0 P2U7 + P1U8 9 0 0 0 0 0 0 0 P2U8 P1U9 0 0 P2U8 + P1U9

10 0 0 0 0 0 0 0 0 P2U9 P1U10 0 P2U9 + P1U1011 0 0 0 0 0 0 0 0 0 P2U10 P1U11 P2U10 + P1U11

0 0 0 0 0 0 0 0 0 0 P2U11 P2U11

The table with numbers plugged in is shown below. The sum of the columns equals the direct runoff hydrograph. The resulting graph shows the unit hydrograph and the storm event or the runoff hydrograph resulting from the total excess rainfall of 15mm occurred within one hour. The area under the unit hydrograph should equal 1.0mm of runoff and the area under the direct runoff hydrograph should equal 15 mm of runoff.

4


Unit Hydrograph


Rainfall (mm) 1 4 6 20 106 30 14 10 4 2 1

Direct Runoff (cms)

1 10 10 0 0 0 0 0 0 0 0 0 0 102 5 5 40 0 0 0 0 0 0 0 0 0 453 0 20 60 0 0 0 0 0 0 0 0 804 0 0 30 200 0 0 0 0 0 0 0 2305 0 0 0 100 1060 0 0 0 0 0 0 11606 0 0 0 0 530 300 0 0 0 0 0 8307 0 0 0 0 0 150 140 0 0 0 0 2908 0 0 0 0 0 0 70 100 0 0 0 1709 0 0 0 0 0 0 0 50 40 0 0 90

10 0 0 0 0 0 0 0 0 20 20 0 4011 0 0 0 0 0 0 0 0 0 10 10 20 0 0 0 0 0 0 0 0 0 0 5 5

0

200

400

600

800

1000

1200

1400

0 2 4 6 8 10 12 1


Run

off (

cms)

4

Unit Hydrograph Runoff Hydrograph

Synthetic Unit Hydrograph If there is no data for the specific catchment to derive the unit hydrograph, it is sometimes possible to construct a synthetic unit hydrograph. This is usually based on empirical functions, which correlate unit hydrograph with some basic morphologic data of the catchment such as area, slope, and land cover together with knowledge of catchments in the region. Two models of synthetic unit hydrograph will be covered in the course. They are Snyder’s method and SCS-dimensionless unit hydrograph. The key properties of a unit hydrograph that will affect design flows are the peak flow rate, the time to peak, and the duration of runoff. In many cases, the exact shape of the unit hydrograph is relatively unimportant as long as the above three properties are reasonably correct. Synthetic unit hydrograph attempt to estimate these three key properties based on information of watershed characteristics.

5


Snyder’s unit hydrograph To develop a unit hydrographs based on Snyder’s method, five inputs are required. They are A (watershed area), L (length of main stream from outlet to divide), Lc (length to centroid of basin), Ct and Cp (model coefficients). These model coefficients can be determined from gauged watersheds in the region and transferred to the ungauged design site. Empirically, Ct values ranges from 0.3 to 6.0. Cp values are in the range of 0.31 to 0.93. Procedure:

1. Estimate basin lag. Basin lag is the time lag from the centroid of excess rainfall hyetograph to the peak runoff. Snyder’s method estimates basin lag as

( ) 3.0catp LLCt = tp (hr), L and Lca (km).

2. Estimate peak discharge ppp tACQ /78.2= Qp (cms), A (square km), tp (hr).

3. Estimate time base of unit hydrograph, which is the time of direct runoff. 8/3 pb tT += Tb (days), tp (hr).

This formula is intended for large watersheds. For small watersheds, the formula will give excessively large time bases. Hence for small and moderate watersheds, the time base (in hours) should be calculated as 3-5 times the basin lag (use Tb=4tp [hrs] in lack of better knowledge).

4. Determine the duration D to which the unit hydrograph corresponds 5.5/ptD = D (hr), tp (hr)

5. Adjust the unit hydrograph to desired duration. In many cases, one is interested in a unit hydrograph with a specific duration. For example, if the design hyetograph is given in time steps of 1 hour, it is desirable to have a unit hydrograph with duration of 1 hour. The duration can be changed using the S-curve method; however, in Snyder’s method the following adjustment is recommended: First, adjust lag time as follows

( )DDtt pp −+= '' 25.0 D’ is the desired rainfall duration tp’ is the corresponding basin lag

6. Calculate the time of rise. Basin lag, tp, is the time from the centroid of excess rainfall to the peak of the unit hydrograph. Hence, the time of rise is calculated as follows.

pR tDT += 2/ 7. To assist in sketching the unit hydrograph, calculate unit hydrograph width at 50% and 75% of Qp.

( ) 08.150 /87.5 −= AQW p

75.1/5075 WW = W75 and W50 (hrs), Qp (cms), A (km2). The endpoints of the intervals defined by W75 and W50 should be placed so that 1/3 appears before TR and 2/3 after. The above information provides sufficient detail to allow a sketching of the unit hydrograph. Adjustments should be made such that the volume of runoff (area under the curve) corresponds to 1 inch (or cm) of runoff. Example The following characteristics are given for a watershed. Develop a 2-hr unit hydrograph for the basin

Area: A = 400 km2 Watershed length L = 45 km Length to center Lc = 25 km Coefficient Ct = 1.257 Coefficient Cp = 0.576

Solution: Step 1: Calculate basin lag (time to peak): ( ) ( ) hrsLLCt ctp 34.102545257.1 3.03.0 =×==

Step 2: Calculated duration of excess rainfall hrs88.15.5

10.345.5

ptD ===

6


Step 3: Adjust lag time to correspond to 2hr excess rain. ( )

( ) hrs

DDtt pp

37.1088.1225.034.10

25.0 ''

=−+=

−+=

Step 4: Calculate the base time of the unit hydrograph hrsdayst

t pb 10329.4

837.103

83 ==+=+=

Step 5: Calculate peak discharge cmst

ACQ

p

pp 77.61

37.10400576.078.278.2

=××

==

Which occurs at time T hrs 37.1137.1012/ =+=+= pR tDStep 6: Compute W50 and W75 ( ) ( ) hrsAQp 44400/77.6187.5/87.5 08.108.1

50 === −−W

W hrs2575.1/4475 ==The unit hydrograph ordinates for a flow of 0.75Qp = 46.3 cms should be plotted at times TR – 25/3 =3.04hrs and TR + 25(2/3) = 28.04hrs. The unit hydrograph ordinates for a flow of 0.50Qp = 30.88 cms has a problem as it results in a negative value. Therefore, it is plotted at times say 2/3(3.04) =2hrs and 44+2=46hrs. This happens because the basin is small.

Snyder's Unit Hydrograph

0

10

20

30

40

50

60

70

0 20 40 60 80 100 120

Time (hrs)

Dis

char

ge (c

ms)

Seven unit hydrograph points (time, discharge) are now available: (0,0), (2, 30.88), (3.04, 46.3), (11.37, 61.77), (28.04, 46.3), (46, 30.88), and (103, 0). The synthetic unit hydrograph can be sketched. Some adjustment may be needed to ensure that the volume of runoff corresponds to 1 cm of net precipitation. S-curve Method: S-curve is the hydrograph produced by a continuous series of effective rainfall at a constant rate. Infinite number of unit hydrographs of this rainfall rate spaced at its duration is summed up to obtain the S-curve. If another S-curve lagged by a given time duration T is subtracted from the original S-curve a hydrograph due to a T hr rainfall can be obtained. Converting the runoff volume to unity a unit hydrograph of T hr duration can be obtained. SCS Dimensionless Hydrograph While peak discharge rates are adequate for many engineering design problems, they are inadequate for design problems where watershed or channel storage is significant. The SCS (US Soil Conservation Service) dimensionless hydrograph is an idealized shape that approximates the flow from an intense storm from a small watershed. The dimensionless hydrograph arbitrarily has units of 100 units of flow for the peak and 100 units of time for the duration of flow. The area under a dimensionless hydrograph has 2,620 square units of runoff. The SCS hydrograph has 19 constant ordinates that represent percentages of flow and time. They can be seen on the figure below. To develop the design hydrograph for a watershed, the peak flow and the runoff volume must be known for the desired return period storm. The design hydrograph is developed from the dimensionless hydrograph by using approximate conversion factors. This allows us to determine the hydrograph from different sized storms by scaling the hydrograph in both space and time. Conversion Factors This dimensionless hydrograph is scaled to create design hydrographs of various storm events that have different peak runoffs and durations. There are three scaling factors used in the dimensionless hydrograph. The first factor is u and it is the ratio of the total runoff volume to the area under the dimensionless hydrograph. Recall that the area under the dimensionless hydrograph has 2,620 units of runoff. So each single unit has a value of u = Q /2620 where Q is the total storm runoff volume. (Generally the peak flow and runoff volume are found from the Curve Number Method, but other methods can also be used)

7


SCS Dimensionless Hydrograph

The second factor is w and is the ratio of the peak runoff for the design storm to the peak flow of 100 on the dimensionless hydrograph. Each unit of flow on the dimensionless hydrograph has a value of w = q/100 where q is the peak runoff (Generally found with the SCS peak flow equation). The third factor is k and is it the value that each unit of time on the dimensionless hydrograph represents in the design hydrograph. On the design hydrograph 1/100 of the peak flow times 1/100 of the duration of the runoff must equal 1/2620 of the flood volume just as it does on the dimensionless hydrograph. Since w is equal to 1/100 of the design peak flow, k must be equal to 1/100 of the design duration, and u is 1/2620 of the design flood volume therefore, w*k = u and k = u/w When runoff rate is measured in m3/s, runoff volume is measured in hectares-meters, and time is measured in minutes. So, k = u(ha-m) * 10,000 (m2/ha) = 167 u W(m3/s) * 60 (sec/min) w The coordinates of the design hydrograph are obtained by multiplying the flow and time ordinates of the dimensionless hydrograph by w and k respectively. The SCS Equivalent Triangular Hydrograph is also shown in the above figure. This triangular hydrograph is defined by the ordinates (0,0), (20,100) and (53.33, 0). Example Suppose a watershed has a peak runoff of 8.2 m3/s and a flood volume 2.9 ha-m derive the design hydrograph. Step 1: Calculate u as: u = Q/2620 = 2.9 / 2620 = 0.00110687 ha-m / unit

8


Step 2: Calculate w as: w = q / 100 = 8.2 / 100 = 0.082 m3/s / unit Step 3: Calculate k as: k = 167 u/w = 167 * 0.00110687 / 0.082 = 2.2542 min/unit Step 4: Multiply ordinates

k = 2.2542 w = 0.082 Design Hydrograph Coordinates

Point Time Ordinate Flow Ordinate k * t w * q a 0 0 0 0 b 2 3 4.5084 0.246 c 6 19 13.5252 1.558 d 8 31 18.0336 2.542 e 12 66 27.0504 5.412 f 14 82 31.5588 6.724 g 16 93 36.0672 7.626 h 18 99 40.5756 8.118 I 20 100 45.084 8.2 j 22 99 49.5924 8.118 k 24 93 54.1008 7.626 l 26 86 58.6092 7.052 m 30 68 67.626 5.576 n 34 46 76.6428 3.772 o 38 33 85.6596 2.706 p 44 21 99.1848 1.722 q 52 11 117.2184 0.902 r 64 4 144.2688 0.328 s 100 0 225.42 0

Resulting design hydrograph.

00.5

11.5

22.5

33.5

44.5

55.5

66.5

77.5

88.5

0 25 50 75 100 125 150 175 200 225

Time (minutes)

Flow

Rat

e (c

ms)

The result is a graph of flow rate versus time for a watershed with a peak flow rate of 8.2 m3/s with total runoff volume equaling 2.9 ha-m. The length of time is 225.4 minutes.

9


Instantaneous Unit Hydrograph A main disadvantage of the unit hydrograph is that it is dependant on the duration of the excess rainfall. Given one unit hydrograph it is difficult to arrive at a unit hydrograph of a different duration. To overcome this difficulty the concept of Instantaneous Unit Hydrograph is proposed. Limiting the duration of a unit hydrograph to zero an Instantaneous Unit Hydrograph is obtained. Instantaneous Unit Hydrograph is the hydrograph resulting from an instantaneous rainfall of one unit uniformly over the basin. Derivation of instantaneous unit hydrograph Consider two S-curves Sa derived from D hr unit hydrograph. The average intensity of excess rainfall of Sa is i=1/D cm/hr. Let there be another S-curve Sb which is leading Sa by a time interval ∆t. Subtracting Sb from Sa and deviding it by i ∆t will provide the ∆t hr unit hydrograph. Limiting ∆t → 0 will bring up the instantaneous unit hydrograph. Thus, mathematically the instantaneous unit hydrograph can be expressed as follows.

( )dtdS

itiSS

ttu ba 1

0lim =

∆−

→∆=

Therefore, it is seen that the slope of the S-curve defines the instantaneous unit hydrograph. The most commonly applied models for the derivation of the unit hydrograph are the Clark’s model and the Nash’s model. Applications of instantaneous unit hydrograph An effective rainfall distributed according to a function of duration t0 results in the runoff hydrograph Q(t) and this hydrograph can be calculated using the equation given below. How the instantaneous unit hydrograph is used in the calculation is shown in the following figure.

( ) ( ) ( ) τττ dItutQt

∫ −=0

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Rational Method to Calculate Peak Discharge The most widely used uncalibrated equation is the Rational Method. Mathematically, the rational method relates the peak discharge (q, m3/sec) to the drainage area (A, ha), the rainfall intensity (i, mm/hr), and the runoff coefficient (C). SI Units q = 0.0028CiA Where q = design peak runoff rate in m3/s C = runoff coefficient

i = rainfall intensity in mm/h for the design return period and for a duration equal to the “time of concentration” of the watershed.

To use the rational method there are a few assumptions. • Rainfall intensity and duration is uniform over the area of study • Storm duration must be equal to or greater than the time of concentration of the watershed. Rational Method - Runoff Coefficients The rational method uses runoff coefficients in the same fashion as the SCS curve number method for estimating runoff volume. They have been determined over the years and primarily focus on urban watershed applications. Below are several tables for different land conditions.

from Schwab (1981)

11


12


from McCuen (1998)

13


Hydrologic Routing

A hydrograph is function of discharge with time. In fact it describes the passage of a wave along the river. As this wave moves down the river its shape gets distorted due to various factors such as channel storage, resistance and lateral addition or withdrawal etc. The process is known as routing and it can be separated in to two categories as follows.

1. Reservoir Routing 2. Channel Routing.

Routing methods can be classified in to (1) Hydrologic routing and (2) Hydraulic routing. Hydraulic routing uses the St. Venant equation and it tries to preserve the hydraulic properties and provides more accurate results while the hydrologic routing employs the continuity equation. Hydrologic routing is much simpler and is used mainly in calculation of related design parameters. Most common application of hydrologic routing is to obtain the design parameters of dams and spillways. In this case the outflow hydrograph is needed to determine the length of spillway and flood control storage. In urban hydrology, the design of detention basins is dependent on an accurate routing of flow. For rivers the interest could be related to the design of an early warning system. In this case we want to predict outflow from a river reach based on a known inflow. Basic principle of continuity can be stated as follows.

I - Q = dS/dt Where,

I = inflow, Q = outflow, S = storage

In finite difference form the continuity equation becomes StQtI ∆=∆−∆ Using suffixes 1 and 2 to denote the conditions at the beginning and end of time interval , the above equation can be written as follows.

t∆

122121

22SSt

QQt

II−=∆

+−∆

+

For a reservoir both initial S and Q are known. Mostly, they are functions of water level (H), which is a known data for a reservoir. If we know the initial condition, we can then solve for Q2 stepwise in time and we need to use numerical or graphical methods. Storage Routing The continuity equation given above is rearranged as follows to use in this method of storage routing.

∆+=

∆−+∆

+2222

21

121 tQ

StQ

StII

For a given time step the left-hand side of the equation is known and the right-hand side contain the unknowns. But, the available data of the reservoir can establish the relationships between reservoir stage (water level) and storage, and between reservoir stage and outflow. This means that the right hand side quantity can be plotted against the reservoir stage. Through this plot the value of right hand side of the equation can graphically point out the corresponding reservoir stage, which in turn will show the outflow. Knowing these quantities the next time step can be solved.

14


The graphical solution looks as the figure given below. The plot contains the set of curves, (1) reservoir stage verses

, (2) reservoir stage verses S

∆−

2tQS

, (3) reservoir stage verses

∆+

2tQS

, and (4) reservoir stage verses .

First step is to locate the water level corresponding to the initial storage and thereby the relevant point on the curve

of stage vs.

Q

∆−

2tQS

as shown in figure. Then, starting from this point, the quantity tII

∆

+2

21 is marked

horizontally. From the end point a vertical line is drawn until it meets the curve stage vs.

∆

+2

tQS . Moving

horizontally the storage, discharge at the end of the time step and the starting point of the next time step on the curve

of stage vs.

∆−

2tQS

can be located.

Channel Routing One way of looking at the channel routing is to see it as a system of interconnected storages with each storage has an inflow hydrograph and an outflow hydrograph. Each storage is considered to be consists of a prism storage and a wedge storage as shown in the figure below.

In a river, storage is not uniquely defined by the outflow. This can be understood if one considers the case of an approaching wave as opposed to a wave just leaving the reach. Equal storage volumes can then occur at totally different outflows. Therefore, channel routing require a method to deal with this situation and most popular method is the Muskingum method. It assumes the following linear relationship.

S = K [xI + (1-x) Q] Where, K = parameter (of dimension time) which is approximately the traveling time for a wave through the

storage unit or the reach x = parameter (no dimensions) in the interval 0<x<0.5. Often around 0.30

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Estimation of K and x In order to find the parameters S is plotted versus [xI + (1-x) Q] for measured events, assuming different values for x. When the data plot almost as a straight line; x, and K are evaluated. When an incorrect value for x is chosen the plotted points will trace a looping curve. This procedure is illustrated in the following example. Example Following inflow and outflow hydrographs were observed in a river reach. Estimate the values of K and x applicable to this reach.

Time (h) 0 6 12 18 24 30 36 42 48 54 60 66 Inflow (m3/s) 5 20 50 50 32 22 15 10 7 5 5 5 Outflow m3/s) 5 6 12 29 38 35 29 23 17 13 9 7

The variation of storage during this time interval is estimated first in [m3/s.h] and then it was compared with the quantity ‘K [xI + (1-x) Q]’ for x = 0.35, 0.30 and 0.25 which has the units [m3/s]. Then the gradient of the plot of S versus [xI + (1-x) Q] will give K in hours. Following table and the figure shows this procedure.

Continuity equation can be written as; ( ) ( ) StQQtII ∆=∆

+−∆

+22 2121

Rearranging the above equation: ( ) ( )[ ] StQIQI ∆=∆

−+−22211

[xI + (1-x) Q] (m3/s) Time

(h) I

(m3/s) Q

(m3/s) I-Q

(m3/s) Mean I-Q

(m3/s)

∆ S (m3/s.h)

S = Σ ∆ S

(m3/s.h) x=0.35 x=0.30 x=0.25

0 5 5 0 0 5.0 5.0 5.0 7.0 42 6 20 6 14 42 10.9 10.2 9.5 26.0 156 12 50 12 38 198 25.3 23.4 21.5 29.5 177 18 50 29 21 375 36.4 35.3 34.3 7.5 45 24 32 38 -6 420 35.9 36.2 36.5 -9.5 -57 30 22 35 -13 363 30.5 31.1 31.8 -13.5 -81 36 15 29 -14 282 24.1 24.8 25.5 -13.5 -81 42 10 23 -13 201 18.5 19.1 19.8 -11.5 -69 48 7 17 -10 132 13.5 14.0 14.5 -9.0 -54 54 5 13 -8 78 10.2 10.6 11.0 -6.0 -36 60 5 9 -4 42 7.6 7.8 8.0 -3.0 -18 66 5 7 2 24 6.3 6.4 6.5

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0

10

20

30

40

0 100 200 300 400 500

[xI + (1-x) Q] (m3/s)

S (m

3/s.

h)

0

10

20

30

40

0 100 200 300 400 500

[xI + (1-x) Q] (m3/s)

S (m

3/s.

h)

5 0

0

10

20

30

40

0

S (m

3/s.

h)

Procedure for cWith known x, K Change in the st

Using the above Where,

For best results t

x = 0.3

100 200 300 400 500

[xI + (1-x) Q] (m3/s)

5

This prelationTherefoestimat= 13.3h

hannel routing a given inflow hydrograph can be routed through

orage according to Muskingum equation is; ( ) ( )( )[ ]121212 1 QQxIIxKSS −−+−=−

equation and the continuity equation Q2 is evaluate

1211202 QCICICQ ++=

tKxKtKxC∆+−

∆+−=

5.05.0

0

tKxKtKxC∆+−

∆+=

5.05.0

1

tKxKtKxKC

∆+−∆−−

=5.05.0

2

he routing interval ( t) should be chosen such tha∆

x = 0.3

lot for x = 0.25 show a more straight line ship compared to x = 0.35 or x = 0.30. re, x = 0.25 is selected and the value of K is

ed from the gradient of the line as K = (400/30) rs.

x = 0.2

the reach using the following method.

d as follows.

t K > ∆ t > xK.

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Hydro Graph