Class lectures on Hydrology by Rabindra Ranjan Saha Lecture 9

Hydrologic cycle,weather and temperature, Rain gauge, precipitation, discharge measurement,stream flow, hydrographs, flood and flood routing. unit hydrographs

1. Lecture 9 Annual runoff volume Methods of estimation   Yield of the river : The total quantity of water that can be expected from a stream in a given period such as a year is called the yield of the rive r. Yield for a period of a year means the total annual runoff volume = ∑ R There are four(4) methods for estimation of yield are listed below :   Correlation of stream flow and rainfall Empirical equations Watershed simulation Rational method 1

2. Lecture 9(contd.) Method- 1: Rainfall –Runoff Correlation   The relation between rainfall and resulting runoff is quite complex due to paucity of data. However , one of the most common method is to correlate runoff –R with rainfall- P values. Plotting of values R against P and drawing a best-fit line can be adopted for very rough estimates . A better method is to fit a linear regression line between R and P and to accept the result if the correlation coefficient is linear unity . The equation for straight line regression between runoff R and rainfall P is given by R = aP + b ........................................... Eq 6-2   P R Linear line 2

3. Lecture 9(contd.) and the values of a and b are given by a = {N (∑ PR) – ( ∑P)( ∑R )} / {N(∑ P 2 ) – (∑ P) 2 } ......... Eq 6-3 b = { (∑R) – a (∑P )}/N .................... Eq 6-4 where, N = number of observation sets R and P. The coefficient of correlation (r) can be calculated as r =[{N (∑ PR) – ( ∑P)( ∑R )}] / √ [{ N(∑ P 2 ) – (∑ P) 2 } × { N (∑ R 2 ) – (∑ R) 2 }] 3

4. 4 Lecture 9(contd.) Example 6-2 The monthly rainfall P and the corresponding runoff R values covering a period of 18 months for a catchment given below in the table. Develop a correlation between R and P. And hence also calculate Correlation coefficient . Month P R Month P R Month P R 1 5 0.5 7 5 0.1 13 2 0.0 2 35 10.0 8 31 12.0 14 22 6.5 3 40 138 9 36 16.0 15 30 9.4 4 30 8.2 10 30 8.0 16 25 7.6 5 15 3.1 11 10 2.3 17 8 1.5 6 10 3.2 12 8 1.6 18 6 0.5 Data Table

5. 5 Solution : Given Rainfall and Runoff data in the table. To be computed correlation between R and P The Equation : R = aP + b .............. Eq-1 T he values of a and b are calculated by Eq-6-3 & 6-4 as given below: a = {N(∑ PR) – ( ∑P)( ∑R)} / {N(∑ P 2 ) – (∑ P) 2 } b = { (∑R) – a (∑P)}/N Lecture 9(contd.)

6. 6 Lecture 9(contd.) Month P P 2 R R 2 PR 1 5 25 0.5 0.25 2.5 2 35 1225 10.0 100.0 350 3 40 1600 13.8 190.44 552 4 30 900 8.2 67.24 246 5 15 225 3.1 9.61 46.5 6 10 100 3.2 10.24 32.0 7 5 25 0.1 0.01 0.5 8 31 961 12.0 144.0 372.0 9 36 1296 16.0 256.0 576.0 10 30 900 8.0 64.0 240.0 11 10 100 2.3 5.29 23.0 12 8 64 1.6 2.56 12.8 13 2 4 0.0 0.0 0.0 14 22 484 6.5 42.25 143.0 15 30 900 9.4 88.36 282.0 16 25 625 7.6 57.76 190.0 17 8 64 1.5 2.25 12.0 18 6 36 0.5 0.25 3.0 Total= 348 9534 104.3 1040.51 3083.3 From the given data we can calculate the required values as provided in the calculated Table below for a & b :

7. 7 Lecture 9(contd.) Putting the respective values for calculation a & b a = {N(∑ PR) – ( ∑P)( ∑R)} / {N(∑ P 2 ) – (∑ P) 2 } a = 18 ( 3083.3) – ( 348)( 104.3 )/ { 18(9534 ) – 121104} a = 0.380 b = { (∑R) – a (∑P)}/N = { 104.3 – 0.380× 348} / 18 b = (-) 1.55 Now putting the values of a & b in Eq - 1 above R = aP + b = 0.380 P + (-1.55) R = 0.380 P – 1.55 This equation is the Co-relation between R &P

8. 8 Calculation of Correlation coefficient (r ) r = { N(∑ PR) – ( ∑P)( ∑R )} / √ [{ N(∑ P 2 ) – (∑ P) 2 } × { N (∑ R 2 )– (∑ R) 2 }] Putting the corresponding values in the above equation r= { ( 18 × 3083.3) (348 × 104.3)}/ √ [ (18 ×9534) – 121104] × [ (18 × 1040.51) – 10878.49] = 0.964 Hence Correlation coefficient (r) = 0.964 Lecture 9(contd.)

9. 9 Lecture 9(contd.) Method -2 Empirical Equations : Khosla ’ s Formula   Establishment of correlation between annual rainfall and runoff : Khosla’s Formula: Khosla (1960) analyzed the rainfall, runoff and temperature data for various catchments in India and USA to arrive at an empirical relationship between runoff and rainfall and arrived a formula which is called Khosla’s Formula. The time period is taken as a month. The relationship for monthly runoff is R m = P m –L m ....................... Eq 6-8 and L m = 0.48 T m for T m > 4.5 C

10. 10 Lecture 9(contd.) where R m = monthly runoff in cm and R m ≥ 0 P m = monthly rainfall in cm   L m = monthly losses in cm T m = mean monthly temperature of the catchment in C   For T m ≤ 4.5 0C, the loss L m may provisionally be assumed as T C 4.5 - 1 - 6.5 L m 2.17 1.78 1.52 (cm) Important : Monthly loss is higher than monthly Rainfall or precipitation then monthly Runoff will be equal to zero, i.e. R m = 0 when L m > P m Monthly Annual runoff = ∑ R m

11. 11 Lecture 9(contd.) Example 6-3 For a catchment,the mean monthly rainfall and temperatures are given. Estimate the annual runoff and annual runoff coefficient by Khosla’s formula.   Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec Temp C 12 16 21 27 31 34 31 29 28 29 19 14 Rainfall ( cm) 4 4 2 2 12 32 29 16 2 1 2 Solution: Given , Rainfall & temp. of January to December is mentioned in the table. To be estimated : Total annual Runoff ∑ R m & Runoff coefficient

12. 12 Lecture 9(contd.) If the loss (L m ) is higher than the P m then R m is taken to be zero. We know, R m = P m –L m and L m = 0.48 T m for T m > 4.5 C Estimation of Runoff for the month of January: We know, R m = P m –L m = 4 – L m L m = 0.48 * 12 C (Values from Data Table ) = 5.76 cm R m = 4 –L m R m = 4 – 5.76 = - 1.76 ; As per condition R m = 0

13. 13 Lecture 9(contd.) Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec Total cm T C 12 16 21 27 31 34 31 29 28 29 19 14 - L m = 0.48T m ( cm) 5.76 7.68 10.08 12.96 17.76 16.32 14.88 13.92 13.44 13.92 9.12 6.72 - Rainfall ( cm ) 4 4 2 2 12 32 29 16.0 2 1 2 106 Runoff (cm ) R m =P m –L m 17.22 15.08 2.56 34.9 Similarly estimation of Runoff for remaining months Estimation Table Annual Runoff coefficient r = (Total Annual Runoff) / (Total Annual Rainfall) r = 34.9/ 106 = 0.329 , say 0.33

14. 14 Lecture 9(contd.) HYDROGRAPH Definition of Hydrograph   A hydrograph is the graphical representation of the discharge flowing in a river at the given location with the passage of time . A hydrograph or a runoff : It is thus a plot between time (on (X-axis), and discharge (on Y-axis) as shown in the following figure-1. Discharge in the river Recessional curve of the hydrograph Peak Flow Peak Flow Time Figure-1 : Typical hydrograph

15. 15 A hydrograph represents discharge fluctuation in the river at a given site over a given time period indicates the peak flow and can help to design the hydraulic structure at site The characteristics region (elements) of hydrographs Peak flow The maximum flow in the river due to any given storm is known as the peak flow. The peak flow will be different for different kinds of storms.   Base flow The gr ound water inflow is called base flow or t he flow in a channel due to soil moisture or ground water is called base flow. Lecture 9(contd.)

16. 16 Lecture 9(contd.) Discharge Q (m 3 /s) Figure 7-2 : Elements of flood hydrograph Lag time T L T C Storm ● A C B ● ● P● ● D T B Time in hours. t pk Q P Elements 1. The rising limb-AB : The joining point A ( the starting point of the rising curve) to the point B (the point of inflection) 2. The crest segment- BP C: between the two points of inflection B with a peak P to the deplection point or the falling limb Point C or depletion curve CD– starting from the point of depletion C.

17. 17 Lecture 9(contd.) 3. The falling limb or depletion curve CD – starting from the point of depletion C.   4. Peak time( t pk )– The time to peak from the starting point A ,   5. T L (Lag time) : The time interval from the centre of mass of rainfall to the centre of mass of hydrograph called lag time(T L ).   6. Q p - peak discharge : The highest discharge flowing in the river is peak discharge.   7. The time base of hydrograph-T B -time from the starting point A to the end of depletion point D .  

18. 18 Lecture 9(contd.) Factors Affecting hydrograph Sl.No Physiographic factors Climatic factors 1. Basin characteristics: Storm characteristics: shape size slope nature of the valley elevation drainage density. a. precipitation b. duration c. magnitude and d. movement of storm 2. Infiltration characteristics: Infiltration characteristics: land use and cover soil type and geological conditions lakes, swamps and other storage 2. Initial loss 3. Channel characteristics: cross-section, roughness and storage capacity. 3. Evapotranspiration .

19. 19 Components of hydrograph   The components of hydrograph are three(3).They are : (1) the rising limb (2) the crest segment (3) the recession limb   (1) The Rising Limb The rising limb of a hydrograph also called concentration curve represents the increase in discharge due to the gradual building up of storage in channels and over the catchments surface. The initial losses and high infiltration losses during the early period of a storm cause the discharge to rise rather slowly in initial periods. Lecture 9(contd.)

20. 20 As the storm continues, more and more flow from distant parts reach the basin outlet. Simultaneously the infiltration losses also decrease with time. Thus under a uniform storm over the catchment the runoff increases rapidly with time. Hence the basin and storm characteristics control the rising limb of the hydrograph. Crest segment The most important part of hydrograph is the crest segment because it contains peak flow. The peak flow occurs when the runoff from various parts of the catchments simultaneously contribute the maximum amount of flow at the outlet of the basin. Generally for large basin/catchments, the peak flow occurs after the cessation of rainfall, the time interval from the center of mass of rainfall to the peak being essentially controlled by basin and storm characteristics. Multiple- peaked complex hydrographs in a basin can occur when two or more storms occur in close succession. Lecture 9(contd.)

21. 21 Lecture 9(contd.) (3) The Recession limb   The recession limb which extends from the point of inflection at the end of the crest segment to the commencement of the natural groundwater flow represents the withdrawal of water from the storage built up in the basin during the earlier phases of the hydrograph. The starting point of the recession limb, i.e. the point of inflection represents the condition of maximum storage. Since the deflection of storage takes place after the creation of rainfall, the shape of this part of the hydrograph is independent of storm characteristics and depends entirely on the basin characteristics. The storage of water in the basin exists as : surface storage interflow storage ground water storage/base flow storage.

22. 22 Lecture 9(contd.) Barnes (1940) showed that the recession of a storage can be expressed as   Q t = Q K r t ............................. Eq -1 Where, Qt = discharge at a time interval t days and Q = initial discharge. K r = recession constant at time interval t days value < unity Eq - 1 can be represented as in the form of exponential decay as   Q t = Q e -at .........................Eq-2 a = - ln K r

23. 23 Where the recession constant K r can be considered to be made up of three components to take care of the three types of storages :   K r = K rs . K ri . K rb .......................Eq-3   Where. K rs = recession constant for surface storage K ri = recession constant for interflow K rb = recession constant for base flow . Lecture 9(contd.)

24. 24 The values of recession constants are as follows for time t in days: K rs = 0.05 to 0.20 K ri = 0.50 to 0.85 K rb = 0.85 to 0.99 If the interflow is not significant K ri can be assumed to be unity. When Eq-1 or Eq-2 is plotted on a semi-log paper with discharge on the log-scale, it plots as a straight line and from this the value of K r can be found. The values of K r can be calculated from storage recession curve   Lecture 9(contd.)

25. 25 Example 7-1 The recession portion of a flood hydrograph is given below. The time is indicated from the arrival of peak. Assuming the interflow component to be negligible, calculate the base flow and surface flow recession coefficients,   Lecture 9(contd.) Time from peak (days) Discharge (m 3 /s) Time from peak (days) Discharge (m 3 /s ) 0.0 90 3.5 5.0 0.5 66 4.0 3.8 1.0 34 4.5 3.0 1.5 20 5.0 2.6 2.0 13 5.5 2.2 2.5 9 6.0 1.8 3.0 6.7 6.5 1.6 7.0 1.5

26. 26 Discharge Q in m 3 /s Recession curve Figure 7-5: Storage recession curve –for the above example Solution: Given Assumption- interflow component negligible. To be calculated base flow and surface flow recession coefficients, The problem will be solved by using recession curve as Figure 7-5. Lecture 9(contd.)

27. 27 Lecture 9(contd.) The given data are plotted on a semi-log paper with discharge on the log-scale as per the above figure 7-5. The straight line AB of the curve indicates base flow. The surface flow terminates at B – 5 days after peak. Since we know, at the time elapse t , discharge Q t Q t = Q K r t , Hence base flow recession coefficient K rb t K rb t = Q t / Q Taking log in both sides;   log K rb t = log (Q t / Q ) t log K rb = log (Q t / Q ) log K rb = 1/t log (Q t / Q )

28. 28 Lecture 9(contd.) The base flow recession is shown by line ABM figure 7-5. From the figure initial discharge at 1 day after the peak ; from curve   Q = 6.60 m 3 / s and time interval of t = 2 days, i.e. 3 days after the peak, the discharge Q t from the curve = 4.0 m 3 /s . Thus putting the values: we get log K rb = 1/t log(Q t / Q )   log K rb = log(Q t / Q ) 1/t K rb = (Q t / Q ) 1/t = ( 4.0/6.6) 1/2 = 0.778, say K rb = 0.78

29. 29 Lecture 9(contd.) Surface runoff is calculated by subtracting base flow from the storage recession curve. Then plot surface runoff curve PA with the surface runoff values as calculated shown in the following table. The above curve shows the surface runoff depletion plot as a straight line. Time after Peak Values from storage recession curve PA (m 3 /sec) Values of Base flow from base flow recession ABM line (m 3 /sec) Surface runoff = col 2 – col 3 (m 3 /sec) 1 2 3 4 0.5 66.0 8.3 57.7 1.0 34.0 8.0 26.0 1.5 20.0 5.5 14.5 2.0 13.0 4.5 8.5 2.5 9.0 4.3 3.7 3.0 6.7 4.35 2.25 3.5 5.0 2.8 2.2 4.0 3.8 2.4 1.6 4.5 3.0 2.3 0.70 5.0

30. 30 Lecture 9(contd.) From the surface runoff curve, initial discharge- Q = 26 m 3 /s at time t=1.0 h and the corresponding Q t = 2.25 m 3 /s at t = 3.0 h i.e. at an time interval 2 hours.   Hence the recession constant for surface storage K rs is given by   log K rs = ½ log(2.25 / 26) = log(2.25 / 26) ½l K rs = 0.29 Ans : K rb = 0.78 and K rs = 0.29