Example: X and Y are two neighboring rainfall stations. Station X has complete records and station Y has some missing values. Find the linear correlation equation between the two series as mentioned in columns and 3 of the following table (8) and then check the correlation by computing both the correlation and regression coefficients, showing the correlation line on an X-Y diagram? Using the derived equation, find the missing data of Y if the observed data at X for the same years are (11, 17 and 166 mm). Table (6): Annual precipitation amounts as recorded by stations X and Y. No. X Y 1 145 5 155 3 19 148 4 3 15 5 5 153 6 15 14 7 167 138 Table (7): Annual precipitation amounts as recorded by stations X and Y. (continued) 8 196 145 9 11 18 1 15 13
11 135 133 Solution: Let us compute (a) and (b) first. To do so, the following table (4) should be completed. Step 1: preparing of table (9): Table (8): Computations of the Components to be Used in the Least Squares Method No. X Y X*Y X Y 1 145 9 4 15 5 155 34875 565 45 3 19 148 81 361 194 4 3 15 3496 59 314 5 5 153 31365 45 349 6 15 14 1 5 196 7 167 138 346 7889 1944 8 196 145 84 38416 15 9 11 18 14336 1544 16384 1 15 13 165 1565 169 11 135 133 17955 185 17689 Total 1935 1567 7937 356849 419 Step : Computation of the slope of the line:
In order to compute (b) in the linear equation we have to follow the following computations: b N * ( X * Y ) ( X * * ( X ) ( X ) = N Y ) 11* 7937 1935 *1567 b = 11* 356849 1935 *1935 37597 33145 b = 395339 37445 445 b = 181114 b =.335 Step : Computation of the intercept of the line: For the computation of (a) in the linear equation we have to follow the following computations: Y b * X a = N 1567.335 *1935 a = 11 1567 43.185 a = 11
1134.81775 a = 11 a =13.165 As a result, the linear equation can be written as: Y = 13.17 +.334 * X Step 3: Computation of the correlation coefficient of the line: For the computation of (r) in the linear equation we have to follow the following computations: r = {[ N * ( N * ( X X * Y ) ( X ) * ( ( X ) ]*[ N * ( Y ) Y ) ) ( Y ) ]} r = 11* 7937 1935 *1567 {[11* 356849 1935 *1935]*[11* 419 1567 *1567 r = 37597 33145 {[3945139 37445]*[465199 455489 r = 445 {[181114]*[971] r = 445 41935.867 r =.96676
Step 4: Computation of the regression coefficient of the line: For the computation of (r ) in the linear equation we have to follow the following computations: r a * ( = Y ) + b * ( X * Y ) * ( Y 1 ( N Y ) 1 N Y ) r 13.165 *1567 +.335 * 7937.991* 455489 = 419.991* 455489 r 161659.555 + 6387.685 38.55 = 419 38.55 r = 818.5 88.495 r =.99 r.93, which is the square of the value of (r) Step 5: Filling the missing values at station Y using the regression equation: The derived equation was: Y = 13.17 +.334 * X
From the observed precipitation values at station X, 11, 17 and 166 mm, we can have the computed values at station Y as follows: Y 1 = 13.17+.334*11 = 17.7 mm Y = 13.17+.334*17 = 14.8 mm Y 3 = 13.17+.334*166 = 141.9 mm Step 6: Drawing of the regression line: The regression line can be represented as shown in the following figure (8). Figure (8): The regression line between the precipitation values of the rainfall stations X and Y
16 15 Series1 Linear (Series1) 14 13 y =.34x + 13.17 R =.935 1 1 15 15 175 5 5 Example: Annual precipitation at rain gauge X and the average annual precipitation at twenty surrounding stations are listed in the following table: a. Examine the consistency of station X b. When did a change in regime occur? Discuss possible causes. c. Adjust the data and determine what difference this makes to the 36-year annual average at station X.
Table (9): Observed annual precipitation values at Station X and computed average annual precipitation for twenty nearby stations Year Annual precipitation (mm) Year Annual precipitation (mm) Gauge X -Station Average Gauge X -Station Average 199 188 64 1974 45 3 1991 185 8 1973 8 36 199 31 386 197 315 34 1989 95 33 1971 3 9 1988 8 33 197 5 37 1987 87 38 1969 9 75 1986 183 31 1968 315 55 1985 34 4 1967 31 55 1984 8 34 1966 338 5 1983 16 35 1965 35 35 Table (9): Observed annual precipitation values at Station X and computed average annual precipitation for twenty nearby stations (continued) Year Annual precipitation (mm) Year Gauge X -Station Average 198 4 33 1964 45 7 1981 3 3 1963 333 3 198 84 31 196 35 1979 95 375 1961 35 75 1978 185 31 196 345 1 1977 69 333 1959 318 1976 1 36 1958 313 5 1975 6 4 1957 4 8 Annual precipitation (mm) Gauge X -Station Average Solution: Let us first compute the cumulative values of the precipitation amounts tabulated in the above-mentioned table (9). To do so, the following table (1) should be completed. Step 1: Preparing of table (1):
Both the observations of the X gauge and the data of the average of the surrounding stations are being accumulated on year basis as shown in the following table (1). Step : Plotting the accumulating data against each other where the accumulated precipitation of X gauge on the ordinate and the accumulated average of the surrounding stations on the abscissa. Step 3: Computing of the linear equations of fitting lines and finding the different slopes. Step 4: Finding the averages of the recent data presented by the recent line for both variables and then, finding the ratio between them. Step 5: Finding the annual means of the records of the whole period for. both variables, which was found as 8.5 mm and 99.64 mm for the station X and the series of average of the stations respectively. Table (1): The cumulative data calculation of the precipitation data of station X and the average of the surrounding stations Cumulative Annual Cumulative Annual precipitation (mm) precipitation (mm) Year Year Gauge -Station -Station Gauge X X Average Average 199 188 64 1974 4579 6 1991 373 49 1973 4859 656 199 683 878 197 5174 69 1989 978 18 1971 5474 719
1988 1186 1538 197 574 756 1987 1473 1918 1969 614 7835 1986 1656 8 1968 639 89 1985 196 68 1967 6639 8345 1984 188 86 1966 6977 857 1983 44 3187 1965 737 885 198 68 3517 1964 7777 975 1981 831 3817 1963 811 9375 198 3115 419 196 8435 9575 1979 341 454 1961 8785 985 1978 3595 485 196 913 16 1977 3864 5138 1959 9448 18 1976 474 5498 1958 9761 157 1975 4334 5898 1957 1161 1787 a. The cumulative values are drawn in figure (9). There is inconsistency in the data after the year of 197. b. The cause of the inconsistency could be a change in the gauge location or in the exposure affected by the conditions surrounding the station such as outcropping of trees or construction of buildings nearby the station, with which a rain gauge samples the rainfall in the particular area. For example, moving a gauge from one part of a roof to another can affect the catch of the gage greatly, even though the gauge is moved only a few meters. Also construction a tall building in the vicinity of the gauge changes wind direction and affects thermal air currents, thus influencing the catch in the rain gauge.
c. If the earlier period is correct, then for the period of 1957-197 the ratio between the average precipitation will be: Gauge X Stations Average average 1.33 This ratio applied to the whole gauge X record gives an annual average over 36 years of 334.79 mm instead of the existing average of 8.5 mm. Figure (9): Double Mass Curve for consistency of Precipitation Data 11 1 9 y = 1.345x - 4464.8 R =.9964 Cumulative X Total 8 7 6 5 4 3 1 y =.744x R =.9988 4 6 8 1 1 Cumulative StationsTotal
Example: an irregular cross section of 3. m width and 1.8 m maximum depth has a maximum depth of water equals to 1.6 m that was measured by a standard current meter. The velocities was rated to be as indicated in the following table (5). Distance from Right Edge [m] Vertical Depth [m] Method Used Velocity-1 [m 3 /s] Velocity- [m 3 /s]..4.5 S.V..6.35.6 D.3.8.45.6 D.36 1 1. D,.8 D.18.1 1. 1.3. D,.8 D..118 1.4 1.4. D,.8 D.7.13 1.6 1.6. D,.8 D.3.13 1.8 1.4. D,.8 D.7.13 1.. D,.8 D.19.111..7. D,.8 D..8.4.65. D,.8 D.18.7.6.5.6 D.36.8 3 Table (5): Current Meter Measurements in a Cross-Section X Vertical Depth 1.8 1.6 1.4 1. 1.8.6.4. Fig. (13): Cross-Section for Current Meter Measur...4.6.8 1. 1. 1.4 1.6 1.8...4.6.8 3. Distance (width) [m] Cross Section Water Level Upper Measurement Points Lower M
Solution: Step 1: Averaging the Velocities by Applying the equations indicated for the methods to be used, as in table (4) taking into consideration the depth of water. Column 4 of table (6). Step : Computation of Partial Areas, by multiplying the average width with the depth. Column 5 of table (6). Step 3: Computation of partial Q for each vertical partial section, by multiplying the partial area with its related average velocity. Column 6 of table (6). Step 4: Finding the average velocity of the flow, by dividing the total of the column 6 to column 5 as follows: V = Q A =.13618.18 =.63 Table (36): Discharge Computation by Current Meter Measurement Distance from Right Edge [m] Vertical Depth [m] Method Used Corrected Velocity [m/s] Partial Areas Discharge Q [m ] [m 3 /s] Column 1 Column Column 3 Column 4 Column 5 Column 6..4.5 S.V.17.5.85.6.35.6 D.3.7.4.8.45.6 D.36.9.34 1 1. D,.8 D.6..1 1. 1.3. D,.8 D.7.6.18 1.4 1.4. D,.8 D.75.8.1 1.6 1.6. D,.8 D.8.3.56 1.8 1.4. D,.8 D.75.8.1 1.. D,.8 D.65.4.156..7. D,.8 D.5.14.7.4.65. D,.8 D.45.13.585.6.5.6 D.36.1.36.8 3 Σ.16.13618
Fig. (14): Daily Streamflow Hydrograph [m3/s] Q [m3/s] 45 4 35 3 5 15 1 5 5 1 15 5 3 35 Time [days] Daily Streamflow [m3/s] 3 Fig. (15): Mass Curve of Annual Streamflow [m3] 5 Mass Curve [m3] Q [m3] 15 1 5 1985 1986 1987 1988 1989 199 1991 199 1993 1994 Time [years]
4 Fig. (16): Mass Curve of Annual Streamflow [m3] 35 3 y = 9.139x - 57766 R =.9455 5 Q [m3] 15 1 5 Mass Curve [m3] Series Linear (Mass Curve [m3]) 1985 1986 1987 1988 1989 199 1991 199 1993 1994 Time [years]
Figure (17): Runoff Accumulation-time Curve for a River Accumulation and its Probability 3 5 15 1 5 1% 3% 5% 7% 9% J F M A M J J A S O N D Date on Which Runoff Accumulation Commences Figure (18): Hydrograph of Streamflow with Baseflow Guage Height (m) 3.6 3.4 3..8 3.6.4. 1.8 1.6 1.4 1. 1.8.6.4. Streamflow [m3/s] Baseflow [m3/s] 4 6 8 1 1 14 16 18 4 Duration (h)