İ.T.Ü. CIVIL ENGINEERING FACULTY HYDRAULICS DIVISION HYDROLOGY Examples –3 Precipitation 1. a) On the river basin illustrated the precipitation heights observed during a strom are given below. Determine the mean precipitation height using the arithmetic mean, Thiessen and isohyetal metods (isohyets will pass every 5mm). b) Determine the local distibution of this precipitation and draw “Precipitation height-area” curve. Solution 1-a) ARITHMETIC AVERAGE METHOD: The simplest method is to take the average of the readings of all instruments. N Pmean P i i 1 N PA PD PE PF PH PI 135,4 22.6mm. 6 6 THIESSEN METHOD: The area is divided such that each point lies in the subregion of the gage to which it is closest. This is done by joining the neighboring gages by straight lines and drawing their perpendicular bisectors (normals at their midpoints). Table 1: Gage Precipitation Pi(mm) Thissen Polygon Area Ai (km2) PiAi A 8.7 233.10 2027.97 B 17.8 644.91 11479.40 C 18.3 481.74 8815.84 D 17.7 186.48 3300.70 E 21.7 85.47 1854.70 F 23.7 828.80 19642.56 G 24.9 160.58 3998.44 H 34.3 297.85 10216.26 I 29.3 903.91 26484.56 J 33.2 297.85 9888.62 K 35.5 248.64 8826.72 Total 4369.33 106535.77 N Pmean P A i 1 N i i A i 1 i 106535.77 24.4mm. 4369.33 ISOHYETAL METHOD: Isohyets (curves of equal precipitation depth) are drawn by joining the points of equal precipitation. Table 2 isohyetals Pi (mm) 35-30 32.5 30-25 27.5 25-20 22.5 20-15 17.5 15-10 12.5 <10 7.5 Total Ai (km2) 1103.34 818.44 1186.22 924.63 300.44 36.26 4369.33 A=Ai (km2) 1103.34 1921.78 3108.00 4032.63 4333.07 4369.33 PiAi 35858.55 22507.10 26689.95 16181.03 3755.50 271.95 105264.08 PiAi 35858.55 58365.65 85055.60 101236.63 104992.13 105264.08 Pmean (mm) 32.5 30.4 27.4 25.1 24.2 24.1 N Pmean P A i i 1 N i A i 1 105264.08 24.1mm. 4369.33 i Precipitation – Area Curve Yağış Yüksekliği-Alan Eğrisi Port(mm) Ortalama Yağış Yüksekliği Mean Precipitation (mm) 34 32 30 28 26 24 22 20 0 1000 2000 3000 4000 5000 22 Area (km) ) Alan,A(km 2. The reservoir surface area of a dam is 32 km2 the heat energy reaching the reservoir within a day is 500 cal/cm2. The albedo of water surface is %10. Compute the daily evaporation volume on the reservoir lake (The heat brought by the incoming river and the heat taken by the out going river will be neglected and it will be assumed that the temperature degree on the lake will not change during the day.) E Hi H0 H L(1 R ) E is the volume of evaporated water. L equals 590 cal/cm3 at normal atmospheric pressure. Hi is the heat that enters the mass (sum of the heat from the sun and the heat of inflow), Ho is the sum of the heat of outflow and the reflected heat, R is the Bowen ratio and H is the heat to change the temperature of the mass of water. Hi = 500 cal/cm2 Albedo = %10 H0 = 500*0.1=50 cal/cm2 H=0 R = Bowen Ratio = 0.2-0.3 L = 590 cal/cm3 in normal atm. pressure E Hi H0 H 500 50 0 0.61cm daily evoporation. L(1 R ) 590(1 0.25) daily evoporation volume = 0.0061m*32*106m2=195200 m3

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# solution 3 - ITU Faculty