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- 1. Question # 01 A Reservoir operator has to be release water from reservoir for being picked up at a distanceof 50 Km for downstream users. The average width of the stream for the anticipator discharge of40km, The mean daily class a tank evaporation for this season is 0.5cm.Estimate the mean daily evaporation from the stream in Acre-feet per day .If the pan co-efficient is0.7. Assuming 15% loss due to seepage , find discharge at the head of canal if the required dischargeat the tail is to be 50 m3/s.SOLUTION : Discharge at tail Qtail = 50m3/s Discharge at head Qhead =? Surface area of stream = A =LxW as L = 50 km= 50 x 1000= 50000 m W = 40 m A = 50000 x 40 = 2000000 Seepage Lossees = 15 %Evaporation Losses : PAN Evaporation E = 0.5 cm PAN Co-efficient Kp = 0.7 LAKE Evaporation El =? El = Kp.E El = 0.7 x 0.5 El = 0.35 = 3.5 x10-3m Or 0.0035mSo Evaporation losses from stream per day E stream = 0.0035 x 2000000 = 7000m3/day = _ 7000_ 24 x 3600 = 0.081m3/sec As 1 Acer-ft = 43560ft 1 meter = 3.282 As = 7000 m3 day = 7000 x (3.28)3 = 247013 ft3 / day As 1 Acer-ft = 43560 ft So = 247013 43560 = 5.6 Acer-ft day
- 2. Question # 02 An Engineer-In-Charge of Reservoir operation has to release water from reservoir toprovide irrigation supplies at a discharge of 40 km for downstream Users. The average widthof the stream for the anticipates discharge is 25 km. The daily mean Class A pan evaporationfor this season is 5mm per day.Estimate the daily evaporation losses from the stream in hector-meter-per day.Solution: Stream Length = 40 km = 40 x 1000 = 40000 m Stream width (W) = 25m Surface Area of stream = 40000 x 25 = 1000000m2 Assuming Kp = 0.7 Pan Evaporation =E = 5mm/day El = Kp E = 0.7 x 5 = 3.5mnm or 0.0035m E Stream = 0.0035 x 1000000m 3 = 3500m3 1 Hector = 10000m2 E Stream = 3500_ 10000 = 0.35 hector-meter
- 3. Question # 03 A small catchment area 150 hectare received a rainfall of 10.5cm in 90 minutes, Dueto a storm draining the catchment was dry before the storm and experienced a runofflasting for 10 hours with an average discharge Value of 2m3/sec. The stream was given dryafter the runoff event.(a) What is the amount of water in acre-feet which was not available to runoff alone tocombined effect of infiltration, evaporation and transpiration.(b) What is the ratio of total and direct runoff to precipitation.Solution : Catchment Area “A” = 150 hectr as 1hectr = 10000m2 = 1500000m2 Precipitation “P” = 10.5cm = 0.105m Total Runoff Volume = A x P = 1500000 x .0.015 = 1575000m3 Time Duration of Rain = 90 min = 90 x 60 = 5400 Sec Total Runoff Discharge Q Rain = 157500 5400 = 29.17m3/Sec Q Runoff = 2m3/sec Runoff time = 10 hrs = 10 x 3600 = 36000 sec Total Runoff Volume = 2 x 36000 = 72000m3(a) For Water Lost : Water Lost = Q Rain – Q Runoff = 157500 – 72000 Water Lost = 85500 m3 as 1m3 = 35.315 ft3 1acre-f t = 43560 ft3So, Water Lost = 85500 x 35.315 43560 Water Lost = 69.3 Acer-Ft(b) Ratio Between Total Runoff to Direct Runoff : Direct Runoff = 72000m3 Total Runoff = 157500m3 Ratio = 72000_ = 0.457 157500 Percentage = 72000 x100 157500 = 45.7%
- 4. Question # 04 For Data given Example 4.4 Find total infiltration during the storm period UsingHorton’s Equation assuming fo = 1.5cm/h and fc = 0.5cm/h.Solution : Total Infiltration “F” = ? Rain = 50mm Time Duration = 2 HRS Initial Infiltration Rate fo = 1.5cm/h Constant Infiltration Rate fc = 0.5cm/h.AS Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet K = [1.5 – 0.5] (1 –e -1x2) + 0.5 x 2 1 = [1] (0.864) + 1 1 = 1.864 cmQuestion # 05 A storm width 10cm precipitation produced a direct runoff as 5.8cm given the time
- 5. distribution of the storm in Table 4.3Estimate the ɸ Index.Solution : Table 4.3 Time Distribution of the Storm.Hour(Time) 1 2 3 4 5 6 7 8International Rainfall (cm) 0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5 Calculation for Φ Index Rainfall (cm) “f” values (cm) 0.45 .50 0.6 0.4 - - - 0.9 0.45 0.4 0.3 1.5 1.05 1.0 0.9 2.3 1.85 1.8 2 1.8 1.35 1.3 1.2 1.6 1.15 1.10 1 1.0 0.55 0.5 0.4 0.5 0.05 - Total 6.45 6.10 5.8
- 6. Question # 07 For 3-Hours duration 225mm of total Rainfall we observed over 3200 km 2 catchmentArea. The infiltration capacity curve for this area can be given by Horton’s Equation(Equation 4.6 & 4.7) in wich fo = 10mm/h and fc = 0.5mm/h. Evaporation and other lossesduring the storm period were observed to be 50mm.Find excess Rainfall over the Catchment(a) Estimate Direct Runoff Volume in m3 & Hectare –m from excess Rainfall(b) Total Run off in Hector-mSolution : Time Duration “t” = 3Hours Rainfall “P” =225mm Area “A” = 3200km2 Initial Infiltration Rate fo = 10mm/h Constant Infiltration Rate fc = 0.5mm/h K = 1h-1 Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet K = [10 – 0.5] (1 –e -1x3) + 0.5 x 3 1 = [9.5] (0.950) + 1.5 = 10.5 mm Excess Rain Fall = Rain – F – Losses = 225 – 10.5 – 50 = 164.5 Direct Runoff = Rainfall Excess x Area of Catchment = 164.5 x 3200 (1000)2 1000 = 526400000m3 = 526400000 10000
- 7. = 52640 Hector – meter Total Runoff = [Rain – Losses other than infiltration ] x A = [225-50] x 3200 (1000)2 m3 1000 = 0.175 x 3200 (1000)2 = 560000000 m3 As 1 Hectr – meter = 560000000 hector – meter 10000 = 56000 hector – meterQuestion # 08 An infiltration Capacity Curve prepared for a Catchment indicates an initial capacityof 2.5cm/h and attains a constant value of 0.5 cm/h after 10 hour of Rainfall. With theHorton’s constant K = 6 day-1Solution : Time Duration “t” = 10 Hours Initial Infiltration Rate fo = 2.5 mm/h Constant Infiltration Rate fc = 0.5 mm/h K = 6 days-1 Or 6/24 = 0.25h-1 As Horton’s Equation Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet K = [2.5 – 0.5] (1 –e -0.25x10) + 0.5 x 10 0.25 = [8] (0.918) + 5 = 7.34 mm = 12.34 cm
- 8. Question # 09 In a project related to rainfall – runoff stadiesw f – curve was plotted to establishedan equation of the form of Horton’s Equation. If F = 8.50 sq Units on the graph with each sqrepresented 1 cm/h on the Vertical and 2 minute on the abscissa and f o = 4.5cm/h fc =1.2cm/h Determine the Horton’s Equation and Calculate “f” for t = 10 minutesSolution : Time Duration “t” = 10 Min Initial Infiltration Rate fo = 4.5 mm/h Constant Infiltration Rate fc = 1.2 mm/h F = 8.50 Units = 8.50 (1cm x 2 h) h 60 As Horton’s Equation Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1 K = [4.5 – 0.5] (1 –e -1x.166) + 1.2 x 0.166 1 = [3.3] (0.152) + 2 = 5.3 mmQuestion # 10 In a storm, total rain fall is 2.29cm and the total infiltration loss is 0.88cm,
- 9. Calculate the rain fall excess. Neglect evaporation during the period.Solution : Rainfall “P” = 2.29cm Infiltration “F” = 0.88 cm Rainfall excess =? Rainfall excess =P–F = 2.29-0.88 = 1.41cmQuestion # 11 Determine the runoff from a catchment of area 2.3k over wich 7.5cm of Rainfalloccurred during 1 day Storm.An infiltration of 0.6 cm/h and attained a constant Value of 0.15cm/h after 12 hours ofRainfall with Horton’s Constant Value of K=3k-1. A class A-pom installed in the catchmentindicates a degree of 2.5 cm in water level on that day. All other losses were found to benegligible.Solution : Catchment Area “A” = 2.3km2 Rainfall “P” = 7.5cm Time Duration “t” = 1 day 24hoursInfiltration Losses : Time Duration “t” = 12 Hours Initial Infiltration Rate fo = 0.6 mm/h Constant Infiltration Rate fc = 0.15 mm/h K = 3 h-1As Horton’s Equation : Total Infiltration F = [ fo – fc ] (1 – e-kt) + fet Let K = 1 K = [0.6 – 0.15] (1 –e -3x12) + 0.15 x 24 1 = [0.15] (1) + (3.6) = 3.75 mmEvaporation Losses : E = 2.5cm Kp = 0.7 El =? El = Kp x E = 0.7 x 2.5 = 1.75cmRainfall Excess :
- 10. = P – F – El = 7.5 -3.75 – 1.75 = 2.00 cmDirect Runoff : = Rainfall Excess x Area of Catchment = _2_ x 2.3 (1000)2 100 = 46000 m3 = 46000 10000 = 4.6 Hector – meter

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