Compute the mean precipitation for the catchment shown by Arithmetic Mean Method.
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Q: the mean precipitation, as computed by the Thiessen
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Q: Determine average rainfall using Arithmetic mean method
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- The total annual precipitation for a region is approximately normally distributed with a mean of 848 mm and a standard deviation of 124 mm. What is the probability of an annual precipitation greater than 500 mm but less than 700 mm? Answr = 0.1145Water flows at a rate of 0.0300m3/? from reservoir 1 to reservoir 2 through threepipes connect in series. If ? = 0.025 and neglecting minor losses, determine thedifference in water surface elevation.Below is a square basin with a single length of 3180 meters. Watershed8 hours for 5 stations at their corners (a, B, C, D) and at their full midpoint (E)Hyetographs of rainfall Figure 4. is also given. Average rainfall of the basin based on information givencalculate its height using the Thiessen method.(T=159)
- Calculate the slope (%) of the watershed to TWO decimal places given the following information: Highest elevation: 401.6m Lowest elevation: 394.1m Length: 597m Area: 460haFor a drainage basin of 600 km2, isohyets drawn for a storm gave the following data: Isohyets (interval) (cm) 15-12 12-9 9-6 6-3 3-1 Inter-isohyetal area (km2) 92 128 120 175 85 Estimate the average depth of precipitation over the catchment.In a watershed, the average annual precipitation for four subbasins was recorded as follows : 90.84,102.27,74.84 and 63.406cm.The area sof the sub-basins were:93264.3,71243.5,108808.2 and 168393.8 ha.Calculatethe averageprecipitationofthetotalwatershedusingThiessenmethod.
- Hydrology Subject: Using Arithmetic Method, Thiesen Polygon Method and Isohyets Method, Find average rainfall over a catchment. The data isThe annual rainfall at the rain-gauge stations with the catchment area of 10km are indicated below. Thiessen polygon measured are also shown in the table below. Stations Recorded rainfall(P1)(cm) Area of polygon(ai)(km2) A 8.8 570 B 7.6 920 C 10.8 720 D 9.2 620 E 13.8 520 F 10.4 550 G 8.5 400 H 10.5 650 I 11.2 500 J 9.5 350 K 7.8 520 L 5.2 250 M 5.6 350 1. Describe mean rainfall over the area using Thiessen polygon method and Compare Thiessen polygon method with arithmetic mean method.Estimate the total excess/effective rainfall in mm for a 5% AEP storm. The temporal pattern distribution for a 45-min rainfall duration (5-min time step) for a South Australian catchment of 5.4 km2 is given below. 20 19.17 16.67 12.5 5 10 8.33 3.33 5 Consider: IL = 11.5 mm and CL = 1.7 mm/hr Total rainfall depth = 30.1 mm Plot the effective rainfall hyetograph of this Question in mm.
- Year Annual rainfall (mm) 1989 2,411.50 1990 2,889.30 1991 1,775.70 1992 1,706.50 1993 2,634.60 1994 2,283.70 1995 1,696.50 1996 2,429.80 1997 2,369.90 1998 2,006.60 1999 2,295.00 2000 2,765.80 2001 1,592.20 2002 2,061.70 2003 1,757.70 2004 2,421.30 2005 3,315.50 2006 2,826.30 2007 2,958.20 2008 3,536.20 2009 1,962.30 2010 2,604.80 2011 2,455.90 2012 2,470.70 2013 2,430.10 2014 1,824.40 2015 2,324.90 2016 1,872.50 2017 2,648.70 2018 1,851.60 Question 1 Calculate the following (i) Mean annual rainfall (ii). Standard Deviation (iii) Coefficient of Variation (iv) Skewness (v) Kurtosis Explain the meaning of the descriptive statistics values (i) to (v)Hyetograph is the relation between? a. Sum of direct and base flow with time b. Precipitation with time c. Effective precipitation with time d. Net and effective precipitation What correct answer?The following stream flow data was collected from a watershed of 280.35 acres. Plot the hydrograph as an x-y scatter plot. Time (hr) Discharge (cfs) 0.0 60 0.5 54 1.0 50 1.5 90 2.0 137 2.5 176 3.0 200 3.5 198 4.0 140 4.5 97 5.0 60 5.5 40 6.0 35 6.5 32 7.0 29 7.5 24 8.0 22 8.5 20 9.0 20 9.5 19 10.0 19 10.5 18 11.0 18 11.5 18 12.0 17