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CVEN9625 - Fundamentals of Water Engineering

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University of New South Wales

   

Fundamentals of Water Engineering (CVEN9625)

   

Added on  2020-03-04

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This solution provides an introduction to water engineering with a focus on hydrology and hydraulics. Topics discussed include; Hydrology: Australian hydrology, catchment processes and energy cycle, meteorological and hydrological measurement techniques, evaporation estimation, design storm and flood estimation. To give you an understanding of the properties of fluids, hydrostatics and the principles of fluid flow based on mass, energy and momentum.

CVEN9625 - Fundamentals of Water Engineering

   

University of New South Wales

   

Fundamentals of Water Engineering (CVEN9625)

   Added on 2020-03-04

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SOLUTIONS TO THE QUESTIONS1.Question 1 The following information is available for a project site: • Latitude = 35°S • Longitude = 152°E • Elevation = 0 m (at sea level) • Average Albedo = 0.2 Jun-Aug Sep-Nov Dec-Feb Mar-May Air Temperature (°C) 17 21 24 20 Relative Humidity (%) 65 70 70 60 Wind Speed (m/s) 2.2 2.4 2.3 2.1 Average sunshine hours 8 8.4 8.7 7.6 Estimate the following using the above information: (a)Potential Evaporation for an open water body using the Penman Monteith equationThis is given by: PE= (D/D+Y)xQet+ (Y/D+Y)xEatWhere Qet= Qs(1-r)-Q1Eat= 0.3(1+0.5U)(Ea-Ed)D= slope of the saturation vapor pressureY= hygrometric constantQ1= long wave radiationr = coefficient relating to vegetation cover (r=0.25)Ta= air temperature n/N= ratio of actual to possible sunshine hours of bright sunshineῥ = density of water (kg/m3)u= wind velocity ea= saturation vapor pressure ed= actual vapor pressureY= psychometric constant Qet= Qs(1-r)-QQ1= 0.95{(8.64x107)/ ῥλ} xꓶx(273.16+Ta)4x(0.53+0.065(Ed-1)0.5x(0.1+0.9))The saturation vapor pressure is determined as follows:
CVEN9625 - Fundamentals of Water Engineering_1
Ea1= 610.78Exp{T/(T+238.3)x17.2694} = 610.78Exp(17/(17+238.3)x17.2694)= 613.14PaEa2= 610.78Exp{21/(21+238.3)x17.2694}= 613.65PaEa3= 610.78Exp{24/(24+238.3)x17.2694}= 614.02PaEa4= 610.78Exp{20/(20+238.3)}x17.2694= 613.52PaThe long and short wavelengths are given as : Qs= 0.2mm and Ql= 0.4mmThe psychometric constant Y= 0.665x10-3PNow, since this is at sea level, Y= 0.665x10-3x 101.3x10-3= 67.3645The actual vapor pressure Ed can be obtained from the following relationship:RH= P/Ps= Ed/EaEd1= RHxEa1= 0.65x613.14= 398.54PaEd2= 0.7x613.65= 429.56PaEd3= 0.7x614.02= 429.81PaEd4= 0.6x613.52=368.11PaNow, Eat values are determined from the equation: Eat1=0.3(1+0.5U)(613.14-398.54)Eat1 = 0.3x2.1x214.6=135.198Eat2= 0.3(1+0.5x2.4)(613.65-429.56)= 121.49Eat3= 0.3(1+0.5x2.3)(614.02-429.81)= 118.82Eat4= 0.3(1+0.5x2.1)(614.52-368.11)= 151.54 Qet= 0.2(1-0.25)-0.4= 0.25The slope of the saturation vapor pressure is given by: D= 4098{0.6108Exp[17.27T/T+237.3]}/(T+237.3)2Tmean= (17+21+24+20)/4= 20.5D= 4098[0.6108Exp(17.27x20.5)/20.5+237.3]/ (20.5+237.3)2D= 4098x0.6108x3.948/(66 460.84)= 0.1487Hence the actual potential evaporation is determined using equation below:
CVEN9625 - Fundamentals of Water Engineering_2

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