Stream cross-sectional areas (A) are required for a number of tasks in water res
ID: 2875809 • Letter: S
Question
Stream cross-sectional areas (A) are required for a number of tasks in water resources engineering, including flood forecasting and reservoir design. Unless electronic sounding devices are available to obtain continuous profiles of the channel bottom, the engineer must rely on discrete depth measurements to compute A. An example of a typical stream cross section is shown in the figure below.
Stream cross-sectional areas (A) are required for a number of tasks in water resources engineering, including flood forecasting and reservoir design. Unless electronic sounding devices are available to obtain continuous profiles of the channel bottom, the engineer must rely on discrete depth measurements to compute A. An example of a typical stream cross section is shown in the figure below Water surface 0 18244 643.6 3.4 28 4. 10 20 Distance from teft bank, m The data points represent locations where a boat was anchored and depth readings taken. Estimate the cross sectional area A from this data using the trapezoidal method of numerical integration.Explanation / Answer
Data below scaled from drawing In Mathcad origin default is 0, reset to 1 in the Math / built-in variables submenu i := 1.. 12 distance depth Data 6.5 13.00 19.5 26 32.5 39 46.5 52.5 59 65.5 72 0 2 2 4 4 7 5.5 4 3.5 2.8 1.4 æ ç ç ç ç ç ç ç ç ç ç ç ç ö ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ := × ft Area 2 segments 2 × h f( 1) + 4 × f( 2) + f( 3) 6 å × Area 2 segments å (width) × (average_height) i := 1, 3.. n - 1 j := 2, 4.. n k := 3, 5.. n - 1 Simpson's 1/3 rule - each segment requires 3 points i ådepthi j ådepthj + 2 × h 235.3 ft2 = × Simpson's rule with summations multiplied through i := 1, 2.. n - 1 j := 2, 3.. 12 or Area 235.3 ft2 = × Area i depthi + depthi+1 2 å × (h) := Area (width) × (ave_height) i := 1, 2.. 11 compute the flow area using Simpson's rule depthi Datai, 2 := label each column separately disti Datai, 1 := distance between sequential data points h := 6.5 × ft n := 12 number of points 72 78.5 1.4 0 ç è ÷ ø Q Simpson's 3/8 rule 3 1.408 103 ´ ft3 sec = Q 3 Area 3 := × V Q Simpson's 1/3 rule 2 1.456 103 ´ ft3 sec = Q 2 Area 2 := × V Q 1.412 10 Simpson's rule 3 ´ ft3 sec Q := Area × V = V 6 ft sec := × If the average flow velocity in the stream is 6 fps what is the flow rate? Area 3 234.731 ft2 = × Area 3 ( 3 × h) i æådepthi è ö ø 3 j ådepthj + × k æådepthk è ö ø + × 3 l ådepthl + 8 × é ê ë ù ú û := i := 1, 4.. n - 2 j := 2, 5.. n - 1 k := 3, 6.. n l := 4, 7.. n - 2 Area 3 segments 3 × h 8 å × (f( 1) + 3 × f( 2) + 3 × f( 3) + f( 4)) Area 3 segments å (width) × (average_height) Simpson's 3/8 rule - each segment requires 4 points Area 2 242.667 ft2 = × Area 2 2 × h i ådepthi 4 j æådepthj è ö ø + × k æådepthk è ö ø + 6 é ê ë ù ú