(5) A hydrology student is wandering through a large wilderness area during his
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(5) A hydrology student is wandering through a large wilderness area during his vacation. He has seen little wildlife during his trek, although he has been following a stream in the region dining most off his hike. Feeling bored with the lack of wild companionship, he decides to rest along an uncharacteristically straight portion of the stream. To pass the time and sharpen his skills, he decides to estimate the volume of water flowing in the river at this point. He estimates the river is 1 ?deep and 21' wide. He steps out a length along the bank which is about 10 yards long. Rummaging through his pack, he finds an orange, which he tosses into the center of the river. Utilizing his trusty digital watch, he determines that the orange takes 35 seconds to traverse the measured distance. As he sits down in anticipation of crunching a few numbers, he realizes that he forgot to pack his calculator for this true wilderness experience. Upon his return a few days later, he asks you to perform the following computations. (a) Determine the velocity of the orange. (b) Compute the discharge Q in cfs, using the continuity equation for streamflow, which is expressed as = VA, where A is the cross-sectional area of the channel. Be sure to include a correction factor of 0.8 to adjust the surface velocity to the average velocity over the vertical profile. (c) Compute the volume that will flow by the point at which the measurements were taken in one day. Express in AF, ft3 and MCM..Explanation / Answer
a.) 10 yards = 30 ft
1 inch = 0.0833 ft
21 inch = 1.75 ft
Area = 0.0833*1.75 = 0.14583 ft^2
v = s/t = 30/35 = 0.857 ft/s
b.) Q = A*v = 0.8*0.857*0.14583 = 0.1 ft^3/s = 0.1 cfs
c.) Volume on 1 day = Q*t = 0.1*24*60*60 = 8640 ft^3
1 ft^3 = 1/(43560) AF
So Volume in AF = 8640/43560 = 0.19835 AF
1 MCM = 10^6 m^3 = 10^6*(3.28084)^3 ft^3
So 1 ft3 = 2.8317 x 10^-8 MCM
So Volume in MCM = 2.8317 x 10^-8*8640 = 2.4465 x 10^-4 MCM
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