Download the data table (file:co_q.csv, "csv" stands for "comma separated values" can be imported into EXCEL). The flowmeter has been calibrated: 1 revolution/s is equivalent to a flow velocity of 2.2 ft/s. Use the data in the table to calculate the flow velocity and put the result into the next column. Then calculate the area for which the measurement is representative (depth * width). Finally calculate the discharge rate for each section and sum them up to obtain the discharge rate in the entire river.
2) (12 points) Flood estimateA construction project is planned for a small non-monitored stream near Oracle Arizona. This stream is typically dry during most of the year, but floods during extreme presipitation events. The planning agency is concerned that their construction site might be flooded during extreme year and wants to know how high the water might be able to rise in the stream. One way is to estimate the maximum amount of precipitation to be expected in a day, assume that all the water makes it into the stream (no evaporation) and then convert the discharge rate into a depth using Manning's equation.
Info you'll need:
b) Calculate the maximum discharge in the stream assuming that all that precipitation comes down and enters the stream in 2 hours. Express the result in metric SI units.
c) Assume that the channel has a V-shape cross section, with a 90o angle at the bottom (draw a little sketch). Let us call the length of the sides of the "V": L and the depth of the channel: h. Then RH = area/wetted perimeter = L2/(2*2*L) = L/4. Remember that Q=A*U, with A being L2/2. Rearange Mannings equation to get L. Then convert L into h, using L2 = 2h2, valid for this particular triangle. So, how deep can the stream get and how far from the center do you need to be in order to avoid flooding?