ENV* K245: Water Resources Engineering Measuring Streamflow |

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We will compare two techniques for estimating streamflow, Q. Both techniques require us to estimate the cross-sectional area, A. Direct measurement also requires estimating the water velocity, v. The Slope-Area Method will also require us to calculate the hydraulic radius (area divided by wetted perimeter) and to estimate the slope and describe the condition of the bottom of the stream.

We will directly measure depth, width, and velocity. We will look at the bottom to guess what it’s made of. We will estimate slope from a map.

Show map |

- cross-section
- Divide the cross-section of the stream into rectangles and/or triangles. Measure the width of each segment with a surveying tape.
- depth
- Measure the depth with a rod, sounding line, or yardstick. Cross-sectional area is then width times depth for a rectangle or one half width times depth for a triangle.
- velocity, v
- This is the linear velocity of a water molecule
- Using the rotating vane flowmeter, measure the average velocity at one or more points within each rectangle or triangle.
- Measure the velocity at various points in the cross section then calculate an average
- At a point, the average velocity is approximately the average of the velocities at 20 percent and 80 percent of the depth, so:
v

_{average}= (v_{0.2z}+ v_{0.8z})/2 - Alternatively, we can measure the surface velocity with a floating stick and a stop watch (and a measuring tape to give us horizontal distance)
- Note that:
v

_{average}= v_{surface}/1.2 - flow is given by:

Q = vA

- If we know the slope and area of a channel we can calculate a quantity known as normal discharge from Manning’s equation:
- Example: What is the normal discharge for a rectangular concrete channel (n=0.013) with a width of 10 feet and a depth of 2 feet, falling with a slope of 2.5 feet per thousand feet (0.0025)?

Q = (1.49/n)AR^{2/3}S^{1/2}

with

n = a roughness factor (about 0.015 for artificial channels like sewers, 0.022 for river bottoms, 0.05 to 0.1 for flooded ground) (see Table 3.4 pp. 135-137; see also p. 118)

A = cross-sectional area

R = hydraulic radius = A/WP

WP = wetted perimeter (If you draw the cross-section, WP is the length of the line that is underwater. For a rectangle, WP = width + 2*depth.)

S = slope

A = 10*2 = 20 ft^{2}

R = (10*2)/(2+10+2) = 1.43 ft

Q = 1.49/0.013*20*1.43^(2/3)*.0025^.5

= 145 ft^{3}/s

ENV* K245 home

Anthony G Benoit
Revised |