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10.4 Uniform Depth Channel Flow 551<br />

Q, cfs<br />

1200<br />

1000<br />

800<br />

600<br />

400<br />

200<br />

Finished concrete<br />

Brickwork Asphalt<br />

Rubble masonry<br />

concrete case,<br />

10.2 fts<br />

Fr <br />

132.2 fts 2 5 ft2 0.804<br />

1 2<br />

The flow is subcritical.<br />

(Ans)<br />

COMMENT The same results would be obtained for the<br />

channel if its size were given in meters. We would use the same<br />

value of n but set k 1 for this SI units situation.<br />

0<br />

0 0.005 0.01 0.015 0.02 0.025 0.03<br />

n<br />

F I G U R E E10.3b<br />

In some instances a trial-and-error or iteration method must be used to solve for the dependent<br />

variable. This is often encountered when the flowrate, channel slope, and channel material are<br />

given, and the flow depth is to be determined as illustrated in the following examples.<br />

E XAMPLE 10.4<br />

Uniform Flow, Determine Flow Depth<br />

GIVEN Water flows in the channel shown in Fig. E10.3a at a<br />

rate of Q 10.0 m 3 s. The canal lining is weedy.<br />

FIND<br />

Determine the depth of the flow.<br />

SOLUTION<br />

In this instance neither the flow area nor the hydraulic radius are<br />

known, although they can be written in terms of the depth, y.<br />

Since the flowrate is given in m 3 /s, we will solve this problem using<br />

SI units. Hence, the bottom width is (12 ft) (1 m/3.281 ft) 3.66 m<br />

and the area is<br />

where A and y are in square meters and meters, respectively. Also,<br />

the wetted perimeter is<br />

so that<br />

y<br />

A y a<br />

tan 40° b 3.66y 1.19y2 3.66y<br />

y<br />

P 3.66 2 a b 3.11y 3.66<br />

sin 40°<br />

R h A P 1.19y2 3.66y<br />

3.11y 3.66<br />

where R h and y are in meters. Thus, with n 0.030 1from Table<br />

10.12, Eq. 10.20 can be written as<br />

Q 10 k n AR2 3<br />

h S 1 2<br />

0<br />

1.0<br />

0.030 11.19y2 3.66y2 a 1.19y2 3.66y<br />

23<br />

3.11y 3.66 b<br />

10.00142 1 2<br />

which can be rearranged into the form<br />

11.19y 2 3.66y2 5 51513.11y 3.662 2 0<br />

where y is in meters. The solution of Eq. 1 can be easily obtained<br />

by use of a simple rootfinding numerical technique or by trial-<br />

(1)<br />

and-error methods. The only physically meaningful root of Eq. 1<br />

1i.e., a positive, real number2 gives the solution for the normal<br />

flow depth at this flowrate as<br />

y 1.50 m<br />

(Ans)<br />

COMMENT By repeating the calculations for various<br />

flowrates, the results shown in Fig. E10.4 are obtained. Note that<br />

the water depth is not linearly related to the flowrate. That is, if<br />

the flowrate is doubled, the depth is not doubled.<br />

y, m<br />

3.0<br />

2.5<br />

2.0<br />

1.5<br />

1.0<br />

0.5<br />

(10, 1.50)<br />

0<br />

0 5 10 15 20 25 30<br />

F I G U R E E10.4<br />

Q, m 3 /s

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