15 Practice Problems

8/8/2019 15 Practice Problems

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Chapter 15

Varied Flow in OpenChannels

Problem 15.1

Water ows in a circular concrete pipe (Mannings q = 0 =012) with a depth that ishalf of the pipe diameter (0.8 m). If the slope is 0.004, nd the ow rate.

Solution

The ow rate is obtained from the Chezy equation.

T =1=0q

DU2 @3k V

1 @2r

The ow area is

D =12

4

0=82

= 0 =251 m2

The wetted perimeter isS =

2 0=8 = 1=26 m

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132 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS

The hydraulic radius is

Uk =DS

=0=2511=26

= 0 =2 m

The ow rate is

T =1

0=012 0=251 0=22 @3 0=0041 @2

= 0 =45 m3 @s


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133

Problem 15.2

A troweled concrete ( q = 0 =012) open channel has a cross-section as shown. Thedischarge is 400 cfs. The drop of the channel is 10 ft in each horizontal mile (5280ft). Find the depth of the ow, k=

Solution

The ow rate in traditional units is

T =1=49

q DU2 @3k V

1 @2r

The slope is

V r =10

5280= 0 =00189

ThusDU

2 @3k =

qT1=49V

1 @2r

=0=012 400

1=49 0=001891 @2= 74 =1 ft 8 @3

The ow area in terms of depth is

D = (10 + k)k

The wetted perimeter is S = 10 + 2 s 2kso the hydraulic radius is

Uk =DS

=(10 + k)k10 + 2s 2k

Thus

DU2 @3k =

(10 + k)5 @3 k5 @3

(10 + 2 s 2k)2 @3 = 74 =1

or(10 + k)k

(10 + 2 s 2k)2 @5 = 13 =24

Solving this equation by iteration gives

k = 3 =26 ft


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Problem 15.3

Find the ow rate in the channel and overbank area that is shown in the followinggure. The slope of the channel is 0.001, and the depth in the overbank area is 2

m. The Mannings q is 0.04 in the overbank area and 0.03 in the main channel.

All the channel sides have a 1:1 slope.

Solution

The discharge is given by the Chezy equation.

T =1=0q

DU2 @3k V

1 @2r

For the overbank area

D = 2 (50 + kr @2) kr= (100 + kr )k r

where kr is the depth in the overbank area. So when kr = 2 m, the area is D = 204m2 =

The wetted perimeter is

S = 2 (50 + s 2kr )= 100 + 2 s 2kr

So when kr = 2 m, the wetted perimeter is S = 105 =6 m.

The hydraulic radius for the overbank area is

Uk =DS

=204

105=7= 1 =93 m

For the main channelD = 10 k f + 2 5(k f 5) + 2 5 5@2

= 20 k f 25


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135

With the main channel depth being 7 m, the ow area is 115 m 2 = The wettedperimeter is

S = 10 + 2 s 2 5= 24 =1 m

so the hydraulic radius of the main channel is

Uk =DS

=11524=1

= 4 =77 m

The ow rate is the sum in each area.

T =1

0=04 204 1=932 @3 0=0011 @2 +

10=03

115 4=772 @3 0=0011 @2

= 250 =0 + 343=5

= 593 =5 m3 @s

Problem 15.4

Water with a depth of 15 cm and a speed of 6 m/s ows through a rectangularchannel. Determine if the ow is critical, subcritical, or supercritical. If appropri-ate, determine the alternative depth.

Solution

The nature of the ow is determined by the Froude number.

I u =Y

s jG

=6 m/s

r 9=8 m/s2

(0=15 m)= 4 =95Since I u A 1, the ow is supercritical. To nd the alternative depth, note that thespeci c energy of subcritical and supercritical ow are the same.

| + Y 2

2j 1 = | + Y 2

2j 2 (1)= 0=15 + 6

2

2 9=8= 1 =99 m

where subscripts 1 and 2 denote sub- and supercritical, respectively. To solve Eq.

(1) for subcritical depth (| 1 ), speed is needed. The continuity principle gives(Y D)1 = ( Y D)2

(Y |z )1 = ( Y |z )2


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so

Y 1 = Y 2| 2| 1

(2)

= 6 0=15| 1

=0=9| 1

Combining Eqs. (1) and (2) gives

| +Y 2

2j 1 = 1 =99 m| 1 +

0=92 @|212 9=8

= 1 =99

or

| 31 1=99|21 + 0 =04133 = 0

We solved for the roots of this cubic equation using a computer program (MathCad).The solution has three roots: | 1 = ( 0=139> 0=15> 1=979 m). Thus the alternatedepth is

| 1 = 1 =979 m


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137

Problem 15.5

Water ows at a uniform rate of 400 cfs through a rectangular channel that hasa slope of 0.007 and a width of 25 ft. The channel sides are concrete with a rough-ness factor of q = 0 =015= Determine depth of ow, and whether the ow is critical,

subcritical, or supercritical.

Solution

The nature of the ow is determined by the Froude number.

I u =Y

s j|

To nd the depth | , we can use Mannings equation.

T =1=49

qDU

3 @2k V

1 @2r

400 =1=490=015

(25 | ) 25|25 + 2 | 3 @2

(0=007)1 @2

To solve for | in Mannings equation, we used a computer program (MathCad) tond a root for an equation of the form i ({) = 0 = The result is

| = 1 =38 ft

Discharge is

T = Y D = Y |z

400 ft3

@s= Y (1=38

25 ft2

)

so Y = 11 =59 ft/s.

The Froude number is

I u =Y

s j|

=11=59 ft/s

p (32=2 ft@s2 ) (1=38 ft)= 1 =74Thus

ow is supercritical


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Problem 15.6

Water ows in a rectangular channel that ends in a free outfall. The channelhas a slope of 0.005, a width of 20 ft, and a depth at the brink of 2 ft. Find thedischarge in the channel.

Solution

A sketch of the situation is

At the brink, the depth is 71% of critical depth.

| 1 =| 2

0=071

=2 ft0=71

= 2 =82 ft

At section 1, the ow is critical, so the Froude number is 1.0.

1=0 =Y

s j| (1)

From continuityY | = t (2)

Combining Eqs. (1) and (2) gives

t =

p j| 3

= r 32=2 ft/s 2 2=823 ft3

= 26 =9 ft 2 @s


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139

Thus

T = tz

= 26=9 ft2 @s(20 ft)= 538 cfs

Problem 15.7

Water ows with an upstream velocity of 6 ft/s and a depth of 12 ft in a rec-tangular open channel. The water passes over a gradual 18-in. upstep. Determinethe depth of the water and the change in surface level downstream of the upstep.

Solution

Assuming no energy losses, the speci c energy is constant across the upstep.

| 1 +Y 212j

= | 2 +Y 222j

+ } (1)

The continuity principle is| 1 Y 1 = | 2 Y 2

SoY 2 = Y 1

| 1| 2

(2)

Combining Eqs. (1) and (2)

| 1 +Y 212j

= | 2 +Y 212j | 1| 2

2

+ }

12 ft + 62 ft 2 @s2

2 32=2 ft@s2 = | 2 + 62 ft2 @s2

2 32=2 ft@s2 12 ft| 2 2

+ 18 @12 ft

So11=06 = | 2 + 80=50

| 22or

| 32 11=06|22 + 80 =5 = 0


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Solving this cubic equation using a computer program (MathCad) gives three roots| 2 = (-2.442, 3.2, 10.30). The negative root is not possible, and the small root(supercritial ow) is unlikely. Thus, the depth of water at section 2 is

| 2 = 10.3 ft

The elevation of the water surface at section 2 is the sum of the depth of the waterand the height of the upstep.

}2 = | 2 + }= 10 =3 ft + 1 =5 ft= 11.8 ft

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