Aeronautical Engineering Chapter 10 Homework Then The Flow Rate Manning's Formula With

Solution 10.12

The velocity and flow rate were worked out in detail in Prob. 4.36:

Problem 10.13

A large pond drains down an asphalt rectangular channel that is 2 ft wide. The channel slope is

0.6 degrees. If the flow is uniform, at a depth of 21 inches, estimate the time to drain 1 acre-foot

of water.

Solution 10.13

From Table 10.1, for asphalt, Manning’s n = 0.016. The Chézy formula is

Problem 10.14

The Chézy formula (10.18) is independent of fluid density and viscosity. Does this mean that

water, mercury, alcohol, and SAE 30 oil will all flow down a given open channel at the same

rate? Explain.

Solution 10.14

The Chézy formula,

2/3

ho

(1.0/ )(R ) S ,

Vn=

appears to be independent of fluid properties, with n

Problem 10.15

The painted–steel channel of Fig. P10.15 is designed, without the barrier, for a flow rate of 6 m3/s at

a normal depth of 1 m. Determine (a) the design slope of the channel and (b) the percentage of

reduction in flow if the painted-steel central barrier is installed.

Solution 10.15

For painted steel, n  0.014. Evaluate the hydraulic radius and then So:

Problem 10.16

Water flows in a brickwork rectangular channel 2 m wide, on a slope of 5 m per km. (a) Find

the flow rate when the normal depth is 50 cm. (b) If the normal depth remains 50 cm, find the

channel width which will triple the flow rate. Comment on this result.

Solution 10.16

The rectangular channel is fairly easy to set up for uniform flow. For brickwork, n = 0.015.

(a) Work out the hydraulic radius and then the uniform flow rate:

Problem 10.17

The trapezoidal channel of Fig. P10.17 is made of brickwork and slopes at 1:500. Determine the

flow rate if the normal depth is 80 cm.

Solution 10.17

For brickwork, n  0.015. Evaluate the hydraulic radius with y = 0.8 m:

Problem 10.18

A V-shaped painted steel channel, similar to Fig. E10.6, has an included angle of 90°. If the

slope, in uniform flow, is 3 m per km, and the depth is 2 m, estimate (a) the flow rate, in m3/s;

and (b) the average wall shear stress. Take y = 2 m.

Solution 10.18

For painted steel, from Table 10.1, n = 0.014. The properties of the 90 vee are easy:

Problem 10.19

Modify Prob. P10.18, the 90 V-channel, to let the surface be clean earth, which erodes if the

average velocity exceeds 6 ft/s. Find the maximum depth that avoids erosion. The slope is still

3 m per km.

Problem 10.18

A V-shaped painted steel channel, similar to Fig. E10.6, has an included angle of 90°. If the

slope, in uniform flow, is 3 m per km, and the depth is 2 m, estimate (a) the flow rate, in m3/s;

and (b) the average wall shear stress. Take y = 2 m.

Solution 10.19

From Table 10.1, for clean earth, n = 0.022. The vee-channel has easy algebra:

Problem 10.20

An unfinished concrete sewer pipe, of diameter 4 ft, is flowing half-full at 39,500 U.S. gallons

per minute. If this is the normal depth, what is the pipe slope, in degrees?

Solution 10.20

Stay in BG units and convert 39,500 gal/min to 88 ft3/s. For unfinished concrete, from

Problem 10.21

An engineer makes careful measurements with a weir (see Sect. 10.7) that monitors a rectangular

unfinished concrete channel laid on a slope of 1. She finds, perhaps with surprise, that when the

water depth doubles from 2 ft 2 inches to 4 ft 4 inches, the normal flow rate more than doubles,

from 200 to 500 ft3/s. (a) Is this plausible? (b) If so, estimate the channel width.

Solution 10.21

(a) Yes, Q always more than doubles for this situation where the depth doubles. Ans. (a)

Problem 10.22

For more than a century, woodsmen harvested trees in Skowhegan, ME, elevation 171 ft, and

floated the logs down the Kennebec River to Bath, ME, elevation 62 ft, a distance of 72 miles.

The river has an average depth of 14 ft and an average width of 400 ft. Assuming uniform flow

and a stony bottom, estimate the travel time required for this trip.

Solution 10.22

The slope is quite small, So = (171-62)/[5280(72)] = 0.000287. For stony cobbles, from Table

10.1, n = 0.035. Compute the hydraulic radius and the average velocity:

Problem 10.23

It is desired to excavate a clean-earth channel as a trapezoidal cross-section with



= 60 (see

Fig. 10.7). The expected flow rate is 500 ft3/s, and the slope is 8 ft per mile. The uniform flow

depth is planned, for efficient performance, such that the flow cross-section is half a hexagon.

What is the appropriate bottom width of the channel?

Solution 10.23

For clean earth, take n = 0.022. For a half-hexagon, from Fig. 10.7 of the text, depth

y = sin(60)b = 0.866b, and

Problem 10.24

A rectangular channel, laid out on a 0.5 slope, delivers a flow rate of 5,000 gal/min in uniform

flow when the depth is 1 ft and the width is 3 ft. (a) Estimate the value of Manning’s factor n.

(b) What water depth will triple the flow rate?

Solution 10.24

We have enough data to figure everything except n in Manning’s formula:

Problem 10.25

The equilateral-triangle channel in Fig. P10.25 has constant slope So and constant Manning

factor n. If y = a/2, find an analytic expression for the flow rate Q.

Solution 10.25

The geometry is a bit awkward, but you only have to do it once.

The width of the water line is a – 2(a/2)cot60 = 0.423 a

Problem 10.26

In the spirit of Fig. 10.6b, analyze a rectangular channel in uniform flow with constant area

A = by, constant slope, but varying width b and depth y. Plot the resulting flow rate Q,

normalized by its maximum value Qmax, in the range 0.2  b/y  4.0, and comment on whether it

is crucial for discharge efficiency to have the channel flow at a depth exactly equal to half the

channel width.

0.423 a

0.577 a

Solution 10.26

The Manning formula for a rectangular channel is:

Problem 10.27

A circular corrugated metal water channel has a slope of 1:800 and a diameter of 6 ft.

(a) Estimate the normal discharge, in gal/min, when the water depth is 4 ft. (b) For this

condition, calculate the average wall shear stress.

Solution 10.27

The geometry of this circular partly-full

Problem 10.28

A new, finished-concrete trapezoidal channel, similar to Fig. 10.7, has b = 8 ft,

yn = 5 ft, and θ = 50º. For this depth, the discharge is 500 ft3/s. (a) What is the slope of the

channel? (b) As years pass, the channel corrodes and n doubles. What will be the new normal

depth for the same flow rate?

Solution 10.28

For finished-concrete, from Table 10.1, n = 0.012. (a) From the formulas of Fig. 10.7,

Problem 10.29

A new, finished-concrete trapezoidal channel, similar to Fig. 10.7, has b = 8 ft,

yn = 5 ft, and θ = 50º. For this depth, the discharge is 500 ft3/s. (a) What is the slope of the

channel? (b) As years pass, the channel corrodes and n doubles. What will be the new normal

depth for the same flow rate?

Solution 10.29

We relate the Shields critical shear stress to our result in Prob. 10.28 above:

Problem 10.30

A clay tile V-shaped channel, with an included angle of 90, is 1 km long and is laid out on a

1:400 slope. When running at a depth of 2 m, the upstream end is suddenly closed while the

lower end continues to drain. Assuming quasi-steady normal discharge, find the time for the

channel depth to drop to 20 cm.

Solution 10.30

We assume quasi-steady uniform flow at any instant. For a control volume enclosing the entire

channel of length L = 1 km, we obtain

Problem 10.31

An unfinished-concrete 6-ft-diameter sewer pipe flows half full. What is the appropriate slope to

deliver 50,000 gal/min of water in uniform flow?

Solution 10.31

For unfinished concrete, from Table 10.1, n  0.022. For a half-full circle,

o

Problem 10.32

Does half a V-shaped channel perform as well as a full V-shaped channel? The answer to Prob.

10.18 is Q = 12.4 m3/s for y = 2m. (Do not reveal this to your friends still working on P10.18.)

For the painted-steel half-V in Fig. P10.32, at the same slope of 3:1000, find the flow area that

gives the same Q and compare with P10.18.

Problem 10.18

A V-shaped painted steel channel, similar to Fig. E10.6, has an included angle of 90°. If the

slope, in uniform flow, is 3 m per km, and the depth is 2 m, estimate (a) the flow rate, in m3/s;

and (b) the average wall shear stress. Take y = 2 m.

Solution 10.32

For painted steel, from Table 10.1, n = 0.014.

The properties of the half-V are easy:

Problem 10.33

Five sewer pipes, each a 2-m-diameter clay tile pipe running half full on a slope of 0.25°, empty

into a single asphalt pipe, also laid out at 0.25°. If the large pipe is also to run half full, what

should be its diameter?

Solution 10.33

First compute the small-pipe flow rate. For clay tile, n = 0.014.

Problem 10.34

A brick rectangular channel with S0 = 0.002 is designed to carry 230 ft3/s of water in uniform

flow. There is an argument over whether the channel width should be 4 or 8 ft. Which design

needs fewer bricks? By what percentage?

Solution 10.34

For brick, take n  0.015. For both designs, A = by and P = b + 2y. Thus

Problem 10.35

In flood stage a natural channel often consists of a deep main channel plus two floodplains, as in

Fig. P10.35. The floodplains are often shallow and rough. If the channel has the same slope

everywhere, how would you analyze this situation for the discharge? Suppose that y1 = 20 ft,

y2 = 5 ft, b1 = 40 ft, b2 = 100 ft, n1 = 0.020, n2 = 0.040, with a slope

of 0.0002. Estimate the discharge in ft3/s.

Solution 10.35

We compute the flow rate in three pieces, with the dashed lines in the figure above serving as

“water walls” which are not counted as part of the perimeter:

Problem 10.36

The Blackstone River in northern Rhode Island normally flows at about 25 m3/s and resembles

Fig. P10.35 with a clean-earth center channel, b1  20 m and y1  3 m. The bed slope is about

2 ft/mi. The sides are heavy brush with b2  150 m. During hurricane Carol in 1954, a record

flow rate of 1000 m3/s was estimated. Use this information to estimate the maximum flood depth

y2 during this event.

Solution 10.36

For heavy brush, n2 = 0.075 and for clean earth, n1 = 0.022, as shown in the figure. Use the same

“zero-perimeter water-wall” scheme as in Prob. 10.35:

Problem 10.37

A triangular channel (see Fig. E10.6) is to be constructed of corrugated metal and will carry

8 m3/s on a slope of 0.005. The supply of sheet metal is limited, so the engineers want to

minimize the channel surface. What is (a) the best included angle



for the channel; (b) the

normal depth for part (a); and (c) the wetted perimeter for part (b).

Solution 10.37

For corrugated metal, take n = 0.022. From Ex. 10.5, for a vee-channel, recall that

Problem 10.38

For the half-Vee channel in Fig. P10.32, let the interior angle of the Vee be



. For a given value

of area, slope, and n, find the value of



for which the flow rate is a maximum. To avoid

cumbersome algebra, simply plot Q versus



for constant A.

Solution 10.38

Write out Manning’s formula and solve for y in terms of A and



:

Problem 10.39

A trapezoidal channel has n = 0.022 and S0 = 0.0003 and is made in the shape of a half-hexagon

for maximum efficiency. What should the length of the side of the hexagon be if the channel is to

carry 225 ft3/s of water? What is the discharge of a semicircular channel of the same cross-

sectional area and the same So and n?

Solution 10.39

The half-hexagon corresponds to Fig. 10.7 with



= 60. Its properties are

Problem 10.40

Using the geometry of Fig. 10.6a, prove that the most efficient circular open channel (maximum

hydraulic radius for a given flow area) is a semicircle.

Solution 10.40

Maximum hydraulic radius means minimum perimeter. Using Eq. 10.20,