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ENGINEERS WITHOUT BORDERS

Introduction to Fluid Mechanics Pipe Flow and

Open Channel Flow

Ross Gordon rgordon_at_rice.edu

Open Channel Flow

Uniform Open Channel Flow is the hydraulic

condition in which the water depth and the

channel cross section do not change over some

reach of the channel Through experimental

observations and calculations, Mannings Equation

was developed to relate flow and channel geometry

to water depth. Knowing the flow in a channel,

you can solve for the water depth. Knowing the

maximum allowable depth, you can solve for the

maximum flow.

Open Channel Flow

Mannings equation is only accurate for cases

where the cross sections of a stream or channel

are uniform. Mannings equation works accurately

for man made channels, but for natural streams

and rivers, it can only be used as an

approximation.

Mannings Equation

Terms to know in the Mannings equation V

Channel Velocity

A Cross sectional area of the

channel P Wetted perimeter of

the channel R Hydraulic

Radius A/P

S Slope of the channel bottom (ft/ft or m/m)

n Mannings roughness coefficient

n 0.015 for

concrete n 0.03 for clean natural channel n

.01 for glass

Yn Normal depth (depth of uniform flow)

Area

Yn

Y

X

Wetted Perimeter

Slope S Y/X

Mannings Equation

V (1/n)R2/3v(S) for the metric system V

(1.49/n)R2/3v(S) for the English system Q

A(k/n)R2/3v(S) k is either 1 or 1.49

As you can see, Yn is not directly a part of

Mannings equation. However, A and R depend on

Yn. Therefore, the first step to solving any

Mannings equation problem, is to solve for the

geometrys cross sectional area and wetted

perimeter

For a rectangular Channel Area A B x

Yn Wetted Perimeter P B 2Yn Hydraulic

Radius A/P R BYn/(B2Yn)

Yn

B

Simple Mannings Example

A rectangular open concrete (n0.015) channel is

to be designed to carry a flow of 2.28 m3/s. The

slope is 0.006 m/m and the bottom width of the

channel is 2 meters.

Determine the

normal depth that will occur in this channel.

First, find A, P and R A 2Yn P 2 2Yn

R 2Yn/(2 2Yn) Next, apply Mannings

equation Q A(1/n)R2/3v(S) ? 2.28

(2Yn)x(1/0.015)x(2Yn/(2 2Yn))2/3xv(0.006)

Solving for Yn Yn 0.47 meters

Yn

2 m

The Trapezoidal Channel

House flooding occurs along Brays Bayou when

water overtops the banks. What flow is allowable

in Brays Bayou if it has the geometry shown below?

Slope S 0.001 ft/ft

25

T 20

Concrete Lined n 0.015

35

A, P and R for Trapezoidal Channels

A Yn(B Yn cot ?) P B (2Yn/sin ? ) R

(Yn(B Yn cot ?)) / (B (2Yn/sin ? ))

Yn

?

B

The Trapezoidal Channel

Slope S 0.0003 ft/ft

25

T 20

Concrete Lined n 0.015

35

A Yn(B Yn cot ?) A 25( 35 25 x cot(20))

2592 ft2 P B (2Yn/sin ? ) P 35 (2 x

25/sin(20)) 181.2 ft R (Yn(B Yn cot ?)) /

(B (2Yn/sin ? )) R 2592 / 181.2 14.3 ft

The Trapezoidal Channel

Slope S 0.0003 ft/ft

25

T 20

Concrete Lined n 0.015

35

A 2592 ft2 R 14.3 ft Q A(1.49/n)R2/3v(S) Q

2592 x (1.49 / .015) x 14.32/3 x v(.0003) Q

Max allowable Flow 26,273 cfs

Mannings Over Different Terrains

S .005 ft/ft

5

5

5

3

3

Grass n.03

Grass n.03

Concrete n.015

Estimate the flow rate for the above

channel? Hint Treat each different portion of

the channel separately. You must find an A, R, P

and Q for each section of the channel that has a

different roughness coefficient.

Mannings Over Different Terrains

S .005 ft/ft

5

5

5

3

3

Grass n.03

Grass n.03

Concrete n.015

The Grassy portions For each section

A 5 x 3 15 ft2 P 5 3 8

ft R 15 ft2/8 ft 1.88 ft Q

15(1.49/.03)1.882/3v(.005)

Q 80.24 cfs per

section ? For both sections Q 2 x 80.24

160.48 cfs

Mannings Over Different Terrains

S .005 ft/ft

5

5

5

3

3

Grass n.03

Grass n.03

Concrete n.015

The Concrete portions

A 5 x

6 30 ft2 P 5 3 3 11 ft R 30

ft2/11 ft 2.72 ft Q 30(1.49/.015)2.722/3v(.005

)

Q 410.6 cfs For the entire channel Q

410.6 129.3 540 cfs

Pipe Flow and the Energy Equation

For pipe flow, the Bernoulli equation alone is

not sufficient. Friction loss along the pipe,

and momentum loss through diameter changes and

corners take head (energy) out of a system that

theoretically conserves energy. Therefore, to

correctly calculate the flow and pressures in

pipe systems, the Bernoulli Equation must be

modified. P1/? V12/2g z1 P2/? V22/2g z2

Hmaj Hmin Major losses Hmaj Major losses

occur over the entire pipe, as the friction of

the fluid over the pipe walls removes energy from

the system. Each type of pipe as a friction

factor, f, associated with it.

Energy line with no losses

Hmaj

Energy line with major losses

1

2

Pipe Flow and the Energy Equation

Minor Losses Hmin Momentum losses in Pipe

diameter changes and in pipe bends are called

minor losses. Unlike major losses, minor losses

do not occur over the length of the pipe, but

only at points of momentum loss. Since Minor

losses occur at unique points along a pipe, to

find the total minor loss throughout a pipe, sum

all of the minor losses along the pipe. Each

type of bend, or narrowing has a loss

coefficient, KL to go with it.

Minor Losses

Major and Minor Losses

Major Losses Hmaj f x (L/D)(V2/2g) f

friction factor L pipe length D pipe

diameter V Velocity g gravity Minor

Losses Hmin KL(V2/2g) Kl sum of loss

coefficients V Velocity g gravity When

solving problems, the loss terms are added to the

system at the second point P1/? V12/2g z1

P2/? V22/2g z2 Hmaj Hmin

Loss Coefficients

Use this table

to find loss coefficients

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

If oil flows from the upper to lower reservoir at

a velocity of 1.58 m/s in the 15 cm diameter

smooth pipe, what is the elevation of the oil

surface in the upper reservoir? Include major

losses along the pipe, and the minor losses

associated with the entrance, the two bends, and

the outlet.

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

Apply Bernoullis equation between points 1 and

2 Assumptions P1 P2 Atmospheric 0 V1

V2 0 (large tank) 0 0 Z1 0 0 130m

Hmaj Hmin Hmaj (fxLxV2)/(Dx2g)(.035 x 197m x

(1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj 5.85m

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

0 0 Z1 0 0 130m 5.85m Hmin Hmin

2KbendV2/2g KentV2/2g KoutV2/2g From Loss

Coefficient table Kbend 0.19 Kent 0.5

Kout 1 Hmin (0.19x2 0.5 1) x

(1.582/2x9.8) Hmin 0.24 m

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

0 0 Z1 0 0 130m Hmaj Hmin 0 0

Z1 0 0 130m 5.85m 0.24m Z1 136.09

meters