8 OPEN CHANNEL FLOW

Fluid Mechanics

Chapter 8 – Open Channel Flow

8 OPEN CHANNEL FLOW 8.1 Classification & Definition ‹

Open channel flows are flows in rivers, streams, artificial channels, irrigation ditches, partially filled pipe etc.

‹

Basically, it is a flow with free surface. (Free surface is a surface with atmospheric pressure) Open Channel Flow

Steady flow Uniform

Unsteady flow

Varied

Gradually varied flow

Varied

Rapidly varied flow

Gradually varied flow

Uniform

Rapidly varied flow

Classifications of Open Channel Flow

8.1.1 1.

Open Channel Geometry

Depth of flow, y: vertical distance from the bottom to surface. Free surface B flow

A

y

Bottom (cross-section)

2

Top width, B: – the width of the channel at the free surface

3

Flow area, A: – cross-sectional area of the flow

4

Wetted perimeter, P: – the length of the channel cross-section in contact with the fluid P.8-1

Fluid Mechanics

5

Hydraulic radius (hydraulic mean depth), R: R

6

Chapter 8 – Open Channel Flow

=

Flow area A = Wetted perimeter P

Average depth (hydraulic average depth), yave: yave

8.1.2

=

Flow area A = Top width B

Rectangular channel

– B

=b y

– A

= b*y

– P

= b+2*y

– R

=

– yave

=y

8.1.3

b

b*y b + 2* y

Trapezoidal channel

– B

= b + 2*m*y y

– A

= y*(b+m*y)

– P

= b+2*y* 1 + m 2

– R

=

– yave

=

m

1

b

y * ( b + m * y) b + 2 * y * 1 + m2

y * ( b + m * y) b + 2*m* y

P.8-2

Fluid Mechanics

8.1.4

Triangular channel

– B –

Chapter 8 – Open Channel Flow

= 2*m*y 2

A

1

= m*y

– P

= 2*y* 1+ m 2

– R

=

– yave

=

8.1.5

y m

m*y 2 * 1 + m2

y 2

Circular channel

– B

= 2 * y * ( D − y)

– A

D 2 * (θ − sin θ ) = 8

– P

θ*D = 2

– R

=

D ⎛ sin θ ⎞ ⎜1 − ⎟ 4⎝ θ ⎠

– yave

=

D * (θ − sin θ) 8 * sin θ 2

y

θ

D

(θ in radian)

P.8-3

Fluid Mechanics

Chapter 8 – Open Channel Flow

8.2 Steady Uniform Flow ‹

For a steady uniform flow – depth is constant along the flow – velocity is constant over the cross-section – time independent

8.2.1 ‹

Manning Equations

In 1890, Manning, an Irish engineer derived a better and more accurate relationship, Manning equation, based on many field measurement. V •

‹

2 1 1 * R 3 *S 2 n

(8.1)

n - Manning’s coefficient, s/m1/3 (can be found in most of the hydraulic handbooks)

To incorporate the continuity equation, Manning equation becomes Q

‹

=

2 1 A 3 2 * R *S = n

(8.2)

As the flow according to Manning equations is for normal steady uniform flow, – the flow is Normal Flow – the depth is Normal Depth

P.8-4

Fluid Mechanics

Chapter 8 – Open Channel Flow

Worked examples:

1.

Water flows in a rectangular, concrete, open channel that is 12 m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate. (n = 0.013)

Answer

By Manning equation, 2 1 1 V = * R 3 *S 2 n

with n = 0.013 S = 0.0028

∴

A P R

hence

= 12* 2.5 m2 = 30 m2 = 12 + 2*2.5 m = 17 m = A/P = 30 / 17 m = 1.765 m V

2 1 1 * (1.765) 3 * (0.0028) 2 0.013 = 5.945 m/s

=

Discharge, Q = A*V = 30*5.945 m3/s = 178.3 m3/s

P.8-5

Fluid Mechanics

2.

Chapter 8 – Open Channel Flow

Water flows in a rectangular, concrete, open channel that is 12 m wide. The channel slope is 0.0028. If the velocity of the flow is 6 m/s, find the depth of the flow. (n = 0.013)

h

12m

Answer

By Manning equation, 2 1 1 V = * R 3 *S 2 n

with V = 6 m/s n = 0.013 S = 0.0028 A P ∴

R

= 12* h m2 = 12 + 2*h m A 12 * h 6*h = = = P 12 + 2 * h 6 + h

6*h = 1.790 6+h h = 2.551 m Depth of the flow = 2.551m

P.8-6

Fluid Mechanics

3.

Chapter 8 – Open Channel Flow

A trapezoidal channel with side slopes of 2/3, a depth of 2 m, a bottom width of 8 m and a channel slope of 0.0009 has a discharge of 56 m3/s. Find the Manning’s n. 14m

2m

1 1.5 3m

8m

3m

Answer

= (14+8)*2/2 m2 = 22 m2 P = 8 + 2* 2 2 + 32 m = 15.211 m A/P = 22 / 15.211 m = 1.446 m A

By Manning equation, Q=

2 1 A * R 3 *S 2 n

Q

= 56 m3/s,

S

= 0.0009

2 1 22 * (1.446) 3 * (0.0009) 2 n = 0.01507

56 = n

P.8-7

Fluid Mechanics

4.

Chapter 8 – Open Channel Flow

Determine the depth in a trapezoidal channel with side slopes of 1 to 1.5, a bottom width of 8 m and a channel slope of 0.0009. The discharge is 56 m3/s and n = 0.017. (8+3*y) m

y

1 1.5 8m

Answer

A P R

= (8+8+3*y)*y/2 m2 = (8+1.5*y)*y m2 = 8 + 2*y* 12 + 1.5 2 m = 8+3.6056*y m = A/P = (8+1.5*y)*y / 8+3.6056*y

By Manning equation, 2 1 A Q = * R 3 *S 2 n Q = 56 m3/s, S = 0.0009

∴ or

(8 + 1.5 * y) * y ⎡ (8 + 1.5 * y) * y ⎤ 56 = *⎢ ⎥ 0.017 ⎣ 8 + 3.6056 * y ⎦

2

3

* (0.0009)

1

2

[(8 + 1.5 * y) * y]5 3 = 31.7333 [8 + 3.6056 * y]2 3 [(1 + 0.1875 * y) * y]5 3 − 3.9667 = 0 [1 + 0.4507 * y]2 3

By trial & error, y = 2.137 m. The depth of the trapezoidal channel is 2.137m.

P.8-8

Fluid Mechanics

5.

Chapter 8 – Open Channel Flow

Water flows in the triangular steel channel shown in the figure below. Find the depth of flow if the channel slope is 0.0015 and the discharge is 0.22 m3/s. (n=0.014)

y

o

60

Answer

A P R

= 2ytan30° * y/2 m2 = y2*tan30° m2 = 2y/cos30° m = A/P = y2*tan30° / 2y/cos30° m = ysin30°/2 m

By Manning equation, 2 1 A Q = * R 3 *S 2 n Q

= 0.22 m3/s, S = 0.0015 2

y 2 tan 30° ⎛ y sin 30° ⎞ 3 *⎜ 0.22 = ⎟ * 0.0015 0.014 2 ⎝ ⎠ 8/3 = y * 0.6338

or

⎛ 0.22 ⎞ y =⎜ ⎟ ⎝ 0.6338 ⎠ = 0.672 m

3

8

m

Depth of the channel is 0.672 m.

P.8-9

Fluid Mechanics

8.2.2 ‹

Chapter 8 – Open Channel Flow

Optimum Hydraulic Cross-sections (REFERENCE ONLY)

From Manning equation, 5

1 A 3* S Q = * 2 n P 3 Hence, Q will be maximum when P is a minimum. ‹

For a given cross-sectional area, A of an open channel, the discharge, Q is maximum when the wetted perimeter, P is minimum. Hence if the wetted perimeter, P for a given flow area is minimised, the area, A will give the least expensive channel to be construct.

‹

This corresponding cross-section is the optimum hydraulic section or the best hydraulic section.

8.2.2.1

Rectangular section width

=b y

depth

=y

b

area, A = by P

= b+2*y A = + 2y y

Hence

dP A =− 2 +2=0 dy y

i.e.

y=

A or 2

b = 2y

Therefore, the optimum rectangular section is

y

2y

P.8-10

Fluid Mechanics

8.2.2.2

Chapter 8 – Open Channel Flow

Trapezoidal section B

= b+2*m*y y

A

= (b+m*y)*y

P

= b+2*y* 1+ m 2

m

1

b

By eliminating b from P, A P = + ( 2 * 1 + m 2 − m) * y y For a minimum value of P, δP = 0, dP dP i.e. =0 =0 and dy dm From

dP = 0, dy

y2

=

A 3

dP 1 = 0, m = dm 3 It implies the side slope of the channel is 60° to horizontal. A y 2 3 = y b = − my = 3y − y 3 3 2 3 4 y+ y = 2 3y and P = 3 3 i.e. P = 3*b

From

The optimum section is given as follow:

o

b

60 b b

P.8-11

Fluid Mechanics

8.2.2.3

Chapter 8 – Open Channel Flow

Other sections

N-side Channel ‹ from the conclusion of the previous two sections – reflection of the rectangular optimum section about the water surface will form a square of side b. – reflection of the trapezoidal optimum section about the water surface will form a regular hexagon of side b. ‹

For a N-side channel, the optimum hydraulic section should be in a form of half a 2N-side regular polygon.

b

φ

⎛ N −1⎞ φ=⎜ ⎟ *180° N ⎝ ⎠

b

b

b b

b

Triangular Section ‹ N = 2, hence φ = 90°

y

o

45

Circular Section ‹ From the result of N-side channel, it can be concluded that the optimum section of a circular channel is a semi-circle. ‹

It is the most optimum section for all the possible open-channel crosssection. D

P.8-12

Fluid Mechanics

Chapter 8 – Open Channel Flow

Worked examples

1. An open channel is to be designed to carry 1m3/s at a slope of 0.0065. The channel material has an n value of 0.011. Find the optimum hydraulic cross-section for a semi-circular section. D

Answer

The optimum circular section is a semi-circular section with diameter D which can discharge 1 m3/s. For a semi-circular section, A = π*D2/8 P = π*D/2 R = A/P = D/4 As n = 0.011, S = 0.0065 and Q = 1 m3/s. Q=

2 1 A * R 3 *S 2 n 2

π * D2 ⎛ D ⎞ 3 * ⎜ ⎟ * 0.0065 = 8 * 0.011 ⎝ 4 ⎠

i.e. 1

2

8/3

D D

1

− 8 * 0.011 3 = * 4 * 0.0065 2 π = 0.951 m

The diameter of this optimum section is 951mm.

P.8-13

Fluid Mechanics

Chapter 8 – Open Channel Flow

2. Find the optimum rectangular section from the last example.

y

2y

Answer

A P R

= 2*y2 = 4*y = A/P = y/2

By Manning equation, Q=

2 1 A * R 3 *S 2 n

1

=

2 3

1* 0.011* 2 2 * 0.0065 = 0.434 m

y8/3 = y

2 3

2* y ⎛ y ⎞ * ⎜ ⎟ * 0.0065 0.011 ⎝ 2 ⎠ 2

The optimum rectangular section has dimension of width 0.868m and depth 0.434m.

P.8-14

Fluid Mechanics

3.

Chapter 8 – Open Channel Flow

Find the optimum triangular section from the last example.

y

o

45

Answer

A P

= y2 = 2 2*y

R

= A/P

= y

2 2

By Manning equation, 2 1 A Q = * R 3 *S 2 n 2 3

y ⎛ y ⎞ *⎜ ⎟ * 0.0065 0.011 ⎝ 2 2 ⎠ 2

1

=

2 3

0.011* (2 2 ) 0.0065 y = 0.614 m The optimum triangular section is a right angle triangle with depth 0.614 m. y8/3 =

P.8-15

Fluid Mechanics

Chapter 8 – Open Channel Flow

8.3 Non-Uniform flow - Specific Energy in Open Channel & Critical Flow energy line 2

v1 /2g fluid surface

E1

hf v22/2g

y1 y2

z1

channel bed horizontal datum

‹

‹

z2

In open channel, the solution of many problems are greatly assisted by the concept of specific energy, i.e. v2 E = +y (8.3) 2g In terms of flow rate, Q, 1 Q 2 E = ( ) +y 2g A

(8.4)

The minimum energy will be given as dE =0 dy

(8.5)

8.3.1 ‹

E2

Let

∴

Rectangular Channel Q = v*y (8.6) b q - the discharge per unit width of a rectangular channel q

E

=

=

q2 2gy 2

+y

(8.7)

P.8-16

Fluid Mechanics ‹

Chapter 8 – Open Channel Flow

By assuming q is constant

or

yc ‹

‹

dE q2 =1=0 (8.8) 3 dy gy y = yc q 2 13 =( ) (8.9) g - critical depth at which the energy is minimum.

The corresponding energy, E is

From (8.6),

Emin =

3 yc 2

v

q . y

=

(8.10)

Substitute into (8.8), v c2 =0 1gy c

v c2 =1 gy c vc = gy c

or ‹

(8.11) (8.12)

Since Froude number, Fr is defined as Fr

=

v gy ave

(8.13)

Hence, the minimum energy is occurred when Fr2 = 1 ‹

(8.14)

For a given discharge, Q, if the flow is such that E is a min., the flow is critical flow. – critical flow - flow with Emin – critical depth, yc - the depth of the critical flow – critical velocity - vc = gy c P.8-17

Fluid Mechanics

Chapter 8 – Open Channel Flow

y

2

v1 /2g

y1

A

y

y1

subcritical or slow

2

vc /2g

c

C yc

supercritical B or fast

o

45

y2 y2

2

v2 /2g

E

‹

If the flow with E > Emin, there are two possible depths (y1, y2). (y1, y2) are called alternate depths.

‹

C divides the curve AB into AC and CB regions. -

AC - subcritical flow region CB - supercritical flow region

Depth of flow Velocity of flow Slope Froude number Other

Subcritical y > yc v < vc Mild S < Sc Fr < 1.0 v2 yc < 2g 2

Critical y = yc v = vc Critical S = Sc Fr = 1.0 v2 yc = 2g 2

Supercritical y < yc v > vc Steep S > Sc Fr > 1.0 v2 yc > 2g 2

P.8-18

Fluid Mechanics

8.3.2

Chapter 8 – Open Channel Flow

Non - Rectangular Channel

‹

If the channel width varies with y, the specific energy must be written Q2 in the form E = (8.15) +y 2gA 2

‹

The minimum energy also occurs where dE = 0 at constant Q dy

‹

Since A = A(y), therefore (8.15) becomes 2Q 2 A −3 dA 12g dy

or

‹

Since

∴ or

dA gA 3 = 2 dy Q

=0

(8.16)

dA = B - the channel width at the free surface, dy gA 3 B = 2 Q BQ 2 13 A =( ) (8.17) g Q vc = A gA 1 =( ) 2 (8.18) B

‹

For a given channel shape, A(y) & B(y), and a given Q, (8.17) & (8.18) have to be solved by trial and error to find the A and then vc.

♦

If a critical channel flow is also moving uniformly (at constant depth), it must correspond to a critical slope, Sc, with yn = yc. This condition can be analysed by Manning formula.

P.8-19

Fluid Mechanics

Chapter 8 – Open Channel Flow

Worked examples:

1. A triangular channel with an angel of 120° made by 2 equal slopes. For a flow rate of 3 m3/s, determine the critical depth and hence the maximum depth of the flow. B

yc o

30

Answer

∴

For critical flow, v2 = g*yave Q2 = g*yave*A2 gA 3 A = (yave = ) B B For critical flow, B = 2*y*cot 30° & A = y2*cot 30° 3g 5 Q2 = y 2 Hence

y

2Q 2 15 =( ) 3g 2 * 32 15 =( ) m 3 * 9.81 = 0.906 m

The maximum depth is 0.906 m. The critical depth, yc = yave =

A ( B * y ) / 2 y 0906 = = = = 0.453m . B B 2 2

P.8-20

Fluid Mechanics

Chapter 8 – Open Channel Flow

2. In the last example, the channel Manning roughness coefficient is 0.012 and the flow rate is 3 m3/s. What is the value of the channel slope if the flow is critical, subcritical or supercritical? B

y o

30

Answer

B A P

= 2 3 *y 1 = *B*y 2 = 4*y

Using Manning equation, 2 1 A Q = * R 3 *S 2 n 1 1 1 B 2 = * ( * B * y) * ( ) 3 * S 2 8 n 2 2 nQ B − 2 3 ( ) Sc1/2 = By 8 2 nQ 3 * y −23 = ( ) 2 4 2 3*y For critical flow, y = y 2 * 0.012 * 3 3 * 0.906 − 2 3 ∴ Sc1/2 = ( ) 4 2 3 * (0.906) 2 Sc = 0.0472 For flow is

critical, S = 0.0472 subcritical, S < 0.0472 supercritical, S > 0.0472

- critical slope

P.8-21

Fluid Mechanics

Chapter 8 – Open Channel Flow

8.4 Frictionless Flow over a Bump supercritical approach flow

y1

v1

2 y2 v

subcritical approach flow ∆h

‹

When fluid is flowing over a bump, the behaviour of the free surface is sharply different according to whether the approach flow is subcritical or supercritical.

‹

The height of the bump can change the character of the results.

‹

Applying Continuity and Bernoulli’s equations to sections 1 and 2,

& ‹

= v2*y2 v1*y1 2 v22 v1 + y1 = + y 2 + ∆h 2g 2g

Eliminating v2 between these two gives a cubic polynomial equation for the water depth y2 over the bump, v12 * y12 y2 – E2*y2 + 2g 3

2

=0

(8.19)

v12 where E2 = + y1 - ∆h (8.20) 2g This equation has one negative and two positive solutions if ∆h is not too large.

P.8-22

Fluid Mechanics ‹

Chapter 8 – Open Channel Flow

The free surface’s behaviour depends upon whether condition 1 is in subcritical or supercritical flow. y1

water depth

y2

∆ hmax 2

1 subcritical bump

∆h

yc

supercritical bump Ec

E2

E1 specific energy

The specific energy E2 is exactly ∆h less than the approach energy, E1, and point 2 will lie on the same leg of the curve as E1. ‹

A subcritical approach, Fr1 < 1, will cause the water level to decrease at the bump.

‹

Supercritical approach flow, Fr1 > 1, causes a water level increase over the bump.

‹

If the bump height reaches ∆hmax = E1 – Ec, the flow at the crest will be exactly critical (Fr = 1).

‹

If the bump > ∆hmax, there are no physical correct solution. That is, a bump to large will choke the channel and cause frictional effects, typically a hydraulic jump.

P.8-23

Fluid Mechanics

Chapter 8 – Open Channel Flow

Worked example:

Water flow in a wide channel approaches a 10 cm high bump at 1.5 m/s and a depth of 1 m. Estimate (a) the water depth y2 over the bump, and (b) the bump height which will cause the crest flow to be critical. Answer

(a) For the approaching flow, v1 15 . = Fr = 9.81 * 1 gy1 = 0.479 ⇒ subcritical For subcritical approach flow, if ∆h is not too large, the water level over the bump will depress and a higher subcritical Fr at the crest. E1 Hence

E2

v12 = + y1 2g = 1.115 m = E1 - ∆h = 1.015 m

15 . 2 . m = + 10 2 * 9.81 = 1.115 – 0.1 m

Substitute E2 into (8.24), y23 – 1.015*y22 + 0.115 = 0 By trial and error, y2 = 0.859 m, 0.451 m and –0.296 m (inadmissible) The second (smaller) solution is the supercritical condition for E2 and is not possible for this subcritical bump. Hence Checking:

y2 = 0.859 m v2 = 1.745 m/s (By continuity) (OK) Fr2 = 0.601 (> Fr1 and < 1)

P.8-24

Fluid Mechanics

(b)

Chapter 8 – Open Channel Flow

By considering per m width of the channel, q = v*y = 1.5*1 m2/s For critical flow, E2 = Emin yc

E2

∆hmax

=

q 2 13 =( ) g 15 . 2 13 =( ) 9.81 = 0.612 m 3 = *0.612 2 = 0.918 m

3 yc 2

m

= E1 - Emin = 1.115 – 0.918 m = 0.197 m

P.8-25

Fluid Mechanics

Chapter 8 – Open Channel Flow

8.5 Hydraulic Jump in Rectangular Channel ‹

A hydraulic jump is a sudden change from a supercritical flow to subcritical flow.

‹

Assumptions: – the bed is horizontal. – the velocity over each cross-section is uniform. – the depth is uniform across the width. – frictionless boundaries. – surface tension effects are neglect. 1

2

critical depth level y2 y1

v

1

v2 eddy currents

‹

Considering the control volume between 1 and 2, the forces are b y12 = ρgb* (8.21a) F31 = ρgy1* *y1 2 2 y22 Similarly F32 = ρgb* (8.21b) 2

‹

By continuity equation, Q = b*y1*v1

‹

‹

= b*y2*v2

By the momentum equation, F1 = F2 = 0 hence F31 – F32 = ρ*Q*(v2 - v1)

(8.22)

(8.23)

Sub. (8.21a, b) and (8.22) into (8.23), then ρgb 2 Q Q ( y1 − y 2 2 ) = ρQ( − ) 2 y 2 b y 1b ρQ 2 y1 − y 2 = ( ) b y1 y 2

(8.24)

P.8-26

Fluid Mechanics ‹

Chapter 8 – Open Channel Flow

In a hydraulic jump, y1 ≠ y2, 2Q 2 y1*y2*(y1+y2) = gb 2 2Q 2 2 2 ∴ y1 y2 + y1y2 = gb 2 2Q 2 y2 2 y2 i.e. ( ) +( )− 2 3 =0 y1 y1 gb y1 Solving (8.25),

1 8Q 2 y2 = [ −1 + 1 + 2 3 ] 2 y1 gb y1 This is the hydraulic jump equation. ‹

(8.26)

Using Froude number, Fr1

2

v12 Q2 = = gy1 gy13b2 1 = [ −1 + 1 + 8Fr12 ] 2

then,

y2 y1

or

1 y1 = [ −1 + 1 + 8Fr2 2 ] 2 y2

‹

(y1,y2) are called conjugate depths.

‹

The energy loss in a jump is given by v22 v12 + y1 = + y2 + h f 2g 2g v − v2 =( 1 ) + ( y1 − y 2 ) 2g 2

i.e. ‹

(8.25)

hf

(8.27) (8.28a)

(8.28b)

2

(8.29)

Sub. (8.22) into above, hf

Q 2 y1 + y 2 = [− ( ) + 1 ](y1 – y2) 2gb 2 y12 y 2 2

(8.29)

P.8-27

Fluid Mechanics ‹

Using (8.25), (8.29) becomes ( y 2 − y1 ) 3 hf = 4 y1 y 2

Chapter 8 – Open Channel Flow

(8.30)

This is the energy loss equation for the hydraulic jump (y2>y1, hf>0). ‹

The power loss in a jump is P = ρghf*Q

‹

This energy loss is useful for getting away with the unwanted energy of a flow. The energy loss is due to the frictional forces amount the eddy currents in the pump. It will increase the temperature of the fluid.

P.8-28

Fluid Mechanics

Chapter 8 – Open Channel Flow

Worked example:

Water flows in a wide channel at q = 10 m2/s and y1 = 1.25 m. If the flow undergoes a hydraulic jump, calculate (a) y2, (b) v2, (c) Fr2, (d) hf,, and (e) the percentage dissipation of the energy. Answer

(a)

v1

Fr1

Since

or

(b)

y2 y1

y2

q y1 10 = m/s = 8 m/s 1.25 v1 = gy1 8 = 2.285 = 9.81 * 125 . 1 = [ −1 + 1 + 8Fr12 ] 2 1 = [ −1 + 1 + 8 * (2.285) 2 ] 2 = 2.77 = 2.77*1.25 m = 3.46 m

=

By Continuity equation, y v2 = v1*( 1 ) y2 125 . = 8* m/s 3.46 = 2.89 m/s

P.8-29

Fluid Mechanics

(c)

(d)

Chapter 8 – Open Channel Flow

v2 gy 2 2.89 = 9.81 * 3.46 = 0.496

Fr2 =

hf

( y 2 − y1 ) 3 = 4 y1 y 2

(3.46 − 125 . )3 = 4 * 3.46 * 125 . = 0.625 m (e)

E1

percentage loss

v12 + y1 = 2g 82 + 1.25 m = 2 * 9.81 = 4.51 m hf * 100% E1 0.625 * 100% = 4.51 = 14 %

=

P.8-30

Fluid Mechanics

Chapter 8 – Open Channel Flow

8.6 Gradually Varied Flow ‹

It is not always possible to have uniform depth across the flow i.e. normal flow with normal depth.

‹

The depth of flow can be changed by the conditions along the channel.

‹

Examples of Gradually Varied Flow are: – backwater curve water surface dy >0 dx

yn

dam

– Downdrop curve water surface yn1 dy dx Sc - Steep (S) So = Sc - Critical (C) So < Sc - Mild (M) So = 0 - Horizontal (H) So < 0 - Adverse (A)

‹

There are three number designators for the type of profile relates to the position of the actual water surface in relation to the position of the water for normal and critical flow in a channel. 1 the surface of stream lies above both normal and critical depth 2 the surface of stream lies between normal and critical depth 3 the surface of stream lies below both normal and critical depth

M1

yn

M2 M3

yc Mild slope

S1

yc

S2 S3

yn Steep slope

P.8-37

Fluid Mechanics

‹

Chapter 8 – Open Channel Flow

Combining the two designators, we have

Slope class sinθ > S

Slope notation Steep (S)

Depth class yc > yn

sinθ = S

Critical (C)

yc = yn

sinθ < S

Mild (M)

yc < yn

sinθ = 0

Horizontal (H)

yn = ∞

sinθ < 0

Adverse (A)

yn = Im

‹

For type S

‹

For type C

Froude number Fr < 1 Fr > 1 Fr > 1 Fr < 1 Fr > 1 Fr < 1 Fr < 1 Fr > 1 Fr < 1 Fr > 1 Fr < 1 Fr > 1

Actual depth

Profile

y >yn; y>yc yc > y > yn yyc yn > y > yc y