BFC21103 Chapter1.pdf

BFC21103 Hydraulics

Chapter 1. Flow in Open Channel

Tan Lai Wai, Wan Afnizan & Zarina Md [email protected]

Updated: September 2014

Learning Outcomes

At the end of this chapter, students should be able to:

i. Define and explain on types and states of flow

ii. Identify types of open channels

iii. Define open channel geometries

BFC21103 Hydraulics Tan et al. ([email protected])

Open channel flow is flow of a liquid in a conduit with a free surface subjected to atmospheric pressure.

Examples: flow of water in rivers, canals, partially full sewers and drains and flow of water over land.

Free surface

Flow

Datumθ x

y

uyA

B

T

Figure. Sketch of open channel geometry


Stormwater Management and Road Tunnel (SMART), Kuala Lumpur, Malaysia

Tahan river rapids

Siberian meandering river

Practical applications:

a. flow depth in rivers, canals and other conveyance conduits,

b. changes in flow depth due to channel controls e.g. weirs, spillways, and gates,

c. changes in river stage during floods,

d. surface runoff from rainfall over land,

e. optimal channel design, and others


1.1 Flow Parameters and Geometric Elementsa. Depth of flow y is the vertical measure of water depth.

Normal depth d is measured normal to the channel bottom.

d = y cos θ

For most applications, d ≈ y when θ ≤ 10%, e.g. cos 1° = 0.9998.

Free surface

Flow Q

Datumθ x

y d

θ

So = bottom slope

Sw = water surface slope


b. Flow or discharge Q is the volume of fluid passing a cross‐section perpendicular to the direction of flow per unit time.

Mean velocity V is the discharge divided by the cross‐sectional area

AQ

V =


c.Wetted perimeter P is the length of channel perimeter that is wetted or covered by flowing water.

A = cross sectional area covered by flowing water

B = bottom width

T = top width

AP

y



d. Hydraulic radius R is the ratio of the flow area A to wetted perimeter P.

B

T

AP

y

PA

R =

e. Hydraulic depth D is the average depth of irregular cross section.

TA

D ==width topareaflow

Channel sectionArea A

Top width T

Wetted perimeter P

By B B + 2y

Table. Open channel geometries

y

B

T

Rectangular

yz

T

Triangular

1 zy2 2zy 212 zy +

By + zy2 B + 2zy 212 zyB ++yz

T

Trapezoidal

1

B

y

T

Circle

2θD ( )θθ sin22

8

2

−D

DθθsinD


Find:

(a) Top surface width T, flow area A, wetted perimeter P, and hydraulic radius R.

(b) If Q = 2.4 m3/s, determine the state of flow.

(c) If longitudinal length L = 50 m, find the cost to construct the channel. Given excavation cost = RM 3/m3 and lining cost = RM 5/m2.

Activity 1.1

3 m

2 m

1 m

60°


5774.060tan1

==o

z

zyBT 2+=

( )( )25774.023+=T

m 309.5=T

212 zyBP ++=

( ) 25774.01223 ++=P

m619.7=P

2zyByA +=

( ) ( )225774.023 +=A

2m 309.8=A

PA

R =

619.7309.8

=R

m091.1=R

(a) Top surface width T, wetted area A, wetted perimeter P and hydraulic radius R.


(b) If Q = 2.4 m3/s, determine the state of flow.

m/s 2888.0309.84.2

===AQ

v

gDV

=Fr

νVR

=Re

(c) If the length of the channel is L = 50 m, find the cost to construct the channel. Given excavation cost = RM 3/m3 and lining cost = RM 5/m2.

Volume of excavation LA ×=∀ channel

( ) 5035774.033 2 ××+×=∀3m81.709=∀

Cost of excavation ∀×= costUnit 81.709m/3 RM 3 ×=

42.2129 RM=


Area of lining LPA ×= channellining

( ) 505774.01323 2lining ×+××+=A

3lining m 41.496=A

Cost of lining liningcost Unit A×= 41.496m/5RM 2 ×=

05.2482RM=

Total cost 05.2482RM42.2129RM += 611.474RM=



Find T, A, P, R, and D

Additional Question for Assignment #1

1.2 m

1.5 m

32 1.2 m

1.5 m

0.3 m


Find:

(a) Flow area A

(b) Wetted perimeter P

(c) Hydraulic radius R

Activity 1.2

3 m4 m2 m1 m

2 m

2 m

1 m A1

A2 A3A4


1.2 Types of Open Channel

• Prismatic and non‐prismatic channels

Prismatic channel is the channel which cross‐sectional shape, size and bottom slope are constant. Most of the man‐made (artificial) channels are prismatic channels over long stretches. Examples of man‐made channels are irrigation canal, flume, drainage ditches, roadside gutters, drop, chute, culvert and tunnel.

All natural channels generally have varying cross‐sections and therefore are non‐prismatic. Examples of natural channels are tiny hillside rivulets, through brooks, streams, rivers and tidal estuaries.


• Rigid and mobile boundary channels

Rigid channels are channels with boundaries that is not deformable. Channel geometry and roughness are constant over time. Typical examples are lined canals, sewers and non‐erodible unlined canals.

Mobile boundary channels are channels with boundaries that undergo deformation due to the continuous process of erosion and deposition due to the flow. Examples are unlined man‐made channels and natural rivers.


Canalsis usually a long and mild‐sloped channel built in the ground, which may be unlined or lined with stoned masonry, concrete, cement, wood or bituminous material.

Griboyedov Canal, St. Petersburg, Russia

Terusan Wan Muhammad Saman, Kedah


This flume diverts water from White River, Washington to generate electricity Bull Run Hydroelectric Project diversion flume

Flumesis a channel of wood, metal, concrete, or masonry, usually supported on or above the surface of the ground to carry water across a depression.


Open‐channel flume in laboratory


Chuteis a channel having steep slopes.

Natural chute (falls) on the left and man‐made logging chute on the right on the Coulonge River, Quebec, Canada


Dropis similar to a chute, but the change in elevation is within a short distance.

The spillway of Leasburg Diversion Dam is a vertical hard basin drop structure designed to dissipate energy


Stormwater seweris a drain or drain system designed to drain excess rain from paved streets, parkinglots, sidewalks and roofs.

Storm drain receiving urban runoff

Storm sewer


1.3 Types and Classification of Open Channel Flows

Open channel flow

Steady flow Unsteady flow

Uniform flow Non‐uniform flow

Gradually‐varied flowRapidly‐varied flow

Various types of open‐channel flow


Open channel flow conditions can be characterised with respect to space (uniform or non‐uniform flows) and time (steady or unsteady flows).

Space ‐ how do the flow conditions change along the reach of an open channel system.

a. Uniform flow ‐ depth of flow is the same at every section of the flow dy/dx = 0

b. Non‐uniform flow ‐ depth of flow varies along the flow dy/dx ≠ 0


a. Uniform flow

b. Non‐uniform flowy1

y2 Depth changes along the channel

y y Constant water depthx

Depth of flow is the same at every section along the channel, 0dd

=xy

Depth of flow varies at different sections along the channel, 0dd

≠xy


Time ‐ how do the flow conditions change over time at a specific section in an open channel system.

c. Steady flow ‐ depth of flow does not change/ constant during the time interval under consideration dy/dt = 0

d. Unsteady flow ‐ depth of flow changes with time dy/dt ≠ 0


c. Steady flow

d. Unsteady flow

y1

Time = t1

y2

Time = t2

y1

t3

t2t1

Depth of flow is the same at every time interval, 0dd

=ty

Depth of flow changes from time to time, 0dd

≠ty

y1 = y2

y1 ≠ y2 ≠ y3


The flow is rapidly varied if the depth changes abruptly over a comparatively short distance. Examples of rapidly varied flow(RVF) are hydraulic jump, hydraulic drop, flow over weir and flow under a sluice gate.

The flow is gradually varied if the depth changes slowly over a comparatively long distance. Examples of gradually varied flow(GVF) are flow over a mild slope and the backing up of flow (backwater).


RVF RVFGVF RVFGVF RVFGVF

Sluice

Hydraulic jump

Flow over weir

Hydraulic drop

Contraction below the sluice


1.4 State of FlowThe state or behaviour of open‐channel flow is governed basically by the viscosity and gravity effects relative to the inertial forces of the flow.

Effect of visco sity ‐ depending on the effect of viscosity relative to inertial forces, the flow may be in laminar, turbulent, or transitional state.

‐ Reynolds number represents the effect of viscosity relative to inertia,

νVR

=Re

where V is the velocity, R is the hydraulic radius of a conduit and ν is the kinematic viscosity (for water at 20°C, ν = 1.004 × 10−6 m2/s, dynamic viscosity μ = 1.002 × 10−3 Ns/m2 and density ρ = 998.2 kg/m3).


Re < 500 → the flow is laminar

500 < Re < 12500 → the flow is transitional

Re > 12500 → the flow is turbulent

The flow is laminar if the viscous forces are dominant relative to inertia. Viscosity will determine the flow behaviour. In laminar flow, water particles move in definite smooth paths.

The flow is turbulent if the inertial forces are dominant than the viscous force. In turbulent flow, water particles move in irregular paths which are not smooth.

νVR

=Re


Effect of gravity ‐ depending on the effect of gravity forces relative to inertial forces, the flow may be subcritical, critical and supercritical.

‐ Froude number represents the ratio of inertial forces to gravity forces,

gDV

=Fr

where V is the velocity, D is the hydraulic depth of a conduit and g is the gravity acceleration (g = 9.81 m/s2).


Fr < 1 , the flow is in subcritical state

Fr = 1 , the flow is in critical state

Fr > 1 , the flow is in supercritical state

gDV <→

gDV =→

gDV >→


1.5 Regimes of Flow

A combined effect of viscosity and gravity may produce any one of the following four regimes of flow in an open channel:

a. subcritical ‐ laminar , when Fr < 1 and Re < 500

b. supercritical ‐ laminar , when Fr > 1 and Re < 500

c. supercritical ‐ turbulent , when Fr > 1 and Re > 12500

d. subcritical ‐ turbulent , when Fr < 1 and Re > 12500


Assignment #1

Q1. [Final Exam Sem I, Session 2010/2011]Justify the difference between:(a) uniform flow and non‐uniform flow(b) state of flow using Reynolds number Re and Froude number Fr.

Q2. [Final Exam Sem I, Session 2008/2009](a) Define

(i) Wetted perimeter(ii) Gradually‐varied flow(iii) Non‐uniform flow(iv) Froude number

(b) Explain the differences between canal and sewer.


Q3. [Final Exam Sem I, Session 2006/2007]Define(a) Reynolds number(b) Froude number(c) Hydraulic radius(d) Prismatic channel(e) Uniform flow

Q4. A discharge of 16.0 m3/s flows with a depth of 2.0 m in a rectangular channel of 4.0 m wide. Determine the state of flow based on(a) Froude number, and(b) Reynolds number.


Q5. A triangular channel of apex angle 120° carries a discharge of 1573 L/s. Calculate the critical depth.

‐ End of Question ‐

THANK YOU


BFC21103 Chapter1.pdf

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