Cc501chapter 4 - Specific Energy

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CC501- HYDRAULICS 2

CHAPTER 4NON-UNIFORM FLOW

IN OPEN CHANNEL



LEARNING OUTCOMECLO1:

Explain clearly the principles and concept ofhydraulics forces.

CLO2:

Determine hydraulics forces and flow in open

channel as well as pump operation by using

appropriate solutions.



INTRODUCTION

Understand the concept ofnon-uniform ow in open

channel-Dene non-uniform ow.

-Dene specic energy.



*DEFINITION

1. UNIFORM FLOW

Where the flow rate of the flow, velocity,depth, flow cross sectional area and the slopeof the base channel is the sae between a

section with other sections

!. O"#N $%&NN#L

& channel that is e'posed to the atosphereand not f(lly et by the solid bo(ndary

)e'* draina+e, river, trench, canal, forin+





*Open-channel flow occurs when a

liquid flows due to gravity. Usually the

flowing liquid has a free surface, as ina channel, flume or partially full pipe.

he liquid is not under pressure, other

than atmospheric pressure. !anyformulae have been developed to

estimate the flow rate in open-

channels, the !anning formula hasbecome widely accepted as the usual

method of estimating flow rate.





*SPECIFIC ENERGY (E)

*he definition of specific energy at any cross-section in anopen channel is the sum of the "inetic energy per unit

weight of the flowing liquid and the potential energy

relative to the bottom of the channel.

*hus an expression for specific energy is as follow#

*E $ y % &'('g - )*+

*here# E is the specific energy in unit m )meter+

- y is the depth of flow above the bottom of the channel in

unit m )meter+- & is the average liquid velocity )$ (+ in m(sec

- g is the acceleration due to gravity $ /.0*m(s'



*Others expression for specific energy is as follow#

*E $ y % &'('g - )*+

- y - is the 123 E4E567

- &'('g - is the 824E23 E4E567

nother form of the equation with ( in place of & is#

E $ y % '(''g -)'+

he way that specific energy varies with depth of flow in

an open channel can be illustrated by considering a

rectangular open channel with bottom width b.

9or such a channel, $ yb, where b is the channel width.1ubstituting for in equation )'+, gives#

E $ y % '()' y'b'g+ -):+



*he parameter q, the flow rate per unit width of

channel, is often used for a

*rectangular channel. he relationship between q

and is thus# q $ (b

*or $ qb. 1ubstituting for in equation ):+ gives#

E $ y % q'()' y'g+ - );+



Example 1:

rectangular open channel with bottom width $ 'm, is

carrying a flow rate of <m:(s, with depth of flow $ *.<m.

cross-section of the channel is shown in the figure below.

3alculate the 1pecific Energy for this open channel.

Soluton:

E $ y % &

'

('g!here" # $ %&A $ '&(2x1)'* $ 1)+,m&-

$= E $ *.< % *.>?'(')/.0*+

): E $ 1)+.2m



*Example 2:

trape@oidal channel has a width of >mand *#* side slopes drain water at 0 m:(s.

Determine the energy of water, if the

water depth is 'm.

*nswer # E $ '.A*:m

*#*

'm

>m



*Example /:

ater flows in a rectangular channelwith a width of <m flow rate 0 m:(s at

a depth of *.Am. 3alculate the value of

the specific energy.

*nswer # E $ *.*:m

*.Am

<m



*SPECIFIC ENERGY GRAPH'+ y &s E" *+ y &s Es

:+ y &s Ey

Es, E" B E



*T0ES O NON3UNIORM LO!

Three T4pe- o5 lo61) Su7crtcal lo6

2) Supercrtcal lo6

/) Crtcal lo6



SUERCRITICAL LO!*ny open channel flow having

depth of flow less than critical

depth ) y C yc + will berepresented by a point on the

lower leg of the graph above.



*1U35223 9O

*ny open channel flow having depth offlow greater than critical depth ) y = yc +

will be represented by a point on the

upper leg of the graph above

*35223 9O

*he flow condition with y $ yc



*CRITICAL DEPTH FLOW (yc)

he parameter, specific energy, can beused to help clarify the meaning of

supercritical, subcritical, and critical flow

in an open channel.

he symbol yc is commonly used for critical

depth and will be so used in this course.

hrough a little application of calculus, an

equation for the critical depth, yc, can be

derived.



* he derivative of E with respect to y, dE(dy, must be

determined from equation )*+, set equal to @ero and

solved for y. his will give an expression for y that

gives either a minimum or maximum value for E. 9rom

inspection of the graph of E vs y in 9igure *, we can

see that it must be a minimum value for E and that

the value of y at that minimum is the critical depth,

yc. his procedureyields the following equation for yc#

yc $ )q

'

(g+

*(:



*3ritical &elocity, &c

&c $ )yc x g+*('

*!inima 1pecific Energy, Emin

Emin $ *.< yc

*3ritical 1lope, 1c

1c $ )qn ( yc<(:+'



Example ; #

>m wide channel drain water 'A m:(s,

determine the depth of the water when the

specific energy would be minimal.

nswer # yc $ *.A;'m



Example <#

ater flows to the flow rate <.;' m:(s

in the rectangular channel width is

;.Am and !anning, n $ A.A*'.

3alculateG

i. 3ritical depth. )yc $ A.<?'m+

ii. he critical velocity.)vc $ '.:>/m(s+

iii. 3ritical slope.)1c $ A.AA*?+ H *#<00iv. !inimum specific energy. )Emin$A.0<0m+



*95OUDE 4U!E5 )9r+

*9r $ &()y x g+*('

*9r = * $= 1upercritical 9low

*9r C * $= 1ubcritical 9low

*9r $ * $= 3ritical 9low



THAT ALL

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Cc501chapter 4 - Specific Energy