EXAMPLE 6 : HYDRAULIC JUMP

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EXAMPLE 6 : HYDRAULIC JUMPA rectangular horizontal channel 2m.wide, carries a flow of 4 m3/s. Thedepth water on the downstream sideof the hydraulic jump is 1m.a)What is the depth upstream?b)What is the loss of head?Q2Q2F1 F2 y1 A1 y 2 A2 gA1gA22 y12 0 . 815y1 2 . 815V1 4/(2*0.311) 6.43 m/s4242y1 (2 y1 ) 1 * ( 2 * 1) g 2 y1g ( 2 * 1)y1 0.311 mV2 4/(2*1) 2.0 m/shloss 2.42-1.20 1.22 m

Example on Critical Flow : 1A flow of 28 m3/sec occurs in an earth-linedtrapezoidal channel having base width 3.0m,side slopes 1V:2H and n 0.022. Calculatethe critical depth and critical slope.y123mAt critical flow Froude number is equal to unityFr2Q 2T 13gAQ 28 m3/s B 3 n 0.022 slope 1V:2HA 3y 2y2 ; T 3 4y ; P 3 2 y(5)1/2R A/P (3y 2y2) / (3 2 y(5)1/2 )From Manning’sA 2 / 3 1/ 2Q R SnSolving for Sc 0.0052( 28 ) 2( 3y 2y 2 ) 3 9 . 81(3 4 y )solving by trial and error yc 1.495m

Example on Critical Flow : 2A channel of rectangular section 3.5m wide,with n 0.014 and S0 0.001 leads from alake whose surface level is 6m above thechannel bed at the lake outlet. Find thedischarge in the channel.The Qmax would result from critical flow at the outlet!!!At reservoir V small thus E y V22g 6mNo head loss between sections.For a rectangular channel at critical flowycEc y c 2Velocity head 2m or velocity Vc 6.264 m/sQmax 87.69 m3/secor yc 4 m

Example on Critical Flow : 3During a major flood events water flowsover the top of a roadway. Determine thehead on the broad-crested weir for adischarge of 300 m3/s if the overflowsection of the roadway is horizontal and150m. long.For a rectangular channel at critical flow q g 22Q Tc1 gA 3corycq 300/150 2 m3/sec per meter 2 g 2yc1/3 0 . 74 m1/ 3

Flow in Open ChannelsDESIGN OF OPEN CHANNELSFOR UNIFORM FLOW

Channel design for Uniform Flow The basic problem is the economical proportioningof the cross section. Channel with given n and S0 for a known Qthe objective is to minimize the area.A 2 / 3 1/ 2Q R Son QnSo5/3 AR2/3A 2/3 KoPIf A is minimum V maximum from continuityR maximum from Manning’sP minimum (R A/P)

Hydraulic Efficiency of Cross-sections Conveyance of the channel sectionQnSo 5/3 AR2/3A 2/3 KoPIt can be shown that the ideal sectionwould beSemicircle

Best Hydraulic SectionQnSo5/3 AR2/3A 2/3 KoPA channel section havingthe least wetted perimeter for a given areahas the maximum conveyance;such a section is known as the best hydraulicsection

Channel design for Uniform FlowHowever, other economical concerns Total volume of excavation Cost of lining Construction techniques Scour in erodible bed Sedimentation for low V Short channels and variable S0may require changes

The best hydraulic section for arectangular channelArea A b* yPerimeter P b 2yybPerimeter must be minimum for given areaP A/y 2ydp/dy -A/y2 2 0A 2 y2b 2y2/ y b 2y

Example 7What are the most efficient dimensions (the best hydraulic section) for aconcrete (n 0.012) rectangular channel to carry 3.5 m3/s at So 0.0006?Given: n 0.012Find b and y.yQ 3.5 m3/s So 0.0006bA 2 / 3 1/ 2Q R Sonbest sectionb 2yQ 2n1/ 3bybyA A 2 / 3 1/ 2Q Q ( ) Son b 2yn P(2y * y 2y * yQ 2y 2yn( y)(8/3So1/ 2( y)8/3)2/3 1.36So1/ 22yQ n2)y22/3( )So2/31/ 2So1/ 2y 1.123 m and b 2.246m

Best Hydraulic SectionsSectionMost efficientTrapezoidalBase depthRectangularWidth 2 x depthTriangularNo specific relationshipCircularSemicircle if openCircle if closed

Precautions Steep slopes cause high velocities which maycreate erosion in erodible (unlined) channels Very mild slopes may result in low velocities whichwill cause silting in channels. (Sedimentation) The proper channel cross-section must haveadequate hydraulic capacity for a minimum cost ofconstruction and maintenance.

Typical Cross Sections The cross-sections of unlined channels arerecommended as trapezoidal in shape with sideslopes depending mainly on the kind of foundationmaterial(considering construction techniques and equipment,and stability of side inclination,the United States Bureau of Reclamation (USBR) andthe Turkish State Hydraulic Works (DSİ) suggeststandard 1.5H:1V side slopes for trapezoidalchannels)

Recommended side slopesMaterialSide Slope (H:V)RockNearly verticalMuck and peat soils¼: 1Stiff clay or earth with concrete lining½:1 to 1:1Earth with stone lining or earth for large channels1:1Firm clay or earth for small ditches1.5:1Loose sandy earth2:1Sandy loam or porous clay3:1

Recommended side slopesKızılkaya Dam

Recommended side slopesFig. 1

DESIGN OF NONERODIBLE CHANNELS For nonerodible channels the designer simplycomputes the dimensions of the channel by auniform-flow formula and then final dimensions onthe basis of hydraulic efficiency, practicability,and economy.Minimum Permissible velocity In the design of lined channels the minimum permissiblevelocity is consideredto avoid deposition if water carries silt or debrisVmin 0.75 m/s(non-silting velocity)

The determination of section dimensions fornonerodible channels, includes the following steps: All necessary information, i.e. thedesign discharge, the Manning’s nand the bed slope are determined. Compute the section factor, Z,from the Manning equation If the expressions for A and R forthe selected shape areA b zy y P b 2y 1 z2substituted in the equation,one obtains 3 unknowns (b, y, z)5/3for trapezoidal sections, and 2[()]b zyyZ AR 2/3 unknowns (b,y) for rectangularsections.b 2y 1 z2 2/3Q, n, So1Q ARn(2/3S 1o / 2)[]Various combinations of b, y and z can be found to satisfythe above section factor Z.The final dimensions are decided on the basis of hydraulicefficiency, practicability and economy.

Methods and Procedures1.2.3.Assume side slope zget the value of b from theexperience curve,Solve for y.Z AR 2/3 [(b zy ) y ] 5 / 3[b 2 y 1 z ]22/3Experience Curves showing bottom width and water depth of lined channels

Best Hydraulic Section: 1) substitute A and R for best hydraulic section in Eq(*),2)Solve for yEx. Trapezoidal sections :z 13A 23yR 24T 3b 3y23y3 y

ChecksIn the proximity of critical depth, flow becomesunstable with excessive wave action, hence it isrecommended that:1)for subcritical flows: y 1.1yc (or Fr 0.86)for supercritical flows: y 0.9yc (or Fr 1.13)2)2) Check the minimum permissible velocity if thewater carries silt. (Vmin 0.75 m/s)

FreeboardThe freeboard, f, is determined by an empirical equationf 0.2 (1 y)where, f is the freeboard (m) y is the water depth (m) or by the curves given in Figure 1 for irrigation canals for theUSBR and DSI practices.

Finalization1)Modify the dimensions for practicability2)Add a proper freeboard to the depth of thechannel section. Recommended freeboard forcanals is given in figure.3)Draw channel cross section and show dimensionsand given parameter.

Open Channel Design Example 1a A trapezoidal channel carrying 11.5 m3/s clear water isbuilt with concrete (nonerodible) channel having a slopeof 0.0016 and n 0.025. Proportion the sectiondimensions.SOLUTION : Q 11.5 m3/s S0 0.0016 n 0.025A1/ 2Q R 2 / 3Son1yzZ b[(b zy ) y ] 5 / 3[b 2 y 1 z ]22/3 7.1875 Assume b 6m and z 2,Solve for y 1.04 m (by trial an error) n *QSo AR2/3 [(b zy ) y ] 5 / 3[b 2 y 1 z ]22/3

Q 11.5 m3/s S0 0.0016 n 0.025b 6m and z 2, y 1.04 mFor given Q from DSİ’s curve Height of lining abovewater surface 0.33mHeight of bank abovewater surface 0.63mCheck stability :Q 2T2Fr 1At critical flow yc 0.692m3gAFor yn Fr 0.48 subcritical and within limitsVelocity V Q/A 1.37 m /sOK for sedimentation.Fr Fr VgDVg( A / T )

Open Channel Design Example 1b A trapezoidal channel carrying 11.5 m3/s clear water isbuilt with concrete (nonerodible) channel having a slopeof 0.0016 and n 0.025. Proportion the sectiondimensions. Use experience curve and z 1.5SOLUTION : Q 11.5 m3/s S0 0.0016 n 0.0251yzb[(b zy ) y ] 5 / 3[b 2 y 1 z ]22/3 7.1875 Take b 2.5m,Solve for y 1.56m (by trial an error)

Q 11.5 m3/s S0 0.0016 n 0.025b 2.5m and z 1.5, y 1.56m Check stability :At critical flow yc 0.692mFor yn Fr 0.47 subcritical and within limitsVelocity V Q/A 1.52 m /sOK for sedimentation.

Open Channel Design Example 1c A trapezoidal channel carrying 11.5 m3/s clear water isbuilt with concrete (nonerodible) channel having a slopeof 0.0016 and n 0.025. Proportion the sectiondimensions. Use best hydraulic section approach!SOLUTION : Q 11.5 m3/s S0 0.0016 n 0.0251yBest Hydraulic Sectionfor Trapezoidal ChannelzbZ n*QSo AR 2/3 7.1875z T A Solve for y 2.03 m (by trial an error)13433yb 233y3 y2R y2

Q 11.5 m3/s S0 0.0016 n 0.025b 2.34m y 2.03m Check stability :At critical flow yc 0.692mFor yn Fr 0.38 subcritical and within limitsVelocity V Q/A 1.49 m /sOK for sedimentation.

discharge of 300 m3/s if the overflow . 5/3 1/ 2 2/3 o R2/3S n A Q Hydraulic Efficiency of Cross-sections Conveyance of the channel section o o K P A AR S Qn 2/3 5/3 2/3 . Vmin 0.75 m/s (non-silting velocity) The determination of section dimensions for nonerodible channels, includes the following steps: .

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