Determination the Invert Level of a Stilling Basin to Control Hydraulic Jump

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1 Global Avane Researh Journal of Agriultural Siene Vol. (4) pp , June, 0 Available online Copyright 0 Global Avane Researh Journals Full Length Researh Paper Determination the Invert Level of a Stilling Basin to Control Hyrauli Jump Farzin Salmasi, Department of Water Engineering, Faulty of Agriulture, Tabriz University, Tabriz, Iran. Salmasi@Tabrizu.a.ir Aepte 04 June, 0 When water is release over the hutes/spillway, the potential energy is onverte into kineti energy at the base of hutes. This energy must be issipate in orer to prevent the possibility of sever souring of ownstream riverbe an the unermining of founation whih may ause failure of spillway an am. For this purpose energy issipators must be use whih perform the energy reution by onverting the kineti energy into turbulene an finally into heat. The issipation of energy an be ahieve by means of several methos suh as stilling basins. The formation of hyrauli jump in the stilling basin will lea to issipation of exess energy. In this stuy we use a simple metho to etermine stilling basin invert elevation in relation to upstream water elevation, esign isharge an ownstream water/ tail water elevation. Some solve examples have been provie to exhibit the apabilities of the esign proeure. Key wors: Stilling Basin, Hyrauli Jump, Energy Dissipator, Chute, Spillway. INTRODUCTION Chutes are use in irrigation anals to onvey water from a higher elevation to a lower elevation. They are useful struture for broken high slope to low slope in agriultural projets. A hute struture may onsist of an inlet, a hute setion, an energy issipator, an an outlet transition. The hute setion may be pipe as in a pipe hute or an open setion as in an open hannel hute. Chutes are similar to rops exept that they arry the water over longer istanes, over flatter slopes, an through greater hanges in grae (U.S.B.R., 978). The hute setion, either pipe or open hannel, generally follows the original groun surfae an onnets to an energy issipator (stilling basin) at the lower en. Stilling pools or baffle outlets are use as energy issipators on hute strutures. The usual setion for an open hannel hute is retangular. In a stilling pool the water flows own the steep slope setion at a veloity greater than the ritial veloity. The abrupt hange in slope where the flat grae of the stilling pool floor meets the steep slope setion fores the water into a hyrauli jump an energy is issipate in the resulting turbulene. The stilling pool is proportione to ontain the jump. For a stilling pool to operate properly the Froue number shoul be between 4.5 an 5. Speial stuies or moel testing are require for strutures with Froue numbers outsie this range. If the Froue number is less than about 4.5 a stable hyrauli jump may not our. If the Froue number is greater than about 0 a stilling pool may not be the best hoie of energy issipator. Stilling pools require tail water to fore the jump to our where the turbulene an be ontaine. Stilling pools usually have a retangular ross setion, parallel walls, an a level floor. The following equations to etermine pool invert elevation apply to stilling basin (Figure.).

2 075 Glo. Av. Res. J. Agri. Si. Figure. Cross setion of a hute with hyraulis parameters ElC ElC + + hv ElC + + q /(g () / Fr () where, g is aeleration ue to gravity, an are the epth of flow before an after of hyrauli jump respetively, El C is energy level at upstream of hute, El C is stilling basin invert elevation, q is flow rate per unit with (q Q/b) where Q is total isharge, b is the hute with, F r is Froue number at setion before hyrauli jump an efine as: veloity (V ) was alulate as V q/. Eq. is energy equation between upstream of hute an ownstream before hyrauli jump (in stilling basin) an Eq. is a result of ombination of momentum an ontinuity equations. Eq. is name onjugate epth ( an ) relation in hyrauli jump. In Eqs., two unknown parameters an El C must be etermine. Note that is a known parameter from (energy level at ownstream/tail water)-el C. So we must solve two non-linear Eqs., by trial an error to obtain two unknown an El C. In this stuy we present a simple metho for etermination of stilling basin invert elevation base on energy at upstream of hute, esign isharge, an tail water elevation (from isharge-hea urve/rating urve). A number of esign examples are presente to emonstrate the appliation of the algorithm too. F r V / g. The average flow ) MATERIAL AND METHODS Solution of system of non-linear equations Refering to a hute ross setion in Figure., we present an example to solve two non-linear Eqs., by trial an error to obtain two unknown an El C. Example. The reservoir level upstream of a 65-m wie spillway for a flow of 400 m /s is at El. 00 m. The ownstream river hanges to water level for this flow is at El m. Determine the invert level of a stilling basin having the same with as the spillway so that a hyrauli jump is forme in the basin. Assume the losses in the spillway are negligible. Given: Q 400 m /s B 65 m Upstream water level El 00 m. Downstream water level El 95.8 m. Determine: Stilling basin invert elevation to form the jump? Solution: Let El C be the invert elevation of the stilling basin. Then, referring to Figure., 95.8 El C. Sine the losses on the spillway fae are negligible, we have:

3 Salmasi. 076 Table. Results of solution non-imensional Equation. an 4. Row H/ K / Row H/ K / q /( g ) ElC / 00 ElC By alulating Froue number at setion before hyrauli jump we have: F F r r V / r g F q ( g ) ( 400/65) /( 9.8 ) 7.8/ /

4 077 Glo. Av. Res. J. Agri. Si. Table. Design stilling basin invert elevation with ifferent hyrauli parameter TW * Q b y y y El C El C (m) (m) (m /s) (m) (m) (m) (m) (m) *TW is tail water elevation or energy at ownstream of anal. Substituting expressions for F r an 95.8 El C into Eq., we will have: 95.8 ElC * +.669/ Finally 0.69 m by trial an error an then, 4.0 m. So stilling basin invert elevation is alulate from: El C m. a) Introuing non-imension parameters Using Equation an, we an emonstrate the following two non-imensional equations (U.S.B.R., 978): H K( K + ) ( K ) 4K q g K ( K + ) () (4) (5) K In Eqs. an 4, H is energy loss in hyrauli jump an is ritial epth. With assuming ifferent values for H/ from 0.05 to 60 with inrement step equal 0., an using Exel Solver, we alulate Eqs. an 4 for / an K. Results of these alulations are presente in Table. The Mirosoft Offie Exel Solver tool uses the Generalize Reue Graient (GRG) nonlinear optimization oe (Mirosoft Offie Exel, 007). Solver is part of a suite of ommans sometimes alle tools. With Solver, you an fin an optimal value for a formula in one ell-alle the target ell-on a worksheet. Solver works with a group of ells that are relate, either iretly or iniretly, to the formula in the target ell. Solver ajusts the values in the hanging ells that you speify-alle the ajustable ells-to proue the result that you speify from the target ell formula. You an apply onstraine to restrit the values that Solver an use in the moel, an the onstraints an refer to other ells that affet the target ell formula. Now we solve example with this metho. q (400/65). 94 g 9.8 H Then from table, for H/., we fin: m K m Finally we alulate stilling basin invert elevation: m

5 Salmasi. 078 Figure. Variation of H/ against / Figure. Variation of / against / El C m. If we fin from table, H/.6 by interpolation, then El C will be exatly the same as in previous example; El C m. b) Regression equations From introue ata in table, we applie regression analysis to fin orrelation between H/ an / an also between / an / in another han. These orrelations are shown in Figure. an. By seleting ifferent regression equations in Exel software, the following equations were obtaine: 0.9 H 0.460, R 0.98 (6).55.50, R To alulate stilling basin invert elevation using Eqs. (6) an (7), we will have : (7)

6 079 Glo. Av. Res. J. Agri. Si. H/ (.6 ) m.55 ( ) m So this example; El C m. Some other examples are presente in table with ifferent hyrauli parameter in esigning stilling basin invert elevation. CONCLUSION A hyrauli jump is the suen transition from a superritial open hannel flow regime to a subritial flow motion. The loation of a jump in stilling basin may be ontrolle by proviing a number of appurtenanes, suh as baffle bloks, sill, rop or rise in the hannel bottom. For a horizontal retangular hannel an negleting bounary frition, the energy an momentum priniples give stilling basin invert elevation. Thus in this stuy we applie three metho to etermine stilling basin invert. They are (a)- energy an momentum equations, (b)- introuing non-imensional parameters an ()- regression equations. Solving an example shows the usefulness of methos (b) an (), beause these methos o not nee any trial an error alulation. b El C El C hute with; epth of flow before hyrauli jump; epth of flow after of hyrauli jump; energy at the upstream of hute; stilling basin inverts elevation; F r V / g ; F r superritial Froue number, g aeleration ue to gravity; H hea loss ue to hyrauli jump; h v veloity hea before hyrauli jump; h v veloity hea after hyrauli jump; K relative epth efine as: K / ; q isharge per unit with; Q isharge; veloity at the toe of the hute; V REFERENCES Mirosoft Offie Exel (007). User help USBR(978). Design of small anal strutures. Unite state epartment of the interior, Bureau of relamation. NOTATION