DESIGN OF A WEIR. The purpose of this exercise is to acquaint with the basic scope of a small weir calculation. It must cover the following topics:

Transcription

1 DESIGN OF A WEIR The purpose of this exercise is to acquaint ith the basic scope of a small eir calculation. It must cover the folloing topics: Calculation part (calculations should be illustrated ith sketches): - designation of spillay idth, - selecting the shape of the spillay (trapezoidal), - calculation of stilling basin (length and depht), - calculation of the seepage in eir s subsoil, - checking the stability of eir s body and stilling basin Draing part: - horizontal projection and cross section of the eir (scale 1:50 or 1:100) Term: 15. January. 01 SDF = Spillay design flood [m /s] Max storage level = maximum storage elevation in the reservoir No. Name River SDF [m /s] Aver. flo rate [m /s] Max storage level [m.a.s.l.] Subsoil Class of hydrostructure decline of ater surface i [-] 1 Alvaro Correa Andres Alcazar Pagona 8,00 6,00 9,50 Gravel IV 0,00050 Adan Gomez Roberto Lopez Pagona 6,00 5,50 9,0 Soft clay III 0,0005 Ana Lopez Val Irene Rivera Pagona 4,00 5,00 9,0 Sandgravel mix II 0, Piotr Prus Marcin Mosiejko Pagona,00 4,50 9,90 Fine II 0, Krzysztof Jednachoski Sergiusz Szreder Pagona 0,00 4,00 9,0 Gravel IV 0, Borys Sajko Irmina Byzdra Pagona 19,00 4,00 9,70 Sandgravel mix IV 0, Samuel Rodriguez Inigo De la Calle Prosna 100,00 0,00 149,00 Gravel II 0,00045 Page 1

2 No. Name River 8 9 Alekra Latiszeska Beata Mikołajczuk Mateusz Samociuk Krzysztof Skrzydłoski SDF [m /s] Aver. flo rate [m /s] Max storage level [m.a.s.l.] Prosna 90,00 18,00 149,00 Prosna 85,00 19,00 148,80 Subsoil Fine Class of hydrostructure Sandgravel mix decline of ater surface i [-] IV 0,00048 III 0, Tomasz Zybała Antoni Żabniak Prosna 80,00 16,00 148,70 Soft clay II 0, Piotr Stolarski Maciej Serin Prosna 75,00 15,00 148,50 Medium III 0, Anna Jaczeska Timothy Apiyo Prosna 70,00 14,00 148,40 Coarse IV 0, Paeł Stanisłaski Maksymilian Mokrzycki Prosna 95,00 14,00 148,40 Coarse IV 0, Karolina Zamiar Wojciech Podleś Pagona 15,00,00 9,00 Coarse IV 0, Krzysztof Tomczak Maciej Zemfler Pagona 16,00,50 9,40 Sandgravel mix III 0, Przemysła Wróbel Tomasz Szymczak Pagona 17,00,00 9,80 Fine III 0, Łukasz Jagalski Remi Mordome Pagona 18,00,50 9,00 Gravel IV 0, Mohammad Ramadan Marta Gosz Pagona 19,00 4,00 9,40 Fine III 0, Daniel Wrzosek Marcin Mosakoski Prosna 100,00 18,00 149,0 Sandgravel mix IV 0, Jędrzej Sznajder Ilga Navitski Prosna 70,00 0,00 150,0 Fine IV 0,0006 Page

3 PROSNA river Valley profile Flo-rate curie (vertical axis: elevation [m a.s.l.]; horizontal axis: flo rate [m /s]) Page

4 PAGONA river Valley profile Flo-rate curie (vertical axis: elevation [m a.s.l.]; horizontal axis: flo rate [m /s]) Page 4

5 . DESIGN OF A WEIR 1. The minimum idth of the spillay may be determined depending on the alloable maximum unit discharge:: b min Q 1 5 q 1 max i t q n max max here:: Q discharge of spillay [[m /s],, -- coefficient of river bed erosion resistance, qmax - maximum flo found in naturally floing river ((stream) prior to its construction [m /sm]] Soil in river bed λ Rocks 1,80 Stones 1,60 Gravel 1,40 Sand gravel mixing 1,0 Coarse 1,0 Medium 1,15 Fine 1,10 Loamy 1,08 Organic soils 1,05 Medium clay 1,10 Heavy clay 1,15 Clay 1,0. Discharge capacity of spillay v M b h g Q o / here: c - idth of the crest [m], M discharge coefficient, b is the idth of the spillay [m], h - is the head from the ater level in the upper pool to the dam crest [m], - St Venant ratio = 1,1, v 0 - velocity of the ater hich flos to the spillay (velocity of approach;) [m/s], g - acceleration due to gravity [m/s ], - side throttling ratio, - submerging coefficient of the spillay v 0 g H / c >,0 H / c = 1,0 H / c = 0,5 inclination,0 1,0 factor m 1 1,86 1,78 1,64-1,70 1,78 1,70 1,55-1,60 1,7 1,64 * 5 1,67 1,55 * 10 1,55 * * H C 1 : m Page 5

6 h 1 0,n b here: n - number of bays of the spillay [-], b - idth of the spillay [m], h - depth of ater above the crest of the spillay[m], ξ - coefficient depending on the shape of pillars and abutments [-]. Submerging coefficient 16 a a a a 4 1 dla 0 0,9 or 0,811 0,9 1, 0 H 0 H 0 dla 0 0 H H here: a - the difference beteen the donstream ater level and the level of crest of spillay. If the donstream ater level is arranged belo the crest of the spillay, then it has no impact on its discharge (spillay is unsubmerged) and = 1. Condition to meet: Q > SDF. Stilling basin Condition to meet: h d h 1 The first step is to calculate the fraction here: q=q/b, =1,1 1 q... h o The second step is to choose the shape of spillay inlet, and to determine the velocity index i.e.=0,9 for rounded shape of inlet) The minimum depth of stilling basin: d min = 0,0 m. The length of stilling basin: l = 5(h - h 1 ) Page 6

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8 4. Seepage To avoid soil piping and ensure the stability of dam the outline of the underground part of the foundation should be so developed to provide a sufficiently long filter path to ensure the proper seepage velocity reduction. In an approximate manner the required (minimum) length of filtration path is determined by the formula l p C h C - factor depending on the type of soil and method of calculation h - the maximum difference in ater levels on the upstream and donstream (max head) Bligh method l l i h i l i length of individual horizontal sections h i length of individual vertical sections Lane method 1 l l i h i l i length of individual horizontal sections h i length of individual vertical sections Condition to meet: l l p Subsoil of dam Bligh C B Lane C L Rocks -,5 Stones -,0 Gravel 7,5 Sand gravel mixing 9 4,0 Coarse 1 5,0 Medium 1 6,0 Fine 15 7,0 Loamy 18 8,5 Organic soils - - Medium clay 8,0 Heavy clay 6,0 Clay - 1,6 With insufficient length of filtration path under the building, e have to increase its length by forcing ater to flo around more developed contour. Here e can use a deck (a aterproof element made of clay or geomembrane) located on the reservoir s bottom and / or sheet pile all at the front of the dam s body. In addition, e can also put the sheet pile all on the end of the stilling basin (at point 7). This solution ill reduce the velocity of seepage, but also ill increase the uplift pressure acting on the slab (stilling basin). Page 8

9 5. Uplift pressure An example of a calculation the uplift pressure acting on the body of the eir. Max storage level=149, ,80 (for average flo rate Q=18 m/s) 144, ,50 (the bottom level) 144,0 144, , , ,7 0.0 h= V = 1,4+0,9 d *,0*10 = 4,5 kn/m V =,1*,0*10 = 6,0 kn/m s V tot = 4,5 + 6,0 = 97,5 kn/m Page 9

10 6. Loads An example of a calculation the ater load and self-eight load acting on the body of the eir. Max storage level=149,00 *h1 h1 148,50 h 1 144,50 P *(h1+h) G centroid of profile area 0 144,0 *(h1+h) h 5 14,7 g 6 p Vd+Vs Vtot Vd+Vs v 4 19,70 Water load h1 h1 h P h _[ kn / m] h h 1 h 0,5 4,5 4,0 4,5 0,8 Self-eight load G A b c 8,6 197,8 _[ kn / m] here: A b is the area of eir s cross section [m ], c is the unit eight of concrete = [kn/m ] Uplift pressure load (as in p.5) V tot 97,5 _[ kn / m] Page 10

11 7. Stability of the body of dam Overturning stability - relation of the moments of the hypothetical point of rotation (point n.6 see the draing in p.6) m M M u o Gg m Pp Vv o Class of hydro structure The basic layout of loads m o I II III IV 1,0 1,15 1,10 1,05 Sliding stability - relation of forces in the foundation n Fu Fp f G V P n o f = tan- friction coefficient beteen soil and concrete - internal friction angle of the soil Class of hydro structure The basic layout of loads n o I II III IV 1,0 1,15 1,10 1,05 To improve the sliding stability of the dam the inclined (toards the reservoir) basis of foundation can be used, teeth or spurs at the front (and rear) all. In order to improve the stability of a shift, dam foundation basis shall be inclined. Then the stability of the shift is checked by formula: V G Ptg G cos Psin V n f f cos n o P cos Gsin P Gtg angle of inclination Page 11

12 8. Stability of stilling basin The minimum thickness of stilling basin must meet the condition: h min d n _, _ min d 0, m c 1 here: c is the unit eight of concrete = [kn/m ] is the unit eight of ater = 10 [kn/m ] n safety factor [-] h maximum value of hydrodynamical pressure acting on the bottom of stilling basin [m] Class of hydro structure The basic layout of loads n I II III IV 1,0 1,15 1,10 1,05 Page 1