--- title: "Gravity dam equilibrium and overtopping discharge coefficient" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:gravity-dam-reservoir-capacity-curve-hrwl" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-reservoir-capacity-curve-hrwl site: BCT Engineering --- # Gravity dam equilibrium and overtopping discharge coefficient Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2072 Chaitra, 3+5+3+5 marks - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/gravity-dam-reservoir-capacity-curve-hrwl - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question A concrete gravity dam shown in the figure below was constructed for the development of a hydropower project. The dam has a vertical upstream face and an inclined downstream face. The highest regulated water level (HRWL) of the dam is fixed at 1 m below the top crest level. At HRWL, the storage capacity of the reservoir created by the dam is 60 million m³. The reservoir capacity curve of the dam is shown in the figure below. In a flood situation the 80 m long dam crest can serve as a spillway to discharge the flood. Assume density of concrete $\gamma_c = 24$ kN/m³ and the friction angle between the dam and foundation $\phi = 43°$. a) Find all main forces acting on the dam when the water level in the reservoir is at HRWL. Give your answer in terms of base width B. b) Find the bottom width B and downstream inclined angle α if the dam is at the state of moment equilibrium with respect to the downstream dam toe. Use a factor of safety against overturning as 1.4. c) Is the dam free from tensile stress? Find the required unit shear resistance (cohesion) if the shear safety factor of the dam is $F_{SF} = 2.5$. d) In a flood event the dam shown in the figure overtopped but did not fail. The outflow discharge over the dam crest was estimated to be 320 m³/s. During this time, the reservoir water level was raised to 722.5 masl. Find the discharge coefficient and give your comments on the value. [Figure: dam with crest width 3 m, vertical upstream face, base at 700 masl, downstream face inclined at angle α, base width B; capacity curve plotting elevation (695 to 740 masl) against volume (0 to 140 million m³).] ## Short answer Base width B≈12.1 m, downstream α≈66°. Heel tensile stress -79 kN/m², so not tension-free. Required cohesion 174 kN/m² for shear safety factor 2.5. Discharge coefficient C=1.01 (C_d≈0.34) is low, indicating possible head overestimation or poor crest shape. ## Answer #### Assumptions The figure values are not given in the question, so these are assumed: crest level **720 masl** (dam height $H_d=20$ m above the base at 700 masl), so HRWL $=719$ masl and the water depth $h=19$ m. Crest width 3 m, vertical u/s face. $\gamma_c=24$ kN/m³, $\gamma_w=9.81$ kN/m³, $\phi=43^\circ$. Uplift is taken as triangular and full (no drainage): $\gamma_wh$ at the heel and zero at the toe. Forces per metre length. ```text 3 m +---+ 720 crest ~|~~~| 719 HRWL | \ | \ alpha | \ 700 +----\ heel <--B--> toe ``` #### a) Main forces at HRWL (in terms of B) Split the section into a rectangle ($3\times20$) and a triangle (base $B-3$, height 20). - $W_1=24\times3\times20=1440$ kN at $B-1.5$ from the toe - $W_2=24\times\frac12(B-3)\times20=240(B-3)$ kN at $\frac23(B-3)$ from the toe - Water thrust $P=\frac12\gamma_wh^2=\frac12\times9.81\times19^2=1,770.7$ kN acting at $h/3=6.33$ m above the base - Uplift $U=\frac12\gamma_whB=\frac12\times9.81\times19\times B=93.20B$ kN acting upward at $\frac23B$ from the toe #### b) Base width B and angle α for $FS_o=1.4$ Moments about the toe: $$ \Sigma M_R=1440(B-1.5)+240(B-3)\times\tfrac23(B-3)=1440(B-1.5)+160(B-3)^2 $$ $$ \Sigma M_O=P\times6.333+U\times\tfrac23B=11,214.5+62.13B^2 $$ Condition $\Sigma M_R=1.4\,\Sigma M_O$ gives, after expanding, a quadratic in $B$: $$ 160B^2-960B+1440+1440B-2160 = 1.4\,(11,214.5+62.13B^2) $$ Solving: **$B=12.07$ m** (check: $FS_o=1.40$). Angle of the d/s face: $\tan\alpha=\dfrac{H_d}{B-3}=\dfrac{20}{9.07}$, so **$\alpha=65.6^\circ$** with the horizontal (the face makes 24.4° with the vertical, slope $\approx0.45$ H : 1 V). #### c) Tension and required shear resistance $W_2=240\times(12.07-3)=2,176$ kN; $U=1,124$ kN. $\Sigma V=1440+2,176-1,124=2,491$ kN/m. $\bar x=\dfrac{\Sigma M_R-\Sigma M_O}{\Sigma V}=3.25$ m from the toe, $e=B/2-\bar x=2.78$ m, and $B/6=2.01$ m. Since $e>B/6$, the resultant is outside the middle third and **the dam is not free from tension**: $\sigma_{heel}=\dfrac{\Sigma V}{B}\left(1-\dfrac{6e}{B}\right)=-79.0$ kN/m² (tensile), and $\sigma_{toe}=491.9$ kN/m². Shear safety factor $F_{SF}=\dfrac{cB+\Sigma V\tan\phi}{P}=2.5$: $$ c=\frac{2.5\,P-\Sigma V\tan43^\circ}{B}=\frac{2.5\times1,770.7-2,491\times0.9325}{12.07}=174.4\ \text{kN/m}^2 $$ **Required unit shear resistance ≈ 174 kN/m².** #### d) Discharge coefficient during overtopping Head over the crest $H=722.5-720=2.5$ m, crest length $L=80$ m, $Q=320$ m³/s. $$ Q=C\,L\,H^{3/2}\Rightarrow C=\frac{320}{80\times2.5^{1.5}}=\frac{320}{80\times3.953}=1.01\ (\text{SI}) $$ In the form $Q=\frac23C_d\sqrt{2g}\,L\,H^{3/2}$: $C_d=0.34$. **Comment:** For a broad-crested or ogee crest, $C$ is about 1.7-2.2 ($C_d\approx0.5$-$0.75$). The value 1.01 is much lower, which means the estimated 320 m³/s is small for a head of 2.5 m. The reasons may be an over-estimated head, a submerged (tail water) flow, or a poor crest shape/rough overtopped surface, or the estimate of discharge is wrong. The check should be repeated with the measured crest level. **Answer: B ≈ 12.1 m, α ≈ 66°; tension occurs at the heel; required shear resistance ≈ 174 kN/m²; C ≈ 1.01 (C_d ≈ 0.34).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the uplift force expressed at HRWL (no drainage, depth h = 19 m)? A. U = gamma_w h B = 186.4 B B. U = (1/2) gamma_w h B = 93.2 B, acting at 2B/3 from the toe C. U = (1/2) gamma_w h^2 = 1770.7 kN D. U = gamma_w h B / 3 Answer: B. Uplift is a triangle from gamma_w h at the heel to zero at the toe, so its force is half base times ordinate, centroid at 2B/3 from the toe. (Key point: Uplift force in terms of B) 2. How is the base width B found in part (b)? A. Set SumM_R = 1.4 SumM_O about the toe, giving a quadratic in B (B = 12.07 m) B. Set SumV = SumH C. Set e = 0 D. Set the shear safety factor to 1.4 Answer: A. Moments about the toe of the weights are equated to 1.4 times the overturning moments of water thrust and uplift, then solved for B. (Key point: Finding B from overturning) 3. What is concluded about tension and how is the required cohesion c found? A. No tension; c = P / B B. Tension at the heel since e = 2.78 m > B/6 = 2.01 m; c = (2.5 P - SumV tan(phi)) / B = 174 kN/m^2 C. Tension at the toe; c = SumV tan(phi) / B D. No tension; c = 0 because phi is large Answer: B. The resultant is outside the middle third, so the heel is in tension (-79 kN/m^2); the shear factor F_SF = (cB + SumV tan phi)/P = 2.5 is solved for c. (Key point: Tension and required cohesion) 4. How is the discharge coefficient found for the overtopping flood, and what does the low value suggest? A. C = Q/(L H^1.5) with H = 2.5 m above the crest; C = 1.01 is low versus 1.7-2.2, so the discharge or head estimate is doubtful B. C = Q/(L H) with H = 722.5 m; the value is normal C. C = Q/(L H^1.5) with H = 22.5 m; the value is high D. C = Q x L x H^1.5; the value is normal Answer: A. Head over the crest is 722.5 - 720 = 2.5 m. C = 320/(80 x 3.953) = 1.01, much lower than typical crest values, so the data or flow conditions need review. 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