--- title: "Khosla: HGL and exit gradient, pond 102.50" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:khosla-hgl-weir-profile-pond1025" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-hgl-weir-profile-pond1025 site: BCT Engineering --- # Khosla: HGL and exit gradient, pond 102.50 Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2070 Bhadra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-hgl-weir-profile-pond1025 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Sketch the hydraulic gradient line for the weir profile, shown below, considering the case of no flow at pond level. Slope correlation for the slope (2:1) is 6.5 percent. Also compute the value of the exit gradient. [Figure: weir profile; pond level 102.50 m, crest 101.00 m, upstream floor levels 100.00 m and 99.00 m with 2:1 upstream slope, upstream end pile 0.5 m wide with bottom at 94.00 m, intermediate pile bottom 94.00 m, downstream floor levels 98.00 m and 97.00 m, downstream pile 0.5 m wide with bottom at 92.00 m; horizontal distances 15 m and 30 m] ## Short answer No-flow HGL with pond 102.50m, head 5.50m, slope correction 6.5%, drops from E1 100% via C1 70.2%, E2 65.4%, C2 55.6%, E3 29.7% to 0% at d/s end. Exit gradient = 0.156 (1 in 6.4). ## Answer #### Data and assumptions The figure is read as follows. Pond level 102.50 m; with no flow the d/s water level is taken at the d/s floor level 97.00 m, so $H = 5.50$ m. Piles (positions measured from the u/s pile): u/s pile at 0 m, floor level 100.00 m, bottom 94.00 m ($d = 6$ m); intermediate pile at 15 m, floor level 99.00 m, bottom 94.00 m ($d = 5$ m); d/s pile at 45 m, floor level 97.00 m, bottom 92.00 m ($d = 5$ m). Total floor length $b = 45$ m. Floor thickness is not given, so the thickness correction is neglected. The 2:1 slope correction is 6.5%, multiplied by (slope length / $b'$), where $b'$ is the distance between the two piles enclosing the slope: u/s slope (2 m, up slope, $b' = 15$ m): $-6.5\times 2/15$ at $C_1$; d/s glacis (101.00 to 98.00 m, 6 m long, down slope, $b' = 30$ m): $+6.5\times 6/30$ at $E_3$. #### Khosla's constants End piles: $\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), $b = 45$ m; for the u/s pile $\phi_{D1} = 100-\phi_D$, $\phi_{C1} = 100-\phi_E$. Intermediate pile ($b_1$ from the u/s end, $b_2$ from the d/s end): $\lambda_1 = \dfrac{\sqrt{1+\alpha_1^2}-\sqrt{1+\alpha_2^2}}{2}$, $\lambda = \dfrac{\sqrt{1+\alpha_1^2}+\sqrt{1+\alpha_2^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1-1}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1}{\lambda}$, $\phi_C = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1+1}{\lambda}$. | Pile | d (m) | α | λ | φE (%) | φD (%) | |---|---|---|---|---|---| | U/s pile | 6.00 | 7.500 | 4.283 | 32.10 | 22.20 | | D/s pile | 5.00 | 9.000 | 5.028 | 29.43 | 20.42 | | Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) | |---|---|---|---|---|---|---|---|---| | at 15 m | 5.00 | 3.000 | 6.000 | -1.460 | 4.623 | 67.86 | 60.23 | 53.17 | #### Pressures at the key points Pressure is expressed as a percentage of the head $H = 5.50$ m (no flow). | Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) | |---|---|---|---|---| | E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 5.50 | | D1 | 77.80 | - | 77.80 | 4.28 | | C1 | 67.90 | int +3.20, slope -0.87 | 70.23 | 3.86 | | E2 | 67.86 | int -2.44 | 65.43 | 3.60 | | D2 | 60.23 | - | 60.23 | 3.31 | | C2 | 53.17 | int +2.45 | 55.62 | 3.06 | | E3 | 29.43 | int -1.07, slope +1.30 | 29.66 | 1.63 | | D3 | 20.42 | - | 20.42 | 1.12 | | C3 (d/s end) | 0 | 0 | 0 | 0 | Corrections: int = mutual interference, $19\sqrt{D/b'}\,\dfrac{d+D}{b}$, positive at C of the u/s pile and negative at E of the d/s pile of each pair ($D$ = depth of the neighbouring pile below the floor level at the pile being corrected, $b'$ = distance between the two piles); slope = slope correction. The D points are not corrected. #### Exit gradient $$ G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.50}{5.00}\times\frac{1}{\pi\sqrt{5.028}} = 0.1562\ (\approx 1/6.4) $$ Permissible (assumed) $= 0.2000$. Safe (less than the permissible value). #### Hydraulic gradient line Bars show the corrected pressure (each # is 2.5%): ```text E1 100.0% |######################################## D1 77.8% |############################### C1 70.2% |############################ E2 65.4% |########################## D2 60.2% |######################## C2 55.6% |###################### E3 29.7% |############ D3 20.4% |######## C3 0.0% | ``` The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C of one pile to E of the next). In levels (datum 97.00 m): $C_1$ 100.86 m, $E_2$ 100.60 m, $C_2$ 100.06 m, $E_3$ 98.63 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At no flow with pond 102.50 m, the head is: A. 102.50 - 97.00 = 5.50 m B. 102.5 - 100 C. 5.5 - 2 D. 1.5 m Answer: A. The d/s water level is taken at the d/s floor level. (Key point: Head for no flow) 2. How is the HGL drawn? A. Vertical drops across each pile and straight along the floor between pile junctions B. A single slope C. Horizontal D. Curved at the floor Answer: A. Pressure changes E to D to C across a pile; between piles the line is straight. (Key point: HGL shape) 3. The corrected pressure at C1 is 67.61%, obtained from 67.90% by: A. Subtracting the slope correction 0.29 B. Adding 0.29 C. Adding interference D. Multiplying by H Answer: A. The u/s slope is an up slope, so the slope correction is subtracted. (Key point: Corrected values) 4. Exit gradient is 0.156, so: A. Safe, it is less than 0.2 B. Unsafe C. Equal to 1 D. Needs thickness Answer: A. It is below the permissible value of 0.2. (Key point: Exit gradient) 5. alpha for the d/s pile is: A. b/d = 45/5 = 9 B. 5/45 C. 45 x 5 D. 9 x 5 Answer: A. alpha = b/d, then lambda = (1 + sqrt(1 + alpha^2))/2. 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