--- title: "Khosla: percentage pressure and piping, pond 503" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:khosla-floor-piping-abcd-503" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-floor-piping-abcd-503 site: BCT Engineering --- # Khosla: percentage pressure and piping, pond 503 Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2076 Baisakh, 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-floor-piping-abcd-503 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question For the figure shown below ignoring floor thickness and slope corrections, find the percentage pressure at the key points of the piles using Khosla's theory. The floor thickness may not be less than 30 cm anywhere. If permissible exit gradient is 0.15, check the floor against piping failure. Also find the thickness of floor at A, B, C and D points. [Figure: pond level 503 m; floor level 500 m (A); 497 m (B, C) and 497 m (D); pile depths 1.5 m and 2.0 m; horizontal distances 6 m and 10 m] ## Short answer Percentage pressures: E1=100%, D1=81.1%, C1=77.2%, E2=31.1%, D2=21.6%, C2=0. Required thickness: A 1.17 m, B 2.57 m, C 1.95 m, D 1.33 m. Exit gradient 0.449 > 0.15, not safe against piping. ## Answer #### Data and assumptions Reading of the figure: pond level 503 m; d/s floor and bed level 497 m, so $H = 6$ m (no flow, no tail water). U/s pile 1.5 m deep at the u/s end (floor top 500 m, point A; pile bottom 498.5 m); d/s pile 2.0 m deep at the d/s end (floor top 497 m, point D; pile bottom 495 m); floor length $b = 6 + 10 = 16$ m. B is at the foot of the drop ($x = 6$ m) and C at $x = 11$ m. Floor thickness and slope corrections are ignored (as stated), but the mutual interference correction still applies. $G = 2.4$; minimum thickness 0.30 m; permissible exit gradient 0.15. #### Khosla's constants $\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 16$ m. For the u/s pile, $\phi_{D1} = 100-\phi_D$ and $\phi_{C1} = 100-\phi_E$. | Pile | d (m) | α | λ | φE (%) | φD (%) | |---|---|---|---|---|---| | U/s pile | 1.50 | 10.667 | 5.857 | 27.12 | 18.88 | | D/s pile | 2.00 | 8.000 | 4.531 | 31.13 | 21.56 | #### Pressures at the key points Pressure is expressed as a percentage of the head $H = 6.00$ m (no flow). | Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) | |---|---|---|---|---| | E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 6.00 | | D1 | 81.12 | - | 81.12 | 4.87 | | C1 | 72.88 | int +4.31 | 77.20 | 4.63 | | E2 | 31.13 | - | 31.13 | 1.87 | | D2 | 21.56 | - | 21.56 | 1.29 | | C2 (d/s end) | 0 | 0 | 0 | 0 | Corrections: int = mutual interference at $C_1$ due to the d/s pile, $19\sqrt{D/b'}\,\dfrac{d+D}{b} = 19\sqrt{5/16}\times\dfrac{1.5+5}{16} = 4.31\%$, with $D = 500-495 = 5$ m and $d = 1.5$ m. At $E_2$ the u/s pile bottom (498.5 m) is above the d/s floor (497 m), so there is no interference correction. #### Floor thickness Unbalanced head $h$ = hydraulic-gradient level at the floor $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$ (the HGL is taken as a straight line from $C_1$ to $E_2$ along the floor). | Location | x (m) | φ (%) | HG level (m) | Floor top (m) | Unbalanced head (m) | Required t (m) | Provided t (m) | Check | |---|---|---|---|---|---|---|---|---| | A (x = 0, u/s) | 0 | 77.20 | 501.63 | 500.00 | 1.63 | 1.17 | - | provide at least 1.17 | | B (x = 6) | 6 | 59.92 | 500.60 | 497.00 | 3.60 | 2.57 | - | provide at least 2.57 | | C (x = 11) | 11 | 45.53 | 499.73 | 497.00 | 2.73 | 1.95 | - | provide at least 1.95 | | D (x = 16, d/s) | 16 | 31.13 | 498.87 | 497.00 | 1.87 | 1.33 | - | provide at least 1.33 | All of these exceed the 0.30 m minimum, so they govern. #### Exit gradient $$ G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{6.00}{2.00}\times\frac{1}{\pi\sqrt{4.531}} = 0.4486\ (\approx 1/2.2) $$ Permissible $= 0.1500$. Not safe against piping; the d/s pile must be deepened or the floor lengthened. **Answer:** Corrected pressures: $E_1 = 100\%$, $D_1 = 81.1\%$, $C_1 = 77.2\%$, $E_2 = 31.1\%$, $D_2 = 21.6\%$, $C_2 = 0$. Floor thickness: A 1.17 m, B 2.57 m, C 1.95 m, D 1.33 m. The exit gradient (0.449) is about three times the permissible 0.15, so the floor is not safe against piping. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Pond level 503 m and d/s bed level 497 m give H = A. 6 m B. 2 m C. 3 m D. 10 m Answer: A. H = 503 - 497 = 6 m. (Key point: Head with no flow) 2. Since floor thickness and slope corrections are ignored, the pressures used are A. Zero everywhere B. Corrected values with interference C. Bligh's values D. The simple Khosla phi values Answer: D. No corrections are applied, so the table gives the simple phi values. (Key point: Corrections ignored) 3. How is the final floor thickness at each of A, B, C, D chosen? A. The larger of the required thickness and 0.30 m B. The smaller of the two C. Always 0.30 m D. Always the required value even if it is below 0.30 m Answer: A. The floor must not be thinner than 0.30 m anywhere, so take the larger value. (Key point: Minimum thickness rule) 4. G_E = 0.4486 and permissible is 0.15. The floor is A. Not safe against piping B. Safe because H/d = 3 C. Safe if thickness is 2.45 m D. Safe against piping Answer: A. G_E exceeds 0.15; the d/s pile must be deepened or the floor lengthened. 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